使用高阶有限差分/初步估计

一维问题

连续问题

本书主要关注对偏微分方程解的有限差分逼近。然而，通过有限差分研究一些一维问题本身就很有用，并且具有指导意义。

考虑以下简单的解题问题：

$-y^{\prime \prime }(x)\;=\;f(x)$    在  $a\;\leq \;x\;\leq \;b$

满足边界条件

$y(a)\;=\;y(b)\;=\;0$.

以下推导的不等式促使我们分析涉及拉普拉斯算子的高维问题。

$\int _{a}^{b}-y(x)y^{\prime \prime }(x)dx\;=\;\mid _{a}^{b}\,-y(x)y^{\prime }(x)+\int _{a}^{b}\left\vert \,y^{\prime }(x)\,\right\vert ^{2}\,dx\;=\;\int _{a}^{b}\left\vert \,y^{\prime }(x)\,\right\vert ^{2}\,dx$

利用不等式()得到

$\int _{a}^{b}\left\vert \,y^{\prime }(x)\,\right\vert ^{2}\,dx\;\geq \;{\frac {\pi ^{2}}{(b-a)^{2}}}\int _{a}^{b}\left\vert \,y(x)\,\right\vert ^{2}\,dx$

并使用该方程和柯西-施瓦茨不等式

$\int _{a}^{b}-y(x)y^{\prime \prime }(x)\,dx\;=\;\int _{a}^{b}y(x)f(x)\,dx\;\leq \;{\sqrt {\int _{a}^{b}\left\vert \,y(x)\,\right\vert ^{2}\,dx\;}}{\sqrt {\int _{a}^{b}\left\vert \,f(x)\,\right\vert ^{2}\,dx\;}}$.

因此

${\frac {\pi ^{2}}{(b-a)^{2}}}\int _{a}^{b}\left\vert \,y(x)\,\right\vert ^{2}\,dx\;\leq \;{\sqrt {\int _{a}^{b}\left\vert \,y(x)\,\right\vert ^{2}\,dx\;}}{\sqrt {\int _{a}^{b}\left\vert \,f(x)\,\right\vert ^{2}\,dx\;}}$.

以及

${\sqrt {\int _{a}^{b}\left\vert \,y(x)\,\right\vert ^{2}\,dx\;}}\;\leq \;{\frac {(b-a)^{2}}{\pi ^{2}}}{\sqrt {\int _{a}^{b}\left\vert \,f(x)\,\right\vert ^{2}\,dx\;}}$.

现在，考虑当  $z(x)$  解决近似问题

$-z^{\prime \prime }(x)\;=\;g(x)$    在  $a\;\leq \;x\;\leq \;b$

满足边界条件  $z(a)\;=\;z(b)\;=\;0$,  其中  $g(x)$  接近  $f(x)$.
由于 $-(y(x)-z(x))^{\prime \prime }\;=\;(f(x)-g(x))$ ，

${\sqrt {\int _{a}^{b}\left\vert \,y(x)-z(x)\,\right\vert ^{2}\,dx\;}}\;\leq \;{\frac {(b-a)^{2}}{\pi ^{2}}}{\sqrt {\int _{a}^{b}\left\vert \,f(x)-g(x)\,\right\vert ^{2}\,dx\;}}$.

上述不等式将被推广到离散和更高维度的模拟。这将使我们能够分析多种类型的方程的有限差分方法的精度。

另一个需要说明的是，如果非零边界条件

$y(a)\;=A,\quad \;y(b)\;=\;B$,

是需要的，那么

$y(x)+{\frac {(B-A)}{(b-a)}}(x-a)+A$

将解决新问题。因此，在大多数情况下，假设零边界条件不会损失一般性。在更高维的情况下，这将把问题分成两个部分。

有限差分法求解

回到()上一节要解决的问题。

$-y^{\prime \prime }(x)\;=\;f(x)$    在  $a\;\leq \;x\;\leq \;b$

满足边界条件

$y(a)\;=\;y(b)\;=\;0$.

这个问题可能比有限差分法更适合用其他方法解决。例如，

令 $g(x)\;=\;\int _{a}^{x}(x-t)f(t)\,dt$，并应用规则()， $g^{\prime }(x)\;=\;(x-x)f(x)+\int _{a}^{x}{\tfrac {\partial }{\partial x}}(x-t)f(t)\,dt\;=\;\int _{a}^{x}f(t)\,dt$  因此   $g^{\prime \prime }(x)\;=\;f(x)$.

那么  $y(x)\;=\;g(x)-{\frac {g(b)}{b-a}}(x-a)$   是该问题的解。

然而，作为一个开始，它将说明有限差分方法，而不会太难理解。另一方面，偏微分方程可能难以用直接的解析方法求解。

从给定区间的划分开始

$a\;=\;x_{0}\;<\;x_{1}\;<\;x_{2}\;<\;\ldots \;<\;x_{n}\;<\;x_{n+1}\;=\;b$.

为了简单起见，假设均匀网格 $h\;=\;(b-a)\,/\,(n+1)$.

也就是说， $x_{i}-x_{i-1}\,=\,h$    对于   $i\,=\,1,\;2\;,\ldots ,\;n+1$.

对于开始，我们用二阶精确差分算子来近似二阶导数

${\frac {d_{h}^{2}}{d_{h}x^{2}}}y(x)\,=\,h^{-2}(y(x-h)-2\,y(x)+y(x+h))$.

定义  $f_{i}\,=\,f(x_{i})$.  将会证明存在唯一的

$y_{0},\;y_{1},\;y_{2},\;,\ldots ,\;y_{n},\;y_{n+1}$   使得   $y_{0}\,=\,y_{n+1}\,=\,0$

能够解以下方程

$-h^{-2}(y_{i-1}-2\,y_{i}+y_{i+1})\;=\;f_{i}$   对于   $i\,=\,1,\;2,\ldots ,\;n$.

此外，最重要的是， $y_{i}$  近似  $y(x)$  按照以下意义

${\sqrt {\textstyle \sum _{i\,=\,1}^{\,n}\left\vert \,y(x_{i})-y_{i}\,\right\vert ^{2}\;}}\;\leq \;{\sqrt {n}}\,B\,h^{2}$

对于某个边界  $B$  独立于  $h$.

在这一点上，我们需要澄清一些内容并引入一些符号。术语有限差分算子用于两种不同的但相关的算子。一个是应用于函数 $y(x)$的差商，即

${\frac {d_{h}^{2}}{d_{h}x^{2}}}y(x)\,=\,h^{-2}(y(x-h)-2\,y(x)+y(x+h))$

而另一个是线性算子

$h^{-2}(y_{i-1}-2\,y_{i}+y_{i+1})$

应用于向量 ${\vec {y}}\,=\,<y_{0}\;\,y_{1}\;\,y_{2}\;\ldots \;y_{n}\;\,y_{n+1}>$.

符号

$({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{i}\,{\vec {y}}\,=\,h^{-2}(y_{i-1}-2\,y_{i}+y_{i+1})$

将用于区分两者。方程()可以写成

$-({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{i}\,{\vec {y}}\,=\,f_{i}$.

线性算子  $({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}$  由以下定义

$({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\;=\;<({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{1}\,{\vec {y}}\;\;\;({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{2}\,{\vec {y}}\;\;\ldots \;\;({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{n}\,{\vec {y}}>$.

那么  $({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}$  可以被认为是

$({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\;=\;<({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{1}\;\;({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{2}\;\;\ldots \;\;({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{n}>$.

这种线性算子的表示方式不同于通常使用的矩阵表示法。这种表示方法的优势在于它允许将估计更轻松地推广到更高阶算子和二维或三维域。

事实上，这种线性算子已经得到了充分的研究，并具有矩阵表示形式

${\begin{bmatrix}-2&1&0&0&\cdots &0&0&0&0\\1&-2&1&0&\cdots &0&0&0&0\\\vdots &\vdots &\ddots &\vdots &\cdots &\vdots &\vdots &\vdots &\vdots \\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots &\vdots \\\vdots &\vdots &\vdots &\vdots &\cdots &\vdots &\ddots &\vdots &\vdots \\0&0&0&0&\cdots &0&1&-2&1\\0&0&0&0&\cdots &0&0&1&-2\\\end{bmatrix}}$,

当应用于向量 $<y_{1}\;\,y_{2}\;\,\ldots \;\,y_{n}>^{T}$时。

该矩阵的特征值、特征向量和逆矩阵是已知的，可以在一些参考资料中找到。如果仅仅是为了这个介绍性示例的缘故，矩阵的细节将用于分析。正如所指出的，正在开发一种可以推广到更高阶算子和域的分析方法。

最后，方程()可以写成

$-({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\,=\,{\vec {f}}$,

其中向量 ${\vec {f}}\,=\,<f_{1}\;\,f_{2}\;\ldots \;f_{n}>$。

最后一点记号，对于向量 ${\vec {y}}\;=\;<y_{0}\,\;y_{1}\,\;y_{2}\,\;\,\ldots \,\;y_{n}\,\;y_{n+1}>$， 

的 ${\vec {y}}$ 的内部点， 表示为 ${\vec {y}}^{\,o}$， 是 $n$ 维向量

${\vec {y}}^{\,o}\;=\;<y_{1}\,\;y_{2}\,\;\,\ldots \,\;y_{n}>$.

换句话说，将证明存在唯一

${\vec {y}}\;=\;<y_{0}\,\;y_{1}\,\;y_{2}\,\;\,\ldots \,\;y_{n}\,\;y_{n+1}>$  其中  $y_{0}\,=\,y_{n+1}\,=\,0$

求解方程()

$-({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\,=\,{\vec {f}}$,

此外，最重要的是， $y_{i}$  近似  $y(x)$  按照以下意义

${\sqrt {\textstyle \sum _{i\,=\,1}^{\,n}\left\vert \,y(x_{i})-y_{i}\,\right\vert ^{2}\;}}\;\leq \;{\sqrt {n}}\,B\,h^{2}$

其中边界 $B$ 与 $h$ 无关，并且 $B$ 的一个很好的估计是

$B\;=\;M\,(\alpha _{n})^{-1}\,h^{2}$

其中  $M\;=\;{\tfrac {1}{12}}\;{\underset {a\;\leq \;x\;\leq \;b}{\max }}(f^{(iv)}(x))$，

$\alpha _{n}\;=\;2\,(1-\cos(\pi \,{\tfrac {1}{n+2}}))\;\to \;{\frac {\pi ^{2}}{(n+2)^{2}}}$,

并且  $B\;\to \;{\frac {M}{\pi ^{2}}}$.

本节的剩余部分是对上述断言()的证明。

证明通过首先证明运算符  $-({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}$  是正定的。

特别是对于  $\alpha _{n}\;=\;2\,(1-\cos(\pi \,{\tfrac {1}{n+2}}))\;\to \;{\frac {\pi ^{2}}{(n+2)^{2}}}$

$-({\vec {y}}^{\,o})^{T}({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\;\geq \;\alpha _{n}\,h^{-2}\,\lVert \,{\vec {y}}^{\,o}\,\rVert _{2}^{2}$.

为了证明上面所说的

${\begin{aligned}-h^{2}({\vec {y}}^{\,o})^{T}({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\;=&\;\textstyle \sum _{i\,=\,1}^{\,n}y_{i}(-y_{i-1}+2\,y_{i}-y_{i+1})\\\;=&\;-\textstyle \sum _{i\,=\,1}^{\,n}y_{i}((y_{i+1}-y_{i})-(y_{i}-y_{i-1}))\end{aligned}}$.

重新排列  分部求和  ()

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

作为

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

现在，令 $m\;=\;n+1$ 得到

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

设定  $v_{i}\;=\;y_{i}$  当  $i\;=\;0,\;1,\;\ldots ,\;n+1$.

设定  $w_{0}\;=\;y_{0}$  以及  $w_{i}\;=\;y_{i}-y_{i-1}$  当  $i\;=\;1,\;2,\;\ldots ,\;n+1$.

由于  $v_{0}\;=\;v_{n+1}\;=\;0$,  该恒等式变为

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

进而得到

$-\sum _{i=1}^{\,n}y_{i}\,((y_{i+1}-y_{i})-(y_{i}-y_{i-1}))\;=\;\sum _{i=1}^{\,n+1}(y_{i}-y_{i-1})(y_{i}-y_{i-1})$.

这证明了等式

$-h^{2}({\vec {y}}^{\,o})^{T}({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\;=\;\sum _{i=1}^{\,n+1}(y_{i}-y_{i-1})^{2}$.

利用 **()**

如果  $v_{0}\,=\,v_{n+1}\,=\,0$  那么

${\sqrt {\textstyle \sum _{\,i\,=\,1}^{\,n+1}(v_{i}-v_{i-1})^{2}}}\geq \,{\sqrt {2\,(1-\cos(\pi \,{\tfrac {1}{n+2}}))}}\,{\sqrt {\textstyle \sum _{\,i\,=\,1}^{\,n}\,v_{i}^{2}}}$

以及

${\sqrt {2\,(1-\cos(\pi \,{\tfrac {1}{n+2}}))}}\;\to \;{\frac {\pi }{n+2}}$.

我们有以下结论

$\textstyle \sum _{\,i\,=\,1}^{\,n+1}(y_{i}-y_{i-1})^{2}\;\geq \;2\,(1-\cos(\pi \,{\tfrac {1}{n+2}}))\,\textstyle \sum _{\,i\,=\,1}^{\,n}\,y_{i}^{2}$,

其中  $2\,(1-\cos(\pi \,{\tfrac {1}{n+2}}))\;\to \;{\frac {\pi ^{2}}{(n+2)^{2}}}$.

这完成了对断言()的证明。

由于该算子是线性且正定的，  ${\vec {y}}$,   ()的解存在且唯一。

接下来要观察的是

$-({\vec {y}}^{\,o})^{T}({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}\;=\;({\vec {y}}^{\,o})^{T}{\vec {f}}\;\leq \;{\sqrt {({\vec {y}}^{\,o})^{T}{\vec {y}}^{\,o}\;}}{\sqrt {({\vec {f}})^{T}{\vec {f}}\;}}$.

并将此与() 结合，    $-({\overrightarrow{y}}^o{)}^T(\frac{d_h^2}{d_h{x}^2}{)}_I\overrightarrow{y}\ge {\alpha }_n$

$\alpha _{n}\,h^{-2}({\vec {y}}^{\,o})^{T}{\vec {y}}^{\,o}\;\leq \;{\sqrt {({\vec {y}}^{\,o})^{T}{\vec {y}}^{\,o}\;}}{\sqrt {({\vec {f}})^{T}{\vec {f}}\;}}$

以及

${\sqrt {({\vec {y}}^{\,o})^{T}{\vec {y}}^{\,o}\;}}\;\leq \;(\alpha _{n})^{-1}h^{2}\,{\sqrt {({\vec {f}})^{T}{\vec {f}}\;}}$.

符号

${\vec {y}}(x)\;=\;<y(x_{0})\;\,y(x_{1})\;\,y(x_{2})\;\,\ldots \;\,y(x_{n})\;\,y(x_{n+1})>$

将用于表示 $y(x)$的精确值，以及符号

${\vec {y}}^{\,o}(x)\;=\;<y(x_{1})\;\,y(x_{2})\;\,\ldots \;\,y(x_{n})>$

用于表示 ${\vec {y}}(x)$的内部点。

向量 $({\vec {y}}(x)-{\vec {y}}\,)$ 满足问题

$-({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,({\vec {y}}(x)-{\vec {y}}\,)\,=\,({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)-{\vec {f}}$,

其中 $(y(a)-y_{0})\;=\;(y(b)-y_{n+1})\;=\;0$.

因此，可以使用估计() 来得到

${\sqrt {({\vec {y}}^{\,o}(x)-{\vec {y}}^{\,o})^{T}({\vec {y}}^{\,o}(x)-{\vec {y}}^{\,o})\;}}\;\leq \;(\alpha _{n})^{-1}h^{2}\,{\sqrt {(({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)-{\vec {f}}\;)^{T}(({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)-{\vec {f}}\;)\;}}$.

回顾()

${\frac {d_{h}^{2}}{d_{h}x^{2}}}\,f(x)\;=\;f^{\prime \prime }(x)+({\tfrac {1}{24}}\,f^{(iv)}(z_{h})+{\tfrac {1}{24}}\,f^{(iv)}(z_{-h}))\,h^{2}$,

其中  $x-h\;<\;z_{-h}\;<\;x$  以及  $x\;<\;z_{h}\;<\;x+h$.

现在，

$({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)\;=\;<{\frac {d_{h}^{2}}{d_{h}x^{2}}}\,y(x_{1})\;\;{\frac {d_{h}^{2}}{d_{h}x^{2}}}\,y(x_{2})\;\;\ldots \;\;{\frac {d_{h}^{2}}{d_{h}x^{2}}}\,y(x_{n})>$

以及

${\vec {f}}\;=\;<y^{\prime \prime }(x_{1})\;\;y^{\prime \prime }(x_{2})\;\;\ldots \;\;y^{\prime \prime }(x_{n})>$.

因此，   $(({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)-{\vec {f}}\;)$    每个分量都具有以下形式

$({\tfrac {1}{24}}\,f^{(iv)}(z_{h})+{\tfrac {1}{24}}\,f^{(iv)}(z_{-h}))\,h^{2}$.

这导致了以下估计

${\sqrt {(({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)-{\vec {f}}\;)^{T}(({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}\,{\vec {y}}(x)-{\vec {f}}\;)\;}}\;\leq \;{\sqrt {n}}\,M\,h^{2}$,

其中

  $M\;=\;{\tfrac {1}{12}}\;{\underset {a\;\leq \;x\;\leq \;b}{\max }}(f^{(iv)}(x))$.

将不等式 () 和 () 结合后，可以得到

$\lVert \,{\vec {y}}^{\,o}(x)-{\vec {y}}^{\,o}\,\rVert _{2}\;\leq \;{\sqrt {n}}\,M\,(\alpha _{n})^{-1}\,h^{4}$.

差分解的概论

在继续讨论使用高阶有限差分算子的影响之前，研究线性方程解的一些一般性很有用。这为设计方法和证明提供了必要的动机。

假设 $L$ 是一个线性算子，它是正定的，这意味着存在一个常数 $\alpha \;>\;0$， 不依赖于 ${\vec {x}}$ 使得

${\vec {x}}^{\,T}L\,{\vec {x}}\;\geq \;\alpha \,{\vec {x}}^{\,T}{\vec {x}}$.

那么正如在()中针对矩阵解释的那样

$\lVert \,L^{-1}\,\rVert _{2}\;\leq \;(\alpha )^{-1}$.

现在，如果精确解 ${\vec {y}}(x)$ 到某个问题由下式给出

$L\,{\vec {y}}(x)\;=\;{\vec {f}}(x)$

除了 ${\vec {f}}(x)$ 仅通过 ${\vec {f}}$， 知道到某个近似程度，那么解 ${\vec {y}}$ 到方程

$L\,{\vec {y}}\;=\;{\vec {f}}$,

是所需精确解 ${\vec {y}}(x)$的近似。  由于

$L\,({\vec {y}}(x)-{\vec {y}})\;=\;({\vec {f}}(x)-{\vec {f}}\,)$,

近似的接近程度可以用以下公式估计

$\lVert \,({\vec {y}}(x)-{\vec {y}})\,\rVert _{2}\;\leq \;\lVert \,L^{-1}\,\rVert _{2}\,\lVert \,({\vec {f}}(x)-{\vec {f}}\,)\,\rVert _{2}\;\leq \;\,(\alpha )^{-1}\lVert \,({\vec {f}}(x)-{\vec {f}}\,)\,\rVert _{2}$.

关于*有限差分方法*，策略将是定义一个*线性算子*  $L_{h}$  使得

${\vec {y}}^{\,T}L_{h}\,{\vec {y}}\;\geq \;\alpha \,{\vec {y}}^{\,T}{\vec {y}}$,

其中  $\alpha$  与  $h$无关。那么对于任何问题的精确解  ${\vec {y}}(x)$，

$\lVert \,({\vec {y}}(x)-{\vec {y}})\,\rVert _{2}\;\leq \;\,(\alpha )^{-1}\lVert \,(L_{h}\,{\vec {y}}(x)-L_{h}\,{\vec {y}})\,\rVert _{2}$.

当  $L_{h}\,{\vec {y}}(x)$  被  ${\vec {f}}$近似到一定程度时，也就是说当

 $\lVert \,L_{h}\,{\vec {y}}(x)-{\vec {f}}\,\rVert _{2}\;\leq \;B_{h}$

然后对于 ${\vec {y}}$,  方程 $L_{h}\,{\vec {y}}\;=\;{\vec {f}}$ 的解是

$\lVert \,({\vec {y}}(x)-{\vec {y}})\,\rVert _{2}\;\leq \;\,(\alpha )^{-1}\lVert \,(L_{h}\,{\vec {y}}(x)-{\vec {f}}\,)\,\rVert _{2}\;\leq \;(\alpha )^{-1}B_{h}$.

三阶估计

在本节中，我们研究了用于二阶导数近似的五点有限差分算子的性质。本节将沿用本章之前章节中介绍的符号。符号  ${\vec {y}}$   和   ${\vec {y}}^{\,o}$   的含义与“有限差分法求解”一节中相同。符号

$({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{i}$   和   $({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{I}$

重新定义为表示五点算子。

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1}\ y\ =$.
${\frac {1}{12}}\,y_{4}-{\frac {1}{3}}\,y_{3}-{\frac {1}{2}}\,y_{2}+{\frac {5}{3}}\,y_{1}-{\frac {11}{12}}\,y_{0}$

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i}\ y\ =$.
${\frac {1}{12}}\,y_{i+2}-{\frac {4}{3}}\,y_{i+1}+{\frac {5}{2}}\,y_{i}-{\frac {4}{3}}\,y_{i-1}+{\frac {1}{12}}\,y_{i-2}$

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

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{n}\ y\ =$.

$-{\frac {11}{12}}\,y_{n+1}+{\frac {5}{3}}\,y_{n}-{\frac {1}{2}}\,y_{n-1}-{\frac {1}{3}}\,y_{n-2}+{\frac {1}{12}}\,y_{n-3}$

五点二阶导数算子在区间端点附近的点处具有三阶精度，由于它是中心差分算子，因此对更内部的点具有四阶精度。

为了使求和过程更容易理解，  $({\frac {d_{h}^{2}}{d_{h}x^{2}}})_{i}$  的表达式被重新排列，使求和过程更容易理解。这些恒等式可以通过简单的系数比较来验证。

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1}\ y\;=\;{\frac {7}{6}}(-y_{2}+2\,y_{1}-y_{0})$

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

$-{\frac {1}{12}}\,(-y_{2}+2\,y_{1}-y_{0})$

$-{\frac {1}{12}}\,(-y_{3}+2\,y_{2}-y_{1})$

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i}\ y\;=\;{\frac {7}{6}}\,(-y_{i+1}+2\,y_{i}-y_{i-1})$

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

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

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

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{n}\ y\;=\;{\frac {7}{6}}(-y_{n+1}+2\,y_{n}-y_{n-1})$

${\begin{aligned}+\;&(-{\frac {1}{12}}\,(y_{n-2}-y_{n-3})+{\frac {1}{3}}\,(y_{n-1}-y_{n-2})-{\frac {5}{12}}\,(y_{n}-y_{n-1})+{\frac {1}{6}}\,(y_{n+1}-y_{n}))\\\end{aligned}}$

$-{\frac {1}{12}}\,(-y_{n}+2\,y_{n-1}-y_{n-2})$

$-{\frac {1}{12}}\,(-y_{n+1}+2\,y_{n}-y_{n-1})$

本节的目的是建立不等式。

$-h^{2}({\vec {y}}^{\,o})^{T}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{I}\;{\vec {y}}\;\geq \;{\frac {2}{3}}\,{\textstyle \sum _{\,i\,=\,1}^{\,n+1}}\,(y_{i}-y_{i-1})^{2}$.

这是对有限差分逼近的精度进行估计时最具技术性的部分。分析的其余部分通过应用“关于差分解的一般性”部分中描述的推理来进行。对五点有限差分估计函数二阶导数精度的估计比较容易，将在单独的部分中介绍。

考虑到 $y_{0}\,=\,0$.   所以 $y_{1}\,=\,(y_{1}-y_{0})$  然后

$-h^{2}y_{1}\,({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1}{\vec {y}}\;={\frac {7}{6}}\,y_{1}(-y_{2}+2\,y_{1}-y_{0})$

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

$-{\frac {1}{12}}\,y_{1}(-y_{2}+2\,y_{1}-y_{0})-{\frac {1}{12}}\,y_{1}(-y_{3}+2\,y_{2}-y_{1})$

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

$-{\frac {1}{12}}\,y_{1}(-y_{2}+2\,y_{1}-y_{0})-{\frac {1}{12}}\,y_{2}(-y_{3}+2\,y_{2}-y_{1})$

$+{\frac {1}{12}}\,(y_{2}-y_{1})^{2}+{\frac {1}{6}}\,(y_{1}-y_{0})^{2}$

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

$-h^{2}y_{i}\,({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i}\,{\vec {y}}\;={\frac {7}{6}}\,y_{i}\,(-y_{i+1}+2\,y_{i}-y_{i-1})$

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

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

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

$+{\frac {1}{12}}\,(y_{i+1}-y_{i})^{2}\,+{\frac {1}{12}}\,(y_{i}-y_{i-1})^{2}$

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

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

考虑到 $y_{n+1}\,=\,0$$-h^{2}y_{n}\,({\partial _{h}^{2} \over \partial _{h}x^{2}})_{n}\,{\vec {y}}\;={\frac {7}{6}}\,y_{n}\,(-y_{n+1}+2\,y_{n}-y_{n-1})$

${\begin{aligned}+\;&y_{n}(-{\frac {1}{12}}\,(y_{n-2}-y_{n-3})+{\frac {1}{3}}\,(y_{n-1}-y_{n-2})-{\frac {5}{12}}\,(y_{n}-y_{n-1})+{\frac {1}{6}}\,(y_{n+1}-y_{n}))\\\end{aligned}}$

$-{\frac {1}{12}}\,y_{n}\,(-y_{n}+2\,y_{n-1}-y_{n-2})-{\frac {1}{12}}\,y_{n}\,(-y_{n+1}+2\,y_{n}-y_{n-1})$

$={\frac {7}{6}}\,y_{n}\,(-y_{n+1}+2\,y_{n}-y_{n-1})$

$-{\frac {1}{12}}\,y_{n-1}\,(-y_{n}+2\,y_{n-1}-y_{n-2})-{\frac {1}{12}}\,y_{n}\,(-y_{n+1}+2\,y_{n}-y_{n-1})$

$+{\frac {1}{6}}\,(y_{n+1}-y_{n})^{2}\,+{\frac {1}{12}}\,(y_{n}-y_{n-1})^{2}$

$-{\frac {1}{12}}\,(y_{n}-y_{n-1})(y_{n-1}-y_{n-2})+{\frac {1}{12}}\,(y_{n+1}-y_{n})(y_{n}-y_{n-1})$

${\begin{aligned}+\;&(y_{n+1}-y_{n})({\frac {1}{12}}\,(y_{n-2}-y_{n-3})-{\frac {1}{3}}\,(y_{n-1}-y_{n-2})+{\frac {1}{3}}\,(y_{n}-y_{n-1})-{\frac {1}{3}}\,(y_{n+1}-y_{n}))\\\end{aligned}}$

这种项的组织方式会导致

$-h^{2}({\vec {y}}^{\,o})^{T}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{I}\;{\vec {y}}\;=\;-h^{2}{\textstyle \sum _{\,i\,=\,1}^{\,n}}\,y_{i}\,({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i}\;{\vec {y}}\;=$

${\textstyle \sum _{\,i\,=\,1}^{\,n}}\,y_{i}\,(-y_{i+1}+2\,y_{i}-y_{i-1})+{\frac {1}{6}}\,{\textstyle \sum _{\,i\,=\,1}^{\,n+1}}\,(y_{i}-y_{i-1})^{2}-{\frac {1}{6}}\,{\textstyle \sum _{\,i\,=\,2}^{\,n-1}}\,(y_{i+1}-y_{i})(y_{i}-y_{i-1})$

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

${\begin{aligned}+\;&(y_{n+1}-y_{n})({\frac {1}{12}}\,(y_{n-2}-y_{n-3})-{\frac {1}{3}}\,(y_{n-1}-y_{n-2})+{\frac {1}{3}}\,(y_{n}-y_{n-1})-{\frac {1}{3}}\,(y_{n+1}-y_{n}))\\\end{aligned}}$

根据 **()**，已知和   ${\textstyle \sum _{\,i\,=\,1}^{\,n}}\,y_{i}\,(-y_{i+1}+2\,y_{i}-y_{i-1})\,=\,{\textstyle \sum _{\,i\,=\,1}^{\,n+1}}\,(y_{i}-y_{i-1})^{2}$。   利用这个等式并将第二个和式的第一项和最后一项移项

$-h^{2}({\vec {y}}^{\,o})^{T}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{I}\;{\vec {y}}\;=$

${\textstyle \sum _{\,i\,=\,1}^{\,n+1}}\,(y_{i}-y_{i-1})^{2}+{\frac {1}{6}}\,{\textstyle \sum _{\,i\,=\,2}^{\,n}}\,(y_{i}-y_{i-1})^{2}-{\frac {1}{6}}\,{\textstyle \sum _{\,i\,=\,2}^{\,n-1}}\,(y_{i+1}-y_{i})(y_{i}-y_{i-1})$

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

${\begin{aligned}+\;&(y_{n+1}-y_{n})({\frac {1}{12}}\,(y_{n-2}-y_{n-3})-{\frac {1}{3}}\,(y_{n-1}-y_{n-2})+{\frac {1}{3}}\,(y_{n}-y_{n-1})-{\frac {1}{6}}\,(y_{n+1}-y_{n}))\\\end{aligned}}$

使用()

$\left\vert \,{\textstyle \sum _{\,i\,=\,3}^{\,n-2}}\,(y_{i+1}-y_{i})(y_{i}-y_{i-1})\,\right\vert \;\leq \;{\frac {1}{2}}\,(y_{3}-y_{2})^{2}\,+\,{\textstyle \sum _{\,i\,=\,4}^{\,n-2}}\,(y_{i}-y_{i-1})^{2}\,+\,{\frac {1}{2}}\,(y_{n-1}-y_{n-2})^{2}$,

$\left\vert \,(y_{3}-y_{2})(y_{2}-y_{1})\,\right\vert \;\leq \;{\frac {1}{2}}\,\alpha _{1}^{2}\,(y_{2}-y_{1})^{2}\,+\,{\frac {1}{2}}\,(y_{3}-y_{2})^{2}\,/\,\alpha _{1}^{2}$,

$\left\vert \,(y_{n}-y_{n-1})(y_{n-1}-y_{n-2})\,\right\vert \;\leq \;{\frac {1}{2}}\,\beta _{1}^{2}\,(y_{n}-y_{n-1})^{2}\,+\,{\frac {1}{2}}\,(y_{n-1}-y_{n-2})^{2}\,/\,\beta _{1}^{2}$,

下一个不等式如下。

$-h^{2}({\vec {y}}^{\,o})^{T}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{I}\;{\vec {y}}\;\geq \;{\textstyle \sum _{\,i\,=\,1}^{\,n+1}}\,(y_{i}-y_{i-1})^{2}-{\frac {1}{6}}\,(y_{1}-y_{0})^{2}-{\frac {1}{6}}\,(y_{n+1}-y_{n})^{2}$

$+\,{\frac {1}{12}}\,(1-1\,/\,\alpha _{1}^{2})(y_{3}-y_{2})^{2}\,+\,{\frac {1}{6}}\,(1-{\frac {1}{2}}\,\alpha _{1}^{2})(y_{2}-y_{1})^{2}$

$+\,{\frac {1}{6}}\,(1-{\frac {1}{2}}\,\beta _{1}^{2})(y_{n}-y_{n-1})^{2}\,+\,{\frac {1}{12}}\,(1-1\,/\,\beta _{1}^{2})(y_{n-1}-y_{n-2})^{2}$

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

${\begin{aligned}+\;&(y_{n+1}-y_{n})({\frac {1}{12}}\,(y_{n-2}-y_{n-3})-{\frac {1}{3}}\,(y_{n-1}-y_{n-2})+{\frac {1}{3}}\,(y_{n}-y_{n-1}))\\\end{aligned}}$

现在，使用()

$\left\vert \,({\frac {1}{12}}\,(y_{4}-y_{3})-{\frac {1}{3}}\,(y_{3}-y_{2})+{\frac {1}{3}}\,(y_{2}-y_{1}))(y_{1}-y_{0})\,\right\vert \;\leq$

${\frac {1}{24}}\,(\alpha _{2}^{2}\,(y_{4}-y_{3})^{2}+{\frac {1}{\alpha _{2}^{2}}}\,(y_{1}-y_{0})^{2})+{\frac {1}{6}}\,(\alpha _{3}^{2}\,(y_{3}-y_{2})^{2}+{\frac {1}{\alpha _{3}^{2}}}\,(y_{1}-y_{0})^{2})$

${\frac {1}{6}}\,(\alpha _{4}^{2}\,(y_{2}-y_{1})^{2}+{\frac {1}{\alpha _{4}^{2}}}\,(y_{1}-y_{0})^{2})$.

取   $\alpha _{1}^{2}\,=\,{\frac {1+{\sqrt {5}}}{2}},\;\alpha _{2}^{2}\,=\,8,$   和   $\alpha _{3}^{2}\,=\,\alpha _{4}^{2}\,=\,2+{\frac {3-{\sqrt {5}}}{4}}$   得出

$-{\frac {1}{6}}\,(y_{1}-y_{0})^{2}+\,{\frac {1}{12}}\,(1-1\,/\,\alpha _{1}^{2})(y_{3}-y_{2})^{2}\,+\,{\frac {1}{6}}\,(1-{\frac {1}{2}}\,\alpha _{1}^{2})(y_{2}-y_{1})^{2}$

$+({\frac {1}{12}}\,(y_{4}-y_{3})-{\frac {1}{3}}\,(y_{3}-y_{2})+{\frac {1}{3}}\,(y_{2}-y_{1}))(y_{1}-y_{0})\;\geq$

$-{\frac {1}{3}}((y_{4}-y_{3})^{2}\,+\,(y_{3}-y_{2})^{2}\,+\,(y_{2}-y_{1})^{2}\,+\,(y_{1}-y_{0})^{2})$.

以相同的方式

$-{\frac {1}{6}}\,(y_{n+1}-y_{n})^{2}+\,{\frac {1}{6}}\,(1-{\frac {1}{2}}\,\beta _{1}^{2})(y_{n}-y_{n-1})^{2}\,+\,{\frac {1}{12}}\,(1-1\,/\,\beta _{1}^{2})(y_{n-1}-y_{n-2})^{2}$

$+(y_{n+1}-y_{n})({\frac {1}{12}}\,(y_{n-2}-y_{n-3})-{\frac {1}{3}}\,(y_{n-1}-y_{n-2})+{\frac {1}{3}}\,(y_{n}-y_{n-1}))\;\geq$

$-{\frac {1}{3}}((y_{n+1}-y_{n})^{2}\,+\,(y_{n}-y_{n-1})^{2}\,+\,(y_{n-1}-y_{n-2})^{2}\,+\,(y_{n-2}-y_{n-3})^{2})$.

所求不等式成立。

$-h^{2}({\vec {y}}^{\,o})^{T}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{I}\;{\vec {y}}\;\geq \;{\frac {2}{3}}\,{\textstyle \sum _{\,i\,=\,1}^{\,n+1}}\,(y_{i}-y_{i-1})^{2}$.

二维域

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

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

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

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

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

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

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

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

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

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

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

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

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

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

离散拉普拉斯算子

为了数值近似 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

${\partial ^{2} \over \partial x^{2}}\ u(x,y)$      以及      ${\partial ^{2} \over \partial y^{2}}\ u(x,y)$

可以在网格上用*差商*来近似。

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

这些*差商*可以通过点的数量或精度阶来选择。在任何一种情况下，它们都将与*定义和基础*章节中差商部分所解释的一样。在使用其他标准（例如相对于阶的系数的大小或差异的最小化）的情况下，通过选择精度阶低于最大阶的差商而产生的可能性在此时未进行分析。

由于在差分算子的符号中包含许多索引很麻烦，因此用于近似二阶导数的差商的相同表达式将被重复用于不同阶的算子。在需要时，将明确哪一个。某些普遍性适用于其中任何一个，并且可以以此为基础进行讨论。

然后*拉普拉斯算子*  $\Delta u(x,y)$  可以在*网格*的*内部*用以下方法*近似*。

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

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

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

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

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

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

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

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

二阶偏导数
${\partial ^{2} \over \partial y^{2}}\ u(x,y)$
可以在网格上用*差商*来近似。
$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})$.

这些差商 由下式给出

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

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

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

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

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

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

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