Burgers 方程的差分方法

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\(1.1\) 引 言

Burgers 方程是描述许多物理现象的模型方程, 如流体力学、非线性声学、气体动力学、交通流动力学问题. Burgers 方程也可以作为流体动力学 Navier-Stokes 方程的简化模型. 近年来, 求解 Burgers 方程的数值方法受到科研人员的广泛关注.

考虑一维非线性 Burgers 方程初边值问题

\[ \begin{aligned} &u_{t}+u u_{x}=\nu u_{x x}, \quad 0<x<L, 0<t \leqslant T, \\ &u(x, 0)=\varphi(x), \quad 0<x<L \\ &u(0, t)=0, \quad u(L, t)=0, \quad 0 \leqslant t \leqslant T \end{aligned} \]

其中 \(\nu\) 为动力黏性系数, \(\varphi(x)\) 为给定函数, \(\varphi(0)=\varphi(L)=0\).

在介绍差分格式之前, 我们先用能量方法给出问题 (1.1)-(1.3) 解的先验估计式.

定理 \(1.1\) 设 \(u(x, t)\) 为问题 (1.1)-(1.3) 的解. 记

\[ E(t)=\int_{0}^{L} u^{2}(x, t) \mathrm{d} x+2 \nu \int_{0}^{t}\left[\int_{0}^{L} u_{x}^{2}(x, s) \mathrm{d} x\right] \mathrm{d} s \]

则有

\[ E(t)=E(0), \quad 0<t \leqslant T \]

证明 用 \(u\) 乘以 \((1.1)\) 的两边, 可得

\[ \left(\frac{1}{2} u^{2}\right)_{t}+\left(\frac{1}{3} u^{3}\right)_{x}=\nu\left[\left(u u_{x}\right)_{x}-u_{x}^{2}\right] \]

将上式两边关于 \(x\) 在区间 \([0, L]\) 上积分, 并利用 (1.3), 得到

\[ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t} \int_{0}^{L} u^{2}(x, t) \mathrm{d} t+\nu \int_{0}^{L} u_{x}^{2}(x, t) \mathrm{d} x=0 \]

可将上式写为

\[ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left\{\int_{0}^{L} u^{2}(x, t) \mathrm{d} x+2 \nu \int_{0}^{t}\left[\int_{0}^{L} u_{x}^{2}(x, s) \mathrm{d} x\right] \mathrm{d} s\right\}=0 \]

即

\[ \frac{\mathrm{d} E(t)}{\mathrm{d} t}=0, \quad 0<t \leqslant T \]

因而

\[ E(t)=E(0), \quad 0<t \leqslant T \]

\(1.2\) 二层非线性差分格式

\(1.2.1\) 记号及引理

为了用差分格式求解问题 \((1.1)-(1.3)\), 将求解区域 \([0, L] \times[0, T]\) 作剖分. 取正整 数 \(m, n\). 将 \([0, L]\) 作 \(m\) 等分, 将 \([0, T]\) 作 \(n\) 等分. 记 \(h=L / m, \tau=T / n ; x_{i}=i h, 0 \leqslant\) \(i \leqslant m ; t_{k}=k \tau, 0 \leqslant k \leqslant n ; \Omega_{h}=\left\{x_{i} \mid 0 \leqslant i \leqslant m\right\}, \Omega_{\tau}=\left\{t_{k} \mid 0 \leqslant k \leqslant n\right\} ; \Omega_{h \tau}=\) \(\Omega_{h} \times \Omega_{\tau} .\) 称在直线 \(t=t_{k}\) 上的所有结点 \(\left\{\left(x_{i}, t_{k}\right) \mid 0 \leqslant i \leqslant m\right\}\) 为第 \(k\) 层结点. 此外, 记 \(x_{i+\frac{1}{2}}=\frac{1}{2}\left(x_{i}+x_{i+1}\right), t_{k+\frac{1}{2}}=\frac{1}{2}\left(t_{k}+t_{k+1}\right)\).

记

\[ \begin{aligned} &\mathcal{U}_{h}=\left\{u \mid u=\left(u_{0}, u_{1}, \cdots, u_{m}\right) \text { 为 } \Omega_{h} \text { 上的网格函数 }\right\}, \\ &\dot{\mathcal{U}}_{h}=\left\{u \mid u \in \mathcal{U}_{h}, u_{0}=u_{m}=0\right\} . \end{aligned} \]

设 \(u \in \mathcal{U}_{h}\), 引进如下记号:

\(\delta_{x} u_{i+\frac{1}{2}}=\frac{1}{h}\left(u_{i+1}-u_{i}\right), \quad \delta_{x}^{2} u_{i}=\frac{1}{h^{2}}\left(u_{i-1}-2 u_{i}+u_{i+1}\right), \quad \Delta_{x} u_{i}=\frac{1}{2 h}\left(u_{i+1}-u_{i-1}\right) .\) 易知

\[ \delta_{x}^{2} u_{i}=\frac{1}{h}\left(\delta_{x} u_{i+\frac{1}{2}}-\delta_{x} u_{i-\frac{1}{2}}\right), \quad \Delta_{x} u_{i}=\frac{1}{2}\left(\delta_{x} u_{i-\frac{1}{2}}+\delta_{x} u_{i+\frac{1}{2}}\right) \]

设 \(u, v \in \mathcal{U}_{h}\), 引进内积、范数及半范数

\[ \begin{aligned} &(u, v)=h\left(\frac{1}{2} u_{0} v_{0}+\sum_{i=1}^{m-1} u_{i} v_{i}+\frac{1}{2} u_{m} v_{m}\right), \\ &\|u\|_{\infty}=\max _{0 \leqslant i \leqslant m}\left|u_{i}\right|, \quad\|u\|=\sqrt{h\left(\frac{1}{2} u_{0}^{2}+\sum_{i=1}^{m-1} u_{i}^{2}+\frac{1}{2} u_{m}^{2}\right)}, \\ &|u|_{1}=\sqrt{h \sum_{i=1}^{m}\left(\delta_{x} u_{i-\frac{1}{2}}\right)^{2}}, \quad\|u\|_{1}=\sqrt{\|v\|^{2}+|u|_{1}^{2}}, \\ &|u|_{2}=\sqrt{h \sum_{i=1}^{m-1}\left(\delta_{x}^{2} u_{i}\right)^{2}}, \quad\|u\|_{2}=\sqrt{\|u\|^{2}+|u|_{1}^{2}+|u|_{2}^{2}} \end{aligned} \]

如果 \(\mathcal{U}_{h}\) 为复空间, 则相应的内积定义为

\[ (u, v)=h\left(\frac{1}{2} u_{0} \bar{v}_{0}+\sum_{i=1}^{m-1} u_{i} \bar{v}_{i}+\frac{1}{2} u_{m} \bar{v}_{m}\right) \]

其中 \(\bar{v}_{i}\) 为 \(v_{i}\) 的共轭.

记

\[ \mathcal{S}_{\tau}=\left\{w \mid w=\left(w_{0}, w_{1}, \cdots, w_{n}\right) \text { 为 } \Omega_{\tau} \text { 上的网格函数 }\right\} \]

没 \(w \in \mathcal{S}_{\tau}\), 引进如下记号:

\[ \begin{aligned} &w^{k+\frac{1}{2}}=\frac{1}{2}\left(w^{k}+w^{k+1}\right), \quad w^{\bar{k}}=\frac{1}{2}\left(w^{k+1}+w^{k-1}\right) \\ &\delta_{t} w^{k+\frac{1}{2}}=\frac{1}{\tau}\left(w^{k+1}-w^{k}\right), \quad \Delta_{t} w^{k}=\frac{1}{2 \tau}\left(w^{k+1}-w^{k-1}\right) . \end{aligned} \]

易知

\[ \Delta_{t} w^{k}=\frac{1}{2}\left(\delta_{t} w^{k-\frac{1}{2}}+\delta_{t} w^{k+\frac{1}{2}}\right) \]

设 \(u=\left\{u_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}\) 为 \(\Omega_{h \tau}\) 上的网格函数, 则 \(v=\left\{u_{i}^{k} \mid 0 \leqslant i \leqslant\right.\) \(m\}\) 为 \(\Omega_{h}\) 上的网格函数, \(w=\left\{u_{i}^{k} \mid 0 \leqslant k \leqslant n\right\}\) 为 \(\Omega_{\tau}\) 上的网格函数.

引理 \(1.1([4,25])\) (a) 设 \(u, v \in \mathcal{U}_{h}\), 则有

\[ -h \sum_{i=1}^{m-1}\left(\delta_{x}^{2} u_{i}\right) v_{i}=h \sum_{i=1}^{m}\left(\delta_{x} u_{i-\frac{1}{2}}\right)\left(\delta_{x} v_{i-\frac{1}{2}}\right)+\left(\delta_{x} u_{\frac{1}{2}}\right) v_{0}-\left(\delta_{x} u_{m-\frac{1}{2}}\right) v_{m} \]

2. 设 \(v \in \stackrel{\circ}{\mathcal{U}}_{h}\), 则有

\[ \begin{aligned} &-h \sum_{i=1}^{m-1}\left(\delta_{x}^{2} u_{i}\right) u_{i}=|u|_{1}^{2} \\ &|u|_{1}^{2} \leqslant\|u\| \cdot|u|_{2} \\ &\|u\|_{\infty} \leqslant \frac{\sqrt{L}}{2}|u|_{1} \\ &\|u\| \leqslant \frac{L}{\sqrt{6}}|u|_{1} \end{aligned} \]

3. 设 \(u \in \stackrel{\circ}{\mathcal{U}}_{h}\), 则有

\[ \|u\|_{\infty}^{2} \leqslant\|u\| \cdot|u|_{1} \]

且对任意 \(\varepsilon>0\), 有

\[ \|u\|_{\infty}^{2} \leqslant \varepsilon|u|_{1}^{2}+\frac{1}{4 \varepsilon}\|u\|^{2} \]

4. 设 \(u \in \mathcal{U}_{h}\), 则

\[ |u|_{1}^{2} \leqslant \frac{4}{h^{2}}\|u\|^{2} \]

5. 设 \(u \in \mathcal{U}_{h}\), 则有

\[ \|u\|_{\infty}^{2} \leqslant 2\|u\| \cdot|u|_{1}+\frac{1}{L}\|u\|^{2} . \]

且对任意 \(\varepsilon>0\), 有

\[ \|u\|_{\infty}^{2} \leqslant \varepsilon|u|_{1}^{2}+\left(\frac{1}{\varepsilon}+\frac{1}{L}\right)\|u\|^{2} . \]

证明 我们仅证明(c)和 \((\mathrm{e})\).

3. 由 \(u_{0}=0\), 当 \(1 \leqslant i \leqslant m-1\) 时, 有

\[ u_{i}^{2}=\sum_{l=1}^{i}\left(u_{l}^{2}-u_{l-1}^{2}\right)=\sum_{l=1}^{i}\left(u_{l}+u_{l-1}\right)\left(u_{l}-u_{l-1}\right)=2 h \sum_{l=1}^{i} u_{l-\frac{1}{2}} \delta_{x} u_{l-\frac{1}{2}} \]

因而

\[ u_{i}^{2} \leqslant 2 h \sum_{l=1}^{i}\left|u_{l-\frac{1}{2}}\right| \cdot\left|\delta_{x} u_{l-\frac{1}{2}}\right| \]

类似地, 注意到 \(u_{m}=0\), 可得

\[ u_{i}^{2} \leqslant 2 h \sum_{l=i+1}^{m}\left|u_{l-\frac{1}{2}}\right| \cdot\left|\delta_{x} u_{l-\frac{1}{2}}\right| \]

将以上两式相加得到

\[ u_{i}^{2} \leqslant h \sum_{l=1}^{m}\left|u_{l-\frac{1}{2}}\right| \cdot\left|\delta_{x} u_{l-\frac{1}{2}}\right| \leqslant \sqrt{h \sum_{l=1}^{m}\left|u_{l-\frac{1}{2}}\right|^{2}} \cdot \sqrt{h \sum_{l=1}^{m}\left|\delta_{x} u_{l-\frac{1}{2}}\right|^{2}} \leqslant\|u\| \cdot|u|_{1} \]

容易得到

\[ \|u\|_{\infty}^{2} \leqslant\|u\| \cdot|u|_{1} \]

对任意的 \(\varepsilon>0\) 有

\[ \|u\|_{\infty}^{2} \leqslant \varepsilon|u|_{1}^{2}+\frac{1}{4 \varepsilon}\|u\|^{2} \]

5. 当 \(i>j\) 时,

\[ \begin{aligned} u_{i}^{2} &=u_{j}^{2}+\sum_{l=j+1}^{i}\left(u_{l}^{2}-u_{l-1}^{2}\right) \\ &=u_{j}^{2}+2 h \sum_{l=j+1}^{i} u_{l-\frac{1}{2}} \delta_{x} u_{l-\frac{1}{2}} \\ & \leqslant u_{j}^{2}+2 h \sum_{l=j+1}^{i}\left|u_{l-\frac{1}{2}}\right| \cdot\left|\delta_{x} u_{l-\frac{1}{2}}\right| \\ & \leqslant u_{j}^{2}+2 h \sum_{l=1}^{m}\left|u_{l-\frac{1}{2}}\right| \cdot\left|\delta_{x} u_{l-\frac{1}{2}}\right| \\ & \leqslant u_{j}^{2}+2\|u\| \cdot|u|_{1} \end{aligned} \]

易知上式对 \(i \leqslant j\) 也是成立的.

记

\[ w_{j}= \begin{cases}1, & 1 \leqslant j \leqslant m-1 \\ \frac{1}{2}, & j=0, m\end{cases} \]

将 \((1.5)\) 乘以 \(h w_{j}\) 并对 \(j\) 从 0 到 \(m\) 求和得到

\[ h \sum_{j=0}^{m} w_{j} u_{i}^{2} \leqslant h \sum_{j=0}^{m} w_{j} u_{j}^{2}+2 h \sum_{j=0}^{m} w_{j}\|u\| \cdot|u|_{1} \]

由上式易得

\[ L\|u\|_{\infty}^{2} \leqslant\|u\|^{2}+2 L\|u\| \cdot|u|_{1} \]

即

\[ \|u\|_{\infty}^{2} \leqslant 2\|u\| \cdot|u|_{1}+\frac{1}{L}\|u\|^{2} . \]

对任意的 \(\varepsilon>0\), 有

\[ \|u\|_{\infty}^{2} \leqslant \varepsilon|u|_{1}^{2}+\left(\frac{1}{\varepsilon}+\frac{1}{L}\right)\|u\|^{2} . \]

下面我们给出几个常用的数值微分公式.

引理 \(1.2\) ([4]) 设 \(c, h\) 为给定的常数, 且 \(h>0\).

1. 如果 \(g(x) \in C^{2}[c-h, c+h]\), 则有

\[ g(c)=\frac{1}{2}[g(c-h)+g(c+h)]-\frac{h^{2}}{2} g^{\prime \prime}\left(\xi_{0}\right), \quad c-h<\xi_{0}<c+h \]

2. 如果 \(g(x) \in C^{2}[c, c+h]\), 则有

\[ g^{\prime}(c)=\frac{1}{h}[g(c+h)-g(c)]-\frac{h}{2} g^{\prime \prime}\left(\xi_{1}\right), \quad c<\xi_{1}<c+h \]

3. 如果 \(g(x) \in C^{2}[c-h, c]\), 则有

\[ g^{\prime}(c)=\frac{1}{h}[g(c)-g(c-h)]+\frac{h}{2} g^{\prime \prime}\left(\xi_{2}\right), \quad c-h<\xi_{2}<c \]

4. 如果 \(g(x) \in C^{3}[c-h, c+h]\), 则有

\[ g^{\prime}(c)=\frac{1}{2 h}[g(c+h)-g(c-h)]-\frac{h^{2}}{6} g^{\prime \prime \prime}\left(\xi_{3}\right), \quad c-h<\xi_{3}<c+h \]

5. 如果 \(g(x) \in C^{4}[c-h, c+h]\), 则有

\(g^{\prime \prime}(c)=\frac{1}{h^{2}}[g(c+h)-2 g(c)+g(c-h)]-\frac{h^{2}}{12} g^{(4)}\left(\xi_{4}\right), \quad c-h<\xi_{4}<c+h\)

6. 如果 \(g(x) \in C^{3}[c, c+h]\), 则有

\[ g^{\prime \prime}(c)=\frac{2}{h}\left[\frac{g(c+h)-g(c)}{h}-g^{\prime}(c)\right]-\frac{h}{3} g^{\prime \prime \prime}\left(\xi_{5}\right), \quad c<\xi_{5}<c+h \]

如果 \(g(x) \in C^{4}[c, c+h]\), 则有

\[ \begin{gathered} g^{\prime \prime}(c)=\frac{2}{h}\left[\frac{g(c+h)-g(c)}{h}-g^{\prime}(c)\right]-\frac{h}{3} g^{\prime \prime \prime}(c)-\frac{h^{2}}{12} g^{(4)}\left(\xi_{6}\right) \\ c<\xi_{6}<c+h \end{gathered} \]

7. 如果 \(g(x) \in C^{3}[c-h, c]\), 则有

\[ g^{\prime \prime}(c)=\frac{2}{h}\left[g^{\prime}(c)-\frac{g(c)-g(c-h)}{h}\right]+\frac{h}{3} g^{\prime \prime \prime}\left(\xi_{7}\right), \quad c-h<\xi_{7}<c \]

如果 \(g(x) \in C^{4}[c-h, c]\), 则有

\[ \begin{gathered} g^{\prime \prime}(c)=\frac{2}{h}\left[g^{\prime}(c)-\frac{g(c)-g(c-h)}{h}\right]+\frac{h}{3} g^{\prime \prime \prime}(c)-\frac{h^{2}}{12} g^{(4)}\left(\xi_{8}\right) \\ c-h<\xi_{8}<c \end{gathered} \]

8. 如果 \(g(x) \in C^{6}[c-h, c+h]\), 则有

\[ \begin{aligned} & \frac{1}{12}\left[g^{\prime \prime}(c-h)+10 g^{\prime \prime}(c)+g^{\prime \prime}(c+h)\right] \\ =& \frac{1}{h^{2}}[g(c+h)-2 g(c)+g(c-h)]+\frac{h^{4}}{240} g^{(6)}\left(\xi_{9}\right), \quad c-h<\xi_{9}<c+h \end{aligned} \]

\(1.2.2\) 差分格式的建立

定义 \(\Omega_{h \tau}\) 上的网格函数 \(U=\left\{U_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}\), 其中

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

在 \(\left(x_{i}, t_{k+\frac{1}{2}}\right)\) 处考虑方程 (1.1), 有

\[ \begin{gathered} u_{t}\left(x_{i}, t_{k+\frac{1}{2}}\right)+u\left(x_{i}, t_{k+\frac{1}{2}}\right) u_{x}\left(x_{i}, t_{k+\frac{1}{2}}\right)=\nu u_{x x}\left(x_{i}, t_{k+\frac{1}{2}}\right) \\ 1 \leqslant i \leqslant m-1, \quad 0 \leqslant k \leqslant n-1 \end{gathered} \]

应用引理 \(1.2\), 有

\[ \begin{aligned} u_{t}\left(x_{i}, t_{k+\frac{1}{2}}\right) &=\delta_{t} U_{i}^{k+\frac{1}{2}}+O\left(\tau^{2}\right) \\ u\left(x_{i}, t_{k+\frac{1}{2}}\right) &=\frac{1}{3}\left[u\left(x_{i-1}, t_{k+\frac{1}{2}}\right)+u\left(x_{i}, t_{k+\frac{1}{2}}\right)+u\left(x_{i+1}, t_{k+\frac{1}{2}}\right)\right]+O\left(h^{2}\right) \\ &=\frac{1}{3}\left(U_{i-1}^{k+\frac{1}{2}}+U_{i}^{k+\frac{1}{2}}+U_{i+1}^{k+\frac{1}{2}}\right)+O\left(\tau^{2}+h^{2}\right) \\ u_{x}\left(x_{i}, t_{k+\frac{1}{2}}\right) &=\frac{1}{2}\left[u_{x}\left(x_{i}, t_{k}\right)+u_{x}\left(x_{i}, t_{k+1}\right)\right]+O\left(\tau^{2}\right) \\ &=\frac{1}{2}\left(\Delta_{x} U_{i}^{k}+\Delta_{x} U_{i}^{k+1}\right)+O\left(\tau^{2}+h^{2}\right) \\ &=\Delta_{x} U_{i}^{k+\frac{1}{2}}+O\left(\tau^{2}+h^{2}\right) \\ u_{x x}\left(x_{i}, t_{k+\frac{1}{2}}\right) &=\frac{1}{2}\left[u_{x x}\left(x_{i}, t_{k}\right)+u_{x x}\left(x_{i}, t_{k+1}\right)\right]+O\left(\tau^{2}\right) \\ &=\frac{1}{2}\left(\delta_{x}^{2} U_{i}^{k}+\delta_{x}^{2} U_{i}^{k+1}\right)+O\left(\tau^{2}+h^{2}\right) \\ &=\delta_{x}^{2} U_{i}^{k+\frac{1}{2}}+O\left(\tau^{2}+h^{2}\right) \end{aligned} \]

将 (1.7)-(1.10) 代入 (1.6), 得到

\[ \begin{gathered} \delta_{t} U_{i}^{k+\frac{1}{2}}+\frac{1}{3}\left(U_{i-1}^{k+\frac{1}{2}}+U_{i}^{k+\frac{1}{2}}+U_{i+1}^{k+\frac{1}{2}}\right) \Delta_{x} U_{i}^{k+\frac{1}{2}}=\nu \delta_{x}^{2} U_{i}^{k+\frac{1}{2}}+R_{i}^{k+\frac{1}{2}} \\ 1 \leqslant i \leqslant m-1, \quad 0 \leqslant k \leqslant n-1 \end{gathered} \]

存在常数 \(c_{1}\) 使得

\[ \left|R_{i}^{k+\frac{1}{2}}\right| \leqslant c_{1}\left(\tau^{2}+h^{2}\right), \quad 1 \leqslant i \leqslant m-1, \quad 0 \leqslant k \leqslant n-1 \]

注意到初边值条件 (1.2)-(1.3), 有

\[ \begin{aligned} U_{i}^{0} &=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1 \\ U_{0}^{k} &=0, \quad U_{m}^{k}=0, \quad 0 \leqslant k \leqslant n \end{aligned} \]

在 (1.11) 中略去小量项 \(R_{i}^{k+\frac{1}{2}}\), 用 \(u_{i}^{k}\) 代替 \(U_{i}^{k}\), 对问题 (1.1)-(1.3) 建立如下差 分格式

\[ \begin{aligned} &\delta_{t} u_{i}^{k+\frac{1}{2}}+\frac{1}{3}\left(u_{i-1}^{k+\frac{1}{2}}+u_{i}^{k+\frac{1}{2}}+u_{i+1}^{k+\frac{1}{2}}\right) \Delta_{x} u_{i}^{k+\frac{1}{2}}=\nu \delta_{x}^{2} u_{i}^{k+\frac{1}{2}}, \\ &\quad 1 \leqslant i \leqslant m-1, \quad 0 \leqslant k \leqslant n-1, \\ &u_{i}^{0}=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1, \\ &u_{0}^{k}=0, \quad u_{m}^{k}=0, \quad 0 \leqslant k \leqslant n . \end{aligned} \]

差分格式 \((1.15)-(1.17)\) 是一个二层非线性差分格式.

\(1.2.3\) 差分格式解的守殂性和有界性

差分格式 (1.15) 中的非线性项可作如下变形

\[ \begin{aligned} & \frac{1}{3}\left(u_{i-1}^{k+\frac{1}{2}}+u_{i}^{k+\frac{1}{2}}+u_{i+1}^{k+\frac{1}{2}}\right) \Delta_{x} u_{i}^{k+\frac{1}{2}} \\ =& \frac{1}{3}\left[u_{i}^{k+\frac{1}{2}} \Delta_{x} u_{i}^{k+\frac{1}{2}}+\left(u_{i+1}^{k+\frac{1}{2}}+u_{i-1}^{k+\frac{1}{2}}\right) \Delta_{x} u_{i}^{k+\frac{1}{2}}\right] \\ =& \frac{1}{3}\left[u_{i}^{k+\frac{1}{2}} \Delta_{x} u_{i}^{k+\frac{1}{2}}+\Delta_{x}\left(u_{i}^{k+\frac{1}{2}} u_{i}^{k+\frac{1}{2}}\right)\right] \end{aligned} \]

没 \(v, w \in \mathcal{U}_{h}\), 定义

\[ \psi(v, w)_{i}=\frac{1}{3}\left[v_{i} \Delta_{x} w_{i}+\Delta_{x}(v w)_{i}\right], \quad 1 \leqslant i \leqslant m-1 \]

则

\[ \begin{aligned} &\frac{1}{3}\left(u_{i-1}^{k+\frac{1}{2}}+u_{i}^{k+\frac{1}{2}}+u_{i+1}^{k+\frac{1}{2}}\right) \Delta_{x} u_{i}^{k+\frac{1}{2}}=\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)_{i} \\ &\frac{1}{3}\left(U_{i-1}^{k+\frac{1}{2}}+U_{i}^{k+\frac{1}{2}}+U_{i+1}^{k+\frac{1}{2}}\right) \Delta_{x} U_{i}^{k+\frac{1}{2}}=\psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)_{i} \end{aligned} \]

于是 (1.15) 可以写为

\[ \delta_{t} u_{i}^{k+\frac{1}{2}}+\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)_{i}=\nu \delta_{x}^{2} u_{i}^{k+\frac{1}{2}}, \quad 1 \leqslant i \leqslant m-1, \quad 0 \leqslant k \leqslant n-1 \]

将 \(u u_{x}\) 写为 \(\frac{1}{3}\left[u u_{x}+\left(u^{2}\right)_{x}\right]\), 可将 \(\psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)_{i}\) 看成为后者在 \(\left(x_{i}, t_{k+\frac{1}{2}}\right)\) 处 的离散化.

算子 \(\psi\) 具有如下结论.

引理 \(1.3\) 设 \(v \in \mathcal{U}_{h}, w \in \stackrel{\circ}{\mathcal{U}}_{h}\), 则有

\[ (\psi(v, w), w)=0 \]

证明

\[ \begin{aligned} &(\psi(v, w), w) \\ =& \frac{1}{3}\left(v \Delta_{x} w+\Delta_{x}(v w), w\right) \\ =& \frac{1}{3}\left[\left(v \Delta_{x} w, w\right)+\left(\Delta_{x}(v w), w\right)\right) \\ =& \frac{1}{3}\left[\left(\Delta_{x} w, v w\right)+\left(\Delta_{x}(v w), w\right)\right] \\ =& 0 . \end{aligned} \]

定理 \(1.2\) 设 \(\left\{u_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}\) 是差分格式 (1.15)-(1.17) 的解. 令

\[ E^{k}=\left\|u^{k}\right\|^{2}+2 \nu \tau \sum_{l=0}^{k-1}\left|u^{l+\frac{1}{2}}\right|_{1}^{2}, \quad 0 \leqslant k \leqslant n \]

则有

\[ E^{k}=E^{0}, \quad 1 \leqslant k \leqslant n \]

证明 注意到差分格式 (1.15) 可写成

\[ \delta_{t} u_{i}^{k+\frac{1}{2}}+\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)_{i}-\nu \delta_{x}^{2} u_{i}^{k+\frac{1}{2}}=0, \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1 \]

用 \(u^{k+\frac{1}{2}}\) 与上式作内积, 可得

\[ \left(\delta_{t} u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)+\left(\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right), u^{k+\frac{1}{2}}\right)-\nu\left(\delta_{x}^{2} u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)=0 . \]

注意到 \(u^{k+\frac{1}{2}} \in \dot{\mathcal{U}}_{h}\), 有

\[ \begin{aligned} &\left(\delta_{t} u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)=\frac{1}{2 \tau}\left(\left\|u^{k+1}\right\|^{2}-\left\|u^{k}\right\|^{2}\right) \\ &\left(\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right), u^{k+\frac{1}{2}}\right)=0 \\ &-\left(\delta_{x}^{2} u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)=\left|u^{k+\frac{1}{2}}\right|_{1}^{2} \end{aligned} \]

因而

\[ \frac{1}{2 \tau}\left(\left\|u^{k+1}\right\|^{2}-\left\|u^{k}\right\|^{2}\right)+\nu\left|u^{k+\frac{1}{2}}\right|_{1}^{2}=0, \quad 0 \leqslant k \leqslant n-1 \]

将上式中的 \(k\) 换为 \(l\), 并对 \(l\) 从 0 到 \(k-1\) 求和, 得

\[ \frac{1}{2 \tau}\left(\left\|u^{k}\right\|^{2}-\left\|u^{0}\right\|^{2}\right)+\nu \sum_{l=0}^{k-1}\left|u^{l+\frac{1}{2}}\right|_{1}^{2}=0, \quad 1 \leqslant k \leqslant n \]

将上式变形即得 (1.18).

由定理 \(1.2\) 易知

\[ \left\|u^{k}\right\| \leqslant\left\|u^{0}\right\|, \quad 1 \leqslant k \leqslant n \]

\(1.2.4\) 差分格式解的存在性和唯一性

我们借助于下列 Browder 定理证明差分格式解的存在性.

定理 \(1.3\) (Browder 定理 \([9,10]\) ) 设 \((H,(\cdot, \cdot))\) 是一个有限维内积空间, \(\|·\|\) 是导出范数算子, \(\Pi: H \rightarrow H\) 是连续的. 进一步假设存在常数 \(\alpha>0\), 对于任意的 \(z \in H\),|z|=$, 有 \(\operatorname{Re}(\Pi(z), z) \geqslant 0\), 则存在 \(z^{*} \in H\) 使得 \(\Pi\left(z^{*}\right)=0\), 且 \(\left\|z^{*}\right\| \leqslant \alpha\).

定理 \(1.4\) 差分格式 (1.15)-(1.17) 存在解.

证明 由 (1.16)-(1.17) 知第 0 层值 \(u^{0}\) 已给定. 设已求得第 \(k\) 层的解 \(u^{k}\). 令

\[ w_{i}=u_{i}^{k+\frac{1}{2}}, \quad 0 \leqslant i \leqslant m \]

则有

\[ u_{i}^{k+1}=2 w_{i}-u_{i}^{k}, \quad 0 \leqslant i \leqslant m \]

由 (1.15), (1.17) 可得关于 \(w=\left(w_{0}, w_{1}, \cdots, w_{m}\right)\) 的方程组

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

对于任意的 \(u, v \in \mathcal{U}_{h}\), 定义内积

\[ (u, v)=h \sum_{i=1}^{m-1} u_{i} v_{i} \]

则 \(\stackrel{\circ}{\mathcal{U}}_{h}\) 为一个内积空间, \(\|u\|=\sqrt{(u, u)}\) 为导出范数.

定义 \(\Pi: \dot{\mathcal{U}}_{h} \rightarrow \dot{\mathcal{U}}_{h}\)

\[ \Pi(w)_{i}=\frac{2}{\tau}\left(w_{i}-u_{i}^{k}\right)+\psi(w, w)_{i}-\nu \delta_{x}^{2} w_{i}, \quad 1 \leqslant i \leqslant m-1 \]

则 \(\Pi(w)\) 为 \(\dot{\mathcal{U}}_{h}\) 上的连续函数, 应用引理 \(1.3\), 可得

\[ \begin{aligned} (\Pi(w), w) &=\frac{2}{\tau}\left[(w, w)-\left(u^{k}, w\right)\right]+(\psi(w, w), w)-\nu\left(\delta_{x}^{2} w, w\right) \\ &=\frac{2}{\tau}\left[\|w\|^{2}-\left(u^{k}, w\right)\right]+\nu|w|_{1}^{2} \\ & \geqslant \frac{2}{\tau}\left(\|w\|^{2}-\left\|u^{k}\right\| \cdot\|w\|\right) \\ & \geqslant \frac{2}{\tau}\left(\|w\|-\left\|u^{k}\right\|\right) \cdot\|w\| \end{aligned} \]

因而当 \(\|w\|=\left\|u^{k}\right\|\) 时, \((\Pi(w), w) \geqslant 0\). 由定理 \(1.3\) 知存在 \(w^{*} \in \dot{\mathcal{U}}_{h}\) 使得 \(\Pi\left(w^{*}\right)=0\), 且 \(\left\|w^{*}\right\| \leqslant\left\|u^{k}\right\|\), 即 \((1.19),(1.20)\) 存在解 \(w^{*}\). 定理 \(1.5\) 记 \(c_{2}=\left\|u^{0}\right\| .\) 因当 \(\tau<\frac{4 \nu^{3}}{c_{2}^{4}}\) 时, 差分格式 (1.15)-(1.17) 的解是唯 一的.

证明 由定理 \(1.4\) 的证明可知只要证明 (1.19)-(1.20) 的解是唯一的.

设 \((1.19)-(1.20)\) 有两个解 \(X, Y \in \dot{U}_{h}\), 即

\[ \begin{aligned} &\frac{2}{\tau}\left(X_{i}-u_{i}^{k}\right)+\psi(X, X)_{i}-\nu \delta_{x}^{2} X_{i}=0, \quad 1 \leqslant i \leqslant m-1 \\ &X_{0}=0, \quad X_{m}=0 \\ &\frac{2}{\tau}\left(Y_{i}-u_{i}^{k}\right)+\psi(Y, Y)_{i}-\nu \delta_{x}^{2} Y_{i}=0, \quad 1 \leqslant i \leqslant m-1 \\ &Y_{0}=0, \quad Y_{m}=0 \end{aligned} \]

令

\[ z=X-Y \]

将 (1.21)-(1.22) 与 (1.23)-(1.24) 相减, 得

\[ \begin{aligned} &\frac{2}{\tau} z_{i}+\psi(X, X)_{i}-\psi(Y, Y)_{i}-\nu \delta_{x}^{2} z_{i}=0, \quad 1 \leqslant i \leqslant m-1 \\ &z_{0}=0, \quad z_{m}=0 \end{aligned} \]

由定理 \(1.2\) 知存在常数 \(c_{2}\) 使得

\[ \|X\| \leqslant c_{2}, \quad\|Y\| \leqslant c_{2} \]

用 \(z\) 与 \((1.25)\) 作内积, 得

\[ \frac{2}{\tau}\|z\|^{2}+(\psi(X, X)-\psi(Y, Y), z)+\nu|z|_{1}^{2}=0 \]

由

\[ \psi(X, X)-\psi(Y, Y)=\psi(X, X)-\psi(X-z, X-z)=\psi(z, X)+\psi(X, z)-\psi(z, z) \]

及引理 \(1.3\) 得

\[ (\psi(X, X)-\psi(Y, Y), z)=(\psi(z, X), z) \]

因而

\[ \begin{aligned} &-(\psi(X, z)-\psi(Y, Y), z) \\ =&-\frac{1}{3} h \sum_{i=1}^{m-1}\left[z_{i} \Delta_{x} X_{i}+\Delta_{x}(z X)_{i}\right] z_{i} \\ =& \frac{1}{3} h \sum_{i=1}^{m-1}\left[X_{i} \Delta_{x}\left(z_{i}^{2}\right)+(z X)_{i} \Delta_{x} z_{i}\right] \end{aligned} \]

\[ \begin{aligned} &\leqslant \frac{1}{3}\left(2\|z\|_{\infty} \cdot\|X\| \cdot|z|_{1}+\|z\|_{\infty} \cdot\|X\| \cdot|z|_{1}\right) \\ &=\|X\| \cdot\|z\|_{\infty} \cdot|z|_{1} \\ &\leqslant c_{2}\|z\|_{\infty} \cdot|z|_{1} \end{aligned} \]

由 \((1.27)\) 得

\[ \frac{2}{\tau}\|z\|^{2}+\nu|z|_{1}^{2} \leqslant c_{2}\|z\|_{\infty} \cdot|z|_{1} \]

由引理 1.1(c) 知, 对任意 \(\varepsilon>0\), 有

\[ \|z\|_{\infty} \leqslant \varepsilon|z|_{1}+\frac{1}{2 \varepsilon}\|z\| . \]

因而

\[ \begin{aligned} \frac{2}{\tau}\|z\|^{2}+\nu|z|_{1}^{2} & \leqslant c_{2}\left(\varepsilon|z|_{1}+\frac{1}{2 \varepsilon}\|z\|\right)|z|_{1} \\ &=c_{2} \varepsilon|z|_{1}^{2}+\frac{c_{2}}{2 \varepsilon}\|z\| \cdot|z|_{1} \\ & \leqslant c_{2} \varepsilon|z|_{1}^{2}+c_{2} \varepsilon|z|_{1}^{2}+\frac{1}{4 c_{2} \varepsilon}\left(\frac{c_{2}}{2 \varepsilon}\right)^{2}\|z\|^{2} \\ &=2 c_{2} \varepsilon|z|_{1}^{2}+\frac{c_{2}}{16 \varepsilon^{3}}\|z\|^{2} \end{aligned} \]

取 \(\varepsilon=\frac{\nu}{2 c_{2}}\), 则有

\[ \frac{2}{\tau}\|z\|^{2} \leqslant \frac{c_{2}^{4}}{2 \nu^{3}}\|z\|^{2} \]

当 \(\tau<\frac{4 \nu^{3}}{c_{2}^{4}}\) 时 \(\|z\|=0\), 即 (1.19)-(1.20) 的解是唯一的.

\(1.2.5\) 差分格式解的收敛性

我们先给出重要的 Gronwall 不等式.

定理 \(1.6\) (a) 设 \(\left\{F^{k}\right\}_{k=0}^{\infty}\) 是一个非负序列, \(c\) 和 \(g\) 是两个非负常数, 且满足

\[ F^{k+1} \leqslant(1+c \tau) F^{k}+\tau g, \quad k=0,1,2, \cdots \]

则有

\[ F^{k} \leqslant e^{c k \tau}\left(F^{0}+\frac{g}{c}\right), \quad k=1,2,3, \cdots \]

2. 设 \(\left\{F^{k}\right\}_{k=0}^{\infty}\) 和 \(\left\{g^{k}\right\}_{k=0}^{\infty}\) 是两个非负序列, \(c\) 为非负常数, 且满足

\[ F^{k+1} \leqslant(1+c \tau) F^{k}+\tau g^{k}, \quad k=0,1,2, \cdots \]

则有

\[ F^{k} \leqslant e^{c k \tau}\left(F^{0}+\tau \sum_{l=0}^{k-1} g^{l}\right), \quad k=0,1,2, \cdots \]

3. 设 \(\left\{F^{k}\right\}_{k=0}^{\infty}\) 是非负序列, \(c\) 和 \(g\) 是两个非负常数, 且满足

\[ F^{k} \leqslant c \tau \sum_{k=0}^{k-1} F^{l}+g, \quad k=0,1,2, \cdots \]

则有

\[ F^{k} \leqslant e^{c k \tau} g, \quad k=0,1,2, \cdots \]

4. 设 \(\left\{F^{k}\right\}_{k=0}^{\infty}\) 是非负序列, \(\left\{g^{k}\right\}_{k=0}^{\infty}\) 是非负单调递增 (不必严格单调), 且满 足

\[ F^{k} \leqslant c \tau \sum_{l=0}^{k-1} F^{l}+g^{k}, \quad k=0,1,2, \cdots \]

则有

\[ F^{k} \leqslant e^{c k \tau} g^{k}, \quad k=0,1,2, \cdots \]

证明 (a)

\[ \begin{aligned} F^{k+1} & \leqslant(1+c \tau) F^{k}+\tau g \\ & \leqslant(1+c \tau)\left[(1+c \tau) F^{k-1}+\tau g\right]+\tau g \\ &=(1+c \tau)^{2} F^{k-1}+[(1+c \tau)+1] \tau g \\ & \leqslant(1+c \tau)^{2}\left[(1+c \tau) F^{k-2}+\tau g\right]+[(1+c \tau)+1] \tau g \\ &=(1+c \tau)^{3} F^{k-2}+\left[(1+c \tau)^{2}+(1+c \tau)+1\right] \tau g \\ & \leqslant \cdots \\ & \leqslant(1+c \tau)^{k} F^{1}+\left[(1+c \tau)^{k-1}+(1+c \tau)^{k-2}+\cdots+1\right] \tau g \\ & \leqslant(1+c \tau)^{k}\left[(1+c \tau) F^{0}+\tau g\right]+\left[(1+c \tau)^{k-1}+(1+c \tau)^{k-2}+\cdots+1\right] \tau g \\ &=(1+c \tau)^{k+1} F^{0}+\left[(1+c \tau)^{k}+(1+c \tau)^{k-1}+\cdots+1\right] \tau g \\ &=(1+c \tau)^{k+1} F^{0}+\frac{(1+c \tau)^{k+1}-1}{c \tau} \cdot \tau g \\ & \leqslant e^{c(k+1) \tau}\left(F^{0}+\frac{g}{c}\right), \quad k=0,1, \cdots \end{aligned} \]

\[ \begin{aligned} F^{k+1} & \leqslant(1+c \tau) F^{k}+\tau g^{k} \\ & \leqslant(1+c \tau)\left[(1+c \tau) F^{k-1}+\tau g^{k-1}\right]+\tau g^{k} \\ &=(1+c \tau)^{2} F^{k-1}+(1+c \tau) \tau g^{k-1}+\tau g^{k} \\ & \leqslant(1+c \tau)^{2}\left[(1+c \tau) F^{k-2}+\tau g^{k-2}\right]+(1+c \tau) \tau g^{k-1}+\tau g^{k} \\ &=(1+c \tau)^{3} F^{k-2}+(1+c \tau)^{2} \tau g^{k-2}+(1+c \tau) \tau g^{k-1}+\tau g^{k} \\ & \leqslant(1+c \tau)^{3}\left[(1+c \tau) F^{k-3}+\tau g^{k-3}\right]+(1+c \tau)^{2} \tau g^{k-2}+(1+c \tau) \tau g^{k-1}+\tau g^{k} \\ &=(1+c \tau)^{4} F^{k-3}+(1+c \tau)^{3} \tau g^{k-3}+(1+c \tau)^{2} \tau g^{k-2}+(1+c \tau) \tau g^{k-1}+\tau g^{k} \\ & \leqslant(1+c \tau)^{k+1} F^{0}+\tau \sum_{l=0}^{k}(1+c \tau)^{k-l} g^{l} \\ & \leqslant(1+c \tau)^{k+1}\left(F^{0}+\tau \sum_{l=0}^{k} g^{l}\right), \quad k=0,1,2, \cdots \end{aligned} \]

3. 易知

\[ F^{0} \leqslant g \]

令

\[ G^{k}=c \tau \sum_{l=0}^{k-1} F^{l}+g, \quad k=0,1,2, \cdots \]

则有

\[ \begin{aligned} &G^{0}=g, \\ &F^{k} \leqslant G^{k}, \quad k=0,1,2, \cdots, \\ &G^{k}=G^{k-1}+c \tau F^{k-1} \leqslant G^{k-1}+c \tau G^{k-1}=(1+c \tau) G^{k-1}, \quad k=1,2,3, \cdots, \end{aligned} \]

递推可得

\[ G^{k} \leqslant(1+c \tau)^{k} G^{0} \leqslant e^{c k \tau} g, \quad k=0,1,2, \cdots \]

因而

\[ F^{k} \leqslant G^{k} \leqslant e^{c k \tau} g, \quad k=0,1,2, \cdots \]

4. 易知

\[ F^{0} \leqslant g^{0} \]

令

\[ G^{k}=c \tau \sum_{l=0}^{k-1} F^{l}+g^{k}, \quad k=0,1,2, \cdots \]

则

\[ \begin{aligned} G^{0} &=g^{0}, \\ F^{k} & \leqslant G^{k}, \quad k=0,1,2, \cdots \\ G^{k} &=c \tau \sum_{l=0}^{k-2} F^{l}+g^{k-1}+c \tau F^{k-1}+\left(g^{k}-g^{k-1}\right) \\ &=G^{k-1}+c \tau F^{k-1}+\left(g^{k}-g^{k-1}\right) \\ & \leqslant(1+c \tau) G^{k-1}+\left(g^{k}-g^{k-1}\right), \quad k=1,2, \cdots . \end{aligned} \]

应用 (b) 之结果, 得

\[ F^{k} \leqslant G^{k} \leqslant e^{c k \tau}\left[G^{0}+\sum_{l=1}^{k}\left(g^{l}-g^{l-1}\right)\right]=e^{c k \tau} g^{k}, \quad k=0,1,2, \cdots \]

记

\[ c_{3}=\max _{0 \leqslant x \leqslant L, 0 \leqslant t \leqslant T}\left|u_{x}(x, t)\right| \]

定理 \(1.7\) 设 \(\left\{U_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}\) 为问题 (1.1)-(1.3) 的解, \(\left\{u_{i}^{k} \mid 0 \leqslant\right.\) \(i \leqslant m, 0 \leqslant k \leqslant n\}\) 为差分格式 \((1.15)-(1.17)\) 的解, 记

\[ e_{i}^{k}=U_{i}^{k}-u_{i}^{k}, \quad 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n \]

则存在常数 \(c_{4}\) 使得

\[ \left\|e^{k}\right\| \leqslant c_{4}\left(\tau^{2}+h^{2}\right), \quad 0 \leqslant k \leqslant n \]

证明 将 \((1.11),(1.13),(1.14)\) 和 \((1.15)-(1.17)\) 相减, 得到误差方程组

\[ \begin{aligned} &\delta_{t} e_{i}^{k+\frac{1}{2}}+\psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)_{i}-\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right)_{i}=\nu \delta_{x}^{2} e_{i}^{k+\frac{1}{2}}+R_{i}^{k+\frac{1}{2}} \\ &\qquad 1 \leqslant i \leqslant m-1, \quad 0 \leqslant k \leqslant n-1 \\ &e_{i}^{0}=0, \quad 1 \leqslant i \leqslant m-1 \\ &e_{0}^{k}=0, \quad e_{m}^{k}=0, \quad 0 \leqslant k \leqslant n \end{aligned} \]

用 \(e^{k+\frac{1}{2}}\) 与 \((1.30)\) 作内积, 得

\[ \begin{aligned} &\left(\delta_{t} e^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right)+\left(\psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)-\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right), e^{k+\frac{1}{2}}\right)+\nu\left|e^{k+\frac{1}{2}}\right|_{1}^{2} \\ =&\left(R^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right), \quad 0 \leqslant k \leqslant n-1 \end{aligned} \]

易知

\[ \left(\delta_{t} e^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right)=\frac{1}{2 \tau}\left(\left\|e^{k+1}\right\|^{2}-\left\|e^{k}\right\|^{2}\right) \]

注意到

\[ \begin{aligned} & \psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)-\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right) \\ =& \psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)-\psi\left(U^{k+\frac{1}{2}}-e^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}-e^{k+\frac{1}{2}}\right) \\ =& \psi\left(e^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)+\psi\left(U^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right)-\psi\left(e^{k+\frac{1}{2}}, e^{k+\frac{1}{2}}\right) \end{aligned} \]

再应用引理 \(1.3\), 可得

\[ \begin{aligned} &-\left(\psi\left(U^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right)-\psi\left(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}\right), e^{k+\frac{1}{2}}\right) \\ =&-\left(\psi\left(e^{k+\frac{1}{2}}, U^{k+\frac{1}{2}}\right), e^{k+\frac{1}{2}}\right) \\ =&-\frac{1}{3} h \sum_{i=1}^{m-1}\left[e_{i}^{k+\frac{1}{2}} \Delta_{x} U_{i}^{k+\frac{1}{2}}+\Delta_{x}\left(e^{k+\frac{1}{2}} U^{k+\frac{1}{2}}\right)_{i}\right] e_{i}^{k+\frac{1}{2}} \\ =&-\frac{1}{3} h \sum_{i=1}^{m-1}\left(e_{i}^{k+\frac{1}{2}}\right)^{2} \Delta_{x} U_{i}^{k+\frac{1}{2}}+\frac{1}{3} h \sum_{i=1}^{m-1} e_{i}^{k+\frac{1}{2}} U_{i}^{k+\frac{1}{2}} \Delta_{x} e_{i}^{k+\frac{1}{2}} \\ =&-\frac{1}{3} h \sum_{i=1}^{m-1}\left(e_{i}^{k+\frac{1}{2}}\right)^{2} \Delta_{x} U_{i}^{k+\frac{1}{2}}-\frac{1}{6} h \sum_{i=0}^{m-1} \frac{U_{i+1}^{k+\frac{1}{2}}-U_{i}^{k+\frac{1}{2}}}{h} e_{i}^{k+\frac{1}{2}} e_{i+1}^{k+\frac{1}{2}} \\ \leqslant & \frac{1}{2} c_{3}\left\|e^{k+\frac{1}{2}}\right\|^{2} \end{aligned} \]

将 (1.34) 和 (1.35) 代入 (1.33), 可得

\[ \begin{aligned} & \frac{1}{2 \tau}\left(\left\|e^{k+1}\right\|^{2}-\left\|e^{k}\right\|^{2}\right) \\ \leqslant & \frac{1}{2} c_{3}\left\|e^{k+\frac{1}{2}}\right\|^{2}+\left\|R^{k+\frac{1}{2}}\right\| \cdot\left\|e^{k+\frac{1}{2}}\right\| \\ \leqslant & \frac{1}{2} c_{3}\left(\frac{\left\|e^{k}\right\|+\left\|e^{k+1}\right\|}{2}\right)^{2}+\left\|R^{k+\frac{1}{2}}\right\| \cdot \frac{\left\|e^{k}\right\|+\left\|e^{k+1}\right\|}{2}, \quad 0 \leqslant k \leqslant n-1 \end{aligned} \]

上式两边约去\(\frac{\left\|e^{k+1}\right\|+\left\|e^{k}\right\|}{2}\), 得到

\[ \frac{1}{\tau}\left(\left\|e^{k+1}\right\|-\left\|e^{k}\right\|\right) \leqslant \frac{c_{3}}{4}\left(\left\|e^{k}\right\|+\left\|e^{k+1}\right\|\right)+\left\|R^{k+\frac{1}{2}}\right\|, \quad 0 \leqslant k \leqslant n-1 \]

即

\[ \begin{aligned} &\left(1-\frac{c_{3}}{4} \tau\right)\left\|e^{k+1}\right\| \leqslant\left(1+\frac{c_{3} \tau}{4}\right)\left\|e^{k}\right\|+\tau\left\|R^{k+\frac{1}{2}}\right\|, \quad 0 \leqslant k \leqslant n-1 . \\ \text { 当 } \frac{c_{3}}{4} \tau \leqslant \frac{1}{3} \text { 时, } \\ \left\|e^{k+1}\right\| & \leqslant\left(1+\frac{3 c_{3} \tau}{4}\right)\left\|e^{k}\right\|+\frac{3}{2} \tau\left\|R^{k+\frac{1}{2}}\right\| \\ \leqslant\left(1+\frac{3 c_{3}}{4} \tau\right)\left\|e^{k}\right\|+\frac{3}{2} c_{1} \sqrt{L} \tau\left(\tau^{2}+h^{2}\right), \quad 0 \leqslant k \leqslant n-1 . \end{aligned} \]

由 Gronwall 不等式 (定理 1.6), 并注意到 \(\left\|e^{0}\right\|=0\), 得

\[ \left\|e^{k+1}\right\| \leqslant e^{\frac{3 c_{3}}{4} k \tau} \cdot \frac{2 c_{1} \sqrt{L}}{c_{3}}\left(\tau^{2}+h^{2}\right)=c_{4}\left(\tau^{2}+h^{2}\right), \quad 0 \leqslant k \leqslant n-1 \]

其中 \(c_{4}=e^{\frac{3 c_{3}}{4} T} \cdot \frac{2 c_{1} \sqrt{L}}{c_{3}}\).

\(1.3\) 三层线性化差分格式

\(1.3.1\) 差分格式的建立 }

在点 \(\left(x_{i}, t_{0}\right)\) 处考虑方程 (1.1), 并注意到 (1.2), 有

\[ u_{t}\left(x_{i}, 0\right)=\nu \varphi^{\prime \prime}\left(x_{i}\right)-\varphi\left(x_{i}\right) \varphi^{\prime}\left(x_{i}\right) \]

Hㄹ

\[ \hat{u}_{i}=\varphi\left(x_{i}\right)+\frac{\tau}{2}\left[\nu \varphi^{\prime \prime}\left(x_{i}\right)-\varphi\left(x_{i}\right) \varphi^{\prime}\left(x_{i}\right)\right], \quad 0 \leqslant i \leqslant m \]

在点 \(\left(x_{i}, t_{\frac{1}{2}}\right)\) 处考虑方程 (1.1), 并应用 Taylor 展开式, 有

\[ \delta_{t} U_{i}^{\frac{1}{2}}+\psi\left(\hat{u}, U^{\frac{1}{2}}\right)_{i}=\nu \delta_{x}^{2} U_{i}^{\frac{1}{2}}+R_{i}^{0}, \quad 1 \leqslant i \leqslant m-1 \]

且存在常数 \(c_{5}\) 使得

\[ \left|R_{i}^{0}\right| \leqslant c_{5}\left(\tau^{2}+h^{2}\right), \quad 1 \leqslant i \leqslant m-1 \]

在点 \(\left(x_{i}, t_{k}\right)\) 处考虑方程 (1.1), 并应用 Taylor 展开式, 得到

\[ \Delta_{t} U_{i}^{k}+\psi\left(U^{k}, U^{\bar{k}}\right)_{i}=\nu \delta_{x}^{2} U_{i}^{\bar{k}}+R_{i}^{k}, \quad 1 \leqslant i \leqslant m-1,1 \leqslant k \leqslant n-1 \]

且存在常数 \(c_{6}\) 使得

\[ \left|R_{i}^{k}\right| \leqslant c_{6}\left(\tau^{2}+h^{2}\right), \quad 1 \leqslant i \leqslant m-1,1 \leqslant k \leqslant n-1 \]

注意到初边值条件 (1.2) 和 (1.3), 有

\[ \begin{aligned} U_{i}^{0} &=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1 \\ U_{0}^{k} &=0, \quad U_{m}^{k}=0, \quad 0 \leqslant k \leqslant n \end{aligned} \]

在 \((1.36),(1.38)\) 中略去小量项, 对问题 (1.1)-(1.3) 建立如下差分格式

\[ \begin{aligned} &\delta_{t} u_{i}^{\frac{1}{2}}+\psi\left(\hat{u}, u^{\frac{1}{2}}\right)_{i}=\nu \delta_{x}^{2} u_{i}^{\frac{1}{2}}, \quad 1 \leqslant i \leqslant m-1 \\ &\Delta_{t} u_{i}^{k}+\psi\left(u^{k}, u^{\bar{k}}\right)_{i}=\nu \delta_{x}^{2} u_{i}^{\bar{k}}, \quad 1 \leqslant i \leqslant m-1,1 \leqslant k \leqslant n-1 \\ &u_{i}^{0}=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1 \\ &u_{0}^{k}=0, \quad u_{m}^{k}=0, \quad 0 \leqslant k \leqslant n \end{aligned} \]

\(1.3.2\) 差分格式解的存在性和唯一性

定理 \(1.8\) 差分格式 (1.42)-(1.45) 的解是存在唯一的.

证明 由 \((1.44)\) 和 \((1.45)\) 知第 0 层的值 \(u^{0}\) 已给定. 由 \((1.42)\) 和 \((1.45)\) 可得 关于第 1 层值 \(u^{1}\) 的线性方程组. 考虑其齐次方程组

\[ \begin{aligned} &\frac{1}{\tau} u_{i}^{1}+\frac{1}{2} \psi\left(\hat{u}, u^{1}\right)_{i}=\frac{1}{2} \nu \delta_{x}^{2} u_{i}^{1}, \quad 1 \leqslant i \leqslant m-1 \\ &u_{0}^{1}=0, \quad u_{m}^{1}=0 \end{aligned} \]

用 \(u^{1}\) 与 \((1.46)\) 作内积, 得

\[ \frac{1}{\tau}\left\|u^{1}\right\|^{2}+\frac{1}{2}\left(\psi\left(\hat{u}, u^{1}\right), u^{1}\right)=\frac{1}{2} \nu\left(\delta_{x}^{2} u^{1}, u^{1}\right) \]

由 \(\left(\psi\left(\hat{u}, u^{1}\right), u^{1}\right)=0\) 和 \(\left(\delta_{x}^{2} u^{1}, u^{1}\right)=-\left|u^{1}\right|_{1}^{2}\) 得

\[ \frac{1}{\tau}\left\|u^{1}\right\|^{2}+\frac{1}{2} \nu\left|u^{1}\right|_{1}^{2}=0 \]

因而 \(\left\|u^{1}\right\|=0\). 方程组 (1.46)-(1.47) 只有零解. (1.42), (1.45) 唯一确定 \(u^{1}\).

现设第 \(k-1\) 层值 \(u^{k-1}\) 和第 \(k\) 层值 \(u^{k}\) 已确定, 则由 \((1.43)\) 和 \((1.45)\) 可得关 于 \(u^{k+1}\) 的线性方程组. 考虑其齐次方程组

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

用 \(u^{k+1}\) 与 \((1.48)\) 作内积, 得

\[ \frac{1}{2 \tau}\left\|u^{k+1}\right\|^{2}+\frac{1}{2}\left(\psi\left(u^{k}, u^{k+1}\right), u^{k+1}\right)=\frac{1}{2} \nu\left(\delta_{x}^{2} u^{k+1}, u^{k+1}\right) \]

由 \(\left(\psi\left(u^{k}, u^{k+1}\right), u^{k+1}\right)=0\) 及 \(\left(\delta_{x}^{2} u^{k+1}, u^{k+1}\right)=-\left|u^{k+1}\right|_{1}^{2}\) 得

\[ \frac{1}{2 \tau}\left\|u^{k+1}\right\|^{2}+\frac{1}{2} \nu\left|u^{k+1}\right|_{1}^{2}=0 \]

因而 \(\left\|u^{k+1}\right\|=0 .\) 方程组 (1.48)-(1.49) 只有零解. 因而 \((1.43)\) 和 (1.45) 唯一确定 \(u^{k+1}\)

由归纳原理, 定理证毕.

\(1.3.3\) 差分格式解的守恒性和有界性

定理 \(1.9\) 设 \(\left\{u_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}\) 为 \((1.42)-(1.45)\) 的解. 则有

\[ \begin{aligned} &\frac{1}{2}\left(\left\|u^{1}\right\|^{2}+\left\|u^{0}\right\|^{2}\right)+\nu \tau\left|u^{\frac{1}{2}}\right|_{1}^{2}=\left\|u^{0}\right\|^{2} \\ &E\left(u^{k+1}, u^{k}\right)=E\left(u^{1}, u^{0}\right), \quad k=1,2,3, \cdots, n-1 \end{aligned} \]

其中

\[ E\left(u^{k+1}, u^{k}\right)=\frac{1}{2}\left(\left\|u^{k+1}\right\|^{2}+\left\|u^{k}\right\|^{2}\right)+2 \nu \tau \sum_{l=1}^{k}\left|u^{\bar{l}}\right|_{1}^{2}, \quad k=0,1, \cdots, n-1 \]

证明 (I) 用 \(u^{\frac{1}{2}}\) 与 \((1.42)\) 作内积, 得

\[ \left(\delta_{t} u^{\frac{1}{2}}, u^{\frac{1}{2}}\right)+\left(\psi\left(\hat{u}, u^{\frac{1}{2}}\right), u^{\frac{1}{2}}\right)=\nu\left(\delta_{x}^{2} u^{\frac{1}{2}}, u^{\frac{1}{2}}\right) \]

由

\[ \begin{aligned} &\left(\delta_{t} u^{\frac{1}{2}}, u^{\frac{1}{2}}\right)=\frac{1}{2 \tau}\left(\left\|u^{1}\right\|^{2}-\left\|u^{0}\right\|^{2}\right) \\ &\left(\psi\left(\hat{u}, u^{\frac{1}{2}}\right), u^{\frac{1}{2}}\right)=0 \\ &\left(\delta_{x}^{2} u^{\frac{1}{2}}, u^{\frac{1}{2}}\right)=-\left|u^{\frac{1}{2}}\right|_{1}^{2} \end{aligned} \]

得

\[ \frac{1}{2 \tau}\left(\left\|u^{1}\right\|^{2}-\left\|u^{0}\right\|^{2}\right)+\nu\left|u^{\frac{1}{2}}\right|_{1}^{2}=0 \]

即

\[ \frac{1}{2}\left(\left\|u^{1}\right\|^{2}+\left\|u^{0}\right\|^{2}\right)+\nu \tau\left|u^{\frac{1}{2}}\right|_{1}^{2}=\left\|u^{0}\right\|^{2} \]

2. 用 \(u^{\bar{k}}\) 与 (1.43) 作内积, 得到

\[ \left(\Delta_{t} u^{k}, u^{\bar{k}}\right)+\left(\psi\left(u^{k}, u^{\bar{k}}\right), u^{\bar{k}}\right)=\nu\left(\delta_{x}^{2} u^{\bar{k}}, u^{\bar{k}}\right), \quad 1 \leqslant k \leqslant n-1 \]

易得

\[ \frac{1}{4 \tau}\left(\left\|u^{k+1}\right\|^{2}-\left\|u^{k-1}\right\|^{2}\right)+\nu\left|u^{\bar{k}}\right|_{1}^{2}=0, \quad 1 \leqslant k \leqslant n-1 \]

或

\[ \frac{1}{2 \tau}\left(\frac{\left\|u^{k+1}\right\|^{2}+\left\|u^{k}\right\|^{2}}{2}-\frac{\left\|u^{k}\right\|^{2}+\left\|u^{k-1}\right\|^{2}}{2}\right)+\nu\left|u^{\bar{k}}\right|_{1}^{2}=0, \quad 1 \leqslant k \leqslant n-1 \]

上式可进一步写为

\[ \frac{1}{2 \tau}\left[E\left(u^{k+1}, u^{k}\right)-E\left(u^{k}, u^{k-1}\right)\right]=0, \quad 1 \leqslant k \leqslant n-1 \]

因而

\[ E\left(u^{k+1}, u^{k}\right)=E\left(u^{1}, u^{0}\right), \quad 1 \leqslant k \leqslant n-1 \]

注 1.1 (1.50) 和 \((1.51)\) 可以统一写为

\(\frac{1}{2}\left(\left\|u^{k+1}\right\|^{2}+\left\|u^{k}\right\|^{2}\right)+\nu \tau\left|u^{\frac{1}{2}}\right|_{1}^{2}+2 \nu \tau \sum_{l=1}^{k}\left|u^{\bar{l}}\right|_{1}^{2}=\left\|u^{0}\right\|^{2}, \quad k=0,1, \cdots, n-1\)

注 \(1.2\) 由 \((1.52)\) 和 \((1.53)\) 可得

\[ \left\|u^{k}\right\| \leqslant\left\|u^{0}\right\|, \quad 1 \leqslant k \leqslant n \]

\(1.3.4\) 差分格式解的收敛性

定理 \(1.10\) 设 \(\left\{U_{i}^{k} \mid 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n\right\}\) 为问题 \((1.1)-(1.3)\) 的解, \(\left\{u_{i}^{k} \mid 0 \leqslant\right.\) \(i \leqslant m, 0 \leqslant k \leqslant n\}\) 为差分格式 \((1.42)-(1.45)\) 的解. 记

\[ e_{i}^{k}=U_{i}^{k}-u_{i}^{k}, \quad 0 \leqslant i \leqslant m, 0 \leqslant k \leqslant n \]

则存在常数 \(c_{7}\), 当 \(\tau^{2}+h^{2} \leqslant \frac{1}{c_{7}}\) 时成立

\[ \begin{aligned} &\left|e^{k}\right|_{1} \leqslant c_{7}\left(\tau^{2}+h^{2}\right), \quad 0 \leqslant k \leqslant n \\ &\left\|e^{k}\right\|_{\infty} \leqslant \frac{\sqrt{L}}{2} c_{7}\left(\tau^{2}+h^{2}\right), \quad 0 \leqslant k \leqslant n \end{aligned} \]

证明 将 \((1.36),(1.38),(1.40)-(1.41)\) 与 \((1.42)-(1.45)\) 相减, 得误差方程组

\[ \begin{aligned} &\delta_{t} e_{i}^{\frac{1}{2}}+\psi\left(\hat{u}, e^{\frac{1}{2}}\right)_{i}=\nu \delta_{x}^{2} e_{i}^{\frac{1}{2}}+R_{i}^{0}, \quad 1 \leqslant i \leqslant m-1 \\ &\Delta_{t} e_{i}^{k}+\psi\left(U^{k}, U^{\bar{k}}\right)_{i}-\psi\left(u^{k}, u^{\bar{k}}\right)_{i}=\nu \delta_{x}^{2} e_{i}^{\bar{k}}+R_{i}^{k} \\ &\quad 1 \leqslant i \leqslant m-1,1 \leqslant k \leqslant n-1 \\ &e_{i}^{0}=0, \quad 1 \leqslant i \leqslant m-1 \\ &e_{0}^{k}=0, \quad e_{m}^{k}=0, \quad 0 \leqslant k \leqslant n . \end{aligned} \]

我们将用数学归纳法证明所要求的结果.

由 \((1.58)-(1.59)\) 得

\[ \left|e^{0}\right|_{1}=0, \quad\left\|e^{0}\right\|_{\infty}=0 \]

故 (1.54) 及 (1.55) 对 \(k=0\) 成立.

1. 用 \(\delta_{t} e^{\frac{1}{2}}\) 与 \((1.56)\) 作内积, 得

\[ \left\|\delta_{t} e^{\frac{1}{2}}\right\|^{2}+\left(\psi\left(\hat{u}, e^{\frac{1}{2}}\right), \delta_{t} e^{\frac{1}{2}}\right)=\nu\left(\delta_{x}^{2} e^{\frac{1}{2}}, \delta_{t} e^{\frac{1}{2}}\right)+\left(R^{0}, \delta_{t} e^{\frac{1}{2}}\right) \]

注意到

\[ e_{i}^{0}=0, \quad 0 \leqslant i \leqslant m \]

有

\[ \frac{1}{\tau^{2}}\left\|e^{1}\right\|^{2}+\frac{1}{2 \tau}\left(\psi\left(\hat{u}, e^{1}\right), e^{1}\right)=-\frac{\nu}{2 \tau}\left|e^{1}\right|_{1}^{2}+\frac{1}{\tau}\left(R^{0}, e^{1}\right) \]

再注意到

\[ \left(\psi\left(\hat{u}, e^{1}\right), e^{1}\right)=0 \]

有

\[ \frac{1}{\tau^{2}}\left\|e^{1}\right\|^{2}+\frac{\nu}{2 \tau}|e|_{1}^{2}=\frac{1}{\tau}\left(R^{0}, e^{1}\right) \leqslant \frac{1}{\tau^{2}}\left\|e^{1}\right\|^{2}+\frac{1}{4}\left\|R^{0}\right\|^{2} \]

再由 \((1.37)\), 得到

\[ \left|e^{1}\right|_{1}^{2} \leqslant \frac{2 \tau}{\nu} \cdot \frac{1}{4}\left\|R^{0}\right\|^{2} \leqslant \frac{\tau}{2 \nu} L c_{5}^{2}\left(\tau^{2}+h^{2}\right)^{2} \]

当 \(\tau \leqslant 2 \nu\) 时

\[ \left|e^{1}\right|_{1}^{2} \leqslant L c_{5}^{2}\left(\tau^{2}+h^{2}\right)^{2} \]

或

\[ \left|e^{1}\right|_{1} \leqslant \sqrt{L} c_{5}\left(\tau^{2}+h^{2}\right) \]

2. 将 \(\Delta_{t} e^{k}\) 与 \((1.57)\) 作内积, 得

\[ \begin{gathered} \left\|\Delta_{t} e^{k}\right\|^{2}+\left(\psi\left(U^{k}, U^{\bar{k}}\right)-\psi\left(u^{k}, u^{\bar{k}}\right), \Delta_{t} e^{k}\right)=\nu\left(\delta_{x}^{2} e^{\bar{k}}, \Delta_{t} e^{k}\right)+\left(R^{k}, \Delta_{t} e^{k}\right), \\ 1 \leqslant k \leqslant n-1 \end{gathered} \]

或

\[ \begin{aligned} &\left\|\Delta_{t} e^{k}\right\|^{2}+\frac{\nu}{4 \tau}\left(\left|e^{k+1}\right|_{1}^{2}-\left|e^{k-1}\right|_{1}^{2}\right) \\ =&-\left(\psi\left(U^{k}, U^{\bar{k}}\right)-\psi\left(u^{k}, u^{\bar{k}}\right), \Delta_{t} e^{k}\right)+\left(R^{k}, \Delta_{t} e^{k}\right), \quad 1 \leqslant k \leqslant n-1 \end{aligned} \]

易知

\[ \left|U^{k}\right|_{1} \leqslant \sqrt{L} c_{3}, \quad\left\|U^{k}\right\|_{\infty} \leqslant \frac{\sqrt{L}}{2}\left|U^{k}\right|_{1} \leqslant \frac{L}{2} c_{3}, \quad 0 \leqslant k \leqslant n \]

设 (1.54) 对 \(k=1,2, \cdots, l\) 成立. 则当 \(c_{7}\left(\tau^{2}+h^{2}\right) \leqslant 1\), 有

\[ \begin{aligned} &\left|u^{k}\right|_{1} \leqslant\left|U^{k}\right|_{1}+\left|e^{k}\right|_{1} \leqslant \sqrt{L} c_{3}+1, \quad 1 \leqslant k \leqslant l \\ &\left\|u^{k}\right\|_{\infty} \leqslant \frac{\sqrt{L}}{2}\left|u^{k}\right|_{1} \leqslant \frac{\sqrt{L}}{2}\left(\sqrt{L} c_{3}+1\right), \quad 1 \leqslant k \leqslant l \end{aligned} \]

注意到

\[ \begin{aligned} & \psi\left(U^{k}, U^{\bar{k}}\right)_{i}-\psi\left(u^{k}, u^{\bar{k}}\right)_{i} \\ =& \psi\left(e^{k}, U^{\bar{k}}\right)_{i}+\psi\left(u^{k}, e^{\bar{k}}\right)_{i} \\ =& \frac{1}{3}\left[e_{i}^{k} \Delta_{x} U_{i}^{\bar{k}}+\Delta_{x}\left(e^{k} U^{\bar{k}}\right)_{i}\right]+\frac{1}{3}\left[u_{i}^{k} \Delta_{x} e_{i}^{\bar{k}}+\Delta_{x}\left(u^{k} e^{\bar{k}}\right)_{i}\right] \\ =& \frac{1}{3}\left[e_{i}^{k} \Delta_{x} U_{i}^{\bar{k}}+\frac{1}{2}\left(\delta_{x} e_{i+\frac{1}{2}}^{k}\right) U_{i+1}^{\bar{k}}+e_{i}^{k} \Delta_{x} U_{i}^{\bar{k}}+\frac{1}{2}\left(\delta_{x} e_{i-\frac{1}{2}}^{k}\right) U_{i-1}^{\bar{k}}\right] \\ &+\frac{1}{3}\left[u_{i}^{k} \Delta_{x} e_{i}^{\bar{k}}+\frac{1}{2}\left(\delta_{x} u_{i+\frac{1}{2}}^{k}\right) e_{i+1}^{\bar{k}}+u_{i}^{k} \Delta_{x} e_{i}^{\bar{k}}+\frac{1}{2}\left(\delta_{x} u_{i-\frac{1}{2}}^{k}\right) e_{i-1}^{\bar{k}}\right] \end{aligned} \]

以及 \((1.63)-(1.65)\), 有

\[ \begin{aligned} &-\left(\psi\left(U^{k}, U^{\bar{k}}\right)-\psi\left(u^{k}, u^{k}\right), \Delta_{t} e^{k}\right) \\ \leqslant & \frac{1}{3}\left(\left\|e^{k}\right\|_{\infty}\left|U^{\bar{k}}\right|_{1}+\left\|U^{\bar{k}}\right\|_{\infty}\left|e^{k}\right|_{1}+\left\|e^{k}\right\|_{\infty}\left|U^{\bar{k}}\right|_{1}\right)\left\|\Delta_{t} e^{k}\right\| \\ &+\frac{1}{3}\left(\left\|u^{k}\right\|_{\infty}\left|e^{\bar{k}}\right|_{1}+\left\|e^{\bar{k}}\right\|_{\infty}\left|u^{k}\right|_{1}+\left\|u^{k}\right\|_{\infty}\left|e^{\bar{k}}\right|_{1}\right)\left\|\Delta_{t} e^{k}\right\| \\ \leqslant & \frac{1}{3}\left(2 \sqrt{L} c_{3}\left\|e^{k}\right\|_{\infty}+\frac{L}{2} c_{3}\left|e^{k}\right|_{1}\right)\left\|\Delta_{t} e^{k}\right\| \\ &+\frac{1}{3}\left(2 \cdot \frac{\sqrt{L}}{2}\left(\sqrt{L} c_{3}+1\right)\left|e^{\bar{k}}\right|_{1}+\left(\sqrt{L} c_{3}+1\right)\left\|e^{\bar{k}}\right\|_{\infty}\right)\left\|\Delta_{t} e^{k}\right\| \\ \leqslant & \frac{1}{3}\left(2 \sqrt{L} c_{3} \frac{\sqrt{L}}{2}\left|e^{k}\right|_{1}+\frac{L}{2} c_{3}\left|e^{k}\right|_{1}\right)\left\|\Delta_{t} e^{k}\right\| \\ &+\frac{1}{3}\left[\sqrt{L}\left(\sqrt{L} c_{3}+1\right)\left|e^{\bar{k}}\right|_{1}+\left(\sqrt{L} c_{3}+1\right) \frac{\sqrt{L}}{2}\left|e^{\bar{k}}\right|_{1}\right]\left\|\Delta_{t} e^{k}\right\| \\ =& \frac{1}{2} L c_{3}\left|e^{k}\right|_{1} \cdot\left\|\Delta_{t} e^{k}\right\|+\frac{1}{2} \sqrt{L}\left(\sqrt{L} c_{3}+1\right)\left|e^{\bar{k}}\right|_{1} \cdot\left\|\Delta_{t} e^{k}\right\| \\ \leqslant & \frac{1}{4}\left\|\Delta_{t} e^{k}\right\|^{2}+\frac{L^{2} c_{3}^{2}}{4}\left|e^{k}\right|_{1}^{2}+\frac{1}{4}\left\|\Delta_{t} e^{k}\right\|^{2}+\frac{L\left(\sqrt{L} c_{3}+1\right)^{2}}{4}\left|e^{\bar{k}}\right|_{1}^{2}, \quad 1 \leqslant k \leqslant l \end{aligned} \]

此外,有

\[ \left(R^{k}, \Delta_{t} e^{k}\right) \leqslant \frac{1}{2}\left\|\Delta_{t} e^{k}\right\|^{2}+\frac{1}{2}\left\|R^{k}\right\|^{2} \]

将以上两式代入 (1.62), 得

\[ \begin{aligned} & \frac{\nu}{4 \tau}\left(\left|e^{k+1}\right|_{1}^{2}-\left|e^{k-1}\right|_{1}^{2}\right) \\ \leqslant & \frac{L^{2} c_{3}^{2}}{4}\left|e^{k}\right|_{1}^{2}+\frac{L\left(\sqrt{L} c_{3}+1\right)^{2}}{4}\left|e^{\bar{k}}\right|_{1}^{2}+\frac{1}{2}\left\|R^{k}\right\|^{2} \\ \leqslant & \frac{L^{2} c_{3}^{2}}{4}\left|e^{k}\right|_{1}^{2}+\frac{L\left(\sqrt{L} c_{3}+1\right)^{2}}{4} \cdot \frac{\left|e^{k+1}\right|_{1}^{2}+\left|e^{k-1}\right|_{1}^{2}}{2}+\frac{1}{2} L c_{6}^{2}\left(\tau^{2}+h^{2}\right)^{2}, \quad 1 \leqslant k \leqslant l \end{aligned} \]

两边乘以 \(\frac{4 \tau}{\nu}\), 并移项, 得

\[ \begin{aligned} \left|e^{k+1}\right|_{1}^{2} \leqslant \mid &\left.e^{k-1}\right|_{1} ^{2}+\frac{L^{2} c_{3}^{2}}{\nu} \tau\left|e^{k}\right|_{1}^{2}+\frac{1}{2 \nu} L\left(\sqrt{L} c_{3}+1\right)^{2} \tau\left(\left|e^{k+1}\right|_{1}^{2}+\left|e^{k-1}\right|_{1}^{2}\right) \\ &+\frac{2}{\nu} L c_{6}^{2} \tau\left(\tau^{2}+h^{2}\right)^{2}, \quad 1 \leqslant k \leqslant l \end{aligned} \]

即

\[ \begin{aligned} &\quad\left[1-\frac{L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} \tau\right]\left|e^{k+1}\right|_{1}^{2} \\ &\leqslant \frac{L^{2} c_{3}^{2}}{\nu} \tau\left|e^{k}\right|_{1}^{2}+\left[1+\frac{L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} \tau\right]\left|e^{k-1}\right|_{1}^{2}+\frac{2}{\nu} L c_{6}^{2} \tau\left(\tau^{2}+h^{2}\right)^{2}, \quad 1 \leqslant k \leqslant l . \\ &\text { 当 } \frac{L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} \tau \leqslant \frac{1}{3} \text { 时, } \\ &\qquad \begin{array}{c} \left|e^{k+1}\right|_{1}^{2} \leqslant \frac{3 L^{2} c_{3}^{2}}{2 \nu} \tau\left|e^{k}\right|_{1}^{2}+\left[1+\frac{3 L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} \tau\right]\left|e^{k-1}\right|_{1}^{2}+\frac{3}{\nu} L c_{6}^{2} \tau\left(\tau^{2}+h^{2}\right)^{2} \\ 1 \leqslant k \leqslant l . \end{array} \\ &\left.\qquad \begin{array}{c} 2 \nu \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \nu \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \nu \end{array}\right) \\ &\qquad \begin{array}{l} 1 \leqslant \end{array} \\ &\left.\qquad \begin{array}{l} 1 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \nu \end{array}\right) \\ &\left.\qquad \begin{array}{l} 1 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 1 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \end{array}\right] \\ &\left.\qquad \begin{array}{l} 1 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 1 \end{array}\right) \\ &\left.\qquad \begin{array}{l} 2 \end{array}\right) \\ &\qquad_{1}{l}_{1}+\frac{L(\sqrt{2})}{2} \\ &\left.\qquad \begin{array}{l} 2 \end{array}\right) \\ &\qquad_{1}{l}_{1}+\frac{2}{2}+\frac{2}{2} \\ &\qquad \begin{array}{l} 2 \end{array} \\ &\qquad_{1}{l}_{1}+\frac{2}{2} \\ &\qquad \end{aligned} \]

易知

\[ \begin{aligned} & \max \left\{\left|e^{k}\right|_{1}^{2},\left|e^{k+1}\right|_{1}^{2}\right\} \\ \leqslant &\left[1+\frac{3 L^{2} c_{3}^{2}+3 L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} \tau\right] \max \left\{\left|e^{k-1}\right|_{1}^{2},\left|e^{k}\right|_{1}^{2}\right\}+\frac{3}{\nu} L c_{6}^{2} \tau\left(\tau^{2}+h^{2}\right)^{2}, 1 \leqslant k \leqslant l \end{aligned} \]

由 Gronwall 不等式, 得

\[ \begin{aligned} & \max \left\{\left|e^{l}\right|_{1}^{2},\left|e^{l+1}\right|_{1}^{2}\right\} \\ \leqslant & \exp \left\{\frac{3 L^{3} c_{3}^{2}+3 L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} T\right\} \\ & \cdot\left(\max \left\{\left|e^{0}\right|_{1}^{2},\left|e^{1}\right|_{1}^{2}\right\}+\frac{2 L c_{6}^{2}}{L^{2} c_{3}^{2}+L\left(\sqrt{L} c_{3}+1\right)^{2}}\left(\tau^{2}+h^{2}\right)^{2}\right) \end{aligned} \]

注意到 \((1.60),(1.61)\) 可得

\[ \begin{aligned} \left|e^{l+1}\right|_{1}^{2} \leqslant & \exp \left\{\frac{3 L^{2} c_{3}^{2}+3 L\left(\sqrt{L} c_{3}+1\right)^{2}}{2 \nu} T\right\} \\ & \cdot\left(L c_{5}^{2}+\frac{2 c_{6}^{2}}{L c_{3}^{2}+\left(\sqrt{L} c_{3}+1\right)^{2}}\right)\left(\tau^{2}+h^{2}\right)^{2} \\ \equiv & c_{7}^{2}\left(\tau^{2}+h^{2}\right)^{2} \end{aligned} \]

其中

\[ c_{7}=\exp \left\{\frac{3 L^{2} c_{3}^{2}+3 L\left(\sqrt{L} c_{3}+1\right)^{2}}{4 \nu} T\right\} \cdot\left(L c_{5}^{2}+\frac{2 c_{6}^{2}}{L c_{3}^{2}+\left(\sqrt{L} c_{3}+1\right)^{2}}\right)^{\frac{1}{2}} \]

即 (1.54) 对 \(k=l+1\) 成立. 由归纳原理知 (1.54) 对 \(k=0,1,2, \cdots, n\) 成立. 由 \((1.54)\) 及引理 1.1(b) 得到 \((1.55)\).

\(1.4\) Hopf-Cole 变换与高阶差分格式

\(1.4.1\)Hopf-Cole 变换

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

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

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

其中

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

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 \(\alpha(s)=(1-s)^{2}\left[5-(1-s)^{2}\right]\).

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

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

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

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

3. 设 \(f(x) \in C^{6}\left[x_{i-1}, x_{i+1}\right]\), 有

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

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

\[ \begin{aligned} &w_{x x x}(0, t)=0, \quad w_{x x x}(L, t)=0, \quad 0 \leqslant t \leqslant T, \\ &w_{x x x x x}(0, t)=0, \quad w_{x x x x x}(L, t)=0, \quad 0 \leqslant t \leqslant T \end{aligned} \]

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

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

存在常数 \(c_{8}\) 使得

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

注意到初值条件

\[ W_{i}^{0}=\tilde{\varphi}\left(x_{i}\right), \quad 0 \leqslant i \leqslant m \]

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

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

\(1.4.3\) 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 \(\|v\|_{\mathcal{A}}\) 和 \(\|v\|\) 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

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

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

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

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

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

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

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

2. 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

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

3. 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

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

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

\[ 2 \leqslant i \leqslant m-2 \]

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

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

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

存在常数 \(c_{11}\) 使得

\[ \left|\hat{R}_{i}^{k}\right| \leqslant c_{11} h^{4}, \quad 1 \leqslant i \leqslant m-1,1 \leqslant k \leqslant n \]

在 (1.95)-(1.97) 中略去小量项, 得到如下计算格式

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

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

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

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, \(\|\cdot\|\) 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:

1. \(T_{\lambda}(w)\) 是 \(H\) 上的连续算子;

2. \(T_{0}(w)=0\) 有唯一解;

3. \(T_{\lambda}(w)=0\) 的一切可能解有一致的界,

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

\[ \frac{2}{\tau}\left((w, w)-\left(u^{k}, w\right)\right)+\lambda(\psi(w, w), w)-\nu\left(\delta_{x}^{2} w, w\right)=0 \]

利用

\[ (\psi(w, w), w)=0, \quad-\left(\delta_{x}^{2} w, w\right)=|w|_{1}^{2} \]

得

\[ \frac{2}{\tau}\left(\|w\|^{2}-\left(u^{k}, w\right)\right)+\nu|w|_{1}^{2}=0 \]

于是

\[ \|w\|^{2} \leqslant\left(u^{k}, w\right) \leqslant\left\|u^{k}\right\| \cdot\|w\| \]

易知

\[ \|w\| \leqslant\left\|u^{k}\right\| \]

条件 (c) 满足. 由 Leray-Schauder 定理. (1.19)-(1.20) 存在解.