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数学物理学报, 2019, 39(6): 1342-1351 doi:

论文

广义超弹性杆波方程的行波解

Traveling Wave Solutions of the Generalized Hyperelastic-Rod Wave Equation

Gu Yongyi1, Yuan Wenjun,2, Wu Yonghong3,4

通讯作者: 袁文俊, E-mail: wjyuan1957@126.com

收稿日期: 2018-08-15  

基金资助: 国家自然科学基金.  11901111
国家自然科学基金.  11271090

Received: 2018-08-15  

Fund supported: the NSFC.  11901111
the NSFC.  11271090

摘要

该文研究了广义超弹性杆波方程.利用行波变换将广义超弹性杆波方程转化为一个复微分方程,并通过弱h,k条件和Fuchs指数证明了该复微分方程的亚纯解属于W类.进一步求出了该复微分方程的所有亚纯解,从而得到了广义超弹性杆波方程的行波解.可将该文的方法应用到一些相关的数学物理方程.

关键词: 广义超弹性杆波方程 ; 微分方程 ; 椭圆函数 ; 亚纯函数

Abstract

In this paper, we study the generalized hyperelastic-rod wave equation. We changed the generalized hyperelastic-rod wave equation into a complex differential equation by using traveling wave transform and show that meromorphic solutions of the complex differential equation belong to the class W by the weak h,k condition and the Fuchs index. Furthermore, we find out all meromorphic solutions of the complex differential equation, then we obtain the traveling wave solutions of the generalized hyperelastic-rod wave equation. We can apply the idea of this study to some related mathematical physics equations.

Keywords: Generalized hyperelastic-rod wave equation ; Differential equation ; Elliptic function ; Meromorphic function

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本文引用格式

. 广义超弹性杆波方程的行波解. 数学物理学报[J], 2019, 39(6): 1342-1351 doi:

Gu Yongyi, Yuan Wenjun, Wu Yonghong. Traveling Wave Solutions of the Generalized Hyperelastic-Rod Wave Equation. Acta Mathematica Scientia[J], 2019, 39(6): 1342-1351 doi:

1 引言

2005年, Coclite等[1-2]首次考虑了如下广义超弹性杆波方程

utuxxt+νux+2κuux+3θu2uxγuxuxxuuxxx=0,
(1.1)

其中ν,κ,θγ为常数,建立了耗散解的存在性,得到了"弱等于强"的唯一性结果,并研究了该问题解的稳定性.

近年来,对广义超弹性杆波方程的研究取得了许多成果[3-6],例如得到广义超弹性杆波方程的整体保守解、几何有限差分格式和整体弱解等.本文运用复方法[7-9]得到广义超弹性杆波方程的行波解.

运用行波变换

u(x,t)=w(z),z=xνt
(1.2)

于方程(1.1),并关于z积分一次,令积分常数为0,得

ww
(1.3)

如果一个亚纯函数g为关于z的有理函数,或为关于e^{\mu z}, \mu \in {\mathbb C}的有理函数,或为椭圆函数,则称g属于W类.

本文主要得到如下定理.

定理1.1  若\theta\neq 0,则方程(1.3)的亚纯解w属于W类.更进一步,当\gamma为一个不属于\mathbb M (定义于第二节)的常数时,方程(1.3)有如下形式的亚纯解

(Ⅰ)有理函数解

W_r(z)=\frac{2(\gamma+2)}{\theta}\frac {1}{(z-z_{0})^2}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中\nu=-\frac{\kappa(\gamma+2)}{3\theta}, z_0\in {\mathbb C}.

(Ⅱ)单周期函数解

W_{s}(z)=\frac{(\gamma+2)\mu^2}{2\theta} \coth^2\frac{\mu (z-z_{0})}{2} -\frac{(\gamma+2)\mu^2}{3\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中

\nu=\frac{\mu^2(-\gamma^2-\gamma+\sqrt{2(-4\gamma^4-8\gamma^3+3\gamma^2+7\gamma+2)}+2)}{18\theta},

\kappa=\frac{\mu^2(-4\gamma^2-10\gamma+\sqrt{2(-4\gamma^4-8\gamma^3+3\gamma^2+7\gamma+2)}-4)}{6(2\gamma+1)}, \quadz_0\in {\mathbb C}.

(Ⅲ)椭圆函数解

W_{d}(z)=\frac{2(\gamma+2)}{\theta}\left(-\wp(z)+\frac{1}{4}\left(\frac{\wp'(z)+A}{\wp(z)-B}\right)^2\right)-\frac{2(\gamma+2)B}{\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中

c_3=\frac{(\kappa^2\gamma^3-3\kappa\theta\nu\gamma^2+9\theta^2\nu^2\gamma+\kappa^2\gamma^2-6\kappa\theta\nu\gamma+27\theta^2\nu^2 -2\kappa^2\gamma+9\kappa\theta\nu)(\kappa(\gamma+2)+3\theta\nu)}{27r(\gamma-1)(\gamma+2)^2(\gamma+1)^3},

c_2=-\frac{(\kappa\gamma-3\theta\nu)(\kappa(\gamma+2)+3\theta\nu)}{3r(\gamma+2)(\gamma+1)^2}, \quad A^2=4B^3-c_2B-c_3.

\gamma=1时,方程(1.3)有如下形式的亚纯解

(Ⅰ)有理函数解

U_r(z)=\frac{6}{\theta}\frac {1}{(z-z_{0})^2}-\frac{\kappa+\theta\nu}{2\theta},

其中\nu=-\frac{\kappa}{\theta}, z_0\in {\mathbb C}.

(Ⅱ)单周期函数解

U_{s}(z)=\frac{3\mu^2}{2\theta} \coth^2\frac{\mu (z-z_{0})}{2} -\frac{\mu^2}{2\theta},

其中\nu=0, \kappa=\mu^2, z_0\in {\mathbb C}.

(Ⅲ)椭圆函数解

U_{d}(z)=\frac{6}{\theta}\left(-\wp(z)+\frac{1}{4}\left(\frac{\wp'(z)+E}{\wp(z)-F}\right)^2\right)-\frac{6F}{\theta}-\frac{\kappa+\theta\nu}{2\theta},

其中\nu=-\frac{\kappa}{\theta}, E^2=4F^3-c_3, c_2=0, c_3为任意常数.

2 基础知识

设集合p\in {\mathbb N}^*:=\{1, 2, 3, \cdots \}, r_j\in {\mathbb M}^*={\mathbb N}^*\cup\{0\}, {\mathbb M}={\mathbb M}^*\cup\{-2, -1\}, r=(r_0, r_1, \cdots, r_p), j=0, 1, \cdots, p.

微分单项式为

K_r[w](z):=\prod\limits_{j=0}^{m}[w^{(j)}(z)]^{r_j},

d(r):=\sum\limits_{j=0}^{p}r_j被称为K_r[w]的次数.

微分多项式为

F(w, w', \cdots, w^{(p)}):=\sum\limits_{r\in J}a_rK_r[w],

其中J为一个有限指标集, a_{r} 为常数. \deg F(w, w', \cdots, w^{(p)}):= \mathop {\max }\limits_{r \in J} \{d(r)\}称为P[w]的全次数.

考虑如下复常微分方程

D(z, w):=F(w, w', \cdots, w^{(p)})-cw^q-d=0,
(2.1)

其中c\neq 0, d为常数, q\in {\mathbb N}^*.

h, ~k\in {\mathbb N}^*,方程(2.1)的亚纯解w至少存在一个极点,若方程(2.1)恰好存在h个不同的亚纯解以z=0k重极点,则称方程(2.1)满足\langle h, k \rangle条件.若将Laurent展式

w(z)=\sum\limits_{\tau=-k}^{\infty}\beta_{\tau}z^{\tau}, \beta_{-k}\neq 0, k>0,
(2.2)

代入方程(2.1)可得h个不同的Laurent展式的奇异部分: \sum\limits_{\tau=-k}^{-1}\beta_{\tau}z^{\tau}, 则称方程(2.1)满足弱\langle h, k \rangle条件.

给定两个复数l_1, ~l_2使得\mbox{Im} \frac{l_1}{l_2}>0, L:=L[2l_1, ~ 2l_2]=\{l~|~l=2al_1+2bl_2, ~a, b\in \mathbb Z\}为同构于{\mathbb Z}\times {\mathbb Z}的离散子集.判别式\Delta=\Delta(b_1, b_2):=b_1^3-27b_2^2 H_n=H_n(L):=\sum\limits_{l\in L\setminus \{0\}}\frac{1}{l^n}.

具有双周期2l_1, ~2l_2的Weierstrass椭圆函数\wp(z):=\wp(z, c_2, c_3)满足方程

(\wp'(z))^2=4 \wp(z)^3-c_2 \wp(z)-c_3,

其中 c_2=60H_4, ~c_3=140H_6,和\Delta(c_2, c_3)\neq 0, 且满足如下加法公式[10]

\wp(z-z_0)= \frac{1}{4}\left[\frac{\wp'(z)+\wp'(z_0)}{\wp(z)-\wp(z_0)}\right]^2-\wp(z)-\wp(z_0).
(2.3)

控制项[11]决定了D(z, w)的亚纯解w的重数k,控制部分\hat{D}(z, w)由所有控制项组成.每个控制项有相同的重数,记为I(k).

控制部分\hat{D}(z, w)关于w的微分定义为

\hat{D}'(z, w)=\lim\limits_{\lambda \rightarrow 0}\frac{\hat{D}(z, w+\lambda v)-\hat{D}(z, w)}{\lambda v},
(2.4)

其为线性算子.方程

P(i)=\lim\limits_{z \rightarrow 0}z^{-i+I(k)}\hat{D}'(z, \beta_{-k}z^{-k})z^{i-k}=0
(2.5)

的根,被称为Fuchs指数.

引理2.1[12-13]  设h, ~p, ~q, ~s\in {\mathbb N}^*, ~\deg F(w, w', \cdots, w^{(p)}) < q, 且方程(2.1)满足\langle h, k \rangle条件,则方程(2.1)的亚纯解w\in W.若对某些参数值使得解w存在,则其他亚纯解的形式是一个单参量族w(z-z_{0}), z_0 \in {\bf C}.并且,每一个在z=0处有极点的椭圆函数解表示为

\begin{array}[b]{rl} w(z)=&\displaystyle\sum\limits_{i=1}^{s-1}\sum\limits_{j=2}^{k}\frac{(-1)^j\beta_{-ij}}{(j-1)!}\frac{d^{j-2}}{dz^{j-2}} \left(\frac{1}{4}\left(\frac{\wp'(z)+C_i}{\wp(z)-D_i}\right)^2 -\wp(z)\right)\\ &+ \displaystyle\sum\limits_{i=1}^{s-1}\frac{\beta_{-i1}}{2}\frac{\wp'(z)+C_i}{\wp(z)-D_i} +\sum\limits_{j=2}^{k}\frac{(-1)^j\beta_{-sj}}{(j-1)!}\frac{d^{j-2}}{dz^{j-2}}\wp (z)+\beta_0, \end{array}
(2.6)

其中\beta_{-ij} (2.2)式给出, \sum\limits_{i=1}^s\beta_{-i1}=0C_i^2=4D_i^3-c_2D_i-c_3.

每一个有理函数解w=R(z)的形式为

R(z)=\sum\limits_{i=1}^{s}\sum\limits_{j=1}^{n}\frac{\beta_{ij}}{(z-z_{i})^{j}}+\beta_0,
(2.7)

它有s(\leq h)个区别的k重极点.

每一个单周期解w=R(\eta)是关于\eta=e^{\mu z} (\mu\in{\bf C})的有理函数,且其形式为

R(\eta)=\sum\limits_{i=1}^{s}\sum\limits_{j=1}^{n}\frac{\beta_{ij}}{(\eta-\eta_{i})^{j}}+ \beta_0,
(2.8)

它有s(\leq h)个区别的k重极点.

2009年, Eremenko等[14]研究了p阶Briot-Bouquet方程

F(w, w^{(p)})=\sum\limits_{j=0}^{q}F_{j}(w)(w^{(p)})^{j}=0,

其中F_{j}(w)为常系数多项式, p\in {\mathbb N}^*.对于p阶Briot-Bouquet方程,有如下引理.

引理2.2[12, 15]  设h, ~p, ~q, ~s \in {\mathbb N}^*, ~\deg F(w, w^{(p)}) < q .若一个p阶Briot-Bouquet方程满足弱\langle h, k \rangle条件,则其所有的亚纯解w \in W, 并且引理2.1的结论成立.

3 理1.1的证明

  将(2.2)式代入方程(1.3),可得h=1, ~k=2, ~\beta_{{-2}}=\frac{2(\gamma+2)}{\theta}, ~\beta_{{-1}}=0, ~\beta_{{0}}=-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta}, ~\beta_{{1}}=0, ~\beta_{{2}}=-\frac{(\kappa\gamma-3\theta\nu)(\kappa\gamma+3\theta\nu+2\kappa)}{30(\gamma+1)^2\gamma\theta}, \beta_3=0,和\beta_4为任意常数.

下面证明方程(1.3)满足\langle h, k\rangle条件.

实际上,由于方程(1.3)满足弱\langle h, k\rangle条件,控制部分\hat{D}(z, w)=ww''+\frac{\gamma-1}{2}(w')^2-\theta w^3,可得I(k)=3k,

\begin{eqnarray} P(i) &=&\lim\limits_{z \rightarrow 0}z^{-i+I(k)}\hat{D}'(z, \beta_{{-k}}z^{-k})z^{i-k}\\& =& \lim\limits_{z \rightarrow 0}z^{-i+3k}[(\beta_{{-k}}z^{-k})\frac{d^2}{dz^2}+(\beta_{{-k}}z^{-k})''+(\gamma-1)w'\frac{d}{dz} -3\theta(\beta_{{-k}}z^{-k})^2]z^{i-k} \\& =&\beta_{{-k}}(i-k)(i-k-1)+\beta_{{-k}}(k+1)k-\beta_{{-k}}(\gamma-1)k(i-k)-3\theta (\beta_{{-k}})^2 =0.\end{eqnarray}
(3.1)

k=2, \beta_{-2}=\frac{2(\gamma+2)}{\theta}代入方程(3.1),可得i=-1, i=2\gamma+4.因此,当\gamma为一个不属于\mathbb M的常数时, i不是非负整数.这意味着,当\gamma为一个不属于\mathbb M的常数时,方程P(i)=0没有非负整数解,即方程(1.3)没有非负整数Fuchs指数.所以, w(z)的Laurent展式能够通过其主要部分唯一确定[11].又由于方程(1.3)满足弱\langle h, k \rangle条件,则可得该方程满足\langle h, k \rangle条件.由引理2.1可得,方程(1.3)的亚纯解w属于W类.

由(2.7)式,推断方程(1.3)在极点z=0处的有理函数解有如下形式

R_{1}(z)=\frac{\beta_{12}}{z^2} +\frac{\beta_{11}}{z}+\beta_{10},

R_{1}(z)代入方程(1.3),可得

R_{1}(z)=\frac{2(\gamma+2)}{\theta}\frac {1}{z^2}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中\nu=-\frac{\kappa(\gamma+2)}{3\theta}.

所以方程(1.3)的有理函数解为

W_r(z)=\frac{2(\gamma+2)}{\theta}\frac {1}{(z-z_{0})^2}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中\nu=-\frac{\kappa(\gamma+2)}{3\theta}, z_0\in {\mathbb C}.

为了得到单周期函数解,令\eta=e^{\mu z},并将w=R(\eta)代入方程(1.3),则可得

\mu^2 (R-\nu)(\eta R'+\eta^2R'')+\frac{\gamma-1}{2}(\mu R'\eta)^2 -\theta R^{3} -\kappa R^{2}=0.
(3.2)

R_{2}(z)=\frac{\beta_{22}}{(\eta-1)^2} +\frac{\beta_{21}}{(\eta-1)}+\beta_{20},

代入方程(3.2),可得

R_{2}(z)=\frac{2(\gamma+2)\mu^2}{\theta}\frac{1}{(\eta-1)^2} +\frac{2(\gamma+2)\mu^2}{\theta}\frac{1}{(\eta-1)}+\frac{(\gamma+2)\mu^2}{6\theta}-\frac{\kappa(\gamma+2) +3\theta\nu}{3(\gamma+1)\theta},
(3.3)

其中\nu=\frac{\mu^2(-\gamma^2-\gamma+\sqrt{2(-4\gamma^4-8\gamma^3+3\gamma^2+7\gamma+2)}+2)}{18\theta}, \kappa=\frac{\mu^2(-4\gamma^2-10\gamma+\sqrt{2(-4\gamma^4-8\gamma^3+3\gamma^2+7\gamma+2)}-4)}{6(2\gamma+1)}.

\eta=e^{\mu z}代入(3.3)式,可得方程(1.3)在极点z=0处的单周期函数解为

\begin{eqnarray*} W_{s0}(z) &=&\frac{2(\gamma+2)\mu^2}{\theta}\frac{1}{(e^{\mu z}-1)^2} +\frac{2(\gamma+2)\mu^2}{\theta}\frac{1}{(e^{\mu z}-1)}+\frac{(\gamma+2)\mu^2}{6\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta}\\&=& \frac{2(\gamma+2)\mu^2}{\theta}\frac{e^{\mu z}}{(e^{\mu z}-1)^2} +\frac{(\gamma+2)\mu^2}{6\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta}\\&=&\frac{(\gamma+2)\mu^2}{2\theta} \coth^2\frac{\mu z}{2} -\frac{(\gamma+2)\mu^2}{3\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta}.\end{eqnarray*}

所以方程(1.3)的单周期函数解为

W_{s}(z)=\frac{(\gamma+2)\mu^2}{2\theta} \coth^2\frac{\mu (z-z_{0})}{2} -\frac{(\gamma+2)\mu^2}{3\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中\nu=\frac{\mu^2(-\gamma^2-\gamma+\sqrt{2(-4\gamma^4-8\gamma^3+ 3\gamma^2+7\gamma+2)}+2)}{18\theta}, \kappa=\frac{\mu^2(-4\gamma^2-10\gamma+\sqrt{2(-4\gamma^4-8\gamma^3+3\gamma^2+7\gamma+2)}-4)}{6(2\gamma+1)}, z_0\in {\mathbb C}.

由引理2.1中的(2.6)式,可将方程(1.3)在极点z=0处的椭圆函数解表示为

W_{d0}(z)=\beta_{-2}\wp(z)+\beta_{0}.

W_{d0}(z)代入方程(1.3),可得

W_{d0}(z)=\frac{2(\gamma+2)}{\theta}\wp(z)-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中

c_3=\frac{(\kappa^2\gamma^3-3\kappa\theta\nu\gamma^2+9\theta^2\nu^2\gamma+\kappa^2\gamma^2-6\kappa\theta\nu\gamma+27\theta^2\nu^2 -2\kappa^2\gamma+9\kappa\theta\nu)(\kappa(\gamma+2)+3\theta\nu)}{27\gamma(\gamma-1)(\gamma+2)^2(\gamma+1)^3},

c_2=-\frac{(\kappa\gamma-3\theta\nu)(\kappa(\gamma+2)+3\theta\nu)}{3\gamma(\gamma+2)(\gamma+1)^2}.

所以方程(1.3)的椭圆函数解为

W_{d}(z)=\frac{2(\gamma+2)}{\theta}\wp(z-z_{0})-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中z_0\in {\mathbb C}.运用椭圆函数的加法公式,可将其改写为

W_{d}(z)=\frac{2(\gamma+2)}{\theta}\left(-\wp(z)+\frac{1}{4}\left(\frac{\wp'(z)+A}{\wp(z)-B}\right)^2\right)-\frac{2(\gamma+2)B}{\theta}-\frac{\kappa(\gamma+2)+3\theta\nu}{3(\gamma+1)\theta},

其中

c_3=\frac{(\kappa^2\gamma^3-3\kappa\theta\nu\gamma^2+9\theta^2\nu^2\gamma+\kappa^2\gamma^2-6\kappa\theta\nu\gamma+27\theta^2\nu^2 -2\kappa^2\gamma+9\kappa\theta\nu)(\kappa(\gamma+2)+3\theta\nu)}{27\gamma(\gamma-1)(\gamma+2)^2(\gamma+1)^3},

c_2=-\frac{(\kappa\gamma-3\theta\nu)(\kappa(\gamma+2)+3\theta\nu)}{3\gamma(\gamma+2)(\gamma+1)^2}, \ \ \ A^2=4B^3-c_2B-c_3.

\gamma=1时,方程(1.3)是二阶Briot-Bouquet微分方程,并且满足弱\langle 1, 2 \rangle条件.故由引理2.2,可知方程(1.3)的亚纯解w属于W类,则得到方程(1.3)如下形式的亚纯解.

由(2.7)式,推断方程(1.3)在极点z=0处的有理函数解有如下形式

U_{1}(z)=\frac{\beta_{32}}{z^2} +\frac{\beta_{31}}{z}+\beta_{30}.

U_{1}(z)代入方程(1.3),可得

U_{1}(z)=\frac{6}{\theta}\frac {1}{z^2}-\frac{\kappa+\theta\nu}{2\theta},

其中\nu=-\frac{\kappa}{\theta}.

所以方程(1.3)的有理函数解为

U_r(z)=\frac{6}{\theta}\frac {1}{(z-z_{0})^2}-\frac{\kappa+\theta\nu}{2\theta},

其中\nu=-\frac{\kappa}{\theta}, z_0\in {\mathbb C}.

为了得到单周期函数解,令\eta=e^{\mu z},并将w=R(\eta)代入方程(1.3),则可得

\mu^2 (R-\nu)(\eta R'+\eta^2R'') -\theta R^{3} -\kappa R^{2}=0.
(3.4)

U_{2}(z)=\frac{\beta_{42}}{(\eta-1)^2} +\frac{\beta_{41}}{(\eta-1)}+\beta_{40} 代入方程(3.4),可得

U_{2}(z)=\frac{6\mu^2}{\theta}\frac{1}{(\eta-1)^2} +\frac{6\mu^2}{\theta}\frac{1}{(\eta-1)}+\frac{\mu^2}{\theta},
(3.5)

其中\nu=0, \kappa=\mu^2.\eta=e^{\mu z}代入(3.5)式,可得方程(1.3)在极点z=0处的单周期函数解为

U_{s0}(z) =\frac{6\mu^2}{\theta}\frac{1}{(e^{\mu z}-1)^2} +\frac{6\mu^2}{\theta}\frac{1}{(e^{\mu z}-1)}+\frac{\mu^2}{\theta}=\frac{6\mu^2}{\theta}\frac{e^{\mu z}}{(e^{\mu z}-1)^2} +\frac{\mu^2}{\theta} =\\ \frac{3\mu^2}{2\theta} \coth^2\frac{\mu z}{2}-\frac{\mu^2}{2\theta}.

所以方程(1.3)的单周期函数解为

U_{s}(z)=\frac{3\mu^2}{2\theta} \coth^2\frac{\mu (z-z_{0})}{2} -\frac{\mu^2}{2\theta},

其中\nu=0, \kappa=\mu^2, z_0\in {\mathbb C}.

由引理2.1中的(2.6)式,可将方程(1.3)在极点z=0处的椭圆函数解表示为

U_{d0}(z)=\beta_{-2}\wp(z)+\beta_{0}.

U_{d0}(z)代入方程(1.3),可得

U_{d0}(z)=\frac{6}{\theta}\wp(z)-\frac{\kappa+\theta\nu}{2\theta},

其中c_2=0, c_3为任意常数.

所以方程(1.3)的椭圆函数解为

U_{d}(z)=\frac{6}{\theta}\wp(z-z_{0})-\frac{\kappa+\theta\nu}{2\theta},

其中z_0\in {\mathbb C}.运用椭圆函数的加法公式,可将其改写为

U_{d}(z)=\frac{6}{\theta}\left(-\wp(z)+\frac{1}{4}\bigg(\frac{\wp'(z)+E}{\wp(z)-F}\right)^2\bigg)-\frac{6F}{\theta}-\frac{\kappa+\theta\nu}{2\theta},

其中\nu=-\frac{\kappa}{\theta}, E^2=4F^3-c_3, c_2=0, c_3为任意常数.

4 计算机模拟

本节将运用一些计算机模拟来描述主要结果,通过以下的图 1图 2,可对有理函数解U_{r}(z)和单周期函数解U_{s}(z)的性质展开分析.

(1)对U_{r}(z),取\theta=6, \kappa=-6\nu=1.

(2)对U_{s}(z),取\mu=1, \theta=\frac{3}{2}, \kappa=1\nu=0.

5 结论

运用复方法,容易得到某些非线性偏微分方程的精确解.本文首先通过弱\langle h, k \rangle条件和Fuchs指数,证明了广义弹性杆波方程的亚纯解属于W类,然后得到该方程的亚纯精确解.结果表明,此方法较为便捷,并且可被应用于其他数学物理方程.

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