11footnotetext:  School of Mathematical Sciences, Fudan University, Shanghai
200433, P.R. China.

#  Inverse scattering transformation for the Fokas-Lenells equation with
nonzero boundary conditions

Yi ZHAO1 and Engui FAN1 Corresponding author and email address:
<EMAIL_ADDRESS>
###### Abstract

In this article, we focus on the inverse scattering transformation for the
Fokas-Lenells (FL) equation with nonzero boundary conditions via the Riemann-
Hilbert (RH) approach. Based on the Lax pair of the FL equation, the
analyticity and symmetry, asymptotic behavior of Jost solutions and scattering
matrix are discussed in detail. With these results, we further present a
generalized RH problem, from which a reconstruction formula between the
solution of the FL equation and the Riemann-Hilbert problem is obtained. The
N-soliton solutions of the FL equation is obtained via solving the RH problem.

Keywords: Fokas-Lenells equation; nonzero boundary conditions; Inverse
scattering transformation, Riemann-Hilbert problem; N-soliton solution.

##  1 Introduction

The nonlinear schrödinger(NLS) equation

$iu_{t}+u_{xx}-2\nu|q|^{2}q=0$ (1)

($\nu=-1$ and $\nu=1$ denote the focusing and defocusing cases), is a
significant mathematical and physical model for the optical fibers, deep water
waves and plasma physics [1]. The NLS equation is one of most integrable
systems, which admits many special features, including the existence of a Lax
pair and the possession of a bi-Hamiltonian formulation [2, 3].

In the late seventies, standard bi-Hamiltonian formulation was used to obtain
an integrable generalization of a given equation [4]. For instance, the
Cammssa-Holm equation is derived from the KdV equation via the bi-Hamiltonian
structure [5], and the same mathematical trick can be applied to the NLS
equation yields Fokas-Lenells (FL) equation [6]

$iu_{t}-\nu u_{tx}+\gamma u_{xx}+\sigma|u|^{2}(u+i\nu u_{x})=0,\quad\sigma=\pm
1.$ (2)

If let $\alpha=\gamma/\nu>0$ and $\beta=\frac{1}{\nu}$ and

$u\rightarrow\beta\sqrt{\alpha}e^{i\beta x}u,\quad\sigma\rightarrow-\sigma,$

then equation (2) can be converted into

$u_{tx}+\alpha\beta^{2}u-2i\alpha\beta u_{x}-\alpha u_{xx}+\sigma
i\alpha\beta^{2}|u|^{2}u_{x}=0,$ (3)

where $\alpha,\ \beta>0$, and here we consider the focusing case of
$\sigma=-1$.

In recent years, much work has been done on the FL equation (3). For example,
Lax pair for FL equation was obtained via the bi-Hamiltonian structure by
Fokas and Lenells [6]. They further considered the initial-boundary problem
for the FL equation (2) on the half-line by using Fokas unified method [7];
The dressing method is applied to obtain an explicit formula for N-bright-
soliton solutions for the equation (2) [8]; The N-dark soliton solution is
obtained in [9]. The bilinear method was used to obtain bright and dark
$N$-soliton solution are obtained via a systematic method [10, 11]. The
Darboux transformation is used to obtain rogue waves of the FL equation [12,
13]. The Deift-Zhou nonlinear steepest decedent method was used to analyze
long-time asymptotic behavior for FL equation with decaying initial value
[14]. Riemann-Hilbert approach was adopted to construct explicit soliton
solutions under zero boundary conditons[15]. As far as we know, the soliton
solutions for the FL equation (3) with nonzero boundary conditions (NZBCs)
have not been reported. In this paper, we apply Riemann-Hilbert (RH) approach
to focus on the inverse transformation of the FL equation (3 with the
following NZBCs

$u(x,t)\sim q_{\pm}e^{-i\beta x+2i\alpha\beta t},\quad x\rightarrow\pm\infty,$
(4)

where $|q_{\pm}|=\sqrt{\frac{2}{\beta}}$.

The inverse scattering transform is an important method to study physically
important nonlinear wave equations with Lax pair such as the NLS equation, the
modified KdV equation, Sine-Gordan equation [16, 17]. As an improved version
of inverse scattering transform, the Riemann-Hilbert method has been widely
adopted to solve nonlinear integrable models [18, 19, 20, 21, 22, 23]

The paper is organized as follows. In Section 2, starting from a Lax pair, a
transformation is introduced to invert the boundary conditions into constants,
and then Jost solutions are obtained. Furthermore, the analyticity, symmetry
and asymptotic behavior associated with eigenfunctions and scattering matrix
are obtained. In Section 3, a generalized RH problem for the FL equation is
constructed, and the distribution of discrete spectrum and residue conditions
associated with RH problem are discussed. Based on these results, we
reconstruct the potential function. In Section 4, we gives the N-soliton
solutions via solving RH problem under reflectionless case.

##  2 The direct scattering with NZBCs

###  2.1 Jost solutions

It is well-known that the FL equation (3) admits a Lax pair

$\begin{split}&\psi_{x}+ik^{2}\sigma_{3}\psi=kU_{x}\psi,\\\
&\psi_{t}+i\eta^{2}\sigma_{3}\psi=\left[\alpha
kU_{x}+\frac{i\alpha\beta^{2}}{2}\sigma_{3}\left(\frac{1}{k}U-U^{2}\right)\right]\psi,\end{split}$
(5)

where

$U=\left(\begin{array}[]{cc}0&u\\\
v&0\end{array}\right),\quad\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right),\quad\eta=\sqrt{\alpha}\left(k-\frac{\beta}{2k}\right),\quad
v=-u^{*}.$

In order to invert the nonzero boundary conditions into the constant boundary
conditions, we introduce a transformation for eigenfunctions and potentials.
Theorem 2.1  By transformation

$\begin{split}&u=qe^{-i\beta x+2i\alpha\beta t},\\\
&\psi=e^{\left(-\frac{1}{2}i\beta x+i\alpha\beta
t\right)\sigma_{3}}\varphi,\end{split}$ (6)

the FL equation (3) becomes

$q_{tx}-i\beta q_{t}+2i\alpha\beta q_{x}-\alpha
q_{xx}+(2\alpha\beta^{2}-\alpha\beta^{3}|q|^{2})q-i\alpha\beta^{2}|q|^{2}q_{x}=0$
(7)

with corresponding boundary conditions

$q\rightarrow q_{\pm},\quad x\rightarrow\pm\infty.$ (8)

And Eq. (7) is the compatibility condition of the Lax pair

$\varphi_{x}=X\phi,\quad\varphi_{t}=T\varphi,$ (9)

where

$\begin{split}X&=-ik^{2}\sigma_{3}+\frac{1}{2}i\beta\sigma_{3}-i\beta
k\sigma_{3}Q+kQ_{x},\\\
T&=-i\eta^{2}\sigma_{3}-\frac{1}{2}i\alpha\beta^{2}\sigma_{3}Q^{2}-\frac{1}{2}i\alpha\beta^{2}q_{0}^{2}\sigma_{3}\\\
&-i\alpha\beta k\sigma_{3}Q+\alpha
kQ_{x}+\frac{i\alpha\beta^{2}}{2k}\sigma_{3}Q.\end{split}$

###### Proof.

Suppose the transformation is

$\begin{split}&u=qe^{Aix+Bit},\ \
\psi=e^{(iCx+iDt)\sigma_{3}}\varphi.\end{split}$

Substituting it into (3) yields

$\begin{split}&q_{xt}+Aiq_{t}+i(B-2\alpha\beta-2\alpha A)q_{x}+(2\alpha\beta
A+\alpha\beta^{2}+\alpha A^{2}-AB)q\\\ -&\alpha
q_{xx}+A\alpha\beta^{2}|q|^{2}q-i\alpha\beta^{2}|q|^{2}q_{x}=0.\end{split}$
(10)

And the Lax pair becomes

$\begin{split}\varphi_{x}=&e^{\left[i\left(\frac{A}{2}-C\right)x+i\left(\frac{B}{2}-D\right)t\right]\hat{\sigma_{3}}}\left(\begin{array}[]{cc}-ik^{2}-iC&k(Aiq+q_{x})\\\
k(Aiq^{*}-q_{x}^{*})&ik^{2}+iC\end{array}\right)\varphi,\\\
\varphi_{t}=&e^{\left[i\left(\frac{A}{2}-C\right)x+i\left(\frac{B}{2}-D\right)t\right]\hat{\sigma_{3}}}\left(\begin{array}[]{cc}-i\eta^{2}+\frac{1}{2}i\alpha\beta^{2}|q|^{2}-iD&\alpha
k(Aiq+q_{x})+\frac{i\alpha\beta^{2}}{2k}q\\\ -\alpha
k(-Aiq^{*}+q_{x}^{*})+\frac{i\alpha\beta^{2}}{2k}q^{*}&i\eta^{2}-\frac{1}{2}i\alpha\beta^{2}|q|^{2}+iD\end{array}\right)\varphi.\end{split}$
(11)

Take $A=2C$ and $B=2D$ in the above equation and consider the limit as
$x\rightarrow\pm\infty$, one gets

$\begin{split}X_{\pm}=&\left(\begin{array}[]{cc}-ik^{2}-\frac{A}{2}i&kAiq_{\pm}\\\
kAiq_{\pm}^{*}&ik^{2}+\frac{A}{2}i\end{array}\right),\quad\\\
T_{\pm}=&\left(\begin{array}[]{cc}-i\eta^{2}+i\alpha\beta-iD&\alpha
kiAq_{\pm}+\frac{i\alpha\beta^{2}}{2k}q_{\pm}\\\ \alpha
kiAq_{\pm}^{*}+\frac{i\alpha\beta^{2}}{2k}q_{\pm}^{*}&i\eta^{2}-i\alpha\beta+iD\end{array}\right).\end{split}$

These two matrices are proportional if

$A=-\beta,\quad B=2\alpha\beta.$

Moreover, $T_{\pm}$ and $X_{\pm}$ are proportional

$T_{\pm}=\frac{\sqrt{\alpha}\eta}{k}X_{\pm}.$

which implies that their eigenvalues are proportional and they share the same
eigenvector matrices. ∎

To diagonalize the matrices $X_{\pm}$ and $T_{\pm}$ and further obtain the
Jost solutions, we need to get the eigenvalues and eigenvector matrices of
them. Direct calculation shows that the eigenvalues of the matrices $X_{\pm}$
are $\pm\frac{ik}{\sqrt{\alpha}}\lambda$; and the eigenvalues of the matrices
$T_{\pm}$ are $\pm i\eta\lambda$, where

$\lambda=\sqrt{\alpha}\big{(}k+\frac{\beta}{2k}\big{)},$

and the corresponding eigenvector matrices are

$Y_{\pm}(k)={\rm I}-\frac{\beta}{2k}Q_{\pm},\ \
\det{Y_{\pm}(k)}=1+\frac{\beta}{2k^{2}}\triangleq\gamma(k).$

With the above results, the matrices can be diagonalized as:

$\begin{split}&X_{\pm}=Y_{\pm}\left(-\frac{ik\lambda\sigma_{3}}{\sqrt{\alpha}}\right)Y_{\pm}^{-1},\\\
&T_{\pm}=Y_{\pm}(-i\eta\sigma_{3})Y_{\pm}^{-1}.\end{split}$ (12)

Thus, the asymptotic Lax pair

$\
\widetilde{\varphi}_{x}=X_{\pm}\widetilde{\varphi},\quad\widetilde{\varphi}_{t}=T_{\pm}\widetilde{\varphi}$
(13)

can be written as

$(Y_{\pm}^{-1}\widetilde{\varphi})_{x}=-\frac{ik\lambda\sigma_{3}}{\sqrt{\alpha}}(Y_{\pm}^{-1}\widetilde{\varphi}),\
\
(Y_{\pm}^{-1}\widetilde{\varphi})_{t}=-i\eta\sigma_{3}(Y_{\pm}^{-1}\widetilde{\varphi}),$

which have a solution

$\widetilde{\varphi}=Y_{\pm}(k)e^{-i\theta(x,t,k)\sigma_{3}}.$ (14)

Therefore, the asymptotic of the eigenfunctions $\varphi_{\pm}$ are

$\varphi_{\pm}\sim Y_{\pm}(k)e^{-i\theta(x,t,k)\sigma_{3}},\quad
x\rightarrow\pm\infty,$ (15)

where

$\theta(x,t,k)=\frac{k\lambda x}{\sqrt{\alpha}}+\lambda\eta t.$

Define the Jost solutions as

$J_{\pm}(x,t,k)=\varphi_{\pm}(x,t,k)e^{i\theta\sigma_{3}},$ (16)

then the Lax pair (9) is changed to

$\begin{split}&J_{\pm,x}-\frac{ik\lambda}{\sqrt{\alpha}}J_{\pm}\sigma_{3}=XJ_{\pm},\\\
&J_{\pm,t}-i\lambda\eta J_{\pm}\sigma_{3}=TJ_{\pm},\end{split}$ (17)

and

$J_{\pm}\sim Y_{\pm}(k),\quad x\rightarrow\pm\infty.$ (18)

###  2.2 Scattering matrix

The functions $\varphi_{\pm}$ are both fundamental matrix solution of the Lax
pair (9), thus there exists a matrix $S(k)$ that only depends on $k$, such
that

$\varphi_{+}(x,t,k)=\varphi_{-}(x,t,k)S(k),$ (19)

where $S(k)$ is called scattering matrix. Columnwise, Eq. (19) reads

$\varphi_{+,1}=s_{11}\varphi_{-,1}+s_{21}\varphi_{-,2},\quad\varphi_{+,2}=s_{12}\varphi_{-,1}+s_{22}\varphi_{-,2}.$
(20)

Since $tr(X)=tr(T)=0$, according to the Abel’s formula, we have

$({\rm det}\ \varphi_{\pm})_{x}=({\rm det}\ \varphi_{\pm})_{t}=0,$

which implies

${\rm det}\ \varphi_{\pm}=\lim_{x\rightarrow\pm\infty}{\rm det}\
\varphi_{\pm}=\gamma(k),$

then the scattering coefficients can be expressed as Wronskians of columns
$\varphi_{\pm}$ in the following way

$\begin{split}&s_{11}(k)=\frac{Wr(\varphi_{+,1},\varphi_{-,2})}{\gamma(k)},\quad
s_{12}(k)=\frac{Wr(\varphi_{+,2},\varphi_{-,2})}{\gamma(k)},\\\
&s_{21}(k)=\frac{Wr(\varphi_{-,1},\varphi_{+,1})}{\gamma(k)},\quad
s_{22}(k)=\frac{Wr(\varphi_{-,1},\varphi_{+,2})}{\gamma(k)}.\end{split}$ (21)

###  2.3 Asymptotic analysis

To properly construct the Riemann-Hilbert problem, we need consider the
asymptotic behavior of eigenfunctions and scattering matrix as
$k\rightarrow\infty$ and $k\rightarrow 0$.

###  2.4 Asymptotic as $k\rightarrow\infty$

Consider a solution of (17) of the form

$J_{\pm}=J_{\pm}^{(0)}+\frac{J_{\pm}^{(1)}}{k}+\frac{J_{\pm}^{(2)}}{k^{2}}+\cdots,$

then substituting the above expansion into (17) and comparing the coefficients
of the same order of $k$, we get the comparison results:
x-part:

$\displaystyle O(k^{0}):J_{\pm,x}^{(0)}+i[\sigma_{3},J_{\pm}^{(2)}]$
$\displaystyle=i\beta\sigma_{3}J_{\pm}^{(0)}-i\beta\sigma_{3}QJ_{\pm}^{(1)}+Q_{x}J_{\pm}^{(1)},$
(22) $\displaystyle O(k^{2}):J_{\pm}^{(0)}\sigma_{3}$
$\displaystyle=\sigma_{3}J_{\pm}^{(0)}.$ (23)

t-part:

$\displaystyle O(k):[\sigma_{3},J_{\pm}^{(1)}]=$
$\displaystyle-\beta\sigma_{3}QJ_{\pm}^{(0)}-iQ_{x}J_{\pm}^{(0)},$ (24)
$\displaystyle O(k^{0}):J_{\pm,t}^{(0)}+i\alpha[\sigma_{3},J_{\pm}^{(2)}]=$
$\displaystyle
i\alpha\beta\sigma_{3}J_{\pm}^{(0)}-\frac{1}{2}i\alpha\beta^{2}Q^{2}\sigma_{3}J_{\pm}^{(0)}-\frac{1}{2}i\alpha\beta^{2}q_{0}^{2}\sigma_{3}J_{\pm}^{(0)}$
$\displaystyle-i\alpha\beta\sigma_{3}QJ_{\pm}^{(1)}+\alpha
Q_{x}J_{\pm}^{(1)}.$ (25)

Based on these results, we derive that $J_{\pm}^{(0)}$ is diagonal and
satisfies

$\displaystyle J^{(0)}_{\pm,x}=$ $\displaystyle
i\nu_{1}\sigma_{3}J_{\pm}^{(0)},$ (26) $\displaystyle J^{(0)}_{\pm,t}=$
$\displaystyle i\nu_{2}\sigma_{3}J_{\pm}^{(0)},$ (27)

where

$\begin{split}\nu_{1}(x,t)=&\beta+\frac{1}{2}\beta^{2}qr-\frac{1}{2}i\beta
qr_{x}+\frac{1}{2}i\beta q_{x}r+\frac{1}{2}q_{x}r_{x},\\\
\nu_{2}(x,t)=&\alpha\beta-\frac{1}{2}\alpha\beta^{2}q_{0}^{2}+\frac{1}{2}i\alpha\beta(q_{x}r-qr_{x})+\frac{1}{2}\alpha
q_{x}r_{x}.\end{split}$

Note that the FL equation (7) admits the conservation law

$\left(q_{x}r_{x}-i\beta qr_{x}+i\beta
q_{x}r+\beta^{2}qr\right)_{t}=\left(\alpha
q_{x}r_{x}+i\alpha\beta(q_{x}r-qr_{x})\right)_{x},$

thus equations (26) and (27) are consistent and are both satisfied if we
define

$J_{\pm}^{(0)}=e^{i\nu\sigma_{3}},\quad\nu=\int_{-\infty}^{x}\left(\beta+\frac{1}{2}\beta^{2}qr-\frac{1}{2}i\beta
qr_{x}+\frac{1}{2}i\beta q_{x}r+\frac{1}{2}q_{x}r_{x}\right)dx^{{}^{\prime}}.$
(28)

Therefore, we obtain the limit of the Jost solutions as $k\rightarrow\infty$:

$J_{\pm}\sim J_{\pm}^{(0)},\quad k\rightarrow\infty.$ (29)

Define

$J_{\pm}=J_{\pm}^{(0)}\mu_{\pm},$ (30)

then we have

$\mu_{\pm}\rightarrow{\rm I},\quad k\rightarrow\infty.$ (31)

In addition, the asymptotic property of $\varphi_{\pm}$ follows:

$\varphi_{\pm}\sim e^{i(\nu-\theta)\sigma_{3}},\quad k\rightarrow\infty.$

Consider the wronskians expression of the scattering coefficients (21), we
find that

$s_{11}(k)\rightarrow 1,\ s_{22}(k)\rightarrow 1,\quad k\rightarrow\infty.$
(32)

###  2.5 Asymptotic as $k\rightarrow 0$

Consider a solution of (17) of the form

$J_{\pm}=\frac{D_{\pm}^{(-1)}}{k}+D_{\pm}^{(0)}+D_{\pm}^{(1)}k+\cdots,$

then we obtain the comparison results:
x-part:

$O(k^{-1}):D_{\pm,x}^{(-1)}-\frac{1}{2}i\beta
D_{\pm}^{(-1)}\sigma_{3}=\frac{1}{2}i\beta\sigma_{3}D_{\pm}^{-1},$ (33)

t-part:

$O(k^{-3}):D_{\pm}^{-1}\sigma_{3}=-\sigma_{3}D_{\pm}^{-1}.$ (34)

It is easy to check that

$D_{\pm,x}^{(-1)}=0,$ (35)

which implies $D_{\pm}^{(-1)}$ is a constant independent on $x$. Taking limit
on $kJ_{\pm}$ yields

$\lim_{x\rightarrow\pm\infty}\lim_{k\rightarrow 0}kJ_{\pm}=\lim_{k\rightarrow
0}\lim_{x\rightarrow\pm\infty}kJ_{\pm}=-\frac{\beta}{2}Q_{\pm},$

thus we know

$D_{\pm}^{(-1)}=-\frac{\beta}{2}Q_{\pm},$

substituting into (17) leads to

$J_{\pm}=-\frac{\beta Q_{\pm}}{2k}+O(1),\quad k\rightarrow 0,$ (36)

then the asymptotic behaviour of functions $\varphi_{\pm}$ and $\mu_{\pm}$
follows. Consider the wronskian expressions of scattering coefficients and
note the boundary conditions, we find that

$\begin{split}&s_{11}=Wr(J_{+,1}e^{-i\theta},J_{-,2}e^{i\theta})=\frac{q_{-}}{\beta
q_{+}}+O(1),\quad k\rightarrow 0,\\\
&s_{22}=Wr(J_{-,1}e^{-i\theta},J_{+,2}e^{i\theta})=\frac{q_{-}^{*}}{\beta
q_{+}^{*}}+O(1),\quad k\rightarrow 0.\end{split}$

###  2.6 Analyticity

Under the transformation (30), we write the Lax pair as

$\begin{split}\mu_{\pm,x}-\frac{ik\lambda}{\sqrt{\alpha}}[\mu_{\pm},\sigma_{3}]=&\hat{X}\mu_{\pm},\\\
\mu_{\pm,t}-i\lambda\eta[\mu_{\pm},\sigma_{3}]=&\hat{T}\mu_{\pm},\end{split}$
(37)

where

$\hat{X}=e^{-i\nu\sigma_{3}}(X+\frac{ik\lambda\sigma_{3}}{\sqrt{\alpha}}-i\nu_{1}\sigma_{3}),$
$\hat{T}=e^{-i\nu\sigma_{3}}(T+i\lambda\eta\sigma_{3}-i\nu_{2}\sigma_{3}).$

Then we obtain a full derivative

$d(e^{i\theta\hat{\sigma}_{3}}\mu_{\pm})=e^{i\theta\hat{\sigma}_{3}}(\hat{X}dx+\hat{T}dt)\mu_{\pm},$
(38)

which implies the solution of Eq. (37) satisfies the following integral
equation:

$\displaystyle\mu_{-}=\mu_{-}^{0}+\int_{-\infty}^{x}e^{-\frac{ik\lambda}{\sqrt{\alpha}}(x-y)\hat{\sigma}_{3}}\hat{X}(y,t,k)\mu_{-}(y,t,k)dy,$
(39)
$\displaystyle\mu_{+}=\mu_{+}^{0}-\int_{x}^{+\infty}e^{-\frac{ik\lambda}{\sqrt{\alpha}}(x-y)\hat{\sigma}_{3}}\hat{X}(y,t,k)\mu_{+}(y,t,k)dy,$
(40)

where

$\mu_{-}^{0}=Y_{-},\quad\mu_{+}^{0}=e^{-i\int_{-\infty}^{+\infty}\nu_{1}(x^{{}^{\prime}})dx^{{}^{\prime}}\sigma_{3}}Y_{+}.$

Thus, the eigenfunctions can be analytically extended in the complex $k$-plane
into the following regions:

$\mu_{-,1},\mu_{+,2}:\quad D^{+},\quad\mu_{+,1},\mu_{-,2}:\quad D^{-},$ (41)

where

$D_{+}=\left\\{k\ |\ arg\
k\in\left(0,\frac{\pi}{2}\right)\cup\left(\pi,\frac{3}{2}\pi\right)\right\\},$
$D_{-}=\left\\{k\ |\ arg\
k\in\left(\frac{\pi}{2},\pi\right)\cup\left(\frac{3}{2}\pi,2\pi\right)\right\\},$

which are shown in Fig.1, and the subscripts 1 and 2 identify matrix columns,
i.e., $\mu_{\pm}=(\mu_{\pm,1,\mu_{\pm,2}})$.
Apparently, the functions $\varphi_{\pm}$ and $\mu_{\pm}$ share the same
analyticity, hence from the wronskians expression of the scattering
coefficients, we know that $s_{11}$ is analytic in $D^{-}$, and $s_{22}$ is
analytic in $D^{+}$.

$+$$-$$-$$+$${\rm Im}k$${\rm Re}k$

Fig. 1. The jump contour in the complex k-plane

###  2.7 Symmetry

To investigate the discrete spectrum and residue conditions in the Riemann-
Hilbert problem, one needs to analyze the symmetric property for the solutions
$\varphi_{\pm}$ and the scattering matrix $S(k)$.
Theorem 2.2.  The Jost eigenfunctions satisfy the following symmetric
relations

$\displaystyle\sigma\varphi_{\pm}^{*}(k^{*})\sigma$
$\displaystyle=-\varphi_{\pm}(k),$ (42)
$\displaystyle\sigma_{1}\varphi_{\pm}^{*}(-k^{*})\sigma_{1}$
$\displaystyle=\varphi_{\pm}(k),$ (43)

and the scattering matrix satisfies

$\displaystyle-\sigma S(k)\sigma$ $\displaystyle=S^{*}(k^{*}),$ (44)
$\displaystyle\sigma_{1}S^{*}(-k^{*})\sigma_{1}$ $\displaystyle=S(k).$ (45)

where

$\sigma=\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right),\quad\sigma_{1}=\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right).$

###### Proof.

We only prove (42) and (44), (43) and (45) can be shown in a similar way. The
functions $\varphi_{\pm}$ are the solutions of the spectral problem

$\varphi_{\pm,x}(k)=X(k)\varphi_{\pm}(k).$ (46)

Conjugating and multiplying left by $\sigma$ on both sides of the equation
gives

$\left(\sigma\varphi_{\pm}^{*}(k^{*})\sigma\right)_{x}=\sigma
X^{*}(k^{*})\varphi_{\pm}^{*}(k^{*})\sigma.$

Note that $\sigma X^{*}(k^{*})=X(k)\sigma$, hence
$\sigma\varphi_{\pm}^{*}(k^{*})\sigma$ is also a solution of (46). By using
relations $\sigma Y_{\pm}^{*}(k^{*})=Y_{\pm}(k)\sigma$ and $\sigma
e^{i\theta\sigma_{3}}\sigma=-e^{-i\theta\sigma_{3}}$, we obtain

$\sigma\varphi_{\pm}^{*}(k^{*})\sigma\sim-
Y_{\pm}(k)e^{-i\theta\sigma_{3}},\quad x\rightarrow\pm\infty,$ (47)

which leads to (42).

For the scattering matrix, conjugating on the both sides of equation (19)
leads to

$\varphi_{+}^{*}(k^{*})=\varphi_{-}^{*}(k^{*})S^{*}(k^{*}),$

substituting (42) into the above formula yields

$\varphi_{+}(k)=-\varphi_{-}(k)\sigma S^{*}(k^{*})\sigma,$

comparing with (19) gives

$S^{*}(k^{*})=-\sigma S(k)\sigma.$

∎

Elementwise, equations (42)-(45) read as

$\begin{split}s_{11}^{*}(k^{*})&=s_{22}(k),\quad
s_{21}^{*}(k^{*})=-s_{12}(k),\\\ s_{11}^{*}(-k^{*})&=s_{22}(k),\quad
s_{21}^{*}(-k^{*})=s_{12}(k).\end{split}$ (48)
$\begin{split}\varphi_{\pm,1}(k)&=\sigma\varphi_{\pm,2}^{*}(k^{*}),\quad\varphi_{\pm,2}(k)=-\sigma\varphi_{\pm,1}^{*}(k^{*}),\\\
\varphi_{\pm,1}(k)&=\sigma_{1}\varphi_{\pm,2}^{*}(-k^{*}),\quad\varphi_{\pm,2}(k)=\sigma_{1}\varphi_{\pm,1}^{*}(-k^{*}),\end{split}$
(49)

which implies

$\varphi_{\pm,2}(k)=-\sigma_{3}\varphi_{\pm,2}(-k),\quad\varphi_{\pm,1}(k)=\sigma_{3}\varphi_{\pm,1}(-k),$
(50)

note the relation (16) and (30), there exists

$\varphi_{\pm}=e^{i\nu\sigma_{3}}\mu_{\pm}e^{-i\theta\sigma_{3}},$ (51)

thus for the eigenfunctions $\mu_{\pm}$, we derive

$\mu_{\pm,2}(k)=-\sigma_{3}\mu_{\pm,2}(-k),\quad\mu_{\pm,1}(k)=\sigma\mu_{\pm,1}(-k).$
(52)

##  3 The inverse scattering with NZBCs

###  3.1 Generalized Riemann-Hilbert problem

We define the two matrices

$M^{+}=\left(\mu_{-,1},\frac{\mu_{+,2}}{s_{22}}\right),\quad k\in D^{+},$
$M^{-}=\left(\frac{\mu_{+,1}}{s_{11}},\mu_{-,2}\right),\quad k\in D^{-},$

which are analytical in $D^{+},\ D^{-}$ respectively, and admit asymptotic

$\begin{split}M^{\pm}&=I+O(1/k),\quad k\rightarrow\infty,\\\
M^{\pm}&=\frac{1+\beta}{k}e^{-i\nu\sigma_{3}}\widetilde{Q}_{\pm}+O(1),\quad
k\rightarrow 0,\end{split}$ (53)

where $\widetilde{Q}_{-}=\left(\begin{array}[]{cc}0&-\frac{1}{q_{-}^{*}}\\\
\frac{1}{q_{-}}&0\end{array}\right)$.

By using (51) (53), we rewrite (20) and get a generalized Riemann-Hilbert
problem

$\displaystyle\star\ M(x,t,k)\ is\ meromorphic\ in\ \rm{C}\setminus\Sigma,$
(54) $\displaystyle\star\ M^{+}(x,t,k)=M^{-}(x,t,k)(I-G(x,t,k)),\quad
k\in\Sigma,$ (55) $\displaystyle\star\ M(x,t,k)\ satisfies\ residue\
conditions\ at\ zeros\ \\{k:s_{11}(k)=s_{22}(k)=0\\},$ (56)
$\displaystyle\star\ {M^{\pm}=I+O(1/k),\quad k\rightarrow\infty,}$ (57)
$\displaystyle\star\
M^{\pm}=\frac{1+\beta}{k}e^{-i\nu\sigma_{3}}\widetilde{Q}_{\pm}+O(1),\quad
k\rightarrow 0,$ (58)

where the jump matrix

$G=\left(\begin{array}[]{cc}0&-e^{-2i\theta}\widetilde{\rho}(k)\\\
e^{2i\theta}\rho(k)&\rho(k)\widetilde{\rho}(k)\end{array}\right),\ \
\rho(k)=\frac{s_{21}}{s_{11}},\quad\widetilde{\rho}(k)=\frac{s_{12}}{s_{22}}.$

and $\Sigma={\rm R}\cup i{\rm R}$ denotes the jump contour in Fig.1.

###  3.2 Discrete spectrum and residue conditions

The discrete spectrum is the set of all values $k\in{\rm C}\setminus\Sigma$
such that eigenfunctions exist in $L^{2}({\rm R})$. As usual, these values are
the zeros of $s_{11}(k)$ in $D^{+}$ and those of $s_{22}(k)$ in $D^{+}$.
Suppose that $s_{11}(k)$ has $N$ simple zeroes in $D^{+}\bigcap\\{{\rm Im}\
k>0\\}$ denoted by $k_{n},\ n=1,2,\ldots,N.$ Owing to the symmetries in (48),
there exists

$s_{11}(k_{n})=0\Longleftrightarrow s_{22}(-k_{n}^{*})=0\Longleftrightarrow
s_{22}(k_{n}^{*})=0\Longleftrightarrow s_{11}(-k_{n})=0,$

thus the discrete spectrum is the set

$\\{\pm k_{n},\pm k_{n}^{*}\\},$ (59)

which distribute in the $k$-plane as shown in Fig.2.

$k_{n}$$\centerdot$$k_{n}^{*}$$\centerdot$$-k_{n}^{*}$$\centerdot$$-k_{n}$$\centerdot$${\rm
Im}k$${\rm Re}k$

Fig. 2. Discrete spectrum

Next we derive the residue conditions will be needed for the RH problem. If
$s_{11}(k)=0$ at $k=k_{n}$ the eigenfunctions $\varphi_{+,1}$ and
$\varphi_{-,2}$ at $k=k_{n}$ must be proportional

$\varphi_{+,1}(k_{n})=b_{n}\varphi_{-,2}(k_{n}),$ (60)

where $b_{n}$ is an arbitrary constant independent on $x,t$. Under the
transformation (51), there exists linear relation for $\mu_{\pm}$

$\mu_{+,1}(k_{n})=b_{n}e^{2i\theta(k_{n})}\mu_{-,2}(k_{n}).$ (61)

Thus we get

${\rm
Res}_{k=k_{n}}\left[\frac{\mu_{+,1}}{s_{11}}\right]=\frac{\mu_{+,1}(k_{n})}{s_{11}^{\prime}(k_{n})}=C_{n}e^{2i\theta(k_{n})}\mu_{-,2}(k_{n}),$
(62)

where $C_{n}=\frac{b_{n}}{s_{11}^{\prime}(k_{n})}$. As for $k=-k_{n}$,
substituting (50) into (60) leads to

$\varphi_{+,1}(-k_{n})=-b_{n}\varphi_{-,2}(-k_{n}),$ (63)

then applying the relation (51) yileds

$\mu_{+,1}(-k_{n})=-b_{n}e^{2i\theta(k_{n})}\mu_{-,2}(-k_{n}),$ (64)

thus we obtain

${\rm
Res}_{k=-k_{n}}\left[\frac{\mu_{+,1}}{s_{11}}\right]=\frac{\mu_{+,1}(-k_{n})}{s_{11}^{{}^{\prime}}(-k_{n})}=-C_{n}e^{2i\theta(k_{n})}\sigma_{3}\mu_{-,2}(k_{n}).$
(65)

Similarly, the residue conditions at $k=\pm k_{n}^{*}$ are

$\displaystyle{\rm Res}_{k=k_{n}^{*}}\left[\frac{\mu_{+,2}}{s_{22}}\right]$
$\displaystyle=\frac{\mu_{+,2}(k_{n}^{*})}{s_{22}^{{}^{\prime}}(k_{n}^{*})}=\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}\mu_{-,1}(k_{n}^{*}),$
(66) $\displaystyle{\rm
Res}_{k=-k_{n}^{*}}\left[\frac{\mu_{+,2}}{s_{22}}\right]$
$\displaystyle=\frac{\mu_{+,2}(-k_{n}^{*})}{s_{22}^{{}^{\prime}}(-k_{n}^{*})}=\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}\sigma_{3}\mu_{-,1}(k_{n}^{*}),$
(67)

where $\widetilde{C}_{n}=-C_{n}^{*}$.
Recall the defination of $M^{\pm}$, there follows

$\begin{split}&{\rm
Res}_{k=k_{n}^{*}}M^{+}=\left(0,\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}\mu_{-,1}(k_{n}^{*})\right),\quad{\rm
Res}_{k=-k_{n}^{*}}M^{+}=\left(0,\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}\sigma_{3}\mu_{-,1}(k_{n}^{*})\right),\\\
&{\rm
Res}_{k=k_{n}}M^{-}=\left(C_{n}e^{2i\theta(k_{n})}\mu_{-,2}(k_{n}),0\right),\quad{\rm
Res}_{k=-k_{n}}M^{-}=\left(-C_{n}e^{2i\theta(k_{n})}\sigma_{3}\mu_{-,2}(k_{n}),0\right).\end{split}$
(68)

###  3.3 Reconstruction formula for the potential

To solve the Riemann-Hilbert problem (55),one needs to regularize it by
subtracting out the asymptotic behaviour and the pole contribution. Hence, we
rewrite Eq. (55) as

$\begin{split}&M^{+}-I-\frac{1+\beta}{k}e^{-i\nu\sigma_{3}}\widetilde{Q}_{-}-\sum_{n=1}^{N}\frac{{\rm
Res}_{k=k_{n}^{*}}M^{+}}{k-k_{n}^{*}}-\sum_{n=1}^{N}\frac{{\rm
Res}_{k=k_{n}}M^{-}}{k-k_{n}}\\\ -&\sum_{n=1}^{N}\frac{{\rm
Res}_{k=-k_{n}^{*}}M^{+}}{k+k_{n}^{*}}-\sum_{n=1}^{N}\frac{{\rm
Res}_{k=-k_{n}}M^{-}}{k+k_{n}}\\\
=&M^{-}-I-\frac{1+\beta}{k}e^{-i\nu\sigma_{3}}\widetilde{Q}_{-}-\sum_{n=1}^{N}\frac{{\rm
Res}_{k=k_{n}^{*}}M^{+}}{k-k_{n}^{*}}-\sum_{n=1}^{N}\frac{{\rm
Res}_{k=k_{n}}M^{-}}{k-k_{n}}\\\ -&\sum_{n=1}^{N}\frac{{\rm
Res}_{k=-k_{n}^{*}}M^{+}}{k+k_{n}^{*}}-\sum_{n=1}^{N}\frac{{\rm
Res}_{k=-k_{n}}M^{-}}{k+k_{n}}-M^{-}G,\end{split}$ (69)

where $\widetilde{Q}_{-}=\left(\begin{array}[]{cc}0&-\frac{1}{q_{-}^{*}}\\\
\frac{1}{q_{-}}&0\end{array}\right)$. Then the Plemelj’s formula shows

$\begin{split}M(x,t,k)=&I+\frac{1+\beta}{k}e^{-i\nu\sigma_{3}}\widetilde{Q}_{-}+\sum_{n=1}^{N}\frac{{\rm
Res}_{k=k_{n}^{*}}M^{+}}{k-k_{n}^{*}}+\sum_{n=1}^{N}\frac{{\rm
Res}_{k=k_{n}}M^{-}}{k-k_{n}}\\\ +&\sum_{n=1}^{N}\frac{{\rm
Res}_{k=-k_{n}^{*}}M^{+}}{k+k_{n}^{*}}+\sum_{n=1}^{N}\frac{{\rm
Res}_{k=-k_{n}}M^{-}}{k+k_{n}}+\frac{1}{2\pi
i}\int_{\Sigma}\frac{M^{-}G(x,t,\xi)}{\xi-k}d\xi,\end{split}$ (70)

and the $(1,2)$-element of $M$ is

$\begin{split}M_{12}=&\frac{1}{k}\bigg{\\{}(1+\beta)e^{-i\nu\sigma_{3}}\widetilde{Q}_{-}+\sum_{n=1}^{N}\big{(}{\rm
Res}_{k=k_{n}^{*}}M^{+}+{\rm Res}_{k=-k_{n}^{*}}M^{+}\big{)}\\\
&-\frac{1}{2\pi
i}\int_{\Sigma}\frac{M^{-}G(x,t,\xi)}{\xi-k}d\xi\bigg{\\}}_{12}+O(1/k^{2}).\end{split}$
(71)

Comparing the $(1,2)$-element on the both sides of Eq. (24) yields

$q_{x}-i\beta q=2ie^{i\nu}J_{\pm,12}^{(1)}.$ (72)

Recall the transformation (6), we know

$u_{x}e^{i\tau(x,t)}=q_{x}-i\beta q,$

where $\tau(x,t)=\beta x+\alpha\beta^{2}q_{0}^{2}t.$ Thus we can write (72) as

$u_{x}=2ie^{i\nu-i\tau}\lim_{k\rightarrow\infty}\left(kJ_{\pm}\right)_{12}=2ie^{2i\nu-i\tau}\lim_{k\rightarrow\infty}\left(k\mu_{\pm}\right)_{12}=2ie^{2i\nu-i\tau}\left(M^{-}_{1}\right)_{12},$
(73)

where

$M^{-}=M^{-}_{0}+\frac{M^{-}_{1}}{k}+\cdots.$

Substituting (71) and (68) into (73), we obtain the reconstruction formula for
potential

$\begin{split}u_{x}=&2ie^{2i\nu-i\tau}\bigg{\\{}-\frac{(1+\beta)e^{-i\nu}}{q_{-}^{*}}+2\sum_{n=1}^{N}\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}\mu_{-,1,1}(k_{n}^{*})\\\
&-\frac{1}{2\pi
i}\int_{\Sigma}\big{(}M^{-}G\big{)}_{12}(\xi)d\xi\bigg{\\}}.\end{split}$ (74)

##  4 Reflectionless potentials

Now we consider the potential $u(x,t)$ for which the reflection coefficient
$\rho(k)$ vanishes identically, that is, $G=0$. In this case, Eq. (74) reads
as

$u_{x}=2ie^{-i\tau}\left(\sum_{n}2e^{2i\nu}\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}\mu_{-,1,1}(k_{n}^{*})-\frac{(1+\beta)e^{i\nu}}{q_{-}^{*}}\right).$
(75)

To obtain the expression of the term $\mu_{-,1,1}(k_{n}^{*})$, we consider the
first and second column of (70) respectively under reflectionless case:

$\begin{split}&\mu_{-,2}(k_{n})=\left(\begin{array}[]{c}-\frac{(1+\beta)e^{-i\nu}}{k_{n}q_{-}^{*}}\\\
1\end{array}\right)+\sum_{j=1}^{N}\frac{\widetilde{C}_{j}e^{-2i\theta(k_{j}^{*})}}{k_{n}-k_{j}^{*}}\mu_{-,1}(k_{j}^{*})+\sum_{j=1}^{N}\frac{\widetilde{C}_{j}e^{-2i\theta(k_{j}^{*})}}{k_{n}+k_{j}^{*}}\sigma_{3}\mu_{-,1}(k_{j}^{*}),\\\
&\mu_{-,1}(k_{n}^{*})=\left(\begin{array}[]{c}1\\\
\frac{(1+\beta)e^{i\nu}}{k_{n}^{*}q_{-}}\end{array}\right)+\sum_{j=1}^{N}\frac{C_{j}e^{2i\theta(k_{j})}}{k_{n}^{*}-k_{j}}\mu_{-,2}(k_{j})-\sum_{j=1}^{N}\frac{C_{j}e^{2i\theta(k_{j})}}{k_{n}^{*}+k_{j}}\sigma_{3}\mu_{-,2}(k_{j}),\end{split}$
(76)

which can be further written as

$\begin{split}&\mu_{-,2}(k_{n})=\left(\begin{array}[]{c}-\frac{(1+\beta)e^{-i\nu}}{k_{n}q_{-}^{*}}\\\
1\end{array}\right)+2\sum_{j=1}^{N}\frac{\widetilde{C}_{j}e^{-2i\theta(k_{j}^{*})}}{k_{n}^{2}-(k_{j}^{*})^{2}}K_{1}\mu_{-,1}(k_{j}^{*}),\\\
&\mu_{-,1}(k_{n}^{*})=\left(\begin{array}[]{c}1\\\
\frac{(1+\beta)e^{i\nu}}{k_{n}^{*}q_{-}}\end{array}\right)+2\sum_{j=1}^{N}\frac{C_{j}e^{2i\theta(k_{j})}}{(k_{n}^{*})^{2}-k_{j}^{2}}K_{2}\mu_{-,2}(k_{j}),\end{split}$
(77)

where

$K_{1}=\left(\begin{array}[]{cc}k_{n}&0\\\ 0&k_{j}^{*}\end{array}\right),\quad
K_{2}=\left(\begin{array}[]{cc}k_{j}&0\\\ 0&k_{n}^{*}\end{array}\right).$

Define

$c_{j}(x,t,k)=\frac{C_{j}}{(k^{*})^{2}-k_{j}^{2}}e^{2i\theta(x,t,k_{j})},$

whose conjugate gives

$c_{j}^{*}(k)=-\frac{\widetilde{C}_{j}}{k^{2}-(k_{j}^{*})^{2}}e^{-2i\theta(k_{j}^{*})}.$

Then (77) reduces to

$\displaystyle\mu_{-,1,1}(k_{n}^{*})$
$\displaystyle=1+2\sum_{j=1}^{N}k_{j}c_{j}(k_{n})\mu_{-,1,2}(k_{j}),$ (78)
$\displaystyle\mu_{-,1,2}(k_{j})$
$\displaystyle=-\frac{(1+\beta)e^{-i\nu}}{k_{j}q_{-}^{*}}-2\sum_{m=1}^{N}k_{j}c_{m}^{*}(k_{j})\mu_{-,1,1}(k_{m}^{*}),$
(79)

substituting (79) into (78) yields

$\mu_{-,1,1}(k_{n}^{*})=1-\frac{(2+2\beta)e^{-i\nu}}{q_{-}^{*}}\sum_{j=1}^{N}c_{j}(k_{n})-4\sum_{j=1}^{N}\sum_{m=1}^{N}k_{j}^{2}c_{j}(k_{n})c_{m}^{*}(k_{j})\mu_{-,1,1}(k_{m}^{*}),\quad
n=1,2,\ldots,N.$ (80)

Introducing notations

$X=(X_{1},X_{2},\ldots,X_{N})^{t},\quad A=(A_{n,m}),\quad
B=(B_{1},B_{2},\ldots,B_{N})^{t}$

with components being

$X_{n}=\mu_{-,1,1}(k_{n}^{*}),\quad
A_{nm}=\sum_{j=1}^{N}4k_{j}^{2}c_{j}(k_{n})c_{m}^{*}(k_{j}),\quad
B_{n}=1-\frac{(2+2\beta)e^{-i\nu}}{q_{-}^{*}}\sum_{j=1}^{N}c_{j}(k_{n}),$

then the system (80) can be written as matrix form

$HX=B,$ (81)

where

$H=I+A=(H_{1},H_{2},\ldots,H_{N}).$

By standard Crammer rule, the system (81) is the solution of

$\mu_{-,1,1}(k_{n}^{*})=X_{n}=\frac{det\ H_{n}^{ext}}{det\ H},$ (82)

where

$H_{n}^{ext}=(H_{1},\ldots,H_{n-1},B,\ldots,H_{N}).$

Note that

$u_{x}v_{x}=q_{x}r_{x}+i\beta q_{x}r-i\beta qr_{x}+\beta^{2}qr,$

then Eq. (28) reduces to

$\nu=\int_{-\infty}^{x}(\beta+\frac{1}{2}u_{x}v_{x})(x^{{}^{\prime}},t)dx^{{}^{\prime}},$
(83)

Therefore, we obtain a compact solution:

$u_{x}=2ie^{-i\tau(x,t)+i\nu}\bigg{(}\frac{{\rm det}\ H^{aug}}{{\rm det}\
H}e^{i\nu}-\frac{1+\beta}{q_{-}^{*}}\bigg{)},$ (84)

where the augmented $(N+1)\times(N+1)$ matrix $H^{aug}$ is

$H^{aug}=\left(\begin{array}[]{cc}0&Y^{t}\\\ B&H\end{array}\right),\quad
Y=(Y_{1},\ldots,Y_{N})^{t},$

and
$Y_{n}=2\widetilde{C}_{n}e^{-2i\theta(k_{n}^{*})}=-2C_{n}^{*}e^{-2i\theta(k_{n}^{*})}$.

##  5 Trace formula and theta condition

Define

$\beta^{-}=s_{11}(k)\prod_{n=1}^{N}\frac{k^{2}-(k_{n}^{*})^{2}}{k^{2}-k_{n}^{2}},\quad\beta^{+}=s_{22}(k)\prod_{n=1}^{N}\frac{k^{2}-k_{n}^{2}}{k^{2}-(k_{n}^{*})^{2}},$
(85)

we see that they are analytic and no zeros in $D^{-}$ and $D^{+}$,
respectively. Moreover, $\beta^{+}\beta^{-}=s_{11}(k)s_{22}(k).$ Note that
${\rm det}\ S(k)=s_{11}s_{22}-s_{21}s_{12}=1$, this implies

$\frac{1}{s_{11}s_{22}}=1-\rho(k)\widetilde{\rho}(k)=1+\rho(k)\rho^{*}(k^{*}),$

thus

$\beta^{+}\beta^{-}=s_{11}s_{22}=\frac{1}{1+\rho(k)\rho^{*}(k^{*})},\quad
k\in\Sigma.$

Taking logarithms leads to

$\log\beta^{+}-(-\log\beta^{-})=-\log[1+\rho(k)\rho(k^{*})],\quad k\in\Sigma,$

then Applying Plemelj formula, we have

$\log\beta^{\pm}=\mp\frac{1}{2\pi
i}\int_{\Sigma}\frac{\log[1+\rho(s)\rho^{*}(s^{*})]}{s-k}ds,\quad k\in
D^{\pm}.$ (86)

Substituting into (85), we obtain the trace formula

$\begin{split}s_{11}(k)&=\exp\bigg{[}\frac{1}{2\pi
i}\int_{\Sigma}\frac{\log[1+\rho(s)\rho^{*}(s^{*})]}{s-k}ds\bigg{]}\prod_{n=1}^{N}\frac{k^{2}-k_{n}^{2}}{k^{2}-(k_{n}^{*})^{2}},\quad
k\in D^{-},\\\ s_{22}(k)&=\exp\bigg{[}-\frac{1}{2\pi
i}\int_{\Sigma}\frac{\log[1+\rho(s)\rho^{*}(s^{*})]}{s-k}ds\bigg{]}\prod_{n=1}^{N}\frac{k^{2}-(k_{n}^{*})^{2}}{k^{2}-k_{n}^{2}},\quad
k\in D^{+}.\end{split}$ (87)

Under reflectionless condition, they reduce to

$s_{11}(k)=\prod_{n=1}^{N}\frac{k^{2}-k_{n}^{2}}{k^{2}-(k_{n}^{*})^{2}},\quad
k\in D^{-};\quad
s_{22}(k)=\prod_{n=1}^{N}\frac{k^{2}-(k_{n}^{*})^{2}}{k^{2}-k_{n}^{2}},\quad
k\in D^{+}.$ (88)

Taking limit as $k\rightarrow 0$ for (88) leads to

$\frac{q_{-}}{\beta
q^{+}}=\exp\bigg{[}-\frac{i}{2\pi}\int_{\Sigma}\frac{\log[1+\rho(s)\rho^{*}(s^{*})]}{s}ds\bigg{]}\exp\bigg{[}4i\sum_{n=1}^{N}arg(k_{n})\bigg{]},\quad
k\in D^{-},$ (89)

note that $\beta$ is a positive constant, then we obtain the theta condition

$\arg\bigg{(}\frac{q_{-}}{q_{+}}\bigg{)}=-\frac{1}{2\pi}\int_{\Sigma}\frac{\log[1+\rho(s)\rho^{*}(s^{*})]}{s}ds+4\sum_{n=1}^{N}arg(k_{n}).$
(90)

under reflectionless condition, we have

$\arg\bigg{(}\frac{q_{-}}{q_{+}}\bigg{)}=4\sum_{n=1}^{N}\arg(k_{n}).$ (91)

##  6 One-soliton solution

As an application of the formula (84)of N-soliton solution, we construct one-
soliton solution for the FL equation, which corresponding to $N=1$. Then Eq.
(84) becomes

$u_{x}=2ie^{-i\tau+i\nu}\bigg{(}-\frac{2C^{*}_{1}e^{-2i\theta(k_{1}^{*})}e^{i\nu}}{1+4k_{1}^{2}|c_{1}(k_{1})|^{2}}+\frac{(4+4\beta)C^{*}_{1}e^{-2i\theta(k_{1}^{*})}c_{1}(k_{1})}{q_{-}^{*}(1+4k_{1}^{2}|c_{1}(k_{1})|^{2})}-\frac{1+\beta}{q_{-}^{*}}\bigg{)},$
(92)

where $k_{1}$ is an eigenvalue, $C_{1}$ is an arbitrary constant and

$\begin{split}\theta(k_{1})&=\frac{k_{1}\lambda_{1}}{\sqrt{\alpha}}x+\lambda_{1}\eta_{1}t,\quad\lambda_{1}=\sqrt{\alpha}\bigg{(}k_{1}+\frac{\beta}{2k_{1}}\bigg{)},\\\
\eta_{1}&=\sqrt{\alpha}\bigg{(}k_{1}-\frac{\beta}{2k_{1}}\bigg{)},\quad
c_{1}(k_{1})=\frac{C_{1}}{(k_{1}^{*})^{2}-k_{1}^{2}}e^{2i\theta(k_{1})}.\end{split}$
(93)

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