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Riemann-Hilbert problems and N-soliton solutions of the nonlocal reverse space-time Chen-Lee-Liu equation

  • Tongshuai Liu ,
  • Tiecheng Xia , *
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  • Department of Mathematics, Shanghai University, Shanghai, 200444, China

*Author to whom any correspondence should be addressed.

Received date: 2022-09-06

  Revised date: 2022-12-14

  Accepted date: 2023-02-02

  Online published: 2023-03-17

Copyright

© 2023 Institute of Theoretical Physics CAS, Chinese Physical Society and IOP Publishing

Abstract

In this paper, the N-soliton solutions to the nonlocal reverse space-time Chen-Lee-Liu equation have been derived. Under the nonlocal symmetry reduction to the matrix spectral problem, the nonlocal reverse space-time Chen-Lee-Liu equation can be obtained. Based on the spectral problem, the specific matrix Riemann-Hilbert problem is constructed for this nonlocal equation. Through solving this associated Riemann-Hilbert problem, the N-soliton solutions to this nonlocal equation can be obtained in the case of the jump matrix as an identity matrix.

Cite this article

Tongshuai Liu , Tiecheng Xia . Riemann-Hilbert problems and N-soliton solutions of the nonlocal reverse space-time Chen-Lee-Liu equation[J]. Communications in Theoretical Physics, 2023 , 75(3) : 035002 . DOI: 10.1088/1572-9494/acb81a

1. Introduction

As we all know, soliton theory plays an important role in nonlinear science [1, 2]. In recent years, nonlocal integrable equations have become a hot topic since Ablowitz and Musslimani proposed a series of integrable nonlocal nonlinear equations by taking specific nonlocal symmetry reductions to the AKNS scattering problem [3]. More and more scholars have carried out research on this vast field, and many significant results have been obtained [4-8].
Many ways have been tried to explore this wide research field. Among those methods, Riemann-Hilbert approach is a highly effective method to solve this kind of nonlocal nonlinear integrable equations. A great number of equations have been analyzed by constructing a corresponding Riemann-Hilbert problem, such as the nonlocal reverse space-time NLS equation [9-11], the nonlocal GI equation [12], the nonlocal mKdV equation [13], and so on [14, 15].
The classical Chen-Lee-Liu (CLL) equation, which also can be called the second type derivative nonlinear Schrödinger equation, is of great importance in mathematical physics. Therefore, there are many significant works on the local CLL equation [16, 17]. The soliton solutions are constructed by means of Hirota bilinear formalism in [18]. Besides, this kind of solution also can be obtained with the help of the Riemann-Hilbert problem [19, 20]. At the same time, the initial-boundary value problem on the half-line to this equation is analyzed in [21]. The higher order rogue wave is derived by Darboux transformation in [22]. But for the nonlocal reverse space-time CLL equation, there is not much research about it up to now. The solutions to the nonlocal CLL equation and the other types of nonlocal derivative nonlinear Schrödinger equations are investigated in [23]. The N-fold Darboux transformation is formulated and various kinds of solutions are obtained in [24, 25]. And the explicit formula of N-soliton solutions derived by the Hirota bilinear method are given in [26]. To our best knowledge, there is no literature on this nonlocal equation by using the Riemann-Hilbert approach. Therefore, it is essential to investigate this equation further.
The rest of this paper is arranged as follows. In section 2, we would like to analyze the spectral problem of the nonlocal CLL equation. Based on the Lax pair, the corresponding Riemann-Hilbert problem is formulated. After that, the solution to the specific Riemann-Hilbert problem is obtained in section 3, and the N-soliton solutions were obtained when the jump matrix as an identity matrix. Finally, we draw a conclusion for this paper and look forward to future research.

2. The construction of Riemann-Hilbert problem

In order to formulate the Riemann-Hilbert problem for the nonlocal reverse space-time CLL equation, we should first consider the following spectral problem with
$\begin{eqnarray}\left\{\begin{array}{l}{\phi }_{x}=(-{\rm{i}}{\lambda }^{2}{\sigma }_{3}+\lambda Q-\displaystyle \frac{{\rm{i}}}{4}{Q}^{2}{\sigma }_{3})\phi ,\\ {\phi }_{t}=(-2{\rm{i}}{\lambda }^{4}{\sigma }_{3}+2{\lambda }^{3}Q-{\rm{i}}{Q}^{2}{\sigma }_{3}{\lambda }^{2}+\left(\displaystyle \frac{1}{2}{Q}^{3}-{\rm{i}}{Q}_{x}{\sigma }_{3}\right)\lambda +\displaystyle \frac{1}{4}\left({Q}_{x}Q-{{QQ}}_{x}\right)-\displaystyle \frac{{\rm{i}}}{8}{Q}^{4}{\sigma }_{3})\phi ,\end{array}\right.\end{eqnarray}$
where $Q=\left(\begin{array}{cc}0 & q(x,t)\\ r(x,t) & 0\end{array}\right)$, ${\sigma }_{3}=\left(\begin{array}{cc}1 & 0\\ 0 & -1\end{array}\right)$ and λ is the spectral parameter.
The compatibility condition of the above system gives the coupled system
$\begin{eqnarray}\left\{\begin{array}{l}{\rm{i}}{q}_{t}+{q}_{{xx}}-{\rm{i}}{{qq}}_{x}r=0,\\ {\rm{i}}{r}_{t}-{r}_{{xx}}-{\rm{i}}{{rr}}_{x}q=0.\end{array}\right.\end{eqnarray}$
Under the reduction r(x, t) = −q*(x, t), where the superscript * denotes the complex conjugation, the above system is converted to the classical CLL equation:
$\begin{eqnarray}{\rm{i}}{q}_{t}+{q}_{{xx}}+{\rm{i}}| q{| }^{2}{q}_{x}=0.\end{eqnarray}$
Under the nonlocal symmetry reduction [3] r(x, t) = σq(−x, −t), the system (2) is compatible and leads to the nonlocal reverse space-time CLL equation:
$\begin{eqnarray}{\rm{i}}{q}_{t}+{q}_{{xx}}-{\rm{i}}{{qq}}_{x}\sigma q(-x,-t)=0,\end{eqnarray}$
which we mainly discussed in this paper.
The formulation of the Riemann-Hilbert problem to the nonlocal integrable is similar to the local case [27]. Without loss of generality, we can assume σ = 1 and q(x, t) decays to zero sufficiently when x → ±∞ or t → ±∞ . Hence, we can see clearly that $\phi \to {{\rm{e}}}^{-{\rm{i}}{\lambda }^{2}{\sigma }_{3}x-2{\rm{i}}{\lambda }^{4}{\sigma }_{3}t}$ when x or t at infinity. It will be natural to express $\phi$ as
$\begin{eqnarray}\phi =\psi {{\rm{e}}}^{{\rm{i}}{\lambda }^{2}{\sigma }_{3}x+2{\rm{i}}{\lambda }^{4}{\sigma }_{3}t}.\end{eqnarray}$
Inserting above equation into the spectral problem, we have
$\begin{eqnarray}\left\{\begin{array}{l}{\psi }_{x}={\rm{i}}{\lambda }^{2}[\psi ,{\sigma }_{3}]+U\psi ,\\ {\psi }_{t}=2{\rm{i}}{\lambda }^{4}[\psi ,{\sigma }_{3}]+V\psi ,\end{array}\right.\end{eqnarray}$
where $U=\lambda Q-\tfrac{{\rm{i}}}{4}{Q}^{2}{\sigma }_{3}$, $V=2{\lambda }^{3}Q-{\rm{i}}{Q}^{2}{\sigma }_{3}{\lambda }^{2}\,+(\tfrac{1}{2}{Q}^{3}-{\rm{i}}{Q}_{x}{\sigma }_{3})\lambda +\tfrac{1}{4}({Q}_{x}Q-{{QQ}}_{x})-\tfrac{{\rm{i}}}{8}{Q}^{4}{\sigma }_{3}$ and [ψ, σ3] = ψσ3σ3ψ. Clearly, UT(−x, −t) = U(x, t). Considering the adjoint scattering equation
$\begin{eqnarray}{K}_{x}={\rm{i}}{\lambda }^{2}[K,{\sigma }_{3}]-{KU}.\end{eqnarray}$
Both the inverse matrix ψ−1(x, t; λ) and ψT(−x, −t; λ) satisfy the above equation. Actually, by utilizing the condition
$\begin{eqnarray}{\left(\psi {\psi }^{-1}\right)}_{x}={\psi }_{x}{\psi }^{-1}+\psi {\left({\psi }^{-1}\right)}_{x}=0\end{eqnarray}$
and the spatial scattering problem of (4), we find
$\begin{eqnarray}{\left({\psi }^{-1}\right)}_{x}={\rm{i}}{\lambda }^{2}[{\psi }^{-1},{\sigma }_{3}]-{\psi }^{-1}U.\end{eqnarray}$
As for the matrix ψT(−x, −t; λ), through direct calculation we get
$\begin{eqnarray}\begin{array}{l}{\left[{\psi }^{{\rm{T}}}(-x,-t;\lambda )\right]}_{x}=-[{\psi }_{x}^{{\rm{T}}}(-x,-t;\lambda )]\\ \quad ={\rm{i}}{\lambda }^{2}[{\psi }^{{\rm{T}}}(-x,-t;\lambda ),{\sigma }_{3}]-{\psi }^{{\rm{T}}}(-x,-t;\lambda ){U}^{{\rm{T}}}(-x,-t)\\ \quad ={\rm{i}}{\lambda }^{2}[{\psi }^{{\rm{T}}}(-x,-t;\lambda ),{\sigma }_{3}]-{\psi }^{{\rm{T}}}(-x,-t;\lambda )U(x,t).\end{array}\end{eqnarray}$
Hence, ψT(−x, −t; λ) = ψ−1(x, t; λ) due to the asymptotic conditions for ψ as well as the uniqueness of solutions.
In this section, we just analyze the spatial scattering problem and under this condition, the time variable t will be omitted. In the scattering problem, the matrix Jost solutions ψ±(x, λ) possess this asymptotic behavior at large infinity:
$\begin{eqnarray}{\psi }_{\pm }(x,\lambda )\to I,x\to \pm \infty .\end{eqnarray}$
Based on Abel's identity as well as the boundary conditions (11), we see that
$\begin{eqnarray}\det {\psi }_{\pm }(x,\lambda )=1\end{eqnarray}$
for all x and λ. For convenience, we introduce the notation
$\begin{eqnarray}E(x,\lambda )={{\rm{e}}}^{-{\rm{i}}{\lambda }^{2}{\sigma }_{3}x}.\end{eqnarray}$
Indeed, both ψE and ψ+E are solutions for the first linear matrix equation of (1), so they are not independent but linearly related by a matrix S(λ):
$\begin{eqnarray}{\psi }_{-}E={\psi }_{+}{ES}(\lambda ),\lambda \in {\mathbb{R}}\cup {\rm{i}}{\mathbb{R}}.\end{eqnarray}$
The matrix S(λ) is called the scattering matrix. Clearly, det S(λ) = 1. Without loss of generality, we assume:
$\begin{eqnarray}S(\lambda )=\left(\begin{array}{cc}{s}_{11} & {s}_{12}\\ {s}_{21} & {s}_{22}\end{array}\right),{S}^{-1}(\lambda )=\left(\begin{array}{cc}{\hat{s}}_{11} & {\hat{s}}_{12}\\ {\hat{s}}_{21} & {\hat{s}}_{22}\end{array}\right).\end{eqnarray}$
Besides, Q has symmetry σ3Qσ3 = −Q, which implies that
$\begin{eqnarray}\psi (x,t;-\lambda )={\sigma }_{3}\psi (x,t;\lambda ){\sigma }_{3}.\end{eqnarray}$
Combing with the scattering condition (14), we can see
$\begin{eqnarray}S(-\lambda )={\sigma }_{3}S(\lambda ){\sigma }_{3},\end{eqnarray}$
which shows
$\begin{eqnarray}\begin{array}{l}{s}_{11}(-\lambda )={s}_{11}(\lambda ),{s}_{12}(-\lambda )=-{s}_{12}(\lambda ),{s}_{21}(-\lambda )\\ \quad =-{s}_{21}(\lambda ),{s}_{22}(-\lambda )={s}_{22}(\lambda ).\end{array}\end{eqnarray}$
Noting that S(λ) is a 2 × 2 matrix with identity determinant, thus the elements of S−1(λ) can be represented by
$\begin{eqnarray}{\hat{s}}_{11}={s}_{22},{\hat{s}}_{12}=-{s}_{12},{\hat{s}}_{21}=-{s}_{21},{\hat{s}}_{22}={s}_{11}.\end{eqnarray}$
This matrix contains scattering data from which we can reconstruct the potential q. The process of reconstruction will rely heavily on the analytical properties of Jost solutions ψ±(x, λ) on the complex λ plane.
Based on the method of variation of parameter and the boundary conditions [28], we obtain the Volterra integral equations for ψ and ψ+:
$\begin{eqnarray}\begin{array}{rcl}{\psi }_{-}(x,\lambda ) & = & I+{\int }_{-\infty }^{x}{{\rm{e}}}^{{\rm{i}}{\lambda }^{2}{\sigma }_{3}(\xi -x)}U(\xi ){\psi }_{-}(\xi ,\lambda ){{\rm{e}}}^{{\rm{i}}{\lambda }^{2}{\sigma }_{3}(x-\xi )}d\,{\xi },\\ {\psi }_{+}(x,\lambda ) & = & I-{\int }_{x}^{\infty }{{\rm{e}}}^{{\rm{i}}{\lambda }^{2}{\sigma }_{3}(\xi -x)}U(\xi ){\psi }_{+}(\xi ,\lambda ){{\rm{e}}}^{{\rm{i}}{\lambda }^{2}{\sigma }_{3}(x-\xi )}d\,{\xi }.\end{array}\end{eqnarray}$
According to the diagonal form of σ3 and the structure of U(x, λ), we find that the first column of ψ±(x, λ) contains the exponential factor ${{\rm{e}}}^{2{\rm{i}}{\lambda }^{2}(x-\xi )}$ and the second column contains the element ${{\rm{e}}}^{2{\rm{i}}{\lambda }^{2}(\xi -x)}$, respectively. In this way, it can be proved that ${[{\psi }_{-}]}_{1}$ and ${[{\psi }_{+}]}_{2}$ are analytic for λD+ and continued for $\lambda \in {D}^{+}\cup {\mathbb{R}}\cup {\rm{i}}{\mathbb{R}}$, while ${[{\psi }_{-}]}_{2}$ and ${[{\psi }_{+}]}_{1}$ are analytic for λD and continued for $\lambda \in {D}^{-}\cup {\mathbb{R}}\cup {\rm{i}}{\mathbb{R}}$, where
$\begin{eqnarray}\begin{array}{rcl}{D}^{+} & = & \left\{\lambda | \arg \lambda \in \left(0,\displaystyle \frac{\pi }{2}\right)\cup (\pi ,\displaystyle \frac{3\pi }{2})\right\},\\ {D}^{-} & = & \left\{\lambda | \arg \lambda \in \left(\displaystyle \frac{\pi }{2},\pi \right)\cup \left(\displaystyle \frac{3\pi }{2},2\pi \right)\right\},\end{array}\end{eqnarray}$
and ${[{\psi }_{\pm }]}_{1}$ denotes the first column of ψ± while ${[{\psi }_{\pm }]}_{2}$ stands for the second column of ψ±.
In order to get the behavior of the Jost solutions ψ±(x, λ) for very large λ, we should consider the following expansion:
$\begin{eqnarray}\psi ={\psi }_{0}+\displaystyle \frac{{\psi }_{1}}{\lambda }+\displaystyle \frac{{\psi }_{2}}{{\lambda }^{2}}+O({\lambda }^{-2}).\end{eqnarray}$
Substituting above expansion into the scattering problem (6) and comparing the same power of both sides, we have
$\begin{eqnarray}\begin{array}{l}O({\lambda }^{2}):{\rm{i}}[{\psi }_{0},{\sigma }_{3}]=0,\\ O(\lambda ):{\rm{i}}[{\psi }_{1},{\sigma }_{3}]+Q{\psi }_{0}=0,\\ O(1):{\psi }_{0x}={\rm{i}}[{\psi }_{2},{\sigma }_{3}]+Q{\psi }_{1}-\displaystyle \frac{{\rm{i}}}{4}{Q}^{2}{\sigma }_{3}{\psi }_{0}.\end{array}\end{eqnarray}$
From the first equation, we find ψ0 is a diagonal matrix. Observing the diagonal element of O(1), we get
$\begin{eqnarray}{\psi }_{0x}=\displaystyle \frac{{\rm{i}}}{4}{Q}^{2}{\sigma }_{3}{\psi }_{0}.\end{eqnarray}$
Similarly, plugging (22) into the time-part linear matrix problem, one can get in the same way
$\begin{eqnarray}{\psi }_{0t}=\left[\displaystyle \frac{1}{4}({{QQ}}_{x}-{Q}_{x}Q)+\displaystyle \frac{{\rm{i}}}{8}{Q}^{4}{\sigma }_{3}\right]{\psi }_{0}.\end{eqnarray}$
To meet the normalization condition of the Riemann-Hilbert problem to be formulated, a new specific matrix P+(x, λ) should be introduced:
$\begin{eqnarray}{P}^{+}(x,\lambda )={\psi }_{0}^{-1}({[{\psi }_{-}]}_{1},{[{\psi }_{+}]}_{2})={\psi }_{0}^{-1}({\psi }_{-}{H}_{1}+{\psi }_{+}{H}_{2}),\end{eqnarray}$
where ${H}_{1}=\left(\begin{array}{cc}1 & 0\\ 0 & 0\end{array}\right)$ and ${H}_{2}=\left(\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right)$. The above definition implies that P+(x, λ) is analytic for λD+ with the asymptotic behavior when λ goes to infinity:
$\begin{eqnarray}{P}^{+}(x,\lambda )\to I,\lambda \in {D}^{+}\to \infty .\end{eqnarray}$
Next, recalling that ${\psi }_{\pm }^{-1}(x,\lambda )$ are solutions of the adjoint scattering problem (7) with the asymptotic behavior ${\psi }_{\pm }^{-1}(x,\lambda )\to I$ as x → ∞ , we can formulate the counterpart of P+(x, λ) in D based on ${\psi }_{\pm }^{-1}(x,\lambda )$. Using the similar approach to P+(x, λ), the adjoint matrix P(x, λ) can be defined:
$\begin{eqnarray}{P}^{-}(x,\lambda )=\left(\begin{array}{c}{\left[{\psi }_{-}^{-1}\right]}^{1}\\ {\left[{\psi }_{+}^{-1}\right]}^{2}\end{array}\right){\psi }_{0}=({H}_{1}{\psi }_{-}^{-1}+{H}_{2}{\psi }_{+}^{-1}){\psi }_{0},\end{eqnarray}$
where ${[{\psi }_{\pm }^{-1}]}^{1}$ denotes the first row of ${\psi }_{\pm }^{-1}$ while ${[{\psi }_{\pm }^{-1}]}^{2}$ stands for the second row of ${\psi }_{\pm }^{-1}$. In addition, P(x, λ) is analytic for λD and has the asymptotic behavior:
$\begin{eqnarray}{P}^{-}(x,\lambda )\to I,\lambda \in {D}^{-}\to \infty .\end{eqnarray}$
According to the definition of P± and the scattering condition (14), through direct calculation
$\begin{eqnarray}{\rm{\det }}{P}^{+}={s}_{11},{\rm{\det }}{P}^{-}={\hat{s}}_{11}.\end{eqnarray}$
Up to now, the two matrix functions P± which are analytic for λD± have been constructed. So, it is natural to formulate the following Riemann-Hilbert problem:
$\begin{eqnarray}{P}^{-}(x,\lambda ){P}^{+}(x,\lambda )=G(x,\lambda ),\lambda \in {\mathbb{R}}\cup {\rm{i}}{\mathbb{R}},\end{eqnarray}$
where
$\begin{eqnarray}G(x,\lambda )=E\left(\begin{array}{cc}1 & {\hat{s}}_{12}\\ {s}_{21} & 1\end{array}\right){E}^{-1}.\end{eqnarray}$
Recalling the symmetry condition ψT(−x, −t; λ) = ψ−1(−x, −t; λ), we get
$\begin{eqnarray}{\left({P}^{+}\right)}^{{\rm{T}}}(-x,-t;\lambda )={P}^{-}(x,t;\lambda ).\end{eqnarray}$
And in this way
$\begin{eqnarray}{G}^{{\rm{T}}}(-x,-t;\lambda )=G(x,t;\lambda ).\end{eqnarray}$

3. Solution to the Riemann-Hilbert problem and N-soliton solution

In this section, we mainly discuss how to solve the Riemann-Hilbert problem in the complex λ plane which is formulated in the previous section. In the case of ${\rm{\det }}{P}^{\pm }\ne 0$ in their respective analytical areas, which is also called the regular Riemann-Hilbert problem, the jump condition can be rewritten into [27]:
$\begin{eqnarray}{\left({P}^{+}\right)}^{-1}(\lambda )-{P}^{-}(\lambda )=\hat{G}(\lambda ){\left({P}^{+}\right)}^{-1}(\lambda ),\lambda \in {\mathbb{R}}\cup {\rm{i}}{\mathbb{R}},\end{eqnarray}$
where
$\begin{eqnarray}\hat{G}=I-G=-E\left(\begin{array}{cc}0 & {\hat{s}}_{12}\\ {s}_{21} & 0\end{array}\right){E}^{-1}.\end{eqnarray}$
With the help of Plemelj formula as well as the boundary conditions, the unique solution to above equation can be provided as follows:
$\begin{eqnarray}{\left({P}^{+}\right)}^{-1}(\lambda )=I+\frac{1}{2\pi {\rm{i}}}{\int }_{-\infty }^{\infty }\frac{\hat{G}(\xi ){\left({P}^{+}\right)}^{-1}(\xi )}{\xi -\lambda }{\rm{d}}\xi ,\lambda \in {D}^{+}.\end{eqnarray}$
But in the more general case, the Riemann-Hilbert problem is non-regular, which means that ${\rm{\det }}{P}^{\pm }(\lambda )=0$ at a finite number of points. According to the condition (30) and the symmetry of Q, it is natural to assume that s11(λ) has 2N simple zeros. Actually, if λk is a simple zero of s11(λ), there must exist another simple zero −λk due to s11(λ) is an even function. Similarly, ${\hat{s}}_{11}(\lambda )$ has 2N simple zeros $\pm {\hat{\lambda }}_{k}$, 1 ≤ kN. In this situation, the kernels of P+(λk) and ${P}^{-}({\hat{\lambda }}_{k})$ contain only a single column vector vk and row vector ${\hat{v}}_{k}$, respectively
$\begin{eqnarray}{P}^{+}({\lambda }_{k}){v}_{k}=0,{\hat{v}}_{k}{P}^{-}(\hat{{\lambda }_{k}})=0,1\leqslant k\leqslant N.\end{eqnarray}$
Taking the x-derivative to the vk equation, we get
$\begin{eqnarray}{P}^{+}(x,{\lambda }_{k})\left(\displaystyle \frac{{\rm{d}}{v}_{k}}{{\rm{d}}x}+{\rm{i}}{\lambda }^{2}{\sigma }_{3}{v}_{k}\right)=0.\end{eqnarray}$
Based on our assumption, vk is the only vector in the kernel of P+(x, λk), thus
$\begin{eqnarray}\displaystyle \frac{{\rm{d}}{v}_{k}}{{\rm{d}}x}+{\rm{i}}{\lambda }^{2}{\sigma }_{3}{v}_{k}={\alpha }_{k}(x){v}_{k}.\end{eqnarray}$
The solution to above equation can be written into
$\begin{eqnarray}{v}_{k}(x)={{\rm{e}}}^{-{\rm{i}}{\lambda }_{k}^{2}{\sigma }_{3}x}{v}_{k0}.\end{eqnarray}$
Then, after the same operation to the ${\hat{v}}_{k}$ equation, we get immediately
$\begin{eqnarray}{\hat{v}}_{k}={\hat{v}}_{k0}{{\rm{e}}}^{{\rm{i}}{\hat{\lambda }}_{k}^{2}{\sigma }_{3}x}.\end{eqnarray}$
Similarly, taking t-derivative to vk and ${\hat{v}}_{k}$ and combining with the spatial dependence, we get the involution condition of vk and ${\hat{v}}_{k}$, respectively,
$\begin{eqnarray}{v}_{k}(x,t)={{\rm{e}}}^{-{\rm{i}}{\lambda }_{k}^{2}{\sigma }_{3}x-2{\rm{i}}{\lambda }_{k}^{4}{\sigma }_{3}t}{v}_{k,0},{\hat{v}}_{k}(x,t)={\hat{v}}_{k,0}{{\rm{e}}}^{{\rm{i}}{\hat{\lambda }}_{k}^{2}{\sigma }_{3}x+2{\rm{i}}{\hat{\lambda }}_{k}^{4}{\sigma }_{3}t}.\end{eqnarray}$
It is noted that the symmetry relations between the classical CLL equation and the nonlocal reverse space-time CLL equation are rather different. For the local or classical CLL equation, which can be obtained from the spectral problem under the reduction r(x, t) = −q*(x, t), the matrix Q possesses the symmetry property: Q = −Q, where the symbol † stands for the Hermitian transpose. And in this way, if λkD+ is a simple zero of $\det {P}^{+}$, we can get the corresponding zero ${\hat{\lambda }}_{k}$ of $\det {P}^{-}$ immediately ${\hat{\lambda }}_{k}={\lambda }_{k}^{* }$. Besides, if vk is the only column vector in the kernels of $\det {P}^{+}$, the corresponding row vector can be derived directly ${\hat{v}}_{k}={v}_{k}^{\dagger }$ through the symmetry relations.
However, in the nonlocal case, we could not get this relation due to the matrix Q no longer having the symmetry besides σ3Qσ3 = −Q. In other words, the nonlocal reduction kills the symmetry property that existed in the local case. Therefore, for the nonlocal reverse space-time CLL equation, the simple zero λk of $\det {P}^{+}$ can be anywhere in D+, and on the other hand, ${\hat{\lambda }}_{k}$ can be anywhere in D, respectively.
Under the above assumption, the non-regular Riemann-Hilbert problem (31) can be transformed into a regular one [29]. Different from the classical integrable equations, the following condition is not satisfied in the nonlocal case:
$\begin{eqnarray}\left\{{\lambda }_{k}| 1\leqslant k\leqslant N\}\cap \{{\hat{\lambda }}_{k}| 1\leqslant k\leqslant N\right\}\ne \varnothing .\end{eqnarray}$
The above conditions are often seen in the nonlocal integrable equations. Therefore, it is of great significance to develop the previous solutions in the literature into the nonlocal case. Indeed, in order to get rid of the zeros structure, we introduce
$\begin{eqnarray}{\rm{\Gamma }}(\lambda )=I+\sum _{j=1}^{N}\left(\displaystyle \frac{{B}_{j}}{\lambda -{\hat{\lambda }}_{j}}-\displaystyle \frac{{\sigma }_{3}{B}_{j}{\sigma }_{3}}{\lambda +{\hat{\lambda }}_{j}}\right),\end{eqnarray}$
where
$\begin{eqnarray}{B}_{j}={w}_{j}{\hat{v}}_{j}\end{eqnarray}$
with
$\begin{eqnarray}{w}_{j,1}={\left({M}^{-1}\right)}_{{jk}}{v}_{j,1},{w}_{j,2}={\left({\hat{M}}^{-1}\right)}_{{jk}}{v}_{j,2}.\end{eqnarray}$
Here, vj,i and wj,i denote the ith element of vj and wj, respectively. When ${\lambda }_{j}={\hat{\lambda }}_{k}$ or ${\lambda }_{j}=-{\hat{\lambda }}_{k}$, the jkth element of matrices M is equal to zero. Otherwise, the jkth element of matrices M and $\hat{M}$ can be represented by
$\begin{eqnarray}\begin{array}{rcl}{M}_{{jk}} & = & \displaystyle \frac{{\hat{v}}_{k}{\sigma }_{3}{v}_{j}}{{\lambda }_{j}+{\hat{\lambda }}_{k}}-\displaystyle \frac{{\hat{v}}_{k}{v}_{j}}{{\lambda }_{j}-{\hat{\lambda }}_{k}},\\ {\hat{M}}_{{jk}} & = & -\displaystyle \frac{{\hat{v}}_{k}{\sigma }_{3}{v}_{j}}{{\lambda }_{j}+{\hat{\lambda }}_{k}}-\displaystyle \frac{{\hat{v}}_{k}{v}_{j}}{{\lambda }_{j}-{\hat{\lambda }}_{k}},1\leqslant j,k\leqslant N.\end{array}\end{eqnarray}$
Obviously, if $\{{\lambda }_{j}={\hat{\lambda }}_{k}\}\cup \{{\lambda }_{j}=-{\hat{\lambda }}_{k}\},1\leqslant j,k\leqslant N,$ we have
$\begin{eqnarray}{\hat{v}}_{k}{v}_{j}=0.\end{eqnarray}$
Defining ${\hat{P}}^{+}(\lambda )={P}^{+}(\lambda ){{\rm{\Gamma }}}^{-1}(\lambda )$ and ${\hat{P}}^{-}(\lambda )={\rm{\Gamma }}(\lambda ){P}^{-}(\lambda )$. Easy to prove that ${\hat{P}}^{\pm }(\lambda )$ are analytic for λD±, respectively. Besides, these two matrices satisfy the normalization condition ${\hat{P}}^{\pm }(\lambda )\to I$ when λ → ∞ . In this way, the non-regular case of (31) becomes one regular Riemann-Hilbert problem:
$\begin{eqnarray}{\hat{P}}^{-}(\lambda ){\hat{P}}^{+}(\lambda )={\rm{\Gamma }}(\lambda )G(\lambda ){{\rm{\Gamma }}}^{-1}(\lambda ),\lambda \in {\mathbb{R}}\cup {\rm{i}}{\mathbb{R}}.\end{eqnarray}$
Based on the form of solution (37) to the regular Riemann-Hilbert problem, the solution to above equation can be derived directly:
$\begin{eqnarray}\begin{array}{l}{\left({\hat{P}}^{+}\right)}^{-1}(\lambda )=I+\frac{1}{2\pi {\rm{i}}}\\ \quad \times {\int }_{-\infty }^{\infty }\frac{{\rm{\Gamma }}(\xi )\hat{G}(\xi ){{\rm{\Gamma }}}^{-1}(\xi ){\left({\hat{P}}^{+}\right)}^{-1}(\xi )}{\xi -\lambda }{\rm{d}}\xi ,\lambda \in {D}^{+}.\end{array}\end{eqnarray}$
As λ → ∞, we have
$\begin{eqnarray}{\hat{P}}^{+}(\lambda )=I+\displaystyle \frac{1}{2\pi {\rm{i}}\lambda }{\int }_{-\infty }^{\infty }{\rm{\Gamma }}(\xi )\hat{G}(\xi ){{\rm{\Gamma }}}^{-1}(\xi ){\left({\hat{P}}^{+}\right)}^{-1}(\xi ).\end{eqnarray}$
The matrix P+(x, t; λ) possesses the following expansion
$\begin{eqnarray}{P}^{+}(x,t;\lambda )=I+\displaystyle \frac{{P}_{1}^{+}(x,t)}{\lambda }+O(\displaystyle \frac{1}{{\lambda }^{2}}),\lambda \to \infty .\end{eqnarray}$
Combing (45) with (48), we find that ${P}_{1}^{+}(x,t)$ in the above equation is
$\begin{eqnarray}\begin{array}{l}{P}_{1}^{+}(x,t)={\sum }_{j=1}^{N}({B}_{j}-{\sigma }_{3}{B}_{j}{\sigma }_{3})\\ \quad +\frac{1}{2\pi {\rm{i}}}{\int }_{-\infty }^{\infty }{\rm{\Gamma }}(\xi )\hat{G}(\xi ){{\rm{\Gamma }}}^{-1}(\xi ){\left({\hat{P}}^{+}\right)}^{-1}(\xi ){\rm{d}}\xi .\end{array}\end{eqnarray}$
The solution to the regular Riamnn-Hilbert problem (37) is not explicitly solvable since the right side of the above equation contains the integral term. In order to obtain the N-soliton solutions, we can consider the non-reflection case, which means that the jump matrix G is an identity matrix so as to remove the integral term. At this time $\hat{G}=0$ and then
$\begin{eqnarray}{P}_{1}^{+}(x,t)=\sum _{j=1}^{N}({B}_{j}-{\sigma }_{3}{B}_{j}{\sigma }_{3}).\end{eqnarray}$
According to (22) and (53), we have
$\begin{eqnarray}{\psi }_{1}={\psi }_{0}{P}_{1}^{+}.\end{eqnarray}$
Recalling the second equation of (23), it is obviously to find $Q={\rm{i}}[{\sigma }_{3},{\psi }_{1}]{\psi }_{0}^{-1}$, which implies
$\begin{eqnarray}q=2{\rm{i}}{\left({\psi }_{0}^{11}\right)}^{2}{\left({P}_{1}^{+}\right)}_{12}=2{\rm{i}}{\left({\psi }_{0}^{11}\right)}^{2}{\left[\displaystyle \sum _{j=1}^{N}({B}_{j}-{\sigma }_{3}{B}_{j}{\sigma }_{3})\right]}_{12}.\end{eqnarray}$
Based on (46) and (47), the solution q can be rewritten into
$\begin{eqnarray}q=4{\rm{i}}{\left({\psi }_{0}^{11}\right)}^{2}{\left(\sum _{j,k=1}^{N}{v}_{j}{\left({M}^{-1}\right)}_{{jk}}{\hat{v}}_{k}\right)}_{12}.\end{eqnarray}$
The rest problem is to find ${\psi }_{0}^{11}$ and then the explicit expression of the N-soliton solutions can be derived. In the case of reflectionless,
$\begin{eqnarray}\psi (x,t;\lambda )={\psi }_{0}(x,t){\rm{\Gamma }}(\lambda ).\end{eqnarray}$
When λ = 0 in the scattering problem, it can be checked that
$\begin{eqnarray}\begin{array}{rcl}{\psi }_{x}(x,t;0) & = & -\displaystyle \frac{{\rm{i}}}{4}{Q}^{2}{\sigma }_{3}\psi (x,t;0),\\ {\psi }_{t}(x,t;0) & = & [\displaystyle \frac{1}{4}({Q}_{x}Q-{{QQ}}_{x})-\displaystyle \frac{{\rm{i}}}{8}{Q}^{4}{\sigma }_{3}]\psi (x,t;0).\end{array}\end{eqnarray}$
Recalling ψ0(x, t) satisfies (24) and (25), we find that the ${\left({\psi }_{0}(x,t)\right)}^{-1}=\psi (x,t;0)$, which indicates:
$\begin{eqnarray}{\left({\psi }_{0}^{11}\right)}^{2}={\left[1-\sum _{j,k=1}^{N}\displaystyle \frac{2{\left({v}_{j}{\left({M}^{-1}\right)}_{{jk}}{\hat{v}}_{k}\right)}_{11}}{{\hat{\lambda }}_{j}}\right]}^{-1}.\end{eqnarray}$
Eventually, the N-soliton solution for nonlocal reverse space-time CLL equation can be expressed as
$\begin{eqnarray}\begin{array}{l}q(x,t)=4{\rm{i}}{\left[1-\sum _{j,k=1}^{N}\displaystyle \frac{2{\left({v}_{j}{\left({M}^{-1}\right)}_{{jk}}{\hat{v}}_{k}\right)}_{11}}{{\hat{\lambda }}_{j}}\right]}^{-1}\\ \quad \times {\left(\sum _{j,k=1}^{N}{v}_{j}{\left({M}^{-1}\right)}_{{jk}}{\hat{v}}_{k}\right)}_{12}.\end{array}\end{eqnarray}$
Here, ${v}_{j},{\hat{v}}_{k}$ , and M follow the previous definition. According to the properties of the determinant, we can rewrite the above solution as a ratio of two determinants as follows:
$\begin{eqnarray}q=4{\rm{i}}\displaystyle \frac{\det {R}_{1}}{\det {R}_{2}},\end{eqnarray}$
with ${R}_{1}=\left(\begin{array}{ccccc}{M}_{11} & {M}_{12} & \cdots & {M}_{1N} & {v}_{\mathrm{1,1}}\\ {M}_{21} & {M}_{22} & \cdots & {M}_{2N} & {v}_{\mathrm{2,1}}\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ {M}_{N1} & {M}_{N2} & \cdots & {M}_{{NN}} & {v}_{N,1}\\ {\hat{v}}_{\mathrm{1,2}} & {\hat{v}}_{\mathrm{2,2}} & \cdots & {\hat{v}}_{N,2} & 0\end{array}\right)$ and ${R}_{2}=\left(\begin{array}{ccccc}{M}_{11} & {M}_{12} & \cdots & {M}_{1N} & {v}_{\mathrm{1,1}}\\ {M}_{21} & {M}_{22} & \cdots & {M}_{2N} & {v}_{\mathrm{2,1}}\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ {M}_{N1} & {M}_{N2} & \cdots & {M}_{{NN}} & {v}_{N,1}\\ \tfrac{2{\hat{v}}_{\mathrm{1,1}}}{{\hat{\lambda }}_{1}} & \tfrac{2{\hat{v}}_{\mathrm{2,1}}}{{\hat{\lambda }}_{2}} & \cdots & \tfrac{2{\hat{v}}_{N,1}}{{\hat{\lambda }}_{N}} & 1\end{array}\right),$ where the subscripts 1 and 2 of vi and ${\hat{v}}_{i}(1\leqslant i\leqslant N)$ represent their first and second elements, respectively. In particular, when N = 1, we denote ${v}_{\mathrm{1,0}}={\left({c}_{11},{c}_{12}\right)}^{{\rm{T}}}$ and ${\hat{v}}_{\mathrm{1,0}}=({\hat{c}}_{11},{\hat{c}}_{12})$. For simplicity, we let ${\theta }_{k}(x,t)\,=-{\rm{i}}{\lambda }_{k}^{2}x-2{\rm{i}}{\lambda }_{k}^{4}t$ and ${\hat{\theta }}_{k}=-{\rm{i}}{\hat{\lambda }}_{k}^{2}x-2{\rm{i}}{\hat{\lambda }}_{k}^{4}t$. Through direct calculation, the explicit expression of one-soliton solution can be derived for the nonlocal reverse space-time integrable CLL equation:
$\begin{eqnarray}q(x,t)=2{\rm{i}}\displaystyle \frac{{c}_{11}{\hat{c}}_{12}{\hat{\lambda }}_{1}({\hat{\lambda }}_{1}^{2}-{\lambda }_{1}^{2}){{\rm{e}}}^{{\theta }_{1}+{\hat{\theta }}_{1}}}{{c}_{11}{\hat{c}}_{11}{\lambda }_{1}^{2}{{\rm{e}}}^{{\theta }_{1}-{\hat{\theta }}_{1}}+{c}_{12}{\hat{c}}_{12}{\lambda }_{1}{\hat{\lambda }}_{1}{{\rm{e}}}^{{\hat{\theta }}_{1}-{\theta }_{1}}},\end{eqnarray}$
where λ1 and ${\hat{\lambda }}_{1}$ are arbitrary values in their analytical domains, respectively. Through analysis, we can get this soliton travels with the velocity $V=-\tfrac{2\mathrm{Im}({\hat{\lambda }}_{1}^{4}-{\lambda }_{1}^{4})}{\mathrm{Im}({\hat{\lambda }}_{1}^{2}-{\lambda }_{1}^{2})}.$ On this trajectory x = Vt, the amplitude of this soliton can be written as
$\begin{eqnarray}| q| =2| {\hat{\lambda }}_{1}^{2}-{\lambda }_{1}^{2}| | \displaystyle \frac{{\hat{\lambda }}_{1}{{\rm{e}}}^{2{At}}}{{c}_{11}{\hat{c}}_{11}{\hat{\lambda }}_{1}^{2}+{c}_{12}{\hat{c}}_{12}{\lambda }_{1}{\hat{\lambda }}_{1}{{\rm{e}}}^{2{\rm{i}}{Bt}}}| ,\end{eqnarray}$
where $A=V\ \mathrm{Im}{\hat{\lambda }}_{1}^{2}+2\mathrm{Im}{\hat{\lambda }}_{1}^{4}$ and $B\,=-V\ \mathrm{Re}({\hat{\lambda }}_{1}^{2}\,-{\lambda }_{1}^{2})-2\mathrm{Re}({\hat{\lambda }}_{1}^{4}-{\lambda }_{1}^{4})$. As usual, Imz and Rez stand for the imaginary and real parts of the arbitrary complex number z, respectively. In order to get the effect of λ1 and ${\hat{\lambda }}_{1}$ on the amplitude, we rewrite A and B as
$\begin{eqnarray*}\displaystyle A=-\displaystyle \frac{8\mathrm{Re}{\lambda }_{1}\mathrm{Im}{\lambda }_{1}\mathrm{Re}{\hat{\lambda }}_{1}\mathrm{Im}{\hat{\lambda }}_{1}({\mathrm{Re}}^{2}{\lambda }_{1}-{\mathrm{Im}}^{2}{\lambda }_{1}-{\mathrm{Re}}^{2}{\hat{\lambda }}_{1}+{\mathrm{Im}}^{2}{\hat{\lambda }}_{1})}{\mathrm{Re}{\lambda }_{1}\mathrm{Im}{\lambda }_{1}-\mathrm{Re}{\hat{\lambda }}_{1}\mathrm{Im}{\hat{\lambda }}_{1}}\end{eqnarray*}$
and
$\begin{eqnarray*}B=-\displaystyle \frac{2\left[{\left(\mathrm{Re}{\lambda }_{1}-\mathrm{Re}{\hat{\lambda }}_{1}\right)}^{2}+{\left(\mathrm{Im}{\lambda }_{1}-\mathrm{Im}{\hat{\lambda }}_{1}\right)}^{2}\right]\left[{\left(\mathrm{Re}{\lambda }_{1}+\mathrm{Re}{\hat{\lambda }}_{1}\right)}^{2}+{\left(\mathrm{Im}{\lambda }_{1}+\mathrm{Im}{\hat{\lambda }}_{1}\right)}^{2}\right]\left(\mathrm{Re}{\lambda }_{1}\mathrm{Im}{\lambda }_{1}+\mathrm{Re}{\hat{\lambda }}_{1}\mathrm{Im}{\hat{\lambda }}_{1}\right)}{\mathrm{Re}{\lambda }_{1}\mathrm{Im}{\lambda }_{1}-\mathrm{Re}{\hat{\lambda }}_{1}\mathrm{Im}{\hat{\lambda }}_{1}}.\end{eqnarray*}$
From above analysis procedure, when B ≠ 0, which means that $\mathrm{Re}{\lambda }_{1}\mathrm{Im}{\lambda }_{1}+\mathrm{Re}{\hat{\lambda }}_{1}\mathrm{Im}{\hat{\lambda }}_{1}\ne 0$, the soliton will collapse periodically and the period is equal to $\tfrac{\pi }{| B| }$. When A ≠ 0, that is, ${\mathrm{Re}}^{2}{\lambda }_{1}-{\mathrm{Im}}^{2}{\lambda }_{1}-{\mathrm{Re}}^{2}{\hat{\lambda }}_{1}+{\mathrm{Im}}^{2}{\hat{\lambda }}_{1}\ne 0,$ the amplitude of the soliton will increase or decrease exponentially, which is decided by the sign of A. On the other hand, the soliton will not collapse in this case of B = 0 and A ≠ 0, but its amplitude always changes with movement. It should be noted that when both A and B are equal to zero, in other words, $\mathrm{Re}{\lambda }_{1}=-\mathrm{Re}{\hat{\lambda }}_{1}$ with $\mathrm{Im}{\lambda }_{1}=\mathrm{Im}{\hat{\lambda }}_{1}$ or ${\lambda }_{1}={\hat{\lambda }}_{1}^{* }$, the amplitude of this soliton will remain unchanged, just like the classical soliton solution of the classical CLL equation. In order to reflect the physical significance of these solutions more intuitively, the 3D graphics are drawn in figure 1 by selecting several sets of parameter values that satisfy the above specific condition.
Figure 1. The 1-soliton solution with ${c}_{11}={c}_{12}={\hat{c}}_{11}={\hat{c}}_{12}=1$, and (a): ${\lambda }_{1}=0.2+0.3{\rm{i}},{\hat{\lambda }}_{1}=0.45-0.5{\rm{i}}$, (b): ${\lambda }_{1}=0.2+0.3{\rm{i}},{\hat{\lambda }}_{1}=0.3-0.2{\rm{i}}$, (c): ${\lambda }_{1}=0.5+0.4{\rm{i}},{\hat{\lambda }}_{1}=0.5-0.4{\rm{i}}$.
In the next, we discuss the dynamic behaviors of the 2-soliton solution to (4). When N = 2 in (63), we can obtain this 2-soliton solution easily. Here, we leave it out because the expression of the 2-soliton solution is rather cumbersome. According to the symmetry, we have known that the spectral parameters λ1, λ2 and ${\hat{\lambda }}_{1},{\hat{\lambda }}_{2}$ are independent of each other. Consequently, the different configurations of pairs $({\lambda }_{1},{\hat{\lambda }}_{1})$ and $({\lambda }_{2},{\hat{\lambda }}_{2})$ will give rise to various novel types of solitons. It should be pointed out that we find that the two-soliton solution is not the nonlinear superposition of two 1-soliton solutions. Only when the parameters are specified this principle is satisfied. Similarly, we choose some sets of parameter values in the 2-soliton solution and plot the corresponding images for them in figure 2, and in this way, we can see the dynamic characteristics of the 2-soliton solution clearly. Obviously, the 2-soliton solution to the nonlocal reverse space-time CLL equation is identical to the 2-soliton solution for the classical case under the condition ${\hat{\lambda }}_{k}={\lambda }_{k}^{* }(k=1,2)$.
Figure 2. Graphics of the 2-soliton solution with ${c}_{11}={c}_{12}={c}_{21}={c}_{22}={\hat{c}}_{11}={\hat{c}}_{12}={\hat{c}}_{21}={\hat{c}}_{22}=1$, and (a): λ1 = −0.2 − 0.3i, ${\hat{\lambda }}_{1}=-0.45+0.5{\rm{i}}$, λ2 = 0.8 + 0.6i, ${\hat{\lambda }}_{2}=0.7-0.5{\rm{i}}$, (b): λ1 = 0.2 + 0.3i, ${\hat{\lambda }}_{1}=0.3-0.2{\rm{i}}$, λ2 = 0.4 + 0.15i, ${\hat{\lambda }}_{2}=0.15-0.4{\rm{i}}$, (c): λ1 = 0.2 + 0.3i, ${\hat{\lambda }}_{1}=0.3-0.2{\rm{i}}$, λ2 = −0.4 − 0.15i, ${\hat{\lambda }}_{2}=0.4+0.15{\rm{i}}$, (d): λ1 = 0.8 + 0.6i, ${\hat{\lambda }}_{1}=0.8-0.6{\rm{i}}$, λ2 = 0.5 + 0.5i, ${\hat{\lambda }}_{2}=0.5-0.5{\rm{i}}$, (e): λ1 = 0.2 + 0.3i, ${\hat{\lambda }}_{1}=0.2-0.3{\rm{i}}$, λ2 = 1 + i, ${\hat{\lambda }}_{2}=1-{\rm{i}}$, (f): λ1 = 1 + i, ${\hat{\lambda }}_{1}=1-{\rm{i}}$, λ2 = 0.2 + 0.3i, ${\hat{\lambda }}_{2}=0.4-0.4{\rm{i}}$.
Even though the solution to the nonlocal reverse space-time CLL equation looks the same as the solution to the local CLL equation solved by the Riemann-Hilbert approach, it is just that the expressions are similar. Actually, the symmetry relations are quite different between the nonlocal and local case, which further lead to the different relationship between λk and ${\hat{\lambda }}_{k}$. In the local situation, λk and ${\hat{\lambda }}_{k}$ are conjugate with each other. However, λk and ${\hat{\lambda }}_{k}$ are completely independent in the nonlocal reverse space-time case. Therefore, we can impose different values for the pair $({\lambda }_{k},{\hat{\lambda }}_{k})$ due to this independence, which will give rise to some novel N-soliton solutions. In particular, the solution (62) is exactly the same as the N-soliton solution to the local CLL equation when ${\hat{\lambda }}_{k}={\lambda }_{k}^{* }$.

4. Conclusion

In this paper, the N-soliton solutions to the nonlocal reverse space-time CLL equation have been derived. From the corresponding spectral problem, we formulate a specific Riemann-Hilbert problem, and the N-soliton solutions can be represented by the solution to the Riemann-Hilbert problem in the case of reflectionless. Compared with the local case, there are many differences in the symmetry relations of the nonlocal reverse space-time CLL equation. As a consequence, the scattering data λk and ${\hat{\lambda }}_{k}$ are independent of each other, which will lead to new types of N-soliton solutions. Indeed, different nonlocal reductions generate different symmetry relations. This paper provides a new method to solve the nonlocal reverse space-time CLL equation. Also, it can be applied to other nonlocal integrable equations. Not only the Riemann-Hilbert approach, but the Hirota bilinear method and the Darboux transformation also apply equally to the nonlocal integrable equations. There are still many problems concerning the nonlocal integrable equations that have not been done and we will carry out further research in the future.

The authors are grateful to the Editors and the Reviewers for their invaluable comments and suggestions, which have greatly improved the quality of this paper. This work has been supported by the National Natural Science Foundation of China under Grant No. 11975145.

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