instructions: AMSTeX 2.1 worked for this file. The line "input amstex" is commented out; it may have to be reinserted or modified for certain systems. There are many lines with over 80 characters. --------------- CUT HERE ------------------- %\input amstex \magnification\magstep1 \documentstyle{amsppt} \hsize 6 truein \vsize 9truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \def\Icirc{\hbox{$I$}\kern-3pt\raise 5pt\hbox{$\circ$}} \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 $$\gather \phi(E_k)=\pi(k+1/2)\quad\text{for\ integers}\ k,\ \text{where}\tag 2\\ \phi(E)=\int_{\{V(x)(\min V)+c\lambda^2. Then F_k(x) is given by several different formulas in overlapping regions, and (2) is the consistency condition for the different formulas to agree. The formulas for F_k are as follows. For x \in [x_{\text{left}}(E_k),x_{\text{right}}(E_k)] not too near the turning points,$$ 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)|\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 $$\noindent with \varepsilon>0 arbitrarily small and N>0 arbitrarily large, and with b_k a real constant satisfying |b_k-1|cS(x)B^{-2}(x)\quad\text{for}\quad |x-x_0|\le c_1B(x_0) \tag 23\\ |V^\prime(x)|>cS(x)B^{-1}(x)\quad\text{for}\ x \in I\quad\text{with}\quad |x-x_0|\ge c_1B(x_0).\tag 24\endgather$$ \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 $|b-1|0$. 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 $-\lambda^{-\varepsilon}C_*^{\varepsilon N}$ 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 $$|\partial_x^\alpha j(x,\lambda)|\le C_*^{\varepsilon N}\,\lambda^{-\frac {\varepsilon N}{s}}\quad \text{for}\quad \alpha=0,1,2\quad \text{and}\quad \lambda^{\varepsilon-2/3} From (17) we conclude that$$\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 010. $$\noindent Whereas A(t) is bounded on the whole real line A_c(t) is bounded on [-1,\infty) but grows rapidly as t \to -\infty. Using A_c(t) in place of A(t), we define a complex analogue of F_0(x,\lambda), namely$$ 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)). $$\noindent Thus F_0(x,\lambda)=\text{Re}[F_c(x,\lambda)]. Note that A_c(\lambda^{2/3}y_N(x,\lambda)) remains bounded by C_*^{\varepsilon N} when 0C_*^{\varepsilon N}. To see this, we write y_N(x,\lambda)=y_0(x)+O(\lambda^{-2}) with y_0(x)>0 for 00. Otherwise, we can repeat the proofs of Lemmas 2 and 3 virtually word for word. Now we return to the proof of Lemma 4. Setting w_0(x)\equiv 1 and w_{N+1}\equiv 0 we obtain from (25) by elementary calculation the formula:$$\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$$\noindent for \lambda^{\varepsilon-2/3}0}\endcases by definition of \tilde u, so \text{Res}_k(u)=\sum\limits_{\ell=0}^k c_{k-\ell}^+ \text{Res}_\ell(\tilde u)=c_k^+. We must therefore show that \text{Res}_k(u)\in \Cal P. \smallskip Lemmas 3 and 5 give$$\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). $$Our estimates for \tilde g_{k\ell}(\tau) and g_k^{\#}(y,\tau) show that |\partial_{y,\tau}^\alpha f_{k+1}(y,\tau)|\le C_{**}^{\alpha k} on Q for all \alpha. The induction is complete, and the Lemma is proven. \qquad\blacksquare \proclaim{Corollary} The conclusion of Lemma 11 holds also for the canonical solution of the transport equation.\endproclaim \medskip \demo{Proof} Express the canonical solution in terms of the elementary solution using the matching coefficients c_k^+. Since c_k^+ \in \Cal P, the matching coefficients depend smoothly on \tau, and the Corollary follows at once from Lemma 11. \qquad \blacksquare \smallskip It remains to verify that F_+(x,\lambda) approximately satisfies the basic differential equation (1). Since we only established the uniqueness of the canonical solution (u_0(x),u_1(x),\ldots,u_k(x),\ldots) for k \le N^{\prime\prime}, we prefer to truncate the sum (5) defining F_+(x,\lambda) below k=N^{\prime\prime}. We will check that the truncated solution still agrees well with F_0(x,\lambda) as in Lemma 5. (This problem arises only from the technicalities of our exposition. Clearly the canonical u_k(x) are defined for all k, by taking N large enough depending on k. We didn't want to bother with the N dependence of u_k.) \vglue 1pc \proclaim{Lemma 12} Set N^{\prime\prime\prime}=[\varepsilon N/500] and define F_{++}(x,\lambda)=\text{Re}\Bigl[\frac {e^{\pm i \pi/4}e^{i\int_0^x(\lambda^2p(t))^{1/2}\,dt}} {(\lambda^2p(x))^{1/4}}\cdot\hfill\break \cdot\sum\limits_{k=0}^{N^{\prime\prime\prime}} \,\lambda^{-k}u_k(x)\Bigr] with (u_0(x),\ldots u_{N^{\prime\prime}} (x)) the canonical solution of the transport equations. Then for \lambda^{\varepsilon - 2/3}0 define p_{\#}(x)=t^2p(x). Then let F_0^{\#}(x,\lambda_{\#}), F_+^{\#}(x,\lambda_{\#}) be the matching approximate solutions for$$ \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>0 on \{|x-x(E)|\hfill\break C.\smallskip \item{(C)} We have |\partial_x^\alpha F_0(x,E,\lambda)|\le C\lambda^{-N^\prime} for -\lambda^{-\varepsilon}C, \alpha\le 2.\smallskip \item{(D)} The functions u_k(x,E) are defined for 0C.\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}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 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: \pageno 38 \magnification \magstep1 \input amstex \documentstyle{amsppt} \hsize 6 truein \vsize 9truein \baselineskip 20 pt \hoffset .15 truein \def\hy{y\kern -8pt\raise 5pt\hbox{\rightharpoonup}} \def\ha{a\kern -8pt\raise 5pt\hbox{\rightharpoonup}} \rightheadtext{} \catcode\@=11 \def\logo@{} \catcode\@=12 \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_{\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 $$\noindent in a small neighborhood of the turning point x_{\text{left}}. To do so, we simply rescale the problem to reduce matters to the setting of the local WKB lemma. Specifically, set z(\tilde E)=(\text{solution\ of}\ V(z)=\tilde E near z = x_{\text{left}}) for |\tilde E-E_0|C_{\#}. Let (u_k^{\text{left}}(x,E)) be the canonical solution to the transport equations for the potential E-V(x) and the turning point x_{\text{left}}(E). Then there exist auxiliary functions Y(x,E), Y_0(x,E), Y_{\#}(x,E) with the following properties. \roster \item"{(I)}" Set F_0^{\text{left}}(x,E)=\lambda_{\text{left}}^{-1/3} (\frac{\partial Y(x,E)}{\partial x})^{-1/2}A(\lambda_{\text{left}}^{2/3} Y(x,E)). Then for |x-x_{\text{left}}(E)|\hfill\break <\lambda_{\text{left}}^{-\varepsilon} B(x_{\text{left}}) we have |(\frac{\partial^2}{\partial x^2}+E-V(x))F_0^ {\text{left}}|\le C_{\#}\lambda_{\text{left}}^{-N^\prime}B^{-3/2} (x_{\text{left}}). Also, for 0 \le \alpha \le 2 and for -\lambda_{\text{left}}^{-\varepsilon} B(x_{\text{left}}) c_{\#}B(x_{\text{left}}) for |E-E_0|1. In each I_\nu, we take (u_k^\nu(x,E))_{0\le k\le N^\prime} to be the solution of the transport equations$$ 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 $$\noindent for x \in I_\nu, |E-E_0|0, since for k_2=0 we have h_{k_2}^j(E)\equiv 1 so that \partial_E^{\beta_2+1}h_{k_2}^j\equiv 0. Hence from (7), (14) and inductive hypothesis we get$$ |\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 $$\noindent with \Cal L=\Bigl(\frac {5}{16}\frac{(V^\prime)^2}{(E-V)^{5/2}}+\frac 14 \frac{V^{\prime\prime}}{(E-V)^{3/2}}\Bigr)+\frac 12 \frac{V^\prime}{(E-V)^{3/2}}\frac {\partial}{\partial x}+\frac {1}{(E-V)^{1/2}}\frac {\partial^2}{\partial x^2}. \noindent Since only negative powers of (E-V) appear in (E-V)^{1/4}\Cal L, and since E-V(x)>c_{\#}S(x) (as we saw in the proof of Lemma 2), we see that (E-V)^{1/4}\Cal L=\sum\limits_{\ell=0}^2 \hat P_\ell (x,E)(\frac{\partial}{\partial x}) ^\ell with |\hat P_\ell|\le C_{\#}S^{-1/4}(x)B^{-2+\ell}(x) in U_{\text{center}}. \noindent Therefore |(E-V)^{1/4}\Cal Lu_{N^\prime}|\le\hfill\break \sum\limits_{\ell=0}^2 C_{\#}S^{-1/4}(x)B^{-2+\ell}(x)|\partial_x^\ell u_{N^\prime}(x,E)|\le \sum\limits_{\ell=0}^2 C_{\#}S^{-1/4}(x)B^{-2+\ell}(x)\cdot\hfill\break C_{\#} \Lambda^{-N^\prime}B^{-\ell}(x) by Lemma 2. That is, |(E-V)^{1/4}\Cal Lu_{N^\prime}(x,E)|\le\hfill\break C_{\#}\Lambda^{-N^\prime}S^{-1/4} (x)B^{-2}(x) on U_{\text{center}}. The Corollary now follows at once from (19).\,\, \blacksquare \medskip We have carried out the task of extending F_+^{\text{left}}(x,E) to an approximate solution of (1) in U_{\text{center}}. Completely analogous to our solution of (1) is another approximate solution defined near x=x_{\text{rt}}(E). This solution has the form F_0^{\text{rt}}(x,E)=\lambda_{\text{rt}}^{-1/3}\Bigl(\frac {-\partial Y^{\text{rt}}(x,E)}{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{rt}} ^{2/3}Y^{\text{rt}}(x,E)) for |x-x_{\text{rt}}(E)|<\lambda_{\text{rt}}^ {-\varepsilon}B(x_{\text{rt}}), |E-E_0|0. A similar formula holds for x=x_{\text{rt}}(E)-\delta, and therefore (26) implies$$ \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|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|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|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|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}}, x0. \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|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 $$V(x)-E_1\ge c_{\#}\Bigl(\frac{S_{\text{left}}}{B_{\text{left}}}\Bigr) (x_{\text{left}}(E_1)-x)\quad\text{for}\ x_{\text{left}}(E_1)-c_{\#}B_{\text{left}}x_{\text{left}}(E)-\lambda_{\text{left}}^KB_{\text{left}}\}. \noindent Estimates (2), (3), (4) together imply$$ 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_{\#}>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>\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$_{\Cal H}=\int \chi^2(F_c^{\text{left}})^2<0$. 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$usolves the ODE and has norm \align \Vert u\Vert_{L^2(I_{\text{center}})}&\le C_{\#}\int_{I_{\text{center}}} \frac{dx}{(E_1-V(x))^{1/2}}\cdot\Vert f_1\Vert_{L^2(I_{\text{center}})}\\ &\le C_{\#}\int_{I_{\text{center}}}\frac{dx}{(E_1-V(x))^{1/2}}\Vert f\Vert, \endalign \noindent provided\Vert H(\cdot,\cdot)\Vert_{\text{Hilbert--Schmidt}}\le \frac 12 \operatornamewithlimits{inf}_{y \in I_{\text{center}}}|H_1(y)|$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$\overline x_1+\hat c_{\#}B_{\text{left}}\le x\le \overline x_2 -\hat c_{\#}B_{\text{right}}, \align \Big|\Bigl(\frac{d}{dx}\Bigr)^m\chi(x)\Big|&\le C_{\#}^m B_{\text{left}}^{-m}\quad\text{in}\ [\overline x_1,\overline x_1+\hat c_{\#} B_{\text{left}}],\\ \Big|\Bigl(\frac{d}{dx}\Bigr)^m\chi(x)\Big|&\le C_{\#}^mB_{\text{right}}^{-m} \quad\text{in}\ [\overline x_2-\hat c_{\#}B_{\text{right}},\overline x_2], \endalign \noindent0\le \chi\le 1$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, $$|_{\Cal H}|= |\int_{I_{\text{center}}}\chi(F_c^{\text{left}})^2\,dx|\le C_{\#}\Lambda^{-N^\prime}\int_{I_{\text{center}}}\frac {dx}{(E_1-V(x))^{1/2}}.\tag 25$$ \noindent To see this, write a partition of unity$\chi=\sum\limits_\nu\chi_\nu$with$\chi_\nu$supported in$\{|x-x_\nu|_{\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 $$\noindent In writing (38), we changed the limits of integration without affecting the order of magnitude of the integral. Combining estimates (29), (36), (37), (38), we obtain the following result. \vglue 1pc \proclaim{Lemma 11} For a real constant b_{\text{left}} of size |b_{\text{left}}|\le C_{\#}(\int_{Vc_{\#}\frac{B_{\text{left}}}{S_{\text{left}}^{1/2}}=c_{\#} \frac{\lambda_{\text{left}}}{S_{\text{left}}}, so$$ \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_{\#}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$E_1^\prime0$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$I_{\text{far\ left}}we have \align (\frac{d^2}{dx^2}+E_1-V(x))F&=\chi_{\text{Airey\ left}}\Bigl\{\Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F_{\text{Airey\ left}}\Bigr\}\\ &\,\,+2 \frac{d\chi_{\text{Airey\ left}}}{dx}\cdot \frac{dF_{\text{Airey\ left}}} {dx}\\ &\,\,+\frac{d^2\chi_{\text{Airey\ left}}}{dx^2}\cdot F_{\text{Airey\ left}}.\endalign \noindent OutsideI_{\text{far\ left}}\cap I_{\text{Airey\ left}}$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$\tilde I_{\text{medium\ right}}we conclude that \align \Bigl(\frac{d^2}{dx^2}+E_1-V(x)\Bigr)F&=\chi_{\text{center}}\Bigl( \frac{d^2}{dx^2}+E_1-V(x)\Bigr)\text{Re}(F_c^{\text{left}})\\ &\,\,\,+\chi_{\text{medium\ right}}\cdot (-1)^{k_1}R(E_1)\Bigl( \frac{d^2}{dx^2}+E_1-V(x)\Bigr)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{Re}(F_c^{\text{right}})\\ &\,\,\,+2 \frac{d\,\chi_{\text{center}}}{dx}\cdot \frac{d}{dx}\{\text{Re} (F_c^{\text{left}}-\Cal A(E_1)F_c^{\text{right}})\}\\ &\,\,+\frac{d^2\chi_{\text{center}}}{dx^2}\cdot\text{Re}(F_c^{\text{left}} -\Cal A(E_1)F_c^{\text{right}})\,\text{on}\ \tilde I_{\text{medium\ right}}. \endalign \noindent The global WKB lemma shows that the first two terms on the right are bounded byC_{\#}\Lambda^{-N^\prime}S_{\text{right}}^{-1/4}B_{\text{right}}^{-2}$,$C_{\#}\lambda_{\text{right}}^{10-\frac 32\varepsilon N^\prime}B_{\text{right}} ^{3/2}$, 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$|\Phi(E)-\pi k|\frac{c_{\#}\Lambda}{(\int_{x_{\text{left}}}^{x_{\text{right}}}\frac{ dx}{S^{1/2}(x)})}$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$|E_\infty-E_0|0$on$\{|E-E_0|0$on$\{|E-E_0|x_{\text{rt}}+ \frac 12\lambda_{\text{rt}}^KB_{\text{rt}}$, 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$xx_{\text{rt}}+ \lambda_{\text{rt}}^KB_{\text{rt}}$is analogous. For$x$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_{\#}\ \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|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_{\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 \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 xx_{\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|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 $$\noindent Combining this with the preceding equation, we get the conclusion of the lemma.\quad\blacksquare \vglue 1pc \demo{Remark} Later, we will see that the universal constants \tilde c(0), \tilde c(1), \tilde c(2) are all zero. This will be important for the normalization of WKB eigenfunctions. The ideas in the proofs of Lemmas 1\ldots4 have used only the asymptotic properties of the Airey function, which are not enough to specify the \tilde c(m). Now suppose we are in the setting of the WKB Eigenfunction Theorem, and let F be a real--valued eigenfunction of H, with L^2--norm 1 and eigenvalue E. We assume |E-E_0|cB^{-1}. Hence we can make the change of variable t=\lambda^{2/3}Y(x,E) to reduce matters to Lemma 4. Define \hat I=\text{Image\ of}\ \{|x-x(E)|<\lambda^{-\varepsilon}B\} under the map x\mapsto t=\lambda^{2/3}Y(x,E), and set$$\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). $$\noindent That is, t_0 is the image of x(E) under x\mapsto t. The WKB Eigenfunction Theorem gives Y=Y_0+\lambda^{-2}Y_1 with |Y_1|\le C_{\#} and Y_0=0 at x=x(E). Hence |t_0|0}\Big\{ \frac{\theta(x)\lambda^{-4/3}(\frac{\partial Y}{\partial x})^{-2}} {(\lambda^{1/3}Y^{1/2})}\Big\}\lambda^{2/3}(\frac{\partial Y}{\partial x}) \,dx+\\ +\sum\limits_{m=0}^2 \tilde c(m)(\lambda^{-2/3}(\frac{\partial Y} {\partial x})^{-1}\frac{d}{dx})^m\Big\{\theta(x)\lambda^{-4/3} (\frac{\partial Y}{\partial x})^{-2}\Big\}\Bigm|_{x=x(E)}+ \text{Error}_3,\endmultline$$ \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 $$\frac{\partial\xi}{\partial x}>c_{\#} B^{-1}\quad\text{for}\quad |x-x(E)|, |\tau|0}\Big\{\theta(x)\Bigl(\frac{\partial Y_0} {\partial x}+\tau \frac{\partial Y_1}{\partial x}\Bigr)^{-2}\Bigr\}\Bigm|_ {x=x(E)+Bg(\xi,\tau)}\xi^{-1/2}\,d\xi.\tag 10$$ 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 $$\split \Big|\partial_\tau\Bigl\{\theta(x)\Bigl(\frac{\partial Y_0}{\partial x}+ \tau\frac{\partial Y_0}{\partial x}\Bigr)^{-2}\Bigr\}\Big|\le C_{\#}B^2,\\ \text{for}\ |x-x(E)|0} \theta(x)\Bigl(\lambda^2(\frac{\partial Y_0}{\partial x})^2Y_0 \Bigr)^{-1/2}\,dx\\ +\sum\limits_{m=0}^2 \tilde c(m)\lambda^{-(\frac{2m+4}{3})} \Bigl((\frac{\partial Y}{\partial x})^{-1}\frac{d}{dx}\Bigr)^m\Bigl\{\theta(x) (\frac{\partial Y}{\partial x})^{-2}\Bigr\}\Bigm|_{x=x(E)}\\ +\text{Error}_4, \quad\text{with}\quad |\text{Error}_4|\le C_{\#}\lambda^{\frac 52 \varepsilon-3}B^2. \endmultline\tag 11$$ 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|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 \{xx_{\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 $$We shall show that \text{Term}\ 1(\theta), \text{Term}\ 2(\theta) are negligibly small. In fact, using a partition of unity we can write \theta=\sum\limits_\nu\theta_\nu with each \theta_\nu supported in \{|x-x_\nu|c_{\#}\lambda^{-\varepsilon}(x_\nu)B(x_\nu), and \text{supp}\ \theta_\nu \subset \text{supp}\ \theta. We then have \text{Term}\ 1(\theta)= \sum\limits_\nu\text{Term}\ 1(\theta_\nu), \text{Term}\ 2(\theta)= \sum\limits_\nu\text{Term}\ 2(\theta_\nu), and |\text{Term}\ 1(\theta_\nu)|, |\text{Term}\ 2(\theta_\nu)|\le C_{\#} (\lambda(x_\nu))^{-2N}\frac{B(x_\nu)}{S^{1/2}(x_\nu)}b_{\text{left}}^2 by the stationary phase Lemma (i.e. Lemma 1 in the section on eigenfunctions and eigenvalues of Schr\"odinger operators.) Hence,$$\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 $$\noindent by the bound for |b_{\text{left}}| in the WKB Eigenfunction Theorem. Regarding \text{Term}\ 3(\theta), we have \Bigl\vert|1-u_{\text{left}} (x,E)|^2-1\Bigr\vert\le C_{\#}\Lambda^{-2} in \text{supp}\ \theta \cap \{x>x_{\text{left}}(E)+c_{\#}B_{\text{left}}\}, by (17), (18^{\text{bis}}). Similarly \Bigl\vert|1-u_{\text{left}}(x,E)|^2-1\Bigr \vert\le C_{\#}\Lambda^{3\varepsilon-2} in \text{supp}\ \theta\cap \{x0} \frac{dx}{(E-V(x))^{1/2}}\le C_{\#}\Lambda^{3\varepsilon-2},\endmultline \tag 25$$ \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|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 $$We apply (33) to compute the inner product$$ with $F$ a Hermite function, $HF=EF$. The first two sums on the right side of (33) do not affect the inner product, since $$=\lambda^{m_1-2m}E^{m_2} = 0 \quad\text{for}\quad m_0\ge 1,$$ \noindent and similarly $$\ =\lambda^{m_1-2m}E^{m_2} = 0\quad\text{for}\quad m_0\ge 1.$$ \noindent Therefore (33) gives \align &=\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 \documentstyle{amsppt} \hsize 6 truein \vsize 9truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \def\bbar{b{\kern -4pt\raise 3pt\hbox{-}}} \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 $I=\{x \in \Bbb R\Bigm||x-x_0|\frac 12 \lambda^KB$, 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)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 $$\noindent since V-E\ge (V(x_0)+\frac 12 cSB^{-2} (x-x_0)^2)-(V(x_0)+c_{\#}\lambda^{-2\varepsilon}S)\ge c_{\#}\lambda^{-2\varepsilon}S for |x-x_0|\sim \lambda^{-\varepsilon}B. If x0. Then we can find formal power series y=\sum\limits_{k,m,s\ge 0} y_{kms}\tau^k\lambda^{-2m}x^s and \tilde\tau=\sum\limits_{k,m\ge 0} \tilde\tau_{km}\tau^k\lambda^{-2m}, satisfying the equation$$ (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}0$ for $s \ge 0$, this allows us to solve for $y_{kms}$ in terms of $y_{00s^\prime}$ and $(s^{\prime\prime} A$. 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 $k+mc_{\#}>0$.\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 $$|\partial_x^\alpha\partial_E^\beta y(x,E)|\le C_{\#}^{\alpha\beta}B^{-\alpha} S^{-\beta}\quad\text{for}\quad |x-x_0|c_{\#}B^{-1}\quad\text{for}\quad |x-x_0|c_{\#}\lambda^2S^{-1}=c_{\#}B^2\quad\text{for}\quad |E-V(x_0)|c_{\#}. Since |\frac{\partial^2y}{\partial x^2}|\le C_{\#}B^{-2} and |\frac{\partial^2y}{\partial x\partial E}|\le C_{\#}B^{-1}S^{-1}, the lower bound for \frac{\partial y}{\partial x} remains valid throughout |x-x_0|\lambda^{-\varepsilon}B} |F(x)|^2\ dx\le C_{\#}^\prime\lambda^{-N}\int_{|x-x_0|<\lambda^{-\varepsilon}B} |F(x)|^2\,dx, hence$$ \Vert\tilde F\Vert_{L^2(\frac 12C_{\#}\lambda^{-\varepsilon}<|y|\,=(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|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/10}S$. Hence in all cases we have$E+C_{\#}\lambda^{-\varepsilon N/10}Sc_{\#}^\prime\lambda^{-\varepsilon})}&\le C_{\#} ^\prime\lambda^{-N}\Vert\tilde F\Vert_{L^2(|y|c_{\#}^\prime\lambda ^{-\varepsilon})} &\le C_{\#}^\prime\lambda^{-N}\Vert\tilde F\Vert_{L^2(|y|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)|\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) $$\noindent for V(x_0)c_1>0. Define \phi_W(E)=\int_{E-W(x)>0}(E-W(x))^{1/2}\,dx, \psi_W(E)=\operatornamewithlimits {lim}_{\delta\to 0+} \bigl[\int_{E-W(x)>\delta}W^{\prime\prime}(x) (E-W(x))^{-3/2}\,dx-q(E)\delta^{-1/2}\Bigr] for small, positive E. Then for 0\hat c_1>0, |(\frac{d}{dx})^\alpha y(x)|\le \hat C_\alpha for |x|<\hat c. Here, \hat c_1, \hat C_\alpha depend only on c_1 and the C_\alpha. Changing variable from x to y=y(x) in the definitions of \phi_W, \psi_W gives$$ \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)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)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$We summarize the results of this section in the next section. %Subject: Section 9 \pageno 156 \magnification \magstep1 \input amstex \documentstyle{amsppt} \hsize 6 truein \vsize 9 truein \baselineskip 20 pt \hoffset .15 truein \rightheadtext{} \catcode\@=11 \def\logo@{} \catcode\@=12 \centerline{\bf {The WKB Theorem on Low Eigenvalues}} \smallskip Let$\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|\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)\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} $$\noindent for V(x_0)C_{\#}\lambda^{-2+4\varepsilon}, then k_{\text{max}}=\overline n. In any case, |k_{\text{max}}-\overline n|\le 1. \item"{(b)}" If 0\le kc_{\#}\lambda S^{-1}, which we leave to the reader. The WKB Theorem on low eigenvalues follows at once from the corollaries to Lemmas 5 and 7 in the preceding section. Finally, our present Lemma 2 just restates Lemma 1 from the preceding section.\qquad\blacksquare %Subject: Section 10 \pageno 159 \magnification\magstep1 \documentstyle{amsppt} \hsize 6 truein \vsize 9truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \def\Icirc{\hbox{I}\kern-3pt\raise 5pt\hbox{\circ}} \centerline{\bf{WKB Theory with Weak Turning Points}} \smallskip In this section, we develop a crude form of WKB Theory that requires only very weak hypotheses on the potential near the turning points. \noindent\underbar{Set--up}. We are given an energy E_0 and a potential V(x) defined on a (possibly unbounded) interval I_{\text{BVP}}. The interval I_{\text{BVP}} is partitioned into subintervals I_{\text{far\ left}}, I_{\text{left}}, I_{\text{center}}, I_{\text{rt}}, I_{\text{far\ rt}} with I_{\text{far\ left}} to the left of I_{\text{left}}, I_{\text{left}} to the left of I_{\text{center}}, etc. Here, I_{\text{far\ left}} and I_{\text{far\ right}} may be empty. On I_{\text{center}} we are given positive weight functions S(x), B(x). Set \lambda(x)=S^{1/2}(x)B(x) and \Lambda=(\int_{I_{\text{center}}}\frac{dx} {\lambda(x)B(x)})^{-1}. We make the following assumptions. \noindent\underbar{Hypotheses} \roster \item"{(H\hat 0)}" I_{\text{center}} is non--empty, and for x,y \in I_{\text{center}} with |x-y|cB(x). \item"{(H\hat 1)}" For x \in I_{\text{center}} we have |(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x) B^{-\alpha}(x) and cS(x)y.$$ \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 $I_\nu=\{|x-x_\nu|c_{\#}|I_\nu|$. 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 $I_\nu= \{|x-x_\nu|0$ 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 $0c_1$. 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}\ 00 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>\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}\ H0$ 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 $|x-y|cB(x)$. \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$\overline 2$)}" $\{x \in I_{\text{BVP}}\mid V(x)cB(x_{\text{left}})$, $\text{dist}(x_{\text{right}},\partial I)>cB(x_{\text{right}})$. \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)