使用高阶有限差分/三阶方法

三阶精确方法

问题陈述

令D为矩形

$D=\{(x,y):0<x<a,\ 0<y<b\}$

并令C为D的边界。

算子

$\Delta u(x,y)={\partial ^{2} \over \partial x^{2}}u(x,y)+{\partial ^{2} \over \partial y^{2}}u(x,y)$

是通常的拉普拉斯算子。问题是，确定一个函数 u(x, y) 使得

${\begin{cases}{\begin{aligned}-\Delta u(x,y)&=f(x,y),\ {\text{for}}\ (x,y)\in D\\\\u(x,y)&=g(x,y),\ {\text{for}}\ (x,y)\in C,\\\end{aligned}}\end{cases}}\quad _{(1.0)}$

被称为泊松问题。

离散逼近

为了数值逼近 u(x, y)，使用网格

$(x_{i},y_{j}),\ 0\leq i\leq m+1,\ 0\leq j\leq n+1$.
其中
$x_{i}=i\,h,\,y_{i}=j\ k$
以及
$h=a\,/\,(m+1),k=b\,/\,(n+1)$

二阶偏导数
${\partial ^{2} \over \partial x^{2}}\ u(x,y)$
可以用差商在网格上进行逼近
$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})$.

这些差商由以下公式给出：

$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{1},y_{j})\ =$.

$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})\ =$.

${-{\frac {1}{12}}\,u(x_{i}+2\,h,y_{j})+{\frac {4}{3}}\,u(x_{i}+h,y_{j})-{\frac {5}{2}}\,u(x_{i},y_{j})+{\frac {4}{3}}\,u(x_{i}-h,y_{j})-{\frac {1}{12}}\,u(x_{i}-2\,h,y_{j}) \over \ h^{2}}$

${\text{for}}\ i\,=\,2,\ 3,\ \ldots \,m-1\quad {\text{and}}$

$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{m},y_{j})\ =$.

${{\frac {11}{12}}\,u(x_{m}+h,y_{j})-{\frac {5}{3}}\,u(x_{m},y_{j})+{\frac {1}{2}}\,u(x_{m}-h,y_{j})+{\frac {1}{3}}\,u(x_{m}-2\,h,y_{j})-{\frac {1}{12}}\,u(x_{m}-3\,h,y_{j}) \over \ h^{2}}$

二阶偏导数
${\partial ^{2} \over \partial y^{2}}\ u(x,y)$
可以用差商在网格上进行逼近
$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})$.

这些差商由以下公式给出：

$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{1})\ =$.

${-{\frac {1}{12}}\,u(x_{i},y_{1}+3\,k)+{\frac {1}{3}}\,u(x_{i},y_{1}+2\,k)+{\frac {1}{2}}\,u(x_{i},y_{1}+k)-{\frac {5}{3}}\,u(x_{i},y_{1})+{\frac {11}{12}}\,u(x_{i},y_{1}-k) \over \ k^{2}}$

$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})\ =$.

${-{\frac {1}{12}}\,u(x_{i},y_{j}+2\,k)+{\frac {4}{3}}\,u(x_{i},y_{j}+k)-{\frac {5}{2}}\,u(x_{i},y_{j})+{\frac {4}{3}}\,u(x_{i},y_{j}-k)-{\frac {1}{12}}\,u(x_{i},y_{j}-2\,k) \over \ k^{2}}$

${\text{for}}\ j\,=\,2,\ 3,\ \ldots \,n-1\quad {\text{and}}$

$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{n})\ =$.

${{\frac {11}{12}}\,u(x_{i},y_{n}+k)-{\frac {5}{3}}\,u(x_{i},y_{n})+{\frac {1}{2}}\,u(x_{i},y_{n}-k)+{\frac {1}{3}}\,u(x_{i},y_{n}-2\,k)-{\frac {1}{12}}\,u(x_{i},y_{n}-3\,k) \over \ k^{2}}$

截断误差

差商   $({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})$   是三阶精度的，具有截断误差

${\begin{aligned}\tau _{x}(x_{i},y_{j})&=({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})-{\partial ^{2} \over \partial x^{2}}\ u(x_{i},y_{j})\\&={h^{3} \over \ 120}\,M{_{x}^{(5)}}(x_{i},y_{j})\end{aligned}}$

其中

$M{_{x}^{(5)}}(x_{1},y_{j})={\frac {67}{6}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{1}+\phi {_{1}^{(1,\,j)}}\,h,\ y_{j})-{\frac {254}{12}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{1}+\phi {_{2}^{(1,\,j)}}\,h,\ y_{j})$ ,

对于    $0<\phi {_{1}^{(1,\,j)}}<2\,,\quad -1<\phi {_{2}^{(1,\,j)}}<3$ ,

$M{_{x}^{(5)}}(x_{i},y_{j})=4\,{\partial ^{5} \over \partial x^{5}}\,u(x_{i}+\phi {_{1}^{(i,\,j)}}\,h,\ y_{j})-4\,{\partial ^{5} \over \partial x^{5}}\,u(x_{i}+\phi {_{2}^{(i,\,j)}}\,h,\ y_{j})$ ,

对于    $-2<\phi {_{1}^{(i,\,j)}}<1\,,\quad -1<\phi {_{2}^{(i,\,j)}}<2\,,$

${\text{for}}\ \ i\ =\ 2,\ 3,\ \ldots \,,\ m-1.$     

当  ${\partial ^{6} \over \partial x^{6}}\,u(x,\,y)$   连续时，这些估计也成立。

$\tau _{x}(x_{i},\,y_{j})\;=\;{h^{4} \over \ 720}M_{x}^{6}(x_{i},\,y_{j}),$   其中

$M_{x}^{6}(x_{i},\,y_{j})\;=\;{\frac {8}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi {_{1}^{(i,\,j)}}\,h,\;y_{j})-{\frac {32}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi {_{2}^{(i,\,j)}}\,h,\;y_{j})$

对于某个    $-1\;<\;\phi {_{1}^{(i,\,j)}}\;<\;1\,,\quad -2\;<\;\phi {_{2}^{(i,\,j)}}\;<\;2,$    

${\text{for}}\ \ i\ =\ 2,\ 3,\ \ldots \,,\ m-1.$

情况为  $i\;=\;m$  是

$M{_{x}^{(5)}}(x_{m},y_{j})={\frac {254}{12}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{m}+\phi {_{1}^{(m,\,j)}}\,h,\ y_{j})-{\frac {67}{6}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{m}+\phi {_{2}^{(m,\,j)}}\,h,\ y_{j})$ ，

对于某些    $-3<\phi {_{1}^{(m,\,j)}}<1\,,\quad -2<\phi {_{2}^{(m,\,j)}}<0$ 。

差商   $({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})$   是三阶精度的，带有截断误差

${\begin{aligned}\tau _{y}(x_{i},y_{i})&=({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{1})-{\partial ^{2} \over \partial y^{2}}\ u(x_{i},y_{1})\\&={k^{3} \over \ 120}\,M{_{y}^{(5)}}(x_{i},y_{1})\end{aligned}}$

其中

$M{_{y}^{(5)}}(x_{i},y_{1})={\frac {67}{6}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{1}+\psi {_{1}^{(i,\,1)}}\,k)-{\frac {254}{12}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{1}+\psi {_{2}^{(i,\,1)}}\,k)$ ，

对于某些    $0<\psi {_{1}^{(i,\,1)}}<2\,,\quad -1<\psi {_{2}^{(i,\,1)}}<3$ ，

$M{_{y}^{(5)}}(x_{i},y_{j})=4\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{j}+\psi {_{1}^{(i,\,j)}}\,k)-4\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{j}+\psi {_{2}^{(i,\,j)}}\,k)$ ，

对于一些    $-2<\psi {_{1}^{(i,\,j)}}<1\,,\quad -1<\psi {_{2}^{(i,\,j)}}<2\,,$

${\text{for}}\ \ j\ =\ 2,\ 3,\ \ldots \,,\ n-1.$     

当  ${\partial ^{6} \over \partial y^{6}}\,u(x,\,y)$   连续时，这些估计也成立。

$\tau _{y}(x_{i},\,y_{j})\;=\;{k^{4} \over \ 720}M_{y}^{6}(x_{i},\,y_{j}),$   其中

$M_{y}^{6}(x_{i},\,y_{j})\;=\;{\frac {8}{3}}{\partial ^{6} \over \partial y^{6}}u(x_{i}+\psi {_{1}^{(i,\,j)}}\,k,\;y_{j})-{\frac {32}{3}}{\partial ^{6} \over \partial y^{6}}u(x_{i}+\psi {_{2}^{(i,\,j)}}\,k,\;y_{j})$

对于一些    $-1\;<\;\psi {_{1}^{(i,\,j)}}\;<\;1\,,\quad -2\;<\;\psi {_{2}^{(i,\,j)}}\;<\;2,$    

${\text{for}}\ \ j\ =\ 2,\ 3,\ \ldots \,,\ n-1.$

当  $j\;=\;n$  时的情况是

$M{_{y}^{(5)}}(x_{i},y_{n})={\frac {254}{12}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{n}+\psi {_{1}^{(i,\,n)}}\,k)-{\frac {67}{6}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{n}+\psi {_{2}^{(i,\,n)}}\,k)$ ,

对于一些    $-3<\psi {_{1}^{(i,\,n)}}<1\,,\quad -2<\psi {_{2}^{(i,\,n)}}<0$ .

然后，拉普拉斯算子  $\Delta u(x,y)$  可以近似为网格的内部

$\Delta _{h,\,k}u(x_{i},y_{j})={\partial _{h}^{2} \over \partial _{h}x^{2}}u(x_{i},y_{j})+{\partial _{k}^{2} \over \partial _{k}y^{2}}u(x_{i},y_{j})$

截断误差

$\tau (x_{i},y_{j})=\Delta _{h,\,k}u(x_{i},y_{j})-\Delta u(x_{i},y_{j})$

由以下给出

$\tau (x_{i},y_{j})=\tau _{x}(x_{i},y_{j})+\tau _{y}(x_{i},y_{j})$.

定义的有限差分运算

对于网格向量

$U=\{U_{i,\,j}\}\quad 0\leq i\leq m+1,\ 0\leq j\leq n+1$

定义有限差分运算

$\Delta _{i,\,j}=({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,U+({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,U$

由以下公式给出。

$({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1,\,j}\ U\ =$.

${-{\frac {1}{12}}\,U_{4,\,j}+{\frac {1}{3}}\,U_{3,\,j}+{\frac {1}{2}}\,U_{2,\,j}-{\frac {5}{3}}\,U_{1,\,j}+{\frac {11}{12}}\,U_{0,\,j} \over \ h^{2}}$

$({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\ U\ =$.

${-{\frac {1}{12}}\,U_{i+2,\,j}+{\frac {4}{3}}\,U_{i+1,\,j}-{\frac {5}{2}}\,U_{i,\,j}+{\frac {4}{3}}\,U_{i-1,\,j}-{\frac {1}{12}}\,U_{i-2,\,j} \over \ h^{2}}$

${\text{for}}\ i\,=\,2,\ 3,\ \ldots \,m-1\quad {\text{and}}$

$({\partial _{h}^{2} \over \partial _{h}x^{2}})_{m,\,j}\ U\ =$.

${{\frac {11}{12}}\,U_{m+1,\,j}-{\frac {5}{3}}\,U_{m,\,j}+{\frac {1}{2}}\,U_{m-1,\,j}+{\frac {1}{3}}\,U_{m-2,\,j}-{\frac {1}{12}}\,U_{m-3,\,j} \over \ h^{2}}$

$({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,1}\ U\ =$.

${-{\frac {1}{12}}\,U_{i,\,4}+{\frac {1}{3}}\,U_{i,\,3}+{\frac {1}{2}}\,U_{i,\,2}-{\frac {5}{3}}\,U_{i,\,1}+{\frac {11}{12}}\,U_{i,\,0} \over \ k^{2}}$

$({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\ U\ =$.

${-{\frac {1}{12}}\,U_{i,\,j+2}+{\frac {4}{3}}\,U_{i,\,j+1}-{\frac {5}{2}}\,U_{i,\,j}+{\frac {4}{3}}\,U_{i,\,j-1}-{\frac {1}{12}}\,U_{i,\,j-2} \over \ k^{2}}$

${\text{for}}\ j\,=\,2,\ 3,\ \ldots \,n-1\quad {\text{and}}$

$({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,n}\ U\ =$.

${{\frac {11}{12}}\,U_{i,\,n+1}-{\frac {5}{3}}\,U_{i,\,n}+{\frac {1}{2}}\,U_{i,\,n-1}+{\frac {1}{3}}\,U_{i,\,n-2}-{\frac {1}{12}}\,U_{i,\,n-3} \over \ k^{2}}$

模拟问题

为了模拟问题(1.0)，令

${\begin{aligned}U_{0,\,j}\ \ &=&g(x_{0},\ y_{j})\,,&\quad 0\leq j\leq n+1\\U_{m+1,\,j}\ \ &=&g(x_{m+1},\ y_{j})\,,&\quad 0\leq j\leq n+1\\U_{i,\,0}\ \ &=&g(x_{i},\ y_{0})\,,&\quad 0\leq i\leq m+1\\U_{i,\,n+1}\ \ &=&g(x_{i},\ y_{n+1)}\,,&\quad 0\leq i\leq m+1\\\end{aligned}}$

然后求解非奇异线性方程组

$-\Delta _{i,\,j}\,U=f(x_{i},\,y_{j})\quad 1\leq i\leq m,\ 1\leq j\leq n$  ;

对于剩下的  $U_{i,\,j}$ 

误差估计

该误差

$U-u=\{U_{i,\,j}-u(x_{i},\,y_{j})\}$ ,

满足

${\lVert U-u\rVert }_{2}\leq {\rho \,a^{2}b^{2} \over \ \pi ^{2}(a^{2}+b^{2})}{\lVert \tau \rVert }_{2}(1+O(h^{2}+k^{2}))$ .

对于

$\tau =\{\tau (x_{i},\,y_{j})\},\quad 1\leq i\leq m,\ 1\leq j\leq n$  ,

以及

$\rho \ \leq \ 1\,/\,(1-(1+{\sqrt {1}}0)\,/\,12)$ .

截断误差估计的证明

该截断误差对于  ${\big (}{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j}){\big )}$  是在假设  $u(x,\;y)$  足够光滑，以至于  ${\partial ^{5} \over \partial x^{5}}\,u(x,\;y)$  是连续的。为了符号方便，令

$g(x)\;=\;u(x,\;y_{j}).$

将  $g(x)$  在它关于  $x_{i}$的泰勒展开式中展开，

${\begin{aligned}g(x_{i}+\Delta x)\;&=\;g(x_{i})+g^{(1)}(x_{i})\Delta x+{\frac {1}{2}}\,g^{(2)}(x_{i})({\Delta x})^{2}+{\frac {1}{6}}\,g^{(3)}(x_{i})({\Delta x})^{3}\\&+{\frac {1}{24}}\,g^{(4)}(x_{i})({\Delta x})^{4}+{\frac {1}{120}}\,g^{(5)}(z)({\Delta x})^{5}\\\end{aligned}}$.

其中 $z$ 是 $x_{i}$ 和 $x_{i}+\Delta x$ 之间的某个数字。那么

${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{1},\,y_{j})\;=\\&{{\big (}-{\frac {1}{12}}g(x_{1}+3\,h)+{\frac {1}{3}}g(x_{1}+2\,h)+{\frac {1}{2}}g(x_{1}+h)-{\frac {5}{3}}g(x_{1})+{\frac {11}{12}}g(x_{1}-h){\big )} \over \ h^{2}}\end{aligned}}$

${\begin{aligned}&=\;g(x_{1}){(-{\frac {1}{12}}+{\frac {1}{3}}+{\frac {1}{2}}-{\frac {5}{3}}+{\frac {11}{12}}) \over \ h^{2}}\\&+\;g^{(1)}(x_{1}){(-{\frac {1}{12}}(3\,h)+{\frac {1}{3}}(2\,h)+{\frac {1}{2}}(h)-{\frac {5}{3}}(0)+{\frac {11}{12}}(-h)) \over \ h^{2}}\\&+\;{\frac {1}{2}}g^{(2)}(x_{1}){(-{\frac {1}{12}}(3\,h)^{2}+{\frac {1}{3}}(2\,h)^{2}+{\frac {1}{2}}(h)^{2}-{\frac {5}{3}}(0)^{2}+{\frac {11}{12}}(-h)^{2}) \over \ h^{2}}\\&+\;{\frac {1}{6}}g^{(3)}(x_{1}){(-{\frac {1}{12}}(3\,h)^{3}+{\frac {1}{3}}(2\,h)^{3}+{\frac {1}{2}}(h)^{3}-{\frac {5}{3}}(0)^{3}+{\frac {11}{12}}(-h)^{3}) \over \ h^{2}}\\&+\;{\frac {1}{24}}g^{(4)}(x_{1}){(-{\frac {1}{12}}(3\,h)^{4}+{\frac {1}{3}}(2\,h)^{4}+{\frac {1}{2}}(h)^{4}-{\frac {5}{3}}(0)^{4}+{\frac {11}{12}}(-h)^{4}) \over \ h^{2}}\\&+\;{\frac {1}{120}}{(-{\frac {1}{12}}g^{(5)}(z_{1})(3\,h)^{5}+{\frac {1}{3}}g^{(5)}(z_{2})(2\,h)^{5}+{\frac {1}{2}}g^{(5)}(z_{3})(h)^{5}-{\frac {5}{3}}(0)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {+{\frac {11}{12}}g^{(5)}(z_{4})(-h)^{5}) \over \ h^{2}}\\\end{aligned}}$

$=\;g^{(2)}(x_{1})+{h^{3} \over \ 120}{\big (}-{\frac {81}{4}}g^{(5)}(z_{1})+{\frac {32}{3}}g^{(5)}(z_{2})+{\frac {1}{2}}g^{(5)}(z_{3})-{\frac {11}{12}}g^{(5)}(z_{4}){\big )}$

其中

${\begin{aligned}&x_{1}\;<\;z_{1}\;<\;x_{1}+3\,h\,,\;x_{1}\;<\;z_{2}\;<\;x_{1}+2\,h\,,\;\\&x_{1}\;<\;z_{3}\;<\;x_{1}+h\,,\;\;{\text{and}}\;\;x_{1}-h\;<\;z_{4}\;<\;x_{1}.\\\end{aligned}}$.

由于

${\frac {67}{6}}{\underset {0\;\leq \;\phi \;\leq \;2}{\min(g^{(5)}(x_{1}\;+\;\phi \,h))}}\;\leq \;{\frac {32}{3}}g^{(5)}(z_{2})+{\frac {1}{2}}g^{(5)}(z_{3})\;\leq \;{\frac {67}{6}}{\underset {0\;\leq \;\phi \;\leq \;2}{\max(g^{(5)}(x_{1}\;+\;\phi \,h))}},$

${\frac {254}{12}}{\underset {-1\;\leq \;\phi \;\leq \;3}{\min(g^{(5)}(x_{1}\;+\;\phi \,h))}}\;\leq \;{\frac {81}{4}}g^{(5)}(z_{1})+{\frac {11}{12}}g^{(5)}(z_{4})\;\leq \;{\frac {254}{12}}{\underset {-1\;\leq \;\phi \;\leq \;3}{\max(g^{(5)}(x_{1}\;+\;\phi \,h))}},$

由*中间值定理*可知

${\frac {32}{3}}g^{(5)}(z_{2})+{\frac {1}{2}}g^{(5)}(z_{3})\;=\;{\frac {67}{6}}\,g^{(5)}(x_{1}\;+\;\phi _{1}\,h)),\quad {\text{for}}\;0\;<\;\phi _{1}\,,\;<\;2\,,$

${\frac {81}{4}}g^{(5)}(z_{1})+{\frac {11}{12}}g^{(5)}(z_{4})\;=\;{\frac {254}{12}}\,g^{(5)}(x_{1}\;+\;\phi _{2}\,h),\quad {\text{for}}\;-1\;<\;\phi _{2}\;<\;3\,.$.

这给出了

${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{1},\,y_{j})\;=\;g^{(2)}(x_{1})+{h^{3} \over \ 120}{\big (}{\frac {67}{6}}\,g^{(5)}(x_{1}+\phi _{1}\,h)-{\frac {254}{12}}\,g^{(5)}(x_{1}+\phi _{2}\,h){\big )}\\&\;=\;{\partial ^{2} \over \partial x^{2}}\,u(x_{1},\,y_{j})+{h^{3} \over \ 120}{\big (}{\frac {67}{6}}{\partial ^{5} \over \partial x^{5}}u(x_{1}+\phi _{1}\,h,\;y_{j})-{\frac {254}{12}}{\partial ^{5} \over \partial x^{5}}u(x_{1}+\phi _{2}\,h,\;y_{j}){\big )}\\\end{aligned}}$

这指的是

$\tau _{x}(x_{1},\,y_{j})\;=\;{h^{3} \over \ 120}M_{x}^{5}(x_{1},\,y_{j}).$

当 $i\;=\;2,\;3,\;\ldots \,,\;m-1$

${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j})\;=\\&{{\big (}-{\frac {1}{12}}g(x_{i}+2\,h)+{\frac {4}{3}}g(x_{i}+h)-{\frac {5}{2}}g(x_{i})+{\frac {4}{3}}g(x_{i}-h)-{\frac {1}{12}}g(x_{i}-2\,h){\big )} \over \ h^{2}}\end{aligned}}$



$=\;g^{(2)}(x_{i})+{h^{3} \over \ 120}{\big (}-{\frac {8}{3}}g^{(5)}(z_{1})+{\frac {4}{3}}g^{(5)}(z_{2})-{\frac {4}{3}}g^{(5)}(z_{3})+{\frac {8}{3}}g^{(5)}(z_{4}){\big )}$

其中

${\begin{aligned}&x_{i}\;<\;z_{1}\;<\;x_{i}+2\,h\,,\;x_{i}\;<\;z_{2}\;<\;x_{i}+h\,,\;\\&x_{i}-h\;<\;z_{3}\;<\;x_{i}\,,\;\;{\text{and}}\;\;x_{i}-2\,h\;<\;z_{4}\;<\;x_{i}.\\\end{aligned}}$.

如同之前推导，将同号项合并，并使用中值定理，

${\frac {4}{3}}g^{(5)}(z_{2})+{\frac {8}{3}}g^{(5)}(z_{4})\;=\;4\,g^{(5)}(x_{i}\;+\;\phi _{1}\,h)),\quad {\text{for}}\;-2\;<\;\phi _{1}\,,\;<\;1\,,$

${\frac {8}{3}}g^{(5)}(z_{1})+{\frac {4}{3}}g^{(5)}(z_{3})\;=\;4\,g^{(5)}(x_{i}\;+\;\phi _{2}\,h),\quad {\text{for}}\;-1\;<\;\phi _{2}\;<\;2\,.$.

这给出了

${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j})\;=\;g^{(2)}(x_{i})+{h^{3} \over \ 120}{\big (}{\frac {67}{6}}\,g^{(5)}(x_{i}+\phi _{1}\,h)-{\frac {254}{12}}\,g^{(5)}(x_{i}+\phi _{2}\,h){\big )}\\&\;=\;{\partial ^{2} \over \partial x^{2}}\,u(x_{i},\,y_{j})+{h^{3} \over \ 120}{\big (}4{\partial ^{5} \over \partial x^{5}}u(x_{i}+\phi _{1}\,h,\;y_{j})-4{\partial ^{5} \over \partial x^{5}}u(x_{i}+\phi _{2}\,h,\;y_{j}){\big )}\\\end{aligned}}$

这指的是

$\tau _{x}(x_{i},\,y_{j})\;=\;{h^{3} \over \ 120}M_{x}^{5}(x_{i},\,y_{j}).$

假设  ${\partial ^{6} \over \partial x^{6}}\,u(x,\;y)$  是连续的，在上述论证中，表达式

${\begin{aligned}&+\;{\frac {1}{120}}{(-{\frac {1}{12}}g^{(5)}(z_{1})(2\,h)^{5}+{\frac {4}{3}}g^{(5)}(z_{2})(h)^{5}-{\frac {5}{2}}(0)^{5}+{\frac {4}{3}}g^{(5)}(z_{3})(-h)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {-{\frac {1}{12}}g^{(5)}(z_{4})(-2\,h)^{5}) \over \ h^{2}}\\\end{aligned}}$

可以替换为

${\displaystyle {\begin{aligned}&+\;{\frac {1}{120}}g^{(5)}(x_{i}){(-{\frac {1}{12}}(2\,h)^{5}+{\frac {4}{3}}(h)^{5}-{\frac {5}{2}}(0)^{5}+{\frac {4}{3}}(-h)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {-{\frac {1}{12}}(-2\,h)^{5}) \over \ h^{2}}\\&\;\\&+\;$

这给出了

${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j})\;=\;\\&\;=\;{\partial ^{2} \over \partial x^{2}}\,u(x_{i},\,y_{j})+{h^{4} \over \ 720}{\big (}{\frac {8}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi _{1}\,h,\;y_{j})-{\frac {32}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi _{2}\,h,\;y_{j}){\big )}\\\end{aligned}}$

当 $-1\;<\;\phi _{1}\;<\;1\,,\quad -2\;<\;\phi _{2}\;<\;2$ 时，

$\tau _{x}(x_{i},\,y_{j})\;=\;{h^{4} \over \ 720}M_{x}^{6}(x_{i},\,y_{j}).$

其他 截断误差 估计也是以相同的方式完成。

结束工作

设 误差

$e=\{e_{i,\,j}\},\quad 1\leq i\leq m,\ 1\leq j\leq n,$

定义为

$e_{i,\,j}=U_{i,\,j}-u(x_{i},y_{j})$ .

$U$ 是 有限差分格式 (xx) 的解，而 $u(x_{i},\,y_{j})$ 是 (1.0) 的解。

由于

${\begin{aligned}-\Delta _{i,\,j}\,e&=-\Delta _{i,\,j}\,U+\Delta _{h,\,k}\,u(x_{i},\,y_{j})\\&=f(x_{i},\,y_{j})+(\Delta \,u(x_{i},\,y_{j})+\tau (x_{i},\,y_{j}))\\&=\tau (x_{i},\,y_{j})\\\end{aligned}}$ 
我们得到

$\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)\ =\ \sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,\tau (x_{i},\,y_{j})\ \leq \ {\lVert e\rVert }_{2}\,{\lVert \tau \rVert }_{2}$ .

接下来将证明算子  $-\Delta _{i,\,j}$  对  $e$,  特别是

$\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)$ 

      $\geq \;\lambda (4a^{-2}\,{(m+1)}^{2}\,(something)+4b^{-2}\,{(n+1)}^{2}\,())\,{\lVert e\rVert }_{2}$ ，

其中

$\lambda \ \geq \ 1-(1+{\sqrt {1}}0)\,/\,12$ 。

从  $\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)\;=$
$\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e-({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)\;=$
$\sum _{j=1}^{n}\sum _{i=1}^{m}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)\;+\;\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)$

首先估计和  $\sum _{i=1}^{m}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)$ 。

工作

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1,\,j}\ e\ =$.

${\frac {1}{12}}\,e_{4,\,j}-{\frac {1}{3}}\,e_{3,\,j}-{\frac {1}{2}}\,e_{2,\,j}+{\frac {5}{3}}\,e_{1,\,j}-{\frac {11}{12}}\,e_{0,\,j}$

$={\frac {7}{6}}(-{\frac {1}{1}}\,e_{2,\,j}+{\frac {2}{1}}\,e_{1,\,j}-{\frac {1}{1}}\,e_{0,\,j})$

${\begin{aligned}+\;&({\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})-{\frac {1}{4}}\,(e_{3,\,j}-e_{2,\,j})+{\frac {5}{12}}\,(e_{2,\,j}-e_{1,\,j})-{\frac {1}{4}}\,(e_{1,\,j}-e_{0,\,j}))\\\end{aligned}}$

${\begin{aligned}=\;&({\frac {1}{12}}\,e_{3,\,j}-{\frac {5}{4}}\,e_{2,\,j}+{\frac {5}{4}}\,e_{1,\,j}-{\frac {1}{12}}\,e_{0,\,j})\\-\;&(-{\frac {1}{12}}\,e_{4,\,j}+{\frac {5}{12}}\,e_{3,\,j}-{\frac {3}{4}}\,e_{2,\,j}-{\frac {5}{12}}\,e_{1,\,j}+{\frac {5}{6}}\,e_{0,\,j})\\\end{aligned}}$

$-h^2(\frac{{\mathrm{\partial }}_h^2}{{\mathrm{\partial }}_h{x}^{2.}}$

${\frac {1}{12}}\,e_{i+2,\,j}-{\frac {4}{3}}\,e_{i+1,\,j}+{\frac {5}{2}}\,e_{i,\,j}-{\frac {4}{3}}\,e_{i-1,\,j}+{\frac {1}{12}}\,e_{i-2,\,j}$

$={\frac {7}{6}}\,(-{\frac {1}{1}}\,e_{i+1,\,j}+{\frac {2}{1}}\,e_{i,\,j}-{\frac {1}{1}}\,e_{i-1,\,j})$

$+{\frac {1}{12}}\,(e_{i+2,\,j}-e_{i+1,\,j})-{\frac {1}{12}}\,(e_{i+1,\,j}-e_{i,\,j})$

$+{\frac {1}{12}}\,(e_{i,\,j}-e_{i-1,\,j})-{\frac {1}{12}}\,(e_{i-1,\,j}-e_{i-2,\,j})$

${\begin{aligned}=\;&({\frac {1}{12}}\,e_{i+2,\,j}&-{\frac {5}{4}}\,e_{i+1,\,j}&+{\frac {5}{4}}\,e_{i,\,j}&-{\frac {1}{12}}\,e_{i-1,\,j})\\-\;&({\frac {1}{12}}\,e_{i+1,\,j}&-{\frac {5}{4}}\,e_{i,\,j}\quad &+{\frac {5}{4}}\,e_{i-1,\,j}&-{\frac {1}{12}}\,e_{i-2,\,j})\\\end{aligned}}$${\text{for}}\ i\,=\,2,\ 3,\ \ldots \,m-1\quad {\text{and}}$

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{m,\,j}\ e\ =$.

$-{\frac {11}{12}}\,e_{m+1,\,j}+{\frac {5}{3}}\,e_{m,\,j}-{\frac {1}{2}}\,e_{m-1,\,j}-{\frac {1}{3}}\,e_{m-2,\,j}+{\frac {1}{12}}\,e_{m-3,\,j}$

${\begin{aligned}=\;&(-{\frac {5}{6}}\,e_{m+1,\,j}+{\frac {5}{12}}\,e_{m,\,j}+{\frac {3}{4}}\,e_{m-1,\,j}+{\frac {1}{4}}\,e_{m-2,\,j}+{\frac {1}{12}}\,e_{m-3,\,j})\\-\;&({\frac {1}{12}}\,e_{m+1,\,j}-{\frac {5}{4}}\,e_{m,\,j}+{\frac {5}{4}}\,e_{m-1,\,j}-{\frac {1}{12}}\,e_{m-2,\,j})\\\end{aligned}}$

$={\frac {7}{6}}(-{\frac {1}{1}}\,e_{m+1,\,j}+{\frac {2}{1}}\,e_{m,\,j}-{\frac {1}{1}}\,e_{m-1,\,j})$

${\begin{aligned}+\;&(-{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})+{\frac {1}{4}}\,(e_{m-1,\,j}-e_{m-2,\,j})-{\frac {5}{12}}\,(e_{m,\,j}-e_{m-1,\,j})+{\frac {1}{4}}\,(e_{m+1,\,j}-e_{m,\,j}))\\\end{aligned}}$

end work

现在，为了使用， 分部求和公式  被陈述如下。

$\sum _{i=1}^{m}w_{i}\,(v_{i}-v_{i-1})=w_{m}\,v_{m}-w_{0}\,v_{0}-\sum _{i=0}^{m-1}v_{i}\,(w_{i+1}-w_{i})$

$Now,\quad -h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\ e\ =v_{i,\,j}-v_{i-1,\,j},\quad {\text{with}}$

$v_{0,\,j}\;=\;(-{\frac {1}{12}}\,e_{4,\,j}+{\frac {5}{12}}\,e_{3,\,j}-{\frac {3}{4}}\,e_{2,\,j}-{\frac {5}{12}}\,e_{1,\,j}+{\frac {5}{6}}\,e_{0,\,j})$

$=\;-{\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})+{\frac {1}{3}}\,(e_{3,\,j}-e_{2,\,j})-{\frac {5}{12}}\,(e_{2,\,j}-e_{1,\,j})-{\frac {5}{6}}\,(e_{1,\,j}-e_{0,\,j})$

${\begin{aligned}v_{i,\,j}\;&=\;({\frac {1}{12}}\,e_{i+2,\,j}-{\frac {5}{4}}\,e_{i+1,\,j}+{\frac {5}{4}}\,e_{i,\,j}-{\frac {1}{12}}\,e_{i-1,\,j})\\\;&=\;{\frac {1}{12}}\,(e_{i+2,\,j}-e_{i+1,\,j})-{\frac {7}{6}}\,(e_{i+1,\,j}-e_{i,\,j})+{\frac {1}{12}}\,(e_{i,\,j}-e_{i-1,\,j})\\\end{aligned}}$

${\text{for}}\ i\,=\,1,\ 2,\ \ldots \,m-1\quad {\text{and}}$.

$=\;-{\frac {5}{6}}\,(e_{m+1,\,j}-e_{m,\,j})-{\frac {5}{12}}\,(e_{m,\,j}-e_{m-1,\,j})+{\frac {1}{3}}\,(e_{m-1,\,j}-e_{m-2,\,j})-{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})$

${\text{So the sum}}\;\sum _{i=1}^{m}e_{i,\,j}\,(-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)=\;\sum _{i=1}^{m}e_{i,\,j}\,(v_{i,\,j}-v_{i-1,\,j})$ $=e_{m,\,j}\,v_{m,\,j}-e_{0,\,j}\,v_{0,\,j}-\sum _{i=0}^{m-1}v_{i,\,j}\,(e_{i+1,\,j}-e_{i,\,j})$

考虑到  $e_{m+1,\,j}\,=\,e_{0,\,j}\,=\,0$  得出以下结论

$=\;-v_{0,\,j}\,(e_{1,\,j}-e_{0,\,j})-\sum _{i=1}^{m-1}v_{i,\,j}\,(e_{i+1,\,j}-e_{i,\,j})-v_{m,\,j}\,(e_{m+1,\,j}-e_{m,\,j})$