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PACS (2003): 2.30.Zz, 3.65.Nk, 43.25.+y;
MSC (2000): 34A55, 34L25, 34L30
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nonlinear potential, generalized nonlinear Schroedinger equation, inverse scattering, nonlinear scatterer
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%January 27, 2004
\input amstex
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\baselineskip 18 pt
\parskip 6 pt
\centerline {\bf INVERSE SCATTERING ON THE LINE FOR}
\centerline {\bf A GENERALIZED NONLINEAR SCHR\"ODINGER EQUATION}
\vskip 10 pt
\centerline {Tuncay Aktosun}
\vskip -8 pt
\centerline {Department of Mathematics and Statistics}
\vskip -8 pt
\centerline {Mississippi State University}
\vskip -8 pt
\centerline {Mississippi State, MS 39762, USA}
\vskip -8 pt
\centerline {aktosun\@math.msstate.edu}
\centerline {Vassilis G. Papanicolaou and Vassilis Zisis}
\vskip -8 pt
\centerline {Department of Mathematics}
\vskip -8 pt
\centerline {National Technical University of Athens}
\vskip -8 pt
\centerline {Zografou Campus}
\vskip -8 pt
\centerline {157 80, Athens, Greece}
\vskip -8pt
\centerline {papanico\@math.ntua.gr}
\noindent {\bf Abstract}: A one-dimensional generalized nonlinear Schr\"{o}dinger equation
is considered, and the corresponding inverse scattering problem is
analyzed when the potential is compactly supported
and depends on the wave function. The unique recovery of the potential is
established from an appropriate set of scattering data.
\vskip 15 pt
\par \noindent {\bf PACS (2003):} 2.30.Zz, 3.65.Nk, 43.25.+y
\vskip -8 pt
\par \noindent {\bf Mathematics Subject Classification (2000):}
34A55, 34L25, 34L30
\vskip -8 pt
\par\noindent {\bf Keywords:} nonlinear potential, generalized nonlinear
Schr\"{o}dinger equation, inverse scattering, nonlinear scatterer
\newpage
\noindent {\bf 1. INTRODUCTION}
\vskip 3 pt
Consider the nonlinear equation
$$-u''+Q(x,u)\,u=k^{2}u,\qquad x\in {\bold R},\tag 1.1$$
where $k$ is a real parameter, the prime denotes the derivative
with respect to the spatial variable $x$, and $Q(x,u)$ has the
form
$$Q(x,u)=\sum_{n=0}^{\infty }q_{n}(x)\,u^{n},\tag{1.2}$$
with each $q_{n}(x)$ being real valued, bounded, measurable,
supported in $[0,b]$ for a fixed $b>0,$ and the series
$$\sum_{n=0}^{\infty }\left(\sup_{x\in [0,b]}
|q_{n}(x)| \right)u^{n}\tag{1.3}$$
being entire in $u$. The assumption that the series in (1.3) is entire
is equivalent to assuming that its radius of convergence is infinite, i.e.
$$\varlimsup_{n\to+\infty}\left(\sup_{x\in [0,b]}|q_{n}(x)|
\right)^{1/n}=0.\tag{1.4}$$
Note that such an assumption is stronger than just assuming that $Q(x,u)$
given in (1.2) is entire in $u$.
In this paper we consider the solution to (1.1) satisfying
$$u(0;k)=\varepsilon,\qquad u^{\prime}(0;k)=-ik\varepsilon.\tag{1.5}$$
As shown in Proposition~2.1, a unique solution to (1.1) exists for
all $x\in{\bold R}$ and $k\in{\bold R}$ when $|\varepsilon|$ is
sufficiently small. The results given in our
paper hold both for real and complex values of
$\varepsilon.$ We suppress the dependence of $u$ on
$\varepsilon$ for simplicity. Since for $n\geq 0$ we assume
$q_{n}(x)=0$ when $x\notin [0,b],$ it follows that for each
$k\in{\bold R}\setminus \{0\}$ the general solution to (1.1) for
$x\notin (0,b)$ is a linear combination of $e^{ikx}$ and
$e^{-ikx}$. Thus, we have
$$u(x;k)=\varepsilon e^{-ikx},\qquad x\leq 0,\tag{1.6}$$
which is equivalent to (1.5), and
$$u(x;k)=A(k;\varepsilon)\,e^{ikx}+B(k;\varepsilon)\,
e^{-ikx},\qquad x\geq b.\tag{1.7}$$
Because of the resemblance with the (linear) Schr\"{o}dinger equation, we
will refer to (1.1) as a generalized nonlinear Schr\"{o}dinger equation
and to $Q(x,u) $ as the (nonlinear) potential.
The expressions for $u(x;k)$ given in (1.6) and (1.7) indicate that
we have a scattering problem in hand, where a plane wave is sent from $%
x=+\infty $ onto the nonhomogeneity $Q(x,u)$, and a part of the wave is
transmitted to $x=-\infty $ and a part is reflected back to $x=+\infty $.
However, contrary to the linear case, there is in general
no energy conservation, i.e. $|B(k;\varepsilon)|^2-|A(k;\varepsilon)|^2$
in general depends on $k.$
The linear part $q_0(x)$ of the potential $Q(x,u)$ in (1.2)
represents the restoring force density in wave propagation
governed by (1.1) in the frequency domain. The higher order
terms in the potential allow the description of nonlinear reaction
of the medium to propagation of
elastic waves.
In analogy with
the direct and inverse scattering problems for the linear Schr\"{o}dinger
equation, the corresponding problems for (1.1) can be formulated as
follows. In the direct problem, given $Q(x,u)$, our task is to determine the
``scattering coefficients'' $A(k;\varepsilon)$ and $B(k;\varepsilon)$ for
sufficiently small $|\varepsilon|.$ On the other hand, in the inverse
scattering problem, given some scattering data related to $A(k;\varepsilon) $
and $B(k;\varepsilon)$, the task is to recover $Q(x,u)$. In our paper, we do
not study the characterization of the scattering data
so that the existence of a corresponding potential is assured;
we only discuss the uniqueness and
recovery aspects of our inverse problem by assuming that there exists
at least one potential corresponding to our scattering coefficients.
The inverse scattering problem
analyzed in this paper is analogous to the recent study by Weder [1],
where the time-dependent Schr\"{o}dinger equation with a nonlinear term is
investigated and both the linear and nonlinear parts of the potential are
recovered from some appropriate scattering data by using a time-domain method.
For other related studies of inverse problems on nonlinear equations using time-domain
methods, we refer the reader to [2-5] and the references therein.
Our paper is organized as follows. In Section~2 we show that the direct
and inverse scattering problems
for the nonlinear equation (1.1) are equivalent to the corresponding
problems for an infinite
number of linear equations, and we analyze
the basic properties of the scattering data for each linear equation
and define the appropriate data set $\Cal D_n$ given in (2.16) for each $n\ge 1.$
We then solve the inverse scattering problem for each $n\ge 1$
recursively. The solution of the inverse problem when
$n=1$ is well known, and in Section~3 we list
the basic facts from the case $n=1$
that are needed later on to solve the inverse problems for
$n\ge 2.$ In Section~4 we show that the solution of
the inverse scattering problem for each
$n\ge 2$ can be obtained by inverting either of the two
integral equations (4.4) and (4.5).
In Section~5 we prove the unique invertibility of (4.4) and (4.5) to recover
$q_{n-1}(x)$ for each $n\ge 2,$ and we summarize the recovery
of the nonlinear potential $Q(x,u)$ in terms of the scattering
data involving $A(k;\varepsilon)$ and $B(k;\varepsilon).$
Finally, in Section~6, we illustrate the direct and inverse problems
for (1.1) with some concrete examples.
\vskip 10 pt
\noindent {\bf 2. PRELIMINARIES}
\vskip 3 pt
It is straightforward to verify that $u(x;k)$ satisfies (1.1) and (1.5)
if and only if it satisfies the integral equation
$$u(x;k)=\varepsilon\, e^{-ikx}+\frac{1}{k}\int_{0}^{x}
\sin\left(k(x-t)\right)\,Q\left(t,u(t;k)\right)\,u(t;k)\,dt.\tag{2.1}$$
\noindent {\bf Proposition 2.1.} {\it There exists a constant $\delta>0$ depending only on
$Q(x,u)$, but not on $k$ (as long as $k$ is real), such that if $|\varepsilon
|\leq \delta $ then a solution $u(x;k)$ to (1.1) satisfying (1.5)
exists for all $x\in {\bold R}$, and it is unique.}
\noindent PROOF: Suppose that a solution $u(x;k)$ to (2.1) ceases to exist,
i.e. blows up, for some $x\in (0,b]$. Then, for any $r$ sufficiently large,
there is an $x_{0}\in (0,b]$ such that $|u(x_{0};k)|=r$ and $|u(x;k)|0$ such that $|Q(x,u)|\leq C$
for $x\in[0,b]$ and $|u|\leq r$. Let us take
$$|\varepsilon |\leq \delta :=(r/2)\,e^{-Cb^{2}}.\tag{2.2}$$
Using $|\sin \theta |\leq |\theta |$ for real $\theta $ and the
realness of $k,$ for $xt.\endcases
\tag{3.7}$$
It is seen from (2.7), the second line in (3.4), and (3.7)
that, when $k\in
{\bold C}^{+}\setminus \{i\kappa _{j}\}_{j=1}^{N}$ the Green's
function $G(x,t;k)$ decays exponentially
as $x\to \pm\infty$, and it is the
only solution to (3.6) having this property.
\vskip 10 pt
\noindent {\bf 4. INVERSE SCATTERING FOR $q_{n-1}(x)$ WITH $n\geq 2$}
\vskip 3 pt
In this section we show that the recovery of $q_{n-1}(x)$ for each $n\geq 2$
is equivalent to inverting the integral equation given in (4.4) below
or the one given in (4.5). Having seen in the previous section
that $q_{0}(x)$ can be recovered uniquely from $\{A_{1}( k) ,B_{1}(k)\}$, we
proceed recursively and prove the unique recovery of $q_{n-1}(x)$
from the data set ${\Cal D}_{n}$ defined in (2.16).
With the help of (2.6)-(2.11), we define
$$y_{n}(x;k):=u_{n}(x;k)-\frac{B_{n}(k)}
{B_{1}(k)}\,u_{1}(x;k),\qquad n\geq 2.\tag{4.1}$$
The relevant properties of $y_{n}(x;k)$ are analyzed next.
\noindent {\bf Proposition 4.1.} {\it Let $y_{n}(x;k)$ be the quantity defined in (4.1).
Then:}
\item{(i)} {\it For each $k\in {\bold C}^{+}\setminus \{i\kappa
_{j}\}_{j=1}^{N}$, where $\{i\kappa
_{j}\}_{j=1}^{N}$ is the set of zeros of $B_{1}(k)$
in ${\bold C}^{+}$, the quantity
$y_{n}(x;k)$ is a solution to (2.9), and
it is the only solution belonging to $L_{2}({\bold R})$ in $x$.
In fact, $y_{n}(\cdot;k)$ decays exponentially as $x\to\pm\infty.$}
\item{(ii)} {\it $y_{n}(x;k)$ satisfies}
$$y_{n}(x;k)=\cases -\displaystyle
\frac{B_{n}(k)}{B_{1}(k)}\,e^{-ikx},
\qquad x\leq 0, \\
\noalign{\medskip}
\left[A_{n}(k)-\displaystyle\frac{B_{n}(k)}{B_{1}(k)}\,A_{1}(k)\right]
\,e^{ikx}, \qquad x\geq b.\endcases\tag{4.2}$$
\item{(iii)} {\it In terms of the nonhomogeneous term $g_{n}(x;k)$
defined in (2.12) and the Green's function $G(x,t;k)$ given in (3.7),
we have}
$$y_{n}(x;k)=-\int_{-\infty }^{\infty }G(x,t;k)\,g_{n}(t;k)\,
dt,\qquad k\in {\bold C
}^{+}\setminus \{i\kappa _{j}\}_{j=1}^{N}.\tag{4.3}$$
\noindent PROOF: First, from Propositions~2.2 and 2.3
and the fact that the
only zeros of $B_{1}(k)$ in ${\bold C}^{+}$ occur at $k=i\kappa _{j}$
for $j=1,\dots,N,$ it
follows that $y_{n}(x;k)$ is well defined for each
$k\in {\bold C}^{+}\setminus \{i\kappa _{j}\}_{j=1}^{N}$.
Note that $y_{n}(x;k)$ solves
(2.9) because $u_{n}(x;k)$ is a solution to the same equation
and $u_{1}(x;k)$ is a solution to the corresponding homogeneous equation. We
obtain (4.2) directly from (2.7), (2.8), (2.10), and (2.11).
For each fixed $k\in {\bold C}^{+}\setminus \{i\kappa_{j}\}_{j=1}^{N}$,
since $e^{-ikx}$ and $e^{ikx}$ decay
exponentially as $x\to-\infty $ and as $x\to +\infty $,
respectively, we get the $L_{2}({\bold R})$-property
of $y_{n}(\cdot;k)$ stated in (i). Note that $y_{n}(x;k)$ is
the only $L_{2}({\bold R})$-solution to (2.9) because
the difference with any other $L_{2}({\bold R})$-solution must
satisfy (2.6); however, as indicated in (i) of Section~3, the
only $L_{2}({\bold R})$-solutions to (2.6) occur
when $k=i\kappa _{j}$ for $j=1,\dots,N$. Thus,
we have proved (i) and (ii). From the properties listed in Proposition~2.3, it
follows that $g_{n}(\cdot;k)$ is bounded and supported in $[0,b]$ and hence
belongs to $L_{2}({\bold R})$. In terms of the operator $L$ defined in
Section~3, $y_{n}(x;k)$ satisfies $(L-k^{2})y_{n}=-g_{n}$, and hence with
the help of (3.5) we obtain (4.3). \qed
In the next theorem we present the main
integral equations from which
$q_{n-1}(x)$ with $n\ge 2$ will be recovered.
\noindent {\bf Theorem 4.2.} {\it
For each $n\geq 2$, the potential $q_{n-1}(x)$ satisfies}
$$\int_{0}^{b}v_{1}(t;k)\,u_{1}(t;k)^{n}\,
q_{n-1}(t)\,dt=E_{n}(k),\qquad k\in
{\bold C},\tag{4.4}$$
$$\int_{0}^{b}u_{1}(t;k)^{n+1}q_{n-1}(t)\,dt=F_{n}(k),
\qquad k\in {\bold C},\tag{4.5}$$
{\it where $E_{n}(k)$ and $F_{n}(k)$ are completely determined by the data
${\Cal D}_{n}$ defined in (2.16), and they are given as}
$$E_{n}(k):=-2ik\,B_{n}(k)-\int_{0}^{b}v_{1}(t;k)\,h_{n}(t;k)\,dt,\tag 4.6$$
$$F_{n}(k):=2ik[B_{1}(k)\,A_{n}(k)-A_{1}(k)\,B_{n}(k)]
-\int_{0}^{b}u_{1}(t;k)\,h_{n}(t;k)\,dt.\tag{4.7}$$
\noindent PROOF: Using (2.7) and (3.4) in (3.7), we get
$$G(x,t;k)=\cases
-\displaystyle\frac{1}{2ik\,B_{1}(k)}\,e^{-ikx}v_{1}(t;k),
\qquad x\leq \min \{0,t\},
\\
\noalign{\medskip}
-\displaystyle\frac{1}{2ik\,B_{1}(k)}\,e^{ikx}u_{1}(t;k),
\qquad x\geq \max \{b,t\}.\endcases\tag{4.8}$$
Using (4.8) on the right hand side of (4.3), we obtain
$$y_{n}(x;k)=\cases
\displaystyle\frac{1}{2ikB_{1}(k)}\,\displaystyle
e^{-ikx}\int_{0}^{b}v_{1}(t;k)\,g_{n}(t;k)\,dt, \qquad
x\leq 0, \\
\noalign{\medskip}
\displaystyle\frac{1}{2ik\,B_{1}(k)}\,
e^{ikx}\int_{0}^{b}u_{1}(t;k)\,g_{n}(t;k)\,dt, \qquad
x\geq b.\endcases\tag{4.9}$$
Comparing (4.9) with (4.2), we see that
$$2ikB_{n}(k)=-\int_{0}^{b} v_1(t;k)\, g_{n}(t;k)\,dt,\qquad n\geq 2,\tag{4.10}$$
$$2ik[B_{1}(k)\,A_{n}(k)-A_{1}(k)\,B_{n}(k)]=
\int_{0}^{b}u_{1}(t;k)\,g_{n}(t;k)\,dt,\qquad n\geq 2.\tag{4.11}$$
With the help of (2.12), we rewrite
(4.10) and (4.11) as (4.4) and
(4.5), respectively. Even though we have derived
(4.4) and (4.5) only for $k\in
{\bold {C}}^{+}\setminus \{i\kappa _{j}\}_{j=1}^{N}$, with the
help analytic extensions indicated in Proposition~2.2, we see that
the former is actually valid for $k\in {\bold{C}}$ and
the latter for $k\in {\bold{C}}\setminus \{0\}$.
Furthermore, the analytic extension to $k=0$ for (4.5) is proved by showing
that the quantity $k[B_{1}(k)\,A_{n}(k)-A_{1}(k)\,B_{n}(k)]$ appearing
in (4.7) is continuous at $k=0$. In order to see this,
with the help of (2.15), as $k\to 0$ in
${\bold{C}}$ we obtain
$$2ik[B_{1}(k)\,A_{n}(k)-A_{1}(k)\,B_{n}(k)]=u_{n}^{\prime
}(x;k)[A_{1}(k)+B_{1}(k)]+O(1).$$
Although each of
$A_{1}(k)$ and $B_{1}(k)$ may have a simple pole at $k=0,$ their
sum $A_{1}(k)+B_{1}(k)$ is continuous at $k=0$ because
(2.8) implies
$$A_{1}(k)+B_{1}(k)=e^{-ikx}u_{1}(x;k)+k\,B_{1}(k)\,\frac{1-e^{-2ikx}}{k},%
\qquad x\geq b,$$
and the right hand side is $O(1)$ as $k\to 0$ in ${\bold{C}}$. \qed
With the help of Theorem~4.2, we see that our inverse scattering problem of
recovery of $q_{n-1}(x)$ for each $n\geq 2$ can be stated as follows:
Given $E_{n}(k)$ or $F_{n}(k)$, determine $q_{n-1}(x)$ for $x\in [0,b]$
by inverting either (4.4) or (4.5).
From (2.13) and (4.6) it is seen that the data
set $\{B_{n}(k),q_{0}(x),\dots ,q_{n-1}(x)\}$
uniquely determines all the
quantities given in (4.4) except for $q_{n-1}(x)$
there, and hence one
can use (4.4) to determine $q_{n-1}(x)$
without needing $A_{n}(k)$ in the data.
\vskip 10 pt
\noindent {\bf 5. UNIQUE RECOVERY OF $q_{n-1}(x)$ WITH $n\geq 2$}
\vskip 3 pt
In this section we show that $q_{n-1}(x)$ with $n\ge 2$ can be recovered
by inverting
either (4.4) or (4.5). We will present the inversion only for (4.5) because
the inversion of (4.4) is similar.
From (3.3) we see that we can write
$$u_{1}(t;k)^{n+1}=e^{-ik(n+1)t}\left[1+\frac{U(t;k)}{k}\right],
\qquad x\in[0,b],\quad
k\in\overline{{\bold C}^{+}}\setminus \{0\}
,\tag 5.1$$
where $U(t;k)$ is uniformly bounded
for $x\in[0,b]$ and $k\in\overline{{\bold C}^{+}};$ i.e.
$$|U(x;k)|\le M,\qquad x\in[0,b],\quad k\in\overline{{\bold C}^{+}}
,\tag 5.2$$
where $M$ is independent of $x$ and $k.$
Let us separate the real and imaginary parts of the complex $k$ variable
in ${\bold C}^{+},$ discretize the former, and keep the latter as
a fixed positive parameter by writing
$$k_{m}=\frac{2\pi m}{(n+1)b}+i\xi ,\qquad m\in {\bold Z}.\tag{5.3}$$
Let us evaluate (4.5) at $k=k_m$ and write the resulting equation
in the operator form as
$K\phi=p,$ or equivalently as
$$\int_0^b K(m,t)\,\phi(t)\,dt=p(m),\qquad m\in {\bold Z},\tag 5.4$$
where we have defined [cf. (5.1) and (5.3)]
$$K(m,t):=e^{-\xi (n+1)t}u_{1}\left( t;k\right)^{n+1}=
e^{-2\pi imt/b}\left[ 1+\frac{U(t;k_{m})}{2\pi
m(n+1)^{-1}b^{-1}+i\xi }\right],\tag 5.5$$
$$\phi(t):=e^{\xi (n+1)t}\,q_{n-1}(t),\tag 5.6$$
$$p(m):=F_n(k_m).\tag 5.7$$
Note that we suppress the dependence on $n$ and $\xi$
because the inversion of (4.5) will be done at fixed values
of $n$ and $\xi.$
When $q_0(x)\equiv 0,$ we have $u_1(t;k)\equiv e^{-ikt}$ and
$U(t;k)\equiv 0;$ in that case the operator
$K$ reduces to $K_0,$ which is given by
[cf. (5.5)]
$$(K_0g)(m)=\int_0^b K_0(m,t)\,\phi(t)\,dt
=\int_0^b e^{-2\pi imt/b}\,\phi(t)\,dt.\tag 5.8$$
The properties of $K$ and $K_0$ are analyzed next.
\noindent {\bf Theorem 5.1.} {\it
The operators $K$ and $K_0$ defined
in (5.4) and (5.8), respectively, satisfy the following:}
\item{(i)} {\it $K_0$ maps $L_2(0,b)$ into $l_2(\bold Z),$ and
its operator norm is given by
$||K_0||=\sqrt{b}.$}
\item{(ii)} {\it $K_0^{-1}$ exists
as a map from $l_2(\bold Z)$ into $L_2(0,b),$ and
its operator norm is given by
$||K_0^{-1}||=1/\sqrt{b}.$}
\item{(iii)} {\it $K-K_0$ maps $L_2(0,b)$ into $l_2(\bold Z).$ The
operator norm of $K-K_0$ satisfies
$||K-K_0||\le \sqrt{s(\xi)},$ where $s(\xi)$
is a monotone decreasing function on
$\xi\in(0,+\infty)$ vanishing at infinity.}
\item{(iv)} {\it There exists $\xi_0>0,$ determined
by $n,$ $b,$ and $\int_0^b |q_0(t)|\,dt$ alone,
such that
the inverse operator $K^{-1}$ exists for any
$\xi>\xi_0.$}
\noindent PROOF: For the proof of (i),
note that $K_0$ is the standard Fourier series
operator and hence it maps $L_2(0,b)$ into $l_2(\bold Z).$
With the help of the completeness relation
$$\displaystyle\sum_{m\in{\bold Z}}e^{2\pi imt/b}\,e^{-2\pi imx/b}=b\cdot
\delta(t-x),\qquad t,x\in\bold R,$$
using the definition of the operator
norm, from (5.8) we obtain $||K_0||=\sqrt{b}.$
As for (ii), the inverse Fourier series
operator $K_0^{-1}$ is given by
$$\left( K_{0}^{-1}h\right)(t)=\displaystyle\frac{1}{b}\,
\displaystyle\sum_{m\in {\bold Z}}h\left( m\right) e^{2\pi imt/b},\tag 5.9$$
and its operator norm is computed in a straightforward manner with the help of
$$\int_0^b e^{2\pi imt/b}\,e^{-2\pi ijt/b}\,dt=b\cdot
\delta_{mj},\qquad m,j\in\bold Z,$$
where $\delta_{mj}$ denotes the Kronecker delta,
yielding $||K_0^{-1}||=1/\sqrt{b}.$
As for (iii), note that
the square of the operator norm of $K-K_0$ is defined as
$$||K-K_{0}||^{2}:=\sup_{||g||=1}\left\| \left(
K-K_{0}\right) g\right\| ^{2}=\sup_{\left\| g\right\| =1}\sum_{m\in {\bold Z}%
}\left| \left[ \left( K-K_{0}\right) g\right] \left( m\right) \right| ^{2},
$$
where $||g||$ denotes the norm of $g$ in $L_{2}(0,b)$. With
the help of (5.3)-(5.5) and (5.8) we get
$$\left\| K-K_{0}\right\|^{2}=\sup_{\left\| g\right\| ^{2}=1}\sum_{m\in {\bold
Z}}\left| \int_{0}^{b}\frac{U(t;k_{m})}{2\pi
m(n+1)^{-1}b^{-1}+i\xi}\,\phi(t)\,dt\right|^{2}.\tag 5.10$$
Using the Cauchy-Schwarz inequality on (5.10), $||g||^{2}=1,$
and (5.2), we obtain
$$\aligned
||K-K_{0}||^{2}&\leq \displaystyle\sum_{m\in {\bold Z}}\int_{0}^{b}\left|
\frac{U(t;k_m)}{2\pi m(n+1)^{-1}b^{-1}+i\xi }\right|^{2}\,dt \\
\noalign{\medskip}
&\leq M^{2}b\displaystyle\sum_{m\in {\bold Z}}\frac{1}{4\pi
^{2}m^{2}(n+1)^{-2}b^{-2}+\xi^{2}},\endaligned$$
and since the summation in the last term can be evaluated
in a closed form, we obtain
$$||K-K_{0}||^{2}\leq \displaystyle\frac{M^{2}b^{2}(n+1)}{2\xi}\,
\coth \left( \frac{(n+1)b\xi}{2}\right),
\qquad \xi>0,\quad n\ge 2.\tag 5.11$$
Note that $\coth(x)$ is monotone decreasing
on $x\in(0,+\infty).$ Using (3.3), (5.2), and (5.11), we see that
$||K-K_{0}||\le \sqrt{s(\xi)},$ where
$$s(\xi):=\displaystyle\frac{b^2\,(n+1)^3}{2\xi}\left(
\int_0^b |q_0(x)|\,dx\right)^2\,
\coth \left( \frac{(n+1)b\xi}{2}\right)
\,\exp\left(2b(n+1) \int_0^b |q_0(x)|\,dx\right).$$
Next, we need to prove (iv).
Writing
$$K=K_0+(K-K_0)=K_0[I+K_0^{-1}(K-K_0)],$$
we see that $K^{-1}$ can be evaluated as a Neumann series
$$K^{-1}=[I+K_0^{-1}(K-K_0)]^{-1}K_0^{-1}=
\displaystyle\sum_{j=0}^\infty (-1)^j
[K_0^{-1}(K-K_0)]^jK_0^{-1},$$
whose convergence is assured by choosing
$||K_0^{-1}||\,||(K-K_0)||<1.$ From (ii) and (iii), we see
that this is achieved by choosing $\xi_0$ as the unique solution
to $s(\xi_0)=b.$ Then, for any $\xi>\xi_0,$
the existence of $K^{-1}$ is assured. \qed
Having shown that $K$ is invertible for sufficiently
large $\xi$ values, we can recover $q_{n-1}(x)$ with the help of
(5.4), (5.6), and (5.7) as
$$q_{n-1}(x)=e^{-\xi(n+1)x}\,(K^{-1}F_n)(x),\qquad
x\in[0,b],\quad n\ge 2.\tag 5.12$$
Let us now summarize the reconstruction
in the solution of the inverse scattering problem for (1.1).
Assume that we are given the scattering data
consisting of $A(k;\varepsilon)$ and $B(k;\varepsilon)$ for
all $k$ in some subinterval
of the positive $k$-axis and for all
$|\varepsilon|\le \delta,$ where $\delta$
is some positive number (no matter how small).
Our aim is to recover the potential $Q(x,u)$ corresponding to
the given scattering data.
\item{(i)} Using (2.5), obtain $A_n(k)$ and $B_n(k)$ for $n\ge 1.$
Note that $kA_n(k)$ and $kB_n(k)$ have entire extensions to $\bold C.$
\item{(ii)} Recover $q_0(x)$ and $u_1(x;k)$ from
$\{A_1(k),B_1(k)\}$ by using any one of
the available inversion methods described in (v) of Section~3.
\item{(iii)} Recover $q_{n-1}(x)$ and $u_n(x;k)$
for $n\ge 2$ by using the data $\Cal D_n$ given in (2.16) in
a recursive way. The recovery of $q_{n-1}(x)$
is achieved by inverting either (4.4) or (4.5). The recovery
of $u_n(x;k)$ amounts to solving the linear equation (2.9) with
the condition in (2.10).
\item{(iv)} Having recovered all the $q_{n-1}(x)$ for $n\ge 1,$ we obtain
$Q(x,u)$ via (1.2).
The uniqueness of the recovery of $Q(x,u)$ from
the data $\{A(k;\varepsilon),B(k;\varepsilon)\}$ is summarized next.
\noindent {\bf Theorem 5.2.} {\it Consider the data
consisting of $A(k;\varepsilon)$ and $B(k;\varepsilon)$
given for
all $k$ in some subinterval
of the positive $k$-axis and
for all $|\varepsilon|\le \delta,$ where $\delta$ is
some positive number (no matter how small). Further
assume that there exists a potential $Q(x,u)$
corresponding to this data. Then $Q(x,u)$ is the only potential
corresponding to that data.}
\vskip 10 pt
\noindent {\bf 6. EXAMPLES}
\vskip 3 pt
In this section we illustrate the direct and inverse problems
for (1.1) in a special case with some explicit examples.
Assume that $Q(x,u)$ given in (1.2) has the form $Q(x,u)=q_2(x)\,u^2,$
i.e. $q_j(x)\equiv 0$ for $j\ge 0$ except when $j=2.$
From (2.6)-(2.8) we get $u_1(x;k)=e^{-ikx}$ for $x\in\bold R,$
$A_1(k)=0,$ $B_1(k)=1.$
From (2.9)-(2.13) with $n=2,$ we obtain $u_2(x;k)=0$ for $x\in\bold R,$
$A_2(k)=B_2(k)=0.$ Using (2.9)-(2.11) with $n=3,$ we see that
$u_3(x;k)$ satisfies
$$u_3''(x;k)+k^2u_3(x;k)=q_2(x)\,e^{-3ikx},\qquad x\in\bold R$$
with the initial conditions $u_3(0;k)=u_3'(0;k)=0.$ Using
variation of parameters, we get
$$u_3(x;k)=\displaystyle\frac{1}{2ik}\,e^{ikx}\displaystyle\int_0^x q_2(t)\,
e^{-4ikt}dt-\displaystyle\frac{1}{2ik}\,e^{-ikx}\displaystyle\int_0^x q_2(t)\,
e^{-2ikt}dt,\qquad x\in\bold R.\tag 6.1$$ Comparing (6.1) with
(2.11) we see that
$$A_3(k)=\displaystyle\frac{1}{2ik}\displaystyle\int_0^b
q_2(t)\,e^{-4ikt}dt,\quad B_3(k)=-\displaystyle\frac{1}{2ik}\displaystyle\int_0^b
q_2(t)\,e^{-2ikt}dt.\tag 6.2$$
Thus, we see that $A_3(k)$ and
$B_3(k)$ are related to each other as $A_3(k)=-2\,B_3(2k).$ With
the help of $\int_{-\infty}^\infty
e^{ik\alpha}dk=2\pi\delta(\alpha),$ the potential $q_2(x)$ is
uniquely recovered from $A_3(k)$ or $B_3(k)$ given in (6.2) as
$$q_2(x)=\displaystyle\frac{4i}{\pi}\displaystyle\int_{-\infty}^\infty
k\,A_3(k)\,e^{4ikx}dk=-\displaystyle\frac{2i}{\pi}\displaystyle\int_{-\infty}^\infty
k\,B_3(k)\,e^{2ikx}dk.\tag 6.3$$
Alternatively, $q_2(x)$ can be recovered by using the method of Section~5
as follows. Let us suppose that we are given $A_3(k).$
Using (2.14) and (4.7) we see that
$F_3(k)=2ikA_3(k).$ Thus, with the help of (5.3) and (5.7) we get
$$p(m)=(im\pi/b-2\xi)\,A_3(i\xi+m\pi/2b),\qquad m\in\bold Z.$$
Since $q_0(x)\equiv 0,$ we have $K=K_0,$ where $K$ and $K_0$ are
the operators appearing in (5.4) and (5.8), respectively.
Thus, for any $\xi>0,$ with the help of
(5.9) and (5.12), we construct $q_2(x)$ explicitly on $[0,b]$ via
$$q_2(x)=\displaystyle\frac{1}{b^2}\,e^{-\xi(n+1)x}\,
\displaystyle\sum_{m\in {\bold Z}} e^{2\pi imx/b}\,(im\pi-2b\xi)\,A_3(i\xi+m\pi/2b).
\tag 6.4$$
In particular, if $q_2(x)$ is a constant on the
interval $[0,b],$ say $q_2(x)=\gamma,$ from (6.2) we get
$$A_3(k)=-\displaystyle\frac{\gamma}{8k^2}\left(1-e^{-4ikb}\right),
\quad
B_3(k)=\displaystyle\frac{\gamma}{4k^2}\left(1-e^{-2ikb}\right).\tag 6.5$$
Conversely, if $A_3(k)$ and $B_3(k)$ are as in (6.5), then
with the help of
$$\int_{-\infty}^\infty \displaystyle\frac{e^{ika}}{k}\,dk=i\pi\,\text{sgn}(a),$$
where $\text{sgn}(a)$ denotes
the signature function, we recover $q_2(x)=\gamma$ via (6.3).
Alternatively, $q_2(x)$ can be recovered by using (6.4).
In another particular case, in which $q_2(x)=e^{\alpha x}$ on
the interval $[0,b],$ where $\alpha$ is a constant, from (6.2) we get
$$A_3(k)=\displaystyle\frac{e^{(\alpha-4ik)b}-1}{2ik(\alpha-4ik)},
\quad
B_3(k)=-\displaystyle\frac{e^{(\alpha-2ik)b}-1}{2ik(\alpha-2ik)}.\tag 6.6$$
Conversely, if $A_3(k)$ and $B_3(k)$ are as in (6.6), then
we recover $q_2(x)=e^{\alpha x}$ via (6.3) with the help of
a contour integration. Alternatively, we
can recover $q_2(x)$ by using (6.4).
\vskip 10 pt
\noindent {\bf Acknowledgment.} The research leading to this
article was supported in part by the National Science Foundation
under grant DMS-0204437 and the Department of Energy under grant
DE-FG02-01ER45951.
\vskip 10 pt
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