## Dynamical Analysis and Traveling Wave Solutions for Generalized (3+1)-Dimensional Kadomtsev-Petviashvili Equation

 基金资助: 国家自然科学基金.  11371267四川省教育厅自然科学重点基金.  2012ZA135

 Fund supported: Sponsored by the NSFC.  11371267the SCNSF.  2012ZA135 Abstract

Dynamical analysis and explicit solutions for generalized (3+1)-dimension Kadomtsev-Petviashvili equation have been carried out. The singular solution is obtained by the ansatz method, the bifurcation phase portraits and corresponding explicit solution are also constructed by the approach of dynamical analysis.

Keywords： Generalized (3+1)-dimension Kadomtsev-Petviashvili equation ; Ansatz method ; Bifurcation analysis ; Phase portraits ; Traveling wave solutions

Zhang Xue, Sun Yuhuai. Dynamical Analysis and Traveling Wave Solutions for Generalized (3+1)-Dimensional Kadomtsev-Petviashvili Equation. Acta Mathematica Scientia[J], 2019, 39(3): 501-509 doi:

## 1 引言

$\begin{equation} (u_{t}+6uu_{x}+u_{xxx})_{x} = \sigma(u_{yy}+u_{zz}), (x, y, z)\in {\Bbb R} ^{3}, t>0. \end{equation}$

$\begin{equation} (u_{t}+auu_{x}+u_{xxx})_{x}+bu_{yy}+cu_{zz} = 0, \end{equation}$

### 2.1 奇异孤子解

$\begin{equation} u(x, y, z, t) = A{\rm csch}^{p}\tau, \end{equation}$

$\begin{equation} u(x, y, z, t) = -\frac{3(k-b-c)}{a}{\rm csch}^{2}\bigg(\sqrt{\frac{k-b-c}{2}}\tau\bigg), \end{equation}$

## 3 分支相图和定性分析

$\begin{equation} u(x, y, z, t) = v(\xi), \quad \xi = x+y+z-kt, \end{equation}$

$\begin{equation} (-k+b+c)v+\frac{a}{2}v^{2}+v'' = 0. \end{equation}$

$v' = y$,由平面动力系统

$\begin{equation} \left\{ \begin{array}{ll} v' = y, \\ y' = (k-b-c)v-\frac{a}{2}v^{2}.\\ \end{array} \right. \end{equation}$

$\begin{equation} H(v, y) = y^{2}-(k-b-c)v^{2}+\frac{a}{3}v^{3} = h, \end{equation}$

$\begin{equation} f(v) = (k-b-c)v-\frac{a}{2}v^{2}, \end{equation}$

$\begin{equation} f'(v) = (k-b-c)-va. \end{equation}$

$\begin{equation} \lambda_{\pm} = \pm\sqrt{f'(v_{i})}. \end{equation}$

(ⅰ)若$f'(v_{i}) > 0$,则奇点$(v_{i}, 0)$是鞍点.

(ⅱ)若$f'(v_{i}) < 0$,则奇点$(v_{i}, 0)$是中心.

(ⅲ)若$f'(v_{i}) = 0$,则奇点$(v_{i}, 0)$是退化的鞍点.由以上结论,借助数学软件maple可得到系统(3.3)的分支相图,如图 14.

### 图 1 ### 图 2 ### 图 3 ### 图 4 $\begin{equation} h^{\ast} = H(v_{1}, 0) = -\frac{4k^{3}}{3a^{2}(2c+1)}. \end{equation}$

(ⅰ)当$h^{\ast} < h < 0$,系统(3.3)有一个周期轨道$L_{2}$和一个特殊的轨道$L_{3}$.

(ⅱ)当$h > 0$或者$h < h^{\ast}$,系统(3.3)没有任何轨道.

(ⅲ)当$h = 0$,系统(3.3)有一个同宿轨道$L_{1}$,两个特殊轨道$L_{1}^{\ast} $$L_{1}^{\star} . (ⅳ)当 h = h^{\ast} ,系统(3.3)有一个特殊轨道 L_{4} . 命题3.2 当 a < 0 , k-b-c < 0 , f'(v_{0}) < 0, f'(v_{1}) > 0 , (v_{0}, 0) 是中心, (v_{1}, 0) 是鞍点. (如图 2所示) (ⅰ)当 0 < h < h^{\ast} ,系统(3.3)有一个周期轨道 L_{6} 和特殊轨道 L_{7} . (ⅱ)当 h < 0 或者 h > h^{\ast} ,系统(3.3)没有任何轨道. (ⅲ)当 h = 0 ,系统(3.3)有一个特殊的轨道 L_{8} . (ⅳ)当 h = h^{\ast} ,系统(3.3)有一个同宿轨道 L_{5} ,两个特殊轨道 L_{5}^{\ast}$$ L_{5}^{\star}$.

(ⅰ)当$h = 0$,方程(1.2)有一个孤立波解和一个奇异孤立波解.

(ⅱ)当$h^{\ast} < h < 0$,方程(1.2)有一个周期波解和一个奇异波解.

(ⅲ)当$h = h^{\ast}$,方程(1.2)有一个周期奇异波解.

(ⅰ)当$h = h^{\ast}$时,方程(1.2)有一个孤立波解和一个奇异孤立波解.

(ⅱ)当$0 < h < h^{\ast}$时,方程(1.2)有一个周期波解和一个奇异波解.

(ⅲ)当$h = 0$时,方程(1.2)有一个周期奇异波解.

## 4 行波解及其联系

(ⅰ)从分支相图 1,可注意到有一条过鞍点(0, 0)的同宿轨道$L_{1}$.两条特殊轨道$L_{1}^{\star} $$L_{1}^{\ast} ,它们在 (v, y) 平面的表达式为 \begin{equation} y = \pm\sqrt{-\frac{a}{3}v^{2}(v-v_{2})}, \end{equation} 其中 v_{2} = \frac{3(k-b-c)}{a} . 把(4.1)式代入系统(3.3)并且沿轨道 L_{1} 积分和沿特殊轨道 L_{1}^{\star}$$ L_{1}^{\ast}$积分可得

(ⅱ)从分支相图 1,可注意到有一条过点$(v_{3}, 0), (v_{4}, 0)$的周期轨道$L_{2} $$(v_{5}, 0) 的周期轨道 L_{3} .它们在 (v, y) 平面的表达式为 \begin{equation} y = \pm\sqrt{-\frac{a}{3}(v-v_{3})(v-v_{4})(v-v_{5})}, \end{equation} 其中 v_{3} < v_{4} < v_{5} . 把(4.2)式代入系统(3.3)并且沿轨道 L_{2}$$ L_{3}$积分可得

(ⅲ)从分支相图 1,可注意到有一条特殊轨道$L_{4}$,它与中心$(v_{1}, 0)$有相同的哈密顿量.它在$(v, y)$平面的表达式为

$\begin{equation} y = \pm\sqrt{-\frac{a(v-v_{1})^{2}(v-v_{6})}{3}}, \end{equation}$

$\begin{equation} y = \pm\sqrt{-\frac{a}{3}(v-v_{1})^{2}(v-v_{7})}, \end{equation}$

(ⅱ)从分支相图 2,可看到过$(v_{8}, 0)$, $(v_{9}, 0)$的轨道$L_{6}$和过点$(v_{10}, 0)$的特殊轨道$L_{7}$.它们在$(v, y)$平面的表达式

$\begin{equation} y = \pm\sqrt{-\frac{a}{3}v^{3}+(k-b-c)v^{2}+h} = \pm{\sqrt{-\frac{a}{3}(s-v_{8})(s-v_{9})(s-v_{10})}}, \end{equation}$

$a > 0 $$k-b-c < 0 时,结果与上面类似,就不一一详细求解. 最后,研究方程行波解之间的联系. (ⅰ)令 h\rightarrow0_{-} ,则 v_{3}\rightarrow\frac{3k}{a} , v_{4}\rightarrow0 , v_{5}\rightarrow0\frac{v_{4}-v_{3}}{v_{5}-v_{3}}\rightarrow 1 因此,可得 u_{3}(x, y, z, t)\rightarrow u_{1}(x, y, z, t)$$ u_{4}(x, y, z, t)\rightarrow u_{2}(x, y, z, t)$.

(ⅱ)令$h\rightarrow h^{\ast}_{+}$,则$v_{3}\rightarrow\frac{2k}{a}$, $v_{4}\rightarrow \frac{2k}{a}$, $v_{5}\rightarrow \frac{-k}{a}$, $\frac{v_{4}-v_{3}}{v_{5}-v_{3}}\rightarrow0$

(ⅲ)令$h\rightarrow h^{\ast}_{-}$,则$v_{8}\rightarrow\frac{-k}{a}$, $v_{9}\rightarrow\frac{2k}{a}$, $v_{10}\rightarrow \frac{2k}{a}$, $\frac{v_{9}-v_{8}}{v_{10}-v_{8}}\rightarrow1$

(ⅳ)令$h\rightarrow 0^{+}$,从而有$v_{8}\rightarrow0$, $v_{9}\rightarrow 0$, $v_{10}\rightarrow \frac{k}{a}$, $\frac{v_{9}-v_{8}}{v_{10}-v_{8}}\rightarrow0$

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