A Five-Component Generalized mKdV Equation and Its Exact Solutions

Abstract

In this paper, a $3\times 3$ spectral problem is proposed and a five-component equation that consists of two different mKdV equations is derived. A Darboux transformation of the five-component equation is presented relating to the gauge transformations between the Lax pairs. As applications of the Darboux transformations, interesting exact solutions, including soliton-like solutions and a solution that consists of rational functions of $e^x$ and t, for the five-component equation are obtained.

1. Introduction

In the study of nonlinear partial differential equations, the theory on solitons is an indispensable part [1]. A lot of systematic methods, such as the inverse scattering method [2,3], are developed to find exact explicit solutions for soliton equations. In the past few decades, as the soliton theory grew vigorously, solitons has been observed in solid physics, fluid physics, plasma physics, laser physics, condensed matter physics, etc. The research on soliton is an important topic in many physics laboratories. Many published articles are about soliton equations and integrability [4], Hamiltonian structures [5,6,7], various forms of solutions and their properties [8,9,10] and so on. Among the many ways to study soliton equations [11,12,13,14,15,16,17], the Darboux transformation method [18,19,20,21,22,23,24,25] is one of the most effective and fruitful tools. Darboux transformations can be used to obtain the rogue-wave solution [26,27], solutions on periodic backgrounds [28,29,30], and so on.

Inspired by the coupled mKdV equation [31,32], the complex modified Korteweg-de Vries equation [33], the coupled nonlinear Schrödinger equation [34,35], and so on, we propose a new five-component nonlinear integrable equation in this paper,

$$\begin{array}{cc}\hfill u_t=& -u_{xxx}-6u^2{u}_x+{\left(wv\right)}_{xx}+{(u{v}^2-rv-\frac{1}{2}u{w}^2)}_x+qv,\hfill \\ \hfill v_t=& v_{xxx}+6v^2{v}_x-{\left(wu\right)}_{xx}-{(u^{2}v+ru-\frac{1}{2}v{w}^2)}_x-qu,\hfill \\ \hfill w_x=& -2uv,\hfill \\ \hfill r_x=& (u^2-v^2)w+(v{u}_x-u{v}_x),\hfill \\ \hfill q_x=& (u^2-v^2)r+\frac{3}{2}{w}_x(u^2+v^2)+\frac{1}{2}{(u^2+v^2)}_{x}w+\frac{1}{6}{\left(w^3\right)}_x-(v{u}_{xx}+u{v}_{xx}),\hfill \end{array}$$

(1)

where $x,t\in \mathbb{R}$. Different from the traditional two-component mKdV equation, the five-component Equation (1) contains two different mKdV equations. That is, when $v=w=r=q=0$ and when $u=w=r=q=0$, the five-component Equation (1) is reduced to, respectively,

$$-u_t=u_{xxx}+6u^2{u}_x,$$

(2)

and

$$v_t=v_{xxx}+6v^2{v}_x.$$

(3)

Of course these are related by the reflection $x^{\prime }=-x$ that replaces right-traveling waves by left-traveling waves. Resorting to the gauge transformations between the Lax pairs, we derive Darboux transformations and exact solutions for (1). As a result, we present several different type of solitons (including soliton solutions and a solution that is a rational function of $e^x$ and t) for (1).

The structure of the paper is as follows. In Section 2, a Lax pair associated with the five-component Equation (1) is proposed, and a Darboux transformation is constructed. In Section 3, as applications of the Darboux transformations, some example solutions for (1) are constructed and illustrated.

2. Darboux Transformation

The five-component nonlinear integrable Equation (1) is associated with the spectral problem

$${\Phi }_x=U\Phi ,{\Phi }_t=V\Phi ,\Phi =\left(\begin{array}{c}\varphi \\ \psi \\ \phi \end{array}\right),U=\left(\begin{array}{ccc}\lambda & u& v\\ -u& 0& 0\\ -v& 0& 0\end{array}\right),$$

(4)

where $\lambda \in \mathbb{C}$ is the spectral parameter independent of x and t, and $u=u(x,t),v=v(x,t)$ are potentials, the expression of V is

$$V=\left(\begin{array}{ccc}\lambda (-u^2+v^2)& -{\lambda }^{2}u+\lambda (vw-u_x)-P& {\lambda }^{2}v+\lambda (-uw+v_x)-Q\\ {\lambda }^{2}u+\lambda (vw-u_x)+P& {\lambda }^3+\lambda (u^2+\frac{1}{2}{w}^2)& {\lambda }^{2}w+\lambda r+q\\ -{\lambda }^{2}v+\lambda (-uw+v_x)+Q& -{\lambda }^{2}w+\lambda r-q& -{\lambda }^3-\lambda (v^2+\frac{1}{2}{w}^2)\end{array}\right)$$

(5)

with

$$\begin{array}{c}\hfill P=2u^3+rv-u{v}^2+\frac{1}{2}u{w}^2-w{v}_x-v{w}_x+u_{xx},\\ \hfill Q=ru+u^{2}v-2v^3-\frac{1}{2}v{w}^2+w{u}_x+u{w}_x-v_{xx}.\end{array}$$

(6)

The soliton Equation (1) is yielded by the zero-curvature equation $U_t-V_x+[U,V]=0$.

Letting $\widehat{\Phi }=T\Phi$ we have

$${\widehat{\Phi }}_x=\widehat{U}\widehat{\Phi },{\widehat{\Phi }}_t=\widehat{V}\widehat{\Phi },$$

(7)

where $\widehat{U}=(T_x+TU)T^{-1}$, $\widehat{V}=(T_t+TV)T^{-1}$. In order to ensure that $\widehat{U},\widehat{V}$ and $U,V$ are the same form except that the old potentials $u,v,w,q,r$ are replaced by the new potentials $\widehat{u},\widehat{v},\widehat{w},\widehat{q},\widehat{r}$ respectively, we suppose that T has the following form,

$$\begin{array}{c}\hfill T=\left(\begin{array}{ccc}\lambda +d& a& b\\ a& \lambda +g& c\\ b& c& \lambda +f\end{array}\right),\end{array}$$

(8)

where

$$d=\frac{ab}{2c}-\frac{ac}{2b}-\frac{bc}{2a},g=\frac{ac}{2b}-\frac{ab}{2c}-\frac{bc}{2a},f=\frac{bc}{2a}-\frac{ac}{2b}-\frac{ab}{2c}.$$

(9)

It is easy to see that

$$\begin{array}{c}\hfill detT={(\lambda +{\lambda }_1)}^2(\lambda -{\lambda }_1),\end{array}$$

(10)

where

$$\begin{array}{c}\hfill {\lambda }_1=-\frac{bc}{2a}-\frac{ab}{2c}-\frac{ac}{2b}.\end{array}$$

(11)

In view of

$$\begin{array}{cc}\hfill {(detT)}_x& =\mathrm{tr}\left(T_x{T}^{*}\right)=\mathrm{tr}[(\widehat{U}T-TU)T^{*}]=\mathrm{tr}(\widehat{U}T{T}^{*})-\mathrm{tr}\left(TU{T}^{*}\right)\hfill \\ \hfill & =(detT)(\mathrm{tr}\widehat{U})-(detT)\left(\mathrm{tr}U\right)=(detT)\mathrm{tr}(\widehat{U}-U)=0,\hfill \\ \hfill {(detT)}_t& =0,\hfill \end{array}$$

where $T^{*}=T^{-1}detT$, we find $detT$ is independent of $(x,t)$. As a result, ${\lambda }_1$ is a constant.

Suppose ${\Phi }_1={({\varphi }_1,{\psi }_1,{\phi }_1)}^T$ is a solution of the Lax pair (4) and (5) when $\lambda ={\lambda }_1$, and denote ${\beta }_1={\psi }_1/{\varphi }_1$ and ${\gamma }_1={\phi }_1/{\varphi }_1$. Assume the quantities $a,b,c,d,f,g$ satisfy

$$\begin{array}{c}\hfill \left(\begin{array}{ccc}{\lambda }_1+d& a& b\\ a& {\lambda }_1+g& c\\ b& c& {\lambda }_1+f\end{array}\right)\left(\begin{array}{c}1\\ {\beta }_1\\ {\gamma }_1\end{array}\right)=0,\end{array}$$

(12)

or equivalently

$$\begin{array}{c}\hfill -\frac{bc}{a}-\frac{ac}{b}+a{\beta }_1+b{\gamma }_1=0,\\ \hfill a+(2{\lambda }_1+\frac{ac}{b}){\beta }_1+c{\gamma }_1=0,\\ \hfill b+c{\beta }_1+(2{\lambda }_1+\frac{bc}{a}){\gamma }_1=0.\end{array}$$

(13)

Computing Equation (13), the expressions of a, b, c are obtained,

$$\begin{array}{c}\hfill a=\frac{-2{\lambda }_1{\beta }_1}{1+{\beta }_1^2+{\gamma }_1^2},b=\frac{-2{\lambda }_1{\gamma }_1}{1+{\beta }_1^2+{\gamma }_1^2},c=\frac{-2{\lambda }_1{\beta }_1{\gamma }_1}{1+{\beta }_1^2+{\gamma }_1^2}.\end{array}$$

(14)

Substituting (14) into (9), the expressions of d, e, f can be written as

$$\begin{array}{c}\hfill d={\lambda }_1+\frac{-2{\lambda }_1}{1+{\beta }_1^2+{\gamma }_1^2},g={\lambda }_1+\frac{-2{\lambda }_1{\beta }_1^2}{1+{\beta }_1^2+{\gamma }_1^2},f={\lambda }_1+\frac{-2{\lambda }_1{\gamma }_1^2}{1+{\beta }_1^2+{\gamma }_1^2}.\end{array}$$

(15)

Theorem 1.

Let the following conditions be satisfied:

1. 

$u,v,w,q,r$ is a known solution of the five-component Equation (1);

2. 

${\lambda }_1\in \mathbb{R}$ is a fixed constant; and

3. 

${\Phi }_1={({\varphi }_1,{\psi }_1,{\phi }_1)}^T$ is a nonzero solution of the Lax pair (4) and (5) when $\lambda ={\lambda }_1$.

Denote ${\beta }_1={\psi }_1/{\varphi }_1$ and ${\gamma }_1={\phi }_1/{\varphi }_1$. Then the new potentials $\widehat{u},\widehat{v},\widehat{w},\widehat{q},\widehat{r}$ given by the Darboux transformation

$$\begin{array}{cc}\hfill \widehat{u}=& u-a,\widehat{v}=v-b,\widehat{w}=w-2c,\hfill \\ \hfill \widehat{r}=& r-ub+va+(g-f)w+(f-g)c,\hfill \\ \hfill \widehat{q}=& q-auw-bvw+b{u}_x+a{v}_x-(u^2+v^2+w^2)c+(g-f)r+(bc-af)v\hfill \\ \hfill & +(ac-bg)u+(2c^2+a^2+g^2-gf)w-2c(a^2+b^2+d^2)\hfill \end{array}$$

(16)

is a new solution of the five-component Equation (1), and the corresponding Darboux matrix is T. In the above equations, the quantities $a,b,c,d,f,g$ are given by (14) and (15).

Proof. 

First of all, we prove $\widehat{U}T=T_x+TU$, where $\widehat{U}{=U|}_{u=\widehat{u},v=\widehat{v}}$. Denote

$$A=\left(\begin{array}{ccc}d& a& b\\ a& g& c\\ b& c& f\end{array}\right),B=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),C=\left(\begin{array}{ccc}0& u& v\\ -u& 0& 0\\ -v& 0& 0\end{array}\right),E=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),$$

so that $T=\lambda E+A$, $U=\lambda B+C$. Denote $\widehat{C}{=C|}_{u=\widehat{u},v=\widehat{v}}$, then $\widehat{U}=\lambda B+\widehat{C}$. Comparing the coefficients of ${\lambda }^j$ in $\widehat{U}T=T_x+TU$ we have

$$\begin{array}{cc}\hfill & \widehat{C}+BA=C+AB,\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill & \widehat{C}A=A_x+AC.\hfill \end{array}$$

(18)

Now we prove that expressions (17) and (18) are true.

A direct computation shows that

$$\widehat{C}+BA=\left(\begin{array}{ccc}d& \widehat{u}+a& \widehat{v}+b\\ -\widehat{u}& 0& 0\\ -\widehat{v}& 0& 0\end{array}\right),C+AB=\left(\begin{array}{ccc}d& u& v\\ -u+a& 0& 0\\ -v+b& 0& 0\end{array}\right).$$

From the first two expressions of the Darboux transformation (16), it is easy to verify that expression (17) holds.

Next, we prove this expression (18) is valid. From Equation (9), it is immediate that

$$\begin{array}{c}\hfill ga+cb+ad=0,ca+fb+bd=0,ab+cg+cf=0.\end{array}$$

(19)

From ${\Phi }_{1,x}=\left(U{|}_{\lambda ={\lambda }_1}\right){\Phi }_1$ we have the Riccati equations

$$\begin{array}{cc}\hfill {\beta }_{1,x}& =-u-{\beta }_1{\lambda }_1-u{\beta }_1^2-v{\beta }_1{\gamma }_1,\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill {\gamma }_{1,x}& =-v-{\gamma }_1{\lambda }_1-u{\beta }_1{\gamma }_1-v{\gamma }_1^2.\hfill \end{array}$$

(21)

According to Equations (14) and (15), we derive the first derivatives of a, b, c, d, g, f with respect to x.

$$\begin{array}{cc}\hfill & a_x=(g-d)u+vc+ad,b_x=cu+(f-d)v+bd,c_x=-bu-av+ab,\hfill \\ \hfill & d_x=2(au+bv)-a^2-b^2,g_x=-2au+a^2,f_x=-2bv+b^2.\hfill \end{array}$$

(22)

Then, we arrive at

$$\begin{array}{cc}\hfill & A_x+AC=\left(\begin{array}{ccc}{d}_x-au-bv& a_x+du& b_x+dv\\ a_x-gu-cv& g_x+au& c_x+av\\ b_x-cu-fv& c_x+bu& f_x+bv\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{ccc}2(au+bv)-a^2-b^2-au-bv& (g-d)u+vc+ad+du& cu+(f-d)v+bd+dv\\ (g-d)u+vc+ad-gu-cv& -2au+a^2+au& -bu-av+ab+av\\ cu+(f-d)v+bd-cu-fv& -bu-av+ab+bu& -2bv+b^2+bv\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{ccc}a(u-a)+b(v-b)& g(u-a)+c(v-b)& c(u-a)+f(v-b)\\ -d(u-a)& -a(u-a)& -b(u-a)\\ -d(v-b)& -a(v-b)& -b(v-b)\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{ccc}a\widehat{u}+b\widehat{v}& g\widehat{u}+c\widehat{v}& c\widehat{u}+f\widehat{v}\\ -d\widehat{u}& -a\widehat{u}& -b\widehat{u}\\ -d\widehat{v}& -a\widehat{v}& -b\widehat{v}\end{array}\right)=\widehat{C}A.\hfill \end{array}$$

Hence, Equation (18) is true, and then $\widehat{U}T=T_x+TU$ is valid.

In the second place, we prove $\widehat{V}T=T_t+TV$, where $\widehat{V}{=V|}_{u=\widehat{u},v=\widehat{v}}$. Denote

$$V_1=\left(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right),V_2=\left(\begin{array}{ccc}0& -u& v\\ u& 0& w\\ -v& -w& 0\end{array}\right),$$

$$V_3=\left(\begin{array}{ccc}-u^2+v^2& vw-u_x& -uw+v_x\\ vw-u_x& u^2+\frac{1}{2}{w}^2& r\\ -uw+v_x& r& -v^2-\frac{1}{2}{w}^2\end{array}\right),V_4=\left(\begin{array}{ccc}0& -P& -Q\\ P& 0& q\\ Q& -q& 0\end{array}\right),$$

so that $T=\lambda E+A$, and $V={\lambda }^3{V}_1+{\lambda }^2{V}_2+\lambda V_3+V_4$. Denote ${\widehat{V}}_2=V_2{|}_{u=\widehat{u},v=\widehat{v}}$, ${\widehat{V}}_3=V_3{|}_{u=\widehat{u},v=\widehat{v}}$, and ${\widehat{V}}_4=V_4{|}_{u=\widehat{u},v=\widehat{v}}$, then $\widehat{V}={\lambda }^3{V}_1+{\lambda }^2{\widehat{V}}_2+\lambda {\widehat{V}}_3+{\widehat{V}}_4$. Comparing the coefficients of ${\lambda }^j$ in $\widehat{V}T=T_t+TV$, we have

$$\begin{array}{cc}\hfill & {\widehat{V}}_2+V_{1}A=V_2+A{V}_1,\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill & {\widehat{V}}_3+{\widehat{V}}_{2}A=V_3+A{V}_2,\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill & {\widehat{V}}_4+{\widehat{V}}_{3}A=V_4+A{V}_3,\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill & {\widehat{V}}_{4}A=A_t+A{V}_4.\hfill \end{array}$$

(26)

So we prove that expressions (23)–(26) are true.

It is easy to see that

$$\begin{array}{cc}\hfill {\widehat{V}}_2+V_{1}A& =\left(\begin{array}{ccc}0& -\widehat{u}& \widehat{v}\\ \widehat{u}+a& g& \widehat{w}+c\\ -\widehat{v}-b& -\widehat{w}-c& -f\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{ccc}0& -(u-a)& v-b\\ (u-a)+a& g& (w-2c)+c\\ -(v-b)-b& -(w-2c)-c& -f\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{ccc}0& -u+a& v-b\\ u& g& w-c\\ -v& -w+c& -f\end{array}\right)=V_2+A{V}_1.\hfill \end{array}$$

So expression (23) holds.

A direct computation shows that

$$\begin{array}{c}\hfill {\widehat{V}}_3+{\widehat{V}}_{2}A=\left(\begin{array}{ccc}-{\widehat{u}}^2+{\widehat{v}}^2-a\widehat{u}+b\widehat{v}& \widehat{v}\widehat{w}-{\widehat{u}}_x-g\widehat{u}+c\widehat{v}& -\widehat{u}\widehat{w}+{\widehat{v}}_x-c\widehat{u}+f\widehat{v}\\ \widehat{v}\widehat{w}-{\widehat{u}}_x+d\widehat{u}+b\widehat{w}& {\widehat{u}}^2+\frac{1}{2}{\widehat{w}}^2+a\widehat{u}+c\widehat{w}& b\widehat{u}+f\widehat{w}+\widehat{r}\\ -\widehat{u}\widehat{w}+{\widehat{v}}_x-d\widehat{v}-a\widehat{w}& -a\widehat{v}-g\widehat{w}+\widehat{r}& -{\widehat{v}}^2-\frac{1}{2}{\widehat{w}}^2-b\widehat{v}-c\widehat{w}\end{array}\right).\end{array}$$

(27)

Substituting (16) and (22) into (27), we obtain

$$\begin{array}{cc}\hfill {\widehat{V}}_3+{\widehat{V}}_{2}A& =\left(\begin{array}{ccc}-u^2+v^2+au-bv& vw-u_x-du-bw& -uw+v_x+dv+aw\\ vw-u_x+gu-cv& u^2+\frac{1}{2}{w}^2-au-cw& av+gw+r\\ -uw+v_x+cu-fv& -bu-fw+r& -v^2-\frac{1}{2}{w}^2+bv+cw\end{array}\right)\hfill \\ \hfill & =V_3+A{V}_2.\hfill \end{array}$$

Therefore, the expression (24) is true.

From Equation (19), it is immediate that

$$a^2+d^2-f^2-c^2=0,a^2+g^2-b^2-f^2=0,c^2+g^2-b^2-d^2=0.$$

(28)

And the following expressions can be derived from Equation (6):

$$\begin{array}{cc}\hfill \widehat{P}=& 2u^3-2a{u}^2+a{v}^2+u{v}^2+\frac{1}{2}(u-a)w^2+(g-d)wv-cwu+rv-br\hfill \\ \hfill & +(ac-bg)w+(cd-ab)v+(2a^2+d^2+b^2-gd)u+(d-g)u_x+(c-w)v_x\hfill \\ \hfill & +u_{xx}-a(a^2+b^2+d^2),\hfill \\ \hfill \widehat{Q}=& -2v^3-u^{2}v-b{u}^2+2b{v}^2+\frac{1}{2}(b-v)w^2+(d-f)wu+cwv+(u-a)r\hfill \\ \hfill & +(af-bc)w+(fd-a^2-2b^2-d^2)v+(ab-cd)u+(w-c)u_x+(f-d)v_x\hfill \\ \hfill & -v_{xx}+b(a^2+b^2+d^2).\hfill \end{array}$$

(29)

Because

$$\begin{array}{c}{\widehat{V}}_4+{\widehat{V}}_{3}A=\left(\begin{array}{c}(-{\widehat{u}}^2+{\widehat{v}}^2)d+(\widehat{v}\widehat{w}-{\widehat{u}}_x)a+(-\widehat{u}\widehat{w}+{\widehat{v}}_x)b\\ (\widehat{v}\widehat{w}-{\widehat{u}}_x)d+({\widehat{u}}^2+\frac{1}{2}{\widehat{w}}^2)a+\widehat{r}b+\widehat{P}\\ (-\widehat{u}\widehat{w}+{\widehat{v}}_x)d+(-{\widehat{v}}^2-\frac{1}{2}{\widehat{w}}^2)b+\widehat{r}a+\widehat{Q}\end{array}\right.\hfill \\ \left.\begin{array}{cc}(-{\widehat{u}}^2+{\widehat{v}}^2)a+(\widehat{v}\widehat{w}-{\widehat{u}}_x)g+(-\widehat{u}\widehat{w}+{\widehat{v}}_x)c-\widehat{P}& (-{\widehat{u}}^2+{\widehat{v}}^2)b+(\widehat{v}\widehat{w}-{\widehat{u}}_x)c+(-\widehat{u}\widehat{w}+{\widehat{v}}_x)f-\widehat{Q}\\ (\widehat{v}\widehat{w}-{\widehat{u}}_x)a+({\widehat{u}}^2+\frac{1}{2}{\widehat{w}}^2)g+\widehat{r}c& (\widehat{v}\widehat{w}-{\widehat{u}}_x)b+({\widehat{u}}^2+\frac{1}{2}{\widehat{w}}^2)c+\widehat{r}f+\widehat{q}\\ (-\widehat{u}\widehat{w}+{\widehat{v}}_x)a+(-{\widehat{v}}^2-\frac{1}{2}{\widehat{w}}^2)c+\widehat{r}g-\widehat{q}& (-\widehat{u}\widehat{w}+{\widehat{v}}_x)b+(-{\widehat{v}}^2-\frac{1}{2}{\widehat{w}}^2)f+\widehat{r}c\end{array}\right),\hfill \end{array}$$

(30)

inserting (16), (22) and (29) into (30), we have

$$\begin{array}{cc}\hfill & {\widehat{V}}_4+{\widehat{V}}_{3}A=\left(\begin{array}{c}(-u^2+v^2)d+(vw-u_x)a+(-uw+v_x)b\\ (-u^2+v^2)a+(vw-u_x)g+(-uw+v_x)c+P\\ (-u^2+v^2)b+(vw-u_x)c+(-uw+v_x)f+Q\end{array}\right.\hfill \\ \hfill & \left.\begin{array}{cc}(vw-u_x)d+(u^2+\frac{1}{2}{w}^2)a+br-P& (-uw+v_x)d+(-v^2-\frac{1}{2}{w}^2)b+ar-Q\\ (vw-u_x)a+(u^2+\frac{1}{2}{w}^2)g+cr& (-uw+v_x)a+(-v^2-\frac{1}{2}{w}^2)c+q+rg\\ (vw-u_x)b+(u^2+\frac{1}{2}{w}^2)c+fr-q& (-uw+v_x)b+(-v^2-\frac{1}{2}{w}^2)f+cr\end{array}\right)\hfill \\ \hfill & =V_4+A{V}_3.\hfill \end{array}$$

So the expression (25) is valid.

In what follows, we show that expression (26) holds. From direct calculations we get

$$\begin{array}{c}\hfill A_t+A{V}_4=\left(\begin{array}{ccc}{d}_t+aP+bQ& a_t-dP-bq& b_t-dQ+aq\\ a_t+gP+cQ& g_t-aP-cq& c_t-aQ+gq\\ b_t+cP+fQ& c_t-bP-fq& f_t-bQ+cq\end{array}\right),\end{array}$$

(31)

$$\begin{array}{c}\hfill {\widehat{V}}_{4}A=\left(\begin{array}{ccc}-\widehat{P}a-\widehat{Q}b& -\widehat{P}g-\widehat{Q}c& -\widehat{P}c-\widehat{Q}f\\ \widehat{P}d+\widehat{q}b& \widehat{P}a+\widehat{q}c& \widehat{P}b+\widehat{q}f\\ \widehat{Q}d-\widehat{q}a& \widehat{Q}a-\widehat{q}g& \widehat{Q}b-\widehat{q}c\end{array}\right).\end{array}$$

(32)

From the equation ${\Phi }_{1,t}=\left(V{|}_{\lambda ={\lambda }_1}\right){\Phi }_1$, we arrive at

$$\begin{array}{cc}\hfill {\beta }_{1,t}=& {\lambda }_1^{2}u+{\lambda }_1(vw-u_x)+P+[{\lambda }_1^3+{\lambda }_1(2u^2-v^2+\frac{1}{2}{w}^2)]{\beta }_1\hfill \\ \hfill & +({\lambda }_1^{2}w+{\lambda }_{1}r+q){\gamma }_1-[{\lambda }_1^{2}v+{\lambda }_1(-uw+v_x)-Q]{\beta }_1{\gamma }_1\hfill \\ \hfill & +[{\lambda }_1^{2}u-{\lambda }_1(vw-u_x)+P]{\beta }_1^2,\hfill \\ \hfill {\gamma }_{1,t}=& -{\lambda }_1^{2}v+{\lambda }_1(-uw+v_x)+Q+(-{\lambda }_1^{2}w+{\lambda }_{1}r-q){\beta }_1\hfill \\ \hfill & +[-{\lambda }_1^3-{\lambda }_1(2v^2-u^2+\frac{1}{2}{w}^2)]{\gamma }_1+[{\lambda }_1^{2}u-{\lambda }_1(vw-u_x)+P]{\beta }_1{\gamma }_1\hfill \\ \hfill & -[{\lambda }_1^{2}v+{\lambda }_1(-uw+v_x)-Q]{\gamma }_1^2.\hfill \end{array}$$

(33)

From (14) and (15) and the equation above, we obtain the first derivatives of a, b, c, d, g, f with respect to t:

$$\begin{array}{cc}\hfill a_t=& 2(d-g)u^3+2c{v}^3+(ag-ad)u^2+(ad-bc)v^2+\frac{1}{2}(d-g)u{w}^2+\frac{1}{2}cv{w}^2\hfill \\ \hfill & +\frac{1}{2}(ag-bc)w^2+(d-g)u{v}^2+c{u}^{2}v-(ab+dc)wu+(dg-g^2-c^2)wv\hfill \\ \hfill & +bq+(d-g)vr-cur+(bg+ac)r+b(a^2+c^2+g^2)w+(c{b}^2+c{d}^2-abf-abd)v\hfill \\ \hfill & +(d-g)(a^2+c^2+g^2)u+(g^2+c^2-dg-cw)u_x+(ab+cd+gw-dw)v_x\hfill \\ \hfill & +(d-g)u_{xx}+c{v}_{xx}+(ag-bc)(a^2+b^2+d^2),\hfill \\ \hfill b_t=& -2c{u}^3+2(f-d)v^3+(ac-bd)u^2+(bd-bf)v^2-\frac{1}{2}cu{w}^2+\frac{1}{2}(f-d)v{w}^2\hfill \\ \hfill & +\frac{1}{2}(ac-bf)w^2+(f-d)u^{2}v-cu{v}^2+(a^2+d^2-fb)wu+(ab+cd)wv\hfill \\ \hfill & -aq+(d-f)ur-cvr+(bc+af)r-a(b^2+c^2+f^2)w-(a^2+b^2+d^2)cu\hfill \\ \hfill & +(f-d)(a^2+b^2+d^2)v+(wd-wf-ab-cd)u_x+(cw+fd-d^2-a^2)v_x\hfill \\ \hfill & -c{u}_{xx}+(f-d)v_{xx}+(ac-bf)(a^2+b^2+d^2),\hfill \\ \hfill c_t=& 2b{u}^3-2a{v}^3+(gc-ab)u^2+(ab-cf)v^2+\frac{1}{2}bu{w}^2-\frac{1}{2}av{w}^2+\frac{1}{2}(gc-cf)w^2\hfill \\ \hfill & -a{u}^{2}v+bu{v}^2-(bc+af)wu+(ac+bg)wv+aru+brv+(gf-f^2-b^2)r\hfill \\ \hfill & +(f-g)q+(f-g)(a^2+c^2+g^2)w-a(a^2+c^2+g^2)v+b(a^2+c^2+g^2)u\hfill \\ \hfill & +(wa-ac-bg)u_x+(bc+af-bw)v_x+b{u}_{xx}-a{v}_{xx}+(gc-cf)(a^2+b^2+d^2),\hfill \\ \hfill d_t=& -2aP-2bQ+2abr+2b(a^2+b^2+d^2)v-2a(a^2+b^2+d^2)u-2bdwu+2adwv\hfill \\ \hfill & -2ad{u}_x+2bd{v}_x+(2a^2+b^2)u^2-(a^2+2b^2)v^2+\frac{1}{2}(a^2-b^2)w^2\hfill \\ \hfill & +(a^2-b^2)(a^2+b^2+d^2),\hfill \\ \hfill g_t=& 2aP+2cq+2cgr+2a(a^2+c^2+g^2)u+2c(a^2+c^2+g^2)w+2agwv-2ag{u}_x\hfill \\ \hfill & +2ac{v}_x-(2a^2+c^2)u^2-(\frac{1}{2}{a}^2+c^2)w^2+(a^2-c^2)v^2-(a^2+2c^2)(a^2+c^2+g^2),\hfill \\ \hfill f_t=& 2bQ-2cq+2fcr-2c(a^2+c^2+g^2)w-2b(a^2+b^2+d^2)v-2bfwu+2bcwv\hfill \\ \hfill & +2bf{v}_x-2bc{u}_x+(c^2-b^2)u^2+(2b^2+c^2)v^2+(c^2+\frac{1}{2}{b}^2)w^2\hfill \\ \hfill & +(b^2+2c^2)(a^2+b^2+d^2).\hfill \end{array}$$

(34)

By calculating the Formulas (14) and (15), we can also deduce

$$\begin{array}{c}\hfill a^2+b^2+d^2=a^2+c^2+g^2=b^2+c^2+f^2={\lambda }_1^2.\end{array}$$

(35)

Substituting (34) and (6) into (31), and inserting (29) and (16) into (32), we derive ${\widehat{V}}_{4}A=A_t+A{V}_4$. Theorem 1 is proved. ☐

3. Exact Solutions

In this section, we derive some examples of exact solutions to (1) by using Darboux transform (16).

(i) Let the seed solution be $u=v=w=r=q=0$, and we have

$$U=\left(\begin{array}{ccc}\lambda & 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),V=\left(\begin{array}{ccc}0& 0& 0\\ 0& {\lambda }^3& 0\\ 0& 0& -{\lambda }^3\end{array}\right).$$

As a result, ${\Phi }_x=U\Phi$ and ${\Phi }_t=V\Phi$ have the general solution:

$$\Phi ={\alpha }_1\left(\begin{array}{c}{e}^{\lambda x}\\ 0\\ 0\end{array}\right)+{\alpha }_2\left(\begin{array}{c}0\\ e^{{\lambda }^{3}t}\\ 0\end{array}\right)+{\alpha }_3\left(\begin{array}{c}0\\ 0\\ e^{-{\lambda }^{3}t}\end{array}\right),$$

which means

$$\begin{array}{cc}\hfill & {\beta }_1=\frac{{\alpha }_2{e}^{{\lambda }_1^{3}t}}{{\alpha }_1{e}^{{\lambda }_{1}x}}=\frac{{\alpha }_2}{{\alpha }_1}{e}^{{\lambda }_1^{3}t-{\lambda }_{1}x},\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & {\gamma }_1=\frac{{\alpha }_3{e}^{-{\lambda }_1^{3}t}}{{\alpha }_1{e}^{{\lambda }_{1}x}}=\frac{{\alpha }_3}{{\alpha }_1}{e}^{-{\lambda }_1^{3}t-{\lambda }_{1}x}.\hfill \end{array}$$

(37)

Without loss of generality, we set ${\alpha }_1=1$. Then Equations (14) and (15) imply

$$\begin{array}{cc}\hfill a& =\frac{-2{\alpha }_2{\lambda }_1{e}^{{\lambda }_1^{3}t-{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill b& =\frac{-2{\alpha }_3{\lambda }_1{e}^{-{\lambda }_1^{3}t-{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill c& =\frac{-2{\alpha }_2{\alpha }_3{\lambda }_1{e}^{-2{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill d& ={\lambda }_1+\frac{-2{\lambda }_1}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill g& ={\lambda }_1+\frac{-2{\alpha }_2^2{\lambda }_1{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill f& ={\lambda }_1+\frac{-2{\alpha }_3^2{\lambda }_1{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}}.\hfill \end{array}$$

(43)

From Darboux transformation (16) we obtain an explicit solution $(\widehat{u},\widehat{v},\widehat{w},\widehat{r},\widehat{q})$ for (1),

$$\begin{array}{cc}\hfill \widehat{u}& =\frac{2{\alpha }_2{\lambda }_1{e}^{{\lambda }_1^{3}t-{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill \widehat{v}& =\frac{2{\alpha }_3{\lambda }_1{e}^{-{\lambda }_1^{3}t-{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill \widehat{w}& =\frac{4{\alpha }_2{\alpha }_3{\lambda }_1{e}^{-2{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}},\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill \widehat{r}& =\frac{4{\alpha }_2{\alpha }_3{\lambda }_1^2{e}^{-4{\lambda }_{1}x}({\alpha }_3^2{e}^{-2{\lambda }_1^{3}t}-{\alpha }_2^2{e}^{2{\lambda }_1^{3}t})}{{(1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x})}^2},\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill \widehat{q}& =\frac{4{\alpha }_2{\alpha }_3{\lambda }_1^3{e}^{-2{\lambda }_{1}x}}{1+{\alpha }_2^2{e}^{2{\lambda }_1^{3}t-2{\lambda }_{1}x}+{\alpha }_3^2{e}^{-2{\lambda }_1^{3}t-2{\lambda }_{1}x}}.\hfill \end{array}$$

(48)

When ${\lambda }_1=-1$, ${\alpha }_2=e^{-1}$, ${\alpha }_3=e$, the solution is illustrated in Figure 1.

(ii) Let $u=v=0$, $w=2$, $r=0$, $q=1$ and ${\lambda }_1=1$. When $\lambda ={\lambda }_1$ we have

$${U|}_{\lambda ={\lambda }_1}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right){,V|}_{\lambda ={\lambda }_1}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 3& 3\\ 0& -3& -3\end{array}\right).$$

Then ${\Phi }_{1,x}=\left(U{|}_{\lambda ={\lambda }_1}\right){\Phi }_1$ and ${\Phi }_{1,t}=\left(V{|}_{\lambda ={\lambda }_1}\right){\Phi }_1$ have the general solution:

$$\begin{array}{cc}\hfill {\Phi }_1=& {\alpha }_1\left(\begin{array}{c}{e}^x\\ 0\\ 0\end{array}\right)+{\alpha }_2\left(\begin{array}{c}0\\ 1+3t\\ -3t\end{array}\right)+{\alpha }_3\left(\begin{array}{c}0\\ 3t\\ 1-3t\end{array}\right).\hfill \end{array}$$

Letting ${\alpha }_1=1$, we have

$$\begin{array}{cc}\hfill {\beta }_1=& {\alpha }_2(1+3t)e^{-x}+3{\alpha }_{3}t{e}^{-x},\hfill \\ \hfill {\gamma }_1=& -3{\alpha }_{2}t{e}^{-x}+{\alpha }_3(1-3t)e^{-x}.\hfill \end{array}$$

(49)

An explicit solutions of the soliton Equation (1) can be derived by solving Equations (14), (15) and Darboux transformation (16). The expressions are as follows:

$$\begin{array}{}\hfill & \widehat{u}=\frac{[2{\alpha }_2(1+3t)+6{\alpha }_{3}t]e^x}{e^{2x}+{\alpha }_2^2(1+6t+18t^2)+{\alpha }_3^2(1-6t+18t^2)+36{\alpha }_2{\alpha }_3{t}^2},\hfill & \mathrm{(50)}\hfill & \widehat{v}=\frac{[-6{\alpha }_{2}t+2{\alpha }_3(1-3t)]e^x}{e^{2x}+{\alpha }_2^2(1+6t+18t^2)+{\alpha }_3^2(1-6t+18t^2)+36{\alpha }_2{\alpha }_3{t}^2},\hfill & \mathrm{(51)}\hfill & \widehat{v}=\frac{[-6{\alpha }_{2}t+2{\alpha }_3(1-3t)]e^x}{e^{2x}+{\alpha }_2^2(1+6t+18t^2)+{\alpha }_3^2(1-6t+18t^2)+36{\alpha }_2{\alpha }_3{t}^2},\hfill & \mathrm{(52)}\hfill & \widehat{v}=\frac{[-6{\alpha }_{2}t+2{\alpha }_3(1-3t)]e^x}{e^{2x}+{\alpha }_2^2(1+6t+18t^2)+{\alpha }_3^2(1-6t+18t^2)+36{\alpha }_2{\alpha }_3{t}^2},\hfill & \mathrm{(53)}\hfill & +(\frac{12{\alpha }_2^2(t+3t^2)-4{\alpha }_2{\alpha }_3(1-18t^2)-12{\alpha }_3^2(t-3t^2)}{{[e^{2x}+{\alpha }_2^2(1+6t+18t^2)+{\alpha }_3^2(1-6t+18t^2)+36{\alpha }_2{\alpha }_3{t}^2]}^2}\hfill & \mathrm{(54)}\hfill & \times [{\alpha }_2^2(1+6t)+12{\alpha }_2{\alpha }_{3}t+{\alpha }_3^2(6t-1)]),\hfill & \mathrm{(55)}\hfill & \widehat{q}=1+\frac{4{\alpha }_2^2(1-3t-9t^2)+12{\alpha }_2{\alpha }_3(1-6t^2)+4{\alpha }_3^2(1+3t-9t^2)}{e^{2x}+{\alpha }_2^2(1+6t+18t^2)+{\alpha }_3^2(1-6t+18t^2)+36{\alpha }_2{\alpha }_3{t}^2}.\hfill & \mathrm{(56)}\end{array}$$

When ${\alpha }_2={\alpha }_3=1$, the solution is illustrated in Figure 2.

4. Conclusions

In this paper, we propose a new five-component nonlinear integrable equation associated with a $3\times 3$ spectral problem. This five-component equation has two different mKdV equations a reduction. By means of Darboux transformations between Lax pairs, we constructed a Darboux transformation for this equations. Finally, as applications of the Darboux transformations, we obtained exact solutions for (1) including a solution that consists of rational functions of $e^x$ and t.