1. Introduction
In this paper, we consider the complex modified Korteweg–de Vries equation (cmKdV) [1] as follows:
$\begin{eqnarray}{u}_{t}+{u}_{{xxx}}+6| u{| }^{2}{u}_{x}=0.\end{eqnarray}$
Here, u = u(x, t) is a complex function of x and t. The cmKdV equation is a typical integrable partial differential equation. It has been studied extensively over the last decades and its mathematical properties are well-documented in the literature. For example, the soliton solutions, the breathers and super-regular breathers, the rogue wave solutions, the existence and stability of solitary wave solutions, periodic traveling waves, exact group invariant solutions and conservation laws have been discussed in [1–11]. On the other hand, it is often suitable for many physical situations, such as the propagation of few-cycle optical pulses in cubic nonlinear media, electromagnetic wave propagation in liquid crystal waveguides and transverse wave propagation in molecular chain models [12–17].The $\bar{\partial }$-dressing method is a powerful tool for constructing and solving integrable nonlinear equations and describing their transformations and reductions. This method was first proposed by Zakharov and Shabat [18], and further developed by Beals, Coifman, Ablowitz, ManBakov, Fokas et al [19–24]. So far, a great deal of integrable equations have been widely discussed by the $\bar{\partial }$-dressing method [25–33]. However, to our knowledge, there is still no research work on the cmKdV equation (1.1 ) by using the $\bar{\partial }$-dressing method. In this paper, we use the $\bar{\partial }$-dressing method to study Lax pairs, and cmKdV hierarchy with source and N-soliton solutions of the cmKdV equation (1.1 ).
The layout of this paper is organized as follows. In section 2 , we obtain the Lax pair with singular dispersion relation by using the $\bar{\partial }$-dressing method. In section 3 , we introduce a cmKdV hierarchy with source, which contains a cmKdV hierarchy for a special case. In section 4 , the N-soliton solutions is constructed. As an application of the N-soliton formula, we discuss one-soliton and two-soliton in section 5 . Finally, the conclusions will be drawn based on the above sections.
2. Spectral problem and Lax pair
2.1. The spatial spectral problem
We consider the 2 × 2 matrix $\bar{\partial }$-problem in the complex k-plane2.1 ) can be written as2.1 ) will be given from (2.2 ) as
$\begin{eqnarray}\bar{\partial }\psi =\psi R,\end{eqnarray}$
with a boundary condition ψ(x, t, k) → I, k → ∞, then a solution of the equation ( $\begin{eqnarray}\psi (k)=I+\displaystyle \frac{1}{2\pi {\rm{i}}}\int \int \displaystyle \frac{\psi (z)R(z)}{z-k}{\rm{d}}z\wedge {\rm{d}}\bar{z}\equiv I+\psi {{RC}}_{k},\end{eqnarray}$
where Ck denotes the Cauchy–Green integral operator acting on the left. The formal solution of $\bar{\partial }$-problem ( $\begin{eqnarray}\psi (k)=I\cdot {\left(I-{{RC}}_{k}\right)}^{-1}.\end{eqnarray}$
For convenience, we define a pairing [26] $\begin{eqnarray*}\begin{array}{rcl}\langle f,g\rangle & = & \displaystyle \frac{1}{2\pi {\rm{i}}}\int \int f(k){g}^{{\rm{T}}}(k){\rm{d}}k\wedge {\rm{d}}\bar{k},\\ \langle f\rangle & = & \langle f,I\rangle =\displaystyle \frac{1}{2\pi {\rm{i}}}\int \int f(k){\rm{d}}k\wedge {\rm{d}}\bar{k},\end{array}\end{eqnarray*}$
which can be shown to possess the following properties $\begin{eqnarray}\begin{array}{rcl}\langle f,g{\rangle }^{{\rm{T}}} & = & \langle g,f\rangle ,\ \langle {fR},g\rangle =\langle f,{{gR}}^{{\rm{T}}}\rangle ,\\ \langle {{fC}}_{k},g\rangle & = & -\langle f,{{gC}}_{k}\rangle .\end{array}\end{eqnarray}$
It is easy to prove that for some matrix functions f(k) and g(k), the operator Ck satisfies $\begin{eqnarray}g(k)[f(k){C}_{k}]{C}_{k}+[g(k){C}_{k}]f(k){C}_{k}=[g(k){C}_{k}][f(k){C}_{k}].\end{eqnarray}$
It is well known that the Lax pairs of nonlinear equations play an important role in the study of integrable systems. Such as the Darboux transformation, inverse scattering transformation, Riemann–Hilbert method, and the algebro-geometric method rely on their Lax pairs. Here we prove that if the transform matrix R(x, t, k) satisfies a simple linear equation, the spatial-time spectral problems of the cmkdv equation can be established from (2.1 ). In particular, we obtain the spatial-time spectral problems of the cmKdV equation.
Let the transform matrix R satisfies
$\begin{eqnarray}{R}_{x}={\rm{i}}k[R,{\sigma }_{3}],\end{eqnarray}$
where ${\sigma }_{3}=\mathrm{diag}(1,-1)$, then the solution ψ of the $\bar{\partial }$-equation ( $\begin{eqnarray}{\psi }_{x}+{\rm{i}}k[{\sigma }_{3},\psi ]=Q\psi ,\end{eqnarray}$
where $\begin{eqnarray}Q=\left(\begin{array}{cc}0 & q\\ -\bar{q} & 0\end{array}\right)=-{\rm{i}}[{\sigma }_{3},\langle \psi R\rangle ].\end{eqnarray}$
Using (
$\begin{eqnarray}{\psi }_{x}={\rm{i}}k\psi R{\sigma }_{3}{C}_{k}{\left(I-{{RC}}_{k}\right)}^{-1}-{\rm{i}}k\psi {\sigma }_{3}{{RC}}_{k}{\left(I-{{RC}}_{k}\right)}^{-1},\end{eqnarray}$
According to the definition of Ck, we can obtain $\begin{eqnarray}{\rm{i}}k\psi {{RC}}_{k}={\rm{i}}\langle \psi R\rangle +{\rm{i}}k(\psi -I).\end{eqnarray}$
Since ${{RC}}_{k}=I-I\cdot (I-{{RC}}_{k})$, then we find $\begin{eqnarray}{{RC}}_{k}{\left(I-{{RC}}_{k}\right)}^{-1}={\left(I-{{RC}}_{k}\right)}^{-1}-I.\end{eqnarray}$
Substituting ( $\begin{eqnarray}{\psi }_{x}={\rm{i}}\langle \psi R\rangle {\sigma }_{3}\psi -{\rm{i}}{\sigma }_{3}k{\left(I-{{RC}}_{k}\right)}^{-1}+{\rm{i}}k\psi {\sigma }_{3}.\end{eqnarray}$
From ( $\begin{eqnarray}k{\left(I-{{RC}}_{k}\right)}^{-1}=\langle \psi R\rangle \psi +k\psi .\end{eqnarray}$
Substituting (2.2. The time spectral problem
Suppose that R satisfies the linear equation
$\begin{eqnarray}{R}_{t}=[R,{\rm{\Omega }}],\end{eqnarray}$
where $\begin{eqnarray}{\rm{\Omega }}={{\rm{\Omega }}}_{p}+{{\rm{\Omega }}}_{s}=4{\rm{i}}{k}^{3}{\sigma }_{3}+\displaystyle \frac{1}{2\pi {\rm{i}}}\int \int \displaystyle \frac{\omega (\xi ){\sigma }_{3}}{\xi -k}{\rm{d}}\xi \wedge {\rm{d}}\bar{\xi },\end{eqnarray}$
which comprises both a polynomial part ${{\rm{\Omega }}}_{p}(k)$ and a singular part ${{\rm{\Omega }}}_{s}(k)$ and $\omega (\xi )$ is a scalar function. Then the solution ψ of the $\bar{\partial }$-equation ( $\begin{eqnarray}\begin{array}{l}{\psi }_{t}+4{\rm{i}}{k}^{3}[{\sigma }_{3},\psi ]=(2{Q}^{3}+2{\rm{i}}k{\sigma }_{3}{Q}_{x}-2{\rm{i}}{{kQ}}^{2}{\sigma }_{3}\\ \quad +4{k}^{2}Q-{Q}_{{xx}})\psi \\ \quad -\displaystyle \frac{1}{2\pi {\rm{i}}}(\psi {\sigma }_{3}{\psi }^{-1}\omega (k){C}_{k})\psi +\psi {{\rm{\Omega }}}_{s}.\end{array}\end{eqnarray}$
As $\omega (k)=0$, the equation (
$\begin{eqnarray}\begin{array}{l}{\psi }_{t}+4{\rm{i}}{k}^{3}[{\sigma }_{3},\psi ]=(2{Q}^{3}+2{\rm{i}}k{\sigma }_{3}{Q}_{x}\\ \quad -2{\rm{i}}{{kQ}}^{2}{\sigma }_{3}+4{k}^{2}Q-{Q}_{{xx}})\psi ,\end{array}\end{eqnarray}$
which together with (We first use the polynomial dispersion relation only ${\rm{\Omega }}={{\rm{\Omega }}}_{p}=4{\rm{i}}{k}^{3}{\sigma }_{3}$. From equations (
$\begin{eqnarray}\begin{array}{l}{\psi }_{t}=4{\rm{i}}[{k}^{3}\psi {{RC}}_{k}{\sigma }_{3}{\left(I-{{RC}}_{k}\right)}^{-1}\\ \quad -{k}^{3}\psi {\sigma }_{3}{\left(I-{{RC}}_{k}\right)}^{-1}]+4{\rm{i}}{k}^{3}\psi {\sigma }_{3}.\end{array}\end{eqnarray}$
Through the following direct computation $\begin{eqnarray}\begin{array}{rcl}k\psi {{RC}}_{k} & = & \langle \psi R\rangle +k(\psi -I),\\ {k}^{2}\psi {{RC}}_{k} & = & \langle k\psi R\rangle +k\langle \psi R\rangle +{k}^{2}(\psi -I),\\ {k}^{3}\psi {{RC}}_{k} & = & \langle {k}^{2}\psi R\rangle +k\langle k\psi R\rangle +{k}^{2}\langle \psi R\rangle +{k}^{3}(\psi -I),\end{array}\end{eqnarray}$
then ( $\begin{eqnarray}\begin{array}{rcl}{\psi }_{t} & = & -4{\rm{i}}{k}^{3}[{\sigma }_{3},\psi ]+4{\rm{i}}[\langle {k}^{2}\psi R\rangle ,{\sigma }_{3}]\psi \\ & & +4{\rm{i}}[\langle k\psi R\rangle ,{\sigma }_{3}](k+\langle \psi R\rangle )\psi \\ & & +4{\rm{i}}[\langle \psi R\rangle ,{\sigma }_{3}]({k}^{2}+\langle k\psi R\rangle +k\langle \psi R\rangle +\langle \psi R{\rangle }^{2})\psi .\end{array}\end{eqnarray}$
By using ( $\begin{eqnarray}\begin{array}{rcl}\langle \psi R{\rangle }_{x} & = & {\rm{i}}[\langle k\psi R\rangle ,{\sigma }_{3}]+Q\langle \psi R\rangle ,\\ \langle k\psi R{\rangle }_{x} & = & {\rm{i}}[\langle {k}^{2}\psi R\rangle ,{\sigma }_{3}]+Q\langle k\psi R\rangle .\end{array}\end{eqnarray}$
Hence, equation ( $\begin{eqnarray}\begin{array}{l}{\psi }_{t}+4{\rm{i}}{k}^{3}[{\sigma }_{3},\psi ]=(4\langle k\psi R{\rangle }_{x}+4k\langle \psi R{\rangle }_{x}\\ \quad +4\langle \psi R{\rangle }_{x}\langle \psi R\rangle +4{k}^{2}Q)\psi .\end{array}\end{eqnarray}$
By means of ( $\begin{eqnarray}\begin{array}{rcl}\langle k\psi R{\rangle }_{x}^{\mathrm{off}} & = & -\displaystyle \frac{{\rm{i}}}{2}{\sigma }_{3}Q\langle \psi R\rangle +\displaystyle \frac{{\rm{i}}}{2}{\sigma }_{3}\langle \psi R{\rangle }_{x},\\ \langle k\psi R{\rangle }_{x}^{\mathrm{diag}} & = & \displaystyle \frac{{\rm{i}}}{2}{\sigma }_{3}{Q}^{2}\langle \psi R\rangle -\displaystyle \frac{{\rm{i}}}{2}{\sigma }_{3}Q\langle \psi R{\rangle }_{x}.\end{array}\end{eqnarray}$
Substituting ( $\begin{eqnarray}\begin{array}{l}{\psi }_{t}+4{\rm{i}}{k}^{3}[{\sigma }_{3},\psi ]=(2{Q}^{3}+2{\rm{i}}k{\sigma }_{3}{Q}_{x}-2{\rm{i}}{{kQ}}^{2}{\sigma }_{3}\\ \quad +4{k}^{2}Q-{Q}_{{xx}})\psi .\end{array}\end{eqnarray}$
In the following, we consider the singular dispersion relation in ( $\begin{eqnarray}{\psi }_{t}=(\psi R{{\rm{\Omega }}}_{s}{C}_{k}-\psi {{\rm{\Omega }}}_{s}){\left(I-{{RC}}_{k}\right)}^{-1}+\psi {{\rm{\Omega }}}_{s}.\end{eqnarray}$
And resorting ( $\begin{eqnarray}\psi R{{\rm{\Omega }}}_{s}{C}_{k}=\psi {{\rm{\Omega }}}_{s}-\displaystyle \frac{1}{2\pi {\rm{i}}}\int \int \displaystyle \frac{\omega (\xi )\psi (\xi ){\sigma }_{3}}{\xi -k}{\rm{d}}\xi \wedge {\rm{d}}\bar{\xi }.\end{eqnarray}$
Hence, we have $\begin{eqnarray}{\psi }_{t}=-\displaystyle \frac{1}{2\pi {\rm{i}}}\int \int \displaystyle \frac{\omega (\xi )\psi (\xi ){\sigma }_{3}}{\xi -k}{\rm{d}}\xi \wedge {\rm{d}}\bar{\xi }{\left(I-{{RC}}_{k}\right)}^{-1}+\psi {{\rm{\Omega }}}_{s}.\end{eqnarray}$
By using the relations $\begin{eqnarray*}\displaystyle \frac{1}{\rho -k}\displaystyle \frac{1}{\xi -\rho }=\displaystyle \frac{1}{\xi -k}\left(\displaystyle \frac{1}{\rho -k}-\displaystyle \frac{1}{\rho -\xi }\right),\end{eqnarray*}$
we find that $\begin{eqnarray*}\displaystyle \frac{1}{k-\xi }{\left(I-{{RC}}_{k}\right)}^{-1}=\displaystyle \frac{1}{k-\xi }{\psi }^{-1}(\xi )\psi (k),\end{eqnarray*}$
by which, then ( $\begin{eqnarray}{\psi }_{t}=-\displaystyle \frac{1}{2\pi {\rm{i}}}(\psi {\sigma }_{3}{\psi }^{-1}\omega (k){C}_{k})\psi +\psi {{\rm{\Omega }}}_{s},\end{eqnarray}$
which together with (2.3. Gauge equivalence
In this section, we prove that there is a gauge equivalence between the cmKdV equation and the Heisenberg chain equation.
The cmKdV equation (
$\begin{eqnarray}{S}_{t}=-\displaystyle \frac{{\rm{i}}}{2}[S,{S}_{{xx}}],\ \ \ \ {S}^{2}=I.\end{eqnarray}$
Making a reversible transformation $g(x,t)$,
$\begin{eqnarray*}\psi (x,t,k)=g(x,t)\varphi (x,t,k),\ \ \ g(x,t)=\psi (k=0).\end{eqnarray*}$
Then we can calculate that $\varphi (x,t,k)$ satisfies the following $\bar{\partial }$-problem, $\begin{eqnarray}\bar{\partial }\varphi ={g}^{-1}\psi R=\varphi R,\varphi (k)\to {g}^{-1},k\to \infty .\end{eqnarray}$
The equation ( $\begin{eqnarray}\varphi (k)={g}^{-1}+\displaystyle \frac{1}{2\pi {\rm{i}}}\int \int \displaystyle \frac{\varphi (\zeta )R(\zeta )}{\zeta -k}{\rm{d}}\zeta \wedge {\rm{d}}\bar{\zeta }={g}^{-1}+\varphi {{RC}}_{k}.\end{eqnarray}$
From ( $\begin{eqnarray}\varphi (k)={g}^{-1}{\left(I-{{RC}}_{k}\right)}^{-1}.\end{eqnarray}$
Deriving equation ( $\begin{eqnarray}\begin{array}{l}{\varphi }_{x}=-{g}^{-1}{g}_{x}\varphi +{\rm{i}}k\varphi R{\sigma }_{3}{C}_{k}{\left(I-{{RC}}_{k}\right)}^{-1}\\ \quad -{\rm{i}}k\varphi {\sigma }_{3}{{RC}}_{k}{\left(I-{{RC}}_{k}\right)}^{-1}.\end{array}\end{eqnarray}$
Similar to the previous calculation, we can simplify the above formula to $\begin{eqnarray}{\varphi }_{x}=-{g}^{-1}\{{g}_{x}+{\rm{i}}[{\sigma }_{3},\langle \psi R\rangle ]g\}\varphi -{\rm{i}}{{kg}}^{-1}{\sigma }_{3}g\varphi +{\rm{i}}k\varphi {\sigma }_{3}.\end{eqnarray}$
We can choose the function g to satisfy the following condition $\begin{eqnarray*}{g}_{x}=-{\rm{i}}[{\sigma }_{3},\langle \psi R\rangle ]g.\end{eqnarray*}$
Let $S={g}^{-1}{\sigma }_{3}g$, then the equation ( $\begin{eqnarray}{\varphi }_{x}=-{\rm{i}}{kS}\varphi +{\rm{i}}k\varphi {\sigma }_{3}.\end{eqnarray}$
The compatible condition leads to the Heisenberg chain equation $\begin{eqnarray}{S}_{t}=-\displaystyle \frac{{\rm{i}}}{2}[S,{S}_{{xx}}],\ \ \ \ {S}^{2}=I.\end{eqnarray}$
3. Recursive operators and cmKdV hierarchy
In this section, we derive the cmKdV equation with a source. First of all, we define the matrix M in the following form2.8 ) and (3.1 ), we can obtain the following proposition.
$\begin{eqnarray}M=\left(\begin{array}{cc}-r & p\\ \bar{p} & r\end{array}\right)=\psi {\sigma }_{3}{\psi }^{-1}.\end{eqnarray}$
By using (Q defined by (
$\begin{eqnarray}\begin{array}{ccc}{Q}_{t}+2{\alpha }_{n}{\sigma }_{3}{{\rm{\Lambda }}}^{n}Q & = & {\rm{i}}[{\sigma }_{3},\langle \omega (k)M(k)\rangle ],{\unicode{x000A0}}n=1,2,\ldots \\ {M}_{x}+{\rm{i}}k[{\sigma }_{3},M] & = & [Q,M].\end{array}\end{eqnarray}$
For the special case when $n=2,{\alpha }_{n}=2{\rm{i}}$, the hierarchy (
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{q}_{x}+{q}_{{xx}}+2| q{| }^{2}q=\ll p\gg ,\\ {p}_{x}+2{\rm{i}}{kp}=2{qr},\ {r}_{x}=-(r\bar{p}+\bar{r}p).\end{array}\end{eqnarray}$
where $\ll p\gg =\,\tfrac{1}{2\pi {\rm{i}}}\int \int \omega (\xi )p(\xi ){\rm{d}}\xi \wedge {\rm{d}}\bar{\xi }$, and $\omega (\xi )$ is a scalar function in the singular dispersion relation. By choosing $\omega (\xi )=0$, the equation (Differentiating the expression of Q with respect to t yields
$\begin{eqnarray}{Q}_{t}=-{\rm{i}}[{\sigma }_{3},\langle \psi R{\rangle }_{t}],\end{eqnarray}$
Because of $\bar{\partial }f(k){C}_{k}=f(k)$, then we have $\begin{eqnarray}\begin{array}{rcl}{(\psi R)}_{t} & = & \bar{\partial }{\psi }_{t}(k)=\bar{\partial }[I\cdot (I-{{RC}}_{k}{)}_{t}^{-1}]\\ & = & \bar{\partial }[\psi {R}_{t}{\left(I-{{RC}}_{k}\right)}^{-1}]{C}_{k}\\ & = & \psi {R}_{t}{\left(I-{{RC}}_{k}\right)}^{-1}.\end{array}\end{eqnarray}$
By using ( $\begin{eqnarray}\begin{array}{l}{Q}_{t}=-{\rm{i}}[{\sigma }_{3},\langle \psi {R}_{t}{\left(I-{{RC}}_{k}\right)}^{-1},I\rangle ]=\\ \,-\,{\rm{i}}[{\sigma }_{3},\,\langle \psi {R}_{t},I\cdot {\left(I+{R}^{T}{C}_{k}\right)}^{-1}\rangle ].\end{array}\end{eqnarray}$
From the $\bar{\partial }$-equation ( $\begin{eqnarray*}\bar{\partial }{\psi }^{-1}=-R{\psi }^{-1},\end{eqnarray*}$
which leads to $\begin{eqnarray*}{\left({\psi }^{-1}\right)}^{T}=I\cdot {\left(I+{R}^{T}{C}_{k}\right)}^{-1}.\end{eqnarray*}$
Therefore, using ( $\begin{eqnarray}{Q}_{t}=-{\rm{i}}[{\sigma }_{3},\langle (\bar{\partial }\psi ){\rm{\Omega }}{\psi }^{-1}+\psi {\rm{\Omega }}\bar{\partial }{\psi }^{-1}\rangle ].\end{eqnarray}$
Taking into the fact that ${{\rm{\Omega }}}_{p}={\alpha }_{n}{k}^{n}{\sigma }_{3},{\alpha }_{n}=\mathrm{const}$ and ${{\rm{\Omega }}}_{s}\to 0$ as $k\to \infty $, the above equation can be further reduced $\begin{eqnarray}\begin{array}{rcl}{Q}_{t} & = & -{\rm{i}}[{\sigma }_{3},\langle \psi {\rm{\Omega }}\bar{\partial }{\psi }^{-1}\rangle ]-{\rm{i}}[{\sigma }_{3},\langle (\bar{\partial }\psi ){\rm{\Omega }}{\psi }^{-1}\rangle ]\\ & = & -{\rm{i}}{\alpha }_{n}[{\sigma }_{3},\langle \bar{\partial }({k}^{n}M(k))\rangle ]+{\rm{i}}[{\sigma }_{3},\langle \omega (k)M(k)\rangle ]\\ & = & -2{\rm{i}}{\alpha }_{n}{\sigma }_{3}\langle \bar{\partial }({k}^{n}M{\left(k\right)}^{{\rm{off}}})\rangle +{\rm{i}}[{\sigma }_{3},\langle \omega (k)M(k)\rangle ].\end{array}\end{eqnarray}$
By virtue of the spectral problem ( $\begin{eqnarray}{M}_{x}+{\rm{i}}k[{\sigma }_{3},M]-[Q,M]=0.\end{eqnarray}$
From ( $\begin{eqnarray}\begin{array}{rcl}{M}_{x}^{\mathrm{diag}} & = & [Q,{M}^{\mathrm{off}}],\\ {M}_{x}^{\mathrm{off}} & = & 2{\rm{i}}k{\sigma }_{3}{M}^{\mathrm{off}}+[Q,{M}^{\mathrm{diag}}],\end{array}\end{eqnarray}$
which lead to $\begin{eqnarray}\begin{array}{rcl}{M}^{\mathrm{diag}} & = & {\sigma }_{3}+{\partial }_{x}^{-1}[Q,{M}^{\mathrm{off}}],\\ {M}^{\mathrm{off}} & = & -{\rm{i}}{\left({\rm{\Lambda }}-k\right)}^{-1}Q,\end{array}\end{eqnarray}$
where $\begin{eqnarray*}{\rm{\Lambda }}\cdot =\,\displaystyle \frac{1}{2}{\rm{i}}{\sigma }_{3}({\partial }_{x}\cdot -[Q,{\partial }_{x}^{-1}[Q,\cdot ]]).\end{eqnarray*}$
The operator Λ usually is called as recursion operator. We expand ${\left({\rm{\Lambda }}-k\right)}^{-1}$ in the series $\begin{eqnarray*}{\left({\rm{\Lambda }}-k\right)}^{-1}=-\displaystyle \sum _{j=1}^{\infty }{k}^{-j}{{\rm{\Lambda }}}^{j-1}.\end{eqnarray*}$
By using $\bar{{\rm{\partial }}}{k}^{n-j}=\pi \delta (k){\delta }_{j,n+1},j\,=\,1,2,\ldots $, we can derive that $\begin{eqnarray*}\displaystyle \sum _{j=1}^{\infty }\langle \bar{\partial }{k}^{n-j}\rangle {{\rm{\Lambda }}}^{j-1}Q=-{{\rm{\Lambda }}}^{n}Q,\end{eqnarray*}$
Substituting it into (4. N-soliton solutions of cmKdV equation
In this section, we will derive the N-soliton solutions of the cmKdV equation (1.1 ).
Suppose that ${k}_{j}$ and ${\bar{k}}_{j}$ are 2N discrete spectrals in complex plane ${\mathbb{C}}$. we choose a spectral transform matrix R as
$\begin{eqnarray}R=\displaystyle \sum _{j=1}^{N}\pi {{\rm{e}}}^{-{\rm{i}}\theta (k){\sigma }_{3}}\left(\begin{array}{cc}0 & -{c}_{j}\delta (k-{k}_{j})\\ {\bar{c}}_{j}\delta (k-{\bar{k}}_{j}) & 0\end{array}\right){{\rm{e}}}^{{\rm{i}}\theta (k){\sigma }_{3}},\end{eqnarray}$
where cj is const and $\theta (k)={kx}+4{k}^{3}t$, then the cmKdV equation ( $\begin{eqnarray}{q}_{N}(x,t)=2{\rm{i}}\displaystyle \frac{\det \ {M}^{\mathrm{aug}}}{\det \ M},\end{eqnarray}$
where ${M}^{\mathrm{aug}}$ is $(N+1)\times (N+1)$ matrices defined by $\begin{eqnarray*}\begin{array}{ccc}{M}^{{\rm{a}}{\rm{u}}{\rm{g}}} & = & \left(\begin{array}{cc}0 & {\boldsymbol{Y}}\\ {\boldsymbol{B}} & M\end{array}\right),{\boldsymbol{Y}}=({Y}_{1},\cdots ,{Y}_{N}),{\unicode{x000A0}}{Y}_{j}=-{c}_{j}{{\rm{e}}}^{\left.-2{\rm{i}}\theta ({k}_{j}\right)},\\ {\boldsymbol{B}} & = & {\left(1,\cdots ,1\right)}^{{\rm{T}}}.\end{array}\end{eqnarray*}$
Substituting (
$\begin{eqnarray}q(x,t)=-2{\rm{i}}\displaystyle \sum _{j=1}^{N}{c}_{j}{{\rm{e}}}^{-2{\rm{i}}\theta ({k}_{j})}{\psi }_{11}({k}_{j}).\end{eqnarray}$
Substituting ( $\begin{eqnarray}{\psi }_{11}(k)=1+\displaystyle \sum _{j=1}^{N}\displaystyle \frac{\bar{{c}_{j}}}{k-{\bar{k}}_{j}}{{\rm{e}}}^{2{\rm{i}}\theta ({\bar{k}}_{j})}{\psi }_{12}({\bar{k}}_{j}),\end{eqnarray}$
$\begin{eqnarray}{\psi }_{12}(k)=-\displaystyle \sum _{m=1}^{N}\displaystyle \frac{{c}_{m}}{k-{k}_{m}}{{\rm{e}}}^{-2{\rm{i}}\theta ({k}_{m})}{\psi }_{11}({k}_{m}).\end{eqnarray}$
Replacing k in ( $\begin{eqnarray}{\psi }_{11}({k}_{n})=1-\displaystyle \sum _{j=1}^{N}\displaystyle \sum _{m=1}^{N}{C}_{j}({k}_{n})\overline{\left.{C}_{m}({k}_{j}\right)}{\psi }_{11}({k}_{m}),{\unicode{x000A0}}n=1,2,\cdots ,N.\end{eqnarray}$
The above equation can be rewritten as $\begin{eqnarray}M{\left({\psi }_{11}({k}_{1}),\cdots ,{\psi }_{11}({k}_{N})\right)}^{{\rm{T}}}={\left(1,\cdots ,1\right)}^{{\rm{T}}},\end{eqnarray}$
where M is N × N matrices defined by $\begin{eqnarray*}\begin{array}{rlc}M & = & I+({A}_{n,m}),{\unicode{x000A0}}{\unicode{x000A0}}{A}_{n,m}=\displaystyle \sum _{j=1}^{N}{C}_{j}({k}_{n})\overline{\left.{C}_{m}({k}_{j}\right)},\\ {C}_{j}(k) & = & \displaystyle \frac{{\bar{c}}_{j}}{k-{k}_{j}}{{\rm{e}}}^{\left.2{\rm{i}}\theta ({\bar{k}}_{j}\right)},{\unicode{x000A0}}m,j=1,2,\cdots ,N.\end{array}\end{eqnarray*}$
According to Cramer's rule, the equation has the following solution $\begin{eqnarray}\psi ({k}_{n})=\displaystyle \frac{det{\unicode{x000A0}}{M}_{n}^{{\rm{a}}{\rm{u}}{\rm{g}}}}{det{\unicode{x000A0}}M},{\unicode{x000A0}}{\unicode{x000A0}}n=1,2,\cdots ,N,\end{eqnarray}$
where $\begin{eqnarray*}{M}_{n}^{\mathrm{aug}}=({{\boldsymbol{M}}}_{1},\cdots ,{{\boldsymbol{M}}}_{n-1},{\boldsymbol{B}},{{\boldsymbol{M}}}_{n+1},\cdots ,{{\boldsymbol{M}}}_{N}).\end{eqnarray*}$
Finally, substituting (5. Application of the N-soliton formula
In the following, we will give the one-soliton and two-soliton solutions for the cmKdV equation (1.1 ).
▶For N = 1, taking k1 = ξ + iη, the formula (4.2 ) gives the one-soliton solution of the cmKdV equation (1.1 )
$\begin{eqnarray}q=2{\rm{i}}\eta {{\rm{e}}}^{-2{\rm{i}}\xi x+{\rm{i}}\phi }{\rm{sech}} 2\eta (x-\alpha ),\end{eqnarray}$
where α = (− 3ξη + (ξ3 + ω1)/η)t + ξ0, φ = − 2(3ξ2η − η3)t + 2ω2t + φ0. The graphic of the one-soliton solution is shown in figure 1.
Figure 1. One-soliton solution of equation ( |
▶For N = 2, the formula (4.2 ) gives the two-soliton solution of the cmKdV equation (1.1 ) that is given by
$\begin{eqnarray}{q}_{2}=2{\rm{i}}\displaystyle \frac{\det \ {M}^{\mathrm{aug}}}{\det \ M},\end{eqnarray}$
where $\begin{eqnarray*}\begin{array}{rcl}M & = & \left(\begin{array}{cc}1+{A}_{11} & {A}_{12}\\ {A}_{21} & 1+{A}_{22}\end{array}\right),\\ {M}^{\mathrm{aug}} & = & \left(\begin{array}{ccc}0 & -{c}_{1}{{\rm{e}}}^{-2{\rm{i}}({k}_{1}x+4{k}_{1}^{3}t)} & -{c}_{2}{{\rm{e}}}^{-2{\rm{i}}({k}_{2}x+4{k}_{2}^{3}t)}\\ 1 & 1+{A}_{11} & {A}_{12}\\ 1 & {A}_{21} & 1+{A}_{22}\end{array}\right),\end{array}\end{eqnarray*}$
and $\begin{eqnarray*}{A}_{i,j}=\displaystyle \frac{{{\rm{\Theta }}}_{1,j}}{-({k}_{i}-{\bar{k}}_{1})({k}_{j}-{\bar{k}}_{1})}+\displaystyle \frac{{{\rm{\Theta }}}_{2,j}}{-({k}_{i}-{\bar{k}}_{2})({k}_{j}-{\bar{k}}_{2})},\end{eqnarray*}$
$\begin{eqnarray*}{\bar{c}}_{i}{c}_{j}={{\rm{e}}}^{{v}_{i}+{v}_{j}},\ {{\rm{\Theta }}}_{i,j}={{\rm{e}}}^{2{\rm{i}}[({\bar{k}}_{i}-{k}_{j})x+4({\bar{k}}_{i}^{3}-{k}_{j}^{3})t]+{v}_{i}+{v}_{j}},\ i,j\,=\,1,2.\end{eqnarray*}$
with vi and vj being two arbitrary constants. The graphic of the two-soliton solution is shown in figure 2.
Figure 2. Two-solion solution of equation ( |
6. Conclusion
By means of the $\bar{\partial }$-dressing method, the spatial and time spectral problems associated with the cmKdV equation are obtained. Furthermore, the gauge equivalence between the cmKdV equation and Heisenberg chain equation is obtained. Then a cmKdV hierarchy with source by means of a recursive operator is proposed. Finally, the N-soliton solutions of the cmKdV equation are constructed based on the $\bar{\partial }$-equation by selecting a special spectral transformation matrix. In particular, the explicit one-soliton and two-soliton solutions are discussed.