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\centerline{\bf{Introduction}}
\smallskip

In [FS1] we announced a precise asymptotic formula for the
ground--state energy of a large atom.  One part of our proof
of that formula is a refined analysis of the eigenvalues and
eigenfunctions of an ordinary differential operator
$\frac{-d^2}{dx^2}+V(x)$.  In particular, we need to improve on
the standard WKB approximations for rather general potentials $V$.
This paper contains our results on individual eigenvalues and
eigenfunctions of ordinary differential operators.  Our later
papers [FS2,3,4,5] will study sums of eigenvalues and sums
of squares of eigenfunctions, and then pass to spherically symmetric
three--dimensional problems by separation of variables.  In [FS6]
we will apply our knowledge of three--dimensional problems to the study of
atoms, which is a $3N$--dimensional problem for large $N$.  Finally,
in [FS7] we study the Hamiltonian flow in a three--dimensional potential
arising in atomic physics, in order to verify a crucial hypothesis in
[FS4,5].

We begin our discussion of ordinary differential operators $\frac{-d^2}{
dx^2}+V(x)$ by reviewing the standard WKB approximation in one dimension.
The approximation applies to large, slowly varying potentials.  A basic
example is
$$
V(x)=\lambda^2V_1(x)\tag 1
$$
\noindent for fixed, smooth $V_1$ and $\lambda >>1$.  For simplicity,
we suppose $V_1^\prime(0)=0$ and $V_1^{\prime\prime}(x)>c>0$ for all
$x$.

According to WKB theory, the eigenvalues $E_k$ of $\frac{-d^2}{dx^2}+
V(x)$ are given approximately as the solutions of
$$\gather
\phi(E_k)=\pi(k+1/2)\quad\text{for\ integers}\ k,\ \text{where}\tag 2\\
\phi(E)=\int_{\{V(x)<E\}}(E-V(x))^{1/2}\,dx.\tag 3\endgather
$$

\noindent Under our assumptions on $V_1$, $\{V(x)<E\}$ is an interval
$[x_{\text{left}}(E),x_{\text{right}}(E)]$.  The endpoints $x_{\text{left}}
(E)$, $x_{\text{right}}(E)$ are called turning points.

For $E_k$ not too near the minimum of the potential, WKB theory also gives
approximate formulas for an eigenfunction $F_k(x)$ corresponding to $E_k$.
In the simplest case, we take $E_k>(\min V)+c\lambda^2$.  Then $F_k(x)$ is
given by
several different formulas in overlapping regions, and (2) is the consistency
condition for the different formulas to agree.  The formulas for $F_k$
are as follows.

For $x \in [x_{\text{left}}(E_k),x_{\text{right}}(E_k)]$ not too near the
turning points,
$$
F_k(x)\approx (E_k-V(x))^{-1/4}\cos(-\frac \pi 4+\int_{x_{\text{left}}(E_k)}
^{x}(E_k-V(t))^{1/2}\,dt).\tag 4
$$

For $x$ near $x_{\text{left}}(E_k)$, a smooth change of variable
$x \to y=y_k(x)$ transforms
$$
\Bigl(-\frac{d^2}{dx^2}+V(x)-E\Bigr)F_k(x)=0\tag 5
$$
\noindent approximately to Airey's equation
$$
\Bigl(\frac{d^2}{dy^2}+\lambda^2y\Bigr)\tilde F_k(y)=0.\tag 6
$$
\noindent In fact, the function
$$
x\mapsto \Bigl(\frac 32 \lambda^{-1}\int_{x_{\text{left}}(E_k)}^x(E_k-V(t))
^{1/2}\,
dt\Bigr)^{2/3}\,\,\text{for}\ x \in [x_{\text{left}}(E_k),x_{\text{left}}(E_k)
+c]\tag 7
$$
\noindent extends to a smooth function $y_k(x)$ on $[x_{\text{left}}(E_k)-
c, x_{\text{left}}(E_k)+c]$ that transforms (5) approximately to (6).

Airey's equation (6) has a special solution $\tilde F_k(y)=\lambda^{-1/3}
A(\lambda^{2/3}y)$, where $A$ is the Airey function, defined by
$$
A(y)=(2\sqrt \pi)^{-1}\operatornamewithlimits{lim}_{\delta\to 0+}
\int_{-\infty}^\infty \exp\{-\delta\xi^2+i\xi y-
\frac i3\xi^3\}\,d\xi.\tag 8
$$
\noindent Pulling back $\tilde F_k(y)$ by the map $x\mapsto y_k(x)$, we
obtain a formula for $F_k(x)$ near $x_{\text{left}}(E)$, namely
$$
F_k(x)\approx \lambda^{-1/3}\Bigl(\frac{dy_k}{dx}\Bigr)^{-1/2}
A(\lambda^{2/3}y_k(x))\quad\text{for}\ |x-x_{\text{left}}(E_k)|<c.\tag 9
$$
\noindent The Airey function is well--understood, so (9) gives a good 
understanding of $F_k$ near $x_{\text{left}}(E_k)$.

Similarly,
$$
F_k(x)\approx \pm\lambda^{-1/3}\Bigl(\frac{-d\tilde y_k}{dx}\Bigr)^{-1/2}
A(\lambda^{2/3}\tilde y_k(x))\,\,\text{for}\ |x-x_{\text{right}}(E_k)|<c,
\tag 10
$$

\noindent where $\tilde y_k$ is a smooth change of variable analogous
to $y_k$.

Outside $[x_{\text{left}}(E_k),x_{\text{right}}(E_k)]$, $F_k(x)$ is
exponentially small unless $x$ is so near turning point that (9) or
(10) applies.  Hence we have an approximate description of $F_k(x)$
for all $x$.

WKB theory also asserts that the $L^2$--norm of the eigenfunction described
by (4), (9), (10) is given by
$$
\Vert F_k\Vert^2\approx \frac 12 \int_{\{V(x)<E_k\}} (E_k-V(x))^{-1/2}\,dx.
\tag 11
$$

A standard reference containing rigorous results to justify (2), (4),
(9), (10), (11) is Erdelyi [E].  For instance, one knows that
$\phi(E_k)=\pi(k+1/2)+O(\lambda^{-1})$ for eigenvalues $E_k$.  Equation (4)
holds modulo an error $O(\lambda^{-1}(E_k-V(x))^{-1/4})$, provided
$x \in [x_{\text{left}}(E_k),x_{\text{right}}(E_k)]$ stays bounded away
from the turning points.  As $x$ approaches $x_{\text{left}}(E_k)$, the
error degrades to $O(\lambda^{-1}(x-x_{\text{left}}(E_k))^{-3/2}
(E_k-V(x))^{-1/4})$, and similarly for $x_{\text{right}}(E_k)$.

The first main purpose of this article is to give more accurate
approximations than the above for the eigenvalues and eigenfunctions
of $\frac{-d^2}{dx^2}+V(x)$.  When $V(x)$ is given by (1), our results
are essentially as follows.
\smallskip

\noindent (A) The eigenvalues $E_k$ satisfy
$$\gather
\phi(E_k)+\frac{1}{48}\psi(E_k)=\pi(k+1/2)+O(\lambda^{-2+\varepsilon}),\quad
\text{with}\tag 12\\
\psi(E)=\operatornamewithlimits{lim}_{\delta\to 0+}
\Bigl[\int_{E-V(x)>\delta}V^{\prime\prime}
(x)(E-V(x))^{-3/2}\,dx-q(E)\delta^{-1/2}\Bigr]\quad \text{and}\tag 13\\
q(E)=\frac{2V^{\prime\prime}(x_{\text{right}}(E))}{V^\prime(x_{\text{right}}
(E))}-\frac{2V^{\prime\prime}(x_{\text{left}}(E))}{V^\prime(x_{\text{left}}
(E))}.\tag 14\endgather
$$

If $E_k>(\min V)+c\lambda^2$, then an eigenfunction $F_k(x)$ corresponding
to $E_k$ will satisfy the following.
\smallskip

\noindent (B) For $x \in [x_{\text{left}}(E_k),x_{\text{right}}(E_k)]$
not too near a turning point, we have
$$\gather
F_k(x)\approx (E_k-V(x))^{-1/4}\cos(-\frac \pi 4+\int_{x_{\text{left}}(E_k)}
^x(E_k-V(t))^{1/2}\,dt+\omega_k(x)),\quad\text{with}\tag 15\\
\omega_k(x)=\operatornamewithlimits{lim}_{\delta\to 0+}\Bigl[\int_
{x_{\text{left}}(E_k)+\delta}^x
\Bigl\{\frac{5}{32}\frac{(V^\prime)^2}{(E_k-V)^{5/2}}+\frac 18
\frac{V^{\prime\prime}}{(E_k-V)^{3/2}}\Bigr\}\,dt\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
-\sum\limits_{\ell=1}^3 q_\ell(E_k)\delta^{-\ell/2}\Bigr]\tag 16\endgather
$$

\noindent and $q_\ell(E_k)$ uniquely specified by demanding the finiteness of
the limit (16). ``Not too near a turning point" means that $|x-x_{\text{left}}
(E_k)|, |x-x_{\text{right}}(E_k)|>\lambda^{\varepsilon-2/3}$.  Equation (15)
holds modulo an error $O(\lambda^{-2}(E_k-V(x))^{-1/4})$, provided $x$
stays bounded away from the turning points.  When $x$ is close to
$x_{\text{left}}(E_k)$, the error degrades to $O(\lambda^{-2}(x-x_{\text{left}}
(E_k))^{-3}(E_k-V(x))^{-1/4})$, and similarly for $x_{\text{right}}(E_k)$.
\smallskip

\noindent (C) Near the turning points, we can describe $F_k$ with great
accuracy by formulas (9) and (10) for suitable smooth coordinate changes
$y_k(x)$, $\tilde y_k(x)$.  In fact,
$$
F_k(x)=\lambda^{-1/3}\Bigl(\frac{dy_k}{dx}\Bigr)^{-1/2}A(\lambda^{2/3}
y_k(x))+O(\lambda^{-N})\ \text{for}\ |x-x_{\text{left}}(E_k)|
<\lambda^{-\varepsilon}\tag 17
$$
\noindent and
$$
F_k(x)=\pm b_k\lambda^{-1/3}\Bigl(-\frac{d\tilde y_k}{dx}\Bigr)^{-1/2}
A(\lambda^{2/3}\tilde y_k(x))+O(\lambda^{-N})\ \text{for}\
|x-x_{\text{right}}(E_k)|<\lambda^{-\varepsilon},\tag 18
$$

\noindent with $\varepsilon>0$ arbitrarily small and $N>0$ arbitrarily
large, and with $b_k$ a real constant satisfying 
$|b_k-1|<C\lambda^{\varepsilon
-2}$.  However, the formula (7) for $y_k(x)$ and its analogue for
$\tilde y_k(x)$ have to be corrected to achieve (17), (18).  This was
known already to Cherry [C].

Away from the intervals where (B) or (C) holds, we know already from
standard WKB theory that $F_k(x)$ is exponentially small.
\smallskip

\noindent (D) The $L^2$--norm of $F_k$ is given by
$$
\Vert F_k\Vert^2=\frac 12 \int_{\{V(x)<E_k\}}(E_k-V(x))^{-1/2}\,dx\cdot
(1+O(\lambda^{\varepsilon-2})).\tag 19
$$

Thus we have improved on the accuracy of standard WKB theory in describing
the eigenvalues and eigenfunctions for potentials of the form (1).

The second main point of this article is to deal with potentials more general
than (1).  We need the added generality in order to understand atoms.  Already
from the Hydrogen atom, we get
$$
V(x)=\frac{\ell(\ell+1)}{x^2}-\frac 1x,\tag 20
$$
\noindent which is certainly not of the form (1).  Hughes [H] extended
the standard WKB theory to potentials relevant to atomic physics.  For
the potential (20), our results yield analogues of (A)$\ldots$(D) above,
with $\ell$ playing the r$\hat{\text{o}}$le of the large parameter $\lambda$.

To capture the behavior of a general potential $V$, we introduce positive
weight functions $S(x)$, $B(x)$, and assume the estimates
$$
\Big|\Bigl(\frac{d}{dx}\Bigr)^{\alpha}V(x)
\Big|\le C_{\alpha}S(x)(B(x))^{-\alpha},
\quad\text{on\ a\ suitable\ interval}\ I.\tag 21
$$

\noindent For instance, if $V(x)$ is given by (1), then (21) holds with
$S(x)=\lambda^2$, $B(x)=1$, $I=[-1,1]$, and our results apply to
eigenvalues $E$ for which $[x_{\text{left}}(E),x_{\text{right}}(E)]
\subset\subset I$.

For the potential (20) we can take $S(x)=\frac 1x$, $B(x)=x$, and (21)
holds on $I=(\frac {\ell(\ell+1)}{10},\infty)$.  
In (21), $S(x)$ controls the size
of $V$, while $B(x)$ controls the lengths of intervals over which $V$
is essentially constant.  Thus, (21) gives a precise meaning to the notion
of a large, slowly varying potential.  A closely related idea underlies the 
Beals--Fefferman pseudodifferential operator calculus [BF].

Our general WKB theorems apply for instance to potentials that satisfy (21) 
and have no
critical points on $I$ except for a single non--degenerate minimum.  More
precisely, for an $x_0 \in I$ we assume:
$$\gather
V^\prime(x_0)=0\tag 22\\
V^{\prime\prime}(x)>cS(x)B^{-2}(x)\quad\text{for}\quad |x-x_0|\le c_1B(x_0)
\tag 23\\
|V^\prime(x)|>cS(x)B^{-1}(x)\quad\text{for}\ x \in I\quad\text{with}\quad
|x-x_0|\ge c_1B(x_0).\tag 24\endgather
$$

\noindent In addition, we make various technical assumptions on $V(x)$,
$S(x)$, $B(x)$.  Under these conditions, we can prove analogues of
(A)$\ldots$(D) above.

We focus on the eigenvalues $E_k$ that are not too close to the minimum
of the potential $V$.  Our results are essentially as follows.
\smallskip

\noindent (A$^\prime$) If $E>V(x_0)+cS(x_0)$ is an eigenvalue of
$-\frac{d^2}{dx^2}+V(x)$, then
$$
\phi(E)+\frac{1}{48}\psi(E)=\pi(k+1/2)+O(\Lambda^{-2})\quad\text{for\
an\ integer}\ k,\tag 25
$$
\noindent with $\phi$ and $\psi$ defined by (3) and (13), and with
$$
\Lambda^{-1}=\int_{x_{\text{left}}(E)}^{x_{\text{right}}(E)}\frac
{dx}{S^{1/2}(x)B^2(x)}.\tag 26
$$

Moreover, an eigenfunction $F(x)$ corresponding to $E$ has the following
properties.
\smallskip

\noindent (B$^\prime$) For $x \in [x_{\text{left}}(E),x_{\text{right}}
(E)]$ not too near the turning points, we have
$$
F(x)\approx (E-V(x))^{-1/4}\cos(-\frac \pi 4+\int_{x_{\text{left}}(E)}
^{x}(E-V(t))^{1/2}\,dt+w(x))\tag 27
$$

\noindent with $w(x)$ as in (16).  ``Not too near the turning points"
means $|x-x_{\text{left}}(E)|>\lambda_{\text{left}}^{\varepsilon-2/3}
B(x_{\text{left}}(E))$ and $|x-x_{\text{right}}(E)|>\lambda_{\text{right}}^
{\varepsilon-2/3}B(x_{\text{right}}(E))$, with
$$
\lambda_{\text{left}}=S^{1/2}(x_{\text{left}}(E))B(x_{\text{left}}(E))
\quad\text{and}\quad \lambda_{\text{right}}=S^{1/2}(x_{\text{left}}(E))
B(x_{\text{left}}(E)).\tag 28
$$

\noindent Equation (27) holds modulo an error $O(\Lambda^{\varepsilon-2}
(E-V(x))^{-1/4})$, provided \hfill\break
$|x-x_{\text{left}}(E)|>cB(x_{\text{left}}(E))$
and $|x-x_{\text{right}}(E)|>cB(x_{\text{right}}(E))$.  When
$|x-x_{\text{left}}(E)|\le cB(x_{\text{left}}(E))$, the error is
degraded to $O(\Lambda^{\varepsilon-2}\cdot\bigl(\frac{x-x_{\text{left}}
(E)}{B(x_{\text{left}}(E))}\bigr)^{-3}\cdot(E-V(x))^{-1/4})$, and 
similarly for $x_{\text{right}}(E)$.
\smallskip

\noindent (C$^\prime$)  For $|x-x_{\text{left}}(E)|<\lambda_{\text{left}}
^{-\varepsilon}B(x_{\text{left}}(E))$ we have
$$
F(x)=\lambda_{\text{left}}^{-1/3}\Bigl(\frac{dY}{dx}\Bigr)^{-1/2}
A(\lambda_{\text{left}}^{2/3}Y)+O(\Lambda^{-N}B^{1/2}(x_{\text{left}}(E))),
\tag 29
$$

\noindent with $Y$ a smooth function of $\frac{x-x_{\text{left}}(E)}
{B(x_{\text{left}}(E))}$.

Similarly, for $|x-x_{\text{right}}(E)|<\lambda_{\text{right}}^{-\varepsilon}
B(x_{\text{right}}(E))$ we have
$$
F(x)=\pm b\lambda_{\text{right}}^{-1/3}\Bigl(\frac{-d\tilde Y}{dx}\Bigr)^{-1/2}
A(\lambda_{\text{right}}^{2/3}\tilde Y)+O(\Lambda^{-N}B^{1/2}
(x_{\text{right}}(E)))\tag 30
$$

\noindent with $\tilde Y$ a smooth function of $\frac {x-x_{\text{right}}(E)}
{B(x_{\text{right}}(E))}$, and with a real constant $b$ satisfying
$|b-1|<C\Lambda^{\varepsilon-2}$.

Outside the regions where (B$^\prime$), (C$^\prime$) hold, $F$ is
exponentially small.  Finally, the $L^2$--norm of $F$ is given by
\smallskip

\noindent  (D$^\prime$)\,\,
$\Vert F\Vert^2=\frac 12 \int_{V(x)<E}\,(E-V(x))^{-1/2}\,dx\cdot
(1+O(\Lambda^{\varepsilon-2}))$.

These are our basic results on the eigenvalues and eigenfunctions of
ordinary differential operators.  Their accuracy is determined by the
number $\Lambda$ in (26).  For $V(x)$ as in (1) one checks that
$\Lambda\sim \lambda$, while for $V(x)$ as in (20) one
gets $\Lambda\sim \ell$. In view of (A$^\prime$), (B$^\prime$), the
uncertainty in the phase at an eigenvalue and the percentage error
in $F(x)$ at a typical point of $[x_{\text{left}}(E),x_{\text{right}}(E)]$
are $O(\Lambda^{\varepsilon-2})$.  These errors are $O(\lambda^{-1})$ for the 
standard WKB approximation applied to potentials (1).  Thus (A$^\prime$)
$\ldots$(D$^\prime$) sharpen the standard approximations.  We will need
both the precision and generality of (A$^\prime$)$\ldots$(D$^\prime$)
in order to deal with atoms in our later papers.

We will investigate what happens to (A$^\prime$)$\ldots$(D$^\prime$) when
the eigenvalue $E$ gets close to the minimum of the potential.  As in 
standard WKB theory for potentials (1), we find that the accuracy of
(B$^\prime$)$\ldots$(D$^\prime$) is degraded for $E$ near $(\text{min}\ V)$,
but (A$^\prime$) retains its full accuracy.  Also, we will work out a very
crude WKB theory that applies, for instance, to the Hydrogen atom (20)
for small $\ell$.  The reader should consult the main body of this
paper for a careful statement of each of our WKB theorems.

Note that (8) differs slightly from the usual Airey function.  In our
normalization, $A(y)$ satisfies
$$\gather
\Bigl(\frac{d^2}{dy^2}+y\Bigr)A(y)=0\\
A(y)\sim\text{Re}\Bigl[e^{-\frac \pi 4 i}\,\frac{e^{\frac 23 iy^{3/2}}}
{y^{1/4}}(1+\sum\limits_{s=1}^\infty c_sy^{-\frac 32 s})\Bigr]\quad
\text{as}\ y \to +\infty\\
A(y)=O(|y|^{-M}), \text{any}\ M, \quad\text{as}\ y \to -\infty.  \endgather
$$
\noindent See [AS].

\noindent We are grateful to Maureen Kirkham for expertly texing our paper.
\hfill\break  Let us now begin the work of solving ordinary differential
equations.


%Section 1:


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\centerline{\bf {Approximate Local Solutions of Ordinary Differential
Equations}}
\smallskip

Let $p \in C^\infty[-1,1]$ with $p(0)=0$, $p^\prime(0) >0$.  Our goal in
this section is to write down in closed form a function $F(x,\lambda)$
defined for $|x| \le c$, $\lambda > C$ that solves the ordinary
differential equation
$$
\frac {\partial^2}{\partial x^2}\, F(x,\lambda)+\lambda^2p(x)F(x,\lambda)=0
\tag 1 $$

\noindent modulo a high power of $1/\lambda$.

Our approximate solution $F$ is patched together from three different
approximations $F_-(x,\lambda)$, $F_0(x,\lambda)$, $F_+(x,\lambda)$
defined on the three regions $\Cal J_-=\{x<-\lambda^{\varepsilon-2/3}\}$,
$\Cal J_0=\{|x|<\lambda^{-\varepsilon}\}$, $\Cal J_+=\{x >+
\lambda^{\varepsilon-2/3}\}$.  Here $\varepsilon > 0$ is a small constant
to be picked (much) later.

In $\Cal J_-$ we simply set $F_-(x,\lambda)=0$.

In $\Cal J_0$ we attempt to transform (1) into Airey's equation
$$
\frac {\partial^2}{\partial y^2}\,A(y,\lambda)+\lambda^2yA(y,\lambda)=0
\tag 2
$$

\noindent by a change of variable $y=y(x,\lambda)$.  The change of
variable makes (1) equivalent to (2) via the transformation law
$F(x,\lambda)=\Bigl(\frac{\partial y(x,\lambda)}{\partial x}\Bigr)^{-1/2}
A(y(x,\lambda),\lambda)$, provided $y$ satisfies the equation
$$
p(x)=\Bigl(\frac{\partial y}{\partial x}\Bigr)^2y+\lambda^{-2}
\{y,x\}\quad \text{with} \quad \{y,x\}={\tsize \frac 12} \frac {\partial_x^3y}
{\partial_xy}-{\tsize \frac 34} \Bigl(
\frac {\partial_x^2 y}{\partial_x y}\Bigr)^2.
\tag 3
 $$

\noindent We construct a formal power series $y$ in $x$ and $\lambda^{-2}$
that solves (3) to infinite order, and then truncate $y$ at degree $N$
to obtain a function $y_N(x,\lambda)$ that solves (3) modulo errors
$O(\lambda^{-\varepsilon N})$ in $\Cal J_0$.  As our solution of (2) we take
$A(y,\lambda)=\lambda^{-1/3}A(\lambda^{2/3}y)$, where $A(t)$ is the 
Airey function defined in the introduction.  Thus our approximate solution 
of (1) in $\Cal J_0$ is
$$
F_0(x,\lambda)=\lambda^{-1/3}\biggl(\frac{\partial y_N}{\partial x}(x,\lambda)
\biggr)^{-1/2}A(\lambda^{2/3}y_N(x,\lambda)), \quad |x|<\lambda^{-\varepsilon}.
\tag 4
$$

\noindent We call this the $\underline {N^{\text {th}}\ \text{Airey\ 
approximation}}$.  Note that $F_-$ and $F_0$ differ by
$O(\lambda^{-M})$ on $\Cal J_- \cap \Cal J_0$ for any $M$, since $A(t)$
decreases exponentially as $t \to -\infty$.

In $\Cal J_+=\{x > \lambda^{\varepsilon-2/3}\}$, our approximate solution of
(1) is
$$
F_+(x,\lambda)=\text{Re}\biggl[\frac {e^{\pm i\,\pi/4}e^{i\int
_0^x(\lambda^2p(t))^{1/2}dt}}{(\lambda^2p(x))^{1/4}}
\,\Bigl(\sum\limits_{k=0}^{N^\prime}u_k(x)
\lambda^{-k}\Bigr)\biggr], \tag 5
$$

\noindent where $u_0(x),u_1(x),\ldots,u_{N^\prime}(x)$ are defined for
$x \in (0,c]$ and satisfy the \underbar {transport}\\ \underbar {equations}:
$$\gather
u_0\equiv 1\tag 6\\
2iu_{k+1}^\prime+({\tsize\frac {5}{16}}(p^\prime)^2p^{-5/2}-{\tsize\frac 14}
p^{\prime\prime}
p^{-3/2})u_k-{\tsize\frac 12} p^\prime p^{-3/2}u_k^\prime+p^{-1/2}u_k^
{\prime\prime}
=0, \quad 0 \le k < N^\prime. \tag 6\endgather
$$

\noindent If we put (5) into (1) and expand formally in decreasing powers
of $\lambda$, we get (6), which is the classical motivation for the
transport equations.  We expect $u_k(x)$ to have a singularity like
$x^{-\frac 32 k}$ at $x=0$.  Hence the successive terms $u_k(x)\lambda^{-k}$
decrease rapidly, provided we take $x >\lambda^{\varepsilon-2/3}$.  This
is the classical motivation for using (5) only on $\Cal J_+$.

The transport equations (6) let us solve successively for each $u_{k+1}$
in terms of $u_k$.  Each time we do this, we get an arbitrary constant of
integration.  There is a unique way to pick all these constants of
integration in order to make (4) and (5) agree on the overlap $\Cal J_0 \cap
\Cal J_+$ modulo a high power of $1/\lambda$.  With the integration
constants picked in this way, we get the \underbar{canonical solution}
$(u_0(x),u_1(x),\ldots,u_{N^\prime}(x))$ to the transport equations.

Thus we produce three excellent approximate solutions of (1) on the
regions $\Cal J_-$, $\Cal J_0$, $\Cal J_+$, which agree closely on the overlaps.
A partition of unity lets us patch these solutions into a single approximate
solution $F(x,\lambda)$ defined in $|x|\le c$, $\lambda > C$.

It will be important to understand how $F_0(x,\lambda)$ and $F_+(x,\lambda)$
depend on $p$.  To keep track of this, we introduce the following
notation.  $\Cal P$ denotes the set of polynomials in the derivatives
$(p^{(m)}(0))_{m\ge 1}$ and in $p^\prime(0)^{-1/6}$.  
By $c_*$, $C_*$, $C^\prime_*$,
etc. with a subscript $*$, we denote positive constants that depend 
only on an upper bound for $\Vert p\Vert_{C^m[-1,1]}$ for some $m$,
and on a positive lower bound for $p^\prime(0)$.  Similarly,
$C_*^{\alpha\beta}$ denotes a constant depending only on $\alpha$, $\beta$,
upper bounds for some $\Vert p\Vert_{C^m[-1,1]}$, and a postive lower
bound for $p^\prime(0)$.

We now indicate in more detail how to carry out the program outlined above.
\vglue 1pc
\proclaim{LEMMA 1}  There is one and only one formal power series\,\,\,
$y(x,\lambda)=\hfil\break \sum\limits_{k,\ell\ge 0}f_{k\ell}x^\ell\lambda^{-2k}$
with $f_{00}=0$, $f_{01}>0$, and satisfying (3) as formal power series.
Moreover, each $f_{k\ell}$ belongs to $\Cal P$.  We have $f_{01}=
(p^\prime(0))^{1/3}$.
\endproclaim
\smallskip
\noindent\underbar{Sketch of proof}:  Set $y_k(x)=
\sum\limits_{\ell\ge 0}f_{k\ell}x^\ell$
so that $y(x,\lambda)=\sum\limits_{k\ge 0}y_k(x)\cdot\lambda^{-2k}$, and
define $z_k(x)$ by setting $(\partial_xy)^2y+\lambda^{-2}\{y,x\}-p(x)=
\sum\limits_{k\ge 0}z_k(x)\lambda^{-2k}$.  By induction on $k$ we show
that $y_k$ can be picked uniquely in terms of $p(x)$ and earlier $y_{k^\prime}$
to make $z_k=0$.  If $k=0$, this amounts to justifying the obvious solution
$y_0=(\frac 32\int_0^xp^{1/2}(t)dt)^{2/3}$ in terms of formal power
series and checking that $f_{0\ell}\in \Cal P$.

If $k>0$ then the contribution of $\lambda^{-2}\{y,x\}$ to $z_k$ is
determined by $y_0\ldots y_{k-1}$, and the contribution of
$(\partial_xy)^2y$ to $z_k$ has the form $2y_0(\frac{\partial y_0}
{\partial x})\,\frac {\partial y_k}{\partial x}+(\frac {\partial}
{\partial x}y_0)^2y_k+[\text{stuff determined by}\ y_0\ldots y_{k-1}]$.
Hence to make $z_k=0$ we must solve the regular singular equation
$2y_0(\frac{\partial y_0}{\partial x})\frac{\partial y_k}{\partial x}
+(\frac{\partial y_0}{\partial x})^2y_k=\text{given formal power series}$.

To solve this we take $y_k=\sum\limits_{\ell\ge0}f_{k\ell}x^\ell$ and
solve successively for $f_{k0}, f_{k1},\ldots$.  It is again easy to check
that each $f_{k\ell} \in \Cal P$. \qquad $\blacksquare$
\medskip
Once we have found $y(x,\lambda)$ by Lemma 1, the truncation $y_N(x,\lambda)$
is defined as
$$
y_N(x,\lambda)=\sum\limits_{0\le k,\ell\le N}f_{k\ell}x^\ell\lambda^{-2k},\
\text{with}\ N>>{\tsize\frac 1\varepsilon}\ \text{to be picked later}. \tag 7
$$

\noindent Thus
$$\split 
p(x)-\Bigl({\tsize\frac {\partial}{\partial x}} y_N(x,\lambda)\Bigr)^
2y_N(x,\lambda)-
\lambda^{-2}\{y_N(x,\lambda),x\}\,\,\, 
\text{vanishes to}\\ \text{order}\ N-3\  \text{at}\
x=0, \lambda^{-2}=0.\endsplit\tag 8$$ 
\vfill\eject

\proclaim{Lemma 2}  For $|x|$, $\lambda^{-1}\le c_*^N$ we have 
$\frac {\partial y_N}{\partial x}>c_*$, so the Airey approximation
(4) is well-defined.  Moreover, $|\frac {\partial^2}{\partial x^2}
F_0(x,\lambda)+\lambda^2p(x)F_0(x,\lambda)|\le C_*^N\,\lambda^{-\frac
{\varepsilon N}{2}}$ for $|x|<\lambda^{-\varepsilon}$ and $\lambda
\ge C_*^{\varepsilon N}$.  For $-\lambda^{-\varepsilon}<x<-\lambda^
{\varepsilon-2/3}$ and $\lambda\ge C_*^{\varepsilon N}$ we have
$|(\frac {\partial}{\partial x})^\alpha F_0(x,\lambda)|\le
c_*^{\varepsilon N}  \lambda^{-\frac{\varepsilon N}{2}}$ for $\alpha=0,1,2$.
\endproclaim
\medskip

\noindent\underbar{Sketch of proof}:  Airey's equation (2) holds
exactly for $A(y,\lambda)=\lambda^{-1/3}A(\lambda^{2/3}y)$, hence
$F_0(x,\lambda)$ satisfies $\frac{\partial^2F_0}{\partial x^2}+
\lambda^2\tilde p(x,\lambda)F_0=0$ with
$$
\tilde p(x,\lambda)=\Bigl(\frac{\partial y_N}{\partial x}\Bigr)^2y_N+
\lambda^{-2}\{y_N,x\}.
$$

\noindent Thus $\frac{\partial^2F_0}{\partial x^2}+\lambda^2p(x)F_0=
\lambda^2 (p(x)-\tilde p(x,\lambda))F_0$.

Regard $(x,\lambda^{-2})$ as the independent variables in $y_N$, $F_0$.  Then
$p(x)-\tilde p(x,\lambda)$ vanishes to order $N-3$ at the origin, and
$|\partial_x^\alpha\partial_{\lambda^{-2}}^\beta [p(x)-\tilde p
(x,\lambda)]|\le C_*^{\alpha\beta N}$  for $|x|$, $|\lambda^{-2}|\le c_*^N$.
Taylor's theorem with remainder yields $|p(x)-\tilde p(x,\lambda)|
\le C_*^N(|x|+
\lambda^{-2})^{N-3}$ for $|x|$, $\lambda^{-2}\le c_*^N$, hence
$$
|p(x)-\tilde p(x,\lambda)|\le C_*^N\lambda^{-\varepsilon(N-3)}\ \text{if}\quad 
|x|<\lambda^{-\varepsilon}\quad \text{and}\quad\lambda \ge C_*^N.
\tag "(8\text{bis})"
$$

\noindent Since the Airey function $A(t)$ is bounded, a look at (4)
gives $|F_0|\le C_*^N\lambda^{-1/3}$, so $|\lambda^2(p(x)-\tilde p
(x,\lambda))F_0| \le C_*^N\,\lambda^{2-\frac 13-\varepsilon(N-3)}$
for $|x|<\lambda^{-\varepsilon}$, $\lambda \ge C_*^N$.\qquad $\blacksquare$
\medskip
Thus, $F_0$ is an excellent approximate solution of the basic ODE (1)
in $\Cal J_0$, and $F_0$ agrees very well with $F_-\equiv 0$ on
$\Cal J_0 \cap \Cal J_-$.

We prepare to construct a solution (5) that agrees very well with $F_0$
in $\Cal J_0 \cap \Cal J_+$.  Clearly the first task is to understand
how $F_0$ behaves in $\Cal J_0 \cap \Cal J_+$.
\vglue 1pc
\proclaim{Lemma 3}  For $\lambda^{\varepsilon-2/3}<x<
\lambda^{-\varepsilon}$ and $\lambda>C_*^{\varepsilon N}$ we have\hfill\break
$F_0(x,\lambda)=\text{Re}\Bigl[\frac{e^{\pm i \pi/4}\,e^{i \int_0^x
(\lambda^2p(t))^{1/2}dt}}{(\lambda^2p(x))^{1/4}}\,\{1+\sum\limits_
{k=1}^N\lambda^{-k}w_k(x)\}\Bigr]+\ \operatornamewithlimits{Error}
(x,\lambda)\hfill\break
\text{where}\ |(\frac{\partial}{\partial x})^\alpha\ 
\operatornamewithlimits{Error}(x,\lambda)|
\le C_*^{\varepsilon N}\lambda^{-\frac{\varepsilon N}{10}}\quad (\alpha
=0,1,2)\
\text{and}\ w_k(x)=
\hfill\break\sum\limits_{-3k\le\ell\le M_k} w_{k\ell}x^{\ell/2}
,\, w_{k\ell} \in \Cal P$.   We have $w_{10}=0$.\endproclaim
\medskip
\noindent\underbar{Sketch of proof}:  We use the asymptotic form of the
Airey function:
$$
A(t)=\text{Re}\Bigl[\frac{e^{\pm i \pi/4}\,e^{\frac 23 it^{3/2}}}{t^{1/4}}
(1+\sum\limits_{s=1}^Mc_st^{-\frac 32 s})\Bigr]+\text{Error}(t)
$$

\noindent with $|\partial_t^\alpha\text{Error}(t)|\le Ct^{-M}$ for
$t>1$ and $\alpha=0,1,2$.  Put $t=\lambda^{2/3}y_N(x,\lambda)$ and
substitute the result in the definition (4) of $F_0$ to obtain the
following.
$$\align
\split
F_0(x,\lambda)&=\lambda^{-1/3}\Bigl(\frac{\partial}{\partial x} y_N(x,\lambda)
\Bigr)
^{-1/2}\Bigl(\lambda^{2/3}y_N(x,\lambda)\Bigr)^{-1/4}\text{Re}\Bigl[
e^{\pm i \pi/4}
e^{i\lambda\cdot\frac 23(y_N(x,\lambda))^{3/2}}\\
&\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad
\cdot\, \Bigl(1+\sum\limits_{s=1}^M c_s(\lambda^{2/3}
y_N(x,\lambda))^{-\frac 32 s}\Bigr)\Bigr]\\
&\qquad+\text{junk}_{10}(x,\lambda).\endsplit\tag 9\endalign
$$

Throughout the proof of Lemma 3, $\text{junk}_s(x,\lambda)$ denotes
a function $j(x,\lambda)$ which satisfies
$$
|\partial_x^\alpha j(x,\lambda)|\le C_*^{\varepsilon N}\,\lambda^{-\frac
{\varepsilon N}{s}}\quad \text{for}\quad \alpha=0,1,2\quad
\text{and}\quad \lambda^{\varepsilon-2/3}
<x<\lambda^{-\varepsilon}, \lambda\ge C_*^{\varepsilon N}.$$
(In particular $\text{Error}(\lambda^{2/3}y_N(x,\lambda))$ contributes
the term $\text{junk}_{10}(x,\lambda)$ in (9), provided we take $M$
large enough.)

To understand how $F_0$ behaves by using (9), we must understand the
three expressions
$$
\exp(\tfrac 23 i\lambda(y_N(x,\lambda))^{3/2}), ({\tsize
\frac {\partial}{\partial x}}y_N
(x,\lambda))^{-1/2}(y_N(x,\lambda))^{-1/4},\ \text{and}\ (y_N(x,\lambda))^
{-\frac 32\,s}.
$$

\noindent We begin with the exponential.

Recall that $y_N(x,\lambda)=\sum\limits_{k,\ell=0}^N 
f_{k\ell}x^\ell\lambda^{-2k}$
with $f_{k\ell} \in \Cal P$, and that $y_0(x)=\sum\limits_{\ell=0}^N
f_{0\ell}x^\ell$ satisfies $(y_0^\prime)^2y_0=p(x)$ to order $\ge N-1$.
It follows that
$$
\frac 23 y_0^{3/2}=\int_0^x(p(t))^{1/2}dt+\text{junk}_2(x,\lambda). \tag 10
$$

\noindent (An error of order $\ge N-1$ is small since we are looking at
the region $|x| \le \lambda^{-\varepsilon}$.)

Also,
$$
(y_N(x,\lambda))^{3/2}=(y_0(x))^{3/2}\cdot\biggl(1+\Bigl\{(\frac{\lambda^{-2}}
{x})(\frac {x}{y_0(x)})(\sum\limits_{k=1}^N\sum\limits_{\ell=0}^N
f_{k\ell}x^\ell\lambda ^{-2(k-1)})\Bigr\}\biggr)^{3/2} \tag 11
$$

\noindent Regarding the factors in curly brackets, we note the following:
$$
\frac {x}{y_0(x)}=(p^\prime{(0)})^{-1/3}\cdot\biggl(1+\Bigl[\sum\limits_
{\ell=1}^N (p^\prime{(0)})^{-1/3}f_{0\ell}x^\ell\Bigr]\biggr)^{-1}.
$$

\noindent Expanding the factor $(1+[\ldots])^{-1}$ in a high-order Taylor
series with remainder, we find that
$$
\frac{x}{y_0(x)}=\sum\limits_{\ell=0}^Nf_\ell^\#x^\ell+\text{junk}_2
(x,\lambda)\quad \text{with}\quad f_\ell^\# \in \Cal P. \tag 12
$$

\noindent A high enough power of $\frac {\lambda^{-2}}{x}$ has the form
$\text{junk}_2(x,\lambda)$, since we are working in the region
$\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$.  Hence substituting
(12) in (11) and then expanding $(1+\{\ldots\})^{3/2}$ in a high-order Taylor
series with remainder, we get
$$\split
(y_N(x,\lambda))^{3/2}=(y_0(x))^{3/2}\cdot\Bigl(1+\sum\limits_{k=1}^N
\lambda^{-2k}\Bigl\{\sum\limits_{\ell=-k}^Mf_{k\ell}^{\#\#}x^\ell\Bigr\}
+\text{junk}_2(x,\lambda)\Bigl)\\ \text{with}\,\, f_{k\ell}^{\#\#}
\in \Cal P.\endsplit
$$

\noindent Putting in (10), we conclude that
$$\multline
{\tsize
\frac 23} \lambda(y_N(x,\lambda))^{3/2}=\int_0^x(\lambda^2p(t))^{1/2}dt+
{\tsize
\frac 23} \lambda(y_0(x))^{3/2}\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^N
f_{k\ell}^{\#\#}x^\ell\lambda^{-2k}\\+\text{junk}_3(x,\lambda). \endmultline
\tag 13
$$

\noindent The proof of (12) shows also \footnotemark"$^{\dag}$"
\footnotetext"$^{\dag}$"{Note that
the factor $(p^\prime(0))^{-1/3}$ in the proof of (13) becomes now
$(p^\prime(0))^{1/2}$.  This still belongs to $\Cal P$, since we
carefully included $(p^\prime(0))^{-1/6}$ in the definition of $\Cal P$.}
that $(\frac{y_0(x)}{x})^{3/2}=\sum\limits_{\ell=0}^N
\tilde f_\ell^\#x^\ell+\text{junk}_2(x,\lambda)$ with $\tilde f_\ell^{\#} \in
\Cal P$.

\noindent Writing the factor $(y_0(x))^{3/2}$ in (13) as
$(\frac{y_0(x)}{x})^{3/2}\cdot x^{3/2}$ and putting in our result for
$(\frac {y_0(x)}{x})^{3/2}$, we obtain from (13) the following:
$$\multline
{\tsize
\frac 23} i\lambda(y_N(x,\lambda))^{3/2}=i\int_0^x(\lambda^2p(t))^{1/2}dt+
\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^Nh_{k\ell}x^{(\ell+\frac 32)}
\lambda^{-(2k-1)}+\hfill\break\\
+\ \text{junk}_4(x,\lambda),\,\, h_{k\ell} \in \Cal P. \endmultline\tag 14
$$

\noindent We prepare to exponentiate both sides of (14). First of all,
$\exp(\text{junk}_4(x,\lambda))=1+\text{junk}_4(x,\lambda)$.  Secondly,
set $X=\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^Nh_{k\ell}
x^{(\ell+3/2)}\lambda^{-(2k-1)}$.  A high enough power of $X$ will be
of the form $\text{junk}_4(x,\lambda)$, so we can evaluate $\exp(X)$ using
a high-order Taylor series with remainder.  If $s\ge 2$ then $X^s$ contains
no terms of the form $x^a\lambda^{-1}$; while for $s=1$, $X^s=X$ contains 
terms of the form $x^a\lambda^{-1}$ only for non-integer $a$.  Consequently,
$\sum\limits_{s=1}^M\frac {X^s}{s!}$ contains no terms of the form
$x^a\lambda^{-1}$ with $a$ an integer.  Moreover, a monomial
$x^{(\ell+3/2)}\lambda^{-(2k-1)}$ with $1 \le k\le N$, $-\ell\le k\le N$
(i.e. the sort that appears in $X$) has the form
$$
x^{\ell^\prime/2}\lambda^{-k^\prime} \ \text{with}\quad 1 \le k^\prime\le 2N,
\,\, 2-k^\prime\le \ell^\prime\le 2N+3.
$$

\noindent Hence $\sum\limits_{s=1}^M \frac {X^s}{s!}=\sum\limits_
{k^\prime=1}^{\overline M} \sum\limits_{\ell^\prime=2-k^\prime}^{\overline M}
h_{k^\prime\ell^\prime}^{\#}x^{\ell^\prime/2}\lambda^{-k^\prime}, h_{k^\prime
\ell^\prime}^{\#} \in \Cal P$, for some large $\overline M$.  
\smallskip

Putting together the conclusions of the preceding paragraph, we obtain
$$\multline
\exp\Bigl(\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^{N}h_{k\ell}x^{(\ell+
\frac 32)}
\lambda^{-(2k-1)}\Bigr)=
\Bigl(1+\sum\limits_{k=1}^{2N}\sum\limits_{\ell=2-k}^{3N}
h_{k\ell}^{\#}x^{\ell/2}\lambda^{-k}\\
+\text{junk}_4(x,\lambda)\Bigr)\endmultline
$$
\noindent with $h_{k\ell}^{\#} \in \Cal P$, and with $h_{1\ell}^{\#}=0$
for $\ell$ even.

Substituting this into (14) gives our basic result on the exponential term,
namely
$$\multline
\exp(
{\tsize\frac 23} i\lambda(y_N(x,\lambda))^{3/2})=\exp(i\int_0^
x(\lambda^2p(t))^{1/2}dt)\cdot\Bigl[1+\sum\limits_{k=1}^{2N}
\sum\limits_{\ell=2-k}^{3N} h_{k\ell}^{\#}
x^{\ell/2}\lambda^{-k}\\
+\text{junk}_4(x,\lambda)\Bigr] \endmultline\tag 15
$$
\noindent with $h_{k\ell}^{\#} \in \Cal P$ and $h_{k\ell}^{\#}=0$ for $k=1$,
$\ell$ even.

Next we study the expression $(\frac{\partial y_N}{\partial x} (x,\lambda)
)^{-\frac 12}(y_N(x,\lambda))^{-\frac 14}$.


Since $(y_0^\prime)^2y_0=p(x)$ to order $\ge N-1$ and both sides vanish to
first order at $x=0$, we have $(y_0^\prime)^2y_0=p(x)\cdot(1+(x^{N-2}g(x))$
with $\,|\partial_x^\alpha g(x)|\le C_*^N$, $\alpha=0,1,2$ for $|x|\le c_*^N$.
We know also that $(\frac {\partial}{\partial x}y_N)^2y_N=(y_0^\prime)^2y_0+
\sum\limits_{k=1}^{3N}\sum\limits_{\ell=0}^{3N}q_{k\ell}x^\ell\lambda^{-2k}$
with $q_{k\ell} \in \Cal P$, since $y_N(x,\lambda)=\sum\limits_{k,\ell=0}^N
f_{k\ell}x^\ell\lambda^{-2k}$.  Thus
$$\align
\bigl({\tsize
\frac {\partial}{\partial x}} y_N\bigr)^2y_N&=(1+x^{N-2}g(x))p(x)+
\lambda^{-2}
\sum\limits_{k=1}^{3N}
\sum\limits_{\ell=0}^{3N}q_{k\ell}x^\ell\lambda^{-2(k-1)}\\
&=p(x)\cdot\Bigl(1+\Bigl[x^{N-2}g(x)+(\frac {\lambda^{-2}}{x})(\frac {x}
{p(x)})(\sum\limits_{k=1}^{3N}\sum\limits_{\ell=0}^{3N}q_{k\ell}x^\ell
\lambda^{-2(k-1)})\Bigr]\Bigr).\tag 16\endalign
$$
\noindent As a function of $(x,\lambda)$, we have $x^{N-2}g(x)=\text{junk}_2
(x,\lambda)$. (Recall we're  working in the region
$|x|\le \lambda^{-\varepsilon}$).  Also $\frac {x}{p(x)}=(p^\prime(0))^{-1}
\cdot\Bigl(1+\Bigl[\sum\limits_{\ell=2}^{M}\frac{p^{(\ell)}(0)}{\ell!
p^\prime(0)}x^{\ell-1}+\text{junk}_1(x,\lambda)\Bigr]\Bigr)^{-1}$ for large $M$,
so that by expanding $(1+X)^{-1}$ in a high-order Taylor series with remainder,
we get
$$
\frac {x}{p(x)}=\sum\limits_{\ell=0}^{N}\tilde p_\ell x^\ell+\text{junk}_2
(x,\lambda) \quad \text{with}\quad \tilde p_\ell \in \Cal P, \quad
\tilde p_0=(p^\prime(0))^{-1}. \tag"{(16\text{bis})}"
$$
\noindent Hence (16) yields
$$
(\frac {\partial}{\partial x}y_N)^2y_N=p(x)\cdot(1+\Bigl[\frac {\lambda^{-2}}
{x}\sum\limits_{k=0}^{4N}\sum\limits_{\ell=0}^{4N}\tilde q_{k\ell}x^\ell
\lambda^{-2k}+\text{junk}_3(x,\lambda)\Bigr]\Bigr) \quad \text{with}\quad
\tilde q_{k\ell} \in \Cal P. \tag 17
$$

We prepare to raise both sides to the power $-\frac 14$.  Since a high
enough power of 
$\frac {\lambda^{-2}}{x}$ has the form $\text{junk}_1(x,\lambda)$
(recall we're working in the region
$\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$), we may 
expand $(1+X)^{-1/4}$ in a high-order Taylor series with remainder, taking
$X=$ the expression in square brackets in (17).  The remainder will have
the form $\text{junk}_3(x,\lambda)$.  Therefore, from (17) we get
$$\multline
\Bigl(\frac{\partial y_N(x,\lambda)}{\partial x}\Bigr)^{-\frac 12}
\bigl(y_N(x,\lambda)\bigr)^{-1/4}=(p(x))^{-1/4}\cdot\Bigl(1+\sum\limits_{k=1}
^{N}\sum\limits_{\ell=-k}^{N} q_{k\ell}^{\#} x^\ell\lambda^{-2k}
+\text{junk}_4(x,\lambda)\Bigr)\\ \quad\text{with}\quad q_{k\ell}^{\#}
\in \Cal P.\endmultline\tag 18
$$
\noindent This is our basic result on $(\frac{\partial}{\partial x}y_N)^{-1/4}
y_N^{-1/2}$.

We turn now to $(
y_N(x,\lambda))^{-\frac 32 s}$.

>From (17) we conclude that
$$\multline
\Bigl(\frac{\partial y_N(x,\lambda)}{\partial x}\Bigr)^{-3s}
\Bigl(y_N(x,\lambda)\Bigr)^{-\frac 32 s}=\Bigl(p(x)\Bigr)^{-\frac 32 s}\cdot
\Bigl(1+\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^Nq_{k\ell}^sx^\ell
\lambda^{-2k}+\text{junk}_4(x,\lambda)\Bigr)\\
\text{with}\ q_{k\ell}^s \in \Cal P.\endmultline \tag 19
$$
\noindent This follows the same way as (18), except that we raise (17)
to the power $-\frac 32 s$ instead of $-\frac 14$.

Also
$$
\lambda^{-s}\Bigl(p(x)\Bigr)^{-\frac 32 s}=\lambda^{-s}x^{-\frac 32 s}
\Bigl(\frac {x}{p(x)}\Bigr)^{\frac 32 s}.
$$
\noindent Equation (16bis) shows that
$$
\Bigl(\frac{x}{p(x)}\Bigr)^{\frac 32 s}=\sum\limits_{\ell=0}^{N}p_\ell^s
x^\ell+\text{junk}_3(x,\lambda)\quad\text{with}\quad p_\ell^s \in \Cal P.
$$
\noindent (To derive this we use a high-order Taylor expansion with
remainder for $(1+X)^{\frac 32 s}$.)
\noindent Hence
$$
\lambda^{-s}\Bigl(p(x)\Bigr)^{-\frac 32 s}=\lambda^{-s}x^{-\frac 32 s}
\Bigl(\sum\limits_{k=0}^N\sum\limits_{\ell=-k}^N\hat q_{k\ell}^sx^\ell
\lambda^{-2k}+\text{junk}_4(x,\lambda)\Bigr)\quad\text{with}\quad
\hat q_{k\ell}^s\in \Cal P. \tag 20
$$
\noindent Since $y_N(x,\lambda)=\sum\limits_{k,\ell=0}^Nf_{k\ell}
x^\ell\lambda^{-2k}$ we have also
$$
\Bigl(\frac{\partial y_N}{\partial x}\Bigr)^{+3s}=\sum\limits_{k,\ell=0}
^{3N}\hat f_{k\ell}^sx^\ell\lambda^{-2k}\quad\text{exactly,}\quad
\hat f_{k\ell}^s \in \Cal P. \tag 21
$$
\noindent Rewriting (19) as
$$\multline
\lambda^{-s}\bigl(y_N(x,\lambda)\bigr)^{-\frac 32 s}\negthinspace
=\negthinspace\Bigl(\frac{\partial y_N}
{\partial x}\Bigr)^{+3s}\cdot\lambda^{-s}\bigl(p(x)\bigr)^{-\frac 32 s}\cdot
\Bigl(1+\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^{N}q_{k\ell}^sx^\ell
\lambda^{-2k}\\
+\text{junk}_4(x,\lambda)\Bigr)\endmultline
$$
\noindent and substituting (20), (21) into the right-hand side, we obtain
$$
\lambda^{-s}(y_N(x,\lambda))^{-\frac 32 s}=\lambda
^{-s}x^{-\frac 32 s}
\cdot\Bigl(\sum\limits_{k=0}^N\sum\limits_{\ell=-k}^Nh_{k\ell}^sx^\ell
\lambda^{-2k}+\text{junk}_5(x,\lambda)\Bigr)\quad\text{with}\quad
h_{k\ell}^s\in \Cal P.
$$
\noindent This is our basic result on $(y_N)^{-\frac 32 s}$.  It gives
at once
$$
\Bigl(1+\sum\limits_{s=1}^Mc_s(\lambda^{2/3}y_N(x,\lambda))^{-\frac 32s}\Bigr)=
\Bigl(1+\sum\limits_{k=1}^N\sum\limits_{\ell=-3k}^N\hat h_{k\ell}
x^{\ell/2}\lambda^{-k}+\text{junk}_6(x,\lambda)\Bigr) \tag 22
$$
\noindent with $\hat h_{k\ell} \in \Cal P$ and $\hat h_{1\ell}=0$ for $\ell$
even.

>From (15), (18), (22) we get
$$\multline
\frac{e^{\pm i \pi/4}\,e^{\frac 23 i\lambda(y_N(x,\lambda))^{\frac 32}}}
{\lambda^{1/2}(\frac{\partial y_N}{\partial x})^{1/2}y_N^{1/4}}\Bigl(
1+\sum\limits_{s=1}^Mc_s(\lambda^{2/3}y_N(x,\lambda))^{-\frac 32s}\Bigr)=\\
{}\\
\frac{e^{\pm i \pi/4}\,e^{i \int_0^x(\lambda^2p(t))^{1/2}dt}}
{(\lambda^2p(x))^{1/4}}\cdot
\Biggl[\Bigl(1+\sum\limits_{k=1}^{2N}\sum\limits_{\ell=2-k}^{3N}h_{k\ell}^{\#}
x^{\ell/2}\lambda^{-k}+\text{junk}_4(x,\lambda)\Bigr)\cdot\\
\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,\,
\Bigl(1+\sum\limits_{k=1}^N\sum\limits_{\ell=-k}^Nq_{k\ell}^{\#}
x^\ell\lambda^{-2k}+\text{junk}_4(x,\lambda)\Bigr)\cdot\\
\Bigl(1+\sum\limits_{k=1}^N\sum\limits_{\ell=-3k}^N\hat h_{k\ell}
x^{\ell/2}\lambda^{-k}+\text{junk}_6(x,\lambda)
\Bigr)\Biggr].\endmultline\tag 23
$$
\noindent The product of the three series has the form $\bigl(1+
\sum\limits_{k=1}^{10N}\sum\limits_{\ell=-3k}^{10N}h_{k\ell}^+
x^{\ell/2}\lambda^{-k}+\text{junk}_7(x,\lambda)\bigr)$, with
$h_{k\ell}^+ \in \Cal P$.  Since $h_{1\ell}^{\#}=0$ and $\hat h_{1\ell}=0$
for even $\ell$, it follows at once that $h_{1\ell}^+=0$ for even $\ell$.
In particular $h_{10}^+=0$.  Now the conclusion of Lemma 3 is immediate
from (9) and (23).\quad\quad$\blacksquare$
\vglue 1pc
\proclaim{Lemma 4}  The functions $w_k(x)$ in the statement of the previous
lemma satisfy the approximate transport equations
$$
|2iw_{k+1}^\prime(x)+\Cal Lw_k(x)|\le C_*^{\varepsilon N}x^{N/32}\quad
\text{for}\quad 0<x<c_*^{\varepsilon N}\quad\text{and}\quad
0\le k\le {\tsize\frac{\varepsilon N}{32}},
$$
\noindent where we set $w_0(x)\equiv 1$ and
$$
\Cal Lw\equiv ({\tsize
\frac {5}{16}}(p^\prime)^2p^{-5/2}-{\tsize\frac 14}p^{\prime\prime}
p^{-3/2})w-{\tsize\frac 12} p^\prime p^{-3/2}w^\prime+p^{-1/2}w^{\prime\prime}.
$$  
\endproclaim
\medskip
\demo{Sketch of Proof}  We start with a seeming digression. 
The Airey function $A(t)$ may be written as the real part of a complex
solution $A_c(t)$ of Airey's equation $\frac {d^2A_c(t)}{dt^2}+tA_c(t)=0$,
having the asymptotic expansion
$$
A_c(t)=\frac{e^{\pm i \pi/4}\,e^{\frac 23 it^{3/2}}}{t^{1/4}}
\Bigl(1+\sum\limits_{s=1}^Mc_st^{-\frac 32 s}\Bigr)+ \ \text{Error}_M(t)
$$
\noindent with
$$
\Big|(\frac {d}{dt})^\alpha\text{Error}_M(t)\Big|\le C_M\,t^{-M}\quad
\text{for}\quad \alpha=0,1,2\quad \text{and}\quad t>10.
$$
\noindent Whereas $A(t)$ is bounded on the whole real line $A_c(t)$ is bounded
on $[-1,\infty)$ but grows rapidly as $t \to -\infty$.  Using $A_c(t)$
in place of $A(t)$, we define a complex analogue of $F_0(x,\lambda)$,
namely
$$
F_c(x,\lambda)=\lambda^{-1/3}\Bigl(\frac{\partial y_N(x,\lambda)}{\partial x}
\Bigr)^{-1/2}A_c(\lambda^{2/3}y_N(x,\lambda)).
$$
\noindent Thus $F_0(x,\lambda)=\text{Re}[F_c(x,\lambda)]$.  Note that
$A_c(\lambda^{2/3}y_N(x,\lambda))$ remains bounded by $C_*^{\varepsilon N}$
when $0<x<c_*^{\varepsilon N}$ and $\lambda>C_*^{\varepsilon N}$.  To see
this, we write $y_N(x,\lambda)=y_0(x)+O(\lambda^{-2})$ with $y_0(x)>0$
for $0<x<c_*^{\varepsilon N}$.  Hence $\lambda^{2/3}y_N(x,\lambda)\ge -1$
and $A_c(\lambda^{2/3}y_N(x,\lambda))$ remains bounded as claimed.

Now we can repeat the proofs of Lemmas 2 and 3 above to derive their
complex analogues:
$$
\Big|{\tsize
\frac{\partial^2}{\partial x^2}}F_c(x,\lambda)+\lambda^2p(x)F_c(x,\lambda)
\Big|\le
C_*^{\varepsilon N}\lambda^{-\frac{\varepsilon N}{2}}\quad\text{for}
\quad 0<x<\lambda^{-\varepsilon}, \lambda\ge C_*^ {\varepsilon N}.\tag 24
$$
$$\split
F_c(x,\lambda)=\frac{e^{\pm i\pi/4}\,e^{i\int_0^x(\lambda^2p(t))^{1/2}dt}}
{(\lambda^2p(x))^{1/4}}\Bigl(1+\sum\limits_{k=1}^{N}\lambda^{-k}w_k(x)\Bigr)
+\text{Error}(x,\lambda)\quad\text{with}\\
\big|\partial_x^\alpha\text{Error}(x,\lambda)\big|\le C_*^{\varepsilon N}
\lambda^{-\frac {\varepsilon N}{10}}\quad \text{for}\,\,
\lambda^{\varepsilon -2/3}<x<\lambda^{-\varepsilon}, \,\,
\lambda\ge C_*^{\varepsilon
N}, \,\,\alpha=0,1,2.\endsplit\tag 25
$$

The proof of Lemma 2 used the boundedness of $A(\lambda^{2/3}y_N(x,\lambda))$,
which we have just verified for $A_c$ when $x>0$.  Otherwise, we can
repeat the proofs of Lemmas 2 and 3 virtually word for word.

Now we return to the proof of Lemma 4.  Setting $w_0(x)\equiv 1$ and
$w_{N+1}\equiv 0$ we obtain from (25) by elementary calculation the
formula:
$$\multline
\Bigl({\tsize
\frac{\partial^2}{\partial x^2}}+\lambda^2p(x)\Bigr)F_c(x,\lambda)=
\frac{e^{\pm i\pi/4}\,e^{i \int_0^x(\lambda^2p(t))^{1/2}dt}}
{(\lambda^2p(x))^{1/4}}\sum\limits_{k=0}^N\lambda^{-k}\{2iw_{k+1}^\prime
(x)+\Cal Lw_k(x)\}\\
+\Bigl(
{\tsize\frac{\partial^2}{\partial x^2}}
+\lambda^2p(x)\Bigr)\text{Error}(x,\lambda),
\endmultline$$
\noindent for $\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon},
\lambda\ge C_*^{\varepsilon N}$.

\noindent The last expression on the right-hand side is dominated by
$C_*^{\varepsilon N}\lambda^{-\frac {\varepsilon N}{2}}$, by virtue of (24).
Therefore,
$$
\Big|\sum\limits_{k=0}^N\lambda^{-k}\{2i\,w_{k+1}^\prime(x)+\Cal L
w_k(x)\}\Big|\le C_*^{\varepsilon N}\lambda^{-\frac{\varepsilon N}{15}}
(\lambda^2p(x))^{1/4}\le C_*^{\varepsilon N}\lambda^{-
\frac{\varepsilon N}{16}}\tag 26
$$

\noindent for $\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$,
$\lambda\ge C_*^{\varepsilon N}$.

\noindent Had we not taken the trouble to remove the real part from the
statement of Lemma 3 by passing to $F_c(x,\lambda)$, the derivation of
(26) would have been more difficult.  For $0<x<c_*^{\varepsilon N}$
and $\tilde\lambda\in [\frac 12,1]$, set $\lambda=\tilde\lambda x^
{-\frac 1\varepsilon}$.  Thus $\lambda^{\varepsilon-2/3}<x<
\lambda^{-\varepsilon}$ and $\lambda\ge C_*^{\varepsilon N}$, so (26) 
holds.  Since $\lambda=\tilde\lambda x^{-\frac 1\varepsilon}$ with
$\tilde\lambda\in [\frac 12,1]$, (26) becomes
$$
\Big|\sum\limits_{k=0}^N\tilde\lambda^{-k}
\Bigl[x^{k/\varepsilon} \{2iw_{k+1}^\prime(x)+\Cal Lw_k(x)\}\Bigr]\Big|
\le C_*^{\varepsilon N} x^{\frac {N}{16}}. \tag 27
$$
\noindent Fix $x \in (0,c_*^{\varepsilon N}]$.  Since (27) holds for all
$\tilde\lambda \in [\frac 12,1]$, it follows that
$|x^{k/\varepsilon}\{2iw_{k+1}^\prime(x)+\Cal Lw_k(x)\}|\le C_*^{\varepsilon N}
x^{\frac {N}{16}}$ for $0\le k\le N$.  The conclusion of Lemma 4 is now
obvious.$\quad\blacksquare$\medskip

Clearly Lemmas 3 and 4 put us in position to pick an exact solution\hfill\break
$(u_0(x),u_1(x),\ldots,u_N^\prime(x))$ of the transport equations to match
$F_0(x,\lambda)$ with\hfill\break $F_+(x,\lambda)$ defined by (5).  Put
$N^\prime=[N/100]$.
\vglue 1pc
\proclaim{Lemma 5}  There is a 
solution $(u_0(x),u_1(x),\ldots,u_{N^\prime}(x))$
of the transport equations, defined for $x \in (0,c_*^{\varepsilon N})$,
with 
$$
\Big|\Bigl({\tsize
\frac {d}{dx}}\Bigr)^\alpha\{u_k(x)-w_k(x)\}\Big|
\le C_*^{\varepsilon N}x^{N/100}\quad\text {for}\quad 0\le \alpha\le N/100,
\,\,x \in (0,c_*^{\varepsilon N}].$$
\medskip
With $F_+(x,\lambda)$ defined by (5) in terms of this solution, we have
$$\multline
\Big|\Bigl({\tsize
\frac{d}{dx}}\Bigr)^\alpha\{F_0(x,\lambda)-F_+(x,\lambda)\}\Big|\le
C_*^{\varepsilon N}\lambda^{-\varepsilon N/200} \quad\text{for}\quad
\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon},\\ 
\lambda\ge C_*^{\varepsilon
N},\,\, \alpha=0,1,2.\endmultline
$$\endproclaim
\medskip
\demo{Proof}  With $\Cal L$ as in Lemma 4, we start by studying 
$v_k=2i\,w_{k+1}^\prime+\Cal Lw_k$.\enddemo
\smallskip

For $M$ as large as we please, we have
$$
p(x)=p^\prime(0)x\cdot \Bigl(1+\sum\limits_{\ell=2}^M \frac{p^{(\ell)}(0)}
{\ell!p^\prime(0)} x^{\ell-1}+x^{M+1}g_M(x)\Bigr)
$$
\noindent with $|\partial_x^\alpha g_M(x)|\le C_*^{M\alpha}$
for $|x|\le 1$.  We can raise this equation to the powers $-5/2$,
$-3/2$, $-1/2$ using a high-order Taylor series with remainder for
$(1+X)^{-m/2}$.  As a result, the coefficients of $\Cal L$ all have the
form $x^{-5/2}(\sum\limits_{\ell=0}^M e_\ell x^\ell+x^{M+1}e_M^{\#}
(x))$ with $e_\ell\in \Cal P$ and $|\partial_x^\alpha e_M^{\#}(x)|\le C_*^
{M\alpha}$ for $|x|\le c_*^M$, all $\alpha$.  The functions $w_k(x)$, to which
we apply $\Cal L$, have the form $\sum\limits_{\ell=-3k}^Nw_{k\ell}
x^{\ell/2}$ with $w_{k\ell} \in \Cal P$.  Therefore,
$$
v_k(x)=\sum\limits_{\ell=-3k-100}^{N+10M}v_{k\ell}x^{\ell/2}+x^{M-3k-100}
\Bigl[g_{k1}^M(x)x^{1/2}+g_{k0}^M(x)\Bigr] \tag 28
$$

\noindent with $v_{k\ell} \in \Cal P$ and $|\partial_x^\alpha g_{k0}^M|$,
$|\partial_x^\alpha g_{k1}^M|\le C_*^{M\alpha k}$ for $|x| \le c_*^M$.

\noindent Lemma 4 shows that $v_{k\ell}=0$ for $k\le \varepsilon N/32$,
$\ell<N/32$ (provided we take $M=1000 N$, which
we now do.)  Hence (28) shows that $v_k(x)$ may be written as
$$
v_k(x)=x^{\overline N}f_k(x^{1/2})\,\,\,\text{with}\,\,\, \overline N=
{\tsize
[\frac {N}{33}}]\,\,\,\text{and}\,\,\, |\partial_y^\alpha f_k(y)|\le
C_*^{\alpha N}\,\,\,\text{for}\,\,\, |y|\le c_*^N,
\,\,\,\text{all}\,\,\,\alpha,\tag"(28${\text{bis}})$"
$$
\noindent for $0\le k\le \varepsilon N/32$.

Now we define $u_k(x)=w_k(x)+U_k(x)$ with $U_k$ picked to make $u_k$ satisfy
the transport equations.  That is, $U_0\equiv 0$, and
$$
2iU_{k+1}^\prime=-\Cal LU_k+v_k. \tag 29
$$

By induction on $k$ we check that $U_0,U_1,\ldots,U_k$ can be picked to
satisfy (29), with $U_k$ of the form
$$
U_k(x)=x^{\overline N-5k}V_k(x^{1/2}) \quad\text{with}\quad |\partial_y^\alpha
V_k(y)|\le C_*^{\alpha kN}\quad\text{for}\quad |y|\le c_*^N,\,\,\,\text{all}
\,\,\, \alpha.
$$
\noindent In fact, for $k=0$ it's clear.  To pass from $k$ to $k+1$, we use
inductive hypothesis and (28${\text{bis}}$) to write (29) as
$$
2i{\tsize
\frac {d}{dx}} U_{k+1}(x)=-\Cal L(x^{\overline N-5k}V_k(x^{1/2}))+
x^{\overline N}f_k(x^{1/2}).\tag 30
$$
\noindent The right-hand side may be written as $x^{\overline N-5(k+1)}h_k
(x^{1/2})$ with $|\partial_y^\alpha h_k(y)|\le C_*^{\alpha kN}$ for
$|y|\le c_*^N$, all $\alpha$.  Putting $U_{k+1}=x^{\overline N-5(k+1)}
V_{k+1}(x^{1/2})$ for unknown $V_{k+1}$, we may therefore rewrite (30) in
terms of $y=x^{1/2}$ as
$$
y^{-1}{\tsize\frac {d}{dy}} \Bigl(y^{2\overline N-10(k+1)}V_{k+1}(y)\Bigr)=
(\text{const.})y^{2\overline N-10(k+1)}h_k(y)
$$

\noindent This has a solution $V_{k+1}$ with $|\partial_y^\alpha V_{k+1}(y)|
\le C_*^{\alpha kN}$, completing the inductive transition.

Our solution $(u_0(x),u_1(x),\ldots u_{N^\prime}(x))$ of the transport
equations has $u_k(x)-w_k(x)=x^{\overline N-5k}V_k(x^{1/2})$ with
$|\partial_y^\alpha V_k(y)|\le C_*^{\alpha kN}$ for $|y|\le c_*^N$,
all $\alpha$.  Therefore by induction on $\alpha=0,1,\ldots,\overline N-5k$
we have
$$\multline
\partial_x^\alpha\{u_k(x)-w_k(x)\}=x^{\overline N-5k-\alpha}V_{k\alpha}
(x^{1/2}) \quad\text{with}\quad |\partial_y^\beta V_{k\alpha}(y)|\le
C_*^{\alpha\beta kN}\\
\text{for}\quad |y|\le c_*^N, \quad\text{all}\quad\beta.\endmultline
$$
\noindent In particular 
$|\partial_x^\alpha\{u_k(x)-w_k(x)\}|\le C_*^Nx^{N/100}$
for $\alpha \le N/100$, $k \le N/100$, which is one of the conclusions of
the Lemma.

By the defining equation (5) for $F_+(x,\lambda)$, and by Lemma 3 for
$F_0(x,\lambda)$ we have
$$\multline
F_0(x,\lambda)-F_+(x,\lambda)=\text{Re}\Big[\frac{e^{\pm i\pi/4}\,
e^{i \int_0^x(\lambda^2p(t))^{1/2}dt}}{(\lambda^2p(x))^{1/4}}
\Bigl(\sum\limits_{k=1}^{N^\prime}\lambda^{-k}(w_k(x)-u_k(x))\\
+\sum\limits_{k=N^\prime+1}^N\lambda^{-k}w_k(x)\Bigr)\Big]
+\text{Error}(x,\lambda)\endmultline
$$

\noindent for $\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$,
$\lambda\ge C_*^{\varepsilon N}$.

\noindent  Hence for $\alpha=0,1,2$ we get
$$\multline
|\partial_x^\alpha\{F(x,\lambda)-F_+(x,\lambda)\}|\le \frac{C_*^{\varepsilon N}
\lambda^{10}}{(p(x))^{10}}\Bigl\{\sum\limits_{k=1}^{N^\prime}\sum\limits
_{\beta=0}^2
|\partial_x^\beta(w_k(x)-u_k(x))|\lambda^{-k}\\
+\sum\limits_{k=N^\prime+1}^N\lambda^{-k}|\partial_x^\beta w_k(x)|\Bigr\}
+|\partial_x^\alpha
\,\text{Error}(x,\lambda)|.\endmultline\tag 31
$$
\noindent The last term on the right is at most $C_*^{-\varepsilon N}
\lambda^{-\varepsilon N/100}$ by Lemma 3, and the double sum in braces
is dominated by
$$
C_*^Nx^{N/100}\le C_*^{\varepsilon N}\lambda^{-\varepsilon N/100}
\qquad\text{for}\quad \lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}.
$$
\noindent The single sum in braces is dominated by
$$
\sum\limits_{k=N^\prime+1}^N
\lambda^{-k}(C_*^{kN}x^{-\frac 32 k})x^{-\beta}\qquad
\text{since}\qquad w_k(x)=\sum\limits_{\ell=-3k}^Nw_{k\ell}
x^{\ell/2}\quad \text{with}\quad w_{k\ell} \in \Cal P.
$$
\noindent Since $\lambda^{-2/3+\varepsilon}<x<\lambda^{-\varepsilon}$, 
this sum is in turn 
dominated by $C_*^{N\varepsilon}(\lambda^{-\varepsilon})^{N^\prime}$.
Hence (31) yields
$$\multline
|\partial_x^\alpha\{F_0(x,\lambda)-F_+(x,\lambda)\}|\le
\frac {C_*^{N\varepsilon}\lambda^{10}}{(p(x))^{10}}\cdot\lambda^{-\varepsilon
N/100}\quad \text{for} \quad 
\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon},
\\
\lambda\ge C_*^{\varepsilon N}, \alpha=0,1,2.\endmultline
$$
\noindent Since also $(p(x))^{-10}\le C_*^{\varepsilon N}\lambda^{\frac
{20}{3}}$ for $\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$, the
final conclusion of Lemma 5 is obvious.$\qquad\blacksquare$
\vglue 1pc
\proclaim{Corollary} The solution of the transport equations constructed
in Lemma 5 may be written as $u_k(x)=x^{-\frac 32 k}f_k(x^{1/2})$ with
$f_k(y)$ smooth in a neighborhood of the origin.\endproclaim
\medskip
\demo{Proof}  We defined $u_k(x)=w_k(x)+x^{\overline N-5k}V_k(x^{1/2})$
with $V_k(y)$ smooth in a neighborhood of the origin.  Since
$w_k(x)=\sum\limits_{\ell=-3k}^Nw_{k\ell}x^{\ell/2}$, the corollary is
obvious.$\qquad\blacksquare$
\vglue 1pc
Next we understand how a general solution of the transport equations
depends on arbitrary constants of integration.
\vglue 1pc
\proclaim{Lemma 6}  Let $(u_0(x),u_1(x),\ldots,u_{N^\prime}(x))$ be a
solution of the transport equations.  Then the general solution
$(\tilde u_0(x),\tilde u_1(x),\ldots,\tilde u_N^\prime(x))$ of the
transport equations is given by $\tilde u_k(x)=\sum\limits_{\ell=0}^k c_\ell
u_{k-\ell}(x)$, where $c_1,\ldots c_{N^\prime}$ are arbitrary
constants and $c_0=1$.  The constants $(c_k)$ are uniquely determined
by the $(u_k)$ and $(\tilde u_k)$.\endproclaim
\medskip
\demo{Proof}  Trivial induction on $N^\prime$. $\qquad\blacksquare$
\vglue 1pc
\proclaim{Corollary} Every solution of the transport equations may be
written as $u_k(x)=x^{-\frac 32 k}f_k(x^{1/2})$ with $f_k(y)$ smooth
in a neighborhood of the origin.\endproclaim
\medskip
\demo{Proof}  Immediate from Lemma 6 and the Corollary to Lemma 5.$\qquad
\blacksquare$
\vglue 1pc
This corollary is not immediately clear by inspection of the transport
equations, since successive integrations of $x^{-\text{power}}$ singularities
might give rise to logarithmic terms.

Lemma 5 constructed a solution of the transport equations to 
match $F_0(x,\lambda)$, but we haven't yet checked the uniqueness of 
$(u_0,u_1,\ldots,u_N)$.  This
is taken care of in the next result.
\vglue 1pc
\proclaim{Lemma 7} Let $(\tilde u_0(x),\tilde u_1(x),\ldots,\tilde u_{N^\prime}
(x))$ be a solution of the transport equations, for which $F_0(x,\lambda)$
defined by (4) matches $F_+(x,\lambda)$ defined by (5) in the sense that
$|F_0(x,\lambda)-F_+(x,\lambda)|\le \tilde C\lambda^{-\varepsilon N/200}$
for $\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$, $\lambda\ge \tilde C$.
(Here $\tilde C$ denotes a positive number independent of $x$ and $\lambda$.)
Then for $0\le k\le \varepsilon N/300$, the function $\tilde u_k$ is exactly
the same as the function $u_k$ constructed in Lemma 5.\endproclaim
\medskip
\demo{Proof}  Let $\tilde C$ denote any positive number independent of $x$
and $\lambda$.  By Lemma 6 we have $\tilde u_k=\sum\limits_{\ell=0}^k
c_\ell u_{k-\ell}$, where $c_0\ldots c_{N^\prime}$ are constants with
$c_0=1$, and $(u_k)$ are the functions constructed in Lemma 5.  We want
to show that $c_k=0$ for $1\le k\le \varepsilon N/300$, for then
$\tilde u_k=u_k$ in that range.  We assume $c_{k_0}\ne 0$ for some $k_0$
between $1$ and $\varepsilon N/300$ and deduce a contradiction.  Take $k_0$ 
to be 
as small as possible.  Thus $\tilde u_k-u_k=\sum\limits_{\ell=k_0}^k c_\ell
u_{k-\ell}$.  By the Corollary to Lemma 5, we have therefore
$$\multline
\tilde u_k-u_k=0\quad\text{for}\quad k<k_0,\,\, \tilde u_{k_0}-u_{k_0}=c_{k_0},
\,\,|\tilde u_k(x)-u_k(x)|\le \tilde C x^{-\frac 32 (k-k_0)}\\
\text{for}\quad k_0<k\le N^\prime.\endmultline
$$
\noindent Hence
$$
\Big|\sum\limits_{k=0}^{N^\prime}\lambda^{-k}(\tilde u_k(x)-u_k(x))-
\lambda^{-k_0}
c_{k_0}\Big|\le \sum\limits_{k=k_0+1}^{N^\prime}
\lambda^{-k}\cdot\tilde C\, x^{-\frac 32 (k-k_0)}\le \tilde C\,\lambda^
{-k_0-\frac 32\varepsilon}$$
\noindent for $\quad \lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$,
and therefore
$$\multline
\Bigg|\text{Re}\Bigl[\frac{e^{\pm i \pi/4}\,e^{i\int_0^x(\lambda^2p(t))^{1/2}
dt}}{(\lambda^2p(x))^{1/4}}\,\Bigl\{\sum\limits_{k=0}^{N^\prime}\lambda^{-k}
(\tilde u_k(x)-u_k(x))-\lambda^{-k_0}c_{k_0}\Bigr\}\Bigr]\Bigg|\le
\frac {\tilde C\,\lambda^{-k_0-\frac {3}{2}\varepsilon}}{(\lambda
^2p(x))^{1/4}}\\
\text{for}\ \lambda^{\varepsilon -2/3}<x<\lambda^{-\varepsilon}.\endmultline
\tag 32
$$

On the other hand, since $F_+(x,\lambda)$ matches $F_0(x,\lambda)$ as in
the hypothesis of the Lemma, both for $(u_k(x))$ and for $(\tilde u_k(x))$,
we have
$$\multline
\Big|\text{Re}\Bigl[
\frac{e^{\pm i\pi/4}\,e^{i \int_0^x(\lambda^2p(t))^{3/2}]dt}}
{\lambda^2p(x))^{1/4}}\Bigl(\sum\limits_{k=0}^{N^\prime}\lambda^{-k}
(\tilde u_k(x)-u_k(x))\Bigr)\Bigr]\Big|\le \tilde C\lambda^{-\frac {\varepsilon N}
{200}}\\
\text{for}\quad \lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}
,\lambda\ge \tilde C.
\endmultline
$$
\noindent  Combining this with (32), we obtain the following
$$\multline
\Big|\text{Re}\Bigl[\frac{e^{\pm 
i \pi/4}\,e^{i \int_0^x(\lambda^2p(t))^{1/2}dt}}
{(\lambda^2p(x))^{1/4}}\cdot\lambda^{-k_0}c_{k_0}\Bigr]\Big|
\le \tilde C\lambda^{-\frac {\varepsilon N}{200}}+
\frac {\tilde C\lambda^{-k_0-\frac 32 \varepsilon}}{(\lambda^2p(x))^{1/4}}\\
\text{for}\quad \lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon},
\lambda\ge \tilde C.\endmultline\tag 33
$$
\noindent For $\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$ we have
$(\lambda^2p(x))^{-\frac 14}\ge \tilde c(\lambda^2\cdot\lambda^{-\varepsilon})
^{-\frac 14} \ge \tilde c\lambda^{-10}$.  Since $k_0\le \frac {\varepsilon N}
{300}$, it follows that the second term on the right-hand side of (33)
dominates the first term on the right.  Thus,
$$\multline
\Big|\text{Re}\Bigl[\frac{e^{\pm i\pi/4}\,e^{\lambda_i \int_0^x(p(t))^{1/2}dt}}
{(\lambda^2p(x))^{1/4}}\cdot\lambda^{-k_0}c_{k_0}\Bigr]\Big|\le
\frac {\tilde C\lambda^{-k_0-\frac 32 \varepsilon}}{(\lambda^2p(x))^{1/4}}\\
\text{for}\quad \lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon},
\lambda\ge \tilde C. \endmultline\tag 34
$$

For small positive $x$, let $I_x=\{\lambda\mid \frac 12 x^{-\frac 1\varepsilon}
<\lambda<x^{-\frac 1\varepsilon}\}$.  Note that $\lambda^{\varepsilon-2/3}
<x<\lambda^{-\varepsilon}$, $\lambda\ge \tilde C$ if $x$ is small enough
and $\lambda \in I_x$.  Thus (34) holds if $x$ is small and $\lambda \in 
I_x$.  Also, $\int_0^x(p(t))^{1/2}dt \ge \tilde c x^{3/2}$, while the
length of $I_x$ is large compared to $x^{-3/2}$ when $x$ is small. 
Therefore, as $\lambda$ varies over $I_x$, $\lambda\int_0^x(p(t))^{1/2}dt$
varies over an interval whose length is large, say more than $100\pi$.
Hence we can pick some $\lambda_x\in I_x$ for which
$$
\pm {\tsize\frac \pi 4}+\lambda_x\int_0^x(p(t))^{1/2}dt+\arg(c_{k_0})\equiv 0
\mod 2\pi.
$$
\noindent Putting this into (34) yields
$\frac {\lambda_x^{-k_0}|c_{k_0}|}{(\lambda_x^2p(x))^{1/4}}\le
\frac {\tilde C\lambda_x^{-k_0-\frac 32 \varepsilon}}{(\lambda_x^2p(x))^{1/4}}$.
Since $|c_{k_0}|\ne 0$, and since $\lambda_x\to \infty$ as $x\to 0$,
this is clearly false for small enough positive $x$.  Thus we have derived
a contradiction, and the Lemma is proven. $\qquad \blacksquare$

Set $N^{\prime\prime}=[\varepsilon N/300]$.
\vglue 1pc
\demo{Definition}  A solution $(u_0(x),\ldots,u_{N^{\prime\prime}}(x))$
of the transport equations is called the \underbar{canonical solution}
(for potential $p(x)$ at the turning point $x=0$) if we can define
$u_{N^{\prime\prime}+1}(x)\ldots u_{N^\prime}(x)$ for which
$(u_0(x),\ldots,u_{N^\prime}(x))$ still solves the transport equations, and
for which
$$
|F_0(x,\lambda)-F_+(x,\lambda)|\le \tilde C\lambda^{-\varepsilon N/200}
\quad \text{for}\quad \lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon},
\lambda\ge \tilde C.
$$
\noindent with $F_0$ defined by (4), $F_+$ defined by (5), and $\tilde C$
independent of $(x,\lambda)$.

Lemmas 5 and 7 show that there is one and only one canonical solution
of the transport equations.  We want to write down the canonical solution
rather explicitly.  Our method is to write down an ``elementary solution"
of the transport equations, whose properties are easily understood,
and then to relate the canonical solution to the elementary one.  To write
down the elementary solution we proceed we follows.

If $u=(u_0(x),\ldots,u_{N^{\prime\prime}}(x))$ solves the transport
equations, then define the \underbar{residue}
$$
\text{Res}_k(u)=\lim\limits_{x\to 0+}\Big(u_k(x)-\sum\limits_{\ell=1}^{3k}
g_{k\ell}x^{-\ell/2}\Big),$$
\noindent  where the $g_{k\ell}$ are uniquely specified
by demanding the finiteness of the limit.  The Corollary to Lemma 6 shows
that $\text{Res}_k(u)$ is well-defined.  If $u=\hfill\break
(u_0(x),u_1(x),\ldots,u_
{N^{\prime\prime}}(x))$ and $\tilde u=(\tilde u_0(x),\ldots,\tilde u_{N^{\prime
\prime}}(x))$ are two solutions of the transport equations related by
$\tilde u_k=\sum\limits_{\ell=0}^k c_\ell u_{k-\ell}$ as in Lemma 6, then
we have
$$
\text{Res}_k(\tilde u)=\sum\limits_{\ell=0}^k c_\ell\text{Res}_{k-\ell}
(u).$$

\noindent Note that $\text{Res}_0(u)\equiv 1$.
\vglue 1pc
\demo{Definition}  A solution $\tilde u=(\tilde u_0(x),\tilde u_1(x),
\ldots,\tilde u_{N^{\prime\prime}}(x))$ of the transport equations is
called the \underbar{elementary solution} (for potential $p(x)$ and
turning point $x=0$) if $\text{Res}_k(\tilde u)=0$ for
$k=1,2,\ldots,N^{\prime\prime}$.
\vglue 1pc
\proclaim{Lemma 8}  There is one and only one elementary solution of
the transport equations.\endproclaim
\medskip

\demo {Proof}  Let $u=(u_0,u_1,\ldots,u_{N^{\prime\prime}})$ be any solution
of the transport equations.  We show that there is one and only one sequence
$c_0,c_1\ldots c_{N^{\prime\prime}}$ with $c_0=1$, for which
$\tilde u_k=\sum\limits_{\ell=0}^kc_\ell u_{k-\ell}$ has residue zero
$k=1,2,\ldots,N^{\prime\prime})$.\enddemo
\smallskip
Since the equations to be solved for the $(c_k)$ may be written as
$c_k+\hfill\break
+\sum\limits_{\ell=0}^{k-1}c_\ell\text{Res}_{k-\ell}(u)=0$,
the $c_k$ are uniquely determined by an obvious induction.$\quad\blacksquare$
\bigskip

The elementary solution is given rather explicitly by the following result.
\vglue 1pc
\proclaim{Lemma 9}  The elementary solution $(u_0(x),u_1(x),\ldots,u_{N^{\prime
\prime}}(x))$ is obtained from the inductive procedure $u_0(x)\equiv 1$,
$$
-2i\,u_{k+1}(x)=\lim_{\delta\to 0+}\Bigl[\int_\delta^x\Cal Lu_k(t)dt+
\sum\limits_{\ell=1}^{3k+3}g_{k\ell}\delta^{-\ell/2}\Bigr],
$$
\noindent with $\Cal L$ as in Lemma 4, and $g_{k\ell}$ uniquely specified
by demanding the finiteness of the limit.  The $g_{k\ell}$ all belong to
$\Cal P$.  We can write $u_k$ in the form $u_k(x)=x^{-\frac 32 k}f_k(x^{1/2})$
with $f_k$ smooth and $(\frac d{dy})^\alpha f_k(0)\in \Cal P$ for all $\alpha$.
\endproclaim
\medskip
\demo{Proof}  By the corollary to Lemma 6 we can write
$$
-2i\, u_{k+1}(t)=\sum\limits_{\ell=1}^{3k+3}q_{k\ell}t^{-\ell/2}+
\tilde q_k(t^{1/2})\tag 35$$
\noindent with $\tilde  q_k(y)$ smooth.
\smallskip
\noindent In particular, $\lim_{t\to0_+}[2iu_{k+1}(t)+
\sum\limits_{\ell=1}^
{3k+3}q_{k\ell}t^{-\ell/2}]=-\tilde q_k(0)$, so  $2i\,\text{Res}_{k+1}(u)=
\hfill\break-\tilde q_k(0)$.  
For the elementary solution the residues are zero,
so
$\tilde q_k(0)=0$.  Now we return to (35) and differentiate, obtaining
$$
-2iu_{k+1}^\prime(t)=\sum\limits_{\ell=1}^{3k+3}(-{\tfrac
\ell 2})q_{k\ell}
t^{-\frac \ell 2-1}+{\tsize
\frac 12} t^{-1/2}\tilde q_k^\prime(t^{1/2}).\tag 36
$$
\noindent 
Integrating from $t=\delta$ to $t=x$, we get
$$
\int_\delta^x(-2i\,u_{k+1}^\prime(t))dt=\sum\limits_{\ell=1}^{3k+3}
q_{k\ell}(x^{-\ell/2}-\delta^{-\ell/2})+\tilde q_k(x^{1/2})-\tilde q_k(\delta^
{1/2}).
$$
\noindent Since $\tilde q_k$ is smooth and vanishes at the origin, it
follows that
$$\multline
\lim_{\delta\to 0+}\Bigl[\int_\delta^x(-2i\,u_{k+1}^\prime(t))dt+\sum\limits_
{\ell=1}^{3k+3}q_{k\ell}\delta^{-\ell/2}\Bigr]=\sum\limits_{\ell=1}^{3k+3}
q_{k\ell}\,x^{-\ell/2}+\tilde q_k(x^{1/2})\\
=-2i\,u_{k+1}(x).\endmultline
$$
\noindent The inductive formula for $-2i\,u_{k+1}(x)$ in the statement of
the Lemma now follows at once, since $-2i\,u_{k+1}^\prime(t)=\Cal Lu_k(t)$
by the transport equation.  The $g_{k\ell}$ are equal to the $q_{k\ell}$.
Again using the Corollary to Lemma 6 
to write $u_k(x)=x^{-\frac 32 k}f_k(x^{1/2})$
with $f_k(y)$ smooth, we see from (35) that the $q_{k\ell}$ are proportional
to the derivatives of $f_{k+1}(y)$ at $y=0$.  Hence the assertion
$g_{k\ell} \in \Cal P$ will follow, once we prove that
$(\frac {d}{dy})^\alpha f_k(0)\in \Cal P$.  So it remains only to verify
the latter assertion.  We proceed by induction on $k$.  For $k=0$ we
have $f_0(y)\equiv 1$ and there is nothing to prove.  Suppose we know
that $(\frac {d}{dy})^\alpha f_k(0)\in \Cal P$ for all $\alpha$.  Then
$$\split
-2i\,u_{k+1}^\prime(t)=\Cal Lu_k(t)=\{({\tsize
\frac {5}{16}}(p^\prime)^2p^{-5/2}-{\tsize\frac 14}
p^{\prime\prime}p^{-3/2})-{\tsize\frac 12} p^\prime p^{-3/2}
{\tsize\frac {d}{dt}}
+p^{-1/2}{\tsize\frac {d^2}{dt^2}}\}\cdot\\
\{t^{-\frac 32 k}f_k(t^{1/2})\}\equiv X.\endsplit
$$
\noindent With $y=t^{1/2}$ we have $\frac {d}{dt}=\frac 12 y^{-1}\frac {d}{dy}$,
$p(t)=g_1(y)$, $p^\prime(t)=g_2(y)$, $p^{\prime\prime}(t)=g_3(y)$,
$y^5/p^{5/2}(t)=g_4(y)$ with $g_i(y)$ smooth functions and
$(\frac {d}{dy})^\alpha g_i(0)\in \Cal P$.  Hence
$$ 
X=\frac {(g_5(y)+g_6(y)\frac {d}{dy}+g_7(y)\frac {d^2}{dy})f_k(y)}
{y^{3k+5}}
$$
\noindent where again $g_i(y)$ are smooth and $(\frac {d}{dy})^\alpha
g_i(0)\in \Cal P$.  By inductive hypothesis, the numerator is smooth,
and its derivatives at $y=0$ all belong to $\Cal P$.  Taylor's theorem
with remainder gives therefore
$$\align
X=\sum\limits_{\ell=1}^{3k+5}\hat q_{k\ell}\,
y^{-\ell}+&\hat q_k(y)\quad\text{with}
\quad \hat q_{k\ell}\in \Cal P,\\
&\hat q_k(y)\quad\text{smooth},\quad ({\tsize\frac {d}{dy}})^\alpha\hat q_k(0)
\in \Cal P.\endalign
$$
\noindent That is,
$$
-2i\,u_{k+1}^\prime(t)=\sum\limits_{\ell=1}^{3k+5}\hat q_{k\ell}\,t^{-\ell/2}
+\hat q_k(t^{1/2}), \quad \text{by\ definition\ of}\  t \ \text{and} \ X.
\tag 37
$$
\noindent Comparing (36) and (37) gives
$$
\sum\limits_{\ell=1}^{3k+5}\hat q_{k\ell}\,t^{-\ell/2}-\sum\limits_{\ell=1}^
{3k+3}(-{\tsize\frac \ell 2})q_{k\ell}\,t^{-\frac \ell 2-1}=
{\tsize\frac 12} t^{-1/2}
\tilde q_k^\prime(t^{1/2})-\hat q_k(t^{1/2}),\tag 38
$$
\noindent where the $\prime$ on the right is taken with respect to $y$.
The right-hand side here is $O(t^{-1/2})$ at the origin, while the left-hand
side is a sum of negative half-integer powers of $t$.  We conclude by
matching like powers of $t$ that
$$
q_{k\ell}=-{\tsize\frac 2 \ell} \hat q_{k\ell+2}\ (\ell=1,\ldots,3k+3)\tag 39
$$
\noindent and that  $\hat q_{k2}=0$.
Hence (38) simplifies to $\hat q_{k1}\,t^{-1/2}=\frac 12 t^{-1/2}\,
\tilde q_k^\prime(t^{1/2})-\hat q_k(t^{1/2})$, i.e.,
$\frac {d}{dy} \tilde q_k(y)=2\hat q_{k1}+2y\,\hat q_k(y)$.
\noindent Since $\tilde q_k(0)=0$, we conclude that
$$
\tilde q_k(y)=2\hat q_{k1}\,y+\int_0^y2s\,\hat q_k(s)\,ds. \tag 40
$$
\noindent Since $\hat q_{k\ell} \in \Cal P$ and $(\frac {d}{dy})^\alpha
\hat q_k(0)\in \Cal P$ for all $\alpha$, equations (39), (40) show
that $q_{k\ell}\in \Cal P$ and $(\frac {d}{dy})^\alpha\tilde q_k(0)\in \Cal P$
for all $\alpha$.  Therefore, (35) allows us to write
$$\multline
u_{k+1}(x)=x^{-\frac 32 (k+1)}f_{k+1}(x^{1/2})\quad\text{with}\\
f_{k+1}(y)=\frac {1} {-2i}\Bigl(\sum\limits_{\ell=1}^{3k+3}
q_{k\ell}\,y^{3k+3-\ell}+y^{3k+3}\tilde q_k(y)\Bigr)\endmultline
$$

\noindent and thus $(\frac {d}{dy})^\alpha f_{k+1}(0)\in \Cal P$ for all $\alpha$.
The induction step is complete, and the Lemma is proven.$\qquad\blacksquare$
\medskip

Now it is trivial to relate the canonical to the elementary solution.
\vglue 1pc
\proclaim{Lemma 10}  Let $u=(u_0(x),u_1(x),\ldots,u_{N^{\prime\prime}}(x))$
be the canonical solution of the transport equations, and let $\tilde u=
(\tilde u_0(x),\ldots,\tilde u_{N^{\prime\prime}}(x))$ be the
 elementary solution.  Then we have $u_k(x)=\sum\limits_{\ell=0}^k
c_{k-\ell}^+\tilde u_\ell(x)$ with $c_0^+=1$ and $c_k^+\in \Cal P$. 
In particular, $c_1^+=0$.  The coefficients $(c_k^+)$ are called \underbar{
matching coefficients}.\endproclaim
\medskip
\demo{Proof}  In view of Lemma 6, the only points to be checked are
that $c_k^+\in \Cal P$ and that $c_1^+=0$.  Now $\text{Res}_\ell(\tilde u)
=\cases 1&\text{if $\ell=0$}\\
0&\text{if $\ell>0$}\endcases$ by definition of $\tilde u$, so
$\text{Res}_k(u)=\sum\limits_{\ell=0}^k c_{k-\ell}^+
\text{Res}_\ell(\tilde u)=c_k^+$.  We must therefore show that
$\text{Res}_k(u)\in \Cal P$.
\smallskip
Lemmas 3 and 5 give
$$\split
u_k(x)=(u_k(x)-w_k(x))+\sum\limits_{-3k\le \ell\le M_k}w_{k\ell}
\,x^{\ell/2}\quad\text{with}\quad w_{k\ell} \in \Cal P,\\
w_{10}\equiv 0, \quad\text{and}\quad \lim_{x\to0^+}(u_k(x)-w_k(x))=0.\endsplit
$$
\noindent Therefore $\lim_{x\to 0^+}[u_k(x)-\sum\limits_{-3k\le \ell\le -1}
w_{k\ell}\,x^{\ell/2}]=w_{k0}$, which shows that \hfill\break
$c_k^+=\text{Res}_k(u)= w_{k0}
\in \Cal P$.  Also, $c_1^+=w_{10}=0$.  The Lemma is proven. $\quad\blacksquare$
\vglue 1pc
\demo{Remark}  It would be interesting to exhibit the first few matching
coefficients beyond $c_1^+$.  In principle, this is routine.  In practice
we don't even know whether the higher $c_k^+$ are all zero.
\medskip

Next, we want a-priori bounds on the $C^\infty$ seminorms of the functions
$f_k$ in Lemma 9.  For later application, we take the potential $p(x)$
to depend smoothly on an additional parameter $\tau$, and investigate how
$f_k(x,\tau)$ depends on $(x,\tau)$ together.
\vglue 1pc
\proclaim{Lemma 11} Let $p(x,\tau)$ be $C^\infty$ on $\{|x|,|\tau|\le 1\}$
with $p(0,\tau)=0$ and $\frac {\partial}{\partial x} p(0,\tau)\ge c_1>0$.
For each fixed $\tau$, let 
$(u_0(x,\tau), u_1(x,\tau),\ldots u_{N^{\prime\prime}}
(x,\tau))$ be the elementary solution to the transport equations for potential
$x\mapsto p(x,\tau)$.  Then define $f_k(y,\tau)$ by $u_k(x,\tau)=x^{-\frac 32 k}
f_k(x^{1/2},\tau)$ as in Lemma 9.  The $C^\infty$ seminorms of $f_k$ can be
bounded a-priori in terms of the $C^\infty$ seminorms of $p(x,\tau)$ and the
constant $c_1$.
\endproclaim
\medskip
\demo{Proof}  We use induction on $k$.  For $k=0$ we have $f_k(y,\tau)\equiv
1$, so
there is nothing to prove.  For the inductive step, we use $C_{**}$, $C_{**}^
\alpha$, etc. to denote constants determined a-priori in terms of the $C^\infty$
seminorms of $p(x,\tau)$ and the lower bound $c_1$ for $\frac {\partial p}
{\partial x}(0,\tau)$.  Thus we must prove that $|\partial_{y,\tau}^\alpha
f_{k+1}|\le C_{**}^\alpha$ for all $\alpha$, given that $|\partial_{y,\tau}^
\alpha f_k|\le C_{**}^\alpha$ for all $\alpha$.\enddemo
\smallskip

Let $Q$ be a small square about the origin with $\frac {p(x,\tau)}{x}
>\frac 12 c_1$ for $x>0$ and $(x^{1/2},\tau)\in Q$.  We will make our
estimates on $Q$.  Note that the size of $Q$ is bounded below by $c_{**}$.

We have $\Cal L u_k(x,\tau)=\Cal L(x^{-\frac 32 k}f_k(x^{1/2},\tau))=
x^{-\frac 32 k-\frac 52}(\hat\Cal L_kf_k)(x^{1/2},\tau)$ with
$$
\hat\Cal L_kw=y^{3k+5}\Bigl[(\tsize
\frac {5}{16}(p^\prime)^2p^{-5/2}-\tsize\frac 14
p^{\prime\prime}p^{-3/2})-{\tsize
\frac 12}p^\prime p^{-3/2}\cdot\tsize\frac {1}{2y}
\tsize\frac {\partial}{\partial y}+p^{-1/2}(\tsize\frac {1}{2y}
\tsize\frac{\partial}
{\partial y})^2\Bigr]y^{-3k}w,
$$
\noindent where $p,p^\prime,p^{\prime\prime}$ are evaluated at
$(y^2,\tau)$.

\noindent The defining formula for $\hat{\Cal L}_k$ may be written in the
form
$$
\hat{\Cal L}_k=a_k^{(2)}(y,\tau)\tfrac{\partial^2}{\partial y^2}+
a_k^{(1)}(y,\tau)\tfrac{\partial}{\partial y}+a_k^{(0)}(y,\tau)
$$
\noindent with $a_k^{(i)}$ smooth on $Q$ and $|\partial_{y,\tau}^\alpha
a_k^{(i)}(y,\tau)|\le C_{**}^{\alpha k}$, $(y,\tau) \in Q$, all
$\alpha$.  Since also $|\partial_{y,\tau}^\alpha f_k(y,\tau)|\le C_{**}^
{\alpha k}$ by inductive hypothesis, we have for $g_k=\hat{\Cal L}_kf_k$
the estimates $|\partial_{y,\tau}^\alpha g_k(y,\tau)|\le C_{**}^{\alpha k}$
on $Q$, all $\alpha$.

We have also $\Cal Lu_k(x,\tau)=x^{-\frac 32 k-\frac 52}g_k(x^{1/2},\tau)$.
Applying Taylor's theorem with remainder to $g_k$, we get smooth functions
$\tilde g_{k\ell}(\tau)$ and a smooth function $\tilde g_k(y,\tau)$
that satisfy $\Cal Lu_k(x,\tau)=\sum\limits_{\ell=1}^{3k+5}\tilde g_{k\ell}
(\tau)x^{-\ell/2}+\tilde g_k(x^{1/2},\tau)$, $|\partial_\tau^\alpha
\tilde g_{k\ell}(\tau)|\le C_{**}^{\alpha k\ell}$ and
$|\partial_{y,\tau}^\alpha\tilde g_k(y,\tau)|\le C_{**}^{\alpha k}$
on $Q$.  Therefore, Lemma 9 says that
$$\split
-2i\, u_{k+1}(x,\tau)=\lim_{\delta \to 0+}\Bigl[\int_\delta^x
\Bigl\{\sum\limits_{\ell=1}^{3k+5} \tilde g_{k\ell}(\tau)t^{-\ell/2}
+\tilde g_k(t^{1/2},\tau)\Bigr\}\,dt\\
+\sum\limits_{\ell=1}^{3k+3} g_{k\ell}(\tau)\delta^{-\ell/2}\Bigr]
\endsplit\tag 41
$$
\noindent for suitable functions $g_{k\ell}(\tau)$.  In particular, the limit
exists, which shows that $\tilde g_{k2}\hfill\break
\equiv 0$.  (Otherwise the
integral would produce a logarithmic singularity in $\delta$, which could
not be remedied by any choice of the $g_{k\ell}(\tau)$.)
\noindent Hence (41) means that
$$\multline
-2i\, u_{k+1}(x,\tau)=\sum\limits_{\ell=3}^{3k+5} (\tfrac {-2}{\ell-2})
x^{-(\ell-2)/2} \tilde g_{k\ell}(\tau)+g_k^{\#}(x^{1/2},\tau) \\
\text{with}\ g_k^{\#}(y,\tau)=2\tilde g_{k1}(\tau)y+\int_0^y \tilde g_k
(s,\tau)\cdot 2sds.\endmultline\tag 42
$$

We have $|\partial_{y,\tau}^\alpha\, g_k^{\#}(y,\tau)|\le C_{**}^{\alpha k}$
on $Q$, by virtue of our estimates for $\tilde g_{k\ell}(\tau)$ and
$\tilde g_k(y,\tau)$.

By (42) we have $u_{k+1}(x,\tau)=x^{-\frac 32 (k+1)}f_{k+1}(x^{1/2},\tau)$
with
$$
f_{k+1}(y,\tau)=\sum\limits_{\ell=3}^{3k+5} \tfrac {1}{i(\ell-2)}
\tilde g_{k\ell}(\tau)y^{(3k+3)-(\ell-2)}-\tfrac {1}{2i}
y^{3k+3}g_k^{\#}(y,\tau).
$$

Our estimates for $\tilde g_{k\ell}(\tau)$ and $g_k^{\#}(y,\tau)$ show that
$|\partial_{y,\tau}^\alpha f_{k+1}(y,\tau)|\le C_{**}^{\alpha k}$ on $Q$
for all $\alpha$.  The induction is complete, and the Lemma is proven.
$\qquad\blacksquare$
\proclaim{Corollary}  The conclusion of Lemma 11 holds also for the
canonical solution of the transport equation.\endproclaim
\medskip
\demo{Proof} Express the canonical solution in terms of the elementary
solution using the matching coefficients $c_k^+$.  Since
$c_k^+ \in \Cal P$, the matching coefficients depend smoothly on $\tau$,
and the Corollary follows at once from Lemma 11. $\qquad \blacksquare$
\smallskip

It remains to verify that $F_+(x,\lambda)$ approximately satisfies the
basic differential equation (1).  Since we only established the uniqueness
of the canonical solution $(u_0(x),u_1(x),\ldots,u_k(x),\ldots)$ for
$k \le N^{\prime\prime}$, we prefer to truncate the sum (5)
defining $F_+(x,\lambda)$ below $k=N^{\prime\prime}$.  We will check
that the truncated solution still agrees well with $F_0(x,\lambda)$ as
in Lemma 5.  (This problem arises only from the technicalities of our
exposition.  Clearly the canonical $u_k(x)$ are defined for all $k$,
by taking $N$ large enough depending on $k$.  We didn't want to bother
with the $N$ dependence of $u_k$.)
\vglue 1pc
\proclaim{Lemma 12}  Set $N^{\prime\prime\prime}=[\varepsilon N/500]$
and define $F_{++}(x,\lambda)=\text{Re}\Bigl[\frac
{e^{\pm i \pi/4}e^{i\int_0^x(\lambda^2p(t))^{1/2}\,dt}}
{(\lambda^2p(x))^{1/4}}\cdot\hfill\break
\cdot\sum\limits_{k=0}^{N^{\prime\prime\prime}}
\,\lambda^{-k}u_k(x)\Bigr]$ with $(u_0(x),\ldots u_{N^{\prime\prime}}
(x))$ the canonical solution of the transport equations.  Then for
$\lambda^{\varepsilon - 2/3}<x<\lambda^{-\varepsilon}$ and $\lambda \ge
C_*^{\varepsilon N}$ we have
$|\partial_x^\alpha\{F_{++}(x,\lambda)-F_0(x,\lambda)\}|\le
C_*^{\varepsilon N}(\lambda x^{3/2})^{-N^{\prime\prime\prime}}
\,\lambda^{10}\quad (\alpha=0,1,2)$.  Also 
$|(\frac{\partial^2}{\partial x^2}+\lambda^2p(x))
F_{++}(x,\lambda)|\hfill\break
\le C_*^{\varepsilon N}(\lambda x^{3/2})^{-N^{\prime\prime
\prime}}\lambda^{10}$ for $\lambda^{\varepsilon 
-\frac 23}<x<c_*^{\varepsilon N}
$, $\lambda \ge C_*^{\varepsilon N}$.\endproclaim
\medskip
\demo{Proof}  For $x$, $\lambda$ as in the Lemma, we have from Lemma 5
$$\split
|\partial_x^\alpha\{F_+(x,\lambda)-F_0(x,\lambda)\}|\le C_*^{\varepsilon N}
\lambda^{-\varepsilon N/200} \le C_*^{\varepsilon N} (\lambda x^{3/2})^
{-N^{\prime\prime\prime}}\lambda^{10}\\
(\alpha=0,1,2).\endsplit
$$
\noindent So the first conclusion of Lemma 12 will follow if we can show that
$$
|\partial_x^\alpha\{F_{++}(x,\lambda)-F_+(x,\lambda)\}|\le C_*^{\varepsilon N}
(\lambda x^{3/2})^{-N^{\prime\prime\prime}}\lambda^{10} \quad
(\alpha=0,1,2).\tag 43
$$
\noindent Now $F_{++}-F_+$ is a sum of real parts of expressions
$$
X=\frac{e^{\pm i \pi/4}}{(\lambda^2p(x))^{1/4}}
e^{i \int_0^x(\lambda^2p(t))^{1/2}dt}\lambda^{-k}u_k(x)\quad
\text{with}\quad N^{\prime\prime\prime} <k\le N^\prime.
$$
\noindent Also, Lemma 5 gives $|\partial_x^\alpha\{u_k(x)-w_k(x)\}|\le
C_*^{\varepsilon N} \quad (\alpha=0,1,2)$, while
$|\partial_x^\alpha w_k(x)|\le C_*^{\varepsilon N}x^{-\frac 32 k-\alpha}
\quad (\alpha=0,1,2)$ \,since $w_k(x)=\sum\limits_{\ell=-3k}^{M_k}
w_{k\ell}\,x^{\ell/2}$, $w_{k\ell} \in \Cal P$.  Hence
$|\partial_x^\alpha u_k(x)|\hfill\break
\le C_*^{\varepsilon N}x^{-\frac 32 k-\alpha}
\quad (\alpha=0,1,2)$.  Putting this into the definition of $X$,
we find that $|\partial_x^\alpha X|\le C_*^{\varepsilon N}
\lambda^{-k}x^{-\frac 32 k}\cdot x^{-3}\lambda^3\quad \text{for}\,
\alpha=0,1,2$.  In the region 
$\lambda^{\varepsilon-2/3}<x<\lambda^{-\varepsilon}$
this implies $|\partial_x^\alpha X|\le C_*^{\varepsilon N}
(\lambda x^{3/2})^{-k}\lambda^{10}\quad (\alpha=0,1,2)$, which proves
(43).

To establish the second conclusion of the Lemma, note that
$$
\Bigl(\frac{\partial^2}{\partial x^2}+\lambda^2p(x)\Bigr)F_{++}(x,\lambda)=
\text{Re}\Bigl[\frac{e^{\pm i\pi/4}\,e^{i\int_0^x(\lambda^2p(t))^{1/2}dt}}
{(\lambda^2p(x))^{1/4}}\,p(x)^{1/2}\Cal Lu_{N^{\prime\prime\prime}}
(x)\lambda^{-N^{\prime\prime\prime}}\Bigr],\tag 44
$$
\noindent because $(u_k(x))$ satisfy the transport equations.

In proving Lemma 11, we showed that $\Cal L\tilde u_k(x)=x^
{-\frac 32 k-\frac 52}g_k(x^{1/2})$ with
$|\partial_y^\alpha g_k(y)|\hfill\break
\le C_*^{\varepsilon N\alpha}$, $0 \le k \le
N^{\prime\prime}$, all $\alpha$ and all $|y|\le c_*^{\varepsilon N}$.
Here $(\tilde u_k(x))$ denotes the elementary solution of the transport
equations.  

Since the canonical and elementary solutions are related by the matching 
coefficients $c_k^+\in \Cal P$, we have the same result for the
$\Cal L u_k(x)$ arising from the canonical solution.  In particular,
$|\Cal L u_k(x)|\le C_*^{\varepsilon N}x^{-\frac 32 k-\frac 52}$ for
$0 \le k \le N^{\prime\prime}$ and all $x \in (0,c_*^{\varepsilon N}]$.
Taking $k=N^{\prime\prime\prime}$ and substituting in (44),
we obtain the second conclusion of the Lemma, since we are in the
region $\lambda^{\varepsilon -2/3} <x$,\,\,\,\, 
$\lambda \ge C_*^{\varepsilon N}$.
$\qquad\blacksquare$

We pause to write down in detail the function $u_1(x)$ for the canonical
solution of the transport equations.  Since the first two matching coefficients
are $c_0^+=1$ and $c_1^+=0$ by Lemma 10, the elementary and canonical
solutions have the same $u_1(x)$.  Therefore, recalling that
$u_0(x)\equiv 1$, we may apply Lemma 9 and the definition of $\Cal L$
to derive:
$$
-2i u_1(x)=\lim_{\delta\to 0+}\Bigl[\int_\delta^x
(\tfrac 5{16}(p^\prime)^2p^{-5/2}-\tfrac 14 p^{\prime\prime}
p^{-3/2})dt-g_1\delta^{-1/2}-g_2\delta^{-1}-g_3\delta^{-3/2}\Bigr]\tag 45
$$
\noindent with $g_1$, $g_2$, $g_3$ uniquely specified by demanding
the finiteness of the limit.  We needn't bother to compute $g_1$,
$g_2$, $g_3$.

Next we rescale our approximate ODE solutions.

Let $F_0(x,\lambda)$, $F_+(x,\lambda)$ be the matching approximate solutions for
$$
\Bigl(\frac{\partial^2}{\partial x^2}+\lambda^2p(x)\Bigr)F=0\quad
\text{given\ by}\, (4), (5).\tag 46
$$
\noindent For $t >0$ define $p_{\#}(x)=t^2p(x)$.  Then let
$F_0^{\#}(x,\lambda_{\#})$, $F_+^{\#}(x,\lambda_{\#})$ be the matching
approximate solutions for
$$
\Bigl(\frac{\partial^2}{\partial x^2}+\lambda_{\#}^2p_{\#}(x)\Bigr)
F^{\#}=0 \quad \text{given\ similarly}.\tag 47
$$

If we set $\lambda_{\#}=t^{-1}\lambda$, then (46) and (47) are the same
equation.  One checks that $y^{\#}(x,\lambda_{\#})=t^{2/3}y(x,t\lambda_{\#})$
solves the analogue of (3) for $p_{\#}(x)$.  Therefore
$F_0^{\#}(x,\lambda_{\#})=F_0(x,\lambda)$, i.e.\ our Airey approximation
really depends only on the potential, not how it is factored.  Since $F_+^{\#}
(x,\lambda_{\#})$ and $F_+(x,\lambda)$ match $F_0^{\#}(x,\lambda_{\#})$ and 
$F_0(x,\lambda)$, it follows that $F_+^{\#}(x,\lambda_{\#})=F_+(x,\lambda)$
as well, and the canonical
solutions $(u_0(x),u_1(x),\ldots,u_{N^{\prime\prime}}(x))$,
$(u_0^{\#}(x),u_1^{\#},\ldots,u_{N^{\prime\prime}}^{\#}(x))$ are related by
$u_k^{\#}(x)=t^{-k}u_k(x)$.  Thus, $\lambda^{-k}u_k(x)$ depends only
on the potential and not how it is factored.

There is a second kind of rescaling for our approximate solutions.  With
$t>0$ we take $p_{\#}(x)=t^2p(tx)$.  The solution of equation (3) for
$p_{\#}(x)$ is $y_{\#}(x,\lambda)=y(tx,\lambda)$.  Thus
$\Bigl(\frac{\partial y_{\#}(x,\lambda)}{\partial x}\Bigr)^{-1/2}
=t^{-1/2}\Bigl(\frac{\partial y}{\partial x}\mid_{tx,\lambda)}\Bigr)^{-1/2}$,
so the Airey approximation
 for $p_{\#}(x)$ is $F_0^{\#}(x,\lambda)=
t^{-1/2}F_0(tx,\lambda)$.

Again, since $F_+^{\#}$ matches $F_0^{\#}$ and $F_+$ matches $F_0$, we can
deduce the scaling laws for $F_+$ and for the canonical solution of
the transport equations we get $F_+^{\#}(x,\lambda)=t^{-1/2}F_+(tx,\lambda)$,
$$
u_k^{\#}(x)=u_k(tx).
$$
\noindent For both notions of rescaling, the elementary solutions of the
transport equations scale in the same way as the canonical solutions.
Details of the rescaling are easily filled in by the reader.

The main application of our local results is to the one-parameter family
of potentials $\tilde p(x,\tau)=p(x+x(\tau))-\tau$, where $x(\tau)$
is the solution of $p(x(\tau))=+\tau$.  If $p(x)$ is $C^\infty$ with
$p^\prime(0)>0$, then $\tilde p(x,\tau)$ is $C^\infty$ with
$\tilde p(0,\tau)=0$, $\frac {\partial\tilde p}{\partial x}
(0,\tau)>c_1>0$  for $|p(0)-\tau|$ small.  For each fixed $\tau$ we apply
the local ODE results of this section to $\tilde p(x,\tau)$.

Quantities belonging to $\Cal P$, such as the matching coefficients $c_k^+$
or the $f_{k\ell}$ in Lemma 1, evidently depend smoothly on $\tau$.  The
$\tau$-dependence of the canonical solutions to the transport equation
is given by the Corollary to Lemma 11.  Therefore, after a small change
of notation, we obtain the following result.
\vglue 1pc
\proclaim{Local WKB Lemma}  Let $\varepsilon$, $N$ be given, and put
$N^\prime=[\varepsilon N/500]$.  Let $p(x)$ be a smooth function
on $[-1,1]$ with $p^\prime(0)>0$.  For $|E-p(0)|<c$ let $x(E)$ denote
the solution of $p(x)=E$, nearest to $x=0$.  Then there exist functions
$y(x,E,s)$ and $u_k(x,E)$ $(0 \le k \le N^\prime)$ with the following
properties.
\smallskip
\item{(A)} The function $y(x,E,s)$ is smooth and satisfies 
$\frac{\partial y}{\partial x}>c>0$ on
$\{|x-x(E)|\hfill\break <c, |E-p(0)|<c, |s|<c\}$.
\smallskip
\item{(B)} The function $F_0(x,E,\lambda)=\lambda^{-1/3}(\partial_xy(x,E,
\lambda^{-2}))^{-1/2}A(\lambda^{2/3}y(x,E,\lambda^{-2}))$ satisfies
$|(\frac{\partial^2}{\partial x^2}+\lambda^2(p(x)-E))F_0(x,E,\lambda)|\le
C\lambda^{-N^\prime}$ for $|x-x(E)|<\lambda^{-\varepsilon}$,
$|E-p(0)|<c$, $\lambda>C$.\smallskip
\item{(C)} 
We have $|\partial_x^\alpha F_0(x,E,\lambda)|\le C\lambda^{-N^\prime}$
for $-\lambda^{-\varepsilon}<x-x(E)<-\lambda^{\varepsilon-2/3}$,
$|E-p(0)|<c$, $\lambda>C$, $\alpha\le 2$.\smallskip
\item{(D)} The functions $u_k(x,E)$ are defined for $0<x-x(E)<c$,
$|E-p(0)|<c$ and have there the form $u_k(x,E)=(x-x(E))^{-\frac 32 k}f_k
((x-x(E))^{1/2},E)$ with $f_k(y,E)$ smooth.\smallskip
\item{(E)} The function $F_+(x,E,\lambda)=\text{Re}
\Bigl[\frac{\exp(\pm i \frac \pi 4+i\lambda\int_{x(E)}^x (p(t)-E)^{1/2}dt)}
{\lambda^{1/2}(p(x)-E)^{1/4}}\Bigl\{1+\hfill\break
\sum\limits_{k=1}^{N^\prime}
\lambda^{-k}u_k(x,E)\Bigr\}\Bigr]$ satisfies
$|(\frac{\partial^2}{\partial x^2}+\lambda^2(p(x)-E))F_+(x,E,\lambda)|\le C
\lambda^{10}\cdot(x-x(E))^{-\frac 32 N^\prime}\lambda^{-N^\prime}$
for $\lambda^{\varepsilon-2/3}<x-x(E)<c$, $|E-p(0)|<c$, $\lambda>C$.\smallskip
\item{(F)} We have $|\partial_x^\alpha\{F_+(x,E,\lambda)-F_0(x,E,\lambda)\}|
\le C\lambda^{10}\cdot(x-x(E))^{-\frac 32 N^\prime}\lambda^{-N^\prime}$
for $\alpha \le 2$, $\lambda^{\varepsilon-2/3}<x-x(E)<\lambda^{-\varepsilon}$,
$|E-p(0)|<c$, $\lambda>C$.\smallskip

More precise information on the functions $y(x,E,s)$ and $u_k(x,E)$ is given
by the following:
\item{(G)} We have $(\partial_xy(x,E,s))^2y(x,E,s)=p(x)-E+(x-x(E))^{N^\prime}
g_0(x,E)+sg_1(x,E,s)$ with $g_0(x,E)$ smooth on $\{|x-x(E)|\le c$,
$|E-p(0)|\le c\}$ and $g_1(x,E,s)$ smooth on
$\{|x-x(E)|,|E-p(0)|,|s|\le c\}$.  For $|E-p(0)|<c$, $|x-x(E)|<\lambda^
{-\varepsilon}$, $s=\lambda^{-2}$, $\lambda>C$ we have
$|(\partial_xy)^2+s\{y,x\}-(p(x)-E)|\le C\lambda^{-N^\prime-2}$, with
$\{y,x\}$ given by (3).\smallskip
\item{(H)} For each fixed $E$, $(u_k(x,E))$ is the canonical solution
of the transport equations for the potential $p(x)-E$ and the turning
point $x(E)$.  In particular $u_0(x,E)\equiv 1$, and
$u_1(x,E)=\frac i2 \lim_{\delta\to 0+} \Bigl[\int_{x(E)+\delta}^x
(\frac 5{16} (p^\prime)^2(p-E)^{-5/2}-\frac 14 p^{\prime\prime}
(p-E)^{-3/2})\,dt-\sum\limits_{\ell=1}^3 q_\ell(E)\delta^{-\ell/2}\Bigr]$,
with $q_\ell(E)$ uniquely specified by demanding the finiteness of the limit.

We can give a positive lower bound for $c$ and positive upper bounds for
$C$ and for the $C^\infty$ seminorms of the functions $y(x,E,s)$, $f_k(y,E)$,
$g_0(x,E)$, $g_1(x,E,s)$ depending only on upper bounds for the $C^\infty$ 
seminorms of $p(x)$ and on a positive lower bound for $p^\prime(0)$.
\endproclaim


%Section 2:


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\centerline{\bf{Approximate Global Solutions of}}
\centerline{\bf{Ordinary Differential Equations}}
\smallskip


In this section, we attempt to write down explicit, highly accurate
approximate solutions of the ordinary differential equation
$$
\Bigl[{{d^2}\over {dx^2}}+(E-V(x))\Bigr]F=0\tag 1
$$
\noindent in an interval $I$ containing two turning points
$x_{\text{left}} < x_{\text{rt}}$.  First we cover $I$ by two
subintervals $I_{\text{left}}$,
$I_{\text{rt}}$, each containing only one of the
turning points.  Next we apply the local theory of the previous
section to construct an accurate approximate solution $F_{\text{left}}(x,E)$
of (1) in a small neighborhood $J_{\text{left}} \subset I_{\text{left}}$ of
the turning point.  We then extend $F_{\text{left}}(x,E)$ from the small
neighborhood $J_{\text{left}}$ to the whole of $I_{\text{left}}$.  It is
quite simple to define the extension:  Near the left endpoint of $J_{\text{
left}}$, $F_{\text{left}}(x,E)$ is very tiny.  Hence we may simply
set $F_{\text{left}}=0$ to be the left of $J_{\text{left}}$.  Near
the right endpoint of $J_{\text{left}}$, $F_{\text{left}}$ is given in terms
of the canonical solution $(u_k)$ of the transport equations.  A glance at
the transport equations shows that the solution $(u_k)$ can be
continued until we get to the second turning point.  Thus $(u_k)$
may be defined in all of $I_{\text{left}} \cap (x_{\text{left}},\infty)$,
and then $F_{\text{left}}(x,E)$ may be defined in terms of $(u_k)$
to the right of $J_{\text{left}}$ by the usual formula.  Thus we have
extended our approximate solution to all of $I_{\text{left}}$.  Of course
we still have to see how $F_{\text{left}}$ behaves on its enlarged
domain, and check that it approximately satisfies (1).  Also, there is an
approximate solution $F_{\text{rt}}(x,E)$ on $I_{\text{rt}}$, completely
analogous to $F_{\text{left}}$.

Finally, we attempt to patch together $F_{\text{left}}$ and $F_{\text{rt}}$
into a single approximate solution of (1) defined on the whole of $I$.
Unless $E$ lies near an eigenvalue of ${{-d^2}\over {dx^2}}+V(x)$,
we expect $F_{\text{left}}$ and $F_{\text{rt}}$ to disagree, so that
the patching cannot be done.  We will define complex approximate solutions
$F_c^{\text{left}}(x,E)$, $F_c^{\text{rt}}(x,E)$ on $I_{\text{left}} \cap
I_{\text{rt}}$ and a complex number $R(E)e^{i\Phi(E)}$ with the following
properties:  $F_{\text{left}}=\text{Re}(F_c^{\text{left}})$,
$F_{\text{rt}}=\text{Re}(F_c^{\text{rt}})$, and $F^{\text{left}}_c$
is very well approximated by  $R(E)e^{i\Phi(E)} F^{\text{rt}}_c
(x,E)$.  Thus, when $\Phi(E)\equiv 0 \mod \pi$ the two solutions
$F_{\text{left}}$, $F_{\text{rt}}$ are nearly proportional, and can
therefore be patched together into an approximate eigenfunction.  It is
natural to conjecture that the true eigenvalues of ${{-d^2}\over{dx^2}}+V(x)$
are well-approximated by the solutions of $\Phi(E)\equiv 0 \mod \pi$, and 
that the true eigenfunctions are nearly proportional to $F_{\text{left}}$
on $I_{\text{left}}$, and to $F_{\text{rt}}$ on $I_{\text{rt}}$.  However,
we postpone discussion of the true eigenvalues and eigenfunctions to
a later section.  Our task here is to construct the approximate solutions
$F_{\text{left}}(x,E)$, $F_{\text{rt}}(x,E)$ and the complex factor
$R(E)\exp(i\Phi(E))$.

We start by formulating our hypotheses on the potential $V(x)$.  They
involve auxiliary functions $S(x)$, $B(x)$ used to capture the size
and smoothness of $V$.  Our hypotheses are as follows.
\roster
\item"{(H0)}" $V(x)$, $S(x)$, $B(x)$ are functions defined on an
interval $I$, and $E_0$ is a real number.

\item"{(H1)}" $S(x),B(x)>0$ on $I$. If $x,y \in I$ and $|x-y|<cB(x)$,
then $cB(x)<B(y)<CB(x)$ and $cS(x)<S(y)<CS(x)$.

\item"{(H2)}" For $x \in I$ and $\alpha \ge 0$ we have
$|({{d}\over {dx}})^\alpha V(x)|\le C_{\alpha}S(x)(B(x))^{-\alpha}$.

\item"{(H3)}" The equation $V(x)=E_0$ has two solutions $x_{\text{left}}<x_
{\text{rt}}$ in $I$.

\item"{(H4)}" We have distance 
$(x_{\text{left}},\partial I)>cB(x_{\text{left}})$
and distance $(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$.

\item"{(H5)}" For $x \in [x_{\text{left}},x_{\text{left}}+
c_1B(x_{\text{left}})]$ we have $-V^\prime(x)
>cS(x_{\text{left}})/B(x_{\text{left}})$.

\item"{(H6)}" For $x \in [x_{\text{rt}}-c_1B(x_{\text{rt}}),x_{\text{rt}}]$
we have $+V^\prime(x)>cS(x_{\text{rt}})/B(x_{\text{rt}})$.

\item"{(H7)}" For $x \in [x_{\text{left}}+c_1B(x_{\text{left}}),x_{\text{rt}}
-c_1B(x_{\text{rt}})]$ we have $E_0-V(x)>cS(x)$.
\endroster

We will denote by $C_{\#}$, $c_{\#}$, $C_{\#}^{\alpha\beta}$ etc. a
positive constant depending only on the constants $c$, $C$, $C_1$,
$C_\alpha$ appearing in (H1)$\ldots$(H7), but not on $V(x)$, $S(x)$,
$B(x)$.
\vglue 1pc
\demo{Example}  Take $E_0=0$, $B(x)\equiv 1$, $S(x)\equiv \lambda^2$,
$I=[-1,1]$, $V(x)=\lambda^2p(x)$.  Then (H1)$\ldots$(H7) hold if $p$ is
smooth and $\{p<0\}=(x_{\text{lt}},x_{\text{rt}})$ with $p^\prime\ne 0$
at $x_{\text{lt}}$, $x_{\text{rt}}$.  Constants $C_{\#}$ depend 
on the $C^\infty$ seminorms of $p(x)$ and on lower bounds for
$|p^\prime(x_{\text{lt}})|$, $|p^\prime(x_{\text{rt}})|$, but are 
independent of $\lambda$.
\medskip

The following quantities play the role of the large parameter $\lambda$
in the  local WKB theory of the previous section.  Set $\lambda(x)=
S^{1/2}(x)B(x)$ and define $\Lambda$ by the equation
$$
\frac 1\Lambda=\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{1}{\lambda(x)}\frac{dx}{B(x)}=\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{S^{1/2}(x)B^2(x)}.
$$

In our elementary example above, $\lambda(x)\equiv\lambda$, $\Lambda=
(\text{const.})\lambda$.  Our approximate solutions will satisfy
(1) modulo a large negative power of $\Lambda$.

Another basic quantity is $\sigma(\beta)=\int_{x_{\text{left}}}^
{x_{\text{right}}}
\frac{\Lambda\,dy}{
\lambda(y)S^\beta(y)B(y)}$.  We note that 
$$
\sigma(\beta_1)\le\hfill\break
(\sigma(\beta))^{\beta_1/\beta}\quad\text{for}\ 0 \le \beta_1\le \beta,
\tag "{(H\"o A)}"
$$
\noindent as follows from H\"older's inequality and the definition of
$\Lambda$.  This implies the useful inequality
$$
\sigma(\beta_1)\sigma(\beta_2)\le \sigma(\beta_1+\beta_2)\tag "{(H\"o B)}"
$$
\noindent We define also $\sigma=\min\limits_{x \in I}S(x)$.

We begin the work of constructing the approximate solutions
$F_{\text{left}}$, $F_{\text{rt}}$.  Let $\varepsilon$, $N$ be
given, and let $N^\prime=[\varepsilon N/500]$ as in the local
WKB lemma.  We change notation slightly  and allow the constants
$c_{\#}$, $C_{\#}$ etc\. to depend on $\varepsilon$, $N$ as well
as on $c$, $c_1$, $C$, $C_\alpha$ in (H0)$\ldots$(H7).  We construct
an approximate solution of
$$
\Bigl(\frac{\partial^2}{\partial z^2}+\tilde E-V(z)\Bigr)\hat F(z,\tilde E)=0
\tag 2
$$
\noindent in a small neighborhood of the turning point
$x_{\text{left}}$.  To do so, we simply rescale the problem to
reduce matters to the setting of the local WKB lemma.  Specifically, set
$z(\tilde E)=(\text{solution\ of}\ V(z)=\tilde E$ near $z = x_{\text{left}}$)
for $|\tilde E-E_0|<c_{\#}S(x_{\text{left}})$.  Then define
$\lambda_{\text{left}}=S^{1/2}(x_{\text{left}})B(x_{\text{left}})$,
$p(x)=-V(z)/S(x_{\text{left}})$, $E=-\tilde E/S(x_{\text{left}})$, where
$z=x_{\text{left}}+c_{\#}B(x_{\text{left}})x$. 

The point is that $p(x)$
satisfies the hypotheses of the local WKB lemma, with constants depending
only on $c$, $c_1$, $C$, $C_\alpha$ in (H0)$\ldots$(H7).  We assume
$\lambda_{\text{left}} \ge C_{\#}$.  Then the local WKB lemma produces
approximate solutions $F_0(x,E,\lambda)$, $F_+(x,E,\lambda)$ to
$(\frac{\partial^2}{\partial x^2}+\lambda^2(p(x)-E))F=0$.

Recall that $F_0$ and $F_+$ are defined in terms of auxiliary functions
$y(x,E,s)$, $u_k(x,E)$.  Conclusion (G) of the local WKB lemma shows that
$y$ can be written as $y(x,E,s)=y_0(x,E)+sy_1(x,E,s)$ with $y_0$, $y_1$
smooth and $y_0(\frac{\partial y_0}{\partial x})^2=p(x)-E$ to
order $\ge N^\prime$ at $x(E)=(\text{solution\ to}\ p(x)=E)$.  We
pull back $F_0$, $F_+$, $y$, $y_0$, $y_1$, $u_k$ as follows.  With
$z=x_{\text{left}}+c_{\#}B(x_{\text{left}})\cdot x$ and
$E=-\tilde E/S(x_{\text{left}})$ as before, define
$$\gather
\hat F_0(z,\tilde E)=
B^{1/2}(x_{\text{left}})\cdot F_0(x,E,\lambda_{\text{left}})\\
\hat F_+(z,\tilde E)=
B^{1/2}(x_{\text{left}})\cdot F_+(x,E,\lambda_{\text{left}})\\
Y(z,\tilde E)=y(x,E,\lambda_{\text{left}}^{-2}), Y_0(z,\tilde E)=y_0(x,E),\\
\qquad\qquad\qquad 
\qquad Y_{\#}(z,\tilde E)=y_1(x,E,\lambda_{\text{left}}^{-2})\\
U_k(z,\tilde E)=\lambda_{\text{left}}^{-k}u_k(x,E).\endgather
$$

\noindent Note that $B^{1/2}(x_{\text{left}})(\frac{\partial y}{\partial x}
(x,E,\lambda_{\text{left}}^{-2}))^{-1/2}=(\frac{\partial Y(z,\tilde E)}
{\partial z})^{-1/2}$, so that
$\hat F_0(z,\tilde E)=\hfill\break
\lambda_{\text{left}}^{-1/3}(\frac{\partial Y(z,\tilde E)}
{\partial z})^{-1/2}A(\lambda_{\text{left}}^{2/3}Y(z,\tilde E))$.  Also,
$Y=Y_0+\lambda_{\text{left}}^{-2}Y_{\#}$ and $\lambda_{\text{left}}^2
Y_0(\frac{\partial Y_0}{\partial z})^2=\hfill\break
S(x_{\text{left}})y_0
(\frac{\partial y_0}{\partial x})^2=S(x_{\text{left}})
(p(x)-E)=\tilde E-V(z)$
to order at least $N^\prime$ at $z=z(\tilde E)$.

Regarding $\hat F_+$, we note that $\frac{B^{1/2}(x_{\text{left}})}
{\lambda^{1/2}_{\text{left}}(p(x)-E)^{1/4}}=\frac{1}{(\tilde E-V(z))^{1/4}}$
and $\lambda_{\text{left}}\int_{x(E)}^x
(p(t)-E)^{1/2}\,dt=\int_{z(\tilde E)}^z(\tilde E-V(t))^{1/2}\,dt$.  Thus,
$$
\hat F_+(z,\tilde E)=\text{Re}\Bigl[\frac{e^{\pm i\frac {\pi}{4}}\,
e^{i\int_{z(\tilde E)}^z(\tilde E-V(t))^{1/2}\,dt}}
{(\tilde E-V(z))^{1/4}}\,\sum\limits_{k=0}^{N^\prime}U_k(z,\tilde E)\Bigr].
$$

Our discussion of rescaling in the previous section shows that for fixed
$\tilde E$,\hfill\break
$(U_k(z,\tilde E))$ is the canonical solution of the transport
equations for the potential $\tilde E-V(z)$ and the turning point
$z=z(E)$.  Hence after an obvious change of notation, we arrive at the
following rescaled version of the local WKB lemma.

\proclaim{Lemma 1}  Suppose $V(x)$, $S(x)$, $B(x)$ satisfy 
(H0)$\ldots$(H7).  For $|E-E_0|<\hfill\break
c_{\#}S(x_{\text{left}})$,
let $x_{\text{left}}(E)$
be the solution of $V(x)=E$ near $x_{\text{left}}$.  Set $\lambda_{\text{left}}
=\hfill\break
S^{1/2}(x_{\text{left}})B(x_{\text{left}})$, and assume
$\lambda_{\text{left}}>C_{\#}$.  Let $(u_k^{\text{left}}(x,E))$ be the
canonical solution to the transport equations for the potential
$E-V(x)$ and the turning point $x_{\text{left}}(E)$.  Then there exist
auxiliary functions $Y(x,E)$, $Y_0(x,E)$, $Y_{\#}(x,E)$ with the
following properties.
\roster
\item"{(I)}" Set $F_0^{\text{left}}(x,E)=\lambda_{\text{left}}^{-1/3}
(\frac{\partial Y(x,E)}{\partial x})^{-1/2}A(\lambda_{\text{left}}^{2/3}
Y(x,E))$.  Then for $|x-x_{\text{left}}(E)|\hfill\break
<\lambda_{\text{left}}^{-\varepsilon}
B(x_{\text{left}})$ we have  $|(\frac{\partial^2}{\partial x^2}+E-V(x))F_0^
{\text{left}}|\le C_{\#}\lambda_{\text{left}}^{-N^\prime}B^{-3/2}
(x_{\text{left}})$.  
Also, for $0 \le \alpha \le 2$ and for $-\lambda_{\text{left}}^{-\varepsilon}
B(x_{\text{left}})<x-x_{\text{left}}(E)<-\lambda_{\text{left}}^{\varepsilon-2/3}
B(x_{\text{left}})$ we have $|\partial_x^\alpha F_0^{\text{left}}|\le C_{\#}
\lambda_{\text{left}}^{-N^\prime}B^{\frac 12-\alpha}(x_{\text{left}})$.
\medskip

\item"{(II)}" Set $F_+^{\text{left}}(x,E)=\text{Re}\Bigl[
\frac{e^{\pm i\frac \pi 4}\,e^{i\int_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt}}
{(E-V(x))^{1/4}}\,\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(x,E)\Bigr]$.
Then for $\lambda_{\text{left}}^{\varepsilon-2/3}B(x_{\text{left}})<x-x_
{\text{left}}(E)<c_{\#}B(x_{\text{left}})$ we have
$|(\frac{\partial^2}{\partial x^2}+E-V(x))F_+^{\text{left}}|\le C_{\#}
\lambda_{\text{left}}^{10}(\lambda_{\text{left}}^{2/3}(x-x_{\text{left}}(E))/
B(x_{\text{left}}))^{-\frac 32 N^\prime}B^{-\frac 32}(x_{\text{left}})$.
Also, for $0 \le \alpha  \hfill\break
\le 2$ and $\lambda_{\text{left}}^{\varepsilon-2/3}
B(x_{\text{left}})<x-x_{\text{left}}(E)<\lambda_{\text{left}}^{-\varepsilon}
B(x_{\text{left}})$ we have $|\partial_x^\alpha\{F_0^{\text{left}}-F_+^
{\text{left}}\}|\le C_{\#}\lambda_{\text{left}}^{10}(\lambda_{\text{left}}^
{2/3}(x-x_{\text{left}}(E))/B(x_{\text{left}}))^{-\frac 32 N^\prime}B^
{\frac 12-\alpha}(x_{\text{left}})$.\medskip

\item"{(III)}" We have $Y(x,E)=Y_0(x,E)+\lambda_{\text{left}}^{-2}Y_{\#}
(x,E)$ with $|\partial_x^\alpha\partial_E^\beta(Y_0,Y_{\#})|
\le  \hfill\break C_{\#}
^{\alpha\beta}B^{-\alpha}(x_{\text{left}})S^{-\beta}(x_{\text{left}})$,
(all $\alpha$, $\beta$) for $|x-x_{\text{left}}(E)|<c_{\#}B(x_{\text{left}})$
 and $\lambda_{\text{left}}^2Y_0(\frac{\partial Y_0}
{\partial x})^2=E-V(x)$ to order at least $N^\prime$ at $x=x_{\text{left}}(E)$.
For $|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\varepsilon}
B(x_{\text{left}})$ we have $|\lambda_{\text{left}}^{2}Y(\frac{\partial Y}
{\partial x})^2+\{Y,x\}-
(E-V(x))|\le C_{\#}\lambda^{-N^\prime}S(x_{\text{left}})$.
\medskip

\item"{(IV)}" The functions $u_k^{\text{left}}(x,E)$ may be written as
$u_k^{\text{left}}(x,E)=\hfill\break
\lambda_{\text{left}}^{-k}(\frac{x-x_{\text{left}}(E)}
{B(x_{\text{left}})})^{-\frac 32 k}\cdot f_k((\frac{x-x_{\text{left}}(E)}
{B(x_{\text{left}})})^{1/2},\frac {E-E_0}{S(x_{\text{left}})})$ with
$|\partial_y^\alpha\partial_s^\beta f_k(y,s)|\le C_{\#}^{\alpha\beta}$
(all $\alpha$, $\beta$) for $|y|$, $|s|<c_{\#}$.\smallskip

\item"{(V)}" We have $u_0(x,E)\equiv 1$, and
$$\multline
u_1(x,E)=\lim_{\delta\to 0+}\Bigl[\int_{x_{\text{left}}(E)+\delta}^x
\Bigl\{\frac{5i}{32}\frac{(V^\prime)^2}{(E-V)^{5/2}}+\frac i8\frac
{V^{\prime\prime}}{(E-V)^{3/2}}\Bigr\}\,dt-\hfill\break\\
\sum\limits_{\ell=1}^3 q_\ell(E)\delta^
{-\ell/2}\Bigr]\endmultline
$$
\noindent with $q_\ell(E)$ uniquely determined by demanding the finiteness
of the limit.\endroster\endproclaim

\demo{Remark}  The subscript in $Y_{\#}$ has nothing to do with our convention
for constants $C_{\#}$.  Note the signs in (V), which are correct for the
potential $E-V(x)$.

Thus, we have succeeded in finding a good approximate
solution of our ODE (1) in $U=\{|E-E_0|<c_{\#}S(x_{\text{left}})$,
$-\lambda_{\text{left}}^{-\varepsilon}B(x_{\text{left}})<x-x_{\text{left}}(E)<
c_{\#}B(x_{\text{left}})\}$.  We simply patch together the two solutions
$F_0^{\text{left}}$, $F_+^{\text{left}}$, with a partition of unity.  The
intersection 
of $U$ with a line $E=\text{const.}$ is the interval which we called
$J_{\text{left}}$ at the beginning of this section.

Our next task is to continue the approximate solution of (1) into a larger
domain by extending the canonical solution of the transport equations.
Thus define $U_{\text{center}}=\{|E-E_0|<c_{\#}^{(0)}\sigma$, $x_{\text{left}}
+c_{\#}^{(1)}B(x_{\text{left}})<x<x_{\text{rt}}-c_{\#}^{(1)}B(x_{\text{rt}})\}$.
The intersection of $U_{\text{center}}$ with a line $E=\text{const.}$\ is an
interval which we call $I_{\text{center}}$.  If we first take $c_{\#}^{(1)}$
small enough and then take $c_{\#}^{(0)}$ small depending on $c_{\#}^{(1)}$,
then the following properties hold:  $I_{\text{center}}\cap J_{\text{left}}\ne
\emptyset$ if $|E-E_0|<c_{\#}^{(0)}\sigma$.  Also $x-x_{\text{left}}(E)>
c_{\#}B(x_{\text{left}})$ for $|E-E_0|<c_{\#}^{(0)}\sigma$ and
$x \in J_{\text{left}} \cap I_{\text{center}}$.  Hence for such $(x,E)$
we have from Lemma 1 (IV) the estimates:
$$\multline
|\partial_x^\alpha\partial_E^\beta u_k^{\text{left}}(x,E)|\le C_{\#}^
{\alpha,\beta}
\lambda_{\text{left}}^{-k}
B^{-\alpha}(x_{\text{left}})S^{-\beta}(x_{\text{left}})
\\ (\text{all}\ \alpha,\beta)\,\, (x,E) \in U \cap U_{\text{center}}.
\endmultline\tag 3
$$
\noindent We continue $(u_k^{\text{left}}(x,E))$ to a solution of the
transport equations in $U_{\text{center}}$.  This can be done, as follows
by inspection of the transport equations and the fact 
that $E-V(x) \ge E_0-V(x)-
c_{\#}^{(0)}\sigma \ge c_{\#}S(x)$ in $U_{\text{center}}$.  We must
estimate $u_k^{\text{left}}(x,E)$ and its derivatives in $U_{\text{center}}$.
This is accomplished by the following result.
\vglue 1pc
\proclaim{Lemma 2}  In $U_{\text{center}}$ we have $|\partial_x^\alpha
\partial_E^\beta u_k^{\text{left}}(x,E)|\le C_{\#}^{\alpha\beta}
\Lambda^{-k}B^{-\alpha}(x)\sigma(\beta)$.\endproclaim
\medskip
\demo{Proof}  
Cover $I_{\text{center}}$ by intervals $I_\nu=\{x \in I \mid |x-x_
\nu|< b_\nu\}$, $1 \le \nu \le \nu_{\text{max}}$, with $b_\nu \sim B(x_\nu)$.
We may arrange the $b_\nu$, $x_\nu$ so that $x_{\nu+1}-x_\nu \sim B(x_\nu)$,
$I_\nu \cap I_{\nu+1}\ne \emptyset$, but $I_\mu \cap I_\nu =\emptyset$
for $|\mu-\nu|>1$.  
In each $I_\nu$, we take $(u_k^\nu(x,E))_{0\le k\le N^\prime}$
to be the solution of the transport equations
$$
u_0^\nu(x,E)\equiv 1 \tag 4
$$
$$\multline
2i \frac{\partial u_{k+1}^\nu(x,E)}{\partial x}+\Bigl(\frac {5}{16}
\frac{(V^\prime(x))^2}{(E-V(x))^{5/2}}+\frac 14 \frac{V^{\prime\prime}
(x)}{(E-V(x))^{3/2}}\Bigr)u_k^\nu(x,E)+\hfill\break\\ +\frac 12
\frac{V^\prime(x)}{(E-V(x))^{3/2}}\frac{\partial}{\partial x}u_k^\nu
(x,E) + \frac{1}{(E-V(x))^{1/2}}\frac{\partial^2}{\partial x^2}u_k^\nu(x,E)=0
\endmultline\tag 4
$$
\noindent for $x \in I_\nu$, $|E-E_0|<c_{\#}^{(0)}\sigma$, with boundary
conditions $u_k^\nu(x_\nu,E)=0$ for $k \ge 1$.  In this region we have
$E-V(x) \ge E_0-V(x)-c_{\#}^{(0)}\sigma \ge c_{\#}S(x_\nu)$.  Therefore
the transport equations take the form
$$
u_0^\nu \equiv 1, \frac{\partial u_{k+1}^\nu}{\partial x}
=\sum\limits_{\ell=0}^2 P_\ell
(x,E)(\frac{\partial}{\partial x})^\ell u_k^\nu, \,u_{k+1}^\nu=0 
\quad \text{at}\quad x=x_\nu,
$$
\noindent with $|\partial_x^\alpha\partial_E^\beta P_\ell|\le C_{\#}
^{\alpha\beta}S^{-\frac 12-\beta}(x_\nu)B^{-2+\ell-\alpha}(x_\nu)$
all $\alpha,\beta$ for $x \in I_\nu$, $|E-E_0|<c_{\#}^{(0)}\sigma$.
\noindent Hence an easy induction on $k$ shows that
$$
|\partial_x^\alpha\partial_E^\beta u_k^\nu
(x,E)|\le C_{\#}^{\alpha\beta}\lambda^{-k}
(x_\nu)B^{-\alpha}(x_\nu)
S^{-\beta}(x_\nu) \quad(\text{all}\ \alpha,\beta)\tag 5
$$
\noindent for $x \in I_\nu$, $|E-E_0|<c_{\#}^{(0)}\sigma$.

Next we relate $(u_k^\nu(x,E))$ to $(u_k^{\nu+1}(x,E))$ on $I_\nu \cap I_
{\nu+1}$.  Lemma 6 in the previous section shows that
$$
u_k^\nu(x,E)=\sum\limits_{\ell=0}^k h_\ell^\nu(E)u_{k-\ell}^{\nu+1}(x,E)\tag 6
$$
\noindent for uniquely determined 
$(h_\ell^\nu(E))$ with $h_0^\nu(E) \equiv 1$.
\noindent In particular, $h_k^\nu(E)=u_k^\nu(x,E)-\sum\limits_{\ell=0}^{k-1}
h_\ell^\nu(E)u_{k-\ell}^{\nu+1}(x,E)$ on $I_\nu \cap I_{\nu+1}$.  Hence
an easy induction on $k$ using (5) shows that
$$
|\partial_E^\beta h_k^\nu(E)|\le C_{\#}^\beta\lambda^{-k}(x_\nu)S^{-\beta}
(x_\nu)\quad  \text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma,\quad \text{all}\ \beta.
\tag 7
$$
\noindent Using (6), we can patch together the $(u_k^\nu(x,E))$ for
$1 \le \nu \le \nu_{\max}$ into a solution $(U_k(x,E))$ of the transport
equations on $U_{\text{center}}$.  We define
$$
U_k(x,E)=\sum\limits_{\ell=0}^k H_\ell^{1\nu}(E)u_{k-\ell}^\nu(x,E)
\quad \text{for}\ x \in I_\nu, |E-E_0|<c_{\#}^{(0)}\sigma, \tag 8
$$
\noindent with
$$
H_\ell^{\mu\nu}(E)=\sum\limits_{\ell_\mu+
\ell_{\mu+1}+\ldots+\ell_{\nu-1}=\ell}\
\prod_{j=\mu}^{\nu-1}\  h_{\ell_j}^j (E) \quad \text{for}\ \mu < \nu
$$
$$
H_\ell^{\mu\nu}(E)=\cases 1&\text{for $\ell=0$}\\
0&\text{for\ $\ell\ne 0$}\endcases \quad \text{for}\ \mu =\nu. 
$$
\noindent On each region $\Omega_\nu=
\{x \in I_\nu, |E-E_0|<c_{\#}^{(0)}\sigma\}$,
equation (8) defines a solution of the transport equations.  Moreover, our
descriptions of $U_k(x,E)$ on $\Omega_\nu$ and on $\Omega_{\nu+1}$
agree on $\Omega_\nu \cap \Omega_{\nu+1}$ by virtue of (6) and the
definition of $H_\ell^{\mu\nu}(E)$.

Next we estimate the global solution $U_k(x,E)$.  To do so, we use the 
inequality
$$
\sum\limits_{\nu=1}^{\nu_{\max}}\frac{\Lambda}{\lambda(x_\nu)S^\beta(x_\nu)}
\le C_{\#}\sigma(\beta),\tag 9
$$
\noindent which follows from the definition of $\sigma(\beta)$
and the geometry of the $x_\nu$.  We begin with the size of the $H_\ell^{\mu\nu}$
.  We have for $\mu<\nu$:
$$\align
\sum\limits_{\ell=0}^{N^\prime}\Lambda^\ell|H_\ell^{\mu\nu}(E)|&\le \sum\limits_
{\ell_\mu+\ldots+\ell_{\nu-1}\le N^{\prime}}\prod_{j=\mu}^{\nu-1}
\Lambda^{\ell_j}|h_{\ell_j}^j(E)|\\
&\le \prod_{j=\mu}^{\nu-1}\Bigl(\sum\limits_{\ell=0}^{N^\prime}\Lambda^\ell
|h_\ell^j(E)|\Bigr).\tag 10\endalign
$$

\noindent For each $j$ we have $\sum\limits_{\ell=0}^{N^\prime}\Lambda^\ell
|h_\ell^j(E)|\le 1+\sum\limits_{\ell=1}^{N^\prime}
 C_{\#}\Lambda^\ell\lambda^{-\ell}
(x_j)$ by (7), and the right--hand side is less than or equal to
$\exp(C_{\#}\Lambda/\lambda(x_j))$.  Putting this into (10), we find that
$$
\sum\limits_{\ell=0}^{N^\prime}\Lambda^\ell |H_{\ell}^{\mu\nu}(E)|\le
\exp\Bigl(\sum\limits_{j=\mu}^{\nu-1}\frac{C_{\#}\Lambda}{\lambda(x_j)}\Bigr).
$$
\noindent The right--hand side is at most $C_{\#}$ by (9) with $\beta=0$
and therefore
$$
|H_\ell^{\mu\nu}(E)|\le C_{\#}\Lambda^{-\ell}\quad (0 \le \ell \le N^\prime)
\quad\text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma, \mu < \nu. \tag 11
$$
\noindent Obviously, (11) holds also for $\mu=\nu$.

Next we control the derivatives of $H_\ell^{\mu\nu}$.  We will show that
$$\split
|\partial_E^\beta H_k^{\mu\nu}(E)|\le C_{\#}^\beta\Lambda^{-k}\sigma(\beta)
\quad (\text{all} \ \beta)\\
\text{for}\ 0 \le k \le N^\prime, |E-E_0|< c_{\#}^{(0)}\sigma, \mu \le \nu.
\endsplit\tag 12
$$
\noindent We use induction on $\beta$.  The case $\beta=0$ is already
known.  We assume (12) for $0 \le \beta \le \overline\beta$ and then
prove (12) for $\beta = \overline\beta+1$.  The case $\mu=\nu$ is trivial,
so suppose $\mu <\nu$.  Differentiating the sum of products defining
$H_\ell^{\mu\nu}$, we get
$$
(\frac{\partial}{\partial E})H_k^{\mu\nu}(E)=\sum\limits_{j=\mu}^{\nu-1}\
\sum\limits_{
k_1+k_2+k_3=k}\ H_{k_1}^{\mu j}(E)\Bigl[\frac{\partial h_{k_2}^j(E)}
{\partial E}\Bigr] H_{k_3}^{j+1\nu}(E),\quad \text{and}\tag 13
$$
\noindent therefore 
$$
(\partial_E^{\overline\beta+1}H_k^{\mu\nu})=\sum\limits_{j=\mu}^{\nu-1}
\sum\Sb k_1+k_2+k_3=k\\ \beta_1+\beta_2+\beta_3=\overline \beta\endSb
\text{coeff}\ (\beta_1\beta_2\beta_3)(\partial_E^{\beta_1}H_{k_1}
^{\mu j})(\partial_E^{\beta_2+1}h_{k_2}^j)(\partial_E^{\beta_3}
H_{k_3}^{j+1\nu}).\tag 14
$$

We use inductive hypothesis (12) together with (7) to estimate the
terms on the right of (14).  Note that we may restrict the sum in (14)
to $k_2>0$, since for $k_2=0$ we have $h_{k_2}^j(E)\equiv 1$ so that
$\partial_E^{\beta_2+1}h_{k_2}^j\equiv 0$.  Hence from (7), (14) and
inductive hypothesis we get
$$
|\partial_E^{\overline\beta+1}H_k^{\mu\nu}|\le C_{\#}^{\overline\beta}\
\sum\limits_{j=\mu}^{\nu-1}\ 
\sum\Sb k_1+k_2+k_3=k\\ \beta_1+\beta_2+\beta_3=\overline\beta\\
k_2>0\endSb\ \Lambda^{-k_1}\sigma(\beta_1)
\lambda^{-k_2}(x_j)S^{-\beta_2-1}(x_j)\Lambda^{-k_3}\sigma(\beta_3).
$$
\noindent Since $\lambda(x_\nu)\ge\Lambda$, this implies by virtue of (9)
and (H\"o B) that
$$\align
|\partial_E^{\overline\beta+1}H_k^{\mu\nu}|&\le \sum\limits_{\beta_1+\beta_2+
\beta_3=\overline\beta} C_{\#}^{\overline\beta}\Lambda^{-k}\sigma(\beta_1)
\sigma(\beta_3)\sum\limits_{j=\mu}^{\nu-1}\Bigl(\sum\limits_{k_2=1}^k
(\frac{\Lambda}{\lambda(x_j)})^{k_2}\Bigr)S^{-(\beta_2+1)}(x_j)\\ 
&\le C_{\#}^{\overline\beta}\Lambda^{-k}\sum\limits_{\beta_1+\beta_2+\beta_3=
\overline\beta}\sigma(\beta_1)\sigma(\beta_3)\sum\limits_{j=\mu}^
{\nu-1} \frac{\Lambda}{\lambda(x_j)S^{\beta_2+1}(x_j)}\\
&\le \frac{C_{\#}^{\overline\beta}}{\Lambda^k}\sum\limits_{\beta_1+\beta_2+
\beta_3=\overline\beta}\sigma(\beta_1)\sigma(\beta_2+1)\sigma(\beta_3)
\le \frac {C_{\#}}{\Lambda^k}\sigma(\overline\beta+1)\endalign
$$
\noindent Thus we have proven (12) for $\beta=\overline\beta+1$, completing
the induction step.  So (12) is proven for all $\beta$. 

Now from (5), (8), (12), (H\"o B) we obtain easily the estimate
$$
|\partial_x^\alpha\partial_E^\beta U_k(x,E)|\le C_{\#}^{\alpha\beta}
\Lambda^{-k}B^{-\alpha}(x)\sigma(\beta)\quad \text{for}\ (x,E) \in U_
{\text{center}}.\tag 15
$$
\noindent To complete the proof of Lemma 2,  we relate $u_k^{\text{left}}(x,E)$
to $U_k(x,E)$.  By Lemma 6 in the previous section, we have
$$
u_k^{\text{left}}(x,E)=\sum\limits_{\ell=0}^k h_\ell(E)U_{k-\ell}
(x,E)\quad \text{in}\ U_{\text{center}}\tag 16
$$
\noindent for a unique set of $(h_\ell(E))$ with $h_0(E)\equiv 1$.

To estimate $h_\ell(E)$, we take for each $E$ some $x_0$ so that $(x_0,E)
\in U \cap U_{\text{center}}$.  Estimate (3) applies at $(x_0,E)$, from
which we deduce the weaker estimate 
$$
|\partial_E^\beta u_k^{\text{left}}(x_0,E)|\le C_{\#}^\beta\Lambda^{-k}
\sigma(\beta).\tag 17
$$
\noindent Since $h_k(E)=u_k^{\text{left}}(x_0,E)-\sum\limits_{\ell=0}^
{k-1}h_\ell(E)U_{k-\ell}(x_0,E)$, an obvious induction on $k$, using
(15) and (17) and (H\"o B), implies the estimate
$$
|\partial_E^\beta h_k(E)|\le C_{\#}^\beta\Lambda^{-k}\sigma(\beta)\quad
\text{for}\ |E-E_0|< c_{\#}^{(0)}\sigma,\ \text{all}\ \beta. \tag 18
$$
\noindent From (15), (16), (18) and (H\"o B) we get
$$\multline
|\partial_x^\alpha\partial_E^\beta u_k^{\text{left}}(x,E)|\le C_{\#}
^{\alpha\beta}\Lambda^{-k}B^{-\alpha}(x)\sigma(\beta)\quad \text{for}\
(x,E) \in U_{\text{center}},\ 0 \le k \le N^\prime,\\
\text{all} \ \alpha,\beta.\endmultline
$$
\noindent This is the conclusion of Lemma 2. $\qquad \blacksquare$
\medskip

It is now trivial to show that
$$
F_+^{\text{left}}(x,E)=\text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4}\,
e^{i\int_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}
\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(x,E)\Bigr]
$$
\noindent approximately solves the ODE (1) in $U_{\text{center}}$.  The
result is as follows.
\vglue 1pc
\proclaim{Corollary} 
$$|(\frac{\partial^2}{\partial x^2}+E-V(x))F_+^{\text{left}}
(x,E)|\le C_{\#}\Lambda^{-N^{\prime}}S^{-1/4}(x)B^{-2}(x)\quad
\text{in}\quad U_{\text{center}}.$$\endproclaim
\medskip
\demo{Proof}  A formal calculation using the transport equations gives
$$\split
(\frac{\partial^2}{\partial x^2}+E-V(x))F_+^{\text{left}}=\text{Re}
\Bigl[e^{\pm i \pi/4} e^{i\int_{x_{\text{left}}(E)}^x (E-V(t))^{1/2}\,dt}\\
\cdot (E-V(x))^{1/4}\Cal Lu_{N^\prime}(x,E)\Bigr]\endsplit\tag 19
$$
\noindent with
$\Cal L=\Bigl(\frac {5}{16}\frac{(V^\prime)^2}{(E-V)^{5/2}}+\frac 14
\frac{V^{\prime\prime}}{(E-V)^{3/2}}\Bigr)+\frac 12
\frac{V^\prime}{(E-V)^{3/2}}\frac {\partial}{\partial x}+\frac
{1}{(E-V)^{1/2}}\frac {\partial^2}{\partial x^2}$.
\noindent Since only negative powers of $(E-V)$ appear in $(E-V)^{1/4}\Cal L$,
and since $E-V(x)>c_{\#}S(x)$ (as we saw in the proof of Lemma 2), we see
that $(E-V)^{1/4}\Cal L=\sum\limits_{\ell=0}^2 \hat P_\ell 
(x,E)(\frac{\partial}{\partial x})
^\ell$ with $|\hat P_\ell|\le C_{\#}S^{-1/4}(x)B^{-2+\ell}(x)$ in
$U_{\text{center}}$.
\noindent Therefore $|(E-V)^{1/4}\Cal Lu_{N^\prime}|\le\hfill\break
\sum\limits_{\ell=0}^2
C_{\#}S^{-1/4}(x)B^{-2+\ell}(x)|\partial_x^\ell u_{N^\prime}(x,E)|\le
\sum\limits_{\ell=0}^2 C_{\#}S^{-1/4}(x)B^{-2+\ell}(x)\cdot\hfill\break
 C_{\#}
\Lambda^{-N^\prime}B^{-\ell}(x)$ by Lemma 2.  That is,
$|(E-V)^{1/4}\Cal Lu_{N^\prime}(x,E)|\le\hfill\break
 C_{\#}\Lambda^{-N^\prime}S^{-1/4}
(x)B^{-2}(x)$ on $U_{\text{center}}$.
The Corollary now follows at once from (19).$\,\, \blacksquare$
\medskip
We have carried out the task of extending $F_+^{\text{left}}(x,E)$ to
an approximate solution of (1) in $U_{\text{center}}$.  Completely analogous
to our solution of (1) is another approximate solution defined near
$x=x_{\text{rt}}(E)$.  This solution has the form
$F_0^{\text{rt}}(x,E)=\lambda_{\text{rt}}^{-1/3}\Bigl(\frac
{-\partial Y^{\text{rt}}(x,E)}{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{rt}}
^{2/3}Y^{\text{rt}}(x,E))$ for $|x-x_{\text{rt}}(E)|<\lambda_{\text{rt}}^
{-\varepsilon}B(x_{\text{rt}})$, $|E-E_0|<c_{\#}S(x_{\text{rt}})$, with
$\lambda_{\text{rt}}=S^{1/2}(x_{\text{rt}})B(x_{\text{rt}})$ and a suitable
function $Y^{\text{rt}}(x,E)$.  For \hfill\break
$-c_{\#}B(x_{\text{rt}})<x-x_{\text{rt}}
(E)<-\lambda_{\text{rt}}^{\varepsilon-2/3}B(x_{\text{rt}})$,
$|E-E_0|<c_{\#}S(x_{\text{rt}})$ we have
$$
F_+^{\text{rt}}(x,E)=\text{Re}\Bigl[\frac{e^{\mp i\frac \pi 4}\,
e^{-i\int_x^{x_{\text{rt}}(E)}(E-V(t))^{1/2}dt}}{(E-V(x))^{1/4}}
\sum\limits_{k=0}^{N^\prime}u_k^{\text{rt}}(x,E)\Bigr]
$$
\noindent where, in an obvious sense $(u_k^{\text{rt}}(x,E))$ is the canonical
solution of the transport equations for the potential $E-V(x)$ and the turning
point $x_{\text{rt}}(E)$.  As in the preceding lemma, $(u_k^{\text{rt}}(x,E))$
continues to a solution of the transport equations in $U_{\text{center}}$,
and the function $F_+^{\text{rt}}(x,E)$ approximately solves the ODE (1) there.
In particular $u_0^{\text{rt}}(x,E)\equiv 1$ and
$$\multline
u_1^{\text{rt}}(x,E)= 
- \operatornamewithlimits{lim}_{\delta\to 0+}\Bigl[
\int_x^{x_{\text{rt}}(E)-\delta}\Bigl\{\frac {5i}{32} \frac
{(V^\prime)^2}{(E-V)^{5/2}}+\frac i8 \frac{V^{\prime\prime}}{(E-V)^{3/2}}
\Bigr\}\,dt\\
-\sum\limits_{\ell=1}^3 q_\ell^{\text{rt}}(E)\delta^{-\ell/2}\Bigr]
\endmultline\tag 20
$$
\noindent with $q_{\ell}^{\text{rt}}(E)$ uniquely specified by demanding the
finiteness of the limit.

Our next task is to compare $(u_k^{\text{left}}(x,E))$ with $(u_k^{\text{rt}}
(x,E))$ in $U_{\text{center}}$.  Both solve the same transport equations, so
by Lemma 6 in the previous section, they are related by the equation
$$
u_k^{\text{left}}(x,E)=\sum\limits_{\ell=0}^k G_\ell(E)u_{k-\ell}^{\text{rt}}
(x,E) \quad (x,E) \in U_{\text{center}} \tag 21
$$
\noindent for uniquely determined coefficients $(G_\ell(E))_{0\le \ell \le
N^\prime}$ with $G_0(E)\equiv 1$.  We can estimate $G_\ell(E)$ by using
Lemma 2 and its analogue for $u_k^{\text{rt}}(x,E)$.  In particular, we have
$$
|\partial_E^\beta u_k^{\text{left}}|, |\partial_E^\beta u_k^{\text{rt}}|\le
C_{\#}^\beta\Lambda^{-k}\sigma(\beta)\quad \text{in}\ U_{\text{center}},\quad
\text{all}\ \beta. \tag 22
$$

Since $G_k(E)=u_k^{\text{left}}(x,E)-\sum\limits_{\ell=0}^{k-1} G_\ell
(E)u_{k-\ell}^{\text{rt}}(x,E)$, an obvious induction on $k$ using (22),
(H\"o B) shows that
$$
|\partial_E^\beta G_k(E)|\le C_{\#}^\beta\Lambda^{-k}\sigma(\beta)\quad
\text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma,\quad \text{all}\ \beta. \tag 23
$$
\noindent As a simple but important consequence of (23), we have
$$
|\sum\limits_{k=1}^{N^\prime}G_k(E)|<\frac {1}{10}\quad \text{for}\
|E-E_0|< c_{\#}^{(0)}\sigma, \tag 24
$$
\noindent provided we have $\Lambda\ge C_{\#}$.

We need an explicit formula for $G_1(E)$.  Since $u_0^{\text{left}}\equiv u_0
^{\text{rt}}\equiv 1$ and $G_0(E)\equiv 1$, equation (21) for $k=1$
becomes
$$
G_1(E)=u_1^{\text{left}}(x,E)-u_1^{\text{rt}}(x,E), \quad (x,E) \in U_
{\text{center}}.
$$

Using the known formulas (20) and Lemma 1 (IV) for $u_1^{\text{left}}$,
$u_1^{\text{rt}}$, we obtain
$$\multline
G_1(E)=\lim_{\delta\to 0+}\ \Bigl[\int_{x_{\text{left}}(E)+\delta}^
{x_{\text{rt}}(E)-\delta}\Bigl\{\frac {5i}{32} \frac{(V^\prime)^2}
{(E-V)^{5/2}}+\frac i8 \frac{V^{\prime\prime}}{(E-V)^{3/2}}\Bigr\}\,dt\\
- \sum\limits_{\ell=1}^3 q_\ell^{\text{total}}(E)\delta^{-\ell/2}\Bigl]
\endmultline \tag 25
$$
\noindent with $q_\ell^{\text{total}}(E)$ uniquely specified by demanding
the finiteness of the limit.  Formula (25) can be simplified, using the
elementary identity $(\frac{5i}{48}\frac{V^\prime}{(E-V)^{3/2}})^\prime=
\frac {5i}{32}\frac{(V^\prime)^2}{(E-V)^{5/2}}+\frac{5i}{48}\frac{V^{\prime
\prime}}{(E-V)^{3/2}}$, which integrates to
$$\split
\int_{x_{\text{left}}(E)+\delta}^{x_{\text{rt}}(E)-\delta}\Bigl\{
\frac{5i}{32}\frac{(V^\prime)^2}{(E-V)^{5/2}}+\frac{5i}{48}
\frac{V^{\prime\prime}}{(E-V)^{3/2}}\Bigl\}\,dt\
=\\
\frac{5i}{48}\frac{V^\prime(x)}{((E-V)(x))^{3/2}}\Bigm|_{x=x_{\text{left}}
(E)+\delta}^{x=x_{\text{rt}}(E)-\delta}. \endsplit\tag 26
$$
\smallskip

For $x=x_{\text{left}}(E)+\delta$ we have $\frac{V^\prime(x)}{(E-V(x))^{3/2}}
=\delta^{-3/2}\Bigl[\bigl(\frac{E-V(x)}{\delta}\bigr)^{-3/2}V^\prime(x)\Bigr]$,
and the quantity in square brackets is a smooth function of $\delta$.  Hence
$\frac{5i}{48}\cdot\hfill\break
\frac{V^\prime(x)}{(E-V(x))^{3/2}}\Bigm|_{x=x_{\text{left}}
(E)+\delta}=a(E)\delta^{-3/2}+b(E)\delta^{-1/2}+O(\delta^{+1/2})$ for small
$\delta>0$.  A similar formula holds for $x=x_{\text{rt}}(E)-\delta$, and
therefore (26) implies
$$
\lim_{\delta\to 0+}\Bigl[\int_{x_{\text{left}}(E)+\delta}^{x_{\text{rt}}
(E)-\delta}\Bigl\{\frac {5i}{32}\frac{(V^{\prime})^2}{(E-V)^{5/2}}
+\frac{5i}{48}\frac{V^{\prime\prime}}{(E-V)^{3/2}}\Bigl\}\,dt-
\sum\limits_{\ell=1}^3 q_\ell^{\text{extra}}(E)\delta^{-\ell/2}\Bigl]=0,
$$
\noindent with $(q_\ell^{\text{extra}}(E))$ uniquely specified by demanding
the finiteness of the limit.  Subtracting this equation from (25), we obtain
$$
G_1(E)=\frac{i}{48}\ \lim_{\delta\to 0+}\ \Bigl[ \int_{x_{\text{left}}(E)+\delta}
^{x_{\text{rt}}(E)-\delta}V^{\prime\prime}(t)(E-V(t))^{-3/2}\,dt-\sum\limits_
{\ell=1}^3 q_\ell^{\text{final}}(E)\delta^{-\ell/2}\Bigl],
$$
\noindent with $q_\ell^{\text{final}}(E)$ uniquely specified by demanding
the finiteness of the limit.  The integral is easily seen to be
$0(\delta^{-1/2})$, so $q_\ell^{\text{final}}(E)=0$ for $\ell=2,3$.  Thus
we obtain our basic formula for $G_1(E)$, namely
$$
G_1(E)=\frac {i}{48}\lim_{\delta\to 0+}\Bigl[\int_{x_{\text{left}}(E)+\delta}^
{x_{\text{rt}}(E)-\delta}V^{\prime\prime}(t)(E-V(t))^{-3/2}\,dt-q(E)\delta^{-1/2}
\Bigl], \tag 27
$$
\noindent with $q(E)$ specified by demanding the finiteness of the limit.

Now we apply equation (21) and our knowledge of the $G_k(E)$ to compare
the approximate ODE solutions $F_+^{\text{left}}(x,E)$ and $F_+^{\text{rt}}
(x,E)$ on their common domain $U_{\text{center}}$.  We will see that they
are real parts of complex approximate solutions $F_c^{\text{left}}(x,E)$,
$F_c^{\text{rt}}(x,E)$ which differ only by a complex factor $\Cal A(E)$
modulo a tiny error.  In fact, define
$$\align
F_c^{\text{left}}(x,E)&=\frac{e^{\pm i\frac \pi 4}\,
e^{i\int_{x_{\text{left}}(E)}
^x(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\sum\limits_{k=0}^{N^\prime}
u_k^{\text{left}}(x,E)\
\text{for}\ (x,E) \in U_{\text{center}}\\
F_c^{\text{rt}}(x,E)&=\frac{e^{\mp i\frac \pi 4}\,e^{-i\int_x^{x_{\text{rt}}(E)}
(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\sum\limits_{k=0}^{N^\prime}u_k^{\text{rt}}
(x,E)\
\text{for}\ (x,E) \in U_{\text{center}}\\
\Cal A_{\text{semiclassical}}(E)&=
\exp\{\pm i\frac \pi 2+i\int_{x_{\text{left}}(E)}
^{x_{\text{rt}}(E)}(E-V(t))^{1/2}dt\}\
\text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma\\
\Cal A_{\text{correction}}(E)&=\sum\limits_{k=0}^{N^\prime}G_k(E)=1+\sum\limits_
{k=1}^{N^\prime}G_k(E)\
\text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma\\
\Cal A(E)&=\Cal A_{\text{semiclassical}}(E)\cdot\Cal A_{\text{correction}}(E)\
\text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma.\endalign
$$

\noindent Note that $|\Cal A_{\text{semiclassical}}(E)|=1$ and
$|1-\Cal A_{\text{correction}}(E)|<\frac {1}{10}$, by (24).  Evidently,
$F_+^{\text{left}}=\text{Re}(F_c^{\text{left}})$ and $F_+^{\text{rt}}
=\text{Re}(F_c^{\text{rt}})$ on $U_{\text{center}}$.  We have 
$$\multline
\Cal A_{\text{correction}}(E)\cdot\sum\limits_{k=0}^{N^\prime}u_k^{\text{rt}}
(x,E)=\sum\limits_{k_1,k_2=0}^{N^\prime}G_{k_1}(E)u_{k_2}^{\text{rt}}(x,E)=\\
\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(x,E)+\sum\limits_{k_1+k_2>N^\prime}
G_{k_1}(E)\cdot u_{k_2}^{\text{rt}}(x,E)\equiv \sum\limits_{k=0}^{N^\prime}
u_k^{\text{left}}(x,E)+u_{\text{error}}(x,E),\endmultline\tag 28
$$
\noindent by virtue of (21).

\noindent To estimate $u_{\text{error}}(x,E)$ we recall that $|G_{k_1}(E)|\le
C_{\#}\Lambda^{-k_1}$ by (23), and  \hfill\break
$|\partial_x^\alpha u_{k_2}^{\text{rt}}
(x,E)|\le C_{\#}^\alpha\Lambda^{-k_2}B^{-\alpha}(x)$ by the analogue of
Lemma 2 for $(u_k^{\text{rt}})$.  Hence
$|\partial_x^\alpha\{G_{k_1}(E)u_{k_2}^{\text{rt}}(x,E)\}|\le
C_{\#}^\alpha\Lambda^{-N^\prime-1}B^{-\alpha}(x)\quad\text{for}\
k_1+k_2>N^\prime$, so
$$
|\partial_x^\alpha u_{\text{error}}(x,E)|\le C_{\#}^\alpha\Lambda^{-N^\prime-1}
B^{-\alpha}(x)\quad\text{in}\ U_{\text{center}}.\tag 29
$$
\noindent Thus $u_{\text{error}}(x,E)$ is extremely small.

\noindent Also
$$
\Cal A_{\text{semiclassical}}(E)\cdot\frac{e^{\mp i\frac \pi 4}\,e^{-i
\int_x^{x_{\text{rt}}(E)}(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}=
\frac{e^{\pm i\frac \pi 4}\,e^{+i\int_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt}}
{(E-V(x))^{1/4}}.
$$

\noindent Multiplying this identity by (28) and recalling the definitions of
$F_c^{\text{left}}$, $F_c^{\text{rt}}$, we get
$$\align
\Cal A(E)F_c^{\text{rt}}(x,E)&=F_c^{\text{left}}(x,E)+
\Bigl[A_{\text{semiclassical}}(E)\cdot\frac{e^{\mp i\frac \pi 4}
\,e^{-i\int_x^{x_{\text{rt}}(E)}(E-V(t))^{1/2}\,dt}}
{(E-V(x))^{1/4}}\cdot\\
&{}\qquad\qquad\qquad\qquad\qquad\qquad\cdot u_{\text{error}}(x,E)\Bigl]\\
&\equiv F_c^{\text{left}}(x,E)+F_{\text{error}}(x,E).
\tag 30\endalign
$$

\noindent Since $|\Cal A_{\text{semiclassical}}(E)|=1$, estimate (29) gives
$$
|F_{\text{error}}(x,E)|\le C_{\#}\Lambda^{-N^\prime-1}(E-V(x))^{-1/4}
\quad\text{in}\quad U_{\text{center}}.\tag 31
$$

\noindent Also 
$$\multline
|\frac{\partial}{\partial x}F_{\text{error}}(x,E)|=
|i(E-V(x))^{1/4}u_{\text{error}}(x,E)+\frac 14 (E-V(x))^{-5/4}V^\prime(x)
\cdot\\ \qquad\qquad\qquad\cdot
u_{\text{error}}(x,E)
+(E-V(x))^{-1/4}\frac{\partial}{\partial x}u_{\text{error}}(x,E)|\\
\le (E-V(x))^{1/4}\{|u_{\text{error}}(x,E)|\cdot(1+(E-V(x))^{-3/2}|V^\prime
(x)|)+(E-V(x))^{-1/2}|\frac{\partial}{\partial x}\cdot\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,
\cdot u_{\text{error}}(x,E)|\}\\
\le C_{\#}(E-V(x))^{1/4}\{\Lambda^{-N^\prime-1}\cdot(1+(E-V(x))^{-3/2}
|V^\prime(x)|)+(E-V(x))^{-1/2}\cdot\\
\qquad\qquad\qquad\qquad\qquad\cdot \Lambda^{-N^\prime-1}B^{-1}(x)\}
\endmultline\tag 31
$$
\noindent by (29).  Since $E-V(x)\ge c_{\#}S(x)$ in $U_{\text{center}}$
and $|V^\prime(x)|\le C_{\#}S(x)B^{-1}(x)$, this gives
$$
|\frac{\partial}{\partial x}F_{\text{error}}(x,E)|\le C_{\#}(E-V(x))^{1/4}
\{\Lambda^{-N^\prime-1}\cdot(1+S^{-1/2}(x)B^{-1}(x))\}.
$$
\noindent Since $S^{1/2}(x)B(x)=\lambda(x)\ge c_{\#}\Lambda>c_{\#}$, this implies
$$
|\frac{\partial}{\partial x} F_{\text{error}}(x,E)|\le
C_{\#}\Lambda^{-N^\prime-1}(E-V(x))^{+\frac 14}\quad\text{in}\ U_{\text{center}}.
\tag 32
$$
\noindent Estimates (31), (32) show that $F_{\text{error}}$ is extremely small.

Note that $F_c^{\text{left}}(x,E)$ and $F_c^{\text{rt}}(x,E)$ are excellent
approximate solutions of (1) in $U_{\text{center}}$.  In fact, the proof of
the Corollary to Lemma 2 shows that
$$\multline
|(\frac{\partial^2}{\partial x^2}+E-V(x))F_c^{\text{left}}|,
|(\frac{\partial^2}{\partial x^2}+E-V(x))F_c^{\text{rt}}|\le
C_{\#}\Lambda^{-N^\prime}S^{-1/4}(x)B^{-2}(x)\\
\text{in}\ 
U_{\text{center}}.\endmultline\tag 33
$$

We have almost completely carried out the program sketched in the introduction
to this section.  What remains is to write the complex factor
$\Cal A(E)$ in the form $R(E)e^{i\Phi(E)}$.  That is, we have to define a
branch of the complex logarithm of $\Cal A(E)$.  This is quite easy.  Let
$\log(1+z)$ be the branch of the logarithm defined on $\{|z|<\frac {1}{10}\}$
and equal to zero when $z=0$.  Then define the complex phase   
$$\multline
\Phi_c(E)=\pm i\frac \pi 2+i\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
(E-V(t))^{1/2}\,dt+
\log(1+\sum\limits_{k=1}^{N^\prime}G_k(E))\\
\text{for}\ |E-E_0|<c_{\#}^{(0)}\sigma.\endmultline
$$
\noindent This makes sense by virtue of (24).  Now we have
$\Cal A(E)=R(E)e^{i\Phi(E)}$ with $R(E)=\exp(\text{Re}\ \Phi_c(E))$,
$\Phi(E)=\text{Im}(\Phi_c(E))$.  We make some straightforward estimates
on $R(E)$, $\Phi(E)$.  From (23), (27) we have $|G_k(E)|\le C_{\#}\Lambda^{-k}$,
$\text{Re}\ G_1(E)=0$.  Therefore $|\text{Re}\ \Phi_c(E)|\le C_{\#}
\Lambda^{-2}$, so $|R(E)-1|\le C_{\#}\Lambda^{-2}$.

\noindent Also,
$$\multline
\Phi(E)=\pm \frac \pi 2+\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
(E-V(t))^{1/2}\,dt+\text{Im}\biggl\{G_1(E)+\Bigl[\log(1+G_1(E))-G_1(E)\Bigr]\\
+\Bigl[\log\Bigl(1+\frac{\sum\limits_{k=2}^{N^\prime}G_k(E)}{1+G_1(E)}\Bigr)\Bigr]
\biggr\}\endmultline
$$
\noindent and the two expressions in square brackets satisfy
$|\partial_E^\beta[\text{Expression}]|\le C_{\#}^\beta\Lambda^{-2}
\sigma(\beta)$ by virtue of (23).  Thus $\Phi(E)=\pm \frac \pi 2+
\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}(E-V(t))^{1/2}\,dt+\text{Im}\
G_1(E)+\Phi_{\text{error}}(E)$, with $|\partial_E^\beta\Phi_{\text{error}}(E)|
\le C_{\#}^\beta\Lambda^{-2}\sigma(\beta)$.  Recalling equation (27)
for $G_1(E)$, we see that
$\Phi(E)=\pm\frac \pi 2+\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}(E-V(t))^{1/2}
\,dt+\frac {1}{48}\lim_{\delta\to 0+}\Bigl[\int_{x_{\text{left}}(E)+\delta}
^{x_{\text{rt}}(E)-\delta}V^{\prime\prime}(t)(E-V(t))^{-3/2}\,dt-q(E)
\delta^{-1/2}\Bigl]+\Phi_{\text{error}}(E)$ with
$q(E)$ uniquely specified by demanding the finiteness
of the limit.





%Section 3:

\pageno 55
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\centerline{\bf{The Global WKB Lemma}}
\smallskip

In this section we assume the following 

\noindent{\bf Hypotheses:}
\roster
\item"{(H0)}" We are given positive numbers $\varepsilon$, $N$, and
we set $N^\prime=[\varepsilon N/500]$.  On an interval $I$, we
are given positive functions $B(x)$, $S(x)$ and a real-valued function
$V(x)$.  We are given a real number $E_0$.
\item"{(H1)}" If $x,y \in I$ and $|x-y|< cB(x)$, then $c<B(y)/B(x)<C$
and $c<S(y)/S(x)<C$.\smallskip
\item"{(H2)}"
For $x \in I$ and $\alpha \ge 0$ we have $|(\frac{d}{dx})^\alpha
V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$.\smallskip
\item"{(H3)}" The equation $V(x)=E_0$ has two solutions $x_{\text{left}}
<x_{\text{rt}}$ in $I$, and they satisfy $\text{dist}(x_{\text{left}},\partial I
)>cB(x_{\text{left}})$,
$\text{dist}(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$.\smallskip
\item"{(H4)}" For $x_{\text{left}}\le x\le x_{\text{left}}+
c_1B(x_{\text{left}})$
we have $-V^\prime(x)>cS(x_{\text{left}})B^{-1}(x_{\text{left}})$, and
for $x_{\text{rt}}-c_1B(x_{\text{rt}})\le x\le x_{\text{rt}}$
we have $+V^\prime(x)>cS(x_{\text{rt}})B^{-1}
(x_{\text{rt}})$.\smallskip
\item"{(H5)}" For $x_{\text{left}}+c_1B(x_{\text{left}})
\le x\le x_{\text{rt}}-c_1
B(x_{\text{rt}})$ we have $E_0-V(x)\ge cS(x)$.\smallskip
\item"{(H6)}" The number $\Lambda$, defined by $\Lambda^{-1}=
\int_{x_{\text{left}}}^{x_{\text{right}}}
S^{-1/2}(x)B^{-2}(x)\,dx$, satisfies $\Lambda\ge C_{\#}^1$,
where $C_{\#}^1$ is a positive number determined entirely by $\varepsilon$,
$N$, $c$, $c_1$, $C$ and finitely many of the $C_\alpha$ in
(H0)$\ldots$(H5).
\endroster

Assuming these hypotheses, we make the following\hfill\break

\noindent{\bf Definitions}. 
Set $\lambda(x)=S^{1/2}(x)B(x)$, $S_{\text{min}}=\text{inf}_
{x \in I}S(x)$, $\sigma(\beta)=\hfill\break
\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{\Lambda\, dy}{\lambda(y)
S^\beta(y)B(y)}$, $B_{\text{left}}=B(x_{\text{left}})$, $S_{\text{left}}
=S(x_{\text{left}})$, $\lambda_{\text{left}}=\lambda(x_{\text{left}})$,
$B_{rt}=B(x_{\text{rt}})$, $S_{rt}=S(x_{\text{
rt}})$, $\lambda_{rt}=\lambda(x_{\text{rt}})$.

For $|E-E_0|< c_{\#}S_{\text{left}}$, define $x_{\text{left}}(E)$
to be the solution of $V(x)=E$ near $x_{\text{left}}$.

For $|E-E_0|<c_{\#}S_{\text{rt}}$, define $x_{\text{rt}}(E)$ to be the
solution of $V(x)=E$ near $x_{\text{rt}}$.

\noindent Define regions

$U_{\text{far\ left}}\ =\{(x,E)\mid|E-E_0|<c_{\#}S_{\text{left}}$,
$x \in I$, $x<x_{\text{left}}(E)-\lambda_{\text{left}}^{\varepsilon-2/3}
B_{\text{left}}\}$

$U_{\text{Airey\ left}}\ =\{(x,E)\mid|E-E_0|<c_{\#}S_{\text{left}}$,
$|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}\}$

$U_{\text{medium\ left}}\ =\{(x,E)\mid|E-E_0|<c_{\#}S_{\text{left}}$,
$x_{\text{left}}(E)+\lambda_{\text{left}}^{\varepsilon -2/3}B_{\text{left}}
<x<x_{\text{left}}(E)+$
${}\qquad\qquad\qquad\qquad\,\, c_{\#}B_{\text{left}}\}$
\smallskip

$U_{\text{center}}\ =\{(x,E)\mid|E-E_0|<c_{\#}^{(0)}S_{\min}$,
$x_{\text{left}}+c_{\#}^{(1)}B_{\text{left}}<x<x_{\text{
rt}}-c_{\#}^{(1)}B_{\text{rt}}\}$

$U_{\text{medium\ rt}}\ =\{(x,E)\mid|E-E_0|<c_{\#}S_{\text{rt}}$,
$x_{\text{rt}}(E)-c_{\#}B_{\text{rt}}<x<x_{\text{rt}}(E)-$
${}\qquad\qquad\qquad\qquad\,\, \lambda_{\text{rt}}
^{\varepsilon-2/3}B_{\text{rt}}\}$

$U_{\text{Airey\ rt}}\ =\{(x,E)\mid|E-E_0|<c_{\#}S_{\text{rt}}$,
$|x-x_{\text{rt}}(E)|<\lambda_{\text{rt}}^{-\varepsilon}B_{\text{rt}}\}$

$U_{\text{far\ rt}}\ =\{(x,E)\mid|E-E_0|<c_{\#}S_{\text{rt}}$,
$x \in I$, $x>x_{\text{rt}}(E)+\lambda_{\text{
rt}}^{\varepsilon-2/3}B_{\text{rt}}\}.$
\medskip

We pick the constants $c_{\#}$, $c_{\#}^{(0)}$, $c_{\#}^{(1)}$ so that
for $|E_1-E_0|<c_{\#}^{(0)} S_{\text{min}}$, the intervals defined
by intersecting the above regions with $\{E=E_1\}$ cover $I$.  The
$c_{\#}$, $c_{\#}^{(0)}$, $c_{\#}^{(1)}$ depend only on $\varepsilon$, $N$,
$c$, $c_1$, $C$ and finitely many of the $C_\alpha$ in hypotheses
(H0)$\ldots$(H5).

For $|E-E_0|<c_{\#}S_{\text{min}}$, let $(u_k^{\text{left}}(x,E))
_{0\le k\le N^\prime}$ be the canonical solution of the transport 
equations for the potential
$E-V(x)$ and the turning point $x_{\text{left}}(E)$.  Similarly, in the
same range of $E$, let $(u_k^{rt}(x,E))$ be the canonical solution of
the transport equations for the potential $E-V(x)$ and the turning
point $x_{rt}(E)$.  Then define
$$
F_c^{\text{left}}(x,E)=\frac{e^{\pm i\, \frac{\pi}{4}}\,e^{i\int_{x_{\text{left}}
(E)}^x\,(E-V(t))^{1/2}dt}}{(E-V(x))^{1/4}}\, \sum\limits_{k=0}^{N^\prime}
u_k^{\text{left}}(x,E)
$$
\noindent and
$$
F_c^{\text{rt}}(x,E)=\frac{e^{\mp i\,\frac{\pi}{4}}\,
e^{-i\int_x^{x_{\text{rt}}(E)}
(E-V(t))^{1/2}dt}}{(E-V(x))^{1/4}}\,\sum\limits_{k=0}^{N^\prime}
u_k^{\text{rt}}(x,E).
$$

Thus, $u_k^{\text{left}}(x,E)$, $u_k^{\text{rt}}(x,E)$, $F_c^{\text{left}}
(x,E)$, $F_c^{\text{rt}}(x,E)$ are all well-defined for $|E-E_0|<c_{\#}
S_{\text{min}}$, $x_{\text{left}}(E)<x<x_{\text{rt}}(E)$.

In terms of these definitions, our conclusions on the construction
of approximate solutions of $(\frac{\partial^2}{\partial x^2}+E-V(x))F=0$
are as follows.
\vglue 1pc
\proclaim{Global WKB Lemma}
Assume $\varepsilon$, $N$, $E_0$, $V(x)$,
$S(x)$, $B(x)$ are given, and that hypotheses (H0)$\ldots$(H6) are
satisfied.  Then there are functions $Y^{\text{left}}(x,E)$ and
$Y^{\text{rt}}(x,E)$ with the following properties.
\smallskip

\noindent (I) \underbar{Satisfying the ODE}
\roster
\item"{(A)}" On $U_{\text{Airey\ left}}$, define $F_{\text{Airey\ left}}
(x,E)=\lambda_{\text{left}}^{-1/3} \Bigl(\frac{\partial Y^{\text{left}}(x,E)}
{\partial x}\Bigr)^{-1/2}\cdot$\hfill\break
$\cdot A(\lambda_{\text{left}}^{2/3}Y^{\text{left}}
(x,E))$.  Then $|(\frac{\partial^2}{\partial x^2}+E-V(x))F_{\text{Airey\ left}}
|\le C_{\#}\lambda_{\text{left}}^{-N^\prime}B_{\text{left}}^{-3/2}$
on $U_{\text{Airey\ left}}$.

\item"{(B)}" On $U_{\text{medium\ left}}$ we have $|(\frac{\partial^2}
{\partial x^2}+E-V(x))F_c^{\text{left}}|\le C_{\#}
\lambda_{\text{left}}^{10}\hfill\break\Bigl(\frac{\lambda_{\text{left}}^{2/3}
(x-x_{\text{left}}(E))}{B_{\text{left}}}\Bigr)^{-3/2\, N^\prime}
B_{\text{left}}^{-3/2}$.

\item"{(C)}" On $U_{\text{center}}$ we have $|(\frac{\partial^2}
{\partial x^2}+E-V(x))F_c^{\text{left,\ right}}|\le C_{\#}
\Lambda^{-N^\prime}S^{-1/4}(x)\negthinspace B^{-2}(x)$.

\item"{(D)}" On $U_{\text{medium\ rt}}$ we have
$|(\frac{\partial^2}{\partial x^2}+E-V(x))F_c^{\text{rt}}|\le C_{\#}
\lambda_{\text{rt}}^{10}\cdot \hfill\break\cdot\Bigl(\frac{\lambda_{
\text{rt}}^{2/3}(x_{\text{rt}}
(E)-x)}{B_{\text{rt}}}\Bigr)^{-\frac 32 N^\prime}\cdot B_{\text{rt}}^{-3/2}$.

\item"{(E)}" On $U_{\text{Airey\ rt}}$, define $F_{\text{Airey\ rt}}(x,E)=
\lambda_{\text{rt}}^{-1/3}(-\frac{\partial Y^{\text{
rt}}(x,E)}{\partial x})^{-1/2}\cdot
\hfill\break \cdot A(\lambda_{\text{rt}}^{2/3}Y^{\text{rt}}(x,E))$.
Then $|(\frac{\partial^2}{\partial x^2}+E-V(x))F_{\text{Airey\ rt}}|\le
C_{\#}\lambda_{\text{rt}}^{-N^\prime}B_{\text{
rt}}^{-3/2}$ on $U_{\text{Airey\ rt}}$.\endroster
\medskip

\noindent (II) \underbar{Behavior of the Solutions}\roster
\item"{(A)}" On $U_{\text{Airey\ left}}$ we can express $Y^{\text{left}}$
as $Y^{\text{left}}(x,E)=Y_0^{\text{left}}(x,E)+\hfill\break
\lambda_{\text{left}}^{-2}
Y_1^{\text{left}}(x,E)$ with $|\partial_x^\alpha\partial_E^\beta Y_i^
{\text{left}}| \le C_{\#}^{\alpha\beta}B_{\text{left}}^{-\alpha}S_{\text{left}}
^{-\beta}$ and $\lambda_{\text{left}}^2 Y_0^{\text{left}}(\frac{\partial Y_0^
{\text{left}}}{\partial x})^2=E-V(x)$ to order $\ge N^\prime$ at $x=
x_{\text{left}}(E)$.
Also $|\lambda_{\text{left}}^2Y^{\text{left}}(\frac 
{\partial Y^{\text{left}}}{\partial x})^2+\{Y^{\text{left}},x\}-(E-V(x))|\le
C_{\#}\lambda_{\text{left}}^{-N^\prime}S_{\text{left}}$ on $U_{\text{Airey\
left}}$.

\item"{(B)}" On $U_{\text{medium\ left}}$ we can express $u_k^{\text{left}}$
as $u_k^{\text{left}}(x,E)=\lambda_{\text{left}}^{-k}
\Bigl(\frac{x-x_{\text{left}}(E)}{B_{\text{left}}}\Bigr)^{-\frac 32 k}
\cdot \cdot f_k^{\text{left}}\Bigl((\frac{x-x_{\text{left}}(E)}{B_{\text{left}}}
)^{1/2}, \frac {E-E_0}{S_{\text{left}}}\Bigr)$ with
$|\partial_y^\alpha\partial_s^\beta f_k^{\text{left}}(y,s)|\le C_{\#}
^{\alpha\beta}$.

\item"{(C)}" On $U_{\text{center}}$ we have $|\partial_x^\alpha\partial_E^
\beta u_k^{\text{left,\ rt}}(x,E)|\le C_{\#}^{\alpha\beta}
\Lambda^{-k}B^{-\alpha}(x)\sigma(\beta)$.

\item"{(D)}" On $U_{\text{medium\ rt}}$ we can express $u_k^{\text{rt}}(x,E)$
as
$$u_k^{\text{rt}}(x,E)=\lambda_{\text{rt}}^{-k}(\frac{x_{\text{rt}}(E)-x}
{B_{\text{rt}}})^{-\tfrac 32 k}f_k^{\text{rt}}((
\tfrac{x_{\text{rt}}(E)-x}{B_{\text{rt}}})^{1/2},\tfrac{E-E_0}{S_{\text{rt}}}
)
$$
with $|\partial_y^\alpha\partial_s^\beta f_k^{{\text{rt}}}(y,s)|\le C_{\#}
^{\alpha\beta}$.


\item"{(E)}" On $U_{\text{Airey\ rt}}$ we can express $Y^{\text{rt}}(x,E)$
as $Y^{\text{rt}}(x,E)=Y_0^{\text{rt}}(x,E)+\hfill\break
\lambda_{\text{rt}}^{-2}
Y_1^{\text{rt}}(x,E)$, with $|\partial_x^\alpha\partial_E^\beta Y_i^{\text{
rt}}|\le C_{\#}^{\alpha\beta}B_{\text{rt}}^{-\alpha}S_{\text{rt}}^{-\beta}$
and $\lambda_{\text{rt}}^2$ $Y_0^{\text{rt}}(\frac{\partial
Y_0^{\text{rt}}}{\partial x})^2=
E-V(x)$ to order $\ge N^\prime$ at $x=x_{\text{rt}}(E)$.  Also
$|\lambda_{\text{rt}}^2Y^{\text{rt}}(\frac{\partial Y^{\text{rt}}}
{\partial x})^2+\{Y^{\text{rt}},x\}-(E-V(x))|\le
C_{\#}\lambda_{\text{rt}}^{-N^\prime}S_{\text{rt}}$ on
$U_{\text{Airey,\ rt}}$.
\endroster

\noindent{(III) \underbar{Patching the Solutions Together}\roster
\item"{(A)}" On $U_{\text{far\ left}} \cap U_{\text{Airey\ left}}$
we have $|\partial_x^\alpha F_{\text{Airey\ left}}|\le C_{\#}
\lambda_{\text{left}}^{-N^\prime}B_{\text{left}}^{\frac 12 -\alpha}$,
$0\le \alpha\le 1$.

\item"{(B)}" On $U_{\text{Airey\ left}}\cap U_{\text{medium\ left}}$
we have $|\partial_x^\alpha\{F_{\text{Airey\ left}}-\text{Re}\ F_c^
{\text{left}}\}|\hfill\break
\le C_{\#}\lambda_{\text{left}}^{10}
\Bigl(\frac{\lambda_{\text{left}}^{2/3}(x-x_{\text{left}}(E))}
{B_{\text{left}}}\Bigr)^{-\frac 32 N^\prime}\cdot B_{\text{left}}
^{\frac 12 -\alpha}$ for $0 \le \alpha \le 1$.

\item"{(C)}" On $U_{\text{center}}$ we can satisfy
$u_k^{\text{left}}(x,E)=\sum\limits_{\ell=0}^k G_\ell(E)u_{k-\ell}^{\text{rt}}
(x,E)$ for a unique sequence $(G_\ell(E))_{0\le \ell \le N^\prime}$
with $G_0(E)=1$.  With $\Cal A(E)=\Bigl[\exp\ i\{\pm \frac \pi 2+
\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}(E-V(t))^{1/2}\,dt\}\Bigr]
\cdot\sum\limits_{k=0}^{N^\prime}G_k(E)$, we have
$$
|F_c^{\text{left}}(x,E)-\Cal A(E)F_c^{\text{rt}}(x,E)|\le C_{\#}
\Lambda^{-N^\prime-1}(E-V(x))^{-1/4} \quad \text{and}
$$
$$
|\tfrac {\partial}{\partial x}\{F_c^{\text{left}}(x,E)-\Cal A(E)F_c^{\text{rt}}
(x,E)\}|\le C_{\#}\Lambda^{-N^\prime-1}(E-V(x))^{+1/4}
$$
\noindent for $(x,E) \in U_{\text{center}}$.

We have $|\partial_E^\beta G_\ell(E)|\le C_{\#}^\beta\Lambda^{-\ell}
\sigma(\beta)\quad \text{for}\quad |E-E_0|<c_{\#}^{(0)}S_{\text{min}}$.

\item"{(D)}" On $U_{\text{medium\ rt}} \cap U_{\text{Airey\ rt}}$
we have $|\partial_x^\alpha\{F_{\text{Airey\ rt}}-\text{Re}\ F_c^{\text{rt}}\}|
\hfill\break \le C_{\#}\lambda_{\text{rt}}^{10}(\frac{\lambda_{\text{rt}}^{2/3}
(x_{\text{rt}}(E)-x)}{B_{\text{rt}}}\Bigr)^{-\frac 32 N^\prime}\cdot
B_{\text{rt}}^{\frac 12 -\alpha}$ for $\alpha=0,1$.

\item"{(E)}" On $U_{\text{Airey\ rt}} \cap U_{\text{far\ rt}}$ we have
$|\partial_x^\alpha F_{\text{Airey\ rt}}|\le C_{\#}\lambda_{\text{rt}}
^{-N^\prime}B_{\text{rt}}^{\frac 12-\alpha}$, $0 \le \alpha \le 1$.
\endroster

\noindent{(IV) \underbar{Explicit Formulas}}
\roster\item"{(A)}" For $|E-E_0|<c_{\#}^{(0)}S_{\text{min}}$ and
$x_{\text{left}}(E)<x<x_{\text{rt}}(E)$ we have $u_0^{\text{left}}
(x,E)=u_0^{\text{rt}}(x,E)=1$ and
$$
u_1^{\text{left}}(x,E)=\lim_{\delta\to 0+} \Bigl[\int_{x_{\text{left}}(E)+\delta}
^x \{\tfrac {5i}{32} \tfrac{(V^\prime)^2}{(E-V)^{5/2}}+ \tfrac i 8
\tfrac {V^{\prime\prime}}{(E-V)^{3/2}}\}\,dt-\sum\limits_{\ell=1}^3
q_\ell^{\text{left}}(E)\delta^{-\ell/2}\Bigr]
$$
$$
u_1^{\text{rt}}(x,E)=-\lim_{\delta\to 0+}\Bigl[\int_x^{x_{\text{rt}}(E)-\delta}
\{\tfrac {5i}{32} \tfrac{(V^\prime)^2}{(E-V)^{5/2}}
+\tfrac i8 \tfrac{V^{\prime\prime}}{(E-V)^{3/2}}\}\,dt-\sum\limits_
{\ell=1}^3 q_\ell^{\text{rt}}(E)\delta^{-\ell/2}\Bigr]
$$
\noindent with $q_\ell^{\text{left}}(E)$, $q_\ell^{\text{rt}}(E)$ uniquely
specified by demanding the finiteness of the limit.

\item"{(B)}" For $|E-E_0|<c_{\#}^{(0)}S_{\text{min}}$ we have $G_0(E)\equiv 1$
and
$$
G_1(E)=\tfrac {i}{48}\lim_{\delta\to 0+} \Bigl[\int_{x_{\text{left}}(E)+\delta}
^{x_{\text{rt}}(E)-\delta} V^{\prime\prime}(t)(E-V(t))^{-3/2}\,dt-q(E)\delta^
{-1/2}\Bigr]
$$
\noindent with $q(E)$ uniquely specified by demanding the finiteness of
the limit.

\item"{(C)}" For $|E-E_0|<c_{\#}^{(0)}S_{\text{min}}$ we have
$\Cal A(E)=R(E)e^{i\varPhi(E)}$ with $|R(E)-1|<C_{\#}\Lambda^{-2}$ and $R(E)$
real,
$$\align
\varPhi(E)=&\pm \tfrac \pi 2+\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
(E-V(t))^{1/2}\,dt+\\&\,\,\,\,+ \tfrac 1{48}\lim_{\delta\to 0+}
\cdot \Bigl[\int_{x_{\text{left}}
(E)+\delta}^{x_{\text{rt}}(E)-\delta} V^{\prime\prime}(t)(E-V(t))
^{-3/2}\,dt-q(E)\delta^{-1/2}\Bigr]+
\varPhi_{\text{extra}}(E)\endalign$$

\noindent with $\varPhi_{\text{extra}}(E)$ real and satisfying
$|\partial_E^\beta\varPhi_{\text{extra}}|\le C_{\#}^\beta
\Lambda^{-2}\sigma(\beta)$ for all $\beta \ge 0$.
\endroster

\noindent{(V)\underbar{What the Constants May Depend On}}

The constants $C_{\#}$, $c_{\#}$, $C_{\#}^\alpha$, $C_{\#}^\beta$,
$C_{\#}^{\alpha\beta}$ in parts (I)$\ldots$(IV) above depend only
on $\varepsilon$, $N$, $c_1$, $c$, $C$ and finitely many of the $C_\alpha$
in the hypotheses (H0)$\ldots$(H5).
\endproclaim
\vglue 1pc
\demo{Proof} All these assertions were proven in the previous section.
$\qquad\blacksquare$
\vglue 1pc
\demo{Remark}  Fix $\varepsilon$, $N$, and set $N^\prime=[\varepsilon N/
500]$.  Suppose $V(x)$ is any smooth potential defined on an interval $I$,
and let $E$ be a real number.  Assume $\{x \in I\mid V(x)<E\}=
(x_{\text{left}}(E), x_{\text{rt}}(E))$ for $x_{\text{left}}(E)$,
$x_{\text{rt}}(E)$ in the interior of $I$, and assume also that
$V^\prime(x)\ne 0$ at $x=x_{\text{left}}(E)$, $x_{\text{rt}}(E)$.  Then
the canonical solutions $u_k^{\text{left}}(x)$, $u_k^{\text{rt}}(x)$
to the transport equations for the potential $E-V(x)$ are well-defined
for $x_{\text{left}}(E)<x<x_{\text{rt}}(E)$.  Hence also the coefficients
$G_k(E)$ with $u_k^{\text{left}}=\sum\limits_{\ell=0}^{k} G_\ell u_{k-\ell}^
{\text{rt}}$ are well-defined, so the complex factor $\Cal A(E)$ of
III(C) above is well-defined.  If $|\sum\limits_{k=1}^{N^\prime} G_k(E)|
<\frac 1{10}$, then the functions $R(E)$, $\varPhi(E)$ in IV(C) are
also well-defined.  The point is that the definition of $R(E)$, 
$\varPhi(E)$ is independent of the choice of weight functions
$S(x)$, $B(x)$ used to describe the problem.


%Subject: Section 4

\pageno 60
\magnification \magstep1
\documentstyle{amsppt}

\hsize 6 truein
\vsize 9truein
\baselineskip 20pt
\hoffset .15truein
\catcode`\@=11
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\def\kbar{k{\kern -4.5pt\raise 3pt\hbox{-}}}

\centerline{\bf{Eigenvalues and Eigenfunctions of Schr\"odinger
Operators}}
\smallskip

In this section we compare the exact eigenvalues and eigenfunctions of
a \hfill\break Schr\"odinger operator with the approximations developed in the
preceding sections.  Recall that the global WKB lemma constructs several
approximate solutions of $(\frac{d^2}{dx^2}+E-V(x))F=0$, which are
mutually consistent if the phase $\Phi(E)$ is an integer multiple of
$\pi$.  We will prove that the eigenvalues of $H=-\frac{d^2}{dx^2}+V(x)$
are very close to the solutions of $\Phi(E)\equiv 0 \mod \pi$, and that
the eigenfunctions of $H$ are very close to constant multiples of the
approximate solutions in the WKB lemma.

Thus let $V(x)$ be defined on an interval $I_{\text{BVP}}$ (possibly
$[0,\infty)$ or $\Bbb R^1$).  We impose Dirichlet or Neumann conditions
at the endpoints, so that $H$ becomes a self--adjoint operator on $L^2$.
Perhaps $H$ has continuous spectrum, so we assume
\roster
\item"{(E1)}" The continuous spectrum of $H$ lies entirely in $[E_\infty,
+\infty)$ for a critical energy $E_\infty$.
\endroster

We fix an energy $E_0\le E_\infty$ and a subinterval $I\subset I_{\text{BVP}}$
on which $V(x)$, $E_0$ satisfy the hypotheses (H0)$\ldots$(H6) of the
global WKB lemma.  We adopt the definitions and notation of the previous
section, and we make the following additional assumptions.

\roster
\item"{(E2)}" If $|E-E_0|\le c_{\#}^{(0)}S_{\text{min}}$ and $E \le E_\infty$,
then $V(x)\ge E$ outside \hfill\break $(x_{\text{left}}(E),x_{\text{rt}}(E))$.
\item"{(E3)}" If $|E-E_0|<c_{\#}^{(0)}S_{\text{min}}$ and $E \le E_\infty$
then for some $\tau \in (0,1)$ we have
$$
\split
\tau(V(x)-E)\exp(2(1-\tau)\int_{x_{\text{right}}(E)}^x(V(t)-E)^{1/2}\,dt)\ge
\lambda_{\text{right}}^N\cdot\text{max}(E-V)\\
\text{for}\ x \in I_{\text{BVP}},\, x>x_{\text{right}}(E)+\lambda_{\text{right}}
^KB_{\text{right}},\endsplit
$$
\noindent and
$$\split
\tau(V(x)-E)\exp(2(1-\tau)\int_x^{x_{\text{left}}(E)} (V(t)-E)^{1/2}
\,dt)\ge \lambda_{\text{left}}^N\cdot \text{max}(E-V)\\
\text{for}\ x \in I_{\text{BVP}}, x<x_{\text{left}}(E)-\lambda_{\text{left}}^K
B_{\text{left}}.\endsplit
$$
\item"{(E4)}" $\text{max}(E_0-V)\le \lambda_{\text{right}}^KS_{\text{right}}$
and $\text{max}(E_0-V)\le \lambda_{\text{left}}^KS_{\text{left}}$
\item"{(E5)}" For $|E-E_0|\le c_{\#}^{(0)}S_{\text{min}}$ we have
$$
\Bigl[\int_{x_{\text{left}}(E)}^{x_{\text{right}}(E)}\frac{dx}
{S^{1/2}(x)B^4(x)}\Bigr]\cdot\bigl[\int_{x_{\text{left}}(E)}^
{x_{\text{right}}(E)}\frac{dx}{(E-V(x))^{1/2}}\Bigr]\le \Lambda^K.
$$
\item"{(E6)}" $\int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}
\frac{dx}{(E_1-V(x))^{1/2}}\le \Lambda^K\cdot\min(S_{\text{left}}^{-1/2}
B_{\text{left}},S_{\text{right}}^{-1/2}B_{\text{right}})\ \text{for}\
|E_1-E_0|<c_{\#}^0S_{\text{min}}$
\item"{(E7)}" $|E_0|\le C_{\#}S_{\text{min}}$.
\endroster

Here $K$ is a large positive number.  We assume that $N$ in the global
WKB lemma has been taken larger than $K\varepsilon^{-10}$.  We broaden
our notion of constants $C_{\#}$, $c_{\#}$, etc. to allow them to
depend on $\tau$, $K$, as well as on the constants in (H0)$\ldots$(H5).

Assumptions (E1), (E2) are very natural, while (E3)$\ldots$(E7) are technical
but are satisfied in most of the examples we care about.

Before we can study the eigenvalues and eigenfunctions of $H$, we need a
crude lower bound for the $L^2$--norm of the approximate ODE solutions
$\text{Re}(F_c^{\text{left}})$, \hfill\break
 $\text{Re}(F_c^{\text{right}})$ in the global
WKB lemma.  (After all, knowing that $(H-E)F$ is small tells us nothing if
$F$ is small also.)  Such a lower bound is easily proved using a standard
stationary phase inequality.
\vglue 1pc
\proclaim{Lemma 1 (Stationary Phase)}  Suppose $\theta(x)$ is
supported in $\{|x-x_0|<\delta\}$ and satisfies $|(\frac{d}{dx})^m\theta(x)|
\le C_m\delta^{-m}$ for $m\ge 0$.  Let $\phi(x)$ be real--valued
and satisfy $|(\frac{d}{dx})^m\phi(x)|\le C_m\delta^{-m} (m\ge 0)$
and $|\frac{d\phi}{dx}|\ge c\delta^{-1}$ on $\{|x-x_0|<\delta\}$.  Then
for $\lambda$ a large real number we have
$$
|\int_{\Bbb R}\theta(x)e^{i\lambda\phi(x)}\,dx|\le C_K\lambda^{-K}\delta,
\quad\text{any}\ K>0.
$$
\endproclaim

\demo{Sketch of proof} Make repeated use of the identity $e^{i\lambda\phi}=
\frac{1}{i\lambda\phi^\prime}\,\frac{d}{dx}e^{i\lambda\phi}$ and integration
by parts.$\qquad\blacksquare$

Our lower bound for $\Vert\text{Re}(F_c^{\text{left}})\Vert$ is as follows.

\vglue 1pc
\proclaim{Lemma 2}  Suppose $|E-E_0|<c_{\#}^{(0)}S_{\text{min}}$ and suppose
$J=\{|x-x_0|<c_{\#}B(x_0)\}$ is contained in $(x_{\text{left}}(E)+
c_{\#}B_{\text{left}}, x_{\text{right}}(E)-c_{\#}B_{\text{right}})$.
Then for $|\omega|=1$ we have
$$
\int_J|\text{Re}(\omega F_c^{\text{left}})|^2\,dx\ge c_{\#}\int_J
\frac{dx}{(E-V(x))^{1/2}}.
$$\endproclaim

\demo{Sketch of Proof}  Let $\chi(x)$ be a smooth cutoff function supported
in $J$.  We have
$$
\int_J\chi(\text{Re}(\omega F_c^{\text{left}}))^2\,dx=\frac 14 \int_J\chi(
\omega F_c^{\text{left}})^2+\frac 14
\int_J\chi(\overline\omega \overline{F_c^{\text{left}}})^2
+\frac 12 \int_J\chi|F_c^{\text{left}}|^2.\tag 1
$$
\noindent The first integral on the right is negligibly small, by virtue
of the stationary phase lemma with $\theta=\frac{\chi(\sum\limits_{k=0}^
{N^\prime}u_k^{\text{left}}(x,E))^2}{(E-V(x))^{1/2}}$, 
$\lambda=\lambda(x_0)$, $\phi(x)=\frac{1}{\lambda(x_0)}\int_{x_0}^{x}
(E-V(t))^{1/2}\,dt$.  The second term on the right of (1) is the complex
conjugate of the first term and so is also negligibly small.  The third
term on the right in (1) has the correct order of magnitude. $\qquad
\blacksquare$
 
\vglue 1pc
\proclaim{Corollary}  $\int_{x_{\text{left}}(E)+c_{\#}B_{\text{left}}}
^{x_{\text{right}}(E)-c_{\#}B_{\text{rt}}}|\text{Re}(\omega F_c^{\text{left}}
)|^2\,dx\ge c_{\#}\int_{x_{\text{left}}(E)}^{x_{\text{right}}(E)}\frac
{dx}{(E-V(x))^{1/2}}$.\endproclaim

\demo{Proof}  On the right we can change the limits of integration to
$x_{\text{left}}(E)+c_{\#}B_{\text{left}}$, $x_{\text{right}}(E)-c_{\#}
B_{\text{right}}$
without changing the order of magnitude of the integral.  The Corollary
then follows by simply dividing $[x_{\text{left}}(E)+c_{\#}B_{\text{left}},
x_{\text{right}}-c_{\#}B_{\text{right}}]$ into intervals $J_\nu$ to which we
can apply Lemma 2. $\qquad\blacksquare$

Of course we have analogous results for $\text{Re}(F_c^{\text{right}})$.

Now we begin to study the eigenfunctions of $H$.  Thus suppose
$F \in \text{Domain}(H)$, $\int_{I_{\text{BVP}}}|F|^2\,dx=1$,
$F$ is real--valued and
$$
-\frac{d^2}{dx^2} F+VF=E_1F\quad\text{for\ some}\ E_1 \quad\text{with}\
|E_1-E_0|<c_{\#}^{(0)}S_{\text{min}}, E_1\le E_\infty.
$$
\noindent We compare $F$ with the approximate solutions in the global WKB
lemma.  To do so, we look separately at each of the following intervals:
$$\align
I_{\text{far\ left}}\ &=\{x \in I_{\text{BVP}}\mid x<x_{\text{left}}(E_1)-
\lambda_{\text{left}}^{\varepsilon-2/3}B_{\text{left}}\}\\
I_{\text{Airey\ left}}\ &=\{x \in I\mid (x,E_1)\in U_{\text{Airey\ left}}\}\\
I_{\text{medium\ left}}\ &=\{x \in I\mid (x,E_1)\in U_{\text{medium\ left}}\}\\
I_{\text{center}}\ &=\{x \in I\mid (x,E_1)\in U_{\text{center}}\}\\
I_{\text{medium\ right}}\ &= \{x \in I\mid (x,E_1)\in U_{\text{medium\ right}}\}\\
I_{\text{Airey\ right}}\ &=\{x \in I\mid (x,E_1)\in U_{\text{Airey\ right}}\}\\
I_{\text{far\ right}}\ &=\{x \in I_{\text{BVP}}\mid x>x_{\text{right}}(E_1)+
\lambda_{\text{right}}^{\varepsilon-2/3}B_{\text{right}}\}\endalign
$$

These intervals cover $I_{\text{BVP}}$.  After understanding how $F$ is
forced to behave on each of these intervals, we check our information
for consistency on their intersections.  We begin with $I_{\text{far\ left}}$.
Our basic tool is Agmon's decay estimate, specialized to the trivial
one--dimensional case.

\vglue 1pc
\proclaim{Lemma 3 (Agmon lemma)} Define an auxiliary function $\varphi(x)$
as follows.
$$\align
\varphi(x)&=\int_x^{x_{\text{left}}(E)}(V(t)-E_1)^{1/2}\,dt\quad\text{if}\
x\le x_{\text{left}}(E_1),\\
\varphi(x)&=0\quad\text{if}\ x_{\text{left}}(E_1)\le x\le x_{\text{right}}(E_1),\\
\varphi(x)&=\int_{x_{\text{right}}(E_1)}^{x}(V(t)-E_1)^{1/2}\,dt\quad\text{if}\
x\ge x_{\text{right}}(E_1).\endalign
$$
\noindent Then we have the estimate
$$\split
\tau\int_{I_{\text{BVP}}\backslash(x_{\text{left}}(E_1),x_{\text{right}}(E_1))}
e^{2(1-\tau)\varphi}\Bigl[(\frac{dF}{dx})^2+(V-E_1)F^2\Bigr]\,dx\\
\le\int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}(E_1-V)F^2\,dx.\endsplit
$$\endproclaim

\demo{Sketch of Proof}  Integrating by parts formally we get
$$\align
0&=\int_{I_{\text{BVP}}}(e^{2(1-\tau)\varphi}F)(-\frac{d^2F}{dx^2}
+(V-E_1)F)\,dx\\
&=\int_{I_{\text{BVP}}}e^{2(1-\tau)\varphi}\Bigl[(\frac{dF}{dx})^2+
(V-E_1)F^2+2(1-\tau)\frac{d\varphi}{dx}(\frac{dF}{dx})F\Bigr]\,dx.\endalign
$$

\noindent This formal process is easily justified:  We approximate $\varphi$
by a smooth function assumed constant near the endpoints of $I_{\text{BVP}}$
and pass to the limit.  Outside $(x_{\text{left}}(E_1),x_{\text{right}}(E_1))$
we have $(\frac{d\varphi}{dx})^2=(V-E_1)$, so the expression in brackets
is greater than or equal to $\tau\{(\frac{dF}{dx})^2+(V-E_1)F^2\}$.  Inside
$(x_{\text{left}}(E_1), x_{\text{right}}(E_1))$ we have $\varphi=\frac
{d\varphi}{dx}=0$.  Therefore
$$\split
0\ge \int_{I_{\text{BVP}}\backslash(x_{\text{left}}(E_1),x_{\text{right}}(E_1))}
\tau e^{2(1-\tau)\varphi}\Bigl\{\Bigl(\frac{dF}{dx}\Bigr)^2+(V-E_1)F^2\Bigl\}
\,dx\\
-\int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}(E_1-V)F^2\,dx,\endsplit
$$
\noindent which proves the lemma. $\qquad\blacksquare$

Let us see how Agmon's function $\varphi(x)$ behaves to the left of
$x_{\text{left}}(E_1)$.  We know that
$$
V(x)-E_1\ge c_{\#}\Bigl(\frac{S_{\text{left}}}{B_{\text{left}}}\Bigr)
(x_{\text{left}}(E_1)-x)\quad\text{for}\
x_{\text{left}}(E_1)-c_{\#}B_{\text{left}}<x<x_{\text{left}}(E_1).
$$

\noindent Therefore $\varphi(x)\ge c_{\#}\lambda_{\text{left}}
(\frac{x_{\text{left}}(E_1)-x}{B_{\text{left}}})^{3/2}$ in that range of $x$.
In particular, in $J_1=\{x_{\text{left}}(E_1)-c_{\#}B_{\text{left}}<x
<x_{\text{left}}(E_1)-\lambda_{\text{left}}^
{\varepsilon-2/3}B_{\text{left}}\}$ we have $\varphi(x)\ge c_{\#}
\lambda_{\text{left}}^{\frac 32\varepsilon}$.  Since $\varphi(x)$ is
decreasing in $I_{\text{far\ left}}$, it follows that $\varphi(x)\ge
c_{\#}\lambda_{\text{left}}^{\frac 32 \varepsilon}$ throughout
$I_{\text{far\ left}}$.

So Lemma 3 implies:
$$\split
\tau\exp(c_{\#}\lambda_{\text{left}}^{\frac 32 \varepsilon})\int_{
I_{\text{far\ left}}}\Bigl(\frac{dF}{dx}\Bigr)^2\,dx\le
\int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}(E_1-V)F^2\,dx\le
\text{max}(E_1-V)\endsplit\tag 2
$$

\noindent and
$$\split
\tau\exp(c_{\#}\lambda_{\text{left}}^{\frac 32 \varepsilon})
\int_{x_{\text{left}}(E_1)-\hat c_{\#}B_{\text{left}}}^
{x_{\text{left}}(E_1)-\frac 12 \hat c_{\#}B_{\text{left}}}
S_{\text{left}}F^2\,dx\le \int_{x_{\text{left}}(E_1)}^
{x_{\text{right}}(E)}(E_1-V)F^2\,dx\\
\le \text{max}(E_1-V)\endsplit\tag 3
$$

\noindent Since $F$ has norm $1$.

We invoke the elementary inequality
$$
c\int_{I_1}F^2\,dx\le \Bigl(\frac{|I_1|}{|I_0|}\Bigr)^2\int_{I_0}
F^2\,dx+|I_1|^2\int_{I_1}\Bigl(\frac{dF}{dx}\Bigr)^2\,dx\tag 4
$$

\noindent for intervals $I_0\subset I_1$, taking $I_0=\{x_{\text{left}}
(E_1)-\hat c_{\#}B_{\text{left}}<x<x_{\text{left}}(E_1)-\frac 12
\hat c_{\#}B_{\text{left}}\}$ and $I_1=I_{\text{far\ left}}\cap
\{x>x_{\text{left}}(E)-\lambda_{\text{left}}^KB_{\text{left}}\}$.

\noindent Estimates (2), (3), (4) together imply
$$
c_{\#}\int_{I_1}F^2\,dx\le \frac{\lambda_{\text{left}}^{2K}\text{max}
(E_1-V)}{\tau\exp(c_{\#}\lambda_{\text{left}}^{\frac 32 \varepsilon})
S_{\text{left}}}+\frac{\lambda_{\text{left}}^{2K}B_{\text{left}}^2
\text{max}(E_1-V)}{\tau\exp(c_{\#}\lambda_{\text{left}}^{\frac 32 
\varepsilon})}
$$
\noindent Hypothesis (E4) gives $\text{max}(E_1-V)\le \lambda_{\text{left}}^K
S_{\text{left}}$, and by definition $B_{\text{left}}^2\cdot S_{\text{left}}
=\lambda_{\text{left}}^2$.  Therefore
$$
\int_{I_1}F^2\,dx\le C_{\#}\lambda_{\text{left}}^{3K+2}\exp(-c_{\#}
\lambda_{\text{left}}^{\frac 32 \varepsilon})\le C_{\#}^\prime
\lambda_{\text{left}}^{-N}.\tag 5
$$

\noindent On the other hand, hypothesis (E3) gives $\tau e^{2\varphi}
(V-E_1)\ge \lambda_{\text{left}}^N\cdot\text{max}(E_1-V)$
in $I_{\text{far\ left}}\backslash I_1$.  Therefore Lemma 3 implies
$$
\text{max}(E_1-V)\cdot\lambda_{\text{left}}^N\int_{I_{\text{far\ left}}
\backslash I_1}F^2\,dx\le \int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}
(E_1-V)F^2\,dx\le \text{max}(E_1-V),
$$

\noindent again becuase $F$ has norm $1$.  Thus, $\int_{I_{\text{far\
left}}\backslash I_1}F^2\,dx\le \lambda_{\text{left}}^{-N}$.  Together
with (5), this gives the following conclusion.

\vglue 1pc
\proclaim{Lemma 4}  We have $\int_{I_{\text{far\ left}}}F^2\,dx
\le C_{\#}\lambda_{\text{left}}^{-N}$. \endproclaim

\noindent  There is of course an analogous result for $I_{\text{far\ right}}$.

Next we study how $F$ behaves on $I_{\text{Airey\ left}}$.  We need some
elementary results on Airey's equation.

\vglue 1pc
\proclaim{Lemma 5}  Suppose $\frac{d^2u}{dy^2}+yu=f$ in an interval
$[-T,+T]$ with $T$ greater than a large universal constant.  Assume
$\int_{-T}^T|f|^2\,dy\le CT^{-2M}$ and $\int_{-T}^T|u|^2\,dy\le C (M\ge 2)$.
Then there is a constant $b$ of size $|b|\le C$ for which $\int_{-\frac 12 T}
^{+\frac 12 T}|u(y)-bA(y)|^2\,dy\le C_MT^{2-2M}$.\endproclaim

\demo{Sketch of Proof}  First we set up a Green's function for Airey's
equation with good bounds.  Let $A_1(y)=A(y)$, and let $A_2(y)$
be another real solution of $\frac{d^2A_2}{dy^2}+yA_2=0$, taken so that
the Wronskian $A_1^\prime A_2-A_2^\prime A_1\equiv \pm 1$.  Both $A_1(y)$
and $A_2(y)$ remain bounded as long as $y>-C$ for some fixed constant $C$.
As $y\to -\infty$ we have $|A_1(y)|\le \frac{C\,e^{-\frac 23|y|^{3/2}}}
{|y|^{1/4}}$, $|A_2(y)|\le \frac{C\,e^{+\frac 23|y|^{3/2}}}{|y|^{1/4}}$.
Hence the Green's function $G(x,y)=A_1(\text{min}(x,y))\cdot
A_2(\text{max}(x,y))$ remains bounded on all of $\Bbb R^2$.  Using the
Green's function, we construct the special solution $u_1(x)=\int_{-T}^T
G(x,y)f(y)\,dy$ of $\frac{d^2u_1}{dy^2}+yu_1=f$ on $[-T,T]$.  Moreover,
$$\split
\int_{-T}^T|u_1(y)|^2\,dy\le CT^2\int_{-T}^T |f(y)|^2\,dy \le
CT^{2-2M}\\
\text{since}\ |G(x,y)|\le C.\endsplit
$$

\noindent The given solution $u$ may be written as $u=u_1+b_1A_1+b_2A_2$
on $[-T,T]$ for constants $b_1$, $b_2$.  Our estimate for $\Vert u_1\Vert^2$
and hypothesis on $\Vert u\Vert^2$ together imply
$$
\int_{-T}^T|b_1A_1(y)+b_2A_2(y)|^2\,dy\le C.\tag 6
$$

\noindent For $T$ greater than a large universal constant, we have
$$
\Big|\int_{-T}^TA_1(y)A_2(y)\,dy\Big|<<\Bigl
(\int_{-T}^TA_1^2(y)\,dy\Bigr)^{1/2}\Bigl(\int_{-T}^T
A_2^2(y)\,dy\Bigr)^{1/2}.
$$

\noindent In fact, the left side grows as a power of $T$, while the right side
grows exponentially.  Thus $A_1$ and $A_2$ are nearly orthogonal in
$L^2[-T,T]$, so (6) implies
$$
\int_{-T}^T|b_1A_1(y)|^2\,dy\le C\tag 7
$$
\noindent and
$$
\int_{-T}^T |b_2A_2(y)|^2\,dy\le C.\tag 8
$$

\noindent Estimate (7) implies $|b_1|\le C$, while (8) implies
$\int_{-T/2}^{+T/2}|b_2A(y)|^2\,dy\le C_MT^{2-2M}$.  (Again we exploit the
exponential growth of $\int_{-T}^T|A_2(y)|^2\,dy$.)

\noindent Now $u(y)-b_1A_1(y)=u_1(y)+b_2A_2(y)$, and the two terms on
the right both have norm squared at most $C_MT^{2-2M}$ in
$L^2[-T/2,+T/2]$.  Thus\TagsOnRight
$$
\int_{-T/2}^{T/2}|u(y)-b_1A(y)|^2\,dy\le
C_MT^{2-2M}, |b_1|\le C.\tag"$\blacksquare$"
$$
\TagsOnLeft

\proclaim{Corollary 1}  Suppose $[\frac{d^2}{dy^2}+W(y)]u=0$ on $[-T,T]$
with $T$ greater than a large universal constant.  Suppose also
$|W(y)-y|\le CT^{-M}$ on $[-T,T]$ and $\int_{-T}^T|u(y)|^2\,dy\hfill\break\le C$.
Then for a constant $b$ of size $|b|\le C_m$ we have
$$
\int_{-T/2}^{+T/2}|u(y)-bA(y)|^2\,dy\le C_M\,T^{2-2M}.
$$\endproclaim

\demo{Proof}  Just write $\frac{d^2u}{dy^2}+yu=[y-W(y)]u\equiv f$ and
apply Lemma 5.$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary 2}  Suppose $[\frac{d^2}{dy^2}+W(y)]u=0$ on
$J=[-c\lambda^{-\varepsilon},+c\lambda^{-\varepsilon}]$ with
$0<\varepsilon<\frac{1}{10}$ and $\lambda$ greater than a large constant
depending on $\varepsilon$.  Suppose $|W(y)-\lambda^2y|\le C\lambda^{2-N^\prime}$
on $J$, and suppose $\int_J|u|^2\,dy\le 1$.  Then for a constant $b_0$
of size $|b_0|\le C\lambda$ we have $\int_{\tilde J}|u(y)-b_0A
(\lambda^{2/3}y)|^2\,dy\le C\lambda^{10-2N^\prime}$, where $\tilde J$
denotes the middle half of $J$.\endproclaim

\demo{Proof}  Rescale to reduce to the previous Corollary.$\qquad\blacksquare$

We apply the preceding Corollary to study our eigenfunction $F$ on the
interval $I_{\text{Airey\ left}}$.

\vglue 1pc
\proclaim{Lemma 6}  For a constant $b$ of size $|b|\le C_{\#}\lambda_{\text{left}}
^{20}B_{\text{left}}^{-1}$ we have
$$
\int_{|x-x_{\text{left}}(E_1)|<c_{\#}\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}}|F(x)-bF_{\text{Airey\ left}}\ (x,E_1)|^2\,dx\le
C_{\#}\lambda_{\text{left}}^{10-2N^\prime}.
$$\endproclaim

\demo{Proof}  We use the change of variable $x\mapsto Y^{\text{left}}(x,E_1)
\equiv y$ to reduce matters to Corollary 2 of Lemma 5.  Recall that a general
ODE $[\frac{d^2}{dx^2}+(E_1-V(x))]F(x)=0$ is transformed by a change
$x\mapsto y(x)$ of independent variable to
$[\frac {d^2}{dy^2}+\hat W(y)]
\hat F(y)=0$ where $F(x)=(\frac{\partial y}{\partial x})^{-1/2}\hat F(y(x))$
and $E_1-V(x)=(\frac{\partial y}{\partial x})^2\hat W(y)+\{y,x\}$ with
$y=y(x)$.  We rewrite this as $(E_1-V(x))=\lambda_{\text{left}}^2(\frac
{\partial y}{\partial x})^2y+\{y,x\}+(\frac{\partial y}{\partial x})^2(
\hat W(y)-\lambda_{\text{left}}^2y)$.  The global WKB lemma tells us that
$$
\Big|-(E_1-V(x))+\lambda_{\text{left}}^2(\frac{\partial y}{\partial x})y
+\{y,x\}\Big|\le C_{\#}\lambda_{\text{left}}^{-N^\prime}S_{\text{left}},
$$

\noindent for $|x-x_{\text{left}}(E_1)|<\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}$, and also that
$$
|\frac{\partial y}{\partial x}|\sim B_{\text{left}}^{-1}\quad\text{for}\
|x-x_{\text{left}}(E_1)|<\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}.
$$

\noindent Therefore for $|x-x_{\text{left}}(E_1)|<\lambda_{\text{left}}^
{-\varepsilon}B_{\text{left}}$ we have $|\hat W(y)-\lambda_{\text{left}}^2y|\le
C_{\#}\lambda_{\text{left}}^{-N^\prime}S_{\text{left}}B_{\text{left}}^2=
C_{\#}\lambda_{\text{left}}^{2-N^\prime}$, with $y=y(x)$.  That is, with
$$
\hat I=\{y(x)\Bigm||x-x_{\text{left}}(E_1)|<\lambda_{\text{left}}^
{-\varepsilon}B_{\text{left}}\},
$$

\noindent we have $|\hat W(y)-\lambda_{\text{left}}^2y|\le C_{\#}
\lambda_{\text{left}}^{2-N^\prime}$ on $\hat I$.  Also
$$\align
\int_{\hat I}|\hat F(y)|^2\,dy&=\int_{|x-x_{\text{left}}(E_1)|<\lambda_{\text{
left}}^{-\varepsilon}B_{\text{left}}}\Big|\Bigl(\frac{\partial y}
{\partial x}\Bigr)^{-1/2}\hat F(y(x))\Big|^2\Bigl(\frac{\partial y}
{\partial x}\Bigr)^2\,dx\\
&=\int_{|x-x_{\text{left}}(E_1)|<\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}}|F(x)|^2\Bigl(\frac{\partial y}{\partial x}\Bigr)^2\,dx\\
&\le \int_{|x-x_{\text{left}}(E_1)|<\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}}|F(x)|^2\,dx\cdot C_{\#}B_{\text{left}}^{-2}\le C_{\#}
B_{\text{left}}^{-2}.\endalign
$$

\noindent Hence on $\hat I$, the function $B_{\text{left}}$ $\hat F(y)$
has norm at most $C_{\#}$ and is annihilated by $\frac{d^2}{dy^2}+
\hat W(y)$ with $|\hat W(y)-\lambda_{\text{left}}^2y|\le C_{\#}
\lambda^{2-N^\prime}$.  Note also that $\hat I$ contains an interval
$\hat J=\{|y|<c_{\#}\lambda_{\text{left}}^{-\varepsilon}\}$.  The
preceding Corollary tells us that $\int_{\hat J_1}|B_{\text{left}}\hat F(y)-
b_0A(\lambda_{\text{left}}^{2/3}y)|^2\,dy\le C_{\#}\lambda_{\text{left}}
^{10-2N^\prime}$ for a constant $b_0$ of size $|b_0|\le C_{\#}
\lambda_{\text{left}}^{10}$.  Here $\hat J_1=\{|y|<c_{\#}\lambda_{\text{left}}
^{-\varepsilon}\}$ for a smaller constant $c_{\#}$ than that used in the
definition of $\hat J$.

Now set $\hat I_1=\{|x-x_{\text{left}}(E_1)|<c_{\#}\lambda_{\text{left}}^
{-\varepsilon}B_{\text{left}}\}$ for  yet a smaller constant $c_{\#}$.
Then the image of $\hat I_1$ under $x\mapsto y(x)$ is contained in
$\hat J_1$.  Therefore 
$$\align
&\int_{\hat I_1}\Big|F(x)-b_0B_{\text{left}}^{-1}\Bigl(\frac{\partial y}
{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{left}}^{2/3}y(x))\Big|^2\,dx=\\
&\int_{\hat I_1}\Big|\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}
\hat F(y(x))-b_0B_{\text{left}}^{-1}\Bigl(\frac{\partial y}{\partial x}
\Bigr)^{-1/2}A(\lambda_{\text{left}}^{2/3}y(x))\Big|^2\,dx\\
&\le \int_{\hat I_1}|\hat F(y(x))-b_0B_{\text{left}}^{-1}A
(\lambda_{\text{left}}^{2/3}y(x))|^2\cdot\Bigl(\frac{\partial y}
{\partial x}\Bigr)\,dx\cdot(C_{\#}B_{\text{left}}^2)\endalign
$$

\noindent (since $(\frac{\partial y}{\partial x})\sim B_{\text{left}}^{-1}$
on $\hat I_1$) 
$$
\le C_{\#}\int_{\hat J_1}|B_{\text{left}}\hat F(y)-b_0A(\lambda_{\text{left}}^
{2/3}y)|^2\,dy\le C_{\#}\lambda_{\text{left}}^{10-2N^\prime}.
$$

We have by definition  $b_0B_{\text{left}}^{-1}(\frac{\partial y}{\partial x}
)^{-1/2}A(\lambda_{\text{left}}^{2/3}y(x))\hfill\break
=(b_0B_{\text{left}}^{-1}
\lambda_{\text{left}}^{1/3})F_{\text{Airey\ left}}(x,E_1)$.  Putting
$b=b_0B_{\text{left}}^{-1}\lambda_{\text{left}}^{1/3}$, we have
$|b|\le C_{\#}\lambda_{\text{left}}^{20}B_{\text{left}}^{-1}$ and
$ \int_{|x-x_{\text{left}}(E_1)|<c_{\#}\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}}|F(x)-bF_{\text{Airey\ left}}(x,E_1)|^2\,dx\le C_{\#}
\lambda_{\text{left}}^{10-2N^\prime}$.  The proof of the Lemma is complete.
$\qquad\blacksquare$

Set $\hat I_{\text{Airey\ left}}\ =\{|x-x_{\text{left}}(E_1)|<c_{\#}
\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\}$, so that Lemma 6 asserts
that
$$
\int_{\hat I_{\text{Airey\ left}}}|F(x)-bF_{\text{Airey\ left}}(x,E_1)|^2\,dx
\le C_{\#}\lambda_{\text{left}}^{10-2N^\prime}.\tag 9
$$

\noindent Of course there is an analogous result asserting that
$$
\int_{\hat I_{\text{Airey\ rt}}}|F(x)-b^\prime F_{\text{Airey\ rt}}(x,E_1)|^2
\,dx\le C_{\#}\lambda_{\text{rt}}^{10-2N^\prime}\tag 10
$$

\noindent with $\hat I_{\text{Airey\ rt}}=\{|x-x_{\text{rt}}(E_1)|<
c_{\#}\lambda_{\text{rt}}^{-\varepsilon}B_{\text{rt}}\}$, and another
constant $b^\prime$ of size $|b^\prime|\le C_{\#}\lambda_{\text{rt}}^{20}
B_{\text{rt}}^{-1}$.  Since $F$ is real, we can take the constants $b$,
$b^\prime$ to be real.

Next we study how $F$ behaves on $I_{\text{medium\ left}}$.  We make use
of the following result.
\vglue 1pc
\proclaim{Lemma 7}  The equation $(\frac{d^2}{dx^2}+E_1-V(x))u_1=f$ on
$I_{\text{medium left}}$ has a solution $u_1$ wth
$$
\Vert u_1\Vert_{L^2(I_{\text{medium\ left}})}\le C_{\#}
S_{\text{left}}^{-1/2}B_{\text{left}}\Vert f\Vert_{L^2(I_{\text{medium\ left}})}.
$$\endproclaim

\demo{Proof}  Let us write $F_c^{\text{left}}(x)$ for
$F_c^{\text{left}}(x,E_1)$. We define an approximate Green's function
$$ 
G(x,y)=\cases F_c^{\text{left}}(x)\overline{F_c^{\text{left}}(y)}
&\text{if $x\le y$}\\
\overline{F_c^{\text{left}}(x)}F_c^{\text{left}}(y)&\text{if
$x\ge y$}\endcases.
$$

\noindent Thus $|G(x,y)|\le \frac{C_{\#}}{[(E_1-V(x))(E_1-V(y))]^{1/4}}$
on $I_{\text{medium\ left}} \times I_{\text{medium\ left}}$, while
$(\partial_x^2+E_1-V(x))G(x,y)=H_1(y)\delta(x-y)+H(x,y)$ with
$|H(x,y)|\le\hfill\break
 C_{\#}\lambda_{\text{left}}^{10}\Bigl(\frac{
\lambda_{\text{left}}^{2/3}(x-x_{\text{left}}(E_1))}{B_{\text{left}}}
\Bigr)^{-\frac 32 N^\prime}B_{\text{left}}^{-3/2}(E_1-V(y))^{-1/4}$.

These estimates are immediate from the global WKB lemma.  Here
$H_1(y)=(\text{const.})\text{Im}(\partial_yF_c^{\text{left}}(y,E_1)\cdot
\overline{F_c^{\text{left}}(y,E_1}))$.  We have therefore
$$\multline
\int_{I_{\text{medium\ left}}\times I_{\text{medium\ left}}}|H(x,y)|^2\,dxdy
\le C_{\#}\lambda_{\text{left}}^{20-3\varepsilon N^\prime}B_{\text{left}}
^{-3}|I_{I_{\text{medium\ left}}}|\cdot\\
\cdot\int_{\text{medium\ left}}\frac{dy}
{(E_1-V(y))^{1/2}}\le C_{\#}\lambda_{\text{left}}^{20-3\varepsilon N^\prime}
B_{\text{left}}^{-1}S_{\text{left}}^{-1/2}
=C_{\#}\lambda_{\text{left}}^{19-3\varepsilon N^\prime},\endmultline
$$

\noindent and
$$\align
\int_{I_{\text{medium\ left}}\times I_{\text{medium\ left}}}
|G(x,y)|^2\,dxdy&\le C_{\#}\Bigl(\int_{I_{\text{medium\ left}}}
\frac{dx}{(E_1-V(x))^{1/2}}\Bigr)^2\\
&\le C_{\#} \Bigl(
S_{\text{left}}^{-1/2}B_{\text{left}}\Bigr)^2.\endalign
$$

\noindent Thus $f\mapsto \Cal E f(x)=\int_{I_{\text{medium\ left}}}
H(x,y)f(y)\,dy$ has Hilbert--Schmidt norm at most $C_{\#}\lambda_
{\text{left}}^{+\frac{19}{2}-\frac 32 \varepsilon N^\prime}$, while
$f \mapsto Gf(x)=\int_{I_{\text{medium\ left}}} G(x,y)f(y)\,dy$ has
Hilbert--Schmidt norm at most $C_{\#} S_{\text{left}}^{-1/2}B_{\text{left}}$.
We next study $H_1(y)$.  Recall that
$$
F_c^{\text{left}}(y)=\frac{e^{\pm i\frac \pi 4}\,e^{i\int_{x_{\text{left}}
(E_1)}^y(E_1-V(t))^{1/2}\,dt}}{(E_1-V(y))^{1/4}}\sum\limits_{k=0}^{N^\prime}
u_k^{\text{left}}(y,E_1)\tag 11
$$
$$\multline
\frac{\partial}{\partial y}F_c^{\text{left}}(y)=e^{\pm i\frac \pi 4}\,
e^{i\int_{x_{\text{left}}(E_1)}^y(E_1-V(t))^{1/2}\,dt}\Bigl\{i(E_1-V(y))^{1/4}
\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(y,E_1)+\\
(E_1-V(y))^{-1/4}\Bigl(\sum\limits_{k=0}^{N^\prime}\frac{\partial}{\partial y}
u_k^{\text{left}}(y,E_1)\Bigr)+\frac 14 (E_1-V(y))^{-\frac 54}V^\prime(y)
\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(y,E_1)\Bigr\}\\
=e^{\pm i\frac \pi 4}\,e^{i\int_{x_{\text{left}}(E)}^y(E_1-V(t))^{1/2}\,dt}
(E_1-V(y))^{1/4}\Bigl\{\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}
(y,E_1)+\frac{\sum\limits_{k=0}^{N^\prime}\frac{\partial}{\partial y}
u_k^{\text{left}}(y,E_1)}{(E_1-V(y))^{1/2}}\\
+\frac{V^\prime(y)\sum\limits_
{k=0}^{N^\prime}u_k^{\text{left}}(y,E_1)}{4(E_1-V(y))^{3/2}}\Bigr\}.
\endmultline\tag 12
$$

\noindent In $I_{\text{medium\ left}}$ we have $\sum\limits_{k=0}^{N^\prime}
u_k^{\text{left}}(y,E_1)=1+O(\lambda_{\text{left}}^{-\frac 32 \varepsilon})$
and
$$
\sum\limits_{k=0}^{N^\prime}\frac{\partial u_k^{\text{left}}}{\partial y}
(y,E_1)=\sum\limits_{k=1}^{N^\prime}O(\lambda_{\text{left}}^{-k}
B_{\text{left}}^{+\frac 32 k}(y-x_{\text{left}}(E_1))^{-\frac 32 k-1}),
$$

\noindent by the global WKB lemma, since $u_0^{\text{left}}\equiv 1$.
The last estimate simplifies to
$$
\Big|\sum\limits_{k=0}^{N^\prime}\frac{\partial u_k^{\text{left}}}
{\partial y}(y,E_1)\Big|\le C_{\#}\lambda_{\text{left}}^{-1}
B_{\text{left}}^{+\frac 32}(y-x_{\text{left}}(E_1))^{-5/2}.
$$

\noindent Since also $E_1-V(y)\sim S_{\text{left}}B_{\text{left}}^{-1}
(y-x_{\text{left}}(E_1))$ in $I_{\text{medium\ left}}$, we get
$$\multline
\bigg|\frac{\sum\limits_{k=0}^{N^\prime}\frac{\partial}{\partial y}
u_k^{\text{left}}(y,E_1)}{(E_1-V(y))^{1/2}}\bigg|\le
\frac{C_{\#}\lambda_{\text{left}}^{-1}B_{\text{left}}^{+3/2}
(y-x_{\text{left}}(E_1))^{-5/2}}{S_{\text{left}}^{1/2}
B_{\text{left}}^{-1/2}(y-x_{\text{left}}(E_1))^{1/2}}=
\frac{C_{\#}\lambda_{\text{left}}^{-1}}{S_{\text{left}}^{1/2}B_{\text{left}}}
\cdot\\
\cdot\Bigl(\frac{y-x_{\text{left}}(E_1)}{B_{\text{left}}}\Bigr)^{-3}\\
=C_{\#}\Bigl(\lambda_{\text{left}}^{2/3}\frac{(y-x_{\text{left}}(E_1))}
{B_{\text{left}}}\Bigr)^{-3}\le C_{\#}\lambda_{\text{left}}^{-3\varepsilon}
\quad\text{in}\ I_{\text{medium\ left}},\endmultline\tag 13
$$

\noindent and
$$\multline
\Bigg|\frac{\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(y,E_1)V^\prime
(y)}{4(E_1-V(y))^{3/2}}\Bigg|\le \frac{C_{\#}S_{\text{left}}B_{\text{left}}
^{-1}}{S_{\text{left}}^{3/2}B_{\text{left}}^{-3/2}(y-x_{\text{left}}(E_1))^
{3/2}}\\
=\frac{C_{\#}}{S_{\text{left}}^{1/2}B_{\text{left}}(\frac{y-x_{\text{left}}
(E_1)}{B_{\text{left}}})^{3/2}}=\frac{C_{\#}}
{\Bigl[\lambda_{\text{left}}^{2/3}(\frac{y-x_{\text{left}}(E_1)}
{B_{\text{left}}})\Bigr]^{3/2}}\le C_{\#}\lambda_{\text{left}}^{-\frac
32 \varepsilon}\\
\text{in}\ I_{\text{medium\ left}}.\endmultline\tag 14
$$

Putting (13) and (14) into (12), we find that
$$
\partial yF_c^{\text{left}}(y,E_1)=e^{\pm i\frac \pi 4}e^{i\int_{x_{\text{left}}
(E_1)}^y(E_1-V(t))^{1/2}\,dt}(E_1-V(y))^{1/4}\cdot (i+\text{error})\tag 15
$$

\noindent with
$$
|\text{error}|<C_{\#}\lambda^{-\frac 32 \varepsilon}\quad\text{in}\
I_{\text{medium\ left}}.\tag 16
$$

\noindent From (11) we get at once
$$
F_c^{\text{left}}(y,E_1)=\frac{e^{\pm i\frac \pi 4}\,e^{i\int_{
x_{\text{left}}(E_1)}^y(E_1-V(t))^{1/2}\,dt}}{(E_1-V(y))^{1/4}}\cdot
(1+\text{error}^\prime)\tag 17
$$

\noindent with
$$
|\text{error}^\prime|<C_{\#}\lambda_{\text{left}}^{-\frac 32 \varepsilon}
\text{in}\ I_{\text{medium\ left}}.\tag 18
$$

\noindent Equations (15), (16), (17), (18) show that
$$\multline
H_1(y)=(\text{const.})\cdot \text{Im}(\partial_yF_c^{\text{left}}(y,E_1)
\cdot\overline{F_c^{\text{left}}(y,E_1)})=(\text{const.})+(\text{error}^
{\prime\prime})\\
\text{with}\ |\text{error}^{\prime\prime}|<C_{\#}\lambda_{\text{left}}
^{-\frac 32 \varepsilon}\quad\text{in}\ I_{\text{medium\ left}}.\endmultline
$$

\noindent All we need from this is $|H_1(y)|>c_{\#}>0$ on $I_{\text{medium\
left}}$.

Now we can complete the proof of the lemma.  With operators $\Cal E$ and
$G$ as defined above, we know that $\Vert G\Vert<C_{\#}S_{\text{left}}
^{-1/2}B_{\text{left}}$ and $\Vert\Cal E\Vert\le C_{\#}\lambda_{\text{left}}
^{\frac{19}{2}-\frac 32 \varepsilon N^\prime}$.  Recall that 
$N^\prime=[\varepsilon N/500]$ and that we pick $N>>\varepsilon^{-10}$.
Thus $\Vert\Cal E\Vert<< \operatornamewithlimits{\inf}_
{y \in I_{\text{medium\ left}}}|H_1(y)|$.  It follows that the equation
$H_1(y)\cdot f_1(y)+\Cal E f_1(y)=f(y)$ on $I_{\text{medium\ left}}$
can be solved by a Neumann series with $\Vert f_1\Vert_{L^2(I
_{\text{medium\ left}})}\le C_{\#}\Vert f\Vert_{L^2(I_{\text{medium\ left}})}$.
Then taking $u_1=Gf_1$, we have
$$
\Vert u_1\Vert_{L^2}\le C_{\#}S_{\text{left}}^{-1/2}B_{\text{left}}\Vert f_1
\Vert_{L^2}\le C_{\#}S_{\text{left}}^{-1/2}B_{\text{left}}\Vert f\Vert_{L^2},
$$

\noindent and
$$\split
\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)u_1(x)=\int\Bigl[
(\partial_x^2+E_1-V(x))G(x,y)\Bigr]f_1(y)\,dy\\
=H_1\cdot f_1+\Cal E f_1=f(x).\endsplit
$$

\noindent Thus we have constructed the desired solution with good bounds.
$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary}  There is an exact solution $F_c(y)=F_c^{\text{left}}
(y,E_1)-F_{\text{error}}(y)$ to $(\frac{d^2}{dx^2}+E_1-V(x))F_c(x)=0$
in $I_{\text{medium\ left}}$, with $\Vert F_{\text{error}}\Vert_{L^2}\le
C_{\#}\lambda_{\text{left}}^{10-\frac 32 \varepsilon N^\prime}S_{\text{left}}^
{-1/2}$.\endproclaim\smallskip

\demo{Proof}  Set $f=(\partial_x^2+E_1-V(x))F_c^{\text{left}}(x,E_1)$ and
take $F_{\text{error}}=u_1$ as in the preceding Lemma.  Clearly $F_c$ is
an exact solution
of the ODE, and $\Vert F_{\text{error}}\Vert_{L^2}\le C_{\#}
S_{\text{left}}^{-1/2}B_{\text{left}}\Vert f\Vert_{L^2}$.  The global WKB
lemma gives
$$\split
|f(x)|\le C_{\#}\lambda_{\text{left}}^{10}\Bigl(\frac
{\lambda_{\text{left}}^{2/3}(x-x_{\text{left}}(E_1))}{B_{\text{left}}}\Bigr)
^{-\frac 32 N^\prime}B_{\text{left}}^{-3/2}\le C_{\#}\lambda_{\text{left}}
^{10-\frac 32 \varepsilon N^\prime}B_{\text{left}}^{-3/2}\\
\text{in}\ I_{\text{medium\ left}}.\endsplit
$$

\noindent so $\Vert f\Vert_{L^2(I_{\text{medium\ left}})}\le C_{\#}
\lambda_{\text{left}}^{10-\frac 32 \varepsilon N^\prime}B_{\text{left}}^{-1}$.
The Corollary follows at once.$\qquad\blacksquare$

Now we can understand our given eigenfunction $F$ in $I_{\text{medium\ left}}$,
by comparing it with the known solutions $F_c$ and $\overline F_c$.
We fix a cutoff $\chi$ supported in the middle half of $I_{\text{medium\ left}}$,
thus safely away from $x_{\text{left}}(E_1)$.  Then we work in the Hilbert
space $\Cal H=L^2(I_{\text{medium\ left}},\chi\,dx)$.  (Say
$0\le \chi\le 1$ everywhere and $\chi=1$ in the middle third of $I_{\text{medium\
left}}$.)

Now $\Vert F_c^{\text{left}}\Vert_{\Cal H}^2=\Vert\overline F_c^{\text{left}}\Vert_{\Cal H}^2\sim S_{\text{left}}^{-1/2}B_{\text{left}}$, while the stationary
phase lemma (lemma 1) shows that
$<F_c^{\text{left}},\overline F_c^{\text{left}}>_{\Cal H}=\int
\chi^2(F_c^{\text{left}})^2<<S_{\text{left}}^{-1/2}B_{\text{left}}$.

Since $\Vert F_{\text{error}}\Vert_{\Cal H}\le \Vert F_{\text{error}}\Vert_
{L^2(I_{\text{medium\ left}})}\le C_{\#}\lambda_{\text{left}}^
{10-\frac 32 \varepsilon N^\prime}S_{\text{left}}^{-1/2}$ and the right--hand
side is small compared to $\Vert F_c^{\text{left}}\Vert_{\Cal H}$, it follows
that the exact solutions $F_c$, $\overline F_c$ have norms in $\Cal H$
of the same order of magnitude as $\Vert F_c^{\text{left}}\Vert_{\Cal H}$,
and $F_c$, $\overline F_c$ are nearly orthogonal.

In particular, $F_c$, $\overline F_c$ are two linearly independent exact
solutions of $(\frac{d^2}{dx^2}+E_1-V(x))F=0$.  Hence our eigenfunction $F$
may be expresed as a linear combination
$$
F=b_1F_c+b_2\overline F_c
$$
\noindent for uniquely determined complex coefficients $b_1$, $b_2$.  Since
$F$ is real, we have $b_2=\overline b_1$ and $F=2\text{Re}(bF_c)$.  Since $F_c$
and $\overline F_c$ are nearly orthogonal in $\Cal H$, we have
$$
1=\Vert F\Vert_{L^2(I_{\text{BVP}})}^2\ge \Vert F\Vert_{\Cal H}^2\ge
c_{\#}(|b_1|^2\Vert F_c\Vert_{\Cal H}^2+|b_2|^2\Vert \overline F_c\Vert_{\Cal H}
^2).
$$

\noindent So $|b_1|\le C_{\#}\Vert F_c\Vert_{\Cal H}^{-1}\le C_{\#}S_{\text{left}}
^{1/4}B_{\text{left}}^{-1/2}$.

Now $F-2\text{Re}(b_1F_c^{\text{left}})=2\text{Re}(b_1F_{\text{error}})$, so
$$\multline
\int_{I_{\text{medium\ left}}}|F-2\text{Re}(b_1F_c^{\text{left}})|^2\,dx\le
C_{\#}|b_1|^2\int_{I_{\text{medium\ left}}}|F_{\text{error}}|^2\,dx\\
\le (C_{\#}S_{\text{left}}^{1/2}B_{\text{left}}^{-1})\cdot
(C_{\#}\lambda_{\text{left}}^{20-3\varepsilon N^\prime}S_{\text{left}}^{-1})
=C_{\#}\lambda_{\text{left}}^{19-3\varepsilon N^\prime},\endmultline
$$

\noindent since $S_{\text{left}}^{1/2}B_{\text{left}}=\lambda_{\text{left}}$.
Thus we have proven the following result.

\vglue 1pc
\proclaim{Lemma 8}  For a constant $b_{\text{medium\ left}}$ of size
$|b_{\text{medium\ left}}|<C_{\#}S_{\text{left}}^{1/4}B_{\text{left}}^{-1/2}$
we have
$$
\int_{I_{\text{medium\ left}}}|F(x)-\text{Re}(b_{\text{medium\ left}}
F_c^{\text{left}}(x,E_1))|^2\,dx\le C_{\#}\lambda_{\text{left}}^
{20-3\varepsilon N^\prime}.
$$
\noindent Here the constant $b_{\text{medium\ left}}$ is complex.\endproclaim

\noindent  There is of course an analogous result on $I_{\text{medium\ right}}$.

Next we study the given eigenfunction $F$ on $I_{\text{center}}$.  The
ideas are analogous to the proof of Lemma 8, but easier.
\vglue 1pc
\proclaim{Lemma 9}  On $I_{\text{center}}$ we can solve the equation
$$
\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)u=f\quad\text{with}\ \Vert u\Vert_{L^2}
\le C_{\#}\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\Vert f\Vert
_{L^2}.
$$\endproclaim

\demo{Proof}  Define the approximate Green's function
$$
G(x,y)=\cases F_c^{\text{left}}(x,E_1)\,\overline{F_c^{\text{left}}(y,E_1)}
&\text{if $x\le y$}\\
\overline{F_c^{\text{left}}(x,E_1)}\,F_c^{\text{left}}(y,E_1) &\text{if
$x \ge y$}.\endcases
$$

\noindent As in the proof of Lemma 7, we have $(\frac{\partial^2}{\partial x^2}
+E_1-V(x))G(x,y)=H_1(y)\delta(x-y)+H(x,y)$, $H_1(y)=(\text{const.})
\text{Im}[\frac{\partial}{\partial y}F_c^{\text{left}}(y,E_1)
\cdot\overline{F_c^{\text{left}}(y,E_1)}]$, and $|G(x,y)|\le
C_{\#}(E_1-V(x))^{-1/4}(E_1-V(y))^{-1/4}$ on $I_{\text{center}}\times
I_{\text{center}}$, $|H(x,y)|\le\hfill\break
 C_{\#}\Lambda^{-N^{\prime}}S^{-1/4}
(x)B^{-2}(x)(E_1-V(y))^{-1/4}$ on $I_{\text{center}}\times
I_{\text{center}}$.

The estimates are immediate from the global WKB lemma.  In particular,
the Hilbert--Schmidt norms of the kernels $H(x,y)$, $G(x,y)$ are
estimated as follows:
$$\align
\int_{I_{\text{center}}\times I_{\text{center}}}|H(x,y)|^2\,dxdy &\le
C_{\#}\Lambda^{-2N^\prime}\Bigl(\int_{I_{\text{center}}}S^{-1/2}
(x)B^{-4}(x)\,dx\Bigr)\\
&{}\qquad\qquad\qquad\qquad
\Bigl(\int_{I_{\text{center}}}\frac
{dy}{(E_1-V(y))^{1/2}}\Bigr)\\
&\le C_{\#}\Lambda^{K-2N^\prime}\quad\text{by\ hypothesis\ (E5)}\ ,
\tag 19\endalign
$$

\noindent and
$$
\Bigl(\int_{I_{\text{center}}\times I_{\text{center}}}|G(x,y)|^2\,dxdy\Bigr)^{1/2}
\le C_{\#}\Bigl(\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\Bigr).
\tag 20
$$

We turn to $H_1(y)$.  The equation (11), (12) hold also in $I_{\text{center}}$.
Moreover we have in $I_{\text{center}}$ the estimates:
$$\align
\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(y,E_1)&=1+O(\Lambda^{-1})\\
\Big|\sum\limits_{k=0}^{N^\prime}\frac{\partial}{\partial y}u_k^{\text{left}}
(y,E_1)\Big|&\le C_{\#}\Lambda^{-1}B^{-1}(y)\tag 21\endalign
$$

\noindent by the global WKB lemma.  Also $E_1-V(y)\ge c_{\#}S(y)$ and
$|V^\prime(y)|\le C_{\#}S(y)B^{-1}(y)$ in $I_{\text{center}}$.  Hence,
$$
\frac{|\sum\limits_{k=0}^{N^\prime}\frac{\partial}{\partial y}u_k^
{\text{left}}(y,E_1)|}{(E_1-V(y))^{1/2}}\le \frac{C_{\#}\Lambda^{-1}
B^{-1}(y)}{S^{1/2}(y)}=\frac{C_{\#}\Lambda^{-1}}{\lambda(y)}\le C_{\#}
\Lambda^{-1}\quad\text{in}\ I_{\text{center}}\tag 22
$$

\noindent and
$$\split
\frac{|V^\prime(y)||\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(y,E_1)|}
{(E_1-V(y))^{3/2}}\le \frac{C_{\#}S(y)B^{-1}(y)}{S^{3/2}(y)}=\frac{C_{\#}}
{S^{1/2}(y)B(y)}=\frac{C_{\#}}{\lambda(y)}\\
\le C_{\#}\Lambda^{-1}\quad\text{in}\ I_{\text{center}}.\endsplit\tag 23
$$

\noindent Putting (21), (22), (23) into (11) and (12), we get
$$\split
\frac{\partial}{\partial y}F_c^{\text{left}}(y,E_1)=
e^{\pm i\frac \pi 4}e^{i\int_{x_{\text{left}}(E_1)}^{y}(E_1-V(t))^{1/2}\,dt}
(E_1-V(y))^{1/4}\{i+\ \text{error}\}\\
\text{with}\ |\text{error}|\le C_{\#}\Lambda^{-1}\endsplit
$$

\noindent and
$$\split
F_c^{\text{left}}(y,E_1)=\frac{e^{\pm i
\frac \pi 4}e^{i\int_{x_{\text{left}}(E_1)}
^y(E_1-V(t))^{1/2}\,dt}}{(E_1-V(y))^{1/4}}\cdot \{1+\text{error}^\prime\}\\
\text{with}\ |\text{error}^\prime|\le C_{\#}\Lambda^{-1}.\endsplit
$$

\noindent Hence $H_1(y)=(\text{const.})\text{Im}[\frac{\partial F_c^
{\text{left}}}{\partial y}\cdot\overline F_c^{\text{left}}]=(\text{const.})+
(\text{error}^{\prime\prime})$ with $|\text{error}^{\prime\prime}|<
C_{\#}\Lambda^{-1}$.  All we need from this is $|H_1(y)|\ge c_{\#}>0$.
Now we solve\hfill\break
$H_1(y)f_1(y)+\int_{I_{\text{center}}}H(y,z)f_1(z)\,dz=f(y)$
by a Neumann series and put $u(x)=\int_{I_{\text{center}}}G(x,y)f_1(y)\,dy$
as in Lemma 7.  Thus $u$ solves the ODE and has norm
$$\align
\Vert u\Vert_{L^2(I_{\text{center}})}&\le C_{\#}\int_{I_{\text{center}}}
\frac{dx}{(E_1-V(x))^{1/2}}\cdot\Vert f_1\Vert_{L^2(I_{\text{center}})}\\
&\le C_{\#}\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\Vert f\Vert,
\endalign
$$

\noindent provided $\Vert H(\cdot,\cdot)\Vert_{\text{Hilbert--Schmidt}}\le
\frac 12 \operatornamewithlimits{inf}_{y \in I_{\text{center}}}|H_1(y)|$
to make the Neumann series converge.  This condition is satisfied by virtue
of (19), since we picked $N>K\varepsilon^{-10}$.$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary}  There is an exact solution $F_0=F_0^{\text{left}}-
F_{\text{error}}$ of $[\frac{d^2}{dx^2}+E_1-V(x)]F_c=0$ in $I_{\text{center}}$, with
$\Vert F_{\text{error}}\Vert_{L^2}^2\le C_{\#}\Lambda^{K-2N^\prime}
\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}$.\endproclaim

\demo{Proof}  Take $F_{\text{error}}=u$ arising from $f=(\frac{\partial^2}
{\partial x^2}+E_1-V(x))F_0^{\text{left}}$ in Lemma 9.  Thus the ODE is 
satisfied exactly,
and $\Vert F_{\text{error}}\Vert_{L^2}\le C_{\#}\int_{I_{\text{center}}}
\frac{dx}{(E_1-V(x))^{1/2}}\Vert f\Vert_{L^2}$.

The global WKB lemma gives $|f(x)|\le C_{\#}\Lambda^{-N^\prime}S^{-1/4}
(x)B^{-2}(x)$ pointwise in $I_{\text{center}}$.  Hence
$\int_{I_{\text{center}}}|f(x)|^2\,dx\le C_{\#}\Lambda^{-2N^\prime}
\Bigl(\int_{I_{\text{center}}}\frac{dx}{S^{1/2}(x)B^4(x)}\Bigr)$,
so \hfill\break
$\Vert F_{\text{error}}\Vert_{L^2(I_{\text{center}})}^2\le
C_{\#}\Lambda^{-2N^\prime}\Bigl(\int_{I_{\text{center}}}\frac
{dx}{S^{1/2}(x)B^4(x)}\Bigr)\Bigl(\int_{I_{\text{center}}}
\frac{dx}{(E_1-V(x))^{1/2}}\Bigr)^2\le\hfill\break
C_{\#}\Lambda^{K-2N^\prime}
\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}$ by hypothesis (E5).
$\qquad\blacksquare$

Now we can repeat the proof of Lemma 8:

Let $\chi(x)$ be a smooth cutoff function supported in
$I_{\text{center}}=[\overline x_1,\overline x_2]$ with $\chi(x)=1$
for $\overline x_1+\hat c_{\#}B_{\text{left}}\le x\le \overline x_2
-\hat c_{\#}B_{\text{right}}$,
$$\align
\Big|\Bigl(\frac{d}{dx}\Bigr)^m\chi(x)\Big|&\le C_{\#}^m
B_{\text{left}}^{-m}\quad\text{in}\ [\overline x_1,\overline x_1+\hat c_{\#}
B_{\text{left}}],\\
\Big|\Bigl(\frac{d}{dx}\Bigr)^m\chi(x)\Big|&\le C_{\#}^mB_{\text{right}}^{-m}
\quad\text{in}\ [\overline x_2-\hat c_{\#}B_{\text{right}},\overline x_2],
\endalign
$$
\noindent $0\le \chi\le 1$ everywhere.

We work in the Hilbert space $\Cal H=L^2(I_{\text{center}},\chi\,dx)$.  The
Corollary of Lemma 2 gives
$$
\Vert F_c^{\text{left}}\Vert_{\Cal H}^2=\Vert \overline F_c^{\text{left}}\Vert_
{\Cal H}^2\ge c\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}.\tag 24
$$

\noindent Moreover,
$$
|<F_c^{\text{left}},\overline F_c^{\text{left}}>_{\Cal H}|=
|\int_{I_{\text{center}}}\chi(F_c^{\text{left}})^2\,dx|\le
C_{\#}\Lambda^{-N^\prime}\int_{I_{\text{center}}}\frac
{dx}{(E_1-V(x))^{1/2}}.\tag 25
$$

\noindent To see this, write a partition of unity $\chi=\sum\limits_\nu\chi_\nu$
with $\chi_\nu$ supported in $\{|x-x_\nu|<c_{\#}B(x_\nu)\}$ and satisfying
natural estimates.  Then we have
$$\multline
|\int_{I_{\text{center}}}\chi(F_c^{\text{left}})^2\,dx|\le
\sum\limits_\nu|\int_{I_{\text{center}}}\chi_\nu(F_c^{\text{left}})^2\,dx|\\
\le C_{\#}\sum\limits_\nu\int_{\text{supp}\ \chi_\nu}
\frac{dx}{(\lambda(x))^{N^\prime}(E_1-V(x))^{1/2}}\endmultline
$$

\noindent as one sees by applying the stationary phase Lemma.  Since
$\lambda(x)\ge \Lambda$, estimate (25) follows.

Now (24), (25) show that $F_c^{\text{left}}$, $\overline F_c^{\text{left}}$
are nearly orthogonal in $\Cal H$.  Also, (24) and the preceding corollary
show that $F_c$, $\overline F_c$ are small perturbations of $F_c^{\text{left}}$,
$\overline F_c^{\text{left}}$.  We conclude that
$$
\Vert F_c\Vert_{\Cal H}^2=\Vert\overline F_c\Vert_{\Cal H}^2\ge c_{\#}
\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}},\tag 26
$$

\noindent and
$$
|<F_c,\overline F_c>_{\Cal H}| \le \frac{1}{10}\,\Vert F_c\Vert_{\Cal H}
\Vert \overline F_c\Vert_{\Cal H}.\tag 27
$$

\noindent In particular, $F_c$ and $\overline F_c$ are linearly independent
solutions of $(\frac{d^2}{dx^2}+E_1-V(x))F\hfill\break=0$.  Hence our given
solution $F$ may be expressed as a linear combination $F=b_1F_c+b_2\overline
F_c$ with uniquely determined (complex) constants $b_1$, $b_2$.  Since $F$
is real we have $b_2=\bbar_1$, so $F=2\text{Re}(b_1F_c)$.  Moreover,
$$\multline
1=\Vert F\Vert_{L^2(I_{\text{BVP}})}^2\ge \Vert F\Vert_{\Cal H}^2\ge c_{\#}
(|b_1|^2\Vert F_c\Vert_{\Cal H}^2+|b_2|^2\Vert\overline F_c\Vert_{\Cal H}^2
)\ \text{(by\ (27))}\\
\ge c_{\#}|b_1|^2\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}
,\endmultline
$$ 
i.e. $|b_1|\le C_{\#}\Bigl(\int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}
\frac{dx}{(E_1-V(x))^{1/2}}\Bigr)^{-1/2}$.

\noindent Here we have changed the limits of integration, but the order
of magnitude of the integral remains unaffected.  Now we have
$$
F=2\text{Re}(b_1F_c)=2\text{Re}(b_1F_c^{\text{left}})-2\text{Re}(b_1
F_{\text{error}}),
$$

\noindent so that
$$\align
&\int_{I_{\text{center}}}|F-2\text{Re}(b_1F_c^{\text{left}})|^2\,dx\le
C_{\#}|b_1|^2\int_{I_{\text{center}}}|F_{\text{error}}(x)|^2\,dx\\
&\,\,\le C_{\#}\Bigl(\int_{I_{\text{center}}}\frac{dx}
{(E_1-V(x))^{1/2}}\Bigr)^{-1}\Vert F_{\text{error}}\Vert_{L^2}^2\\
&\,\,\le C_{\#}\Bigl(\int_{I_{\text{center}}}\frac{dx}
{(E_2-V(x))^{1/2}}\Bigr)^{-1}\cdot C_{\#}\Lambda^{K-2N^\prime}
\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\\
&\,\,=C_{\#}\Lambda^{K-2N^\prime}.\endalign
$$

\noindent Thus we have proven the following result.

\vglue 1pc
\proclaim{Lemma 10}  For a constant $b_{\text{center}}^{\text{left}}$
of size $|b_{\text{center}}^{\text{left}}|\le\hfill\break C_{\#}
\Bigl(\int_{x_{\text{left}}(E_1)}^{x_{\text{right}}(E_1)}\frac{dx}
{(E_1-V(x))^{1/2}}\Bigr)^{-1/2}$ we have
$$
\int_{I_{\text{center}}}|F(x)-\text{Re}(b_{\text{center}}
^{\text{left}}F_c^{\text{left}}(x,E_1))|^2\,dx\le C_{\#}\Lambda^{K-2N^\prime}
.
$$\endproclaim

Of course, there is an analogous result for $F_c^{\text{right}}$, namely
$$
\int_{I_{\text{center}}}|F(x)-\text{Re}(b_{\text{center}}^
{\text{right}}F_c^{\text{right}}(x,E_1))|^2\,dx\le C_{\#}\Lambda^{K-2N^\prime}.
\tag 28
$$

\noindent for another constant $b_{\text{center}}^{\text{right}}$ of
size $|b_{\text{center}}^{\text{right}}|\le C_{\#}
\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}(E_1)}\frac{dx}
{(E_1-V(x))^{1/2}}\Bigr)^{-1/2}$.

Lemmas 4, 6, 8, 10 tell us how $F$ looks on $I_{\text{far\ left}}$,
$\hat I_{\text{Airey\ left}}$, $I_{\text{medium\ left}}$, $I_{\text{center}}$.
The analogous results for the right--hand solutions tell us how $F$ looks
on $I_{\text{far\ right}}$, $\hat I_{\text{Airey\ right}}$,
$I_{\text{medium\ right}}$,
$I_{\text{center}}$.  These intervals cover $I_{\text{BVP}}$, the domain
of $F$.  Since $I_{\text{medium\ left}}$ overlaps both
$\hat I_{\text{Airey\ left}}$ and $I_{\text{center}}$, we have two
different descriptions of $F$ on each overlap.  Naturally, these
descriptions must be consistent.  We will deduce from this that the constants
$b$ in Lemma 6, $b_{\text{medium\ left}}$ from Lemma 8, and $b_{\text{center}}^
{\text{left}}$ from Lemma 10 are nearly the same.  Thus we can describe
$F$ on $I_{\text{far\ left}}\hfill\break
\cup \hat I_{\text{Airey\ left}}\cup
I_{\text{medium\ left}} \cup I_{\text{center}}$ using a single unknown
constant $b$.  Analogously, we can describe $F$ on $I_{\text{far\ right}}
\cup \hat I_{\text{Airey\ right}}\cup I_{\text{medium\ right}}\cup
I_{\text{center}}$ using another single unknown constant $b^\prime$.
On $I_{\text{center}}$, we have two different descriptions of $F$, which
must be consistent.  This tells us that the phase $\Phi(E_1)$ must be
nearly a multiple of $\pi$, and that $b$ is approximately $\pm b^\prime$. 
We carry out this plan in the paragraphs below.

>From lemmas 6, 8, 10 we get
$$
\int_{\hat I_{\text{Airey\ left}}}|F-b_{\text{Airey\ left}}F_{\text{Airey\ left}}
|^2\,dx\le C_{\#}\lambda_{\text{left}}^{10-2N^\prime}\tag 29
$$
$$
|b_{\text{Airey\ left}}|\le C_{\#}\lambda_{\text{left}}^{20}
B_{\text{left}}^{-1}, b_{\text{Airey\ left}}\,\,\text{\underbar{real}}
\tag 30
$$
$$
\int_{I_{\text{medium\ left}}}|F-\text{Re}(b_{\text{medium\ left}}F_c^
{\text{left}})|^2\,dx\le C_{\#}\lambda_{\text{left}}^{20-3\varepsilon N^\prime}
\tag 31
$$
$$
|b_{\text{medium\ left}}|\le C_{\#}S_{\text{left}}^{1/4}B_{\text{left}}^{-1/2}
\tag 32
$$
$$
\int_{I_{\text{center}}}|F-\text{Re}(b_{\text{center}}^{\text{left}}
F_c^{\text{left}})|^2\,dx \le C_{\#}\Lambda^{K-2N^\prime}\tag 33
$$
$$
|b_{\text{center}}^{\text{left}}|\le C_{\#}\Bigl(\int_{x_{\text{left}}(E_1)}
^{x_{\text{right}}(E_1)}\frac{dx}{(E_1-V(x))^{1/2}}\Bigr)^{-1/2}.\tag 34
$$

The global WKB lemma gives
$$
|F_{\text{Airey\ left}}-\text{Re}(F_c^{\text{left}})|\le C_{\#}
\lambda_{\text{left}}^{10-\frac 32 \varepsilon N^\prime}B_{\text{left}}^
{1/2}\ \text{on}\ \hat I_{\text{Airey\ left}}\cap I_{\text{medium\ left}},
$$

\noindent hence
$$\align
&\int_{\hat I_{\text{Airey\ left}}\cap I_{\text{medium\ left}}}|b_
{\text{Airey\ left}}F_{\text{Airey\ left}}-\text{Re}(b_{\text{Airey\ left}}
F_c^{\text{left}})|^2\,dx\\
&\,\, \le \Bigl(C_{\#}\lambda_{\text{left}}^{20-3\varepsilon N^\prime}
|b_{\text{Airey\ left}}|^2B_{\text{left}}\Bigr)|I_{\text{Airey\ left}}
\cap I_{\text{medium\ left}}|\\
&\,\,\le C_{\#}\lambda_{\text{left}}^{20-3\varepsilon N^\prime}B_{\text{left}}^2
|b_{\text{Airey\ left}}|^2\le C_{\#}\lambda_{\text{left}}^
{20-3\varepsilon N^\prime}B_{\text{left}}^2\cdot C_{\#}\lambda_{\text{left}}
^{40}B_{\text{left}}^{-2}\\
&\,\,\ \text{(by\ (30))}\ = C_{\#}\lambda^{60-3\varepsilon N^\prime}.\endalign
$$

\noindent Combining this inequality with (29), (31), we get
$$
\int_{\hat I_{\text{Airey\ left}}\cap I_{\text{medium\ left}}}|\text{Re}
((b_{\text{Airey\ left}}-b_{\text{medium\ left}})F_c^{\text{left}})|^2\,dx
\le C_{\#}\lambda_{\text{left}}^{60-3\varepsilon N^\prime}.
$$

\noindent Introduce a smooth cutoff $\chi$ $(0\le \chi\le 1)$ supported
in the middle half of $\hat I_{\text{Airey\ left}}\hfill\break
\cap I_{\text{medium\ left}}$.
(Thus $x-x_{\text{left}}(E_1)\sim \lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}$ in $\text{supp}\ \chi$).  The preceding inequality implies
$$
\int_{\Bbb R}\chi |\text{Re}((b_{\text{Airey\ left}}-b_{\text{medium\ left}})
F_c^{\text{left}}|^2\,dx\le C_{\#}\lambda_{\text{left}}^{60-3\varepsilon
N^\prime}.\tag 35
$$

\noindent As in the proof of Lemma 2, the stationary phase Lemma shows
that the integral on the left dominates
$$\align
c_{\#}&\int_{\Bbb R}\frac{\chi\,dx}{(E_1-V(x))^{1/2}}\cdot
|b_{\text{Airey\ left}}-b_{\text{medium\ left}}|^2\\
&\,\, \ge \frac{c_{\#}\,\lambda_{\text{left}}^{-\varepsilon/2}B_{\text{left}}}
{S_{\text{left}}^{1/2}}\cdot|b_{\text{Airey\ left}}-b_{\text{medium\ left}}|^2.
\endalign
$$

\noindent Therefore $|b_{\text{Airey\ left}}-b_{\text{medium\ left}}|^2\le
C_{\#} \lambda_{\text{left}}^{63-3\varepsilon N^\prime}S_{\text{left}}^{1/2}
B_{\text{left}}^{-1}$.  Since $S_{\text{left}}^{1/2}B_{\text{left}}^{-1}=
\hfill\break\lambda_{\text{left}}B_{\text{left}}^{-2}$, this yields
$|b_{\text{Airey\ left}}-b_{\text{medium\ left}}|^2\le C_{\#}
\lambda_{\text{left}}^{66-3\varepsilon N^\prime}B_{\text{left}}^{-2}$, hence
$$\align
&\int_{I_{\text{medium\ left}}}|b_{\text{Airey\
left}}\text{Re}(F_c^{\text{left}})-\text{Re}(b_{\text{medium\
left}}F_c^{\text{left}})|^2\,dx\\
&\,\,\,\le C_{\#}\lambda_{\text{left}}^{66-3\varepsilon
N^\prime}B_{\text{left}}^{-2} \int_{I_{\text{medium\ left}}}|F_c|^2\,dx\\
&\qquad\qquad\qquad\qquad\qquad\le
C_{\#}\lambda_{\text{left}}^ {66-3\varepsilon
N^\prime}B_{\text{left}}^{-2}\int_{I_{\text{medium\ left}}}
\frac{dx}{(E_1-V(x))^{1/2}}\\
&\,\,\,\le C_{\#}\lambda_{\text{left}}^{66-3\varepsilon
N^\prime}B_{\text{left}}^{-2} \cdot S_{\text{left}}^{-1/2}B_{\text{left}}\le
C_{\#}\lambda_{\text{left}}^ {65-3\varepsilon N^\prime},\ \text{since}\
S_{\text{left}}^{1/2}B_{\text{left}} =\lambda_{\text{left}}.\endalign
$$

\noindent Combining this with (31), we get
$$
\int_{I_{\text{medium\ left}}}|F-b_{\text{Airey\
left}}\text{Re}(F_c^{\text{left}})|^2\,dx\le C_{\#}\lambda_{\text{left}}^
{65-3\varepsilon N^\prime}.\tag 36
$$

Next, from (36) and (33) we get
$$
\int_{I_{\text{medium\ left}}\cap I_{\text{center}}}|\text{Re}
((b_{\text{Airey\ left}}-b_{\text{center}}^{\text{left}})F_c^{\text{left}})|^2
\,dx\le C_{\#}\Lambda^{65-3\varepsilon N^\prime+K}.
$$

Since $|I_{\text{medium\ left}}\cap I_{\text{center}}|\ge c_{\#}
B_{\text{left}}$, $\text{dist}(I_{\text{medium\ left}}\cap I_{\text{center}},
x_{\text{left}}(E_1))\ge\hfill\break c_{\#}B_{\text{left}}$, lemma 2 shows that
the integral on the left is at least \hfill\break
 $c_{\#}|b_{\text{Airey\ left}}
-b_{\text{center}}^{\text{left}}|^2\int_{I_{\text{medium\ left}}\cap
I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\ge\hfill\break c_{\#}
|b_{\text{Airey\ left}}-b_{\text{center}}^{\text{left}}|^2S_{\text{left}}^{-1/2}
B_{\text{left}}$.  Hence $|b_{\text{Airey\ left}}-b_{\text{center}}^{\text{
left}}|^2\le\hfill\break
 C_{\#}\Lambda^{K+65-3\varepsilon N^\prime}S_{\text{left}}^{1/2}
B_{\text{left}}^{-1}$, so that
$$\align
&\int_{I_{\text{center}}}|\text{Re}(b_{\text{Airey}}^{\text{left}}F_c^
{\text{left}}-b_{\text{center}}^{\text{left}}F_c^{\text{left}})|^2\,dx\\
&\,\,\,\,
\le C_{\#}\Lambda^{K+65-3\varepsilon N^\prime}S_{\text{left}}^{1/2}
B_{\text{left}}^{-1}\int_{I_{\text{center}}}|F_c^{\text{left}}|^2\,dx\\
&\,\,\,\,\le C_{\#}\Lambda^{65+K-3\varepsilon N^\prime}S_{\text{left}}^{1/2}
B_{\text{left}}^{-1}\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\\
&\,\,\,\,\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}\quad\text{by (E6)}.
\endalign
$$

\noindent Combining this with (33), we have
$$
\int_{I_{\text{center}}}|F-b_{\text{Airey\ left}}\text{Re}
(F_c^{\text{left}})|^2\,dx\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}.\tag 
37
$$

\noindent Since also $\int_{I_{\text{center}}}|F|^2\,dx\le 1$, we get
$$\split
b_{\text{Airey\ left}}^2\int_{I_{\text{center}}}(\text{Re}\
F_c^{\text{left}})^2\,dx\le C_{\#} \quad\text{since}\ N^\prime=[\varepsilon
N/500],\\
N >> K\varepsilon^{-10}.\endsplit
$$

\noindent Lemma 2 shows that this amounts to $c_{\#}b_{\text{Airey\ left}}^2
\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\le C_{\#}$, i.e.
$$
|b_{\text{Airey\ left}}|\le C_{\#}\Bigl(\int_{x_{\text{left}}(E_1)}
^{x_{\text{right}}(E_1)}\frac{dx}{(E_1-V(x))^{1/2}}\Bigr)^{-1/2}.\tag 38
$$
\noindent In writing (38), we changed the limits of integration without
affecting the order of magnitude of the integral.  Combining estimates (29),
(36), (37), (38), we obtain the following result.

\vglue 1pc
\proclaim{Lemma 11}  For a real constant $b_{\text{left}}$ of size
$|b_{\text{left}}|\le C_{\#}(\int_{V<E_1}\frac{dx}{(E_1-V(x))^{1/2}})^{-1/2}$,
we have
$$
\int_{\hat I_{\text{Airey\ left}}}|F(x)-b_{\text{left}}F_{\text{Airey\ left}}
(x,E_1)|^2\,dx\le C_{\#}\lambda_{\text{left}}^{10-2N^\prime}\quad\text{and}
\tag 39
$$
$$\int_{I_{\text{medium\ left}}\cup I_{\text{center}}}|F(x)-b_{\text{left}}
\text{Re}(F_c^{\text{left}}(x,E_1))|^2\,dx\le C_{\#}\Lambda^{2K+65-3
\varepsilon N^\prime}.\tag 40
$$\endproclaim
\bigskip

\noindent Of course there is an analogous result for $F_c^{\text{right}}$:
For a real constant  $b_{\text{right}}$ of size $|b_{\text{right}}|
\le C_{\#}(\int_{V<E_1}\frac{dx}{(E_1-V(x))^{1/2}})^{-1/2}$ we have
$$
\int_{\hat I_{\text{Airey\ right}}}|F(x)-b_{\text{right}}F_{\text{Airey\ right}}
(x,E_1)|^2\,dx\le C_{\#}\lambda_{\text{right}}^{10-2N^\prime}\quad\text{and}
\tag 41
$$
$$
\int_{I_{\text{medium\ right}}\cup I_{\text{center}}}|F(x)-b_{\text{right}}
\text{Re}(F_c^{\text{right}}
(x,E_1))|^2\,dx\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}.
\tag 42
$$

Comparing (40) and (42), we see that
$$
\int_{I_{\text{center}}}|\text{Re}(b_{\text{left}}F_c^{\text{left}}-
b_{\text{right}}F_c^{\text{right}})|^2\,dx\le C_{\#}
\Lambda^{2K+65-3\varepsilon N^\prime}.\tag 43
$$

\noindent From the global WKB lemma we recall that
$$
|F_c^{\text{left}}-\Cal A(E_1)F_c^{\text{right}}|\le C_{\#}
\Lambda^{-N^\prime-1}(E_1-V(x))^{-1/4}\ \text{on}\ I_{\text{center}},
$$
\noindent so
$$\align
\int_{I_{\text{center}}}|&\text{Re}(b_{\text{left}}F_c^{\text{left}}-
b_{\text{left}}\Cal A(E_1)F_c^{\text{right}})|^2\,dx\\&\le 
C_{\#}\Lambda^{-2N^\prime-2}\int_{I_{\text{center}}}\frac{dx}
{(E_1-V(x))^{1/2}}|b_{\text{left}}|^2\\
&\le C_{\#}\Lambda^{-2N^\prime-2}\quad\text{by\ the\ estimate\ on}\
|b_{\text{left}}|\ \text{in\ Lemma\ 11}.\endalign
$$

\noindent Combining this with (43) gives
$$
\int_{I_{\text{center}}}|\text{Re}((b_{\text{right}}-b_{\text{left}}
\Cal A(E_1))F_c^{\text{right}})|^2\,dx\le 
C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}.
$$

\noindent The Corollary to Lemma 2 shows that the left--hand side dominates
$c_{\#}|b_{\text{right}}-b_{\text{left}}\Cal A(E_1)|^2\int_{
I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}$.  Hence we obtain:
$$
|b_{\text{right}}-b_{\text{left}}\Cal A(E_1)|^2\le C_{\#}
\Lambda^{2K+65-3\varepsilon N^\prime}\Bigl(\int_{V<E_1}\frac{dx}
{(E_1-V(x))^{1/2}}\Bigr)^{-1}.\tag 44
$$

\noindent We next derive lower bounds for $|b_{\text{right}}|$, $|b_{\text{
left}}|$, to allow us to deduce from (44) that the complex number
$\Cal A(E)$ has small imaginary part.

Note that $\lambda_{\text{left}}^2(\frac{\partial Y^{\text{left}}}{\partial x}
)^{2}Y^{\text{left}}=(E_1-V(x))+\text{Error}$ with $|\text{Error}|\le
C_{\#}\lambda_{\text{left}}^{-2}S_{\text{left}}=C_{\#}B_{\text{left}}^{-2}$
in $\hat I_{\text{Airey\ left}}$, by the global WKB lemma.

Also $C_{\#}B_{\text{left}}^{-2}\ge (\frac{\partial Y^{\text{left}}}{\partial x}
)^2\ge c_{\#}B_{\text{left}}^{-2}$, in $\hat I_{\text{Airey\ left}}$
again by the global WKB lemma.

Hence
$$\align
&|F_{\text{Airey\ left}}(x,E_1)|=\Big|\lambda_{\text{left}}^{-1/3}\Bigl(\frac
{\partial Y^{\text{left}}}{\partial
x}\Bigr)^{-1/2}A(\lambda_{\text{left}}^{2/3} Y^{\text{left}})\Big|\\
&\,\,\le C_{\#}\lambda_{\text{left}}^{-1/3}\Big|\frac{\partial Y^{\text{left}}}
{\partial x}\Big|^{-1/2}(1+\lambda_{\text{left}}^{2/3}|Y^{\text{left}}|
)^{-1/4}\\
&\,\,=C_{\#}\Bigl[\lambda_{\text{left}}^{4/3}\Bigl(\frac{\partial
Y^{\text{left}}} {\partial x}\Bigr)^{+2}+\lambda_{\text{left}}^2
\Bigl(\frac{\partial
Y^{\text{left}}} {\partial x}\Bigr)^2|Y^{\text{left}}|\Bigr]^{-1/4}\\
&\,\,\le C_{\#}[\lambda_{\text{left}}^{4/3}B_{\text{left}}^{-2}
+|E_1-V(x)+\text{Error}|]^{-1/4}\\
&\,\,\le C_{\#}[\lambda_{\text{left}}^{4/3}B_{\text{left}}^{-2}+|E_1-V(x))|]
^{-1/4}\\
&\noindent (\text{since}\ |\text{Error}|\le C_{\#}B_{\text{left}}^{-2}<<
\lambda_{\text{left}}^{4/3}B_{\text{left}}^{-2})\\
&\,\,\le C_{\#}|E_1-V(x)|^{-1/4}\quad\text{in}\ \hat I_{\text{Airey\ left}}.
\endalign
$$

\noindent This crude estimate is enough to show that
$$\split
\int_{\hat I_{\text{Airey\ left}}}|F_{\text{Airey\ left}}
(x,E_1)|^2\,dx\le C_{\#}\int_{\hat I_{\text{Airey\ left}}}\frac{dx}
{|E_1-V(x)|^{1/2}}\le C_{\#}S_{\text{left}}^{-1/2}B_{\text{left}}.
\endsplit\tag"(44 bis)"
$$

\noindent Hence (39) yields
$$\align
\int_{\hat I_{\text{Airey\ left}}}&|F(x)|^2\,dx\le C_{\#}
\lambda_{\text{left}}^{10-2N^\prime}+|b_{\text{left}}|^2\int_{
\hat I_{\text{Airey\ left}}}|F_{\text{Airey\ left}}(x,E_1)|^2\,dx\\
&\quad \le C_{\#}\lambda_{\text{left}}^{10-2N^\prime}+C_{\#}
S_{\text{left}}^{-1/2}B_{\text{left}}|b_{\text{left}}|^2\\
&\quad \le C_{\#}\lambda_{\text{left}}^{10-2N^\prime}+C_{\#}
\int_{V<E_1}\frac{dx}{(E_1-V(x))^{1/2}}\cdot|b_{\text{left}}|^2.\tag 45
\endalign
$$

\noindent Similarly,
$$
\int_{\hat I_{\text{Airey\ right}}}|F(x)|^2\,dx\le C_{\#}
\lambda_{\text{right}}^{10-2N^\prime}+C_{\#}\int_{V<E_1}
\frac{dx}
{(E_1-V(x))^{1/2}}\cdot |b_{\text{right}}|^2.\tag 46
$$

\noindent From (40) we get
$$\align
&\int_{I_{\text{medium\ left}}\cup I_{\text{center}}}|F(x)|^2\,dx
\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+|b_{\text{left}}|^2
\int_{I_{\text{medium\ left}}\cup I_{\text{center}}}|F_c^{\text{left}}|^2\,dx
\\
&\,\,\,\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}+C_{\#}
\int_{V<E_1}\frac{dx}{(E_1-V(x))^{1/2}}|b_{\text{left}}|^2.\tag 47\endalign
$$

\noindent Similarly,
$$\split
\int_{I_{\text{center}}\cup I_{\text{medium\ right}}}|F(x)|^2\,dx\le
C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}\\+C_{\#}
\int_{V<E_1}\frac{dx}{(E_1-V(x))^{1/2}}|b_{\text{right}}|^2.\endsplit\tag 48
$$

\noindent Lemma 4 and its analogue for $I_{\text{far\ right}}$ give
$$
\int_{I_{\text{far\ left}}\cup I_{\text{far\ right}}}|F(x)|^2\,dx\le
C_{\#}\lambda_{\text{left}}^{-N}+C_{\#}\lambda_{\text{right}}^{-N}.\tag 49
$$

\noindent Since the regions of integration in (45), (46), (47), (48), (49)
cover $I_{\text{BVP}}$, we may add these inequalities to obtain
$$
\int_{I_{\text{BVP}}}|F(x)|^2\,dx\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}
+C_{\#}\int_{V<E_1}\frac{dx}{(E_1-V(x))^{1/2}}\cdot(|b_{\text{left}}|^2+
|b_{\text{right}}|^2).
$$

\noindent Here the left--hand side is $1$, and the first term on the right
is much smaller than $1$.  Therefore the second term on the right is at
least $1/2$, which proves that
$$
|b_{\text{left}}|^2+|b_{\text{right}}|^2\ge c_{\#}\Bigl(\int_{V<E_1}
\frac{dx}{(E_1-V(x))^{1/2}}\Bigr)^{-1}.\tag 50
$$

\noindent From (44) and (50), we get
$$
|b_{\text{right}}-b_{\text{left}}\Cal A(E_1)|^2\le C_{\#}
\Lambda^{2K+65-3\varepsilon N^\prime}(|b_{\text{left}}|^2+|b_{\text{right}}|^2).
\tag 51
$$

\noindent From the global WKB lemma we recall that
$$
\Cal A(E_1)=R(E_1)e^{i\Phi(E_1)}\quad\text{with}\ |R(E_1)-1|<\frac{1}{10}.
\tag 52
$$

\noindent From (51) and (52) we get
$$
\frac 12 |b_{\text{right}}|<|b_{\text{left}}|<2|b_{\text{right}}|, \quad
\text{and\ then}\tag 53
$$
$$
|\text{Im}\ \Cal A(E_1)|^2\le C_{\#}\Lambda^{2K+65-3\varepsilon N^\prime}.
$$

\noindent Hence for a suitable integer  $k$, we have
$$
|\Phi(E_1)-\pi k_1|\le C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}.
\tag 54
$$

\noindent From (51), (52), (53), (54) we get also
$$
|b_{\text{right}}-b_{\text{left}}R(E_1)e^{i \pi k_1}|\le C_{\#}
\Lambda^{K+33-\frac 32 \varepsilon N^\prime}|b_{\text{left}}|.\tag 55
$$

Therefore
$$\multline
\int_{\hat I_{\text{Airey\ right}}}|b_{\text{right}}F_{\text{Airey\ right}}
-R(E_1)(-1)^{k_1}b_{\text{left}}F_{\text{Airey\ right}}|^2\,dx\le\\
C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\int_{\hat I_{\text{Airey\
right}}}|F_{\text{Airey\ right}}|^2\,dx\cdot |b_{\text{left}}|^2\le
C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\endmultline
$$

\noindent by (44 bis) and the upper bound for $|b_{\text{left}}|$ in Lemma 11.
Combining this with (41), we get
$$
\int_{\hat I_{\text{Airey\ right}}}|F(x)-(-1)^{k_1}R(E_1)b_{\text{left}}
F_{\text{Airey\ right}}|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}.
\tag 56
$$

\noindent Similarly,
$$\align
&\int_{I_{\text{medium\ right}}\cup I_{\text{center}}}|b_{\text{right}}\text{Re}
(F_c^{\text{right}})-(-1)^{k_1}R(E_1)b_{\text{left}}\text{Re}(F_c^{\text{right}})|
^2\,dx\\
&\,\,\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\int_{I_{\text{medium\
right}}\cup I_{\text{center}}}|F_c^{\text{right}}|^2\,dx\cdot
|b_{\text{left}}|^2\quad \text{(by\ (55))}\\
&\,\,\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\int_{V<E_1}
\frac{dx}{(E_1-V(x))^{1/2}}|b_{\text{left}}|^2\\
&\,\,\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\ \text{(by\ lemma\ 11)}.
\endalign
$$

\noindent Combining this with (42) gives
$$\split
\int_{I_{\text{center}}\cup I_{\text{medium\ right}}}|F(x)-(-1)^{k_1}
R(E_1)b_{\text{left}}\text{Re}(F_c^{\text{right}}(x,E_1))|^2\,dx\\
\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}.\endsplit\tag 57
$$

\noindent Estimates (39), (40), (49), (56), (57) give an accurate
description of $F(x)$ in terms of a single unknown real constant.  We
record these results as follows.

\vglue 1pc
\proclaim{Lemma 12}  For a real constant $b$ of size $c_{\#}(\int_{V<E_1}
\frac{dx}{(E_1-V)^{1/2}})^{-1/2}\le\hfill\break |b|\le C_{\#}(\int_{V<E_1}
\frac{dx}{(E_1-V)^{1/2}})^{-1/2}$, we have
$$\align
&\int_{I_{\text{far\ left}}}|F(x)|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon 
N^\prime}\\
&\int_{\hat I_{\text{Airey\ left}}}|F(x)-bF_{\text{Airey\ left}}(x,E_1)
|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\\
&\int_{I_{\text{medium\ left}}}|F(x)-b\text{Re}(F_c^{\text{left}}(x,E_1))|^2
\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\\
&\int_{I_{\text{center}}}|F(x)-b\text{Re}(F_c^{\text{left}}(x,E_1))|^2\,dx\le
C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\\
&\int_{I_{\text{center}}}|F(x)-(-1)^{k_1}R(E_1)b\text{Re}
(F_c^{\text{right}}(x,E_1))|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}
\\
&\int_{I_{\text{medium\ right}}}|F(x)-(-1)^{k_1}R(E_1)b\text{Re}
(F_c^{\text{right}}(x,E_1))|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}
\\
&\int_{\hat I_{\text{Airey\ right}}}|F(x)-(-1)^{k_1}R(E_1)bF_{\text{Airey\
right}}(x,E_1)|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N^\prime}\\
&\int_{I_{\text{far\ right}}}|F(x)|^2\,dx\le C_{\#}\Lambda^{2K+66-3\varepsilon N
^\prime}.\endalign
$$
Here $k_1$ is an integer, and $|\Phi(E_1)-\pi k_1|\le C_{\#}
\Lambda^{K+33-\frac 32 \varepsilon N^\prime}$.\endproclaim

Of course the real constant $b$ is determined (up to sign) modulo a tiny
error by the requirement that $\int_{I_{\text{BVP}}}|F|^2\,dx=1$.  We
postpone the calculation of $b$ to a later section.

Lemma 12 gives precise control of the eigenfunction corresponding to a
given eigenvalue.  We must now understand the eigenvalues.  Lemma 12 gives
us strong information on the eigenvalues, but there is more to do.

Our next task is to show that there is at most one eigenvalue $E_1$
whose phase $\Phi(E_1)$ lies near a given multiple of $\pi$.  To show
this, we check that the functions $F_{\text{Airey\ left}}(x,E_1)$,
$\text{Re}(F_c^{\text{left}}(x,E_1))$, $\text{Re}(F_c^{\text{right}}
(x,E_1))$, $F_{\text{Airey\ right}}(x,E_1)$ and the factor $R(E_1)$
change very little under a change in $E_1$ which produces only a small
shift in the phase $\Phi(E_1)$.  Hence by Lemma 12, if $F_1$, $F_2$
are eigenfunctions whose eigenvalues $E_1$, $E_2$ both have phase
near $\pi k_1$, then $F_1$ must be nearly proportional to $F_2$.
If $E_1\ne E_2$, then $F_1$ and $F_2$ are orthogonal and therefore
far from proportional.  This contradiction shows that there can be at
most one eigenvalue $E_1$ with $\Phi(E_1)$ near $\pi k_1$.  We now
carry out the details.

\vglue 1pc
\proclaim{Lemma 13}  For $|E-E_0|<c_{\#}S_{\text{min}}$ we have
$$
\int_{\hat I_{\text{Airey\ left}}}|\frac{\partial}{\partial E}
F_{\text{Airey\ left}}(x,E)|^2\,dx\le C_{\#}\Bigl(\int_{V<E}
\frac{dy}{(E-V(y))^{1/2}}\Bigr)^3\tag 58
$$
$$
\int_{I_{\text{medium\ left}}}|\frac{\partial}{\partial E}F_c^{\text{left}}
(x,E)|^2\,dx\le C_{\#}\Bigl(\int_{V<E}\frac{dy}{
(E-V(y))^{1/2}}\Bigr)^3\tag 59
$$
$$
\int_{I_{\text{center}}}|\frac{\partial}{\partial E}F_c^{\text{left}}
(x,E)|^2\,dx\le C_{\#}\Bigl(\int_{V<E}\frac{dy}{(E-V(y))^{1/2}}\Bigr)^3
\tag 60
$$
$$
\int_{I_{\text{medium\ right}}}|\frac{\partial}{\partial E}\{R(E)F_c^
{\text{right}}(x,E)\}|^2\,dx\le C_{\#}\Bigl(\int_{V<E}\frac{dy}
{(E-V(y))^{1/2}}\Bigr)^3\tag 61
$$
$$
\int_{\hat I_{\text{Airey\ right}}}|\frac{\partial}{\partial E}
\{R(E)F_{\text{Airey\ right}}(x,E)\}|^2\,dx\le C_{\#}
\Bigl(\int_{V<E}\frac{dy}{(E-V(y))^{1/2}}\Bigr)^3.\tag 62
$$
\noindent Here the intervals $\hat I_{\text{Airey\ left}}\ldots\hat I_
{\text{Airey\ right}}$ are defined in terms of the energy $E$.\endproclaim

\demo{Proof}  First we estimate $R(E)=|1+\sum\limits_{k=1}^{N^\prime}
G_k(E)|$.  The global WKB lemma gives $|R(E)-1|\le C_{\#}
\Lambda^{-1}$, and
$$\align
c_{\#}|\frac{\partial}{\partial E}G_k(E)|\negthinspace&\le\negthinspace\Lambda^{-k}
\int_{V<E}\frac{\Lambda\,dy}{\lambda(y)S(y)B(y)}\negthinspace\le
\negthinspace \Lambda^{-k}\negthinspace
\int_{V<E}\frac{\Lambda}{\lambda(y)}\cdot\frac{1}{S^{1/2}(y)B(y)}
\cdot \frac{dy}{S^{1/2}(y)}\\
&\le\negthinspace \Lambda^{-k-1}\negthinspace
\int_{V<E}\frac{dy}{S^{1/2}(y)}\negthinspace\le \negthinspace
\frac{C_{\#}}{\Lambda^{k+1}}\int_{V<E}
\frac{dy}{(E-V(y))^{1/2}}\,\,\text{by\ (E7)}.
\tag"(62 bis)"\endalign$$

\noindent Hence $|\frac{\partial}{\partial E}R(E)|\le C_{\#}\Lambda^{-2}
\int_{V<E}\frac{dy}{(E-V)^{1/2}}$, so
$$\split
\Big|\frac{\partial R(E)}{\partial E}\Big|^2\int_{\hat I_{\text{Airey\ right}}}
|F_{\text{Airey\ right}}(x,E)|^2\,dx\le \Bigl[C_{\#}
\Lambda^{-2}\int_{V<E}
\frac{dy}{(E-V)^{1/2}}\Bigr]^2\\
\cdot\Bigl[C_{\#}\int_{V<E}\frac{dy}{(E-V)^{1/2}}\Bigr]\endsplit
$$

\noindent by virtue of the analogue of (44 bis) for $F_{\text{Airey\ right}}$.
This shows that (62) follows from the analogue of (58) for $F_{\text{Airey\
right}}$.  Similarly,
$$\multline
\Big|\frac{\partial R(E)}{\partial E}\Big|^2\int_{I_{\text{medium\ right}}}
|F_c^{\text{right}}(x,E)|^2\,dx\le\\ \Bigl[C_{\#}\Lambda^{-2}
\int_{V<E}\frac{dy}{(E-V)^{1/2}}\Bigr]^2\int_{I_{\text{medium\ right}}}
\frac{C_{\#}}{(E-V)^{1/2}}\,dy,\endmultline
$$

\noindent which reduces (61) to the analogue of (59) for $F_c^{\text{right}}$.
So it is enough to prove (58), (59), (60).

To prove (58), note the estimates
$$\gather
|A(t)|\le C,\,\, |A^\prime(t)|\le C(1+|t|)^{1/4}\quad\text{for}\ t \in \Bbb R,\\
\Big|\frac{\partial Y_{\text{left}}}{\partial E}\Big|\le C_{\#}
S_{\text{left}}^{-1},\,\, \Big|\frac{\partial^2Y_{\text{left}}}{\partial
x\,\partial E}\Big|\le
C_{\#}B_{\text{left}}^{-1}S_{\text{left}}^{-1}\quad\text{in}\ U_{\text{Airey\
left}},\\
|Y_{\text{left}}|\le C_{\#},\,\, c_{\#}B_{\text{left}}^{-1}<\Bigl(\frac
{\partial Y_{\text{left}}}{\partial x}\Bigr)<C_{\#}B_{\text{left}}^{-1}
\quad\text{in}\ U_{\text{Airey\ left}}.\endgather
$$

\noindent The estimates in $U_{\text{Airey\ left}}$ are consequences of the
global WKB lemma.

Now recall $F_{\text{Airey\ left}}(x,E)=\lambda_{\text{left}}^{-1/3}
\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-1/2}
A(\lambda_{\text{left}}^{2/3}Y_{\text{left}})$, so that
$$\align
&\Big|\frac{\partial}{\partial E}F_{\text{Airey\ left}}(x,E)\Big|=\Big|
-\frac 12 \lambda_{\text{left}}^{-1/3}\Bigl(\frac{\partial Y_{\text{left}}}
{\partial x}\Bigr)^{-3/2}\Bigl(\frac{\partial^2Y_{\text{left}}}
{\partial x\,\partial E}\Bigr)A(\lambda_{\text{left}}^{2/3}
Y_{\text{left}})\\
&\qquad\qquad\qquad\qquad\qquad\qquad+\lambda_{\text{left}}^{-1/3}
\cdot \lambda_{\text{left}}^{2/3}\frac{\partial Y_{\text{left}}}{\partial E}
\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-1/2}A^\prime(\lambda
^{2/3}Y_{\text{left}})\Big|\\
&\le C_{\#}[\lambda_{\text{left}}^{-1/3}B_{\text{left}}^{+3/2}
(B_{\text{left}}^{-1}S_{\text{left}}^{-1})+
\lambda_{\text{left}}^{-1/3}B_{\text{left}}^{+1/2}(1+\lambda_{\text{left}}^
{2/3})^{1/4}\cdot\lambda_{\text{left}}^{2/3}S_{\text{left}}^{-1}]\\
&=C_{\#}[\lambda_{\text{left}}^{-1/3}B_{\text{left}}^{1/2}S_{\text{left}}^{-1}
+\lambda_{\text{left}}^{1/2}B_{\text{left}}^{1/2}S_{\text{left}}^{-1}]
\le C_{\#}\lambda_{\text{left}}^{1/2}B_{\text{left}}^{1/2}S_{\text{left}}^{-1}
\quad\text{in}\ U_{\text{Airey\ left}}.\endalign
$$

\noindent Therefore 
$$\split
\int_{\hat I_{\text{Airey\ left}}}|\frac{\partial}
{\partial E}F_{\text{Airey\ left}}(x,E)|^2\,dx\le C_{\#}
\lambda_{\text{left}}B_{\text{left}}S_{\text{left}}^{-2}|
\hat I_{\text{Airey\ left}}|\\
\le C_{\#}S_{\text{left}}^{-3/2}B_{\text{left}}^3\quad\text{by}\
|\hat I_{\text{Airey\ left}}|<C_{\#}B_{\text{left}}\ \text{and\ by\
definition\ of}\ \lambda_{\text{left}}.\endsplit\tag 63
$$

\noindent On the other hand, $(\int_{V<E}\frac{dx}{(E-V)^{1/2}})^3\ge
c_{\#}(\frac{B_{\text{left}}}{S_{\text{left}}^{1/2}})^3$, so (63) implies
(58).

Next we prove (59).  By definition of $F_c^{\text{left}}$ we have
$$\multline
\frac{\partial}{\partial E}F_c^{\text{left}}(x,E)=\exp\Bigl[\pm i\frac \pi 4
+i\int_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt\Bigr]\cdot\\
\bigg\{\frac{\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(x,E)}
{(E-V(x))^{1/4}}\cdot \frac i2 \int_{x_{\text{left}}(E)}^x(E-V(t))^{-1/2}\,dt
-\frac 14 \frac{\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}(x,E)}
{(E-V(x))^{5/4}}\\
+\frac{\sum\limits_{k=0}^{N^\prime}\frac{\partial u_k^{\text{left}}}
{\partial E}(x,E)}{(E-V(x))^{1/4}}\biggr\}.\endmultline\tag 64$$
\noindent In $I_{\text{medium\ left}}$ we have $\Big|
\frac{\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}}
{(E-V)^{5/4}}\Big|\le
\frac{C_{\#}}{(E-V)^{5/4}}\le \frac{C_{\#}}{(E-V(x))^{1/4}}
\cdot\frac{C_{\#}}{S_{\text{left}}\lambda_{
\text{left}}^{\varepsilon-2/3}}$ and\hfill\break
 $\frac{\sum\limits_{k=1}^{N^\prime}
|\frac{\partial u_k^{\text{left}}}{\partial E}|}
{(E-V(x))^{1/4}}\le\hfill\break\frac
{C_{\#}S_{\text{left}}^{-1}}{(E-V(x))^{1/4}}$ by the global WKB lemma.
So (64) shows that
$$
\Big\vert\frac{\partial F_c^{\text{left}}}{\partial E}\Big\vert\le C_{\#}
(E-V(x))^{-1/4}\Bigl\{\int_{V<E}\frac{dy}{(E-V)^{1/2}}+
S_{\text{left}}^{-1}\lambda_{\text{left}}^{2/3-\varepsilon}\Bigr\}
\quad\text{in}\ I_{\text{medium\ left}}.
$$
\noindent Now the integral in curly brackets dominates $S_{\text{left}}^{-1/2}
B_{\text{left}}=S_{\text{left}}^{-1}\lambda_{\text{left}}$, which
dominates the $2^{\text{nd}}$ term in curly brackets.  Hence
$|\frac{\partial F_c^{\text{left}}}{\partial E}|\le\hfill\break
 C_{\#}(E-V(x))^{-1/4}
\{\int_{V<E}\frac{dy}{(E-V)^{1/2}}\}$ in $I_{\text{medium\ left}}$,
from which (59) is obvious.

To prove (60), we return to (64), and estimate the right--hand side on
$I_{\text{center}}$.  On $I_{\text{center}}$ the global WKB lemma shows
that
$$\align
\Big|&\sum\limits_{k=0}^{N^\prime}u_k^{\text{left}}\Big|\le C_{\#},
\Big|\sum\limits_{k=1}^{N^\prime} \frac{\partial u_k^{\text{left}}}
{\partial E}\Big|\le \sum\limits_{k=1}^{N^\prime}C_{\#}\Lambda^{-k}\int_{V<E}
\frac{\Lambda\,dy}{\lambda(y)S(y)B(y)}\\
&\le C_{\#}\int_{V<E}\frac{dy}{\lambda(y)S(y)B(y)}=
C_{\#}\int_{V<E}\frac{dy}{\lambda(y)[S^{1/2}(y)B(y)]S^{1/2}(y)}\\
&\le C_{\#}\Lambda^{-2}\int_{V<E}\frac{dy}{S^{1/2}(y)}\le C_{\#}
\Lambda^{-2}\int_{V<E}\frac{dy}{(E-V(y))^{1/2}}\quad\text{by\ (E7)}.
\endalign
$$

\noindent Hence (64) implies
$$
\Big|\frac{\partial F_c^{\text{left}}}{\partial E}\Big|\le \frac{C_{\#}}
{(E-V(x))^{1/4}}\Bigl\{\int_{V<E}\frac{dy}{(E-V)^{1/2}}+\frac{1}
{(E-V(x))}\Bigr\}\quad\text{on}\ I_{\text{center}},
$$

\noindent and since
$$\align
\int_{V<E}\frac{dy}{(E-V)^{1/2}}&\ge  c_{\#}\frac{B(x)}{S^{1/2}(x)}
\quad\text{(by\ (E7))}\\
&=\frac{c_{\#}\lambda(x)}{S(x)}\ge \frac{c_{\#}\lambda(x)}{(E-V(x))},
\endalign
$$

\noindent the first term in curly brackets dominates.  So we have
$$
\Big|\frac{\partial F_c^{\text{left}}}{\partial E}\Big|\le
\frac{C_{\#}}{(E-V(x))^{1/4}}\int_{V<E}\frac{dy}{(E-V)^{1/2}}\quad\text{on}\
I_{\text{center}},
$$
\noindent from which (60) is obvious. $\qquad\blacksquare$
\smallskip

As $E_1$ varies, the intervals $I_{\text{far\ left}}$, $\hat I_{\text{
Airey\ left}}$, etc. in Lemma 12 vary.  However, for a given energy
$\overline E_1$, we form the interval $\{|E_1-\overline E_1|<\tilde S\}=
\Cal T$, with $\tilde S=\text{min}(\lambda_{\text{left}}^{-\varepsilon}
S_{\text{left}},\lambda_{\text{right}}^{-\varepsilon}S_{\text{right}},
c_{\#}^0S_{\text{min}})$.  Then we can find fixed intervals
$\overline I_{\text{far\ left}}$, $\overline I_{\text{Airey\ left}}$,
$\overline I_{\text{medium\ left}}$, $\overline I_{\text{center}}$,
$\overline I_{\text{medium\ right}}$, $\overline I_{\text{Airey\ right}}$,
$\overline I_{\text{far\ right}}$ which cover $I_{\text{BVP}}$,
do not depend on $E_1$, but satisfy $\overline I_{\text{far\ left}}\subset
I_{\text{far\ left}}$, $\overline I_{\text{Airey\ left}}\subset \hat I_{
\text{Airey\ left}}$, etc. for $E_1\in \Cal T$.

The conclusions of Lemmas 12 and 13 hold a fortiori with $I_{\text{far\ left}}
\ldots I_{\text{far\ right}}$ replaced by $\overline I_{\text{far\ left}}\ldots
\overline I_{\text{far\ right}}$.

For $E_1\in \Cal T$ we note that $\int_{V<E_1}\frac{dy}{(E_1-V)^{1/2}}$
is comparable to $\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dy}
{S^{1/2}(y)}$, which does not depend on $E_1$.

Now suppose $E_1,E_2\in \Cal T\cap (-\infty,E_\infty]$
are distinct eigenvalues of $-\frac{d^2}
{dx^2}+V(x)$.  Let $F_1$, $F_2$ denote the corresponding (real)
eigenfunctions of norm $1$.  Lemma 12 shows precisely how $F_1$, $F_2$
behave, in terms of two real constants $b_1$, $b_2$ of size\hfill\break
$\sim (\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})^{-1/2}$.
Lemma 12, Lemma 13 and the fundamental theorem of calculus show that
$$\split
\Bigl(\int_{\overline I_{\text{Airey\ left}}\cup \overline I_{\text{medium\
left}}\cup \overline I_{\text{center}}\cup \overline I_{\text{medium\ right}}
\cup \overline I_{\text{Airey\ right}}}|b_1^{-1}F_1-b_2^{-1}
F_2|^2\,dx\Bigr)^{1/2}\\
\le C_{\#}|E_1-E_2|\Bigl(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{3/2}\\
+C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}\Bigl(\int_
{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{+1/2}
\endsplit
$$
\noindent and that
$$\split
\Bigl(\int_{\overline I_{\text{far\ left}}\cup \overline I_{\text{far\ right}}}
|b_1^{-1}F_1|^2+|b_2^{-1}F_2|^2\,dx\Bigr)^{1/2}\le \Bigl(\int_{x_{\text{left}}}
^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{+1/2}\\
\cdot C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}.\endsplit
$$
\noindent Since $\overline I_{\text{far\ left}},\ldots,\overline I_{\text{
far\ right}}$ cover $I_{\text{BVP}}$, we get:
$$\split
\Bigl(\int_{I_{\text{BVP}}}|F_1-(b_1b_2^{-1})F_2|^2\,dx\Bigr)^{1/2}\le
C_{\#}|E_1-E_2|\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{S^{1/2}(x)}\Bigr)\\
+C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}.\endsplit
$$
\noindent Since $F_1$ and $F_2$ are eigenfunctions for distinct eigenvalues,
they are orthogonal on $I_{\text{BVP}}$, so the left--hand side is at least
one.  Consequently,
$$
|E_1-E_2|\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}\Bigr)\ge c_{\#}\ \text{if}\ E_1, E_2 \in \Cal T=\{
|E-\overline E_1|<\tilde S\}\cap (-\infty,E_\infty].
$$
\noindent That is, if $E_1$, $E_2$ are distinct eigenvalues in $\{|E-
\overline E_0|<c_{\#}^0S_{\text{min}}\}\cap (-\infty,E_\infty]$, then
$$
|E_1-E_2|\ge \text{min}\Bigl\{\lambda_{\text{left}}^{-\varepsilon}
S_{\text{left}},\lambda_{\text{left}}^{-\varepsilon}S_{\text{right}},
c_{\#}\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}\Bigr)^{-1}\Bigr\}\tag 65
$$
\noindent We have $\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}>c_{\#}\frac{B_{\text{left}}}{S_{\text{left}}^{1/2}}=c_{\#}
\frac{\lambda_{\text{left}}}{S_{\text{left}}}$, so
$$
\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}\Bigr)^{-1}\le C_{\#}\frac{S_{\text{left}}}
{\lambda_{\text{left}}}<<c_{\#}\frac{S_{\text{left}}}
{\lambda_{\text{left}}^\varepsilon}.
$$
\noindent Similarly for $\frac{S_{\text{right}}}{\lambda_{\text{right}}
^\varepsilon}$.  Therefore the right--hand side of (65) is comparable to
\hfill\break
$(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})
^{-1}$, and we get
$$\split
|E_1-E_2|\ge c_{\#}\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{S^{1/2}(x)}\Bigr)^{-1}\ \text{for}\ E_1, E_2\ \text{distinct}\\
\text{eigenvalues\ in}\ \{|E-E_0|<c_{\#}^0S_{\text{min}}\}
\cap (-\infty,E_\infty].\endsplit\tag 66
$$
\noindent This is our basic estimate for the gap between eigenvalues.  Let
us check what it tells us about the phase difference.  Recall that
$$
\Phi(E)=\pm \frac \pi 2+\int_{V<E}(E-V(t))^{1/2}\,dt+\text{Im}\ \log
(1+\sum\limits_{k=1}^{N^\prime}G_k(E))\tag"{(66 bis)}"
$$
\noindent for a suitable branch of the logarithm.  We know that
$$\split
\Big|\sum\limits_{k=1}^{N^\prime}G_k(E)\Big|<\frac{1}{10}\ \text{and}\
\Big|\frac{\partial G_k}{\partial E}\Big|\le \frac{C_{\#}}{\Lambda^2}
\int_{V<E}\frac{dx}{(E-V)^{1/2}}\\
\text{by\ (62 bis)}.\endsplit\tag 67
$$
\noindent Hence
$$\align
\frac{d\Phi}{dE}&\ge \frac 12 \int_{V<E}(E-V(t))^{-1/2}\,dt-\frac{C_{\#}}
{\Lambda^2}\int_{V<E}(E-V)^{-1/2}\,dt\\
&\ge c_{\#}\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}\quad\text{for}\ |E-E_0|\le c_{\#}^0S_{\text{min}}.
\endalign
$$
\noindent For any two energies $E_1$, $E_2$ in that range we have
therefore $|\Phi(E_1)-\Phi(E_2)|\ge (c_{\#}\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})\cdot|E_1-E_2|$.  Estimate (66)
now implies the following result.

\vglue 1pc
\proclaim{Lemma 14}  Let $E_1$, $E_2$ be distinct eigenvalues of
$-\frac{d^2}{dx^2}+V(x)$ with $|E_i-E_0|<c_{\#}^0S_{\text{min}}\,\,(i=1,2)$
and $E_i\le E_\infty$.
Then $|\Phi(E_1)-\Phi(E_2)|\ge c_{\#}$.\endproclaim

In particular, there is at most one eigenvalue with a phase near $\pi k_1$
for a given integer $k_1$.

Lemma 12 tells us what an eigenfunction looks like, once we know the
eigenvalue.  It says also that eigenvalues have phase very close to
multiples of $\pi$.  Lemma 14 says that or given $k_1$ there is at most
one eigenvalue $E_1$ with $\Phi(E_1)$ near $k_1$.  It remains to show
that there is \underbar{at least} one eigenvalue $E_1$ with $\Phi(E_1)$
near $k_1$.

\vglue 1pc
\proclaim{Lemma 15}  Suppose $|E_1-E_0|<\frac{c_{\#}^0}{2}S_{\text{min}}$
with $\Phi(E_1)=\pi k_1$ for an integer $k_1$.  Then there is an energy
$E_1^\prime$ in the spectrum of $-\frac{d^2}{dx^2}+V(x)$ with
$|E_1^\prime-E_0|<c_{\#}^0S_{\text{min}}$ and $|\Phi(E_1^\prime)-\pi k_1|
\le C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}$.  In particular,
if $E_\infty\ge E_1+C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}
(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})^{-1}$,
then $E_1^\prime$ is an eigenvalue and $E_1^\prime\le E_\infty$.\endproclaim

\demo{Proof}  We use the global WKB lemma to construct a non zero function $F$
in the domain of $H=-\frac{d^2}{dx^2}+V(x)$ satisfying
$$\split
\Vert(H-E_1)F\Vert_{L^2(I_{\text{BVP}})}\le C_{\#}\Bigl(\int_{x_{\text{left}}}
^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{-1}\Vert F\Vert_{L^2(
I_{\text{BVP}})}\\
\cdot \Lambda^{K+33-\frac 32 \varepsilon N^\prime}.\endsplit\tag 68
$$
\noindent Elementary spectral theory will then show that some $E_1^\prime$
can be found in the spectrum of $H$, with
$$
|E_1^\prime-E_1|\le C_{\#}\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{S^{1/2}(x)}\Bigr)^{-1}\cdot \Lambda^{K+33 -\frac 32 \varepsilon
N^\prime}.\tag 69
$$
\noindent From (66 bis) and (67) we have $\frac{d\Phi}{dE}\le
C_{\#}(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)})$ for $|E-E_0|\le c_{\#}^0S_{\text{min}}$.  We will
check that $E_1^\prime$ lies in this range.  That is a consequence of 
the estimate
$$
\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}\ge c_{\#}\Bigl(\frac{S_{\text{min}}}{\Lambda}\Bigr)^{-1}
\tag 70
$$
\noindent and (69).

To check (70), pick $\overline x \in [x_{\text{left}},x_{\text{right}}]$
with $S(\overline x)\sim S_{\text{min}}$.  We may take
$|\overline x-x_{\text{left}}|>c_{\#}B_{\text{left}}$ and 
$|\overline x-x_{\text{right}}|>c_{\#}B_{\text{right}}$.  Then we have
$$\split
\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\ge
\frac{B(\overline x)}{S^{1/2}(\overline x)}=c_{\#}
\frac{\lambda(\overline x)}{S(\overline x)}\ge \frac{c_{\#}\Lambda}
{S_{\text{min}}},\\ \text{proving\ (70)}.\endsplit
$$
\noindent Now we know that $|\frac{d\Phi}{dE}|\le
C_{\#}(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})$
for $E$ between $E_1$ and $E_1^\prime$.  Thus
$$\split
|\Phi(E_1^\prime)-\Phi(E_1)|\le C_{\#}\Bigl(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)|E_1^\prime-E_1|\\
\le C_{\#}\Lambda^{K+33-\varepsilon N^\prime}\quad\text{by\ (69)}.\endsplit
$$
\noindent That is, $|\Phi(E_1^\prime)-\pi k_1|\le C_{\#}\Lambda^{K+33-
\frac 32 \varepsilon N^\prime}$.

Furthermore, if $E_\infty\ge E_1+C_{\#}\Lambda^{K+33-\frac 32 \varepsilon
N^\prime}(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)})^{-1}$, then $E_1^\prime<E_\infty$.  Since the part of
the spectrum of $H$ below $E_\infty$ consists entirely of eigenvalues,
$E_1^\prime$ must be an eigenvalue of $H$.

Thus the proof of the lemma is reduced to constructing a nonzero\hfill\break
$F \in \text{Domain}(H)$ satisfying (68).

Let $\chi_{\text{Airey\ left}}$, $\chi_{\text{medium\ left}}$,
$\chi_{\text{center}}$, $\chi_{\text{medium\ right}}$,
$\chi_{\text{Airey\ right}}$ be smooth functions with the following
properties:
$$\multline
\chi_{\text{Airey\ left}}+\chi_{\text{medium\ left}}+
\chi_{\text{center}}+\chi_{\text{medium\ right}}+
\chi_{\text{Airey\ right}}=1\\
\text{on}\
\hat I_{\text{Airey\ left}} \cup I_{\text{medium\ left}}
\cup I_{\text{center}} \cup I_{\text{medium\ right}}\cup
\hat I_{\text{Airey\ right}}.\endmultline\tag 71
$$
$$\multline
\Big|\Bigl(\frac{d}{dx}\Bigr)^\alpha\chi_{\text{Airey\ left}}\Big|
\le
C_{\#}^\alpha(\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}})^{-\alpha}\
\text{and\ similarly\ for}\ \chi_{\text{Airey\ right}}\endmultline
\tag 72
$$
$$
\Big|\Bigl(\frac{d}{dx}\Bigr)^\alpha\chi_{\text{medium\ left}}\Big|\le
C_{\#}^\alpha(\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}})^{-\alpha}
\text{and\ similarly\ for}\ \chi_{\text{medium\ right}}\tag 73
$$
$$
\Big|\Bigl(\frac{d}{dx}\Bigr)^\alpha\chi_{\text{center}}(x)\Big|\le C_{\#}^
\alpha B^{-\alpha}(x)\tag 74
$$
\noindent and with $\chi_{\text{Airey\ left}}$ supported in $I_{\text{Airey\
left}}$,
$\chi_{\text{medium\ left}}$ supported in $I_{\text{medium\ left}}$,
$\chi_{\text{center}}$ supported in $I_{\text{center}}$, $\chi_{\text{medium\
right}}$ supported in $I_{\text{medium\ right}}$, $\chi_{\text{Airey\ right}}$
supported in $I_{\text{Airey\ right}}$.  We may assume $\chi_{\text{Airey\
left}}>0$ in $\hat I_{\text{left}}$, 
and similarly for $\hat I_{\text{Airey\ right}}$.
\noindent We define
$$\multline
F=\Bigl[\chi_{\text{Airey\ left}}F_{\text{Airey\ left}}(x,E_1)+(\chi_
{\text{medium\
left}}+\chi_{\text{center}})\text{Re}(F_c^{\text{left}}(x,E_1))\Bigr]\\
+(-1)^{k_1}R(E_1)\Bigl[\chi_{\text{medium\ right}}\cdot\text{Re}
(F_c^{\text{right}}(x,E_1))+\chi_{\text{Airey\ right}}F_{\text{Airey\ right}}
(x,E_1)\Bigr].\endmultline
$$
\noindent In view of the support properties of the $\chi$'s, we know that $F$
vanishes in a neighborhood of the endpoints of $I_{\text{BVP}}$, hence
belongs to the domain of $H$.  The Corollary to Lemma 2 shows us that
$$
\int_{I_{\text{center}}}|F|^2\,dx\ge c_{\#}\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}.\tag 75
$$
\noindent We estimate $(\frac{d^2}{dx^2}+E_1-V(x))F$ on each of the intervals
$I_{\text{far\ left}}$ $I_{\text{Airey\ left}}\ldots\hfill\break
I_{\text{Airey\ right}}$
$I_{\text{far\ right}}$.  In $I_{\text{far\ left}}$ we have
$$\align
(\frac{d^2}{dx^2}+E_1-V(x))F&=\chi_{\text{Airey\
left}}\Bigl\{\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F_{\text{Airey\ left}}\Bigr\}\\
&\,\,+2 \frac{d\chi_{\text{Airey\ left}}}{dx}\cdot \frac{dF_{\text{Airey\
left}}} {dx}\\
&\,\,+\frac{d^2\chi_{\text{Airey\ left}}}{dx^2}\cdot F_{\text{Airey\
left}}.\endalign
$$
\noindent Outside $I_{\text{far\ left}}\cap I_{\text{Airey\ left}}$ the
right--hand side is identically zero, while inside $I_{\text{far\ left}}\cap
I_{\text{Airey\ left}}$ it is dominated by $C_{\#}\lambda_{\text{left}}^
{-N^\prime+2\varepsilon}B_{\text{left}}^{-3/2}$, by the global WKB lemma.

Hence
$$\align
\int_{I_{\text{far\
left}}}\Big|\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F\Big|^2\,dx&\le C_{\#}
\lambda_{\text{left}}^{-2N^\prime+4\varepsilon}B_{\text{left}}^{-3}
|I_{\text{Airey\ left}}|\\
&\le C_{\#}\lambda_{\text{left}}^{-2N^\prime+4\varepsilon}
B_{\text{left}}^{-2}.\endalign
$$
\noindent Hypothesis (E6) yields
$$
\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}
\le \Lambda^K\frac{B_{\text{left}}}{S_{\text{left}}^{1/2}}=\Lambda^K
\frac{B_{\text{left}}^2}{\lambda_{\text{left}}},\tag "{(75 bis)}"
$$
\noindent so that $\lambda_{\text{left}}^{-2N^\prime+4\varepsilon}
B_{\text{left}}^{-2}\le C_{\#}\Lambda^{-2N^\prime+K+4\varepsilon}
(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)})^{-1}$.  Hence
$$
\int_{I_{\text{far\ left}}}\Bigl|\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F\Big|^2
\,dx\le
C_{\#}\Lambda^{K-2N^\prime+4\varepsilon}\Bigl(\int_{x_{\text{left}}}
^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{-1}.\tag 76
$$
\noindent An analogous result holds for $I_{\text{far\ right}}$.

On $\tilde I_{\text{Airey\ left}}\equiv \text{supp}\ \chi_{\text{Airey\ left}}$
we have all the $\chi$'s identically zero except $\chi_{\text{Airey}}$
and $\chi_{\text{medium\ left}}$.  Thus
$$\align
&\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F=\chi_{\text{Airey\ left}}
\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F_{\text{Airey\ left}}\\
&\,\,+\chi_{\text{medium\ left}}\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)
\text{Re}(F_c^{\text{left}})\\
&\,\,+2 \frac{d\chi_{\text{Airey\ left}}}{dx}\frac{d}{dx}\Bigl(F_
{\text{Airey\ left}}-\text{Re}(F_c^{\text{left}})\Bigr)\\
&\,\,+\frac{d^2}{dx^2}\chi_{\text{Airey\ left}}\cdot (F_{\text{Airey\ left}}
-\text{Re}(F_c^{\text{left}})\Bigr)\quad\text{on}\ \tilde I_{\text{Airey\
left}}.\tag 77\endalign
$$
\noindent Here we have used $\frac{d}{dx}\chi_{\text{medium\ left}}=-
\frac{d}{dx}\chi_{\text{Airey\ left}}$ and $\frac{d^2\chi_{\text{medium\
left}}}{dx^2}=-\frac{d^2\chi_{\text{Airey\ left}}}{dx^2}$, which hold on
$\tilde I_{\text{Airey\ left}}$ since $\chi_{\text{Airey\ left}}+
\chi_{\text{medium\ left}}\equiv 1$ there.

The global WKB lemma shows that the terms on the right of (77) are dominated by
$C_{\#}\lambda_{\text{left}}^{-N^\prime}B_{\text{left}}^{-3/2}$,
$C_{\#}\lambda_{\text{left}}^{10-\frac 32 \varepsilon
N^\prime}B_{\text{left}}^{-3/2}$, $C_{\#}\lambda_{\text{left}}^{10-\frac 32
\varepsilon N^\prime+\varepsilon} B_{\text{left}}^{-3/2}$, $C_{\#}
\lambda_{\text{left}}^{10-\frac 32 \varepsilon
N^\prime+2\varepsilon}B_{\text{left}}^{-3/2}$ respectively, in view of our
estimates on the derivatives of the $\chi$'s.  Hence
$$
\Big|\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F\Big|\le C_{\#}
\lambda_{\text{left}}^{10-\frac 32 \varepsilon N^\prime+2\varepsilon}
B_{\text{left}}^{-3/2}\quad\text{on}\ \tilde I_{\text{Airey\ left}},
$$
\noindent so that as in the proof of (76) we have
$$\split
\int_{\tilde I_{\text{Airey\ left}}}\Big|\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)
F\Big|^2\,dx\le C_{\#}\lambda_{\text{left}}^{20-3\varepsilon
N^\prime+4\varepsilon}B_{\text{left}}^{-2}\\
\le C_{\#}\Lambda^{21-3\varepsilon N^\prime+K}\Bigl(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{-1}.\endsplit\tag 78
$$
\noindent There is an analogous estimate for $\tilde I_{\text{Airey\ right}}
\equiv \text{supp}\ \chi_{\text{Airey\ right}}$.  Note that\hfill\break
$\hat I_{\text{Airey\ left}}
\subset \tilde I_{\text{Airey\ left}}$, and
similarly for $\hat I_{\text{Airey\ right}}$.  Next we investigate\hfill\break
$I_{\text{medium\ left}}\backslash \tilde I_{\text{Airey\ left}}\equiv
\tilde I_{\text{medium\ left}}$.  Here all the $\chi$'s are identically
zero except for $\chi_{\text{medium\ left}}$, $\chi_{\text{center}}$.  So
$\chi_{\text{medium\ left}}+\chi_{\text{center}}=1$ on $\tilde I_{\text{medium\
left}}$, and thus $F=\text{Re}(F_c^{\text{left}})$ on $\tilde I_{\text{medium\
left}}$.  Hence the global WKB lemma gives $|(\frac{d^2}{dx^2}
+E_1-V(x))F|\le C_{\#}\lambda_{\text{left}}^{10-\frac 32 \varepsilon N^\prime}
B_{\text{left}}^{-3/2}$ on $\tilde I_{\text{medium\ right}}$, so that
as in the proofs of (77), (78) we get

$$
\int_{\tilde I_{\text{medium\
left}}}\Big|\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F\Big|^2\,dx\le C_{\#}
\Lambda^{20-3\varepsilon
N^\prime+K}\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}
\Bigr)^{-1}.\tag 79
$$
\noindent For $\tilde I_{\text{medium\ right}}\equiv I_{\text{medium\
right}}\backslash \tilde I_{\text{Airey\ right}}$, the situation is different,
since the matching of $F_c^{\text{left}}$ with $F_c^{\text{right}}$ comes
into play. We argue as follows.

\noindent Since $\Cal A(E_1)=R(E_1)e^{i\Phi(E_1)}$ with $\Phi(E_1)=\pi k_1$,
we have $\Cal A(E_1)=(-1)^{k_1}R(E_1)$.  On $\tilde I_{\text{medium\ right}}$,
all the $\chi$'s are identically zero except for $\chi_{\text{center}}$
and $\chi_{\text{medium\ right}}$.  Hence
$$
F=\chi_{\text{center}}\text{Re}(F_c^{\text{left}})+\chi_{\text{medium\ right}}
\cdot (-1)^{k_1}R(E_1)\text{Re}(F_c^{\text{right}})\quad\text{in}\
\tilde I_{\text{medium\ right}}.
$$
\noindent Using also $\chi_{\text{center}}+\chi_{\text{medium\ right}}
\equiv 1$ in $\tilde I_{\text{medium\ right}}$ we conclude that
$$\align
\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F&=\chi_{\text{center}}\Bigl(
\frac{d^2}{dx^2}+E_1-V(x)\Bigr)\text{Re}(F_c^{\text{left}})\\
&\,\,\,+\chi_{\text{medium\ right}}\cdot (-1)^{k_1}R(E_1)\Bigl(
\frac{d^2}{dx^2}+E_1-V(x)\Bigr)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{Re}(F_c^{\text{right}})\\
&\,\,\,+2 \frac{d\,\chi_{\text{center}}}{dx}\cdot \frac{d}{dx}\{\text{Re}
(F_c^{\text{left}}-\Cal A(E_1)F_c^{\text{right}})\}\\
&\,\,+\frac{d^2\chi_{\text{center}}}{dx^2}\cdot\text{Re}(F_c^{\text{left}}
-\Cal A(E_1)F_c^{\text{right}})\,\text{on}\ \tilde I_{\text{medium\ right}}.
\endalign
$$
\noindent The global WKB lemma shows that the first two terms on the right
are bounded by
$C_{\#}\Lambda^{-N^\prime}S_{\text{right}}^{-1/4}B_{\text{right}}^{-2}$,
$C_{\#}\lambda_{\text{right}}^{10-\frac 32\varepsilon N^\prime}B_{\text{right}}
^{3/2}$, while the last two terms on the right are supported in
$I_{\text{center}}\cap \tilde I_{\text{medium\ right}}$ and are dominated
by $C_{\#}\Lambda^{-N^\prime-1}S_{\text{right}}^{-1/4}B_{\text{right}}^{-2}$
and $C_{\#}\Lambda^{-N^\prime-1}S_{\text{right}}^{+1/4}B_{\text{right}}^{-1}$
there.  Hence $|(\frac{d^2}{dx^2}+E_1-V(x))F|\le\hfill\break
C_{\#}\Lambda^{10-\frac 32
\varepsilon N^\prime}\lambda_{\text{right}}^{1/2}B_{\text{right}}^{-3/2}$
in $\tilde I_{\text{medium\ right}}$, so that
$$\multline
\int_{\tilde I_{\text{medium\ right}}}\Big|\Bigl(\frac{d^2}{dx^2}
+E_1-V(x)\Bigr)F\Big|^2\,dx\le C_{\#}\Lambda^{20-3\varepsilon N^\prime}
\lambda_{\text{right}}B_{\text{right}}^{-3}|\tilde I_{\text{medium\ right}}|\\
\le C_{\#}\Lambda^{20-3\varepsilon
N^\prime}\lambda_{\text{right}}B_{\text{right}}^{-2}\le C_
{\#}\Lambda^{20-3\varepsilon N^\prime+K}\Bigl(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{-1},\endmultline\tag 80
$$
\noindent by virtue of the analogue of (75 bis) for $\lambda_{\text{right}}$,
$B_{\text{right}}$.

Next we look at $\tilde I_{\text{center}}=I_{\text{center}}\backslash
(I_{\text{medium\ left}}\cup I_{\text{medium\ right}})$.  Here
$\chi_{\text{center}}\equiv 1$ and the other $\chi$'s are identically zero.
Hence $F=\text{Re}(F_c^{\text{left}})$ on $\tilde I_{\text{center}}$.  The
global WKB lemma therefore tells us that $|(\frac{d^2}{dx^2}+E_1-V(x))F|\le
C_{\#}\Lambda^{-N^\prime}S^{-1/4}(x)B^{-2}(x)$ on $\tilde I_{\text{center}}$,
hence $\int_{\tilde I_{\text{center}}}|(\frac{d^2}{dx^2}+E_1-V(x))F|^2\,dx
\le C_{\#}\Lambda^{-2N^\prime}\int_{I_{\text{center}}}\frac{dx}
{S^{1/2}(x)B^4(x)}$. \hfill\break Hypothesis (E5) shows that the right--hand side
is dominated by\hfill\break $C_{\#}\Lambda^{-2N^\prime+K}(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})^{-1}$, so we get
$$
\int_{\tilde I_{\text{center}}}\Big|\Bigl(\frac{d^2}{dx^2}
+E_1-V(x)\Bigr)F\Big|^2\,dx\le C_{\#}\Lambda^{40+K-3\varepsilon N^\prime}
\Bigl(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^
{-1}.\tag 81
$$

Estimates (76), (78) and their analogues for $I_{\text{far\ right}}$,
$\tilde I_{\text{Airey\ right}}$, together with estimates (79), (80),
(81) show that
$$
\int_{I_{\text{BVP}}}\Big|\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F\Big|^2\,dx\le
C_{\#}\Lambda^{40+K-3\varepsilon N^\prime}\Bigl(\int_{x_{\text{left}}}
^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\Bigr)^{-1}.\tag 82
$$
\noindent The reason is that the regions of integration in those estimates
cover $I_{\text{BVP}}$.

Now from (75) and (82), we see that
$$
\Vert(H-E_1)F\Vert^2\le \Vert F\Vert^2\cdot\frac
{C_{\#}\Lambda^{40+K-3\varepsilon N^\prime}(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})^{-1}}
{(\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)})},
$$
\noindent which implies (68).  The proof of the lemma is complete.$\qquad
\blacksquare$

We summarize our knowledge of the eigenvalues of $E$ in the following result.
\vglue 1pc
\proclaim{Lemma 16}  Let $J=[E_0-\frac 14c_{\#}^0S_{\text{min}},E_0+\frac 14
c_{\#}^0S_{\text{min}}]\cap (-\infty,E_\infty]$, $J(k)=\{E\mid|E-E_0|<
c_{\#}^0S_{\text{min}}$ and $|\Phi(E)-\pi k|<C_{\#}\Lambda^{K+33-\frac 32
\varepsilon N^\prime}\}$, $\Cal K=\{k \in \Bbb Z\mid J(k)\cap J\ne \emptyset\}$.
Then $\Cal K=\{k \in \Bbb Z\mid k_{\text{min}}\le k\le k_{\text{max}}\}$
for integers $k_{\text{min}}<k_{\text{max}}$.  Every eigenvalue in $J$
belongs to one of the $J(k)$ $(k \in \Cal K)$.  For $k_{\text{min}}\le
k<k_{\text{max}}$, there is exactly one eigenvalue 
in $J(k)$.  Also there
is exactly one eigenvalue $\le E_\infty$ in $J(k_{\text{max}})$, unless
$|E_\infty-E_0|<c_{\#}^0S_{\text{min}}$ and $|\Phi(E_\infty)-\pi k_{\text{max}}
|<C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}$, in which case
there is at most one eigenvalue $\le E_\infty$ in $J(k_{\text{max}})$.
\endproclaim

\demo{Proof}  First we check that $\Cal K=\{k \in \Bbb Z\mid k_{\text{min}}
\le k\le k_{\text{max}}\}$ with $k_{\text{min}}<k_{\text{max}}$.  In fact,
$\Cal K=\{k \in \Bbb Z\mid \text{distance}\ (\pi k,\Phi(J))<C_{\#}
\Lambda^{K+33-\frac 32 \varepsilon N^\prime}\}$.  Since
$0<c_{\#}\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}<\frac{d\Phi(E)}{dE}<C_{\#}\int_{x_{\text{left}}}^
{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}$ for $E \in J$, and since\hfill\break
$(\text{length}\ J)\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}
{S^{1/2}(x)}
\ge \frac 14 c_{\#}^0S_{\text{min}}
\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{dx}{S^{1/2}(x)}\ge
c_{\#}\Lambda$ by (70), we know that $\Phi(J)$ is a finite interval
of length $\ge c_{\#}\Lambda$.  Hence the desired form of $\Cal K$ is clear.

Next we check that every eigenvalue $E \in J$ belongs to $J(k)$ for
some $k \in \Cal K$.  In fact, lemma 12 shows that $|\Phi(E)-\pi k|<C_{\#}
\Lambda^{K+33-\frac 32 \varepsilon N^\prime}$ for an integer $k$.  Thus,
$E \in J(k)$.  Also, $k \in \Cal K$ since $E\in J(k)\cap J$.  So $E$ belongs
to $J(k)$ for some $k \in \Cal K$, as asserted.

Next we note that every $J(k)$ contains at most one eigenvalue $\le E_\infty$,
as follows immediately from Lemma 14.

It remains only to show that a given $J(k)$ $(k \in \Cal K)$ contains at
least one eigenvalue $\le E_\infty$, unless $k=k_{\text{max}}$, $|E_\infty-E_0|<
c_{\#}^0S_{\text{min}}$
and $|\Phi(E_\infty)-\pi k|<C_{\#}\Lambda^{K+33-\frac 32\varepsilon N^\prime}$.
Our argument is as follows.  Given $k \in \Cal K$ there is by definition
some $E_1 \in J$ with $|\Phi(E_1)-\pi k|<C_{\#}\Lambda^{K+33-\frac 32
\varepsilon N^\prime}$.

Now $\frac{d\Phi}{dE}\ge c_{\#}\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{S^{1/2}(x)}$ on $J^+=[E_0-\frac{c_{\#}^0}{2}S_{\text{min}},
E_0+\frac{c_{\#}^0}{2}S_{\text{min}}]$, and the distance from $E_1 \in J$
be the boundary of $J^+$ is at least $\frac 14 c_{\#}^0S_{\text{min}}
>\frac{c_{\#}\Lambda}{(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{
dx}{S^{1/2}(x)})}$ by (70).  Hence there is some $\hat E_1 \in J^+$
with $\Phi(\hat E_1)=\pi k$.  Lemma 15 implies that there is a point
$E\in J(k)\cap \text{spectrum}(H)$.  If $J(k)$ lies to the left of
$E_\infty$, then $E$ must be an eigenvalue, by virtue of assumption (E1),
and $E\le E_\infty$.
Hence it is enough to show that $J(k)$ lies to the left of $E_\infty$
unless $|E_\infty-E_0|<c_{\#}^0S_{\text{min}}$ and $|\Phi(E_\infty)-
\pi k|<C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}$ and
$k=k_{\text{max}}$.  This is proved as follows.

Since $k \in \Cal K$, we know that $J(k)$ intersects $J \subset (-\infty,
E_\infty]$.  Also since $\frac{d\Phi}{dE}>0$ on $\{|E-E_0|<c_{\#}^0
S_{\text{min}}\}$, we know that $J(k)$ is an interval.  Hence
either $J(k)$ lies to the left of $E_\infty$, or else $E_\infty \in J(k)$.
We need only analyze the case $E_\infty \in J(k)$.  Let $k^\prime \in \Cal K$
be different from $k$.  The same argument just used for $k$ shows that
either $J(k^\prime)$ lies to the left of $E_\infty$, or else
$E_\infty \in J(k^\prime)$.  We cannot have $E_\infty \in J(k^\prime)$,
since $E_\infty$ belongs to $J(k)$, and $J(k)\cap J(k^\prime)=\emptyset$.
Thus $J(k^\prime)$ lies to the left of $E_\infty$, while $J(k)$ contains
$E_\infty$.  Since $J(k^\prime)$ and $J(k)$ are disjoint intervals, it
follows that $J(k^\prime)$ lies to the left of $J(k)$.  Since
$\frac{d\Phi}{dE}>0$ on $\{|E-E_0|<c_{\#}^0S_{\text{min}}\}$, it follows
that $k^\prime<k$.

In other words, $k$ is the maximal element of $\Cal K$, $k=k_{\text{max}}$.
Also since $E_\infty\in J(k)$, we have by definition
$|E_\infty-E_0|<c_{\#}^0S_{\text{min}}$ and $|\Phi(E_\infty)-\pi k|<
C_{\#}\Lambda^{K+33-\frac 32 \varepsilon N^\prime}$.  This is what we
were supposed to show.  The proof of the lemma is complete.$\qquad\blacksquare$





\pageno 100
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\centerline{\bf{Minor Improvements}}
\smallskip

The hypotheses in the preceding section have become two baroque.  We
begin this section by introducing a reasonably simple hypothesis that
implies the complicated assumption (E3).  In fact, suppose
$$
\text{For}\ x \in I_{\text{BVP}}, x<x_{\text{left}}-\frac 12
\lambda_{\text{left}}^KB_{\text{left}},\,\, \text{we\ have}\,\, V(x)-E_\infty\ge
\frac{100}{|x-x_{\text{left}}|^2}. \tag "{(E3)$^\prime$}"
$$
\noindent Similarly, for $x \in I_{\text{BVP}}$, $x>x_{\text{rt}}+
\frac 12\lambda_{\text{rt}}^KB_{\text{rt}}$, we have $V(x)-E_\infty
\ge \frac{100}{|x-x_{\text{rt}}|^2}$. \hfill\break
\noindent We check that this and (E4) imply (E3),
with $\tau=\frac 12$.  It is enough to study $x<x_{\text{left}}-\lambda
_{\text{left}}^KB_{\text{left}}$, since the argument for $x>x_{\text{rt}}+
\lambda_{\text{rt}}^KB_{\text{rt}}$ is analogous.  For $x<x_{\text{left}}
-\lambda_{\text{left}}^KB_{\text{left}}$, we have
$$\multline
\int\limits_x^{x_{\text{left}}(E)}(V(t)-E)^{1/2}\ dt\ge
\int\limits_x^{x_{\text{left}}-\frac 12\lambda_{\text{left}}^KB_{\text{left}}}
\frac {10}{|t-x_{\text{left}}|}\ dt+\\
\int\limits_{x_{\text{left}}(E)-c_{\#}
B_{\text{left}}}^{x_{\text{left}}(E)}
\cdot\Bigl(c_{\#}\frac{S_{\text{left}}}
{B_{\text{left}}}(x_{\text{left}}(E)-t)\Bigr)^{1/2}\ dt\\
\ge 10 \ln \Bigl(\frac{|x-x_{\text{left}}|}{\frac 12\lambda_{\text{left}}^K
B_{\text{left}}}\Bigr)+c_{\#}S_{\text{left}}^{1/2}B_{\text{left}}=
10 ln \Bigl(\frac{|x-x_{\text{left}}|}{\frac 12\lambda_{\text{left}}^K
B_{\text{left}}}\Bigr)+c_{\#}\lambda_{\text{left}}.\endmultline
$$
\noindent Hence with $\tau=\frac 12$ we have
$\tau(V(x)-E)\text{exp}(2(1-\tau)\int\limits_x^{x_{\text{left}}(E)}
(V(t)-E)^{1/2}\ dt)\ge \frac{50}{|x-x_{\text{left}}|^2}
\Bigl(\frac{|x-x_{\text{left}}|}{\frac 12\lambda_{\text{left}}^KB_{\text{left}}}
\Bigr)^{10}e^{c_{\#}\lambda_{\text{left}}}$.  For $x \le x_{\text{left}}-
\lambda_{\text{left}}^KB_{\text{left}}$, the right--hand side is
smallest when $|x-x_{\text{left}}|=\lambda_{\text{left}}^KB_{\text{left}}$.
Hence
$\tau(V(x)-E)\text{exp}(2(1-\tau)\int\limits_x^{x_{\text{left}}(E)}(V(t)-E)^{1/2}
\ dt) \ge \frac{c_{\#}e^{c_{\#}\lambda_{\text{left}}}}{\lambda_{\text{left}}^K
B_{\text{left}}^2}=\frac{c_{\#}e^{c_{\#}\lambda_{\text{left}}}S_{\text{left}}}
{\lambda_{\text{left}}^{K+2}}=\Bigl[\frac{c_{\#}e^{c_{\#}\lambda_{\text{left}}}}{\lambda_{\text{left}}^
{2K+N+2}}\Bigr] \lambda_{\text{left}}^N\cdot
(\lambda_{\text{left}}^KS_{\text{left}})\ge\hfill\break \lambda_{\text{left}}^N
\Bigl[\frac{c_{\#}e^{c_{\#}\lambda_{\text{left}}}}{\lambda_{\text{left}}
^{2K+N+2}}\Bigr]\cdot\ \text{MAX}\ (E-V)$ by (E4).

For $\lambda_{\text{left}}\ge C_{\#}$, the factor in brackets is more
than $1$, so the preceding estimate implies (E3).  We have
$\lambda_{\text{left}}\ge \Lambda\ge C_{\#}$, so (E3) is proven.

Also, assumptions (E3)$^\prime$ and (E2) imply (E1) on the continuous
spectrum of $H$.  In fact, (E2) shows that $V$ is bounded below
on $I_{\text{BVP}}$, while (E3$^\prime$) gives $V-E_\infty\ge 0$ outside
a bounded subinterval $I^\prime \subset I_{\text{BVP}}$.  Letting 
$-M$ be a lower bound for $V-E_\infty$ on $I_{\text{BVP}}$, we cut $I^\prime$
into finitely many subintervals $I_\nu$ of length $< 10^{-10}M^{-1/2}$.
If $F$ has average zero on each of the $I_\nu$, then $\int\limits_{I_\nu}
(F^\prime)^2+(V-E_\infty)F^2\ge \int\limits_{I_\nu}(F^\prime)^2-MF^2\ge 0$,
so $\int\limits_{I^\prime}(F^\prime)^2+(V-E_\infty)F^2\ge 0$.

This shows that the quadratic form $<(H-E_\infty)F,F>$ is non--negative
on a finite--codimension subspace of $L^2(I_{\text{BVP}})$, which implies
that $H_\infty$ has no continuous spectrum in $(-\infty,E_\infty)$.  Thus
(E1) follows from (E2), (E3$^\prime$).

One last task for this section is to estimate the derivatives of the 
semiclassical phase
$$
\phi_{\text{sc}}(E)=\int\limits_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
(E-V(t))^{1/2}\ dt.
$$

\noindent Under the hypotheses of the global WKB lemma, we have the
following result.
\vglue 1pc
\proclaim{Lemma} $|(\frac{d}{dE})^\beta \phi_{\text{sc}}(E)|\le C_{\#}
\int\limits_{x_{\text{left}}}^{x_{\text{rt}}} S^{\frac 12-\beta}(x)\ dx$
for $|E-E_0|<c_{\#}^0S_{\text{min}}$.
\endproclaim\smallskip\smallskip
\demo{Sketch of Proof}  Write a partition of unity $1=\theta_{\text{left}}(x)+
\theta_{\text{rt}}(x)+\sum\limits_\nu\theta_\nu(x)$ in $[x_{\text{left}}-c_{\#}
B_{\text{left}}, x_{\text{rt}}+c_{\#}B_{\text{rt}}]$, with
$$\gather
\theta_{\text{left}}\ \text{supported\ in}\,\, |x-x_{\text{left}}|<c_{\#}
B_{\text{left}},\,\, |(\frac {d}{dx})^\alpha\theta_{\text{left}}|\le
C_{\#}^\alpha B_{\text{left}}^{-\alpha},\\
\theta_{\text{rt}}\ \text{supported\ in}\,\,|x-x_{\text{rt}}|<c_{\#}
B_{\text{rt}},\,\,|(\frac{d}{dx})^\alpha\theta_{\text{rt}}|\le
C_{\#}B_{\text{rt}}^{-\alpha},\\
\theta_\nu\ \text{supported\ in}\,\,|x-x_\nu|<c_{\#}B(x_\nu),\,\,|(\frac{d}{dx}
)^\alpha\theta_\nu|\le C_{\#}^\alpha B^{-\alpha}(x_\nu),\\
|x_{\text{left}}-x_\nu|, |x_{\text{rt}}-x_\nu|>c_{\#}\ \text{diam}(\text{supp}\
\theta_\nu).\endgather
$$
\noindent Then
$$\multline
\phi_{\text{sc}}(E)=\int\limits_{-\infty}^\infty \theta_{\text{left}}(t)\cdot
(E-V(t))^{1/2}_+\ dt+\int\limits_{-\infty}^\infty\theta_{\text{rt}}(t)\cdot
(E-V(t))_+^{1/2}\ dt\\
+ \sum\limits_\nu\int\limits_{-\infty}^\infty\theta_\nu(t)\cdot (E-V(t))^{1/2}
\ dt \equiv \phi_{\text{left}}(E)+\phi_{\text{rt}}(E)+\sum\limits_\nu
\phi_\nu(E).\endmultline
$$
\noindent We have $E-V(t)=(G_{\text{left}}(t,E))^2\cdot(t-x_{\text{left}}(E))$
for $|E-E_0|<c_{\#}^0S_{\text{min}}$, $t \in \text{supp}\ \theta_{\text{left}}$,
with $|\partial_t^\alpha\partial_E^\beta G_{\text{left}}|\le C_{\#}^{\alpha\beta}
S_{\text{left}}^{\frac 12-\beta}B_{\text{left}}^{-\frac 12 -\alpha}$,
in view of our hypotheses on non--degeneracy of $V^\prime(x)$ at
$x_{\text{left}}$.  Hence, setting $t=x_{\text{left}}(E)+s$, we have
$\phi_{\text{left}}(E)=\int\limits_0^\infty\theta_{\text{left}}(x_{\text{left}}
(E)+s)\cdot G_{\text{left}}(x_{\text{left}}(E)+s,E)\cdot s^{1/2}\ ds
\equiv \int\limits_0^\infty F_{\text{left}}(E,s)\cdot s^{1/2}\ ds$,
with $F_{\text{left}}(E,s)$ supported in $0 \le s \le c_{\#}B_{\text{left}}$,
$|\partial_E^\beta F_{\text{left}}|\le C_{\#}^\beta S_{\text{left}}
^{\frac 12-\beta}B_{\text{left}}^{-\frac 12}$.
Hence $|(\frac {d}{dE})^\beta\phi_{\text{left}}(E)|\le \int\limits_0^
{c_{\#}B_{\text{left}}}C_{\#}^\beta S_{\text{left}}^{\frac 12-\beta}
B_{\text{left}}^{-1/2}s^{1/2}\ ds \le 
C_{\#}^\beta S_{\text{left}}^{\frac 12-\beta}B_{\text{left}}\le \hfill\break
C_{\#}^\beta
\int\limits_{|x-x_{\text{left}}(E)|<c_{\#}B_{\text{left}}}S^{1/2-\beta}(x)\ dx$.
Similarly $|(\frac{d}{dE})^\beta\phi_{\text{rt}}(E)|\le\hfill\break
C_{\#}^\beta \int\limits_{|x-x_{\text{rt}}(E)|<c_{\#}B_{\text{rt}}}S^{1/2-\beta}
(x)\ dx$.  Trivially, $(\frac{d}{dE})^\beta\phi_\nu(E)=\hfill\break
c_\beta\int\theta_\nu(x)(E-V(x))^{\frac 12-\beta}\ dx$ so
$|(\frac{d}{dE})^\beta\phi_\nu(E)|\le C_{\#}^\beta\int\limits_{\text{supp}\
\theta_\nu} S^{\frac 12-\beta}(x)\ dx$.  Summing our estimates for
$(\frac{d}{dE})^\beta\phi_\nu(E)$, $(\frac{d}{dE})^\beta\phi_{\text{left}}(E)$,
$(\frac{d}{dE})^\beta\phi_{\text{rt}}(E)$, we get the conclusion of the Lemma.
\qquad $\blacksquare$

In the next section, we summarize what we know so far about eigenvalues and
eigenfunctions of $-\frac{d^2}{dx^2}+V(x)$.



%Subject: Section 6

\pageno 103
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\centerline{\bf{The WKB Theorems}}
\smallskip

\noindent\underbar{Set-up}.  We are given the following:  A
potential $V(x)$ defined 
on a (possibly unbounded) interval $I_{\text{BVP}}$; two positive
functions $S(x)$, $B(x)$ defined on a subinterval $I \subset I_{\text{BVP}}$;
two real numbers $E_0\le E_\infty$; positive numbers $\varepsilon<\frac{1}
{100}$, $K>1$ and $N>K\varepsilon^{-10}$.
We define $N^\prime=[\varepsilon N/500]$ and
$N^{\prime\prime}=\frac 32 \varepsilon N^\prime-K-33$.

Our goal is to understand the eigenvalues and eigenfunctions of the
self--adjoint operator $H=\frac{-d^2}{dx^2}+V(x)$ on $L^2(I_{\text{BVP}})$,
with Dirichlet or Neumann conditions at the endpoints.

\noindent{\bf{Hypotheses}}

\noindent\underbar{Assumptions on $V(x)$, $S(x)$, $B(x)$ in $I$}

\roster
\item"(Hyp0)" If $x,y \in I$ and $|x-y|<cB(x)$, then $c<\frac{B(y)}
{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$.
\item"(Hyp1)" For $x \in I$ and $\alpha \ge 0$ we have
$|(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$.
\item"(Hyp2)" The equation $V(x)=E_0$ has two solutions
$x_{\text{left}}<x_{\text{rt}}$ in $I$, and they satisfy
$\text{dist}(x_{\text{left}},\partial I)>cB(x_{\text{left}}),
\text{dist}(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$.
\item"(Hyp3)" For $x_{\text{left}}\le x\le x_{\text{left}}+c_1
B(x_{\text{left}})$ we have $-V^\prime(x)>cS(x_{\text{left}})B^{-1}
(x_{\text{left}})$, and for $x_{\text{rt}}-c_1B(x_{\text{rt}})\le x\le
x_{\text{rt}}$ we have $+V^\prime(x)>cS(x_{\text{rt}})B^{-1}(x_{\text{rt}})$.
\item"(Hyp4)" For $x_{\text{left}}+c_1B(x_{\text{left}})\le x\le
x_{\text{rt}}-c_1B(x_{\text{rt}})$ we have $cS(x)<E_0-V(x)<CS(x)$.
\endroster

To state the remaining hypotheses, we establish some notation.  Set
$\lambda(x)=S^{1/2}(x)B(x)$ for $x \in I$.  Then set
$$
B_{\text{left}}=B(x_{\text{left}}),\quad S_{\text{left}}=S(x_{\text{left}}),
\quad\lambda_{\text{left}}=\lambda(x_{\text{left}}).
$$
$$
B_{\text{rt}}=B(x_{\text{rt}}),\quad S_{\text{rt}}=S(x_{\text{rt}})
,\quad\lambda_{\text{rt}}=\lambda(x_{\text{rt}}).
$$

For $|E-E_0|<cS_{\text{left}}$, let $x_{\text{left}}(E)$ be the solution
of $V(x)=E$ nearest to $x_{\text{left}}$, and for $|E-E_0|<cS_{\text{rt}}$,
let $x_{\text{rt}}(E)$ be the solution of $V(x)=E$ nearest to $x_{\text{rt}}$.
Define $S_{\text{min}}=\operatornamewithlimits{inf}_{x_{\text{left}}<x<x_
{\text{rt}}}S(x)$ and $\Lambda=\Bigl(\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac
{dx}{S^{1/2}(x)B^2(x)}\Bigr)^{-1}$.

Our remaining hypotheses are as follows.

\noindent\underbar{Assumptions on $V(x)$ in all of $I_{\text{BVP}}$}

\roster
\item"{(Hyp5)}" If $|E-E_0|<c_2S_{\text{min}}$ and $E\le E_\infty$, then
$V(x)>E$ \quad for all\hfill\break
$x \in I_{\text{BVP}}\backslash [x_{\text{left}}(E),
x_{\text{rt}}(E)]$.
\item"{(Hyp6)}" If $x \in I_{\text{BVP}}$ satisfies
$x<x_{\text{left}}-\frac 12 \lambda_{\text{left}}^KB_{\text{left}}$ then
$V(x)\ge E_\infty+\frac{100}{|x-x_{\text{left}}|^2}$, and if
$x \in I_{\text{BVP}}$ satisfies $x>x_{\text{rt}}+\frac 12 \lambda_{\text{rt}}
^KB_{\text{rt}}$, then $V(x)\ge E_\infty+\frac{100}{|x-x_{\text{rt}}|^2}$.
\endroster
\smallskip
\noindent\underbar{Technical Assumptions}
\roster
\item"{(Hyp7)}" $\operatornamewithlimits{max}_{x \in I} S(x)\le
\lambda_{\text{left}}^KS_{\text{left}}$ and $\operatornamewithlimits
{max}_{x \in I}S(x)\le \lambda_{\text{rt}}^KS_{\text{rt}}$
\item"{(Hyp8)}" $\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{1/2}(x)}\le \Lambda^K\cdot\min(S_{\text{left}}^{-1/2}B_{\text{left}},
S_{\text{rt}}^{-1/2}B_{\text{rt}})$
\item"{(Hyp9)}" $[\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{1/2}(x)B^{4}(x)}]\cdot[\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{1/2}(x)}]\le \Lambda^K$
\endroster
\smallskip
\noindent\underbar{WKB Condition}

\roster
\item"{(Hyp10)}" $\Lambda$ is bounded below by a positive constant
depending only on $\varepsilon$, $K$, $N$,\hfill\break
and on $c$, $C$, $c_1$, $c_2$, $C_\alpha$
in (Hyp0)$\ldots$(Hyp4).
\endroster

\smallskip
\noindent\underbar{Definitions and Basic Properties of Phases}

Assume hypotheses (Hyp0)$\ldots$(Hyp10).  For $|E-E_0|<c_{\#}S_{\text{min}}$,
define
$$\gather
\phi(E)=\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}(E-V(x))^{1/2}\,
dx\quad \text{and}\\
\psi(E)=\operatornamewithlimits{lim}_{\delta\to 0+} \Bigl[
\int_{x_{\text{left}}(E)+\delta}^{x_{\text{rt}}(E)-\delta}V^{\prime\prime}
(x)(E-V(x))^{-\frac 32}\,dx-q(E)\delta^{-1/2}\Bigr]\endgather
$$

\noindent with $q(E)$ uniquely specified by demanding the finiteness of
the limit.
\proclaim{Lemma 1}  For $|E-E_0|<c_{\#}S_{\text{min}}$ we have
$$\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\phi(E)\Big|\le C_{\#}^\beta
\int_{x_{\text{left}}}^{x_{\text{rt}}}S^{\frac 12-\beta}(x)\,dx
\quad\text{and}
$$
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\psi(E)\Big|\le C_{\#}^\beta
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{\frac 12+\beta}
(x)B^2(x)}.
$$
\noindent Also $c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{1/2}(x)}<\frac{d\phi(E)}{dE}<C_{\#}\int_{x_{\text{left}}}^{
x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}$.

The constants $c_{\#}$, $C_{\#}$, $C_{\#}^\beta$ depend only on
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$,  $c_2$, $C_\alpha$ in the
hypotheses (Hyp 0)$\ldots$(Hyp 4).
\endproclaim

\proclaim{WKB Eigenvalue Theorem}  If (Hyp0)$\ldots$(Hyp10) hold, then
there is a function $\Phi(E)$ on $[E_0-c_{\#}S_{\text{min}},E_0+c_{\#}
S_{\text{min}}]$ and there are numbers $E_{k_{\text{min}}},E_{k_{\text{min}}+1},
\ldots,E_{k_{\text{max}}}\hfill\break \le E_\infty$ with the following 
properties.
\roster
\item"{(A)}" $\Phi(E)=\pm\frac \pi 2+\phi(E)+\frac{1}{48}\psi(E)+\phi_
{\text{error}}(E)$, with
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\phi_{\text{error}}\Big|\le C_{\#}
^\beta\Lambda^{-1}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{\frac 12+\beta}(x)B^2(x)}, \quad\text{all}\ \beta\ge 0.
$$
\item"{(B)}" $\{k_{\text{min}},k_{\text{min}}+1,\ldots,k_{\text{max}}\}$
is exactly the set of integers $k$ for which $|\Phi(E)-\pi k|<C_{\#}
\Lambda^{-N^{\prime\prime}}$ for some $E \in [E_0-\frac 14
c_{\#}S_{\text{min}},E_0+\frac 14c_{\#}S_{\text{min}}]\cap (-\infty,E_{\infty}]$.
\item"{(C)}" If $k_{\text{min}}\le k<k_{\text{max}}$, then $E_k$ is an 
eigenvalue of $H$.
\item"{(D)}" If $k=k_{\text{max}}$, then $E_k$ is either an eigenvalue of $H$
or equal to $E_\infty$.
\item"{(E)}" For $k_{\text{min}}\le k\le k_{\text{max}}$ we have
$|E_k-E_0|<c_{\#}S_{\text{min}}$ and $|\Phi(E_k)-\pi k|<C_{\#}
\Lambda^{-N^{\prime\prime}}$.
\item"{(F)}" Every eigenvalue of $H$ in the interval $[E_0-\frac 14
c_{\#}S_{\text{min}},E_0+\frac 14c_{\#}S_{\text{min}}]\hfill\break
 \cap (-\infty,E_\infty]$
is one of the $E_k$ $(k_{\text{min}}\le k\le k_{\text{max}})$.
\endroster
\endproclaim

\noindent The constants $c_{\#}$, $C_{\#}^\beta$ depend only on $\varepsilon$,
$K$, $N$, $c$, $C$, $c_1$ $c_2$, $C_\alpha$ in (Hyp0)$\ldots$(Hyp5).

\demo{Remark} Perhaps $E_{k_{\text{min}}}$ or $E_{k_{\text{max}}}$ or both lie
slightly outside 
$$
[E_0-\frac 14c_{\#}S_{\text{min}},E_0+\frac 14 c_{\#}
S_{\text{min}}].$$ 

We prepare to describe the eigenfunctions of $H$.  Given an energy $E$
with $|E-E_0|<c_{\#}S_{\text{min}}$, define intervals:
$$
I_{\text{far\ left}}^E=I_{\text{BVP}}\cap (-\infty,x_{\text{left}}(E)
-\lambda_{\text{left}}^{\varepsilon-2/3}B_{\text{left}}]
$$
$$ 
I_{\text{Airey\ left}}^E=[x_{\text{left}}(E)-\lambda_{\text{left}}^
{-\varepsilon}B_{\text{left}},x_{\text{left}}(E)+\lambda_{\text{left}}^
{-\varepsilon} B_{\text{left}}]
$$
$$
I_{\text{medium\ left}}^E=[x_{\text{left}}(E)+\lambda_{\text{left}}^{
\varepsilon-\frac 23}B_{\text{left}}, x_{\text{left}}(E)+c_{\#}
B_{\text{left}}]
$$
$$
I_{\text{center}}^E=[x_{\text{left}}(E)+\frac 12 c_{\#}B_{\text{left}},
x_{\text{rt}}(E)-\frac 12 c_{\#}B_{\text{rt}}]
$$
$$
I_{\text{medium\ rt}}^E=[x_{\text{rt}}(E)-c_{\#}B_{\text{rt}},
x_{\text{rt}}(E)-\lambda_{\text{rt}}^{\varepsilon-2/3}B_{\text{rt}}]
$$
$$
I_{\text{Airey\ rt}}^E=[x_{\text{rt}}(E)-\lambda_{\text{rt}}^{-\varepsilon}
B_{\text{rt}},x_{\text{rt}}(E)+\lambda_{\text{rt}}^{-\varepsilon}B_{\text{rt}}]
$$
$$
I_{\text{far\ rt}}^E=[x_{\text{rt}}(E)+\lambda_{\text{rt}}^{\varepsilon-2/3}
B_{\text{rt}},\infty)\cap I_{\text{BVP}}.
$$
\noindent Also define regions
$$
U_{\text{Airey\ left}}=\{(x,E)\Bigm| |E-E_0|<c_{\#}S_{\text{min}},
|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\},
$$
$$
U_{\text{Airey\ rt}}=\{(x,E)\Bigm| |E-E_0|<c_{\#}S_{\text{min}},|x-x_{\text{rt}}
(E)|<\lambda_{\text{rt}}^{-\varepsilon}B_{\text{rt}}\}
$$
$$
U_{\text{oscil}}=\{(x,E)\Bigm| |E-E_0|<c_{\#}S_{\text{min}},x_{\text{left}}(E)<
x<x_{\text{rt}}(E)\}.
$$
\vglue 1pc
\proclaim{WKB Eigenfunction Theorem}  Assume (Hyp0)$\ldots$(Hyp10).  Then
there exist real--valued functions $Y_{\text{left}}(x,E)$ on
$U_{\text{Airey\ left}}$, $Y_{\text{rt}}(x,E)$ on $U_{\text{Airey\ rt}}$,
and complex--valued functions $u_{\text{left}}(x,E)$, $u_{\text{rt}}(x,E)$
on $U_{\text{oscil}}$, for which the following hold.\endproclaim
\smallskip
\noindent\underbar{Properties of the Auxiliary Functions}
\roster
\item"{(1)}" On $U_{\text{Airey\ left}}$\ we can write
$Y_{\text{left}}=Y_0^{\text{left}}+\lambda_{\text{left}}^{-2}
Y_1^{\text{left}}$, with $|\partial_x^\alpha\partial_E^\beta Y_i^{\text{left}}|
\le C_{\#}^{\alpha\beta}B_{\text{left}}^{-\alpha}S_{\text{left}}^{-\beta}$
and $\lambda_{\text{left}}^2(\frac{\partial Y_0^{\text{left}}}{\partial x})^2Y_0^
{\text{left}}=E-V(x)$ to order at least $N^\prime$ at $x=x_{\text{left}}(E)$.
More precisely, $|\lambda_{\text{left}}^2(\frac{\partial Y_{\text{left}}}
{\partial x})^2\cdot Y_{\text{left}}+\{Y_{\text{left}},x\}-
(E-V(x))|\le C_{\#}\lambda_{\text{left}}^{-N^\prime}
S_{\text{left}}$ on $U_{\text{Airey}}^{\text{left}}$. 
Similarly, on
$U_{\text{Airey\ rt}}$ we can write $Y_{\text{rt}}=Y_0^{\text{rt}}+
\lambda_{\text{rt}}^{-2}Y_1^{\text{rt}}$, with $|\partial_x^\alpha
\partial_E^\beta Y_i^{\text{rt}}|\le C_{\#}^{\alpha\beta}
B_{\text{rt}}^{-\alpha}S_{\text{rt}}^{-\beta}$ and $\lambda_{\text{rt}}^2
(\frac{\partial Y_0^{\text{rt}}}{\partial x})^2Y_0^{\text{rt}}=E-V(x)$
to order at least $N^\prime$ at $x=x_{\text{rt}}(E)$.  More precisely,
$|\lambda_{\text{rt}}^2
(\frac{\partial Y_{\text{rt}}}{\partial x})^2Y_{\text{rt}}+\{Y_{\text{rt}},x\}-
(E-V(x))|\le C_{\#}\lambda_{\text{rt}}^{-N^\prime}S_{\text{rt}}$ on
$U_{\text{Airey}}^{\text{rt}}$.

\item"{(2)}" For $|E-E_0|<c_{\#}S_{\text{min}}$ and $x \in I_{\text{medium\
left}}^E$ we have $|\partial_x^\alpha u_{\text{left}}|\le \hfill\break
C_{\#}^\alpha
\lambda_{\text{left}}^{-1}(\frac{x-x_{\text{left}}(E)}{B_{\text{left}}})
^{-\frac 32}\cdot(x-x_{\text{left}}(E))^{-\alpha}$, and
$|\text{Re}\ u_{\text{left}}|\le \hfill\break
C_{\#}\lambda_{\text{left}}^{-2}
(\frac{x-x_{\text{left}}(E)}{B_{\text{left}}})^{-3}$.  Similarly, for
$|E-E_0|<c_{\#}S_{\text{min}}$ and $x \in I_{\text{medium\ rt}}^E$ we have
$|\partial_x^\alpha u_{\text{rt}}|\le C_{\#}^\alpha\lambda_{\text{rt}}^{-1}
(\frac{x_{\text{rt}}(E)-x}{B_{\text{rt}}})^{-\frac 32}\cdot (x_{\text{rt}}
(E)-x)^{-\alpha}$ and $|\text{Re}\ u_{\text{rt}}|\le C_{\#}
\lambda_{\text{rt}}^{-2}(\frac{x_{\text{rt}}(E)-x}{B_{\text{rt}}})^{-3}$.
\item"{(3)}" For $|E-E_0|<c_{\#}S_{\text{min}}$ and $x \in I_{\text{center}}^E$
we have
$$
|\partial_x^\alpha u_{\text{left}}|\le C_{\#}^\alpha\Lambda^{-1}B^{-\alpha}
(x)\quad\text{and}\quad |\text{Re}\ u_{\text{left}}|\le C_{\#}\Lambda^{-2}.
$$
Similarly, on the same region we have $|\partial_x^\alpha u_{\text{rt}}|\le
C_{\#}^\alpha\Lambda^{-1}B^{-\alpha}(x)$ and $|\text{Re}\ u_{\text{rt}}|
\hfill\break\le C_{\#}\Lambda^{-2}$.
\endroster

\smallskip
\noindent\underbar{Description of Eigenfunctions}

Let $F(x)$ be a real--valued eigenfunction of $H$ with eigenvalue $E$ and
norm $1$.  Assume $|E-E_0|<c_{\#}S_{\text{min}}$ and $E\le E_\infty$.  Then
there exist real constants $b_{\text{left}}$, $b_{\text{rt}}$ for which we
have:
\roster
\item"{(4)}" $\int_{I_{\text{far\ left}}^E}|F(x)|^2\,dx\le\Lambda^
{-N^{\prime\prime}}$ and $\int\limits_{I_{\text{far\ rt}}^E}|F(x)|^2\,dx\le
\Lambda^{-N^{\prime\prime}}$
\item"{(5)}" $\int_{I_{\text{Airey\ left}}^E}|F(x)-b_{\text{left}}
\lambda_{\text{left}}^{-1/3}\Bigl(\frac{\partial Y_{\text{left}}(x,E)}
{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E))|^2\,dx
\le \hfill\break
\Lambda^{-N^{\prime\prime}}$ and $\int\limits_{I_{\text{Airey\ rt}}^E}
|F(x)-b_{\text{rt}}\lambda_{\text{rt}}^{-1/3}\Bigl(\frac{-\partial Y_{\text{rt}}
(x,E)}{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{rt}}^{2/3}
Y_{\text{rt}}(x,E))|^2\,dx\hfill\break\le
\Lambda^{-N^{\prime\prime}}$
\item"{(6)}"$$\split
\int_{I_{\text{medium\ left}}^E\cup I_{\text{center}}^E}
\Big|F(x)-b_{\text{left}}\text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{i\int_
{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\\
\cdot(1+u_{\text{left}}(x))\Bigr]\Big|^2\,dx\le \Lambda^{-N^{\prime\prime}}
\endsplit
$$
\noindent and
$$\split
\int_{I_{\text{medium\ rt}}^E\cup I_{\text{center}}^E}
|F(x)-b_{\text{rt}}\text{Re}\Bigl[\frac{e^{\mp i\frac \pi 4}\,
e^{-i\int_x^{x_{\text{rt}}(E)}(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\\
\cdot (1+u_{\text{rt}}(x))\Bigr]|^2\,dx\le \Lambda^{-N^{\prime\prime}}
\endsplit
$$
\item"{(7)}" $c_{\#}<(|b_{\text{left}}|^2+|b_{\text{rt}}|^2)
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}<C_{\#}$, and
\item"{(8)}" $\Bigl||b_{\text{left}}|\cdot|b_{\text{rt}}|^{-1}-1\Bigr|
\le C_{\#}\Lambda^{-2}$.
\endroster

\newpage
\noindent\underbar{What the Constants may Depend on}
\smallskip

The constants $c_{\#}$, $C_{\#}$, $C_{\#}^\alpha$ above depend only on 
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$ in
(Hyp0)$\ldots$(Hyp5).

\vglue 1pc
\demo{Proofs} (Hyp0)$\ldots$(Hyp10) imply the assumptions used in the
section on eigenfunctions and eigenvalues, namely (H0)$\ldots$(H6)
and (E1)$\ldots$(E7).  This follows from the discussion in the preceding
section.  The estimates for $\frac{d^\beta\phi(E)}{dE^\beta}$
in Lemma 1 were proved in the preceding section, and the estimates for
$\frac{d^\beta}{dE^\beta}\psi(E)$ are immediate consequences of the
global WKB lemma and the fact that $\psi(E)=48\ \text{Im}(G_1(E))$.

The WKB Eigenvalue Theorem is immediate from Lemma 16 in the section
on eigenvalues and eigenfunctions.  We define $E_k$ to be the unique
eigenvalue  $\le E_\infty$ in $J(k)$, assuming an eigenvalue exists there;
and we set $E_k=E_\infty$ otherwise.  (See Lemma 16 for the definition of
$J(k)$). This proves (B)$\ldots$(F) of the WKB Eigenvalue Theorem.  Part (A)
is already contained in the global WKB lemma.

The WKB Eigenfunction Theorem is immediate from Lemma 12 in the section 
on eigenfunctions and eigenvalues.  We set $u_{\text{left}}=\sum\limits_
{k=1}^{N^\prime}u_k^{\text{left}}(x,E)$, $u_{\text{rt}}=\sum\limits_
{k=1}^{N^\prime}u_k^{\text{rt}}(x,E)$, $b_{\text{left}}=b$ and
$b_{\text{rt}}=(-1)^{k_1}R(E_1)b$.  (See Lemma 12 for the notation.)

Parts (1), (2), (3) of the Eigenfunction Theorem are contained in the
global WKB lemma, and parts (4), (5), (6), (7) come from Lemma 12.  Part (8)
asserts that $|R(E_1)-1|\le C_{\#}\Lambda^{-2}$, which is contained in
the global WKB lemma.$\qquad \blacksquare$


%Subject: Section 7


\pageno 109
\magnification \magstep1
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\centerline{\bf{Normalizing the WKB Eigenfunctions}}
\smallskip

The goal of this section is to compute the constants $b_{\text{left}}$,
$b_{\text{rt}}$ in the statement of the WKB Eigenfunction Theorem.
The first step is to give a close approximation for integrals of the form
$\int_{-\infty}^\infty\theta(t)A^2(t)\,dt$, which we accomplish in the
next few lemmas.
\vglue 1pc

\proclaim{Lemma 1}  Suppose $|(\frac{d}{dy})^m\theta(y)|\le C_mR^{-m}\,\,
(m\ge 0)$ and $\theta$ of compact support and $|K(y)|\le \frac{C^\prime}{
|y|^M}$, where $M$, $R>10$.  Then
$$
\int_1^\infty\theta(y)K(y)\,dy=\sum\limits_{m=0}^{M-3}c(m)\theta^{(m)}
(0)+\roman{Error}
$$
\noindent with $|\roman{Error}|\le \roman{Const.}(C^\prime,C_m,M)\cdot
R^{-(M-2)}$, $c(m)$ depending only on $K(y)$, and 
$|c(m)|\le\roman{Const.}(C^\prime,M)$.
\endproclaim
\vglue 1pc
\demo{Proof} Taylor's theorem gives $\theta(y)=\sum\limits_{m=0}^{M-1}
\frac{1}{m!}\theta^{(m)}(0)y^m+\chi(y)\cdot y^M$ with $|\chi(y)|\le
\text{Const.}(C_M)R^{-M}$, so
$$\align
\int_1^\infty\theta(y)K(y)\,dy&=\sum\limits_{m=0}^{M-1}\theta^{(m)}(0)
\cdot \Bigl[\frac{1}{m!}\int_1^RK(y)y^m\,dy\Bigr]+\int_1^R\chi(y)y^M
K(y)\,dy\\
&\qquad\qquad  +\int_R^\infty \theta(y)K(y)\,dy.\endalign
$$

\noindent The last two terms on the right are dominated by
$\text{Const.}(C^\prime,C_m,M)R^{-(M-1)}$, and the term
$\theta^{(M-1)}(0)\cdot\Bigl[\frac{1}{(M-1)!}\int_1^RK(y)y^{M-1}\,dy\Bigr]$
is dominated by \hfill\break
$\text{Const.}(C_{M-1},C^\prime)R^{-(M-1)}\ell nR\le
\text{Const.}(C_{M-1},C^\prime)R^{-(M-2)}$.  Hence
$$
\split
\int_1^\infty\theta(y)K(y)\,dy=\sum\limits_{m=0}^{M-2}\theta^{(m)}(0)\cdot
\Bigl[\frac{1}{m!}\int_1^\infty K(y)y^m\,dy\Bigr]\\
-\sum\limits_{m=0}^{M-2}\theta^{(m)}(0)\Bigl[\frac{1}{m!}
\int_R^\infty K(y)y^m\,dy\Bigr]+\text{Error}
\endsplit
$$
\noindent with $|\text{Error}|\le \text{Const.}(C^\prime,C_m,M)\cdot
R^{-(M-2)}$.  The second sum on the right is also dominated by
$\text{Const.}(C^\prime,C_m,M)R^{-(M-2)}$, as is the term $m=M-2$
in the first sum.
Hence the lemma is proven, with $c(m)=\frac{1}{m!}\int_1^\infty y^mK(y)\,dy$.
$\qquad\blacksquare$

\vglue 1pc
\proclaim{Lemma 2}  If $\theta(y)$ has compact support and $|(\frac{d}{dy})^m
\theta(y)|\le C_mR^{-m}$ (all $m$) with $R\ge 10$, then
$$ 
\int_1^\infty e^{\frac 43 iy^{\frac 32}}y^{-s}\theta(y)\,dy=
\sum\limits_{m=0}^{M-1}c(s,m)\theta^{(m)}(0)+\roman{Error}
$$
\noindent with $c(s,m)$ universal constants, $|\roman{Error}|<
\roman{Const.}(C_m,M)\cdot R^{-M}$ and $M$ as large as we please.
\endproclaim

\vglue 1pc
\demo{Proof}  Integration by parts gives the formula
$$\split
\int_1^\infty e^{\frac 43 iy^{3/2}}y^{-s}\theta(y)\,dy=
c_1(s)\theta(1)+c_2(s)\int_1^\infty e^{\frac 43 iy^{3/2}}y^{-(s+
\frac 32)}\theta(y)\,dy\\
+c_3(s)\int_1^\infty e^{\frac 43 iy^{\frac 32}}y^{-(s+1/2)}\theta^\prime(y)
\,dy.\endsplit
$$

\noindent The integrals on the right are analogous to the left--hand
side, with $s$ increased by at least $1/2$.  Hence repeated use of
this formula yields an identity of the form
$$\split
\int_1^\infty e^{\frac 43 iy^{3/2}}y^{-s}\theta(y)\,dy=
\sum\limits_{j=1}^{j_{\text{max}}}c(s,j)\theta^{(m_j)}(1)\\
+\sum\limits_{j=1}^{j_{\text{max}}^\prime}c^\prime(s,j)\int_1^
\infty e^{\frac 43 i y^{3/2}}y^{-s_j}\theta^{(m_j^\prime)}(y)\,dy\endsplit
$$

\noindent with all the $s_j \ge M+2$.  Lemma 1 (with $M+2$ in place of $M$)
applies to the integrals on the right, so we get
$$\int_1^\infty e^{\frac 43 iy^{3/2}}y^{-s}\theta(y)\,dy=\sum\limits_
{j=1}^{j_{\text{max}}}c(s,j)\theta^{(m_j)}(1)+\sum\limits_{m=0}^{M-1}
\tilde c(s,m)\theta^{(m)}(0)+\text{Error},$$

\noindent with $|\text{Error}|\le \text{Const.}(C_m,M)R^{-M}$.

The terms $c(s,j)\theta^{(m_j)}(1)$ with $m_j \ge M$ are dominated by
$\text{Const.}(C_m,M)R^{-M}$, and for $m_j\le M-1$
Taylor's theorem gives $\theta^{(m_j)}(1)=
\sum\limits_{m_j\le m\le M-1}\frac{\theta^{(m)}(0)}{(m-m_j)!}+\text{error}$,
with $|\text{error}|\le \text{Const.}(C_m,M)R^{-M}$.  Therefore, the previous
equation for our integral implies
$$\split
\int_1^\infty e^{\frac 42 iy^{3/2}}y^{-s}\theta(y)\,dy=\sum\limits_{m=0}^{M-1}
c(s,m)\theta^{(m)}(0)+\text{Error}\\
\text{with}\quad |\text{Error}|\le \text{Const.}(C_m,M)R^{-M}.\endsplit
$$
\TagsOnRight
$$
\tag"$\blacksquare$"
$$\TagsOnLeft

Of course, Lemma 2 has an analogue with $e^{\frac 43 iy^{3/2}}$ replaced by
$e^{-\frac 43 iy^{3/2}}$.
\vglue 1pc
\proclaim{Lemma 3} Suppose $\theta(y)$ has compact support and satisfies
$|(\frac{d}{dy})^m\theta(y)|\le C_mR^{-m}$ (all $m$) with $R \ge 10$.
Suppose $\Cal A(y)$ satisfies
$$\split
\Big|\Cal A(y)-\text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{\frac 23 iy^{3/2}}}
{y^{1/4}}\,(1+\sum\limits_{k=1}^M c_k\,y^{-\frac 32 k})\Bigr]\chi_{y>1}\Big|
\le C_M^\prime(1+|y|)^{-M}\\ (\text{any}\ M).\endsplit
$$
\noindent Then $\int_{-\infty}^\infty\Cal A^2(y)\theta(y)\,dy=\frac 12
\int_1^\infty\Big|\frac{1+\sum\limits_{k=1}^Mc_ky^{-\frac 32 k}}{y^{1/4}}\Big|^2
\theta(y)\,dy+\sum\limits_{m=0}^{M-1}c(m)\theta^{(m)}(0)+\roman{Error}$,
where the $c(m)$ are independent of $\theta$, $M$ is as large as we please,
and $|\roman{Error}|\le \roman{Const.}(M,C_M^\prime,C_m)\cdot R^{-M}$.
\endproclaim

\vglue 1pc
\demo{Proof}  Set $H_M(y)=\Cal A^2(y)-\Bigl(\text{Re}\Bigl[\frac
{e^{\pm i\frac \pi 4}\,e^{\frac 23 iy^{3/2}}}{y^{1/4}}\,(1+\sum\limits_{k=1}^
{M}c_ky^{-\frac 32 k})\Big]\chi_{y>1}\Bigr)^2$.\hfill\break
Our hypothesis implies
$$
|H_M(y)|\le C_M(1+|y|)^{-M}\quad, M\  \text{as\ large\ as\ we\ please}.
$$

\noindent Lemma 1 shows that
$$\split
\int_1^\infty H_M(y)\theta(y)\,dy=\sum\limits_{m=0}^{M-3}c_1(m)\theta^{(m)}
(0)+\text{Error}_1,\\
|\text{Error}_1|\le \text{Const.}(M,C_M^\prime,C_m)R^{-(M-2)}\endsplit
$$

\noindent and that
$$\split
\int_{-\infty}^{-1}H_M(y)\theta(y)\,dy=\sum\limits_{m=0}^{M-3}
c_2(m)\theta^{(m)}(0)+\text{Error}_2,\\
|\text{Error}_2|\le \text{Const.}(M,C_M^\prime,C_m)R^{-(M-2)}.
\endsplit
$$

\noindent Taylor's theorem gives
$$\split
\int_{-1}^{1}H_M(y)\theta(y)\,dy=\sum\limits_{m=0}^{M-3}c_3(m)\theta^{(m)}
(0)+\text{Error}_3,\\
|\text{Error}_3|\le \text{Const.}(M,C_M^\prime,C_m)R^{-(M-2)}.\endsplit
$$

\noindent Adding the last three equations, we get
$$\split
\int_{-\infty}^{\infty}H_M(y)\theta(y)\,dy=\sum\limits_{m=0}^{M-3}
c_4(m)\theta^{(m)}(0)+\text{Error}_4,\\
|\text{Error}_4|\le \text{Const.}(M,C_{M}^\prime,C_m)R^{-(M-2)}.
\endsplit
$$

\noindent Therefore, the lemma follows if we can show that
$$\multline
Q\equiv \int_1^\infty\Bigl(\text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{\frac
23 iy^{3/2}}}{y^{1/4}}(1+\sum\limits_{k=1}^Mc_ky^{-\frac 32 k})\Bigr]\Bigr)^2
\theta(y)\,dy\\
=\frac 12 \int_1^\infty\Big|\frac{1+\sum\limits_{k=1}^Mc_ky^{-\frac 32 k}}
{y^{1/4}}\Big|^2\theta(y)\,dy+\sum\limits_{m=0}^{M-3}c_5(m)\theta^{(m)}
(0)+\text{Error}_5\endmultline\tag1
$$

\noindent with $|\text{Error}_5|\le \text{Const.}(M,C_M^\prime,C_m)\cdot
R^{-(M-2)}$.

We expand $Q$ using $(\text{Re}(\zeta))^2=(\frac 12\zeta+\frac 12\overline\zeta)
^2=\frac 12|\zeta|^2+\frac 14 \zeta^2+\frac 14\overline\zeta^2$.  Thus
$$\align
Q&=\frac 12\int_1^\infty\bigg|\frac{1+\sum\limits_{k=1}^{M}c_ky^{-\frac 32 k}}
{y^{1/4}}\bigg|^2\theta(y)\,dy+\frac 14 e^{\pm i\frac \pi 2}
\int_1^\infty\frac{e^{\frac 43 iy^{3/2}}}{y^{1/2}}\\
&\qquad\qquad\qquad\qquad\qquad \Bigl(1+\sum\limits_{k=1}^Mc_ky^{-\frac 32 k}
\Bigr)^2\theta(y)\,dy\\
&\,\,+\frac 14 e^{\mp i\frac \pi 2}\int_1^\infty
\frac{e^{-\frac 43 iy^{3/2}}}{y^{1/2}}\Bigl(1+\sum\limits_{k=1}^M\overline c_k
y^{-\frac 32 k}\Bigr)^2\theta(y)\,dy.\endalign
$$

The last two terms on the right are sums of integrals to which we can
apply Lemma 2 and its analogue for $e^{-\frac 43 iy^{3/2}}$, and the desired
equation (1) follows easily. 
\TagsOnRight
$$
\tag"$\blacksquare$"
$$\TagsOnLeft
\smallskip

We apply Lemma 3 to the Airey function.
\vglue 1pc
\proclaim{Lemma 4}  Suppose $\theta(y)$ has compact support and satisfies
$|(\frac {d}{dy})^m\theta(y)|\le C_mR^{-m}$ $(m \ge 0)$ with $R \ge 10$.

For universal constants $\tilde c(0)$, $\tilde c(1)$, $\tilde c(2)$ we have
$$
\int_{-\infty}^\infty A^2(y)\theta(y)\,dy=\frac 12 \int_0^\infty
\frac{\theta(y)\,dy}{y^{1/2}}+\sum\limits_{m=0}^2 \tilde c(m)\theta^{(m)}(0)
+\text{Error},$$

\noindent with $|\text{Error}|\le \text{Const.}(C_m)\cdot R^{-5/2}$.


\endproclaim
\demo{Proof}  The Airey function $A(y)$ satisfies the hypothesis of the
previous lemma, with $c_1=\pm \frac {5}{48}i$.  Since $c_1$ is purely
imaginary, we have  $|1+\sum\limits_{k=1}^Mc_ky^{-\frac 32 k}|^2=1+
O(y^{-3})$ for $y \ge 1$.  Hence
$$
\frac 12 \operatornamewithlimits{\int}_1^\infty\Big|\frac{1+\sum\limits_{k=1}^M
c_ky^{-\frac 32 k}}{y^{1/4}}\Big|^2\theta(y)\,dy=\frac 12\int_1^\infty
\frac{\theta(y)\,dy}{y^{1/2}}
+\int_1^\infty K(y)\theta(y)\,dy
$$

\noindent with $|K(y)|\le C y^{-7/2}$.  Taylor's theorem gives
$$\multline
\int_1^\infty K(y)\theta(y)\,dy=\int_1^RK(y)\Bigl
[\theta(0)+y\theta^\prime(0)+\frac 12
y^2\theta^{\prime\prime}(0)+O(R^{-3}y^3)\Bigr]\,dy\\
+\int_R^\infty K(y)O(1)\,dy
=\sum\limits_{m=0}^2\theta^{(m)}(0)\cdot \frac{1}{m!}\int_1^\infty
y^mK(y)\,dy\\
+\Big\{-\sum\limits_{m=0}^2 \theta^{(m)}(0)\\
\cdot \frac 1{m!}\int_R^\infty y^mK(y)\,dy
+\int_1^RK(y)\cdot O(R^{-3}y^3)\,dy+\int_R^\infty K(y)O(1)\,dy\Big\}.
\endmultline
$$

\noindent The terms in curly brackets are all dominated by $R^{-5/2}$, so that
$$
\frac 12 \int_1^\infty \bigg|\frac{1+\sum\limits_{k=1}^Mc_k y^{-\frac 32 k}}
{y^{1/4}}\bigg|^2\theta(y)\,dy=\frac 12 \int_1^\infty\frac{\theta(y)\,dy}
{y^{1/2}}+\sum\limits_{m=0}^2c_1(m)\theta^{(m)}(0)+\text{Error}_1
$$

\noindent with $|\text{Error}_1|\le \text{Const.}(C_m)\cdot R^{-5/2}$.  The
previous lemma therefore gives  \hfill\break
$\int_{-\infty}^\infty A^2(y)\theta(y)\,dy=\frac
12 \int_1^\infty \frac{\theta(y)\,dy}{y^{1/2}}+\sum\limits_{m=0}^2 c_2(m)\theta^
{(m)}(0)+\text{Error}_2$, with $|\text{Error}_2|\le \text{Const.}(C_m)\cdot
R^{-5/2}$.  Taylor's theorem shows that 
$$\split
\frac 12 \int_0^1\frac
{\theta(y)\,dy}{y^{1/2}}=\sum\limits_{m=0}^2\theta^{(m)}(0)\Big[
\frac 12 \int_0^1\frac{y^{m-\frac 12}}{m!}\,dy\Big]+\text{Error}_3,\\ 
\text{with}|\text{Error}_3|\le \text{Const.}(C_m)R^{-3}. \endsplit
$$

\noindent Combining this with the preceding equation, we get the conclusion
of the lemma.$\quad\blacksquare$
\vglue 1pc
\demo{Remark}  Later, we will see that the universal constants
$\tilde c(0)$, $\tilde c(1)$, $\tilde c(2)$ are all zero.  This will be
important for the normalization of WKB eigenfunctions.  The ideas in
the proofs of
Lemmas 1$\ldots$4 have used only the asymptotic properties of the Airey
function, which are not enough to specify the $\tilde c(m)$.

Now suppose we are in the setting of the WKB Eigenfunction Theorem, and let $F$
be a real--valued eigenfunction of $H$, with $L^2$--norm $1$ and eigenvalue
$E$.  We assume $|E-E_0|<c_{\#}S_{\text{min}}$ and $E\le E_\infty$, so that
$F$ is described by the WKB Eigenfunction Theorem.  Our goal is to
compute $\int_{I_{\text{BVP}}}\theta(x)
F^2(x)\,dx$ for suitable smooth functions $\theta$. 

Let us start with the case of $\theta$ supported near a turning point.  Thus
we assume $\text{supp}\ \theta\subset I_{\text{Airey\ left}}^E=
\{|x-x_{\text{left}}
(E)|\le \lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\}$, and we
assume the estimates $|(\frac {d}{dx})^\alpha\theta(x)|\le C^\alpha
(\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}})^{-\alpha}$.

For the next few paragraphs, we delete the subscript ``left".  According to
the WKB Eigenfunction Theorem, we have in\ $\text{supp}\,\,\,\, \theta$
$F(x)=\hfill\break
b\lambda^{-1/3}(\frac{\partial Y}{\partial x})^{-1/2}A
(\lambda^{2/3}Y)+\text{Error}$, with $\int|\text{Error}|^2\,dx<
\Lambda^{-N^{\prime\prime}}$.

Thus 
$$
F^2(x)\le (1+\Lambda^{-\frac 1{10}N^{\prime\prime}})(b\lambda^{-1/3}
(\frac{\partial Y}{\partial x})^{-1/2}A(\lambda^{2/3}Y))^2+\Lambda^{\frac 12
N^{\prime\prime}}|\text{Error}|^2
$$
\noindent and 
$$
F^2(x)\ge (1-\Lambda^{-\frac 1{10}
N^{\prime\prime}})(b\lambda^{-1/3}(\frac{\partial Y}{\partial x})^{-1/2}
A(\lambda^{2/3}Y))^2-\Lambda^{\frac 12 N^{\prime\prime}}|\text{Error}|^2.
$$

\noindent Estimate (44 bis) in the section on Schr\"odinger operators,
together with the results on the order of magnitude of $|b|$ in the WKB
Eigenfunction Theorem, shows that
$$
|\int|\theta(x)|\Lambda^{-\frac 1{10}N^{\prime\prime}}(b\lambda^{-1/3}
(\frac{\partial Y}{\partial x})^{-1/2}A(\lambda^{2/3}Y))^2\,dx|\le
C_{\#}\Lambda^{-\frac{1}{10}N^{\prime\prime}}.
$$

\noindent Hence
$$
\int|\theta(x)|\,|F^2(x)-(b\lambda^{-1/3}(\frac{\partial Y}{\partial x})^
{-1/2}A(\lambda^{2/3}Y))^2|\,dx\le C_{\#}^\prime\Lambda^{-N^{\prime\prime}/10},
\quad \text{so}
$$
$$\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx=b^2\int_{I_{\text{BVP}}}
\theta(x)\lambda^{-2/3}(\frac{\partial Y}{\partial x})^{-1}A^2(\lambda^{2/3}
Y)\,dx+\text{Error}_1, \tag 2
$$
$$
|\text{Error}_1|\le C_{\#}\Lambda^{-\frac 1{10}N^{\prime\prime}}.\tag 3
$$

In $\text{supp}\ \theta$, we have $\frac{\partial Y}{\partial x}>cB^{-1}$.
Hence we can make the change of variable $t=\lambda^{2/3}Y(x,E)$ to reduce
matters to Lemma 4.

Define $\hat I=\text{Image\ of}\ \{|x-x(E)|<\lambda^{-\varepsilon}B\}$
under the map $x\mapsto t=\lambda^{2/3}Y(x,E)$, and set
$$\align
&\theta^{\#}(t)=\theta(x)\lambda^{-4/3}(\frac{\partial Y}{\partial x})^{-2}
\qquad \text{for}\quad t=\lambda^{2/3}Y(x,E),\,\, |x-x(E)|<\lambda^{
-\varepsilon}B;\\
&\theta^{\#}(t)=0  \qquad \qquad\qquad\qquad\quad\,\,\text{for}
\quad t\notin \hat I.\endalign 
$$

\noindent We have
$$\multline
\int_{I_{\text{BVP}}}\theta(x)\lambda^{-2/3}(\frac{\partial Y}{\partial x})^
{-1}A^2(\lambda^{2/3}Y)\,dx=\\
\int\limits_{|x-x(E)|<\lambda^{-\varepsilon}B}\theta(x)\lambda^{-4/3}
(\frac{\partial Y}{\partial x})^{-2}A^2(\lambda^{2/3}Y)\cdot (\lambda^{2/3}
\frac{\partial Y}{\partial x}\,dx)\\
=\int_{-\infty}^\infty \theta^{\#}(t)A^2(t)\,dt.\endmultline\tag 4
$$

Note that $\text{supp}\ \theta^{\#}\subset\subset \hat I$, hence $\theta^{\#}$
is smooth on the whole real line.  To estimate the derivatives of $\theta^{\#}$
we argue as follows.  From the definitions of $t$, $\theta^{\#}$, and
from the WKB Eigenfunction Theorem, we see that
$$\split
\lambda^{-2/3}t\ \text{is\ a\ smooth\ function\ of}\ \frac{x-x(E)}{B},\
\text{with\ first\ derivative}\\ \text{bounded\ below}.\endsplit\tag 5
$$

$$
\theta(x)\cdot\lambda^{-4/3}(\frac{\partial Y}{\partial x})^{-2}\quad
\text{has\ the\ form}\ \lambda^{-4/3}B^2\cdot(\text{smooth\ function\ of}\
\frac{x-x(E)}{B\lambda^{-\varepsilon}}).\tag 6
$$

Hence $\theta^{\#}(t)$ has the form $\lambda^{-4/3}B^2\cdot(\text{smooth\
function\ of}\ \lambda^{-\frac 23+\varepsilon}t)$. 
The $C^\infty$--seminorms of the smooth
functions in (5), (6) are bounded a--priori by the WKB Eigenfunction Theorem.
Hence we have $\theta^{\#}(t)=\lambda^{-4/3}B^2\cdot \theta^{\#\#}(t)$,
with $|(\frac{d}{dt})^m\theta^{\#\#}|\hfill\break
\le C_{\#}^m\lambda^{(\varepsilon-\frac 23)
m}$.  Applying Lemma 4 to $\theta^{\#\#}$, with $R=\lambda^{2/3-\varepsilon}$,
we get
$$
\int_{-\infty}^\infty\theta^{\#}(t)A^2(t)\,dt=\frac 12 \int_0^\infty
\frac{\theta^{\#}(t)\,dt}{t^{1/2}}+\sum\limits_{m=0}^2 \tilde c(m)
\,(\frac{d}{dt})^m\,\theta^{\#}(0)+\text{Error}_2\tag 7
$$
\noindent with $|\text{Error}_2|\le C_{\#}\lambda^{-4/3}B^2\cdot
(\lambda^{2/3-\varepsilon})^{-5/2}=C_{\#}\lambda^{\frac 52 \varepsilon-3}
B^2$.

On the right--hand side of (7), we want to replace
$$
(\frac{d}{dt})^m\theta^{\#}(0)\quad \text{by}\quad (\frac{d}{dt})^m
\theta^{\#}(t_0),\quad\text{with}\quad t_0=\lambda^{2/3}Y(x(E),E).
$$

\noindent That is, $t_0$ is the image of $x(E)$ under $x\mapsto t$.

The WKB Eigenfunction Theorem gives $Y=Y_0+\lambda^{-2}Y_1$ with
$|Y_1|\le C_{\#}$ and $Y_0=0$ at $x=x(E)$.  Hence
$|t_0|<C_{\#}\lambda^{2/3}\cdot\lambda^{-2}=C_{\#}\lambda^{-4/3}$, so
$$\multline
\Big|(\frac{d}{dt})^m\theta^{\#}(t_0)-(\frac{d}{dt})^m\theta^{\#}(0)\Bigr|
\le |t_0|\cdot\text{max}\Big|(\frac{d}{dt})^{m+1}\theta^{\#}\Big|\\
\le C_{\#}^m\lambda^{-4/3}\cdot\lambda^{-4/3}B^2\cdot\lambda^{(\varepsilon
-\frac 23)(m+1)}\le C_{\#}^m\lambda^{-\frac {10}{3}+(m+1)\varepsilon}B^2.
\endmultline
$$

\noindent Putting this into (7), we get
$$
\int_{-\infty}^\infty\theta^{\#}(t)A^2(t)\,dt=\frac 12 \int_0^\infty
\frac{\theta^{\#}(t)\,dt}{t^{1/2}}+\sum\limits_{m=0}^2 \tilde c(m)
(\frac{d}{dt})^m\theta^{\#}(t_0)+\text{Error}_3\tag 8
$$

\noindent with $|\text{Error}_3|\le C_{\#}\lambda^{\frac 52 \varepsilon-3}
B^2$.

We change variable from $t$ back to $x$, obtaining:
$$\multline
\int_{-\infty}^\infty\theta^{\#}(t)A^2(t)\,dt=\frac 12 \int_{Y>0}\Big\{
\frac{\theta(x)\lambda^{-4/3}(\frac{\partial Y}{\partial x})^{-2}}
{(\lambda^{1/3}Y^{1/2})}\Big\}\lambda^{2/3}(\frac{\partial Y}{\partial x})
\,dx+\\
+\sum\limits_{m=0}^2 \tilde c(m)(\lambda^{-2/3}(\frac{\partial Y}
{\partial x})^{-1}\frac{d}{dx})^m\Big\{\theta(x)\lambda^{-4/3}
(\frac{\partial Y}{\partial x})^{-2}\Big\}\Bigm|_{x=x(E)}+
\text{Error}_3,\endmultline
$$

\noindent that is,
$$\multline
\int_{-\infty}^\infty\theta^{\#}(t)A^2(t)\,dt=\frac 12 \int_{Y>0}
\theta(x)\Bigl(\lambda^2(\frac{\partial Y}{\partial x})^2Y\Bigr)^{-1/2}\,dx\\
+\sum\limits_{m=0}^2\tilde c(m)\lambda^{-(\frac{2m+4}{3})}\Bigl((\frac
{\partial Y}{\partial x})^{-1}\frac{d}{dx}\Bigr)^m\Big\{\theta(x)
(\frac{\partial Y}
{\partial x})^{-2}\Big\}\Bigm|_{x=x(E)}+\text{Error}_3,
\endmultline\tag 9
$$

\noindent with $|\text{Error}_3|\le C_{\#}\lambda^{\frac 52 \varepsilon-3}
B^2$.

In (9) we would like to change $Y$ to $Y_0$, where we recall from the WKB
Eigenfunction Theorem that $Y=Y_0+\lambda^{-2}Y_1$.  To see the effect of
this change, define
$$
\Cal T(\tau)=\int_{Y_0+\tau Y_1>0}\theta(x)\lambda^{-1}\Bigl(\frac{\partial Y_0}
{\partial x}+\tau\frac{\partial Y_1}{\partial x}\Bigr
)^{-1}(Y_0+\tau Y_1)^{-1/2}\,
dx,
$$

\noindent for $|\tau|\le c_{\#}$.  The integral on the right--hand side
of (9) is $\Cal T(\lambda^{-2})$, and we want to replace it by $\Cal T(0)$.
So we study the dependence of $\Cal T(\tau)$ on $\tau$.  This is most easily
done by introducing a new independent variable $\xi =Y_0(x,E)+\tau Y_1(x,E)$.
We regard $\xi$ as a function of $x$ and $\tau$.
>From the properties of $Y_0$, $Y_1$
asserted in the WKB Eigenfunction Theorem, we see that
$$
|(\frac{\partial}
{\partial x})^\alpha(\frac{\partial}{\partial\tau})^\gamma\,\xi|\le C_{\#}
^{\alpha\gamma}B^{-\alpha}
$$
\noindent and
$$\frac{\partial\xi}{\partial x}>c_{\#}
B^{-1}\quad\text{for}\quad |x-x(E)|, |\tau|<c_{\#},
$$

\noindent and also that $\xi=0$ when
$x=x(E)$, $\tau=0$.  The implicit function theorem lets us write
$x=x(E)+Bg(\xi,\tau)$ for the solution of $Y_0+\tau Y_1=\xi$, with
$|\partial_\xi^\alpha\partial_\tau^\gamma g(\xi,\tau)|\le C_{\#}^{\alpha\gamma}$
for $|\xi|$, $|\tau|\le c_{\#}^\prime$.

Writing $\Cal T(\tau)$ in terms of the new integration variable $\xi$, we
get the formula
$$
\Cal T(\tau)=\lambda^{-1}\int_{\xi>0}\Big\{\theta(x)\Bigl(\frac{\partial Y_0}
{\partial x}+\tau \frac{\partial Y_1}{\partial x}\Bigr)^{-2}\Bigr\}\Bigm|_
{x=x(E)+Bg(\xi,\tau)}\xi^{-1/2}\,d\xi.\tag 10
$$

Now $|\partial_x^\alpha\partial_\tau^\gamma(\frac{\partial Y_0}
{\partial x}+\tau\frac{\partial Y_1}{\partial x})^{-2}|\le C_{\#}
^{\alpha\gamma}B^{2-\alpha}$ and $|\partial_x\theta(x)|\le C_{\#}
\lambda^{+\varepsilon}B^{-1}$.

Hence $$
\Big|\partial_x\Bigl\{\theta(x)\Bigl(\frac{\partial Y_0}{\partial x}+\tau\frac
{\partial Y_1}{\partial x}\Bigr)^{-2}\Bigr\}\Big|\le C_{\#}\lambda^\varepsilon B
$$

\noindent and 
$$\split
\Big|\partial_\tau\Bigl\{\theta(x)\Bigl(\frac{\partial Y_0}{\partial x}+
\tau\frac{\partial Y_0}{\partial x}\Bigr)^{-2}\Bigr\}\Big|\le C_{\#}B^2,\\
\text{for}\
|x-x(E)|<c_{\#}B,\quad  |\tau|<c_{\#}.\endsplit
$$ 

\noindent Therefore 
$$\split
\frac{\partial}{\partial\tau}\Big[\Big\{\theta(x)\Bigl(\frac{\partial Y_0}
{\partial x}+\tau\frac{\partial Y_1}{\partial x}\Bigr)^{-2}\Big\}\Bigm|_
{x=x(E)+Bg(\xi,\tau)}\Big]\\
=\Big[\partial_x\Bigl\{\theta(x)\Bigl(\frac{\partial Y_0}{\partial x}+\tau\frac
{\partial Y_1}{\partial x}\Bigr)^{-2}\Bigr\}\Big]\cdot B
\frac{\partial g(\xi,\tau)}{\partial\tau}\\
+\Bigl[\partial_\tau\Bigl\{\theta(x)\Bigl(\frac{\partial Y_0}{\partial x}+\tau
\frac{\partial Y_1}{\partial x}\Bigr)^{-2}\Bigr\}\Bigr
]\quad\text{is\ dominated\ by}
\quad C_{\#}\lambda^{\varepsilon} B^2.\endsplit
$$

The integrand in (10) thus has $\tau$--derivative at most $\lambda^
\varepsilon B^2\xi^{-1/2}$.  That integrand is supported in $\{|\xi|<c_{\#}\}$
because of the $\theta$--factor.  Therefore from (10) we get
$|\frac{d}{dt}\Cal T(\tau)|\le C_{\#}\lambda^{\varepsilon-1}B^2$,
for $|\tau|<c_{\#}$.

Hence $|\Cal T(\lambda^{-2})-\Cal T(0)|\le C_{\#}\lambda^{\varepsilon-3}
B^2$, so (9) implies
$$\multline
\int_{-\infty}^\infty\theta^{\#}(t)A^2(t)\,dt=\frac 12 \int_{Y_0>0}
\theta(x)\Bigl(\lambda^2(\frac{\partial Y_0}{\partial x})^2Y_0
\Bigr)^{-1/2}\,dx\\
+\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-(\frac{2m+4}{3})}
\Bigl((\frac{\partial Y}{\partial x})^{-1}\frac{d}{dx}\Bigr)^m\Bigl\{\theta(x)
(\frac{\partial Y}{\partial x})^{-2}\Bigr\}\Bigm|_{x=x(E)}\\
+\text{Error}_4, \quad\text{with}\quad |\text{Error}_4|\le 
C_{\#}\lambda^{\frac 52 \varepsilon-3}B^2. \endmultline\tag 11
$$

Recall from the WKB Eigenfunction Theorem that
$\lambda^2(\frac{\partial Y_0}{\partial x})^2Y_0=E-V(x)$ to order
$\ge N^\prime$ at $x=x(E)$.

Hence in $\text{supp}\ \theta$, $Y_0>0$ is equivalent to $E-V(x)>0$.
Also, Taylor's theorem with remainder and our estimates
for the $x$--derivatives of $Y_0$, $V(x)$ yield
$$\multline
\Big|\lambda^2(\frac{\partial Y_0}{\partial x})^2Y_0-(E-V(x))\Big|\le C_{\#}
S\cdot\Bigl(\frac{x-x(E)}{B}\Bigr)^{N^\prime}\\
\le C_{\#}S\cdot\Bigl(\frac{x-x(E)}{B}\Bigr)^{N^\prime-1}\cdot\Bigl(\frac{
E-V(x)}{S}\Bigr)\le C_{\#}\lambda^{-\varepsilon(N^\prime-1)}(E-V(x))
\endmultline
$$

\noindent for $|x-x(E)|<\lambda^{-\varepsilon}B$.  Thus
$$
\Bigl|\frac{\lambda^2(\frac{\partial Y_0}{\partial x})^2Y_0}{E-V(x)}
-1\Bigr|\le C_{\#}\lambda^{-\varepsilon(N^\prime-1)}\quad \text{in}\
\quad \text{supp}\ \theta.
$$

\noindent Hence  
$$
\frac 12\int_{Y_0>0}\theta(x)\Bigl(\lambda^2(\frac{\partial Y_0}{\partial x}
)^2
Y_0\Bigr)^{-1/2}\,dx=\frac 12 \int_{E-V(x)>0}\theta(x)(E-V(x))^{-1/2}\,dx
+\text{Error}_5,\tag 12
$$
$$\align
|\text{Error}_5|&\le C_{\#}\lambda^{-\varepsilon(N^\prime-1)}\int_{E-V(x)>0}
|\theta(x)|(E-V(x))^{-1/2}\,dx\\
&\le C_{\#}\lambda^{-\varepsilon(N^\prime-1)}
S^{-1/2}B.\tag 13\endalign
$$

Since $S^{-1/2}B=\lambda^{-1}B^2$, (13) may be rewritten as $|\text{Error}_5|
\le C_{\#}\lambda^{-1-\varepsilon(N^\prime-1)}B^2\hfill\break
\le C_{\#}\lambda^{-100}
B^2$, since we take $N>100\varepsilon^{-10}$, $N^\prime=[\varepsilon N/500]$,
$0 \le \varepsilon<1/10$.  Therefore, (11) and (12) yield
$$\multline
\int_{-\infty}^\infty\theta^{\#}(t)A^2(t)\,dt=\frac 12 \int_{E-V(x)>0}
\theta(x)(E-V(x))^{-1/2}\,dx\\
+\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-(\frac{2m+4}{3})}
\Bigl((\frac{\partial Y}{\partial x})^{-1}\frac{d}{dx}\Bigr)^m
\Bigl\{\theta(x)(\frac{\partial Y}
{\partial x})^{-2}\Bigr\}\Bigm|_{x=x(E)}+\text{Error}_6\endmultline
$$
\noindent with $|\text{Error}_6|\le C_{\#}\lambda^{\frac 52 \varepsilon-3}
B^2$.  Putting this into (2), (3), (4) we get
$$\multline
b^{-2}\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx=\frac 12
\int_{E-V(x)>0}\theta(x)(E-V(x))^{-1/2}\,dx\\
+\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-(\frac{2m+4}{3})}
\Bigl((\frac{\partial Y}{\partial x})^{-1}\frac {d}{dx}\Bigr)^m
\Bigl\{\theta(x)(\frac{\partial Y}{\partial x})^{-2}\Bigr\}\Bigm|_
{x=x(E)}+\text{Error}_7\endmultline\tag 14
$$
\noindent where $|\text{Error}_7|\le C_{\#}\lambda^{\frac 52 \varepsilon-3}
B^2+C_{\#}\Lambda^{-\frac{1}{10}N^{\prime\prime}}b^{-2}$.

The WKB Eigenfunction Theorem gives $b^{-2}\le C_{\#}
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}$,
and hypothesis (Hyp 8) dominates the right--hand side by $C_{\#}
\Lambda^KS^{-1/2}B=C_{\#}\Lambda^K\lambda^{-1}B^2$.  Hence
$$
|\text{Error}_7|\le C_{\#}B^2\cdot(\lambda^{\frac 52 \varepsilon-3}
+\Lambda^{K-\frac{1}{10}N^{\prime\prime}}\lambda^{-1}).
$$
\noindent We record our results in the following lemma.
\vglue 1pc

\proclaim{Lemma 5} In the setting of the WKB Eigenfunction Theorem, let $F$
be a real--valued eigenfunction of $H$, with $L^2$--norm $1$ and eigenvalue
$E$ satisfying $|E-E_0|<c_{\#}S_{\text{min}}$, $E\le E_\infty$.  Suppose
$\theta(x)$ is supported in $\{|x-x_{\text{left}}(E)|<\lambda_{\text{left}}
^{-\varepsilon}B_{\text{left}}\}$ and satisfies
$\Bigl|(\frac{d}{dx})^m\theta(x)\Bigr|\le C_{\#}^m(\lambda_{\text{left}}
^{-\varepsilon}B_{\text{left}})^{-m}$.  Then we have
$$\multline
b_{\text{left}}^{-2}\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx=\frac 12
\int_{E-V(x)>0}\frac{\theta(x)\,dx}{(E-V(x))^{1/2}}+\sum\limits_
{m=0}^2\tilde c(m)\lambda_{\text{left}}^{-(\frac{2m+4}{3})}\cdot\\
\cdot \Bigl((\frac{\partial Y_{\text{left}}}{\partial x})^{-1}
\frac{d}{dx}\Bigr)^m\Bigl\{\theta(x)(\frac{\partial Y_{\text{left}}}
{\partial x})^{-2}\Bigr\}\Bigm|_{x=x_{\text{left}}(E)}+\roman{Error},
\endmultline
$$
\noindent with 
$$
|\roman{Error}|\le C_{\#}B_{\text{left}}^2\cdot(\lambda_{\text{left}}^
{\frac 52\varepsilon-3}+\Lambda^{K-\frac{N^{\prime\prime}}{10}}
\lambda_{\text{left}}^{-1}).
$$
\noindent The coefficients $\tilde c(m)$ are as in Lemma 4.\endproclaim

Of course, there is an analogue of Lemma 5 for $\theta(x)$ supported near
$x_{\text{rt}}(E)$.  Once we know that the $\tilde c(m)$ are all zero,
Lemma 5 will simplify.

Next suppose $\theta(x)$ is supported in $I_\ell=[x_{\text{left}}+c_{\#}
\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}},x_{\text{rt}}
-c_{\#}B_{\text{rt}}]$ and satisfies
$$
\Bigl|(\frac{d}{dx})^m\theta(x)\Bigr|\le C_{\#}^m(\lambda^{-\varepsilon}
(x)B(x))^{-m}.\tag 15
$$
\noindent Again we want to compute $\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx$
for eigenfunctions $F$ as in Lemma 5.

The WKB Eigenfunction Theorem shows that
$$
F=b_{\text{left}}\text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{
i\int_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}
(1+u_{\text{left}}(x,E))\Bigr]+\text{Error}_8\tag 16
$$
\noindent in $\text{supp}\ \theta$, with
$$
|\text{Re}\ u_{\text{left}}(x,E)|\le C_{\#}\Lambda^{3\varepsilon-2}
\quad\text{in}\quad \text{supp}\ \theta,\tag 17
$$
$$\split
\Bigl\vert(\frac{d}{dx})^mu_{\text{left}}(x,E)\Bigr\vert\le C_{\#}^m
\lambda_{\text{left}}
^{\frac 32 \varepsilon-1}(\lambda_{\text{left}}^{-\varepsilon}B_{\text
{left}})^{-m}\\
\text{in}\quad \text{supp}\ \theta\cap \{x<x_{\text{left}}(E)+c_{\#}
B_{\text{left}}\}\endsplit\tag 18
$$
$$\split
\Bigl|(\frac{d}{dx})^mu_{\text{left}}(x,E)\Bigr|\le C_{\#}^m\Lambda^{-1}
(B(x))^{-m}\\
\text{in}\quad \text{supp}\ \theta\cap \{x>x_{\text{left}}(E)+c_{\#}
B_{\text{left}}\}.\endsplit\tag"$(18^{\text{bis}})$"
$$
\noindent and
$$\int_{\text{supp}\ \theta}|\text{Error}_8|^2\,dx\le \Lambda^{-N^{\prime
\prime}}.\tag 19
$$
\noindent From (16) we get
$$
F^2\le b_{\text{left}}^2(\text{Re}[\text{etc}])^2(1+\Lambda^{-\frac{1}
{10}N^{\prime\prime}})+\Lambda^{\frac 12 N^{\prime\prime}}|\text{Error}_8
|^2$$
\noindent and
$$
F^2\ge b_{\text{left}}^2(\text{Re}[\text{etc}])^2(1-\Lambda^{-\frac{1}{10}
N^{\prime\prime}})-\Lambda^{\frac 12 N^{\prime\prime}}|\text{Error}_8|^2,
$$
\noindent so that
$$\multline
\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx=\\
b_{\text{left}}^2\int_{I_{\text{BVP}}}\theta(x)\Bigl(\text{Re}
\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{i\int_{\text{left}(E)}^x(E-V(t))^{1/2}
\,dt}}{(E-V(x))^{1/4}}(1+u_{\text{left}}(x,E))\Bigr]\Bigr)^2\,dx\\
+\text{Error}_9\endmultline\tag 20
$$
\noindent with
$$\split
|\text{Error}_9|\le C_{\#}b_{\text{left}}^2\Lambda^{-\frac{1}{10}N
^{\prime\prime}}\int_{I_{\text{BVP}}}|\theta(x)|\frac{dx}
{(E-V(x))^{1/2}}+\Lambda^{-\frac 12 N^{\prime\prime}}\\
\le C_{\#}\Lambda^{-\frac{1}{10}N^{\prime\prime}}.\endsplit\tag 21
$$
\noindent The last inequality uses our estimate of $|b_{\text{left}}|$
from the WKB Eigenfunction Theorem.

Now we expand $(\text{Re}(\zeta))^2=(\frac{\zeta}{2}+
\frac{\overline\zeta}{2})^2=\frac{\zeta^2}{4}+\frac{\overline\zeta^2}
{4}+\frac{|\zeta|^2}{2}$ to write:
$$\align
b_{\text{left}}^2&\int_{I_{\text{BVP}}}\theta(x)\Bigl(\text{Re}
\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{i\int_{x_{\text{left}}(E)}^x
(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\,\,(1+u_{\text{left}}(x,E))\Bigr]\Bigr)^2\,
dx=\\
&\frac 14 b_{\text{left}}^2\int_{I_{\text{BVP}}}\frac
{\theta(x)\,e^{\pm i\frac \pi 2}\,e^{+2i\int_{x_{\text{left}}(E)}
^x(E-V(t))^{1/2}
\,dt}}{(E-V(x))^{1/2}}\,\,(1+u_{\text{left}}(x,E))^2\,dx\\
&+\frac 14 b_{\text{left}}^2\int_{I_{\text{BVP}}}\frac
{\theta(x)e^{\mp i\frac \pi 2}\,e^{-2i\int_{x_{\text{left}}(E)}^x
(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/2}}\,\,(1+\overline{u_{\text{left}}(x,E)})^2\,
dx\\
&+\frac 12 b_{\text{left}}^2\int_{I_{\text{BVP}}}\frac{\theta(x)}
{(E-V(x))^{1/2}}\,|1+u_{\text{left}}(x,E)|^2\,dx\\
&\equiv \text{Term}\ 1(\theta)+\text{Term}\ 2(\theta)+\text{Term}\ 3(\theta).
\tag 22\endalign
$$

We shall show that $\text{Term}\ 1(\theta)$, $\text{Term}\ 2(\theta)$ are
negligibly small.  In fact, using a partition of unity we can write
$\theta=\sum\limits_\nu\theta_\nu$ with each $\theta_\nu$ supported in
$\{|x-x_\nu|<c_{\#}\lambda^{-\varepsilon}(x_\nu)B(x_\nu)\}$,
satisfying $\Big\vert\Bigl(\frac{d}{dx}\Bigr)^m\theta_\nu\Big\vert\le
C_{\#}^m(\lambda^{-\varepsilon}(x_\nu)B(x_\nu))^{-m}$, $x_{\nu+1}-x_\nu
>c_{\#}\lambda^{-\varepsilon}(x_\nu)B(x_\nu)$, and $\text{supp}\ \theta_\nu
\subset \text{supp}\ \theta$.  We then have $\text{Term}\ 1(\theta)=
\sum\limits_\nu\text{Term}\ 1(\theta_\nu)$, $\text{Term}\ 2(\theta)=
\sum\limits_\nu\text{Term}\ 2(\theta_\nu)$, and
$|\text{Term}\ 1(\theta_\nu)|$, $|\text{Term}\ 2(\theta_\nu)|\le C_{\#}
(\lambda(x_\nu))^{-2N}\frac{B(x_\nu)}{S^{1/2}(x_\nu)}b_{\text{left}}^2$ by
the stationary phase Lemma (i.e. Lemma 1 in the section on eigenfunctions
and eigenvalues of Schr\"odinger operators.)  Hence,
$$\split
|\text{Term}\ 1(\theta)|, |\text{Term}\ 2(\theta)|\le b_{\text{left}}^2
\Lambda^{-N}\sum\limits_\nu\frac{B(x_\nu)\lambda^{-\varepsilon}(x_\nu)}
{(S(x_\nu))^{1/2}}\le C_{\#}\Lambda^{-N}b_{\text{left}}^2
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}\\
\le C_{\#}\Lambda^{-N}\endsplit\tag 23
$$
\noindent by the bound for $|b_{\text{left}}|$ in the WKB Eigenfunction
Theorem.

Regarding $\text{Term}\ 3(\theta)$, we have $\Bigl\vert|1-u_{\text{left}}
(x,E)|^2-1\Bigr\vert\le C_{\#}\Lambda^{-2}$ in $\text{supp}\ \theta
\cap \{x>x_{\text{left}}(E)+c_{\#}B_{\text{left}}\}$,
by (17), (18$^{\text{bis}}$). 
Similarly $\Bigl\vert|1-u_{\text{left}}(x,E)|^2-1\Bigr
\vert\le C_{\#}\Lambda^{3\varepsilon-2}$ in
$\text{supp}\ \theta\cap \{x<x_{\text{left}}(E)+c_{\#}B_{\text{left}}\}$
by (17), (18) and the fact that $\lambda_{\text{left}}\ge c_{\#}\Lambda$.

\noindent Hence, 
$$
\text{Term}\ 3(\theta)=\frac 12 \int_{I_{\text{BVP}}}b_{\text{left}}^2
 \frac{\theta(x)\,dx}{(E-V(x))^{1/2}}+\text{Error}_{10},\tag 24
$$
\noindent with
$$\multline
|\text{Error}_{10}|\le C_{\#}\Lambda^{-2}b_{\text{left}}^2\int_{\text{supp}\
\theta} \frac{dx}{(E-V(x))^{1/2}}\\
+C_{\#}\Lambda^{3\varepsilon-2}b_{\text{left}}^2
\int_{\text{supp}\ \theta\cap\{x<x_{\text{left}}(E)+c_{\#}B_{\text{left}}\}}
\frac{dx}{(E-V(x))^{1/2}}\\
\le C_{\#}\Lambda^{3\varepsilon-2}b_{\text{left}}^2\int_{E-V(x)>0}
\frac{dx}{(E-V(x))^{1/2}}\le C_{\#}\Lambda^{3\varepsilon-2},\endmultline
\tag 25
$$
\noindent again using our bound on $|b_{\text{left}}|$.

Putting (23), (24), (25) into (22) and then comparing the result with
(20), (21), we obtain:
$$
\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx=\frac 12 b_{\text{left}}^2
\int_{E-V(x)>0}\frac{\theta(x)\,dx}{(E-V(x))^{1/2}}+\text{Error}_{11},
$$
\noindent with $|\text{Error}_{11}|\le C_{\#}\Lambda^{3\varepsilon-2}$.
Here we wrote the region of integration as $\{E-V(x)>0\}$, which is harmless 
since $E-V>0$ on $\text{supp}\ \theta$.

We record this result as a Lemma.

\vglue 1pc
\proclaim{Lemma 6}  Let $F$ be as in Lemma 5, and suppose $\theta(x)$ is
supported in $[x_{\text{left}}(E)+c_{\#}\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}},x_{\text{rt}}(E)-c_{\#}B_{\text{rt}}]$, satisfying
$\Bigl\vert(\frac{d}{dx})^m\theta(x)\Bigr\vert\le C_{\#}^m(\lambda^{-
\varepsilon}(x)B(x))^{-m}$.  Then
$$
\int_{I_{\text{BVP}}}\theta(x)F^2(x)\,dx=\frac 12 b_{\text{left}}^2
\int_{E-V(x)>0} \frac{\theta(x)\,dx}{(E-V(x))^{1/2}}+\roman{Error}
$$
\noindent with $|\text{Error}|\le C_{\#}\Lambda^{3\varepsilon-2}$.
\endproclaim

Of course there is an analogue of Lemma 6 with the roles of $x_{\text{left}}
(E)$, $x_{\text{rt}}(E)$ interchanged.  At last we are ready to dispose
of $\tilde c(m)$.
\vglue 1pc
\proclaim{Lemma 7}  The constants $\tilde c(0)$, $\tilde c(1)$, $\tilde c(2)$
in Lemma 4 are all zero.\endproclaim
\demo{Proof}  We apply Lemmas 5 and 6 to the harmonic oscillator, and compare
the results with the known behavior of the Hermite functions.  Specifically,
let $H=-\frac {d^2}{dx^2}+\lambda^2x^2$ on the whole line, and put
$E_0=\lambda^2$, $E_\infty=10 \lambda^2$, $I=[-10,+10]$, $B(x)\equiv 1$,
$S(x)\equiv \lambda^2$.  Then our assumptions (Hyp0)$\ldots$(Hyp10) hold
if $\lambda$ is bigger than a (large) universal constant.    The
eigenfunctions of $H$ are the Hermite functions, and there are many
eigenvalues $E$ with $|E-E_0|<c_{\#}^0S_{\text{min}}$, i.e. $|E-\lambda^2|
<<\lambda^2$.  Let $F$ and $E$ be as in Lemmas 5 and 6.  Suppose
$\theta(x)$ is an even $C^\infty$ function of polynomial growth on the
line.  Using a partition of unity we can write
$$
\theta(x)=\theta_{\text{far\ left}}\ +\theta_{\text{Airey\ left}}\
+\theta_{\text{med.\ left}}\ +\theta_{\text{center}}\ +
\theta_{\text{med.\ rt}}\ +\theta_{\text{Airey\ rt}}\ +\theta_{\text{far\
rt}}
$$
\noindent with:
$$\align
&\text{supp}\ \theta_{\text{far\ left}}\ \subset(-\infty,x_{\text{left}}
(E)-c_{\#}\lambda^{-\varepsilon}],\,\, \theta_{\text{far\ rt}}(x)=\theta
_{\text{far\ left}}(-x)\\
&\text{supp}\ \theta_{\text{Airey\ left}} \subset \{|x-x_{\text{left}}(E)|
<\lambda^{-\varepsilon}\},\,\, \theta_{\text{Airey\ rt}}(x)=\theta_
{\text{Airey\ left}}(-x)\\
&\text{supp}\ \theta_{\text{med.\ left}} \subset [x_{\text{left}}(E)+
c_{\#}\lambda^{-\varepsilon},x_{\text{left}}(E)+c_{\#}],\,\,
\theta_{\text{med\ rt}}(x)=\theta_{\text{med\ left}}(-x)\\
&\text{supp}\ \theta_{\text{center}} \subset [x_{\text{left}}(E)+
\frac{c_{\#}}{2}, x_{\text{rt}}(E)-\frac{c{\#}}{2}],\,\, \theta_{\text{center}}
(x)= \theta_{\text{center}}(-x).\endalign
$$

\noindent Since $F$ is either even or odd, we have
$$\multline
\int_{-\infty}^{\infty}\theta(x)F^2(x)\,dx=2\int_{-\infty}^\infty
\theta_{\text{far\ left}}F^2\,dx+2\int_{-\infty}^\infty\theta
_{\text{Airey\ left}}F^2\,dx\\
+2\int_{-\infty}^\infty \theta_{\text{med\ left}}F^2\,dx +
\int_{-\infty}^\infty \theta_{\text{center}}F^2\,dx.
\endmultline
$$

The last two integrals on the right are computed by Lemma 6, and the
integral involving $\theta_{\text{Airey\ left}}$ is computed by Lemma 5.

Since $\int_{-\infty}^\infty F^2\,dx=1$, the rapid decrease of the Hermite
functions in\hfill\break 
$(-\infty,x_{\text{left}}(E)-c_{\#}\lambda^{-\varepsilon}]$
implies that $|\int_{-\infty}^\infty\theta_{\text{far\ left}}F^2\,dx|
\le \lambda^{-10}$.  Consequently, we have the formula
$$\align
\int_{-\infty}^\infty&\theta(x)F^2(x)\,dx=b_{\text{left}}^2\cdot \frac 12
\int_{E-\lambda^2x^2>0}\frac{\theta(x)\,dx}{(E-\lambda^2x^2)^{1/2}}\\
&+2\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-\frac{(2m+4)}{3}}
\Bigl((\frac{\partial Y_{\text{left}}}{\partial x})^{-1}\frac{d}{dx}\Bigr)^m
\Bigl\{\theta(x)(\frac{\partial Y_{\text{left}}}{\partial x})^{-2}\Bigr\}
\Bigm|_{x=x_{\text{left}}(E)}\cdot b_{\text{left}}^2\\
&+\text{Error}\tag 26\endalign
$$
\noindent with
$$
|\text{Error}|\le C_{\#}\lambda^{3\varepsilon-2}+C_{\#}b_{\text{left}}^2
\lambda^{\frac 52 \varepsilon-3}.\tag 27
$$

The WKB Eigenfunction Theorem gives $b_{\text{left}}^{-2}\sim
\int\limits_{E-\lambda^2x^2>0}\frac{dx}{(E-\lambda^2x^2)^{1/2}}\sim
\lambda^{-1}$, so (27) shows that
$$
|\text{Error}|\le C_{\#}\lambda^{3\varepsilon-2}.\tag 28
$$

Equations (26), (28) hold for $\theta(x)$ an even $C^\infty$ function of
polynomial growth.  Let us test (26), (28) in the case $\theta(x)=\text{
polynomial}$.  The computation of the moments $\int_{-\infty}^\infty x^{2m}
F^2(x)\,dx$ is an elementary exercise using raising and lowering operators.
We provide  details for the convenience of the reader.  We can express
$H=\frac 12 (a_0a_1+a_1a_0)$ with $[a_0,a_1]$ a scalar multiple of the
identity, using $a_0=+\frac {d}{dx}+\lambda x$, $a_1=-\frac{d}{dx}+\lambda x$,
or by using $a_0=-\frac{d}{dx}+\lambda x$, $a_1=+\frac{d}{dx}+\lambda x$.
As operators we have $x=\frac{1}{2\lambda}(a_0+a_1)$, so $x^{2m}=
(2\lambda)^{-2m}\sum\limits_{i_1\ldots i_{2m}=0}^1 a_{i_1}a_{i_2}\ldots
a_{i_{2m}}$.

Each monomial $a_{i_1}a_{i_2}\ldots a_{i_{2m}}$ may be written as a linear
combination of terms of the form:
$$\align
a_1^{2m_0}[a_0,a_1]^{m_1}H^{m_2}\quad&\text{with}\quad m_0+m_1+m_2=m,
\quad m_0\ne 0\tag 29\\
a_0^{2m_0}[a_0,a_1]^{m_1}H^{m_2}\quad&\text{with}\quad m_0+m_1+m_2=m,
\quad m_0\ne 0 \tag 30\\
[a_0,a_1]^{m_1}H^{m_2}\quad&\text{with}\quad m_1+m_2=m\tag 31\endalign
$$
In fact, (29) arises from monomials with more $a_1$'s than the $a_0$'s;
(30) arises from monomials with an excess of $a_0$'s; and (31)
arises from monomials with as many $a_1$'s as $a_0$'s.

Hence as operators we have
$$\align
x^{2m}=&(2\lambda)^{-2m}\sum\Sb m_0+m_1+m_2=m\\ m_0\ge 1\endSb
\text{coeff}^\prime(m_0,m_1,m_2)a_1^{2m_0}[a_0,a_1]^{m_1}\,H^{m_2}\\
&+(2\lambda)^{-2m}\sum\Sb m_0+m_1+m_2=m\\ m_0\ge 1\endSb
\text{coeff}^{\prime\prime}(m_0,m_1,m_2)a_0^{2m_0}[a_0,a_1]^{m_1}\,H^{m_2}\\
&+(2\lambda)^{-2m}\sum\limits_{m_1+m_2=m} \text{coeff}(m_1,m_2)
[a_0,a_1]^{m_1}\,H^{m_2}.\tag 32\endalign
$$
\noindent This holds with the same coefficients whether we take
$a_0=\frac{d}{dx}+\lambda x$ and $a_1=-\frac{d}{dx}+\lambda x$ or instead
we set $a_0=-\frac{d}{dx}+\lambda x$ and $a_1=+\frac{d}{dx}-\lambda x$.  Thus
there are two versions of (32), one with $[a_0,a_1]=2\lambda$, and
the other with $[a_0,a_1]=-2\lambda$.

Averaging together the two versions of (32) cancels the terms
$[a_0,a_1]^{m_1}H^{m_2}$ with $m_1$ odd, yielding
$$\align
x^{2m}=&\sum\Sb m_0+m_1+m_2=m\\ m_0\ge 1\endSb \text{coeff}^\prime
(m_0,m_1,m_2)a_1^{2m_0}\,\lambda^{m_1-2m}\,H^{m_2}\\
&+\sum\Sb m_0+m_1+m_2=m\\ m_0\ge 1\endSb \text{coeff}^{\prime\prime}
(m_1,m_1,m_2)a_0^{2m_0}\,\lambda^{m_1-2m}\,H^{m_2}\\
&+\sum\Sb m_1+m_2=m\\ m_1\ \text{even}\endSb \text{coeff}(m_1,m_2)\,
\lambda^{m_1-2m}\,H^{m_2}.\tag 33\endalign
$$

We apply (33) to compute the inner product $<x^{2m}F,F>$ with $F$ a
Hermite function, $HF=EF$.

The first two sums on the right side of (33) do not affect the inner
product, since
$$<a_1^{2m_0}\,\lambda^{m_1-2m}H^{m_2}F,F> =\lambda^{m_1-2m}E^{m_2}
<a_1^{2m_0}F,F> = 0 \quad\text{for}\quad m_0\ge 1,
$$
\noindent and similarly
$$
<a_0^{2m_0}\,\lambda^{m_1-2m}H^{m_2}F,F>\ =\lambda^{m_1-2m}E^{m_2}
<a_0^{2m_0}F,F> = 0\quad\text{for}\quad m_0\ge 1.
$$
\noindent Therefore (33) gives
$$\align
<x^{2m}F,F>&=\sum\Sb m_1+m_2=m\\ m_1\ \text{even}\endSb \text{coeff}
(m_1,m_2)\lambda^{m_1-2m}E^{m_2}\\
&=\sum\Sb m_1+m_2=m\\ m_1\ \text{even}\endSb \text{coeff}(m_1,m_2)\lambda
^{-m_1}(\frac{E}{\lambda^2})^{m_2}.\endalign
$$

In particular, for $\frac{E}{\lambda^2} \sim 1$ we get
$$
\int_{-\infty}^\infty x^{2m}F^2(x)\,dx=\text{coeff}(m)(\frac{E}{\lambda^2})^m
+O(\lambda^{-2})\quad\text{for\ each}\ m.\tag 34
$$
\noindent On the other hand, from (26), (28) we get trivially
$$
\int_{-\infty}^\infty x^{2m}F^2(x)\,dx=\int_{E-\lambda^2x^2>0} (\frac 12 
b_{\text{left}}^2)\frac{x^{2m}\,dx}{(E-\lambda^2x^2)^{1/2}}+O(\lambda^{-1/3}),
$$
\noindent since $b_{\text{left}}^2 \sim \lambda$ and $\frac{\partial Y_{\text
{left}}}{\partial x}$ satisfies $|\partial_x^\alpha Y_{\text{left}}|\le
C_{\#}^\alpha$, $\frac{\partial Y_{\text{left}}}{\partial x}\ge c_{\#}$
by the WKB Eigenfunction Theorem.

The change of variable $x=\frac{E^{1/2}}{\lambda}\overline x$ gives
$$\int_{E-\lambda^2x^2>0}\frac{x^{2m}\,dx}{(E-\lambda^{2}x^2)^{1/2}}=
\hat{\text{coeff}}(m)\lambda^{-1}\cdot\Bigl(\frac{E}{\lambda^2}\Bigr)^m,\tag 36
$$

\noindent so the previous equation implies
$$
\int_{-\infty}^\infty x^{2m}F^2(x)\,dx=\Bigl(\frac 12b_{\text{left}}^2
\lambda^{-1}\Bigr)\Bigl(\frac{E}{\lambda^2}\Bigr)^{m}\hat{\text{coeff}}
(m)+O(\lambda^{-1/3})\quad\text{for\ each}\ m.
$$

\noindent Comparing this with (34) and recalling that $\frac{E}{\lambda^2}\sim
1$, we get
$$
\Bigl(\frac 12b_{\text{left}}^2\lambda^{-1}\Bigr)\hat{\text{coeff}}(m)
=\text{coeff}(m)+O(\lambda^{-1/3}) \quad\text{for\ each}\ m.\tag 37
$$

Taking $m=0$ and recalling that $\int_{-\infty}^\infty F^2\,dx=1$, we get
$\hat{\text{coeff}}(0)=\pi$ by (36), and $\text{coeff}(0)=1+O(\lambda^{-2})$
by (34).  Hence (37) for $m=0$ gives $(\frac 12b_{\text{left}}^2\lambda^{-1})
=\frac 1\pi+O(\lambda^{-1/3})$.  So (37) may be rewritten in the form
$$
\text{coeff}(m)=\frac 1\pi \hat{\text{coeff}}(m)+O(\lambda^{-1/3}).
$$

\noindent Here, $\text{coeff}(m)$ and $\hat{\text{coeff}}(m)$ are universal
constants, while $\lambda$ may be taken arbitrarily large.  Consequently,
$$
\text{coeff}(m)=\frac 1\pi \hat{\text{coeff}}(m).
$$

\noindent This equation and (34) yield
$$
\int_{-\infty}^\infty x^{2m}F^2(x)\,dx=\frac 1\pi \hat{\text{coeff}}
(m)\cdot\Bigl(\frac{E}{\lambda^2}\Bigr)^m+O(\lambda^{-2}),
$$
\noindent which is equivalent to
$$
\int_{-\infty}^\infty x^{2m}F^2(x)\,dx=\frac \lambda\pi\int_{
E-\lambda^2x^2>0}\frac{x^{2m}\,dx}{(E-\lambda^2x^2)^{1/2}}+O(\lambda^{-2})
\quad\text{for\ each}\ m,
$$

\noindent by virtue of (36).  This implies
$$
\int_{-\infty}^\infty\theta(x)F^2(x)\,dx=\frac{\lambda}{\pi}\int_{
E-\lambda^2x^2>0}\frac{\theta(x)\,dx}{(E-\lambda^2x^2)^{1/2}}+O(\lambda^{-2})
$$

\noindent for $\theta(x)=\sum\limits_{k=0}^{10}a_k(E,\lambda)x^{2k}$, provided
$|a_k(E,\lambda)|=O(1)$.  Comparing this equation with (26), (28), we
conclude that
$$\multline
\frac \lambda\pi\int_{E-\lambda^2x^2>0}\frac{\theta(x)\,dx}
{(E-\lambda^2x^2)^{1/2}}=\frac 12 b_{\text{left}}^2\int_{E-\lambda^2x^2>0}
\frac{\theta(x)\,dx}{(E-\lambda^2x^2)^{1/2}}\\
+2\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-\frac{(2m+4)}{3}}b_{\text{left}}^2
\Bigl(\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-1}
\frac{d}{dx}\Bigr)^m\Bigl\{\theta(x)\Bigl(\frac{\partial Y_{\text{left}}}
{\partial x}\Bigr)^{-2}\Bigr\}\Bigm|_{x=x_{\text{left}}(E)}\\
+O(\lambda^{3\varepsilon-2}).\endmultline\tag"{(37a)}"
$$

We first apply (37a) to $\theta_0(x)=(x^2-\frac{E}{\lambda^2})^4$, which
vanishes to $4^{\text{th}}$ order at $x=x_{\text{left}}(E)=-
(\frac{E}{\lambda^2})^{1/2}$, so that the second term on the right in (37a)
equals zero.  Thus, (37a) for $\theta_0$ yields
$$
b_{\text{left}}^2=\frac{2\lambda}{\pi}+O(\lambda^{3\varepsilon-1}),
\quad\text{since}\ \int_{E-\lambda^2x^2>0}\frac{\theta_0(x)\,dx}
{(E-\lambda^2x^2)^{1/2}}\sim \lambda^{-1}.\tag"{(37b)}"
$$

\noindent Putting (37b) into (37a), we get
$$\split
2b_{\text{left}}^2\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-\frac{(2m+4)}
{3}}\Bigl(\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-1}
\frac{d}{dx}\Bigr)^m \Bigl\{\theta(x)\Bigl(\frac{\partial Y_{\text{left}}}
{\partial x}\Bigr)^{-2}\Big\}\Bigm|_{x=x_{\text{left}}(E))}\\
=O(\lambda^{3\varepsilon-2})\endsplit
$$

\noindent for $\theta(x)=\sum\limits_{k=0}^{10}a_k(E,\lambda)x^{2k}$
with $|a_k(E,\lambda)|=O(1)$.  That is,

$$\split
\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-(\frac{2m+4}{3})}\Bigl((\frac
{\partial Y_{\text{left}}}{\partial x})^{-1}\frac{d}{dx}\Bigr)^m
\Bigl\{\theta(x)\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-2}
\Bigr\}\Bigm|_{x=x_{\text{left}}(E)}\\
=O(\lambda^{3\varepsilon-3}).\endsplit\tag 38
$$
\noindent We use (38) for special $\theta(x)$ picked to make the computation
of $Y_{\text{left}}(x,E)$ unnecessary.  First take $\theta(x)\equiv 1$
in (38) to obtain
$$
\tilde c(0)\lambda^{-4/3}(\frac{\partial Y_{\text{left}}}{\partial x}
)^{-2}\Bigm|_{x=x_{\text{left}}(E)}=O(\lambda^{-2}).
$$
\noindent Since $\frac{\partial Y_{\text{left}}}{\partial x}\le C_{\#}$
near $x_{\text{left}}(E)$, this implies $\tilde c(0)=0$.  Next take
$\theta(x)=(\frac{E}{\lambda^2}-x^2)$.  At $x=x_{\text{left}}(E)$
we have $\theta=0$, hence $(\frac{\partial Y_{\text{left}}}{\partial x})^{-1}
\frac{d}{dx}
\Bigl\{\theta(x)\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-2}
\Bigr\}\Bigm|_{x=x_{\text{left}}(E)}=\Bigl(\frac{\partial Y_{\text{left}}}
{\partial x}\Bigr)^{-3}\frac{d\theta}{dx}\Bigm|_{x=x_{\text{left}}(E)}
=-2x \Bigl(\frac{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-3}\Bigm|_
{x=x_{\text{left}}(E)}=\hfill\break
+2(\frac{E}{\lambda^2})^{1/2}\cdot \Bigl(\frac
{\partial Y_{\text{left}}}{\partial x}\Bigr)^{-3}\Bigm|_{x=x_{\text{left}}(E)}$.
For this $\theta$, (38) yields
$$
\tilde c(1)\lambda^{-2}\cdot\Bigl[2(\frac{E}{\lambda^2})^{1/2}\cdot
\Bigl(\frac{\partial Y_{\text{left}}}{\partial x}
\Bigm|_{x=x_{\text{left}}(E)}\Bigr)^{-3}\Bigr]=O(\lambda^{-8/3}),\tag 39
$$
\noindent since we already know that $\tilde c(0)=0$.  The quantity in
brackets has order of magnitude $1$, so (39) implies $\tilde c(1)=0$.
Finally, take $\theta(x)=(\frac{E}{\lambda^2}-x^2)^2$.  At
$x=x_{\text{left}}(E)$ we have $\theta=0$, $\theta^\prime=0$, $\theta^
{\prime\prime}=\frac{8E}{\lambda^2}$, and therefore
$$
\Bigl((\frac{\partial Y_{\text{left}}}{\partial x})^{-1}\frac{d}{dx}\Bigr)^
{2}\Bigl\{\theta(x)(\frac{\partial Y_{\text{left}}}{\partial x})^{-2}
\Bigr\}\Bigm|_{x=x_{\text{left}}(E)}=\Bigl\{(\frac{\partial Y_{\text{left}}}
{\partial x})^{-4}\cdot \frac{8E}{\lambda^2}\Bigr\}\Bigm|_{x=x_{\text{left}}
(E)}.
$$ 
\noindent Since $\tilde c(0)=\tilde c(1)=0$, equation (38) for this $\theta$
yields
$$
\Bigl[\frac{8E}{\lambda^2}(\frac{\partial Y_{\text{left}}}{\partial x})^{-4}
\Bigm|_{x=x_{\text{left}}(E)}\Bigr]\cdot \tilde c(2)\lambda^{-8/3}=
O(\lambda^{2\varepsilon-3}).\tag 40
$$

The quantity in brackets has order of magnitude $1$, so (40) implies
that $\tilde c(2)=0$ as well.  The proof of the lemma is complete.
$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary}  Under the hypotheses of Lemma 5 we have
$$
\int_{I_{\text{BVP}}} \theta(x)F^2(x)\,dx=\frac 12 b_{\text{left}}^2
\int_{E-V(x)>0} \frac{\theta(x)\,dx}{(E-V(x))^{1/2}}+\roman{Error},\tag 41
$$
\noindent with $|\text{Error}|\le C_{\#}\Lambda^{\frac 52 \varepsilon-2}$.
\endproclaim
\demo{Proof}  Since $\tilde c(0)=\tilde c(1)=\tilde c(2)=0$, Lemma 5 gives
(41), with $|\text{Error}|\le \hfill\break
C_{\#}b_{\text{left}}^2B_{\text{left}}^2
\lambda_{\text{left}}^{-1}(\lambda_{\text{left}}^{\frac 52 \varepsilon-2}
+\Lambda^{K-\frac{N^{\prime\prime}}{10}})$.

>From the WKB Eigenfunction Theorem we get
$$
b_{\text{left}}^{-2}\ge c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{1/2}(x)}\ge  c_{\#}\frac{B_{\text{left}}}
{S_{\text{left}}^{1/2}}
= c_{\#}\frac{B_{\text{left}}^2}{\lambda_{\text{left}}},
$$

\noindent and therefore $b_{\text{left}}^2B_{\text{left}}^2
\lambda_{\text{left}}^{-1}\le C_{\#}$.  Putting this into our previous
estimate for the error, we get
$|\text{Error}|\le C_{\#}(\lambda_{\text{left}}^{\frac 52 \varepsilon-2}+
\Lambda^{K-\frac{N^{\prime\prime}}{10}})$.  Since $\lambda_{\text{left}}
\ge \Lambda$ and $K-\frac{N^{\prime\prime}}{10}<-2$, the Corollary follows.
$\qquad\blacksquare$

Now we can compute the normalization constants $b_{\text{left}}$, $b_{\text{rt}}$ in the WKB Eigenfunction Theorem.  We write a partition of unity
$$
1=\theta_{\text{far\ left}}\ +\theta_{\text{Airey\ left}}\ +\theta
_{\text{medium\ left}}\ +\theta_{\text{center}}\ +\theta_{\text{medium\ rt}}\
+\theta_{\text{Airey\ rt}}\ +\theta_{\text{far\ rt}},$$
\noindent with:
$$
\text{supp}\ \theta_{\text{far\ left}}\ \subset 
(-\infty,x_{\text{left}}(E)-
c_{\#}\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}],\,\,
|\theta_{\text{far\ left}}|\le C_{\#};
$$ 
\noindent $\theta_{\text{Airey\ left}}$
satisfying the hypotheses of Lemma 5;
$$
\text{supp}\ \theta_{\text{medium\ left}}\ \subset [x_{\text{left}}
(E)+c_{\#}\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}},x_{\text{left}}
(E)+c_{\#}B_{\text{left}}],
$$

\noindent $\theta_{\text{medium\ left}}$ satisfying the hypotheses of Lemma 6;
$$
\text{supp}\ \theta_{\text{center}}\subset [x_{\text{left}}
(E)+\frac{c_{\#}}{2}
B_{\text{left}}, x_{\text{rt}}(E)-\frac{c_{\#}}{2}B_{\text{rt}}],
$$
\noindent $\theta_{\text{center}}$ satisfying the hypotheses of Lemma 6;
$$\align
&\theta_{\text{medium\ rt}}\ \quad \text{analogous\ to}\ \theta_{\text{medium\
left}}\ ;\\
&\theta_{\text{Airey\ rt}}\ \quad \text{analogous\ to}\ \theta_{\text{Airey\
left}}\ ;\\
&\theta_{\text{far\ rt}}\  \quad \text{analogous\ to}\ 
\theta_{\text{far\ left}}\ .\endalign
$$

The WKB Eigenfunction Theorem gives
$$
|\int_{I_{\text{BVP}}} \theta_{\text{far\ left}}\ F^2\,dx|,
|\int_{I_{\text{BVP}}} \theta_{\text{far\ rt}}\
F^2\,dx|\le C_{\#}\Lambda^{-100}. \tag 42
$$

The Corollary to Lemma 7 gives
$$
\int_{I_{\text{BVP}}}\theta_{\text{Airey\ left}}\ F^2\,dx=\frac 12
b_{\text{left}}^2\int_{E-V(x)>0} \frac{\theta_{\text{Airey\ left}}\ (x)
\,dx}{(E-V(x))^{1/2}}+\ \text{Error}\tag 43
$$
\noindent with $|\text{Error}|\le C_{\#}\Lambda^{\frac 52 \varepsilon-2}$.

Analogously we have
$$
\int_{I_{\text{BVP}}}\theta_{\text{Airey\ rt}}\ F^2\,dx=\frac 12 b_{\text{rt}}^2
\int_{E-V(x)>0} \frac{\theta_{\text{Airey\ rt}}\ (x)\,dx}{(E-V(x))^{1/2}}
+\ \text{Error}\ ,\tag 44
$$
\noindent with $|\text{Error}|\le C_{\#}\Lambda^{\frac 52 \varepsilon-2}$.
We want to change $b_{\text{rt}}$ to $b_{\text{left}}$.  The WKB
Eigenfunction Theorem gives $|b_{\text{rt}}|=|b_{\text{left}}|\cdot
(1+\text{error})$ with $|\text{error}| \le C_{\#}\Lambda^{-2}$.  Hence
$$\align
\Bigl|&\frac 12 b_{\text{rt}}^2\int_{E-V(x)>0} \frac{\theta_{\text{Airey\
rt}}\ (x)\,dx}{(E-V(x))^{1/2}}-\frac 12 b_{\text{left}}^2 \int_
{E-V(x)>0} \frac{\theta_{\text{Airey\ rt}}\, (x)\,dx}{(E-V(x))^{1/2}}\Bigr|\\
&\le C_{\#}\Lambda^{-2}b_{\text{left}}^2\int_{x_{\text{left}}(E)}
^{x_{\text{rt}}(E)} \frac{dx}{(E-V(x))^{1/2}}\le C_{\#}\Lambda^{-2}
b_{\text{left}}^2 \int_{x_{\text{left}}}^{x_{\text{rt}}} \frac {dx}
{S^{1/2}(x)}\\
&\le C_{\#}\Lambda^{-2},\endalign
$$
\noindent by the bound on $b_{\text{left}}$ given by the WKB Eigenfunction
Theorem.  Putting this into (44), we get
$$\split
\int_{I_{\text{BVP}}}\theta_{\text{Airey\ rt}}\ F^2\,dx=\frac 12
b_{\text{left}}^2\int_{E-V(x)>0}\frac{\theta_{\text{Airey\ rt}}\, (x)\,dx}
{(E-V(x))^{1/2}}+\text{Error},\\
\text{with}\  |\text{Error}|\le C_{\#}\Lambda^{\frac 52 \varepsilon-2}.
\endsplit\tag 45
$$

Lemma 6 yields 
$$\split
\int_{I_{\text{BVP}}}\theta_{\text{medium\ left}} F^2\,dx=\frac 12
b_{\text{left}}^2\int_{E-V(x)>0} \frac{\theta_{\text{medium\ left}}\
(x)\,dx}{(E-V(x))^{1/2}}+\text{Error},\\
\text{with}\ |\text{Error}|\le C_{\#}\Lambda^{3\varepsilon-2}.\endsplit
\tag 46
$$

\noindent Analogously, we have
$$\split
\int_{I_{\text{BVP}}}\theta_{\text{medium\ rt}}\ F^2\,dx=\frac 12
b_{\text{rt}}^2\int_{E-V(x)>0}\frac{\theta_{\text{medium\ rt}}\,(x)\,dx}
{(E-V(x))^{1/2}}\ +\text{Error}\\
\text{with}\ |\text{Error}|\le C_{\#}\Lambda^{3\varepsilon-2}.\endsplit
\tag 47
$$

As in the proof of (45), we have
$$\multline
\Big|\frac 12 b_{\text{rt}}^2\int_{E-V(x)>0}\frac{\theta_{\text{medium\ rt}}\,
(x)\,dx}{(E-V(x))^{1/2}}-\frac 12 b_{\text{left}}^2\int_{E-V(x)>0}
\frac{\theta_{\text{medium\ rt}}\,(x)\,dx}{(E-V(x))^{1/2}}\Big|\\
\le C_{\#}\Lambda^{-2}b_{\text{left}}^2\int_{x_{\text{left}}(E)}^
{x_{\text{rt}}(E)} \frac {dx}{(E-V(x))^{1/2}}\le C_{\#}
\Lambda^{-2}b_{\text{left}}^2 \int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac {dx}{S^{1/2}(x)}\le C_{\#}\Lambda^{-2},\endmultline
$$
\noindent so that (47) implies
$$\split
\int_{I_{\text{BVP}}}\theta_{\text{medium\ rt}}\, F^2\,dx=\frac 12
b_{\text{left}}^2\int_{E-V(x)>0} \frac {\theta_{\text{medium\ rt}}\, (x)\,dx}
{(E-V(x))^{1/2}} +\ \text{Error},\\
\text{with}\ |\text{Error}|\le C_{\#}\Lambda^{3\varepsilon-2}.\endsplit
\tag 48
$$
\noindent Finally, Lemma 6 yields
$$\split
\int_{I_{\text{BVP}}}\theta_{\text{center}}\, F^2\,dx=\frac 12
b_{\text{left}}^2\int_{E-V(x)>0} \frac{\theta_{\text{center}}\, (x)\,dx}
{(E-V(x))^{1/2}}+\ \text{Error},\\
\text{with}\ |\text{Error}|\le C_{\#}\Lambda^{3\varepsilon-2}.\endsplit
\tag 49
$$

Adding (42), (43), (45), (46), (48), (49) and recalling that
$\theta_{\text{Airey\ left}}\ +\theta_{\text{medium\ left}}\hfill\break
+\theta_{\text{center}}\ +\theta_{\text{medium\ rt}}+
\theta_{\text{Airey\ rt}}\ =1$
on $\{E-V(x)>0\}$, we obtain the formula:
$$\split
\int_{I_{\text{BVP}}}F^2\,dx=\frac 12 b_{\text{left}}^2
\int_{E-V(x)>0}\frac{dx}{(E-V(x))^{1/2}}+\ \text{Error},\\
\text{with}\ |\text{Error}| \le C_{\#}\Lambda^{3\varepsilon-2}.\endsplit
$$
\noindent Since $\int_{I_{\text{BVP}}}F^2\,dx=1$, this means that
$|b_{\text{left}}|=\Bigl(\frac 12 \int_{E-V(x)>0}\frac {dx}
{(E-V(x))^{1/2}}\Bigr)^{-1/2}\cdot (1+\ \text{Error})$,
with $|\text{Error}|\le C_{\#}\Lambda^{3\varepsilon-2}$.  The WKB
Eigenfunction Theorem tells us that
$\Bigl||b_{\text{rt}}/b_{\text{left}}|-1\Bigr|\le C_{\#}\Lambda^{-2}$,
so we get the same result for $b_{\text{rt}}$ as for $b_{\text{left}}$.

This completes our computation of the normalizing contants.  We record
our result in the following theorem.
\vglue 1pc
\proclaim{WKB Normalization Theorem}  The constants $b_{\text{left}}$,
$b_{\text{rt}}$ in the WKB Eigenfunction Theorem satisfy
$$
\Big|b_{\text{left}}^2\cdot \Bigl(\frac 12 \int_{E-V(x)>0}\frac {dx}
{(E-V(x))^{1/2}}\Bigr)-1\Big|\le C_{\#}\Lambda^{3\varepsilon-2}\quad
\text{and}
$$
$$
\Big|b_{\text{rt}}^2\cdot \Bigl(\frac 12 \int_{E-V(x)>0}\frac {dx}
{(E-V(x))^{1/2}}\Bigr)-1\Big|\le C_{\#}\Lambda^{3\varepsilon-2},
$$
\noindent with $C_{\#}$ depending only on $\varepsilon$, $K$, $N$, $c$,
$C$, $c_1$, $c_2$, $C_\alpha$ in (Hyp 0)$\ldots$(Hyp 5).
\endproclaim



%Subject: Section 8

\pageno 133
\magnification \magstep1
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\hsize 6 truein
\vsize 9truein
\baselineskip 20pt
\hoffset .15truein
\catcode`\@=11
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\def\kbar{k{\kern -4.5pt\raise 3pt\hbox{-}}}

\centerline{\bf{Eigenvalues near the Minimum of the
Potential}}
\smallskip

In this section we study the eigenvalues of $H=-\frac
{d^2}{dx^2}+V(x)$ near the minimum of the potential $V$.
We shall see that the WKB formulas for eigenvalues $E$
remain highly accurate as $E$ nears $\text{min}\ V$,
even though the WKB description of eigenfunctions loses
precision.  This phenomenon is familiar for the harmonic
oscillator, all of whose eigenvalues are perfectly
described by the semiclassical approximation.  To prove
our results, we transform $H$ to the harmonic oscillator
by a (formal power series) change of variable $y=y(x)$.
The precise statement of the problem is as follows.

Let $K,\varepsilon,N>0$ be given with $\varepsilon N>
100$.  Let $V(x)$ be a potential defined on a (possibly
unbounded) interval $I_{\text{BVP}}$.  Let $S,B>0$
be given numbers, and let $x_0 \in I_{\text{BVP}}$
be given.  Define $\lambda=S^{1/2}B$.  Let $E_\infty$
be a given energy, with $E_\infty\ge V(x_0)$.  We make
the following assumptions.
\roster
\item"{(H1*)}" $|(\frac{d}{dx})^\alpha V(x)|\le C_\alpha
SB^{-\alpha}$ in $I=\{x \in \Bbb R\Bigm||x-x_0|<cB\}
\subset I_{\text{BVP}}$
\item"{(H2*)}" $\frac{d^2}{dx^2}V\ge cSB^{-2}$ in $I$
\item"{(H3*)}" $V^\prime(x_0)=0$
\item"{(H4*)}" For $x \in I_{\text{BVP}}\backslash I$
we have $V(x)\ge \text{min}\{E_\infty,V(x_0)+c
\lambda^{-2\varepsilon}S\}$
\item"{(H5*)}" For $x \in I_{\text{BVP}}$ with $|x-x_0|
>\frac 12 \lambda^KB$, we have $V(x)\ge E_\infty+
\frac{1000}{|x-x_0|^2}$
\item"{(H6*)}" $\lambda$ is bounded below by a constant
depending only on $C_\alpha$, $c$, $\varepsilon$, $K$
in (H1*)$\ldots$(H5*), and on $N$.
\endroster

Let $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_{\text{BVP}})$
with Dirichlet or Neumann conditions.  Our goal is
to understand the eigenvalues $E$ of $H$ in the range
$V(x_0)\le E\le \text{min}(E_\infty,V(x_0)+c\lambda^
{-2\varepsilon} S)$.  We begin with two elementary
observations.  First of all, the continuous spectrum
of $H$ is contained in $[E_\infty,\infty)$,  since
$V\ge E_\infty$ outside a bounded subinterval of 
$I_{\text{BVP}}$.  Secondly, for $E\le \text{min}
(E_\infty,V(x_0)+c\lambda^{-2\varepsilon}S)$ an eigenvalue
of $H$, the eigenfunction $F(x)$ is strongly concentrated
on $\{|x-x_0|<\lambda^{-\varepsilon}B\}$.  This will
make it possible to study the eigenvalues of $H$ by
Taylor--expanding $V(x)$ about $x=x_0$.  Specifically,
we have the following result.

\vglue 1pc
\proclaim{Lemma 1}  Let $F$ be an eigenfunction of $H$
with $L^2$--norm $1$ and with eigenvalue $E\le \text{min}
(E_\infty,V(x_0)+c_{\#}\lambda^{-2\varepsilon}S)$.  Then
$$
\int_{I_{\text{BVP}}\cap\{|x-x_0|>\lambda^{-\varepsilon}
B\}}|F|^2\,dx\le C_{\#}\lambda^{-N}.
$$
\noindent In this section, $c_{\#}$, $C_{\#}$, $C_{\#}^
{\alpha\beta}$ etc. denote constants depending only on
$K$, $\varepsilon$, $N$, $C_\alpha$, $c$ in
(H1*)$\ldots$(H5*).
\endproclaim

\demo{Sketch of Proof}  We follow the discussion of
the Agmon lemma in the section on eigenvalues and
eigenfunctions of Schr\"odinger operators.  We have
$\{x \in I_{\text{BVP}}\mid V(x)<E\}=(x_-,x_+)$
with $|x_\pm-x_0|<c_{\#}\lambda^{-\varepsilon}B$,
and the proof of the Agmon lemma gives us
$$\split
\int_{I_{\text{BVP}}\backslash(x_-,x_+)}\frac {e^\varphi}
{2} (|\frac{dF}{dx}|^2+(V-E)|F|^2)\,dx\\
\le \int_{x_-}^{x_+} (E-V)|F|^2\,dx\endsplit\tag 1
$$

\noindent with $\varphi(x)=\int_{x}^{x_-}(V-E)^{1/2}$
for $x \le x_-$, $\varphi(x)=\int_{x_+}^x(V-E)^{1/2}$
for $x \ge x_+$. \enddemo

We establish lower bounds for $\varphi$.  Say $x<x_0-
\lambda^{-\varepsilon}B$.  Then
$$\align
\varphi(x)\ge\int_{x_0-\lambda^{-\varepsilon}B}^{x_-}
(V-E)^{1/2}
\,dy&>c_{\#}\int_{x_0-\lambda^{-\varepsilon}B}^
{x_0-\frac 12 \lambda^{-\varepsilon}B}(\lambda^{-2
\varepsilon}S)^{1/2}\,dy\\
&\ge c_{\#}\lambda^{-2\varepsilon}BS^{1/2}=c_{\#}
\lambda^{1-2\varepsilon},
\endalign
$$
\noindent since $V-E\ge (V(x_0)+\frac 12 cSB^{-2}
(x-x_0)^2)-(V(x_0)+c_{\#}\lambda^{-2\varepsilon}S)\ge
c_{\#}\lambda^{-2\varepsilon}S$ for $|x-x_0|\sim
\lambda^{-\varepsilon}B$.

If $x<x_0-\lambda^KB$, then
$$\align
\varphi(x)&\ge \int_x^{x_0-\lambda^KB}(V-E)^{1/2}\,dy+
\int_{x_0-\lambda^{-\varepsilon}B}^{x_-}(V-E)^{1/2}\,dy\\
&\ge \int_x^{x_0-\lambda^KB}(V-E)^{1/2}\,dy+c_{\#}
\lambda^{1-2\varepsilon}\endalign
$$
\noindent  (by the previous chain of inequalities), and
the integral on the right exceeds\hfill\break $\int_x^{x_0-\lambda^KB}
\frac{10}{|y-x_0|}\,dy=10\ \ell n(\frac{|x-x_0|}
{\lambda^KB})$.  Hence:
$$\gather
\varphi(x)\ge c_{\#}\lambda^{1-2\varepsilon}\quad\text{if}
\quad x<x_0-\lambda^{-\varepsilon}B\\
V(x)-E\ge
c_{\#}\lambda^{-2\varepsilon}S\quad\text{if}\quad
|x-x_0|\sim \lambda^{-\varepsilon}B\\
\varphi(x)\ge c_{\#}\lambda^{1-2\varepsilon}+10\ \ell n
(\frac{|x-x_0|}{\lambda^KB})\quad \text{if}\quad
x<x_0-\lambda^KB.\endgather
$$
\noindent In $(x_-,x_+)$ we have $E-V\le C_{\#}S$, so
(1) yields the following estimates:
$$
\int_{x_0-2\lambda^{-\varepsilon}B}^{x_0-\lambda^{-
\varepsilon}B}|F|^2\,dy\le C_{\#}\lambda^{2\varepsilon}
\exp(-c_{\#}\lambda^{1-2\varepsilon})\int_{x_-}^{x_+}
|F|^2\,dy\tag 2
$$
$$
\int_{I_{\text{BVP}}\cap(-\infty,x_0-\lambda^{-\varepsilon}
B)}\Big|\frac{dF}{dy}\Big|^2\,dy\le C_{\#}S\text{exp}
(-c_{\#}\lambda^{1-2\varepsilon})\int_{x_-}^{x_+}
|F|^2\,dy\tag 3
$$
$$\split
\int_{I_{\text{BVP}}\cap(-\infty,x_0-\lambda^KB)}
\Bigl[\Bigl(\frac{|y-x_0|}{\lambda^KB}\Bigr)^{10}\text{exp}
(c_{\#}\lambda^{1-2\varepsilon}\Bigr)\Bigr](V-E)|F|^2\,dy\\
\le C_{\#}S\int_{x_-}^{x_+}|f|^2\,dy.\endsplit\tag 4
$$
\noindent From (2) and (3) we deduce
$$
\int_{I_{\text{BVP}}\cap[x_0-\lambda^KB,x_0-\lambda^{
-\varepsilon}B]}|F|^2\,dy\le C_{\#}\lambda^{-N}\int_
{x_-}^{x_+}|F|^2\,dy.\tag 5
$$
\noindent From (4) and the inequalities
$$\align
\Bigl(\frac{|y-x_0|}{\lambda^kB}\Bigr)^{10}(V(y)-E)&\ge
\Bigl(\frac{|y-x_0|}{\lambda^KB}\Bigr)^{10}(V(y)-E_\infty)
\ge \Bigl(\frac{|y-x_0|}{\lambda^KB}\Bigr)^{10}\cdot
\frac{100}{|y-x_0|^2}\\
&\ge\frac{100}{(\lambda^KB)^2}\endalign
$$
\noindent valid in $I_{\text{BVP}}\cap(-\infty,x_0-\lambda
^KB]$, we conclude that
$$\align
\int_{I_{\text{BVP}}\cap(-\infty,x_0-\lambda^{-K}B]}
|F|^2\,dy&\le C_{\#}\exp(-c_{\#}\lambda^{1-2\varepsilon})
\lambda^{2K}SB^2\int_{x_-}^{x_+}|F|^2\,dy\\
&\le
C_{\#}\lambda^{-N}\int_{x_-}^{x_+}|F|^2\,dy,\quad
\text{since}\ SB^2=\lambda^2.\endalign
$$
\noindent Putting this together with (5) gives
$$
\int_{I_{\text{BVP}}\cap(-\infty,x_0-\lambda^{-\varepsilon}
B]}|F|^2\,dy\le C_{\#}\lambda^{-N},\quad\text{since}\
\int_{x_-}^{x_+}|F|^2\,dy\le \int_{I_{\text{BVP}}}|F|^2
\,dy=1.
$$
\noindent Similarly, $\int_{I_{\text{BVP}}\cap[x_0+
\lambda^{-\varepsilon}B,+\infty)}|F|^2\,dy\le C_{\#}
\lambda^{-N}$, and therefore\hfill\break $\int_{I_{\text{BVP}}
\backslash \{|x-x_0|<\lambda^{-\varepsilon}B\}}
|F|^2\,dy\le C_{\#}\lambda^{-N}$, which is the conclusion
of the Lemma. \TagsOnRight
$$
\tag"$\blacksquare$"
$$\TagsOnLeft

Next we show that $H=-\frac{d^2}{dx^2}+V(x)$ can be
transformed to the harmonic oscillator by a (formal power
series) change of variables $y=y(x)$.  The basic result
is as follows.
\vglue 1pc
\proclaim{Lemma 2}  Let $W(x)$ be a smooth function satisfying
$|(\frac{d}{dx})^\alpha W(x)|\le C_\alpha$ and $W(0)\hfill\break=0$,
$W^\prime(0)=0$, $W^{\prime\prime}(0)\ge c_1>0$.
Then we can find formal power series $y=\sum\limits_{k,m,s\ge 0}
y_{kms}\tau^k\lambda^{-2m}x^s$ and $\tilde\tau=\sum\limits_{k,m\ge 0}
\tilde\tau_{km}\tau^k\lambda^{-2m}$, satisfying the equation
$$
(y^2-\tilde\tau)\Bigl(\frac{\partial y}{\partial x}\Bigr)^2+\lambda^{-2}
\Bigl\{-\frac 12 \frac{\partial_x^3y}{\partial_xy}+\frac 34
\Bigl(\frac{\partial_x^2y}{\partial_xy}\Bigr)^2\Bigr\}=W(x)-\tau
\tag 6 $$
\noindent with $y_{001}>0$ (so that $\frac{1}{\partial_xy}$
makes sense as a formal power series).

The coefficients $y_{kms}$ and $\tilde\tau_{km}$ are uniquely
determined, and are bounded a--priori in terms of $c_1$ and the 
$C_{\alpha}$.\endproclaim
\demo{Proof}  Define the \underbar{strength} of a monomial
$C\tau^k\lambda^{-2m}x^s$ to be $k+m$.  By induction on
$A\ge 0$ we will prove\smallskip
\noindent\underbar{Assertion A}:  We can find coefficients
$(y^{\text{correct}}_{kms})$ and $(\tilde\tau_{km}^{\text{correct}})$
so that (6) holds modulo terms of strength $>A$ if and only if
for $k+m\le A$ we have $y_{kms}=y_{kms}^{\text{correct}}$ (all $s$)
and $\tilde\tau_{km}=\tilde\tau_{km}^{\text{correct}}$.


We set $y_{km}(x)=\sum\limits_{s\ge 0}y_{kms}x^s$, and begin
by proving Assertion 0.  The terms of strength $0$ in (6)
give us the following equation:
$$
(y_{00}^2-\tilde\tau_{00})(\frac{dy_{00}}{dx})^2=W(x).
\tag 7
$$
\noindent Since $y_{001}>0$, $(\frac{dy_{00}}{dx})^2$ is
a formal power series with non--vanishing constant term.
On the other hand $W(x)$ vanishes to second order at $0$,
so $y_{00}^2-\tilde\tau_{00}$ is a formal power series
starting with $ax^2$.  Therefore $2y_{00}\frac{
dy_{00}}{dx}$ has no constant term.  Since $\frac{dy_{00}}
{dx}$ has nonzero constant term, it follows that $y_{00}$
has no constant term.  Thus $y_{00}^2$ is a formal power
series starting with $ax^2$, and since $y_{00}^2-\tilde
\tau_{00}$ also starts with an $x^2$--term, we get
$\tilde\tau_{00}=0$.  Equation (7) therefore becomes
$$
y_{00}^2\,\Bigl(\frac{dy_{00}}{dx}\Bigr)^2=W(x).\tag 8
$$
\noindent Our assumptions on $W$ let us write $W(x)=
(F(x))^2$ with $F$ smooth, $F(0)=0$, $F^\prime(0)>0$,
and thus (8) amounts to $y_{00}\frac{dy_{00}}{dx}=\pm
F(x)$.  Since $y_{001}>0$ and $F^\prime(0)\hfill\break
>0$, we must
have the ``$+$" sign, so (7) amounts to $\tilde\tau_{00}
=0$ and $y_{00}\frac{dy_{00}}{dx}=F(x)$, which in turn
means that $\frac 12(y_{00})^2=\int_0^xF(t)\,dt$.  The
right--hand side has the form $ax^2+\text{higher\ terms}$
with $a>0$, and therefore has a smooth square root $g(x)$.
Thus (7) means that $\tilde\tau_{00}=0$ and $y_{00}=\pm
\sqrt 2 g(x)$.  The sign is uniquely specified by
requiring $y_{001}>0$.  Thus (7) holds if and only if
$\tilde\tau_{00}=0$ and $y_{00s}=y_{00s}^{\text{correct}}$
(all $s$).  We have proven Assertion 0.\smallskip

Next we assume Assertion (A-1) and deduce Assertion A
$(A\ge 1)$.  With $y_{kms}=y_{kms}^{\text{correct}}$,
$\tilde\tau_{km}=\tilde \tau_{km}^{\text{correct}}$ for $k+m\le A-1$,
but with $y_{kms}$, $\tilde\tau_{km}$ arbitrary 
for $k+m\ge A$, we examine the terms of
strength $A$ in (6).  Thus we fix $k$, $m$ with $k+m=A$.
The $\tau^k\lambda^{-2m}$--terms in (6) are as follows:
$$\split
2y_{00}\Bigl(\frac{dy_{00}}{dx}\Bigr)^2y_{km}+2y_{00}^2
\Bigl(\frac{dy_{00}}{dx}\Bigr)\Bigl(\frac{dy_{km}}
{dx}\Bigr)-\tilde\tau_{km}\Bigl(\frac{dy_{00}}{dx}\Bigr)^2\\
+f_{km}=g_{km},\endsplit\tag"(9)$_{km}$"\endsplit
$$
\noindent where $f_{km}$ is determined by $(y_{k^\prime
m^\prime s}^{\text{correct}})$, $(\tilde\tau_{k^\prime
m^\prime}^{\text{correct}})$ $(k^\prime+m^\prime\le A-1)$,
and $g_{km}$ is the $\tau^k\lambda^{-2m}$--term in
$W(x)-\tau$.  In fact, $f_{km}=\sum\limits_{s\ge 0}
f_{kms}x^s$ with $f_{kms}$ a polynomial in $(y_{k^\prime
m^\prime s^\prime}^{\text{correct}})$, $(\tilde\tau_{k^\prime
m^\prime s^\prime}^{\text{correct}})$ $(k^\prime+m^\prime
\le A-1)$, $y_{001}^{-1}$.  Hence to solve (6) modulo terms
of strength $>A$, we must solve $(9)_{km}$ for $k+m=A$.  Since
$y_{000}=0$, $y_{001}>0$, the constant term in (9)$_{km}$ is
$$
-\tilde\tau_{km}(y_{001})^2=\text{constant\ term\ in}\ (g_{km}-
f_{km}),\tag"(9$\text{bis}$)"
$$
\noindent which uniquely specifies $\tilde\tau_{km}$.  Once
$\tilde\tau_{km}$ is specified, (9)$_{km}$ takes the form
$$
2y_{00}\Bigl(\frac{dy_{00}}{dx}\Bigr)^2y_{km}+2y_{00}^2\Bigl(
\frac{dy_{00}}{dx}\Bigr)\Bigl(\frac{dy_{km}}{dx}\Bigr)=
\sum\limits_{s\ge 1}h_{kms}x^s\tag"(10)$_{km}$"
$$
\noindent with $h_{kms}$ specified by $(y_{k^\prime m^\prime
s^\prime}^{\text{correct}})$, $(\tilde\tau_{k^\prime m^\prime}
^{\text{correct}})$ for $k^\prime+m^\prime\le A-1$.
To solve (10)$_{km}$ for $y_{km}=\sum\limits_{s\ge 0}
y_{kms}x^s$, we solve successively for the coefficients
$y_{kms}$.  The terms of degree $(s+1)$ in (10)$_{km}$ are 
as follows
$$
(y_{001})^3(2+2s)y_{kms}+(\text{polynomial\ in}\ y_{00s^\prime}
\quad\text{and}\,\, y_{kms^{\prime\prime}}\,\,\text{with}
\,\, s^{\prime\prime}<s)
=h_{km(s+1)}.
$$

Since $(y_{001})^3(2+2s)>0$ for $s \ge 0$, this allows us to
solve for $y_{kms}$ in terms of $y_{00s^\prime}$ and $(s^{\prime\prime}
<s)$.  Hence we obtain inductively the unique sequence
$(y_{kms})_{s\ge 0}$ that solves (10)$_{km}$.  Thus $\tilde\tau_
{km}$ and $y_{kms}$ $(k+m=A)$ can be specified in one and only one
way to make (6) hold modulo terms of strength $>A$.  We have
succeeded in deducing Assertion A from Assertion (A-1).  By
induction, Assertion A holds for all $A$, so there is one and
only one formal power series solution of (6).  To obtain the
a--priori bounds on the coefficients $\tilde\tau_{km}$,
$y_{kms}$ we reexamine the preceding argument.  For terms of
strength $0$, we saw that $\tilde\tau_{00}=0$ and
$y_{00s}=\frac{1}{s!}(\pm\sqrt 2)(\frac{d}{dx})^sg(x)\Bigm|
_{x=0}$ with $g^2(x)=\int_0^xW^{1/2}(t)\,dt$.

It follows that $\tilde\tau_{00}$, $y_{00s}$ can be bounded
a--priori in terms of $C_\alpha$ and $c_1$ in the statement
of the lemma.  Also, $y_{001}$ can be bounded below.  Assuming
we have bounded $y_{kms}$, $\tilde\tau_{km}$ for $k+m<A$,
we look at (9)$_{km}$ and observe that the coefficients of
the power series $f_{km}$ and $g_{km}$ are bounded a--priori
in terms of $C_\alpha$, $c_1$, for $k+m=A$.  (For $f_{km}$
this follows from the fact that $f_{km}=\sum\limits_sf_{kms}x^s$
with $f_{kms}$ a polynomial in quantities which we have already
bounded a--priori.  For $g_{km}$ it is immediate from the
definitions.)

Therefore (9bis) shows that $\tilde\tau_{km}$ is bounded
a--priori in terms of $C_\alpha$, $c_1$.  This in turn shows
that the coefficients $h_{kms}$ in (10)$_{km}$ are bounded
a--priori.  Hence it follows by induction on $s$ that
$y_{kms}$ (which we obtain by solving (10)$_{km}$) is bounded
a--priori in terms of $C_\alpha$, $c_1$.  

Assuming a--priori bounds for $\tilde\tau_{km}$ and $y_{kms}$
$(k+m<A)$, we have derived a--priori bounds for
$\tilde\tau_ {km}$ and $y_{kms}$ when $k+m=A$.  By
induction on $A=k+m$, the $\tilde\tau_{km}$ and $y_{kms}$
are all bounded a--priori in terms of $C_\alpha$ and
$c_1$.  The proof of the lemma is complete.
$\qquad\blacksquare$

Note that $y_{000}=0$ and $\tilde\tau_{00}=0$, as we saw in
the proof of Lemma 2.
\vglue 1pc
\proclaim{Corollary}  With $W$, $y_{kms}$ $\tilde\tau_{km}$
as in Lemma 2, set
$$
y(x,E,\lambda)=\sum\limits_{k,m,s=0}^Ny_{kms}\Bigl(\frac{E}
{\lambda^2}\Bigr)^k\lambda^{-2m}x^s
$$
\noindent and
$$
\tilde E(E,\lambda)=\lambda^2\sum\limits_{k,m=0}^N\tilde\tau_{km}
\Bigl(\frac{E}{\lambda^2}\Bigr)^k\lambda^{-2m}.
$$
\noindent Then for $|x|$, $|\frac{E}{\lambda^2}|\le C_2\lambda^
{-\varepsilon}$ and $\lambda\ge C_2$ we have
$$
\Bigl(\frac{\partial y}{\partial x}\Bigr)^2(\lambda^2y^2-\tilde E)
+\Bigl\{-\frac 12 \frac{\partial_x^3y}{\partial_xy}+\frac 34\Bigl(\frac
{\partial_x^2y}{\partial_xy}\Bigr)^2\Bigr\}=\lambda^2W(x)-E
+\text{Error}(x,E,\lambda)
$$
\noindent with $|\text{Error}(x,E,\lambda)|\le C_3\lambda^{-\varepsilon
N/2}$, $|\frac{\partial}{\partial E}\text{Error}(x,E,\lambda)|\le
C_3\lambda^{-\frac{\varepsilon N}{2}-2}$.  The constant $C_3$ depends
only on $C_2$ and on $c_1$, $C_\alpha$ in the hypothesis of Lemma 2.
\endproclaim
\demo{Proof}  Set $f(x,\tau,\xi)=\lambda^{-2}\text{Error}(x,E,\lambda)$
with $\tau=\frac{E}{\lambda^2}$ and $\xi=\lambda^{-2}$.  Immediately
from the definitions, we see that the $C^\infty$ seminorms of $f$ on
$\{|x|,|\tau|\le c\}$ are bounded a--priori in terms of the $C_\alpha$,
the $|y_{kms}|$, the $|\tilde
\tau_{km}|$, and a lower bound for $y_{001}$.
The a--priori bounds in the statement of Lemma 2 shows that
$|\partial_{x,\tau,\xi}^\alpha f|\le C_\alpha^\prime$ for $|x|,
|\tau|,|\xi|\le c^\prime$ with $C_\alpha^\prime$, $c^\prime$ determined
by $c_1$, $C_\alpha$.  On the other hand, the formal power series
equation (6) shows that $f(x,\tau,\xi)=0$ to order at least $N-5$
at $(x,\tau,\xi)=(0,0,0)$.  Therefore, Taylor's theorem with remainder
gives
$$
|f(x,\tau,\xi)|\le C^\prime\lambda^{-\varepsilon(N-5)}\quad\text{and}\quad
|\frac{\partial}{\partial\tau}f(x,\tau,\xi)\Bigm|\le C^\prime\lambda^
{-\varepsilon(N-6)}$$
\noindent for $|x|,|\tau|,|\xi|\le C_2\lambda^{-\varepsilon}$, with
$C^\prime$ determined by $C_2$, $c_1$, $C_\alpha$.  This is stronger
than the conclusion of the corollary.$\qquad\blacksquare$
\smallskip

The previous lemma applies to potentials $\lambda^2W(x)$.  A simple
rescaling lets us apply that result to study $V(x)$.

\vglue 1pc
\proclaim{Lemma 3}  Let $V(x)$, $S$, $B$ etc. satisfy (H1$^*$)$\ldots$
(H6$^*$).  Then there exist coefficients $\hat y_{ks}$, $\hat\tau_k$ for which
the functions 
$$
y(x,E)=\sum\limits_{k,s=0}^N \hat y_{ks}\Bigl(\frac{E-V(x_0)}{S}\Bigr)^k
\Bigl(\frac{x-x_0}{B}\Bigr)^s
$$
\noindent and
$$
\tilde E(E)=\lambda^2\sum\limits_{k=0}^N \hat\tau_k\Bigl(\frac{E-V(x_0)}
{S}\Bigr)^k
$$
\noindent satisfy the equation
$$\split
\Bigl(\frac{\partial y}{\partial x}\Bigr)^2(\lambda^2y^2-\tilde E)+\Bigl\{
-\frac 12 \frac{\partial_x^3y}{\partial_xy}+\frac 34 \Bigl(\frac{\partial_x^2 y}
{\partial_xy}\Bigr)^2\Bigr\}=V(x)-E\\
+\text{Error}(x,E),
\endsplit\tag 11
$$
\noindent with $|\text{Error}(x,E)|\le C_{\#}\lambda^{-\varepsilon N/2}
S, |\frac{\partial}{\partial E}\ \text{Error}(x,E)|\le
C_{\#}\lambda^{-\varepsilon N/2}$ for $|x-x_0|<C_{\#}\lambda^{-\varepsilon} B$,
$|E-V(x_0)|<C_{\#}\lambda^{-2\varepsilon} S$.

Moreover, we have the a--priori bounds $|\hat y_{ks}|$, $|\hat\tau_k|\le C_{\#}$,
$|\hat y_{00}|\le C_{\#}\lambda^{-2}$, $|\hat\tau_0|\le C_{\#}\lambda^{-2}$,
$\hat y_{01}>c_{\#}>0$.\endproclaim

\demo{Proof}  Apply Lemma 2 to $W(x)=\frac{V(x_0+Bx)-V(x_0)}{S}$.
With $y_{kms}$, $\tilde\tau_{km}$ as in Lemma 2, put $\hat y_{ks}=
\sum\limits_{m=0}^Ny_{kms}\lambda^{-2m}$ and $\hat\tau_k=\sum\limits_{m=0}^
N \tilde \tau_{km}\lambda^{-2m}$.  The estimates asserted for
$\text{Error}(x,E)$ are then immediate consequences of the corollary to Lemma
2.  From Lemma 2 we get $|y_{kms}||\tilde\tau_{km}|\le C_{\#}$, which shows
that $|\hat y_{ks}|,|\hat\tau_k|\le C_{\#}$.  Since $y_{000}=\tilde\tau_{00}=0$,
we get also $|\hat y_{00}|, |\hat\tau_{0}|\le C_{\#}\lambda^{-2}$.  Since
$y_{001}>c_{\#}>0$ and $|y_{0m1}|\le C_{\#}$ for $1\le m\le N$ we get
$\hat y_{01}>c_{\#}>0$.$\qquad\blacksquare$

\vglue 1pc
\demo{Remark}  In Lemma 3, we can take $\hat y_{ks}$, and $\hat\tau_k$ to
depend only on $N$, $S$, $B$ and the power series of $V(x)$ at $x=x_0$.  That's
clear from the proof of Lemma 3, since the $y_{kms}$ and $\tilde\tau_{km}$
provide the unique solution to a formal power series equation (6) and are
consequently determined by the power series of $W(x)$ at $x=0$.
\vglue 1pc
\proclaim{Corollary}  We have the following a--priori bounds
$$
|\partial_x^\alpha\partial_E^\beta y(x,E)|\le C_{\#}^{\alpha\beta}B^{-\alpha}
S^{-\beta}\quad\text{for}\quad |x-x_0|<B,\,\,\, |E-V(x_0)|<S;\tag 12
$$
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\tilde E(E)\Big|\le C_{\#}^\beta\lambda^2
S^{-\beta}\quad\text{for}\quad |E-V(x_0)|<c_{\#}S;\tag 13
$$
$$
\frac{\partial y}{\partial x}>c_{\#}B^{-1}\quad\text{for}\quad |x-x_0|<c_{\#}B,
\,\,\,|E-V(x_0)|<c_{\#}S;\tag 14
$$
$$
\Bigl(\frac{d}{dE}\Bigr)\tilde
E>c_{\#}\lambda^2S^{-1}=c_{\#}B^2\quad\text{for}\quad |E-V(x_0)|<c_{\#}S.
\tag 15
$$\endproclaim

{\sl Proof}. The upper bounds for $|\partial_x^\alpha\partial_E^\beta y|$ and
$|(\frac{d}{dy})^\beta\tilde E|$ are immediate from the definitions and
$|\hat y_{ks}|, |\hat\tau_k|\le C_{\#}$.  The lower bound for $\frac
{\partial y}{\partial x}$ holds at $x=x_0$, $E=V(x_0)$ since $\hat
y_{01}>c_{\#}$.  Since $|\frac{\partial^2y}{\partial x^2}|\le C_{\#}B^{-2}$
and $|\frac{\partial^2y}{\partial x\partial E}|\le C_{\#}B^{-1}S^{-1}$, the
lower bound for $\frac{\partial y}{\partial x}$ remains valid throughout
$|x-x_0|<c_{\#}B$, $|E-V(x_0)|<c_{\#}S$.

To estimate $(\frac{d}{dE})\tilde E$,
we differentiate (11) with respect to $E$ at the point $x=x_0$, $E=V(x_0)$.  Thus
$$\split
2\Bigl(\frac{\partial y}{\partial x}\Bigr)
\Bigl(\frac{\partial y}{\partial E\partial x}\Bigr)
(\lambda^2y^2-\tilde E)+\frac{\partial}{\partial E}\Bigl\{-\frac 12
\frac{\partial_x^3y}{\partial_xy}+\frac 34 \Bigl(\frac{\partial_x^2y}
{\partial_xy}\Bigr)^2\Bigr\}\\
+\Bigl(\frac{\partial y}{\partial x}\Bigr)^2\cdot 2\lambda^2y\Bigl(\frac
{\partial y}{\partial E}\Bigr)-\Bigl(\frac{\partial y}{\partial x}\Bigr)^2\frac
{d\tilde E}{dE}=-1+\frac{\partial}{\partial E}\ \text{Error}(x,E).
\endsplit\tag 16
$$
\noindent At $x=x_0$, $E=V(x_0)$ we have $y=\hat y_{00}$ and $\tilde E=
\lambda^2\hat\tau_0$, so $|y|\le C_{\#}\lambda^{-2}$ and
$|\lambda^2y^2-\tilde E|\le C_{\#}$.  From (12) we get
$$\split
|2\frac{\partial y}{\partial E\partial x}
\Bigl(\frac{\partial y}{\partial x}\Bigr)(\lambda^2y^2-\tilde E)|\le
(C_{\#}B^{-1})\cdot(C_{\#}B^{-1}S^{-1})=C_{\#}\lambda^{-2}\\ \text{at}\
x=x_0, E=V(x_0).\endsplit\tag 17
$$
\noindent Also (12) and (14) show that
$$
\Big|\frac{\partial}{\partial E}\Bigl\{-\frac 12
\frac{\partial_x^3y}{\partial_xy} +\frac 34
\Bigl(\frac{\partial_x^2y}{\partial_xy}\Bigr)^2\Bigr\}\Big|\le C_{\#}
B^{-2}S^{-1}=C_{\#}\lambda^{-2}.\tag 18
$$
\noindent From (12) and the fact that $|y|\le C_{\#}\lambda^{-2}$ at $x=x_0$,
$E=V(x_0)$ we conclude that
$$\split
\Big|\Bigl(\frac{\partial y}{\partial x}\Bigr)^2\cdot 2\lambda^2y
\Bigl(\frac{\partial y}{\partial E}\Bigr)\Big|\le (C_{\#}B^{-1})^2\cdot
C_{\#}S^{-1}=C_{\#}\lambda^{-2}\ \text{at} \ x=x_0,\\
E=V(x_0).\endsplit\tag 19
$$
\noindent Putting (17), (18), (19), and the bound $|\frac{\partial}{\partial E}
\text{Error}(x,E)|\le C_{\#}\lambda^{-\varepsilon N/2}$ into (16), we see that
$|(\frac{\partial y}{\partial x})^2\frac{d\tilde E}{dE}-1|\le C_{\#}
\lambda^{-2}$ at $x=x_0$, $E=V(x_0)$.  So $\frac{d\tilde E}{dE}\ge
c_{\#}(\frac{\partial y}{\partial x})^{-2}\ge c_{\#}
B^2$ at $x=x_0$, $E=V(x_0)$, by (12).  That is $\frac{d\tilde E}
{dE}\ge c_{\#}B^2=c_{\#}\lambda^2S^{-1}$
at $x=x_0$, $E=V(x_0)$.  Since $|(\frac{d}
{dE})^2\tilde E|\le C_{\#}\lambda^2S^{-2}$ by (13), the estimate
$\frac{d}{dE}\tilde E\ge c_{\#}\lambda^2S^{-1}$ remains valid for $|E-V(x_0)|
\le  c_{\#}S$.  The proof of the corollary is complete. $\qquad\blacksquare$
\smallskip

Lemma 3 says that $x\mapsto y(x,E)$ transforms the equation $[\frac{d^2}
{dx^2}+(E-V(x))]F=0$ to
$$\split
\Bigl[\frac{d^2}{dy^2}+(\tilde E-\lambda^2y^2)+\text{Error}\Bigr]
\tilde F(y)=0\quad\text{with}\quad |\text{Error}|\le C_{\#}\lambda
^{-\varepsilon N/10}\\
\text{for}\ |y|<C_{\#}\lambda^{-\varepsilon}\endsplit\tag
20 $$
\noindent by taking
$$
F(x)=\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\tilde F(y),\ \text{
which\ defines}\ \tilde F(y)\ \text{for}\ |y|<C_{\#}\lambda^{-\varepsilon}.
\tag 21
$$
\noindent Hence we can use our knowledge of the harmonic oscillator to
locate the eigenvalues of $-\frac{d^2}{dx^2}+V(x)=H$.  This is achieved
in the next two lemmas.

\vglue 1pc
\proclaim{Lemma 4}  Suppose $E$ is an eigenvalue of $H$ with $|E-V(x_0)|
\le c_{\#}\lambda^{-2\varepsilon}S$ and $E\le E_{\infty}$.
Then with $\tilde E(E)$ as in the previous lemma we have
$$
|\tilde E(E)-(2k+1)\lambda|\le C_{\#}\lambda^{-\varepsilon N/11}\tag 22
$$
\noindent for an integer $k\ge 0$.  Moreover, no other eigenvalue $E^\prime$
of $H$ with $E^\prime\le E_{\infty}$ and
$|E^\prime-V(x_0)|\le c_{\#}\lambda^{-2\varepsilon}S$
satisfies $|\tilde E(E^\prime)-(2k+1)\lambda|\le
C_{\#}\lambda^{-\varepsilon N/11}$ for the same
$k$.\endproclaim

\demo{Proof}  Let $F$ be the eigenfunction associated to $E$, and define
$\tilde F$ by (21).  From (20) we get
$$
\Big\Vert\Bigl[\frac{d^2}{dy^2}+(\tilde E-\lambda^2y^2)\Bigr]\tilde F\Big\Vert_
{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}\le C_{\#}^\prime\lambda^
{-\varepsilon N/10}\Vert\tilde F\Vert_{L^2(|y|\le C_{\#}
\lambda^{-\varepsilon})}.\tag 23
$$
\noindent Lemma 1 gives $\int_{|x-x_0|>\lambda^{-\varepsilon}B}
|F(x)|^2\ dx\le C_{\#}^\prime\lambda^{-N}\int_{|x-x_0|<\lambda^{-\varepsilon}B}
|F(x)|^2\,dx$, hence
$$
\Vert\tilde F\Vert_{L^2(\frac 12C_{\#}\lambda^{-\varepsilon}<|y|<C_{\#}
\lambda^{-\varepsilon})}\le C_{\#}^\prime\lambda^{-N}\Vert\tilde F\Vert_
{L^2(|y|<\frac 12C_{\#}\lambda^{-\varepsilon})}^2\tag 24
$$
\noindent by virtue of (12), (14).  For $|x-x_0|<B$ we have $|\tilde E
-\lambda^2y^2|\le C_{\#}\lambda^2$ by (12), (13), so (24) and (23) imply
$$
\Big\Vert\frac{d^2}{dy^2}\tilde F\Big
\Vert_{L^2(\frac 12 C_{\#}\,\lambda^{-\varepsilon}
<|y|<C_{\#}\lambda^{-\varepsilon})}\le C_{\#}^\prime\lambda^{-\varepsilon
N/10}\,\Vert\tilde F\Vert_{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}.\tag 25
$$
\noindent From (24), (25) we conclude that
$$
\Big\Vert\frac{d}{dy}\tilde F\Big\Vert_{L^2(\frac 23C_{\#}\lambda^{-\varepsilon}
<|y|<C_{\#}\lambda^{-\varepsilon})}\le C_{\#}^\prime\,\lambda^{-\varepsilon N/11}
\,\Vert\tilde F\Vert_{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}.\tag 26
$$

Now let $\chi(y)$ be a cutoff function satisfying $|(\frac{d}{dy})^m\chi|\le
C_{\#}^m\lambda^{\varepsilon m}$ and $\chi(y)=1$ for $|y|\le \frac 23
C_{\#}\lambda^{-\varepsilon}$, $\chi(y)=0$ for $|y|\ge C_{\#}\lambda^{
-\varepsilon}$.  We can pick $C_{\#}$ large so that $\chi(y(x,E))=1$ when
$|x-x_0|<\lambda^{-\varepsilon}B$.  Estimates (23), (24), (26) yield
$$
\Big\Vert\Bigl[\frac{d^2}{dy^2}+(\tilde E-\lambda^2y^2)\Bigr](\chi \tilde F)\Big
\Vert_
{L^2(\Bbb R)}\le C_{\#}^\prime\,\lambda^{-\varepsilon N/11}\,\Vert\tilde F
\Vert_{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}.\tag 27
$$

Also, (24) gives $\Vert\tilde F\Vert_{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}
\le 2\Vert\chi\tilde F\Vert_{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}$,
so that (27) implies
$$
\Big\Vert\Bigl[\frac{d^2}{dy^2}+(\tilde E-\lambda^2y^2)\Bigr](\chi \tilde F)\Big
\Vert_
{L^2(\Bbb R)}\le C_{\#}^\prime\lambda^{-\varepsilon N/11}\Vert\chi \tilde F\Vert_
{L^2(\Bbb R)}.\tag 28
$$
\noindent For $|x-x_0|<\lambda^{-\varepsilon}B$ we have $\chi(y)=1$ and
$F(x)=(\frac{\partial y}{\partial x})^{-1/2}\tilde F(y)$.  Hence $\chi\tilde F$
cannot vanish identically.  From (28) and the known spectrum of the harmonic
oscillator, we get
$$
\vert\tilde E-(2k+1)\lambda|\le C_{\#}^\prime\lambda^{-\varepsilon N/11}
\,\, \text{for\ an\ integer}\ k\ge 0\tag 29
$$

\noindent and also $
\chi\tilde F=a\tilde f_k+\tilde g_k\quad\text{with}\quad \tilde f_k\quad$ the
$k^{\text{th}}$ eigenfunction of $-\frac{d^2}{dy^2}+\lambda^2y^2$,

\noindent $a$ a
constant, and $\Vert\tilde g_k\Vert_{L^2(\Bbb R)}\le C_{\#}^\prime
\lambda^{-\varepsilon N/11}\Vert\chi \tilde F\Vert_{L^2(\Bbb R)}$.  In
$J=\{|x-x_0|<\lambda^{-\varepsilon}B\}$ we have
$$
F(x)=\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\chi(y)\tilde F(y)=
\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\cdot a\tilde f_k(y)+
\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\tilde g_k(y),
$$
\noindent and therefore
$$\multline
\Big\Vert F(x)-a\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\tilde f_k
(y(x,E))\Big\Vert_{L^2(J)}\le \operatornamewithlimits{max}_{x \in J}
\Big|\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1}\Big|\Vert\Bigl(
\frac{\partial y}{\partial x}\Bigr)^{1/2}
\tilde g_k(y(x,E))\Vert_{L^2(J)}\\
\le \operatornamewithlimits{max}_{x \in J}\Big|\Bigl(\frac{\partial y}{\partial
x}\Bigr)^{-1}\Big|\Vert\tilde g_k\Vert_{L^2(\Bbb R)}\le
\operatornamewithlimits{max}_{x \in J}\Big|\Bigl(\frac{\partial y}
{\partial x}\Bigr)^{-1}\Big|\cdot C_{\#}^\prime\lambda^{-\varepsilon N/11}
\Vert \tilde F\Vert_{L^2(|y|\le C_{\#}\lambda^{-\varepsilon})}.
\endmultline\tag 30
$$
\noindent On the other hand,
$$\multline
\Vert F\Vert_{L^2(I_{\text{BVP}})}\ge \Vert\Bigl(\frac{\partial y}{
\partial x}\Bigr)^{-1/2}\tilde F(y(x,E))\Vert_{L^2(|x-x_0|<
C_{\#}^{\prime\prime}\lambda^{-\varepsilon}B)}\\
\ge
\operatornamewithlimits{min}_{|x-x_0|<C_{\#}^{\prime\prime}\lambda
^{-\varepsilon}B}
\Big|\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1}\Big|\cdot\Big\Vert
\Bigl(\frac{\partial y}{\partial x}\Bigr)^{1/2}\tilde F(y(x,E))\Big\Vert_{
L^2(|x-x_0|<C_{\#}^{\prime\prime}\lambda^{-\varepsilon}B)}\\
\ge \operatornamewithlimits{min}_{|x-x_0|<C_{\#}^{\prime\prime}
\lambda^{-\varepsilon}B}\Big|\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1}
\Big|\cdot\Vert\tilde F\Vert_{L^2(|y|<C_{\#}\lambda^{-\varepsilon})}
\endmultline
$$

\noindent if we take $C_{\#}^{\prime\prime}$ large enough so that the image
of $\{|x-x_0|<C_{\#}^{\prime\prime}\lambda^{-\varepsilon}B\}$ under
$x\mapsto y(x,E)$ covers $\{|y|<C_{\#}\lambda^{-\varepsilon}\}$.  Comparing
this to (30), and noting that $(\frac{\partial y}{\partial x})$ has
a constant order of magnitude by (12), (14), we conclude that
$$
\Vert F(x)-a\cdot\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\tilde f_k
(y(x,E))\Vert_{L^2(J)}\le C_{\#}^\prime\,\lambda^{-\varepsilon N/11}\,
\Vert F\Vert_{L^2(I_{\text{BVP}})}.
$$

\noindent Lemma 1 shows that $\Vert F\Vert_{L^2(I_{\text{BVP}})}\le 2\Vert F
\Vert_{L^2(J)}$, and therefore
$$
\Vert F(x)-a\cdot\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}
\tilde f_k(y(x,E))\Vert_{L^2(J)}\le C_{\#}^\prime\lambda^{-\varepsilon N/11}
\Vert F\Vert_{L^2(J)}.\tag 31
$$
\noindent Thus in $L^2(J)$, $F$ is nearly proportional to $f_E(x)=(\frac
{\partial y}{\partial x})^{-1/2}\tilde f_k(y(x,E))$.  Next we investigate
$\frac{\partial}{\partial E}f_E$ for $E$ satisfying (29)
and  $|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}\cdot S$. 
We have: $$\multline
\Big\Vert\frac{\partial}{\partial E}f_E\Big\Vert_{L^2(J)}\le \Big\Vert\Bigl(\frac{
\partial y}{\partial x}\Bigr)^{-3/2}\Bigl(\frac{\partial^2y}{\partial x
\partial E}\Bigr)\tilde f_k(y(x,E))\Big\Vert_{L^2(J)}\\
+\Big\Vert\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\Bigl(\frac{\partial y}
{\partial E}\Bigr)\tilde f_k^\prime(y(x,E))\Big\Vert_{L^2(J)}\\
\le \operatornamewithlimits{max}_{x \in J}\Big|\Bigl(\frac{\partial y}
{\partial x}\Bigr)^{-1}\Bigl(\frac{\partial^2y}{\partial x\partial E}\Bigr)
\Big|\cdot\Big\Vert\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\tilde f_k
(y(x,E))\Big\Vert_{L^2(J)}\\
+\operatornamewithlimits{max}_{x \in J}\Big|\Bigl(\frac{\partial y}{\partial
x}\Bigr)^{-1}\Bigl(\frac{\partial y}{\partial E}\Bigr)\Big|\cdot\Big\Vert\Bigl(
\frac{\partial y}{\partial x}\Bigr)^{1/2}\tilde f_k^\prime(y(x,E))\Big\Vert_{
L^2(J)}\\
\le \operatornamewithlimits{max}_{x \in J}\Big|
\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1}\Bigl(\frac{\partial^2y}
{\partial x\partial E}\Bigr)\Big|\cdot\Big\Vert f_E\Big\Vert_{L^2(J)}+
\operatornamewithlimits{max}_{x \in J}\Big|\Bigl(\frac{\partial y}
{\partial x}\Bigr)^{-1}\Bigl(\frac{\partial y}{\partial E}\Bigr)\Big|\cdot
\Vert\tilde f_k^\prime\Vert_{L^2(\Bbb R)}\endmultline\tag 32
$$

\noindent (by definition of $f_E$ and a change of variable from $x$ to $y$)
$$
\le C_{\#}^\prime S^{-1}\Vert f_E\Vert_{L^2(J)}+C_{\#}^\prime BS^{-1}\Vert
\tilde f_k^\prime\Vert_{L^2(\Bbb R)}.
$$
\noindent We have
$$\split
\Vert\tilde f_k^\prime\Vert_{L^2(\Bbb R)}^2\le\, <(-\frac{d^2}
{dy^2}+\lambda^2y^2)\tilde f_k,\tilde f_k>\,=(2k+1)\lambda\,
\Vert\tilde f_k\Vert_{L^2(\Bbb R)}^2\\
\le C_{\#}\lambda^2\,\Vert\tilde f_k\Vert^2_{L^2(\Bbb R)}\endsplit
$$
\noindent by (29) and (13).  Thus, (32) implies
$$
\Big\Vert\frac{\partial}{\partial E}f_E\Big\Vert_{L^2(J)}\le C_{\#}^\prime
S^{-1}\Vert f_E\Vert_{L^2(J)}+C_{\#}^\prime BS^{-1}\lambda\Vert \tilde f_k
\Vert_{L^2(\Bbb R)}.\tag 33
$$

\noindent On the other hand,
$$\align
\Vert f_E\Vert_{L^2(J)}&\ge \operatornamewithlimits{min}_{x \in J}
\Big|\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1}\Big|\cdot\Big\Vert
\Bigl(\frac{\partial y}{\partial x}\Bigr)^{1/2}\tilde f_k(y(x,E))\Big\Vert_
{L^2(J)}\\
&\ge c_{\#}^\prime B\Vert\tilde f_k\Vert_{L^2(|y|<c_{\#}^\prime\lambda^
{-\varepsilon})}\quad\text{by\ (12)}. \tag 34\endalign
$$

\noindent Since we are assuming $|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S$
for a small enough $c_{\#}$, Lemma 3 shows that $|\tilde E|\le c_{\#}^
{\prime\prime}\lambda^{2-2\varepsilon}$ with small
$c_{\#}^{\prime\prime}$.  We can make
$c_{\#}^{\prime\prime}$ as small as we please by taking
$c_{\#}$ small.  By (29) we have $|(2k+1)\lambda|\le
c_{\#}^{\prime\prime} \lambda^{2-2\varepsilon}$.  If we
take $c_{\#}^{\prime\prime}$ small enough depending on
$c_{\#}^\prime$, then $|(2k+1)\lambda|\le 
c_{\#}^{\prime\prime}\lambda^{2-2\varepsilon}$ implies
that $\tilde f_k$ is heavily concentrated in
$\{|y|<c_{\#}^\prime\lambda^{-\varepsilon}\}$. Hence (34)
implies $\Vert f_E\Vert_{L^2(J)}\ge c_{\#}^\prime
B\Vert\tilde f_k \Vert_{L^2(\Bbb R)}$.  Putting this into
(33), we get $$ \Big\Vert\frac{\partial}{\partial
E}f_E\Big\Vert_{L^2(J)}\le C_{\#}^\prime\lambda
S^{-1}\Vert f_E\Vert_{L^2(J)}.\tag 35 $$

\noindent We proved (35) for any $E$ satisfying
$$
|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S\ \text{and}\ |\tilde E(E)-
(2k+1)\lambda|\le C_{\#}\lambda^{-\varepsilon N/11},
\text{and}\ E\le E_{\infty}.\tag 36
$$

\noindent In particular, if $E$, $E^\prime$ both satisfy (36), then
$$
\Vert f_E-f_{E^\prime}\Vert_{L^2(J)}\le C_{\#}^\prime\lambda S^{-1}
|E-E^\prime|\,\,\,\,\Vert f_E\Vert_{L^2(J)},\tag 37
$$
\noindent provided
$$
\lambda S^{-1}|E-E^\prime|<c_{\#}.\tag 38
$$

\noindent Since $\frac{d\tilde E}{dE}\sim \lambda^2S^{-1}=B^2$ by (13), (15),
condition (36) for $E$ and $E^\prime$ implies $c_{\#}B^2|E-E^\prime|\le C_{\#}
^\prime\lambda^{-\varepsilon N/11}$.  This implies (38), and therefore (37)
holds.  Moreover, from (37) we get
$$\split
\Vert f_E-f_{E^\prime}\Vert_{L^2(J)}\le
C_{\#}^\prime\lambda S^{-1}\cdot B^{-2}
\lambda^{-\varepsilon N/11}\,\Vert f_E\Vert_{L^2(J)}\\
=C_{\#}\lambda^{-1-\varepsilon N/11}\,\Vert
f_E\Vert_{L^2(J)}.\endsplit $$

\noindent Thus $f_E$ and $f_{E^\prime}$ are nearly proportional.  On the
other hand, we have seen that the eigenfunctions $F$, $F^\prime$
corresponding to $E$, $E^\prime$ are nearly proportional to $f_E$,
$f_{E^\prime}$
respectively, in $L^2(J)$.  Therefore $F$ and $F^\prime$ are nearly
proportional in $L^2(J)$.  Lemma 1 shows that $F$ and $F^\prime$ are almost
entirely concentrated in $J$, hence $F$ and $F^\prime$ are nearly proportional
in $L^2(I_{\text{BVP}})$.  If $E\ne E^\prime$, then $F$ and $F^\prime$
 would have to be
orthogonal, and so couldn't be nearly proportional.  Hence $E^\prime=E$,
proving the uniqueness assertion in Lemma 4.  This and estimate (29) complete
the proof.$\qquad\blacksquare$
\vglue 1pc
\proclaim{Lemma 5}  Let $k$ be an integer satisfying $0\le k\le c_{\#}\lambda^
{1-2\varepsilon}$.  Suppose either $E_\infty>V(x_0)+\lambda^{-2\varepsilon}S$,
or else $E_\infty\le V(x_0)+\lambda^{-2\varepsilon}S$ and $(2k+1)\lambda+
C_{\#}\lambda^{-\varepsilon N/11}\le \tilde E(E_\infty)$.  Then there
is an eigenvalue $E^\prime$ of $H$ satisfying $|E^\prime-V(x_0)|\le c_{\#}
\lambda^{-2\varepsilon}S$, and $|\tilde E(E^\prime)-(2k+1)\lambda|<C_{\#}
\lambda^{-\varepsilon N/11}$.\endproclaim
\demo{Proof}  We start by finding an $E$ with $|E-V(x_0)|<c_{\#}\lambda^{
-2\varepsilon}S$, $\tilde E(E)=(2k+1)\lambda$, and $E+C_{\#}\lambda^{-
\varepsilon N/10}S\le E_\infty$.  In fact, we know that $|\tilde E|\le C_{\#}$
when $E=V(x_0)$ by the bound for $\hat\tau_0$ in Lemma 3.  We know also
that $\frac{d\tilde E}{dE}\ge c_{\#}\lambda^2S^{-1}=c_{\#}B^2$
for $|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S$.  \enddemo

Therefore, for $0\le k<c_{\#}\lambda^{1-2\varepsilon}$, we can find $E$ 
satisfying
$$
|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S\quad\text{and}\quad
\tilde E(E)=(2k+1)\lambda.\tag 39
$$
\noindent It remains to check that $E+C_{\#}\lambda^{-\varepsilon N/10}S
<E_\infty$.  This is clear from (39) in case $E_\infty\ge V(x_0)+
\lambda^{-2\varepsilon}S$.  In the alternate case $E_\infty\le V(x_0)+
\lambda^{-2\varepsilon}S$, $(2k+1)\lambda+C_{\#}\lambda^{-\varepsilon N/11}
\le \tilde E(E_\infty)$, we use (39) to get $\tilde E(E)+C_{\#}
\lambda^{-\varepsilon N/11}\le \tilde E(E_\infty)$.  Then from (13), (15)
we conclude that $E_\infty-E\ge c_{\#}\lambda^{-\varepsilon N/11}
(\frac{\lambda^2}{S})^{-1}>C_{\#}\lambda^{-\varepsilon N/10}S$.
Hence in all cases we have $E+C_{\#}\lambda^{-\varepsilon N/10}S<E_\infty$,
as claimed.

Lemma 3 shows that the equation
$$
\Bigl[\frac{d^2}{dy^2}+(\tilde E(E)-\lambda^2y^2)\Bigr]\tilde F(y)=0\tag 40
$$

\noindent is transformed into
$$\split
\Bigl[\frac{d^2}{dx^2}+(E-V(x))+\text{Error}\Bigr]F(x)=0\ \text{with}\
|\text{Error}|<C_{\#}\lambda^{-\varepsilon N/10}S\\
\text{for}\ |x-x_0|<C_{\#}\lambda^{-\varepsilon}B,\endsplit\tag 41
$$

\noindent by setting
$$
F(x)=\Bigl(\frac{\partial y}{\partial x}\Bigr)^{-1/2}\tilde F(y(x,E))\quad\text
{for}\ |x-x_0|<C_{\#}\lambda^{-\varepsilon}S.\tag 42
$$

\noindent Since $\tilde E(E)=(2k+1)\lambda$, equation (40) is satisfied on
the whole real line for a suitable Hermite function $\tilde F$.  Hence,
defining $F$ by (42), we get
$$
\Big\Vert\Bigl[\frac{d^2}{dx^2}+(E-V(x))\Bigr]F\Big\Vert_{L^2(|x-x_0|<C_{\#}
\lambda^{-\varepsilon}B)}\le C_{\#}\lambda^{-\varepsilon N/10} S\Vert F\Vert
_{L^2(|x-x_0|\le C_{\#}\lambda^{-\varepsilon} B)}\tag 43
$$

\noindent by virtue of (41).  Moreover, if $0\le k\le c_{\#}\lambda^{1-2\
\varepsilon}$, then the Hermite function $\tilde F$ is concentrated almost
entirely in $\{|y|<c_{\#}^\prime\lambda^{-\varepsilon}\}$.  We can make
$c_{\#}^\prime$ small by taking $c_{\#}$ small.

\noindent In particular, we have
$$\align
\Vert\tilde F\Vert_{L^2(|y|>c_{\#}^\prime\lambda^{-\varepsilon})}&\le C_{\#}
^\prime\lambda^{-N}\Vert\tilde
F\Vert_{L^2(|y|<c_{\#}^\prime\lambda^{-\varepsilon})}\quad\text{and}\\
\Big\Vert\frac{d}{dy}\tilde
F(y)\Big\Vert_{L^2(|y|>c_{\#}^\prime\lambda
^{-\varepsilon})} &\le
C_{\#}^\prime\lambda^{-N}\Vert\tilde
F\Vert_{L^2(|y|<c_{\#}^\prime
\lambda^{-\varepsilon})}.\endalign $$

\noindent From these estimates, (12), and the fact that $c_{\#}B^{-1}
<\frac{\partial y}{\partial x}<C_{\#}B^{-1}$ for $|x-x_0|<c_{\#}B$, we see that
$$
\Vert F\Vert_{L^2(\lambda^{-\varepsilon}B<|x-x_0|<C_{\#}\lambda^{-\varepsilon}
B)}\le C_{\#}\lambda^{-N}\Vert F\Vert_{L^2(|x-x_0|<\lambda^{-\varepsilon}B)}
\tag 44
$$
\noindent \text{and}
$$
\Big\Vert\frac{d}{dx}F\Big\Vert_{L^2(\lambda^{-\varepsilon}B<|x-x_0|<C_{\#}\lambda
^{-\varepsilon}B)}\le C_{\#}\lambda^{-N}B^{-1}\Vert F\Vert_{L^2(|x-x_0|
<\lambda^{-\varepsilon}B)}.\tag 45
$$

Introduce a cutoff function $\chi(x)$ satisfying $\Big|\Bigl(\frac{d}{dx}
\Bigr)^\alpha\chi\Big|\le C_{\#}^\alpha(\lambda^{-\varepsilon}B)^{-\alpha}$,
$\chi(x)=1$ for $|x-x_0|\le \lambda^{-\varepsilon}B$, $\chi(x)=0$ for
$|x-x_0|>2\lambda^{-\varepsilon}B$.  From (43), (44), (45) we get
$$
\Big\Vert\Bigl[\frac{d^2}{dx^2}+(E-V(x))\Bigr](\chi F)\Big
\Vert_{L^2(I_{\text{BVP}})}
\le C_{\#}\lambda^{-\frac{\varepsilon N}{10}}S\Vert \chi F\Vert_{L^2(
I_{\text{BVP}})}.\tag 46
$$

\noindent A glance at the definition (42) with $\tilde F$ a Hermite function
shows that $\chi F$ doesn't vanish identically.  Hence, (46) implies that
the interval $\Cal J=[E-C_{\#}\lambda^{-\varepsilon N/10}S,E+C_{\#}
\lambda^{-\varepsilon N/10}S]$ contains some point $E^\prime$ in the
spectrum of $H$.  (Note that $\chi F$ satisfies the boundary conditions for $H$
since $\text{supp}\ \chi \subset\subset I_{\text{BVP}}$.)  Since the
continuous spectrum of $H$ is contained in $[E_\infty,\infty)$, and since
$E_\infty>E+C_{\#}\lambda^{-\frac{\varepsilon N}{10}}S$, it follows that
$\Cal J$ contains none of the continuous spectrum of $H$.  Thus $E^\prime$
is an eigenvalue of $H$.  Since $|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S$
and $|E^\prime-E|<C_{\#}\lambda^{-\frac{\varepsilon N}{10}}S$, we get
$|E^\prime-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S$.  We have
$|E^\prime-E|\le C_{\#}\lambda^{-\frac{\varepsilon N}{10}}S$ and therefore
$|\tilde E(E^\prime)-(2k+1)\lambda|=|\tilde E(E^\prime)-\tilde E(E)|\le
C_{\#}\lambda^2S^{-1}|E^\prime-E|\le C_{\#}\lambda^{2-\varepsilon N/10}\le
C_{\#}\lambda^{-\varepsilon N/11}$.  The proof of the lemma is complete.
$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary} There is a finite sequence
$E_0,E_1,\ldots,E_{k_{\text{max}}}$ with the following properties.
\roster
\item"{(a)}" The integers $k=0,1,\ldots,k_{\text{max}}$ are precisely those
for which we can find some $E$ to satisfy
$$
|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S,\,\, E\le E_\infty,\,\, |\tilde E(E)-
(2k+1)\lambda|\le C_{\#}\lambda^{-\frac{\varepsilon N}{11}}.
$$
\item"{(b)}" If $0\le k<k_{\text{max}}$ then $E_k$ is an eigenvalue of $H$.
\item"{(c)}" Either $E_{k_{\text{max}}}$ is an eigenvalue of $H$ or else
$E_{k_{\text{max}}}=E_\infty$.
\item"{(d)}" Every eigenvalue $E$ of $H$ satisfying $|E-V(x_0)|<c_{\#}
\lambda^{-2\varepsilon}S$ and $E\le E_{\infty}$
is one of the $E_k$ $(0\le k\le k_{\text{max}})$.
\item"{(e)}" For $0\le k\le k_{\text{max}}$ we have $|E_k-V(x_0)|\le 2c_{\#}
\lambda^{-2\varepsilon}S$ and $|\tilde E(E_k)-(2k+1)\lambda|\le C_{\#}
\lambda^{-\varepsilon N/11}$.
\endroster\endproclaim

\demo{Sketch of Proof}  Define $k_{\text{max}}$ by (a), recalling that
$E\mapsto \tilde E$ is monotone with $\frac{d\tilde E}{dE}\sim \lambda^2S^{-1}$.
If $0\le k\le k_{\text{max}}$ then either the hypotheses of Lemma 5 are
satisfied, or else $k=k_{\text{max}}$ and
$|E_\infty-V(x_0)|<\lambda^{-2\varepsilon}S$, $|\tilde E(E_\infty)-(2k+1)
\lambda|\le
C_{\#}\lambda^{-\varepsilon N/11}$.  There can be at most one eigenvalue $E$
with $|E-V(x_0)|<\lambda^{-2\varepsilon}S$, $|\tilde E(E)-(2k+1)\lambda|\le
C_{\#}\lambda^{-\varepsilon N/11}$, $ E\le E_\infty$
by virtue of Lemma 4.  Also from
Lemma 5 there exists such an eigenvalue, except perhaps in the case:
$k=k_{\text{max}}$, $|E_\infty-V(x_0)|<\lambda^{-2\varepsilon}S$,
$|\tilde E(E_\infty)-(2k+1)\lambda|\le C_{\#}\lambda^{-\varepsilon N/11}$.
Hence we can define $E_k=\text{the\ unique\ eigenvalue}\ E$ as above
if one exists, $E_k=E_\infty$ if no suitable eigenvalue exists.  Properties
(a), (b), (c), (e) are thus satisfied, and (d) follows from Lemma 4.
We have seen this argument before in establishing the WKB Eigenvalue
Theorem.$\qquad\blacksquare$

It remains to give an explicit formula for $\tilde E(E)$ modulo a small
error.  We shall see that $\tilde E(E)=\frac{2}{\pi}\lambda(\phi(E)+\frac
{1}{48}\psi(E))+\text{(small\ error)}$, where the phases $\phi$ and
$\psi$ are defined as in the WKB Eigenvalue Theorem.  Thus the above
Corollary extends the range of $E$ under which the WKB Eigenvalue Theorem
locates the eigenvalues of $H$ with good precision.  Details are as follows.

For $V(x_0)<E<V(x_0)+c_{\#}S$, the set $\{x\mid |x-x_0|<c_{\#}^\prime B,
V(x)<E\}$ is an interval $(x_{\text{left}}(E), x_{\text{rt}}(E))$ with
$V(x_{\text{left}}(E))=V(x_{\text{rt}}(E))=E$.  We define
$$
\phi(E)=\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}(E-V(x))^{1/2}\,dx\quad
\text{and}\tag 47
$$
$$\split
\psi(E)=\operatornamewithlimits{lim}_{\delta_{\text{left}},\delta_{\text{rt}}
\to 0+}\, \Bigl[\int_{x_{\text{left}}(E)+\delta_{\text{left}}}^
{x_{\text{rt}}(E)-\delta_{\text{rt}}}V^{\prime\prime}(x)(E-V(x))^{-3/2}
\,dx-
q_{\text{left}}(E)\delta_{\text{left}}^{-1/2}\\
-q_{\text{rt}}(E)\delta_{\text{rt}}^{-1/2}\Bigr]\endsplit\tag 48
$$

\noindent with $q_{\text{left}}(E)$, $q_{\text{rt}}(E)$ uniquely 
specified by demanding the finitness of the limit.  Equivalently,
$$
\psi(E)=\operatornamewithlimits{lim}_{\delta\to 0+}
\Bigl[\operatornamewithlimits{\int} \Sb |x-x_0|<c_{\#}B\\
E-V(x)>\delta\endSb V^{\prime\prime}(x)(E-V(x))^{-3/2}\,dx-q(E)\delta^{-1/2}
\Bigr]
$$

\noindent with $q(E)$ uniquely specified by demanding the finiteness of
the limit.  Our goal is to compare $\frac{\tilde E(E)}{2\lambda}$
with $\phi(E)+\frac{1}{48}\psi(E)$.  We begin by studying the smoothness
of $\phi$ and $\psi$.

\vglue 1pc
\proclaim{Lemma 6}  The phases $\phi(E)$ and $\psi(E)$ satisfy
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\phi(E)\Big|\le C_{\#}^\beta
\lambda S^{-\beta},\,\, \Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\psi(E)\Big|\le
C_{\#}^\beta\lambda^{-1}S^{-\beta}\quad(\beta\ge 0)
$$

\noindent for $V(x_0)<E<V(x_0)+c_{\#}S$.\endproclaim

\demo{Proof}  Rescaling reduces matters to the following problem.  Suppose
$|(\frac{d}{dx})^\alpha W(x)|\le C_\alpha$, $W(0)=W^\prime(0)=0$,
$W^{\prime\prime}(0)>c_1>0$.  Define
$\phi_W(E)=\int_{E-W(x)>0}(E-W(x))^{1/2}\,dx$, $\psi_W(E)=\operatornamewithlimits
{lim}_{\delta\to 0+} \bigl[\int_{E-W(x)>\delta}W^{\prime\prime}(x)
(E-W(x))^{-3/2}\,dx-q(E)\delta^{-1/2}\Bigr]$ for small, positive $E$.

Then for $0<E<c_2$ we have $|(\frac{d}{dE})^\beta\phi_W(E)|,|(\frac{d}{dE})
^\beta\psi_W(E)|\le C_\beta^\prime$, with $c_2$, $C_\beta^\prime$ depending
only on $c_1$ and the $C_\alpha$.  This implies Lemma 6 by virtue of the
scaling relations $\phi(E)=\lambda\phi_W(\frac{E-V(x_0)}{S})$,
$\psi(E)=\lambda^{-1}\psi_W(\frac{E-V(x_0)}{S})$ for
$W(x)=\frac{V(x_0+Bx)-V(x_0)}{S}$.  The rest of our proof studies
$\phi_W(E)$, $\psi_W(E)$.  We can write $W(x)=(y(x))^2$ for a smooth
function $y(x)$ satisfying $y(0)=0$, $y^\prime(0)>\hat c_1>0$,
$|(\frac{d}{dx})^\alpha y(x)|\le \hat C_\alpha$ for $|x|<\hat c$.  Here,
$\hat c_1$, $\hat C_\alpha$ depend only on $c_1$ and the $C_\alpha$.
Changing variable from $x$ to $y=y(x)$ in the definitions of $\phi_W$,
$\psi_W$ gives
$$
\phi_W(E)=\int_{E-y^2>0}(E-y^2)^{1/2}\Cal T_1(y)\,dy\tag 49
$$
$$
\psi_W(E)=\lim\Bigl[\int_{E-y^2>\delta}(E-y^2)^{-3/2}\Cal T_2(y)\,dy-q(E)
\delta^{-1/2}\Bigr]\tag 50
$$
\noindent with $\Cal T_i(y)$ defined by $dx=\Cal T_1(y)\,dy$, $V^{\prime
\prime}\,dx=\Cal T_2(y)\,dy$.  The $C^\infty$--seminorms of $\Cal T_i$
are bounded a--priori in terms of $c_1$, $C_\alpha$.  In (49), (50),
we may replace $\Cal T_i(y)$ by $\frac 12[\Cal T_i(y)+\Cal T_i(-y)]$
without affecting the answer.  Hence we may suppose $\Cal T_i(y)$ is
even, and therefore $\Cal T_i(y)=F_i(y^2)$ for smooth functions $F_i$.
The $C^\infty$--seminorms of the $F_i$ are bounded a--priori in terms of
$c_1$, $C_\alpha$.  Changing variable from $y$ to $t=y^2/E$, we get
$$
\phi_W(E)=c\int_0^1E^{1/2}\cdot(1-t)^{1/2}F_1(Et)\cdot E^{1/2}\,\frac
{dt}{t^{1/2}}\quad\text{and}\tag 51
$$
$$
\psi_W(E)=c\lim_{\delta\to 0+} \Bigl[\int_0^{1-\delta}E^{-3/2}\cdot
(1-t)^{-3/2}F_2(Et)\cdot E^{1/2}\frac{dt}{t^{1/2}}-\tilde q(E)\delta^{-1/2}
\Bigr].\tag 52
$$

\noindent The desired a--priori estimates for $\phi_W(E)$ are immediate
from (51), while $\psi_W(E)$ requires a little more work.

Using the elementary fact that $\int_0^{1-\delta}(1-t)^{-3/2}\Bigl.\frac{dt}
{t^{1/2}}=\frac{2t^{1/2}}{(1-t)^{1/2}}\Bigr]_0^{1-\delta}=
\frac{2(1-\delta)^{1/2}}{\delta^{1/2}}=2\delta^{-1/2}+o(1)$,
so that $\lim_{\delta\to
0+}\Bigl[\int_0^{1-\delta}(1-t)^{-3/2}\,\frac{dt}{t^{1/2}}
-2\delta^{-1/2}\Bigr]=0$, we can rewrite (52) in the form
$$ \psi_W(E)=c\lim_{\delta\to
0}\Bigl[\int_0^{1-\delta}(1-t)^{-3/2}
\frac{F_2(Et)-F_2(E)}{E}\,\frac{dt}{t^{1/2}}-\hat
q(E)\delta^{-1/2}\Bigr] \tag 53 $$

\noindent with $\hat q(E)$ uniquely specified by the finiteness of the
limit.  In (53) the integrand has only a weak singularity at $t=1$, so the
limit is finite with $\hat q(E)=0$.  Hence (53) is equivalent to
$$
\psi_W(E)=c^\prime\int_0^1(1-t)^{-1/2}\frac{F_2(Et)-F_2(E)}{Et-E}\frac
{dt}{t^{1/2}},
$$

\noindent which we rewrite as
$$
\psi_W(E)=c^\prime\int_0^1\int_0^1(1-t)^{-1/2}F_2^\prime(sEt+(1-s)E)\,ds
\frac{dt}{t^{1/2}},\tag 54
$$

The derived a--priori estimates for $\psi_W(E)$ are immediate from (54),
completing the proof of the Lemma. $\qquad\blacksquare$
\smallskip

In comparing $\frac{\pi}{2}\frac{\tilde E(E)}{\lambda}$ with $\phi(E)+
\frac{1}{48}\psi(E)$, we make an extra assumption on $V(x)$, which we
remove later.  The extra assumption is as follows.
$$
E_\infty\ge V(x_0)+c_{\#}S.\tag"{(H7$^*$)}"
$$
\vglue 1pc
\proclaim{Lemma 7}  Under hypotheses (H1$^*$)$\ldots$(H7$^*$), we have
$$
\Bigl|\frac{\pi}{2\lambda}\tilde E(E)-(\phi(E)+\frac{1}{48}\psi(E))\Big|\le
C_{\#}\lambda^{-2+4\varepsilon}\quad\text{for}\ V(x_0)<E<V(x_0)+
\lambda^{-2\varepsilon}S.\tag 55
$$\endproclaim

\demo{Proof}  Let $E_1<\ldots<E_{j_{\text{max}}}$ be the eigenvalues of $H$
in the interval
$$
\Cal T=\{|E-(V(x_0)+\hat c_{\#}\lambda^{-2\varepsilon}S)|< \Hat{\Hat c}_{\#}
\lambda^{-2\varepsilon}S\}\quad\text{with}\ \Hat{\Hat c}_{\#}<<\hat c_{\#}
<<1.
$$

We will compute the $E_j$ by using the results of this section, and
independently by applying the WKB Eigenvalue Theorem.  Our proof is based
on comparing the results of these two computations of the $E_j$.


First of all, Lemma 4 shows that
$$
|\frac{\pi}{2\lambda}\tilde E(E_j)-\pi(k_j+1/2)|\le C_{\#}\lambda^
{-\frac {\varepsilon N}{11}-1}\quad\text{for\ integers}\ k_j.\tag 56
$$

\noindent Lemma 4 shows that the $k_j$ are all distinct.  Since $\tilde E(E)$
is increasing, we must have $k_{j+1}\ge k_j$.  Thus $k_{j+1}\ge k_j+1$
since the $k_j$ are all distinct.  We could not have $k_{j+1}\ge k_j+2$.
Otherwise, Lemma 5 would give us an eigenvalue $E^\prime$ with $|\frac{\pi}
{2\lambda}\tilde E(E^\prime)-\pi(k_j+1)|\le C_{\#}\lambda^{-\frac{\varepsilon
N}{11}-1}$, and monotonicity of $\tilde E(\cdot)$ would imply
$E_j<E^\prime<E_{j+1}$, contradicting the fact that $E_{j+1}$ is the
next eigenvalue after $E_j$.

Thus $k_{j+1}=k_j+1$, so that $k_j=j+m$, with $m$ an integer independent
of $j$.  Equation (56) becomes
$$
\Big|\frac{\pi}{2\lambda}\tilde E(E_j)-\pi(j+m+\frac 12)\Big|\le C_{\#}
\lambda^{-\frac{\varepsilon N}{11}-1}\quad (1\le j\le j_{\text{max}}).\tag 57
$$

On the other hand, we can analyze $E_1\ldots E_{j_{\text{max}}}$ using the
WKB Eigenvalue Theorem.  In fact, set $\hat I_{\text{BVP}}=I_{\text{BVP}}$,
$\hat I=\{|x-x_0|<\lambda^{-\varepsilon}B\}$, $\hat V(x)=V(x)-V(x_0)$,
$\hat S(x)=\lambda^{-2\varepsilon}S$ and $\hat B(x)=\lambda^{-\varepsilon}B$
for $x \in \hat I$, $\hat E_{\infty}=E_\infty-V(x_0)$, $\hat E_0=\hat c_{\#}
\lambda^{-2\varepsilon}S$.  These satisfy the hypotheses (Hyp0)$\ldots$
(Hyp10) of the WKB Eigenvalue Theorem, and the constants $c_1$, $C_\alpha$
etc. in (Hyp0)$\ldots$(Hyp5) depend only on the constants $c_1$, $C_{\alpha}$,
in our present hypotheses (H1$^*$)$\ldots$(H7$^*$).  In particular, we find
that $S_{\text{min}}=\lambda^{-2\varepsilon}S$, $\lambda(x)=\lambda^
{1-2\varepsilon}$ for $x \in \hat I$, and $\Lambda=\lambda^{1-2\varepsilon}$.
The WKB Eigenvalue Theorem therefore implies that the eigenvalues of $H$
which lie in $\Cal T$ may be written as $E_{p_{\text{min}}}^+<\ldots<
E_{p_{\text{max}}}^+$ with
$$
\Big|\phi(E_k^+)+\frac{1}{48}\psi(E_k^+)-\pi(k+\frac 12)\Big|\le C_{\#}
\Lambda^{-2}=C_{\#}\lambda^{-2+4\varepsilon} (p_{\text{min}}\le k\le
p_{\text{max}}).\tag 58
$$

The eigenvalues in $\Cal T$ are given as $E_1<\ldots<E_{j_{\text{max}}}$
and as $E_{p_{\text{min}}}^+<\ldots<E_{p_{\text{max}}}^+$.
Thus $E_k^+=E_{j-m^\prime}$ with $m^\prime=p_{\text{min}}-1$, so
(58) becomes
$$
\Big|\phi(E_j)+\frac{1}{48}\psi(E_j)-\pi(j+m^\prime+\frac 12)\Big|\le C_{\#}
\lambda^{-2+4\varepsilon} \quad (1\le j\le j_{\text{max}}).\tag 59
$$

Now set $f(E)=\frac{\pi}{2\lambda}\tilde E(E)-(\phi(E)+\frac{1}{48}
\psi(E))+\pi(m^\prime-m)$.  Estimates (57) and (59) imply
$$
|f(E_j)|\le C_{\#}\lambda^{-2+4\varepsilon}\quad (1\le j\le j_{\text{max}}).
\tag 60
$$

\noindent Lemma 6 and estimate (13) show that
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\Bigl\{\frac{\pi}{2\lambda}\tilde E(E)-
(\phi(E)+\frac{1}{48}\psi(E))\Bigr\}\Big|\le C_{\#}^\beta
\lambda S^{-\beta}\,\,\text{for}\,\, V(x_0)<E<V(x_0)+c_{\#}S.\tag 61
$$

\noindent Take $\beta=0$, $E=E_j$, and compare this estimate with (60),
to conclude that $|m-m^\prime|\le C_{\#}\lambda$.  This and (61) show that
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta f(E)\Big|\le C_{\#}^\beta\lambda
S^{-\beta}\quad\text{for}\ V(x_0)<E<V(x_0)+c_{\#}S.\tag 62
$$

\noindent By (62) and Taylor's theorem with remainder, we can write
$$
f(E)=P(E)+\text{error}(E),\tag 63
$$
\noindent with $P(E)$ a polynomial of degree $N$, and
$$
|\text{error}(E)|\le C_{\#}\lambda\Bigl(\frac{E-V(x_0)}{S}\Bigr)^N\le
C_{\#}\lambda^{-2+4\varepsilon}\ \text{for}\ V(x_0)<E<V(x_0)+\lambda^
{-2\varepsilon}S.\tag 64
$$

\noindent From (60), (63), (64) and $E_j \in \Cal T$, we get
$$
|P(E_j)|\le C_{\#}\lambda^{-2+4\varepsilon}.\tag 65
$$

\noindent Any $E$ in $\Cal T$ lies within $C_{\#}\frac{S}{\lambda}$ of an
$E_j$, by (57) and the fact that $\frac{d}{dE}(\frac{\pi}{2\lambda}
\tilde E(E))\sim \lambda S^{-1}$ in $\Cal T$ by (13).  Hence we can find
$E_{j_0},E_{j_1},\ldots,E_{j_N} \in \Cal T$ with $|E_{j_{\nu+1}}-
E_{j_\nu}|\sim \frac{\text{diam}\ \Cal T}{N}$.  Using (65) for the $E_{j_\nu}$
and applying the Lagrange interpolation formula, we obtain
$$
|P(E)|\le C_{\#}\lambda^{-2+4\varepsilon}\,\, \text{for}\,\, |E-V(x_0)|<
\lambda^{-2\varepsilon}S.\tag 66
$$

\noindent Estimates (63), (64), (66) together show that
$$
|f(E)|\le C_{\#}\lambda^{-2+4\varepsilon}\ \text{for}\ V(x_0)<E<V(x_0)+
\lambda^{-2\varepsilon}S.\tag 67
$$

\noindent Let us specialize (67) to $E\to V(x_0)+$.  Lemma 3 shows that
$|\tilde E(V(x_0))|=|\lambda^2\hat\tau_0|\le C_{\#}$, and Lemma 6 gives
$|\psi(E)|\le C_{\#}\lambda^{-1}$.  Immediately from the definition we
get $\phi(E)\to 0$ as $E\to V(x_0)+$.  Hence for $E$ very slightly greater
than $V(x_0)$, the definition of $f(E)$ gives $|f(E)+\pi(m-m^\prime)|\le
C_{\#}\lambda^{-1}$.  Comparing this with (67) and recalling that
$m$, $m^\prime$ are integers, we conclude that $m-m^\prime=0$.  Therefore,
(67) says that
$$
\Big|\frac{\pi}{2\lambda}\tilde E(E)-(\phi(E)+\frac{1}{48}\psi(E))\Big|\le
C_{\#}\lambda^{-2+4\varepsilon}\,\, \text{for}\,\,
V(x_0)<E<V(x_0)+\lambda^{-2\varepsilon}S,
$$
\noindent which is the conclusion of the Lemma.$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary} The conclusion of Lemma 7 holds also without the
extra assumption (H7$^*$).\endproclaim

\demo{Proof}  Suppose we modify the potential $V(x)$ in $\{|x-x_0|>c_{\#}B\}$,
without changing it in $\{|x-x_0|\le c_{\#}B\}$.  The remark
following the proof of Lemma 3 shows that $\tilde E(E)$
will not change.  Also $\phi(E)$, $\psi(E)$ will not
change, provided $V(x_0)<E<V(x_0)+c_{\#}^\prime S$.
Therefore the conclusion of Lemma 7 is unaffected by
changing $V(x)$ in $\{|x-x_0|>c_{\#}B\}$.  We can make the
change in $V$ so that $V(x)\ge
V(x_0)+\frac{c_{\#}S}{B^2}(x-x_0)^2$ globally, and we then
change $E_\infty$ to the value $V(x_0)+c_{\#}^\prime S$
for a small $c_{\#}^\prime>0$. Hypotheses
(H1$^*$)$\ldots$(H7$^*$) hold for the modified $V(\cdot)$,
$E_\infty$, and therefore the conclusion of Lemma 7 holds
for the original $V$.$\qquad\blacksquare$

We summarize the results of this section in the 
next section.






%Subject: Section 9

\pageno 156
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\centerline{\bf {The WKB Theorem on Low Eigenvalues}}
\smallskip

Let $\varepsilon,K,N>0$ be given, with $\varepsilon
N\ge 100$.  Let $V(x)$ be a potential defined on
a (possibly unbounded) interval $I_{\text{BVP}}$.  Let $S$,
$B$ be positive numbers, and let $x_0 \in I_{\text{BVP}}$
be given.  Define $\lambda=S^{1/2}B$.  Let $E_\infty$
be a given energy, with $E_\infty>V(x_0)$.  We make
the following assumptions.
\roster
 \item"{(H0$^*$)}" $I=\{|x-x_0|<cB\}\subset I_{\text{BVP}}$
\item"{(H1$^*$)}" $|(\frac{d}{dx})^\alpha V(x)|\le
C_\alpha SB^{-\alpha}$ in $I$
\item"{(H2$^*$)}" $\frac{d^2}{dx^2}V\ge c^\prime SB^{-2}$
in $I$
\item"{(H3$^*$)}" $V^\prime(x_0)=0$
\item"{(H4$^*$)}" For $x \in I_{\text{BVP}}\backslash I$
we have $V(x)\ge \text{min}\{E_\infty,V(x_0)+c^{\prime
\prime}\lambda^{-2\varepsilon} S\}$.
\item"{(H5$^*$)}" For $x \in I_{\text{BVP}}$ with
$|x-x_0|>\frac 12 \lambda^KB$, we have $V(x)\ge
E_\infty +\frac{1000}{|x-x_0|^2}$.
\item"{(H6$^*$)}" $\lambda$ is bounded below by a positive
constant depending only on $c$, $c^\prime$, $c^{\prime
\prime}$, $C_\alpha$ in (H0$^*$)$\ldots$(H1$^*$), and on
$\varepsilon$, $K$, $N$.\endroster

Let $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_{\text{BVP}})$
with Dirichlet or Neumann conditions at the endpoints.

For $V(x_0)<E<V(x_0)+\lambda^{-2\varepsilon}S$, define
$x_{\text{left}}(E)<x_{\text{rt}}(E)$ to be the
two values of $x \in I$ at which $V(x)=E$.  Then define
$$
\phi(E)=\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
(E-V(x))^{1/2}\,dx
$$
$$\align
\psi(E)&=\lim_{\delta\to 0+}\Bigl[\operatornamewithlimits{\int} \Sb x \in I\\
E-V(x)>\delta  \endSb V^{\prime\prime}(x)(E-V(x))^{-3/2}\,dx
-q(E)\delta^{-1/2}\Bigr]\\
&=\lim_{\delta_{\text{left}},\delta_{\text{rt}}\to 0+}
\Bigl[\int_{x_{\text{left}}(E)+\delta_{\text{left}}}
^{x_{\text{rt}}(E)-\delta_{\text{rt}}} V^{\prime\prime}
(x)(E-V(x))^{-3/2}\,dx-q_{\text{left}}(E)\delta_{\text{
left}}^{-1/2}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\qquad\qquad\,\,\,\,\,
-q_{\text{rt}}(E)\delta_{\text{rt}}^{-1/2}
\Bigr]\endalign
$$
\noindent with $q(E)$, $q_{\text{left}}(E)$, $q_{\text{
rt}}(E)$ uniquely specified by demanding the finiteness
of the limits.

\vglue 1pc
\proclaim{Lemma 1}  The phases $\phi(E)$, $\psi(E)$
satisfy the estimates
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta \phi(E)\Big|\le
C_{\#}^\beta\lambda S^{-\beta}
$$
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\psi(E)\Big|\le
C_{\#}^\beta\lambda^{-1}S^{-\beta}
$$
$$
\frac{d}{dE}\phi(E)\ge c_{\#}\lambda S^{-1}
$$
\noindent for $V(x_0)<E<V(x_0)+\lambda^{-2\varepsilon}S$.

\noindent The constants $c_{\#}$, $C_{\#}^\beta$ depend
only on $c$, $c^\prime$,$ c^{\prime\prime}$, $C_{\alpha}$,
$\varepsilon$, $K$, $N$ in hypotheses (H0$^*$)$\ldots$
(H6$^*$).\endproclaim

\vglue 1pc
\proclaim{WKB Theorem on Low Eigenvalues} Assume
(H0$^*$)$\ldots$(H6$^*$).  Then there is a finite
sequence $E_0,E_1,\ldots,E_{k_{\max}}$ with the
following properties.
\roster
\item"{(a)}" Let $w=\phi(E_*)+\frac{1}{48}\psi(E_*)$
with $E_*=\text{min}\{E_\infty,V(x_0)+c_{\#}\lambda^{-2\varepsilon}S\}$,
and let  $\overline n$ be the largest integer
with $\pi(\overline n+1/2)\le  w$.  If
$\operatornamewithlimits{min}_{k \in \Bbb Z}|w-
\pi(k+1/2)|>C_{\#}\lambda^{-2+4\varepsilon}$, then
$k_{\text{max}}=\overline n$.  In any case,
$|k_{\text{max}}-\overline n|\le 1$.
\item"{(b)}" If $0\le k<k_{\text{max}}$, then $E_k$
is an eigenvalue of $H$.
\item"{(c)}" Either $E_{k_{\text{max}}}=E_\infty$ or
else $E_{k_{\text{max}}}$ is an eigenvalue of $H$.
\item"{(d)}" Every eigenvalue $E$ of $H$ satisfying
$E\le E_\infty$,
$|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S$ is one of
the $E_k$.
\item"{(e)}" For $0 \le k\le k_{\text{max}}$ we have
$V(x_0)<E_k<V(x_0)+2c_{\#}\lambda^{-2\varepsilon}S$
and $|\phi(E_k)+\frac{1}{48}\psi(E_k)-\pi(k+1/2)|\le
C_{\#}\lambda^{-2+4\varepsilon}$.
\endroster
\noindent The constants $c_{\#}$, $C_{\#}$ depend only on
$\varepsilon$, $K$, $N$, $c$, $c^\prime$,
$c^{\prime\prime}$, $C_\alpha$ in hypotheses
(H0$^*$)$\ldots$(H6$^*$).\endproclaim

\vglue 1pc
\proclaim{Lemma 2}  Assume (H0$^*$)$\ldots$(H6$^*$).
Let $F(x)$ be an eigenfunction of $H$ whose eigenvalue
$E$ satisfies
$V(x_0)<E<V(x_0)+c_{\#}\lambda^{-2\varepsilon}S$ and $E\le
E_\infty$.  Then
$$
\int_{I_{\text{BVP}}\backslash\{|x-x_0|
<\lambda^{-\varepsilon}B\}}|F(x)|^2\,dx\le
C_{\#}\lambda^{-N}\int_{\{|x-x_0|<\lambda^{-\varepsilon}
B\}}|F(x)|^2\,dx.
$$
\noindent The constants $c_{\#}$, $C_{\#}$ depend only 
on $\varepsilon$, $K$, $N$, $c$, $c^\prime$,
$c^{\prime\prime}$, $C_\alpha$ in the hypotheses
(H0$^*$)$\ldots$(H6$^*$).\endproclaim
\demo{Proofs of the results}  Lemma 6 of the preceding
section contains the present Lemma 1, except for the
trivial estimate $\frac{d\phi}{dE}>c_{\#}\lambda S^{-1}$,
which we leave to the reader.  The WKB Theorem on low
eigenvalues follows at once from the corollaries to
Lemmas 5 and 7 in the preceding section.  Finally, our
present Lemma 2 just restates Lemma 1 from the preceding
section.$\qquad\blacksquare$


%Subject: Section 10

\pageno 159
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\centerline{\bf{WKB Theory with Weak Turning Points}}
\smallskip

In this section, we develop a crude form of WKB
Theory that requires only very weak hypotheses on the
potential near the turning points.

\noindent\underbar{Set--up}.  We are given an energy
$E_0$ and a potential $V(x)$ defined on a (possibly
unbounded) interval $I_{\text{BVP}}$.  The interval $I_{\text{BVP}}$
is partitioned into subintervals
 $I_{\text{far\ left}}$, $I_{\text{left}}$,
$I_{\text{center}}$, $I_{\text{rt}}$, $I_{\text{far\ rt}}$
with $I_{\text{far\ left}}$ to the left of
$I_{\text{left}}$, $I_{\text{left}}$ to the left of
$I_{\text{center}}$, etc. Here, $I_{\text{far\ left}}$ and 
$I_{\text{far\ right}}$ may be empty.  On $I_{\text{center}}$ we
are given positive weight functions $S(x)$, $B(x)$.  Set
$\lambda(x)=S^{1/2}(x)B(x)$ and
$\Lambda=(\int_{I_{\text{center}}}\frac{dx}
{\lambda(x)B(x)})^{-1}$.  We make the following
assumptions.

\noindent\underbar{Hypotheses}

\roster
\item"{(H$\hat 0$)}" $I_{\text{center}}$ is non--empty,
and for $x,y \in I_{\text{center}}$ with $|x-y|<cB(x)$
we have $c<B(y)/B(x)<C$ and $c<S(y)/S(x)<C$, and
$|I_{\text{center}}|>cB(x)$.

\item"{(H$\hat 1$)}" For $x \in I_{\text{center}}$ we
have $|(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x)
B^{-\alpha}(x)$ and $cS(x)<E_0-V(x)<CS(x)$.

\item"{(H$\hat 2$)}" $\Lambda$ is bounded below by a
large number depending only on $c$, $C$, $C_\alpha$ in
(H$\hat0$), (H$\hat1$).

\item"{(H$\hat 3$)}" $I_{\text{left}}$, $I_{\text{rt}}$
are non--empty.  If $I_{\text{center}}=[x_{\text{left}},
x_{\text{rt}}]$, then we have $|I_{\text{left}}|\le
\underline{C}B(x_{\text{left}})$, $|I_{\text{rt}}|\le
\underline{C}B(x_{\text{rt}})$, $\lambda(x_{\text{left}})
\le \underline{C}$, $\lambda(x_{\text{rt}})\le 
\underline{C}$.

\item"{(H$\hat 4$)}" If $I_{\text{left}}=[x_{\text{far\
left}},x_{\text{left}}]$, then we have $|V(x)-E_0|\le
\underline{C}|I_{\text{left}}|^{-1}(x-\hat x_{\text{far\
left}})^{-1}$ in $I_{\text{left}}$.  Here
$\hat x_{\text{far\ left}}\le x_{\text{far\ left}}$ with
strict inequality unless $I_{\text{far\ left}}=\emptyset$.

\item"{(H$\hat 5$)}" For $x \in I_{\text{rt}}$ we have
$|V(x)-E_0|\le \underline{C}|I_{\text{rt}}|^{-2}$.

\item"{(H$\hat 6$)}" For $x \in I_{\text{far\ left}}$
we have $V(x)-E_0\ge \underline{c}|I_{\text{left}}|^{-2}$.
$V(x)$ is $C^\infty$ in the interior of $I_{\text{far\
left}}$.

\item"{(H$\hat 7$)}" For $x \in I_{\text{far\ rt}}$
we have $V(x)-E_0\ge
\frac{-10^{-9}}{(x-x_{\text{rt}})^2}$.\endroster

Our goal is to understand the eigenfunctions and
eigenvalues of $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_
{\text{BVP}})$ with Dirichlet boundary conditions.

\noindent Denote by $C_{\#}$ a constant depending on $c$,
$C$, $C_\alpha$ in (H$\hat 0$), (H$\hat 1$).
\noindent Denote by $C_*$, $c_*$ etc. constants
depending only on $c$, $C$, $C_\alpha$, $\underline{c}$,
$\underline{C}$ in (H$\hat 0$)$\ldots$(H$\hat 7$).

Note that $I_{\text{left}}$ and $I_{\text{rt}}$ don't
play completely analogous r$\hat{\text{o}}$les in our
hypotheses.

We start by studying $I_{\text{center}}$.

\vglue 1pc
\proclaim{Lemma 1}  Let $I_\nu=\{|x-x_\nu|<c_{\#}B(x_\nu)
\}\subset I_{\text{center}}$.  Then the equation
$[\frac{d^2}{dx^2}+E_0-V(x)]u=f$ has a solution on
$I_\nu$, with $\Vert u\Vert_{L^\infty(I_\nu)}\le
C_{\#}\frac{B^2(x_\nu)}{\lambda(x_\nu)}\Vert f\Vert_
{L^\infty(I_\nu)}$ and $\Vert\frac{du}{dx}\Vert_{
L^\infty(I_\nu)}\le C_{\#}B(x_\nu)\Vert f\Vert_
{L^\infty(I_\nu)}$.\endproclaim

\demo{Proof}  Put $F_c(x)=\frac{\exp(i\int_{x_{\text{left
}}}^x(E_0-V(t))^{1/2}\,dt)}{(E_0-V(x))^{1/4}}$
on $I_\nu$.  Then
$$\split
\Big|\Bigl[\frac{d^2}{dx^2}+(E_0-V(x))\Bigr]F_c(x)\Big|=
\Big|\frac{5}{16}\frac{(V^\prime)^2}{(E_0-V(x))^{9/4}}
+\frac 14\frac{V^{\prime\prime}}{(E_0-V(x))^{5/4}}
\Big|\\
\le C_{\#}B^{-2}(x_\nu)S^{-1/4}(x_\nu),\endsplit
$$

\noindent while $|F_c(x)|\le C_{\#}S^{-1/4}(x_\nu)$. 
Using $F_c$ we construct the approximate Green's
function
$$
G(x,y)=F_c(x)\overline{F_c(y)}\ \text{for}\ x \le y,\quad
G(x,y)=\overline{F_c(x)}F_c(y)\ \text{for}\ x>y.
$$

\noindent Thus
$$
|G(x,y)|\le C_{\#}S^{-1/2}(x_\nu),
$$

\noindent and as distributions we have
$$
\Bigl[\frac{\partial^2}{\partial x^2}+(E_0-V(x))\Bigr]
G(x,y)=H(y)\delta(x-y)+K(x,y)
$$

\noindent with $c\, H(y)=\text{Im}(F_c^\prime(y)
\overline{F_c(y)})$ and $|K(x,y)|\le C_{\#}B^{-2}(x_\nu)
S^{-1/2}(x_\nu)$.  We have $F_c^\prime\overline{F_c}(y)
=i+\frac{(\text{const.})\ V^\prime(y)}{(E_0-V(y))^{3/2}}
=i+\Cal E(y)$, with $|
\Cal E(y)|\le C_{\#}S^{-1/2}
(x_\nu)B^{-1}(x_\nu)=C_{\#}/\lambda(x_\nu)$.  This 
implies $|H(y)|>c_{\#}$ on $I_\nu$.

Now set $u(x)=\int_{I_\nu}G(x,y)h(y)\,dy$ with $h$ to
be determined.  Thus
$$
\Bigl[\frac{d^2}{dx^2}+E_0-V(x)\Bigr]u(x)=H(x)h(x)+
\int_{I_\nu}K(x,y)h(y)\,dy,
$$

\noindent so $u$ solves our ODE provided we take $h$
to solve
$$
H(x)h(x)+\int_{I_\nu}K(x,y)h(y)\,dy=f(x).
$$

As an operator on $L^\infty(I_\nu)$, $Th(x)=\int_{I_\nu}
K(x,y)h(y)\,dy$ has norm\hfill\break
 $\operatornamewithlimits{
ess\ sup}_{x \in I_\nu}[\int_{I_\nu}|K(x,y)|\,dy]\le
C_{\#}B^{-1}(x_\nu)S^{-1/2}(x_\nu)=C_{\#}/\lambda(x_\nu)$.
Since $|H(x)|\ge c_{\#}$, we can solve the integral
equation by a Neumann series, obtaining also the bound
$\Vert h\Vert_{L^\infty(I_\nu)}\le C_{\#}\Vert f\Vert_
{L^\infty(I_\nu)}$.  To estimate $u$, we note that
$$\align
|u(x)|&\le \int_{I_\nu}|G(x,y)|\,dy\cdot\Vert h\Vert_
{L^\infty(I_\nu)}\le C_{\#}B(x_\nu)S^{-1/2}
(x_\nu)\Vert h\Vert_{L^\infty(I_\nu)}\\
&\le C_{\#}B(x_\nu)S^{-1/2}(x_\nu)\Vert f\Vert_{
L^\infty(I_\nu)}=\frac{C_{\#}B^2(x_\nu)}{\lambda(x_\nu)}
\Vert f\Vert_{L^\infty(I_\nu)}.\endalign
$$

\noindent Similarly,
$$\split
\Big|\frac{du(x)}{dx}\Big|\le \int_{I_\nu}
\Big|\frac{\partial}{\partial x}G(x,y)\Big|\,dy\cdot
\Vert h\Vert_{L^\infty(I_\nu)}\le C_{\#}
S^{-1/4}(x_\nu)\int_{I_\nu}\Big|\frac{dF_c(x)}
{dx}\Big|\,dx\\
\cdot\Vert f\Vert_{L^\infty(I_\nu)}.\endsplit
$$

We saw that $|F_c^\prime(y)\overline{F_c(y)}|=
|i+\Cal E(y)|<2$, so $|F_c^\prime(y)|\le
2(E_0-V(y))^{1/4}\le C_{\#}S^{+1/4}(x_\nu)$.  Therefore
the previous estimate becomes $|\frac{du(x)}{dx}|\le\hfill\break
C_{\#}B(x_\nu)\Vert f\Vert_{L^\infty(I_\nu)}$,
which completes the proof.$\qquad\blacksquare$

\vglue 1pc
\proclaim{Lemma 2}  Let $I_\nu=\{|x-x_\nu|<c_{\#}B(
x_\nu)\}\subset I_{\text{center}}$.  Then we can find
a complex--valued solution of $[\frac{d^2}{dx^2}+E_0
-V(x)]F_\nu(x)=0$ on $I_\nu$, of the form
$$
F_\nu(x)=(E_0-V(x))^{-1/4}\exp(i\int_{x_{\text{left}}}
^x(E_0-V(t))^{1/2}\,dt)-\text{Error}_\nu(x),
$$
\noindent with $|\text{Error}_\nu|\le \frac{C_{\#}
S^{-1/4}(x_\nu)}{\lambda(x_\nu)}$ and
$|\frac{d}{dx}\text{Error}_\nu(x)|\le \frac{C_{\#}
S^{+1/4}(x_\nu)}{\lambda(x_\nu)}$.\endproclaim

\demo{Proof} Let $f=[\frac{d^2}{dx^2}+(E_0-V(x))]\{
(E_0-V(x))^{-1/4}\exp(i\int_{x_{\text{left}}}^x
(E_0-V(t))^{1/2}\,dt)\}$, and let $\text{Error}_\nu(x)$
be the function $u(x)$ given by the previous lemma.  At
the start of the proof of Lemma 1, we saw that
$\Vert f\Vert_{L^\infty(I_\nu)}\le C_{\#}
B^{-2}(x_\nu)S^{-1/4}(x_\nu)$, hence by lemma 1,
$$
\Vert\text{Error}_\nu\Vert_{L^\infty(I_\nu)}\le
C_{\#}\frac{ B^{2}(x_\nu)}{\lambda(x_\nu)}\cdot B^{-2}(x_\nu)S^{-1/4}(x_\nu)=
\frac{C_{\#}S^{-1/4}(x_\nu)}{\lambda(x_\nu)}
$$

\noindent and
$$\split
\Vert\frac{d}{dx}\ \text{Error}_\nu\Vert_{L^\infty(I_\nu)}
\le C_{\#}B(x_\nu)\cdot C_{\#}B^{-2}(x_\nu)S^{-1/4}
(x_\nu)\\
=\frac{C_{\#}S^{+1/4}(x_\nu)}{S^{1/2}(x_\nu)B(x_\nu)}=
\frac{C_{\#}S^{+1/4}(x_\nu)}{\lambda(x_\nu)}.\endsplit
$$

\noindent Also
$(E_0-V(x))^{-1/4}\exp(i\int_{x_{\text{left}}}^x(E_0-V(t))
^{1/2}\,dt)-\text{Error}_\nu(x)$ solves our ODE.
$\qquad\blacksquare$

\vglue 1pc
\proclaim{Lemma 3}  Let $I_\nu$ be as in the previous
lemma.  Suppose $\theta_\nu \in C_0^\infty(I_\nu)$
with $|(\frac{d}{dx})^\alpha\theta_\nu(x)|\le
C_{\#}^\alpha B^{-\alpha}(x_\nu)$.  Then

\roster
\item"{(a)}"
$\Big|\int_{I_\nu}\theta_\nu\exp(\pm 2i\int_
{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\,dx\Big|\le C_{\#}
\frac{B(x_\nu)}{\lambda(x_\nu)}$.
\item"{(b)}" Let $F_\nu(x)$ be the ODE solution
given by the previous lemma.  For $A$ a (complex)
constant, we have
$$\split
\Big|\int_{I_\nu}\theta_\nu(x)|\text{Re}(AF_\nu(x))|^2\,
dx-\frac 12|A|^2\int_{I_\nu}\frac{\theta_\nu(x)\,dx}
{(E_0-V(x))^{1/2}}\Big|\\
\le C_{\#}\frac{|A|^2S^{-1/2}(x_\nu)B(x_\nu)}
{\lambda(x_\nu)}.\endsplit
$$
\item"{(c)}" $F_\nu$ and $\overline F_\nu$ are linearly
independent functions.
\endroster
\endproclaim

\demo{Proof}  $|\int_{I_\nu}\theta_\nu(x)\exp
(\pm 2i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\,dx|=\hfill\break
(\text{const.})
\Big|\int_{I_\nu}\frac
{\theta_\nu(x)}{(E_0-V(x))^{1/2}}\frac{d}{dx}
\Bigl\{\exp(\pm 2i\int_{x_{\text{left}}}^x(E_0-V(t))^
{1/2}\,dt)\Bigr\}\,dx\Big|\hfill\break
=(\text{const.}) \Big|\int_{I_\nu}\exp(\pm 2i
\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\frac{d}{dx}
\Big\{\frac{\theta_\nu(x)}{(E_0-V(x))^{1/2}}\Bigr\}
\,dx\Big|\hfill\break
\le C_{\#}\int_{I_\nu}\Bigl\{\frac{|\theta_\nu^\prime|}
{(E_0-V)^{1/2}}+\frac{|\theta_\nu||V^\prime|}
{(E_0-V)^{3/2}}\Bigr\}\,dx\le C_{\#}\int_{I_\nu}
S^{-1/2}(x_\nu)B^{-1}(x_\nu)\,dx\hfill\break
=\frac{C_{\#}B(x_\nu)}{\lambda(x_\nu)}$, proving (a).

To prove (b), we write
$F_\nu(x)=(1+\text{Error}_\nu(x))\cdot(E_0-V(x))^{-1/4}
\cdot\exp(i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,
dt)$ with $|\text{Error}_\nu(x)|\le \frac{C_{\#}}
{\lambda(x_\nu)}$ by virtue of Lemma 2.  Hence
$$\multline
[\text{Re}(AF_\nu)]^2=\frac 12 |A|^2|F_\nu|^2+\frac 14
A^2F_\nu^2+\frac 14 \overline A\,\overline F_\nu^2=\\
\frac 12 |A|^2\cdot(E_0-V(x))^{-1/2}\cdot|1+\text{Error}
_\nu(x)|^2+\frac 14 A^2\cdot(E_0-V(x))^{-1/2}\cdot
(1+\text{Error}_\nu(x))^2\\
\cdot\exp(2i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\\
+\frac 14 \overline A^2\cdot(E_0-V(x))^{-1/2}\cdot
(1+\overline{\text{Error}_\nu(x)})^2\cdot\exp
(-2i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt).
\endmultline
$$

\noindent This implies
$$\multline
\Big|\int_{I_\nu}\theta_\nu\Bigl[\text{Re}(AF_\nu)\Bigr]
^2\,dx
-\frac 12 \int_{I_\nu}\theta_\nu|A|^2(E_0-V(x))^{-1/2}\,
dx\Big|\le\\
\sum\limits_{\pm} C_{\#}|A|^2\Big|\int_{I_\nu}\frac
{\theta_\nu(x)}{(E_0-V(x))^{1/2}}\,e^{\pm 2i\int_
{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt}\,dx\Big|\\
+C_{\#}|A|^2\int_{I_\nu}\frac{|\theta_\nu(x)|}
{(E_0-V(x))^{1/2}} |\text{Error}_\nu(x)|\,dx,\endmultline
$$
\noindent since $\Big||1+\text{Error}_\nu|^2-1\Big
|, |(1+\text{Error}_\nu)^2-1|,
|(1+\overline{\text{Error}_\nu})^2-1 |\le
C_{\#}|\text{Error}_\nu|$.

The sum over $\pm$ may be estimated by part (a), and
the last term on the right is estimated by using
$|\text{Error}_\nu(x)|\le \frac{C_{\#}}{\lambda(x_\nu)}$.


The result is
$$\split
\Big|\int_{I_\nu}\theta_\nu\Bigl[\text{Re}(AF_\nu)\Big]^2\,dx-
\frac 12 |A|^2\int_{I_\nu}\frac{\theta_\nu\,dx}{(E_0-V)
^{1/2}}\Big|\le C_{\#}\frac{|A|^2S^{-1/2}(x_\nu)B(x_\nu)}{\lambda
(x_\nu)},\\
\text{which\ is\ (b)}.\endsplit
$$

\noindent Finally (c) follows from (b)
since $F_\nu$, $\overline F_\nu$ linearly dependent
implies $\text{Re}(AF_\nu)=0$ for a complex number $A$
with $|A|=1$.  From (b) we would then learn that
$|\int_{I_\nu}\frac{\theta_\nu\,dx}{(E_0-V)^{1/2}}|
\le C_{\#}\frac{S^{-1/2}(x_\nu)B(x_\nu)}{\lambda(x_\nu)}$,
which is plainly false. $\qquad\blacksquare$

\vglue 1pc
\proclaim{Lemma 4}  Let
$I_\nu=\{|x-x_\nu|<c_{\#}B(x_\nu)\}$ and $I_{\nu+1}
=\{|x-x_{\nu+1}|<c_{\#}B(x_{\nu+1})\}$ be subintervals of
$I_{\text{center}}$, with $|I_\nu\cap I_{\nu+1}|
>c_{\#}|I_\nu|$.  Let $F_\nu(x)$, $F_{\nu+1}(x)$
be the solutions of $[\frac{d^2}{dx^2}+(E_0-V(x))]F=0$
given by Lemma 2 on $I_\nu$, $I_{\nu+1}$ respectively.
Then on $I_\nu\cap I_{\nu+1}$ we have
$$
F_{\nu+1}(x)=A_\nu^1 F_\nu(x)+A_\nu^2\overline{F_\nu
(x)}\quad\text{with}\,\, |A_\nu^1-1|,\,\, |A_\nu^2|\le
\frac{C_{\#}}{\lambda(x_\nu)}.\tag 1
$$\endproclaim

\demo{Proof}  Since $F_\nu$, $\overline F_\nu$ are two
independent solutions of the ODE for $F_{\nu+1}$,
we can represent $F_{\nu+1}$ by (1), and the only
problem is to bound the $A_\nu^i$.  We will integrate
(1) against $g_\nu^\pm(x)=(E_0-V(x))^{+1/4}\theta_\nu
(x)\text{exp}(\pm i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,
dt)$.  Here we take $\theta_\nu\in C_0^\infty(I_\nu\cap
I_{\nu+1})$ with $\theta_\nu\ge 0$, \
$\operatornamewithlimits{max}_{x \in
I_\nu}\theta_\nu(x)=1$, and
$|(\frac{d}{dx})^\alpha\theta_\nu|\le
C_{\#}^\alpha B^{-\alpha}(x_\nu)$.  Lemma 2 gives
$$
\int_{I_\nu}g_\nu^-F_\nu\,dx=\int_{I_\nu}\theta_\nu(x)\,dx
+0(\int_{I_\nu}|g_\nu^-(x)||\text{Error}_\nu(x)|\,dx),
$$

\noindent with the last integral dominated by
$C_{\#}\frac{B(x_\nu)}{\lambda(x_\nu)}$.  Thus,
$|\int_{I_\nu}g_\nu^-F_\nu\,dx-\int_{I_\nu}\theta_\nu(x)\,
dx|\le C_{\#}B(x_\nu)/\lambda(x_\nu)$.  Similarly,
$|\int_{I_\nu}g_\nu^-F_{\nu+1}\,dx-\int_{I_\nu}\theta_\nu
(x)\,dx|\le C_{\#}B(x_\nu)/\lambda(x_\nu)$ and
$|\int_{I_\nu}g_\nu^+\overline F_\nu\,dx-\int_{I_\nu}
\theta_\nu(x)\,dx|\le C_{\#}B(x_\nu)/\lambda(x_\nu)$.
Also
$$\align
\Big|\int_{I_\nu}g_\nu^+F_\nu\,dx\Big|&\le\Big|\int_{I_\nu}\theta_\nu(x)
\exp(+2i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\,dx\Big|\\
&\,\,\,+\int_{I_\nu}|g_\nu^+(x)||\text{Error}_\nu(x)|\,dx.
\endalign
$$

\noindent Lemmas (2) and (3a) show that both terms on the
right are dominated by $C_{\#}\frac{B(x_\nu)}
{\lambda(x_\nu)}$, so
$|\int_{I_\nu}g_\nu^+F_\nu\,dx|\le C_{\#}B(x_\nu)/
\lambda(x_\nu)$.  Similarly, $|\int_{I_\nu}g_\nu^+
F_{\nu+1}\,dx|\le C_{\#}B(x_\nu)/\lambda(x_\nu)$ and
$|\int_{I_\nu}g_\nu^-\overline F_\nu\,dx|\le C_{\#}
B(x_\nu)/\lambda(x_\nu)$.

\noindent Using these estimates, we can integrate
(1) against $g_\nu^\pm(x)\,dx$ on $I_\nu$, to obtain
the following equations
$${}\cases 
X_\nu=A_\nu^1P_{11}^\nu+A_\nu^2P_{12}^\nu&{}\\
Y_\nu=A_\nu^1P_{21}^\nu+A_\nu^2P_{22}^\nu&{}\endcases\tag 2
$$

\noindent with 
$$\gather
|X_\nu-\int_{I_\nu}\theta_\nu\,dx|\le C_{\#}
B(x_\nu)/\lambda(x_\nu)\\
|Y_\nu|\le C_{\#}B(x_\nu)/\lambda(x_\nu)\\
|P_{11}^\nu-\int_{I_\nu}\theta_\nu\,dx|,\,\,
|P_{22}^\nu-\int_{I_\nu}\theta_\nu\,dx|\le
C_{\#}B(x_\nu)/\lambda(x_\nu)\\
|P_{12}^\nu|, |P_{21}^\nu|\le C_{\#}B(x_\nu)/
\lambda(x_\nu).\endgather
$$

Dividing (2) by $\int_{I_\nu}\theta_\nu\,dx\sim B(x_\nu)$,
we obtain equations (2) with $|X_\nu-1|$, $|Y_\nu|$, $|P_{11}^\nu-1|$,
$|P_{22}^\nu-1|$, $|P_{12}^\nu|$, $|P_{21}^\nu|\le \frac{C_{\#}}
{\lambda(x_\nu)}$.  Hence  $|A_\nu^1-1|$, $|A_\nu^2|\le
\frac{C_{\#}}{\lambda(x_\nu)}$ as asserted.$\qquad
\blacksquare$

\vglue 1pc
\proclaim{Lemma 5}  Let $u$ be a real--valued solution of
$[-\frac{d^2}{dx^2}+(E_0-V(x))]u=0$ on
$I_{\text{center}}$.
For a (complex) constant $Q$ we can express $u$ as

$$\multline
u(x)=\text{Re}\Bigl[Q(E_0-V(x))^{-1/4}\Bigl\{e^{i
\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt}+\text{Error}
(x)\Bigr\}\Bigr]\quad\text{with}\\
|\text{Error}(x)|\le C_{\#}/\Lambda
\quad\text{and}\,\, |\frac{d}{dx}\text{Error}(x)|\le C_{\#}
S^{1/2}(x)/\Lambda.\endmultline\tag 3
$$

\noindent Equivalently,
$$\multline
u(x)=\text{Re}\Bigl[Q(E_0-V(x))^{-1/4}(1+\text{Error}
(x))\exp(i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)
\Bigr]\quad\text{with}\\
|\text{Error}(x)|\le C_{\#}/\Lambda\quad\text{and}\,\,
|\frac{d}{dx}\text{Error}(x)|\le C_{\#}S^{1/2}
(x)/\Lambda.\endmultline\tag 4
$$\endproclaim

\demo{Proof}  Cover $I_{\text{center}}$ by $I_\nu=
\{|x-x_\nu|<c_{\#}B(x_\nu)\}$ so that $|I_\nu \cap
I_{\nu+1}|\ge c_{\#}|I_\nu|$.  For each $I_\nu$,
let $F_\nu$ be the ODE solution given by lemma 2 on
$I_\nu$.  Since $F_\nu$, $\overline F_\nu$ are
independent solutions of the ODE for $u$ on $I_\nu$,
we can express $u$ on $I_\nu$ as $u=\text{Re}
[Q_\nu F_\nu]$.  The complex constant $Q_\nu$ is
uniquely determined.  On $I_{\nu+1}\cap I_\nu$
we have $F_{\nu+1}=A_\nu^1F_\nu+A_\nu^2\overline F_\nu$
as in lemma 4, hence on $I_{\nu+1}\cap I_\nu$
$u=\text{Re}[Q_{\nu+1}F_{\nu+1}]=\hfill\break
\text{Re}[Q_{\nu+1}(A_\nu^1F_\nu+A_\nu^2\overline
F_\nu)]=\text{Re}[Q_{\nu+1}
A_\nu^1F_\nu+Q_{\nu+1}A_\nu^2\overline
F_\nu]=\hfill\break
\text{Re}[(Q_{\nu+1}A_\nu^1+\overline Q_{\nu+1} \overline
A_\nu^2)F_\nu]$.  Since $Q_\nu$ is uniquely determined,
this implies $Q_\nu=Q_{\nu+1} A_\nu^1+\overline
Q_{\nu+1}\overline A_\nu^2$.  Lemma 4 gives $|A_\nu^1-1|$,
$|A_\nu^2|\le \frac{C_{\#}} {\lambda(x_\nu)}$, hence
$|Q_\nu-Q_{\nu+1}|\le
\frac{C_{\#}|Q_{\nu+1}|}{\lambda(x_\nu)}$. Equivalently,
$Q_{\nu+1}=Q_\nu\exp(b_\nu)$ with $b_\nu$ a complex number
and $|b_\nu|\le \frac{C_{\#}} {\lambda(x_\nu)}$.  Since
$\sum\limits_\nu\frac{C_{\#}} {\lambda(x_\nu)}\le
\frac{C_{\#}^\prime}{\Lambda}$, it follows that
$Q_\nu=Q\exp(b_\nu^\prime)$ for a complex number $Q$
independent of $\nu$, and with $|b_\nu^\prime|\le
C_{\#}/\Lambda$.  Equivalently, $|Q_\nu-Q|\le
\frac{C_{\#}|Q|}{\Lambda}$.

On $I_\nu$ we have
$$\split
u(x)=\text{Re}[Q_\nu
F_\nu(x)]=\text{Re}[Q_\nu(E_0-V(x))^{-1/4}
\exp(i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\\
+Q_\nu w_\nu(x)]\endsplit $$

\noindent with $|w_\nu(x)|\le \frac{C_{\#}S^{-1/4}
(x_\nu)}{\lambda(x_\nu)}$, $|\frac{dw_\nu(x)}
{dx}|\le \frac{C_{\#}S^{+1/4}(x_\nu)}
{\lambda(x_\nu)}$, by Lemma 2.  Hence
$$\multline
u=\text{Re}\Bigl[Q(E_0-V(x))^{-1/4}\exp(i\int_{x_{\text{left}}}
^x(E_0-V(t))^{1/2}\,dt)+\Cal E_\nu(x)\Bigr]\ \text{on}\,\,
I_\nu,\\
\text{with}\ \Cal E_\nu(x)=(Q_\nu-Q)(E_0-V(x))^{-1/4}
\exp(i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)+Q_\nu
w_\nu(x).\endmultline\tag 5
$$

\noindent Our bounds on $|Q_\nu-Q|$ and on $|w_\nu(x)|$,
$|\frac{d}{dx}w_\nu(x)|$ gives us the estimates
$$|\Cal E_\nu(x)|\le C_{\#}S^{-1/4}(x_\nu)|Q|/\Lambda,\,\,
|\frac{d}{dx}\Cal E_\nu(x)|\le C_{\#}S^{+1/4}
(x_\nu)|Q|/\Lambda.\tag 6
$$

Replacing $\Cal E_\nu(x)$ by its real part, we
preserve (5) and (6).  Now $\Cal E_\nu(x)$
is uniquely specified on $I_\nu$ by (5) and $\text{Im}\
\Cal E_\nu\equiv 0$, so $\Cal E_{\nu+1}(x)=\Cal E_\nu(x)$
on $I_\nu\cap I_{\nu+1}$.  Therefore, (5) and (6) become
$$
u(x)=\text{Re}\Bigl[Q(E_0-V(x))^{-1/4}\exp(i
\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)+\Cal E(x)\Bigr]\ \text{on}\
I_{\text{center}}\tag 7
$$

\noindent with
$$
|\Cal E(x)|\le \frac{C_{\#}S^{-1/4}(x)|Q|}{\Lambda},\,\,
|\frac{d}{dx}\Cal E(x)|\le \frac{C_{\#}S^{1/4}(x)|Q|}
{\Lambda}.\tag 8
$$

Writing $\text{Error}(x)=\frac{\Cal E(x)}{Q}\cdot
(E_0-V(x))^{+1/4}$, we get (3) with
$|\text{Error}(x)|\le \frac{C_{\#}}{\Lambda}$,
$|\frac{d}{dx}\ \text{Error}(x)|\le
\frac{C_{\#}S^{+1/4}(x)|Q|}{\Lambda}\cdot
\frac{S^{+1/4}(x)}{|Q|}+\frac{C_{\#}S^{-1/4}(x)|Q|}
{\Lambda}\cdot\frac{S^{+1/4}(x)B^{-1}(x)}{|Q|}\le\hfill\break
\frac{C_{\#}}{\Lambda}\{S^{1/2}(x)+B^{-1}(x)\}=
\frac{C_{\#}S^{1/2}(x)}{\Lambda}\{1+\lambda^{-1}
(x)\}\le \frac{C_{\#}^\prime S^{1/2}(x)}{\Lambda}$.
These are the desired estimates for (3).

We can also put $\text{Error}(x)=\frac{\Cal E(x)}{Q}
(E_0-V(x))^{+1/4}\exp(-i\int_{x_{\text{left}}}^x
(E_0-V(t))^{1/2}\,dt)$, so that (4) holds, together
with the estimate $|\text{Error}(x)|\le C_{\#}/\Lambda$
and
$$\multline
\Big|\frac{d}{dx}\ \text{Error}(x)\Big|\le
\frac{C_{\#}S^{+1/4}(x)|Q|}{\Lambda}\cdot\frac
{S^{+1/4}(x)}{|Q|}+\frac{C_{\#}S^{-1/4}(x)|Q|}{\Lambda}
\cdot\frac{S^{+1/4}(x)B^{-1}(x)}{|Q|}\\
\,+ \frac{C_{\#}S^{-1/4}(x)|Q|}{\Lambda}\cdot
\frac{S^{3/4}(x)}{|Q|}\le\negthinspace
\frac{C_{\#}S^{1/2}(x)} {\Lambda}\{1+\lambda^{-1}(x)\}\le
\negthinspace\frac{C_{\#}S^{1/2}(x)} {\Lambda}\endmultline
$$
\noindent as asserted.  The proof of Lemma 5 is
complete.$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary 1}  In the setting of Lemma 5, the
number of zeroes of $u(x)$ on $I_{\text{center}}$ differs
by at most $C_{\#}$ from $\frac
1\pi\int_{I_{\text{center}}}(E_0-V(t))^{1/2}\,dt$.
\endproclaim

\demo{Proof}  Equation (4) shows that the zeros of $u(x)$
on $I_{\text{center}}$ are the points of
$I_{\text{center}}$ where
$$
f(x)=(\arg Q)+\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}
\,dt+\text{Im}\ \ell n(1+\text{Error}(x))\equiv 0\
\text{mod}\ \pi,
$$

\noindent with
$|\text{Error}(x)|\le\frac{C_{\#}}{\Lambda}$,$|\frac{d}
{dx}\ \text{Error}(x)|\le \frac{C_{\#}S^{1/2}(x)}
{\Lambda}$, and with a branch of $\ell n(1+\zeta)$
defined on $\{|\zeta|<\frac 12\}$ and yielding\
$\ell n 1=0$.  The function $g(x)=\ \text{Im}\ \ell n
(1+\text{Error}(x))$ satisfies $|g(x)|\le \frac{C_{\#}}
{\Lambda}$, $|\frac{d}{dx}g(x)|\le C_{\#}
\frac{S^{1/2}(x)}{\Lambda}$, while $\frac{d}{dx}
(f-g)=(E_0-V(x))^{1/2}\ge c_{\#}S^{1/2}(x)$.  If
$x_k \in I_{\text{center}}$ satisfies $(f-g)(x_k)=\pi k$,
then $|f(x_k)-\pi k|\le \frac{C_{\#}}{\Lambda}$, and
$f^\prime(x)\ge c_{\#}S^{1/2}(x_k)$ for $|x-x_k|<c_{\#}
B(x_k)$.  Hence there is an $x_k^\prime \in
I_{\text{center}}$ with $f(x_k^\prime)=\pi k$
and $|x_k-x_k^\prime|<\frac{(C_{\#}/\Lambda)}{c_{\#}
S^{1/2}(x_k)}=\frac{C_{\#}B(x_k)}{\lambda(x_k)\Lambda}$,
unless $x_k$ lies within $C_{\#}B(x_k)/(\lambda (x_k)
\Lambda)$ of an endpoint of $I_{\text{center}}$, i.e.
unless $|x_k-x_{\text{left,\ rt}}|\le \frac{
C_{\#}B(x_{\text{left\. rt}})}{\lambda(x_{\text{left}\
,rt})\Lambda}$.  So to every $x_k$ not too near
an endpoint of $I_{\text{center}}$ there corresponds
an $x_k^\prime$.  Different $k$ give rise to different
$f(x_k^\prime)$, hence to distinct $x_k^\prime$.  Since
$(f-g)^\prime>0$ there is at most one $x_k$ for each
$k$.  Hence the number of solutions of $f(x)\equiv 0
\mod \pi$ in $I_{\text{center}}$ is at least as great as
the number of solutions of $f(x)-g(x)\equiv 0\mod \pi$
in $I_{\text{center}}\backslash\operatornamewithlimits
{\bigcup}_{\text{left,\ rt}}\{|x-x_{\text{left,\ rt}}|
\le \frac{C_{\#}B(x_{\text{left\, rt}})}{\lambda(x_{
\text{left,\ rt}})\Lambda}\}\equiv I_{\text{center}}
\backslash \operatornamewithlimits{\bigcup}_
{\text{left,\ rt}} J_{\text{left,\ rt}}$.

The roles of $f$ and $f-g$ can be reversed in the above
argument, so the number of solutions of $f(x)\equiv 0
\mod \pi$ in $I_{\text{center}}$ differs from the
corresponding number of solutions for $f-g$ by at most
the number of solutions of $f$ or $f-g\equiv 0\mod \pi$
in $J_{\text{left}}\cup J_{\text{rt}}$.  In
$J_{\text{left}}$ we have $0<f^\prime$, $(f-g)^\prime
<C_{\#}S^{1/2}(x_{\text{left}})$, and
$|J_{\text{left}}|\le \frac{C_{\#}B(x_{\text{left}})}
{\lambda(x_{\text{left}})\Lambda}$.  Hence on
$J_{\text{left}}$, $f$ and $f-g$ are strictly 
increasing and vary by at most $\frac{C_{\#}}{\Lambda}$.
So there is at most one solution of $f(x)\equiv 0
\mod \pi$ in $J_{\text{left}}$.  Similarly, there is
at most one solution of $f-g\equiv 0\mod \pi$ in
$J_{\text{left}}$, and the same argument applies to
$J_{\text{rt}}$.  Thus, the number of solutions of
$f\equiv 0\mod \pi$ and of $f-g\equiv 0\mod \pi$ in
$I_{\text{center}}$ differ by at most $4$.  (We could
have done better.)  The number of solutions of $f-g\equiv
0\mod \pi$ differs from $\frac 1\pi
\int_{I_{\text{center}}}(E_0-V(x))^{1/2}\,dx$ by at most
$20$ while we have seen that the number of solutions
of $f\equiv 0\mod \pi$ is equal to the number of zeroes
of $u$.  Hence $|(\text{Number\ of\ zeroes\ of}\
u\ \text{in}\ I_{\text{center}})-\frac 1\pi
\int_{I_{\text{center}}}(E_0-V)^{1/2}|\le 24$.
$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary 2}  In the setting of Lemma 5 we have
$$
\int_{I_{\text{center}}}u^2\,dx\ge c_{\#}|Q|^2\int_{I_{\text{center}}}
\frac{dx}{(E_0-V(x))^{1/2}}.
$$\endproclaim

\demo{Proof}  For $\theta_\nu\ge 0$ as in lemma 3 we have
$$\split
\int_{I_\nu}\theta_\nu u^2\,dx=\int_{I_\nu}\theta_\nu(x)\Bigl(\text{Re}
\Bigl[Q(E_0-V(x))^{-1/4}\exp(i\int_{x_{\text{left}}}^x(E_0-V(t))^{1/2}\,dt)\\
\cdot (1+\text{Error}(x))\Bigr]\Bigr)^2\,dx\endsplit
$$

\noindent with $|\text{Error}(x)|\le \frac{C_{\#}}{\Lambda}$.

The proof of Lemma 3(b) shows that the right--hand side differs from\hfill\break
$\frac 12 |Q|^2 \int_{I_\nu}\theta_\nu(x)(E_0-V(x))^{-1/2}\,dx$
by at most $\frac{C_{\#}|Q|^2}{\Lambda}S^{-1/2}(x_\nu)B(x_\nu)$.
Covering $I_{\text{center}}$ by $I_\nu$, we see that
$$\split
\int_{I_{\text{center}}}(\sum\limits_\nu\theta_\nu)u^2\,dx\ge \frac 12
|Q|^2\int_{I_{\text{center}}}(\sum\limits_\nu\theta_\nu)(E_0-V(x))^{-1/2}\,dx\\
-\frac{C_{\#}|Q|^2}{\Lambda}\sum\limits_\nu S^{-1/2}(x_\nu)B(x_\nu).
\endsplit
$$

We can arrange the $\theta_\nu$, $I_\nu$ so that each $x \in I_{\text{center}}$
lies in at most $C_{\#}$ of the $I_\nu$, and so that
$\sum\limits_\nu\theta_\nu=1$ on 
$\hbox{$I$}\kern-4pt\raise 8pt\hbox{$\circ$}=[x_{\text{left}}
+c_{\#}B(x_{\text{left}})$,
$x_{\text{rt}}-c_{\#}B(x_{\text{rt}})]\subset I_{\text{center}}$.  Therefore
the preceding inequality gives
$$
\int_{I_{\text{center}}}u^2\,dx\ge \frac 12 \int_{\lower 3pt\Icirc}
(E_0-V)^{-1/2}
\,dx\cdot|Q|^2-\frac{C_{\#}}{\Lambda}\int_{I_{\text{center}}}(E_0-V)^{-1/2}
\,dx\cdot|Q|^2.
$$

Since $c_{\#}\int_{I_{\text{center}}}(E_0-V)^{-1/2}\,dx\le \int_{\lower 3pt
\Icirc}
(E_0-V)^{-1/2}\,dx$, the last inequality implies
$\int_{I_{\text{center}}}u^2\,dx\ge |Q|^2\cdot(\frac 12 c_{\#}
-\frac{C_{\#}}{\Lambda})\int_{I_{\text{center}}}(E_0-V(x))^{-1/2}\,dx$,
which yields the desired corollary.$\qquad\blacksquare$

Now we study $-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{left}}$, $I_{\text{rt}}$,
$I_{\text{far\ left}}$, $I_{\text{far\ rt}}$.

\vglue 1pc
\proclaim{Lemma 6}
\roster
\item"{(a)}" Let $H_{\text{left}}=-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{left}}$
with the following boundary conditions.  If $I_{\text{far\ left}}$ is
non--empty, then we use Neumann conditions. \hfill\break If 
$I_{\text{far\ left}}$ is empty, then we impose a Dirichlet condition at
the left endpoint and a Neumann condition at the right endpoint.
\item"{(b)}" Let $H_{\text{rt}}=-\frac{d^2}{dx^2}+V(x)$ on
$I_{\text{rt}}$ with Neumann conditions.
\item"{(c)}" Let $H_{\text{far\ left}}=-\frac{d^2}{dx^2}+V(x)$ on
$I_{\text{far\ left}}$ with Dirichlet boundary condition on the left
and Neumann condition on the right endpoint. (We consider
$H_{\text{far\ left}}$ only if $I_{\text{far\ left}}$ is
non--empty). \item"{(d)}" Let $H_{\text{far\
right}}=-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{far\ right}}$
with Dirichlet boundary condition on the right and Neumann
boundary condition at the left endpoint.  (We consider
$H_{\text{far right}}$ only if $I_{\text{far\ rt}}$ is
non--empty.) \endroster \noindent Then all of the above
operators have at most $C_{*}$ negative
eigenvalues.\endproclaim

\demo{Proof} For $H_{\text{left}}$, note first of all that $H_{\text{left}}$
is bounded below.  After passing to a subspace of codimension $0$ or $1$ in
the domain of $H_{\text{left}}$, we may assume a Dirichlet boundary
condition at the left endpoint.  Then after rescaling, we have only to
prove that $-\frac{d^2}{dx^2}-\underline{C}x^{-1}$ on $[0,1]$ with
Dirichlet condition at $0$ and Neumann condition at $1$ has at most
$C_{*}$ negative eigenvalues.  Here $C_{*}$ depends only on $\underline C$.
This bound is trivial.

After rescaling $H_{\text{right}}$, we are left with the operator
$-\frac{d^2}{dx^2}-\underline{C}$ on $[0,1]$ with Neumann boundary conditions.
We must prove that this operator has at most $C_*$ negative eigenvaleus,
which is trivial.

$H_{\text{far\ left}}$ has no eigenvalues $< E_0$, since $V(x)-E_0\ge 0$
on $I_{\text{far\ left}}$.

For $H_{\text{far\ right}}$, we again impose a Dirichlet boundary condition
at the left endpoint after passing to a subspace of codimension at most $1$.

Now we see that $H_{\text{far\ right}}$ has no eigenvalues $<E_0$ by
virtue of the elementary inequality $\int_0^\infty|\frac
1y \int_0^y f(t)\,dt|^2\,dy
\le 10^5 \int_0^\infty|f(y)|^2\,dy$.$\qquad\blacksquare$

\vglue 1pc
\proclaim{Lemma 7}  The number of eigenvalues $<E_0$ for $-\frac{d^2}{dx^2}
+V(x)$ on $I_{\text{BVP}}$ with Dirichlet boundary conditions differs
by at most $C_*$ from $\frac 1\pi \int_{I_{\text{center}}}(E_0-V(x))^{1/2}\,
dx$.\endproclaim

\demo{Proof}  We compare our eigenvalue problem with separate problems on
$I_{\text{far\ left}}$, $I_{\text{left}}$, $I_{\text{center}}$,
$I_{\text{right}}$, $I_{\text{far\ right}}$.  To get a lower bound on
the number of eigenvalues $<E_0$, we impose Dirichlet conditions at all
the endpoints.

To get an upper bound on the number of eigenvalues $<E_0$, we impose
Dirichlet conditions at the endpoints of $I_{\text{BVP}}$ and Neumann
conditions at all the other endpoints.  Lemma 6 shows that all the intervals
except $I_{\text{center}}$ contribute at most $C_*$ eigenvalues to the
upper and lower bounds.  Hence it is enough to study $H_{\text{center}}=
-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{center}}$, with Dirichlet or Neumann
boundary conditions.  Sturm--Liouville theory shows that the number of
eigenvalues $<E_0$ for $H_{\text{center}}$ differs by at most $C_*$
from the number of zeroes of $u$ on $I_{\text{center}}$.  Here $u$ is the
real--valued solution of $[-\frac{d^2}{dx^2}+E_0-V(x)]u=0$ with a
Dirichlet or Neumann boundary condition at one endpoint of $I_{\text{center}}$.
Hence Lemma 7 follows from Corollary 1 to Lemma
{5.\hfill$\blacksquare$} \TagsOnRight
\smallskip

We turn to the study of an eigenfunction on $I_{\text{left}}$,
$I_{\text{right}}$.

\vglue 1pc
\proclaim{Lemma 8}  Suppose $|u^{\prime\prime}(x)|\le C_1x^{-1}|u(x)|$
on $(0,1)$, and let $\hat I\subset (0,1)$ be an interval with $|\hat I|>c_1$.
Then  $\operatornamewithlimits{max}_{x \in (0,1)} |u(x)|\le \hat C
\operatornamewithlimits{max}_{x \in \hat I}|u(x)|$ with
$\hat C$ depending only on $c_1$, $C_1$.\endproclaim

\demo{Proof}  Let $x_0$, $x_1$ be respectively the midpoint and right
endpoint of $\hat I$.  Thus $x_0$, $x_1-x_0>\frac 12 c_1$.  By the mean--value
theorem we can find an $\overline x$ between $x_0$ and $x_1$ with
$|u^\prime(\overline x)|=|\frac{u(x_1)-x(x_0)}{x_1-x_0}|\le 2
\operatornamewithlimits{max}_{\hat I}|u|/(\frac 12 c_1)$.  Also
$|u(\overline x)|\le \operatornamewithlimits{max}_{\hat I}|u|$ since 
$\overline x \in \hat I$.
\medskip

Now $\frac{d}{dx}\pmatrix u^\prime(x)\\ u(x)\endpmatrix =\pmatrix 0&W(x)\\
1&0\endpmatrix \pmatrix u^\prime(x)\\ u(x)\endpmatrix$ with\hfill\break
$W(x)=\cases \frac{u^{\prime\prime}}{u(x)}&\text{if $u(x)\ne 0$}\\
0&\text{if $u(x)=0$}\endcases$.  Hypothesis implies $|W(x)|\le C_1x^{-1}$, so
\hfill\break
$\Big|\frac{d}{dx}\pmatrix u^\prime(x)\\ u(x)\endpmatrix\Big|\le (1+C_1x^{-1})
\Big|\pmatrix u^\prime(x)\\ u(x)\endpmatrix\Big|$, 
so $\frac{d}{dx}\Bigl[x^{(1+C_1)}\Big|\pmatrix u^\prime(x)\\ u(x)\endpmatrix
\Big|\,\Bigr]\ge 0\ge\hfill\break \frac{d}{dx}\Bigl[x^{-(1+C_1)}\Big|
\pmatrix u^\prime(x)\\ u(x)\endpmatrix\Big|\,\Bigr]$.  Therefore
$$\align
&\Big|\pmatrix u^\prime(x)\\ u(x)\endpmatrix\Big|\le
\Big|\pmatrix u^\prime(\overline x)\\ u(\overline x)\endpmatrix\Big|
\cdot (x/\overline x)^{(1+C_1)}\,\quad\text{for}\ \overline x\le x<1,\ 
\text{and}\\
&{}\\
&\Big|\pmatrix u^\prime(x)\\ u(x)\endpmatrix\Big| \le \Big|
\pmatrix u^\prime(\overline x)\\ u(\overline x)\endpmatrix\Big|\cdot
(\overline x/x)^{(1+C_1)}\,\,\quad\text{for}\ 0<x\le \overline x.\endalign
$$

Substituting our estimates for $|u(\overline x)|$, $|u^\prime(\overline x)|$
into the last two estimates, we see that
$$
|u^\prime(x)|, |u(x)|\le \hat C_1\operatornamewithlimits{max}_{\hat I}
|u|\quad\text{for}\ x_*\le x<1,\tag 9
$$

\noindent where $\hat C_1$ depends only on $c_1$, $C_1$ provided $x_*>0$
depends only on $c_1$, $C_1$.  We pick $x_*$ so that
$$
x_*C_1\int_0^1(\ell n\frac 1t)\,dt=\frac {1}{10}.
$$

For a constant $\hat C_2>>\hat C_1$ depending only on $c_1$, $C_1$, we
claim that
$$|u(x)|\le \hat C_2\operatornamewithlimits{max}_{\hat I}|u|\quad\text{for}\
0<x\le x_*.\tag 10
$$

To see (10), let $x_{\text{min}}$ be the smallest number $y\ge 0$ for
which $\operatornamewithlimits{max}_{[y,x_*]}|u|\le \hat C_2
\operatornamewithlimits{max}_{\hat I}|u|$.  Either (10) holds or else
$0<x_{\text{min}}<x_*$,
$\operatornamewithlimits{max}_{[x_{\text{min}},
x_*]}|u|\le \hat C_2\operatornamewithlimits{max}_{\hat
I}|u|$ and $|u(x_{\text{min}})|=\hat C_2\text{max}_{\hat
I}|u|$.  We assume the latter case and derive a
contradiction.  Since $|u^\prime(x_*)|\le \hat C_1
\operatornamewithlimits{max}_{\hat I}|u|$ by (9), we have
for $x \in [x_{\text{min}},x_*]$ the estimate $$\align
|u^\prime(x)|&\le
|u^\prime(x_*)|+\int_x^{x_*}|u^{\prime\prime}(t)|\,dt\le
|u^\prime(x_*)|+\int_x^{x_*}C_1t^{-1}|u(t)|\,dt\\ &\le
\hat C_1\operatornamewithlimits{max}_{\hat I}|u|+(\hat C_2
\operatornamewithlimits{max}_{\hat
I}|u|)\int_x^{x_*}C_1t^{-1}\,dt\\ &=(\hat C_1+C_1\hat C_2\
\ell n(x_*/x))\cdot\operatornamewithlimits{max}_ {\hat
I}|u|.\endalign $$

Since also $|u(x_*)|\le \hat C_1\operatornamewithlimits{max}_{\hat I}
|u|$ by (9), it follows that 
$$\align
|u(x_{\text{min}})|&\le |u(x_*)|+\int_{x_{\text{min}}}^{x_*}|u^\prime
(x)|\,dx\\
&\le \hat C_1\operatornamewithlimits{max}_{\hat I}|u|+
(\operatornamewithlimits{max}_{\hat I}|u|)\int_{x_{\text{min}}}^{x_*}
(\hat C_1+C_1\hat C_2\ \ell n(x_*/x))\,dx\\
&\le (2\hat C_1+x_*C_1\hat C_2\int_0^1\ell n(1/t)\,dt)\cdot
\operatornamewithlimits{max}_{\hat I}|u|.\endalign
$$

\noindent We have $2\hat C_1<\frac{1}{10}\hat C_2$ since we picked
$\hat C_2>>\hat C_1$, and $x_*C_1\hat C_2\int_0^1 \ell n(1/t)\,dt=
\frac{1}{10}\hat C_2$ by our choice of $x_*$.  Hence the previous
estimate implies $|u(x_{\text{min}})|\le \frac 15 \hat C_2
\operatornamewithlimits{max}_{\hat I}|u|$, which contradicts
$|u(x_{\text{min}})|=\hat C_2\operatornamewithlimits{max}_{\hat I}
|u|$.  This contradiction proves (10).  The conclusion of lemma 8 is
contained in (9), (10).$\qquad\blacksquare$

\vglue 1pc
\proclaim{Corollary}  If $[\frac{d^2}{dx^2}+(E_0-V(x))]=0$ on $I_{\text{BVP}}$
and $u$ is expressed as in lemma 5 on $I_{\text{center}}$, then we have
$$\gather
\operatornamewithlimits{max}_{I_{\text{left}}}|u|\le C_*|Q|(E_0-V(x_{\text{
left}}))^{-1/4}\quad\text{and}\\
\operatornamewithlimits{max}_{I_{\text{rt}}}|u|\le C_*|Q|\cdot
(E_0-V(x_{\text{right}}))^{-1/4}.\endgather
$$\endproclaim

\demo{Proof}  Take $J=I_{\text{left}}\cup \hat J$ with $\hat J=
[x_{\text{left}},x_{\text{left}}+c_{\#}B(x_{\text{left}})]$.
Then $\hat J\subset J$, $|\hat J|>c_*|J|$, and $|V(x)-E_0|\le
\frac{C_*|J|^{-1}}{|x-\text{min}\ J|}$ on $J$.  Rescaling
this situation brings us to the setting of lemma 8, which implies the
estimate $\operatornamewithlimits{max}_J|u|\le C_*\operatornamewithlimits{max}
_{\hat J}|u|$.  Lemma 5 shows that $\operatornamewithlimits{max}_{\hat J}
|u|\le C_{\#}|Q|\cdot(E_0-V(x_{\text{left}}))^{-1/4}$, so we obtain
the desired estimate for $\operatornamewithlimits{max}_{I_{\text{left}}}
|u|$.  The estimate for $\operatornamewithlimits{max}_{I_{\text{right}}}|u|$
is analogous and easier, since on $I_{\text{right}}$ we make a stronger
assumption on $|V-E_0|$.$\qquad\blacksquare$

We next study how $u$ behaves on $I_{\text{far\ left}}$.  (We will not need
to understand $I_{\text{far\ right}}$.)

\vglue 1pc
\proclaim{Lemma 9}  Let $u(x)$ be a real--valued solution of $[\frac{d^2}
{dx^2}+E_0-V(x)]u=0$ on $I_{\text{BVP}}$ with Dirichlet conditions, with
$u$ described by Lemma 5 on $I_{\text{center}}$.  Then for
$x \in I_ {\text{far\ left}}$ we have
$$
|u(x)|\le
\frac{C_{*}|Q|}{(E_0-V(x_{\text{left}}))^{1/4}}\exp
(-c_*(x_{\text{left}}-x)/B(x_{\text{left}})).\tag 11 $$
\endproclaim

\demo{Proof}  The corollary of Lemma 8 shows that (11) holds for 
$x \in I_{\text{left}}$, in particular for $x =\min J_{\text{left}}=
x_{\text{far\ left}}$.  Now set $w=\pm u$ and 
$$
f(x)=\frac{C_{*}|Q|}
{(E_0-V(x_{\text{left}}))^{1/4}}\exp(-\tilde c_*(x_{\text{left}}-x)
/B(x_{\text{left}}))-w(x)$$

\noindent with $\tilde c_*>0$ to be picked in a moment.  We know that
$f(x_{\text{far\ left}})\ge 0$, and also that
$\lim_{x\to\min I_{\text{BVP}}}f(x)\ge 0$ since $u$ satisfies Dirichlet
boundary conditions.

Also, $f(x)$ is twice differentiable in the interior of $J_{\text{far\ left}}$,
so either $f(x)\ge 0$ throughout $J_{\text{far\ left}}$ or else $f(x)$
has a strictly negative interior minimum at some point $x_*\in I_{\text{far\
left}}$.  We assume the latter case and derive a contradiction.
Thus we assume $f(x_*)<0$, $f^\prime(x_*)=0$, $f^{\prime\prime}
(x_*)\ge 0$.  In particular $w(x_*)\ge \frac{C_*|Q|}{(E_0-V(x_{\text{left}}
))^{1/4}}\exp(-\tilde c_*(x_{\text{left}}-x_*)/B(x_{\text{left}}))>0$ since
$f(x_*)<0$.  Also $V(x)-E_0\ge c_*B^{-2}(x_{\text{left}})$ on
$I_{\text{far\ left}}$, by our basic assumptions on $V(x)$.  Hence
$$
w^{\prime\prime}(x_*)=(V(x_*)-E_0)w(x_*)\ge \frac
{c_*B^{-2}(x_{\text{left}})|Q|}{(E_0-V(x_{\text{left}}))^{1/4}}
\exp(-\tilde c_*(x_{\text{left}}-x_*)/B(x_{\text{left}}))
$$

\noindent whereas
$$\split
\Bigl(\frac{C_*|Q|}{(E_0-V(x_{\text{left}}))^{1/4}}
\exp(-\tilde c_*(x_{\text{left}}-x)/B(x_{\text{left}}))\Bigr)^{\prime\prime}=
\frac{C_*|\tilde c_*|^2B^{-2}(x_{\text{left}})|Q|}
{(E_0-V(x_{\text{left}}))^{1/4}}\cdot\\
\cdot\exp\Bigl(\frac{-\tilde c_*(x_{\text{left}}
-x_*)}{B(x_{\text{left}})}\Bigr)\quad \text{at}\ x=x_*.\endsplit
$$

\noindent Taking $\tilde c_*$ small enough and subtracting, we find that
$f^{\prime\prime}(x_*)<0$, which contradicts our previous assumption
$f^{\prime\prime}(x_*)\ge 0$.

This contradiction shows that $f(x)\ge 0$ in $I_{\text{far\ left}}$ for
either choice of sign in the definitions of $f$, $w$.  That is
$|u(x)|\le \frac{C_*|Q|}{(E_0-V(x_{\text{left}}))^{1/4}}\exp
(-\tilde c_*(x_{\text{left}}-x_*)/B(x_{\text{left}}))$, as asserted.
$\qquad\blacksquare$

We summarize our knowledge of the eigenfunctions and eigenvalues in the
following result.

\vglue 1pc
\proclaim{Theorem 1}  Under the assumptions (H$\hat 0$)$\ldots$(H$\hat 7$)
we have
$$
|(\text{Number\ of\ eigenvalues\ of}\ H<E_0)-\frac 1\pi 
\int_{I_{\text{center}}}(E_0-V(t))^{1/2}\,dt|\le C_*.
$$
\noindent Suppose $E_0$ is an eigenvalue of $H$, and suppose $u$ is the
corresponding eigenfunction, with $u$ real--valued and having $L^2$--norm
$1$ on $I_{\text{BVP}}$.  Then
$$|u(x)|^2\le C_*\Bigl(\int_{I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x))^{-1/2}\ \text{for}\ x \in I_{\text{center}}
$$
$$\split
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x_{\text{left}}))^{-1/2}\exp\Bigl(\frac{-c_*(x_{\text{left}}-x)}
{B(x_{\text{left}})}\Bigr)\\
\text{for}\ x \in I_{\text{BVP}}, x \le x_{\text{left}}=
\text{min}\ I_{\text{center}}.\endsplit
$$
$$
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x_{\text{right}}))^{-1/2}\quad\text{for}\ x \in I_{\text{right}}.
$$\endproclaim

As an application of this theorem, we study the following situation.
\smallskip
\noindent{\underbar{Set--up}}. $V(x)$ is a potential defined on a (possibly
unbounded) interval $I_{\text{BVP}}$.  We are given a subinterval
$I \subset I_{\text{BVP}}$ and weight functions $S(x)$, $B(x)>0$ defined
on $I$.  Set $\lambda(x)=S^{1/2}(x)B(x)$.  We are given an energy $E_0$.

We make the following assumptions.

\noindent{\bf{Hypotheses}}
\roster
\item"{(H$\overline 0$)}" For $x,y \in I$ with $|x-y|<cB(x)$ we have
$c<\frac{B(y)}{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$, and $|I|>cB(x)$.
\item"{(H$\overline 1$)}" For $x \in I$ we have $|(\frac{d}{dx})^\alpha
V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$.
\item"{(H$\overline 2$)}" $\{x \in I_{\text{BVP}}\mid V(x)<E_0\}
=(x_{\text{left}},x_{\text{right}})\subset I$ with
$\text{dist}(x_{\text{left}},\partial I)>cB(x_{\text{left}})$,
$\text{dist}(x_{\text{right}},\partial I)>cB(x_{\text{right}})$.
\item"{(H$\overline 3$)}" In $[x_{\text{left}},x_{\text{left}}+c_1
B(x_{\text{left}})]$ we have $-V^\prime(x)\ge c S(x_{\text{left}})/
B(x_{\text{left}})$, and in $[x_{\text{right}}-c_1B(x_{\text{right}}),
x_{\text{right}}]$ we have $+V^\prime(x)\ge cS(x_{\text{right}})/B
(x_{\text{right}})$.
\item"{(H$\overline 4$)}" In $[x_{\text{left}}+c_1B(x_{\text{left}}),
x_{\text{right}}-c_1B(x_{\text{right}})]$ we have $cS(x)<E_0-V(x)<CS(x)$.
\item"{(H$\overline 5$)}" In $I_{\text{BVP}}^{\text{interior}}\cap
(-\infty,x_{\text{left}})$, $V(x)$ is decreasing and $C^\infty$.
\item"{(H$\overline 6$)}" $\Lambda=(\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{\lambda(x)B(x)})^{-1}$ is bigger than a large positive number
depending only on $c$, $C$, $c_1$, $C_\alpha$ in (H$\overline 0$)$\ldots$
(H$\overline 4$).
\endroster

Let $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_{\text{BVP}})$ with Dirichlet
boundary conditions.

\vglue 1pc
\proclaim{Theorem 2}  Under assumptions (H$\overline 0$)$\ldots$
(H$\overline 6$) above, we have
$$
|\text{(Number\ of\ eigenvalues\ of}\ H<E_0)-\frac 1\pi \int_{I_{\text{BVP}}}
(E_0-V(t))_+^{1/2}\,dt|\le C_*.
$$
\noindent If $E_0$ is an eigenvalue of $H$, and if $u(x)$ (real--valued,
with norm $1$ in $L^2(I_{\text{BVP}})$) is the corresponding eigenfunction,
then
$$
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{BVP}}}(E_0-V(t))_+^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x))^{-1/2}\quad\text{for}\ x_{\text{left}}<x<x_{\text{right}}
$$
$$\split
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{BVP}}}(E_0-V(t))_+^{-1/2}\,dt\Bigr)^{-1}
\Bigl[S(x_{\text{left}})\lambda^{-2/3}(x_{\text{left}})\Bigr]^{-1/2}\cdot\\
\cdot\exp\Bigl(-c_*\lambda^{2/3}(x_{\text{left}})\frac{(x_{\text{left}}-x)}
{B(x_{\text{left}})}\Bigr)\quad
\roman{for}\ x \in I_{\text{BVP}}\cap (-\infty,x_{\text{left}}].\endsplit
$$
\noindent The constants $c_*$, $C_*$ depend only on $c$, $C$, $c_1$,
$C_\alpha$, in (H$\overline 0$)$\ldots$(H$\overline 4$).\endproclaim

\demo{Proof}  We define intervals $\check I_{\text{far\ left}}$,
$\check I_{\text{left}}$, $\check I_{\text{center}}$, $\check I_{\text{right}}$,
$\check I_{\text{far\ right}}$ and weight functions $\check S(x)$,
$\check B(x)$ so that theorem 1 applies.  With a large constant $\check C$
to be picked later, we define the intervals and weights as follows.
$$\align
\check I_{\text{far\ left}}&=I_{\text{BVP}}\cap (-\infty,x_{\text{left}}
-\frac{\check CB(x_{\text{left}})}{\lambda^{2/3}(x_{\text{left}})}]\\
\check I_{\text{left}}&=\{|x-x_{\text{left}}|<
\frac{\check C
B(x_{\text{left}})}{\lambda^{2/3}(x_{\text{left}})}\}\\
\check I_{\text{center}}&=[x_{\text{left}}+\frac{\check CB(x_{\text{left}})}
{\lambda^{2/3}(x_{\text{left}})}, x_{\text{right}}-
\frac{\check C B(x_{\text{right}})}{\lambda^{2/3}(x_{\text{right}})}]\\
\check I_{\text{right}}&=\{|x-x_{\text{right}}|\le \frac
{\check CB(x_{\text{right}})}{\lambda^{2/3}(x_{\text{right}})}\}\\
\check I_{\text{far\ right}}&=I_{\text{BVP}}\cap [x_{\text{right}}
+\check C\frac{B(x_{\text{right}})}{\lambda^{2/3}(x_{\text{right}})},\infty).
\endalign
$$

\noindent For $x \in \check I_{\text{center}}$, we take $\check B(x)=
\text{min}\{|x-x_{\text{left}}|, |x-x_{\text{right}}|, B(x)\}$ and set
$\check S(x)=\frac{S(x_{\text{left}})}{B(x_{\text{left}})}|x-x_{\text{left}}|$
if $|x-x_{\text{left}}|<cB(x_{\text{left}})$, $\check S(x)=\frac
{S(x_{\text{right}})}{B(x_{\text{right}})}|x-x_{\text{right}}|$ if
$|x-x_{\text{right}}|<cB(x_{\text{right}})$, $\check S(x)=S(x)$ otherwise.

We check that hypotheses (H$\hat 0$)$\ldots$(H$\hat 7$) hold.  (H$\hat 0$)
and (H$\hat 1$) are obvious.  To check (H$\hat 2$) we note that
$$\align
\int_{\check I_{\text{center}}}
\frac{dx}{\check\lambda(x)\check B(x)}
&\le \int_{x_{\text{left}}+
\frac{\check CB(x_{\text{left}})}
{\lambda^{2/3}(x_{\text{left}})}}
^{x_{\text{left}}+cB(x_{\text{left}})}
\frac{dx}{\frac{S^{1/2}(x_{\text{left}})}
{B^{1/2}(x_{\text{left}})}
|x-x_{\text{left}}|^{5/2}}\\
&\,\,+\int_{x_{\text{left}}+cB(x_{\text{left}})}^
{x_{\text{right}}-cB(x_{\text{right}})}
\frac{dx}{\lambda(x)B(x)}\\
&\,\,+\int_{x_{\text{right}}-cB(x_{\text{right}})}^
{x_{\text{right}}-\frac{\check
CB(x_{\text{right}})}{\lambda^{2/3}(x_{\text{right}})}}
\frac{dx}{\frac{S^{1/2}(x_{\text{right}})}{B^{1/2}
(x_{\text{right}})}|x-x_{\text{right}}|^{5/2}}.
\endalign
$$

\noindent The first integral on the right is $\sim \frac{1}
{\frac{S^{1/2}(x_{\text{left}})}{B^{1/2}(x_{\text{left}})}
(\frac{\check CB(x_{\text{left}})}{\lambda^{2/3}(x_{\text{left}})})^{3/2}}
=\frac{1}{(\check C)^{3/2}}$.  The last integral on the right is
$\sim \frac{1}{(\check C)^{3/2}}$ as well, by a similar calculation.  The second
integral on the right is at most $\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{\lambda(x)B(x)}=\Lambda^{-1}$.
Hence $\int_{\check
I_{\text{center}}}\frac{dx}{\check\lambda(x)\check B(x)}
\le \frac{C_*}{(\check C)^{3/2}}+\Lambda^{-1}$, with $C_*$
depending only on $c$, $C$, $c_1$, $C_\alpha$ in
(H$\overline 0$)$\ldots$(H$\overline 7$). Hence (H$\hat
2$) holds if we take $\check C$ large enough, which we now
do. 

To check (H$\hat 3$) we note that $\check I_{\text{center}}=[
\check x_{\text{left}},\check x_{\text{right}}]$ with
$\check x_{\text{left}}=x_{\text{left}}+\frac{\check CB(x_{\text{left}})}
{\lambda^{2/3}(x_{\text{left}})}$, $\check S(\check x_{\text{left}})=
\frac{S(x_{\text{left}})\cdot \check C}{\lambda^{2/3}(x_{\text{left}})}$,
$\check B(\check x_{\text{left}})=\frac{\check CB(x_{\text{left}})}
{\lambda^{2/3}(x_{\text{left}})}$ so that
$\check \lambda(\check x_{\text{left}})=\check S^{1/2}(\check x_{\text{left}})
\check B(\check x_{\text{left}})=(\check C)^{3/2}$.  This verifies the
desired a--priori bound on $\check\lambda(x_{\text{left}})$ and
shows that $|\check I_{\text{left}}|=2\check B(\check x_{\text{left}})$.

Analogous arguments work at $x_{\text{right}}$, completing the proof of 
(H$\hat 3$).  To check (H$\hat 4$), (H$\hat 5$), note that
$|V(x)-E_0|\le C_*\frac{S(x_{\text{left}})}{B(x_{\text{left}})}
|x-x_{\text{left}}|\le C_*\check S(\check x_{\text{left}})$ in
$\check I_{\text{left}}$, and $\check S(\check x_{\text{left}})=
(\check\lambda(\check x_{\text{left}}))^2(\check B(\check x_{\text{left}}))^{-2}
\le C_*(\check B(\check x_{\text{left}}))^{-2}\le C_*|\check I_{\text{left}}|
^{-2}$.  Here we used $\check\lambda(\check x_{\text{left}})\le C_*$, which
we verified before.  Thus $\operatornamewithlimits{max}_{\check I_{\text{left}}}
|V(x)-E_0|\le C_*|\check I_{\text{left}}|^{-2}$.  Similarly for $\check I_
{\text{right}}$, completing the proofs of (H$\hat 4$), (H$\hat 5$).

To check (H$\hat 6$), we use the fact that $V(x)$ is decreasing for 
$x<x_{\text{left}}$.  Hence for $x \in \check I_{\text{far\ left}}$ we have
$V(x)-E_0\ge V(x_{\text{left}}-\frac{\check CB(x_{\text{left}})}
{\lambda^{2/3}(x_{\text{left}})})-E_0\ge \frac{c_*S(x_{\text{left}})}
{B(x_{\text{left}})}\cdot \frac{\check CB(x_{\text{left}})}{\lambda^{2/3}
(x_{\text{left}})}=c_*\check S(\check x_{\text{left}})=
c_*(\check\lambda(\check x_{\text{left}}))^2(\check B(\check x_{\text{left}}))
^{-2}\ge c_*|\check I_{\text{left}}|^{-2}$ since $\check\lambda(\check x_{
\text{left}})=(\check C)^{3/2}$ and $|\check I_{\text{left}}|\sim
\check B(\check x_{\text{left}})$.  Also $V(x)$ is $C^\infty$ on
$\check I_{\text{far\ left}}$ which completes the proof of (H$\hat 6$).

To check (H$\hat 7$) we just note that $\{V<E_0\}=(x_{\text{left}},
x_{\text{right}})$, so $V(x)-E_0\ge 0$ in $I_{\text{far\ right}}$.
Thus (H$\hat 0$)$\ldots$(H$\hat 7$) hold, so we can apply Theorem 1.

The number of eigenvalues of $H<E_0$ therefore differs by at most $C_*$
from $\int_{\check I_{\text{center}}}(E_0-V(t))^{1/2}\,dt$.  Also
$\int_{\check I_{\text{left}}}|E_0-V(t)|^{1/2}\,dt
\le \int_{\check I_{\text{left}}}(C_*|\check I_{\text{left}}|^{-2})^{1/2}\,dt
\le C_*$, and similarly for $\check I_{\text{right}}$.  Thus
$\int_{\check I_{\text{center}}}(E_0-V(t))^{1/2}\,dt$
and $\int_{I_{\text{BVP}}}
(E_0-V(t))_+^{1/2}\,dt$ differ by at most $C_*$.  This proves the part of
Theorem 2 on eigenvalues.  Regarding eigenfunctions, the conclusions of
Theorem 2 are obvious consequences of those of Theorem 1, once we note
that $\int_{\check I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt$ has
the same order of magnitude as $\int_{I_{\text{BVP}}}(E_0-V(t))_+^{-1/2}\,dt$,
and that $\check B(\check x_{\text{left}})=\frac{\check CB(x_{\text{left}})}
{\lambda^{2/3}(x_{\text{left}})}$.$\qquad\blacksquare$

When we apply our ODE results to three--dimensional
potentials arising from an atom of nuclear charge $Z$,
Theorem 2 will be used for angular momenta in the range
(Large Const.) $< \ell  < Z^\varepsilon$, and Theorem 1
will be used for angular momenta $\ell\le$ (Large
Const.).  For angular momenta  $\ell \ge Z^\varepsilon$,
we use the much more refined WKB theory of the previous
sections.


\pageno 178
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\centerline{\bf{References}}
\smallskip

\widestnumber\key{FS6}

\ref\key AS 
\by M. Abramowitz and I. Stegun
\book Handbook of Math\. Functions
\bookinfo section 10.4. Nat. Bureau of Standards, Applied
Math Series no. 55
\yr 1964
\endref

\ref\key BF 
\by R. Beals and C. Fefferman
\paper Spatially Inhomogeneous Pseudodifferential Operators I
\jour Comm. Pure Appl. Math.
\vol 27
\yr 1974
\pages 1--24
\endref

\ref\key C 
\by T. Cherry
\paper Uniform Asymptotic Formulae for Functions with Transition\hfill\break
Points
\jour Transactions of the AMS
\vol 68
\yr 1950
\pages 224--257
\endref

\ref\key E 
\by A. Erd\'elyi
\book Asymptotic Expansions
\publ Dover
\yr 1956
\endref

\ref\key FS1 
\by C. Fefferman and L. Seco
\paper The Ground--State Energy of a Large Atom
\jour Bull. A.M.S.
\yr 1990
\pages {}
\endref

\ref\key FS2 
\bysame %C. Fefferman and L. Seco
\paper The Density in a One--Dimensional Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS3 
\bysame %C. Fefferman and L. Seco
\paper The Eigenvalue Sum for a One--Dimensional Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS4 
\bysame %C. Fefferman and L. Seco
\paper The Density in a Three--Dimensional Radial Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS5 
\bysame %C. Fefferman and L. Seco
\paper The Eigenvalue Sum for a Three--Dimensional Radial Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS6
\bysame %C. Fefferman and L. Seco
\paper On the Dirac and Schwinger Corrections to the Ground--State
Energy of an Atom
\jour Advances in Math
\toappear
\endref

\ref\key FS7
\bysame % C. Fefferman and L. Seco
\paper Aperiodicity of the Hamiltonian Flow in the Thomas--Fermi Potential
\jour Advances in Math
\toappear
\endref

\ref\key H
\by W. Hughes
\paper An Atomic Energy Lower Bound that Agrees with Scott's Correction
\jour Advances in Math
\vol 79
\yr 1990
\pages 213--270
\endref












\end