數值方法資格考試問題和解答（馬里蘭大學）/2009 年 8 月 667

問題 4

給定二點邊值問題

${\begin{cases}-u''+\alpha u=f(x),\quad 0\leq x\leq 1,\quad \alpha >0\\u'(0)=A,\\u'(1)=B\end{cases}}\!\,$

問題 4a

基於等距點上的分段線性元素，為該問題建立有限元逼近。確定適當範數下的收斂速度

解答 4a

令 $V=\{v\in H^{1}[0,1]\}\!\,$

找到 $u\in V\!\,$ 使得對於所有 $v\in V\!\,$

$\int _{0}^{1}-u''v+\alpha \int _{0}^{1}uv=\int _{0}^{1}fv\!\,$

或者在分部積分後，幷包含初始條件

$\int _{0}^{1}u'v'+\alpha \int _{0}^{1}uv=\int _{0}^{1}fv+Bv(1)-Av(0)\!\,$

離散變分形式

$V_{h}=\{v\in H^{1}[0,1]:v=\!\,$ 分段線性 $\}\!\,$

$\{\phi _{i}\}_{i=0}^{N+1}\!\,$ 是 $V_{h}\!\,$ 的基底； $h={\frac {1}{N+2}}\!\,$

當 $i=1,2,\ldots ,N\!\,$

$\phi _{i}={\begin{cases}{\frac {x-x_{i-1}}{h}}{\mbox{ for }}x\in [x_{i-1},x_{i}]\\{\frac {x_{i+1}-x}{h}}{\mbox{ for }}x\in [x_{i},x_{i+1}]\\0{\mbox{ otherwise}}\end{cases}}\!\,$

$\phi _{i}'={\begin{cases}{\frac {1}{h}}{\mbox{ for }}x\in [x_{i-1},x_{i}]\\-{\frac {1}{h}}{\mbox{ for }}x\in [x_{i},x_{i+1}]\\0{\mbox{ otherwise}}\end{cases}}\!\,$

$\int \phi _{i}\phi _{j}={\begin{cases}{\frac {2}{3}}h{\mbox{ for }}i=j\\{\frac {1}{6}}h{\mbox{ for }}|i-j|=1\\0{\mbox{ otherwise}}\end{cases}}\!\,$

$\int \phi _{i}'\phi _{j}'={\begin{cases}{\frac {2}{h}}{\mbox{ for }}i=j\\-{\frac {1}{h}}{\mbox{ for }}|i-j|=1\\0{\mbox{ otherwise}}\end{cases}}\!\,$

$\phi _{0}={\begin{cases}{\frac {x_{1}-x}{h}}{\mbox{ for }}x\in [x_{0},x_{1}]\\0{\mbox{ otherwise }}\end{cases}}\!\,$

$\phi _{N+1}={\begin{cases}{\frac {x-x_{N}}{h}}{\mbox{ for }}x\in [x_{N},x_{N+1}]\\0{\mbox{ otherwise }}\end{cases}}\!\,$

找到 $u_{h}\in V_{h}\!\,$ 使得對於所有的 $v\in V_{h}\!\,$

$\int u_{h}'v_{h}'+\alpha \int _{0}^{1}u_{h}v_{h}=\int _{0}^{1}fv_{h}+Bv(1)-Av(0)\!\,$

由於 $\{\phi _{i}\}_{i=0}^{N+1}\!\,$ 構成一個基

$u_{h}=\sum _{i=0}^{N+1}u_{h_{i}}\phi _{i}\!\,$

因此我們有方程組

對於 $j=0,1,\ldots ,N+1\!\,$

$\sum _{i=0}^{N+1}u_{h_{i}}\int _{0}^{1}\phi _{i}'\phi _{j}'+\alpha \sum _{i=0}^{N+1}u_{h_{i}}\int _{0}^{1}\phi _{i}\phi _{j}=\int _{0}^{1}f\phi _{j}+B\phi _{j}(1)-A\phi _{j}(0)\!\,$

${\begin{bmatrix}\alpha {\frac {1}{3}}h+{\frac {1}{h}}&{\frac {\alpha }{6}}h-{\frac {1}{h}}&&\\{\frac {\alpha }{6}}-{\frac {1}{h}}&\alpha {\frac {2}{3}}h+{\frac {2}{h}}&{\frac {\alpha }{6}}h-{\frac {1}{h}}&\\\ddots &\ddots &\ddots &\\&{\frac {\alpha }{6}}h-{\frac {1}{6}}&\alpha {\frac {2}{3}}h+{\frac {2}{h}}&{\frac {\alpha }{6}}h-{\frac {1}{h}}\\&&{\frac {\alpha }{6}}h-{\frac {1}{h}}&\alpha {\frac {1}{3}}h+{\frac {1}{h}}\end{bmatrix}}{\begin{bmatrix}u_{h_{0}}\\u_{h_{1}}\\\vdots \\u_{h_{N}}\\u_{h_{N+1}}\end{bmatrix}}={\begin{bmatrix}\int _{0}^{1}f\phi _{0}-A\\\int _{0}^{1}f\phi _{1}\\\vdots \\\int _{0}^{1}f\phi _{N}\\\int _{0}^{1}f\phi _{N+1}+B\end{bmatrix}}\!\,$