$V_{\pm}(t,x,\lambda)=-U(t,x,\lambda)-\mathrm{i}G_{\pm}(t,x,\lambda)=\mathrm{i}\lambda\sigma_{3}+H(t,x)-\mathrm{i}G_{\pm}(t,x,\lambda)$
(2.7), $G_{\pm}(t,x,\lambda)=G(t,x,\lambda\pm\mathrm{i}0)$ has the form
(1.13), $G_{\pm}(t,x,\lambda)=G(t,x,z)$ for
$z=\lambda\in{\mathbb{R}}\setminus[-\Lambda,\Lambda]$ and
$G(t,x,\lambda):=\dfrac{1}{4}\,{\rm
p.v.}\int\limits_{-\infty}^{\infty}\dfrac{F(t,x,s)n(s)}{s-\lambda}\textup{d}s$
for $\lambda\in[-\Lambda,\Lambda]$. The compatibility conditions of the
systems (4.25), (4.27) imply (2.9), i.e.,
$U_{x}-(V_{\pm})_{t}+[U,V_{\pm}]=U_{x}-V_{t}+[U,V]\mp\dfrac{\pi
n}{4}\big{(}F_{t}-[U,F]\big{)}=0$, $\lambda\in{\mathbb{R}}$. Using these
conditions, we obtain the equation
$H_{x}(t,x)+H_{t}(t,x)-\dfrac{1}{4}\int\limits_{-\infty}^{\infty}[\sigma_{3},F(t,x,s)]n(s)\textup{d}s-\dfrac{\mathrm{i}}{4}\,\int\limits_{-\infty}^{\infty}\dfrac{(F_{t}(t,x,s)+[\mathrm{i}s\sigma_{3}+H(t,x),F(t,x,s)])n(s)}{s-\lambda\mp\mathrm{i}0}\textup{d}s=0$
that yields the MB equations in the matrix form (4.22), (4.23) which are
equivalent to the MB equations (1.5). Thus, the matrices
$\Phi(t,x,\lambda\pm\mathrm{i}0)$ satisfy the systems of the equations (4.25),
(4.27) which coincide with the AKNS systems (1.8), (1.12), and the
compatibility condition (2.3), where $U(t,x,\lambda)$, $V(t,x,\lambda)$ have
the form (2.4), (2.5), is fulfilled. Hence the functions ${\mathcal{E}}(t,x)$,
${\mathcal{N}}(t,x,\lambda)$ and ${\mathcal{\rho}}(t,x,\lambda)$ satisfy the
MB equations (1.5).

Let us proof that the initial and boundary conditions (1.6), (1.7) are
fulfilled. Since $t=0\leq x$ for all $x\in[0,L)$ ($L\leq\infty$), then for
$t=0$ as in the proof of Theorem 3 we obtain that ${\mathcal{E}(0,x)\equiv
0}$, $\rho(0,x,\lambda)\equiv 0$ and $\mathcal{N}(0,x,\lambda)\equiv-1$ or
$H(0,x)\equiv 0$ and $F(0,x,\lambda)\equiv-\sigma_{3}$. Taking into account
the form the solution $M(t,x,z)$ (3.5) and Theorem 1, the solution of the
basic problem RH0 for $x=0$ has the form

$M(t,0,z)\\!=\\!\begin{cases}\begin{pmatrix}\Psi[1](t,0,z)e^{\mathrm{i}zt}&\dfrac{\Phi[2](t,z)}{a(z)}e^{-\mathrm{i}zt}\end{pmatrix},\;z\in{\mathbb{C}}_{+}\setminus[E,\operatorname{Re}E),\\\
\begin{pmatrix}\dfrac{\Phi[1](t,z)}{a(z)}e^{\mathrm{i}zt}&\Psi[2](t,0,z)e^{-\mathrm{i}zt}\end{pmatrix},\;z\in{\mathbb{C}}_{-}\setminus(\operatorname{Re}E,\overline{E}],\end{cases}$

where $\Psi(t,0,z)$ satisfies the $t$-equation
$\Psi_{t}=(-\mathrm{i}z\sigma_{3}-H(t,0))\Psi$ and $\Psi(0,0,z)=I$, and
$\Phi(t,z)=e^{-\mathrm{i}\operatorname{Re}E\,t\sigma_{3}}M_{0}(z)e^{-\mathrm{i}w(z)\,t\sigma_{3}}$
(recall that $\Phi(t,\lambda)=\Phi_{0}(t,0,\lambda)$ (2.22), where
$\Phi_{0}(t,x,\lambda)$ is the background solution of the AKNS system (1.8),
(1.9) for which ${\mathcal{E}}_{bg}(t,0)={\mathcal{E}}_{in}(t)$ is the input
signal (1.7), as shown in Section 2) satisfies the $t$-equation
$\Phi_{t}=(-\mathrm{i}z\sigma_{3}-H(t,0))\Phi$ with
$H(t,0)=H_{0}(t)=\dfrac{1}{2}\begin{pmatrix}0&{\mathcal{E}}_{in}(t)\\\
-\overline{{\mathcal{E}}_{in}(t)}\end{pmatrix}$ (2.15). Thus, using (4.19) and
(4.20), we obtain that
$\mathcal{E}(t,0)=-4\mathrm{i}\lim\limits_{z\to\infty}\big{(}z(M(t,0,z)-I)\big{)}_{12}=2\big{(}H_{0}(t)\big{)}_{12}={\mathcal{E}}_{in}(t)$
or $H(t,0)=H_{0}(t)$.

Since $F(t,x,\lambda)$ satisfies the equation (4.23), then just as in the
proof of Theorem 2 we get (4.21), i.e.,
$F(t,x,\lambda)=-M(t,x,\lambda+\mathrm{i}0)e^{-\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(0,x,\lambda+\mathrm{i}0)\sigma_{3}M(0,x,\lambda+\mathrm{i}0)e^{\mathrm{i}\lambda
t\sigma_{3}}M^{-1}(t,x,\lambda+\mathrm{i}0)$,
$\lambda\in{\mathbb{R}}\setminus\\{\operatorname{Re}E\\}$. The function
$F(t,x,\lambda)$ as well as $M(t,x,\lambda+\mathrm{i}0)=M_{+}(t,x,\lambda)$ is
continuous in $\lambda\in{\mathbb{R}}\setminus\\{\operatorname{Re}E\\}$ and
bounded near $\lambda=\operatorname{Re}E$. Thus, we can appropriately define
$M_{+}(t,x,\operatorname{Re}E)$ (see the proof of Theorem 2) so that
$F(t,x,\lambda)$ is continuous in $\lambda\in{\mathbb{R}}$.

The proof of the uniqueness of $\Phi(t,x,z)$ is similar to the proof of the
uniqueness of $M(t,x,z)$ in Theorem 2. ∎

Theorems 4, 5 and 6 give the following corollary.

######  Corollary 1.

The function $H(t,x)$ (4.19) and the electric field envelope
${\mathcal{E}}(t,x)$ can also be defined by

$H(t,x)=-\mathrm{i}[\sigma_{3},m^{(2)}(t,x)],\quad
m^{(2)}(t,x)=\dfrac{1}{2\pi\mathrm{i}}\int\limits_{\Upsilon}[W^{(2)}(t,x,s)+I][J^{(2)}(t,x,s)-I]\textup{d}s,$
(4.28)

and

${\mathcal{E}}(t,x)=\dfrac{2}{\pi}\int\limits_{\Upsilon}\Big{(}[W^{(2)}(t,x,s)+I][I-J^{(2)}(t,x,s)]\Big{)}_{12}\textup{d}s,$
(4.29)

where $W^{(2)}(t,x,z)$ is a solution of a singular integral equation of the
form (4.12), $J^{(2)}(t,x,z)$ has the form (4.11) and (4.15) for
$\Lambda<\infty$ and $\Lambda=\infty$ respectively, and $\Upsilon=L_{1}\cup
L_{3}\cup\overline{L_{1}}\cup\overline{L_{3}}\cup[\lambda_{1},\lambda_{2}]\cup[E,\overline{E}]$
(see Fig. 2) and $\Upsilon=L_{2}\cup\overline{L_{2}}$ (see Fig. 3, 4) for
$\Lambda<\infty$ and $\Lambda=\infty$ respectively.

###### Proof.

Using the proof of Theorem 5, in particular, the formula (4.13), we obtain
that

$M^{(2)}(t,x,z)=I+\dfrac{m^{(2)}(t,x)}{z}+O(z^{-2}),\;M^{(2)}_{t}(t,x,z)=\dfrac{m^{(2)}_{t}(t,x)}{z}+O(z^{-2}),\;z\to\infty,\;z\in{\mathbb{C}}\setminus\Upsilon,$

where $m^{(2)}(t,x)$ has the form (4.28). Since
$m(t,x)=\lim\limits_{z\to\infty}z(M(t,x,z)-I)=\lim\limits_{z\to\infty}z(M^{(2)}(t,x,z)-I)=m^{(2)}(t,x)$,
where $M(t,x,z)=M^{(2)}(t,x,z)(G^{(2)}(t,x,z))^{-1}\delta^{-\sigma_{3}}(z)$
and $M(t,x,z)=M^{(2)}(t,x,z)(G^{(2)}(t,x,z))^{-1}$ for $\Lambda<\infty$ and
$\Lambda=\infty$ respectively (see the proof of Theorem 5), then, using the
formula (4.19), we obtain (4.28) and (4.29). ∎

## Appendix. Analysis of the phase function

Consider the phase function $\theta$ (3.4):

$\theta(t,x,z)=zt-\eta(z)x=\tau\left(z-\xi\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-z}\textup{d}s\right)\\!,\;\;\xi=\dfrac{x}{4\tau},\;\;\tau=t-x\;\;\left(x=\dfrac{4\xi}{1+4\xi}\,t\right)\\!,$
(4.30)

where $t$ and $x$ as well as $\tau$ and $\xi$ are parameters. In what follows,
we use the same notation $\theta$ for the function of the parameters $x$, $t$
or $\tau$, $\xi$. We will study the function
$\operatorname{Re}(\mathrm{i}\theta(t,x,z))$ to obtain its signature table
characterizing a behavior of the exponents $e^{\pm 2\mathrm{i}\theta(t,x,z)}$.
It has the form

$\operatorname{Re}(\mathrm{i}\theta(t,x,z))=\nu\\!\left(x-t+\dfrac{x}{4}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{|s-z|^{2}}\right)\\!=\tau\nu\,\big{(}\xi\Pi(\lambda,\nu)-1\big{)}=\dfrac{x\nu}{4}\\!\left(\Pi(\lambda,\nu)-\dfrac{1}{\xi}\right)\\!,$
(4.31)

where $z=\lambda+\mathrm{i}\nu$ and

$\Pi(\lambda,\nu):=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-\lambda)^{2}+\nu^{2}}.$
(4.32)

The function $\theta(t,x,z)$ is analytic for
$z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$ and therefore the functions
$\operatorname{Re}(\mathrm{i}\theta(t,x,z))=-\operatorname{Im}\theta(t,x,z)$
and
$\operatorname{Im}(\mathrm{i}\theta(t,x,z))=\operatorname{Re}\theta(t,x,z)$
are harmonic in $z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$. It has the
continuous boundary values
$\theta_{\pm}(t,x,\lambda)=\theta(t,x,\lambda\pm\mathrm{i}0)=\lambda
t-\eta_{\pm}(\lambda)x$ on $[-\Lambda,\Lambda]$ (recall that the contour
$[-\Lambda,\Lambda]$ is oriented from $-\Lambda$ to $\Lambda$), where
$\eta_{\pm}(\lambda)$ are define in (2.20), $z=\lambda+\mathrm{i}\nu$, which
can be represented in the form

$\theta_{\pm}(t,x,\lambda)=\tau\left(\lambda-\xi\,{\rm
p.v.}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)}{s-\lambda}\textup{d}s\right)\mp\dfrac{\pi\mathrm{i}}{4}n(\lambda)x,\quad\lambda\in[-\Lambda,\Lambda].$

Since the equality $\operatorname{Re}(\mathrm{i}\theta(t,x,z))=0$ is satisfied
if $\nu=\operatorname{Im}z=0$ or

$\Pi(\lambda,\nu)=\dfrac{1}{\xi},$ (4.33)

where $\Pi(\lambda,\nu)$ is defined by (4.32), then the curve defined by the
equality $\operatorname{Re}(\mathrm{i}\theta(t,x,z))=0$ divides the complex
plane into two domains ${\mathbb{C}}_{\pm}$ (i.e., this curve is $\nu=0$) when
$\xi<0$ ($t<x$) or $t=x$, and into four domains whose common boundary is
defined by the equalities $\nu=0$ and (4.33) when $\xi>0$ ($t>x$).

Since

$\dfrac{\textup{d}\theta}{\textup{d}z}=\tau\left(1-\xi\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-z)^{2}}\right)=\tau\left(1-\xi\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-\lambda)^{2}-2\mathrm{i}(s-\lambda)\nu-\nu^{2}}\right)=0$

($z\in{\mathbb{C}}\setminus[-\Lambda,\Lambda]$) if
$\dfrac{1}{\xi}=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-z)^{2}}=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)(s-\overline{z})^{2}\textup{d}s}{|s-z|^{4}}=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)[(s-\lambda)^{2}-\nu^{2}]\textup{d}s}{[(s-\lambda)^{2}+\nu^{2}]^{2}}+2\,\mathrm{i}\nu\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)(s-\lambda)\textup{d}s}{[(s-\lambda)^{2}+\nu^{2}]^{2}}=I_{1}(\lambda,\nu)+2\,\mathrm{i}\nu\,I_{2}(\lambda,\nu)$,
then $\dfrac{d\theta}{dz}=0$ if $I_{1}(\lambda,\nu)=\dfrac{1}{\xi}$, and
$\nu=0$ or $I_{2}(\lambda,\nu)=0$.

First, consider the case when $\nu=0$ and $I_{1}(\lambda,\nu)=\dfrac{1}{\xi}$.
In this case, the derivative $\textup{d}\theta/\textup{d}z$ takes the form

$\dfrac{\textup{d}\theta}{\textup{d}z}=\tau\left(1-\xi\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-\lambda)^{2}}\right)=\tau\big{(}1-\xi
I_{1}(\lambda,0)\big{)},\qquad z=\lambda\notin[-\Lambda,\Lambda],$ (4.34)

where $I_{1}(\lambda,0)=\Pi(\lambda,0)$. Hence (since the derivative (4.34)
does not exist for $\lambda\in[-\Lambda,\Lambda]$), for $\xi>0$ ($\tau>0$) the
stationary points of the function
$\mathrm{i}\theta(t,x,z)=\mathrm{i}\theta(\tau,\xi,z)$ are located on the real
line outside the interval $[-\Lambda,\Lambda]$ and they are the points of the
intersection of the real line $\nu=0$ and the level line (4.33). Since
$I_{1}(\lambda,0)=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-\lambda)^{2}}>0$,
then for $\xi<0$ ($\tau<0$) the derivative $\dfrac{d\theta}{dz}<0$ and hence
there are no stationary points.

Now, consider the case when ${\nu\neq 0}$ and
${I_{2}(\lambda,\nu)=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)(s-\lambda)\textup{d}s}{[(s-\lambda)^{2}+\nu^{2}]^{2}}=0}$,
${I_{1}(\lambda,\nu)=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)[(s-\lambda)^{2}-\nu^{2}]\textup{d}s}{[(s-\lambda)^{2}+\nu^{2}]^{2}}=\dfrac{1}{\xi}}$.
In this case ${\xi>0}$. If $\lambda\neq 0$, ${\nu\neq 0}$, then
${I_{2}(\lambda,\nu)\neq 0}$. If $\lambda=0$ and ${\nu\neq 0}$, then
$I_{2}(0,\nu)=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)s}{[s^{2}+\nu^{2}]^{2}}\textup{d}s=\left(\int\limits_{-\Lambda}^{0}+\int\limits_{0}^{\Lambda}\right)\dfrac{n(s)s}{[s^{2}+\nu^{2}]^{2}}\textup{d}s=0$
when $n(s)$ is an even function ($n(-s)=n(s)$), for example, when $n(\lambda)$
has the form (2.17). Then the points $z=\mathrm{i}\nu\neq 0$ will be
stationary if

${I_{1}(0,\nu)=\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)[s^{2}-\nu^{2}]\textup{d}s}{[s^{2}+\nu^{2}]^{2}}=\dfrac{1}{\xi}}.$
(4.35)

Obviously,
$\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)s^{2}\textup{d}s}{[s^{2}+\nu^{2}]^{2}}<\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\nu^{2}\textup{d}s}{[s^{2}+\nu^{2}]^{2}}$,
i.e., $I_{1}(0,\nu)<0$, for $|\nu|>\Lambda$. Since $n(s)$ is even,

$I_{1}(0,\nu)=2\int\limits_{0}^{\Lambda}\dfrac{n(s)[s^{2}-\nu^{2}]\textup{d}s}{[s^{2}+\nu^{2}]^{2}}.$

Consider the function $\chi(s):=\dfrac{s^{2}}{[s^{2}+\nu^{2}]^{2}}$. Since
$\chi^{\prime}(s)=\dfrac{2s(\nu^{2}-s^{2})}{[s^{2}+\nu^{2}]^{3}}=0$ for $s=0$
and $s^{2}=\nu^{2}$, and
$\chi^{\prime\prime}(s)=\dfrac{2\nu^{4}-16\nu^{2}s^{2}+6s^{4}}{[s^{2}+\nu^{2}]^{4}}$
is positive for $s=0$ and negative for $s^{2}=\nu^{2}$ ($\nu\neq 0$), then
$\chi(s)$ attains a strict maximum at $s=\pm\nu$. Consequently,
$\dfrac{n(s)s^{2}}{[s^{2}+\nu^{2}]^{2}}<\dfrac{n(s)\nu^{2}}{[s^{2}+\nu^{2}]^{2}}$,
$s^{2}\neq\nu^{2}$ ($\nu\neq 0$, $|\nu|\leq\Lambda$), and hence
$\int\limits_{0}^{\Lambda}\dfrac{n(s)s^{2}\textup{d}s}{[s^{2}+\nu^{2}]^{2}}<\int\limits_{0}^{\Lambda}\dfrac{n(s)\nu^{2}\textup{d}s}{[s^{2}+\nu^{2}]^{2}}$,
i.e., $I_{1}(0,\nu)<0$, for $|\nu|\leq\Lambda$, $\nu\neq 0$. In particular,
for $n(\lambda)$ of the form (2.17)
$I_{1}(0,\nu)=-\dfrac{1}{\Lambda^{2}+\nu^{2}}<0$ for all $\nu$. Thus, in the
present case the function $\mathrm{i}\theta(t,x,z)$ does not have stationary
points.

Since $n(\lambda)\geq 0$ and
$\int\limits_{-\infty}^{\infty}n(\lambda)\textup{d}\lambda=1$, then

$0<\Pi(\lambda,\nu)\leq\sup\limits_{s\in[-\Lambda,\Lambda]}\dfrac{1}{(s-\lambda)^{2}+\nu^{2}}\int\limits_{-\Lambda}^{\Lambda}n(s)\textup{d}s=\dfrac{1}{\inf\limits_{s\in[-\Lambda,\Lambda]}(s-\lambda)^{2}+\nu^{2}}.$

Thus,
$\dfrac{1}{\xi}=\Pi(\lambda,\nu)\leq\begin{cases}\dfrac{1}{\nu^{2}},&|\lambda|\leq\Lambda,\\\
\dfrac{1}{(\Lambda-|\lambda|)^{2}+\nu^{2}},&|\lambda|>\Lambda.\end{cases}$.
Hence, $|\nu|\leq\sqrt{\xi}$ for $\lambda\in{\mathbb{R}}$ and
$(|\lambda|-\Lambda)^{2}+\nu^{2}\leq\xi$ when
$\lambda\in(-\infty,-\Lambda)\cup(\Lambda,\infty)$. Since
$\Pi(\lambda,\nu)=\Pi(\lambda,-\nu)$ for any $\nu$, the function
$\Pi(\lambda,\nu)$ and consequently the level line (4.33) are symmetric with
respect to the real axis $\nu=0$. If $n(-\lambda)=n(\lambda)$ for all
$\lambda$ ($n(\lambda)$ is an even function), then $\Pi(\lambda,\nu)$ and the
level line (4.33) are also symmetric with respect to the imaginary axis
$\lambda=0$. Moreover,

$\dfrac{1}{\xi}=\Pi(\lambda,\nu)\geq\inf\limits_{s\in[-\Lambda,\Lambda]}\dfrac{1}{(s-\lambda)^{2}+\nu^{2}}=\dfrac{1}{(\Lambda+|\lambda|)^{2}+\nu^{2}}\geq\dfrac{1}{2(\Lambda^{2}+\lambda^{2})+\nu^{2}},$

and, therefore, $\xi\leq 2(\Lambda^{2}+\lambda^{2})+\nu^{2}$.

Note that $\lim\limits_{\nu\to 0}\Pi(\lambda,\nu)=\lim\limits_{\nu\to
0}\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-\lambda)^{2}+\nu^{2}}=+\infty$
for $\lambda\in[-\Lambda,\Lambda]$. Taking into account the signatures of the
derivatives

$\dfrac{\partial\Pi}{\partial\lambda}=2\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)(s-\lambda)\textup{d}s}{[(s-\lambda)^{2}+\nu^{2}]^{2}},\qquad\dfrac{\partial\Pi}{\partial\nu}=-2\nu\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{[(s-\lambda)^{2}+\nu^{2}]^{2}},$

we conclude that when $\Lambda<\infty$, for each $\xi\in{\mathbb{R}}_{+}$
there exists the smooth and closed level line (4.33), which is symmetric with
respect to the real axis $\nu=0$, and we denote this curve (level line) by
$\gamma_{\xi}$ (see Figure 5). Since the derivative
$\dfrac{\partial\widetilde{\Pi}}{\partial\lambda}=\dfrac{\partial\Pi}{\partial\lambda}$,
where $\widetilde{\Pi}(\lambda,\nu,\xi)=\Pi(\lambda,\nu)-\dfrac{1}{\xi}$,
$\xi\in{\mathbb{R}}_{+}$, exists and is not equal to zero for
$\nu\in{\mathbb{R}}$ and $|\lambda|>\Lambda$, then in some neighborhoods of
the stationary points (recall that they are the points of the intersection of
the real line $\nu=0$ and the level line (4.33)) the level line can be
determined by the smooth function $\lambda=\lambda_{-}(\nu,\xi)$ for
$\lambda<-\Lambda$ and $\lambda=\lambda_{+}(\nu,\xi)$ for $\lambda>\Lambda$.
We denote the stationary points by $\lambda_{\pm}=\lambda_{\pm}(\xi)$
($\lambda_{-}<-\Lambda$ and $\lambda_{+}>\Lambda$). At a large $\xi$, the
level line (4.33) goes to the vicinity of the infinity in the $z$-plane, and
at a small $\xi$, it envelops the segment $[-\Lambda,\Lambda]$ ($|\nu|\to+0$
and $|\lambda|\to\Lambda+0$ for $\xi\to+0$). As $\Lambda\to+\infty$ the
stationary points tend to $\pm\infty$. When $\Lambda=+\infty$ the real line
$\nu=0$ becomes an asymptote for the level line (4.33)
($\lambda_{\pm}(\nu,\xi)\to\pm\infty$ as $\nu\to 0$), which consists of the
two curves $\gamma^{\pm}_{\xi}$ symmetric with respect to the real axis (see
Figure 6). As $\nu\to\pm 0$ the curves $\gamma^{\pm}_{\xi}$ tend to the real
line from ${\mathbb{C}}_{\pm}$ respectively.

Taking into account (4.31), we obtain that

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

for $0\leq t<x$, i.e., $\xi<0$, and for $t=x$, $t,x>0$, and
$\operatorname{Re}(\mathrm{i}\theta(0,0,z))\equiv 0$. This case ($t\leq x$,
$t,x\geq 0$) is studied in Theorem 3.

As shown above, in the case $\tau=t-x>0$ and, accordingly, $\xi>0$, the curve
defined by the equality $\operatorname{Re}(\mathrm{i}\theta(t,x,z))=0$, where
$z=\lambda+\mathrm{i}\nu$ and $\Lambda<\infty$, consists of the real line
${\mathbb{R}}$ and the closed level line $\gamma_{\xi}$ defined by the
equality $\Pi(\lambda,\nu)=1/\xi$ (the level line (4.33)). For any fixed
$\xi>0$, the interval $[-\Lambda,\Lambda]$ is located inside the oval
$\gamma_{\xi}$, and $\gamma_{\xi}$ intersects the real line at the stationary
points $\lambda_{\pm}=\lambda_{\pm}(\xi)$ ($\lambda_{-}<-\Lambda$ and
$\lambda_{+}>\Lambda$) which are simple ones because
$\dfrac{\partial^{2}\theta}{\partial\lambda^{2}}=-2\xi\int\limits_{-\Lambda}^{\Lambda}\dfrac{n(s)\textup{d}s}{(s-\lambda)^{3}}$
is negative (i.e., strictly negative) for $\lambda<-\Lambda$ and positive for
$\lambda>\Lambda$. When $\Lambda\to+\infty$ ($\xi>0$ is fixed) the stationary
points $\lambda_{\pm}$ tend to $\pm\infty$. For $\tau>0$ ($\xi>0$) and
$\Lambda=+\infty$, the real line $\nu=0$ is an asymptote for the level line
(4.33) consisting of the two curves $\gamma^{\pm}_{\xi}$, symmetric with
respect to the real line.

It is easy to verify that for $t>x$ ($\xi>0$) the signature table of
$\operatorname{Re}(\mathrm{i}\theta(t,x,z))$ (4.31) has the form presented in
Figures 5, 6. The signature table shows that the exponents
$e^{-2\mathrm{i}\theta(t,x,z)}$ and $e^{2\mathrm{i}\theta(t,x,z)}$ vanish
exponentially on the subintervals $(E,\operatorname{Re}E)$ and
$(\operatorname{Re}E,\overline{E})$ respectively when $z$ and the interval
$(E,\overline{E})$ lie inside of $\gamma_{\xi}$ (when $\Lambda<\infty$) or
between the curves $\gamma^{\pm}_{\xi}$ (when $\Lambda=\infty$). They increase
in $\tau$ unboundedly on the subintervals $(E,\operatorname{Re}E)$ and
$(\operatorname{Re}E,\overline{E})$ respectively if $z$ and the interval
$(E,\overline{E})$ lie outside of $\gamma_{\xi}$ (when $\Lambda<\infty$) or on
those parts of the subintervals $(E,\operatorname{Re}E)$ and
$(\operatorname{Re}E,\overline{E})$ respectively that lie above the curve
$\gamma^{+}_{\xi}$ and below the curve $\gamma^{-}_{\xi}$ if $z$ also lies in
the corresponding regions (when $\Lambda=\infty$).

Figure 5: The curve ${\mathbb{R}}\cup\gamma_{\xi}$ defined by
$\operatorname{Re}(\mathrm{i}\theta(t,x,z))=0$ and the signature table of
$\operatorname{Re}(\mathrm{i}\theta)$ when $t>x$ ($\xi>0$) and
$\Lambda<\infty$ Figure 6: The curve
${\mathbb{R}}\cup\gamma^{+}_{\xi}\cup\gamma^{-}_{\xi}$ defined by
$\operatorname{Re}(\mathrm{i}\theta(t,x,z))=0$ and the signature table of
$\operatorname{Re}(\mathrm{i}\theta)$ when $t>x$ ($\xi>0$) and
$\Lambda=\infty$

## Acknowledgments

The author would like to thank Vladimir Kotlyarov (B. Verkin ILTPE of NAS of
Ukraine) for useful discussions and comments.

This work was partially supported by the National Academy of Sciences of
Ukraine (project 0121U111968 “Nonstandard nonlocal and peakon integrable
equations: asymptotics and inverse scattering transform”).

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