# Inverse scattering transform for the Tzitzéica equation

Linlin Wanga, Junyi Zhub
a College of Science, Henan University of Engineering,
Zhengzhou 451191, Henan, China
b School of Mathematics and Statistics, Zhengzhou University,
Zhengzhou 450001, Henan, China
<EMAIL_ADDRESS>
###### Abstract

The inverse scattering transform is extended to investigate the Tzitzéica
equation. A set of sectionally analytic eigenfunctions and auxiliary
eigenfunctions are introduced. We note that in this procedure, the auxiliary
eigenfunctions play an important role. Besides, the symmetries of the analytic
eigenfunctions and scattering data are discussed. The asymptotic behaviors of
the Jost eigenfunctions are derived systematically. A Riemann-Hilbert problem
is constructed to study the inverse scattering problem. Lastly, some novel
exact solutions are obtained for reflectionless potentials.

Keywords: The Tzitzéica equation; Lax pair; The inverse spectral transform;
Vanishing boundary condition; Explicit solutions.

##  1 Introduction

The inverse scattering transform method is an effective tool to study the
integrable nonlinear equation with sufficient decay boundary condition or non-
zero boundary condition. Besides, some novel and interesting properties for
solutions can be given by the inverse scattering transform. There are so many
works involving these problems. We here only refer to some of them [1, 2, 3,
4, 5, 6, 7, 8, 9, 10]. In contrast to $2\times 2$ matrix linear spectral
problems, the $3\times 3$ or higher order matrix problems are more difficult
to be investigated by the inverse scattering transform method [11, 12, 13,
14].

In the present paper, we will study the Tzitzéica equation by the inverse
scattering transform

$u_{xt}=e^{u}-e^{-2u},$ (1.1)

where $u=u(x,t)$ and has sufficient decay for all $t$ as $|x|\to\infty$. Here,
the subscripted variables $x$ and $t$ in Eq. (1.1) denote the corresponding
partial differentiation. We note that the Tzitzéica equation (1.1) can be
rewritten as other forms [15, 16]

$(\ln h)_{xt}=h-h^{-2},$ (1.2) $u_{xx}-u_{tt}=e^{u}-e^{-2u},$ (1.3)

It can be directly calculated that Eq.(1.1) is equivalent to Eq. (1.2) with
the transform $\ln h=u$. The Tzitzéica equation (1.1) was initially proposed
as the model describing special surfaces in differential geometry, where the
ratio $k/d^{4}$ is constant. The parameter $k$ is the Gauss curvature of the
surface and $d$ is the distance from the origin to the tangent plane at a
fixed point. In [17, 18, 19], the Tzitzéica equation can be reduced from the
two-dimensional Toda lattice equation via the conjugate nets. Furthermore, the
relations between soliton theory and affine differential geometry were
presented in [17].

It is shown that Eq. (1.1) plays an important role in many fields, such as
geometry [20], soliton theory [21, 19, 22, 23] and gas dynamics [24]. The
Tzitzéica equation was studied by many methods, such as the Darboux
transformation [25, 26], Bäcklund transformation [27], the algebro-geometric
approach [28, 29, 30], the Hirota bilinear method [31, 32], the inverse
scattering method [19, 33] and the dressing method [31]. In addition, there
are some other results for the Tzitzéica equation, such as the discretization
[34], ultradiscretization [35], symmetries [36], conservation laws [21] and
others [37, 38, 39, 40, 41, 42].

This paper is to obtain the explicit solution of the Tzitzéica equation
through inverse scattering transforms. To this end, a Riemann-Hilbert (RH)
problem with the corresponding six jump matrices and six jump conditions is
constructed. We note that it is difficult to construct a set of analyticity
eigenfunctions in every domains. Thus, introducing the auxiliary
eigenfunctions and the adjoint problem, we obtain the piecewise meromorphic
function. To avoid the overly cumbersome work, we consider the symmetry
conditions associated with the Tzitzéica equation. By using the Cauchy
projector, the RH problem is transformed to a system of matrix algebraic-
integral equations. Based on the algebraic-integral equations, some explicit
solutions of Eq.(1.1) for the case of reflectionless potentials are given.

The paper is organized as follows. In section 2, the direct problem of the
Tzitzéica equation Eq. (1.1) is considered. In section 3, the discrete
spectrum is discussed in detail. In section 4, the inverse problem is
investigated. In the last section, based on the RH problem, some exact
solutions of Eq. (1.1) are constructed.

##  2 Spectral analysis

In this section, we will consider the direct scattering problem associated
with Eq. (1.1).

###  2.1 Preliminaries: Lax pair, scattering matrix

The Tzitzéica equation (1.1) can be derived through the compatibility
condition of the following Lax pair

$\psi_{0x}=U_{0}(x,\lambda,u)\psi_{0},$ (2.1a)
$\psi_{0t}=V_{0}(t,\lambda,u)\psi_{0},$ (2.1b)

where

$U_{0}(x,\lambda,u)=\left(\begin{matrix}0&0&\lambda e^{u}\\\ e^{-u}&u_{x}&0\\\
0&1&0\end{matrix}\right),\quad
V_{0}(t,\lambda,u)=\left(\begin{matrix}u_{t}&e^{-u}&0\\\ 0&0&e^{u}\\\
\lambda^{-1}&0&0\end{matrix}\right).$ (2.2)

Since the trace of the matrices $U_{0}$ and $V_{0}$ are not zero, we make the
following eigenfunction transformation

$\left(\begin{array}[]{c}\psi_{1}\\\ \psi_{2}\\\
\psi_{3}\end{array}\right)=\left(\begin{matrix}e^{-u}&0&0\\\ 0&\nu&0\\\
0&0&\nu^{2}\end{matrix}\right)\left(\begin{array}[]{c}\psi_{10}\\\
\psi_{20}\\\ \psi_{30}\end{array}\right),$ (2.3)

and have the second Lax pair:

$\psi_{x}=U(x,\lambda,u)\psi,$ (2.4a) $\psi_{t}=V(t,\lambda,u)\psi,$ (2.4b)

with

$U(x,\lambda,u)=\left(\begin{matrix}-u_{x}&0&\nu\\\ \nu&u_{x}&0\\\
0&\nu&0\end{matrix}\right),\quad
V(t,\lambda,u)=\left(\begin{matrix}0&\frac{e^{-2u}}{\nu}&0\\\
0&0&\frac{e^{u}}{\nu}\\\ \frac{e^{u}}{\nu}&0&0\end{matrix}\right),$ (2.5)

$\nu$ is the spectral parameter. Thus, the determinant of the fundamental
matrix solution of $(2.4)$ is $x$ and $t$ independent. In the following, we
will discuss the spectral analysis from the spectral problem (2.4). Under the
vanishing boundary condition $u{\to}0,x{\to}\pm\infty$, we have

$\displaystyle U\to U_{\pm}=\left(\begin{matrix}0&0&\nu\\\ \nu&0&0\\\
0&\nu&0\end{matrix}\right),\quad x\to\pm\infty,$ (2.6) $\displaystyle V\to
V_{\pm}=\left(\begin{matrix}0&\frac{1}{\nu}&0\\\ 0&0&\frac{1}{\nu}\\\
\frac{1}{\nu}&0&0\end{matrix}\right),\quad x\to\pm\infty.$

For convenience, we introduce an invertible matrix $A$, such that

$A^{-1}U_{\pm}A=\nu\sigma,\quad A^{-1}V_{\pm}A=\nu^{-1}\sigma^{-1},$ (2.7)

where

$\displaystyle\sigma=\left(\begin{matrix}1&0&0\\\ 0&\alpha&0\\\
0&0&\alpha^{2}\end{matrix}\right),\quad\quad
A=\left(\begin{matrix}1&\alpha^{2}&\alpha\\\ 1&\alpha&\alpha^{2}\\\
1&1&1\end{matrix}\right).$ (2.8)

Here, $\lambda=\nu^{3},\alpha=e^{\frac{2}{3}\pi i}$. Now we introduce the
matrix solutions $\psi_{\pm}(x,t,\nu)$ of (2.4a) satisfying the following
condition

$\displaystyle\psi_{\pm}(x,t,\nu)=Ae^{\nu
x\sigma+\nu^{-1}\sigma^{-1}t}+o(1),\quad x\to\pm\infty.$ (2.9)

Thus

$\det\psi_{\pm}(x,t,\nu)=3(\alpha-\alpha^{2})=\gamma.$ (2.10)

For the sake of analyticity, we introduce new Jost solutions
$\mu_{\pm}(x,t,\nu)$ defined by

$\psi_{\pm}(x,t,\nu)=\mu_{\pm}(x,t,\nu)e^{\nu\sigma x+\nu^{-1}\sigma^{-1}t},$
(2.11)

with their corresponding asymptotic behaviour

$\mu_{\pm}(x,t,\nu)\to A,\quad x\to\pm\infty.$ (2.12)

$x$$t$$(x,t)$$\gamma_{1}$$x$$t$$(x,t)$$\gamma_{2}$ Figure 1: Domain
$\gamma_{1}:(x,+\infty),\gamma_{2}:(-\infty,x)$.

In addition, they are linked by the scattering matrix $s(\nu)=(s_{ij})$,
namely

$\psi_{-}(x,t,\nu)=\psi_{+}(x,t,\nu)s(\nu).$ (2.13)

###  2.2 Analyticity of the Jost function

In order to study the analytic properties of the matrix Jost functions, we
need to rewrite the spectral equation. Inserting the representations (2.11)
into the Lax pair (2.4), we get

$(A^{-1}\mu_{\pm})_{x}(x,t,\nu)=\nu[\sigma,A^{-1}\mu_{\pm}(x,t,\nu)]+\Omega
A^{-1}\mu_{\pm}(x,t,\nu),$ (2.14a)
$(A^{-1}\mu_{\pm})_{t}(x,t,\nu)=\nu^{-1}[\sigma^{-1},A^{-1}\mu_{\pm}(x,t,\nu)]+\tilde{\Omega}A^{-1}\mu_{\pm}(x,t,\nu),$
(2.14b)

where

$\displaystyle\Omega(x,t,\nu)=A^{-1}\Delta{U}(x,t,\nu)A,\quad\Delta{U}(x,t,\nu)=U(x,t,\nu)-U_{\pm}(x,t,\nu),$
$\displaystyle\tilde{\Omega}(x,t,\nu)=A^{-1}\Delta{V}(x,t,\nu)A,\quad\Delta{V}(x,t,\nu)=V(x,t,\nu)-V_{\pm}(x,t,\nu),$

that is

$\displaystyle\Omega(x,t,\nu)=\frac{1}{3}(\alpha-\alpha^{2})u_{x}\left(\begin{matrix}0&1&-1\\\
-1&0&1\\\ 1&-1&0\end{matrix}\right),$
$\displaystyle\tilde{\Omega}(x,t,\nu)=\frac{1}{3\nu}\left(\begin{matrix}2e^{u}+e^{-2u}-3&\alpha(e^{-2u}-e^{u})&\alpha^{2}(e^{-2u}-e^{u})\\\
\alpha(e^{-2u}-e^{u})&\alpha^{2}(e^{-2u}+2e^{u}-3)&e^{-2u}-e^{u}\\\
\alpha^{2}(e^{-2u}-e^{u})&e^{-2u}-e^{u}&\alpha(e^{-2u}+2e^{u}-3)\end{matrix}\right).$

In the analysis that follows, we choose $t=0$ and the expressions of
$\mu_{\pm}(x,0,\nu)$ can be given from Eq. (2.14)

$\mu_{\pm}(x,0,\nu)=A+\int_{\pm\infty}^{x}Ae^{\nu\hat{\sigma}(x-\xi)}\Omega(\xi)A^{-1}\mu_{\pm}(\xi)d\xi,$
(2.15)

Theorem 1. If $u(x)\in L^{1}(-\infty,a)$ or $L^{1}(a,+\infty)$ for any
constant $a\in\mathbb{R}$, the columns of $\mu_{\pm}(x,0,\nu)$ can be
analytically extended into the corresponding regions of the complex
$\nu$-plane:

$\displaystyle\mu_{+1}(x,0,\nu):~{}\nu\in D_{3}\cup
D_{4},\quad\mu_{-1}(x,0,\nu):~{}\nu\in D_{1}\cup D_{6},$ (2.16)
$\displaystyle\mu_{+2}(x,0,\nu):~{}\nu\in D_{1}\cup
D_{2},\quad\mu_{-2}(x,0,\nu):~{}\nu\in D_{4}\cup D_{5},$
$\displaystyle\mu_{+3}(x,0,\nu):~{}\nu\in D_{5}\cup
D_{6},\quad\mu_{-3}(x,0,\nu):~{}\nu\in D_{2}\cup D_{3},$

where the subscript $j$ refer to the columns and $D_{n}(n=1,2,\cdots,6)$ are
defined by

$\displaystyle D_{1}=\Bigg{\\{}\nu\in\mathbb{C}|\Re l_{2}<\Re l_{3}<\Re
l_{1}\Bigg{\\}},\ D_{2}=\Bigg{\\{}\nu\in\mathbb{C}|\Re l_{2}<\Re l_{1}<\Re
l_{3}\Bigg{\\}},$ (2.17) $\displaystyle D_{3}=\Bigg{\\{}\nu\in\mathbb{C}|\Re
l_{1}<\Re l_{2}<\Re l_{3}\Bigg{\\}},\ D_{4}=\Bigg{\\{}\nu\in\mathbb{C}|\Re
l_{1}<\Re l_{3}<\Re l_{2}\Bigg{\\}},$ $\displaystyle
D_{5}=\Bigg{\\{}\nu\in\mathbb{C}|\Re l_{3}<\Re l_{1}<\Re l_{2}\Bigg{\\}},\
D_{6}=\Bigg{\\{}\nu\in\mathbb{C}|\Re l_{3}<\Re l_{2}<\Re l_{1}\Bigg{\\}},$

namely

$D_{n}=\Bigg{\\{}\nu\in\mathbb{C}:\arg\nu\in\bigg{(}\frac{(n-1)\pi}{3},\frac{n\pi}{3}\bigg{)}\Bigg{\\}},\quad
n=1,\ldots,6.$

Here $l_{i}(\nu)(i=1,2,3)$ are the diagonal entries of matrices $\nu\sigma$.
Proof Setting $J_{\pm}(x,0,\nu)=A^{-1}\mu_{\pm}(x,0,\nu)$, (2.15) can be
rewritten in the column form

$J_{\pm 1}(x,0,\nu)=\left(\begin{array}[]{c}1\\\ 0\\\
0\end{array}\right)+\int_{\pm\infty}^{x}e^{\nu\sigma(x-\xi)}\Omega(\xi)e^{-\nu(x-\xi)}J_{\pm
1}(\xi,t,\nu)d\xi,$ (2.18a) $J_{\pm 2}(x,0,\nu)=\left(\begin{array}[]{c}0\\\
1\\\
0\end{array}\right)+\int_{\pm\infty}^{x}e^{\nu\sigma(x-\xi)}\Omega(\xi)e^{-\alpha\nu(x-\xi)}J_{\pm
2}(\xi,t,\nu)d\xi,$ (2.18b) $J_{\pm 3}(x,0,\nu)=\left(\begin{array}[]{c}0\\\
0\\\
1\end{array}\right)+\int_{\pm\infty}^{x}e^{\nu\sigma(x-\xi)}\Omega(\xi)e^{-\alpha^{2}\nu(x-\xi)}J_{\pm
3}(\xi,t,\nu)d\xi.$ (2.18c)

We consider the integral equation of $J_{+1}(x,0,\nu)$, that is

$J_{+1}(x,0,\nu)=\left(\begin{array}[]{c}1\\\ 0\\\
0\end{array}\right)+\int_{+\infty}^{x}\kappa_{+}(x,\xi,\nu)J_{+1}(\xi,\nu)d\xi,$
(2.19)

where
$\kappa_{+}(x,\xi,\nu)=e^{\nu(x-\xi)\sigma}e^{-\nu(x-\xi)}A^{-1}\tilde{U}A$.
Then $\|\kappa_{+}(x,\xi,\nu)\|\leq 6\rho|u_{x}|,\nu\in D_{3}\cup D_{4}$,
where $\rho=\max\\{\|A\|,\|A\|^{-1}\\},\|\Delta{U}\|=2|u_{x}|$.

Furthermore, we define the Neumann series as

$J_{+1}(x,0,\nu)=\displaystyle\sum_{j=0}^{\infty}C^{(j)}(x,0,\nu),\quad\|C^{(j)}(x,0,\nu)\|\leq\frac{\mu^{j}(x)}{j!},$

where

$\displaystyle\mu(x)=6\rho\int_{+\infty}^{x}|u_{\xi}|d\xi,\quad
C^{(0)}(x,0,\nu)=\left(\begin{array}[]{c}1\\\ 0\\\ 0\end{array}\right),$
$\displaystyle
C^{(j+1)}(x,0,\nu)=\int_{+\infty}^{x}e^{\nu\sigma(x-\xi)}\Omega(\xi)e^{-\nu(x-\xi)}C^{(j)}(\xi,t,\nu)d\xi.$

This prove that $J_{+1}(x,0,\nu)$ is analytic in domain $\nu\in D_{3}\cup
D_{4}$. The others can be proved similarly. $\square$

Under the same hypotheses as in Theorem 1, the scattering coefficients can be
analytically extended from $\Sigma$ (the contour $\Sigma$ is defined as the
boundary of domain $D$ i.e
$\Sigma=\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}\cup\Sigma_{4}\cup\Sigma_{5}\cup\Sigma_{6}$
and $\Sigma_{j}=D_{j}\cap D_{j+1},D_{6+1}=D_{1}$) to the following regions:

$s_{11}:\nu\in D_{1}\cup D_{6},\quad s_{22}:\nu\in D_{4}\cup D_{5},\quad
s_{33}:\nu\in D_{2}\cup D_{3},$ (2.21a) $r_{11}:\nu\in D_{3}\cup D_{4},\quad
r_{22}:\nu\in D_{1}\cup D_{2},\quad r_{33}:\nu\in D_{5}\cup D_{6},$ (2.21b)

where $s_{ij}=(s(\nu))_{ij}$ and $r_{ij}=(s^{-1}(\nu))_{ij}$.

$\pi$$0$$\frac{\pi}{2}$$\frac{3\pi}{2}$$\frac{4\pi}{3}$$\frac{\pi}{3}$$\frac{2\pi}{3}$$\frac{5\pi}{3}$
Figure 2: Domain $D^{+}$, $D^{-}$ and their boundary curve $\partial D$.

Next, we define six solutions $M_{n}(x,0,\nu)$ of (2.14), which take the
following form

$(M_{n})_{ij}(x,0,\nu)=\delta_{ij}+\int_{\gamma_{ij}^{n}}\Big{(}e^{\nu\hat{\sigma}(x-x^{\prime})}\Omega(x^{\prime},0,\nu)M_{n}(x^{\prime},0,\nu)dx^{\prime}\Big{)}_{ij},$
(2.22)

where the contours $\gamma_{ij}^{n}$, $(n=1,\ldots,6,i,j=1,2,3)$ are defined
as

$\gamma_{ij}^{n}=\left\\{\begin{array}[]{ll}\gamma_{1},&\Re(l_{i})>\Re(l_{j}),\\\
\gamma_{2},&\Re(l_{i})<\Re(l_{j}),\end{array}\right.$

and $\gamma_{2}$: $(-\infty,x)$, $\gamma_{1}$: $(x,+\infty).$

For concretely

$\displaystyle\gamma^{1}=\left(\begin{matrix}\gamma_{2}&\gamma_{1}&\gamma_{1}\\\
\gamma_{2}&\gamma_{2}&\gamma_{2}\\\
\gamma_{2}&\gamma_{1}&\gamma_{2}\end{matrix}\right),\quad\gamma^{2}=\left(\begin{matrix}\gamma_{2}&\gamma_{1}&\gamma_{2}\\\
\gamma_{2}&\gamma_{2}&\gamma_{2}\\\
\gamma_{1}&\gamma_{1}&\gamma_{2}\end{matrix}\right),$
$\displaystyle\gamma^{3}=\left(\begin{matrix}\gamma_{2}&\gamma_{2}&\gamma_{2}\\\
\gamma_{1}&\gamma_{2}&\gamma_{2}\\\
\gamma_{1}&\gamma_{1}&\gamma_{2}\end{matrix}\right),\quad\gamma^{4}=\left(\begin{matrix}\gamma_{2}&\gamma_{2}&\gamma_{2}\\\
\gamma_{1}&\gamma_{2}&\gamma_{1}\\\
\gamma_{1}&\gamma_{2}&\gamma_{2}\end{matrix}\right),$
$\displaystyle\gamma^{5}=\left(\begin{matrix}\gamma_{2}&\gamma_{2}&\gamma_{1}\\\
\gamma_{1}&\gamma_{2}&\gamma_{1}\\\
\gamma_{2}&\gamma_{2}&\gamma_{2}\end{matrix}\right),\quad\gamma^{6}=\left(\begin{matrix}\gamma_{2}&\gamma_{1}&\gamma_{1}\\\
\gamma_{2}&\gamma_{2}&\gamma_{1}\\\
\gamma_{2}&\gamma_{2}&\gamma_{2}\end{matrix}\right).$

We note that $\mu_{\pm}(x,t,\nu)$ have the same analytic properties as
$\mu_{\pm}(x,0,\nu)$, and $M_{n}(x,t,\nu)$ can be defined in the same way as
in (2.22), where $e^{\nu\hat{\sigma}(x-x^{\prime})}$ should be replaced by
$e^{\nu\hat{\sigma}(x-x^{\prime})+\nu^{-1}\hat{\sigma}t}$. Supposing that

$\begin{aligned}
M_{n}(x,t,\nu)=\mu_{+}(x,t,\nu)e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}S_{n}(\nu),\\\
M_{n}(x,t,\nu)=\mu_{-}(x,t,\nu)e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}T_{n}(\nu),\end{aligned}\quad\nu\in
D_{n},$ (2.23)

we have

$\displaystyle S_{n}(\nu)=s(\nu)T_{n}(\nu),$ (2.24) $\displaystyle
S_{n}(\nu)=\lim_{x\to\infty}e^{-\nu\hat{\sigma}x-\nu^{-1}\hat{\sigma}^{-1}t}M_{n}(x,t,\nu),$
$\displaystyle
T_{n}(\nu)=\lim_{x\to-\infty}e^{-\nu\hat{\sigma}x-\nu^{-1}\hat{\sigma}^{-1}t}M_{n}(x,t,\nu),$

where $S_{n}$ can be given in terms of the entries of $s(\nu)=(s_{ij}(\nu))$
and $s^{-1}(\nu)=(r_{ij}(\nu))$ as follows

$\displaystyle S_{1}(\nu)=\left(\begin{matrix}s_{11}(\nu)&0&0\\\
s_{21}(\nu)&\frac{1}{r_{22}(\nu)}&\frac{-r_{23}(\nu)}{s_{11}(\nu)}\\\
s_{31}(\nu)&0&\frac{r_{22}(\nu)}{s_{11}(\nu)}\end{matrix}\right),\quad
S_{2}(\nu)=\left(\begin{matrix}\frac{r_{22}(\nu)}{s_{33}(\nu)}&0&s_{13}(\nu)\\\
\frac{-r_{21}(\nu)}{s_{33}(\nu)}&\frac{1}{r_{22}(\nu)}&s_{23}(\nu)\\\
0&0&s_{33}(\nu)\end{matrix}\right),$ $\displaystyle
S_{3}(\nu)=\left(\begin{matrix}\frac{1}{r_{11}(\nu)}&\frac{-r_{12}(\nu)}{s_{33}(\nu)}&s_{13}(\nu)\\\
0&\frac{r_{11}(\nu)}{s_{33}(\nu)}&s_{23}(\nu)\\\
0&0&s_{33}(\nu)\end{matrix}\right),\quad
S_{4}(\nu)=\left(\begin{matrix}\frac{1}{r_{11}(\nu)}&s_{12}(\nu)&\frac{-r_{13}(\nu)}{s_{22}(\nu)}\\\
0&s_{22}(\nu)&0\\\
0&s_{32}(\nu)&\frac{r_{11}(\nu)}{s_{22}(\nu)}\end{matrix}\right),$
$\displaystyle
S_{5}(\nu)=\left(\begin{matrix}\frac{r_{33}(\nu)}{s_{22}(\nu)}&s_{12}(\nu)&0\\\
0&s_{22}(\nu)&0\\\
-\frac{r_{31}(\nu)}{s_{22}(\nu)}&s_{32}(\nu)&\frac{1}{r_{33}(\nu)}\end{matrix}\right),\quad
S_{6}(\nu)=\left(\begin{matrix}s_{11}(\nu)&0&0\\\
s_{21}(\nu)&\frac{r_{33}(\nu)}{s_{11}(\nu)}&0\\\
s_{31}(\nu)&\frac{-r_{32}(\nu)}{s_{11}(\nu)}&\frac{1}{r_{33}(\nu)}\end{matrix}\right).$

###  2.3 Adjoint problem and auxiliary eigenfunctions

It is remarked that, for the Tzitzéica equation (1.1), traditional inverse
scattering transform may fail to construct a set of analytic eigenfunctions in
any given domain. One efficient technique is to introduce the adjoint Lax pair
[3, 11]

$\tilde{\psi}_{x}=\tilde{U}(x,\nu,u)\tilde{\psi},$ (2.25a)
$\tilde{\psi}_{t}=\tilde{V}(x,\nu,u)\tilde{\psi},$ (2.25b)

where

$\tilde{U}(x,\nu,u)=U^{\ast}(x,-\nu^{\ast},u),\quad\tilde{V}(x,\nu,u)=V^{\ast}(x,-\nu^{\ast},u),$

and the star denotes the complex conjugate.

We note that if $\phi(x,t,\nu)$ and $\varphi(x,t,\nu)$ are two arbitrary
vector eigenfunctions of the adjoint problem (2.24), then

$\frac{T_{1}}{3\gamma^{\ast}}[\phi\times\varphi](x,t,\nu),\quad
T_{1}=\left(\begin{matrix}0&1&0\\\ 1&0&0\\\ 0&0&1\end{matrix}\right),$ (2.26)

is an eigenfunction of the linear system (2.4). Here the symbol $\times$
denotes the usual cross product.

Suppose $\tilde{\psi}_{\pm}(x,t,\nu)$ defined similarly as in (2.9) are
fundamental matrix solutions of the (2.24a), where $\tilde{A}$ should be
$\tilde{A}=A^{\ast}$, then there exists an invertible $3\times 3$ matrix
$\tilde{s}(\nu)$ such that

$\tilde{\psi}_{-}(x,t,\nu)=\tilde{\psi}_{+}(x,t,\nu)\tilde{s}(\nu).$

Introducing the modified adjoint eigenfunction $\tilde{\mu}_{\pm}(x,t,\nu)$
defined by

$\tilde{\psi}_{\pm}=\tilde{\mu}_{\pm}(x,t,\nu)e^{-\nu\sigma^{-1}x-\nu^{-1}\sigma
t},$

and proceeding the same analysis for $\mu_{\pm}(x,t,\nu)$, we know that
different collum of $\tilde{\mu}_{\pm}(x,t,\nu)$ can be analyticity extended
into the following domains

$\displaystyle\tilde{\mu}_{+1}(x,t,\nu):~{}\nu\in D_{1}\cup
D_{6},\quad\tilde{\mu}_{-1}(x,t,\nu):~{}\nu\in D_{3}\cup D_{4},$ (2.27)
$\displaystyle\tilde{\mu}_{+2}(x,t,\nu):~{}\nu\in D_{2}\cup
D_{3},\quad\tilde{\mu}_{-2}(x,t,\nu):~{}\nu\in D_{5}\cup D_{6},$
$\displaystyle\tilde{\mu}_{+3}(x,t,\nu):~{}\nu\in D_{4}\cup
D_{5},\quad\tilde{\mu}_{-3}(x,t,\nu):~{}\nu\in D_{1}\cup D_{2}.$

Then the associated scattering coefficients can be similarly extended in the
following region

$\tilde{s}_{11}:\nu\in D_{3}\cup D_{4},\quad\tilde{s}_{22}:\nu\in D_{5}\cup
D_{6},\quad\tilde{s}_{33}:\nu\in D_{1}\cup D_{2},$ (2.28a)
$\tilde{r}_{11}:\nu\in D_{1}\cup D_{6},\quad\tilde{r}_{22}:\nu\in D_{2}\cup
D_{3},\quad\tilde{r}_{33}:\nu\in D_{4}\cup D_{5}.$ (2.28b)

Using (2.26) and the solutions $\tilde{\psi}_{\pm}$ of the adjoint problem
(2.24), we construct six novel solutions for the Lax pair (2.4) as

$\chi_{1}(x,t,\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{-3}\times\tilde{\psi}_{+1}](x,t,\nu),$
(2.29a)
$\chi_{2}(x,t,\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{+2}\times\tilde{\psi}_{-3}](x,t,\nu),$
(2.29b)
$\chi_{3}(x,t,\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{-1}\times\tilde{\psi}_{+2}](x,t,\nu),$
(2.29c)
$\chi_{4}(x,t,\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{+3}\times\tilde{\psi}_{-1}](x,t,\nu),$
(2.29d)
$\chi_{5}(x,t,\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{-2}\times\tilde{\psi}_{+3}](x,t,\nu),$
(2.29e)
$\chi_{6}(x,t,\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{+1}\times\tilde{\psi}_{-2}](x,t,\nu).$
(2.29f)

Here, $\chi_{1}(x,t,\nu),\cdots,\chi_{6}(x,t,\nu)$ are called the auxiliary
eigenfunctions. Due to the Tzitzéica equation has six different domains of
analyticity, we need to define six auxiliary eigenfunctions. Furthermore,
$\chi_{n}(x,t,\nu)$ is analytic in domain $\nu\in D_{n},n=1,\cdots,6$.

In addition, for $\nu\in\Sigma$, the Jost functions $\psi_{\pm}$ in (2.9) and
the solutions $\tilde{\psi}_{\pm}$ of the adjoint problem (2.24) satisfy the
following relations

$\psi_{\pm 1}(\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{\pm
2}\times\tilde{\psi}_{\pm 3}](x,t,\nu),$ (2.30a) $\psi_{\pm
2}(\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{\pm
1}\times\tilde{\psi}_{\pm 2}](x,t,\nu),$ (2.30b) $\psi_{\pm
3}(\nu)=\frac{T_{1}}{3\gamma^{\ast}}[\tilde{\psi}_{\pm
3}\times\tilde{\psi}_{\pm 1}](x,t,\nu),$ (2.30c) $\tilde{\psi}_{\pm
1}(\nu)=\frac{T_{1}}{3\gamma}[\psi_{\pm 2}\times\psi_{\pm 3}](x,t,\nu),$
(2.30d) $\tilde{\psi}_{\pm 2}(\nu)=\frac{T_{1}}{3\gamma}[\psi_{\pm
1}\times\psi_{\pm 2}](x,t,\nu),$ (2.30e) $\tilde{\psi}_{\pm
3}(\nu)=\frac{T_{1}}{3\gamma}[\psi_{\pm 3}\times\psi_{\pm 1}](x,t,\nu).$
(2.30f)

Based on the above facts, the scattering matrices $s(\nu)$ and
$\tilde{s}(\nu)$ are related by

$\tilde{s}(\nu)=\Gamma\bigg{(}s^{-1}(\nu)\bigg{)}^{T}\Gamma,\quad\Gamma=\left(\begin{matrix}1&0&0\\\
0&0&1\\\ 0&1&0\end{matrix}\right).$ (2.31)

For all $\nu\in\Sigma$, the Jost functions and the auxiliary eigenfunctions
obey the following equations

$\psi_{+1}(\nu)=\frac{1}{r_{22}(\nu)}[\chi_{2}(\nu)+\psi_{+2}(\nu)r_{21}(\nu)]=\frac{1}{r_{33}(\nu)}[\chi_{5}(\nu)+\psi_{+3}(\nu)r_{31}(\nu)],$
(2.32a)
$\psi_{+2}(\nu)=\frac{1}{r_{11}(\nu)}[\chi_{3}(\nu)+\psi_{+1}(\nu)r_{12}(\nu)]=\frac{1}{r_{33}(\nu)}[\chi_{6}(\nu)+\psi_{+3}(\nu)r_{32}(\nu)],$
(2.32b)
$\psi_{+3}(\nu)=\frac{1}{r_{22}(\nu)}[\chi_{1}(\nu)+\psi_{+2}(\nu)r_{23}(\nu)]=\frac{1}{r_{11}(\nu)}[\chi_{4}(\nu)+\psi_{+1}(\nu)r_{13}(\nu)],$
(2.32c)
$\psi_{-1}(\nu)=\frac{1}{s_{33}(\nu)}[\chi_{2}(\nu)+\psi_{-3}(\nu)s_{31}(\nu)]=\frac{1}{s_{22}(\nu)}[\chi_{5}(\nu)+\psi_{-2}(\nu)s_{21}(\nu)],$
(2.32d)
$\psi_{-2}(\nu)=\frac{1}{s_{33}(\nu)}[\chi_{3}(\nu)+\psi_{-3}(\nu)s_{32}(\nu)]=\frac{1}{s_{11}(\nu)}[\chi_{6}(\nu)+\psi_{-1}(\nu)s_{12}(\nu)],$
(2.32e)
$\psi_{-3}(\nu)=\frac{1}{s_{11}(\nu)}[\chi_{1}(\nu)+\psi_{-1}(\nu)s_{13}(\nu)]=\frac{1}{s_{22}(\nu)}[\chi_{4}(\nu)+\psi_{-2}(\nu)s_{23}(\nu)].$
(2.32f)

###  2.4 symmetries

First symmetry. It is readily verified that if $\psi(x,t,\nu)$ is a
fundamental matrix solution of the Lax pair (2.4), so is
$T_{1}\bigg{(}\psi^{\dagger}(x,t,-\nu^{\ast})\bigg{)}^{-1}$, where $T_{1}$ is
defined in (2.25).

For all $\nu\in\Sigma$, the Jost functions satisfy the symmetry

$T_{1}\bigg{(}\psi_{\pm}^{\dagger}(x,t,-\nu^{\ast})\bigg{)}^{-1}\frac{\Gamma}{3}=\psi_{\pm}(x,t,\nu),$

where

$\bigg{(}\psi^{-1}_{\pm}(x,t,\nu)\bigg{)}^{T}=\frac{1}{\det\psi_{\pm}(x,t,\nu)}\bigg{(}\psi_{\pm
2}\times\psi_{\pm 3},\psi_{\pm 3}\times\psi_{\pm 1},\psi_{\pm
1}\times\psi_{\pm 2}\bigg{)}(x,t,\nu).$

This symmetry implies a condition between the scattering matrix and its
inverse

$s^{-1}(\nu)=\Gamma s^{\dagger}(-\nu^{\ast})\Gamma^{-1},\quad\nu\in\Sigma,$
(2.33)

or, in component form

$r_{11}(\nu)=s^{\ast}_{11}(-\nu^{\ast}),\quad
r_{12}(\nu)=s^{\ast}_{31}(-\nu^{\ast}),\quad
r_{13}(\nu)=s^{\ast}_{21}(-\nu^{\ast}),$ (2.34a)
$r_{21}(\nu)=s^{\ast}_{13}(-\nu^{\ast}),\quad
r_{22}(\nu)=s^{\ast}_{33}(-\nu^{\ast}),\quad
r_{23}(\nu)=s^{\ast}_{23}(-\nu^{\ast}),$ (2.34b)
$r_{31}(\nu)=s^{\ast}_{12}(-\nu^{\ast}),\quad
r_{32}(\nu)=s^{\ast}_{32}(-\nu^{\ast}),\quad
r_{33}(\nu)=s^{\ast}_{22}(-\nu^{\ast}).$ (2.34c)

The Schwarz reflection principle of (2.33) gives to the following results

$r_{11}(\nu)=s^{\ast}_{11}(-\nu^{\ast}),\quad\nu\in D_{3}\cup D_{4},$ (2.35a)
$r_{22}(\nu)=s^{\ast}_{33}(-\nu^{\ast}),\quad\nu\in D_{1}\cup D_{2},$ (2.35b)
$r_{33}(\nu)=s^{\ast}_{22}(-\nu^{\ast}),\quad\nu\in D_{5}\cup D_{6}.$ (2.35c)

The Jost functions and the auxiliary eigenfunctions obey the following
symmetry relations

$\psi^{\ast}_{+1}(-\nu^{\ast})=\frac{T_{1}}{3\gamma
r_{33}(\nu)}[\chi_{6}(\nu)\times\psi_{+3}(\nu)]=\frac{T_{1}}{3\gamma
r_{22}(\nu)}[\psi_{+2}(\nu)\times\chi_{1}(\nu)],$ (2.36a)
$\psi^{\ast}_{-1}(-\nu^{\ast})=\frac{T_{1}}{3\gamma
s_{33}(\nu)}[\chi_{3}(\nu)\times\psi_{-3}(\nu)]=\frac{T_{1}}{3\gamma
s_{22}(\nu)}[\psi_{-2}(\nu)\times\chi_{4}(\nu)],$ (2.36b)
$\psi^{\ast}_{+2}(-\nu^{\ast})=\frac{T_{1}}{3\gamma
r_{22}(\nu)}[\chi_{2}(\nu)\times\psi_{+2}(\nu)]=\frac{T_{1}}{3\gamma
r_{11}(\nu)}[\psi_{+1}(\nu)\times\chi_{3}(\nu)],$ (2.36c)
$\psi^{\ast}_{-2}(-\nu^{\ast})=\frac{T_{1}}{3\gamma
s_{22}(\nu)}[\chi_{5}(\nu)\times\psi_{-2}(\nu)]=\frac{T_{1}}{3\gamma
s_{11}(\nu)}[\psi_{-1}(\nu)\times\chi_{6}(\nu)],$ (2.36d)
$\psi^{\ast}_{+3}(-\nu^{\ast})=\frac{T_{1}}{3\gamma
r_{11}(\nu)}[\chi_{4}(\nu)\times\psi_{+1}(\nu)]=\frac{T_{1}}{3\gamma
r_{33}(\nu)}[\psi_{+3}(\nu)\times\chi_{5}(\nu)],$ (2.36e)
$\psi^{\ast}_{-3}(-\nu^{\ast})=\frac{T_{1}}{3\gamma
s_{11}(\nu)}[\chi_{1}(\nu)\times\psi_{-1}(\nu)]=\frac{T_{1}}{3\gamma
s_{33}(\nu)}[\psi_{-3}(\nu)\times\chi_{2}(\nu)].$ (2.36f)

Besides, the auxiliary eigenfunctions and the Jost functions satisfy another
symmetry conditions

$\chi^{\ast}_{1}(x,t,-\nu^{\ast})=\frac{T_{1}}{3\gamma}[\psi_{-3}\times\psi_{+1}](x,t,\nu),\quad\nu\in
D_{3},$ (2.37a)
$\chi^{\ast}_{2}(x,t,-\nu^{\ast})=\frac{T_{1}}{3\gamma}[\psi_{+2}\times\psi_{-3}](x,t,\nu),\quad\nu\in
D_{2},$ (2.37b)
$\chi^{\ast}_{3}(x,t,-\nu^{\ast})=\frac{T_{1}}{3\gamma}[\psi_{-1}\times\psi_{+2}](x,t,\nu),\quad\nu\in
D_{1},$ (2.37c)
$\chi^{\ast}_{4}(x,t,-\nu^{\ast})=\frac{T_{1}}{3\gamma}[\psi_{+3}\times\psi_{-1}](x,t,\nu),\quad\nu\in
D_{6},$ (2.37d)
$\chi^{\ast}_{5}(x,t,-\nu^{\ast})=\frac{T_{1}}{3\gamma}[\psi_{-2}\times\psi_{+3}](x,t,\nu),\quad\nu\in
D_{5},$ (2.37e)
$\chi^{\ast}_{6}(x,t,-\nu^{\ast})=\frac{T_{1}}{3\gamma}[\psi_{+1}\times\psi_{-2}](x,t,\nu),\quad\nu\in
D_{4}.$ (2.37f)

Second symmetry. We note that if $\psi_{\pm}(x,t,\nu)$ is a solution of the
Lax pair (2.4), so is $\psi_{\pm}(x,t,\alpha^{2}\nu)$.

For all $\nu\in\Sigma$, the Jost eigenfunctions satisfy the symmetry

$\psi_{\pm}(\nu)=\alpha\sigma\psi_{\pm}(\alpha^{2}\nu)\Delta,\quad\Delta=\left(\begin{matrix}0&0&1\\\
1&0&0\\\ 0&1&0\end{matrix}\right).$ (2.38)

In component form, (2.38) yields

$\psi_{\pm 1}(\nu)=\alpha\sigma\psi_{\pm 2}(\alpha^{2}\nu),\quad\nu\in\Re
l_{1}\lessgtr\Re l_{j},\quad j=2,3,$ (2.39a) $\psi_{\pm
2}(\nu)=\alpha\sigma\psi_{\pm 3}(\alpha^{2}\nu),\quad\nu\in\Re
l_{2}\lessgtr\Re l_{j},\quad j=1,3,$ (2.39b) $\psi_{\pm
3}(\nu)=\alpha\sigma\psi_{\pm 1}(\alpha^{2}\nu),\quad\nu\in\Re
l_{3}\lessgtr\Re l_{j},\quad j=1,2.$ (2.39c)

The scattering matrix satisfies the symmetry

$s(\alpha^{2}\nu)=\Delta s(\nu)\Delta^{-1},\quad\nu\in\Sigma.$ (2.40)

Componentwise, we have

$s_{11}(\nu)=s_{22}(\alpha^{2}\nu),\quad
s_{12}(\nu)=s_{23}(\alpha^{2}\nu),\quad s_{13}(\nu)=s_{21}(\alpha^{2}\nu),$
(2.41a) $s_{21}(\nu)=s_{32}(\alpha^{2}\nu),\quad
s_{22}(\nu)=s_{33}(\alpha^{2}\nu),\quad s_{23}(\nu)=s_{31}(\alpha^{2}\nu),$
(2.41b) $s_{31}(\nu)=s_{12}(\alpha^{2}\nu),\quad
s_{32}(\nu)=s_{13}(\alpha^{2}\nu),\quad s_{33}(\nu)=s_{11}(\alpha^{2}\nu).$
(2.41c)

Beside, the analyticity properties of the scattering matrix imply that

$\displaystyle s_{11}(\nu)=s_{22}(\alpha^{2}\nu),~{}\nu\in D_{1}\cup
D_{6},\quad s_{22}(\nu)=s_{33}(\alpha^{2}\nu),~{}\nu\in D_{4}\cup D_{5},$
(2.42a) $\displaystyle s_{33}(\nu)=s_{11}(\alpha^{2}\nu),~{}\nu\in D_{2}\cup
D_{3},$ $\displaystyle r_{11}(\nu)=r_{22}(\alpha^{2}\nu),~{}\nu\in D_{3}\cup
D_{4},\quad r_{22}(\nu)=r_{33}(\alpha^{2}\nu),~{}\nu\in D_{1}\cup D_{2},$
(2.42b) $\displaystyle r_{33}(\nu)=r_{11}(\alpha^{2}\nu),~{}\nu\in D_{5}\cup
D_{6}.$

Finally, taking (2.31), (2.37) and (2.39) into consideration, we find

$\chi_{1}(\nu)=\alpha\sigma\chi_{5}(\alpha^{2}\nu)=\alpha^{2}\sigma^{2}\chi_{3}(\alpha\nu),\quad\nu\in
D_{1},$ (2.43a)
$\chi_{2}(\nu)=\alpha\sigma\chi_{6}(\alpha^{2}\nu)=\alpha^{2}\sigma^{2}\chi_{4}(\alpha\nu),\quad\nu\in
D_{2}.$ (2.43b)

Third symmetry. It is the fact that $\psi_{\pm}(x,t,\nu)$ and
$\psi^{\ast}_{\pm}(x,t,\nu^{\ast})$ are both the solutions of the system
(2.4).

For all $\nu\in\Sigma$, the Jost eigenfunctions satisfy the symmetry

$\psi_{\pm}(x,t,\nu)=\psi^{\ast}_{\pm}(x,t,\nu^{\ast})\Gamma,$ (2.44)

or equivalently

$\psi_{\pm 1}(x,t,\nu)=\psi^{\ast}_{\pm 1}(x,t,\nu^{\ast}),\quad\nu\in\Re
l_{1}\lessgtr\Re l_{j},~{}j=2,3,$ (2.45a) $\psi_{\pm
2}(x,t,\nu)=\psi^{\ast}_{\pm 3}(x,t,\nu^{\ast}),\quad\nu\in\Re
l_{2}\lessgtr\Re l_{j},~{}j=1,3,$ (2.45b) $\psi_{\pm
3}(x,t,\nu)=\psi^{\ast}_{\pm 2}(x,t,\nu^{\ast}),\quad\nu\in\Re
l_{3}\lessgtr\Re l_{j},~{}j=1,2.$ (2.45c)

In the same way, the scattering matrices admit the symmetry

$s^{\ast}(\nu^{\ast})=\Gamma s(\nu)\Gamma.$ (2.46)

Componentwise, we have

$s^{\ast}_{11}(\nu^{\ast})=s_{11}(\nu),\quad
s^{\ast}_{12}(\nu^{\ast})=s_{13}(\nu),\quad
s^{\ast}_{13}(\nu^{\ast})=s_{12}(\nu),$ (2.47a)
$s^{\ast}_{21}(\nu^{\ast})=s_{31}(\nu),\quad
s^{\ast}_{22}(\nu^{\ast})=s_{33}(\nu),\quad
s^{\ast}_{23}(\nu^{\ast})=s_{32}(\nu),$ (2.47b)
$s^{\ast}_{31}(\nu^{\ast})=s_{21}(\nu),\quad
s^{\ast}_{32}(\nu^{\ast})=s_{23}(\nu),\quad
s^{\ast}_{33}(\nu^{\ast})=s_{22}(\nu).$ (2.47c)

In addition, the analyticity properties of the scattering matrix give the
following results

$\displaystyle s_{11}(\nu)=s^{\ast}_{11}(\nu^{\ast}),~{}\nu\in D_{1}\cup
D_{6},\quad s_{22}(\nu)=s^{\ast}_{33}(\nu^{\ast}),~{}\nu\in D_{4}\cup D_{5},$
(2.48a) $\displaystyle s_{33}(\nu)=s^{\ast}_{22}(\nu^{\ast}),~{}\nu\in
D_{2}\cup D_{3},$ $\displaystyle
r_{11}(\nu)=r^{\ast}_{11}(\nu^{\ast}),~{}\nu\in D_{3}\cup D_{4},\quad
r_{22}(\nu)=r^{\ast}_{33}(\nu^{\ast}),~{}\nu\in D_{1}\cup D_{2},$ (2.48b)
$\displaystyle r_{33}(\nu)=r^{\ast}_{22}(\nu^{\ast}),~{}\nu\in D_{5}\cup
D_{6}.$

The auxiliary eigenfunctions satisfy the symmetries

$\chi_{1}(\nu)=-\frac{\gamma}{\gamma^{\ast}}\chi^{\ast}_{6}(\nu^{\ast}),\quad\nu\in
D_{1},$ (2.49a)
$\chi_{2}(\nu)=-\frac{\gamma}{\gamma^{\ast}}\chi^{\ast}_{5}(\nu^{\ast}),\quad\nu\in
D_{2},$ (2.49b)
$\chi_{3}(\nu)=-\frac{\gamma}{\gamma^{\ast}}\chi^{\ast}_{4}(\nu^{\ast}),\quad\nu\in
D_{3}.$ (2.49c)

The matrix spectral function $s(\nu)$ has the following properties:
(i)
$s(\nu)=I+o(\frac{1}{\nu}),~{}s^{-1}(\nu)=I+o(\frac{1}{\nu}),\quad\nu\longrightarrow\infty,$
(ii) $\det s(\nu)=1,~{}\det s^{-1}(\nu)=1.$

##  3 Discrete spectrum

###  3.1 Discrete spectrum

In the following, we switch our attention to discuss the discrete spectrum for
the Tzitzéica equation. To this end, we introduce the following six $3\times
3$ matrices

$\mathbf{G}_{1}(x,t,\nu)=\bigg{(}\psi_{-1}(x,t,\nu),\psi_{+2}(x,t,\nu),\chi_{1}(x,t,\nu)\bigg{)},\quad\nu\in
D_{1},$ (3.1a)
$\mathbf{G}_{2}(x,t,\nu)=\bigg{(}\chi_{2}(x,t,\nu),\psi_{+2}(x,t,\nu),\psi_{-3}(x,t,\nu)\bigg{)},\quad\nu\in
D_{2},$ (3.1b)
$\mathbf{G}_{3}(x,t,\nu)=\bigg{(}\psi_{+1}(x,t,\nu),\chi_{3}(x,t,\nu),\psi_{-3}(x,t,\nu)\bigg{)},\quad\nu\in
D_{3},$ (3.1c)
$\mathbf{G}_{4}(x,t,\nu)=\bigg{(}\psi_{+1}(x,t,\nu),\psi_{-2}(x,t,\nu),\chi_{4}(x,t,\nu)\bigg{)},\quad\nu\in
D_{4},$ (3.1d)
$\mathbf{G}_{5}(x,t,\nu)=\bigg{(}\chi_{5}(x,t,\nu),\psi_{-2}(x,t,\nu),\psi_{+3}(x,t,\nu)\bigg{)},\quad\nu\in
D_{5},$ (3.1e)
$\mathbf{G}_{6}(x,t,\nu)=\bigg{(}\psi_{-1}(x,t,\nu),\chi_{6}(x,t,\nu),\psi_{+3}(x,t,\nu)\bigg{)},\quad\nu\in
D_{6}.$ (3.1f)

We note that the above matrix function are analytic in their respective
domain. From (2.31), we know that

$\det\Big{(}\mathbf{G}_{1}(x,t,\nu)\Big{)}=s_{11}(\nu)r_{22}(\nu)\gamma,\quad\nu\in
D_{1},$ (3.2a)
$\det\Big{(}\mathbf{G}_{2}(x,t,\nu)\Big{)}=s_{33}(\nu)r_{22}(\nu)\gamma,\quad\nu\in
D_{2},$ (3.2b)
$\det\Big{(}\mathbf{G}_{3}(x,t,\nu)\Big{)}=r_{11}(\nu)s_{33}(\nu)\gamma,\quad\nu\in
D_{3},$ (3.2c)
$\det\Big{(}\mathbf{G}_{4}(x,t,\nu)\Big{)}=r_{11}(\nu)s_{22}(\nu)\gamma,\quad\nu\in
D_{4},$ (3.2d)
$\det\Big{(}\mathbf{G}_{5}(x,t,\nu)\Big{)}=r_{33}(\nu)s_{22}(\nu)\gamma,\quad\nu\in
D_{5},$ (3.2e)
$\det\Big{(}\mathbf{G}_{6}(x,t,\nu)\Big{)}=s_{11}(\nu)s_{33}(\nu)\gamma,\quad\nu\in
D_{6}.$ (3.2f)

Thus, we can easily observe that the columns of $\mathbf{G}_{1}(x,t,\nu)$
become linearly dependent at the zeros of $s_{11}(\nu)$ and $r_{22}(\nu)$.

According to the symmetry conditions about the scattering matrices discussed
in section 2.4, we can locate the distribution of the discrete spectrum. For
convenience, we let $\nu_{0}\in D_{2}$ be a discrete eigenvalue of the
scattering problem. It is easy to see that $\nu_{0}$ can be divided into two
types of eigenvalue.
1\. The first type is that $\nu_{0}=-\nu_{0}^{\ast}$.
2\. The second type is that $\nu_{0}\neq-\nu_{0}^{\ast}$.

When $\nu_{0}=-\nu_{0}^{\ast}$, the symmetry conditions about the scattering
matrices imply that

$\displaystyle s_{33}(\nu_{0})=0\Leftrightarrow
s_{22}(\nu_{0}^{\ast})=0\Leftrightarrow s_{11}(\alpha\nu_{0}^{\ast})=0$
$\displaystyle\Leftrightarrow s_{11}(\alpha^{2}\nu_{0})=0\Leftrightarrow
s_{22}(\alpha\nu_{0})=0\Leftrightarrow s_{33}(\alpha^{2}\nu_{0}^{\ast})=0$
$\displaystyle\Leftrightarrow
r_{11}(\alpha^{2}\nu_{0}^{\ast})=0\Leftrightarrow
r_{11}(\alpha\nu_{0})=0\Leftrightarrow r_{22}(\nu_{0})=0$
$\displaystyle\Leftrightarrow r_{22}(\alpha\nu_{0}^{\ast})=0\Leftrightarrow
r_{33}(\nu_{0}^{\ast})=0\Leftrightarrow r_{33}(\alpha^{2}\nu_{0})=0.$ (3.3)

When $\nu_{0}\neq-\nu_{0}^{\ast}$, we can derive

$\displaystyle s_{33}(\nu_{0})=0\Leftrightarrow
s_{22}(\nu_{0}^{\ast})=0\Leftrightarrow s_{11}(\alpha\nu_{0}^{\ast})=0$
$\displaystyle\Leftrightarrow s_{11}(\alpha^{2}\nu_{0})=0\Leftrightarrow
s_{22}(\alpha\nu_{0})=0\Leftrightarrow s_{33}(\alpha^{2}\nu_{0}^{\ast})=0$
$\displaystyle\Leftrightarrow r_{11}(-\alpha^{2}\nu_{0})=0\Leftrightarrow
r_{11}(-\alpha\nu_{0}^{\ast})=0\Leftrightarrow r_{22}(-\nu_{0}^{\ast})=0$
$\displaystyle\Leftrightarrow r_{22}(-\alpha\nu_{0})=0\Leftrightarrow
r_{33}(-\nu_{0})=0\Leftrightarrow r_{33}(-\alpha^{2}\nu_{0}^{\ast})=0.$ (3.4)

In addition, for $\nu_{0}=-\nu_{0}^{\ast},\nu_{0}\in D_{2}$, Jost functions
and the auxiliary eigenfunctions admit the following statements
(1): $\chi_{2}(\nu_{0})=\textbf{0},$
(2): $\chi_{4}(\alpha\nu_{0})=\textbf{0},$
(3): $\chi_{6}(\alpha^{2}\nu_{0})=\textbf{0},$
(4): There exists a constant $b_{2}$ such that
$\psi_{+2}(-\nu_{0}^{\ast})=b_{2}\psi_{-3}(-\nu_{0}^{\ast}),$
(5): There exists a constant $b_{4}$ such that
$\psi_{+3}(-\alpha^{2}\nu_{0}^{\ast})=b_{4}\psi_{-1}(-\alpha^{2}\nu_{0}^{\ast}),$
(6): There exists a constant $b_{6}$ such that
$\psi_{+1}(-\alpha\nu_{0}^{\ast})=b_{6}\psi_{-2}(-\alpha\nu_{0}^{\ast}).$

Similarly, when $\nu_{0}\neq-\nu_{0}^{\ast},\nu_{0}\in D_{2}$, the following
statements are equivalent
(1): $\chi_{1}(\alpha\nu_{0}^{\ast})=\textbf{0},$
(2): $\chi_{3}(\alpha^{2}\nu_{0}^{\ast})=\textbf{0},$
(3): $\chi_{5}(\nu_{0}^{\ast})=\textbf{0},$
(4): There exists a constant $b_{1}$ such that
$\psi_{+1}(-\alpha^{2}\nu_{0})=b_{1}\psi_{-3}(-\alpha^{2}\nu_{0}),$
(5): There exists a constant $b_{3}$ such that
$\psi_{+2}(-\alpha\nu_{0})=b_{3}\psi_{-1}(-\alpha\nu_{0}),$
(6): There exists a constant $b_{5}$ such that
$\psi_{+3}(-\nu_{0})=b_{5}\psi_{-2}(-\nu_{j})$.

In what follows, we assume that the zeros of $s_{jj}(\nu)$ and
$r_{jj}(\nu)(j=1,2,3)$ are simple.

(i) If $\nu_{0}$ is an eigenvalue of the first type, we can directly get
$\chi_{1}(\alpha\nu_{0}^{\ast})=\chi_{2}(\nu_{0})=\textbf{0}$. As a result, we
observe that

$\psi_{+2}(\nu_{0})=f_{0}\psi_{-3}(\nu_{0}),\quad\psi_{+1}(\alpha^{2}\nu_{0}^{\ast})=\hat{f}_{0}\psi_{-3}(\alpha^{2}\nu_{0}^{\ast}),$
(3.5a)
$\psi_{+2}(\alpha\nu_{0}^{\ast})=\check{f}_{0}\psi_{-1}(\alpha\nu_{0}^{\ast}),\quad\psi_{+3}(\alpha^{2}\nu_{0})=\breve{f}_{0}\psi_{-1}(\alpha^{2}\nu_{0}),$
(3.5b)
$\psi_{+3}(\nu_{0}^{\ast})=\acute{f}_{0}\psi_{-2}(\nu_{0}^{\ast}),\quad\psi_{+1}(\alpha\nu_{0})=\grave{f}_{0}\psi_{-2}(\alpha\nu_{0}),$
(3.5c)

where $f_{0},\hat{f}_{0},\check{f}_{0},\breve{f}_{0},\acute{f}_{0}$, and
$\grave{f}_{0}$ are the associated proportionality constants.

(ii) Suppose $\nu_{0}$ is an eigenvalue of the second type. In this case, we
find that

$\chi_{2}(\nu_{0})=\hat{c}_{0}\psi_{-3}(\nu_{0}),\quad\psi_{+2}(-\nu_{0}^{\ast})=\hat{d}_{0}\chi_{2}(-\nu_{0}^{\ast}),$
(3.6a)
$\chi_{1}(\alpha\nu_{0}^{\ast})=c_{0}\psi_{-1}(\alpha\nu_{0}^{\ast}),\quad\psi_{+2}(-\alpha\nu_{0})=d_{0}\chi_{1}(-\alpha\nu_{0}),$
(3.6b)
$\chi_{3}(\alpha^{2}\nu_{0}^{\ast})=\check{c}_{0}\psi_{-3}(\alpha^{2}\nu_{0}^{\ast}),\quad\psi_{+1}(-\alpha^{2}\nu_{0})=\check{d}_{0}\chi_{3}(-\alpha^{2}\nu_{0}),$
(3.6c)
$\chi_{4}(\alpha\nu_{0})=\breve{c}_{0}\psi_{-2}(\alpha\nu_{0}),\quad\psi_{+1}(-\alpha\nu_{0}^{\ast})=\breve{d}_{0}\chi_{4}(-\alpha\nu_{0}^{\ast}),$
(3.6d)
$\chi_{5}(\nu_{0}^{\ast})=\acute{c}_{0}\psi_{-2}(\nu_{0}^{\ast}),\quad\psi_{+3}(-\nu_{0})=\acute{d}_{0}\chi_{5}(-\nu_{0}),$
(3.6e)
$\chi_{6}(\alpha^{2}\nu_{0})=\grave{c}_{0}\psi_{-1}(\alpha^{2}\nu_{0}),\quad\psi_{+3}(-\alpha^{2}\nu_{0}^{\ast})=\grave{d}_{0}\chi_{6}(-\alpha^{2}\nu_{0}^{\ast}),$
(3.6f)

where, $d_{0},\hat{d}_{0},\check{d}_{0},\breve{d}_{0},\acute{d}_{0}$,
$\grave{d}_{0}$, $c_{0},\hat{c}_{0},\check{c}_{0},\breve{c}_{0},\acute{c}_{0}$
and $\grave{c}_{0}$ are the associated proportionality constants.

###  3.2 Symmetries of the norming constants

To obtain the residue conditions for the inverse problem, we rewrite the
equations (3.5) and (3.6) in terms of the modified eigenfunctions. Setting
$\\{\zeta_{j}\\}_{j=1}^{N_{1}}$ be the set of all eigenvalues of the first
type, then from (2.11) and (3.5), we have

$\mu_{+2}(\zeta_{j})=f_{j}\mu_{-3}(\zeta_{j})e^{(\alpha^{2}-\alpha)\zeta_{j}x+(\alpha-\alpha^{2})\ddot{\zeta_{j}}t},$
(3.7a)
$\mu_{+1}(\alpha^{2}\zeta_{j}^{\ast})=\hat{f}_{j}\mu_{-3}(\alpha^{2}\zeta_{j}^{\ast})e^{(\alpha-\alpha^{2})\zeta_{j}^{\ast}x+(\alpha^{2}-\alpha)\ddot{\zeta_{j}}^{\ast}t},$
(3.7b)
$\mu_{+2}(\alpha\zeta_{j}^{\ast})=\check{f}_{j}\mu_{-1}(\alpha\zeta_{j}^{\ast})e^{(\alpha-\alpha^{2})\zeta_{j}^{\ast}x+(\alpha^{2}-\alpha)\ddot{\zeta_{j}}^{\ast}t}$
(3.7c)
$\mu_{+3}(\alpha^{2}\zeta_{j})=\breve{f}_{j}\mu_{-1}(\alpha^{2}\zeta_{j})e^{(\alpha^{2}-\alpha)\zeta_{j}x+(\alpha-\alpha^{2})\ddot{\zeta_{j}}t}$
(3.7d)
$\mu_{+3}(\zeta_{j}^{\ast})=\acute{f}_{j}\mu_{-2}(\zeta_{j}^{\ast})e^{(\alpha-\alpha^{2})\zeta_{j}^{\ast}x+(\alpha^{2}-\alpha)\ddot{\zeta_{j}}^{\ast}t},$
(3.7e)
$\mu_{+1}(\alpha\zeta_{j})=\grave{f}_{j}\mu_{-2}(\alpha\zeta_{j})e^{(\alpha^{2}-\alpha)\zeta_{j}x+(\alpha-\alpha^{2})\ddot{\zeta_{j}}t}.$
(3.7f)

In order to simplify the calculation, we introduce the modified auxiliary
eigenfunctions

$m_{n}(x,t,\nu)=e^{-\alpha^{2}\nu x-\alpha\nu^{-1}t}\chi_{n}(\nu),\quad
n=1,4,$ (3.8a) $m_{n}(x,t,\nu)=e^{-\nu x-\nu^{-1}t}\chi_{n}(\nu),\quad n=2,5,$
(3.8b) $m_{n}(x,t,\nu)=e^{-\alpha\nu x-\alpha^{2}\nu^{-1}t}\chi_{n}(\nu),\quad
n=3,6.$ (3.8c)

Let $\\{\nu_{j}\\}_{j=1}^{N_{2}}$ be the eigenvalues of the second type, we
obtain that

$\displaystyle m_{2}(\nu_{j})=$
$\displaystyle\hat{c}_{j}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}\mu_{-3}(\nu_{j}),$
(3.9a) $\displaystyle\mu_{+2}(-\nu_{j}^{\ast})=$
$\displaystyle\hat{d}_{j}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t}m_{2}(-\nu_{j}^{\ast}),$

$\displaystyle m_{1}(\alpha\nu_{j}^{\ast})=$ $\displaystyle
c_{j}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t}\mu_{-1}(\alpha\nu_{j}^{\ast}),$
(3.9b) $\displaystyle\mu_{+2}(-\alpha\nu_{j})=$ $\displaystyle
d_{j}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}m_{1}(-\alpha\nu_{j}),$

$\displaystyle m_{3}(\alpha^{2}\nu_{j}^{\ast})=$
$\displaystyle\check{c}_{j}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t}\mu_{-3}(\alpha^{2}\nu_{j}^{\ast}),$
(3.9c) $\displaystyle\mu_{+1}(-\alpha^{2}\nu_{j})=$
$\displaystyle\check{d}_{j}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}m_{3}(-\alpha^{2}\nu_{j}),$

$\displaystyle m_{4}(\alpha\nu_{j})=$
$\displaystyle\breve{c}_{j}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}\mu_{-2}(\alpha\nu_{j}),$
(3.9d) $\displaystyle\mu_{+1}(-\alpha\nu_{j}^{\ast})=$
$\displaystyle\breve{d}_{j}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t}m_{4}(-\alpha\nu_{j}^{\ast}),$

$\displaystyle m_{5}(\nu_{j}^{\ast})=$
$\displaystyle\acute{c}_{j}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t}\mu_{-2}(\nu_{j}^{\ast}),$
(3.9e) $\displaystyle\mu_{+3}(-\nu_{j})=$
$\displaystyle\acute{d}_{j}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}m_{5}(-\nu_{j}),$

$\displaystyle m_{6}(\alpha^{2}\nu_{j})=$
$\displaystyle\grave{c}_{j}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}\mu_{-1}(\alpha^{2}\nu_{j}),$
(3.9f) $\displaystyle\mu_{+3}(-\alpha^{2}\nu_{j}^{\ast})=$
$\displaystyle\grave{d}_{j}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t}m_{6}(-\alpha^{2}\nu_{j}^{\ast}),$

where $\ddot{\nu}_{j}=1/\nu_{j}$.

The norming constants in Eqs.(3.7) and (3.9) admit the following symmetry
relations

$f_{j}=\grave{f}_{j}=\breve{f}_{j},\quad\hat{f}_{j}=\acute{f}_{j}=\check{f}_{j},\quad
f_{j}=\acute{f}^{\ast}_{j},\quad$ (3.10a)
$c_{j}=\check{c}_{j}=\acute{c}_{j},\quad\hat{c}_{j}=\breve{c}_{j}=\grave{c}_{j},\quad
c^{\ast}_{j}=-\frac{\gamma^{\ast}}{\gamma}\hat{c}_{j},$ (3.10b)
$d_{j}=\check{d}_{j}=\acute{d}_{j},\quad\hat{d}_{j}=\breve{d}_{j}=\grave{d}_{j},$
(3.10c) where
$f_{j}=-f^{\ast}_{j}\frac{\dot{r_{22}}(\nu_{j})}{\dot{s_{33}}(\nu_{j})},\quad
d_{j}=-\frac{c^{\ast}_{j}}{s^{\ast}_{33}(-\nu^{\ast}_{j})},\quad\hat{d}_{j}=\frac{\gamma^{\ast}}{\gamma
s_{33}(-\nu^{\ast}_{j})}c_{j}.$ (3.10d)

Here the dot denotes the derivative with respect to the parameter $\nu$.

###  3.3 Trace formula

In this section, we we consider the associated trace formula. Assume that
$s_{33}(\nu)$ has the simple zeros
$\Big{\\{}\zeta_{j}:\zeta_{j}=-\zeta_{j}^{\ast},\zeta_{j}\in
D_{2}\Big{\\}}_{j=1}^{N_{1}}$ and
$\Big{\\{}\nu_{j}:\nu_{j}\neq-\nu_{j}^{\ast},\nu_{j}\in
D_{2}\Big{\\}}_{j=1}^{N_{2}}$.

According to the symmetry of scattering data (3.1) and (3.1), we define

$\displaystyle\bar{s}_{11}(\nu)=\prod_{j=1}^{N_{1}}\frac{(\nu+\alpha^{2}\zeta_{j})(\nu+\alpha\zeta_{j}^{\ast})}{(\nu-\alpha^{2}\zeta_{j})(\nu-\alpha\zeta_{j}^{\ast})}\prod_{j=1}^{N_{2}}\frac{(\nu+\alpha\nu_{j}^{\ast})(\nu+\alpha^{2}\nu_{j})(\nu-\nu_{j}^{\ast})(\nu-\nu_{j})}{(\nu-\alpha\nu_{j}^{\ast})(\nu-\alpha^{2}\nu_{j})(\nu+\nu_{j}^{\ast})(\nu+\nu_{j})}s_{11}(\nu),$
(3.11a)
$\displaystyle\bar{s}_{22}(\nu)=\prod_{j=1}^{N_{1}}\frac{(\nu+\alpha\zeta_{j})(\nu+\zeta_{j}^{\ast})}{(\nu-\alpha\zeta_{j})(\nu-\zeta_{j}^{\ast})}\prod_{j=1}^{N_{2}}\frac{(\nu+\alpha\nu_{j})(\nu-\alpha^{2}\nu_{j})(\nu+\nu_{j}^{\ast})(\nu-\alpha^{2}\nu_{j}^{\ast})}{(\nu-\alpha\nu_{j})(\nu+\alpha^{2}\nu_{j})(\nu-\nu_{j}^{\ast})(\nu+\alpha^{2}\nu_{j}^{\ast})}s_{22}(\nu).$
(3.11b)

Here, $\bar{s}_{11}(\nu)$ and $\bar{s}_{22}(\nu)$ are analytic in $\nu\in
D_{1}\cup D_{6}$ and $\nu\in D_{4}\cup D_{5}$, respectively, whereas they have
no zeros. After a few calculation, we find that the analytic scattering
coefficients of the scattering matrix take the following form,

$\displaystyle
s_{11}(\nu)=\prod_{j=1}^{N_{1}}\frac{(\nu-\alpha^{2}\zeta_{j})(\nu-\alpha\zeta_{j}^{\ast})}{(\nu+\alpha^{2}\zeta_{j})(\nu+\alpha\zeta_{j}^{\ast})}\prod_{j=1}^{N_{2}}\frac{(\nu-\alpha\nu_{j}^{\ast})(\nu-\alpha^{2}\nu_{j})(\nu+\nu_{j}^{\ast})(\nu+\nu_{j})}{(\nu+\alpha\nu_{j}^{\ast})(\nu+\alpha^{2}\nu_{j})(\nu-\nu_{j}^{\ast})(\nu-\nu_{j})}$
(3.12a) $\displaystyle\times\exp\bigg{(}\frac{1}{2\pi
i}\int_{-\infty}^{+\infty}\frac{\log(1-s_{12}(\eta)s_{12}(-\eta)-s_{13}(\eta)s_{13}(-\eta))}{\eta-\nu}d\eta\bigg{)},$
$\displaystyle
s_{22}(\nu)=\prod_{j=1}^{N_{1}}\frac{(\nu-\alpha\zeta_{j})(\nu-\zeta_{j}^{\ast})}{(\nu+\alpha\zeta_{j})(\nu+\zeta_{j}^{\ast})}\prod_{j=1}^{N_{2}}\frac{(\nu-\alpha\nu_{j})(\nu+\alpha^{2}\nu_{j})(\nu-\nu_{j}^{\ast})(\nu+\alpha^{2}\nu_{j}^{\ast})}{(\nu+\alpha\nu_{j})(\nu-\alpha^{2}\nu_{j})(\nu+\nu_{j}^{\ast})(\nu-\alpha^{2}\nu_{j}^{\ast})}$
(3.12b) $\displaystyle\times\exp\bigg{(}\frac{1}{2\pi
i}\int_{-\infty}^{+\infty}\frac{\log(1-s_{12}(\eta)s_{12}(-\eta)-s_{13}(\eta)s_{13}(-\eta))}{\eta-\alpha\nu}d\eta\bigg{)}.$

##  4 Inverse problem

###  4.1 Riemann-Hilbert problem

In this section, the piecewise meromorphic function $\textbf{M}(x,t,\nu)$ in
six regions for the Tzitzéica equation were presented. According to the
analyticities of the Jost functions and the auxiliary eigenfunctions, we
define six piecewise meromorphic functions as follows

$\displaystyle\mathbf{M}_{1}(x,t,\nu)=\Bigg{(}\mu_{-1},\frac{\mu_{+2}}{r_{22}},\frac{m_{1}}{s_{11}}\Bigg{)},\quad\nu\in
D_{1},$ (4.1)
$\displaystyle\mathbf{M}_{2}(x,t,\nu)=\Bigg{(}\frac{m_{2}}{s_{33}},\frac{\mu_{+2}}{r_{22}},\mu_{-3}\Bigg{)},\nu\in
D_{2},$
$\displaystyle\mathbf{M}_{3}(x,t,\nu)=\Bigg{(}\frac{\mu_{+1}}{r_{11}},\frac{m_{3}}{s_{33}},\mu_{-3}\Bigg{)},\quad\nu\in
D_{3},$
$\displaystyle\mathbf{M}_{4}(x,t,\nu)=\Bigg{(}\frac{\mu_{+1}}{r_{11}},\mu_{-2},\frac{m_{4}}{s_{22}}\Bigg{)},\quad\nu\in
D_{4},$
$\displaystyle\mathbf{M}_{5}(x,t,\nu)=\Bigg{(}\frac{m_{5}}{s_{22}},\mu_{-2},\frac{\mu_{+3}}{r_{33}}\Bigg{)},\quad\nu\in
D_{5},$
$\displaystyle\mathbf{M}_{6}(x,t,\nu)=\Bigg{(}\mu_{-1},\frac{m_{6}}{s_{11}},\frac{\mu_{+3}}{r_{33}}\Bigg{)},\quad\nu\in
D_{6}.$

It is readily verified that $\mathbf{M}_{n}(x,t,\nu)$ satisfy the following
jump conditions

$\mathbf{M}_{+}(x,t,\nu)=\mathbf{M}_{-}(x,t,\nu)J(x,t,\nu),\quad\nu\in\Sigma,$
(4.2)

where the matrices $\mathbf{M}_{+}(x,t,\nu),\mathbf{M}_{-}(x,t,\nu)$ and
$J(x,t,\nu)$ admit the following definitions

$\mathbf{M}(x,t,\nu)=\left\\{\begin{array}[]{ll}\mathbf{M}_{+}(x,t,\nu),&\nu\in
D_{1}\cup D_{3}\cup D_{5},\\\ \mathbf{M}_{-}(x,t,\nu),&\nu\in D_{2}\cup
D_{4}\cup D_{6},\end{array}\right.$ (4.3)
$\mathbf{J}(x,t,\nu)=\left\\{\begin{array}[]{ll}J_{1}=e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}\bigg{(}S_{2}^{-1}(\nu)S_{1}(\nu)\bigg{)},&\arg\nu=\frac{\pi}{3},\\\
J_{2}=e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}\bigg{(}S_{2}^{-1}(\nu)S_{3}(\nu)\bigg{)},&\arg\nu=\frac{2\pi}{3},\\\
J_{3}=e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}\bigg{(}S_{4}^{-1}(\nu)S_{3}(\nu)\bigg{)},&\arg\nu=\pi,\\\
J_{4}=e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}\bigg{(}S_{4}^{-1}(\nu)S_{5}(\nu)\bigg{)},&\arg\nu=\frac{4\pi}{3},\\\
J_{5}=e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}\bigg{(}S_{6}^{-1}(\nu)S_{5}(\nu)\bigg{)},&\arg\nu=\frac{5\pi}{3},\\\
J_{6}=e^{\nu\hat{\sigma}x+\nu^{-1}\hat{\sigma}^{-1}t}\bigg{(}S_{6}^{-1}(\nu)S_{1}(\nu)\bigg{)},&\arg\nu=0.\end{array}\right.$
(4.4)

We note that the jump conditions (4.2) can be rewritten as

$\displaystyle\mathbf{M}_{+}(x,t,\nu)-\mathbf{M}_{-}(x,t,\nu)$ (4.5)
$\displaystyle=\mathbf{M}_{-}(x,t,\nu)\bigg{(}e^{\nu\sigma
x+\nu^{-1}\sigma^{-1}t}\mathbf{\hat{J}}(x,t,\nu)e^{-\nu\sigma
x-\nu^{-1}\sigma^{-1}t}\bigg{)},\quad\nu\in\Sigma,$

where the jump matrices $\hat{J}_{1},\ldots,\hat{J}_{6}$ can be derived as

$\displaystyle\mathbf{\hat{J}}_{1}=\left(\begin{matrix}0&0&-\rho_{1}(\nu)\\\
0&0&0\\\
\rho_{2}(\nu)&0&-\rho_{1}(\nu)\rho_{2}(\nu)\end{matrix}\right),\nu\in\Sigma_{1},\quad\mathbf{\hat{J}}_{2}=\left(\begin{matrix}-\rho^{\ast}_{1}(-\nu^{\ast})\rho^{\ast}_{2}(-\nu^{\ast})&-\rho^{\ast}_{2}(-\nu^{\ast})&0\\\
\rho^{\ast}_{1}(-\nu^{\ast})&0&0\\\
0&0&0\end{matrix}\right),\nu\in\Sigma_{2},$ (4.6)
$\displaystyle\mathbf{\hat{J}}_{3}=\left(\begin{matrix}0&0&0\\\
0&-\rho^{\ast}_{3}(\nu^{\ast})\rho_{3}(\nu)&\rho_{3}(\nu)\\\
0&-\rho^{\ast}_{3}(\nu^{\ast})&0\end{matrix}\right),\nu\in\Sigma_{3},\quad\mathbf{\hat{J}}_{4}=\left(\begin{matrix}0&0&\rho_{2}(-\nu)\\\
0&0&0\\\
-\rho_{1}(-\nu)&0&-\rho_{1}(-\nu)\rho_{2}(-\nu)\end{matrix}\right),\nu\in\Sigma_{4},$
$\displaystyle\mathbf{\hat{J}}_{5}=\left(\begin{matrix}-\rho^{\ast}_{1}(\nu^{\ast})\rho^{\ast}_{2}(\nu^{\ast})&\rho^{\ast}_{1}(\nu^{\ast})&0\\\
-\rho^{\ast}_{2}(\nu^{\ast})&0&0\\\
0&0&0\end{matrix}\right),\nu\in\Sigma_{5},\quad\mathbf{\hat{J}}_{6}=\left(\begin{matrix}0&0&0\\\
0&-\rho_{3}(-\nu)\rho^{\ast}_{3}(-\nu^{\ast})&-\rho^{\ast}_{3}(-\nu^{\ast})\\\
0&\rho_{3}(-\nu)&0\end{matrix}\right),\nu\in\Sigma_{6}.$

To find the normalization condition of the Riemann-Hilbert problem, we note
that, from Eq. (2.14), asymptotic behaviors ${\mu_{\pm}(x,t,\nu)}$ take the
following form

$\mu_{\pm}(x,t,\nu)=\left(\begin{matrix}1&\alpha^{2}&\alpha\\\
1&\alpha&\alpha^{2}\\\
1&1&1\end{matrix}\right)+\frac{1}{\nu}\left(\begin{matrix}\eta-\frac{2}{3}u_{x}&\alpha\eta-\frac{2}{3}\alpha
u_{x}&\alpha^{2}\eta-\frac{2}{3}\alpha^{2}u_{x}\\\
\eta+\frac{1}{3}u_{x}&\eta+\frac{1}{3}u_{x}&\eta+\frac{1}{3}u_{x}\\\
\eta+\frac{1}{3}u_{x}&\alpha^{2}\eta+\frac{1}{3}\alpha^{2}u_{x}&\alpha\eta+\frac{1}{3}\alpha
u_{x}\end{matrix}\right)+o(\frac{1}{\nu^{2}}),\nu\longrightarrow\infty,$ (4.7)

where $\eta=\frac{1}{3}\int_{\pm\infty}^{x}u_{\xi}^{2}d\xi$, which implies
that

$\mathbf{M}_{\pm}(x,t,\nu)\rightarrow A,\quad\nu\rightarrow\infty.$

###  4.2 The case of no poles

In this case, matrix functions $\mathbf{M}_{\pm}(x,t,\nu)$ are the analytic in
their respective domains.Applying the Cauchy operator $\frac{1}{2\pi
i}\int_{\Sigma}[\frac{f}{\zeta-\nu}]d\zeta$ to the equation (4.5), we have the
following integral representation

$\mathbf{M}_{\pm}(x,t,\nu)=A+\frac{1}{2\pi
i}\int_{\Sigma_{2}\cup\Sigma_{4}\cup\Sigma_{6}}\frac{\bigg{(}M_{n}\bar{J}_{n}\bigg{)}(\zeta)}{\zeta-\nu}d\zeta+\frac{1}{2\pi
i}\int_{\Sigma_{1}\cup\Sigma_{3}\cup\Sigma_{5}}\frac{\bigg{(}M_{n+1}\bar{J}_{n}\bigg{)}(\zeta)}{\zeta-\nu}d\zeta,$
(4.8)

where

$\zeta\in\Sigma_{n},\quad\bar{J}_{n}=e^{\zeta\sigma
x+\zeta^{-1}\sigma^{-1}t}\hat{J}_{n}e^{-\zeta\sigma
x-\zeta^{-1}\sigma^{-1}t}.$

###  4.3 The case of poles

In this case, we assume that $s_{33}(\nu)$ has the simple zeros
$\\{\zeta_{j}:\zeta_{j}=-\zeta_{j}^{\ast}\\}_{j=1}^{N_{1}}$ and
$\\{\nu_{j}:\nu_{j}\neq-\nu_{j}^{\ast}\\}_{j=1}^{N_{2}}$ in $D_{2}$. The
meromorphic matrices defined in (4.1) satisfy the following residue conditions

$\mathbf{M}_{1,\alpha\zeta_{j}^{\ast}}(x,t,\nu)=\check{F}_{j}\bigg{(}\textbf{0},\mu_{-1}(\alpha\zeta_{j}^{\ast}),\textbf{0}\bigg{)},\mathbf{M}_{2,\zeta_{j}}(x,t,\nu)=F_{j}\bigg{(}\textbf{0},\mu_{-3}(\zeta_{j}),\textbf{0}\bigg{)},$
(4.9a)
$\mathbf{M}_{3,\alpha^{2}\zeta_{j}^{\ast}}(x,t,\nu)=\hat{F}_{j}\bigg{(}\mu_{-3}(\alpha^{2}\zeta_{j}^{\ast}),\textbf{0},\textbf{0}\bigg{)},\mathbf{M}_{4,\alpha\zeta_{j}}(x,t,\nu)=\grave{F}_{j}\bigg{(}\mu_{-2}(\alpha\zeta_{j}),\textbf{0},\textbf{0}\bigg{)},$
(4.9b)
$\mathbf{M}_{5,\zeta_{j}^{\ast}}(x,t,\nu)=\acute{F}_{j}\bigg{(}\textbf{0},\textbf{0},\mu_{-2}(\zeta_{j}^{\ast})\bigg{)},\mathbf{M}_{6,\alpha^{2}\zeta_{j}}(x,t,\alpha^{2}\zeta_{j})=\breve{F}_{j}\bigg{(}\textbf{0},\textbf{0},\mu_{-1}(\alpha^{2}\zeta_{j})\bigg{)},$
(4.9c)
$\mathbf{M}_{1,-\alpha\nu_{j}}(x,t,\nu)=D_{j}\bigg{(}\textbf{0},\frac{m_{1}}{s_{11}}(-\alpha\nu_{j}),\textbf{0}\bigg{)},\mathbf{M}_{1,\alpha\nu_{j}^{\ast}}(x,t,\nu)=C_{j}\bigg{(}\textbf{0},\textbf{0},\mu_{-1}(\alpha\nu_{j}^{\ast})\bigg{)},$
(4.10a)
$\mathbf{M}_{2,-\nu_{j}^{\ast}}(x,t,\nu)=\hat{D}_{j}\bigg{(}\textbf{0},\frac{m_{2}}{s_{33}}(-\nu_{j}^{\ast}),\textbf{0}\bigg{)},\mathbf{M}_{2,\nu_{j}}(x,t,\nu)=\hat{C}_{j}\bigg{(}\mu_{-3}(\nu_{j}),\textbf{0},\textbf{0}\bigg{)},$
(4.10b)
$\mathbf{M}_{3,-\alpha^{2}\nu_{j}}(x,t,\nu)=\check{D}_{j}\bigg{(}\frac{m_{3}}{s_{33}}(-\alpha^{2}\nu_{j}),\textbf{0},\textbf{0}\bigg{)},\mathbf{M}_{3,\alpha^{2}\nu_{j}^{\ast}}(x,t,\nu)=\check{C}_{j}\bigg{(}\textbf{0},\mu_{-3}(\alpha^{2}\nu_{j}^{\ast}),\textbf{0}\bigg{)},$
(4.10c)
$\mathbf{M}_{4,-\alpha\nu_{j}^{\ast}}(x,t,\nu)=\breve{D}_{j}\bigg{(}\frac{m_{4}}{s_{22}}(-\alpha\nu_{j}^{\ast}),\textbf{0},\textbf{0}\bigg{)},\mathbf{M}_{4,\alpha\nu_{j}}(x,t,\nu)=\breve{C}_{j}\bigg{(}\textbf{0},\textbf{0},\mu_{-2}(\alpha\nu_{j})\bigg{)},$
(4.10d)
$\mathbf{M}_{5,-\nu_{j}}(x,t,\nu)=\acute{D}_{j}\bigg{(}\textbf{0},\textbf{0},\frac{m_{5}}{s_{22}}(-\nu_{j})\bigg{)},\mathbf{M}_{5,\nu_{j}^{\ast}}(x,t,\nu)=\acute{C}_{j}\bigg{(}\mu_{-2}(\nu_{j}^{\ast}),\textbf{0},\textbf{0}\bigg{)},$
(4.10e)
$\mathbf{M}_{6,-\alpha^{2}\nu_{j}^{\ast}}(x,t,\nu)=\grave{D}_{j}\bigg{(}\textbf{0},\textbf{0},\frac{m_{6}}{s_{11}}(-\alpha^{2}\nu_{j}^{\ast})\bigg{)},\mathbf{M}_{6,\alpha^{2}\nu_{j}}(x,t,\alpha^{2}\nu_{j})=\grave{C}_{j}\bigg{(}\textbf{0},\mu_{-1}(\alpha^{2}\nu_{j}),\textbf{0}\bigg{)},$
(4.10f)

where

$\check{F}_{j}(x,t,\nu)=\frac{\check{f}_{j}}{\dot{r_{22}}(\alpha\zeta_{j}^{\ast})}e^{(\alpha-\alpha^{2})\zeta_{j}^{\ast}x+(\alpha^{2}-\alpha)\ddot{\zeta_{j}}^{\ast}t},F_{j}(x,t,\nu)=\frac{f_{j}}{\dot{r_{22}}(\zeta_{j})}e^{(\alpha^{2}-\alpha)\zeta_{j}x+(\alpha-\alpha^{2})\ddot{\zeta_{j}}t},$
(4.11a)
$\hat{F}_{j}(x,t,\nu)=\frac{\hat{f}_{j}}{\dot{r_{11}}(\alpha^{2}\zeta_{j}^{\ast})}e^{(\alpha-\alpha^{2})\zeta_{j}^{\ast}x+(\alpha^{2}-\alpha)\ddot{\zeta_{j}}^{\ast}t},\grave{F}_{j}(x,t,\nu)=\frac{\grave{f}_{j}}{\dot{r_{11}}(\alpha\zeta_{j})}e^{(\alpha^{2}-\alpha)\zeta_{j}x+(\alpha-\alpha^{2})\ddot{\zeta_{j}}t},$
(4.11b)
$\acute{F}_{j}(x,t,\nu)=\frac{\acute{f}_{j}}{\dot{r_{33}}(\zeta_{j}^{\ast})}e^{(\alpha-\alpha^{2})\zeta_{j}^{\ast}x+(\alpha^{2}-\alpha)\ddot{\zeta_{j}}^{\ast}t},\breve{F}_{j}(x,t,\nu)=\frac{\breve{f}_{j}}{\dot{r_{33}}(\alpha^{2}\zeta_{j})}e^{(\alpha^{2}-\alpha)\zeta_{j}x+(\alpha-\alpha^{2})\ddot{\zeta_{j}}t},$
(4.11c)
$D_{j}(x,t)=\frac{d_{j}s_{11}(-\alpha\nu_{j})}{\dot{r_{22}}(-\alpha\nu_{j})}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t},C_{j}(x,t)=\frac{c_{j}}{\dot{s_{11}}(\alpha\nu_{j}^{\ast})}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t},$
(4.12a)
$\hat{D}_{j}(x,t)=\frac{\hat{d}_{j}s_{33}(-\nu^{\ast}_{j})}{\dot{r_{22}}(-\nu_{j}^{\ast})}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t},\hat{C}_{j}(x,t)=\frac{\hat{c}_{j}}{\dot{s_{33}}(\nu_{j})}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t},$
(4.12b)
$\check{D}_{j}(x,t)=\frac{\check{d}_{j}s_{33}(-\alpha^{2}\nu_{j})}{\dot{r_{11}}(-\alpha^{2}\nu_{j})}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t},\check{C}_{j}(x,t)=\frac{\check{c}_{j}}{\dot{s_{33}}(\alpha^{2}\nu_{j}^{\ast})}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t},$
(4.12c)
$\breve{D}_{j}(x,t)=\frac{\breve{d}_{j}s_{22}(-\alpha\nu^{\ast}_{j})}{\dot{r_{11}}(-\alpha\nu_{j}^{\ast})}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t},\breve{C}_{j}(x,t)=\frac{\breve{c}_{j}}{\dot{s_{22}}(\alpha\nu_{j})}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t},$
(4.12d)
$\acute{D}_{j}(x,t)=\frac{\acute{d}_{j}(x,t)s_{22}(-\nu_{j})}{\dot{r_{33}}(-\nu_{j})}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t},\acute{C}_{j}(x,t)=\frac{\acute{c}_{j}}{\dot{s_{22}}(\nu_{j}^{\ast})}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t},$
(4.12e)
$\grave{D}_{j}(x,t)=\frac{\grave{d}_{j}s_{11}(-\alpha^{2}\nu_{j}^{\ast})}{\dot{r_{33}}(-\alpha^{2}\nu_{j}^{\ast})}e^{(\alpha-1)\nu_{j}^{\ast}x+(\alpha^{2}-1)\ddot{\nu_{j}}^{\ast}t},\grave{C}_{j}(x,t)=\frac{\grave{c}_{j}}{\dot{s_{11}}(\alpha^{2}\nu_{j})}e^{(\alpha^{2}-1)\nu_{j}x+(\alpha-1)\ddot{\nu_{j}}t}.$
(4.12f)

It is remarked that

$\displaystyle\acute{F}_{j}(x,t,\nu)=F_{j}^{\ast}(x,t,\nu),\check{F}_{j}(x,t,\nu)=\alpha
F_{j}^{\ast}(x,t,\nu),\grave{F}_{j}(x,t,\nu)=\alpha F_{j}(x,t,\nu),$
$\displaystyle\hat{F}_{j}(x,t,\nu)=\alpha^{2}F_{j}^{\ast}(x,t,\nu),\breve{F}_{j}(x,t,\nu)=\alpha^{2}F_{j}(x,t,\nu),$
$\displaystyle\grave{D}_{j}(x,0,\nu)=\alpha^{2}\hat{D}_{j}(x,t,\nu)=\alpha\breve{D}_{j}(x,t,\nu),{D}_{j}(x,t,\nu)=\alpha\acute{D}_{j}(x,t,\nu)=\alpha^{2}\check{D}_{j}(x,t,\nu),$

and

$\displaystyle{C}_{j}(x,t,\nu)=\alpha\acute{C}_{j}(x,t,\nu)=\alpha^{2}\check{C}_{j}(x,t,\nu),\grave{C}_{j}(x,t,\nu)=\alpha^{2}\hat{C}_{j}(x,t,\nu)=\alpha\breve{C}_{j}(x,t,\nu),$
$\displaystyle\acute{D}_{j}(x,t,\nu)=(\acute{C}_{j}(x,t,\nu))^{\ast},\hat{C}_{j}(x,t,\nu)=(\hat{D}_{j}(x,t,\nu))^{\ast},$
$\displaystyle\hat{D}_{j}(x,t,\nu)=\acute{C}_{j}(x,t,\nu),\hat{C}_{j}(x,t,\nu)=(\acute{C}_{j}(x,t,\nu))^{\ast}.$

The key to solving the RH problem of (4.2) is to convert it into a mixed
system of algebraic-integral equations. We similarly apply $\frac{1}{2\pi
i}\int_{\Sigma}[\frac{f}{\zeta-\nu}]d\zeta$ to both sides of the jump
condition (4.5). Taking into account the symmetries properties of the involved
functions, we can easily obtain

$\displaystyle m_{1}^{\pm}(x,t,\nu)=$
$\displaystyle\left(\begin{array}[]{c}1\\\\[-2.0pt] 1\\\\[-2.0pt]
1\end{array}\right)+\sum_{j=1}^{N_{1}}\left\\{\frac{\hat{F}_{j}\mu_{-3}(\alpha^{2}\zeta_{j}^{\ast})}{\nu-\alpha^{2}\zeta_{j}^{\ast}}+\frac{\grave{F}_{j}\mu_{-2}(\alpha\zeta_{j})}{\nu-\alpha\zeta_{j}}\right\\}+\sum_{j=1}^{N_{2}}\left\\{\frac{\hat{C}_{j}\mu_{-3}(\nu_{j})}{\nu-\nu_{j}}\right.$
(4.13)
$\displaystyle\left.+\frac{\check{D}_{j}}{\nu+\alpha^{2}\nu_{j}}\frac{m_{3}}{s_{33}}(-\alpha^{2}\nu_{j})+\frac{\acute{C}_{j}\mu_{-2}(\nu_{j}^{\ast})}{\nu-\nu_{j}^{\ast}}+\frac{\breve{D}_{j}}{\nu+\alpha\nu_{j}^{\ast}}\frac{m_{4}}{s_{22}}(-\alpha\nu_{j}^{\ast})\right\\}$
$\displaystyle+\frac{1}{2\pi
i}\int_{\Sigma}\frac{(M^{-}\bar{J}_{n})_{1}(\zeta)}{\zeta-z}d\zeta,$
$\displaystyle m_{2}^{+}(x,t,\nu)=$
$\displaystyle\left(\begin{array}[]{c}\alpha^{2}\\\\[-1.0pt]
\alpha\\\\[-1.0pt]
1\end{array}\right)+\sum_{j=1}^{N_{1}}\left\\{\frac{\check{F}_{j}\mu_{-1}(\alpha\zeta_{j}^{\ast})}{\nu-\alpha\zeta_{j}^{\ast}}+\frac{F_{j}\mu_{-3}(\zeta_{j})}{\nu-\zeta_{j}}\right\\}+\sum_{j=1}^{N_{2}}\left\\{\frac{D_{j}\frac{m_{1}}{s_{11}}(-\alpha\nu_{j})}{\nu+\alpha\nu_{j}}\right.$
(4.14)
$\displaystyle\left.+\frac{\hat{D}_{j}}{\nu+\nu_{j}^{\ast}}\frac{m_{2}}{s_{33}}(-\nu_{j}^{\ast})+\frac{\check{C}_{j}\mu_{-3}(\alpha^{2}\nu_{j}^{\ast})}{\nu-\alpha^{2}\nu_{j}^{\ast}}+\frac{\grave{C}_{j}\mu_{-1}(\alpha^{2}\nu_{j})}{\nu-\alpha^{2}\nu_{j}}\right\\}$
$\displaystyle+\frac{1}{2\pi
i}\int_{\Sigma}\frac{(M^{-}\bar{J}_{n})_{2}(\zeta)}{\zeta-z}d\zeta,$
$\displaystyle m_{2}^{-}(x,t,\nu)=$
$\displaystyle\left(\begin{array}[]{c}\alpha^{2}\\\\[-1.0pt]
\alpha\\\\[-1.0pt]
1\end{array}\right)+\sum_{j=1}^{N_{1}}\left\\{\frac{\check{F}_{j}\mu_{-1}(\alpha\zeta_{j}^{\ast})}{\nu-\alpha\zeta_{j}^{\ast}}+\frac{F_{j}\mu_{-3}(\zeta_{j})}{\nu-\zeta_{j}}\right\\}+\sum_{j=1}^{N_{2}}\left\\{\frac{D_{j}\frac{m_{1}}{s_{11}}(-\alpha\nu_{j})}{\nu+\alpha\nu_{j}}\right.$
(4.15)
$\displaystyle\left.+\frac{\hat{D}_{j}}{\nu+\nu_{j}^{\ast}}\frac{m_{2}}{s_{33}}(-\nu_{j}^{\ast})+\frac{\check{C}_{j}\mu_{-3}(\alpha^{2}\nu_{j}^{\ast})}{\nu-\alpha^{2}\nu_{j}^{\ast}}+\frac{\grave{C}_{j}\mu_{-1}(\alpha^{2}\nu_{j})}{\nu-\alpha^{2}\nu_{j}}\right\\}$
$\displaystyle+\frac{1}{2\pi
i}\int_{\Sigma}\frac{(M^{-}\bar{J}_{n})_{2}(\zeta)}{\zeta-z}d\zeta,$
$\displaystyle m_{3}^{+}(x,t,\nu)=$
$\displaystyle\left(\begin{array}[]{c}\alpha\\\\[-1.0pt]
\alpha^{2}\\\\[-1.0pt]
1\end{array}\right)+\sum_{j=1}^{N_{1}}\left\\{\frac{\acute{F}_{j}\mu_{-2}(\zeta_{j}^{\ast})}{\nu-\zeta_{j}^{\ast}}+\frac{\breve{F}_{j}\mu_{-1}(\alpha^{2}\zeta_{j})}{\nu-\alpha^{2}\zeta_{j}}\right\\}+\sum_{j=1}^{N_{2}}\left\\{\frac{C_{j}\mu_{-1}(\alpha\nu_{j}^{\ast})}{\nu-\alpha\nu_{j}^{\ast}}\right.$
(4.16)
$\displaystyle\left.+\frac{\breve{C}_{j}\mu_{-2}(\alpha\nu_{j})}{\nu-\alpha\nu_{j}}+\frac{\acute{D}_{j}}{\nu+\nu_{j}}\frac{m_{5}}{s_{22}}(-\nu_{j})+\frac{\grave{D}_{j}}{\nu+\alpha^{2}\nu_{j}^{\ast}}\frac{m_{6}}{s_{11}}(-\alpha^{2}\nu_{j}^{\ast})\right\\}$
$\displaystyle+\frac{1}{2\pi
i}\int_{\Sigma}\frac{(M^{-}\bar{J}_{n})_{3}(\zeta)}{\zeta-z}d\zeta,$
$\displaystyle m_{3}^{-}(x,t,\nu)=$
$\displaystyle\left(\begin{array}[]{c}\alpha\\\\[-1.0pt]
\alpha^{2}\\\\[-1.0pt]
1\end{array}\right)+\sum_{j=1}^{N_{1}}\left\\{\frac{\acute{F}_{j}\mu_{-2}(\zeta_{j}^{\ast})}{\nu-\zeta_{j}^{\ast}}+\frac{\breve{F}_{j}\mu_{-1}(\alpha^{2}\zeta_{j})}{\nu-\alpha^{2}\zeta_{j}}\right\\}+\sum_{j=1}^{N_{2}}\left\\{\frac{C_{j}\mu_{-1}(\alpha\nu_{j}^{\ast})}{\nu-\alpha\nu_{j}^{\ast}}\right.$
(4.17)
$\displaystyle\left.+\frac{\breve{C}_{j}\mu_{-2}(\alpha\nu_{j})}{\nu-\alpha\nu_{j}}+\frac{\acute{D}_{j}}{\nu+\nu_{j}}\frac{m_{5}}{s_{22}}(-\nu_{j})+\frac{\grave{D}_{j}}{\nu+\alpha^{2}\nu_{j}^{\ast}}\frac{m_{6}}{s_{11}}(-\alpha^{2}\nu_{j}^{\ast})\right\\}$
$\displaystyle+\frac{1}{2\pi
i}\int_{\Sigma}\frac{(M^{-}\bar{J}_{n})_{3}(\zeta)}{\zeta-z}d\zeta.$

To finish the potential reconstruction about the scattering data, we consider
the expansion of (4.13) at $\nu\to\infty$, and compare it with (4.7). And the
$O(\nu^{-1})$ terms imply the representation of the potential matrix about the
eigenfunctions at certain eigenvalues, which can be reduced by taking
$\nu=-\alpha^{2}\nu_{j},\nu_{j}^{\ast}$ in (4.14), $\nu=\alpha\zeta_{j}$ in
(4.15), $\nu=\alpha^{2}\zeta_{j}^{\ast}$ in (4.16) and
$\nu=\nu_{j},-\alpha\nu_{j}^{\ast}$ in (4.17). Then the potential can be
reconstructed as

$\displaystyle
U_{x}=\sum_{j=1}^{N_{1}}\Bigg{\\{}\hat{F}_{j}\bigg{(}\mu_{-23}-\mu_{-13}\bigg{)}(\alpha^{2}\zeta_{j}^{\ast})+\grave{F}_{j}\bigg{(}\mu_{-22}-\mu_{-12}\bigg{)}(\alpha\zeta_{j})\Bigg{\\}}$
(4.18)
$\displaystyle+\sum_{j=1}^{N_{2}}\Bigg{\\{}\hat{C}_{j}\bigg{(}\mu_{-23}-\mu_{-13}\bigg{)}(\nu_{j})+\check{D}_{j}\bigg{(}\frac{m_{23}}{s_{33}}-\frac{m_{13}}{s_{33}}\bigg{)}(-\alpha^{2}\nu_{j})$
$\displaystyle+\acute{C}_{j}\bigg{(}\mu_{-22}-\mu_{-12}\bigg{)}(\nu_{j}^{\ast})+\breve{D}_{j}\bigg{(}\frac{m_{24}}{s_{22}}-\frac{m_{14}}{s_{22}}\bigg{)}(-\alpha\nu_{j}^{\ast})\Bigg{\\}}$
$\displaystyle-\frac{1}{2\pi
i}\int_{\Sigma}\Big{(}(M^{-}\bar{J}_{n})_{21}-(M^{-}\bar{J}_{n})_{11}\Big{)}(\zeta)d\zeta.$

##  5 Explicit solutions

In the reflectionless case, it is convenient to write (4.17) as fractional
equation

$U_{x}=\frac{\det(G^{\prime}+b^{\ast}Y^{T})}{\det
G^{\prime}}-\frac{\det(G+bY^{T})}{\det G},$

where

$\displaystyle G=I-F,\quad G^{\prime}=I-F^{\prime},\quad
b=(b_{1},\ldots,b_{2N_{1}+4N_{2}})^{T},\quad
Y=(y_{1},\ldots,y_{2N_{1}+4N_{2}})^{T},$

and $F_{ij},b_{j},y_{j}$ are given by (A.1)-(A.2), (A.4)-(A.9)in the appendix.
(1)$I$-explicit solution.
Assuming $N_{1}=1$ and $N_{2}=0$, and setting
$\zeta_{1}=ie^{s},s\in\mathbb{R},c_{0},d_{0}\in\mathbb{R}$, we then find the
following explicit solution

$u(x,t,\nu)=\partial^{-1}\Bigg{\\{}-\displaystyle\frac{6e^{s}\sinh(c)\sin(d-\frac{\pi}{3})}{\sinh^{2}(c)+2\cos(d+\frac{2\pi}{3})\cosh(c)+2\cos(2d-\frac{2\pi}{3})}\Bigg{\\}},$

where

$c=c_{0}+\sqrt{3}(e^{s}x+e^{-s}t),d=d_{0}.$

(2)$II$-explicit solution.
Choosing $N_{1}=0$ and $N_{2}=1$, and supposing
$\nu_{1}=e^{\epsilon_{1}+i\eta_{1}},\epsilon_{0},\eta_{0},\epsilon_{1},\eta_{1}\in\mathbb{R},$
we get another explicit solution for the Eq. (1.1) as

$u(x,t,\nu)=\partial^{-1}\Bigg{\\{}3\sqrt{3}e^{\epsilon_{1}}\displaystyle\frac{A}{B}\Bigg{\\}},$
(5.1)

where

$\begin{split}A=&-\omega\sin
2b+e^{2\xi-2\tau}\tan\eta_{1}\omega\sinh(2\tau+2a)\\\ &-\frac{3}{8}\omega
e^{2\xi-\tau+\epsilon_{0}}[\sin(s+\gamma)\cos b\sinh(\tau+a)-\sin
b\cos(s+\gamma)\cosh(\tau+a)]\\\ &+\frac{1}{2}e^{6\xi-3\tau}[\sin b\cos
s\cosh(3a+3\tau)+\sin s\cos b\sinh(3a+3\tau)]\\\
&-\frac{1}{2}e^{2\xi-\tau}[\sin s\cos b\sinh(a+\tau)+\sin 3b\cos
s\cosh(a+\tau)],\end{split}$ $\begin{split}B=&\Bigg{[}\cos
b-e^{2\xi-\tau}\cosh(a+\tau)\Bigg{]}^{2}\Bigg{\\{}e^{4\xi-2\tau}\cosh^{2}(a+\tau)+\cos^{2}b-3\frac{\cos^{2}(\eta_{1}-\frac{\pi}{3})}{\cos\eta_{1}\cos(\eta_{1}+\frac{\pi}{3})}\Bigg{\\}},\end{split}$

with

$\displaystyle
a=\epsilon_{0}+e^{\epsilon_{1}}\Bigg{(}\frac{\sqrt{3}}{2}\sin\eta_{1}-\frac{3}{2}\cos\eta_{1}\Bigg{)}x-e^{-\epsilon_{1}}\Bigg{(}\frac{3}{2}\cos\eta_{1}-\frac{\sqrt{3}}{2}\sin\eta_{1}\Bigg{)}t,$
$\displaystyle
b=\eta_{0}+e^{\epsilon_{1}}\Bigg{(}\frac{3}{2}\sin\eta_{1}+\frac{\sqrt{3}}{2}\cos\eta_{1}\Bigg{)}x-e^{-\epsilon_{1}}\Bigg{(}\frac{\sqrt{3}}{2}\cos\eta_{1}+\frac{3}{2}\sin\eta_{1}\Bigg{)}t,$
$\displaystyle
s=\eta_{1}+\frac{2}{3}\pi,\quad\tau=\frac{1}{2}\ln\Bigg{\\{}\frac{\tan\eta_{1}}{\tan(\eta_{1}+\frac{\pi}{3})}\Bigg{\\}},\quad\xi=\frac{1}{2}\ln\Bigg{\\{}\frac{\tan\eta_{1}}{\tan(\eta_{1}-\frac{\pi}{3})}\Bigg{\\}},$
$\displaystyle\gamma=-\frac{2\pi
i}{3}-\frac{i}{2}\ln\Bigg{\\{}\displaystyle\frac{e^{i(4\eta_{1}-\frac{2\pi}{3})}+3e^{-4i\eta_{1}}+2e^{\frac{2\pi
i}{3}}}{2+e^{-i(4\eta_{1}+\frac{2\pi}{3})}+3e^{i(4\eta_{1}+\frac{2\pi}{3})}}\Bigg{\\}},$
$\displaystyle\omega=\frac{\cos^{3}(\eta_{1}-\frac{\pi}{3})}{\cos\eta_{1}\cos(\eta_{1}+\frac{\pi}{3})}.$

## Appendix

##  Appendix A Reflectionless potentials

Through calculating, we can give the coefficients $b_{j}(x,t)$ and
$y_{j}(x,t)$ as follows

$b_{j}(x,t,\nu)=\left\\{\begin{array}[]{ll}\alpha^{2},j=1,\ldots,N_{1},\\\
\alpha,j=N_{1}+1,\ldots,2N_{1},\\\ \alpha,j=2N_{1}+1,\ldots,2N_{1}+N_{2},\\\
\alpha^{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},\\\
\alpha^{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},\\\
\alpha,j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},\end{array}\right.$ (A.1)
$y_{j}(x,t,\nu)=\left\\{\begin{array}[]{ll}\grave{F}_{j},j=1,\ldots,N_{1},\\\
\hat{F}_{j-N_{1}},j=N_{1}+1,\ldots,2N_{1},\\\
\hat{C}_{j-2N_{1}},j=2N_{1}+1,\ldots,2N_{1}+N_{2},\\\
\acute{C}_{j-2N_{1}-N_{2}},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},\\\
\check{D}_{j-2N_{1}-2N_{2}},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},\\\
\breve{D}_{j-2N_{1}-3N_{2}},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}.\end{array}\right.$
(A.2)

For convenience, these definitions are given

$F_{n}^{(1)}(x,t,\nu)=F_{n}(x,t,\nu)\frac{\alpha^{2}}{\nu-\zeta_{n}},F_{n}^{(2)}(x,t,\nu)=\check{F}_{n}(x,t,\nu)\frac{\alpha^{2}}{\nu-\alpha\zeta_{n}^{\ast}},$
(A.3a)
$F_{n}^{(3)}(x,t,\nu)=\breve{F}_{n}(x,t,\nu)\frac{\alpha}{\nu-\alpha^{2}\zeta_{n}},F_{n}^{(4)}(x,t,\nu)=\acute{F}_{n}(x,t,\nu)\frac{\alpha}{\nu-\zeta_{n}^{\ast}},$
(A.3b)
$C_{n}^{(1)}(x,t,\nu)=\grave{C}_{n}(x,t,\nu)\frac{\alpha^{2}}{\nu-\alpha^{2}\nu_{n}},C_{n}^{(2)}(x,t,\nu)=\check{C}_{n}(x,t,\nu)\frac{\alpha^{2}}{\nu-\alpha^{2}\nu_{n}^{\ast}},$
(A.3c)
$C_{n}^{(3)}(x,t,\nu)=\breve{C}_{n}(x,t,\nu)\frac{\alpha}{\nu-\alpha\nu_{n}},C_{n}^{(4)}(x,t,\nu)=C_{n}(x,t,\nu)\frac{\alpha}{\nu-\alpha\nu_{n}^{\ast}},$
(A.3d)
$D_{n}^{(1)}(x,t,\nu)=D_{n}(x,t,\nu)\frac{\alpha^{2}}{\nu+\alpha\nu_{n}},D_{n}^{(2)}(x,t,\nu)=\hat{D}_{n}(x,t,\nu)\frac{\alpha^{2}}{\nu+\nu_{n}^{\ast}},$
(A.3e)
$D_{n}^{(3)}(x,t,\nu)=\acute{D}_{n}(x,t,\nu)\frac{\alpha}{\nu+\nu_{n}},D_{n}^{(4)}(x,t,\nu)=\grave{D}_{n}(x,t,\nu)\frac{\alpha}{\nu+\alpha^{2}\nu_{n}^{\ast}}.$
(A.3f)

Additionally, these terms $F_{ij}$ can be derived from (A.3)

$F_{ij}(x,t,\nu)=F_{j}^{(1)}(\alpha\zeta_{i})~{}(i,j=1,\ldots,N_{1}),$ (A.4a)
For $i=1,\ldots,N_{1},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}(x,t,\nu)=F_{j-N_{1}}^{(2)}(\alpha\zeta_{i}).$ (A.4b) For
$i=1,\ldots,N_{1},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}}^{(1)}(\alpha\zeta_{i}).$ (A.4c) For
$i=1,\ldots,N_{1},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}-N_{2}}^{(2)}(\alpha\zeta_{i}).$ (A.4d) For
$i=1,\ldots,N_{1},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-2N_{2}}^{(1)}(\alpha\zeta_{i}).$ (A.4e) For
$i=1,\ldots,N_{1},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-3N_{2}}^{(2)}(\alpha\zeta_{i}).$ (A.4f)

For $i=N_{1}+1,\ldots,2N_{1},j=1,\ldots,N_{1}$,

$F_{ij}(x,t,\nu)=F_{j}^{(3)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.5a) For
$i=N_{1}+1,\ldots,2N_{1},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}(x,t,\nu)=F_{j-N_{1}}^{(4)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.5b)
For $i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}}^{(3)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.5c)
For $i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}-N_{2}}^{(4)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$
(A.5d) For $i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-2N_{2}}^{(3)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$
(A.5e) For $i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-3N_{2}}^{(4)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$
(A.5f)

For $i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=1,\ldots,N_{1}$,

$F_{ij}(x,t,\nu)=F_{j}^{(3)}(\nu_{i-2N_{1}}).$ (A.6a) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}(x,t,\nu)=F_{j-N_{1}}^{(4)}(\nu_{i-2N_{1}}).$ (A.6b) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}}^{(3)}(\nu_{i-2N_{1}}).$ (A.6c) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}-N_{2}}^{(4)}(\nu_{i-2N_{1}}).$ (A.6d) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-2N_{2}}^{(3)}(\nu_{i-2N_{1}}).$ (A.6e) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-3N_{2}}^{(4)}(\nu_{i-2N_{1}}).$ (A.6f)

For $i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=1,\ldots,N_{1}$,

$F_{ij}(x,t,\nu)=F_{j}^{(1)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$ (A.7a) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}(x,t,\nu)=F_{j-N_{1}}^{(2)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$ (A.7b) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}}^{(1)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$ (A.7c) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}-N_{2}}^{(2)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.7d) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-2N_{2}}^{(1)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.7e) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-3N_{2}}^{(2)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.7f)

For $i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=1,\ldots,N_{1}$,

$F_{ij}(x,t,\nu)=F_{j}^{(1)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$ (A.8a) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}(x,t,\nu)=F_{j-N_{1}}^{(2)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$ (A.8b)
For $i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}}^{(1)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$ (A.8c)
For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}-N_{2}}^{(2)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.8d) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-2N_{2}}^{(1)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.8e) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-3N_{2}}^{(2)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.8f)

For $i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=1,\ldots,N_{1}$,

$F_{ij}(x,t,\nu)=F_{j}^{(3)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.9a) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}(x,t,\nu)=F_{j-N_{1}}^{(4)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$
(A.9b) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}}^{(3)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$
(A.9c) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}(x,t,\nu)=C_{j-2N_{1}-N_{2}}^{(4)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$
(A.9d) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-2N_{2}}^{(3)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$
(A.9e) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}(x,t,\nu)=D_{j-2N_{1}-3N_{2}}^{(4)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$
(A.9f)

For $i,j=1,\ldots,N_{1}$,

$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}F_{j}^{(1)}(\alpha\zeta_{i}).$ (A.10a) For
$i=1,\ldots,N_{1},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}F_{j-N_{1}}^{(2)}(\alpha\zeta_{i}).$
(A.10b) For $i=1,\ldots,N_{1},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}C_{j-2N_{1}}^{(1)}(\alpha\zeta_{i}).$
(A.10c) For $i=1,\ldots,N_{1},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}C_{j-2N_{1}-N_{2}}^{(2)}(\alpha\zeta_{i}).$
(A.10d) For $i=1,\ldots,N_{1},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}D_{j-2N_{1}-2N_{2}}^{(1)}(\alpha\zeta_{i}).$
(A.10e) For $i=1,\ldots,N_{1},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}D_{j-2N_{1}-3N_{2}}^{(2)}(\alpha\zeta_{i}).$
(A.10f)

For $i=N_{1}+1,\ldots,2N_{1},j=1,\ldots,N_{1}$,

$F_{ij}^{\prime}(x,t,\nu)=\alpha
F_{j}^{(3)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.11a) For
$i=N_{1}+1,\ldots,2N_{1},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
F_{j-N_{1}}^{(4)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.11b) For
$i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
C_{j-2N_{1}}^{(3)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.11c) For
$i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
C_{j-2N_{1}-N_{2}}^{(4)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.11d) For
$i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
D_{j-2N_{1}-2N_{2}}^{(3)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.11e) For
$i=N_{1}+1,\ldots,2N_{1},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
D_{j-2N_{1}-3N_{2}}^{(4)}(\alpha^{2}\zeta_{i-N_{1}}^{\ast}).$ (A.11f)

For $i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=1,\ldots,N_{1}$,

$F_{ij}^{\prime}(x,t,\nu)=\alpha F_{j}^{(3)}(\nu_{i-2N_{1}}).$ (A.12a) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha F_{j-N_{1}}^{(4)}(\nu_{i-2N_{1}}).$ (A.12b)
For $i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha C_{j-2N_{1}}^{(3)}(\nu_{i-2N_{1}}).$ (A.12c)
For $i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha C_{j-2N_{1}-N_{2}}^{(4)}(\nu_{i-2N_{1}}).$
(A.12d) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha D_{j-2N_{1}-2N_{2}}^{(3)}(\nu_{i-2N_{1}}).$
(A.12e) For
$i=2N_{1}+1,\ldots,2N_{1}+N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha D_{j-2N_{1}-3N_{2}}^{(4)}(\nu_{i-2N_{1}}).$
(A.12f)

For $i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=1,\ldots,N_{1}$,

$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}F_{j}^{(1)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.13a) For $i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}F_{j-N_{1}}^{(2)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.13b) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}C_{j-2N_{1}}^{(1)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.13c) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}C_{j-2N_{1}-N_{2}}^{(2)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.13d) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}D_{j-2N_{1}-2N_{2}}^{(1)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.13e) For
$i=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}D_{j-2N_{1}-3N_{2}}^{(2)}(\nu_{i-2N_{1}-N_{2}}^{\ast}).$
(A.13f)

For $i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=1,\ldots,N_{1}$,

$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}F_{j}^{(1)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.14a) For $i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}F_{j-N_{1}}^{(2)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.14b) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}C_{j-2N_{1}}^{(1)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.14c) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}C_{j-2N_{1}-N_{2}}^{(2)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.14d) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}D_{j-2N_{1}-2N_{2}}^{(1)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.14e) For
$i=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha^{2}D_{j-2N_{1}-3N_{2}}^{(2)}(-\alpha^{2}\nu_{i-2N_{1}-2N_{2}}).$
(A.14f)

For $i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=1,\ldots,N_{1}$,

$F_{ij}^{\prime}(x,t,\nu)=\alpha
F_{j}^{(3)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.15a) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=N_{1}+1,\ldots,2N_{1}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
F_{j-N_{1}}^{(4)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.15b) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+1,\ldots,2N_{1}+N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
C_{j-2N_{1}}^{(3)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.15c) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+N_{2}+1,\ldots,2N_{1}+2N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
C_{j-2N_{1}-N_{2}}^{(4)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.15d) For
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+2N_{2}+1,\ldots,2N_{1}+3N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
D_{j-2N_{1}-2N_{2}}^{(3)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.15e)
Finally, for
$i=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2},j=2N_{1}+3N_{2}+1,\ldots,2N_{1}+4N_{2}$,
$F_{ij}^{\prime}(x,t,\nu)=\alpha
D_{j-2N_{1}-3N_{2}}^{(4)}(-\alpha\nu_{i-2N_{1}-3N_{2}}^{\ast}).$ (A.15f)

## Acknowledgments

The projects 11871440, 11931017 and 11801144 are all supported by the National
Natural Science Foundation of China.

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