非线性耦合波动方程组的Du Fort Frankel格式

Du Fort Frankel Scheme for Nonlinear Cou-pled Wave Equations

摘要: 为了克服波动方程经典显式差分法条件稳定的限制，本文将建立和分析一维非线性耦合波动方程组的Du Fort-Frankel (DFF)格式。在耦合波动方程组经典显式差分格式的基础上，对二阶中心差分算子提出了一类改进的差分公式，从而建立了具有更好稳定性的DFF格式。运用了能量分析法证明了由当前算法得到的数值解在无穷范数意义下有的收敛阶。最后，数值结果验证了格式的有效性和理论结果的正确性。

Abstract: In order to overcome the limitation of the stability condition of the classical explicit difference method for wave equations, this paper is concerned with the development and analysis of an unconditionally stable Du Fort Frankel (DFF) scheme for coupled wave equations. Based on the classical explicit difference scheme for coupled wave equations, a DFF scheme with better stability is derived by improving the central difference operator. By using the discrete energy method, it is shown that numerical solutions obtained by the current method are convergent with an order of in maximum norm. Finally, numerical results verify the validity of the scheme and the correctness of the theoretical results.

1. 引言

本文考虑如下一维非线性耦合波动方程组：

$u_{tt}-a{u}_{xx}=f\left(u\left(x,t\right),v\left(x,t\right),x,t\right)$, $x_0\le x\le x_l$, $0\le t\le T$, (1a)

$v_{tt}-b{v}_{xx}=g\left(u\left(x,t\right),v\left(x,t\right),x,t\right)$, $x_0\le x\le x_l$, $0\le t\le T$, (1b)

$u\left(x,0\right)={\phi }_1\left(x\right),v\left(x,0\right)={\phi }_2\left(x\right)$, $x_0\le x\le x_l$, (1c)

$u_t\left(x,0\right)={\varphi }_1\left(x\right),v_t\left(x,0\right)={\varphi }_2\left(x\right)$, $x_0\le x\le x_l$,(1d)

$u\left(x_0,t\right)={\chi }_1\left(t\right),u\left(x_l,t\right)={\gamma }_1\left(t\right)$, $0\le t\le T$, (1e)

$v\left(x_0,t\right)={\chi }_2\left(t\right),v\left(x_l,t\right)={\gamma }_2\left(t\right)$, $0\le t\le T$, (1f)

当 $f\left(u\left(x,t\right),v\left(x,t\right),x,t\right)=-{\delta }^2\sin \left(u-v\right)$， $g\left(u\left(x,t\right),v\left(x,t\right),x,t\right)=\sin \left(u-v\right)$ 时，方程(1a)~(1f)被称作非线性耦合sine-Gordon方程组(文献 [1] [2] [3] [4])。它在物理、生物等各个领域都有重要的应用。比如它能描绘脱氧核糖核酸(DNA)的开放状态和光脉冲在光纤波导中的传播等。因此，非线性耦合sine-Gordon方程组已经得到了充分的研究。具体见文献 [1] [2] [3] [4]。当 $f\left(u\left(x,t\right),v\left(x,t\right),x,t\right)=a_{1}u+b_1{u}^3+c_{1}u{v}^2$， $g\left(u\left(x,t\right),v\left(x,t\right),x,t\right)=a_{2}v+b_2{v}^3+c_2{u}^{2}v$ 时，方程(1a)~(1f)被称作Klein-Gordon方程组(文献 [5])。它由Segal首次提出，其在描述带电介子在电磁场中的运动有着极其重要的作用。它的研究见文献 [5] 及其被引文献。

在文献 [6] 中，吴等研究了此方程组的经典显式差分格式。虽然此格式有易于计算、耗时少等优点，但太依赖于稳定的条件。陈等在文献 [7] 中对一类非线性延迟波动方程建立了DFF差分格式。受到抛物方程的DFF差分法和文献 [7] 工作的启发，本文改进方程组(1a)~(1f)的经典显式差分格式，对其建立了DFF格式。运用能量分析法证明了该格式的收敛性。最后，也用数值算例验证了理论的正确性和算法的性能。

2. 差分格式

2.1. 记号

为了运用差分方法求解问题(1a)~(1f)，将区域 $\Omega =\left\{\left(x,t\right)|x_0\le x\le x_l,0\le t\le T\right\}$ 剖分。将空间区间 $\left[x_0,x_l\right]$ 作m等分(m为整数)，记空间步长h ( $h=\left(x_l-x_0\right)/m$ )。在时间方向上，将区间 $\left[0,T\right]$ 作n等分(n为整数)，记时间步长 $h_t$ ( $h_t=T/n$ )。记 $x_i=x_0+ih$， $t_k=k{h}_t$， $0\le i\le m$， $0\le k\le n$，i, k均为整数。记网格剖分区域 ${\Omega }_h=\left\{\left(x_i,t_k\right)|0\le i\le m,0\le k\le n\right\}$，定义网格函数空间 $u_h=\left\{u|u=\left\{u_i|0\le i\le m\right\},u_0=u_m=0\right\}$。对任意 $u_i^k\in u_h$，引进如下记号：

${\delta }_t^2{u}_i^k=\frac{1}{h_t^2}\left(u_i^{k+1}-2u_i^k+u_i^{k-1}\right)$, ${\delta }_{\hat{t}}{u}_i^k=\frac{1}{2}\left({\delta }_t{u}_i^{k+\frac{1}{2}}+{\delta }_t{u}_i^{k-\frac{1}{2}}\right)$, ${\delta }_t{u}_i^{k-\frac{1}{2}}=\frac{1}{h_t}\left(u_i^k-u_i^{k-1}\right)$,

${\delta }_x^2{u}_i^k=\frac{1}{h^2}\left(u_{i+1}^k-2u_i^k+u_{i-1}^k\right)$, $\left(u,v\right)=h{\displaystyle \underset{i=1}{\overset{m-1}{\sum }}{u}_i{v}_i}$, ${\left(u,v\right)}_1=h{\displaystyle \underset{i=1}{\overset{m}{\sum }}\left({\delta }_x{u}_{i-\frac{1}{2}}\right)\left({\delta }_x{v}_{i-\frac{1}{2}}\right)}$,

$\Vert u\Vert =\sqrt{\left(u,u\right)}$, ${\left|u\right|}_1=\sqrt{{\left(u,u\right)}_1}$, ${\Vert u\Vert }_{\infty }=\underset{0\le i\le m}{\max }\left|u_i\right|$.

2.2. DFF差分格式的建立

方程(1a)~(1f)在节点 $\left(x_i,t_k\right)$ 处的精确解为 $u\left(x_i,t_k\right)$， $v\left(x_i,t_k\right)$。即记 $U_i^k=u\left(x_i,t_k\right)$， $V_i^k=v\left(x_i,t_k\right)$。在节点的数值解为 $u_i^k$， $v_i^k$。即 $u_i^k\approx U_i^k$， $v_i^k\approx V_i^k$。

由泰勒展式可知

$u_{tt}\left(x_i,t_k\right)={\delta }_t^2{U}_i^k-\frac{h_t^2}{12}\frac{{\partial }^{4}u\left(x_i,{\eta }_{ik}\right)}{\partial t^4}$, $u_{xx}\left(x_i,t_k\right)={\delta }_x^2{U}_i^k-\frac{h^2}{12}\frac{{\partial }^{4}u\left({\xi }_{ik},t_k\right)}{\partial x^4}$,

$v_{tt}\left(x_i,t_k\right)={\delta }_t^2{V}_i^k-\frac{h_t^2}{12}\frac{{\partial }^{4}v\left(x_i,{\eta }_{ik}\right)}{\partial t^4}$, $v_{xx}\left(x_i,t_k\right)={\delta }_x^2{V}_i^k-\frac{h^2}{12}\frac{{\partial }^{4}v\left({\xi }_{ik},t_k\right)}{\partial x^4}$.

在节点 $\left(x_i,t_k\right)$ 处考虑微分方程(1a)~(1b)，将用上述差分公式 ${\delta }_t^2{U}_i^k$， ${\delta }_t^2{V}_i^k$， ${\delta }_x^2{U}_i^k$， ${\delta }_x^2{V}_i^k$ 离散 $u_{tt}\left(x_i,t_k\right)$， $v_{tt}\left(x_i,t_k\right)$， $u_{xx}\left(x_i,t_k\right)$， $v_{xx}\left(x_i,t_k\right)$ 可得

${\delta }_t^2{U}_i^k-a{\delta }_x^2{U}_i^k=f\left(U_i^k,V_i^k,x_i,t_k\right)+{\left({\eta }_1\right)}_i^k$, $1\le i\le m-1$, $1\le k\le n-1$,(2)

${\delta }_t^2{V}_i^k-b{\delta }_x^2{V}_i^k=g\left(U_i^k,V_i^{k_1},x_i,t_k\right)+{\left({\eta }_2\right)}_i^k$, $1\le i\le m-1$, $1\le k\le n-1$,(3)

其中，

${\left({\eta }_1\right)}_i^k=\left(h_t^2\frac{{\partial }^{4}u\left(x_i,{\eta }_{ik}\right)}{\partial t^4}-a{h}^2\frac{{\partial }^{4}u\left({\xi }_{ik},t_k\right)}{\partial x^4}\right)/12$, $1\le i\le m-1$, $1\le k\le n-1$, (4)

${\left({\eta }_2\right)}_i^k=\left(h_t^2\frac{{\partial }^{4}v\left(x_i,{\eta }_{ik}\right)}{\partial t^4}-b{h}^2\frac{{\partial }^{4}v\left({\xi }_{ik},t_k\right)}{\partial x^4}\right)/12$, $1\le i\le m-1$, $1\le k\le n-1$. (5)

在(2) (3)中用 $u_i^k$ 代替 $U_i^k$，用 $v_i^k$ 代替 $V_i^k$，略去小量项 ${\left({\eta }_1\right)}_i^k$， ${\left({\eta }_2\right)}_i^k$，得到如下经典显式差分格式

${\delta }_t^2{u}_i^k-a{\delta }_x^2{u}_i^k=f\left(u_i^k,v_i^{k_1},x_i,t_k\right)$, $1\le i\le m-1$, $1\le k\le n-1$, (6a)

${\delta }_t^2{v}_i^k-b{\delta }_x^2{v}_i^k=g\left(u_i^k,v_i^k,x_i,t_k\right)$, $1\le i\le m-1$, $1\le k\le n-1$. (6b)

格式(6a)~(6b)就是文献 [1] 中的显格式，它要求网格比 $r_x=h_t/h<1$。

为了得到稳定性更好的格式，我们对差分算子 ${\delta }_x^2{U}_i^k$ 进行如下改进

$\begin{array}{c}{\delta }_x^2{U}_i^k=\frac{1}{h^2}\left(U_{i+1}^k-2U_i^k+U_{i-1}^k\right)\\ =\frac{1}{h^2}\left[U_{i+1}^k-2\left(\frac{U_i^{k+1}+U_i^{k-1}}{2}-\frac{h_t^2}{2}\frac{{\partial }^{2}u\left(x_i,{\varsigma }_{ik}\right)}{\partial t^2}\right)+U_{i-1}^k\right]\\ =\frac{1}{h^2}\left(U_{i+1}^k+U_{i-1}^k\right)-\frac{2}{h^2}{U}_i^k-\frac{1}{h^2}\left(U_i^{k+1}+U_i^{k-1}\right)+\frac{2}{h^2}{U}_i^k+\frac{h_t^2}{h^2}\frac{{\partial }^{2}u\left(x_i,{\varsigma }_{ik}\right)}{\partial t^2}\\ ={\delta }_x^2{U}_i^k-\frac{h_t^2}{h^2}{\delta }_t^2{U}_i^k+\frac{h_t^2}{h^2}\frac{{\partial }^{2}u\left(x_i,{\varsigma }_{ik}\right)}{\partial t^2}\end{array}$ (7)

将(7)式代入(2)式中得

$\left(1+a{r}_x^2\right){\delta }_t^2{U}_i^k-a{\delta }_x^2{U}_i^k=f\left(U_i^k,V_i^k,x_i,t_k\right)+{\left(R_1\right)}_i^k$, $1\le i\le m-1$, $1\le k\le n-1$. (8)

同理可得

$\left(1+b{r}_x^2\right){\delta }_t^2{V}_i^k-b{\delta }_x^2{V}_i^k=g\left(U_i^k,V_i^k,x_i,t_k\right)+{\left(R_2\right)}_i^k$, $1\le i\le m-1$, $1\le k\le n-1$. (9)

截断误差为

${\left(R_1\right)}_i^k={\left({\eta }_1\right)}_i^k+a{r}_x^2\frac{{\partial }^{2}u\left(x_i,{\varsigma }_{ik}\right)}{\partial t^2}$, $1\le i\le m-1$, $1\le k\le n-1$, (10)

${\left(R_2\right)}_i^k={\left({\eta }_2\right)}_i^k+b{r}_x^2\frac{{\partial }^{2}v\left(x_i,{\varsigma }_{ik}\right)}{\partial t^2}$, $1\le i\le m-1$, $1\le k\le n-1$. (11)

显然，此格式是三层显式差分格式。第0层的数值解是已知的。为了启动的计算，还必须算出第一层的数值解。用带有积分型余项的泰勒公式对 $U_i^1$， $V_i^1$ 在节点 $\left(x_i,t_0\right)$ 处泰勒展开可得

$\begin{array}{c}{U}_i^1=u\left(x_i,t_0\right)+h_t\frac{\partial u}{\partial t}\left(x_i,t_0\right)+\frac{h_t^2}{2}\frac{{\partial }^{2}u}{\partial t^2}\left(x_i,t_0\right)+\frac{h_t^3}{6}\frac{{\partial }^{3}u}{\partial t^3}\left(x_i,t_0\right)\\ +\frac{h_t^4}{24}\frac{{\partial }^{4}u}{\partial t^4}\left(x_i,t_0\right)+\frac{h_t^5}{24}{\displaystyle {\int }_0^1\frac{{\partial }^{5}u}{\partial t^5}\left(x_i,\lambda ht\right)}{\left(1-\lambda \right)}^4\text{d}\lambda \\ =u_i^1+{\left[R_1\right]}_i^0,\end{array}$ (12)

$\begin{array}{c}{V}_i^1=v\left(x_i,t_0\right)+h_t\frac{\partial v}{\partial t}\left(x_i,t_0\right)+\frac{h_t^2}{2}\frac{{\partial }^{2}v}{\partial t^2}\left(x_i,t_0\right)+\frac{h_t^3}{6}\frac{{\partial }^{3}v}{\partial t^3}\left(x_i,t_0\right)\\ +\frac{h_t^4}{24}\frac{{\partial }^{4}v}{\partial t^4}\left(x_i,t_0\right)+\frac{h_t^5}{24}{\displaystyle {\int }_0^1\frac{{\partial }^{5}v}{\partial t^5}\left(x_i,\lambda ht\right)}{\left(1-\lambda \right)}^4\text{d}\lambda \\ =v_i^1+{\left[R_2\right]}_i^0,\end{array}$ (13)

其中

$u_i^1={\phi }_1\left(x_i\right)+h_t{\varphi }_1\left(x_i\right)+\frac{h_t^2}{2\left(1+a{r}_x^2\right)}\left[a{\delta }_x^2{u}_i^0+f\left(u_i^0,v_i^0,x_i,t_k\right)\right]$, $0\le i\le m$,

$v_i^1={\phi }_2\left(x_i\right)+h_t{\varphi }_2\left(x_i\right)+\frac{h_t^2}{2\left(1+b{r}_x^2\right)}\left[b{\delta }_x^2{v}_i^0+f\left(u_i^0,v_i^0,x_i,t_k\right)\right]$, $0\le i\le m$,

${\left[R_1\right]}_i^0=\frac{h_t^3}{6}\frac{{\partial }^{3}u}{\partial t^3}\left(x_i,t_0\right)+\frac{h_t^4}{24}\frac{{\partial }^{4}u}{\partial t^4}\left(x_i,t_0\right)+\frac{h_t^5}{24}{\displaystyle {\int }_0^1\frac{{\partial }^{5}u}{\partial t^5}\left(x_i,\lambda ht\right)}{\left(1-\lambda \right)}^4\text{d}\lambda$, $0\le i\le m$,

${\left[R_2\right]}_i^0=\frac{h_t^3}{6}\frac{{\partial }^{3}v}{\partial t^3}\left(x_i,t_0\right)+\frac{h_t^4}{24}\frac{{\partial }^{4}v}{\partial t^4}\left(x_i,t_0\right)+\frac{h_t^5}{24}{\displaystyle {\int }_0^1\frac{{\partial }^{5}v}{\partial t^5}\left(x_i,\lambda ht\right)}{\left(1-\lambda \right)}^4\text{d}\lambda$, $0\le i\le m$.

在(8)~(9)，(12)~(13)中略去小量项 ${\left(R_1\right)}_i^k$， ${\left(R_2\right)}_i^k$，用 $u_i^k$ 代替 $U_i^k$，用 $v_i^k$ 代替 $V_i^k$ 可得DFF格式如下

$\left(1+a{r}_x^2\right){\delta }_t^2{u}_i^k-a{\delta }_x^2{u}_i^k=f\left(u_i^k,v_i^{k_1},x_i,t_k\right)$, $1\le i\le m-1$, $1\le k\le n-1$, (14a)

$\left(1+b{r}_x^2\right){\delta }_t^2{v}_i^k-b{\delta }_x^2{v}_i^k=g\left(u_i^k,v_i^{k_1},x_i,t_k\right)$, $1\le i\le m-1$, $1\le k\le n-1$, (14b)

$u_i^0={\phi }_1\left(x\right),v_i^0={\phi }_2\left(x\right)$, $0\le i\le m$, (14c)

$u_0^k={\chi }_1\left(t\right),v_0^k={\chi }_2\left(t\right)$, $0<k\le n$, (14d)

$u_m^k={\gamma }_1\left(t\right),v_m^k={\gamma }_2\left(t\right)$, $0<k\le n$, (14e)

$u_i^1={\phi }_1\left(x_i\right)+h_t{\varphi }_1\left(x_i\right)+\frac{h_t^2}{2\left(1+a{r}_x^2\right)}\left[a{\delta }_x^2{u}_i^0+f\left(u_i^0,v_i^0,x_i,t_k\right)\right]$, $0\le i\le m$, (14f)

$v_i^1={\phi }_2\left(x_i\right)+h_t{\varphi }_2\left(x_i\right)+\frac{h_t^2}{2\left(1+b{r}_x^2\right)}\left[b{\delta }_x^2{v}_i^0+f\left(u_i^0,v_i^0,x_i,t_k\right)\right]$, $0\le i\le m$. (14g)

2.3. 差分格式的收敛性分析

设问题(1a)~(1f)在节点 $\left(x_i,t_k\right)$ 的精确解为 $U_i^k$， $V_i^k$， $u_i^k$， $v_i^k$ 为差分格式(14a)~(14g)的数值解。令 $U_i^k-u_i^k=e_i^k$， $V_i^k-v_i^k={\tilde{e}}_i^k$， $F_i^k=f\left(U_i^k,V_i^{k_1},x_i,t_k\right)-f\left(u_i^k,v_i^{k_1},x_i,t_k\right)$， $G_i^k=g\left(U_i^k,V_i^{k_1},x_i,t_k\right)-g\left(u_i^k,v_i^{k_1},x_i,t_k\right)$。用(8)~(9)，(10)~(11)，(12)~(13)式依次减去(14a)~(14g)式可得到如下误差方程

$\left(1+a{r}_x^2\right){\delta }_t^2{e}_i^k-a{\delta }_x^2{e}_i^k=F_i^k+{\left[R_1\right]}_i^k$, $1\le i\le m-1$, $1\le k\le n-1$, (15a)

$\left(1+b{r}_x^2\right){\delta }_t^2{\tilde{e}}_i^k-b{\delta }_x^2{\tilde{e}}_i^k=G_i^k+{\left[R_2\right]}_i^k$, $1\le i\le m-1$, $1\le k\le n-1$, (15b)

$e_i^0=0,{\tilde{e}}_i^0=0$, $0\le i\le m$,(15c)

$e_0^k=0,{\tilde{e}}_0^k=0$, $0<k\le n$, (15d)

$e_m^k=0,{\tilde{e}}_m^k=0$, $0<k\le n$, (15e)

$e_i^1={\left[R_1\right]}_i^0,{\tilde{e}}_i^1={\left[R_2\right]}_i^0$, $0\le i\le m$. (15f)

为了研究上述显式差分格式的收敛性，引入如下引理和假设。

引理2.1 [8] 设 $w\in u_h$，则有下列不等式成立

$\left(-{\delta }_x^{2}w,w\right)={\Vert {\delta }_{x}w\Vert }^2$, ${\Vert w\Vert }_{\infty }\le \frac{\sqrt{x_l-x_0}}{2}{\left|w\right|}_1$,

$\Vert w\Vert \le \frac{x_l-x_0}{\sqrt{6}}{\left|w\right|}_1$, ${\left|w\right|}_1^2\le \frac{4}{h^2}{\Vert w\Vert }^2$.

另外，根据(10) (11)可假设，存在常数 $c_1$， $c_2$，使得

${\Vert {\left(R_1\right)}_i^k\Vert }^2\le c_1{\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)}^2$, $1\le i\le m-1$, $1\le k\le n-1$, (16)

${\Vert {\left(R_2\right)}_i^k\Vert }^2\le c_2{\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)}^2$, $1\le i\le m-1$, $1\le k\le n-1$. (17)

成立。

假设函数 $f\left(u,v,x,t\right)$， $g\left(u,v,x,t\right)$ 满足如下局部Lipschitz条件：

设u，v为问题方程(1a)~(1f)的精确解，且存在正常数 $c_3$， ${\epsilon }_0$，当 $\text{|}{\varsigma }_i\text{|}<{\epsilon }_0$，(i = 1, 2时)，函数 $f\left(u,v,x,t\right)$， $g\left(u,v,x,t\right)$ 满足如下不等式

$\left|f\left(u+{\varsigma }_1,v+{\varsigma }_2,x,t\right)-f\left(u,v,x,t\right)\right|\le c_3\left(\left|{\varsigma }_1\right|+\left|{\varsigma }_2\right|\right)$, (18)

$\left|g\left(u+{\varsigma }_1,v+{\varsigma }_2,x,t\right)-g\left(u,v,x,t\right)\right|\le c_3\left(\left|{\varsigma }_1\right|+\left|{\varsigma }_2\right|\right)$. (19)

其中 $c_3$ 为Lipschitz常数。

引理2.2 设 $H^k=\left(1+a{r}_x^2\right){\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2+a\left({\delta }_x{e}^k,{\delta }_x{e}^{k-1}\right)+\left(1+b{r}_x^2\right){\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2+b\left({\delta }_x{\tilde{e}}^k,{\delta }_x{\tilde{e}}^{k-1}\right)$，有

${\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2\text{+}{\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2+a{\left|e^{k-\frac{1}{2}}\right|}_1^2\text{+}b{\left|{\tilde{e}}^{k-\frac{1}{2}}\right|}_1^2\le H^k$, (20)

${\left|e^k\right|}_1^2\le 2\max \left(r_x^2,a^{-1}\right)H^k$, ${\left|{\tilde{e}}^k\right|}_1^2\le 2\max \left(r_x^2,b^{-1}\right)H^k$, (21)

${\left|e^k\right|}_1^2+{\left|{\tilde{e}}^k\right|}_1^2\le \max \left(2r_x^2,2a^{-1},2b^{-1}\right)H^k$. (22)

证明：根据 $\alpha \beta ={\left[\frac{\alpha +\beta }{2}\right]}^2-{\left[\frac{\alpha -\beta }{2}\right]}^2$，有

$a\left({\delta }_x{e}^k,{\delta }_x{e}^{k-1}\right)=a{\left|e^{k-\frac{1}{2}}\right|}_1^2-\frac{a{h}_t{}^2}{4}{\Vert {\delta }_x{\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2$, (23)

$b\left({\delta }_x{\tilde{e}}^k,{\delta }_x{\tilde{e}}^{k-1}\right)=b{\left|{\tilde{e}}^{k-\frac{1}{2}}\right|}_1^2-\frac{b{h}_t{}^2}{4}{\Vert {\delta }_x{\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2$. (24)

再由引理2.1，可得

$\begin{array}{c}{H}^k=\left(1+a{r}_x^2\right){\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2+a{\left|e^{k-\frac{1}{2}}\right|}_1^2-\frac{a{h}_t^2}{4}{\Vert {\delta }_x{\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2+\left(1+b{r}_x^2\right){\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2+b{\left|{\tilde{e}}^{k-\frac{1}{2}}\right|}_1^2-\frac{b{h}_t^2}{4}{\Vert {\delta }_x{\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2\\ \ge {\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2+{\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2+a{\left|e^{k-\frac{1}{2}}\right|}_1^2+b{\left|{\tilde{e}}^{k-\frac{1}{2}}\right|}_1^2.\end{array}$

即(20)成立。

应用恒等式 $e_i^k=\frac{h_t}{2}\frac{e^k-e^{k-1}}{h_t}+\frac{e^k+e^{k-1}}{2}$ 以及三角不等式，可以得到

${\left|e^k\right|}_1^2\le 2{\left|e^{k-\frac{1}{2}}\right|}_1^2+2\frac{h_t^2}{4}{\left|{\delta }_t{e}_i{}^{k-\frac{1}{2}}\right|}_1^2\le 2{\left|e^{k-\frac{1}{2}}\right|}_1^2+2\frac{h_t^2}{4h^2}{h}^2{\left|{\delta }_t{e}^{k-\frac{1}{2}}\right|}_1^2$.

对上述不等式再应用引理2.1可得

${\left|e^k\right|}_1^2\le 2r_x^2{\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2+2a^{-1}a{\left|e^{k-\frac{1}{2}}\right|}_1^2\le \max \left(2r_x^2,2a^{-1}\right)\left({\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2+a{\left|e^{k-\frac{1}{2}}\right|}_1^2\right)\le 2\max \left(r_x^2,a^{-1}\right)H^k$.

同理可得

${\left|{\tilde{e}}^k\right|}_1^2\le \max \left(2r_x^2,2b^{-1}\right)\left({\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2+b{\left|{\tilde{e}}^{k-\frac{1}{2}}\right|}_1^2\right)\le 2\max \left(r_x^2,b^{-1}\right)H^k$.

即(21)得证。

(22)式可由(21)式直接得出。

引理2.3 [8] Gronwall不等式：设 $\left\{P^k|k\ge 0\right\}$ 为非负序列，且满足 $P^{k+1}\le \left(1+c\tau \right)P^k+\tau q$， $k=0,1,2,\cdots$，其中c和q为非负常数，则有

$P^k\le e^{ck\tau }\left(P^0+\frac{q}{c}\right)$, $k=0,1,2,\cdots$.

定理2.1设问题(1a)~(1f)在节点 $\left(x_i,t_k\right)$ 处的精确解为 $U_i^k$， $V_i^k$。DFF格式(14a)~(14g)的数值解为 $u_i^k$， $v_i^k$。记 $L=x_l-x_0$。假设存在常数 $c_4\ge 0$，使得 $H^1\le c_4{\left(h_t^2+h^2+{\left(h_t/h\right)}^2\right)}^2$。当步长满足如下条件

$h\le \sqrt{\frac{2{\epsilon }_0}{3c_7\sqrt{L}}}$, $h_t\le \sqrt{\frac{2{\epsilon }_0}{3c_7\sqrt{L}}}$, $\frac{h_t}{h}\le \sqrt{\frac{2{\epsilon }_0}{3c_7\sqrt{L}}}$, $c_5{h}_t\le \frac{1}{3}$

时，有如下误差估计

$\underset{0\le k\le N}{\max }\left({\left|e^k\right|}_1,{\left|{\tilde{e}}^k\right|}_1\right)\le c_7\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)$, (25)

$\underset{0\le k\le N}{\max }\left({\Vert e^k\Vert }_{\infty },{\Vert {\tilde{e}}^k\Vert }_{\infty }\right)\le c_8\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)$, (26)

其中，

$c_5=1+\frac{4}{3}{\left(c_{3}L\right)}^2\max \left(r_x^2,a^{-1},b^{-1}\right)$, $c_6=\max \left(2r_x^2,2a^{-1},2b^{-1}\right)e^{3c_{5}T}\left[c_4+\frac{1}{2c_5}\left(c_1+c_2\right)\right]$,

$c_7=\max \left(\sqrt{2r_x^2{c}_4},\sqrt{2a^{-1}{c}_4},\sqrt{2b^{-1}{c}_4},\sqrt{c_6}\right)$, $c_8=\frac{\sqrt{L}}{2}{c}_7$.

证明：用数学归纳法证明(25)成立，当 $k=0,1$ 时，式(25)显然成立。假设当 $k=0,1,\cdots ,l$ 时，(25)式成立。下面证明当 $k=l+1$ 时，(25)也成立。

由引理2.1可得

$\underset{0\le k\le N}{\max }\left({\Vert e^k\Vert }_{\infty },{\Vert {\tilde{e}}^k\Vert }_{\infty }\right)\le \frac{\sqrt{L}}{2}\underset{0\le k\le N}{\max }\left({\left|e^k\right|}_1,{\left|{\tilde{e}}^k\right|}_1\right)\le \frac{\sqrt{L}}{2}{c}_7\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)$.

又由 $h\le \sqrt{\frac{2{\epsilon }_0}{3c_7\sqrt{L}}}$， $h_t\le \sqrt{\frac{2{\epsilon }_0}{3c_7\sqrt{L}}}$， $\frac{h_t}{h}\le \sqrt{\frac{2{\epsilon }_0}{3c_7\sqrt{L}}}$ 可得

$\underset{0\le k\le N}{\max }\left({\Vert e^k\Vert }_{\infty },{\Vert {\tilde{e}}^k\Vert }_{\infty }\right)\le {\epsilon }_0$.

于是，根据(18)，(19)有

${\Vert F^k\Vert }^2\le 2c_3^2\left({\Vert e^k\Vert }^2+{\Vert {\tilde{e}}^k\Vert }^2\right)$, ${\Vert G^k\Vert }^2\le 2c_3^2\left({\Vert e^k\Vert }^2+{\Vert {\tilde{e}}^k\Vert }^2\right)$, $0\le k\le l$.(27)

接下来对(15a) (15b)两边分别与 $2{\delta }_{\hat{t}}{e}^k$， $2{\delta }_{\hat{t}}{\tilde{e}}^k$ 作内积，根据三角不等式可得

$\begin{array}{c}\frac{H^{k+1}-H^k}{h_t}=2\left(F^k,{\delta }_{\hat{t}}{e}^k\right)+2\left(R_1^k,{\delta }_{\hat{t}}{e}^k\right)+2\left(G^k,{\delta }_{\hat{t}}{\tilde{e}}^k\right)+2\left(R_2^k,{\delta }_{\hat{t}}{\tilde{e}}^k\right)\\ \le {\Vert F^k\Vert }^2\text{+}{\Vert {\delta }_{\hat{t}}{e}^k\Vert }^2+{\Vert R_1^k\Vert }^2+{\Vert {\delta }_{\hat{t}}{e}^k\Vert }^2+{\Vert G^k\Vert }^2+{\Vert {\delta }_t{\tilde{e}}^k\Vert }^2+{\Vert R_2^k\Vert }^2+{\Vert {\delta }_t{\tilde{e}}^k\Vert }^2\end{array}$, $0\le k\le l$. (28)

将(27)代入(28)中以及应用不等式 ${\left(\frac{a+b}{2}\right)}^2\le \frac{a^2+b^2}{2}$ 和引理2.1可得

$\begin{array}{c}\frac{H^{k+1}-H^k}{h_t}\le 4c_3^2\left({\Vert e^k\Vert }^2+{\Vert {\tilde{e}}^k\Vert }^2\right)+\left({\Vert {\delta }_t{e}^{k+\frac{1}{2}}\Vert }^2+{\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2\right)+\left({\Vert {\delta }_t{\tilde{e}}^{k+\frac{1}{2}}\Vert }^2+{\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2\right)+{\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2\\ \le \frac{2}{3}{\left(c_{3}L\right)}^2\left({\left|e^k\right|}_1^2+{\left|{\tilde{e}}^k\right|}_1^2\right)+\left({\Vert {\delta }_t{e}^{k+\frac{1}{2}}\Vert }^2+{\Vert {\delta }_t{e}^{k-\frac{1}{2}}\Vert }^2\right)+\left({\Vert {\delta }_t{\tilde{e}}^{k+\frac{1}{2}}\Vert }^2+{\Vert {\delta }_t{\tilde{e}}^{k-\frac{1}{2}}\Vert }^2\right)+{\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2.\end{array}$ (29)

对(29)式应用引理2.2得

$\begin{array}{c}\frac{H^{k+1}-H^k}{h_t}\le \frac{2}{3}{\left(c_{3}L\right)}^2\max \left(2r_x^2,2a^{-1},2b^{-1}\right)H^k+H^k+H^{k+1}+{\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2\\ \le c_5\left(H^k+H^{k+1}\right)+{\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2\end{array}$. (30)

对(30)式两边同乘 $h_t$，再将 $H^{k+1}$ 都放在不等号左边，当 $c_5{h}_t\le \frac{1}{3}$ 时，整理式(30)可得

$\begin{array}{c}{H}^{k+1}\le \frac{1+c_5{h}_t}{1-c_5{h}_t}{H}^k+\frac{h_t}{1-c_5{h}_t}\left({\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2\right)\\ \le \left(1+3c_5{h}_t\right)H^k+\frac{3}{2}{h}_t\left({\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2\right)\\ \le \left(1+3c_5{h}_t\right)H^k+\frac{3}{2}{h}_t\underset{0\le k\le N}{\max }\left({\Vert R_1^k\Vert }^2+{\Vert R_2^k\Vert }^2\right)\end{array}$, $k=0,1,\cdots ,l$. (31)

对(31)式应用引理2.3，取 $k=l$ 有，得到

$H^{l+1}\le e^{3c_{5}T}\left[H^1+\frac{3}{2}\frac{1}{3c_5}\underset{0\le k\le N}{\max }\left({\Vert R_1^l\Vert }^2+{\Vert R_2^l\Vert }^2\right)\right]$. (32)

将(16)(17)代入(32)式得到

$H^{l+1}\le e^{3c_{5}T}\left[c_4+\frac{1}{2c_5}\left(c_1+c_2\right)\right]{\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)}^2$. (33)

最后，将(33)式代入(22)式，有

${\left|e^{l+1}\right|}_1^2+{\left|{\tilde{e}}^{l+1}\right|}_1^2\le c_6{\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)}^2$.

即证 $\underset{0\le k\le N}{\max }\left({\left|e^{l+1}\right|}_1^2,{\left|{\tilde{e}}^{l+1}\right|}_1^2\right)\le {\left|e^{l+1}\right|}_1^2+{\left|{\tilde{e}}^{l+1}\right|}_1^2\le c_6{\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)}^2$。显然(25)式对 $k=l+1$ 时也是成立的。所以，由归纳法可证(25)式成立。

对(25)式运用引理2.1可直接证明(26)式成立。

3. 数值实验

算例 考虑如下非线性耦合sine-Gordon方程组：

$u_{tt}-u_{xx}=-{\delta }^2\sin \left(u-v\right)$

$v_{tt}-b{v}_{xx}=-\sin \left(u-v\right)$, $\left(x,t\right)\in \left(0,1\right]\times \left(0,1\right]$

初边值条件由精确解 $u\left(x,t\right)=\frac{c^2}{4\left(e^2-1\right)d^2}\sin \left(2d\left(x-et\right)\right)$， $v\left(x,t\right)=\frac{c^2}{4\left(e^2-1\right)d^2}\sin \left(2d\left(x-et\right)\right)-2d\left(x-et\right)$ 确定。

这里取 $b=2,c=2,d=1.5,e=\sqrt{\left(1-b{c}^2\right)/\left(1-c^2\right)}$。

定义 $E_{\infty }\left(h,h_t\right)=\max \left\{\underset{0\le i\le l,0\le k\le N}{\max }\left|u\left(x_i,t_k\right)-u_i^k\right|,\underset{0\le i\le l,0\le k\le N}{\max }\left|v\left(x_i,t_k\right)-v_i^k\right|\right\}$， $\text{order}={\text{log}}_2\left(E_{\infty }\left(2h,2h_t\right)/E_{\infty }\left(h,h_t\right)\right)$。

表1为格式(3a)~(3g)在 $h_t=h^2$ 时取不同步长时得到的数值解的最大误差，收敛阶，CPU时间(秒)。从表中可以看到当 $h_t=h^2$ 时，数值解在最大范数意义下有 $O\left(h^2\right)$ 的收敛阶。从而，表1表明了定理2.1的正确性。

Table 1. Difference method (3a)~(3f) numerical results for this problem

表1. 差分方法(3a)~(3f)求解该问题得到的数值结果( $h_t=h^2$ )

4. 结论

本文受抛物方程的Du Fort-Frankel差分法的启发，对非线性耦合波动方程组建立了Du Fort-Frankel差分格式。运用能量分析法，证明了它在最大范数意义下有 $O\left(h_t^2+h^2+\frac{h_t^2}{h^2}\right)$ 的收敛阶。数值结果验证了理论结果的正确性。

基金项目

国家自然科学基金项目(No. 11861047)；江西省自然科学基金(20202BABL201005)；江西省杰出青年基金(20212ACB211006)。