本文研究如下一维非线性Klein-Gordon方程Neumann的边值问题的数值解

$ \begin{align} &\frac{\partial^2 u}{\partial t^2}-\alpha\frac{\partial^2u}{\partial x^2}+g(u) = f(x, t), \quad a<x<b, \; 0<t\leq T, \end{align} $

(1.1)

$ \begin{align} &u(x, 0) = \varphi(x), \; \frac{\partial u}{\partial t}(x, 0) = \psi(x), \quad a\leq x\leq b, \end{align} $

(1.2)

$ \begin{align} &\frac{\partial u}{\partial x}(a, t) = 0, \; \frac{\partial u}{\partial x}(b, t) = 0, \quad 0\leq t\leq T, \end{align} $

(1.3)

其中$ \alpha>0 $为常数, $ f(x, t) $, $ g(u) $为满足相容性条件的光滑函数.

Klein-Gordon方程是相对论量子力学和量子场论中用于描述零自旋粒子的自由运动方程, 关于它的数值解法已有不少研究结果.文献[1]基于样条基函数提出了一个数值格式; 四阶紧格式在文献[2]中进行了研究; 基于变分迭代法的数值格式及边界元方法可参见文献[3, 4]; 在文献[5]中导出了以三层样条差分格式逼近非线性Klein-Gordon方程; 无界域上的问题的数值研究可参见文献[6]; 文献[7]中提出了一个基于有限差分和匹配法的新的数值格式; 而文献[8]提出了微分积分法.所有这些文献中的研究都是针对Dirichlet边界条件, 对于Neumann边界条件下的高阶差分格式还没有很好的结果.近年来, 具有Neumann边界条件的热方程的高阶差分格式已有一些研究结果如文献[9, 10, 11].文献[12]研究了Cahn-Hilliard方程Neumann边界条件下的三层线性化紧格式; 文献[13]中建立了哈密尔顿非线性波方程Neumann边界条件下的高阶显格式, 该方法空间方向基于紧格式, 时间方向基于Runge-Kutta-Nystrom方法.通过分析以上文献, 了解到Klein-Gordon方程Neumann边值问题的无条件稳定的高阶差分格式, 目前还没有这方面的结果, 其主要困难是边界点的处理.本文利用Klein-Gordon方程及边界条件可得到在边界处的三阶导数和五阶导数的函数值, 从而建立边界点和内点处的两点和三点紧差分格式, 构造一个紧格式, 并证明差分格式关于时间$ 2 $阶收敛, 关于空间$ 4 $阶收敛.

设问题(1.1)–(1.3)存在光滑解$ u $, 记$ \Omega = [a, b]\times[0, T] $, 本文假设

H1: $ u\in C^{(4, 3)}(\Omega) $, 且存在常数$ C_0 $, 使$ \forall(x, t)\in(\Omega) $有

$ \begin{align*} \max(|\frac{\partial^4 u}{\partial t^4}|, |\frac{\partial^4 u}{{\partial x^2}{\partial t^2}}|, |\frac{\partial^4 u}{{\partial x}{\partial t^3}}|, |\frac{\partial^3u}{\partial x^3}|, |u|)\leq C_0. \end{align*} $

H2:函数$ g $一阶可导, 且存在正常数$ C $, 使得当$ |s|\leq C_0+1 $时, 有

$ \max\{|g(s)|, | g'(s)|, |g''(s)|\}\leq C. $

H3: $ \forall(x, t)\in\Omega $, 且存在常数$ C_1 $, 使$ \max{|\frac{\partial f}{\partial x}(x, t)| = C_1} $.

由(1.1)式中的方程可得

$ \begin{align} \alpha\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2}+g(u)-f(x, t). \end{align} $

(2.1)

将(2.1)式两边关于$ x $的$ k\; (k = 3, 4, 5) $阶导数, 且由边界条件(1.3)可得

$ \begin{align*} &\frac{\partial^3u}{\partial x^3}(a, t) = -\frac{1}{\alpha}\frac{\partial f}{\partial x}(a, t), \; \frac{\partial^3 u}{\partial x^3}(b, t) = -\frac{1}{\alpha}\frac{\partial f}{\partial x}(b, t), \\ &\frac{\partial^5 u}{\partial x^5}(a, t) = -\left[\frac{1}{\alpha}\frac{\partial^3 f}{\partial x^3}(a, t)+\frac{g'(u)}{\alpha^2}\frac{\partial f}{\partial x}(a, t) +\frac{1}{\alpha^2}\frac{\partial^3 f}{\partial t^2\partial x}(a, t)\right], \\ &\frac{\partial^5 u}{\partial x^5}(b, t) = -\left[\frac{1}{\alpha}\frac{\partial^3 f}{\partial x^3}(b, t)+\frac{g'(u)}{\alpha^2}\frac{\partial f}{\partial x}(b, t) +\frac{1}{\alpha^2}\frac{\partial^3 f}{\partial t^2\partial x}(b, t)\right]. \end{align*} $

取正整数$ m, n $, 记空间和时间步长分别为$ h = \dfrac{b-a}{m} $, $ \tau = \dfrac{T}{n} $, 记$ x_i = a+ih, \; 0\le i\le m $, $ t_k = k\tau, \; 0\le k\le n $.定义

$ \Omega_h = \{x_i|0\le i\le m\}, \; \; \Omega_{h\tau} = \{(x_i, t_k)|0\le i\le m, 0\le k\le n\}, $

称$ (x_i, t_k) $为节点, 并设$ \{v_i^k|0\le i\le m, 0\le k\le n\} $为$ \Omega_{h\tau} $上的网格函数, 引进下面的记号

$ \begin{align*} & v_i^{k+1/2} = \frac{1}{2}(v_i^k+v_i^{k+1}), \quad v_i^{\bar{k}} = \frac{1}{2}(v_i^{k+1}+v_i^{k-1}), \; \; \quad\delta_t v_i^{k+1/2} = \frac{1}{\tau}(v_i^{k+1}-v_i^k), \\ &\delta_xv_{i+1/2}^k = \frac{1}{h}(v_{i+1}-v_i^k), \quad \delta_x^2 v_i^k = \frac{1}{h^2}(v_{i+1}^k-2v_i^k+v_{i-1}^k), \quad \delta_t^2 v_i^k = \frac{1}{\tau^2}(v_i^{k+1}-2v_i^k+v_i^{k-1}), \\ & D_{\hat{t}}v_i^k = \frac{1}{2\tau}(v_i^{k+1}-v_i^{k-1}). \end{align*} $

记$ V_h = \{ v|v = \{v_i|0\le i\le m\}\; \mbox{为}\; \Omega_h\; \mbox{上的网格函数}\} $, 设$ v\in V_h $, 引进下列网格函数的模与半模

$ \begin{align*} \|v\| = \sqrt{\frac{h}{2}(v_0^2+ 2{\sum\limits_{i = 1}^{m-1}} v_i^2+v_m^2)}, \; |v|_1 = {\sqrt{\frac{1}{h}{\sum\limits_{i = 1}^{m-1}}(v_i-v_{i-1})^2}}, \; \|v\|_{\infty} = {\max\limits_{0\le i\le m}}|v_i|. \end{align*} $

引理 2.1 ^[3]   记$ \zeta(s) = (1-s)^3[5-3(1-s)^2], s\in[0, 1] $.

(1) 若$ g(x)\in C^6[x_0, x_1] $, 则有

$ \begin{align*} &\frac{1}{6}[5g''(x_0)+g''(x_1)]-\frac{2}{h}\left[ \frac{g(x_1)-g(x_0)}{h}-g'(x_0)\right]\\ = &-\frac{h}{6}g'''(x_0)+\frac{h^3}{90}g^{(5)}(x_0) +\frac{h^4}{180}\int_0^1g^{(6)}(x_0+sh)\zeta(s)ds. \end{align*} $

(2) 若$ g(x)\in C^6[x_{m-1}, x_m] $, 则有

$ \begin{align*} &\frac{1}{6}[g''(x_{m-1})+5g''(x_m)]-\frac{2}{h}\left[ g'(x_m)-\frac{g(x_m)-g(x_{m-1})}{h}\right]\\ = &-\frac{h}{6}g'''(x_m)+\frac{h^3}{90}g^{(5)}(x_m) +\frac{h^4}{180}\int_0^1g^{(6)}(x_m-sh)\zeta(s)ds. \end{align*} $

(3) 若$ g(x)\in C^6[x_{i-1}, x_{i+1}], 1\leq i\leq m-1 $, 则有

$ \begin{align*} &\frac{1}{12}[g''(x_{i-1})+10g''(x_i)+g''(x_{i+1})]-\frac{1}{h^2}[g(x_{i-1})-2g(x_i)+g(x_{i+1})]\\ = &\frac{h^4}{360}\int_0^1[g^{(6)}(x_i+sh)+g^{(6)}(x_i-sh)]\zeta(s)ds. \end{align*} $

引理 2.2 ^[5]  设$ v\in V_h $, 则有$ |v|_1^2\le \dfrac{4}{h^2}\|v\|^2 $, 且对任意$ \varepsilon\!>\!0 $, 有

$ \|v\|_{\infty}^2\le \varepsilon|v|_1^2+\left(\dfrac{1}{\varepsilon}+\dfrac{1}{b-a}\right)\|v\|^2. $

引理 2.3^[14]   设$ h $和$ c $为两个常数, 且$ h>0 $, 若$ g(x)\in C^2[c-h, c+h] $, 则有

$ \begin{align*} &g(c) = \dfrac{1}{2}[g(c-h)+g(c+h)]-\dfrac{h^2}{2}g''(\xi_0), \xi_0\in(c-h, c+h). \end{align*} $

引理 2.4 ^[15]  设$ \{v_i^k|0\le i\le m, 0\le k\le n\} $为$ \Omega_{h\tau} $上的网格函数, 则

$ \|v^k\|^2\le 2k\tau^2 \sum\limits_{l = 0}^{k-1}\|\delta_tv^{l+1/2}\|^2+2\|v^0\|^2. $

引理 2.5  设$ \{v_i^k|0\le i\le m, 0\le k\le n\} $为$ \Omega_{h\tau} $上的网格函数, 记

$ E^k = \frac{5}{6}\|\delta_tv^{k+1/2}\|^2+\frac{h}{6}\sum\limits_{i = 0}^{m-1}(\delta_tv_i^{k+1/2})(\delta_tv_{i+1}^{k+1/2})+\frac{\alpha}{2}(|v^{k+1}|_1^2+|v^{k}|_1^2), \; 0\le k\le n-1, $

则当$ 0\le k\le n-1 $时, 有

$ \frac{2}{3}\|\delta_tv^{k+1/2}\|^2+\frac{\alpha}{2}(|v^{k+1}|_1^2+|v^{k}|_1^2)\le E^k\le \|\delta_tv^{k+1/2}\|^2+\frac{\alpha}{2}(|v^{k+1}|_1^2+|v^{k}|_1^2). $

证  由不等式

$ h\sum\limits_{i = 0}^{m-1}\big|(\delta_{t}v_{i}^{k+\frac{1}{2}})(\delta_{t}v_{i+1}^{k+\frac{1}{2}})\big|\leq \frac{h}{2}\sum\limits_{i = 0}^{m-1}\big[(\delta_{t}v_{i}^{k+\frac{1}{2}})^{2}+(\delta_{t}v_{i+1}^{k+\frac{1}{2}})^{2}\big] = \|\delta_{t}v^{k+\frac{1}{2}}\|^{2}, $

易得引理2.5.

记$ U_{i}^{k} = u(x_{i}, t_{k}), \; 0\leq i\leq m, \; 0\leq k\leq n $, 则$ U = [U_i^k|0\leq i\leq m, 0\leq k\leq n] $为定义在$ \Omega_{h\tau} $上的网格函数.再设$ g\in{V_h} $, 定义算子$ \mathscr{P} $:

$ \begin{align*} (\mathscr{P}g)_i\! = \!\left\{\begin{aligned}\! &\frac{1}{12}(g_{i-1}\!+\!10g_i\!+\!g_{i+1}), 1\leq i\leq m-1, \\ \!&\frac{1}{6}(5g_0+g_1), \quad i = 0, \\ \!&\frac{1}{6}(g_{m-1}+5g_m), \quad i = m. \end{aligned}\right. \end{align*} $

令$ v = \frac{\partial^2 u}{\partial x^2} $, 记$ V_i^k = v(x_i, t_k), 0\leq i\leq m, \; 0\leq k\leq n $处考虑方程(1.1), 则有

$ \begin{align} \frac{\partial^2 u}{\partial t^2}(x, t)- \alpha v(x, t)+g{u(x, t)} = f(x, t), \quad a<x<b, \; 0<t\leq T, \end{align} $

(3.1)

在结点$ (x_i, t_k) $处考虑方程(3.1), 并对两边作用算子$ \mathscr{P} $, 且$ 1\!\leq\! i\!\leq\!m-1\!, \; 0\!\leq\! k\!\leq\!n-1\! $, 可得

$ \begin{align} \mathscr{P}\frac{\partial^2 u}{\partial t^2}(x_i, t_k)- \alpha \mathscr{P}V(x_i, t_k)+\mathscr{P}g{U(x_i, t_k)} = \mathscr{P}f(x_i, t_k). \end{align} $

(3.2)

根据引理2.1, 边界条件及作用算子$ \mathscr{P} $可得

$ \begin{align} \mathscr{P}V_0^k = &\frac{2}{h}\delta_x U_{1/2}^k+\frac{h}{6\alpha}\frac{\partial f}{\partial x}(x_0, t_k)-\frac{h^3}{90\alpha}\frac{\partial {f^3}}{\partial {x^3}}(x_0, t_k)+\frac{1}{\alpha}g'(U_0^k)\frac{\partial f}{\partial x}(x_0, t_k) \\ & +\frac{1}{\alpha}\frac{\partial f^3}{\partial t^2 \partial x}(x_0, t_k)+O(h^4), \; 0\leq k\leq n-1, \end{align} $

(3.3)

$ \begin{align} \mathscr{P}V_m^k = &-\frac{2}{h}\delta_x U_{m-1/2}^k-\frac{h}{6\alpha}\frac{\partial f}{\partial x}(x_m, t_k)+\frac{h^3}{90\alpha}\frac{\partial {f^3}}{\partial {x^3}}(x_m, t_k)+\frac{1}{\alpha}g'(U_m^k)\frac{\partial f}{\partial x}(x_m, t_k)\\ & +\frac{1}{\alpha}\frac{\partial f^3}{\partial t^2 \partial x}(x_m, t_k)+O(h^4), \; 0\leq k\leq n-1, \end{align} $

(3.4)

$ \begin{align} \mathscr{P}V_i^k = &\delta_xU_i^k+O(h^4), \quad 0\leq i\leq m, \; 1\leq k\leq n-1. \end{align} $

(3.5)

由Taylor展开知

$ \begin{align} &\frac {\partial^2 u}{\partial t^2} = \delta_t^2 U_i^k-\frac{\tau^2}{12}\frac{\partial^4}{\partial t^4}(x_i, \eta_{ik}), t_{k-1}<\eta_{ik}<t_{k+1}, \end{align} $

(3.6)

$ \begin{align} &\frac{\partial^2 u}{\partial t^2}(x_i, t_k) = \frac{1}{2}[\frac{\partial^2 u}{\partial x^2}(x_i, t_{k-1})+\frac{\partial^2 u}{\partial x^2}(x_i, t_{k+1})]-\frac{\tau^2}{2}\frac{\partial^4 u}{\partial x^2\partial t^2}(x_i, \xi_{ik}), t_{k-1}<\xi_{ik}<t_{k+1}. \end{align} $

(3.7)

将(3.6)–(3.7)式带入到(3.2)式, 且$ 0\leq i\leq m-1, 0\leq k \leq n-1, $得到

$ \begin{align} \mathscr{P}\frac{\partial^2 u}{\partial x^2}(x_i, t_k) -\!\frac{\alpha}{2}(\mathscr{P}V_i^{k+1}\!+\!\mathscr{P}V_i^{k-1})\!+\!\mathscr{P}g(U_i^k) \! = \!\mathscr{P}f(x_i, t_k). \end{align} $

(3.8)

由(3.3)–(3.5)式求得$ \mathscr{P}V_i^{k+1}, \mathscr{P}V_i^{k-1} $, 将其带入到(3.8)式, 结合(3.7)–(3.8)式和引理2.3得

$ \begin{align} &\mathscr{P}\delta_t^2 U_i^k \!-\!\alpha\delta_x^2 U_i^{\overline{k}}+\mathscr{P}g(U_i^k) = \mathscr{P}f(x_i, t_k)+R_i^k, 0\leq i\leq m-1, 0\leq k\leq n-1, \end{align} $

(3.9)

$ \begin{align} &\mathscr{P}\delta_t^2 U_0^k \!-\!\frac{2\alpha}{h}\delta_x U_{1/2}^{\overline{k}}\!-\!\frac{h}{6}\frac{\partial f}{\partial x}(x_0, t_k)+\frac{h^3}{90}[\!\frac{\partial^3f}{\partial x^3}(x_0, t_k)\!+\!\frac{1}{\alpha}g'(U_0^k)\frac{\partial f}{\partial x}(x_0, t_k)\!+\frac{1}{\alpha}\frac{\partial^3 f}{\partial t^2 \partial x}(x_0, t_k)]\\ &+\mathscr{P}g(U_0^k) = \mathscr{P}f(x_0, t_k)+R_0^k, 0\leq k\leq n-1, \end{align} $

(3.10)

$ \begin{align} &\!\mathscr{P}\delta_t^2 U_m^k\! \!-\!\frac{2\alpha}{h}\delta_x U_{1/2}^{\overline{k}}\!+\!\frac{h}{6}\frac{\partial f}{\partial x}(x_m, t_k)\!-\!\frac{h^3}{90}[\frac{\partial^3f }{\partial x^3}(x_m, t_k)+\!\frac{1}{\alpha}g'(U_0^k)\!\frac{\partial f}{\partial x}(x_m, t_k)\!+\!\frac{1}{\alpha}\frac{\partial^3 f}{\partial t^2 \partial x}(x_m, t_k)]\!\\ &+\mathscr{P}g(U_m^k) = \mathscr{P}f(x_m, t_k)+R_m^k, 0\leq k\leq n-1, \end{align} $

(3.11)

其中$ |R_i^k|\!\leq\! M_1(\tau^2+h^4), 0\!\leq \! i \!\leq\! m-1, 0\leq\! k\! \leq\! n-1\!, M_1 $为正常数, 由初始条件有$ u(x_i, 0)\! = \!\varphi(x_i), 0\!\leq\! i\!\leq\! m $, 并由(1.1)式且根据带积分余项的Taylor展开式以及初始条件得到

$ \begin{align} &U_i^1 = \varphi(x_i)+\tau\psi(x_i)+\frac{\tau^2}{2}[\alpha\varphi''(x_i)+f(x_i, 0)-g(\varphi(x_i)]+r_i^1, 0 \leq i\leq m, \end{align} $

(3.12)

其中

$ r_i^1 = \frac{1}{2} \int_{0}^{\tau}(\tau-t)^2\dfrac{\partial^3 u}{\partial t^3}(x_i, t)dt, 0\leq i\leq m. $

易知$ |r_i^1|\leq M_2\tau^3, M_2 $为正常数, 令

$ \begin{align*} &G(U_i^k, x_i, t_k) = \frac{h}{6}\frac{\partial f}{\partial x}(x_i, t_k)-\frac{h^3}{90}\frac{\partial^3 f}{\partial x^3}(x_i, t_k)+\frac{1}{\alpha}g'(U_i^k)\frac{\partial f}{\partial x}(x_i, t_k)+\frac{1}{\alpha}\frac{\partial^3 f}{\partial t^2\partial x}(x_i, t_k), \\ &0\leq i\leq m, 0\leq k\leq n. \end{align*} $

在(3.9)–(3.12)式中分别略去小量项$ R_i^k, r_i^1 $, 并用$ u_i^k $代替$ U_i^k $, 可得如下差分格式

$ \begin{align} &\mathscr{P}\delta_t^2u_i^k-\alpha\delta_x^2u_{i}^{\overline {k}}+\mathscr{P}g(u_i^k) = \mathscr{P}f(x_i, t_k), \quad 1\leq i\leq m-1, 1\leq k\leq n-1, \end{align} $

(3.13)

$ \begin{align} &\mathscr{P}\delta_t^2u_0^k-\frac{2\alpha}{h}\delta_x u_{1/2}^{\overline {k}}+\mathscr{P}g(u_0^k) = \mathscr{P}f(x_0, t_k)+G(u_0^k, x_0, t_k), \quad 1\leq k\leq n-1, \end{align} $

(3.14)

$ \begin{align} &\mathscr{P}\delta_t^2u_m^k+\frac{2\alpha}{h}\delta_x u_{m-1/2}^{\overline {k}}+\mathscr{P}g(u_m^k) = \mathscr{P}f(x_m, t_k)+G(u_m^k, x_m, t_k), \quad 1\leq k\leq n-1, \end{align} $

(3.15)

$ \begin{align} &u_i^0 = \varphi(x_i), \quad 1\leq i\leq m, \end{align} $

(3.16)

$ \begin{align} &u_i^1 = \varphi(x_i)+\tau\psi(x_i)+\frac{\tau^2}{2}[\alpha\varphi''(x_i)+f(x_i, 0)-g(\varphi(x_i))], \quad 1\leq i\leq m. \end{align} $

(3.17)

定理1   设$ u(x_i, t_k) $是问题(1.1)–(1.3)式的解, $ \{u_i^k|0\leq i\leq m, 0\leq k\leq n\} $是差分格式的解(3.13)–(3.17)的解, 记$ e_i^k = u(x_i, t_k)-u_i^k $, 则当$ \tau\leq \sqrt\frac{C_0+1}{2\tilde{c}}, h\leq \sqrt\frac{C_0+1}{2\tilde{c}} $时, 有

$ \begin{equation} \|e^k\|_{\infty}\leq \tilde{c}(\tau^2+h^4), 1\leq k\leq n, \end{equation} $

(4.1)

其中

$ \begin{align*} \tilde{c} = \max\{\sqrt{\varepsilon M_3^2+M_2^2(\frac{1}{\varepsilon}+\frac{1}{b-a})}, \sqrt{e^{MT}(1+\frac{\alpha}{2}+2T)M^2[\frac{2\varepsilon}{\alpha}+3T^2(\frac{1}{\varepsilon}+\frac{1}{b-a})]}\}. \end{align*} $

证   将(3.9)–(3.12)式分别与(3.13)–(3.17)式相减, 且$ 1\leq k\leq n-1, $得到误差方程

$ \begin{align} &\mathscr{P}\delta_t^2e_i^k -\alpha\delta_x^2e_{i}^{\overline {k}} = R_i^k-\mathscr{P}g(U_i^k)-\mathscr{P}g(u_i^k), \quad 1\leq i\leq m-1, \end{align} $

(4.2)

$ \begin{align} &\mathscr{P}\delta_t^2e_0^k -\frac{2\alpha}{h}\delta_x e_{1/2}^{\overline {k}} = R_0^k-\frac{h^3}{90\alpha}[g'(U_0^k)-g'(u_0^k)\frac{\partial f}{\partial x}(x_0, t_k)-\mathscr{P}[g(U_0^k)-g(u_0^k)], \end{align} $

(4.3)

$ \begin{align} &\!\mathscr{P}\delta_t^2e_m^k\! +\frac{2\alpha}{h}\delta_x e_{m-1/2}^{\overline {k}}\! = \!R_m^k-\!\frac{h^3}{90\alpha}[g'(U_m^k)-g'(u_m^k)\frac{\partial f}{\partial x}(x_m, t_k)\!-\!\mathscr{P}[g(U_m^k)-g(u_m^k)], \end{align} $

(4.4)

$ \begin{align} &e_i^0 = \varphi(x_i), e_i^1 = r_i^1, \quad 1\leq i\leq m. \end{align} $

(4.5)

令

$ M_3 = \!\dfrac{1}{6}\max\{|\dfrac{\partial^4 u}{\partial x\partial t^3}|, a \leq x \leq b, 0\leq t\leq T\}, $

则有

$ \delta_x e_{i+1/2}^l \leq M_3\tau^3, 0\leq i\leq m-1, $

从而得到$ \!|e_i^1|_1^2\leq c_1^2(b-a)\tau^6\! $, 由$ \!|r_1^1|\leq M_2(\tau^2+h^4)\! $可知$ \|e^1\|^2\leq M_2^2(b-a)\tau^6 $.由引理2.2得到

$ \begin{equation} \|e^1\|_\infty^2\leq \varepsilon|e|_1^2+(\frac{1}{\varepsilon}+\frac{1}{b-a})\|e^k\|^2\leq [\varepsilon |e^k|_1^2+M_2^2(\frac{1}{\varepsilon}+\frac{1}{b-a})](b-a)\tau^6. \end{equation} $

(4.6)

显然, (4.1)式对于$ k = 1 $成立, 现假设(4.1)式对于$ 1\leq k\leq l $成立, 其中$ 1\leq l\leq \!n-1\! $.下面证明(4.2)式对$ k = l+1 $也成立, 用$ 2h D_{\widehat{t}}e_i^k, hD_{\widehat{t}}e_0^k, hD_{\widehat{t}}e_m^k $, 分别乘方程(4.2)–(4.4)三式, 并关于$ i $从1到$ m-1 $求和, 且将三式相加可得$ J_1+J_2 = J_3 $, 其中

$ \begin{align*} J_1 = &2h\sum\limits_{i = 1}^{m-1}(\mathscr{P}{\delta}_t^2{ e_i^k})(D_{\hat{t}}e_i^k) +h(\mathscr{P}{\delta_t^2}{ e_0^k})(D_{\hat{t}}e_0^k)+h(\mathscr{P}{\delta_t^2}{ e_m^k})(D_{\hat{t}}e_m^k)\\ \leq& \frac{5}{6\tau}(\|\delta_t e^{k+1/2}\|^2-\|{\delta_t}e^{k-1/2}\|^2) +\frac{h}{6\tau}\sum\limits_{i = 0}^{m-1}(\delta_t e_i^{k+1/2}\delta_t e_{i+1}^ {k+1/2}-\delta_t e_i^{k-1/2}\delta_t e_{i+1}^ {k-1/2}) , \\ J_2 = & -2h \sum\limits_{i = 1}^{m-1} \alpha ({\delta_x^2}{ e_i^{\overline{k}}})D_{\hat{t}}e_i^k-2\alpha\delta_x e_{1/2}^{\bar{k}}D_{\hat{t}}e_m^k+2\alpha\delta_x e_{m-1/2}^{\bar{k}} D_{\hat{t}} e_m^k\leq\frac{\alpha}{2\tau}(|e^{k+1}|_1^2-|e^{k-1}|_1^2), \\ J_3 = & 2h \sum\limits_{i = 1}^{m-1}R_i^k D_{\hat{t}} e_i^k +h R_0^k D_{\hat{t}} e_0^k +hR_m^k D_{\hat{t}} e_m^k + 2h \sum\limits_{i = 1}^{m-1} a_i^k (D_{\hat{t}} e_0^k) + hb_i^k(D_{\hat{t}} e_0^k) + hc_i^k(D_{\hat{t}} e_m^k) \\ \leq& (\frac{1}{2}+\beta)(\|\delta_t e^{k+1/2}\|^2+\|\delta_t e^{k-1/2}\|^2)+2\beta\|e^k\|+\|R^k\|. \end{align*} $

上述$ \beta $为常数, 其中$ a_i^k, b_i^k, c_i^k $如下

$ \begin{align*} & a_i^k = -\mathscr{P}[g(U_i^k)-g(u_i^k)], \\ & b_i^k = -\frac{h^3}{90\alpha} [g'(U_0^k)-g'(u_0^k)]\frac{\partial f}{\partial x}(x_0, t_k)- \mathscr{P}[g(U_0^k)-g(u_0^k)], 0\leq k\leq n-1, \\ &c_i^k = \frac{h^3}{90\alpha} [g'(U_m^k)-g'(u_m^k)]\frac{\partial f}{\partial x}(x_m, t_k)- \mathscr{P}[g(U_m^k)-g(u_m^k)], 0\leq k\leq n-1. \end{align*} $

令

$ E^k = \dfrac{5}{6}\|\delta_t e^{k+1/2}\|^2+\dfrac{h}{6}\sum\limits_{i = 0}^{m-1}\delta_t e_i^{k+1/2}\delta_t e_{i+1}^{k+1/2} +\dfrac{\alpha}{2}(|e^{k+1}|_1^2+|e^k|_1^2), $

则$ J_1+J_2 = J_3 $可写为

$ \begin{equation} E^k-E^{k-1}\leq (\frac{3}{4}+\frac{3\beta}{2})(E^k+E^{k-1})+3T\beta\tau^2 \sum _{l = 0}^{k-1} E^l+\tau\|R^k\|^2. \end{equation} $

(4.7)

记$ \lambda = \frac{3}{4}+\frac{3\beta}{2} $, 且当$ \lambda\tau\leq \frac{1}{2} $时, 则(4.7)式可表示为

$ \begin{equation*} E^k\leq\frac{1+\lambda\tau}{1-\lambda\tau}E^{k-1}+\frac{3T\beta\tau^2}{1-\lambda\tau}\sum \limits_{l = 0}^{k-1}E^l+2\tau\|R^k\|^2. \end{equation*} $

令$ F^k = \max (E^k, E^{k-1}, E^{k-2}, \cdots, E^1, E^0) $, 则有

$ F^k\leq[1+2(2\lambda+3kT^2\beta)\tau]F^{k-1}+2\tau\|R^k\|^2. $

记$ M_4 = 2(2\lambda+3kT^2\beta), M_4 $为正整数, 故$ F^k\leq(1+M_4\tau)F^{k-1}+2\tau\|R^k\|^2 $, 由Granwall不等式可得到

$ F^k\leq e^{M_4 k\tau}(F^0+2\tau\sum\limits_{l = 1}^{k-1}\|F^k\|^2). $

由$ F^k $的定义可知$ F^0 = E^0 $, 则有

$ E^k\leq e^{M_4 k\tau}(E^0+2\tau\sum\limits_{l = 1}^{k-1}\|R^l\|^2). $

取$ M = \max(M_1, M_2, M_3, M_4) $, 并由引理2.5可得

$ \begin{align*} E^k&\leq e^{MT}[\frac{5}{6}\|\delta_t e^{1/2}\|^2+\frac{h}{6}\sum\limits_{i = 0}^{m-1}\delta_t e_i^{1/2}\delta_t e_{i+1}^{1/2}+\frac{\alpha}{2}(|e^1|_1^2+|e^0|_1^2)+2\tau\sum\limits_{l = 1}^{k-1}\|R^k\|^2]\\ &\leq e^{MT}(1+\frac{\alpha}{2}+2T)M^2(\tau^2+h^4)^2. \end{align*} $

由此可得

$ \begin{equation} E^k\leq e^{MT}(1+\frac{\alpha}{2}+2T)M^2(\tau^2+h^4)^2. \end{equation} $

(4.8)

由引理2.4和引理2.5可知

$ \begin{equation} \|e^{k+1}\|^2\leq \frac{3T^2}{2}e^{MT}(1+\frac{\alpha}{2}+2T)M^2(\tau^2+h^4)^2, 0\leq k\leq n-1. \end{equation} $

(4.9)

由(4.8)–(4.9)两式和引理2.2可知, 对于任意的$ \varepsilon>0 $, 有

$ \begin{equation*} \|e^{k+1}\|_\infty^2\leq \varepsilon|e^{k+1}|_1^2+(\frac{1}{\varepsilon}+\frac{1}{b-a})\|e^{k+1}\|^2\leq\widetilde{c}(\tau^2+h^4)^2, \end{equation*} $

其中$ \widetilde{c} $在定理中已经说明, 因此上述结论对于$ l = k+1 $也成立由归纳原理知(4.1)式成立.

类似讨论差分格式的收敛性可以得到差分格式(3.13)–(3.17)式关于初值的稳定性.

定理2   设$ u_i^k, v_i^k $分别是差分格式(3.13)–(3.17)的解, 记$ \varepsilon_i^k = u_i^k-v_i^k $, 假设条件H1, H2, H3成立, 则当$ h, \tau $充分小时, 且$ \; 1\leq k\leq n-1 $时, 有

$ \begin{eqnarray*} &&\!\frac{2}{3}\|\delta_t{\varepsilon^{k+1/2}}\|^2 \!+\!\frac{\alpha}{2}(|\varepsilon^{k+1}|_1^2\!+\!|\varepsilon^k|_1^2)\\ &\leq& e^{B_1 k\tau}[\|\delta_t{\varepsilon^{1/2}}\|^2\!+\!\frac{\alpha}{2}(|\varepsilon^1|_1^2+|\varepsilon^0|_1^2) \!+\!B_2\|{\varepsilon^0}\|^2\!+\!2\tau\sum\limits_{l = 1}^{k-1}\|\varepsilon^l\|^2], \end{eqnarray*} $

且$ B_1, B_2 $是与$ h, \tau $无关的正的常数, 证明略.

本节利用构造的差分格式(3.13)–(3.17)计算下面的定解问题

$ \begin{eqnarray*} \left\{ \begin{array}{lll} \frac{\partial^2 u}{\partial t^2}-\frac{\partial^2 u}{\partial x^2}+u^2 = -[(x^2-1)^2+4x]\cos{t}+(x^2-1)^4\cos^2{t}, -1<x<1, 0<t<1, \\ u(x, 0) = (x^2-1)^2, \; \frac{\partial u}{\partial t}(x, 0) = 0, \; \; \; \; 0\leq x\leq 1, \\ \frac{\partial u}{\partial x}(-1, t) = 0, \; \frac{\partial u}{\partial x}(1, t) = 0, \; \; \; \; 0\leq t\leq 1. \end{array} \right. \end{eqnarray*} $

该问题的精确解为$ u(x, t) = (x^2-1)^2\cos{t} $.计算结果见表 6.1和表 6.2.表 6.1给出了在不同步长时某些节点处的误差, 表 6.2给出了不同步长时数值解的最大误差和误差比, 其中

$ E_i^k = u(x_i, t_k)-u_i^k, \; \; E_{\infty}(h, \tau) = \max\limits_{\begin{subarray}{l} 0\leq i\leq m, 0\leq k\leq n\end{subarray}}\Big|u(x_{i}, t_{k})-u_{i}^{k}\Big|. $

由表 6.2可以看出, 差分格式在无穷范数下的收敛阶为$ O(\tau^2+h^4) $, 这和理论分析结果一致.