非线性发展方程的有限差分方法

$1.4$ Hopf-Cole 变换与高阶差分格式

$1.4.1$ Hopf-Cole 变换

令
$\begin{aligned} &u(x, t)=-2 \nu \frac{w_{x}(x, t)}{w(x, t)} \end{aligned}\tag{1.66}$

则有

$\begin{aligned} &u_{t}=-2 \nu\left(\frac{w_{x}}{w}\right)_{t}=-2 \nu \frac{w_{x t} w-w_{x} w_{t}}{w^{2}}=-2 \nu\left(\frac{w_{t}}{w}\right)_{x} \\ &u_{x}=-2 \nu\left(\frac{w_{x}}{w}\right)_{x} \\ &u_{x x}=-2 \nu\left(\frac{w_{x}}{w}\right)_{x x} \end{aligned}$

将上式代入 (1.1), 得到

$-2 \nu\left(\frac{w_{t}}{w}\right)_{x}+\left(-2 \nu \frac{w_{x}}{w}\right)\left(-2 \nu\left(\frac{w_{x}}{w}\right)_{x}\right)=\nu\left[-2 \nu\left(\frac{w_{x}}{w}\right)_{x x}\right]$

即

$\left(\frac{w_{t}}{w}\right)_{x}-\nu\left[\left(\frac{w_{x}}{w}\right)^{2}\right]_{x}=\nu\left(\frac{w_{x}}{w}\right)_{x x} \text {, }$

或

$\left[\frac{w_{t}}{w}-\nu\left(\frac{w_{x}}{w}\right)^{2}-\nu\left(\frac{w_{x}}{w}\right)_{x}\right]_{x}=0$

上式又可以写成

$\left(\frac{w_{t}-\nu w_{x x}}{w}\right)_{x}=0$

因而

$\frac{w_{t}-\nu w_{x x}}{w}=q(t)$

可以将上式写成

$w_{t}-q(t) w=\nu w_{x x}$

上式两边同乘以 $e^{-\int_{0}^{t} q(s) \mathrm{d} s}$ , 则可得到

$\left[w e^{-\int_{0}^{t} q(s) \mathrm{d} s}\right]_{t}=\nu\left[w e^{-\int_{0}^{t} q(s) \mathrm{d} s}\right]_{x x}$

$\tilde{w}(x, t)=w(x, t) e^{-\int_{0}^{t} q(s) \mathrm{d} s}$

则

$-2 \nu \frac{\widetilde{w}_{x}}{\widetilde{w}}=-2 \nu \frac{w_{x}}{w}=u(x, t)$

即对于任意 $q(t)$ , 不影响 $u(x, t)$ . 因而取 $q(t)=0$ . 于是得到如下等价问题

$\begin{aligned} &w_{t}-\nu w_{x x}=0, \quad 0<x<L, 0<t \leqslant T \\ &w(x, 0)=\tilde{\varphi}(x), \quad 0 \leqslant x \leqslant L, \\ &w_{x}(0, t)=0, \quad w_{x}(L, t)=0, \quad 0 \leqslant t \leqslant T \end{aligned}\tag{1.67-1.69}$

其中

$\widetilde{\varphi}(x)=e^{-\frac{1}{2 \nu} \int_{0}^{x} \varphi(s) \mathrm{d} s} .\tag{1.70}$

称 (1.66) 为 Hopf-Cole 变换.

$1.4.2$ 差分格式的建立

下面给出几个带积分余项的数值微分公式.

引理 $1.4([20]) \quad$ 记 $\alpha(s)=(1-s)^{2}\left[5-(1-s)^{2}\right]$ .

(a) 设 $g(x) \in C^{6}\left[x_{0}, x_{1}\right]$ , 有

$\begin{aligned} & \frac{5}{6} g^{\prime \prime}\left(x_{0}\right)+\frac{1}{6} g^{\prime \prime}\left(x_{1}\right)-\frac{2}{h}\left[\frac{g\left(x_{1}\right)-g\left(x_{0}\right)}{h}-g^{\prime}\left(x_{0}\right)\right] \\ =&-\frac{h}{6} g^{\prime \prime \prime}\left(x_{0}\right)+\frac{h^{3}}{90} g^{(5)}\left(x_{0}\right)+\frac{h^{4}}{180} \int_{0}^{1} g^{(6)}\left(x_{0}+s h\right) \alpha(s) d s \\ =&-\frac{h}{6} g^{\prime \prime \prime}\left(x_{0}\right)+\frac{h^{3}}{90} g^{(5)}\left(x_{0}\right)+\frac{h^{4}}{240} g^{(6)}\left(x_{0}+\theta_{0} h\right), \quad \theta_{0} \in(0,1) \end{aligned}$

(b) 设 $g(x) \in C^{6}\left[x_{m-1}, x_{m}\right]$ , 有

$\begin{aligned} & \frac{1}{6} g^{\prime \prime}\left(x_{m-1}\right)+\frac{5}{6} g^{\prime \prime}\left(x_{m}\right)-\frac{2}{h}\left[g^{\prime}\left(x_{m}\right)-\frac{g\left(x_{m}\right)-g\left(x_{m-1}\right)}{h}\right] \\ =& \frac{h}{6} g^{\prime \prime \prime}\left(x_{m}\right)-\frac{h^{3}}{90} g^{(5)}\left(x_{m}\right)+\frac{h^{4}}{180} \int_{0}^{1} g^{(6)}\left(x_{m}-s h\right) \alpha(s) d s \\ =& \frac{h}{6} g^{\prime \prime \prime}\left(x_{m}\right)-\frac{h^{3}}{90} g^{(5)}\left(x_{m}\right)+\frac{h^{4}}{240} g^{(6)}\left(x_{m}-\theta_{m} h\right), \quad \theta_{m} \in(0,1) \end{aligned}$

$\begin{aligned} & \frac{1}{12}\left[g^{\prime \prime}\left(x_{i-1}\right)+10 g^{\prime \prime}\left(x_{i}\right)+g^{\prime \prime}\left(x_{i+1}\right)\right]-\frac{1}{h^{2}}\left[g\left(x_{i+1}\right)-2 g\left(x_{i}\right)+g\left(x_{i-1}\right)\right] \\ =& \frac{h^{4}}{360} \int_{0}^{1}\left[g^{(6)}\left(x_{i}+s h\right)+g^{(6)}\left(x_{i}-s h\right)\right] \alpha(s) d s \\ =& \frac{h^{4}}{240} g^{(6)}\left(x_{i}+\theta_{i} h\right), \quad \theta_{i} \in(-1,1) \end{aligned}$

设 $v \in \mathcal{U}_{h} .$ 定义平均值算子 $\mathcal{A}$ :

$\mathcal{A} v_{i}= \begin{cases}\frac{5}{6} v_{0}+\frac{1}{6} v_{1}, & i=0, \\ \frac{1}{12}\left(v_{i-1}+10 v_{i}+v_{i+1}\right), & 1 \leqslant i \leqslant m-1 \\ \frac{5}{6} v_{m}+\frac{1}{6} v_{m-1}, & i=m .\end{cases}$

设 $(1.67)-(1.70)$ 存在解 $w(x, t) \in C^{6,4}([0, L] \times[0, T])$ . 定义网格函数

$U_{i}^{k}=u\left(x_{i}, t_{k}\right), \quad w_{i}^{k}=W\left(x_{i}, t_{k}\right), \quad 0 \leqslant i \leqslant m, \quad 0 \leqslant k \leqslant n$

由 (1.67) 和 (1.69) 可得

$\begin{aligned} &w_{x x x}(0, t)=0, \quad w_{x x x}(L, t)=0, \quad 0 \leqslant t \leqslant T, \\ &w_{x x x x x}(0, t)=0, \quad w_{x x x x x}(L, t)=0, \quad 0 \leqslant t \leqslant T \end{aligned}\tag{1.71-1.72}$

在点 $\left(x_{i}, t_{k+\frac{1}{2}}\right)$ 处考虑方程 (1.67), 有

$w_{t}\left(x_{i}, t_{k+\frac{1}{2}}\right)-\nu w_{x x}\left(x_{i}, t_{k+\frac{1}{2}}\right)=0, \quad 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n-1$

应用引理 $1.2$ , 可得

$w_{t}\left(x_{i}, t_{k+\frac{1}{2}}\right)-\frac{\nu}{2}\left[w_{x x}\left(x_{i}, t_{k}\right)+w_{x x}\left(x_{i}, t_{k+1}\right)\right]=\left(R_{x t} w\right)_{i}^{k+\frac{1}{2}}$

$0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n-1$

其中

$\left(R_{x t} w\right)_{i}^{k+\frac{1}{2}}=-\frac{\nu \tau^{2}}{8} \frac{\partial^{4} w}{\partial x^{2} \partial t^{2}}\left(x_{i}, t_{k}+\eta_{i}^{k} \tau\right), \quad \eta_{i}^{k} \in(0,1), 0 \leqslant i \leqslant m$

用算子 $\mathcal{A}$ 作用上述等式的两边, 得

$\begin{gathered} \mathcal{A} w_{t}\left(x_{i}, t_{k+\frac{1}{2}}\right)-\frac{\nu}{2}\left[\mathcal{A} w_{x x}\left(x_{i}, t_{k}\right)+\mathcal{A} w_{x x}\left(x_{i}, t_{k+1}\right)\right]=\mathcal{A}\left(R_{x t} w\right)_{i}^{k+\frac{1}{2}} \\ 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1 \end{gathered}$

应用引理 $1.4$ 并注意到 (1.71)-(1.72), 得到

$\begin{aligned} &\mathcal{A} \delta_{t} W_{0}^{k+\frac{1}{2}}-\nu\left(\frac{2}{h} \delta_{x} W_{\frac{1}{2}}^{k+\frac{1}{2}}\right)=R_{0}^{k+\frac{1}{2}}, \quad 0 \leqslant k \leqslant n-1 \\ &\mathcal{A} \delta_{t} W_{i}^{k+\frac{1}{2}}-\nu \delta_{x}^{2} W_{i}^{k+\frac{1}{2}}=R_{i}^{k+\frac{1}{2}}, \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1 \\ &\mathcal{A} \delta_{t} W_{m}^{k+\frac{1}{2}}-\nu\left(-\frac{2}{h} \delta_{x} W_{m-\frac{1}{2}}^{k+\frac{1}{2}}\right)=R_{m}^{k+\frac{1}{2}}, \quad 0 \leqslant k \leqslant n-1 \end{aligned}\tag{1.73-1.75}$

存在常数 $c_{8}$ 使得

$\left|R_{i}^{k+\frac{1}{2}}\right| \leqslant c_{8}\left(\tau^{2}+h^{4}\right), \quad 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n-1 \tag{1.76}$

注意到初值条件

$W_{i}^{0}=\tilde{\varphi}\left(x_{i}\right), \quad 0 \leqslant i \leqslant m. \tag{1.77}$

在 (1.73)-(1.75) 中略去小量项, 对 (1.67)-(1.69) 建立如下差分格式

$\begin{aligned} &\mathcal{A} \delta_{t} w_{0}^{k+\frac{1}{2}}-\nu\left(\frac{2}{h} \delta_{x} w_{\frac{1}{2}}^{k+\frac{1}{2}}\right)=0, \quad 0 \leqslant k \leqslant n-1 \\ &\mathcal{A} \delta_{t} w_{i}^{k+\frac{1}{2}}-\nu \delta_{x}^{2} w_{i}^{k+\frac{1}{2}}=0, \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1, \\ &\mathcal{A} \delta_{t} w_{m}^{k+\frac{1}{2}}-\nu\left(-\frac{2}{h} \delta_{x} w_{m-\frac{1}{2}}^{k+\frac{1}{2}}\right)=0, \quad 0 \leqslant k \leqslant n-1 \\ &w_{i}^{0}=\widetilde{\varphi}\left(x_{i}\right), \quad 0 \leqslant i \leqslant m \end{aligned}\tag{1.78-1.81}$

$1.4.3$ 差分格式解的存在性和唯一性}

引理 $1.5$ 设 $v=\left(v_{0}, v_{1}, \cdots, v_{m}\right) \in \mathcal{U}_{h}$ , 则有

$\begin{aligned} &(\mathcal{A} v, v)=\|v\|^{2}-\frac{h^{2}}{12}|v|_{1}^{2} \\ &\frac{2}{3}\|v\|^{2} \leqslant(\mathcal{A} v, v) \leqslant\|v\|^{2} \end{aligned}$

证明由

$(\mathcal{A} v)_{i}= \begin{cases}v_{0}+\frac{h}{6} \delta_{x} v_{\frac{1}{2}}, & i=0 \\ v_{i}+\frac{h^{2}}{12} \delta_{x}^{2} v_{i}, & 1 \leqslant i \leqslant m-1 \\ v_{m}-\frac{h}{6} \delta_{x} v_{m-\frac{1}{2}}, & i=m\end{cases}$

得到

$\begin{aligned} (\mathcal{A} v, v) &=h\left[\frac{1}{2}\left(\mathcal{A} v_{0}\right) v_{0}+\sum_{i=1}^{m-1}\left(\mathcal{A} v_{i}\right) v_{i}+\frac{1}{2}\left(\mathcal{A} v_{m}\right) v_{m}\right] \\ &=h\left[\frac{1}{2}\left(v_{0}+\frac{h}{6} \delta_{x} v_{\frac{1}{2}}\right) v_{0}+\sum_{i=1}^{m-1}\left(v_{i}+\frac{h^{2}}{12} \delta_{x}^{2} v_{i}\right) v_{i}+\frac{1}{2}\left(v_{m}-\frac{h}{6} \delta_{x} v_{m-\frac{1}{2}}\right) v_{m}\right] \\ &=h\left(\frac{1}{2} v_{0}^{2}+\sum_{i=1}^{m-1} v_{i}^{2}+\frac{1}{2} v_{m}^{2}\right) \end{aligned}$

$\begin{aligned} &+\frac{h^{2}}{12}\left[\left(\delta_{x} v_{\frac{1}{2}}\right) v_{0}+h \sum_{i=1}^{m-1}\left(\delta_{x}^{2} v_{i}\right) v_{i}-\left(\delta_{x} v_{m-\frac{1}{2}}\right) v_{m}\right] \\ =&\|v\|^{2}-\frac{h^{2}}{12} \cdot h \sum_{i=0}^{m-1}\left(\delta_{x} v_{i+\frac{1}{2}}\right)^{2} \\ =&\|v\|^{2}-\frac{h^{2}}{12}|v|_{1}^{2} . \end{aligned}$

易知

$(\mathcal{A} v, v) \leqslant\|v\|^{2}$

由

$|v|_{1}^{2} \leqslant \frac{4}{h^{2}}\|v\|^{2}$

易得

$(\mathcal{A} v, v) \geqslant\|v\|^{2}-\frac{h^{2}}{12} \cdot \frac{4}{h^{2}}\|v\|=\frac{2}{3}\|v\|^{2}$

定义 $\mathcal{U}_{h}$ 上的范数

$\|v\|_{\mathcal{A}}=\sqrt{(\mathcal{A} v, v)}$

由引理 $1.5$ 知 $\Vert v\Vert _{\mathcal{A}}$ 和 $\Vert v\Vert$ 等价.

定理 $1.11$ 差分格式 (1.78)-(1.81) 的解是存在唯一的.

证明第 0 层的值 $w^{0}$ 是由（1.81）给定. 设已得到第 $k$ 层的值 $w^{k}$ , 则由 (1.78)-(1.80) 可得关于第 $k+1$ 层值 $w^{k+1}$ 的线性方程组. 考虑其齐次方程组

$\begin{aligned} &\frac{1}{\tau} \mathcal{A} w_{0}^{k+1}-\nu \frac{1}{h} \delta_{x} w_{\frac{1}{2}}^{k+1}=0 \\ &\frac{1}{\tau} \mathcal{A} w_{i}^{k+1}-\nu \frac{1}{2} \delta_{x}^{2} w_{i}^{k+1}=0, \quad 1 \leqslant i \leqslant m-1 \\ &\frac{1}{\tau} \mathcal{A} w_{m}^{k+\frac{1}{2}}-\nu\left(-\frac{1}{h} \delta_{x} w_{m-\frac{1}{2}}^{k+1}\right)=0 \end{aligned}\tag{1.82-1.84}$

用 $\frac{1}{2} h w_{0}^{k+1}$ 与 $(1.82)$ 相乘, 用 $h w_{i}^{k+1}$ 与 $(1.83)$ 相乘, 用 $\frac{1}{2} h w_{m}^{k+\frac{1}{2}}$ 与 $(1.84)$ 相乘, 并将所得结果相加, 得

$\frac{1}{\tau}\left(\mathcal{A} w^{k+1}, w^{k+1}\right)-\frac{1}{2} \nu\left[w_{0}^{k+\frac{1}{2}} \delta_{x} w_{\frac{1}{2}}^{k+1}+h \sum_{i=1}^{m-1} w_{i}^{k+1} \delta_{x}^{2} w_{i}^{k+1}-w_{m}^{k+1} \delta_{x} w_{m-\frac{1}{2}}^{k+1}\right]=0$

由上式易得

$\frac{1}{\tau}\left\|w^{k+1}\right\|_{\mathcal{A}}^{2}+\frac{1}{2} \nu\left|w^{k+1}\right|_{1}^{2}=0$

因而

$w^{k+1}=0$

即 $(1.82)-(1.84)$ 只有零解. 于是 $(1.78)-(1.80)$ 唯一确定 $w^{k+1}$ .

$1.4.4$ 差分格式解的收敛性

定理 $1.12$ 设 $\left\{W_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}$ 是问题 (1.67)-(1.69) 的解, $\left\{w_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}$ 是差分格式 $(1.78)-(1.81)$ 的解. 令

$e_{i}^{k}=W_{i}^{k}-w_{i}^{k}, \quad 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n$

则存在常数 $c_{9}, c_{10}$ 使得

$\begin{aligned} &\left\|e^{k}\right\| \leqslant c_{9}\left(\tau^{2}+h^{4}\right), \quad 0 \leqslant k \leqslant n \\ &\left|e^{k}\right|_{1} \leqslant c_{10}\left(\tau^{2}+h^{4}\right), \quad 0 \leqslant k \leqslant n \\ &\left\|e^{k}\right\|_{\infty} \leqslant \frac{\sqrt{L}}{2} c_{10}\left(\tau^{2}+h^{4}\right), \quad 0 \leqslant k \leqslant n \end{aligned}\tag{1.85-1.87}$

证明将 (1.73)-(1.75), (1.77) 和 (1.78)-(1.81) 相减, 得到误差方程组

$\begin{aligned} &\mathcal{A} \delta_{t} e_{0}^{k+\frac{1}{2}}-\nu \cdot \frac{2}{h} \delta_{x} e_{\frac{1}{2}}^{k+\frac{1}{2}}=R_{0}^{k+\frac{1}{2}}, \quad 0 \leqslant k \leqslant n-1 \\ &\mathcal{A} \delta_{t} e_{i}^{k+\frac{1}{2}}-\nu \delta_{x}^{2} e_{i}^{k+\frac{1}{2}}=R_{i}^{k+\frac{1}{2}}, \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1, \\ &\mathcal{A} \delta_{t} e_{m}^{k+\frac{1}{2}}-\nu\left(-\frac{2}{h} \delta_{x} e_{m-\frac{1}{2}}^{k+\frac{1}{2}}\right)=R_{m}^{k+\frac{1}{2}}, \quad 0 \leqslant k \leqslant n-1 \\ &e_{i}^{0}=0, \quad 0 \leqslant i \leqslant m . \\ &\text { (I) 用 } \frac{1}{2} h e_{0}^{k+\frac{1}{2}} \text { 与 }(1.88) \text { 相乘, 用 } h e_{i}^{k+\frac{1}{2}} \text { 与 }(1.89) \text { 相乘, 用 } \frac{1}{2} h e_{m}^{k+\frac{1}{2}} \text { 与 } \end{aligned}\tag{1.88-1.91}$

相乘, 并将结果相加, 得到

$\begin{aligned} & \frac{1}{2 \tau}\left(\left\|e^{k+1}\right\|_{\mathcal{A}}^{2}-\left\|e^{k}\right\|_{\mathcal{A}}^{2}\right)-\nu\left(e_{0}^{k+\frac{1}{2}} \delta_{x} e_{\frac{1}{2}}^{k+\frac{1}{2}}+h \sum_{i=1}^{m-1} e_{i}^{k+\frac{1}{2}} \delta_{x}^{2} e_{i}^{k+\frac{1}{2}}-e_{m}^{k+\frac{1}{2}} \delta_{x} e_{m-\frac{1}{2}}^{k+\frac{1}{2}}\right) \\ =&\left(R^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right), \quad 0 \leqslant k \leqslant n-1 \end{aligned}$

即

$\frac{1}{2 \tau}\left(\left\|e^{k+1}\right\|_{\mathcal{A}}^{2}-\left\|e^{k}\right\|_{\mathcal{A}}^{2}\right)+\nu\left|e^{k+\frac{1}{2}}\right|_{1}^{2}=\left(R^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right), \quad 0 \leqslant k \leqslant n-1$

对上式右端用 Cauchy-Schwarz 不等式, 并应用引理 $1.5$ , 得

$\begin{aligned} & \frac{1}{2 \tau}\left(\left\|e^{k+1}\right\|_{\mathcal{A}}^{2}-\left\|e^{k}\right\|_{\mathcal{A}}^{2}\right) \\ \leqslant &\left\|R^{k+\frac{1}{2}}\right\| \cdot\left\|e^{k+\frac{1}{2}}\right\| \\ \leqslant & \sqrt{\frac{3}{2}}\left\|R^{k+\frac{1}{2}}\right\| \cdot\left\|e^{k+\frac{1}{2}}\right\|_{\mathcal{A}} \\ \leqslant & \sqrt{\frac{3}{2}}\left\|R^{k+\frac{1}{2}}\right\| \cdot \frac{\left\|e^{k+1}\right\|_{\mathcal{A}}+\left\|e^{k}\right\|_{\mathcal{A}}}{2} \end{aligned}$

两边约去 $\frac{1}{2}\left(\left\|e^{k+1}\right\|_{\mathcal{A}}+\left\|e^{k}\right\|_{\mathcal{A}}\right),$ 得到

$\frac{1}{\tau}\left(\left\|e^{k+1}\right\|_{\mathcal{A}}-\left\|e^{k}\right\|_{\mathcal{A}}\right) \leqslant \sqrt{\frac{3}{2}}\left\|R^{k+\frac{1}{2}}\right\|, \quad 0 \leqslant k \leqslant n-1$

因而

$\left\|e^{k+1}\right\|_{\mathcal{A}} \leqslant\left\|e^{0}\right\|_{\mathcal{A}}+\sqrt{\frac{3}{2}} \tau \sum_{l=0}^{k}\left\|R^{l+\frac{1}{2}}\right\|, \quad 0 \leqslant k \leqslant n-1$

由 $(1.76)$ 和 $(1.91)$ , 得到

$\left\|e^{k+1}\right\|_{A} \leqslant \sqrt{\frac{3}{2}}(k+1) \tau \sqrt{L} c_{8}\left(\tau^{2}+h^{4}\right) \leqslant \sqrt{\frac{3}{2}} T \sqrt{L} c_{8}\left(\tau^{2}+h^{4}\right), \quad 0 \leqslant k \leqslant n-1$

再次应用引理 $1.5$ , 得

$\left\|e^{k}\right\| \leqslant \sqrt{\frac{3}{2}}\left\|e^{k}\right\|_{\mathcal{A}} \leqslant \frac{3}{2} T \sqrt{L} c_{8}\left(\tau^{2}+h^{2}\right), \quad 1 \leqslant k \leqslant n$

(II) 用 $\frac{1}{2} h \delta_{t} e_{0}^{k+\frac{1}{2}}$ 与 $(1.88)$ 相乘, 用 $h \delta_{t} e_{i}^{k+\frac{1}{2}}$ 与 (1.89) 相乘, 用 $\frac{1}{2} h \delta_{t} e_{m}^{k+\frac{1}{2}}$ 与 (1.90) 相乘, 并将结果相加, 得

$\left\|\delta_{t} e^{k+\frac{1}{2}}\right\|_{\mathcal{A}}^{2}-\nu\left[\left(\delta_{x} e_{\frac{1}{2}}^{k+\frac{1}{2}}\right) \delta_{t} e_{0}^{k+\frac{1}{2}}+h \sum_{i=1}^{m-1}\left(\delta_{x}^{2} e_{i}^{k+\frac{1}{2}}\right) \delta_{t} e_{i}^{k+\frac{1}{2}}-\left(\delta_{x} e_{m-\frac{1}{2}}^{k+\frac{1}{2}}\right) \delta_{t} e_{m}^{k+\frac{1}{2}}\right]$

$=\left(R^{k+\frac{1}{2}}, \delta_{t} e^{k+\frac{1}{2}}\right), \quad 0 \leqslant k \leqslant n-1$

即

$\begin{aligned} &\left\|\delta_{t} e^{k+\frac{1}{2}}\right\|_{\mathcal{A}}^{2}+\nu \cdot \frac{1}{2 \tau}\left(\left|e^{k+1}\right|_{1}^{2}-\left|e^{k}\right|_{1}^{2}\right) \\ =&\left(R^{k+\frac{1}{2}}, \delta_{t} e^{k+\frac{1}{2}}\right) \\ \leqslant &\left\|R^{k+\frac{1}{2}}\right\| \cdot\left\|\delta_{t} e^{k+\frac{1}{2}}\right\| \\ \leqslant & \frac{2}{3}\left\|\delta_{t} e^{k+\frac{1}{2}}\right\|^{2}+\frac{3}{8}\left\|R^{k+\frac{1}{2}}\right\|^{2} \\ \leqslant &\left\|\delta_{t} e^{k+\frac{1}{2}}\right\|_{\mathcal{A}}^{2}+\frac{3}{8}\left\|R^{k+\frac{1}{2}}\right\|^{2}, \quad 0 \leqslant k \leqslant n-1 \end{aligned}$

因而

$\frac{1}{2 \tau}\left(\left|e^{k+1}\right|_{1}^{2}-\left|e^{k}\right|_{1}^{2}\right) \leqslant \frac{3}{8 \nu}\left\|R^{k+\frac{1}{2}}\right\|^{2}, \quad 0 \leqslant k \leqslant n-1$

递推可得

$\left|e^{k+1}\right|_{1}^{2} \leqslant\left|e^{0}\right|_{1}^{2}+\frac{3}{4 \nu} \tau \sum_{l=0}^{k}\left\|R^{l+\frac{1}{2}}\right\|^{2} \leqslant \frac{3}{4 \nu}(k+1) \tau L c_{8}^{2}\left(\tau^{2}+h^{4}\right)^{2}, \quad 0 \leqslant k \leqslant n-1$

于是

$\left|e^{k}\right|_{1} \leqslant \frac{1}{2} \sqrt{\frac{3}{\nu} T L} c_{8}\left(\tau^{2}+h^{4}\right), \quad 1 \leqslant k \leqslant n$

(III) 由 $(1.86)$ 和引理 1.1(b) 得

$\left\|e^{k}\right\|_{\infty} \leqslant \frac{\sqrt{L}}{2}\left|e^{k}\right|_{1} \leqslant \frac{\sqrt{L}}{2} c_{10}\left(\tau^{2}+h^{4}\right), \quad 1 \leqslant k \leqslant n$

$1.4.5$ 原问题解的计算

设 $g(x) \in C^{5}\left[x_{0}, x_{m}\right]$ , 则存在常数 $\hat{c}_{1}, \hat{c}_{2}, \cdots, \hat{c}_{6}$ 使得

$g^{\prime}\left(x_{1}\right)=\hat{c}_{1} \delta_{x} g_{\frac{1}{2}}+\hat{c}_{2} \delta_{x} g_{\frac{3}{2}}+\hat{c}_{3} \delta_{x} g_{\frac{5}{2}}+\hat{c}_{4} \delta_{x} g_{\frac{7}{2}}+O\left(h^{4}\right)\tag{1.92}$

$g^{\prime}\left(x_{i}\right)=\hat{c}_{5} \delta_{x} g_{i-\frac{3}{2}}+\hat{c}_{6} \delta_{x} g_{i-\frac{1}{2}}+\hat{c}_{6} \delta_{x} g_{i+\frac{1}{2}}+\hat{c}_{5} \delta_{x} g_{i+\frac{3}{2}}+O\left(h^{4}\right)$

$2 \leqslant i \leqslant m-2, \tag{1.93}$

$g^{\prime}\left(x_{m-1}\right)=\hat{c}_{1} \delta_{x} g_{m-\frac{1}{2}}+\hat{c}_{2} \delta_{x} g_{m-\frac{3}{2}}+\hat{c}_{3} \delta_{x} g_{m-\frac{5}{2}}+\hat{c}_{4} \delta_{x} g_{m-\frac{7}{2}}+O\left(h^{4}\right)\tag{1.94}$

由变换 (1.66) 有

$u\left(x_{i}, t_{k}\right)=-2 \nu \frac{w_{x}\left(x_{i}, t_{k}\right)}{w\left(x_{i}, t_{k}\right)}$

利用 $(1.92)-(1.94)$ 可得

$\begin{gathered} U_{1}^{k}=-\frac{2 \nu}{W_{1}^{k}}\left(\hat{c}_{1} \delta_{x} W_{\frac{1}{2}}^{k}+\hat{c}_{2} W_{\frac{3}{2}}^{k}+\hat{c}_{3} \delta_{x} W_{\frac{5}{2}}^{k}+\hat{c}_{4} \delta_{x} W_{\frac{7}{2}}^{k}\right)+\hat{R}_{1}^{k}, \quad 1 \leqslant k \leqslant n \\ U_{i}^{k}=-\frac{2 \nu}{W_{i}^{k}}\left(\hat{c}_{5} \delta_{x} W_{i-\frac{3}{2}}^{k}+\hat{c}_{6} \delta_{x} W_{i-\frac{1}{2}}^{k}+\hat{c}_{6} \delta_{x} \hat{W}_{i+\frac{1}{2}}^{k}+\hat{c}_{5} \delta_{x} W_{i+\frac{3}{2}}^{k}\right)+\hat{R}_{i}^{k} \\ 2 \leqslant i \leqslant m-2,1 \leqslant k \leqslant n . \\ U_{m-1}^{k}=-\frac{2 \nu}{W_{m-1}^{k}}\left(\hat{c}_{4} \delta_{x} W_{m-\frac{7}{2}}^{k}+\hat{c}_{3} \delta_{x} W_{m-\frac{5}{2}}^{k}+\hat{c}_{2} \delta_{x} W_{m-\frac{3}{2}}^{k}+\hat{c}_{1} \delta_{x} W_{m-\frac{1}{2}}^{k}\right)+\hat{R}_{m-1}^{k} \\ 1 \leqslant k \leqslant n \end{gathered}\tag{1.95-1.97}$

存在常数 $c_{11}$ 使得

$\left|\hat{R}_{i}^{k}\right| \leqslant c_{11} h^{4}, \quad 1 \leqslant i \leqslant m-1,1 \leqslant k \leqslant n. \tag{1.98}$

在 (1.95)-(1.97) 中略去小量项, 得到如下计算格式

$\begin{gathered} u_{1}^{k}=-\frac{2 \nu}{w_{1}^{k}}\left(\hat{c}_{1} \delta_{x} w_{\frac{1}{2}}^{k}+\hat{c}_{2} \delta_{x} w_{\frac{3}{2}}^{k}+\hat{c}_{3} \delta_{x} w_{\frac{5}{2}}^{k}+\hat{c}_{4} \delta_{x} w_{\frac{7}{2}}^{k}\right), \quad 1 \leqslant k \leqslant n \\ u_{i}^{k}=-\frac{2 \nu}{w_{i}^{k}}\left(\hat{c}_{5} \delta_{x} w_{i-\frac{3}{2}}^{k}+\hat{c}_{6} \delta_{x} w_{i-\frac{1}{2}}^{k}+\hat{c}_{6} \delta_{x} w_{i+\frac{1}{2}}^{k}+\hat{c}_{5} \delta_{x} w_{i+\frac{3}{2}}^{k}\right) \\ 2 \leqslant i \leqslant m-2,1 \leqslant k \leqslant n, \\ u_{m-1}^{k}=-\frac{2 \nu}{w_{m-1}^{k}}\left(\hat{c}_{4} \delta_{x} w_{m-\frac{7}{2}}^{k}+\hat{c}_{3} \delta_{x} w_{m-\frac{5}{2}}^{k}+\hat{c}_{2} \delta_{x} w_{m-\frac{3}{2}}^{k}+\hat{c}_{1} \delta_{x} w_{m-\frac{1}{2}}^{k}\right) \\ 1 \leqslant k \leqslant n \end{gathered}\tag{1.99-1.101}$

利用定理 $1.12$ 的结果可以证明

$\sqrt{h \sum_{i=1}^{m-1}\left(U_{i}^{k}-u_{i}^{k}\right)^{2}} \leqslant c_{12}\left(\tau^{2}+h^{4}\right), \quad 1 \leqslant k \leqslant n$

$1.5$ 小结与延拓

本章讨论了 Burgers 方程的差分方法. 首先证明了问题 (1.1)-(1.3) 的解满足能量守恒性. 接着在 $1.2$ 节和 $1.3$ 节分别介绍了二层非线性差分格式和三层线性化差分格式. 证明了差分格式解的存在性、唯一性、有界性和收敛性. 三层线性化差分格式的有关结果主要取材于 [30].

对于问题 (1.1)-(1.3) 可建立如下二层线性化差分格式

$\begin{aligned} &\delta_{t} u_{i}^{k+\frac{1}{2}}+\frac{1}{2}\left(u_{i}^{k} \Delta_{x} u_{i}^{k+1}+u_{i}^{k+1} \Delta_{x} u_{i}^{k}\right)=\nu \delta_{x}^{2} u_{i}^{k+\frac{1}{2}}, \\ &\quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1 \\ &u_{i}^{0}=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1 \\ &u_{0}^{k}=0, \quad u_{m}^{k}=0, \quad 0 \leqslant k \leqslant n . \end{aligned}\tag{1.102-1.104}$

可以证明差分格式 (1.102)-(1.104) 是唯一可解的, 在无穷范数下关于时间步长和空间步长均是二阶收玫的.

文 [39] 研究了二维 Burgers 方程的二阶差分方法.

应用 Hopf-Cole 变换可将 Burgers 方程的初边值问题 (1.1)-(1.3) 变为线性的热传导方程的初边值问题 (1.67)-(1.69). 对 $(1.67)-(1.69)$ , 我们建立了紧致差分格式 $(1.78)-(1.81) .$ 证明了 (1.78)-(1.81) 解的存在性和唯一性以及解在无穷范数下关于时间步长 2 阶、空间步长 4 阶的收敛性. 如果在 (1.78)-(1.81) 中用单位算子 $\mathcal{I}$ 代替平均值算子 $\mathcal{A}$ , 得到如下格式

$\begin{aligned} &\delta_{t} w_{0}^{k+\frac{1}{2}}-\nu\left(\frac{2}{h} \delta_{x} w_{\frac{1}{2}}^{k+\frac{1}{2}}\right)=0, \quad 0 \leqslant k \leqslant n-1 \\ &\delta_{t} w_{i}^{k+\frac{1}{2}}-\nu \delta_{x}^{2} w_{i}^{k+\frac{1}{2}}=0, \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1 \\ &\delta_{t} w_{m}^{k+\frac{1}{2}}-\left(-\frac{2}{h} \delta_{x} w_{m-\frac{1}{2}}^{k+\frac{1}{2}}\right)=0, \quad 0 \leqslant k \leqslant n-1 \\ &w_{i}^{0}=\widetilde{\varphi}\left(x_{i}\right), \quad 0 \leqslant i \leqslant m \end{aligned}$

该差分格式是唯一可解的, 在无穷范数下关于时间步长和空间步长均是二阶收敛的.

我们借助于 Browder 定理证明了非线性方程组 (1.19)-(1.20) 解的存在性. 与 Browder 定理相伴的还有一个 Leray-Schauder 定理 [43]. 设 $H$ 是一个有限维内积空间, $\Vert\cdot\Vert$ 是导出范数. 考虑 $H \rightarrow H$ 的算子 $T_{\lambda}(w)$ , 其中 $\lambda \in[0,1]$ 为参数. 如果 $T_{\lambda}(w)$ 满足如下条件:

(a) $T_{\lambda}(w)$ 是 $H$ 上的连续算子;

(b) $T_{0}(w)=0$ 有唯一解;

则对任意 $\lambda \in[0,1], T_{\lambda}(w)=0$ 存在解. 特别地, $T_{1}(w)=0$ 存在解.

现在用上述结论来证明定理 $1.4$ , 即证明 $(1.19)-(1.20)$ 存在解.

令 $H=\dot{\mathcal{U}}_{h} .$ 对任意的 $w \in \stackrel{\circ}{\mathcal{U}}_{h}$ , 定义

$\begin{aligned} &T_{\lambda}(w)_{i}=\frac{2}{\tau}\left(w_{i}-u_{i}^{k}\right)+\lambda \psi(w, w)_{i}-\nu \delta_{x}^{2} w_{i}, \quad 1 \leqslant i \leqslant m-1 \\ &T_{\lambda}(w)_{0}=0 \\ &T_{\lambda}(w)_{m}=0 \end{aligned}$

易知 (a) $T_{\lambda}(w)$ 是连续的; (b) $T_{0}(w)_{i}=0, i=1,2, \cdots, m-1$ 是一个严格对角占优的三对角线性方程组, 故有唯一解. 现在来检验 (c). 设 $w$ 是 $T_{\lambda}(w)=0$ 可能的解. 用 $w$ 和 $T_{\lambda}(w)=0$ 作内积, 得