Rogue wave solutions of the (2+1)-dimensional derivative nonlinear Schrödinger equation

Abstract

In this paper, we focus on the construction of rogue wave solutions for the (2+1)-dimensional derivative nonlinear Schrödinger equation. The N-order generalized Darboux transformation is obtained, and the determinant form of N-order rogue waves is also presented by taking limit on the classical Darboux transformation. On the plane wave solution background, two different kinds of rogue wave solutions (linear rogue wave and parabolic rogue wave) are constructed successively. The characteristics of two types of rogue waves are analyzed by some figures and physical qualities.

Appendix: Choose the special solution

Firstly, we need to solve the linear system (2) with the seed solution by the gauge transformation method:

$$\begin{aligned} \varPsi =M\varPhi , M=diag\left( e^{-\frac{i}{2}\theta },e^{\frac{i}{2}\theta }\right) , \end{aligned}$$

(5)

where

$$\begin{aligned} \varPsi =\left( \begin{array}{c} \varphi _{1} \\ \varphi _{2} \end{array}\right) , \varPhi =\left( \begin{array}{ll} \phi _{1} \\ \phi _{2} \end{array} \right) . \end{aligned}$$

(6)

Then, matrix $\varPhi$ satifies the following linear system:

$$\begin{aligned}&\varPhi _{x}=N\varPhi ,\end{aligned}$$

(7a)

$$\begin{aligned}&\varPhi _{t}=\lambda ^{2}\varPhi _{y}-\nu N\varPhi , \end{aligned}$$

(7b)

where

$$\begin{aligned} N= & {} \frac{i}{2}\left( \begin{array}{cc} \lambda ^{2}+\mu &{} 2ia\lambda \\ -2ia\lambda &{} -\lambda ^{2}-\mu \end{array}\right) . \end{aligned}$$

And (7a) could be rewrited as

$$\begin{aligned}&2 a \lambda \phi _2 -i \left( \lambda ^2+\mu \right) \phi _1+2 \phi _{1x}=0,\end{aligned}$$

(8a)

$$\begin{aligned}&\quad -2 a \lambda \phi _1+i \left( \lambda ^2+\mu \right) \phi _2+2 \phi _{2x}=0. \end{aligned}$$

(8b)

By some calculations, i.e., (8a)+(8b) and (8a)-(8b), we could have

$$\begin{aligned} 2 \left( \phi _1+\phi _2\right) _{x}+\left[ 2 a \lambda +i \left( \lambda ^2+\mu \right) \right] \left( \phi _2-\phi _1\right) =0,\end{aligned}$$

(9a)

$$\begin{aligned} 2 \left( \phi _1-\phi _2\right) _{x}+\left[ 2 a \lambda -i \left( \lambda ^2+\mu \right) \right] \left( \phi _1+\phi _2\right) =0. \end{aligned}$$

(9b)

For convenience, $\phi _2-\phi _1$ and $\phi _2+\phi _1$ are marked as R and K, respectively. Therefore, from (9a) and (9b), we could deriving that

$$\begin{aligned} K&=a_1(t,y) e^{\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}\\&\quad +a_2(t,y) e^{-\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}},\\ R&=\frac{\sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}{2 a \lambda +i \left( \lambda ^2+\mu \right) }\nonumber \\&\quad \times \left[ a_2(t,y)e^{-\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}\right. \\&\quad \left. -\,a_1(t,y) e^{\frac{1}{2}x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}\right] . \end{aligned}$$

Next, inserting K and R into $\phi _2-\phi _1=R$ and $\phi _2+\phi _1=K$, we could obtained

$$\begin{aligned} \phi _{1}= & {} \frac{\sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}+2 a \lambda +i \lambda ^2+i \mu }{2 \left[ 2 a \lambda +i \left( \lambda ^2+\mu \right) \right] }\\&\times e^{\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}a_1(t,y)\\&- \frac{\sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}-2 a \lambda -i \lambda ^2-i \mu }{2 \left[ 2 a \lambda +i \left( \lambda ^2+\mu \right) \right] }\\&\times e^{-\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}a_2(t,y),\\ \phi _{2}= & {} \frac{\sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}+2 a \lambda +i \lambda ^2+i \mu }{2 \left[ 2 a \lambda +i \left( \lambda ^2+\mu \right) \right] }\\&\times e^{-\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}a_2(t,y)\\&-\frac{\sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}-2 a \lambda -i \lambda ^2-i \mu }{2 \left[ 2 a \lambda +i \left( \lambda ^2+\mu \right) \right] }\\&\times e^{\frac{1}{2} x \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}a_1(t,y). \end{aligned}$$

Now, simultaneous Eq. (7b) and $\phi _{i}(i=1,2)$, $a_1(t,y)$ and $a_2(t,y)$ could be derived:

$$\begin{aligned} a_1(t,y)&=f\left( \lambda ^2 t+y\right) e^{-\frac{1}{2} \nu t \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}},\\ a_2(t,y)&=g\left( \lambda ^2 t+y\right) e^{\frac{1}{2} \nu t \sqrt{-4 a^2 \lambda ^2-\left( \lambda ^2+\mu \right) ^2}}, \end{aligned}$$

where $f\left( \lambda ^2 t+y\right)$ and $g\left( \lambda ^2 t+y\right)$ are arbitrary complex functions about $\lambda ^2 t+y$. The different solutions could be derived by changing $f\left( \lambda ^2 t+y\right)$ and $g\left( \lambda ^2 t+y\right)$. Then, $\varphi _{1}$ and $\varphi _{2}$ could be derived.