#  IBVP for the Maxwell-Bloch equations with an arbitrary inhomogeneous
broadening and periodic boundary function

Maria Filipkovska
Friedrich-Alexander-Universität Erlangen-Nürnberg, Cauerstrasse 11, 91058
Erlangen, Germany
B. Verkin Institute for Low Temperature Physics and Engineering
of the National Academy of Sciences of Ukraine, 47 Nauky Ave., 61103 Kharkiv,
Ukraine
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

The initial-boundary value problem (IBVP) for the Maxwell-Bloch equations with
an arbitrary inhomogeneous broadening and periodic boundary condition is
studied. This IBVP describes the propagation of an electromagnetic wave
generated by periodic pumping in a resonant medium with distributed two-level
atoms. We extended the inverse scattering transform method in the form of the
matrix Riemann-Hilbert problem for solving the considered IBVP. Using the
system of Ablowitz-Kaup-Newell-Segur equations equivalent to the system of the
Maxwell-Bloch (MB) equations, we construct the associated matrix Riemann-
Hilbert (RH) problem. Theorems on the existence, uniqueness and smoothness
properties of a solution of the constructed RH problem are proved, and it is
shown that a solution of the considered IBVP is uniquely defined by the
solution of the associated RH problem. It is proved that the RH problem
provides the causality principle. The representation of a solution of the MB
equations in terms of a solution of the associated RH problem are given. The
significance of this method also lies in the fact that, having studied the
asymptotic behavior of the constructed RH problem and equivalent ones, we can
obtain formulas for the asymptotics of a solution of the corresponding IBVP
for the MB equations.

Key words: integrable nonlinear PDEs, Maxwell-Bloch equations, inverse
scattering transform, Riemann-Hilbert problem, singular integral equation,
inhomogeneous broadening, periodic boundary function

MSC2020: 35F31, 35Q15, 37K15, 34L25, 35Q60

##  1 Introduction

Consider the Maxwell-Bloch (MB) equations written in the form

$\mathcal{E}_{t}+\mathcal{E}_{x}=\Omega\int\limits_{-\infty}^{\infty}n(s)\rho(t,x,s)\textup{d}s,\quad\rho_{t}+2\mathrm{i}\lambda\rho=\mathcal{N}\mathcal{E},\quad{\mathcal{N}}_{t}=-\frac{1}{2}(\overline{\mathcal{E}}\rho+\mathcal{E}\overline{\rho}),$
(1.1)

where $x\in(0,L)$, $L\leq\infty$, $t\in{\mathbb{R}}_{+}=(0,+\infty)$,
$\lambda\in{\mathbb{R}}$, ${\mathcal{E}}={\mathcal{E}}(t,x)$ and
$\rho=\rho(t,x,\lambda)$ are complex valued functions,
${\mathcal{N}}={\mathcal{N}}(t,x,\lambda)$ is a real function, subindexes mean
partial derivatives in $t$ and $x$, and $\overline{\phantom{Z}}$ means a
complex conjugation. A Hermitian conjugation will be denoted by ∗. It follows
from the second and third equations of the system (1.1) that
$\frac{\partial}{\partial
t}({\mathcal{N}}^{2}(t,x,\lambda)+|\rho(t,x,\lambda)|^{2})=0$. This relation
gives the well-known normalization condition

${\mathcal{N}}^{2}(t,x,\lambda)+|\rho(t,x,\lambda)|^{2}\equiv 1.$ (1.2)

The system of Maxwell-Bloch (MB) equations is a system of integrable nonlinear
PDEs which can be solved using the inverse scattering transform (IST) method.
The MB equations were originally proposed by Lamb [15, 16]. Ablowitz, Kaup and
Newell [1] have studied the coherent pulse propagation through a resonant two
level optical medium and shown that the Maxwell-Bloch equations describing
this phenomenon can be solved by the inverse scattering method. Zakharov
introduced in the paper [21] the concepts of spontaneous and causal solutions
of the MB equations that initiates the systematic application of the IST
method to laser problems. This work was further evolved by Gabitov, Zakharov
and Mikhailov [9, 10, 11] which have used the Marchenko integral equations and
obtained a new version of IST method. They presented a description of general
solutions of the MB equations and gave their classification. Generally, there
are many pioneering works relating to the Maxwell-Bloch equations and only a
small part of them are cited here. In this paper, we extend the IST method in
the form of the matrix Riemann-Hilbert (RH) problem for solving the initial-
boundary value problem (IBVP) for the MB equations with periodic boundary
condition. In [7, 8] Fokas proposed the unified transform method for solving
IBVPs for linear and for integrable nonlinear PDEs, the main idea of which is
the simultaneous spectral analysis of the both Lax operators (whose
compatibility condition is provided by the satisfaction of corresponding
nonlinear PDEs) to construct a unifying transform for solving certain IBVPs.
We use the similar idea, namely, to construct the associated matrix RH
problem, we use compatible solutions of the systems of Ablowitz-Kaup-Newell-
Segur (AKNS) equations (in general, we consider the three systems), which are
equivalent to the MB equations in the sense that their compatibility
conditions are provided by the satisfaction of the corresponding MB equations.
Notice that the compatibility condition of the AKNS system is the equivalent
of the Lax representation, i.e., the compatibility condition of the Lax pair.
Then we obtain the representation of a solution of the MB equations in terms
of a solution of the associated RH problem and prove the theorems on the
existence, uniqueness and smoothness properties of a solution of the
constructed RH problem.

It is known that the compatibility condition of the AKNS equations is the
equivalent of the Lax representation for the considering integrable

The MB equations use to described different physical phenomenon including
self-induced transparency [1, 16], superfluorescence [9, 11, 11] and others
related to the problems on the propagation of an electromagnetic wave in a
medium with distributed two-level atoms [16, 2, 18]. Short reviews of the
physical meaning of the MB equations and the application of the inverse
scattering transform method can be found in [1, 2, 11, 13], and in [20, 12]
for the reduced MB equations. We consider the problem on the propagation of an
electromagnetic wave in a resonant medium with distributed two-level atoms. In
(1.1), ${\mathcal{E}}(t,x)$ is the complex envelope of an electric field,
${\mathcal{N}}={\mathcal{N}}(t,x,\lambda)$ and $\rho=\rho(t,x,\lambda)$ are
entries of the density matrix of a quantum two-level atom subsystem:
$\begin{pmatrix}{\mathcal{N}}(t,x,\lambda)&\rho(t,x,\lambda)\\\
\overline{\rho}(t,x,\lambda)&-{\mathcal{N}}(t,x,\lambda)\end{pmatrix}$, the
parameter $\lambda$ is the deviation of the transition frequency of the given
two-level atom from the mean frequency $\Omega$, and $L$ is the length of the
attenuator. The weight function $n(\lambda)$ ($n(\lambda)\geq 0$,
$\lambda\in{\mathbb{R}}$) normalized by the condition
$\int\limits_{-\infty}^{\infty}n(\lambda)\textup{d}\lambda=1$ describes the
inhomogeneous broadening of the medium. Consider general initial and boundary
conditions for the mixed problem for the MB equations:

$\displaystyle{\mathcal{E}}(0,x)={\mathcal{E}}_{0}(x),\quad\rho(0,x,\lambda)=\rho_{0}(x,\lambda),\quad{\mathcal{N}}(0,x,\lambda)={\mathcal{N}}_{0}(x,\lambda),\quad
0<x<L\leq\infty,$ (1.3)
$\displaystyle{\mathcal{E}}(t,0)={\mathcal{E}}_{in}(t),\quad 0<t<\infty,$
(1.4)

where ${\mathcal{E}}_{0}(x)$, ${\mathcal{N}}_{0}(x,\lambda)$,
$\rho_{0}(x,\lambda)$ and ${\mathcal{E}}_{in}(t)$ are infinitely
differentiable for $x\in(0,L)$, $\lambda\in{\mathbb{R}}$ and
$t\in{\mathbb{R}}_{+}$. Taking into account (1.2), a function
${\mathcal{N}}_{0}(x,\lambda)$ is defined by $\rho_{0}(x,\lambda)$:

${\mathcal{N}}_{0}(x,\lambda)=-\sqrt{1-|\rho_{0}(x,\lambda)|^{2}},\quad
0<x<L\leq\infty.$

Here we choose the positive branch of the square root. This means that we
consider the problem for an attenuator (self-induced transparency).

In what follows, we put $\Omega=1$ in (1.1) and consider the IBVP for the
Maxwell-Bloch equations

${\mathcal{E}}_{t}+{\mathcal{E}}_{x}=\int\limits_{-\infty}^{\infty}n(s)\rho(t,x,s)\textup{d}s,\quad\rho_{t}+2\mathrm{i}\lambda\rho={\mathcal{N}}{\mathcal{E}},\quad{\mathcal{N}}_{t}=-\dfrac{1}{2}(\overline{{\mathcal{E}}}\rho+{\mathcal{E}}\overline{\rho}),$
(1.5)

where $x\in(0,L)$, $L\leq\infty$, $t\in{\mathbb{R}}_{+}=(0,\infty)$ and
$\lambda\in{\mathbb{R}}$ is a parameter, with the trivial initial conditions

${\mathcal{E}}(0,x)={\mathcal{E}}_{0}(x)=0,\;\rho(0,x,\lambda)=\rho_{0}(x,\lambda)=0,\;{\mathcal{N}}(0,x,\lambda)={\mathcal{N}}_{0}(x,\lambda)=-1,\;x\in(0,L),$
(1.6)

and the periodic boundary function (input signal)

$\mathcal{E}(t,0)=\mathcal{E}_{in}(t)=A_{0}e^{\mathrm{i}\omega_{0}t},\quad
t\in{\mathbb{R}}_{+},$ (1.7)

where $A_{0},\,\omega_{0}>0$. Given the form of boundary and initial data, it
is natural to assume that the function $\mathcal{E}(t,x)$ is bounded in the
vicinity of $(0,0)$ (as well as its partial derivatives with respect to $t$
and $x$).

It is known that the MB equations (1.5) is equivalent to the overdetermined
system of two linear differential equations (see, e.g., [1, 11, 2])

$\displaystyle w_{t}+\mathrm{i}\lambda\sigma_{3}w=-H(t,x)w,$ (1.8)
$\displaystyle
w_{x}-\mathrm{i}\lambda\sigma_{3}w+\mathrm{i}G(t,x,\lambda)w=H(t,x)w,$ (1.9)

where

$\sigma_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},\qquad
H(t,x)=\frac{1}{2}\begin{pmatrix}0&\mathcal{E}(t,x)\\\
-\overline{\mathcal{E}(t,x)}&0\end{pmatrix},$
$G(t,x,\lambda)=\dfrac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)}{s-\lambda}\textup{d}s,\qquad
F(t,x,s)=\begin{pmatrix}{\mathcal{N}}(t,x,s)&\rho(t,x,s)\\\
\overline{\rho(t,x,s)}&-{\mathcal{N}}(t,x,s)\end{pmatrix},$

and the symbol “$\rm{p.v.}$” denotes the principal value of the Cauchy
integral. The relationship of $\rho$ and $\mathcal{N}$ with the fundamental
matrix of the equation (1.8) was indicated in [17].

The following notation will be used in the paper: $[A,B]$ is the commutator of
the matrices $A$ and $B$, i.e., $[A,B]=AB-BA$; $I$ denotes the identity
operator (matrix); $A[i]$ denotes the $i$-th column of a matrix $A$.

The $t$-equation (1.8) and the $x$-equation (1.9) are compatible if and only
if ${\mathcal{E}}(t,x)$, $\rho(t,x,\lambda)$ and ${\mathcal{N}}(t,x,\lambda)$
satisfy the MB equations (1.5) (cf. [2]) which can be written in the matrix
form (see, e.g., [14]):

$\displaystyle\frac{\partial H(t,x)}{\partial t}+\frac{\partial
H(t,x)}{\partial
x}-\frac{1}{4}\int\limits_{-\infty}^{\infty}[\sigma_{3},F(t,x,s)]n(s)\textup{d}s=0,$
(1.10) $\displaystyle\frac{\partial F(t,x,\lambda)}{\partial
t}+[\mathrm{i}\lambda\sigma_{3}+H(t,x),F(t,x,\lambda)]=0.$ (1.11)

In particular, the equation (1.11) is a matrix form of the second and third
equations from (1.5). It is easy to prove that for any fixed $t$, $x$ the
matrix function $F(t,x,\lambda)=\pm
w(t,x,\lambda)\sigma_{3}w^{-1}(t,x,\lambda)$ (or $F(t,x,\lambda)=\pm
w(t,x,\lambda)\sigma_{3}w^{*}(t,x,\lambda)$, where ∗ denotes a Hermitian
conjugation), where $w(t,x,\lambda)$ is a solution of the $t$-equation (1.8)
satisfying the initial condition $w(0,x,\lambda)\equiv I$, is a unique
solution of the matrix linear DE (1.11) satisfying the initial condition
$F(0,x,\lambda)\equiv\pm\sigma_{3}$. The sign plus corresponds to a model of
amplifier, and the sign minus to a model of attenuator which is the subject of
this paper. In the absence of the electric field (${\mathcal{E}}(t,x)\equiv
0$) one has $F(t,x,\lambda)\equiv\pm\sigma_{3}$ that means an unstable/stable
medium (amplifier/attenuator) under consideration.

Since the equation (1.9) cannot be used to construct an appropriate associated
RH problem, we introduce the $x^{+}$-equation and $x^{-}$-equation (the upper
and lower bank equations) in addition to $x$-equation (1.9):

$w_{x}-\mathrm{i}\lambda\sigma_{3}w+\mathrm{i}G_{\pm}(t,x,\lambda)w=H(t,x)w,$
(1.12)

where

$G_{\pm}(t,x,\lambda)=\frac{1}{4}\,\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s=\frac{1}{4}{\rm
p.v.}\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)}{s-\lambda}\textup{d}s\pm\frac{\pi\mathrm{i}}{4}F(t,x,\lambda)n(\lambda).$
(1.13)

The $t$-equation (1.8) and the $x^{\pm}$-equation (1.12) are compatible iff
${\mathcal{E}}(t,x)$, $\rho(t,x,\lambda)$ and ${\mathcal{N}}(t,x,\lambda)$
satisfy the MB equations (1.5) (or the MB equations in the matrix form (1.10),
(1.11)).

In what follows, we suppose that $n(\pm\Lambda)=0$, where
$\Lambda\leq+\infty$, and $n(\lambda)=0$ on
${\mathbb{R}}\setminus[-\Lambda,\Lambda]$ if $\Lambda<\infty$ (i.e.,
$n(\lambda)$ has a compact support $[-\Lambda,\Lambda]$). Also, in the case if
$\Lambda=+\infty$, we denote $[-\Lambda,\Lambda]:={\mathbb{R}}$.

In [5] the IBVP (1.5)–(1.7) have been studied in the case when $n(\lambda)$ is
the Dirac delta-function, i.e., without inhomogeneous broadening (an
unbroadened medium). An application of the IST method to the IBVP (1.1),
(1.3)-(1.4) with a smooth and fast decreasing input signal
${\mathcal{E}}_{in}(t)$ was given in [14], using the simultaneous spectral
analysis of equations (1.8), (1.12) and the RH problem. The present paper
shows that the IST method in the form of matrix Riemann-Hilbert problem can be
used for the case of periodic boundary condition and give an integral
representation for ${\mathcal{E}}(t,x)$, ${\mathcal{N}}(t,x,\lambda)$ and
$\rho(t,x,\lambda)$ through the solution of a singular integral equation.
Also, it shows that the RH problem allow to study long time asymptotic
behavior of the solution of the mixed problem for MB equations, choosing as an
example the problem when ${\mathcal{E}}(t,0)=A_{0}e^{\mathrm{i}\omega_{0}t}$,
which is interesting itself. Later, having studied the asymptotic behavior of
the constructed RH problem and equivalent ones, we will be able to obtain
formulas for the asymptotics of a solution of the considered IBVP.

The paper has the following structure. In Section 1, the problem statement and
its physical interpretation are given. In Section 2, the basic (compatible)
solutions of the systems of the AKNS equations equivalent to the system of the
MB equations are constructed and their properties are stated. These solutions
will be used to construct the matrix Riemann-Hilbert problem associated with
the IBVP (1.5)–(1.7) which is given in Section 3. Then, in Section 4, we
obtain the theorems on the existence, uniqueness and smoothness of a solution
of the associated RH problem and prove that this RH problem generates a
solution of the IBVP (1.5)–(1.7). We give the representation of the solution
of the considered IBVP in terms of the solution of the associated RH problem,
and also the representation of the electric field envelope in terms of a
solution of an integral equation and an equivalent associated RH problem.
Also, we prove that the constructed RH problem provides the causality
principle. Thus, the problem on the propagation of an electromagnetic wave
generated by periodic input signal in a stable medium with distributed two-
level atoms (attenuator) is solved. The presented results will be used later
to obtain the asymptotics of a solution of the considered IBVP. In Appendix,
the analysis of the phase function is carried out, the results of which are
used in proving the theorems.

##  2 Basic solutions of the systems of the AKNS equations

Let

$\displaystyle W_{t}$ $\displaystyle=U(t,x,\lambda)W,$ (2.1) $\displaystyle
W_{x}$ $\displaystyle=V(t,x,\lambda)W$ (2.2)

be a system of matrix differential equations. The system of the equations
(2.1), (2.2) is _compatible_ if ${W_{tx}\equiv W_{xt}}$ for any solution of
these equations, i.e., the following condition holds:

$U_{x}-V_{t}+[U,V]=0,\qquad[U,V]=UV-VU.$ (2.3)

The $t$-equation (1.8) and the $x$-equation (1.9) can be rewritten in the form
of the $t$-equation (2.1) and the $x$-equation (2.2), respectively, where

$U(t,x,\lambda)=-\mathrm{i}\lambda\sigma_{3}-H(t,x),$ (2.4)
$V(t,x,\lambda)=\mathrm{i}\lambda\sigma_{3}+H(t,x)-\mathrm{i}G(t,x,\lambda).$
(2.5)

Accordingly, the $x^{\pm}$-equations (1.12) can be rewritten in the form

$W_{x}=V_{\pm}(t,x,\lambda)W,$ (2.6)

where

$V_{\pm}(t,x,\lambda)=V(t,x,\lambda)\pm\dfrac{\pi}{4}F(t,x,\lambda)n(\lambda)=\mathrm{i}\lambda\sigma_{3}+H(t,x)-\mathrm{i}G_{\pm}(t,x,\lambda),$
(2.7)

$G_{\pm}(t,x,\lambda)$ (1.13) are the limits of the function

$G(t,x,z)=\dfrac{1}{4}\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)\textup{d}s}{s-z}$
(2.8)

as $z\to\lambda\pm\mathrm{i}0$ ($z=\lambda+\mathrm{i}\nu$), i.e.,
$G_{\pm}(t,x,\lambda)=\lim\limits_{\begin{subarray}{c}\varepsilon\to 0\\\
\varepsilon>0\end{subarray}}G(t,x,\lambda\pm\mathrm{i}\varepsilon)$. The
system (2.1), (2.2) with $U$ and $V$ of the form (2.4), (2.5) is commonly
referred to as the _system of Ablowitz-Kaup-Newell-Segur (AKNS) equations_ for
the MB equations. Accordingly, the systems (1.8), (1.12) can be rewritten in
the form the AKNS systems (2.1), (2.6), where $U$, $V_{\pm}$ have the form
(2.4), (2.7).

In general, we have the AKNS system (1.8), (1.9) or (2.1), (2.2) with $U$, $V$
of the form (2.4), (2.5), and the two AKNS systems (1.8), (1.12) or (2.1),
(2.6) with $U$, $V_{\pm}$ of the form (2.4), (2.7).

The _compatibility condition of the AKNS system (the $t$\- and $x$-equations)_
(1.8), (1.9) has the form (2.3), and the _compatibility conditions of the AKNS
systems (the $t$\- and $x^{\pm}$-equations)_ (1.8), (1.12) take the form

$U_{x}(t,x,\lambda)-(V_{\pm})_{t}(t,x,\lambda)+[U(t,x,\lambda),V_{\pm}(t,x,\lambda)]=U_{x}(t,x,\lambda)-V_{t}(t,x,\lambda)+\\\
+[U(t,x,\lambda),V(t,x,\lambda)]\mp\frac{\pi
n(\lambda)}{4}\big{(}F_{t}(t,x,\lambda)-[U(t,x,\lambda),F(t,x,\lambda)]\big{)}=0,$
(2.9)

where $U$, $V$ and $V_{\pm}$ are defined by (2.4), (2.5) and (2.7), and
$(V_{\pm})_{t}:=\partial V_{\pm}/\partial t$. If
${F_{t}-[U,F]}=F_{t}+[\mathrm{i}\lambda\sigma_{3}+H,F]=0$ (that is, (1.11)
holds), then the compatibility conditions (2.9) coincides with (2.3).
Conversely, if (2.3) holds, then $F_{t}+[\mathrm{i}\lambda\sigma_{3}+H,F]=0$
as well as $U_{x}-(V_{\pm})_{t}+[U,V_{\pm}]=0$. Each of the compatibility
conditions gives the same MB equations. The upper and lower bank equations
(1.12) will allow us to obtain solutions of the AKNS systems that have
analytic continuation to the upper and lower complex half-planes
${\mathbb{C}}_{\pm}=\\{z\in{\mathbb{C}}\mid\;\pm\operatorname{Im}z>0\\}$
(except for the cuts). We will use them for a construction of the matrix RH
problem which generates a solution of the considered IBVP.

Let us suppose that a solution of the MB equations (1.5) exists, then the AKNS
systems (1.8), (1.12) (as well as the AKNS system (1.8), (1.9)) are compatible
and we can define their “basic” solutions with the properties enabling to
obtain the matrix RH problem associated with the IBVP (1.5)–(1.7).

First, find the “background” solution of the AKNS system (1.8), (1.9), which
will be used below. We seek the background solution in the form

$\Phi_{0}(t,x,\lambda)=e^{\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}M_{0}(\lambda)e^{-\mathrm{i}(\alpha(\lambda)t+\beta(\lambda)x)\sigma_{3}},\qquad\lambda\in{\mathbb{R}},$
(2.10)

where $\alpha_{0}$, $\beta_{0}$, $\alpha(\lambda)$, $\beta(\lambda)$ and
$M_{0}(\lambda)$ are to be determined. The logarithmic derivative of
$\Phi_{0}$ in $t$ takes the form

$(\Phi_{0})_{t}(t,x,\lambda)\Phi_{0}^{-1}(t,x,\lambda)=\mathrm{i}\alpha_{0}\sigma_{3}-\mathrm{i}\alpha(\lambda)e^{\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}M_{0}(\lambda)\sigma_{3}M_{0}^{-1}(\lambda)e^{-\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}.$

Let

$M_{0}(\lambda):=\begin{pmatrix}a(\lambda)&b(\lambda)\\\
b(\lambda)&a(\lambda)\end{pmatrix}$ (2.11)

be a matrix with the unit determinant: $a^{2}(\lambda)-b^{2}(\lambda)=1$. Then
$M_{0}(\lambda)\sigma_{3}M_{0}^{-1}(\lambda)=(a^{2}+b^{2})\sigma_{3}-2\mathrm{i}ab\sigma_{2}$,
where $\sigma_{2}=\begin{pmatrix}0&-\mathrm{i}\\\ \mathrm{i}&0\end{pmatrix}$
is a Pauli matrix. Hence

$(\Phi_{0})_{t}(t,x,\lambda)\Phi_{0}^{-1}(t,x,\lambda)=\mathrm{i}\left(\alpha_{0}-\alpha(\lambda)[a^{2}(\lambda)+b^{2}(\lambda)]\right)\sigma_{3}-\\\
-2\alpha(\lambda)a(\lambda)b(\lambda)e^{\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}\sigma_{2}e^{-\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}.$

On the other hand, we need to obtain an equation of the form

$(\Phi_{0})_{t}\Phi_{0}^{-1}=-\mathrm{i}\lambda\sigma_{3}-H_{0}(t,x),\qquad
H_{0}(t,x)=\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}_{bg}(t,x)\\\
-\overline{{\mathcal{E}}_{bg}(t,x)}\end{pmatrix}.$

Comparing expressions for the logarithmic derivatives, we find

$\alpha(\lambda)=\frac{\lambda+\alpha_{0}}{a^{2}(\lambda)+b^{2}(\lambda)},\qquad{\mathcal{E}}_{bg}(t,x)=-4\mathrm{i}\alpha(\lambda)a(\lambda)b(\lambda)e^{2\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)}.$

Since ${\mathcal{E}}_{bg}(t,x)$ is independent of $\lambda$, then
$\alpha(\lambda)a(\lambda)b(\lambda)$ have to be a constant. To achieve this,
we put

$a(\lambda)=\frac{\varkappa(\lambda)+\varkappa^{-1}(\lambda)}{2},\quad
b(\lambda)=\frac{\varkappa(\lambda)-\varkappa^{-1}(\lambda)}{2},\quad\varkappa(\lambda)=\left(\dfrac{\lambda-\overline{E}}{\lambda-E}\right)^{1/4},$
(2.12)

where the branch cut for the function
$\varkappa(z)=(z-\overline{E})^{1/4}(z-E)^{-1/4}$, analytic in
$z\in{\mathbb{C}}\setminus[E,\overline{E}]$, is taken along $[E,\overline{E}]$
and the branch is fixed by the asymptotics

$\varkappa(z)=1+(E-\overline{E})/(4z)+O(z^{-2}),\qquad z\to\infty.$

Then $(a(\lambda)+b(\lambda))^{2}=\varkappa^{2}(\lambda)$ and

$a^{2}(\lambda)+b^{2}(\lambda)=\frac{\lambda-\operatorname{Re}E}{w(\lambda)},\quad
a(\lambda)b(\lambda)=\frac{\mathrm{i}\operatorname{Im}E}{2w(\lambda)},\quad
w(\lambda)=\sqrt{(\lambda-E)(\lambda-\overline{E})},$ (2.13)

where we fix the branch of the root $w(z)=\sqrt{(z-E)(z-\overline{E})}$ by
taking the branch cut along $[E,\overline{E}]$ and by asymptotics

$w(z)=z-\operatorname{Re}E+O(z^{-1}),\qquad z\to\infty,$

and thus $w(z)$ is analytic for $z\in{\mathbb{C}}\setminus[E,\overline{E}]$.
Hence,
${\mathcal{E}}_{bg}(t,x)=2\operatorname{Im}E\dfrac{\lambda+\alpha_{0}}{\lambda-\operatorname{Re}E}e^{2\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)}$
and it is independent of $\lambda$ iff $\alpha_{0}=-\operatorname{Re}E$.
Therefore

$\alpha(\lambda)=w(\lambda),\quad{\mathcal{E}}_{bg}(t,x)=2\operatorname{Im}Ee^{2\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)},\quad\alpha_{0}=-\operatorname{Re}E.$

Comparing ${\mathcal{E}}_{bg}(t,0)$ with
${\mathcal{E}}_{in}(t)=A_{0}e^{\mathrm{i}\omega_{0}t}$, we obtain
$2\operatorname{Im}E=A_{0}$, $2\operatorname{Re}E=-2\alpha_{0}=-\omega_{0}$.
The last formulas mean that

$E=-\frac{\omega_{0}}{2}+\mathrm{i}\frac{A_{0}}{2}.$ (2.14)

Thus, the matrix function $\Phi_{0}(t,x,\lambda)$ satisfies the equation
$(\Phi_{0})_{t}=-(\mathrm{i}\lambda\sigma_{3}+H_{0}(t,x))\Phi_{0}$, and the
matrix $H_{0}(t):=H_{0}(t,0)$ is defined by the input signal
${\mathcal{E}}_{in}(t)$ (1.7):

$H_{0}(t):=H_{0}(t,0)=\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}_{bg}(t,0)\\\
-\overline{{\mathcal{E}}_{bg}(t,0)}\end{pmatrix}=\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}_{in}(t)\\\
-\overline{{\mathcal{E}}_{in}(t)}\end{pmatrix}.$ (2.15)

Further we will use the MB equations. Using the equation
$\rho_{bg\,t}+2\mathrm{i}\lambda\rho_{bg}={\mathcal{N}}_{0}{\mathcal{E}}_{bg}$,
we find that if $\rho_{bg}=\mathrm{i}c_{0}\mathcal{E}_{bg}$, then
$\mathcal{N}_{bg}=-2c_{0}(\lambda-\operatorname{Re}E)$. Taking into account
the normalization condition

$\mathcal{N}_{bg}^{2}+|\rho_{bg}|^{2}=4c_{0}^{2}[(\lambda-\operatorname{Re}E)^{2}+\operatorname{Im}^{2}E]=1,$

we obtain $c_{0}=\dfrac{1}{2w(\lambda)}$ and
$\rho_{bg}=\dfrac{\mathrm{i}\mathcal{E}_{bg}}{2w(\lambda)}$. The first MB
equation gives

$\int\limits_{-\infty}^{\infty}n(s)\rho_{bg}(t,x,s)\textup{d}s={\mathcal{E}}_{bg\,t}+{\mathcal{E}}_{bg\,x}=2\mathrm{i}(\alpha_{0}+\beta_{0})\mathcal{E}_{bg}=2\mathrm{i}(\beta_{0}-\operatorname{Re}E)\mathcal{E}_{bg},$

and hence

$\beta_{0}=\operatorname{Re}E+\dfrac{1}{4}\int\limits_{-\infty}^{\infty}\dfrac{n(s)}{w(s)}\textup{d}s.$
(2.16)

Thus, we obtain the periodic solution in the form of a plane wave for the MB
equations (1.5):

$\displaystyle{\mathcal{E}}_{bg}(t,x)=2\operatorname{Im}E\,e^{2\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)},\quad\rho_{bg}(t,x,\lambda)=\dfrac{\mathrm{i}\operatorname{Im}E}{w(\lambda)}\,e^{2\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)},\quad\mathcal{N}_{bg}(t,x,\lambda)=-\frac{\lambda-\operatorname{Re}E}{w(\lambda)},$

where $\operatorname{Im}E=A_{0}/2$,
$\alpha_{0}=-\operatorname{Re}E=\omega_{0}/2$, $\beta_{0}$ has the form
(2.16), and
${\mathcal{E}}_{bg}(t,0)=A_{0}e^{\mathrm{i}\omega_{0}t}={\mathcal{E}}_{in}(t)$
is the input signal (1.7). The third MB equation is evidently fulfilled. The
functions $\rho_{bg}(t,x,z)$ and $\mathcal{N}_{bg}(t,x,z)$ are analytic for
$z\in{\mathbb{C}}\setminus[E,\overline{E}]$, i.e., they are analytic in the
same domain as $w(z)$. Note that $\rho_{bg}(t,x,\lambda)$ and the partial
derivative $(\mathcal{N}_{bg})_{\lambda}(t,x,\lambda)$ are discontinuous in
$\lambda$ at the point $\operatorname{Re}E$. To obtain an infinitely
differentiable solution on the whole line ${\mathbb{R}}$, we have to redefine
the function $w(z)$: let us draw a branch cut connecting the points $E$ and
$\overline{E}$ via infinity without intersection of the real line and fix the
branch of $w(z)$ by the condition $w(0)=|E|$. In this case we obtain the
infinitely differentiable solution of the MB equations with respect to all
$\lambda\in{\mathbb{R}}$.

Now let us consider the logarithmic derivative of $\Phi_{0}(t,x,\lambda)$ with
respect to $x$:

$\displaystyle(\Phi_{0})_{x}\Phi_{0}^{-1}$
$\displaystyle=\mathrm{i}\beta_{0}\sigma_{3}-\mathrm{i}\beta(\lambda)e^{\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}M_{0}(\lambda)\sigma_{3}M_{0}^{-1}(\lambda)e^{-\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}$
$\displaystyle=\mathrm{i}\left(\beta_{0}-\beta(\lambda)[a^{2}(\lambda)+b^{2}(\lambda)]\right)\sigma_{3}-2\beta(\lambda)a(\lambda)b(\lambda)e^{\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}\sigma_{2}e^{-\mathrm{i}(\alpha_{0}\,t+\beta_{0}\,x)\sigma_{3}}.$

Denote $G_{0}(t,x,\lambda)=\dfrac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\dfrac{F_{0}(t,x,s)n(s)}{s-\lambda}\textup{d}s$,
$F_{0}(t,x,s)\\!=\\!\begin{pmatrix}{\mathcal{N}}_{bg}(t,x,s)&\rho_{bg}(t,x,s)\\\
\overline{\rho_{bg}(t,x,s)}&-{\mathcal{N}}_{bg}(t,x,s)\end{pmatrix}$. Since we
need to obtain the $x$-equation of the AKNS system, i.e., the equation (1.9),
then

$(\Phi_{0})_{x}\Phi_{0}^{-1}=\mathrm{i}\lambda\sigma_{3}+H_{0}-\mathrm{i}G_{0}=\mathrm{i}\left(\lambda-\frac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\frac{\mathcal{N}_{bg}(t,x,s)n(s)}{s-\lambda}\textup{d}s\right)\sigma_{3}+{}\\\
+\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}_{bg}(t,x)\\\
-\overline{{\mathcal{E}}_{bg}(t,x)}&0\end{pmatrix}-\frac{\mathrm{i}}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\begin{pmatrix}0&\rho_{bg}(t,x,s)\\\
\overline{\rho_{bg}(t,x,s)}&0\end{pmatrix}\frac{n(s)}{s-\lambda}\textup{d}s=\\\
=\mathrm{i}\left(\lambda+\frac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\frac{(s-\operatorname{Re}E)n(s)}{w(s)(s-\lambda)}\textup{d}s\right)\sigma_{3}+{}\\\
+\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}_{bg}(t,x)\\\
-\overline{{\mathcal{E}}_{bg}(t,x)}&0\end{pmatrix}\left[1+\frac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\dfrac{n(s)}{w(s)(s-\lambda)}\textup{d}s\right].$

Using the relations
$s-\operatorname{Re}E=\lambda-\operatorname{Re}E+s-\lambda$, (2.13), and
(2.16) and comparing the expressions for the logarithmic derivative, we obtain

$\beta(\lambda)=-w(\lambda)\left(1+\frac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\dfrac{n(s)}{w(s)(s-\lambda)}\textup{d}s\right),\qquad\lambda\in{\mathbb{R}}.$

Next, we introduce some preliminary information.

We supposed that $n(\pm\Lambda)=0$, where $\Lambda\leq+\infty$, and if
$\Lambda<\infty$, then $n(\lambda)=0$ on
${\mathbb{R}}\setminus[-\Lambda,\Lambda]$. Recall that in the case if
$\Lambda=+\infty$, we set $[-\Lambda,\Lambda]={\mathbb{R}}$ for convenience.
Thus, we consider the two cases: $\Lambda=+\infty$ and $0<\Lambda<+\infty$.

Below, we use the ordinary Hölder condition and the Hölder condition in a
neighborhood of infinity [19]. As usual, a scalar function $f(\zeta)$
satisfies the Hölder condition on a set $Z\subset{\mathbb{C}}$ or
$\subset{\mathbb{R}}$ if there exist positive constants $C$, $\mu$ such that
$|f(\zeta_{1})-f(\zeta_{2})|\leq C|\zeta_{1}-\zeta_{2}|^{\mu}$ for any
$\zeta_{1},\,\zeta_{2}\in Z$. The scalar function $f(\zeta)$
($\zeta\in{\mathbb{C}}$ or $\in{\mathbb{R}}$) satisfies the _Hölder condition
in a neighborhood of infinity_ if there exist positive constants $C$, $\mu$
such that $|f(\zeta_{1})-f(\zeta_{2})|\leq
C\big{|}\zeta_{1}^{-1}-\zeta_{2}^{-1}\big{|}^{\mu}$ for sufficiently large
$|\zeta_{1}|,\,|\zeta_{2}|$ (for any $|\zeta_{1}|,\,|\zeta_{2}|$ from the
neighborhood of infinity). Notice that if
$\lim\limits_{|\zeta|\to\infty}f(\zeta)=c<\infty$ and
$f(\zeta)=c+O(|\zeta|^{-\mu})$, where $\mu=const>0$, for sufficiently large
$|\zeta|$ ($|\zeta|\to\infty$), then $f(\zeta)$ is said to satisfy the Hölder
condition at the point $\zeta=\infty$ [19]. It is known that if the function
$f(\zeta)$ is define on an arc $L$ and satisfies the Hölder with the index
$\mu>1$ on it, then $f(\zeta)=const$ on $L$ (see, e.g., [19]). Thus, in what
follows, we assume that $0<\mu\leq 1$ in the Hölder conditions.

In what follows, we assume that the function $n(\lambda)$ (recall that,
according to the problem statement, $n(\lambda)\geq 0$ and
$\int\limits_{-\infty}^{\infty}n(\lambda)\textup{d}\lambda=1$, and thus
$n(\lambda)$ is absolutely integrable on any interval in ${\mathbb{R}}$)
satisfies the following conditions:

  1. 1)

$n(\lambda)$ satisfies the Hölder condition on ${\mathbb{R}}$ and in a
neighborhood of infinity, and ${n(\pm\Lambda)=0}$ where $\Lambda=+\infty$
($n(\lambda)=O(|\lambda|^{-\mu})$, $|\lambda|\to+\infty$, $0<\mu\leq 1$);

  2. 2)

$n(\lambda)$ has a compact support $[-\Lambda,\Lambda]$, $\Lambda<\infty$ and
one of the following conditions holds:

    1. 2.1)

$n(\lambda)$ satisfies the Hölder condition on $[-\Lambda,\Lambda]$ and
$n(\pm\Lambda)=0$;

    2. 2.2)

$n(\lambda)=const$ on $(-\Lambda,\Lambda)$ and $n(\pm\Lambda)=0$; in this case

$n(\lambda)=\begin{cases}\dfrac{1}{2\Lambda},&|\lambda|<\Lambda,\\\
0,&|\lambda|\geq\Lambda.\end{cases}$ (2.17)

In general, $n(\lambda)$ can be a singular generalized function, for example,
the delta function. The case when $n(\lambda)=\delta(\lambda)$, where
$\delta(\lambda)$ is the Dirac delta function, was considered in [5].

Introduce the function

$\eta(z)=z+\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-z}\textup{d}s,\qquad\Lambda\leq\infty,$
(2.18)

where $\Lambda<\infty$ if it is supposed that $n(\lambda)$ has the compact
support $[-\Lambda,\Lambda]$, and $\Lambda=+\infty$ otherwise. The function
$\eta(z)$ is analytic for $z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$, and,
hence, the function $\operatorname{Im}\eta(z)$ is harmonic in
${\mathbb{C}}\setminus[-\Lambda,\Lambda]$. Since

$\operatorname{Im}\eta(z)=\operatorname{Im}z\left(1+\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{(s-\operatorname{Re}z)^{2}+\operatorname{Im}^{2}z}\textup{d}s\right),$
(2.19)

then

$\mathop{\mathrm{sgn}}(\operatorname{Im}\eta(z))=\mathop{\mathrm{sgn}}(\operatorname{Im}z).$

The function $\eta(z)$ has the boundary (limiting) values
$\eta_{\pm}(\lambda)=\eta(\lambda\pm\mathrm{i}0)$ ($z=\lambda+\mathrm{i}\nu$),
continuous on $[-\Lambda,\Lambda]$:

$\eta_{\pm}(\lambda)=\lambda+\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s=\lambda+\dfrac{1}{4}{\rm
p.v.}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-\lambda}\textup{d}s\pm\frac{\pi\mathrm{i}}{4}n(\lambda),\quad\lambda\in[-\Lambda,\Lambda],\;\Lambda\leq\infty,$
(2.20)

and the jump
$\eta(\lambda+\mathrm{i}0)-\eta(\lambda-\mathrm{i}0)=\dfrac{\pi\mathrm{i}}{2}n(\lambda)$
across $[-\Lambda,\Lambda]$.

In the formula (2.20), for case 1) and for case 2) with condition 2)2.1) (see
above) we have:

${\rm
p.v.}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-\lambda}\textup{d}s=\begin{cases}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)-n(\lambda)}{s-\lambda}\textup{d}s+n(\lambda)\ln\dfrac{\Lambda-\lambda}{\Lambda+\lambda},&\lambda\in(-\Lambda,\Lambda),\;\Lambda<\infty,\\\
\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s\mp\Lambda}\textup{d}s,&\lambda=\pm\Lambda,\;\Lambda<\infty,\\\
\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)-n(\lambda)}{s-\lambda}\textup{d}s,&\lambda\in{\mathbb{R}},\;\Lambda=+\infty;\end{cases}$

for case 2)2.2), i.e., $n(\lambda)$ of the form (2.17), we have:

${\rm
p.v.}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-\lambda}\textup{d}s=\begin{cases}\dfrac{1}{2\Lambda}\ln\dfrac{\Lambda-\lambda}{\Lambda+\lambda},&\lambda\in(-\Lambda,\Lambda),\\\
0,&\lambda=\pm\Lambda.\end{cases}$

Notice that, generally, instead of condition 2)2.1) one can require that the
function $n(\lambda)$ _belong to the class $H^{*}$ on_ $[-\Lambda,\Lambda]$
[19], which means the following: $n(\lambda)$ satisfies the Hölder condition
on any closed interval in $(-\Lambda,\Lambda)$ and near the endpoints
$\mp\Lambda$ can be represented as
$n(\lambda)=\dfrac{n_{i}(\lambda)}{|\lambda\pm\Lambda|^{\alpha_{i}}}$,
$i=1,2$, respectively, where $0\leq\alpha_{i}<1$ and $n_{1}(\lambda)$,
$n_{2}(\lambda)$ are defined and satisfy the Hölder condition on sufficiently
small parts of $[-\Lambda,\Lambda]$ that are adjacent to the endpoints
$-\Lambda$, $\Lambda$, respectively (the contour $[-\Lambda,\Lambda]$ is
oriented from $-\Lambda$ to $\Lambda$). Then the function $\eta(z)$ (2.18) is
analytic on ${\mathbb{C}}\setminus[-\Lambda,\Lambda]$, continuous up to
$(-\Lambda,\Lambda)$ and has weak singularities at the points $\pm\Lambda$
(the singularities of the type $O\big{(}(z\mp\Lambda)^{-\nu}\big{)}$,
$0<\nu<1$). Near the endpoints $\pm\Lambda$, the Cauchy integral in (2.18)
admits the estimate
$\left|\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)ds}{s-z}\right|<\dfrac{const}{|z\mp\Lambda|^{\nu_{i}}}$,
$\alpha_{i}<\nu_{i}<1$, $i=1,2$, where $\alpha_{1}$, $\alpha_{2}$ were defined
above (cf. [19]). A function $\eta(z)$ will be used when constructing the RH
problem as well as the basic solutions of the AKNS systems (see below). In
order for the RH problem to have no singularities, including weak
singularities, at the points $\pm\Lambda$, we will not use this requirement
and keep condition 2)2.1) for $n(\lambda)$, although in the case of a more
general RH problem it can be used.

Since we consider the problem on the propagation of an electromagnetic wave in
the stable medium (i.e., the problem for an attenuator), we can assume that
${\mathcal{N}}(t,x,\lambda)+1\to 0$ fast enough as $x\to\infty$ so that
$[{\mathcal{N}}(t,x,\lambda)+1]\in L^{1}({\mathbb{R}}_{+})$ with respect to
$x$ for all $t\in{\mathbb{R}}_{+}$, $\lambda\in{\mathbb{R}}$ (or, given the
relationship (1.2) between $\rho(t,x,\lambda)$ and
${\mathcal{N}}(t,x,\lambda)$, we can assume that $\rho(t,x,\lambda)\to 0$ fast
enough as $x\to\infty$ for all $t\in{\mathbb{R}}_{+}$,
$\lambda\in{\mathbb{R}}$).

The next known lemma will be used below.

######  Lemma 1 ([3, 14]).

Let the system of the equations (2.1), (2.2) be compatible (the condition
(2.3) holds) for all $t,x,\lambda\in{\mathbb{R}}$. Assume that
$W(t,x,\lambda)$ is a matrix satisfying the $t$-equation (2.1) for all $x$,
and $W(t_{0},x,\lambda)$ satisfies the $x$-equation (2.2) for some $t=t_{0}$,
$t_{0}\leq\infty$. Then $W(t,x,\lambda)$ satisfies the $x$-equation for all
$t$. The same result is true, if one replaces $t$ by $x$ and vice versa.

We will seek the two pairs $Y_{+}$, $Z_{+}$ and $Y_{-}$, $Z_{-}$ of “basic”
solutions of the AKNS systems (1.8), (1.12) (i.e., the systems (2.1), (2.6)
with $U$, $V_{\pm}$ of the form (2.4), (2.7)).

The first “basic” solutions of the AKNS systems (1.8), (1.12) we seek in the
form:

$Y_{\pm}(t,x,\lambda)=W_{\pm}(t,x,\lambda)\Phi(t,\lambda),\qquad\lambda\in{\mathbb{R}},$
(2.21)

where $W_{\pm}(t,x,\lambda)$ satisfy the $x^{\pm}$-equations (1.12)
(respectively) for each $t\in{\mathbb{R}}_{+}$ and $W_{\pm}(t,0,\lambda)=I$,
and $\Phi(t,\lambda)$ satisfies the $t$-equation (1.8) for $x=0$ with the
matrix $H(t,0)=H_{0}(t)$ (2.15) defined by the input signal (1.7), i.e.,

$\Phi(t,\lambda)=\Phi_{0}(t,0,\lambda)=e^{-\mathrm{i}\operatorname{Re}E\,t\sigma_{3}}M_{0}(\lambda)e^{-\mathrm{i}w(\lambda)\,t\sigma_{3}},$
(2.22)

where $M_{0}(\lambda)$, $w(\lambda)$ are defined by (2.11), (2.12), and
(2.13), and $E=-\omega_{0}/2+\mathrm{i}\,A_{0}/2$ is determined by the
boundary data (1.7). The second “basic” solutions we seek in the form

$Z_{\pm}(t,x,\lambda)=\Psi(t,x,\lambda)w_{\pm}(x,\lambda),\qquad\lambda\in{\mathbb{R}},$
(2.23)

where $\Psi(t,x,\lambda)$ satisfies the $t$-equation (1.8) for each
$x\in(0,L)$, $L\leq\infty$, and $\Psi(0,x,\lambda)=I$, and
$w_{\pm}(x,\lambda)$ satisfy the $x^{\pm}$-equations (1.12) (respectively) for
$t=0$ and the initial conditions
$w_{\pm}(L,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)L\sigma_{3}}=I$,
$L\leq\infty$ (if $L=\infty$, then
$w_{\pm}(L,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)L\sigma_{3}}:=\lim\limits_{x\to\infty}w_{\pm}(x,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}$).

Under the conditions imposed on the function $n(\lambda)$ above,
$Y_{+}(t,x,\lambda)=Y_{-}(t,x,\lambda)$ and
$Z_{+}(t,x,\lambda)=Z_{-}(t,x,\lambda)$ (similarly for $W_{\pm}(t,x,\lambda)$
and $w_{\pm}(x,\lambda)$) for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$.

In the theorem below, for convenience, we assume that $L=\infty$, i.e.,
$x\in{\mathbb{R}}_{+}$. For the case when $x\in(0,L)$, $L<\infty$, the theorem
remains valid and the proof is similar. Recall that $A[i]$ denotes the $i$-th
column of a matrix $A$.

######  Theorem 1.

Let the functions ${\mathcal{E}}(t,x)$, $\rho(t,x,\lambda)$,
$\mathcal{N}(t,x,\lambda)$ be a solution of the IBVP (1.5)–(1.7) and have the
following properties: ${\mathcal{E}}(t,x)\in
C^{1}({\mathbb{R}}_{+}\times{\mathbb{R}}_{+})$, the partial derivatives of
$\rho(t,x,\lambda)$ and ${\mathcal{N}}(t,x,\lambda)$ with respect to $t$ and
$\lambda$ are continuous for
$(t,x,\lambda)\in{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}$,
and ${\mathcal{N}}(t,x,\lambda)+1\to 0$ fast enough as $x\to\infty$ so that
$[{\mathcal{N}}(t,x,\lambda)+1]\in L^{1}({\mathbb{R}}_{+})$ with respect to
$x$ for all $t\in{\mathbb{R}}_{+}$, $\lambda\in{\mathbb{R}}$. Suppose
$n(\lambda)$ satisfies conditions 1), 2) mentioned above.

Then there exist the solutions $Y_{\pm}(t,x,\lambda)$ of the form (2.21) and
$Z_{\pm}(t,x,\lambda)$ of the form (2.23), where
$Y_{+}(t,x,\lambda)=Y_{-}(t,x,\lambda)$ and
$Z_{+}(t,x,\lambda)=Z_{-}(t,x,\lambda)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$
($t,\,x\in[0,\infty)$), for the systems of the $t$\- and $x^{\pm}$-equations
(1.8), (1.12), which have the following properties:

  * •

$\det Y_{\pm}(t,x,\lambda)\equiv 1$ and $\det Z_{\pm}(t,x,\lambda)\equiv 1$
($\lambda\in{\mathbb{R}}$);

  * •

$Y_{\pm}(t,x,\lambda)$ and $Z_{\pm}(t,x,\lambda)$ are continuously
differentiable in $t$ and $x$;

  * •

the functions $Y_{+}(t,x,\lambda)$ and $Y_{-}(t,x,\lambda)$ have the analytic
continuations $Y^{+}(t,x,z)$ and $Y^{-}(t,x,z)$ for
$z\in{\mathbb{C}}_{+}^{cl}\setminus([-\Lambda,\Lambda]\cup[E,\operatorname{Re}E])$
and
$z\in{\mathbb{C}}_{-}^{cl}\setminus([-\Lambda,\Lambda]\cup[\operatorname{Re}E,\overline{E}])$
respectively, where
${\mathbb{C}}_{\pm}^{cl}={\mathbb{C}}_{\pm}\cup{\mathbb{R}}$,
$\Lambda\leq\infty$, ($Y^{+}(t,x,z)=Y^{-}(t,x,z)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$), which
are continuous up to
$([-\Lambda,\Lambda]\setminus\\{\operatorname{Re}E\\})\cup[E,\operatorname{Re}E)$
and
$([-\Lambda,\Lambda]\setminus\\{\operatorname{Re}E\\})\cup(\operatorname{Re}E,\overline{E}]$
and have the singularities of the type $(z-E)^{-1/4}$ and
$(z-\overline{E})^{-1/4}$ at the points $E$ and $\overline{E}$ respectively,
are bounded in the neighborhood of the point $\operatorname{Re}E$, and also,
for fixed $t$ and $x$,

$\displaystyle Y^{-}[1](t,x,z)e^{\mathrm{i}(zt-\eta(z)x)}=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\quad z\to\infty,\;z\in{\mathbb{C}}_{-}^{cl};$ (2.24)
$\displaystyle Y^{+}[2](t,x,z)e^{-\mathrm{i}(zt-\eta(z)x)}=\begin{pmatrix}0\\\
1\end{pmatrix}+O(z^{-1}),\quad z\to\infty,\;z\in{\mathbb{C}}_{+}^{cl}.$ (2.25)

  * •

the functions $Z_{+}(t,x,\lambda)$ and $Z_{-}(t,x,\lambda)$ have the analytic
continuations $Z^{+}(t,x,z)$ and $Z^{-}(t,x,z)$ for
$z\in{\mathbb{C}}_{+}^{cl}\setminus[-\Lambda,\Lambda]$ and
$z\in{\mathbb{C}}_{-}^{cl}\setminus[-\Lambda,\Lambda]$ respectively, where
$\Lambda\leq\infty$, ($Z^{+}(t,x,z)=Z^{-}(t,x,z)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$) which
are continuous up to $[-\Lambda,\Lambda]$, and

$\displaystyle Z^{+}[1](t,x,z)e^{\mathrm{i}(zt-\eta(z)x)}=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\quad z\to\infty,\;z\in{\mathbb{C}}_{+}^{cl};$ (2.26)
$\displaystyle Z^{-}[2](t,x,z)e^{-\mathrm{i}(zt-\eta(z)x)}=\begin{pmatrix}0\\\
1\end{pmatrix}+O(z^{-1}),\quad z\to\infty,\;z\in{\mathbb{C}}_{-}^{cl}.$ (2.27)

###### Proof.

Since ${\mathcal{E}}(t,x)$, $\rho(t,x,\lambda)$ and $\mathcal{N}(t,x,\lambda)$
satisfy the MB equations (1.5), then the AKNS systems (the $t$\- and
$x^{\pm}$-equations) (1.8), (1.12) are compatible, i.e., the compatibility
conditions (2.9) hold.

From Lemma 1 it follows that if the matrices $W_{\pm}(t,x,\lambda)$,
$\Phi(t,\lambda)$ and $\Psi(t,x,\lambda)$, $w_{\pm}(x,\lambda)$ from (2.21)
and (2.23) exist, then the matrix functions
$W_{\pm}(t,x,\lambda)\Phi(t,\lambda)$ (2.21) and
$\Psi(t,x,\lambda)w_{\pm}(x,\lambda)$ (2.23) are solutions of the compatible
AKNS systems (1.8), (1.12). Indeed, since $W_{\pm}(t,x,\lambda)$ are solutions
of the $x^{\pm}$-equations (1.12) (respectively) for all $t$, then the product
$W_{\pm}(t,x,\lambda)\Phi(t,\lambda)$ are also solutions of the
$x^{\pm}$-equations for all $t$. Further, since $\Phi(t,\lambda)$ is a
solution of the $t$-equation (1.8) for $x=0$, then
$W_{\pm}(t,0,\lambda)\Phi(t,\lambda)=\Phi(t,\lambda)$ are also solutions of
$t$-equation for $x=0$ and therefore, by Lemma 1,
$Y_{\pm}(t,x,\lambda)=W_{\pm}(t,x,\lambda)\Phi(t,\lambda)$ are solutions of
the $t$-equation for all $x$. Thus, $Y_{\pm}(t,x,\lambda)$ (2.21) are
solutions of the equations (1.8), (1.12) for all $t$, $x$. The proof for
$Z_{\pm}(t,x,\lambda)$ (2.23) is similar.

The matrix $\Phi(t,\lambda)$ was obtained above. The existence of the matrices
$W_{\pm}(t,x,\lambda)$, $\Psi(t,x,\lambda)$, and $w_{\pm}(x,\lambda)$ as well
as the properties of the solutions (2.21), (2.23) are proved below.

Note that the function (2.8), that is,

$G(t,x,z)=\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{F(t,x,s)n(s)\textup{d}s}{s-z},\qquad\Lambda\leq\infty,$

is analytic for $z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$, and for
$\lambda\in[-\Lambda,\Lambda]$ it has the continuous boundary values
$G_{\pm}(t,x,\lambda)=\dfrac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{F(t,x,s)n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s$
on the left (“+”) and right (“-”) of the contour $[-\Lambda,\Lambda]$ oriented
from the left to the right.

First consider the case when $\Lambda<\infty$ and
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$. For
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ the $x^{\pm}$-equations
(1.12) (or (2.6)) are the same and take the form

$W_{x}=(\mathrm{i}\lambda\sigma_{3}+H(t,x)-\mathrm{i}G(t,x,\lambda))W.$ (2.28)

We rewrite the equation (2.28) in the form

$W_{x}=\left(\mathrm{i}\left[\eta(\lambda)-\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{[{\mathcal{N}}(t,x,s)+1]n(s)}{s-\lambda}\textup{d}s\right]\sigma_{3}+\widehat{H}(t,x,\lambda)\right)W,\quad\lambda\in\mathbb{R}\setminus[-\Lambda,\Lambda],$

where $\eta(\lambda)$ has the form (2.18)
($z=\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$), and

$\widehat{H}(t,x,\lambda)=\begin{pmatrix}0&\dfrac{1}{2}{\mathcal{E}}(t,x)-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-\lambda}\textup{d}s\\\
-\dfrac{1}{2}\overline{\mathcal{E}(t,x)}-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\overline{\rho(t,x,s)}n(s)}{s-\lambda}\textup{d}s&0\end{pmatrix}.$

Further let us put $\widehat{W}=e^{-\mathrm{i}\alpha\sigma_{3}}W$, where

$\alpha=\alpha(t,x,\lambda)=-\dfrac{1}{4}\int\limits_{0}^{x}\int\limits_{-\Lambda}^{\Lambda}\dfrac{[{\mathcal{N}}(t,y,s)+1]n(s)}{s-\lambda}\textup{d}s\textup{d}y=-\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\widehat{{\mathcal{N}}}(t,x,s)n(s)}{s-\lambda}\textup{d}s,\quad\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda],$
$\widehat{\mathcal{N}}(t,x,s)=\int\limits_{0}^{x}[{\mathcal{N}}(t,y,s)+1]\textup{d}y.$
(2.29)

For $\widehat{W}$ the equation (2.28) takes the form

$\widehat{W}_{x}=\big{[}\mathrm{i}\eta(\lambda)\sigma_{3}+\widetilde{H}(t,x,\lambda)\big{]}\widehat{W},\qquad\widetilde{H}(t,x,\lambda)=e^{-\mathrm{i}\alpha\sigma_{3}}\widehat{H}(t,x,\lambda)e^{\mathrm{i}\alpha\sigma_{3}}.$

This equation is equivalent to the matrix integral equation:

$\widetilde{W}(t,x,\lambda)=I+\int\limits_{0}^{x}e^{\mathrm{i}\eta(\lambda)(x-y)\sigma_{3}}\widetilde{H}(t,y,\lambda)\widetilde{W}(t,y,\lambda)e^{-\mathrm{i}\eta(\lambda)(x-y)\sigma_{3}}\textup{d}y,$
(2.30)

where
$\widetilde{W}(t,x,\lambda)=\widehat{W}(t,x,\lambda)e^{-\mathrm{i}\eta(\lambda)x\sigma_{3}}=e^{-\mathrm{i}\alpha(t,x,\lambda)\sigma_{3}}W(t,x,\lambda)e^{-\mathrm{i}\eta(\lambda)x\sigma_{3}}$.
From (2.30) we obtain the vector integral equations (where
$\widetilde{W}[i](t,y,\lambda)$ is the $i$-th column of the matrix
$\widetilde{W}(t,y,\lambda)$):

$\widetilde{W}[1](t,x,\lambda)=\begin{pmatrix}1\\\
0\end{pmatrix}+\int\limits_{0}^{x}\begin{pmatrix}0&\widetilde{H}_{12}(t,y,\lambda)\\\
\widetilde{H}_{21}(t,y,\lambda)e^{-2\mathrm{i}\eta(\lambda)(x-y)}&0\end{pmatrix}\widetilde{W}[1](t,y,\lambda)\textup{d}y,$
(2.31) $\widetilde{W}[2](t,x,\lambda)=\begin{pmatrix}0\\\
1\end{pmatrix}+\int\limits_{0}^{x}\begin{pmatrix}0&\widetilde{H}_{12}(t,y,\lambda)e^{2\mathrm{i}\eta(\lambda)(x-y)}\\\
\widetilde{H}_{21}(t,y,\lambda)&0\end{pmatrix}\widetilde{W}[2](t,y,\lambda)\textup{d}y.$
(2.32)

Let us prove that there exists a solution $\widetilde{W}(t,x,\lambda)$ of the
integral equation (2.30) and it is continuously differentiable in $t$, $x$ and
has an analytic continuation to ${\mathbb{C}}\setminus[-\Lambda,\Lambda]$. To
do this, consider the Neumann series
$\sum\limits_{n=0}^{\infty}\widetilde{W}_{n}(t,x,\lambda)$, where

$\widetilde{W}_{0}(t,x,\lambda)\equiv
I,\quad\widetilde{W}_{n}(t,x,\lambda)=\int\limits_{0}^{x}e^{\mathrm{i}\eta(\lambda)(x-y)\sigma_{3}}\widetilde{H}(t,y,\lambda)\widetilde{W}_{n-1}(t,y,\lambda)e^{-\mathrm{i}\eta(\lambda)(x-y)\sigma_{3}}\textup{d}y,$

$n=1,2,...$, and $t\in{\mathbb{R}}_{+}$, $x\in{\mathbb{R}}_{+}$,
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$. Denote
$\Upsilon(t,x,\lambda)=\begin{pmatrix}0&-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-\lambda}\textup{d}s\\\
-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\overline{\rho(t,x,s)}n(s)}{s-\lambda}\textup{d}s&0\end{pmatrix}$,
then ${\widehat{H}(t,x,\lambda)=H(t,x)+\Upsilon(t,x,\lambda)}$.

Recall that $|{\mathcal{N}}(t,x,\lambda)|\leq 1$ and $|\rho(t,x,\lambda)|\leq
1$ for all
$(t,x,\lambda)\in{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}$ due
to (1.2) and that ${\mathcal{N}}(t,x,\lambda)+1\to 0$ fast enough as
$x\to\infty$. Therefore,
$0\leq\widehat{\mathcal{N}}(t,x,\lambda)\leq\widehat{K}=const$ for
$t,x\in{\mathbb{R}}_{+}$, $\lambda\in{\mathbb{R}}$, where
$\widehat{\mathcal{N}}$ is defined in (2.29). It follows from the properties
of the functions $n$, $\mathcal{N}$, $\widehat{\mathcal{N}}$ and $\rho$ that
$\alpha$, $\Upsilon$, $\alpha_{t}$ and $\Upsilon_{t}$ are continuous for
$(t,x,\lambda)\in{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}\setminus[-\Lambda,\Lambda]$,
and also that $\alpha$, $\Upsilon$ are bounded for all
$t,\,x\in{\mathbb{R}}_{+}$ and
$\lambda\in(-\infty,-\Lambda_{0}]\cup[\Lambda_{0},+\infty)$, where
${\Lambda_{0}>\Lambda}$ is an arbitrary number. Obviously,
$\|e^{\pm\mathrm{i}\eta(\lambda)(x-y)\sigma_{3}}\|\leq C_{1}=const$ and
$\|e^{\pm\mathrm{i}\alpha(t,x,\lambda)\sigma_{3}}\|\leq C_{2}=const$ for
$t,x\in{\mathbb{R}}_{+}$ and
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$. Here we use the operator
(matrix) norm subordinate to the vector norm
$\|x\|_{\infty}=\mathop{\sup}\limits_{1\leq k\leq 2}|x_{k}|$,
$x\in{\mathbb{C}}^{2}$. In general, by $\|\cdot\|$ we denote any suitable norm
(such that $\|I\|=1$), unless it is specified exactly which norm is
considered.

Since ${\mathcal{E}}(t,x)\in C^{1}({\mathbb{R}}_{+}\times{\mathbb{R}}_{+})$ is
bounded near $(0,0)$, i.e., bounded on any compact $[0,L_{0}]\times[0,T_{0}]$
in $[0,\infty)\times[0,\infty)$, then there exists a constant
$C_{3}=C_{3}(\Lambda_{0})$ such that $\|\widehat{H}(t,x,\lambda)\|\leq C_{3}$
for all $t\in[0,T_{0}]$, $x\in[0,L_{0}]$ and
$\lambda\in(-\infty,-\Lambda_{0}]\cup[\Lambda_{0},+\infty)$, where $T_{0}>0$,
$L_{0}>0$ and $\Lambda_{0}>\Lambda$ are arbitrary numbers. In addition, we
obtain that ${\mathcal{E}}(t,x)\in L^{1}[a,b]$, $0\leq a\leq b<\infty$, with
respect to $x$ for each $t\in{\mathbb{R}}_{+}$ and with respect to $t$ for
each $x\in{\mathbb{R}}_{+}$

It follows from the above that
$\|\widetilde{W}_{0}(t,x,\lambda)\|\equiv\|I\|=1$ and
$\|\widetilde{W}_{n}(t,x,\lambda)\|\leq
C\int\limits_{0}^{x}\|\widetilde{W}_{n-1}(t,y,\lambda)\|\textup{d}y$, where
$C=C_{1}^{2}C_{2}^{2}C_{3}$, $t\in[0,T_{0}]$, ${T_{0}>0}$, $x\in[0,L_{0}]$,
$L_{0}>0$, and $\lambda\in(-\infty,-\Lambda_{0}]\cup[\Lambda_{0},+\infty)$,
$\Lambda_{0}>\Lambda$, and using the method of mathematical induction (MMI) it
is easy to prove that
$\|\widetilde{W}_{n}(t,x,\lambda)\|\leq\dfrac{C^{n}x^{n}}{n!}$. Consequently,
$\|\widetilde{W}(t,x,\lambda)\|\leq\sum\limits_{n=0}^{\infty}\|\widetilde{W}_{n}(t,x,\lambda)\|\leq\sum\limits_{n=0}^{\infty}\dfrac{C^{n}x^{n}}{n!}=e^{Cx}$
and hence the series
$\sum\limits_{n=0}^{\infty}\widetilde{W}_{n}(t,x,\lambda)$ converges
absolutely and uniformly for all $t\in[0,T_{0}]$, $x\in[0,L_{0}]$ and
$\lambda\in(-\infty,-\Lambda_{0}]\cup[\Lambda_{0},+\infty)$, and in view of
the arbitrariness of $L_{0}$, $T_{0}$ and $\Lambda_{0}$ it converges locally
uniformly for all $t,\,x\in{\mathbb{R}}_{+}$ and
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$. Consequently, the sum
$\widetilde{W}(t,x,\lambda)=\sum\limits_{n=0}^{\infty}\widetilde{W}_{n}(t,x,\lambda)$
is a solution of the equation (2.30), at that, the function
$\widetilde{W}(t,x,\lambda)$ is continuous on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}\setminus[-\Lambda,\Lambda]$
(since $W_{n}(t,x,\lambda)$ is continuous on this set) and continuously
differentiable in $x$ (since $\widetilde{W}(t,x,\lambda)$ is a solution of the
integral equation (2.30)). Moreover, since $\eta(\lambda)$,
$\alpha(t,x,\lambda)$ and $\Upsilon(t,x,\lambda)$ (and therefore
$\widehat{H}(t,x,\lambda)$) have analytic continuations to
${{\mathbb{C}}\setminus[-\Lambda,\Lambda]}$, then $\widetilde{W}(t,x,\lambda)$
has an analytic continuation $\widetilde{W}(t,x,z)$ for
$z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$. From the properties of the
functions $\mathcal{E}$, $\mathcal{N}$, $\rho$ as a solution of the considered
IBVP (1.5)–(1.7) it follows that $\mathcal{E}_{t}(t,x)$,
$\mathcal{N}_{t}(t,x,\lambda)$ and $\rho_{t}(t,x,\lambda)$ are bounded for
$(t,x)$ close to $(0,0)$ and $\lambda$ from any compact in
${\mathbb{R}}\setminus[-\Lambda,\Lambda]$. Then
${\alpha_{t}(t,x,\lambda)=-\dfrac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\widehat{\mathcal{N}}_{t}(t,x,s)n(s)}{s-\lambda}\textup{d}s}$,
${\widehat{H}_{t}(t,x,\lambda)=H_{t}(t,x)+\Upsilon_{t}(t,x,\lambda)}$, where
$\Upsilon_{t}=\left(\begin{smallmatrix}0&-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho_{t}(t,x,s)n(s)}{s-\lambda}\textup{d}s\\\
-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\overline{\rho_{t}(t,x,s)}n(s)}{s-\lambda}\textup{d}s&0\end{smallmatrix}\right)$,
and
$\widetilde{H}_{t}(t,x,\lambda)=e^{-\mathrm{i}\alpha(t,x,\lambda)\sigma_{3}}\big{(}\widehat{H}_{t}(t,x,\lambda)+\mathrm{i}\alpha_{t}(t,x,\lambda)[\widehat{H}(t,x,\lambda),\sigma_{3}]\big{)}e^{\mathrm{i}\alpha(t,x,\lambda)\sigma_{3}}$
are bounded on any compact in
${\mathbb{R}}_{+}\times[0,\infty)\times{\mathbb{R}}\setminus[-\Lambda,\Lambda]$.
Denote $(\widetilde{W}_{n})_{t}=\dfrac{\partial\widetilde{W}_{n}}{\partial
t}$. Obviously, $(\widetilde{W}_{n})_{t}(t,x,\lambda)$ ($n=0,1,...$) are
continuous on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}\setminus[-\Lambda,\Lambda]$.
In the same way as above, we prove that the series
$\sum\limits_{n=0}^{\infty}(\widetilde{W}_{n})_{t}$ converges absolutely and
locally uniformly on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}\setminus[-\Lambda,\Lambda]$.
Consequently, $\widetilde{W}(t,x,\lambda)$ is continuously differentiable in
$t$.

Now consider the case when $\Lambda\leq\infty$ and
$\lambda\in[-\Lambda,\Lambda]$ (recall that we set
$[-\Lambda,\Lambda]={\mathbb{R}}$ for $\Lambda=\infty$).

The $x^{\pm}$-equations (1.12) (for $W_{\pm}(t,x,\lambda)$ from (2.21)), as
above, can be written in the form

$(W_{\pm})_{x}=\left(\mathrm{i}\left[\eta_{\pm}(\lambda)-\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{[{\mathcal{N}}(t,x,s)+1]n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s\right]\sigma_{3}+\widehat{H}_{\pm}(t,x,\lambda)\right)W_{\pm},$

where $\lambda\in[-\Lambda,\Lambda]$, $\Lambda\leq\infty$,
$\eta_{\pm}(\lambda)$ are the boundary values defined in (2.20), and

$\widehat{H}_{\pm}(t,x,\lambda)=\begin{pmatrix}0&\dfrac{1}{2}{\mathcal{E}}(t,x)-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s\\\
-\dfrac{1}{2}\overline{{\mathcal{E}}(t,x)}-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\overline{\rho(t,x,s)}n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s&0\end{pmatrix}.$

Let us put $\widehat{W}_{\pm}=e^{-\mathrm{i}\alpha_{\pm}\sigma_{3}}W_{\pm}$,
where

$\alpha_{\pm}=\alpha_{\pm}(t,x,\lambda)=-\frac{1}{4}\int\limits_{0}^{x}\int\limits_{-\Lambda}^{\Lambda}\dfrac{[{\mathcal{N}}(t,y,s)+1]n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s\textup{d}y=-\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\widehat{{\mathcal{N}}}(t,x,s)n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s,$

and $\widehat{\mathcal{N}}(t,x,s)$ has the form (2.29). Since
$(\alpha_{\pm})_{x}=-\dfrac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{[{\mathcal{N}}(t,x,s)+1]n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s$,
then

$(\widehat{W}_{\pm})_{x}=\big{[}\mathrm{i}\eta_{\pm}(\lambda)\sigma_{3}+\widetilde{H}_{\pm}(t,x,\lambda)\big{]}\widehat{W}_{\pm},\qquad\widetilde{H}_{\pm}(t,x,\lambda)=e^{-\mathrm{i}\alpha_{\pm}\sigma_{3}}\widehat{H}_{\pm}(t,x,\lambda)e^{\mathrm{i}\alpha_{\pm}\sigma_{3}}.$

These equations are equivalent to the matrix integral equations:

$\widetilde{W}_{\pm}(t,x,\lambda)=I+\int\limits_{0}^{x}e^{\mathrm{i}\eta_{\pm}(\lambda)(x-y)\sigma_{3}}\widetilde{H}_{\pm}(t,y,\lambda)\widetilde{W}_{\pm}(t,y,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)(x-y)\sigma_{3}}\textup{d}y,$
(2.33)

where
$\widetilde{W}_{\pm}(t,x,\lambda)=\widehat{W}_{\pm}(t,x,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}=e^{-\mathrm{i}\alpha_{\pm}(t,x,\lambda)\sigma_{3}}W_{\pm}(t,x,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}$
($\widetilde{W}_{\pm}(t,0,\lambda)=\widehat{W}_{\pm}(t,0,\lambda)=W_{\pm}(t,0,\lambda)=I$).
From (2.33) we obtain the vector integral equations

$\widetilde{W}_{\pm}[1](t,x,\lambda)=\begin{pmatrix}1\\\
0\end{pmatrix}+\int\limits_{0}^{x}\begin{pmatrix}0&\widehat{H}_{\pm
12}e^{-2\mathrm{i}\alpha_{\pm}}\\\ \widehat{H}_{\pm
21}e^{2\mathrm{i}\alpha_{\pm}}e^{-2\mathrm{i}\eta_{\pm}(\lambda)(x-y)}&0\end{pmatrix}\widetilde{W}_{\pm}[1](t,y,\lambda)\textup{d}y,$
(2.34) $\widetilde{W}_{\pm}[2](t,x,\lambda)=\begin{pmatrix}0\\\
1\end{pmatrix}+\int\limits_{0}^{x}\begin{pmatrix}0&\widehat{H}_{\pm
12}e^{-2\mathrm{i}\alpha_{\pm}}e^{2\mathrm{i}\eta_{\pm}(\lambda)(x-y)}\\\
\widehat{H}_{\pm
21}e^{2\mathrm{i}\alpha_{\pm}}&0\end{pmatrix}\widetilde{W}_{\pm}[2](t,y,\lambda)\textup{d}y.$
(2.35)

Let us show that solutions of the integral equations (2.33) exist and they
satisfy the required properties. Solutions of the equations (2.33) can be
represented as the Neumann series
$\widetilde{W}_{\pm}(t,x,\lambda)=\sum\limits_{n=0}^{\infty}\widetilde{W}_{\pm,\,n}(t,x,\lambda)$,
where

$\widetilde{W}_{\pm,\,0}(t,x,\lambda)\equiv
I,\quad\widetilde{W}_{\pm,\,n}(t,x,\lambda)=\int\limits_{0}^{x}e^{\mathrm{i}\eta_{\pm}(\lambda)(x-y)\sigma_{3}}\widetilde{H}_{\pm}(t,y,\lambda)\widetilde{W}_{\pm,\,n-1}(t,y,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)(x-y)\sigma_{3}}\textup{d}y.$

Introduce the matrix
$\Upsilon_{\pm}(t,x,\lambda)=\begin{pmatrix}0&-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s\\\
-\dfrac{\mathrm{i}}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\overline{\rho(t,x,s)}n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s&0\end{pmatrix}$,
then $\widehat{H}_{\pm}(t,x,\lambda)=H(t,x)+\Upsilon_{\pm}(t,x,\lambda)$.

Obviously, $n(\lambda)$ is bounded on ${\mathbb{R}}$ and hence
$\operatorname{Im}\eta_{\pm}(\lambda)=\pm\dfrac{\pi}{4}n(\lambda)$ are bounded
on ${\mathbb{R}}$. Consequently, for $y\in[0,x]$, $x\in{\mathbb{R}}_{+}$ we
obtain

$|e^{\mathrm{i}\eta_{+}(\lambda)(x-y)}|=|e^{-\mathrm{i}\eta_{-}(\lambda)(x-y)}|=e^{-\pi/4\,n(\lambda)(x-y)}\leq
1,$
$|e^{-\mathrm{i}\eta_{+}(\lambda)(x-y)}|=|e^{\mathrm{i}\eta_{-}(\lambda)(x-y)}|=e^{\pi/4\,n(\lambda)(x-y)}\leq
e^{\pi/4\,n(\lambda)x}\leq e^{Nx},\quad N=const>0.$

Since $|\rho(t,x,\lambda)|\leq 1$ for all $(t,x,\lambda)$, then, due to the
properties of $n(\lambda)$ indicated in conditions 1) and 2), where 2)2.1)
holds, the function $\rho(t,x,\lambda)n(\lambda)$ satisfies the Hölder
condition uniformly with respect to $t$, $x$ for
$\lambda\in[-\Lambda,\Lambda]$ ($\Lambda\leq\infty$) and, if $\Lambda=\infty$
(i.e., $[-\Lambda,\Lambda]={\mathbb{R}}$), in a neighborhood of infinity. If
condition 2)2.2) holds, i.e., $n(\lambda)$ has the form (2.17), then
$\rho(t,x,\lambda)n(\lambda)=\frac{1}{2\Lambda}\rho(t,x,\lambda)$ for
$\lambda\in(-\Lambda,\Lambda)$. Note that $\rho(t,x,\lambda)$ satisfies the
Lipschitz condition in $\lambda$ on $[-\Lambda,\Lambda]$. Moreover,
$\rho(t,x,\pm\Lambda)n(\pm\Lambda)\equiv 0$ ($\Lambda\leq\infty$) and
$\rho(t,x,\lambda)n(\lambda)$ belongs to $L^{1}[-\Lambda,\Lambda]$ with
respect to $\lambda$ under both conditions 1) and 2). Consequently, there
exists the Cauchy integral
$\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-z}\textup{d}s$
which is analytic on ${\mathbb{C}}\setminus[-\Lambda,\Lambda]$ and continuous
up to $[-\Lambda,\Lambda]$ (i.e., continuously extendible to
$[-\Lambda,\Lambda]$ from the left and from the right:
$\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s<\infty$,
$\lambda=\operatorname{Re}z\in[-\Lambda,\Lambda]$) as a function of the
variable $z$ with the parameters $t$, $x$, and
$\int\limits_{-\Lambda}^{\Lambda}\dfrac{\rho(t,x,s)n(s)}{s-z}\textup{d}s=O(z^{-1})$,
$z\to\infty$. The same is true for the integral with $\overline{\rho(t,x,s)}$.
Thus, there exists the function $\Upsilon(t,x,z)$ ($\widehat{H}(t,x,z)$
respectively), analytic in $z$ on ${\mathbb{C}}\setminus[-\Lambda,\Lambda]$
and continuous up to $[-\Lambda,\Lambda]$, which has the boundary values
$\Upsilon_{\pm}(t,x,\lambda)$ ($\widehat{H}_{\pm}(t,x,\lambda)$ respectively)
from the left/right of $[-\Lambda,\Lambda]$ and the asymptotics
$\Upsilon(t,x,z)=O(z^{-1})$ as $z\to\infty$, and evidently this function
coincides with the analytic continuation of $\Upsilon(t,x,\lambda)$ mentioned
above for the case $\Lambda<\infty$,
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$. Note that
($\Upsilon_{\pm}(t,x,\lambda)=O(|\lambda|^{-\mu})$, $|\lambda|\to+\infty$)

$\|\Upsilon_{\pm}(t,x,\lambda)\|\leq\upsilon,\quad\upsilon=const>0,\quad(t,x,\lambda)\in{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times[-\Lambda,\Lambda],\;\Lambda\leq\infty.$

Recall that $0\leq\widehat{\mathcal{N}}(t,x,\lambda)\leq\widehat{K}=const$ for
all $(t,x,\lambda)$ due to the properties of ${\mathcal{N}}(t,x,\lambda)$. In
the same way as above, using the properties of ${\mathcal{N}}(t,x,\lambda)$
(and, accordingly, $\widehat{\mathcal{N}}(t,x,\lambda)$) and the properties of
$n(\lambda)$ indicated in conditions 1) and 2), we obtain that the function

$\alpha(t,x,z)=-\dfrac{1}{4}\int\limits_{0}^{x}\int\limits_{-\Lambda}^{\Lambda}\dfrac{[{\mathcal{N}}(t,y,s)+1]n(s)}{s-z}\textup{d}s\textup{d}y=-\frac{1}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\widehat{{\mathcal{N}}}(t,x,s)n(s)}{s-z}\textup{d}s,\quad\Lambda\leq\infty,$

is analytic for $z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$ and continuous
up to ${\mathbb{C}}$ (evidently, this function coincides with the analytic
continuation of $\alpha(t,x,\lambda)$ mentioned above for the case
$\Lambda<\infty$, $\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$), and
$\alpha(t,x,z)=O(z^{-1})$ as $z\to\infty$. Its boundary values
$\alpha_{\pm}(t,x,\lambda)=\alpha(t,x,\lambda\pm\mathrm{i}0)$ are equal to
$\alpha_{\pm}(t,x,\lambda)=-\dfrac{1}{4}\,{\rm
p.v.}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\widehat{{\mathcal{N}}}(t,x,s)n(s)}{s-\lambda}\textup{d}s\mp\dfrac{\pi\mathrm{i}}{4}\widehat{\mathcal{N}}(t,x,\lambda)n(\lambda)$,
$\lambda\in[-\Lambda,\Lambda]$. In addition,

$\operatorname{Im}\alpha(t,x,z)=-\dfrac{\operatorname{Im}z}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{\widehat{\mathcal{N}}(t,x,s)n(s)}{(s-\lambda)^{2}+\operatorname{Im}^{2}z}\textup{d}s,\quad
z\to\infty.$

Since
$\operatorname{Im}\alpha_{\pm}(t,x,\lambda)=\mp\dfrac{\pi}{4}\widehat{\mathcal{N}}(t,x,\lambda)n(\lambda)$,
then

$|e^{\mathrm{i}\alpha_{+}(t,x,\lambda)}|=|e^{-\mathrm{i}\alpha_{-}(t,x,\lambda)}|=e^{\pi\widehat{\mathcal{N}}(t,x,\lambda)n(\lambda)/4}\leq
e^{\pi\widehat{K}n(\lambda)/4},$
$|e^{-\mathrm{i}\alpha_{+}(t,x,\lambda)}|=|e^{\mathrm{i}\alpha_{-}(t,x,\lambda)}|=e^{-\pi\widehat{\mathcal{N}}(t,x,\lambda)n(\lambda)/4}\leq
1.$

Thus, $\|e^{\pm\mathrm{i}\alpha_{\pm}(t,x,\lambda)\sigma_{3}}\|\leq e^{K}$,
where $K=const>0$, and
$\|e^{\pm\mathrm{i}\eta_{\pm}(\lambda)(x-y)\sigma_{3}}\|\leq e^{Nx}$ for each
$t\in{\mathbb{R}}_{+}$, $y\in[0,x]$, $x\in{\mathbb{R}}_{+}$,
$\lambda\in[-\Lambda,\Lambda]$ ($\Lambda\leq\infty$). Note that
$e^{\pm\mathrm{i}\alpha(t,x,z)}=1+O(z^{-1})$, $z\to\infty$
($e^{\pm\mathrm{i}\alpha_{\pm}(t,x,\lambda)}\to 1$, $|\lambda|\to\infty$).

It follows from the above that $\|W_{\pm,\,0}(t,x,\lambda)\|\equiv\|I\|=1$ and

$\|\widetilde{W}_{\pm,\,n}(t,x,\lambda)\|\leq
e^{2(NL_{0}+K)}\int\limits_{0}^{x}(\|H(t,y)\|+\upsilon)\,\|\widetilde{W}_{\pm,\,n-1}(t,y,\lambda)\|\textup{d}y,\quad
n=1,2,...,$

for all $t\in(0,T_{0})$, $x\in(0,L_{0})$ and
$\lambda\in(-\Lambda_{0},\Lambda_{0})$, where $T_{0},\,L_{0},\,\Lambda_{0}>0$
($\Lambda_{0}\leq\Lambda$ for $\Lambda<\infty$ and $\Lambda_{0}<\infty$ for
$\Lambda=\infty$) are arbitrary numbers. Then
$\|\widetilde{W}_{\pm,\,1}(t,x,\lambda)\|\leq
e^{2(NL_{0}+K)}\int\limits_{0}^{x}(\|H(t,y)\|+\upsilon)\textup{d}y=e^{2(NL_{0}+K)}S(t,x)$,
where $S(t,x)=\int\limits_{0}^{x}(\|H(t,y)\|+\upsilon)\textup{d}y$, and using
the MMI we obtain
$\|\widetilde{W}_{\pm,\,n}(t,x,\lambda)\|\leq\dfrac{e^{2n(NL_{0}+K)}S_{t}^{n}(x)}{n!}$.
Consequently,
$\|\widetilde{W}_{\pm}(t,x,\lambda)\|=\big{\|}\sum\limits_{n=0}^{\infty}\widetilde{W}_{\pm,\,n}(t,x,\lambda)\big{\|}\leq\exp\\{e^{2(NL_{0}+K)}S(t,x)\\}$.
Due to the properties of $\mathcal{E}(t,x)$, the function $S(t,x)$ is bounded
on any compact in ${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}$. Thus, the series
$\sum\limits_{n=0}^{\infty}\widetilde{W}_{\pm,\,n}(t,x,\lambda)$ converge
absolutely and locally uniformly for
$(t,x,\lambda)\in{\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times[-\Lambda,\Lambda]$,
$\Lambda\leq\infty$. Since the functions
$\widetilde{W}_{\pm,\,n}(t,x,\lambda)$ are continuous and the series converge
locally uniformly on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times[-\Lambda,\Lambda]$, then the
functions (the sums of the series) $\widetilde{W}_{\pm}(t,x,\lambda)$ are
continuous on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times[-\Lambda,\Lambda]$,
$\Lambda\leq\infty$. The continuous differentiability of
$\widetilde{W}_{\pm}(t,x,\lambda)$ with respect to $t$ and $x$ is proved in
the same way as for $\widetilde{W}(t,x,\lambda)$,
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ (see above).

It also follows from the above that there exist the function (analytic
continuation) $\widetilde{W}(t,x,z)$, analytic in $z$ on
${\mathbb{C}}\setminus[-\Lambda,\Lambda]$ and continuous up to the boundary
$[-\Lambda,\Lambda]$, which satisfies the integral equation (2.30) for
$z=\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$, and
have the boundary values $\widetilde{W}_{\pm}(t,x,\lambda)$ satisfying the
integral equations (2.33) for $\lambda\in[-\Lambda,\Lambda]$,
$\Lambda\leq\infty$ (also, $\widetilde{W}(t,x,z)$ is continuously
differentiable in $t$, $x$ for each $z$). For $\Lambda<\infty$, we extend the
functions $\widetilde{W}_{\pm}(t,x,\lambda)$ by defining them equal to
$\widetilde{W}_{\pm}(t,x,\lambda):=\widetilde{W}(t,x,\lambda)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ (accordingly, in a similar
way we define all the boundary values included in
$\widetilde{W}_{\pm}(t,x,\lambda)$, i.e.,
$\alpha_{\pm}(t,x,\lambda):=\alpha(t,x,\lambda)$,
$\eta_{\pm}(\lambda):=\eta(\lambda)$, etc., for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$). Thus,
the function
$W(t,x,z)=e^{\mathrm{i}\alpha(t,x,z)\sigma_{3}}\widetilde{W}(t,x,z)e^{\mathrm{i}\eta(z)x\sigma_{3}}$
is analytic in $z$ on ${\mathbb{C}}\setminus[-\Lambda,\Lambda]$ and continuous
up to $[-\Lambda,\Lambda]$, it is continuously differentiable in $t$, $x$ (for
each $z\in{\mathbb{C}}$) and its boundary values
$W_{\pm}(t,x,\lambda)=e^{\mathrm{i}\alpha_{\pm}(t,x,\lambda)\sigma_{3}}\widetilde{W}_{\pm}(t,x,\lambda)e^{\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}$
($W_{+}(t,x,\lambda)=W_{-}(t,x,\lambda)=W(t,x,\lambda)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$) satisfy
the $x^{\pm}$-equations (1.12) for each $t\in{\mathbb{R}}_{+}$ and
$W_{\pm}(t,0,\lambda)=I$. Denote by
$\widetilde{W}^{\pm}(t,x,z):=\begin{cases}\widetilde{W}(t,x,z),\quad
z\in{\mathbb{C}}_{\pm}=\\{z\in{\mathbb{C}}\mid\;\pm\operatorname{Im}z>0\\},\\\
\widetilde{W}_{\pm}(t,x,z),\quad z=\lambda\in{\mathbb{R}}\end{cases}$
($W^{+}(t,x,z)=W^{-}(t,x,z)=W(t,x,z)$,
$z=\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$) and
introduce the notation $W^{\pm}(t,x,z)$,
$z\in{\mathbb{C}}_{\pm}\cup{\mathbb{R}}$ in a similar way. Since
$e^{\pm\mathrm{i}\eta(z)x}\to 0$ for $z\in{\mathbb{C}}_{\pm}$ (respectively),
$x\in{\mathbb{R}}_{+}$ as $z\to\infty$ (or $x\to+\infty$), then the columns
$\widetilde{W}^{-}[1](t,x,z)$ and $\widetilde{W}^{+}[2](t,x,z)$ are bounded in
$z\in{\mathbb{C}}_{-}^{cl}={\mathbb{C}}_{-}\cup{\mathbb{R}}$ and
$z\in{\mathbb{C}}_{+}^{cl}={\mathbb{C}}_{+}\cup{\mathbb{R}}$, respectively.
Further, taking into account

$\displaystyle
W^{-}[1](t,x,z)e^{-\mathrm{i}\eta(z)x}=e^{\mathrm{i}\alpha(t,x,z)\sigma_{3}}\widetilde{W}^{-}[1](t,x,z),$
$\displaystyle
W^{-}[2](t,x,z)e^{-\mathrm{i}\eta(z)x}=e^{\mathrm{i}\alpha(t,x,z)\sigma_{3}}\widetilde{W}^{-}[2](t,x,z)e^{-2\mathrm{i}\eta(z)x},$

where $\widetilde{W}^{-}[2]e^{-2\mathrm{i}\eta\,x}=\begin{pmatrix}0\\\
e^{-2\mathrm{i}\eta\,x}\end{pmatrix}+\int\limits_{0}^{x}\begin{pmatrix}0&\widehat{H}_{12}e^{-2\mathrm{i}\alpha}e^{-2\mathrm{i}\eta\,y}\\\
\widehat{H}_{21}e^{2\mathrm{i}\alpha}e^{-2\mathrm{i}\eta\,x}&0\end{pmatrix}\widetilde{W}^{-}[2]\textup{d}y$
(obviously, $\widehat{H}(t,x,z)=\widehat{H}_{\pm}(t,x,z)$,
$\alpha(t,x,z)=\alpha_{\pm}(t,x,z)$ and $\eta(z)=\eta_{\pm}(z)$ for
$z=\lambda\in{\mathbb{R}}$), we obtain that
$W^{-}(t,x,z)e^{-\mathrm{i}\eta(z)x}$ is bounded in
$z\in{\mathbb{C}}_{-}^{cl}$, and as $z\to\infty$, $z\in{\mathbb{C}}_{-}$, it
has the asymptotics of the form

$W^{-}[1](t,x,z)e^{-\mathrm{i}\eta(z)x}=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\quad
W^{-}[2](t,x,z)e^{-\mathrm{i}\eta(z)x}=\begin{pmatrix}0\\\
e^{-2\mathrm{i}\eta(z)x}\end{pmatrix}+O(z^{-1}),$

and for fixed $x$ as $z\to\infty$, $z\in{\mathbb{C}}_{-}$, it has the
asymptotics

$W^{-}[1](t,x,z)e^{-\mathrm{i}\eta(z)x}=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\quad
W^{-}[2](t,x,z)e^{-\mathrm{i}\eta(z)x}=O(z^{-1}).$

It can be proved similarly that $W^{+}(t,x,z)e^{\mathrm{i}\eta(z)x}$ is
bounded in $z\in{\mathbb{C}}_{+}^{cl}$, and for fixed $x$ as $z\to\infty$,
$z\in{\mathbb{C}}_{+}$, it has the asymptotic behavior

$W^{+}[1](t,x,z)e^{\mathrm{i}\eta(z)x}=\begin{pmatrix}e^{2\mathrm{i}\eta(z)x}\\\
0\end{pmatrix}+O(z^{-1})=O(z^{-1}),\quad
W^{+}[2](t,x,z)e^{\mathrm{i}\eta(z)x}=\begin{pmatrix}0\\\
1\end{pmatrix}+O(z^{-1}).$

Note that if the symbol $O(\cdot)$ is present in a matrix expression, then it
denotes a matrix of the appropriate size whose entries have the indicated
order.

The functions $w(z)$ and $\varkappa(z)$ are analytic in
${\mathbb{C}}\setminus[E,\overline{E}]$, and, hence, the function
$M_{0}(z)=\begin{pmatrix}a(z)&b(z)\\\ b(z)&a(z)\end{pmatrix}$ defined by
$a(z)$ and $b(z)$ of the form (2.12) is also analytic in
${\mathbb{C}}\setminus[E,\overline{E}]$. Note that since $\varkappa(z)\to 1$
and $\varkappa^{-1}(z)\to 1$ as $z\to\infty$, then $M_{0}(z)\to I$ as
$z\to\infty$. Thus, the function $\Phi(t,\lambda)$ (2.22) has the analytic
continuation $\Phi(t,z)$ for $z\in{\mathbb{C}}\setminus[E,\overline{E}]$ (the
orientation on $[E,\overline{E}]$ is chosen from up to down) which is
continuous up to the boundary and has the singularities of the type
$(z-E)^{-1/4}$, $(z-\overline{E})^{-1/4}$ at the points $E$ and
$\overline{E}$. Since ${\det M_{0}(z)\equiv 1}$, then ${\det\Phi(t,z)\equiv
1}$. Taking into account that $w(z)=z-\operatorname{Re}E+O(z^{-1})$,
$z\to\infty$, for each fixed $t$ we obtain the asymptotics:

$\Phi[1](t,z)e^{\mathrm{i}zt}=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\quad\Phi[2](t,z)e^{-\mathrm{i}zt}=\begin{pmatrix}0\\\
1\end{pmatrix}+O(z^{-1}),\quad z\to\infty,\quad z\in{\mathbb{C}}.$

It follows from the above that the functions $Y_{\pm}(t,x,\lambda)$ have the
analytic continuations $Y^{\pm}(t,x,z)$ for
$z\in{\mathbb{C}}_{\pm}^{cl}\setminus([-\Lambda,\Lambda]\cup[E,\overline{E}])$,
where ${\mathbb{C}}_{\pm}^{cl}={\mathbb{C}}_{\pm}\cup{\mathbb{R}}$
($Y^{+}(t,x,z)=Y^{-}(t,x,z)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$) which
are continuous up to
$([-\Lambda,\Lambda]\setminus\\{\operatorname{Re}E\\})\cup[E,\operatorname{Re}E)$
and
$([-\Lambda,\Lambda]\setminus\\{\operatorname{Re}E\\})\cup(\operatorname{Re}E,\overline{E}]$
respectively, bounded in the neighborhood of the point $\operatorname{Re}E$,
have the singularities of the type $(z-E)^{-1/4}$ and
$(z-\overline{E})^{-1/4}$ respectively, and in addition, for fixed $t$ and
$x$,

$\displaystyle
Y^{-}[1](t,x,z)e^{\mathrm{i}(zt-\eta(z)x)}=W^{-}(t,x,z)e^{-\mathrm{i}\eta(z)x}\Phi[1](t,z)e^{\mathrm{i}zt}$
$\displaystyle=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\;z\to\infty,\;z\in{\mathbb{C}}_{-}^{cl};$
$\displaystyle
Y^{+}[2](t,x,z)e^{-\mathrm{i}(zt-\eta(z)x)}=W^{+}(t,x,z)e^{\mathrm{i}\eta(z)x}\Phi[2](t,z)e^{-\mathrm{i}zt}$
$\displaystyle=\begin{pmatrix}0\\\
1\end{pmatrix}+O(z^{-1}),\;z\to\infty,\;z\in{\mathbb{C}}_{+}^{cl}.$

To prove that there exists the solution $\Psi(t,x,\lambda)$ of the
$t$-equation (1.8) (for each $x$) satisfying the initial condition
${\Psi(0,x,\lambda)\equiv I}$, we prove that there exists a solution
$\widehat{\Psi}(t,x,\lambda)=\Psi(t,x,\lambda)e^{\mathrm{i}\lambda
t\sigma_{3}}$ of the equivalent integral equation

$\widehat{\Psi}(t,x,\lambda)=I-\int\limits_{0}^{t}e^{-\mathrm{i}\lambda(t-\tau)\sigma_{3}}H(\tau,x)\widehat{\Psi}(\tau,x,\lambda)e^{\mathrm{i}\lambda(t-\tau)\sigma_{3}}d\tau.$
(2.36)

We represent the solution of (2.36) as the Neumann series
$\widehat{\Psi}=\sum\limits_{n=0}^{\infty}\widehat{\Psi}_{n}$, where
${\widehat{\Psi}_{0}(t,x,\lambda)\equiv I}$,
$\widehat{\Psi}_{n}(t,x,\lambda)=-\int\limits_{0}^{t}e^{-\mathrm{i}\lambda(t-\tau)\sigma_{3}}H(\tau,x)\widehat{\Psi}_{n-1}(\tau,x,\lambda)e^{\mathrm{i}\lambda(t-\tau)\sigma_{3}}\textup{d}\tau$,
and prove that the series converges. Since
$|e^{\pm\mathrm{i}\lambda(t-\tau)}|\leq 1$, then
$\|\widehat{\Psi}_{n}(t,x,\lambda)\|\leq\int\limits_{0}^{t}\|H(\tau,x)\|\,\|\widehat{\Psi}_{n-1}(\tau,x,\lambda)\|\textup{d}\tau$.
Hence,
$\|\widehat{\Psi}_{1}(t,x,\lambda)\|\leq\int\limits_{0}^{t}\|H(\tau,x)\|\textup{d}\tau=:M(t,x)$,
and by the MMI we obtain
$\|\widehat{\Psi}_{n}(t,x,\lambda)\|\leq\dfrac{M^{n}(t,x)}{n!}$. Consequently,
$\|\widehat{\Psi}(t,x,\lambda)\|\leq\sum\limits_{n=0}^{\infty}\|\widehat{\Psi}_{n}(t,x,\lambda)\|\leq
e^{M(t,x)}$. Since (due to the properties of $\mathcal{E}(t,x)$) $M(t,x)$ is
bounded on any compact in ${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}$, then the
series $\sum\limits_{n=0}^{\infty}\widehat{\Psi}_{n}$ converges absolutely and
uniformly in $\lambda$ on ${\mathbb{R}}$ and in $(t,x)$ on any compact in
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}$. Since the functions
$\widehat{\Psi}_{n}(t,x,\lambda)$ are continuous and the series converges
locally uniformly on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}$, then the sum of
the series $\widehat{\Psi}(t,x,\lambda)$ is continuous on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}$, and since
$\widehat{\Psi}(t,x,\lambda)$ is a solution of the integral equation (2.36),
then it is continuously differentiable in $t$. Obviously,
$\widehat{\Psi}_{n}(t,x,\lambda)$ are continuously differentiable in $x$. It
is easily verified that the series
$\sum\limits_{n=0}^{\infty}(\widehat{\Psi}_{n})_{x}(t,x,\lambda)$ converges
absolutely and locally uniformly on
${\mathbb{R}}_{+}\times{\mathbb{R}}_{+}\times{\mathbb{R}}$, and hence
$\widehat{\Psi}(t,x,\lambda)$ is continuously differentiable in $x$. The
function $\Psi(t,x,\lambda)=\widehat{\Psi}(t,x,\lambda)e^{-\mathrm{i}\lambda
t\sigma_{3}}$ has the analytic continuation $\Psi(t,x,z)$ for
$z\in{\mathbb{C}}$. Taking into account that $\widehat{\Psi}(t,x,\lambda)$ has
the integral representation (2.36) and that $e^{\pm\mathrm{i}zt}\to 0$ for
$z\in{\mathbb{C}}_{\pm}$ (respectively) as $z\to\infty$
($t\in{\mathbb{R}}_{+}$), we obtain that the functions
$\Psi(t,x,z)e^{\mathrm{i}zt}$ and $\Psi(t,x,z)e^{-\mathrm{i}zt}$ are bounded
in $z\in{\mathbb{C}}_{+}^{cl}$ and $z\in{\mathbb{C}}_{-}^{cl}$, respectively,
and have the asymptotics

$\Psi(t,x,z)e^{\mathrm{i}zt}=\begin{pmatrix}1&0\\\
0&e^{2\mathrm{i}zt}\end{pmatrix}+O(z^{-1}),\quad
z\to\infty,\;z\in{\mathbb{C}}_{+}^{cl},$
$\Psi(t,x,z)e^{-\mathrm{i}zt}=\begin{pmatrix}e^{-2\mathrm{i}zt}&0\\\
0&1\end{pmatrix}+O(z^{-1}),\quad z\to\infty,\;z\in{\mathbb{C}}_{-}^{cl}.$

Similar results can be obtained by using the integral representation for
$\Psi(t,x,\lambda)$ by a transformation operator (cf. [6]).

Since $H(0,x)\equiv 0$ and $F(0,x,\lambda)\equiv-\sigma_{3}$, then for $t=0$
the $x^{\pm}$-equations (1.12) take the form
$w_{x}=\mathrm{i}\eta_{\pm}(\lambda)\sigma_{3}$ ($\lambda\in{\mathbb{R}}$),
where $\eta_{\pm}(\lambda)=\eta(\lambda)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ ($\Lambda<\infty$). These
equations have the solutions
$w_{\pm}(x,\lambda)=e^{\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}$ satisfying
the initial conditions
$\lim\limits_{x\to\infty}w_{\pm}(x,\lambda)e^{-\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}=I$
(recall that we prove the theorem, assuming that $L=\infty$, $x\in(0,L)$).
Obviously, $\det w_{\pm}(x,\lambda)\equiv 1$. Due to the properties of
$\eta(z)$, there exists the function (analytic continuation)
$w(x,z)=e^{\mathrm{i}\eta(z)x\sigma_{3}}$, analytic in $z$ on
${\mathbb{C}}\setminus[-\Lambda,\Lambda]$ and continuous up to
$[-\Lambda,\Lambda]$, which satisfies the $x^{\pm}$-equations (1.12) for
$z=\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ and have the boundary
values $w_{\pm}(x,\lambda)$ satisfying (1.12) for
$\lambda\in[-\Lambda,\Lambda]$, and in addition,
$w[1](x,z)e^{-\mathrm{i}\eta(z)x}=\begin{pmatrix}1\\\ 0\end{pmatrix}$ and
$w[2](x,z)e^{\mathrm{i}\eta(z)x}=\begin{pmatrix}0\\\ 1\end{pmatrix}$.

It follows from the above that the functions $Z_{\pm}(t,x,\lambda)$ have the
analytic continuations $Z^{\pm}(t,x,z)$ for
$z\in{\mathbb{C}}_{\pm}^{cl}\setminus[-\Lambda,\Lambda]$ respectively
($Z^{+}(t,x,z)=Z^{-}(t,x,z)$ for
$\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$, $\Lambda<\infty$), which
are continuous up to $[-\Lambda,\Lambda]$ and

$\displaystyle
Z^{+}[1](t,x,z)e^{\mathrm{i}(zt-\eta(z)x)}=\Psi(t,x,z)e^{\mathrm{i}zt}w[1](x,z)e^{-\mathrm{i}\eta(z)x}$
$\displaystyle=\begin{pmatrix}1\\\
0\end{pmatrix}+O(z^{-1}),\;z\to\infty,\;z\in{\mathbb{C}}_{+}^{cl};$
$\displaystyle
Z^{-}[2](t,x,z)e^{-\mathrm{i}(zt-\eta(z)x)}=\Psi(t,x,z)e^{-\mathrm{i}zt}w[2](x,z)e^{\mathrm{i}\eta(z)x}$
$\displaystyle=\begin{pmatrix}0\\\
1\end{pmatrix}+O(z^{-1}),\;z\to\infty,\;z\in{\mathbb{C}}_{-}^{cl}.$

Since $Y_{\pm}(t,x,\lambda)$ and $Z_{\pm}(t,x,\lambda)$ are solutions of the
compatible systems (2.1), (2.6), where $U$, $V_{\pm}$ have the form (2.4),
(2.7), then $\det Y_{\pm}(t,x,\lambda)\equiv 1$ and $\det
Z_{\pm}(t,x,\lambda)\equiv 1$. Indeed, let $W_{\pm}(t,x,\lambda)$ be solutions
of the compatible systems (2.1), (2.6), respectively. Then their determinants
$X_{\pm}(t,x,\lambda)=\det W_{\pm}(t,x,\lambda)$ satisfy the systems

$\displaystyle(X_{\pm})_{t}$ $\displaystyle=\mathop{\mathrm{tr}}\nolimits
U(t,x,\lambda)X_{\pm},$ (2.37) $\displaystyle(X_{\pm})_{x}$
$\displaystyle=\mathop{\mathrm{tr}}\nolimits V_{\pm}(t,x,\lambda)X_{\pm}.$
(2.38)

and we obtain the Liouville (Ostrogradsky-Liouville) formulas:
$X_{\pm}(t,x,\lambda)=X_{\pm}(t_{0},x,\lambda)\exp\left\\{\int\limits_{t_{0}}^{t}\mathop{\mathrm{tr}}\nolimits
U(\tau,x,\lambda)d\tau\right\\}$,
$X_{\pm}(t,x,\lambda)=X_{\pm}(t,x_{0},\lambda)\exp\left\\{\int\limits_{x_{0}}^{x}\mathop{\mathrm{tr}}\nolimits
V_{\pm}(t,y,\lambda)dy\right\\}$, where $t_{0}\geq 0$ is an arbitrary number.
From the compatibility of the systems (2.1), (2.6), i.e., from the relations
(2.9), it follows that $\mathop{\mathrm{tr}}\nolimits
U_{x}=\mathop{\mathrm{tr}}\nolimits(V_{\pm})_{t}$ and the systems (2.37),
(2.38) are also compatible. Due to the compatibility of the systems (2.37),
(2.38), their solutions have the form

$X_{\pm}(t,x,\lambda)=X_{\pm}(t_{0},x_{0},\lambda)\exp\left\\{\int\limits_{t_{0}}^{t}\mathop{\mathrm{tr}}\nolimits
U(\tau,x,\lambda)d\tau+\int\limits_{x_{0}}^{x}\mathop{\mathrm{tr}}\nolimits
V_{\pm}(t_{0},y,\lambda)dy\right\\}.$

Since $\mathop{\mathrm{tr}}\nolimits
U(t,x,\lambda)=\mathop{\mathrm{tr}}\nolimits V_{\pm}(t,x,\lambda)=0$ for all
$(t,x,\lambda)$, then $\det W_{\pm}(t,x,\lambda)=\det
W_{\pm}(t_{0},x_{0},\lambda)$. Thus, choosing $t_{0}=0$ and $x_{0}=0$, we
obtain $\det Y_{\pm}(t,x,\lambda)=\det Y_{\pm}(0,0,\lambda)=\det
W_{\pm}(0,0,\lambda)\det\Phi(0,\lambda)=1$ and $\det Z_{\pm}(t,x,\lambda)=\det
Z_{\pm}(0,0,\lambda)=\det\Psi(0,0,\lambda)\det w_{\pm}(0,\lambda)=1$ for all
$(t,x,\lambda)$. ∎

##  3 Formulation of the Riemann-Hilbert problem

Since $Y_{\pm}$ (2.21) and $Z_{\pm}$ (2.23) are the solutions of the
compatible AKNS systems (the $t$\- and $x^{\pm}$-equations) (1.8), (1.12),
they are linearly dependent, namely, they satisfy the “scattering relations”

$Y_{\pm}(t,x,\lambda)=Z_{\pm}(t,x,\lambda)T_{\pm}(\lambda),\qquad\lambda\in{\mathbb{R}}.$
(3.1)

Since the transition matrices (scattering matrices) $T_{\pm}(\lambda)$ are
independent of $t$ and $x$, they can be presented in the form (as shown above,
$w_{\pm}(x,\lambda)=e^{\mathrm{i}\eta_{\pm}(\lambda)x\sigma_{3}}$)

$T_{\pm}(\lambda)=(Z_{\pm}(0,0,\lambda))^{-1}Y_{\pm}(0,0,\lambda)=(w_{\pm}(0,\lambda))^{-1}\Phi(0,\lambda)=\Phi(0,\lambda)=M_{0}(\lambda).$

Hence,

$T_{\pm}(\lambda)=T(\lambda)=\begin{pmatrix}a(\lambda)&b(\lambda)\\\
b(\lambda)&a(\lambda)\end{pmatrix}=\frac{1}{2}\begin{pmatrix}\varkappa(\lambda)+\varkappa^{-1}(\lambda)&\varkappa(\lambda)-\varkappa^{-1}(\lambda)\\\
\varkappa(\lambda)-\varkappa^{-1}(\lambda)&\varkappa(\lambda)+\varkappa^{-1}(\lambda)\end{pmatrix},$

where $a(\lambda)$, $b(\lambda)$ and $\varkappa(\lambda)$ are defined in
(2.12).

The spectral functions $a(\lambda)$, $b(\lambda)$ have the analytic
continuations $a(z)$, $b(z)$ to ${\mathbb{C}}\setminus[E,\overline{E}]$ which
have the following asymptotics at infinity:

$\displaystyle a(z)=$ $\displaystyle 1+O(z^{-2}),\quad z\to\infty,\quad
z\in{\mathbb{C}},$ $\displaystyle b(z)=$ $\displaystyle O(z^{-1}),\quad
z\to\infty,\quad z\in{\mathbb{C}}.$

It was assumed that the interval $[E,\overline{E}]$ is oriented downward. For
$z\in(E,\overline{E})$ the functions $a(z)$, $b(z)$ and $\varkappa(z)$ satisfy
the relations

$\varkappa_{-}(z)=\mathrm{i}\varkappa_{+}(z),\qquad
a_{-}(z)=\mathrm{i}b_{+}(z),\qquad b_{-}(z)=\mathrm{i}a_{+}(z),$

where the signs “$-$” and “$+$” in subscripts denote nontangential boundary
values from the left and right of an oriented contour, and in what follows
this notation is also used. It is easy to see that $a(z)$ has no zeros and,
hence, a discrete spectrum is empty.

The “reflection coefficient”

$r(z):=\dfrac{b(z)}{a(z)}=\dfrac{\varkappa(z)-\varkappa^{-1}(z)}{\varkappa(z)+\varkappa^{-1}(z)}=\dfrac{z-\operatorname{Re}E-w(z)}{\mathrm{i}\operatorname{Im}E}=\dfrac{\mathrm{i}\big{(}w(z)-z+\operatorname{Re}E\big{)}}{\operatorname{Im}E},$
(3.2)

where $w(z)=\sqrt{(z-E)(z-\overline{E})}$ (2.13), is analytic in
$z\in{\mathbb{C}}\setminus[E,\overline{E}]$, satisfies the symmetry relation
$\overline{r(\bar{z})}=-r(z)$, and

$r(z)=O(z^{-1}),\quad z\to\infty.$

Since $w(\lambda)\in C^{\infty}({\mathbb{R}})$, then $r(\lambda)$ is
infinitely differentiable as a function of the real variable $\lambda$. Note
that if we, as above, take the cut along an arc $\gamma$ connecting $E$ and
$\overline{E}$ via infinity (on the extended complex plane
${\mathbb{C}}\cup\\{\infty\\}$) without intersection of the real axis and fix
the branch of $w(z)$ by the condition $w(0)=|E|$ or
$w(\operatorname{Re}E)=\operatorname{Im}E$, then $r(z)$ is analytic on
${\mathbb{C}}\setminus\gamma$, including ${\mathbb{R}}$. The same holds for
$a(z)$ and $b(z)$, where we fix the branch of $\varkappa(z)$ by the condition
$\varkappa(\operatorname{Re}E)=e^{\mathrm{i}\pi/4}$.

Denote $\phi=\arg(\lambda-\overline{E})$, $\lambda\in{\mathbb{R}}$, then
$\varkappa^{4}(\lambda)=\dfrac{\lambda-\overline{E}}{\lambda-E}=e^{\mathrm{i}2\phi}$,
where $-\pi<2\phi\leq\pi$, and $\varkappa(\lambda)=e^{\mathrm{i}\phi/2}$.
Since
$r(\lambda)=\dfrac{e^{\mathrm{i}\phi/2}-e^{-\mathrm{i}\phi/2}}{e^{\mathrm{i}\phi/2}+e^{-\mathrm{i}\phi/2}}=\mathrm{i}\tan(\phi/2)$,
$\lambda\in{\mathbb{R}}$, and $-\pi/2<\phi\leq\pi/2$, then

$1-r^{2}(\lambda)=1+|r(\lambda)|^{2}=1+\tan^{2}(\phi/2)=\dfrac{1}{\cos^{2}(\phi/2)}\leq
2,$

and, therefore, $|r(\lambda)|\leq 1$ for $\lambda\in{\mathbb{R}}$.

Due to the relations $a_{-}(z)=\mathrm{i}b_{+}(z)$,
$b_{-}(z)=\mathrm{i}a_{+}(z)$ for $z\in(E,\overline{E})$, the jump of $r(z)$
is

$h(z):=r_{-}(z)-r_{+}(z)=\dfrac{a_{+}^{2}(z)-b_{+}(z)^{2}}{a_{+}(z)b_{+}(z)}=\dfrac{1}{a_{+}(z)b_{+}(z)}=-\dfrac{2\mathrm{i}\,w_{+}(z)}{\operatorname{Im}E},\qquad
z\in(E,\overline{E}).$

Since for $z=\operatorname{Re}E+\mathrm{i}\nu\in(E,\overline{E})$,
$w_{\pm}(\operatorname{Re}E+\mathrm{i}\nu)=\lim\limits_{\varepsilon\to
0}\sqrt{(\pm\varepsilon+\mathrm{i}\nu)^{2}+\operatorname{Im}^{2}E}=\sqrt{\operatorname{Im}^{2}E-\nu^{2}}$,
where $\nu=\operatorname{Im}z\in(\operatorname{Im}E,-\operatorname{Im}E)$,
then
$h(z)=-2\mathrm{i}\,\sqrt{\operatorname{Im}^{2}E-\nu^{2}}/\operatorname{Im}E$.
Thus, for $z\in(E,\overline{E})$ we introduce the function

$h(z)=r_{-}(z)-r_{+}(z)=-\dfrac{2\mathrm{i}\sqrt{\operatorname{Im}^{2}E-\operatorname{Im}^{2}z}}{\operatorname{Im}E},\;\operatorname{Im}z\in(\operatorname{Im}E,-\operatorname{Im}E),\;\operatorname{Re}z=\operatorname{Re}E.$
(3.3)

It satisfies the symmetry relation
$\overline{h(\bar{z})}=\overline{h(z)}=-h(z)$.

Note that the scattering relations (3.1) can be rewritten in the form

$\displaystyle\dfrac{Y_{\pm}[1](t,x,\lambda)}{a(\lambda)}$
$\displaystyle=Z_{\pm}[1](t,x,\lambda)+r(\lambda)Z_{\pm}[2](t,x,\lambda),\qquad\lambda\in{\mathbb{R}},$
$\displaystyle\dfrac{Y_{\pm}[2](t,x,\lambda)}{a(\lambda)}$
$\displaystyle=r(\lambda)Z_{\pm}[1](t,x,\lambda)+Z_{\pm}[2](t,x,\lambda),\qquad\lambda\in{\mathbb{R}},$

and also, it follows from (3.1) that

$\displaystyle
a(\lambda)=W(Y_{\pm}[1](t,x,\lambda),Z_{\pm}[2](t,x,\lambda))=W(Z_{\pm}[1](t,x,\lambda),Y_{\pm}[2](t,x,\lambda)),$
$\displaystyle
b(\lambda)=W(Z_{\pm}[1](t,x,\lambda),Y_{\pm}[1](t,x,\lambda))=W(Y_{\pm}[2](t,x,\lambda),Z_{\pm}[2](t,x,\lambda)),$

where $W(f,g)=\det(f,g)$ is the Wronskian (Wronski determinant) of the
solutions $f$, $g$. In addition, $h(z)=\dfrac{\mathrm{i}}{a_{-}(z)a_{+}(z)}$,
$z\in(E,\overline{E})$.

Denote

$\theta(t,x,z)=zt-\eta(z)x,$ (3.4)

where $\eta(z)$ is defined by (2.18), and introduce the contour
$\Sigma={\mathbb{R}}\cup(E,\overline{E})$ oriented from the left to the right
for the real line ${\mathbb{R}}$ and downward for the interval
$(E,\overline{E})$ (see Fig. 1).

Figure 1: The oriented contour $\Sigma$

The analysis of the phase function $\theta(t,x,z)$ (3.4) is performed in
Appendix.

Using the scattering relations (3.1), the relations obtained for $a(z)$,
$b(z)$, $r(z)$ and $h(z)$, and Theorem 1, we obtain the following. _The
matrix_

$M(t,x,z)\\!=\\!\begin{cases}\begin{pmatrix}Z^{+}[1](t,x,z)e^{\mathrm{i}\theta(t,x,z)}&\dfrac{Y^{+}[2](t,x,z)}{a(z)}e^{-\mathrm{i}\theta(t,x,z)}\end{pmatrix},\;z\in{\mathbb{C}}_{+}\setminus[E,\operatorname{Re}E),\\\
\begin{pmatrix}\dfrac{Y^{-}[1](t,x,z)}{a(z)}e^{\mathrm{i}\theta(t,x,z)}&Z^{-}[2](t,x,z)e^{-\mathrm{i}\theta(t,x,z)}\end{pmatrix},\;z\in{\mathbb{C}}_{-}\setminus(\operatorname{Re}E,\overline{E}],\end{cases}$
(3.5)

_is a solution of the following basic matrix Riemann-Hilbert problem (the
problem of conjugation of boundary values on a contour $\Sigma$), where the
reflection coefficient $r(z)$ and the function $h(z)$ are uniquely determined
by the initial and boundary conditions (1.6), (1.7):_

Basic problem RH0. Given a contour $\Sigma={\mathbb{R}}\cup(E,\overline{E})$
(see Fig. 1) and functions $r(z)$ (3.2) ($z\in{\mathbb{R}}$) and $h(z)$ (3.3)
($z\in(E,\overline{E})$), find a $2\times 2$ matrix function $M(t,x,z)$
satisfying the following conditions:

  * •

$M(t,x,z)$ is analytic in
$z\in{\mathbb{C}}\setminus(\Sigma\cup\\{E,\overline{E}\\})$ and has the
continuous boundary values $M_{\pm}(t,x,z)$ in
$z\in\Sigma\setminus\\{\operatorname{Re}E\\}$;

  * •

$M(t,x,z)$ is bounded near the point of the contour self-intersection
$\operatorname{Re}E$ and has the singularities of the type $(z-E)^{-1/4}$,
$(z-\overline{E})^{-1/4}$ at the endpoints $E$, $\overline{E}$;

  * •

$M(t,x,z)$ satisfies the boundary condition (which is also called the jump
condition) $M_{-}(t,x,z)=M_{+}(t,x,z)J(t,x,z)$,
$z\in\Sigma\setminus\\{\operatorname{Re}E\\}$, where $J(t,x,z)$ is defined as

$\displaystyle J(t,x,z)=$
$\displaystyle\begin{pmatrix}1&h(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},\;$ $\displaystyle z\in(E,\operatorname{Re}E),$
$\displaystyle=$
$\displaystyle\begin{pmatrix}1-r^{2}(z)e^{2\mathrm{i}\,(\theta_{-}(t,x,z)-\theta_{+}(t,x,z))}&-r(z)e^{-2\mathrm{i}\theta_{+}(t,x,z)}\\\
r(z)e^{2\mathrm{i}\theta_{-}(t,x,z)}&1\end{pmatrix},\;$ $\displaystyle
z=\lambda\in{\mathbb{R}},$ (3.6) $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0\\\
h(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},\;$ $\displaystyle
z\in(\operatorname{Re}E,\overline{E});$

  * •

$M(t,x,z)$ satisfies the normalization condition $M(t,x,z)=I+O(z^{-1})$,
$z\to\infty$.

######  Remark 1.

The jump matrix $J(t,x,z)$ for $z=\lambda\in{\mathbb{R}}$ takes two different
form:

$\displaystyle J(t,x,\lambda)=$
$\displaystyle\begin{pmatrix}1-r^{2}(\lambda)e^{-\pi
n(\lambda)\,x}&-r(\lambda)e^{-2\mathrm{i}\theta_{+}(t,x,\lambda)}\\\
r(\lambda)e^{2\mathrm{i}\theta_{-}(t,x,\lambda)}&1\end{pmatrix},$
$\displaystyle\lambda\in[-\Lambda,\Lambda]\qquad(\Lambda\leq\infty),$
$\displaystyle=$
$\displaystyle\begin{pmatrix}1-r^{2}(\lambda)&-r(\lambda)e^{-2\mathrm{i}\theta(t,x,\lambda)}\\\
r(\lambda)e^{2\mathrm{i}\theta(t,x,\lambda)}&1\end{pmatrix},$
$\displaystyle\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]\qquad(\Lambda<\infty),$

because $\eta_{+}(\lambda)=\eta_{-}(\lambda)$ and
$\theta_{+}(t,x,\lambda)=\theta_{-}(t,x,\lambda)$ for
$\lambda\notin[-\Lambda,\Lambda]$, and
$\theta_{-}(t,x,\lambda)-\theta_{+}(t,x,\lambda)=(\eta_{+}(\lambda)-\eta_{-}(\lambda))x$,
$\eta_{+}(\lambda)-\eta_{-}(\lambda)=\pi\mathrm{i}n(\lambda)/2$ for
$\lambda\in[-\Lambda,\Lambda]$ (recall that for the case when
$\Lambda=+\infty$ we set $[-\Lambda,\Lambda]={\mathbb{R}}$).

##  4 Solvability of the IBVP for the MB equations and the RH problem

######  Theorem 2.

Let the functions ${\mathcal{E}}(t,x)$, $\rho(t,x,\lambda)$ and
$\mathcal{N}(t,x,\lambda)$ be a solution of the IBVP for the MB equations
(1.5) with the initial and boundary conditions (1.6), (1.7) and have the
properties indicated in Theorem 1, and let $a(z)$ and $b(z)$ be the
corresponding spectral functions. Then there exists a unique matrix $M(t,x,z)$
($M=(M_{ij})_{1\leq i,j\leq 2}$) which solves the basic RH problem RH0, and
this matrix defines the complex electric field envelope ${\mathcal{E}}(t,x)$
and the density matrix
$F(t,x,\lambda)=\left(\begin{smallmatrix}\mathcal{N}(t,x,\lambda)&\rho(t,x,\lambda)\\\
\overline{\rho(t,x,\lambda)}&-\mathcal{N}(t,x,\lambda)\end{smallmatrix}\right)$
by the formulas

${\mathcal{E}}(t,x)=-4\mathrm{i}\lim\limits_{z\to\infty}\big{(}zM(t,x,z)\big{)}_{12},$
(4.1) $F(t,x,\lambda)=-M(t,x,\lambda+\mathrm{i}0)e^{-\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(0,x,\lambda+\mathrm{i}0)\sigma_{3}M(0,x,\lambda+\mathrm{i}0)e^{\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(t,x,\lambda+\mathrm{i}0)$ (4.2)

(more precisely, the function $F(t,x,\lambda)$ is defined by the formula (4.2)
for $\lambda\in{\mathbb{R}}\setminus\\{\operatorname{Re}E\\}$ and defined
appropriately at the point $\lambda=\operatorname{Re}E$ so that
$F(t,x,\lambda)$ is continuous in $\lambda\in{\mathbb{R}}$).

###### Proof.

The existence of the solution $M(t,x,z)$ (3.5) of the basic RH problem RH0 was
proved in Section 3. Let us prove the uniqueness of the solution. First we
show that if $M(t,x,z)$ is a solution of this RH problem, then $\det
M(t,x,z)\equiv 1$. It follows from the properties of $M(t,x,z)$ that the
function $\det M(t,x,z)$ is analytic for
$z\in{\mathbb{C}}\setminus\Sigma^{cl}$, where
$\Sigma^{cl}=\Sigma\cup\\{E,\overline{E}\\}$, and has continuous boundary
values on $\Sigma\setminus\\{\operatorname{Re}E\\}$, and $\det
M_{-}(t,x,z)=\det M_{+}(t,x,z)\det J(t,x,z)=\det M_{+}(t,x,z)$,
$z\in\Sigma\setminus\\{\operatorname{Re}E\\}$, since $\det J(t,x,z)\equiv 1$.
Hence, $\det M(t,x,z)$ is analytic for all $z\in{\mathbb{C}}$ except for the
points $E$, $\overline{E}$, $\operatorname{Re}E$ which are removable
singularities by virtue of the properties of $M(t,x,z)$. Thus, after
appropriate modifications (definitions) at these points, $\det M(t,x,z)$ is
analytic for $z\in{\mathbb{C}}$, and since $\det M(t,x,z)=1+O(z^{-1})$ as
${z\to\infty}$ (by virtue of the normalization condition for $M(t,x,z)$), then
by the Liouville theorem $\det M(t,x,z)\equiv 1$. Hence, there exists
$M^{-1}(t,x,z)$ and it is analytic on ${\mathbb{C}}\setminus\Sigma^{cl}$ and
has continuous boundary values on $\Sigma\setminus\\{\operatorname{Re}E\\}$.
Now suppose that there is another solution $\hat{M}(t,x,z)$ of the basic RH
problem RH0. Consider the function $N(t,x,z):=\hat{M}(t,x,z)M^{-1}(t,x,z)$
(cf. [4, p. 189]). Note that
$N_{-}(t,x,z)=\hat{M}_{+}(t,x,z)J(t,x,z)J^{-1}(t,x,z)M_{+}^{-1}(t,x,z)=N_{+}(t,x,z)$,
$z\in\Sigma\setminus\\{\operatorname{Re}E\\}$, and $N(t,x,z)\to I$ as
$z\to\infty$. Using the same arguments as for the proof that $\det
M(t,x,z)\equiv 1$, we obtain that $N(t,x,z)\equiv I$. Consequently,
$\hat{M}(t,x,z)\equiv M(t,x,z)$, which proves the uniqueness of the solution.

Since $\mathcal{E}$, $\rho$ and $\mathcal{N}$ is a solution of the IBVP for
the MB equations, then the matrix function

$\Phi(t,x,z)=M(t,x,z)e^{-\mathrm{i}\theta(t,x,z)\sigma_{3}},$ (4.3)

where $\theta(t,x,z)$ is defined in (3.4), satisfies the $t$-equation (1.8),
namely, $\Phi_{t}=-(\mathrm{i}z\sigma_{3}+H(t,x))\Phi$,
$z\in{\mathbb{C}}\setminus\Sigma^{cl}$, and hence the matrix $M(t,x,z)$
satisfies the equation $M_{t}=-\mathrm{i}z[\sigma_{3},M]-HM$. Taking into
account that $M(t,x,z)=I+O(z^{-1})$ as $z\to\infty$, we can represent
$M(t,x,z)$ as $M(t,x,z)=I+\dfrac{m(t,x)}{z}+o(z^{-1})$, $z\to\infty$, where
$m(t,x)=\lim\limits_{z\to\infty}z(M(t,x,z)-I)$. Substituting this expansion
into the equation, we obtain that the equality
$H(t,x)+\mathrm{i}[\sigma_{3},m(t,x)]=0$ holds. Hence,
${\mathcal{E}}(t,x)=-4\mathrm{i}m_{12}(t,x)=-4\mathrm{i}\lim\limits_{z\to\infty}\big{(}zM(t,x,z)\big{)}_{12}$
(as above, the subindex denotes $(1,2)$-entry of the corresponding matrix).

Further, since ${\mathcal{E}}$, $\rho$, $\mathcal{N}$ is a solution of the MB
equations (1.5), then the equation (1.11) is satisfied, i.e.,
$F_{t}+[\mathrm{i}\lambda\sigma_{3}+H,F]=0$. This equation can be represented
in the form

$F_{t}=[U,F],$ (4.4)

where $U(t,x,\lambda)$ is the matrix function (2.4) (note that since
$U^{*}=-U$, the equation (4.4) can be represented as $F_{t}=UF+FU^{*}$). The
initial condition (1.6) for the MB equations implies
$F(0,x,\lambda)\equiv-\sigma_{3}$. Furthermore, for $z=\lambda+\mathrm{i}0$
($\lambda\neq\operatorname{Re}E$) the matrix function $\Phi(t,x,z)$ (4.3)
satisfies the $t$-equation (1.8), which can written as (2.1), i.e.,
${W_{t}=UW}$, with the same matrix function $U(t,x,\lambda)$, and
$\det\Phi(t,x,z)=\det M(t,x,z)\equiv 1$. Consequently, the matrix function
(the Cauchy matrix)
$W(t,x,\lambda)=\Phi(t,x,\lambda+\mathrm{i}0)\Phi^{-1}(0,x,\lambda+\mathrm{i}0)$
is a solution of the $t$-equation (2.1) (where $U$ has the form (2.4)) and
satisfies the initial condition $W(0,x,\lambda)\equiv I$. Then the function
$F(t,x,\lambda)=-W(t,x,\lambda)\sigma_{3}W^{-1}(t,x,\lambda)$ is a unique
solution of the initial value problem (4.4),
$F(0,x,\lambda)\equiv-\sigma_{3}$. Since
$W(t,x,\lambda)=M(t,x,\lambda+\mathrm{i}0)e^{-\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(0,x,\lambda+\mathrm{i}0)$, then $F(t,x,\lambda)$ has the
form (4.2). Due to the properties of $\Phi(t,x,z)$, the function
$F(t,x,\lambda)$ is continuous in
$\lambda\in{\mathbb{R}}\setminus\\{\operatorname{Re}E\\}$ and bounded in the
neighborhood of $\lambda=\operatorname{Re}E$. Thus, we can define it
appropriately at this point so that $F(t,x,\lambda)$ becomes continuous in
$\lambda$ on ${\mathbb{R}}$. Namely, define the matrix $M(t,x,z)$ (3.5) for
$z=\operatorname{Re}E+\mathrm{i}0$ as
$M(t,x,\operatorname{Re}E+\mathrm{i}0)=M_{+}(t,x,\operatorname{Re}E):=\begin{pmatrix}Z^{+}[1](t,x,\operatorname{Re}E+\mathrm{i}0)e^{\mathrm{i}\theta_{+}(t,x,\operatorname{Re}E)}&\dfrac{Y^{+}[2](t,x,\operatorname{Re}E+\mathrm{i}0)}{a(\operatorname{Re}E)}e^{-\mathrm{i}\theta_{+}(t,x,\operatorname{Re}E)}\end{pmatrix}$,
where
$Z^{+}[1](t,x,\operatorname{Re}E+\mathrm{i}0)=Z_{+}[1](t,x,\operatorname{Re}E)$
and
$Y^{+}[2](t,x,\operatorname{Re}E+\mathrm{i}0)=Y_{+}[2](t,x,\operatorname{Re}E)$
($Z_{+}(t,x,\lambda)$ (2.23) and $Y_{+}(t,x,\lambda)$ (2.21),
$\lambda\in{\mathbb{R}}$, are the solution of the $t$\- and
$x^{\pm}$-equations), and
$a(\operatorname{Re}E)=(\varkappa(\operatorname{Re}E)+\varkappa^{-1}(\operatorname{Re}E))/2$,
$\varkappa(\operatorname{Re}E)=e^{\mathrm{i}\pi/4}$ (in this case
$r(\operatorname{Re}E)=\mathrm{i}$, and $r(\lambda)$ is infinitely
differentiable as a function of the real variable $\lambda$). Then
$F(t,x,\lambda)$ defined by (4.2) is continuous in $\lambda\in{\mathbb{R}}$. ∎

######  Theorem 3.

For $t\leq x$ and $t,x\geq 0$, a solution of the IBVP (1.5)–(1.7) is trivial,
i.e., ${\mathcal{E}(t,x)\equiv 0}$, $\rho(t,x,\lambda)\equiv 0$, and
$\mathcal{N}(t,x,\lambda)\equiv-1$.

###### Proof.

First, note that

$\mathop{\mathrm{sgn}}[\operatorname{Re}(\mathrm{i}\theta(t,x,z))]=\mathop{\mathrm{sgn}}\operatorname{Im}z$

for $0\leq t<x$ and $t=x$, $t,x>0$, and $\theta(0,0,z)\equiv 0$ (see
Appendix). The analyticity of $r(z)$ outside the interval $[E,\overline{E}]$
allows one to apply the following transformation:
$M^{(1)}(t,x,z)=M(t,x,z)G^{(1)}(t,x,z)$, where

$G^{(1)}(t,x,z)=\begin{cases}\begin{pmatrix}1&-r(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},\quad
z\in{\mathbb{C}}_{+}\setminus[E,\operatorname{Re}{E}),\\\
\begin{pmatrix}1&0\\\ -r(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},\quad
z\in{\mathbb{C}}_{-}\setminus(\operatorname{Re}{E},\overline{E}].\end{cases}$
(4.5)

Then we obtain the equivalent RH problem

$M^{(1)}_{-}(t,x,z)=M^{(1)}_{+}(t,x,z)J^{(1)}(t,x,z),\qquad
z\in\Sigma\setminus\\{\operatorname{Re}E\\},$

with the jump matrix

$J^{(1)}(t,x,z)=(G^{(1)}_{+}(t,x,z))^{-1}J(t,x,z)G^{(1)}_{-}(t,x,z),\qquad
z\in\Sigma,$

where $G^{(1)}_{\pm}(t,x,z)$ are nontangential boundary values of
$G^{(1)}(t,x,z)$ on the left and right of the contour $\Sigma$ respectively.
Due to the factorization
$J(t,x,z)=G^{(1)}_{+}(t,x,z)(G^{(1)}_{-}(t,x,z))^{-1}$ of the jump matrix (• ‣
3) and the jump relations $r_{-}(z)-r_{+}(z)=h(z)$ for $z\in(E,\overline{E})$,
the jump matrix $J^{(1)}(t,x,z)\equiv I$ for $z\in\Sigma$. Since
$G^{(1)}(t,x,z)=I+O(z^{-1})$ as $z\to\infty$, $t\leq x$, and the points $E$,
$\operatorname{Re}E$, $\overline{E}$ are removable singularities of
$M^{(1)}(t,x,z)$, then $M^{(1)}(t,x,z)\equiv I$ in view of the Liouville
theorem. Hence,
$m^{(1)}(t,x)=\lim\limits_{z\to\infty}z(M^{(1)}(t,x,z)-I)\equiv 0$. Since
$(G^{(1)}(t,x,z))^{-1}\to I$ as $z\to\infty$, then

${\mathcal{E}}(t,x)=-4\mathrm{i}\lim\limits_{z\to\infty}\big{(}zM(t,x,z)\big{)}_{12}=-4\mathrm{i}\lim\limits_{z\to\infty}\left(zM^{(1)}(t,x,z)(G^{(1)}(t,x,z))^{-1}\right)_{12}\equiv
0.$

Since $M(t,x,\lambda+\mathrm{i}0)\equiv I$, then
$F(t,x,\lambda)\equiv-\sigma_{3}$ (see (4.2)) and hence
${\rho(t,x,\lambda)\equiv 0}$ and ${\mathcal{N}(t,x,\lambda)\equiv-1}$. ∎

Note that, since by Theorem 3 in the “causality region” $t\leq x$ ($t,x\geq
0$) the solution of the IBVP (1.5)–(1.7) is trivial, the basic RH problem RH0
associated with the IBVP (1.5)–(1.7) provides the causality principle.

Below, we will prove that the basic problem RH0 has a unique solution
$M(t,x,z)$, infinitely differentiable in $t$, $x$, and that the matrix
$M(t,x,z)$ generates solutions of the AKNS systems (1.8), (1.12) and, as a
result, a solution of the MB equations, and this solution satisfied the
initial and boundary conditions. Also, an integral representation of the
solution of the MB equations through the solution of a singular integral
equation will be given.

######  Theorem 4.

For each $t\in{\mathbb{R}}_{+}$ and each $x\in(0,L)$, $L\leq\infty$, the basic
problem RH0 has a unique solution $M(t,x,z)$, and this solution is continuous
in $(t,x)\in{\mathbb{R}}_{+}\times(0,L)$.

###### Proof.

Let $t\in{\mathbb{R}}_{+}$ and $x\in(0,L)$ be an arbitrary fixed numbers. The
boundary condition $M_{-}(t,x,z)=M_{+}(t,x,z)J(t,x,z)$ on
$\Sigma\setminus\\{\operatorname{Re}E\\}$ can be rewritten in the form of the
jump condition $M_{+}(t,x,z)-M_{-}(t,x,z)=M_{+}(t,x,z)[I-J(t,x,z)]$,
$z\in\Sigma\setminus\\{\operatorname{Re}E\\}$. Since $M(\infty)=I$, then a
solution of the basic problem RH0, if it exists, has the form

$M(t,x,z)=\frac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\frac{M_{+}(t,x,s)[I-J(t,x,s)]}{s-z}\textup{d}s+I,\qquad
z\in{\mathbb{C}}\setminus\Sigma.$ (4.6)

Passing to the limit in the equality (4.6) as $z\to\zeta_{+}$, where
${\zeta_{+}=\lim\limits_{z\to\zeta,\,z\in+\,side}z}$, ${\zeta\in\Sigma}$ and
“$+$ side” denotes a positive side (a boundary value on the left) of the
oriented contour $\Sigma$ (we exclude the points of self-intersection of the
contour), and carrying out elementary transformations, we obtain the following
singular integral equation for $M_{+}(t,x,\zeta)$:

$M_{+}(t,x,\zeta)-I=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\dfrac{[M_{+}(t,x,s)-I][I-J(t,x,s)]}{s-\zeta_{+}}\textup{d}s+\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\dfrac{I-J(t,x,s)}{s-\zeta_{+}}\textup{d}s,\quad\zeta\in\Sigma.$

Denote $W(t,x,\zeta):=M_{+}(t,x,\zeta)-I$ and introduce the singular integral
operator $\mathbf{K}$ and operator function $\Pi$ defined by

$\mathbf{K}[W](t,x,\zeta):=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\dfrac{W(t,x,s)[I-J(t,x,s)]}{s-\zeta_{+}}\textup{d}s,$
$\Pi(t,x,\zeta):=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\dfrac{I-J(t,x,s)}{s-\zeta_{+}}\textup{d}s.$

We will prove that for any fixed $t$, $x$ there exists a unique solution
$W(t,x,\zeta)$ of the singular integral equation (cf. [14])

$W(t,x,\zeta)-\mathbf{K}[W](t,x,\zeta)=\Pi(t,x,\zeta),\quad\zeta\in\Sigma,$
(4.7)

which is considered in the space of operator ($2\times 2$ matrix) functions
$W(t,x,\zeta)=:W(\zeta)\in L^{2}(\Sigma)$. The operator $\displaystyle
C_{\pm}[f](\zeta):=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\dfrac{f(s)}{s-\zeta_{\pm}}\textup{d}s\in
B(L^{2}(\Sigma))$, where $B(L^{2}(\Sigma))$ is the (Banach) space of bounded
linear operators from $L^{2}(\Sigma)$ into $L^{2}(\Sigma)$, since $\Sigma$ is
a Carleson curve (see, e.g., [23] and references therein, or [24]). The
function $I-J(t,x,s)$ belongs to $L^{2}(\Sigma)$ with respect to $s$. Thus,
$\Pi(t,x,\zeta)\in L^{2}(\Sigma)$ as a function of the variable $\zeta$. Note
that $I-J(t,x,s)\in L^{\infty}(\Sigma)$ in $s$. Since
$\operatorname{Re}J(t,x,z)$ is positive definite for $z\in{\mathbb{R}}$, $\det
J(t,x,z)\equiv 1$, the contour $\Sigma$ is symmetric with respect to the real
axis ${\mathbb{R}}$ and $J^{-1}(t,x,z)=J^{*}(t,x,\bar{z})$ for
$z\in\Sigma\setminus{\mathbb{R}}$ (thus, the contour $\Sigma$ and the matrix
$J^{-1}(t,x,z)$ satisfy the Schwartz reflection principle), then it follows
from [22, Theorem 9.3, p. 984] that there exists the operator
$(I-\mathbf{K})^{-1}\in B(L^{2}(\Sigma))$. Consequently, the singular integral
equation (4.7) has a unique solution $W(t,x,\zeta)$ for any fixed $t$ and $x$,
which belongs to $L^{2}(\Sigma)$ with respect to $\zeta$. Therefore, there
exists the solution $M(t,x,z)$ of the basic problem RH0, which can be obtained
by the formula

$M(t,x,z)=\frac{1}{2\pi\mathrm{i}}\int\limits_{\Sigma}\frac{[W(t,x,s)+I][I-J(t,x,s)]}{s-z}\textup{d}s+I,\qquad
z\in{\mathbb{C}}\setminus\Sigma.$ (4.8)

Since the operator $I-\mathbf{K}\in B(L^{2}(\Sigma))$ and the operator
function $\Pi\in L^{2}(\Sigma)$ depends continuously on the parameters
$(t,x)\in{\mathbb{R}}_{+}\times(0,L)$, then the inverse operator
$(I-\mathbf{K})^{-1}$ and the solution $W(t,x,\zeta)$ of the equation (4.7)
also depend continuously on $(t,x)$. Taking into account the form (4.8) of the
solution $M(t,x,z)$, we obtain that $M(t,x,z)$ is continuous in
$(t,x)\in{\mathbb{R}}_{+}\times(0,L)$.

The proof of the uniqueness of the solution $M(t,x,z)$ is carried out in the
same way as for Theorem 2. ∎

######  Theorem 5.

For any $(t,x)\in{\mathbb{R}}_{+}\times(0,L)$, $L\leq\infty$, a solution of
the basic problem RH0 is infinitely differentiable in $t$ and $x$.

###### Proof.

By Theorem 4, the basic problem RH0 has a unique solution $M(t,x,z)$ which can
be obtained by the formula (4.8).

In order to the solution $M(t,x,z)$, as well as the solution $W(t,x,\zeta)$ of
the singular integral equation (4.7), are differentiable in $t$ and $x$, it is
necessary that the integrals that arise when differentiating the solution
$M(t,x,z)$ and equation (4.7) are convergent. The matrix $J(t,x,z)$ and its
derivatives in $t$ and $x$ are responsible for the decrease of the integrands.

On unbounded parts of the contour $\Sigma$, i.e., for $z\in{\mathbb{R}}$ with
$|z|>>1$ ($|z|$ is sufficiently large), the jump matrix has the form (see
Remark 1)

$J(t,x,z)=\begin{pmatrix}1-r^{2}(z)&-r(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
r(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix}$

if $\Lambda<\infty$ ($z\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$), and

$J(t,x,z)=\begin{pmatrix}1-r^{2}(z)e^{-\pi
n(z)\,x}&-r(z)e^{-2\mathrm{i}\theta_{+}(t,x,z)}\\\
r(z)e^{2\mathrm{i}\theta_{-}(t,x,z)}&1\end{pmatrix}$

if $\Lambda=\infty$ ($z\in{\mathbb{R}}$).

Since $r(z)=O(z^{-1})$ and $n(z)\to 0$ as $z\to\pm\infty$ ($z\in{\mathbb{R}}$)
and $\operatorname{Re}(2\mathrm{i}\theta_{\pm}(t,x,z))=\pm\pi n(z)x/2$
($z\in{\mathbb{R}}$), then $J(t,x,z)=I+O(z^{-1})$ as $z\to\pm\infty$ for
$\Lambda\leq\infty$.

Consider the case when $t\leq x$. In this case
$M(t,x,z)=(G^{(1)}(t,x,z))^{-1}$, where $G^{(1)}$ is defined in (4.5) (see the
proof of Theorem 3), and since $\theta(t,x,z)$ is infinitely differentiable in
the parameters $t$ and $x$, then $M(t,x,z)$ is also infinitely differentiable
in the parameters.

Now, consider the case when $t>x$.

Consider $\Lambda<\infty$. Choose any numbers
$\lambda_{1},\,\lambda_{2}\in{\mathbb{R}}$ with $|\lambda_{i}|$ sufficiently
large and such that $\lambda_{1}<-\Lambda$, $\lambda_{1}<\operatorname{Re}E$,
$\lambda_{2}>\Lambda$, and perform the following “$\delta$-transformation”:

$M^{(1)}(t,x,z)=M(t,x,z)\delta^{\sigma_{3}}(z),\qquad\delta^{\sigma_{3}}(z)=\begin{pmatrix}\delta(z)&0\\\
0&\delta^{-1}(z)\end{pmatrix},$

where $\delta(z)$ is a solution of the following problem of conjugation of
boundary values (RH problem):

  * •

$\delta(z)$ is analytic in
${\mathbb{C}}\setminus\big{(}(-\infty,\lambda_{1}]\cup[\lambda_{2},+\infty)\big{)}$.

  * •

$\delta_{+}(z)=\delta_{-}(z)(1-r^{2}(z))$

on $(-\infty,\lambda_{1}]\cup[\lambda_{2},+\infty)$.

  * •

$\delta(z)\to 1$ as $z\to\infty$.

Note that $r^{2}(\lambda)<0$ for $\lambda\in{\mathbb{R}}$, since
$\operatorname{Im}w(\lambda)=0$ on ${\mathbb{R}}$ and $\operatorname{Im}E>0$.
A unique solution of the problem given above is the function

$\delta(z)=\exp\left\\{\frac{1}{2\pi\mathrm{i}}\int\limits_{s\in(-\infty,\lambda_{1}]\cup[\lambda_{2},+\infty)}\frac{\ln(1-r^{2}(s))ds}{s-z}\right\\},\quad
z\in{\mathbb{C}}\setminus\big{(}(-\infty,\lambda_{1}]\cup[\lambda_{2},+\infty)\big{)}.$

For $M^{(1)}(t,x,z)$ the boundary condition has the form
$M^{(1)}_{-}(t,x,z)=M^{(1)}_{+}(t,x,z)J^{(1)}(t,x,z)$ with
$J^{(1)}(t,x,z)=\delta_{+}^{-\sigma_{3}}(z)J(t,x,z)\delta_{-}^{\sigma_{3}}(z)$
which after applying factorizations takes the form

$\displaystyle J^{(1)}(t,x,z)=$ $\displaystyle\begin{pmatrix}1&0\\\
\dfrac{r(z)}{1-r^{2}(z)}\delta_{+}^{2}(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix}\\!\\!\begin{pmatrix}1&-\dfrac{r(z)}{1-r^{2}(z)}\delta_{-}^{-2}(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
z\in(-\infty,\lambda_{1}]\cup[\lambda_{2},+\infty),$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1-r^{2}(z)e^{2\mathrm{i}(\theta_{-}(t,x,z)-\theta_{+}(t,x,z))}&-r(z)\delta^{-2}(z)e^{-2\mathrm{i}\theta_{+}(t,x,z)}\\\
r(z)\delta^{2}(z)e^{2\mathrm{i}\theta_{-}(t,x,z)}&1\end{pmatrix},\;z\in(\lambda_{1},\lambda_{2}),$
$\displaystyle=$
$\displaystyle\begin{pmatrix}1&h(z)\delta^{-2}(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},\qquad\qquad\qquad\qquad\qquad\qquad\quad
z\in(E,\operatorname{Re}E),$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0\\\
h(z)\delta^{2}(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},\qquad\qquad\qquad\qquad\qquad\qquad\qquad
z\in(\operatorname{Re}E,\overline{E}).$ (4.9)

We introduce the smooth curves $L_{1}$, $L_{3}$ and $\overline{L_{1}}$,
$\overline{L_{3}}$, symmetric to $L_{1}$, $L_{3}$ with respect to the real
axis, the additional jump contour $\widetilde{\Sigma}=L_{1}\cup
L_{3}\cup\overline{L_{1}}\cup\overline{L_{3}}$ (the points $\lambda_{1}$,
$\lambda_{2}$ do not belong to $\widetilde{\Sigma}$), the domains $D_{i}$ and
the domains $\overline{D_{i}}$, symmetric to $D_{i}$ with respect to the real
axis, $i=1,2,3$, which are displayed in Fig. 2. Thus, the complex $z$-plane is
decomposed into the parts
${\mathbb{C}}=\widetilde{\Sigma}\cup\Sigma\cup(\cup_{i=1}^{3}D_{i})\cup(\cup_{i=1}^{3}\overline{D_{i}})$.

Figure 2: The contours $\Sigma={\mathbb{R}}\cup(E,\overline{E})$ and
$\widetilde{\Sigma}=L_{1}\cup L_{3}\cup\overline{L_{1}}\cup\overline{L_{3}}$
and the domains $D_{i}$, $\overline{D_{i}}$, $i=1,2,3$

Further, we transform the matrix $M^{(1)}(t,x,z)$ into

$M^{(2)}(t,x,z)=M^{(1)}(t,x,z)G^{(2)}(t,x,z),$

where

$\displaystyle G^{(2)}(t,x,z)=$ $\displaystyle\begin{pmatrix}1&0\\\
\dfrac{r(z)}{1-r^{2}(z)}\delta^{2}(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},$
$\displaystyle z\in D_{1}\cup D_{3},$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&\dfrac{r(z)}{1-r^{2}(z)}\delta^{-2}(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},$ $\displaystyle z\in\overline{D_{1}}\cup\overline{D_{3}},$
$\displaystyle=$ $\displaystyle I,$ $\displaystyle z\in
D_{2}\cup\overline{D_{2}}.$ (4.10)

Since $e^{2\mathrm{i}\theta(t,x,z)}\to 0$ and
$e^{-2\mathrm{i}\theta(t,x,z)}\to 0$ as $z\to\infty$ for
$\operatorname{Im}z>0$ and $\operatorname{Im}z<0$ respectively (see the case
$t>x$, $\Lambda<\infty$ in Appendix and Fig. 5), then $G^{(2)}(t,x,z)\to I$
and hence $M^{(2)}(t,x,z)\to I$ as $z\to\infty$. For $M^{(2)}(t,x,z)$ the jump
(boundary) condition has the form
$M^{(2)}_{-}(t,x,z)=M^{(2)}_{+}(t,x,z)J^{(2)}(t,x,z)$,
${z\in\Sigma\cup\widetilde{\Sigma}\setminus\\{\operatorname{Re}E,\lambda_{1},\lambda_{2}\\}}$,
with the jump matrix
${J^{(2)}(t,x,z)=(G_{+}^{(2)}(t,x,z))^{-1}J^{(1)}(t,x,z)G_{-}^{(2)}(t,x,z)}$
defined on the new contour $\Sigma\cup\widetilde{\Sigma}$ whose orientation
remains the same on $\Sigma={\mathbb{R}}\cup(E,\overline{E})$, and all
branches of $\widetilde{\Sigma}$ (i.e., $L_{i}$, $\overline{L_{i}}$, $i=1,3$)
are oriented from the left to the right. It is easy to verify that

$\displaystyle J^{(2)}(t,x,z)$ $\displaystyle=I,\qquad$ $\displaystyle
z\in(-\infty,\lambda_{1})\cup(\lambda_{2},+\infty),$
$\displaystyle=\begin{pmatrix}1&0\\\
\dfrac{r(z)}{1-r^{2}(z)}\delta^{2}(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},\qquad$
$\displaystyle z\in L_{1}\cup L_{3},\vskip 3.0pt plus 1.0pt minus 1.0pt$
$\displaystyle=\begin{pmatrix}1&-\dfrac{r(z)}{1-r^{2}(z)}\delta^{-2}(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},\qquad$ $\displaystyle
z\in\overline{L_{1}}\cup\overline{L_{3}},$ $\displaystyle=$ $\displaystyle
J^{(1)}(t,x,z),\qquad$ $\displaystyle
z\in(\lambda_{1},\lambda_{2})\cup(E,\overline{E})\setminus\\{\operatorname{Re}E\\}.$
(4.11)

Introduce the contour
$\Upsilon=\widetilde{\Sigma}\cup[\lambda_{1},\lambda_{2}]\cup[E,\overline{E}]=L_{1}\cup
L_{3}\cup\overline{L_{1}}\cup\overline{L_{3}}\cup[\lambda_{1},\lambda_{2}]\cup[E,\overline{E}]$.
As in the proof of Theorem 4, we obtain that for any fixed $t$, $x$ there
exists a unique solution $W^{(2)}(t,x,\zeta)$ of the singular integral
equation (the equation is considered in the space of operator functions
$W^{(2)}(t,x,\zeta)\in L^{2}(\Upsilon)$ with respect to $\zeta$;
$W^{(2)}(t,x,\zeta):=M^{(2)}_{+}(t,x,\zeta)-I$)

$W^{(2)}(t,x,\zeta)-\mathbf{K}^{(2)}[W^{(2)}](t,x,\zeta)=\Pi^{(2)}(t,x,\zeta),\quad\zeta\in\Upsilon,$
(4.12)

where

$\displaystyle\mathbf{K}^{(2)}[W^{(2)}](t,x,\zeta):=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Upsilon}\dfrac{W^{(2)}(t,x,s)[I-J^{(2)}(t,x,s)]}{s-\zeta_{+}}\textup{d}s,$
$\displaystyle\Pi^{(2)}(t,x,\zeta):=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Upsilon}\dfrac{I-J^{(2)}(t,x,s)}{s-\zeta_{+}}\textup{d}s,$

and then we find $M^{(2)}(t,x,z)$ by the formula

$M^{(2)}(t,x,z)=I+\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Upsilon}\frac{[W^{(2)}(t,x,s)+I][I-J^{(2)}(t,x,s)]}{s-z}\textup{d}s,\qquad
z\in{\mathbb{C}}\setminus\Upsilon.$ (4.13)

Since $e^{\pm 2\mathrm{i}\theta(t,x,z)}\to 0$ as $z\to\infty$,
$z\in{\mathbb{C}}_{\pm}$, respectively (see the case $t>x$, $\Lambda<\infty$
in Appendix), then for sufficiently large $z\in\widetilde{\Sigma}$ (i.e., on
the parts of the jump contour $\Upsilon$ contained in a neighborhood of
infinity), the matrix $I-J^{(2)}(t,x,z)$ decreases exponentially and its
derivatives of any order in $t$ and $x$ decrease fast enough so that the
integrals that arise when differentiating are convergent. Consequently, the
solution $W^{(2)}(t,x,\zeta)$ of the equation (4.12) and the function
$M^{(2)}(t,x,z)$ (4.13) are infinitely differentiable in $t$ and $x$. Since
$M^{(2)}(t,x,z)$ and $G^{(2)}(t,x,z)$ are infinitely differentiable in $t$,
$x$, then $M^{(1)}(t,x,z)$ and $M(t,x,z)$ also have this property.

Consider $\Lambda=\infty$ (as before, $t>x$).

We introduce the smooth curves $L_{2}=L_{2}(\xi)$ and
$\overline{L_{2}}=\overline{L_{2}}(\xi)$ and the domains
$D_{2}=D_{2}(\xi)\subset{\mathbb{C}}_{+}\setminus[E,\operatorname{Re}{E})$,
$D_{4}=D_{4}(\xi)\subset{\mathbb{C}}_{+}$ and
$\overline{D_{2}}=\overline{D_{2}}(\xi)\subset{\mathbb{C}}_{-}\setminus(\operatorname{Re}{E},\overline{E}]$,
$\overline{D_{4}}=\overline{D_{4}}(\xi)\subset{\mathbb{C}}_{-}$, symmetric
with respect to ${\mathbb{R}}$, where $\xi=\dfrac{x}{4(t-x)}>0$ (since $t>x$),
which are such that all sufficiently large $z$ belong to $L_{2}(\xi)$ and
$\overline{L_{2}}(\xi)$ (i.e., the parts of $L_{2}$ and $\overline{L_{2}}$
contained in a neighborhood of infinity) contains in the domain bounded (from
above and below) by the curves $\gamma^{\pm}_{\xi}$ defined by the equality
$\int\limits_{-\infty}^{\infty}\dfrac{n(s)\textup{d}s}{|s-z|^{2}}=\dfrac{1}{\xi}$
($\operatorname{Re}(\mathrm{i}\theta)=0$ on $\gamma^{\pm}_{\xi}$; see the case
$t>x$, $\Lambda=\infty$ in Appendix and Fig. 6), the interval
$[E,\overline{E}]$ lies in the domain bounded by $L_{2}$ and
$\overline{L_{2}}$, the domains $D_{2}$ and $\overline{D_{2}}$ are bounded by
$L_{2}$, ${\mathbb{R}}\cup[E,\operatorname{Re}{E})$ and $\overline{L_{2}}$,
${\mathbb{R}}\cup(\operatorname{Re}{E},\overline{E}]$ respectively (see Fig.
3, 4). Then complex $z$-plane is decomposed into the parts
${\mathbb{C}}=L_{2}\cup\overline{L_{2}}(\xi)\cup\Sigma\cup D_{2}\cup
D_{4}\cup\overline{D_{2}}\cup\overline{D_{4}}$. Define the contour
$\widetilde{\Sigma}=L_{2}\cup\overline{L_{2}}$, where $L_{2}$ and
$\overline{L_{2}}$ are oriented from the left to the right. The orientation on
$\Sigma$ remains the same as above.

Figure 3: The contours $\Sigma={\mathbb{R}}\cup(E,\overline{E})$ and
$\widetilde{\Sigma}=L_{2}\cup\overline{L_{2}}$, the curves
$\gamma^{\pm}_{\xi}$ and the domains $D_{i}$, $\overline{D_{i}}$, $i=2,4$, in
the case when the interval $[E,\overline{E}]$ lies inside the domain bounded
by the curves $\gamma^{+}_{\xi}$, $\gamma^{-}_{\xi}$ Figure 4: The contours
$\Sigma={\mathbb{R}}\cup(E,\overline{E})$ and
$\widetilde{\Sigma}=L_{2}\cup\overline{L_{2}}$, the curves
$\gamma^{\pm}_{\xi}$ and the domains $D_{i}$, $\overline{D_{i}}$, $i=2,4$, in
the case when the interval $[E,\overline{E}]$ intersects $\gamma^{+}_{\xi}$,
$\gamma^{-}_{\xi}$

Further, we transform the matrix $M(t,x,z)$ into

$M^{(2)}(t,x,z)=M(t,x,z)G^{(2)}(t,x,z),$

where

$\displaystyle G^{(2)}(t,x,z)=$
$\displaystyle\begin{pmatrix}1&-r(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},$ $\displaystyle z\in D_{2},$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0\\\
-r(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},$ $\displaystyle
z\in\overline{D_{2}},$ $\displaystyle=$ $\displaystyle I,$ $\displaystyle z\in
D_{4}\cup\overline{D_{4}}.$ (4.14)

Then ${M^{(2)}_{-}(t,x,z)=M^{(2)}_{+}(t,x,z)J^{(2)}(t,x,z)}$,
$z\in\widetilde{\Sigma}\cup\Sigma\setminus\\{\operatorname{Re}E\\}$, where
$J^{(2)}(t,x,z)=(G_{+}^{(2)}(t,x,z))^{-1}J(t,x,z)G_{-}^{(2)}(t,x,z)$,
$z\in\widetilde{\Sigma}\cup\Sigma$, has the following form:

$\displaystyle J^{(2)}(t,x,z)$ $\displaystyle=I,\qquad$ $\displaystyle
z\in\Sigma,$
$\displaystyle=\begin{pmatrix}1&-r(z)e^{-2\mathrm{i}\theta(t,x,z)}\\\
0&1\end{pmatrix},$ $\displaystyle z\in L_{2},\vskip 3.0pt plus 1.0pt minus
1.0pt$ $\displaystyle=\begin{pmatrix}1&0\\\
r(z)e^{2\mathrm{i}\theta(t,x,z)}&1\end{pmatrix},$ $\displaystyle
z\in\overline{L_{2}}.$ (4.15)

As above, we consider the singular integral equation (4.12), where
$\Upsilon=\widetilde{\Sigma}=L_{2}\cup\overline{L_{2}}$, and find
$M^{(2)}(t,x,z)$ ($G^{(2)}(t,x,z)\to I$ and $M^{(2)}(t,x,z)\to I$ as
$z\to\infty$) by the formula (4.13). Since $e^{-2\mathrm{i}\theta(t,x,z)}\to
0$ and $e^{2\mathrm{i}\theta(t,x,z)}\to 0$ as $z\to\infty$ for $z$ from the
domains bounded by ${\mathbb{R}}$, $\gamma^{+}_{\xi}$ and ${\mathbb{R}}$,
$\gamma^{-}_{\xi}$ respectively (see Fig. 6 in Appendix), then for
sufficiently large $z\in\widetilde{\Sigma}$ (i.e., on the parts of the jump
contour $\Upsilon$ contained in a neighborhood of infinity), the matrix
$I-J^{(2)}(t,x,z)$ decreases exponentially and its derivatives of any order in
$t$ and $x$ decrease fast enough. Consequently, the solution
$W^{(2)}(t,x,\zeta)$ of the singular integral equation, the function
$M^{(2)}(t,x,z)$ and, hence, the function $M(t,x,z)$ are infinitely
differentiable in $t$ and $x$. ∎

######  Theorem 6.

Let

$\Phi(t,x,z):=M(t,x,z)e^{-\mathrm{i}\theta(t,x,z)\sigma_{3}},$ (4.16)

where $M(t,x,z)$ is a solution of the basic problem RH0. Then $\Phi(t,x,z)$ is
a unique solution of the AKNS system

$\displaystyle\Phi_{t}=-\big{(}\mathrm{i}z\sigma_{3}+H(t,x)\big{)}\Phi,$
(4.17)
$\displaystyle\Phi_{x}=\big{(}\mathrm{i}z\sigma_{3}+H(t,x)-\mathrm{i}G(t,x,z)\big{)}\Phi,\qquad
G(t,x,z)=\dfrac{1}{4}\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)\textup{d}s}{s-z},$
(4.18)

where $t\in{\mathbb{R}}_{+}$, $x\in(0,L)$, $L\leq\infty$, and
$z\in{\mathbb{C}}\setminus\Sigma^{cl}$,
$\Sigma^{cl}=\Sigma\cup\\{E,\overline{E}\\}$. In the system (4.17), (4.18),
the function $H(t,x)$ is defined by

$H(t,x)=\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}(t,x)\\\
-\overline{\mathcal{E}(t,x)}&0\end{pmatrix}=-\mathrm{i}[\sigma_{3},m(t,x)],$
(4.19)

where

$m(t,x)=\lim\limits_{z\to\infty}z(M(t,x,z)-I)\qquad({\mathcal{E}}(t,x)=-4\mathrm{i}\lim\limits_{z\to\infty}\big{(}zM(t,x,z)\big{)}_{12}),$
(4.20)

the function $F(t,x,\lambda)$ is defined by

$F(t,x,\lambda)=\begin{pmatrix}{\mathcal{N}}(t,x,\lambda)&\rho(t,x,\lambda)\\\
\overline{\rho(t,x,\lambda)}&-{\mathcal{N}}(t,x,\lambda)\end{pmatrix}=\\\
=-M(t,x,\lambda+\mathrm{i}0)e^{-\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(0,x,\lambda+\mathrm{i}0)\sigma_{3}M(0,x,\lambda+\mathrm{i}0)e^{\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(t,x,\lambda+\mathrm{i}0),\quad\lambda\in{\mathbb{R}},$
(4.21)

where $M(t,x,\lambda+\mathrm{i}0)=M_{+}(t,x,\lambda)$ is defined appropriately
at the point $\lambda=\operatorname{Re}E$ so that $F(t,x,\lambda)$ is
continuous in $\lambda\in{\mathbb{R}}$, and these functions satisfy the
Maxwell-Bloch equations in the matrix form

$\displaystyle
H_{t}(t,x)+H_{x}(t,x)-\dfrac{1}{4}\int\limits_{-\infty}^{\infty}[\sigma_{3},F(t,x,s)]n(s)ds=0,$
(4.22) $\displaystyle
F_{t}(t,x,\lambda)+[\mathrm{i}\lambda\sigma_{3}+H(t,x),F(t,x,\lambda)]=0,$
(4.23)

where $t\in{\mathbb{R}}_{+}$, $x\in(0,L)$, $\lambda\in{\mathbb{R}}$, and the
initial and boundary conditions

$F(0,x,\lambda)\equiv-\sigma_{3},\quad H(0,x)\equiv 0,\quad
H(t,0)=\dfrac{1}{2}\begin{pmatrix}0&\mathcal{E}_{in}(t)\\\
-\overline{\mathcal{E}_{in}(t)}&0\end{pmatrix},$

that is, the functions ${\mathcal{E}}(t,x)$, $\rho(t,x,\lambda)$ and
$\mathcal{N}(t,x,\lambda)$ satisfy the MB equations (1.5) and the initial and
boundary conditions (1.6) and (1.7).

###### Proof.

First, note that $\Phi(t,x,z)$ (4.16) is infinitely differentiable in $t$ and
$x$ since $M(t,x,z)$ (see Theorem 5) and $\theta(t,x,z)$ (3.4) have this
property and that the solution $M(t,x,z)$ of the basic problem RH0 is unique
(see Theorem 4). Since $M(t,x,z)$ satisfies the normalization condition
$M(t,x,z)=I+O(z^{-1})$, $z\to\infty$, and can be represented in the form
(4.8), then it has the asymptotics

$M(t,x,z)=I+m(t,x)z^{-1}+O(z^{-2}),\qquad
z\to\infty\quad(z\in{\mathbb{C}}\setminus\Sigma),$

where $m(t,x)$ has the form (4.20), i.e.,
$m(t,x)=\lim\limits_{z\to\infty}z(M(t,x,z)-I)$ (the limit at $\infty$ is
nontangential). In addition,

$M_{t}(t,x,z)=m_{t}(t,x)z^{-1}+O(z^{-2}),\qquad
z\to\infty\quad(z\in{\mathbb{C}}\setminus\Sigma).$

Since $M(t,x,z)$ is a solution of the basic problem RH0 and
$\Phi(t,x,z)=M(t,x,z)e^{-\mathrm{i}\theta(t,x,z)\sigma_{3}}$, then
$\Phi(t,x,z)$ is analytic in $z\in{\mathbb{C}}\setminus\Sigma^{cl}$, its
boundary values $\Phi_{\pm}(t,x,z)$ are continuous in
$z\in\Sigma\setminus\\{\operatorname{Re}E\\}$, and $\Phi(t,x,z)$ satisfies the
boundary condition

$\Phi_{-}(t,x,z)=\Phi_{+}(t,x,z)e^{\mathrm{i}\theta_{+}(t,x,z)\sigma_{3}}J(t,x,z)e^{-\mathrm{i}\theta_{-}(t,x,z)\sigma_{3}},\qquad
z\in\Sigma\setminus\\{\operatorname{Re}E\\},$ (4.24)

where the matrix
$e^{\mathrm{i}\theta_{+}(t,x,z)\sigma_{3}}J(t,x,z)e^{-\mathrm{i}\theta_{-}(t,x,z)\sigma_{3}}$
is independent on $t$ for all $z\in{\mathbb{C}}$ and is independent on $x$ for
$z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$ (since
$\theta_{+}(t,x,z)=\theta_{-}(t,x,z)$ for
$z\in(E,\overline{E})\setminus\\{\operatorname{Re}E\\}$ and
$z=\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ and
$\theta_{-}(t,x,\lambda)-\theta_{+}(t,x,\lambda)=\pi\mathrm{i}n(\lambda)x/2$
for $\lambda\in[-\Lambda,\Lambda]$). This implies

$\frac{\partial\Phi_{-}(t,x,z)}{\partial
t}\Phi^{-1}_{-}(t,x,z)=\frac{\partial\Phi_{+}(t,x,z)}{\partial
t}\Phi^{-1}_{+}(t,x,z)$

for $z\in\Sigma\setminus\\{\operatorname{Re}E\\}$. Consequently, the
logarithmic derivative $\Phi_{t}(t,x,z)\Phi^{-1}(t,x,z)$ is analytic in
$z\in{\mathbb{C}}\setminus\\{\operatorname{Re}E,E,\overline{E}\\}$ and the
points $\operatorname{Re}E$, $E$, $\overline{E}$ are removable singularities
due to the properties of $M(t,x,z)$. Thus, we can define
$\Phi_{t}(t,x,z)\Phi^{-1}(t,x,z)$ at these points so that it becomes analytic
in $z\in{\mathbb{C}}$. Using the above asymptotic expansions for $M(t,x,z)$
and $M_{t}(t,x,z)$ and (4.16), we obtain

$\Phi_{t}(t,x,z)\Phi^{-1}(t,x,z)=-\mathrm{i}z\sigma_{3}+\mathrm{i}[\sigma_{3},m(t,x)]+O(z^{-1}),\quad
z\to\infty.$

Hence, by the Liouville theorem,

$\Phi_{t}(t,x,z)\Phi^{-1}(t,x,z)=-\mathrm{i}z\sigma_{3}-H(t,x),$

where $H(t,x)=-\mathrm{i}[\sigma_{3},m(t,x)]$ and, therefore, it has the
structure $H(t,x)=\begin{pmatrix}0&q(t,x)\\\ p(t,x)&0\end{pmatrix}$ (since
$H\sigma_{3}+\sigma_{3}H=0$). Using the symmetries of the contour $\Sigma$ and
the jump matrix $J(t,x,z)$, namely,
$\sigma_{2}\overline{J(t,x,\bar{z})}\sigma_{2}=(J^{*}(t,x,\bar{z}))^{-1}=J(t,x,z)$
for $z\in\Sigma\setminus{\mathbb{R}}$ and $J^{*}(t,x,z)=J(t,x,z)$ for
$z=\lambda\in{\mathbb{R}}$, we find that the matrix $H(t,x)$ is anti-
Hermitian, i.e., $H(t,x)=-H^{*}(t,x)$, and therefore
$q(t,x)=-\overline{p(t,x)}=-2\mathrm{i}m_{12}(t,x)$, where
$m(t,x)=(m_{ij}(t,x))_{1\leq i,j\leq 2}$ is defined by (4.20). We set
$q(t,x):=\mathcal{E}(t,x)/2$, then $\mathcal{E}(t,x)$ has the form (4.1),
i.e.,
${\mathcal{E}}(t,x)=-4\mathrm{i}m_{12}(t,x)=-4\mathrm{i}\lim\limits_{z\to\infty}\big{(}zM(t,x,z)\big{)}_{12}$.
Also, we denote $U(t,x,z):=\Phi_{t}(t,x,z)\Phi^{-1}(t,x,z)$. Thus, it is
proved that the matrix function $\Phi(t,x,z)$ satisfies the $t$-equation
(4.17) which becomes the $t$-equation (1.8) or (2.1) (i.e., ${W_{t}=UW}$),
where $U(t,x,\lambda)=-(\mathrm{i}\lambda\sigma_{3}+H(t,x))$ (2.4), as
$z=\lambda+\mathrm{i}0$.

Using (4.24), we obtain that

$\frac{\partial\Phi_{-}(t,x,z)}{\partial
x}\Phi^{-1}_{-}(t,x,z)=\frac{\partial\Phi_{+}(t,x,z)}{\partial
x}\Phi^{-1}_{+}(t,x,z)$

for $z\in\Sigma\setminus[-\Lambda,\Lambda]$, $z\neq\operatorname{Re}E$.
Consequently, the logarithmic derivative $\Phi_{x}(t,x,z)\Phi^{-1}(t,x,z)$ is
analytic for
$z\in{\mathbb{C}}\setminus([-\Lambda,\Lambda]\cup\\{\operatorname{Re}E,E,\overline{E}\\})$
and the points $E$, $\overline{E}$, as well as $\operatorname{Re}E$ if
$\operatorname{Re}E\not\in[-\Lambda,\Lambda]$, are removable singularities.
Thus, we can modify $\Phi_{x}(t,x,z)\Phi^{-1}(t,x,z)$ so that it becomes
analytic in $z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$. Taking into account
asymptotic expansions for $M(t,x,z)$ and $M_{x}(t,x,z)$ and the formulas
(4.16), (2.18), we obtain

$\Phi_{x}(t,x,z)\Phi^{-1}(t,x,z)=\mathrm{i}z\sigma_{3}+H(t,x)+O(z^{-1}),\quad
z\to\infty\quad(z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]).$

Note that in (4.24), for $z=\lambda\in[-\Lambda,\Lambda]$ the matrix
$\widetilde{J}(x,\lambda):=e^{\mathrm{i}\theta_{+}(t,x,\lambda)\sigma_{3}}J(t,x,\lambda)e^{-\mathrm{i}\theta_{-}(t,x,\lambda)\sigma_{3}}$
can be represented as
$\widetilde{J}(x,\lambda)=\begin{pmatrix}1&-r(\lambda)\\\
0&1\end{pmatrix}e^{\frac{\pi n(\lambda)x\sigma_{3}}{2}}\begin{pmatrix}1&0\\\
r(\lambda)&1\end{pmatrix}$. Thus, using the boundary condition (4.24) for
$z=\lambda\in{\mathbb{R}}$, i.e.,
$\Phi_{-}(t,x,\lambda)=\Phi_{+}(t,x,\lambda)\widetilde{J}(x,\lambda)$, where
$\Phi_{\pm}(t,x,\lambda)=\Phi(t,x,\lambda\pm\mathrm{i}0)$, we obtain that
$\Phi_{x}(t,x,z)\Phi^{-1}(t,x,z)$ has the following jump across
$[-\Lambda,\Lambda]$:

$\Phi_{x}(t,x,\lambda+\mathrm{i}0)\Phi^{-1}(t,x,\lambda+\mathrm{i}0)-\Phi_{x}(t,x,\lambda-\mathrm{i}0)\Phi^{-1}(t,x,\lambda-\mathrm{i}0)=\\\
=-\Phi(t,x,\lambda+\mathrm{i}0)\widetilde{J}_{x}(x,\lambda)\widetilde{J}^{-1}(x,\lambda)\Phi^{-1}(t,x,\lambda+\mathrm{i}0)=\\\
=-\dfrac{\pi
n(\lambda)}{2}\Phi(t,x,\lambda+\mathrm{i}0)\begin{pmatrix}1&-r(\lambda)\\\
0&1\end{pmatrix}\sigma_{3}\begin{pmatrix}1&r(\lambda)\\\
0&1\end{pmatrix}\Phi^{-1}(t,x,\lambda+\mathrm{i}0),$

and we denote
$F(t,x,\lambda):=-\Phi(t,x,\lambda+\mathrm{i}0)\begin{pmatrix}1&-r(\lambda)\\\
0&1\end{pmatrix}\sigma_{3}\begin{pmatrix}1&r(\lambda)\\\
0&1\end{pmatrix}\Phi^{-1}(t,x,\lambda+\mathrm{i}0)$. Thus, the function
$\Phi_{x}(t,x,z)\Phi^{-1}(t,x,z)$ can be found by the formula

$\Phi_{x}(t,x,z)\Phi^{-1}(t,x,z)=\mathrm{i}z\sigma_{3}+H(t,x)+\dfrac{1}{4\mathrm{i}}\int\limits_{[-\Lambda,\Lambda]}\dfrac{n(s)F(t,x,s)}{s-z}\textup{d}s,\quad
z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda].$

Since $J^{*}(t,x,\lambda)=J(t,x,\lambda)$, $\lambda\in{\mathbb{R}}$, then
$\widetilde{J}^{*}(x,\lambda)=\widetilde{J}(x,\lambda)$,
$\lambda\in{\mathbb{R}}$. This yields $\dfrac{\pi
n(\lambda)}{2}F(t,x,\lambda)=(\Phi^{*}(t,x,\lambda+\mathrm{i}0))^{-1}\Phi^{*}_{x}(t,x,\lambda+\mathrm{i}0)-(\Phi^{*}(t,x,\lambda-\mathrm{i}0))^{-1}\Phi^{*}_{x}(t,x,\lambda-\mathrm{i}0)=\dfrac{\pi
n(\lambda)}{2}F^{*}(t,x,\lambda)$, and hence the matrix $F(t,x,\lambda)$ is
Hermitian, i.e., $F(t,x,\lambda)=F^{*}(t,x,\lambda)$. In addition, since
$\det\Phi(t,x,z)=\det M(t,x,z)\equiv 1$, then
$\mathop{\mathrm{tr}}\nolimits(\Phi_{x}(t,x,\lambda\pm\mathrm{i}0)\Phi^{-1}(t,x,\lambda\pm\mathrm{i}0))=(\det\Phi(t,x,\lambda\pm\mathrm{i}0))_{x}\equiv
0$, and therefore $\mathop{\mathrm{tr}}\nolimits F(t,x,\lambda)\equiv 0$.
Consequently, $F(t,x,\lambda)$ has the structure:

$F(t,x,\lambda):=\begin{pmatrix}{\mathcal{N}}(t,x,\lambda)&{\mathcal{\rho}}(t,x,\lambda)\\\
\overline{{\mathcal{\rho}}(t,x,\lambda)}&-{\mathcal{N}}(t,x,\lambda)\end{pmatrix}.$

Thus, $\Phi(t,x,z)$ satisfies the $x$-equation (4.18).

It is proved that the matrix function $\Phi(t,x,z)$ satisfies the $t$\- and
$x$-equations (4.17) and (4.18), that is,

$\displaystyle\Phi_{t}$ $\displaystyle=U(t,x,z)\Phi,\qquad
U(t,x,z)=-\mathrm{i}z\sigma_{3}-H(t,x)$ (4.25) $\displaystyle\Phi_{x}$
$\displaystyle=V(t,x,z)\Phi,\qquad V(t,x,z)=-U(t,x,z)-\mathrm{i}G(t,x,z),$
(4.26)

where
$G(t,x,z)=\dfrac{1}{4}\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)}{s-z}ds$,
$z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$, and
$\Phi_{\pm}(t,x,z)=\Phi(t,x,\lambda\pm\mathrm{i}0)$ for
$z=\lambda\in{\mathbb{R}}$ satisfy the $t$-equation (4.25) and the
$x^{\pm}$-equations

$\Phi_{x}=V_{\pm}(t,x,\lambda)\Phi,$ (4.27)

where