## Periodic Wave Solutions, Solitary Wave Solutions and Their Relationship for Generalized Symmetric Regularized Long Wave Equation with Two Nonlinear Terms

Ling Xingqian,1, Zhang Weiguo,2

 基金资助: 国家自然科学基金.  11471215

 Fund supported: the NSFC.  11471215

Abstract

In this paper, we study the exact solitary wave, periodic wave solutions and their evolutionary relationship with Hamilton energy for generalized symmetric regularized long wave equation. By the planar dynamical system method, we make detailed qualitative analysis on this equation. Then we use the first integral method to obtain the exact solutions for the equation. Furthermore, by establishing the corresponding relationships between the solitary wave, periodic wave solutions and Hamilton energy h, we reveal that the value of Hamilton energy h plays an important role in the appearance of solitary wave and periodic wave solutions. The evolution processes from the periodic wave solution to the solitary wave solution and the kink wave solution are also given in this paper.

Keywords： Generalized symmetric regularized long wave equation ; Planar dynamical system ; Solitary wave solution ; Periodic wave solution ; Evolutionary relationship

Ling Xingqian, Zhang Weiguo. Periodic Wave Solutions, Solitary Wave Solutions and Their Relationship for Generalized Symmetric Regularized Long Wave Equation with Two Nonlinear Terms. Acta Mathematica Scientia[J], 2021, 41(3): 603-628 doi:

## 1 前言

$\begin{eqnarray} \left\{\begin{array}{ll} { } u_{xxt}-u_t = (v-\frac{1}{2}u^2), \\ v_t+u_x = 0 \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} { } u_{xxt}-u_t = \rho_x+\frac{1}{3}(u^3)_x, \\ \rho_t+u_x = 0 \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} u_{xxt}-u_t = (\rho+b_2u^2+b_3u^3)_x, \\ b_i = {\rm constant}, \quad b_3\geq0, \quad i = 2, 3, \\ \rho_t+u_x = 0 \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} u_{xxt}-u_t = (b_1\rho+b_2 u^2+b_3 u^3)_x, \\ b_i = {\rm constant}, \quad i = 1, 2, 3, \\ \rho_t+u_x = 0 \end{array}\right. \end{eqnarray}$

$$$u(\xi) = u(x-ct), \qquad \rho(\xi) = \rho(x-ct)$$$

### 2 定性分析

$$$\left\{\begin{array}{ll} -cu'''+cu' = b_{1}\rho'+2b_{2}uu'+3b_{3}u^{2}u', \\ -c\rho'+u' = 0. \end{array}\right.$$$

$$$u''+(\frac{b_{1}}{c^{2}}-1)u+\frac{b_{2}}{c}u^{2}+\frac{b_{3}}{c}u^{3} = k,$$$

$$$U''+\frac{b_{3}}{c}(U^{3}+mU+n) = 0,$$$

$x = U(\xi)$, $y = U'(\xi)$, 则方程(2.3)等价于如下平面动力系统

$$$\left\{\begin{array}{ll} { }\frac{{\rm d}x}{{\rm d}\xi} = y = P(x, y), \\ { } \frac{{\rm d}y}{{\rm d}\xi} = -\frac{b_{3}}{c}(x^{3}+mx+n) = Q(x, y). \end{array}\right.$$$

$(x, y)$平面上, 系统(2.4)的有限远奇点个数取决于$f(x) = x^{3}+mx+n = 0$实根的个数, 记$f(x) = 0$实根的判别式为$\Delta = (\frac{n}{2})^2+(\frac{m}{3})^3,$根据代数学易知, 当$\Delta>0$时, 方程$f(x) = 0$存在一个实根和两个复根; $\Delta = 0$时, 方程$f(x) = 0$存在一个单根和一组重根；当$\Delta<0$时, 方程$f(x) = 0$存在三个互异的实根. 由于本文着重讨论方程(1.4)的孤波解、周期波解及它们随Hamilton能量的演化关系, 当方程$f(x) = 0$仅有一个根时, 方程(2.3)没有有界解. 故本文以下始终假设$m<0 $$\Delta>0 . 设系统(2.4)存在三个有限远奇点 P_{i}(x_{i}, 0)(i = 1, 2, 3) , 不妨令 x_1<x_2<x_3 , 其中 x_i$$ f(x) = 0$的实根且有

$$$\left\{\begin{array}{ll} x_1+x_2+x_3 = 0, \\ x_1x_2+x_1x_3+x_2x_3 = m, \\ x_1x_2x_3 = -n. \end{array}\right.$$$

$$$H(x, y)\triangleq\frac{y^2}{2}+\frac{b_3}{c}(x^4+2mx^2+4nx) = h,$$$

### 2.1 系统(2.4)的有限奇点

(1) 在$b_3>0$情形

a) $n>0$

(i) 当$\triangle<0$时, 方程$f(x) = 0$有三个不同的实根$x_1$, $x_2 $$x_3 , 且有 x_1<-x_3<-x_2<0<x_2<x_3 , 对应系统(2.4)有三个不同的有限远奇点 P_i(x_i, 0)(i = 1, 2, 3) , 对应Jacobi矩阵 J(x_i, 0)(i = 1, 2, 3) 的行列式分别为 因此, P_1$$ P_3$为中心, $P_2$为鞍点.

(ii) 当$\triangle = 0$时, 方程$f(x) = 0$存在1个单根$x_1$和1组重根$x_{2, 3}$, 且$x_1<0<x_{2, 3}$. 对应系统(2.4)有2个不同的有限远奇点$P_1(x_1, 0) $$P_{2, 3}(x_{2, 3}, 0) . 根据(2.5)式可知 则有对应的Jacobi矩阵 J(x_i, 0)(i = 1, 2, 3) 的行列式分别为 P_1 为中心, P_2, _3 为尖点. b) n<0 (i) 当 \triangle<0 , 方程 f(x) = 0 有3个不同的实根 x_1 , x_2$$ x_3$, 且有$x_1<x_2<0<-x_2<-x_1<x_3$, 对应系统(2.4)有三个不同的有限远奇点$P_i(x_i, 0)(i = 1, 2, 3)$. 类似a)(i), 可得Jacobi矩阵$J(x_i, 0)(i = 1, 2, 3)$的行列式分别为

$P_1 $$P_3 为中心, P_2 为鞍点. (ii) 当 \triangle = 0 时, 方程 f(x) = 0 的3个实根满足 x_2 = x_1<0<x_3 . 故对应系统(2.4)有2个不同的有限远奇点 P_1, _2(x_1, _2, 0)$$ P_3(x_3, 0)$. 类似a)(ii), 可得Jacobi矩阵$J(x_i, 0)(i = 1, 2, 3)$的行列式分别为

$P_1, _2$为尖点, $P_3$为中心.

c) 当$n = 0$, $\triangle <0$时, 方程$f(x) = 0$有三个实根$-x_1 = x_3 = \sqrt{-m} $$x_2 = 0 . 因此, 对应系统(2.4)有2个不同的有限远奇点 P_1, _3$$ P_2$, Jacobi矩阵的行列式为

$P_1 $$P_3 为中心, P_2 为鞍点. (2) 在 b_3<0 情况下 方程 f(x) = 0 有三个实根 x_1 , x_2$$ x_3$, 对应系统存在3个有限远奇点$P_i(d_i, 0) $$(i = 1, 2, 3) . 类似于对(1)的讨论, 我们可以得到以下结论. a) n>0 (i) 当 \triangle<0 时, 实根 x_1 , x_2$$ x_3$互不相同且满足$x_1<-x_3<-x_2<0<x_2<x_3$, 此时对应Jacobi矩阵$J(x_i, 0) $$(i = 1, 2, 3) 的行列式满足 {\rm det}J(x_1, 0)<{\rm det}J(x_{2, 3}, 0) = 0 . P_1$$ P_3$为鞍点, $P_2$是中心.

(ii) 当$\triangle = 0$时, 实根$x_2 $$x_3 重合, 且有 x_1<0<x_2, _3 , 此时对应Jacobi矩阵 J(x_i, 0)\ (i = 1, 2, 3) 的行列式满足 {\rm det}J(x_1, 0)<{\rm det}J(x_{2, 3}, 0) = 0 . P_1 是鞍点, P_2, _3 是尖点. b) n<0 (i) 当 \triangle<0 时, 实根 x_1 , x_2$$ x_3$互不相同且满足$x_1<x_2<0<-x_2<x_1<x_3$, 此时对应Jacobi矩阵$J(x_i, 0)\ (i = 1, 2, 3)$的行列式满足${\rm det}J(x_3, 0)<{\rm det}J(x_1, 0)<0<{\rm det}J(x_2, 0)$.$P_1 $$P_3 是鞍点, P_2 是中心. (ii) 当 \triangle = 0 时, 实根 x_1$$ x_2$重合, 且有$x_{1, 2}<0<x_{3}$, 此时对应Jacobi矩阵$J(x_i, 0)(i = 1, 2, 3)$的行列式满足${\rm det}J(x_3, 0)<{\rm det}J(x_{1, 2}, 0) = 0$.$P_{1, 2}$是尖点, $P_3$是鞍点.

c) 当$n = 0$, $\triangle<0$时, 方程$f(x) = 0$有三个互不相同的实根$-x_1 = x_3 = \sqrt{-m} $$x_2 = 0 . 故我们可以得到Jacobi矩阵的行列式为 P_1$$ P_3$为鞍点, $P_2$是中心.

### 2.2 系统(2.4)的全局相图和方程(1.4)有界孤波解的存在性

$b_3>0$情形, 根据系统(2.4)的全局相图图 2.1图 2.5, 我们可以得到以下性质和定理.

### 图 2.5

$\rm(1) $$\Delta<0 时, 系统(2.4)在 n>0 , n = 0$$ n<0$情形下分别存在两条同宿轨和无数条闭轨(见图 2.1, 图 2.2图 2.3).

$\rm(2) $$\Delta = 0 时, 系统(2.4)在 n>0$$ n<0$情形分别存在一条同宿轨和无数条闭轨(见图 2.4图 2.5).

$b_3<0$情形下, 类似, 由全局相图图 2.6图 2.10, 我们可以得到以下性质和结论.

### 图 2.10

$\rm(1) $$\Delta<0 时, 系统(2.4)在 n = 0 , n>0$$ n<0$情形下分别存在一条同宿轨和无数条闭轨(见图 2.7图 2.8), 在$n = 0$情形下存在两条异宿轨和无数条闭轨(见图 2.6).

$\rm(2) $$\Delta = 0 时, 系统(2.4)在 n>0$$ n<0$情形下均不存在有界轨线(见图 2.9图 2.10).

$\rm(1) $$\Delta<0 时, 方程(1.4)在 n = 0 , n>0 , n<0 情形下均存在两个钟状孤波解和和无数个周期行波解(分别对应于图 2.1, 图 2.2图 2.3中的同宿轨和闭轨). \rm(2)$$ \Delta = 0$时, 方程(1.4)在$n>0$, $n<0$情形下均存在一个钟状孤波解和无数个周期行波解(分别对应于图 2.4图 2.5中的同宿轨和闭轨).

$\rm(1) $$\Delta<0 时, 方程(1.4)在 n>0 , n<0 情形下均存在一个钟状孤波解和和无数个周期行波解(分别对应于图 2.7图 2.8中的同宿轨和闭轨). 在 n = 0 情形下存在两个扭状孤波解和无数个周期行波解(对应于图 2.6的异宿轨和闭轨). \rm(2)$$ \Delta = 0$时, 方程(1.4)在$n>0$, $n<0$情形下均不存在有界行波解.

### 3 方程(1.4)孤波解的求解

$$$y^2 = -\frac{b_{3}}{2c}(x^4+2mx^2+4nx)+2h = -\frac{b_{3}}{2c}(x^4+2mx^2+4nx-\frac{4hc}{b_{3}}),$$$

$$$\int\frac{{\rm d}x}{\sqrt{\pm F_h(x)}} = \pm\sqrt{\frac{|b_{3}|}{2c}}(\xi-\xi_0),$$$

### 3.1 图 2.1–图 2.3对应的钟状孤波解

$$$-F_h(x) = -(x-x_2)^2(x^2+2x_2x+3x^2_2+2m) = -(x-x_2)^2(x-r_1)(x-r_2),$$$

$$$r_1 = -x_2-\sqrt{-2m-2x^2_2}, \quad r_2 = -x_2+\sqrt{-2m-2x^2_2}.$$$

$$$\pm\sqrt{\frac{b_3}{2c}}(\xi-\xi_0) = \int\frac{{\rm d}x}{\sqrt{-(x-x_2)^2(x-r_1)(x-r_2)}} = \int\sqrt{\frac{r_2-x}{x-r_1}}\frac{{\rm d}x}{(x-x_2)(r_2-x)}.$$$

$x\in(r_1, x_2)\cup(x_2, r_2)$, 令$t = \sqrt{\frac{r_2-x}{x-r_1}}$, 即有$x = r_1+\frac{r_2-r_1}{t^2+1}$, ${\rm d}x = \frac{-2(r_2-r_1)^2}{(t^2+1)^2}{\rm d}t$. 将它们代入(3.5)式, 有

$\begin{eqnarray} \pm\sqrt{\frac{|b_3|}{2c}}(\xi-\xi_0)& = &-2\int\frac{{\rm d}x}{(r_2-x_2)-(x_2-r_1)t^2}{}\\ && = -\frac{1}{\sqrt{(x_2-r_1)(r_2-x_2)}}\left(\ln\left|t+\sqrt{\frac{r_2-x_2}{x_2-r_1}}\right|-\ln\left|t+\sqrt{\frac{r_2-x_2}{x_2-r_1}}\right|\right). {\qquad} \end{eqnarray}$

$$$t = \mp\sqrt{\frac{r_2-x_2}{x_2-r_1}}\cdot\frac{e^{\sqrt{\frac{b_3}{2c}(r_2-x_2)(x_2-r_1)}(\xi-\xi_0)}+1}{e^{\sqrt{\frac{b_3}{2c}(r_2-x_2)(x_2-r_1)}(\xi-\xi_0)}-1},$$$

$x\in(x_2, r_2)$时, 有

$$$t = \mp\sqrt{\frac{r_2-x_2}{x_2-r_1}}\cdot\frac{e^{\sqrt{\frac{b_3}{2c}(r_2-x_2)(x_2-r_1)}(\xi-\xi_0)}-1}{e^{\sqrt{\frac{b_3}{2c}(r_2-x_2)(x_2-r_1)}(\xi-\xi_0)}+1}.$$$

$$$x(\xi) = x_2-\frac{2(x_2-r_2)(x_2-r_1)}{4x_2\pm(r_2-r_1)\cosh\sqrt{\frac{b_3}{2c}(x_2-r_1)(r_2-x_2)}(\xi-\xi_0)}.$$$

$$$U^\pm_{S1}(\xi) = x_2-\frac{2(m+3x^2_2)}{2x_2\pm\sqrt{-2m-2x^2_2}\cosh\sqrt{-\frac{b_3}{c}(m+3x^2_2)}(\xi-\xi_0)}.$$$

$$$\left\{\begin{array}{ll} { } u^\pm_{S1}(x, t) = U^\pm_{S1}(\xi)-\frac{b_2}{3b_3}, \\ { } \rho^\pm_{S1}(x, t) = \frac{u^\pm_{S1}(\xi)}{c}, \end{array}\right.$$$

(3.11)式中$U^\pm_{S1}$由(3.10)式给定, 对应于全局相图图 2.2图 2.3中同宿轨.

$$$\widetilde{U}^\pm_{S1}(\xi) = \pm\sqrt{-2m}\cdot {\rm sech}\sqrt{\frac{-b_3m}{c}}(\xi-\xi_0),$$$

$$$\left\{\begin{array}{ll} { } \widetilde{u}^\pm_{S1}(x, t) = \widetilde{U}^\pm_{S1}(\xi)-\frac{b_2}{3b_3}, \\ { }\widetilde{\rho}^\pm_{S1}(x, t) = \frac{\widetilde{u}^\pm_{S1}(\xi)}{c}, \end{array}\right.$$$

(3.13)式中的$\widetilde{U}^\pm_{S1}(\xi)$由(3.12)式给出, 对应于相图 2.1中的同宿轨.

### 3.2 图 2.7、图 2.8同宿轨线对应的孤波解

$$$F_h(x) = (x-x_3)^2(x-\eta_1)(x-\eta_2),$$$

$$$\left\{\begin{array}{ll} { } u^+_{S4}(x, t) = U^+_{S4}(\xi)-\frac{b_2}{3b_3}, \\ { }\rho^+_{S4}(x, t) = \frac{u^+_{S4}(\xi)}{c}, \end{array}\right.$$$

(3.28)中$U^+_{S4}$由(3.24)式给出, 对应相图 2.4中同宿轨.

$\rm(2) $$n<0 时, \rm{Hamilton} 能量为 h = H(x_1, 0) , 方程(1.4)有孤波解 $$\left\{\begin{array}{ll} { } u^-_{S4}(x, t) = U^-_{S4}(\xi)-\frac{b_2}{3b_3}, \\ { } \rho^-_{S4}(x, t) = \frac{u^-_{S4}(\xi)}{c}, \end{array}\right.$$ (3.29)式中 U^-_{S4} 由(3.27)式给出, 对应相图 2.5中同宿轨. ### 3.4 相图 2.6异宿轨线对应的扭状孤波解 此时 b_3<0 , n = 0 , 由于 r_1$$ f(x) = x^3+mx = x(x^2+m)$的根, 有$x^2_1 = -m$, $h = H(x_1, 0) = \frac{b_3}{4c}(x^4_1+2mx^2_1) = -\frac{b_3}{4c}m^2$, 此时有

$$$F_h(x) = x^4+2mx^2-\frac{4c}{b_3}h = x^4+2mx^2+m^2 = (x+\sqrt{-m})^2(x-\sqrt{-m})^2.$$$

$$$\pm\sqrt{\frac{-b_3}{2c}}(\xi-\xi_0) = \int\frac{{\rm d}x}{\sqrt{(x+\sqrt{-m})^2(x-\sqrt{-m})^2}}.$$$

$$$\pm\sqrt{-\frac{b_3}{2c}}(\xi-\xi_0) = \int\frac{{\rm d}x}{(x+\sqrt{-m})(\sqrt{-m}-x)} = \frac{-1}{2\sqrt{-m}}\ln{\frac{\sqrt{-m}-x}{\sqrt{-m}+x}}.$$$

$$$U^\pm_{S5}(\xi) = x^\pm_{S5}(\xi) = \pm\sqrt{-m}\tanh\left(\sqrt{\frac{b_3 m}{2c}}(\xi-\xi_0)\right),$$$

$$$\left\{\begin{array}{ll} { } u^\pm_{S5}(x, t) = U^\pm_{S5}(\xi)-\frac{b_2}{3b_3}, \\ { }\rho^\pm_{S5}(x, t) = \frac{u^\pm_{S5}(x, t)}{c}. \end{array}\right.$$$

### 4.1 图 2.1和图 2.6中闭轨对应的精确周期波解

(1) 当$b_3>0$, $n = 0$, $\Delta<0$, 对应于图 2.1. 系统(2.4)在同宿轨线$L_1 $$L_2 所围区域内部的同一闭轨上有相等的Hamilton能量, 记为 其中 (\eta_1, 0) 为周期轨线与 x 轴的交点, 并且 \mid\eta_1\mid<\sqrt{-m} . 此时 f(x) = x(x^2+m) = 0 的三个解分别为 x_1 = -\sqrt{-m} , x_2 = 0$$ x_3 = \sqrt{-m}$, 可得$H(x_1, 0) = H(x_3, 0)<H(x_2, 0)$.

### 图 4.1

$$$-F_{h_1}(x) = -(x^4+2mx^2-\frac{4ch_1}{b_3}) = m^2+\frac{4ch_1}{b_3}-(x^2+m)^2.$$$

$\begin{eqnarray} \pm\sqrt{\frac{b_3}{2c}}{\rm d}\xi& = &\frac{{\rm d}x}{\sqrt{m^2+\frac{4ch_1}{b_3}-(x^2+m)^2}} {}\\ & = &\frac{{\rm d}x}{\sqrt{-\left(x^2+m+\sqrt{m^2+\frac{4ch_1}{b_3}}\right)\left(x^2+m-\sqrt{m^2+\frac{4ch_1}{b_3}}\right)}}. \end{eqnarray}$

$A^2_1 = \sqrt{m^2+\frac{4ch_1}{b_3}}-m, \quad B^2_1 = -\sqrt{m^2+\frac{4ch_1}{b_3}}-m$, 作变换$t = \frac{x}{A_1} = \frac{U(\xi)}{A_1}$, 则(4.2)式可转化为

$$$\pm\sqrt{\frac{b_3}{2c}}{\rm d}\xi = \frac{{\rm d}x}{\sqrt{-(x^2-A^2_1)(x^2-B^2_1)}} = \frac{{\rm d}t}{A_1\sqrt{(1-t^2)\cdot\left(t^2-\left(\frac{B_1}{A_1}\right)^2\right)}}.$$$

$$$U^\pm_{P1}(\xi) = \pm\sqrt{\frac{-2m}{2-k^2_1}}\cdot dn\left(\sqrt{\frac{-b_3 m}{c(2-k^2_1)}}(\xi-\xi_0), k_1\right),$$$

### 图 4.2

$$$F_{h2}(x) = x^4+2mx^2-\frac{4ch_2}{b_3} = (x+m)^2-\left(m^2+\frac{4ch_2}{b_3}\right).$$$

$A^2_2 = \sqrt{m^2+\frac{4ch_2}{b_3}}-m$, $B^2_2 = -\sqrt{m^2+\frac{4ch_2}{b_3}}-m$, 作变换$t = \frac{x}{A_2} = \frac{U(\xi)}{A_2}$, 则(3.2)式可转化为

$$$\pm\sqrt{-\frac{b_3}{2c}}{\rm d}\xi = \frac{{\rm d}t}{B_2\sqrt{(1-t^2)\cdot\left(1-\left(\frac{A_2}{B_2}\right)^2 t^2\right)}}.$$$

$$$\left\{\begin{array}{ll} { } u^\pm_{P1}(\xi) = U^\pm_{P1}(\xi)-\frac{b_2}{3b_3}, \\ { }\rho^\pm_{P1}(\xi) = \frac{u^\pm_{P1}(\xi)}{c}, \end{array}\right.$$$

$\rm(2) $$b_3<0 时, 方程(1.4)有周期波解 $$\left\{\begin{array}{ll} { } u^\pm_{P2}(\xi) = U^\pm_{P2}(\xi)-\frac{b_2}{3b_3}, \\ { } \rho^\pm_{P2}(\xi) = \frac{u^\pm_{P2}(\xi)}{c}, \end{array}\right.$$ 其中 U^\pm_{P2} 图 2.6中对称异宿轨所围闭轨对应的方程(2.3)的周期解, 由(4.7)式给出. ### 4.2 图 2.2–2.3周期轨线对应的精确周期波解 b_3>0 , n\neq0 时, \triangle<0 , 由于在同宿轨线内包围的同一周期轨线上点有相等的Hamilton能量, 有 $$H(x, y) = H(\eta_3, 0) = h_3,$$ 其中 (\eta_3, 0) 为周期轨线与 x 轴的交点, H(x, y) 由(2.6)式给定. 据图 2.22.3可知 P(d_i, 0)$$ (i = 1, 2, 3)$为系统(2.4)的3个奇点, 且横坐标满足$x_1<x_2<x_3$.

$b_3>0$情况, 易证$n>0$, $H(x_1, 0)<H(x_3, 0)<H(x_2, 0)$; $n<0$, $H(x_3, 0)<H(x_1, 0)<H(x_2, 0)$. 当Hamilton能量$h_3$满足

$$$\max{\{H(x_1, 0), H(x_3, 0)\}}<h_3<H(x_2, 0),$$$

$$$F_h(x) = (x-d_1)(x-d_2)(x-d_3)(x-d_4).$$$

(i) 当$d_1<U<d_2$, 作变换

$$$U_{P3}(\xi) = \frac{d_4(d_2-d_1)sn^2\left(\frac{1}{2}\sqrt{\frac{b_3}{2c}(d_4-d_2)(d_3-d_1)}(\xi-\xi_0), k_3\right)+d_1(d_4-d_2)} {(d_2-d_1)sn^2\left(\frac{1}{2}\sqrt{\frac{b_3}{2c}(d_4-d_2)(d_3-d_1)}(\xi-\xi_0), k_3\right)+(d_4-d_2)}.$$$

(ii) 当$d_3<U<d_4$时, 作变换

$$$U_{P4}(\xi) = \frac{d_1(d_4-d_3)sn^2\left(\frac{1}{2}\sqrt{\frac{b_3}{2c}(d_4-d_2)(d_3-d_1)}(\xi-\xi_0), k_3\right)+d_4(d_3-d_1)} {(d_4-d_3)sn^2\left(\frac{1}{2}\sqrt{\frac{b_3}{2c}(d_4-d_2)(d_3-d_1)}(\xi-\xi_0), k_3\right)+(d_3-d_1)}.$$$

$$$\left\{\begin{array}{ll} { } u_{P3}(\xi) = U_{P3}(\xi)-\frac{b_2}{3b_3}, \\ { }\rho_{P3}(\xi) = \frac{u_{P3}(\xi)}{c} \end{array}\right.$$$

$$$\left\{\begin{array}{ll} { } u_{P4}(\xi) = U_{P4}(\xi)-\frac{b_2}{3b_3}, \\ { }\rho_{P4}(\xi) = \frac{u_{P4}(\xi)}{c}, \end{array}\right.$$$

$$$\left\{\begin{array}{ll} { } u_{P5}(\xi) = U_{P5}(\xi)-\frac{b_2}{3b_3}, \\ { } \rho(\xi) = \frac{u_{P5}(\xi)}{c}, \end{array}\right.$$$

$\rm(2) $$n>0 , 方程(1.4)有周期波解 $$\left\{\begin{array}{ll} { } u_{P6}(\xi) = U_{P6}(\xi)-\frac{b_2}{3b_3}, \\ { } \rho_{P6}(\xi) = \frac{u_{P6}(\xi)}{c}, \end{array}\right.$$ 其中 U_{P6}(\xi)\in(x_2, x_3) 为方程(2.3)的有界解, 由(4.20)式给出. ### 4.4 图 2.1–图 2.5其余周期轨线对应的周期波解 b_3>0 时, 在图 2.1图 2.5上还存在: (i)包围同宿轨线 L_i\;(i = 1, \cdots , 8) 的周期轨线; (ii) L_7$$ L_8$包围着的周期轨线; (iii)$L_3 $$L_6 内部位于中心点附近包围中心的周期轨线. 这些周期轨线可归为一类, 其特点是与 x 轴的交点仅有两个. b_3>0 , \Delta<0 , n<0 的情况为例( b_3<0 的其他情况类似). 如图 4.4所示, 当Hamilton能量 h_5 的取值范围为: h_5>H(x_2, 0)$$ h_5$值介于$H(x_1, 0) $$H(x_3, 0) 之间时, (3.2)式取函数 $$-F_h(x) = -x^4-2mx^2-4nx+\frac{4ch_5}{b_3},$$ ### 图 4.4 图 4.4 b_3>0, n<0, \Delta<0 时, h_5 对应 -F_h(x) 函数曲线和周期轨线示意图 -F_h(x)$$ x$轴有2个不同的交点$\gamma_1<\gamma_2$. 此时(4.23)式可表示为

$$$-F_h(x) = (\gamma_1-x)(x-\gamma_2)((x-\alpha)^2+\beta^2),$$$

$$$(x-\alpha)^2+\beta = \varepsilon_1(x-\sigma_1)^2+(1-\varepsilon_1)(x-\sigma_2)^2,$$$

$$$(\gamma_1-x)(x-\gamma_2) = \varepsilon_2(x-\sigma_1)^2-(1+\varepsilon_2)(x-\sigma_2)^2,$$$

$$$(\gamma_1-\gamma_2)^2\lambda^2-4(\beta^2+(\alpha-\gamma_1)(\alpha-\gamma_2))\lambda-4\beta^2 = 0$$$

$$$\widetilde{U}^\pm_{P7}(\xi) = \pm\sqrt{\frac{-2m\widetilde{k^2_1}}{2\widetilde{k^2_1}-1}}\cdot cn\left(\sqrt{\frac{-b_3 m}{c(2\widetilde{k^2_1}-1)}}(\xi-\xi_0), \widetilde{k_1}\right),$$$

(2) 对比(4.23)式和(4.24)式, 可得$2\alpha+\gamma_1+\gamma_2 = 0$, $2m = \beta^2-3\alpha^2+\gamma_1\gamma_2$以及$4n = 2\alpha(\alpha^2+\beta^2-\gamma_1\gamma_2)$.

$\beta^2+(\alpha-\gamma_1)(\alpha-\gamma_2)<0$, $k_5^2>\frac{1}{2}$时, 有$\beta^2+(\alpha-\gamma_1)(\alpha-\gamma_2) = \beta^2+3\alpha^2+\gamma_1\gamma_2 = 2(m+3\alpha^2)<0$, 即$\left(\frac{\gamma_1+\gamma_2}{2}\right)<\frac{-m}{3}$, 且$\gamma_2<x_1 = \frac{-d-\sqrt{-3d^2-4m}}{2}<x_2 = d<x_3 = \frac{-d+\sqrt{-3d^2-4m}}{2}<\gamma_1$. 此时(4.28)式对应于系统(2.4)的包围同宿轨线$L_i(i = 1, 2, \cdots , 8)$的周期轨线.

$\beta^2+(\alpha-\gamma_1)(\alpha-\gamma_2)>0$, $k_5^2<\frac{1}{2}$时, 得$\left(\frac{\gamma_1+\gamma_2}{2}\right)>\frac{-m}{3}$, 且$d_1 = d<d_2 = \frac{-d-\sqrt{-3d^2-4m}}{2}<\gamma_2<\gamma_1<x_3 = \frac{-d+\sqrt{-3d^2-4l}}{2}<\gamma_1$. 此时, (4.28)式对应于系统(2.4)的$L_7$, $L_8$包围着的周期轨线, 以及$L_3 $$L_6 内部位于中心点附近包围中心的周期轨线. 综上, 我们可以得到如下定理. 定理4.4 若条件(1): b_3>0 , \Delta<0 , \rm{Hamilton} 能量 h>H(x_2, 0)$$ h\in(H(x_3, 0), $$H(x_1, 0)) 成立, 或条件(2): b_3>0 , m<0 , \Delta = 0 , 且 \rm{Hamilton} 能量 h\neq H(x_2, 0) 成立, 则方程(2.3)有周期波解 U^\pm_{P7} , 由(4.28)式给定, 从而方程(1.4)有周期波解 $$\left\{\begin{array}{ll} { } u^\pm_{P7}(\xi) = U^\pm_{P7}(\xi)-\frac{b_2}{3b_3}, \\ { } \rho^\pm_{P7}(\xi) = \frac{u^\pm_{P7}(\xi)}{c}. \end{array}\right.$$ ### 5 方程(1.4)孤波解和周期波解的演化关系 文中第3节已求出了方程(1.4)的两种形式的钟状孤波解, 第4节又求出了方程(1.4)在不同参数条件下的精确周期波解. 本节我们将研究方程(1.4)的孤波解与周期波解的演化关系, 揭示方程(1.4)对应的Hamilton系统的能量变化对解波形的影响. 本节我们假定 \Delta<0 . ### 5.1 方程(1.4)的周期波解、孤波解与Hamilton能量取值的关系 因为平面动力系统(2.4)是Hamilton系统, 在同一闭轨上点的Hamilton能量相等. 在参数不变 b_3\neq0, m, n 条件下, 对任一实数 h , 全局相图中具有能量 C_h = \{(x, y)\in {{\Bbb R}} \times {{\Bbb R}} |H(x, y) = h\} 与(3.2) 式的 F_{h}(x) ( -F_{h}(x) )函数曲线一一对应. 下面我们以全局相图 2.2为例, 讨论(1.4)式的行波解与其对应的Hamilton能量取值的关系. 此时 b_3>0 , 我们取 -F_h(x) = -x^4-2mx^2-4nx+\frac{4ch}{b_3} . 下面我们分3种情况进行讨论, 分别是: a) h = H(x_2, 0) 时同宿轨线; b) h>H(x_2, 0) 时包围同宿轨线的周期轨线; c) h<H(x_2, 0) 时包含于同宿轨线的周期轨线. 据在第3节和第4节中讨论的孤波解, 周期波解与全局相图中同宿轨线和周期轨线的对应关系, 可以推得, 当 b_3>0 , \Delta<0 , n<0 时, Hamilton能量 h 与对应的孤波解、周期波解有如下关系: (1) 当 h\leq H(x_3, 0) 时, 系统(2.4)没有出现有界轨线, 即方程(1.4)无有界行波解. (2) 当 h\in(H(x_3, 0), H(x_1, 0)] 时, 系统(2.4)有以 P_3 为中心且包含于同宿轨线内的周期轨线, 此时方程(1.4)存在对应的周期波解 (u^+_{P7}(\xi), v^+_{P7}(\xi)) , 由(4.32)式给出, 此时模数 k_5^2<\frac{1}{2} . (3) 当 h\in(H(x_1, 0), H(x_2, 0)) 时, 系统(2.4)有以 P_1$$ P_3$和为中心且包含于同宿轨线内的周期轨线. 此时方程(1.4) 存在对应的周期波解$(u_{P3}(\xi), v_{P3}(\xi)) $$(u_{P4}(\xi), v_{P4}(\xi)) , 分别由(4.15) 式和(4.16) 式给出. (4) 当 h = H(x_2, 0) 时, 系统(2.4)出现了两个同宿轨线, 方程(1.4)存在对应的两个钟状孤波解 (u^\pm_{S1}(\xi), v^\pm_{S1}(\xi)) , 由(3.11)式给出. (5) 当 h>H(x_2, 0) 时, 系统(2.4)出现了包围同宿轨线 L_5$$ L_6$的周期轨线. 此时方程(1.4)存在对应的周期波解$(u^+_{P7}(\xi), v^+_{P7}(\xi))$, 由(4.32)式给出, 此时模数$k_5^2>\frac{1}{2}$.

#### 5.2.1 相图 2.1对称同宿轨内周波解与孤波解的演变关系

$b_3>0$, $\Delta<0$, $n = 0$时, 全局相图 2.1有对称同宿轨$L_1 $$L_2 , L_1$$ L_2$对应的方程的孤波解$(\widetilde{u}^\pm_{S1}(\xi), \widetilde{v}^\pm_{S1}(\xi))$在(3.13)式中已给出. 由定理4.1知, 方程(1.4)的周期波解$(u^\pm_{P1}(\xi), v^\pm_{P1}(\xi)) $$U^\pm_{P1}(\xi) 对应的Hamilton能量 h\in\left(-\frac{b_3m^2}{4c}, 0\right) , k^2_1 = \frac{2\sqrt{b_3 m^2+4ch}}{\sqrt{b_3m^2+4ch}-\sqrt{b_3}m}$$ (m<0)$, 而孤波解中$U^\pm_{P1}(\xi)$对应的Hamilton能量此时为$H(x_2, 0) = 0$, 当$h\rightarrow 0^{-}$时, $k^2_1 = \frac{2\sqrt{b_3 m^2}}{\sqrt{b_3m^2}-\sqrt{b_3}m}\rightarrow1(m<0)$, 所以方程(1.4)的周期波解$(u^\pm_{P1}(\xi), v^\pm_{P1}(\xi))$在Hamilton能量$h\rightarrow 0^{-}$时有

$\begin{eqnarray} \lim\limits_{h\rightarrow0^-}U^\pm_{P1}(\xi) & = &\lim\limits_{k_1\rightarrow1}\pm\sqrt{\frac{-2m}{2-k^2_1}}dn(\sqrt{\frac{-b_3m}{c(2-k^2_1)}}\xi, k_1) {}\\ & = &\pm\sqrt{-2m}{\rm sech}\sqrt{\frac{-b_3m}{c}}\xi = \widetilde{U}^\pm_{S1}(\xi), \end{eqnarray}$

$$$\lim\limits_{h\rightarrow0^-}(u^\pm_{P1}(\xi), v^\pm_{P1}(\xi)) = (\widetilde{u}^\pm_{S1}(\xi), \widetilde{v}^\pm_{S1}(\xi)).$$$

$L_9 $$L_{10} 包围的闭轨对应的周期波解由定理4.1结论(2)给出. 当 b_3<0 , \Delta<0 , n = 0 时, 由定理4.1结论(2)知, 方程(1.4)的周期波解 (u^\pm_{P2}(\xi), v^\pm_{P2}(\xi)) 对应的Hamilton能量 h\in\left(0, -\frac{b_3m^2}{4c}\right) , k^2_2 = \frac{b_3m+\sqrt{4cb_3h_3+3b^2_3 m^2}}{b_3m-\sqrt{4cb_3h_3+3b^2_3 m^2}} , 则当 h\rightarrow -\frac{b_3 m^2}{4c} 时, k^2_2 = \frac{b_3l+\sqrt{12b_3h_3+3b^2_3l^2}}{b_3l-\sqrt{12b_3h_3+3b^2_3l^2}}\rightarrow1 , 所以方程(1.4)的周期波解 (u^\pm_{P2}(\xi), v^\pm_{P2}(\xi)) 在Hamilton能量 h\rightarrow -\frac{b_3 m^2}{4c} , 有 \begin{eqnarray} \lim\limits_{h\rightarrow {-\frac{b_3m^2}{4c}}}U^\pm_{P2}(\xi) & = &\lim\limits_{k_2\rightarrow1}\pm\sqrt{\frac{-2k_2^2m}{1+k^2_2}}sn(\sqrt{\frac{b_3m}{c(1+k^2_2)}}\xi, k_2) {}\\ & = &\pm\sqrt{-m}\tan h(\sqrt{\frac{b_3m}{2c}}\xi) = U^\pm_{S5}(\xi), \end{eqnarray} $$\lim\limits_{h\rightarrow -\frac{b_3m^2}{4c}}(u^\pm_{P2}(\xi), v^\pm_{P2}(\xi)) = (u^\pm_{S5}(\xi), v^\pm_{S5}(\xi)).$$ 即当Hamilton能量 h\rightarrow -\frac{b_3 m^2}{4c} , 方程(1.4)的周期波解 (u^\pm_{P2}(\xi), v^\pm_{P2}(\xi)) 演变为扭状孤波解 (u^\pm_{S5}(\xi), v^\pm_{S5}(\xi)) . #### 5.2.3 非对称同宿轨内周期波解和孤波解的演变关系 这里仅讨论图 2.3图 2.8两种情形, 其余情形类似. (1) 在相图 2.3情形 此时 b_3>0 , \Delta<0 , n<0 . 当Hamilton能量 h = H(x_2, 0) , 有同宿轨线 L_5$$ L_6$, 它们对应的方程(1.4)的钟状孤波解$(u^\pm_{S1}(\xi), v^\pm_{S1}(\xi))$由(3.11)式给出. 当Hamilton能量$h\in(H(x_1, 0), H(x_2, 0))$, 系统(2.4)存在同宿轨线$L_5 $$L_6 包围着的周期轨线, 相应的周期波解 (u_{P3}(\xi), v_{P3}(\xi))$$ (u_{P4}(\xi), v_{P4}(\xi))$分别由(4.15) 式和(4.16)式给出.

$$$x_1 = \frac{-x_2-\sqrt{-3x_2^2-4m}}{2}, \qquad x_2, \qquad x_3 = \frac{-x_2+\sqrt{-3x_2^2-4m}}{2}.$$$

(2) 在相图 2.8情形

$h\rightarrow H^+(x_2, 0)$时, 有

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