Burgers 方程的差分方法

第 1章 Burgers 方程的差分方法

$1.1$ 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

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

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

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

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

且对任意 $\varepsilon>0$ , 有

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

(d) 设 $u \in \mathcal{U}_{h}$ , 则

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

(e) 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 $(\mathrm{e})$ .

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

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

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

(e) 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 $h>0$ .

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

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

(b) 如果 $g(x) \in C^{2}[c, c+h]$ , 则有

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

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

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

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

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

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

(f) 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

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

(g) 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

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

(h) 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|=\alpha $, 有$ \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

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

(b) 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

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

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

则有

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

(d) 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)

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

(b)

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

$F^{0} \leqslant g$

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

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

(d) 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

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

(II) 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.

(I) 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

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

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

(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

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

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

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

(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

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

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

$2 \leqslant i \leqslant m-2$

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

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:

(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

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

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.

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Burgers 方程的差分方法

第 1章 Burgers 方程的差分方法

引 言

二层非线性差分格式

记号及引理

差分格式的建立

差分格式解的守殂性和有界性

差分格式解的存在性和唯一性

差分格式解的收敛性

三层线性化差分格式

差分格式的建立 }

差分格式解的存在性和唯一性

差分格式解的守恒性和有界性

差分格式解的收敛性

Hopf-Cole 变换与高阶差分格式

Hopf-Cole 变换

差分格式的建立}

差分格式解的存在性和唯一性}

差分格式解的收敛性

原问题解的计算

小结与延拓

推荐阅读更多精彩内容

$1.1$ 引言

$1.2$ 二层非线性差分格式

$1.2.1$ 记号及引理

$1.2.2$ 差分格式的建立

$1.2.3$ 差分格式解的守殂性和有界性

$1.2.4$ 差分格式解的存在性和唯一性

$1.2.5$ 差分格式解的收敛性

$1.3$ 三层线性化差分格式

$1.3.1$ 差分格式的建立 }

$1.3.2$ 差分格式解的存在性和唯一性

$1.3.3$ 差分格式解的守恒性和有界性

$1.3.4$ 差分格式解的收敛性

$1.4$ Hopf-Cole 变换与高阶差分格式

$1.4.1$ Hopf-Cole 变换

$1.4.2$ 差分格式的建立}

$1.4.3$ 差分格式解的存在性和唯一性}

$1.4.4$ 差分格式解的收敛性

$1.4.5$ 原问题解的计算

$1.5$ 小结与延拓