Problems & Solutions
Question
考虑如下热传导方程的初边值问题：
$$ \left\{\begin{array}{l} \frac{\partial u}{\partial t}-\frac{\partial^{2} u}{\partial x^{2}}+u=f(x, t), x \in(0,1), t \in(0, T] \\ u(x, 0)=\varphi(x), x \in[0,1], \\ u(0, t)=\alpha(t), u(1, t)=\beta(t), t \in[0, T] . \end{array}\right. $$
已知空间步长 $h=1/m$，时间步长 $\tau=T/n$，网格节点定义为 $x_{i}=i h$（$0 \leqslant i \leqslant m$）和 $t_{k}=k \tau$（$0 \leqslant k \leqslant n$）。请设计一个两层隐式差分格式，使其截断误差达到 $O(\tau^{2}+h^{2})$，并推导该截断误差的具体表达式。
$$ \left\{\begin{array}{l} \frac{\partial u}{\partial t}-\frac{\partial^{2} u}{\partial x^{2}}+u=f(x, t), x \in(0,1), t \in(0, T] \\ u(x, 0)=\varphi(x), x \in[0,1], \\ u(0, t)=\alpha(t), u(1, t)=\beta(t), t \in[0, T] . \end{array}\right. $$
已知空间步长 $h=1/m$，时间步长 $\tau=T/n$，网格节点定义为 $x_{i}=i h$（$0 \leqslant i \leqslant m$）和 $t_{k}=k \tau$（$0 \leqslant k \leqslant n$）。请设计一个两层隐式差分格式，使其截断误差达到 $O(\tau^{2}+h^{2})$，并推导该截断误差的具体表达式。
Answer
差分格式：$\left\{\begin{array}{l}
\frac{1}{\tau}\left(u_{i}^{k+1}-u_{i}^{k}\right)-\frac{1}{2 h^{2}}\left(u_{i-1}^{k}-2 u_{i}^{k}+u_{i+1}^{k}+u_{i-1}^{k+1}-2 u_{i}^{k+1}+u_{i+1}^{k+1}\right) \\
\quad+\frac{1}{2}\left(u_{i}^{k}+u_{i}^{k+1}\right)=f\left(x_{i}, t_{k+1 / 2}\right), \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1, \\
u_{i}^{0}=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1, \\
u_{0}^{k}=\alpha\left(t_{k}\right), u_{m}^{k}=\beta\left(t_{k}\right), \quad 0 \leqslant k \leqslant n .
\end{array}\right.$；截断误差：$R_{i}^{k}=\left[\frac{1}{24} \frac{\partial^{3} u\left(x_{i}, \eta_{i}^{k}\right)}{\partial t^{3}}-\frac{1}{8} \frac{\partial^{4} u\left(x_{i}, \tilde{\eta}_{i}^{k}\right)}{\partial x^{2} \partial t^{2}}+\frac{1}{8} \frac{\partial^{2} u\left(x_{i}, \bar{\eta}_{i}^{k}\right)}{\partial t^{2}}\right] \tau^{2} -\frac{1}{24}\left[\frac{\partial^{4} u\left(\xi_{i}^{k}, t_{k}\right)}{\partial x^{4}}+\frac{\partial^{4} u\left(\xi_{i}^{k+1}, t_{k+1}\right)}{\partial x^{4}}\right] h^{2}$
Final answer
差分格式：$\left\{\begin{array}{l}
\frac{1}{\tau}\left(u_{i}^{k+1}-u_{i}^{k}\right)-\frac{1}{2 h^{2}}\left(u_{i-1}^{k}-2 u_{i}^{k}+u_{i+1}^{k}+u_{i-1}^{k+1}-2 u_{i}^{k+1}+u_{i+1}^{k+1}\right) \\
\quad+\frac{1}{2}\left(u_{i}^{k}+u_{i}^{k+1}\right)=f\left(x_{i}, t_{k+1 / 2}\right), \quad 1 \leqslant i \leqslant m-1,0 \leqslant k \leqslant n-1, \\
u_{i}^{0}=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant m-1, \\
u_{0}^{k}=\alpha\left(t_{k}\right), u_{m}^{k}=\beta\left(t_{k}\right), \quad 0 \leqslant k \leqslant n .
\end{array}\right.$；截断误差：$R_{i}^{k}=\left[\frac{1}{24} \frac{\partial^{3} u\left(x_{i}, \eta_{i}^{k}\right)}{\partial t^{3}}-\frac{1}{8} \frac{\partial^{4} u\left(x_{i}, \tilde{\eta}_{i}^{k}\right)}{\partial x^{2} \partial t^{2}}+\frac{1}{8} \frac{\partial^{2} u\left(x_{i}, \bar{\eta}_{i}^{k}\right)}{\partial t^{2}}\right] \tau^{2} -\frac{1}{24}\left[\frac{\partial^{4} u\left(\xi_{i}^{k}, t_{k}\right)}{\partial x^{4}}+\frac{\partial^{4} u\left(\xi_{i}^{k+1}, t_{k+1}\right)}{\partial x^{4}}\right] h^{2}$
Explanation
1. 取时间中间点 $t_{k+1/2}=(t_k+t_{k+1})/2$，在点 $(x_i, t_{k+1/2})$ 处考虑原方程：$\frac{\partial u(x_i, t_{k+1/2})}{\partial t}-\frac{\partial^2 u(x_i, t_{k+1/2})}{\partial x^2}+u(x_i, t_{k+1/2})=f(x_i, t_{k+1/2})$。
2. 对各项进行 Taylor 展开：
- 时间导数项：$\frac{\partial u(x_i, t_{k+1/2})}{\partial t}=\frac{1}{\tau}[u(x_i, t_{k+1})-u(x_i, t_k)]-\frac{\tau^2}{24}\frac{\partial^3 u(x_i, \eta_i^k)}{\partial t^3}$，其中 $t_k<\eta_i^k<t_{k+1}$。
- 空间二阶导数项：$\frac{\partial^2 u(x_i, t_{k+1/2})}{\partial x^2}=\frac{1}{2}[\frac{\partial^2 u(x_i, t_k)}{\partial x^2}+\frac{\partial^2 u(x_i, t_{k+1})}{\partial x^2}]-\frac{\tau^2}{8}\frac{\partial^4 u(x_i, \tilde{\eta}_i^k)}{\partial x^2 \partial t^2}$，进一步将空间二阶导数用中心差商近似：$\frac{1}{2h^2}[u(x_{i-1}, t_k)-2u(x_i, t_k)+u(x_{i+1}, t_k)+u(x_{i-1}, t_{k+1})-2u(x_i, t_{k+1})+u(x_{i+1}, t_{k+1})]-\frac{h^2}{24}[\frac{\partial^4 u(\xi_i^k, t_k)}{\partial x^4}+\frac{\partial^4 u(\xi_i^{k+1}, t_{k+1})}{\partial x^4}]-\frac{\tau^2}{8}\frac{\partial^4 u(x_i, \tilde{\eta}_i^k)}{\partial x^2 \partial t^2}$，其中 $x_{i-1}<\xi_i^k, \xi_i^{k+1}<x_{i+1}$，$t_k<\tilde{\eta}_i^k<t_{k+1}$。
- $u$ 项：$u(x_i, t_{k+1/2})=\frac{1}{2}[u(x_i, t_k)+u(x_i, t_{k+1})]-\frac{\tau^2}{8}\frac{\partial^2 u(x_i, \bar{\eta}_i^k)}{\partial t^2}$，其中 $t_k<\bar{\eta}_i^k<t_{k+1}$。
3. 将上述展开式代入方程，忽略截断误差，结合初边值条件 $u(x_i, 0)=\varphi(x_i)$（$1\leqslant i\leqslant m-1$）、$u(0, t_k)=\alpha(t_k)$ 和 $u(1, t_k)=\beta(t_k)$（$0\leqslant k\leqslant n$），得到差分格式。
4. 整理截断误差项，得到 $R_i^k$ 的表达式。
2. 对各项进行 Taylor 展开：
- 时间导数项：$\frac{\partial u(x_i, t_{k+1/2})}{\partial t}=\frac{1}{\tau}[u(x_i, t_{k+1})-u(x_i, t_k)]-\frac{\tau^2}{24}\frac{\partial^3 u(x_i, \eta_i^k)}{\partial t^3}$，其中 $t_k<\eta_i^k<t_{k+1}$。
- 空间二阶导数项：$\frac{\partial^2 u(x_i, t_{k+1/2})}{\partial x^2}=\frac{1}{2}[\frac{\partial^2 u(x_i, t_k)}{\partial x^2}+\frac{\partial^2 u(x_i, t_{k+1})}{\partial x^2}]-\frac{\tau^2}{8}\frac{\partial^4 u(x_i, \tilde{\eta}_i^k)}{\partial x^2 \partial t^2}$，进一步将空间二阶导数用中心差商近似：$\frac{1}{2h^2}[u(x_{i-1}, t_k)-2u(x_i, t_k)+u(x_{i+1}, t_k)+u(x_{i-1}, t_{k+1})-2u(x_i, t_{k+1})+u(x_{i+1}, t_{k+1})]-\frac{h^2}{24}[\frac{\partial^4 u(\xi_i^k, t_k)}{\partial x^4}+\frac{\partial^4 u(\xi_i^{k+1}, t_{k+1})}{\partial x^4}]-\frac{\tau^2}{8}\frac{\partial^4 u(x_i, \tilde{\eta}_i^k)}{\partial x^2 \partial t^2}$，其中 $x_{i-1}<\xi_i^k, \xi_i^{k+1}<x_{i+1}$，$t_k<\tilde{\eta}_i^k<t_{k+1}$。
- $u$ 项：$u(x_i, t_{k+1/2})=\frac{1}{2}[u(x_i, t_k)+u(x_i, t_{k+1})]-\frac{\tau^2}{8}\frac{\partial^2 u(x_i, \bar{\eta}_i^k)}{\partial t^2}$，其中 $t_k<\bar{\eta}_i^k<t_{k+1}$。
3. 将上述展开式代入方程，忽略截断误差，结合初边值条件 $u(x_i, 0)=\varphi(x_i)$（$1\leqslant i\leqslant m-1$）、$u(0, t_k)=\alpha(t_k)$ 和 $u(1, t_k)=\beta(t_k)$（$0\leqslant k\leqslant n$），得到差分格式。
4. 整理截断误差项，得到 $R_i^k$ 的表达式。