## The Integrability of the KdV-Shallow Water Waves Equation

Hao Xiaohong,1, Cheng Zhilong,2

 基金资助: 安徽省自然科学研究项目.  KJ2016A071

Received: 2017-04-10

 Fund supported: the Natural Science Foundation of Anhui Province.  KJ2016A071

Abstract

In this paper, the binary Bell polynomials to construct bilinear forma, bilinear Bäcklund transformation, Lax pair of the KdV-shallow water waves equation. Through bilinear Bäcklund transformation, some soliton solutions are presented. Moreover, the infinite conservation laws are also derived by Bell polynomials, all conserved densities and fluxes are given with explicit recursion formulas.

Keywords： Bäcklund transformation ; Lax pair ; Infinite conservation laws

Hao Xiaohong, Cheng Zhilong. The Integrability of the KdV-Shallow Water Waves Equation. Acta Mathematica Scientia[J], 2019, 39(3): 451-460 doi:

## 1 引言

1996年, Lambert, Gilson和Nimmo建立了Bell多项式与Hirota双线性算子之间的关系[14],通过转换关系得到双线性Bäcklund变换,这个方法很有效地避免了求Bäcklund变换过程中使用交换公式繁琐的计算,简洁实用.并且直接对其做线性化还可以得到方程的Lax对[9, 12],最后将级数展开式代入计算求解得到方程的无穷守恒律.

$$$r_{t}-\alpha r_{xxt}-4\alpha r r_{t}-2 \alpha r_{x}\int_x^{\infty}r_{t}{\rm d}x+ \alpha r_x+\beta r_{xxx}+6\beta rr_x=0,$$$

$\begin{array}{l}&&{\cal Y}_{x}(v)=v_{x}, {\cal Y}_{2x}(v, w)=w_{2x}+v_{x}^{2}, {\cal Y}_{x, t}(v, w)=w_{xt}+v_{x}v_{t}, \\&&{\cal Y}_{2x, t}(v, w)=v_{2x, t}+w_{2x}v_{t}+2w_{xt}v_x+v_x^2v_t, \\&&{\cal Y}_{3x}(v, w)=v_{3x}+3v_{x}w_{2x}+v_{x}^{3}, \cdots .\end{array}$

$$${\cal Y}_{n_{1}x_{1}, \cdots, n_{l}x_{l}}(v=\ln F/G, w=\lnFG)=(FG)^{-1}D_{x_{1}}^{n_{1}}\cdots D_{x_{l}}^{n_{l}}F\cdot G,$$$

$\begin{array}{ll}(F)^{-2}D_{x_{1}}^{n_{1}}\cdots D_{x_{l}}^{n_{l}}F\cdot F = {\cal Y}_{n_{1}x_{1}, \cdots, n_{l}x_{l}}(0, q=2\ln F)\\ = \left\{\begin{array}{ll} 0, &n_{1}+n_{2}+ \cdots+n_{l} \ \mbox{为奇数, } \\ P_{n_{1}x_{1}, \cdots, n_{l}x_{l}}(q), & n_{1}+n_{2}+ \cdots+n_{l}\ \mbox{为偶数.} \end{array}\right.\end{array}$

$\begin{array}{l}&&P_{2x}(q)=q_{2x}, \qquad P_{x, t}(q)=q_{xt}, \\&&P_{4x}(q)=q_{4x}+3q_{2x}^{2}, \qquadP_{6x}(q)=q_{6x}+15q_{2x}q_{4x}+15q_{2x}^3 \cdots .\end{array}$

$\begin{array}{l} (FG)^{-1}D_{x_{1}}^{n_{1}}\cdots D_{x_{l}}^{n_{l}}F\cdot G\\ = {\cal Y}_{n_{1}x_{1}, \cdots, n_{l}x_{l}}(v, w)|_{v=\ln F/G, w=\ln FG} ={\cal Y}_{n_{1}x_{1}, \cdots, n_{l}x_{l}}(v, v+q)|_{v=\ln F/G, q=2\ln G}\\ = \sum\limits_{n_{1}+\cdots+n_{l}=even}^{}\sum\limits_{r_{1} =0}^{n_{1}}\cdots\sum\limits_{r_{l}=0}^{n_{l}}\prod\limits_{i=1}^{l}(n_{i}, r_{i})' P_{r_{1}x_{1}, \cdots, r_{l}x_{l}}(q)Y_{(n_{1}-r_{1})x_{1}, \cdots, (n_{l}-r_{l})x_{l}}(v), \end{array}$

$$$Y_{n_{1}x_{1}, \cdots, n_{l}x_{l}}(v)|_{v=\ln\psi}=\frac{\psi_{n_{1}x_{1}, \cdots, n_{l}x_{l}}}{\psi},$$$

$$$\eta=\frac{(q_{x_{k}}' -q_{x_{k}})}{2},$$$

$$$C(q', q)=E(q' )-E(q)=0$$$

$$$\eta_{x_{k}}+f(\eta)=0$$$

$$$\partial_{x_{1}}F_{1}(\eta)+\cdots+\partial_{x_{l}}F_{l}(\eta)=0.$$$

### 3 一类广义浅水波KdV方程

$r(x, t)=u_x(x, t)$,方程(1.1)化成

$$$u_{xt}-\alpha u_{xxxt}-4\alpha u_xu_{xt}-2\alpha u_{xx}u_{t}+ \alpha u_{xx}+\beta u_{xxxx}+6 \beta u_xu_{xx}=0,$$$

### 3.1 双线性表达式

$\begin{array}{l}&&(D_{x}^4+D_{x}D_{z})G\cdot G=0, \\&&[D_{x}D_{t}+\alpha(-\frac{2}{3}D_{t}D_{x}^{3}+\frac{1}{3}D_{t}D_{z}+D_{x}^{2})+\betaD_{x}^{4}]G\cdot G=0, \end{array}$

为了得到方程(3.1)的双线性化表达式,首先引入一个辅助变量$q$,且令

$$$u=cq_{x},$$$

$$$q_{2x, t}+\alpha(-\frac{2}{3}q_{4x, t}-2cq_{3x}q_{xt}-4cq_{2x}q_{2x, t}-\frac{1}{3}q_{4x, t}+q_{3x})+\beta(q_{5x}+6cq_{2x}q_{3x})=0,$$$

$$$E(q)\equivq_{xt}+\alpha\big(-\frac{2}{3}(q_{3x, t}+3cq_{2x}q_{xt})-\frac{1}{3}\partial_x^{-1}\partial_t(q_{4x}+3cq_{2x}^2)+q_{2x}\big)+\beta(q_{4x}+3cq_{2x}^2)=0,$$$

$$$E(q)\equivq_{xt}+\alpha\big(-\frac{2}{3}(q_{3x, t}+3q_{2x}q_{xt})-\frac{1}{3}\partial_x^{-1}\partial_t(q_{4x}+3q_{2x}^2)+q_{2x}\big)+\beta(q_{4x}+3q_{2x}^2)=0.$$$

$$$q_{4x}+3q_{2x}^2=-q_{xz},$$$

$$$E(q)\equivq_{xt}+\alpha\big(-\frac{2}{3}(q_{3x, t}+3q_{2x}q_{xt})+\frac{1}{3}q_{tz}+q_{2x}\big)+\beta(q_{4x}+3q_{2x}^2)=0.$$$

$\begin{array}{l}&&P_{4x}(q)+P_{xz}(q)=0, \\&&P_{xt}(q)+\alpha(-\frac{2}{3}P_{3x, t}(q)+\frac{1}{3}P_{tz}(q)+P_{2x}(q))+\beta P_{4x}(q)=0.\end{array}$

$$$q=2\ln G\Longleftrightarrow u=cq_x=2(\ln G)_x$$$

$\begin{array}{l}&&(D_{x}^4+D_{x}D_{z})G\cdot G=0, \\&&[D_{x}D_{t}+\alpha(-\frac{2}{3}D_{t}D_{x}^{3}+\frac{1}{3}D_{t}D_{z}+D_{x}^{2})+\betaD_{x}^{4}]G\cdot G=0. \end{array}$

### 3.2 Bäcklund变换,孤子解与Lax对

$\begin{array}{l}&&(D_{x}^2-\lambda)F\cdot G=0, \\&&[(1-3\alpha\lambda)D_{t}+(\alpha+3\beta\lambda)D_{x}-\alphaD_{x}^{2}D_{t}+\beta D_{x}^{3}]F\cdotG=0, \end{array}$

$G$即为方程(3.11)的另解,其中$\lambda$为任意参数.

令$q$$q' 为方程(3.6)的两个不同的解 $$q=2\ln F, \qquad q' =2\ln G,$$ 相应地,引入两个新的变量 $$w=\frac{q' +q}{2}=\ln (FG), \qquadv=\frac{q' -q}{2}=\ln(\frac{F}{G}),$$ 则二场条件为 \begin{array}{l}E(q' )-E(q) = E(w+v)-E(w-v)\\ = 2v_{xt}+\alpha[-2v_{3x, t}-4w_{2x}v_{x, t}-4w_{x, t}v_{2x} \\ -4\partial_x^{-1}(w_{2x}v_{2x, t}+w_{2x, t}v_{2x})+2v_{xx}]+\beta(2v_{4x}+12w_{xx}v_{xx})\\ = 2\partial_x[{\cal Y}_t(v)-\alpha{\cal Y}_{2x, t}{(v, w)}+\beta {\cal Y}_{3x}{(v, w)}]+R{(v, w)}=0, \end{array} 其中 这个二场条件可以认为是在适当限制条件下便于求得Bäcklund变换. 为了将二场条件(3.15)写成一对限制条件,可加入一个限制条件,则R(v, w)可以表示成{\cal Y}-多项式的x-导数结合形式.可选 $${\cal Y}_{2x}{(v, w)}=w_{2x}+v_x^2=\lambda,$$ 其中\lambda为任意参数.由(3.16)式, R(v, w)可化为 \begin{array}{l}R{(v, w)} = \alpha[2\lambdav_{xt}-4w_{2x}v_{xt}+4w_{2x, t}v_x-4\partial_x^{-1}(w_{2x}v_{2x, t}+w_{2x, t}v_{2x})+2v_{xx}]\\+6\beta(w_{2x}v_{xx}-v_xw_{3x}-v_x^2v_{2x})\\ = -6\alpha\lambda v_{xt}+2\alpha v_{2x}+6\beta\lambda v_{2x}, \end{array} 此处应用w_{2x, t}=-2v_xv_{xt}$$w_{2x}=\lambda-v_x^2.$

$\begin{array}{l}&&{\cal Y}_{2x}(v, w)-\lambda=0, \\&&\partial_x[{(1-3\alpha\lambda)}{\cal Y}_{t}{(v)}+(\alpha+3\beta\lambda){\cal Y}_{x}{(v)}-\alpha{\cal Y}_{2x, t}{(v, w)}+\beta{\cal Y}_{3x}{(v, w)}]=0, \end{array}$

$\begin{array}{l}&&(D_{x}^2-\lambda)F\cdot G=0, \\&&[(1-3\alpha\lambda)D_{t}+(\alpha+3\beta\lambda)D_{x}-\alphaD_{x}^{2}D_{t}+\beta D_{x}^{3}]F\cdotG=0, \end{array}$

$$$G_1=e^{\frac{\xi_1}{2}}+e^{-\frac{\xi_1}{2}}, \quad\xi_1=k_1x-\frac{\alpha k_1+\beta k_1^3}{1-\alphak_1^2}t+\xi_1^{(0)},$$$

$$$G_1=e^{-\frac{\xi_1}{2}}(1+e^{\xi_1}),$$$

$$$r=2\ln[1+e^{k_1x-\frac{\alpha k_1+\beta k_1^3}{1-\alphak_1^2}t+\xi_1^{(0)}}]_{xx},$$$

$$$G_2=(k_1-k_2)(e^{\frac{\xi_1+\xi_2}{2}}+e^{-\frac{\xi_1+\xi_2}{2}})-(k_1+k_2)(e^{\frac{\xi_1-\xi_2}{2}}+e^{-\frac{\xi_1-\xi_2}{2}}), \quad\lambda=\frac{k_2^2}{4},$$$

$$$G_2=1+e^{\eta_1}+e^{\eta_2}+e^{\eta_1+\eta_2+A_{12}},$$$

$$$r=2\ln [1+e^{\eta_1}+e^{\eta_2}+e^{\eta_1+\eta_2+A_{12}}]_{xx}.$$$

$\begin{array}{l}G_3 = a(e^{\frac{\xi_1+\xi_2+\xi_3}{2}}+e^{-(\frac{\xi_1+\xi_2+\xi_3}{2}})+b(e^{\frac{-\xi_1+\xi_2+\xi_3}{2}}+e^{-\frac{-\xi_1+\xi_2+\xi_3}{2}})\\+c(e^{\frac{\xi_1-\xi_2+\xi_3}{2}}+e^{-\frac{\xi_1-\xi_2+\xi_3}{2}})+d(e^{\frac{\xi_1+\xi_2-\xi_3}{2}}+e^{-\frac{\xi_1+\xi_2-\xi_3}{2}}), \end{array}$

$\xi_j=k_j x-\frac{\alpha k_j+\beta k_j^3}{1-\alpha k_j^2}t+\xi_j^{(0)}\ (j=1, 2, 3)$.

$\begin{array}{l}&&G_n=\sum\limits_{\epsilon={\pm1}}\prod \limits_{1\leqj <l}^{n}\epsilon_l(\epsilon_jk_j-\epsilon_lk_l)e^{\frac{1}{2}\sum \limits _{j=1}^{n}\epsilon_j\xi_j}, \\&&\xi_j=k_j x-\frac{\alpha k_j+\beta k_j^3}{1-\alphak_j^2}t+\xi_j^{(0)}\ (j=1, 2, \cdots, n), \quad\lambda=\frac{k_n^2}{4}, \end{array}$

$\begin{array}{l}&&L\psi=\psi_{xx}+(u_{x}-\lambda)\psi=0, \\&&M\psi=(1-3\alpha\lambda-u_{x})\psi_{t}+(\alpha+3\beta\lambda+3\betau_{x}-2\alpha u_{t})\psi_{x}-\alpha\psi_{xxt}+\beta\psi_{xxx}=0, \end{array}$

由Hopf-Cole变换$v=\ln \psi$,由公式(2.8)和(2.9)可得

$\begin{array}{l}&&{\cal Y}_{x}(v)=\frac{\psi_{x}}{\psi}, \qquad{\cal Y}_{t}=\frac{\psi_{t}}{\psi}, \qquad{\cal Y}_{2x}(v, w)=q_{2x}+\frac{\psi_{2x}}{\psi}, \\&&{\cal Y}_{3x}(v, w)=3q_{2x}\frac{\psi_{x}}{\psi}+\frac{\psi_{3x}}{\psi}, \qquad{\cal Y}_{2x, t}(v, w)=2q_{xt}\frac{\psi_{x}}{\psi}+q_{2x}\frac{\psi_{t}}{\psi}+\frac{\psi_{2x, t}}{\psi}.\end{array}$

$\begin{array}{l}L\psi=0, M\psi=0, \end{array}$

$\begin{array}{l}&&L=\partial_{x}^2+q_{2x}-\lambda, \\&&M=(1-3\alpha\lambda-q_{2x})\partial_{t}+(\alpha+3\beta\lambda+3\betaq_{2x}-2\alpha q_{xt})\partial_{x}-\alpha\partial_t\partial_{x}^2+\beta \partial_{x}^3, \end{array}$

$q_{x}$代替$u$

$\begin{array}{l}&&L\psi=\psi_{xx}+(u_{x}-\lambda)\psi=0, \\&&M\psi=(1-3\alpha\lambda-u_{x})\psi_{t}+(\alpha+3\beta\lambda+3\betau_{x}-2\alpha u_{t})\psi_{x}-\alpha\psi_{xxt}+\beta\psi_{xxx}=0.\end{array}$

$$$[L, M]=q_{2x, t}+\alpha(-q_{4x, t}-4 q_{2x}q_{2x, t}-2 q_{3x}q_{xt}+q_{3x})+\beta(q_{5x}+6q_{2x}q_{3x})=0.$$$

### 3.3 无穷守恒律

$$$I_{n, t}+F_{n, x}=0, n=1, 2, 3, \cdots ,$$$

\begin{array}{l}&&I_{1}=-\frac{1}{2}u_{x}, \qquadI_{2}=-\frac{1}{2}I_{1, x}=\frac{1}{4}u_{2x}, \\& &\cdots\cdots\\&&I_{n}=-\frac{1}{2}\bigg[I_{n-1, x}+\sum\limits_{k=1}^nI_{k}I_{n-1-k}\bigg], \ n=2, 3, 4, \cdots. \end{array}

$\begin{array}{l}&&F_{1}=-\frac{1}{2}[\alpha(u_{x}-u_{xxt}-2\partial_x^{-1}(u_xu_{xt})-2u_tu_x)+\beta(u_{3x}+3u_x^2)], \\&&\cdots\cdots\\&&F_{n}=\alpha\bigg[I_n- I_{n, xt}+4\partial_x^{-1}\bigg(\sum\limits_{k=1}^{n}I_{k}I_{n+1-k, t}\bigg)+4\partial_x^{-1}\bigg(\sum\limits_{k=1}^nI_{k, t}\bigg)I_{n+1-k}\\&& +4\partial_x^{-1}\bigg(\sum\limits_{i+j+k=n}I_{i}I_{j, t}I_{k}\bigg)\bigg]\\&& +\beta\bigg(I_{n, 2x}-2\sum\limits_{i+j+k=n}I_{i}I_{j}I_{k}-6\sum\limits_{k=1}^nI_kI_{n+1-k}\bigg), \ n=2, 3, 4, \cdots .\end{array}$

首先从公式(3.19)出发,由关系$\partial_x{\cal Y}_t(v)=\partial_t{\cal Y}_x(v)=v_{xt}$,则(3.19)式可化为

$\begin{array}{l}&&{\cal Y}_{2x}(v, w)-\lambda=0, \\&&\partial_t(1-3\alpha\lambda){\cal Y}_{x}+\partial_x[{(\alpha+3\beta\lambda){\cal Y}_{x}{(v)}-\alpha{\cal Y}_{2x, t}{(v, w)}+\beta{\cal Y}_{3x}{(v, w)}}]=0.\end{array}$

$$$\eta=\frac{q_{x}' -q_{x}}{2},$$$

$$$v_{x}=\eta, w_{x}=q_{x}+\eta,$$$

$$$\eta_{x}+\eta^{2}+q_{2x}=\lambda$$$

$$$(1-4\alpha\lambda)\eta_{t}+\partial_{x}\{(\alpha+6\beta\lambda)\eta-\alpha\eta_{xt}+4\alpha[\partial_x^{-1}(\eta\eta_t)]\eta+\beta\eta_{2x}-2\beta\eta^3\}=0,$$$

$$$\tilde{\eta}_x +\tilde{\eta}^2+2\varepsilon\tilde{\eta}+q_{xx}=0$$$

$$$\tilde{\eta}_t+\partial_x\{\alpha\tilde{\eta}-\alpha\tilde{\eta}_{xt}+4\alpha\partial_x^{-1}(\tilde{\eta}\tilde{\eta}_t)\varepsilon+4\alpha\partial_x^{-1}(\varepsilon\tilde{\eta}_t+\tilde{\eta}\tilde{\eta}_t)\tilde{\eta}+\beta\tilde{\eta}_{2x}-2\beta\tilde{\eta}^3-6\beta\varepsilon\tilde{\eta}^2\}=0.$$$

$$$\tilde{\eta}=\sum\limits_{n=1}^{\infty}I_{n}(q, q_{x}, \cdots)\varepsilon^{-n}$$$

$\begin{array}{l}&&I_{1}=-\frac{1}{2}u_{x}, \qquadI_{2}=-\frac{1}{2}I_{1, x}=\frac{1}{4}u_{2x}, \\&&I_{3}=-\frac{1}{2}(I_{2, x}+I_{1}^{2})=-\frac{1}{8}(u_{xxx}+u_x^2), \\&&I_{4}=-\frac{1}{2}(I_{3, x}+2I_{1}I_{2})=\frac{1}{16}(u_{4x}+4u_xu_{2x}), \\&&\cdots\cdots\\&&I_{n}=-\frac{1}{2}[I_{n-1, x}+\sum\limits_{k=1}^nI_{k}I_{n-1-k}], n=2, 3, 4, \cdots .\end{array}$

$\begin{array}{l}&&\sum\limits_{n=1}^{\infty}I_{n, t}\varepsilon^{-n}+\partial_{x}\bigg[\alpha\sum\limits_{n=1}^{\infty}I_{n}\varepsilon^{-n}-\alpha\sum\limits_{n=1}^{\infty}I_{n, xt}\varepsilon^{-n}+4\alpha\partial_x^{-1}\bigg(\sum\limits_{n=1}^{\infty}I_{n}\varepsilon^{-n}\sum\limits_{n=1}^{\infty}I_{n, t}\varepsilon^{-n}\bigg)\cdot\varepsilon\\&&+4\alpha\partial_x^{-1}\bigg(\varepsilon\cdot\sum\limits_{n=1}^{\infty}I_{n, t}\varepsilon^{-n}+\sum\limits_{n=1}^{\infty}I_{n}\varepsilon^{-n}\sum\limits_{n=1}^{\infty}I_{n, t}\varepsilon^{-n}\bigg)\sum\limits_{n=1}^{\infty}I_{n}\varepsilon^{-n}+\beta\sum\limits_{n=1}^{\infty}I_{n, 2x}\varepsilon^{-n}\\&&-2\beta\bigg(\sum\limits_{n=1}^{\infty}I_{n}\varepsilon^{-n}\bigg)^{3}-6\beta\varepsilon\bigg(\sum\limits_{n=1}^{\infty}I_{n}\varepsilon^{-n}\bigg)^2\bigg]=0, \end{array}$

$$$I_{n, t}+F_{n, x}=0, n=1, 2, 3, \cdots,$$$

$\begin{array}{l}&&F_{1}=-\frac{1}{2}[\alpha(u_{x}-u_{xxt}-2\partial_x^{-1}(u_xu_{xt})-2u_tu_x)+\beta(u_{3x}+3u_x^2)], \\&&F_{2}=\frac{1}{4}[\alpha(u_{2x}-u_{3x, t}-4(u_xu_{xt})-2u_tu_{2x})+\beta(u_{4x}+6u_xu_{2x})], \\&&\cdots\cdots\\&&F_{n}=\alpha\bigg[I_n- I_{n, xt}+4\partial_x^{-1}\bigg(\sum\limits_{k=1}^{n}I_{k}I_{n+1-k, t}\bigg)+4\partial_x^{-1}\bigg(\sum\limits_{k=1}^nI_{k, t}\bigg)I_{n+1-k}\\&& +4\partial_x^{-1}\bigg(\sum\limits_{i+j+k=n}I_{i}I_{j, t}I_{k}\bigg)\bigg]\\&& +\beta\bigg(I_{n, 2x}-2\sum\limits_{i+j+k=n}I_{i}I_{j}I_{k}-6\sum\limits_{k=1}^nI_kI_{n+1-k}\bigg), \ n=2, 3, 4, \cdots .\end{array}$

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