偏微分方程算法之二维初边值问题（紧交替方向隐格式）

news2024/10/6 18:20:46

一、研究对象

二、理论推导

2.1 二维紧差分格式

2.2 紧交替方向格式

2.2.1 紧Peaceman-Rachford格式

2.2.2 紧D’Yakonov格式

2.2.3 紧Douglas格式

三、算例实现

四、结论

一、研究对象

继续以二维抛物型方程初边值问题为研究对象：

$\left\{\begin{matrix} \frac{\partial u(x,y,t)}{\partial t}-(\frac{\partial^{2}u(x,y,t)}{\partial x^{2}}+\frac{\partial^{2}u(x,y,t)}{\partial y^{2}})=f(x,y,t),(x,y)\in \Omega=[0,a]\times [0,b],0<t\leqslant T,\\ u(x,y,0)=\varphi(x,y),(x,y)\in \Omega,\space\space\space\space(1)\\ u(0,y,t)=g_{1}(y,t),u(a,y,t)=g_{2}(y,t),0\leqslant y\leqslant b,0<t\leqslant T,\\ u(x,0,t)=g_{3}(x,t),u(x,b,t)=g_{4}(x,t),0\leqslant x\leqslant a,0<t\leqslant T \end{matrix}\right.$

为了确保连续性，公式（1）中的相关函数满足：

$g_{1}(0,t)=g_{3}(0,t),g_1(b,t)=g_4(0,t),$

$g_{2}(0,t)=g_{3}(a,t),g_{2}(b,t)=g_{4}(a,t)$

$\varphi(x,0)=g_{3}(x,0),\varphi(x,b)=g_{4}(x,0)$

二、理论推导

2.1 二维紧差分格式

从Crank-Nicolson法分析，剖分、离散过程与偏微分方程算法之二维初边值问题（交替方向隐（ADI）格式）-CSDN博客https://blog.csdn.net/L_peanut/article/details/137767096?spm=1001.2014.3001.5501中介绍的一样，节点处离散方程为：

$\frac{\partial u}{\partial t}|_{(x_{i},y_{j},t_{k+\frac{1}{2}})}-(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}})|_{(x_{i},y_{j},t_{k+\frac{1}{2}})}=f(x_{i},y_{j},t_{k+\frac{1}{2}}) \space\space\space\space (2)$

将一维抛物型方程紧差分方法的思路进行推广，引入两个中间函数 $v(x,y,t),w(x,y,t)$ ，满足

$v(x,y,t)=\frac{\partial^{2}u(x,y,t)}{\partial x^{2}},w(x,y,t)=\frac{\partial^{2}u(x,y,t)}{\partial y^{2}}$

则有 $\frac{\partial u}{\partial t}|_{(x_{i},y_{j},t_{k+\frac{1}{2}})}-(v+w)|_{(x_{i},y_{j},t_{k+\frac{1}{2}})}=f(x_{i},y_{j},t_{k+\frac{1}{2}})$

$\frac{u(x_{i-1},y,t)-2u(x_{i},y,t)+u(x_{i+1},y,t)}{\Delta x^{2}}=\frac{\partial^{2}u}{\partial x^{2}}|_{(x_{i},y,t)}+\frac{\Delta x^{2}}{12}\frac{\partial^{4}}{\partial x^{4}}|_{(x_{i},y,t)}+O(\Delta x^{4})$

$v(x_{i},y_{j},t)=\frac{\partial ^{2}u}{\partial x^{2}}|_{(x_{i},y_{j},t)}=\frac{u(x_{i-1},y_{j},t)-2u(x_{i},y_{j},t)+u(x_{i+1},y_{j},t)}{\Delta x^{2}}-\frac{\Delta x^{2}}{12}\frac{\partial^{4}u}{\partial x^{4}}|_{(x_{i},y_{j},t)}+O(\Delta x^{4})$

$=\frac{\delta^{2}_{x}u(x_{i},y_{j},t)}{\Delta x^{2}}-\frac{\Delta x^{2}}{12}\frac{\partial^{2}v}{\partial x^{2}}|_{(x_{i},y_{j},t)}+O(\Delta x^{4})$

$=\frac{\delta^{2}_{x}u(x_{i},y_{j},t)}{\Delta x^{2}}-\frac{\Delta x^{2}}{12}(\frac{v(x_{i-1},y_{j},t)-2v(x_{i},y_{j},t)+v(x_{i+1},y_{j},t)}{\Delta x^{2}})+O(\Delta x^{4})$

即

$\frac{\delta^{2}_{x}u(x_{i},y_{j},t)}{\Delta x^{2}}=\frac{1}{12}(v(x_{i-1},y_{j},t)+10v(x_{i},y_{j},t)+v(x_{i+1},y_{j},t))+O(\Delta x^{4}) \space\space\space\space(3)$