1. Introduction
As one of the current research focuses, the nonlinear evolution equations (NLEEs) with zero/nonzero boundary conditions have been brought to the forefront of nonlinear systems in the past decades [1–4]. Many effective methods have emerged to get the solutions for NLEEs. In 1967, Gardner et al first proposed the inverse scattering transforms (IST) and applied it to the Korteweg–de Vries (KdV) equation [5]. The classical IST methods are generally studied on the basis of the Gel’fand-Levitan-Marchenkoo integral equation. In 1987, Zakharov et al appropriately predigested the IST method by issuing the Riemann–Hilbert (RH) formula [6]. Subsequently, more and more researchers used this method to obtain the soliton solutions of many NLEEs [7–13]. Thereby, the research on the RH formula made important progress in the field of integrable systems, which is remain popular all around the world [14–30].
In recent years, the study of bound-state (BS) solitons and rogue wave (RW) solutions based on the RH method have drawn much attention, thereby more and more BS solitons and RW solutions of the NLEEs have been found. In 1972, Zakharov and Shabat derived the multiple-poles (MPs) soliton solutions of the nonlinear Schrödinger (NLS) equation [31]. Thereafter the increasing MPs solitons of different nonlinear integrable equations have been solved, for example, the modified KdV equation [32], the sine-Gordon (sG) equation [33], the Sasa-Satsuma (SS) equation [34], the Wadati-Konno-Ichikawa (WKI) equation [35], and the complex modified KdV equation [36]. In addition, they discussed the asymptotic for MPs solitons [37, 38]. Most importantly, using the original IST method to solve the BS solitons, which requires not only a lot of calculations but also some complex constraints [32, 33]. However, the RH problem with MPs can be directly expressed by employing the residue theorem and Laurent’s series [35, 36], which not only simplifies the calculation but also obtains the BS solitons. Thereafter, Bilman and Miller found that the robust IST can be applied to solve the higher-order RW solutions of the focusing NLS equation [39]. Simultaneously, this method is used to solve breather wave solutions, rational W-type soliton solutions, and so on. Afterward, the robust IST is used to solve more nonlinear integrable equations of RW, such as the fifth-order NLS equation [40], the sixth-order NLS equation [41], the Hirota equation [42], the quartic NLS equation [43], and the generalized NLS equation [44].
In this paper, we mainly study the inhomogeneous fifth-order NLS equation [45]1 ) and then obtained the RW solutions based on the generalized DT [45]. In 2019, Feng et al studied the determinant representation of the N-fold DT based on Lax pair. Moreover, the higher-order solitary wave, breather wave and RW solutions in equation (1 ) are obtained by using the N-fold DT [46]. In 2020, Yang et al discussed equation (1 ) with NZBCs in detail. For the inverse scattering problem, they discussed simple zeros and double zeros cases of scattering coefficients and further obtained their exact solutions [47]. However, the BS solitons with zero boundary conditions (ZBCs) and the RW with nonzero boundary conditions (NZBCs) of equation (1 ) have not been analyzed. Therefore, we will use the RH problem to obtain the BS solitons of the inhomogeneous fifth-order NLS equation (1 ) with ZBCs. Then the RH problem of equation (1 ) with NZBCs is discussed. Finally, based on the obtained RH problem and the DT method, the RW solution of equation (1 ) with NZBCs is obtained.
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{q}_{t}-{\rm{i}}\epsilon {q}_{{xxxxx}}-10{\rm{i}}\epsilon | q{| }^{2}{q}_{{xxx}}\\ \quad -20{\rm{i}}\epsilon {q}_{x}{q}^{* }{q}_{{xx}}-30{\rm{i}}\epsilon | q{| }^{4}{q}_{x}\\ \quad -10{\rm{i}}\epsilon {(| {q}_{x}{| }^{2}q)}_{x}+{q}_{{xx}}+2q| q{| }^{2}-{\rm{i}}{q}_{x}=0,\end{array}\end{eqnarray}$
where q = q(x, t) represents the complex functions of x and t, ε is the perturbation parameter, and superscript ‘*’ means the complex conjugate. In 2015, Chen first constructed the generalized Darboux transformation (DT) of the inhomogeneous fifth-order NLS equation (Equation (1 ) satisfies the following Lax pair
$\begin{eqnarray}{{\rm{\Psi }}}_{x}=U{\rm{\Psi }},\qquad {{\rm{\Psi }}}_{t}=V{\rm{\Psi }},\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}U & = & -{\rm{i}}\lambda {\sigma }_{3}+Q,\\ V & = & -16{\rm{i}}{\lambda }^{5}\epsilon {\sigma }_{3}+16{\lambda }^{4}\epsilon Q-8{\rm{i}}{\lambda }^{3}\epsilon ({Q}^{2}+{Q}_{x}){\sigma }_{3}\\ & & +4{\lambda }^{2}\epsilon (2{Q}^{3}+{Q}_{x}Q-{{QQ}}_{x}-{Q}_{{xx}})\\ & & -2{\rm{i}}{\lambda }^{2}{\sigma }_{3}-{\rm{i}}\lambda {\sigma }_{3}+2\lambda Q-2{\rm{i}}\lambda \epsilon (3{Q}^{4}+6{Q}^{2}{Q}_{x}\\ & & +{Q}_{x}^{2}-{{QQ}}_{{xx}}-{Q}_{{xx}}Q-{Q}_{{xxx}}){\sigma }_{3}\\ & & -{\rm{i}}({Q}^{2}+{Q}_{x}){\sigma }_{3}+Q+\epsilon (6{Q}^{5}-6{Q}^{3}{Q}_{x}\\ & & +6{Q}^{2}{Q}_{x}Q-6{Q}_{x}{{QQ}}_{x}-4{Q}_{x}^{2}Q\\ & & -2{{QQ}}_{{xx}}Q-8{Q}^{2}{Q}_{{xx}}-{Q}_{x}{Q}_{{xx}}\\ & & +{Q}_{{xx}}{Q}_{x}-{Q}_{{xxx}}Q+{{QQ}}_{{xxx}}+{Q}_{{xxxx}}),\end{array}\end{eqnarray}$
with $\begin{eqnarray}{\sigma }_{3}=\left(\begin{array}{cc}1 & 0\\ 0 & -1\end{array}\right),\qquad Q=\left(\begin{array}{cc}0 & q\\ -{q}^{* } & 0\end{array}\right).\end{eqnarray}$
The structure of this paper is as follows: In section 2 , we construct the RH problem of equation (1 ) with ZBCs, and then derive the BS soliton with a higher-order pole. In section 3 , we construct the RH problem of the inhomogeneous fifth-order NLS equation (1 ) with NZBCs by means of the robust IST. Then the modified DT method is applied to solve the RH problem and obtain the exact breather wave and RW solutions of equation (1 ) with NZBCs. In the last section, we give some conclusions.
2. The IST with ZBCs and BS solution
We will study the BS soliton q(x, t) of the inhomogeneous fifth-order NLS equation (1 ) with ZBCs through infinity under the following conditions
$\begin{eqnarray}\mathop{\mathrm{lim}}\limits_{x\to \pm \infty }q(x,t)=0.\end{eqnarray}$
Next, we will express the IST and BS soliton of equation (1 ) with ZBCs through the research of the RH problem.
2.1. The structure of the RH problem with ZBCs
Let x → ± ∞ , we rewrite the Lax pair (2 ) into the following form
$\begin{eqnarray}\begin{array}{rcl}{{\rm{\Psi }}}_{x} & = & {U}_{0}{\rm{\Psi }}=-{\rm{i}}\lambda {\sigma }_{3}{\rm{\Psi }},\\ {{\rm{\Psi }}}_{t} & = & {V}_{0}{\rm{\Psi }}=(16{\lambda }^{4}\epsilon +2\lambda +1){U}_{0}{\rm{\Psi }},\end{array}\end{eqnarray}$
which satisfies the basic matrix solutions ${{\rm{\Psi }}}_{\pm }^{{fd}}(x,t,\lambda )$ and given by $\begin{eqnarray}\begin{array}{rcl}{{\rm{\Psi }}}_{\pm }^{{fd}}(x,t,\lambda ) & = & {{\rm{e}}}^{-{\rm{i}}{\rm{\Xi }}(x,t,\lambda ){\sigma }_{3}},\\ {\rm{\Xi }}(x,t,\lambda ) & = & \lambda [x+(16{\lambda }^{4}\epsilon +2\lambda +1)t].\end{array}\end{eqnarray}$
We get the Jost solutions $\Psi$±(x, t, λ) $\begin{eqnarray}{{\rm{\Psi }}}_{\pm }(x,t,\lambda )\to {{\rm{e}}}^{-{\rm{i}}{\rm{\Xi }}(x,t,\lambda ){\sigma }_{3}},\qquad {as}\qquad x\to \pm \infty .\end{eqnarray}$
Then, the modified Jost solutions μ±(x, t, λ) are expressed as $\begin{eqnarray}{\mu }_{\pm }(x,t,\lambda )={{\rm{\Psi }}}_{\pm }(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}{\rm{\Xi }}(x,t,\lambda ){\sigma }_{3}},\end{eqnarray}$
which fruit in ${\mu }_{\pm }(x,t,\lambda )\to {\mathbb{I}}$ (x → ± ∞ ), and satisfy the Volterra integral equations $\begin{eqnarray}\begin{array}{l}{\mu }_{-}(x,t,\lambda )=\\ \quad {\mathbb{I}}+{\displaystyle \int }_{-\infty }^{x}{{\rm{e}}}^{{\rm{i}}\lambda {\sigma }_{3}(\xi -x)}Q(y,t){\mu }_{-}(y,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\lambda {\sigma }_{3}(\xi -x)}{\rm{d}}\xi ,\\ {\mu }_{+}(x,t,\lambda )=\\ \quad {\mathbb{I}}-{\displaystyle \int }_{x}^{+\infty }{{\rm{e}}}^{{\rm{i}}\lambda {\sigma }_{3}(\xi -x)}Q(y,t){\mu }_{+}(y,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\lambda {\sigma }_{3}(\xi -x)}{\rm{d}}\xi .\end{array}\end{eqnarray}$
Let ${{\mathbb{C}}}_{+}\,=\,\{\lambda \in {\mathbb{C}}| \mathrm{Im}\lambda \,\gt \,0\}$, ${{\mathbb{C}}}_{-}=\{\lambda \in {\mathbb{C}}| \mathrm{Im}\lambda \lt 0\}$. Obviously, μ−,1 and μ+,2 are analytical for ${{\mathbb{C}}}_{+}$ and continuously extend to ${{\mathbb{C}}}_{+}\,\bigcup \,{\mathbb{R}}$, μ+,1 and μ−,2 are analytical for ${{\mathbb{C}}}_{-}$ and continuously extended to ${{\mathbb{C}}}_{-}\,\bigcup \,{\mathbb{R}}$ (where μ±,j represents the j-th column of matrix μ±).
$\Psi$+(x, t, λ) and $\Psi$−(x, t, λ) are both solutions representing the Lax pair (2 ). Therefore, $\Psi$+(x, t, λ) and $\Psi$−(x, t, λ) can be connected by the scattering matrix $S{(\lambda )=({s}_{{ij}}(\lambda ))}_{2\times 2}$ in the following form
$\begin{eqnarray}{{\rm{\Psi }}}_{+}(x,t,\lambda )={{\rm{\Psi }}}_{-}(x,t,\lambda )S(\lambda ),\qquad \lambda \in {\mathbb{R}},\end{eqnarray}$
or $\begin{eqnarray}\begin{array}{l}{\mu }_{+}(x,t,\lambda )={\mu }_{-}(x,t,\lambda ){{\rm{e}}}^{-{\rm{i}}{\rm{\Xi }}(x,t,\lambda ){\sigma }_{3}}S(\lambda )\\ \times \,{{\rm{e}}}^{{\rm{i}}{\rm{\Xi }}(x,t,\lambda ){\sigma }_{3}},\qquad \lambda \in {\mathbb{R}},\end{array}\end{eqnarray}$
where μ±(x, t, λ) and S(λ) have the following symmetries $\begin{eqnarray}\begin{array}{rcl}{\mu }_{\pm }(x,t,\lambda ) & = & {\sigma }_{2}{\mu }_{\pm }^{* }(x,t,{\lambda }^{* }){\sigma }_{2},\\ S(\lambda ) & = & {\sigma }_{2}{S}^{* }({\lambda }^{* }){\sigma }_{2},\end{array}\end{eqnarray}$
with ${\sigma }_{2}=\left(\begin{array}{cc}0 & -{\rm{i}}\\ {\rm{i}} & 0\end{array}\right)$. Furthermore, we have $\begin{eqnarray}\begin{array}{rcl}{s}_{11}(\lambda ) & = & {s}_{22}^{* }({\lambda }^{* }),\\ {s}_{12}(\lambda ) & = & -{s}_{21}^{* }({\lambda }^{* }),\end{array}\end{eqnarray}$
sjk(j, k = 1, 2) that can be shown as $\begin{eqnarray}\begin{array}{rcl}{s}_{11}(\lambda ) & = & \mathrm{Wr}({{\rm{\Psi }}}_{+,1},{{\rm{\Psi }}}_{-,2}),\\ {s}_{12}(\lambda ) & = & \mathrm{Wr}({{\rm{\Psi }}}_{+,2},{{\rm{\Psi }}}_{-,2}),\\ {s}_{21}(\lambda ) & = & \mathrm{Wr}({{\rm{\Psi }}}_{-,1},{{\rm{\Psi }}}_{+,1}),\\ {s}_{22}(\lambda ) & = & \mathrm{Wr}({{\rm{\Psi }}}_{-,1},{{\rm{\Psi }}}_{+,2}),\end{array}\end{eqnarray}$
and s11 is analytical for ${{\mathbb{C}}}_{-}$, and s22 is analytical for ${{\mathbb{C}}}_{+}$. In addition, s11, s22 → 1 (λ → ± ∞ ) in ${{\mathbb{C}}}_{-},{{\mathbb{C}}}_{+}$, respectively.Next, we will structure the RH problem. First, we need to consider the piecewise meromorphic matrices
$\begin{eqnarray}M(x,t,\lambda )=\left\{\begin{array}{l}\left[{\mu }_{-,1},\displaystyle \frac{{\mu }_{+,2}}{{s}_{22}}\right],\qquad \lambda \in {{\mathbb{C}}}_{+},\\ \left[\displaystyle \frac{{\mu }_{+,1}}{{s}_{11}},{\mu }_{-,2}\right],\qquad \lambda \in {{\mathbb{C}}}_{-}.\end{array}\right.\end{eqnarray}$
Then, we obtain the following RH problem.
$M(x,t,\lambda )$ solve the following RH problem
$\begin{eqnarray}\left\{\begin{array}{l}M(x,t,\lambda ){becomes}\,{analytical}\,{for}\,{\mathbb{C}}\setminus {\mathbb{R}},\\ {M}_{+}(x,t,\lambda )={M}_{-}(x,t,\lambda )J(x,t,\lambda ),\qquad \lambda \in {\mathbb{R}},\\ M(x,t,\lambda )\to {\mathbb{I}},\qquad \lambda \to \infty ,\end{array}\right.\end{eqnarray}$
where the jump matrix $J(x,t,\lambda )$ is $\begin{eqnarray}J(x,t,\lambda )=\left(\begin{array}{cc}1 & r(\lambda ){{\rm{e}}}^{-2{\rm{i}}{\rm{\Xi }}(x,t,\lambda )}\\ {r}^{* }({\lambda }^{* }){{\rm{e}}}^{2{\rm{i}}{\rm{\Xi }}(x,t,\lambda )} & 1+| r(\lambda ){| }^{2}\end{array}\right),\end{eqnarray}$
with $r(\lambda )=\tfrac{{s}_{12}}{{s}_{22}}$.From equations (13 ) and (14 ), we get ${M}_{+}(\lambda )\,={\sigma }_{2}{M}_{-}^{* }({\lambda }^{* }){\sigma }_{2}$. Letting1 ) with ZBCs is shown by the following formula
$\begin{eqnarray}\begin{array}{l}M(x,t,\lambda )={\mathbb{I}}+\displaystyle \frac{1}{\lambda }{M}^{(1)}\left(x,t,\lambda \right)+O\left(\displaystyle \frac{1}{{\lambda }^{2}}\right),\\ \lambda \to \infty ,\end{array}\end{eqnarray}$
then q(x, t) of the inhomogeneous fifth-order NLS equation ( $\begin{eqnarray}q(x,t)=2{\rm{i}}{M}_{12}^{(1)}(x,t,\lambda )=\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }2{\rm{i}}\lambda {M}_{12}(x,t,\lambda ).\end{eqnarray}$
2.2. BS soliton with a higher-order pole
Generally, we assume that λ satisfying s22(λ) = 0 in ${{\mathbb{C}}}_{+}$ and λ* satisfying s11(λ*) = 0 in ${{\mathbb{C}}}_{-}$ (where λ and λ* are exactly discrete spectral points). Without thinking of simple poles, we suppose that s22(λ) has N higher-order poles λn, n = 1, 2, 3, ⋯ ,N in ${{\mathbb{C}}}_{+}$, which means14 ) of the scattering matrix S(λ), we get the following results
$\begin{eqnarray}\begin{array}{rcl}{s}_{22}(\lambda ) & = & {\left(\lambda -{\lambda }_{1}\right)}^{{n}_{1}}{\left(\lambda -{\lambda }_{2}\right)}^{{n}_{2}}{\left(\lambda -{\lambda }_{3}\right)}^{{n}_{3}}\times \cdots \\ & & \times {\left(\lambda -{\lambda }_{N}\right)}^{{n}_{N}}{s}_{22}^{(0)}(\lambda ),\\ {s}_{11}({\lambda }^{* }) & = & {\left(\lambda -{\lambda }_{1}^{* }\right)}^{{n}_{1}}{\left(\lambda -{\lambda }_{2}^{* }\right)}^{{n}_{2}}{\left(\lambda -{\lambda }_{3}^{* }\right)}^{{n}_{3}}\\ & & \times \cdots \times {\left(\lambda -{\lambda }_{N}^{* }\right)}^{{n}_{N}}{s}_{11}^{(0)}({\lambda }^{* }),\end{array}\end{eqnarray}$
where ${s}_{22}^{(0)}(\lambda )={s}_{11}^{(0)}({\lambda }^{* })\ne 0$ ($\forall \lambda \in {{\mathbb{C}}}_{+}$). According to the symmetric relation ( $\begin{eqnarray}{s}_{22}({\lambda }_{n})={s}_{11}({\lambda }_{n}^{* })=0.\end{eqnarray}$
Therefore, the relevant λn and ${\lambda }_{n}^{* }$ is represented as1 ) with ZBCs. In this section, we will discuss the situation with an N-th order pole. This means that s22(λ) has an N-th order zero point on the upper half-plane (UHP) (that is ${s}_{22}(\lambda )\,={\left(\lambda -{\lambda }_{0}\right)}^{N}{s}_{22}^{(0)}(\lambda )$ $({\rm{Im}}\lambda \gt 0,N\gt 1,{s}_{22}^{(0)}({\lambda }_{0})\ne 0)$). Similarly, we obtain that M11(x, t, λ) has an N-th order pole at $\lambda ={\lambda }_{0}^{* }$, and M12(x, t, λ) has an N-th order pole at λ = λ0. Based on the normalization condition for M(x, t, λ), we write the RH problem as follows
$\begin{eqnarray}{\rm{\Upsilon }}=\{{\lambda }_{n},{\lambda }_{n}^{* }\}{}_{n=1}^{N},\end{eqnarray}$
and its distributions are given in figure 1. In the case of reflectionless potential (that is r(λ) = 0), we obtain the soliton solution of the inhomogeneous fifth-order NLS equation ( $\begin{eqnarray}\begin{array}{rcl}{M}_{11}(x,t,\lambda ) & = & 1+\sum _{n=1}^{N}\displaystyle \frac{{F}_{n}(x,t)}{{\left(\lambda -{\lambda }_{0}^{* }\right)}^{n}},\\ {M}_{12}(x,t,\lambda ) & = & \sum _{n=1}^{N}\displaystyle \frac{{G}_{n}(x,t)}{{\left(\lambda -{\lambda }_{0}\right)}^{n}}.\end{array}\end{eqnarray}$
Simultaneously, defining $\begin{eqnarray}\begin{array}{rcl}{{\rm{e}}}^{-2{\rm{i}}{\rm{\Xi }}(x,t,\lambda )} & = & \sum _{s=0}^{+\infty }{f}_{s}(x,t){\left(\lambda -{\lambda }_{0}\right)}^{s},\\ {{\rm{e}}}^{2{\rm{i}}{\rm{\Xi }}(x,t,\lambda )} & = & \sum _{s=0}^{+\infty }{f}_{s}^{* }(x,t){\left(\lambda -{\lambda }_{0}^{* }\right)}^{s},\\ {M}_{11}(x,t,\lambda ) & = & \sum _{s=0}^{+\infty }{\zeta }_{s}(x,t){\left(\lambda -{\lambda }_{0}\right)}^{s},\\ {M}_{12}(x,t,\lambda ) & = & \sum _{s=0}^{+\infty }{\xi }_{s}(x,t){\left(\lambda -{\lambda }_{0}^{* }\right)}^{s},\\ r(\lambda ) & = & {r}_{0}(\lambda )+\sum _{m=1}^{N}\displaystyle \frac{{r}_{m}}{{\left(\lambda -{\lambda }_{0}\right)}^{m}},\\ {r}^{* }({\lambda }^{* }) & = & {r}_{0}^{* }({\lambda }^{* })+\sum _{m=1}^{N}\displaystyle \frac{{r}_{m}^{* }}{{\left(\lambda -{\lambda }_{0}^{* }\right)}^{m}},\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}{f}_{s}(x,t) & = & \mathop{\mathrm{lim}}\limits_{\lambda \to {\lambda }_{0}}\displaystyle \frac{1}{s!}\displaystyle \frac{{\partial }^{s}}{\partial {\lambda }^{s}}{{\rm{e}}}^{-2{\rm{i}}{\rm{\Xi }}(x,t,\lambda )},\\ {\zeta }_{s}(x,t) & = & \mathop{\mathrm{lim}}\limits_{\lambda \to {\lambda }_{0}}\displaystyle \frac{1}{s!}\displaystyle \frac{{\partial }^{s}}{\partial {\lambda }^{s}}{M}_{11}(x,t,\lambda ),\\ {\xi }_{s}(x,t) & = & \mathop{\mathrm{lim}}\limits_{\lambda \to {\lambda }_{0}^{* }}\displaystyle \frac{1}{s!}\displaystyle \frac{{\partial }^{s}}{\partial {\lambda }^{s}}{M}_{12}(x,t,\lambda ),\\ {r}_{m} & = & \mathop{\mathrm{lim}}\limits_{\lambda \to {\lambda }_{0}}\displaystyle \frac{1}{(N-m)!}\displaystyle \frac{{\partial }^{N-m}}{\partial {\lambda }^{N-m}}[{\left(\lambda -{\lambda }_{0}\right)}^{N}r(\lambda )],\\ s & = & 0,1,2,3,\cdots ,\qquad \qquad m=1,2,3,4,\cdots ,N,\end{array}\end{eqnarray}$
and r0(λ) means analytical for the UHP.
Figure 1. Depicts the contours and the discrete spectrum of the RH problem on complex λ-plane, ${{\mathbb{C}}}_{+}$ (yellow) and ${{\mathbb{C}}}_{-}$ (white). |
On the basis of the theorem 1, (24 ) and (25 ), then we collect the correlation coefficients of ${\left(\lambda -{\lambda }_{0}\right)}^{-n}$ and ${\left(\lambda -{\lambda }_{0}^{* }\right)}^{-n}$ to get
$\begin{eqnarray}\begin{array}{rcl}{F}_{n}(x,t) & = & -\sum _{m=n}^{N}\sum _{s=0}^{m-n}{r}_{m}^{* }{f}_{m-n-s}^{* }(x,t){\xi }_{s}(x,t),\\ {G}_{n}(x,t) & = & \sum _{m=n}^{N}\sum _{s=0}^{m-n}{r}_{m}{f}_{m-n-s}(x,t){\zeta }_{s}(x,t),\end{array}\end{eqnarray}$
with n = 1, 2, 3, ⋯ ,N.Similarly, putting (24 ) into (26 ), we have28 ) into (27 ), we obtain
$\begin{eqnarray}\begin{array}{rcl}{\xi }_{s}(x,t) & = & \sum _{l=1}^{N}\left(\begin{array}{c}l+s-1\\ s\end{array}\right)\displaystyle \frac{{\left(-1\right)}^{s}{G}_{l}(x,t)}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{l+s}},\,\,s=0,1,2,\cdots ,\\ {\zeta }_{s}(x,t) & = & \left\{\begin{array}{l}1+\sum _{l=1}^{N}\displaystyle \frac{{F}_{l}(x,t)}{{\left({\lambda }_{0}-{\lambda }_{0}^{* }\right)}^{l}},\,\,\,s=0,\\ \sum _{l=1}^{N}\left(\begin{array}{c}l+s-1\\ s\end{array}\right)\displaystyle \frac{{\left(-1\right)}^{s}{F}_{l}(x,t)}{{\left({\lambda }_{0}-{\lambda }_{0}^{* }\right)}^{l+s}},\,\,s=1,2,3,\cdots .\end{array}\right.\end{array}\,\end{eqnarray}$
Putting ( $\begin{eqnarray}\begin{array}{rcl}{F}_{n}(x,t) & = & -\sum _{m=n}^{N}\sum _{s=0}^{m-n}\sum _{l=1}^{N}\left(\begin{array}{c}l+s-1\\ s\end{array}\right)\\ & & \times \displaystyle \frac{{\left(-1\right)}^{s}{r}_{m}^{* }{f}_{m-n-s}^{* }(x,t){G}_{l}}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{l+s}},\\ {G}_{n}(x,t) & = & \sum _{m=n}^{N}{r}_{m}{f}_{m-n}(x,t)\\ & & +\sum _{m=n}^{N}\sum _{s=0}^{m-n}\sum _{l=1}^{N}\left(\begin{array}{c}l+s-1\\ s\end{array}\right)\\ & & \times \displaystyle \frac{{\left(-1\right)}^{s}{r}_{m}{f}_{m-n-s}(x,t){F}_{l}}{{\left({\lambda }_{0}-{\lambda }_{0}^{* }\right)}^{l+s}},\\ n & = & 1,2,3,\cdots ,N.\end{array}\end{eqnarray}$
Subsequently, we can conclude the following theorem.Based on the ZBCs at infinity provided in (
$\begin{eqnarray}q(x,t)=2{\rm{i}}\left(\displaystyle \frac{\det ({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }}+| \eta \rangle \langle {{\rm{Y}}}_{0}| )}{\det ({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }})}-1\right),\end{eqnarray}$
where $\langle {Y}_{0}| \,=\,{[1,0,0,\cdots ,0]}_{1\times N}$ and $\begin{eqnarray}\begin{array}{rcl}{\rm{\Omega }} & = & {[{{\rm{\Omega }}}_{{nl}}]}_{N\times N}=\left[-\sum _{m=n}^{N}\sum _{s=0}^{m-n}\left(\begin{array}{c}l+s-1\\ s\end{array}\right)\right.\\ & & {\left.\times \displaystyle \frac{{\left(-1\right)}^{s}{r}_{m}^{* }{f}_{m-n-s}^{* }(x,t)}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{l+s}}\right]}_{N\times N},\\ | \eta \rangle & = & {\left[{\eta }_{1},{\eta }_{2},{\eta }_{3},\cdots ,{\eta }_{N}\right]}^{\top },\qquad \,\,\,{\eta }_{n}=\sum _{m=n}^{N}{r}_{m}{f}_{m-n}(x,t),\\ n,l & = & 1,2,3,\cdots ,N.\end{array}\end{eqnarray}$
First, we introduce
$\begin{eqnarray}\begin{array}{rcl}| F\rangle & = & {\left[{F}_{1},{F}_{2},{F}_{3},\cdots ,{F}_{N}\right]}^{\top },\\ | G\rangle & = & {\left[{G}_{1},{G}_{2},{G}_{3},\cdots ,{G}_{N}\right]}^{\top },\end{array}\end{eqnarray}$
where the superscript ”$\top $” means transposition. Subsequently, we can transform equations ( $\begin{eqnarray}| F\rangle ={\rm{\Omega }}| G\rangle ,\qquad \qquad | G\rangle =| \eta \rangle -{{\rm{\Omega }}}^{* }| F\rangle .\end{eqnarray}$
Then, we obtain $\begin{eqnarray}\begin{array}{rcl}| F\rangle & = & {\rm{\Omega }}{\left({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }}\right)}^{-1}| \eta \rangle ,\\ | G\rangle & = & {\left({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }}\right)}^{-1}| \eta \rangle .\end{array}\end{eqnarray}$
Putting ( $\begin{eqnarray}\begin{array}{l}{M}_{11}(x,t,\lambda )=1+\sum _{n=1}^{N}\displaystyle \frac{{F}_{n}(x,t)}{{\left(\lambda -{\lambda }_{0}^{* }\right)}^{n}}\\ \quad =\,\displaystyle \frac{\det \left({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }}+| \eta \rangle \langle {\rm{Y}}(\lambda )| {\rm{\Omega }}\right)}{\det ({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }})},\\ {M}_{12}(x,t,\lambda )=\sum _{n=1}^{N}\displaystyle \frac{{G}_{n}(x,t)}{{\left(\lambda -{\lambda }_{0}\right)}^{n}}\\ \quad =\displaystyle \frac{\det \left({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }}+| \eta \rangle \langle {{\rm{Y}}}^{* }({\lambda }^{* })| \right)}{\det \left({\mathbb{I}}+{{\rm{\Omega }}}^{* }{\rm{\Omega }}\right)}-1,\end{array}\end{eqnarray}$
where $\langle Y(\lambda )| \,=\,\left[\tfrac{1}{(\lambda -{\lambda }_{0}^{* })},\tfrac{1}{{\left(\lambda -{\lambda }_{0}^{* }\right)}^{2}},\cdots ,\tfrac{1}{{\left(\lambda -{\lambda }_{0}^{* }\right)}^{N}}\right]$. According to expression (When N = 2, λ = λ0 is the second-order zero point of s22, then r(λ) is rewritten as follows1 ) is
$\begin{eqnarray}r(\lambda )={r}_{0}(\lambda )+\displaystyle \frac{{r}_{1}}{\lambda -{\lambda }_{0}}+\displaystyle \frac{{r}_{2}}{{\left(\lambda -{\lambda }_{0}\right)}^{2}},\end{eqnarray}$
and Ω11, Ω12, Ω21 and Ω22 are expressed as $\begin{eqnarray}\begin{array}{rcl}{{\rm{\Omega }}}_{11} & = & -\displaystyle \frac{{r}_{1}^{* }{f}_{0}^{* }}{{\lambda }_{0}^{* }-{\lambda }_{0}}-\displaystyle \frac{{r}_{2}^{* }{f}_{1}^{* }}{{\lambda }_{0}^{* }-{\lambda }_{0}}+\displaystyle \frac{{r}_{2}^{* }{f}_{0}^{* }}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{2}},\\ {{\rm{\Omega }}}_{12} & = & -\displaystyle \frac{{r}_{1}^{* }{f}_{0}^{* }}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{2}}-\displaystyle \frac{{r}_{2}^{* }{f}_{1}^{* }}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{2}}+\displaystyle \frac{2{r}_{2}^{* }{f}_{0}^{* }}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{3}},\\ {{\rm{\Omega }}}_{21} & = & -\displaystyle \frac{{r}_{2}^{* }{f}_{0}^{* }}{{\lambda }_{0}^{* }-{\lambda }_{0}},\qquad \qquad {{\rm{\Omega }}}_{22}=-\displaystyle \frac{{r}_{2}^{* }{f}_{0}^{* }}{{\left({\lambda }_{0}^{* }-{\lambda }_{0}\right)}^{2}},\end{array}\end{eqnarray}$
$| \eta \rangle ={\left[{\eta }_{1},{\eta }_{2}\right]}^{\top }$ defined as $\begin{eqnarray}{\eta }_{1}={r}_{1}{f}_{0}+{r}_{2}{f}_{1},\qquad \qquad {\eta }_{2}={r}_{2}{f}_{0},\end{eqnarray}$
and ⟨Y0∣ = [1, 0]. Furthermore, on the basis of the theorem 2, let r1 = r2 = 1, λ0 = a + bi, we obtain that the second-order BS soliton solution of the inhomogeneous fifth-order NLS equation ( $\begin{eqnarray}q(x,t)=-\displaystyle \frac{32{\rm{i}}{b}^{3}\left({\varpi }_{1}{{\rm{e}}}^{-2{\rm{i}}\left(a+b{\rm{i}}\right){\varpi }_{2}}+{\varpi }_{3}{{\rm{e}}}^{{\varpi }_{4}}\right)}{\left({\varpi }_{5}+{{\rm{e}}}^{4b{\varpi }_{6}}\right){{\rm{e}}}^{4b{\varpi }_{6}}+258{b}^{8}},\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}{\varpi }_{1} & = & \left(2560\,{\rm{i}}{a}^{4}{b}^{5}\epsilon -15360\,{\rm{i}}{a}^{2}{b}^{7}\epsilon +2560\,{\rm{i}}{b}^{9}\epsilon \right.\\ & & -10240\,{a}^{3}{b}^{6}\epsilon +10240\,{{ab}}^{8}\epsilon +128\,{\rm{i}}{{ab}}^{5}\\ & & \left.+32\,{\rm{i}}{b}^{5}-128\,{b}^{6}\right)t+32\,{\rm{i}}{b}^{5}x-16\,{b}^{5},\\ {\varpi }_{2} & = & 16\,{b}^{4}t\epsilon -64\,{\rm{i}}{{ab}}^{3}t\epsilon -96\,{a}^{2}{b}^{2}t\epsilon +64\,{\rm{i}}{a}^{3}{bt}\epsilon \\ & & +16\,{a}^{4}t\epsilon +2\,{\rm{i}}{bt}+2\,{at}+t+x,\\ {\varpi }_{3} & = & \left(-160\,{\rm{i}}{a}^{4}b\epsilon +960\,{\rm{i}}{a}^{2}{b}^{3}\epsilon -160\,{\rm{i}}{b}^{5}\epsilon \right.\\ & & \left.-640\,{a}^{3}{b}^{2}\epsilon +640\,{{ab}}^{4}\epsilon -8\,{\rm{i}}{ab}-2\,{\rm{i}}b-8\,{b}^{2}\right)t\\ & & -2\,{\rm{i}}{bx}+2\,{\rm{i}}-b,\\ {\varpi }_{4} & = & \left(-32\,{\rm{i}}{a}^{5}\epsilon +320\,{\rm{i}}{a}^{3}{b}^{2}\epsilon -160\,{\rm{i}}{{ab}}^{4}\epsilon +480\,{a}^{4}b\epsilon \right.\\ & & -960\,{a}^{2}{b}^{3}\epsilon +96\,{b}^{5}\epsilon -4\,{\rm{i}}{a}^{2}+4\,{\rm{i}}{b}^{2}\\ & & \left.-2\,{\rm{i}}a+24\,{ab}+6\,b\right)t-2\,{\rm{i}}{ax}+6\,{bx},\\ {\varpi }_{5} & = & \left(1\,638\,400\,{a}^{8}{b}^{6}{\epsilon }^{2}+6553600\,{a}^{6}{b}^{8}{\epsilon }^{2}\right.\\ & & +9830400\,{a}^{4}{b}^{10}{\epsilon }^{2}+6553600\,{a}^{2}{b}^{12}{\epsilon }^{2}\\ & & +1638400\,{b}^{14}{\epsilon }^{2}+163840\,{a}^{5}{b}^{6}\epsilon -327680\,{a}^{3}{b}^{8}\epsilon \\ & & -491520\,{{ab}}^{10}\epsilon +40960\,{a}^{4}{b}^{6}\epsilon \\ & & -245760\,{a}^{2}{b}^{8}\epsilon +40960\,{b}^{10}\epsilon \\ & & \left.+4096\,{a}^{2}{b}^{6}+4096\,{b}^{8}+2048\,{{ab}}^{6}+256\,{b}^{6}\right){t}^{2}\\ & & +\left(40\,960\,{a}^{4}{b}^{6}\epsilon -245760\,{a}^{2}{b}^{8}\epsilon \right.\\ & & \left.+40960\,{b}^{10}\epsilon +2048\,{{ab}}^{6}+512\,{b}^{6}\right){xt}\\ & & +(81920\,{a}^{3}{b}^{7}\epsilon -81920\,{{ab}}^{9}\epsilon -20480\,{a}^{4}{b}^{5}\epsilon \\ & & +122\,880\,{a}^{2}{b}^{7}\epsilon -20480\,{b}^{9}\epsilon +1024\,{b}^{7}\\ & & \left.-1024\,{{ab}}^{5}-256\,{b}^{5}\right)t+256\,{b}^{6}{x}^{2}\\ & & +64\,{b}^{6}-256\,{b}^{5}x+96\,{b}^{4},\\ {\varpi }_{6} & = & \left(80\,{a}^{4}\epsilon -160\,{a}^{2}{b}^{2}\epsilon +16\,{b}^{4}\epsilon +4\,a+1\right)t+x.\end{array}\end{eqnarray}$
The BS soliton solution (39 ) means the interaction between two soliton solutions, in which the high peak is caused by the interaction of two solitons with related eigenvalues. The relevant evolution process for the solutions (39 ) at different coefficient ε are counseled in Figure 2. We can find that the change of parameter ε affects the phase for the two solitons in Figure 2.
Figure 2. The density plots of the second-order the BS soliton solution ( |
3. The IST with NZBCs and RW
We will study the RW solution q(x, t) of the inhomogeneous fifth-order NLS equation (1 ) with NZBCs through infinity under the following conditions
$\begin{eqnarray}\mathop{\mathrm{lim}}\limits_{x\to \pm \infty }q(x,t)=B{{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)},\end{eqnarray}$
where ω and B > 0 expressions real constants, ν = (30B4ω − 20B2ω3 + ω5)ε + 2B2 − ω2 + ω.3.1. The structure of the RH problem with NZBCs
According to the robust IST, we obtain the RH problem of the inhomogeneous fifth-order NLS equation (1 ) with NZBCs. For ∀ t (x → ± ∞ ), q(x, t) → Bei(ωx+νt). Based on the standard transformation ${\rm{\Psi }}(x,t)={{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)\tfrac{{\sigma }_{3}}{2}}\psi (x,t)$, the Lax pair (2 ) will be converted to
$\begin{eqnarray}{\psi }_{x}=X\psi ,\qquad {\psi }_{t}=T\psi ,\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}X & = & -{\rm{i}}\left(\lambda +\displaystyle \frac{\omega }{2}\right){\sigma }_{3}+{Q}_{1},\\ {Q}_{1} & = & \left(\begin{array}{cc}0 & q{{\rm{e}}}^{-{\rm{i}}(\omega x+\nu t)}\\ -{q}^{* }{{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)} & 0\end{array}\right),\\ T & = & {{\rm{e}}}^{-{\rm{i}}\left(\omega x+\nu t\right)\displaystyle \frac{{\sigma }_{3}}{2}}\left(V-\displaystyle \frac{{\rm{i}}\nu }{2}{\sigma }_{3}\right){{\rm{e}}}^{{\rm{i}}\left(\omega x+\nu t\right)\displaystyle \frac{{\sigma }_{3}}{2}}.\end{array}\end{eqnarray}$
By the NZBCs (41 ), we can rewrite the above Lax pair (42 ) into the following form44 ), we can get the following solution
$\begin{eqnarray}{\psi }_{\pm x}={X}_{\pm }{\psi }_{\pm },\qquad {\psi }_{\pm t}={T}_{\pm }{\psi }_{\pm },\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}{X}_{\pm } & = & -{\rm{i}}\left(\lambda +\displaystyle \frac{\omega }{2}\right){\sigma }_{3}+{Q}_{0},\\ {Q}_{0} & = & \left(\begin{array}{cc}0 & B\\ -B & 0\end{array}\right),\\ {T}_{\pm } & = & \delta (\lambda ){X}_{\pm }=\left[16\,{\lambda }^{4}\epsilon -8\,{\lambda }^{3}\omega \,\epsilon +\left(-8\,{B}^{2}\epsilon +4\,{\omega }^{2}\epsilon \right){\lambda }^{2}\right.\\ & & +\left(12\,{B}^{2}\omega \,\epsilon -2\,{\omega }^{3}\epsilon +2\right)\lambda \\ & & \left.+6\,{B}^{4}\epsilon -12\,{B}^{2}{\omega }^{2}\epsilon +{\omega }^{4}\epsilon -\omega +1\right]{X}_{\pm }.\end{array}\end{eqnarray}$
According to the Lax pair ( $\begin{eqnarray}\begin{array}{rcl}{\psi }_{\pm }(\lambda ,x,t) & = & Y(\lambda ){{\rm{e}}}^{-{\rm{i}}\theta (\lambda ,x,t){\sigma }_{3}},\\ Y(\lambda ) & = & n(\lambda )\left(\begin{array}{cc}1 & \displaystyle \frac{{\rm{i}}\left(\lambda +\tfrac{\omega }{2}-\rho \right)}{B}\\ \displaystyle \frac{{\rm{i}}\left(\lambda +\tfrac{\omega }{2}-\rho \right)}{B} & 1\end{array}\right),\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}n{\left(\lambda \right)}^{2} & = & \displaystyle \frac{\lambda +\tfrac{\omega }{2}+\rho }{2\rho },\\ \theta (\lambda ,x,t) & = & \rho (\lambda )[x+\delta (\lambda )t],\end{array}\end{eqnarray}$
and $\rho {\left(\lambda \right)}^{2}={\left(\lambda +\tfrac{\omega }{2}\right)}^{2}+{B}^{2}$, which is a two-sheeted Riemann surface for λ with branch points being $\lambda =-\tfrac{\omega }{2}\pm {\rm{i}}B$, ρ(λ) = λ + O(λ−1), and $\det (Y(\lambda ))=1$ for $\lambda \ne -\tfrac{\omega }{2}\pm {\rm{i}}B$. The branch cut of ρ(λ) expressed by η = η− ∪ η+ with ${\eta }_{-}=\left[-{\rm{i}}B-\tfrac{\omega }{2},-\tfrac{\omega }{2}\right]$ and ${\eta }_{+}=\left[{\rm{i}}B-\tfrac{\omega }{2},-\tfrac{\omega }{2}\right]$, and η means oriented upward (see figure 3).
Figure 3. The contour ${{\rm{\Sigma }}}_{0}={\mathbb{R}}\cup \eta $ of the basic RH problem. |
We assume that Φ±(λ, x, t) are also the solution of the Lax pair (44 ) and Φ±(λ, x, t) → ψ±(λ, x, t) (x → ± ∞ ). Then, taking transformation
$\begin{eqnarray}{\mu }_{\pm }(\lambda ,x,t)={{\rm{\Phi }}}_{\pm }(\lambda ,x,t){{\rm{e}}}^{{\rm{i}}\theta (\lambda ,x,t){\sigma }_{3}},\end{eqnarray}$
with $\begin{eqnarray}{\mu }_{\pm }(\lambda ,x,t)\to Y(\lambda ),\qquad x\to \pm \infty .\end{eqnarray}$
Then we can immediately calculate that μ± satisfy
$\begin{eqnarray}\begin{array}{l}{\left({Y}^{-1}{\mu }_{\pm }\right)}_{x}-{\rm{i}}\rho \left[{Y}^{-1}{\mu }_{\pm },{\sigma }_{3}\right]={Y}^{-1}({Q}_{1}-{Q}_{0}){\mu }_{\pm },\\ {\left({Y}^{-1}{\mu }_{\pm }\right)}_{t}-{\rm{i}}\rho \delta \left[{Y}^{-1}{\mu }_{\pm },{\sigma }_{3}\right]={Y}^{-1}(T-{T}_{\pm }){\mu }_{\pm },\end{array}\end{eqnarray}$
and satisfy the following Volterra integral equations $\begin{eqnarray}\begin{array}{rcl}{\mu }_{-}(x,t,\lambda ) & = & Y+{\displaystyle \int }_{-\infty }^{x}Y{{\rm{e}}}^{{\rm{i}}\rho {\sigma }_{3}(\xi -x)}\\ & & \times \left[{Y}^{-1}({Q}_{1}-{Q}_{0}){\mu }_{-}(\xi ,t,\lambda )\right]{{\rm{e}}}^{-{\rm{i}}\rho {\sigma }_{3}(\xi -x)}{\rm{d}}\xi ,\\ {\mu }_{+}(x,t,\lambda ) & = & Y-{\displaystyle \int }_{x}^{+\infty }Y{{\rm{e}}}^{{\rm{i}}\rho {\sigma }_{3}(\xi -x)}\\ & & \times \left[{Y}^{-1}({Q}_{1}-{Q}_{0}){\mu }_{+}(\xi ,t,\lambda )\right]{{\rm{e}}}^{-{\rm{i}}\rho {\sigma }_{3}(\xi -x)}{\rm{d}}\xi .\end{array}\end{eqnarray}$
Firstly, we will use the robust IST to structure the RH problem [39]. Let us assume $q(x,t)-B{{\rm{e}}}^{(\omega x+\nu t)}\in {L}^{1}({\mathbb{R}})$ and μ± = [μ±1, μ±2]. Since μ−1 include e−2iρ(ξ−x), it is verified that μ−1 means analytical on ${{\mathbb{C}}}_{+}\setminus {\eta }_{+}$ (where ${{\mathbb{C}}}_{+}=\{\lambda :\mathrm{Im}\lambda \gt 0\}$). Using a similar method, we can find that μ−2 means analytical on ${{\mathbb{C}}}_{-}\setminus {\eta }_{-}$ (where ${{\mathbb{C}}}_{-}=\{\lambda :\mathrm{Im}\lambda \lt 0\}$). In summary, μ−1 and μ+2 are analytically continuous to ${{\mathbb{C}}}_{+}\setminus {\eta }_{+}$, while μ+1 and μ−2 are analytically continuous to ${{\mathbb{C}}}_{-}\setminus {\eta }_{-}$.
Since Φ±(λ, x, t) satisfy (42 ) for $\lambda \in {{\rm{\Sigma }}}_{0}\setminus \left\{-\tfrac{\omega }{2}\pm {\rm{i}}B\right\}$, we can give the scattering relation by S(λ)42 ) is obtained
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{+}(\lambda ,x,t)={{\rm{\Phi }}}_{-}(\lambda ,x,t)S(\lambda ),\\ \lambda \in {{\rm{\Sigma }}}_{0}\setminus \left\{-\displaystyle \frac{\omega }{2}\pm {\rm{i}}B\right\},\end{array}\end{eqnarray}$
the scattering matrix is shown as $\begin{eqnarray}S(\lambda )=\left(\begin{array}{cc}{S}_{11}(\lambda ) & {S}_{12}(\lambda )\\ {S}_{21}(\lambda ) & {S}_{22}(\lambda )\end{array}\right),\qquad \det (S(\lambda ))=1,\end{eqnarray}$
where ${S}_{11}(\lambda )={S}_{22}^{* }({\lambda }^{* })$, ${S}_{12}(\lambda )=-{S}_{21}^{* }({\lambda }^{* })$. Furthermore, the Beals-Coifman (BC) simultaneous solution of the Lax pair ( $\begin{eqnarray}{\phi }^{{BC}}(x,t,\lambda )=\left\{\begin{array}{l}\left[{{\rm{\Phi }}}_{-,1}(\lambda ,x,t),\displaystyle \frac{{{\rm{\Phi }}}_{+,2}\left(\lambda ,x,t\right)}{{S}_{22}\left(\lambda \right)}\right],\,\,\lambda \in {{\mathbb{C}}}_{+}\setminus {\eta }_{+},\\ \left[\displaystyle \frac{{{\rm{\Phi }}}_{+,1}\left(\lambda ,x,t\right)}{{S}_{11}\left(\lambda \right)},{{\rm{\Phi }}}_{-,2}\left(\lambda ,x,t\right)\right],\,\,\lambda \in {{\mathbb{C}}}_{-}\setminus {\eta }_{-}.\end{array}\right.\end{eqnarray}$
Set ${M}^{{BC}}(x,t,\lambda )={\phi }^{{BC}}(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}\theta {\sigma }_{3}}$, the jumping curve of MBC(x, t, λ) is ${\mathbb{R}}\cup \eta $. According to the similar calculation shown in [48, 49], we derive another similar solution of (42 ) for the smaller λ (here we assume that ϵ). In order that this solution has no singularities, we define1 ) with NZBCs is obtained as follows.
$\begin{eqnarray}\phi (x,t,\lambda )=\left\{\begin{array}{l}{\phi }^{{BC}}(x,t,\lambda ),\qquad \lambda \in {D}_{+}\cup {D}_{-},\\ {\phi }^{{\rm{in}}}(x,t,\lambda ),\qquad \,\,\lambda \in {D}_{0},\end{array}\right.\end{eqnarray}$
where φBC(x, t, λ) means the BC simultaneous solution. φin(x, t, λ) represents a complete function, which is redefined as φ(x, t, λ)φ(L, 0, λ)−1. D0 represents an open disk with a boundary of Σ+ ∪ Σ− and a radius of ϵ. It is worth noting that we choose the appropriate ϵ to further make S11(λ), S22(λ) is not equal to zero outside the disk. Concurrently, the branch cut η is included in this disk. In addition, ${D}_{\pm }=\{\lambda \in {\mathbb{C}}:| \lambda | \geqslant \varepsilon ,\mathrm{Im}\lambda \gtrless 0\}$ and Σ = ( − ∞ , − ϵ] ∪ [ϵ, + ∞ ) ∪ Σ+ ∪ Σ− are displayed in figure 4. Let $M(x,t,\lambda )=\phi (x,t,\lambda ){{\rm{e}}}^{{\rm{i}}\theta {\sigma }_{3}}$, the RH problem of the inhomogeneous fifth-order NLS equation ($M(x,t,\lambda )$ solves the RH problem
$\begin{eqnarray}\left\{\begin{array}{l}M(x,t,\lambda ){becomes}\,{analytical}\,{for}\,{\mathbb{C}}\setminus \{{\rm{\Sigma }}\cup \eta \},\\ {M}_{+}(x,t,\lambda )=\left\{\begin{array}{l}{M}_{-}(x,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\theta {\sigma }_{3}}J(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}\theta {\sigma }_{3}},\qquad \,\lambda \in {\rm{\Sigma }},\\ {M}_{-}(x,t,\lambda ){{\rm{e}}}^{2{\rm{i}}{\rho }_{+}(\lambda )[x+\delta (\lambda )t]{\sigma }_{3}},\qquad \qquad \lambda \in \eta ,\end{array}\right.\\ M(x,t,\lambda )\to {\mathbb{I}},\qquad \lambda \to \infty ,\end{array}\right.\end{eqnarray}$
where the jump matrix $J(x,t,\lambda )$ is $\begin{eqnarray}J(x,t,\lambda )=\left\{\begin{array}{l}\left[{{\rm{\Phi }}}_{-,1}(L,0,\lambda ),\displaystyle \frac{{{\rm{\Phi }}}_{+,2}\left(L,0,\lambda \right)}{{S}_{22}(\lambda )}\right],\,\,\lambda \in {{\rm{\Sigma }}}_{+},\\ \left[\displaystyle \frac{{{\rm{\Phi }}}_{+,1}\left(L,0,\lambda \right)}{{S}_{11}(\lambda )},{{\rm{\Phi }}}_{-,2}(L,0,\lambda )\right],\,\,\lambda \in {{\rm{\Sigma }}}_{-},\\ \left[\begin{array}{cc}1 & R(\lambda )\\ {R}^{* }({\lambda }^{* }) & 1+| R(\lambda ){| }^{2}\end{array}\right],\qquad \,\,\lambda \in (-\infty ,-\varepsilon ]\cup [\varepsilon ,+\infty ),\end{array}\right.\end{eqnarray}$
and L expresses a fixed real number, $R(\lambda )=\tfrac{{S}_{12}(\lambda )}{{S}_{22}(\lambda )}$ and the corresponding contour is displayed in figure 4. Then, we can deduce that the solution of the inhomogeneous fifth-order NLS equation ( $\begin{eqnarray}q(x,t)=\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }2{\rm{i}}\lambda {M}_{12}(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)}.\end{eqnarray}$
Figure 4. Definitions of the regions D±, D0 and Σ±, η. |
3.2. RW of the inhomogeneous fifth-order NLS equation
We will use the modified DT for theorem 1 to obtain the higher-order RW of the inhomogeneous fifth-order NLS equation (1 ). We make the following specification transformation42 ) and $\phi (L,0,\lambda )={\mathbb{I}}$ (λ ∈ D0). The T is expressed as
$\begin{eqnarray}\widetilde{\phi }(x,t,\lambda )=\left\{\begin{array}{l}{\boldsymbol{T}}(x,t,\lambda )\phi (x,t,\lambda ),\,\,\,\,\lambda \in {D}_{+}\cup {D}_{-},\\ {\boldsymbol{T}}(x,t,\lambda )\phi (x,t,\lambda ){\boldsymbol{T}}{\left(L,0,\lambda \right)}^{-1},\qquad \lambda \in {D}_{0},\end{array}\right.\end{eqnarray}$
where φ(x, t, λ) satisfies the above Lax pair ( $\begin{eqnarray}{\boldsymbol{T}}(x,t,\lambda )={\mathbb{I}}+\displaystyle \frac{{\boldsymbol{H}}(x,t)}{\lambda -\varsigma }+\displaystyle \frac{{\boldsymbol{Y}}(x,t)}{\lambda -{\varsigma }^{* }},\end{eqnarray}$
for ∀λ ∈ D0 where H(x, t) and Y(x, t) can be written as $\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{H}}(x,t) & = & \displaystyle \frac{4{\beta }^{2}\left(1-{\vartheta }^{* }(x,t)\right){\boldsymbol{s}}(x,t){{\boldsymbol{s}}}^{\top }(x,t){\sigma }_{2}+2{\rm{i}}\beta { \mathcal N }(x,t){\sigma }_{2}{{\boldsymbol{s}}}^{* }(x,t){{\boldsymbol{s}}}^{\top }(x,t){\sigma }_{2}}{4{\beta }^{2}| 1-\vartheta (x,t){| }^{2}+{{ \mathcal N }}^{2}(x,t)},\\ {\boldsymbol{Y}}(x,t) & = & \displaystyle \frac{4{\beta }^{2}\left(\vartheta (x,t)-1\right){\sigma }_{2}{{\boldsymbol{s}}}^{* }(x,t){{\boldsymbol{s}}}^{\dagger }(x,t)-2{\rm{i}}\beta { \mathcal N }(x,t){\boldsymbol{s}}(x,t){{\boldsymbol{s}}}^{\dagger }(x,t)}{4{\beta }^{2}| 1-\vartheta (x,t){| }^{2}+{{ \mathcal N }}^{2}(x,t)},\end{array}\end{eqnarray}$
where $\beta =\mathrm{Im}(\varsigma )$, s(x, t) = φ(x, t)c, ${ \mathcal N }(x,t)={{\boldsymbol{s}}}^{\dagger }(x,t){\boldsymbol{s}}(x,t)$, $\vartheta (x,t)={{\boldsymbol{s}}}^{\top }(x,t){\sigma }_{2}{{\boldsymbol{s}}}^{{\prime} }(x,t)$, ${\boldsymbol{c}}={({c}_{1},{c}_{2})}^{\top }$, and c1 and c2 represent arbitrary constants. Then the homologous jump condition of $\widetilde{M}(x,t,\lambda )=\widetilde{\phi }(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}\rho {\sigma }_{3}}$ changes at λ ∈ Σ+ ∪ Σ−, and J(λ) is rewritten as $\begin{eqnarray}\widetilde{J}(x,t,\lambda )=\left\{\begin{array}{l}{\boldsymbol{T}}(L,0,\lambda )J(x,t,\lambda ),\qquad \,\,\,\lambda \in {{\rm{\Sigma }}}_{+},\\ J(x,t,\lambda ){\boldsymbol{T}}{\left(L,0,\lambda \right)}^{-1},\qquad \lambda \in {{\rm{\Sigma }}}_{-}.\end{array}\right.\end{eqnarray}$
Then, we get $\widetilde{q}(x,t)$ from the new RH problem $\widetilde{M}(x,t,\lambda )$, namely $\begin{eqnarray}\begin{array}{rcl}\widetilde{q}(x,t) & = & \mathop{\mathrm{lim}}\limits_{\lambda \to \infty }2{\rm{i}}\lambda {\widetilde{M}}_{12}(x,t,\lambda ){{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)}\\ & = & q(x,t)+2{\rm{i}}({{\boldsymbol{H}}}_{12}-{{\boldsymbol{H}}}_{21}^{* }){{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)}.\end{array}\end{eqnarray}$
Furthermore, when we change c into the form ϵ−1c∞, where ${{\boldsymbol{c}}}_{\infty }\in {{\mathbb{C}}}^{2}\setminus \{0\}$ represents a fixed vector, and set ϵ → 0, the matrix T(x, t, λ) is also represented as a limit process, that is $\begin{eqnarray}{{\boldsymbol{T}}}_{\infty }(x,t,\lambda )={\mathbb{I}}+\displaystyle \frac{{{\boldsymbol{H}}}_{\infty }(x,t)}{\lambda -\varsigma }+\displaystyle \frac{{{\boldsymbol{Y}}}_{\infty }(x,t)}{\lambda -{\varsigma }^{* }},\end{eqnarray}$
where ${{\boldsymbol{H}}}_{\infty }(x,t)={\mathrm{lim}}_{\varepsilon \to 0}{\boldsymbol{H}}(x,t)$ and ${{\boldsymbol{Y}}}_{\infty }(x,t)={\mathrm{lim}}_{\varepsilon \to 0}{\boldsymbol{Y}}(x,t)$.Given the vector s(x, t), we can solve the inhomogeneous fifth-order NLS equation (1 ) by the DT method. We regard φbg(x, t, λ) = ψ±(x, t, λ), representing the basic solutions, then we have the following results
$\begin{eqnarray}\begin{array}{l}{\phi }_{{bg}}^{{\rm{in}}}(x,t,\lambda )=\left(x+\delta \left(\lambda \right)t\right)\displaystyle \frac{\sin \left(\theta \left(\lambda ,x,t\right)\right)}{\theta (\lambda ,x,t)}{X}_{\pm }\\ \quad +\,\cos (\theta (\lambda ,x,t)){\mathbb{I}}.\end{array}\end{eqnarray}$
Furthermore, we get $\begin{eqnarray}\begin{array}{l}{\boldsymbol{s}}(x,t,\varsigma )={\phi }_{{bg}}^{{\rm{in}}}(x,t,\varsigma ){\boldsymbol{c}}={\rm{i}}\iota (x,t,\varsigma )\\ \quad \times \left[-\left(\varsigma +\displaystyle \frac{\omega }{2}\right){\sigma }_{3}{\boldsymbol{c}}+B{\sigma }_{2}{\boldsymbol{c}}\right]+\chi (x,t,\varsigma ){\boldsymbol{c}},\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}\iota (x,t,\varsigma ) & = & \left(x+\delta \left(\varsigma \right)t\right)\displaystyle \frac{\sin \left(\theta \left(\varsigma ,x,t\right)\right)}{\theta \left(\varsigma ,x,t\right)},\\ \chi (x,t,\varsigma ) & = & \cos (\theta (\varsigma ,x,t)),\end{array}\end{eqnarray}$
and $\begin{eqnarray}\begin{array}{rcl}{ \mathcal N }(x,t) & = & {{\boldsymbol{s}}}^{\dagger }(x,t){\boldsymbol{s}}(x,t)=\left[| \iota {| }^{2}\left(\varsigma +\displaystyle \frac{\omega }{2}\right)\left({\varsigma }^{* }+\displaystyle \frac{\omega }{2}\right)\right.\\ & & \left.+| \chi {| }^{2}+| \iota {| }^{2}{B}^{2}\right]{{\boldsymbol{c}}}^{\dagger }{\boldsymbol{c}}\\ & & +\left[{\rm{i}}{\iota }^{* }\chi \left({\varsigma }^{* }+\displaystyle \frac{\omega }{2}\right)-{\rm{i}}\iota {\chi }^{* }\left(\varsigma +\displaystyle \frac{\omega }{2}\right)\right]{{\boldsymbol{c}}}^{\dagger }{\sigma }_{3}{\boldsymbol{c}}\\ & & +\left({\rm{i}}\iota B{\chi }^{* }-{\rm{i}}{\iota }^{* }B\chi \right){{\boldsymbol{c}}}^{\dagger }{\sigma }_{2}{\boldsymbol{c}}\\ & & +{\rm{i}}| \iota {| }^{2}({\varsigma }^{* }-\varsigma )B{{\boldsymbol{c}}}^{\dagger }{\sigma }_{1}{\boldsymbol{c}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\vartheta (x,t) & = & {{\boldsymbol{s}}}^{\top }(x,t){\sigma }_{2}{{\boldsymbol{s}}}^{{\prime} }(x,t)=\left[{\iota }^{{\prime} }\chi \left(\varsigma +\displaystyle \frac{\omega }{2}\right)\right.\\ & & \left.-\iota {\chi }^{{\prime} }\left(\varsigma +\displaystyle \frac{\omega }{2}\right)+\iota \chi \right]{{\boldsymbol{c}}}^{\top }{\sigma }_{1}{\boldsymbol{c}}+\left[\iota {\iota }^{{\prime} }{\left(\varsigma +\displaystyle \frac{\omega }{2}\right)}^{2}\right.\\ & & \left.+{\iota }^{2}\left(\varsigma +\displaystyle \frac{\omega }{2}\right)+\iota {\iota }^{{\prime} }{B}^{2}+\chi {\chi }^{{\prime} }\right]{{\boldsymbol{c}}}^{\top }{\sigma }_{2}{\boldsymbol{c}}\\ & & +\left({\rm{i}}{\iota }^{{\prime} }\chi B-{\rm{i}}\iota {\chi }^{{\prime} }B\right){{\boldsymbol{c}}}^{\top }{\boldsymbol{c}}-{\iota }^{2}B{{\boldsymbol{c}}}^{\top }{\sigma }_{3}{\boldsymbol{c}}.\end{array}\end{eqnarray}$
Therefore, the solutions of equation (1 ) are66 ), (68 ) and (69 ). Furthermore, letting c = c∞ϵ−1 with ϵ → 0, the solution (70 ) can be rewritten as follows66 ), (68 ) and (69 ) with c replaced by c∞, respectively.
$\begin{eqnarray}\begin{array}{rcl}\widetilde{q}(x,t) & = & \left[B+2{\rm{i}}\left({{\boldsymbol{H}}}_{12}-{{\boldsymbol{H}}}_{21}^{* }\right)\right]{{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)}\\ & = & \left(B+\displaystyle \frac{8{\beta }^{2}\left[\left(1-{\vartheta }^{* }\right){s}_{1}^{2}-\left(1-\vartheta \right){s}_{2}^{{* }^{2}}\right]+8\beta { \mathcal N }{s}_{1}{s}_{2}^{* }}{4{\beta }^{2}| 1-\vartheta {| }^{2}+{{ \mathcal N }}^{2}}\right)\\ & & \times {{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)},\end{array}\end{eqnarray}$
where ${\boldsymbol{s}}={({s}_{1},{s}_{2})}^{\top }$, ${ \mathcal N }$ and ϑ are provided in expressions ( $\begin{eqnarray}\begin{array}{l}{\widetilde{q}}_{\infty }(x,t)=\left[B-\displaystyle \frac{8{\beta }^{2}\left({\vartheta }_{\infty }^{* }{s}_{\infty 1}^{2}-{\vartheta }_{\infty }{s}_{\infty 2}^{{* }^{2}}\right)-8\beta {{ \mathcal N }}_{\infty }{s}_{\infty 1}{s}_{\infty 2}^{* }}{4{\beta }^{2}| {\vartheta }_{\infty }{| }^{2}+{{ \mathcal N }}_{\infty }^{2}}\right]\\ \,\times \,{{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)},\end{array}\end{eqnarray}$
where s∞, ${{ \mathcal N }}_{\infty }$ and ϑ∞ are provided in expressions (According to the spectral analysis theorem, the properties of the corresponding solutions will change with the change of ς. When $\varsigma =-\tfrac{\omega }{2}+{\rm{i}}{\ell }B$ with ∣ℓ∣ > 1, it will be expressed as the temporal-spatial periodic breather wave, which can be verified by figure 5. When ∣ℓ∣ < 1, it becomes a spatial periodic breather wave and can be verified from figure 6.
Figure 5. The temporal-spatial periodic breather wave solutions ( |
Figure 6. The spatial periodic breather wave solutions ( |
We obtain the RW of equation (1 ) by making $\varsigma =-\tfrac{\omega }{2}\pm {\rm{i}}B$. For convenience, we only calculate the case of $\varsigma =-\tfrac{\omega }{2}+{\rm{i}}B$ ($\varsigma =-\tfrac{\omega }{2}-{\rm{i}}B$ can perform similar calculations). Furthermore, we obtain
$\begin{eqnarray}{\boldsymbol{s}}(x,t)=\left(\begin{array}{c}{c}_{1}+B({c}_{1}+{c}_{2})(x+\delta (\varsigma )t)\\ {c}_{2}-B({c}_{1}+{c}_{2})(x+\delta (\varsigma )t)\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}{{\boldsymbol{s}}}^{{\prime} }(x,t)=\left(\begin{array}{c}-\displaystyle \frac{1}{3}{\rm{i}}B\left[B{\left(x+\delta t\right)}^{3}+3{\rm{i}}{\left(x+\delta t\right)}^{{\prime} }\right]({c}_{1}+{c}_{2})-{\rm{i}}{c}_{1}(x+\delta t)\left[B(x+\delta t)+1\right]\\ \displaystyle \frac{1}{3}{\rm{i}}B\left[B{\left(x+\delta t\right)}^{3}+3{\rm{i}}{\left(x+\delta t\right)}^{{\prime} }\right]({c}_{1}+{c}_{2})-{\rm{i}}{c}_{2}(x+\delta t)\left[B(x+\delta t)-1\right]\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{ \mathcal N } & = & | {c}_{1}{| }^{2}+| {c}_{2}{| }^{2}+2\mathrm{Re}\{B{\left({c}_{1}-{c}_{2}\right)}^{* }({c}_{1}+{c}_{2})(x+\delta t)\}\\ & & +2{B}^{2}| {c}_{1}+{c}_{2}{| }^{2}| x+\delta t{| }^{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\vartheta & = & -\displaystyle \frac{1}{3}B\left[2B{\left(x+\delta t\right)}^{3}-3{\rm{i}}{\left(x+\delta t\right)}^{{\prime} }\right]{\left({c}_{1}+{c}_{2}\right)}^{2}\\ & & +2{c}_{1}{c}_{2}(x+\delta t)-B{\left(x+\delta t\right)}^{2}({c}_{1}^{2}-{c}_{2}^{2}).\end{array}\end{eqnarray}$
Then we can find that the first-order RW can be deduced at c1 + c2 = 0.| a | (a) For c1 = − c2 = 1, we obtain the first-order RW solution as (see figure 7) $\begin{eqnarray}\widetilde{q}(x,t)=\left(B+\displaystyle \frac{4{B}^{2}[{\left(x+\delta t\right)}^{* }-(x+\delta t)]-4B}{{B}^{2}\left[1+2(x+\delta t)+2{\left(x+\delta t\right)}^{* }+4| x+\delta t{| }^{2}\right]+1}\right){{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)}.\end{eqnarray}$ |
| b | (b) For ${{\boldsymbol{c}}}_{\infty }={\left(c,-c\right)}^{\top }$, we obtain the first-order RW solution with its (see figure 8) $\begin{eqnarray}\widetilde{q}(x,t)=\left(\displaystyle \frac{B\left[4{B}^{2}| x+\delta t{| }^{2}+4B{\left(x+\delta t\right)}^{* }-4B(x+\delta t)-3\right]}{4{B}^{2}| x+\delta t{| }^{2}+1}\right){{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)}.\,\end{eqnarray}$ |
| c | (c) For c1 + c2 ≠ 0, e.g. ${c}_{1}=c+\tfrac{\varpi }{2}$, ${c}_{2}=-c+\tfrac{\varpi }{2}$ with $c\in {\mathbb{C}}\setminus \{0\}$ and ϖ ≪ 1. Let $x=\tfrac{\overline{x}}{| \varpi | },t=\tfrac{\overline{t}}{| \varpi | }$, if $(\overline{x},\overline{t})\in {{\mathbb{R}}}^{2}$ is fixed, and then we obtain $\begin{eqnarray}{\boldsymbol{s}}(x,t)=\left(\begin{array}{c}c+{{\rm{e}}}^{\mathrm{iarg}(\varpi )}B(\overline{x}+\delta \overline{t})\\ -c-{{\rm{e}}}^{\mathrm{iarg}(\varpi )}B(\overline{x}+\delta \overline{t})\end{array}\right)+O(\varpi ),\end{eqnarray}$ $\begin{eqnarray}\begin{array}{l}{ \mathcal N }=2| c{| }^{2}+4\mathrm{Re}\{{{Bc}}^{* }{{\rm{e}}}^{\mathrm{iarg}(\varpi )}(\overline{x}+\delta \overline{t})\}\\ \quad +2{B}^{2}({\overline{x}}^{2}+| \delta {| }^{2}{\overline{t}}^{2})+O(\varpi ),\end{array}\end{eqnarray}$ $\begin{eqnarray}\begin{array}{rcl}\vartheta & = & \displaystyle \frac{1}{| \varpi | }\left[-\displaystyle \frac{2}{3}{B}^{2}{{\rm{e}}}^{2\mathrm{iarg}(\varpi )}{\left(\overline{x}+\delta \overline{t}\right)}^{3}\right.\\ & & \left.-2{cB}{{\rm{e}}}^{\mathrm{iarg}(\varpi )}{\left(\overline{x}+\delta \overline{t}\right)}^{2}-2{c}^{2}(\overline{x}+\delta \overline{t})\right]+O(1).\end{array}\end{eqnarray}$ Further, we can get $\widetilde{q}(x,t)\approx 1$, unless the leading term in ϑ proportional to $\tfrac{1}{| \varpi | }$ is cancelled, these terms will form a cubic equation of ${\mathfrak{n}}=\overline{x}+\delta \overline{t}$, and its three roots are ${\mathfrak{n}}=0$ and $\tfrac{1}{2}c{\left(B\varpi \right)}^{-1}(-3\pm {\rm{i}}\sqrt{3})$, respectively (see figure 9). |
| d | (d) For ${{\boldsymbol{c}}}_{\infty }={\left(\mathrm{1,1}\right)}^{\top }$, we can obtain the second-order RW (see figure 10) $\begin{eqnarray}\widetilde{q}(x,t)=\left(B-12B\displaystyle \frac{{\mathfrak{P}}}{{\mathfrak{Q}}}\right){{\rm{e}}}^{{\rm{i}}(\omega x+\nu t)},\end{eqnarray}$ |
Figure 7. The first-order RW solutions ( |
Figure 8. The first-order RW solutions ( |
Figure 9. The first-order RW solutions ( |
Figure 10. The second-order RW solutions ( |
where
$\begin{eqnarray}\begin{array}{rcl}{\mathfrak{P}} & = & 16{B}^{5}{\left(x+\delta t\right)}^{3}{\left(x+\delta t\right)}^{* 2}-16{B}^{5}{\left(x+\delta t\right)}^{2}{\left(x+\delta t\right)}^{* 3}\\ & & +24{\rm{i}}{B}^{3}{\left(x+\delta t\right)}^{{\prime} }{\left(x+\delta t\right)}^{* }\\ & & -6{\rm{i}}{B}^{2}{\left(x+\delta t\right)}^{* ^{\prime} }-16{B}^{4}{\left(x+\delta t\right)}^{3}{\left(x+\delta t\right)}^{* }\\ & & +48{B}^{4}{\left(x+\delta t\right)}^{2}{\left(x+\delta t\right)}^{* 2}-6{\rm{i}}{B}^{2}{\left(x+\delta t\right)}^{{\prime} }\\ & & -16{B}^{4}(x+\delta t){\left(x+\delta t\right)}^{* 3}-24{\rm{i}}{B}^{3}(x+\delta t){\left(x+\delta t\right)}^{* ^{\prime} }\\ & & -12{B}^{3}{\left(x+\delta t\right)}^{2}{\left(x+\delta t\right)}^{* }\\ & & +12{B}^{3}(x+\delta t){\left(x+\delta t\right)}^{* 2}+4{B}^{3}{\left(x+\delta t\right)}^{3}\\ & & -4{B}^{3}{\left(x+\delta t\right)}^{* 3}-24{\rm{i}}{B}^{4}{\left(x+\delta t\right)}^{2}{\left(x+\delta t\right)}^{* ^{\prime} }\\ & & -24{\rm{i}}{B}^{4}{\left(x+\delta t\right)}^{* 2}{\left(x+\delta t\right)}^{{\prime} }+24{B}^{2}(x+\delta t){\left(x+\delta t\right)}^{* }\\ & & -9B(x+\delta t)\\ & & +9B{\left(x+\delta t\right)}^{* }-3,\\ {\mathfrak{Q}} & = & 64{B}^{6}{\left(x+\delta t\right)}^{3}{\left(x+\delta t\right)}^{* 3}+96{\rm{i}}{B}^{5}{\left(x+\delta t\right)}^{3}{\left(x+\delta t\right)}^{* ^{\prime} }\\ & & -96{\rm{i}}{B}^{5}{\left(x+\delta t\right)}^{* 3}{\left(x+\delta t\right)}^{{\prime} }\\ & & -48{B}^{4}{\left(x+\delta t\right)}^{3}{\left(x+\delta t\right)}^{* }+144{B}^{4}{\left(x+\delta t\right)}^{2}{\left(x+\delta t\right)}^{* 2}\\ & & -48{B}^{4}(x+\delta t){\left(x+\delta t\right)}^{* 3}\\ & & +144{B}^{4}{\left(x+\delta t\right)}^{{\prime} }{\left(x+\delta t\right)}^{* ^{\prime} }-72{\rm{i}}{B}^{3}(x+\delta t){\left(x+\delta t\right)}^{* ^{\prime} }\\ & & +72{\rm{i}}{B}^{3}{\left(x+\delta t\right)}^{* }{\left(x+\delta t\right)}^{{\prime} }\\ & & +108{B}^{2}(x+\delta t){\left(x+\delta t\right)}^{* }+9.\end{array}\end{eqnarray}$
4. Conclusions
The present work studied the BS soliton and RW solutions of the inhomogeneous fifth-order NLS equation (1 ) with ZBCs and NZBCs by the RH method. In this context, the RH problem of equation (1 ) is constructed, and an N-th order BS solitons of equation (1 ) with ZBCs are obtained by the residue theorem and Laurent’s series. Also, some dynamic behaviors of the second-order BS soliton solutions are analyzed for equation (1 ) in the form of images. It is manifested that parameters can change the shape and size of the two soliton waves (figure 2). In the meantime, the RH problem of equation (1 ) with NZBCs is constructed by robust IST. Then the RW solutions of equation (1 ) are obtained via the modified DT. The graphs of the temporal-spatial periodic breather waves and the spatial periodic breather waves are drawn, which revealed that parameter ς had a certain influence on the breather wave solutions (figure 5 and figure 6). Finally, the first-order and the second-order RW are obtained by modulating parameters in equation (1 ).
Although the exact solutions of equation (1 ) with NZBCs are derived in [47], the RW solutions of equation are studied by DT in [46]. However, in this paper, we mainly study the BS solitons with ZBCs and the RW solutions with NZBCs of equation (1 ), the obtained solutions are of more extensive significance and richer content. In addition, the proposed method in this paper can be further extended to identify some other nonlinear systems, and the method can be optimized to improve the results in the future.