Using High Order Finite Differences/Third Order Method

A Third Order Accurate Method

Statement of the Problem

Let D be the rectangle

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

and let C be the boundary of D.

The operator

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

is the usual Laplacian. The problem, determine a function u(x, y) such that

$\{\begin{array}{c}\begin{array}{cc}-\mathrm{\Delta }u(x,y)& =f(x,y),\text{for}(x,y)\in D\\ \\ u(x,y)& =g(x,y),\text{for}(x,y)\in C,\\ \end{array}\end{array}{}_{(1.0)}$

is called a Poisson problem.

discrete approximation

To approximate u(x, y) numerically, use the grid

$(x_i,y_j),0\le i\le m+1,0\le j\le n+1$.
with
$x_i=ih,y_i=jk$
and
$h=a/(m+1),k=b/(n+1)$

The second partial derivative
$\frac{{\partial }^2}{\partial x^2}u(x,y)$
can be approximated on the grid by difference quotients
$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2})u(x_i,y_j)$.

These difference quotients are given by

$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2})u(x_1,y_j)=$.

$\frac{-\frac{1}{12}u(x_1+3h,y_j)+\frac{1}{3}u(x_1+2h,y_j)+\frac{1}{2}u(x_1+h,y_j)-\frac{5}{3}u(x_1,y_j)+\frac{11}{12}u(x_1-h,y_j)}{h^2}$

$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2})u(x_i,y_j)=$.

$\frac{-\frac{1}{12}u(x_i+2h,y_j)+\frac{4}{3}u(x_i+h,y_j)-\frac{5}{2}u(x_i,y_j)+\frac{4}{3}u(x_i-h,y_j)-\frac{1}{12}u(x_i-2h,y_j)}{h^2}$

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

$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2})u(x_m,y_j)=$.

$\frac{\frac{11}{12}u(x_m+h,y_j)-\frac{5}{3}u(x_m,y_j)+\frac{1}{2}u(x_m-h,y_j)+\frac{1}{3}u(x_m-2h,y_j)-\frac{1}{12}u(x_m-3h,y_j)}{h^2}$

The second partial derivative
$\frac{{\partial }^2}{\partial y^2}u(x,y)$
can be approximated on the grid by difference quotients
$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2})u(x_i,y_j)$.

These difference quotients are given by

$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2})u(x_i,y_1)=$.

$\frac{-\frac{1}{12}u(x_i,y_1+3k)+\frac{1}{3}u(x_i,y_1+2k)+\frac{1}{2}u(x_i,y_1+k)-\frac{5}{3}u(x_i,y_1)+\frac{11}{12}u(x_i,y_1-k)}{k^2}$

$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2})u(x_i,y_j)=$.

$\frac{-\frac{1}{12}u(x_i,y_j+2k)+\frac{4}{3}u(x_i,y_j+k)-\frac{5}{2}u(x_i,y_j)+\frac{4}{3}u(x_i,y_j-k)-\frac{1}{12}u(x_i,y_j-2k)}{k^2}$

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

$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2})u(x_i,y_n)=$.

$\frac{\frac{11}{12}u(x_i,y_n+k)-\frac{5}{3}u(x_i,y_n)+\frac{1}{2}u(x_i,y_n-k)+\frac{1}{3}u(x_i,y_n-2k)-\frac{1}{12}u(x_i,y_n-3k)}{k^2}$

truncation errors

The difference quotients   $(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2})u(x_i,y_j)$   are third order accurate with truncation errors:

$\begin{array}{cc}{\tau }_x(x_i,y_j)& =(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2})u(x_i,y_j)-\frac{{\partial }^2}{\partial x^2}u(x_i,y_j)\\ & =\frac{h^3}{120}M{}_x^{(5)}(x_i,y_j)\end{array}$

with

$M{}_x^{(5)}(x_1,y_j)=\frac{67}{6}\frac{{\partial }^5}{\partial x^5}u(x_1+\varphi {}_1^{(1,j)}h,y_j)-\frac{254}{12}\frac{{\partial }^5}{\partial x^5}u(x_1+\varphi {}_2^{(1,j)}h,y_j)$ ,

for some    $0<\varphi {}_1^{(1,j)}<2,-1<\varphi {}_2^{(1,j)}<3$ ,

$M{}_x^{(5)}(x_i,y_j)=4\frac{{\partial }^5}{\partial x^5}u(x_i+\varphi {}_1^{(i,j)}h,y_j)-4\frac{{\partial }^5}{\partial x^5}u(x_i+\varphi {}_2^{(i,j)}h,y_j)$ ,

for some    $-2<\varphi {}_1^{(i,j)}<1,-1<\varphi {}_2^{(i,j)}<2,$

$\text{for}i=2,3,\dots ,m-1.$     

When  $\frac{{\partial }^6}{\partial x^6}u(x,y)$   is continuous, these estimates also hold.

${\tau }_x(x_i,y_j)=\frac{h^4}{720}{M}_x^6(x_i,y_j),$   with

$M_x^6(x_i,y_j)=\frac{8}{3}\frac{{\partial }^6}{\partial x^6}u(x_i+\varphi {}_1^{(i,j)}h,y_j)-\frac{32}{3}\frac{{\partial }^6}{\partial x^6}u(x_i+\varphi {}_2^{(i,j)}h,y_j)$

for some    $-1<\varphi {}_1^{(i,j)}<1,-2<\varphi {}_2^{(i,j)}<2,$    

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

The case for  $i=m$  is

$M{}_x^{(5)}(x_m,y_j)=\frac{254}{12}\frac{{\partial }^5}{\partial x^5}u(x_m+\varphi {}_1^{(m,j)}h,y_j)-\frac{67}{6}\frac{{\partial }^5}{\partial x^5}u(x_m+\varphi {}_2^{(m,j)}h,y_j)$ ,

for some    $-3<\varphi {}_1^{(m,j)}<1,-2<\varphi {}_2^{(m,j)}<0$ .

The difference quotients   $(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2})u(x_i,y_j)$   are third order accurate with truncation errors:

$\begin{array}{cc}{\tau }_y(x_i,y_i)& =(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2})u(x_i,y_1)-\frac{{\partial }^2}{\partial y^2}u(x_i,y_1)\\ & =\frac{k^3}{120}M{}_y^{(5)}(x_i,y_1)\end{array}$

with

$M{}_y^{(5)}(x_i,y_1)=\frac{67}{6}\frac{{\partial }^5}{\partial y^5}u(x_i,y_1+\psi {}_1^{(i,1)}k)-\frac{254}{12}\frac{{\partial }^5}{\partial y^5}u(x_i,y_1+\psi {}_2^{(i,1)}k)$ ,

for some    $0<\psi {}_1^{(i,1)}<2,-1<\psi {}_2^{(i,1)}<3$ ,

$M{}_y^{(5)}(x_i,y_j)=4\frac{{\partial }^5}{\partial y^5}u(x_i,y_j+\psi {}_1^{(i,j)}k)-4\frac{{\partial }^5}{\partial y^5}u(x_i,y_j+\psi {}_2^{(i,j)}k)$ ,

for some    $-2<\psi {}_1^{(i,j)}<1,-1<\psi {}_2^{(i,j)}<2,$

$\text{for}j=2,3,\dots ,n-1.$     

When  $\frac{{\partial }^6}{\partial y^6}u(x,y)$   is continuous, these estimates also hold.

${\tau }_y(x_i,y_j)=\frac{k^4}{720}{M}_y^6(x_i,y_j),$   with

$M_y^6(x_i,y_j)=\frac{8}{3}\frac{{\partial }^6}{\partial y^6}u(x_i+\psi {}_1^{(i,j)}k,y_j)-\frac{32}{3}\frac{{\partial }^6}{\partial y^6}u(x_i+\psi {}_2^{(i,j)}k,y_j)$

for some    $-1<\psi {}_1^{(i,j)}<1,-2<\psi {}_2^{(i,j)}<2,$    

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

The case for  $j=n$  is

$M{}_y^{(5)}(x_i,y_n)=\frac{254}{12}\frac{{\partial }^5}{\partial y^5}u(x_i,y_n+\psi {}_1^{(i,n)}k)-\frac{67}{6}\frac{{\partial }^5}{\partial y^5}u(x_i,y_n+\psi {}_2^{(i,n)}k)$ ,

for some    $-3<\psi {}_1^{(i,n)}<1,-2<\psi {}_2^{(i,n)}<0$ .

The Laplacian  $\mathrm{\Delta }u(x,y)$  then can be approximated on the interior of the grid by

${\mathrm{\Delta }}_{h,k}u(x_i,y_j)=\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_i,y_j)+\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}u(x_i,y_j)$

The truncation error

$\tau (x_i,y_j)={\mathrm{\Delta }}_{h,k}u(x_i,y_j)-\mathrm{\Delta }u(x_i,y_j)$

is given by

$\tau (x_i,y_j)={\tau }_x(x_i,y_j)+{\tau }_y(x_i,y_j)$.

finite difference operations defined

For the grid vector

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

define the finite difference operations

${\mathrm{\Delta }}_{i,j}=(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}U+(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}U$

by the following.

$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{1,j}U=$.

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

$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}U=$.

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

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

$(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{m,j}U=$.

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

$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,1}U=$.

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

$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}U=$.

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

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

$(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,n}U=$.

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

simulation of the problem

To simulate the problem  (1.0)   let

$\begin{array}{cccc}{U}_{0,j}& =& g(x_0,y_j),& 0\le j\le n+1\\ U_{m+1,j}& =& g(x_{m+1},y_j),& 0\le j\le n+1\\ U_{i,0}& =& g(x_i,y_0),& 0\le i\le m+1\\ U_{i,n+1}& =& g(x_i,y_{n+1)},& 0\le i\le m+1\\ \end{array}$

Then solve the non-singular linear system

$-{\mathrm{\Delta }}_{i,j}U=f(x_i,y_j)1\le i\le m,1\le j\le n$  ;

for the remaining  $U_{i,j}$ 

error estimation

The error

$U-u=\{U_{i,j}-u(x_i,y_j)\}$ ,

satisfies

${\Vert U-u\Vert }_2\le \frac{\rho a^2{b}^2}{{\pi }^2(a^2+b^2)}{\Vert \tau \Vert }_2(1+O(h^2+k^2))$ .

for

$\tau =\{\tau (x_i,y_j)\},1\le i\le m,1\le j\le n$  ,

and

$\rho \le 1/(1-(1+\sqrt{1}0)/12)$ .

proof of truncation error estimates

The truncation error estimates for  $(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_i,y_j))$  are done under the assumption that  $u(x,y)$  is sufficiently smooth so that  $\frac{{\partial }^5}{\partial x^5}u(x,y)$  is continuous. For notational convenience let

$g(x)=u(x,y_j).$

Expand  $g(x)$  in it's Taylor expansion about  $x_i$,

$\begin{array}{cc}g(x_i+\mathrm{\Delta }x)& =g(x_i)+g^{(1)}(x_i)\mathrm{\Delta }x+\frac{1}{2}{g}^{(2)}(x_i)(\mathrm{\Delta }x{)}^2+\frac{1}{6}{g}^{(3)}(x_i)(\mathrm{\Delta }x{)}^3\\ & +\frac{1}{24}{g}^{(4)}(x_i)(\mathrm{\Delta }x{)}^4+\frac{1}{120}{g}^{(5)}(z)(\mathrm{\Delta }x{)}^5\\ \end{array}$.

where  $z$  is some number between  $x_i$  and  $x_i+\mathrm{\Delta }x$. Then

$\begin{array}{cc}& \frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_1,y_j)=\\ & \frac{(-\frac{1}{12}g(x_1+3h)+\frac{1}{3}g(x_1+2h)+\frac{1}{2}g(x_1+h)-\frac{5}{3}g(x_1)+\frac{11}{12}g(x_1-h))}{h^2}\end{array}$

$\begin{array}{cc}& =g(x_1)\frac{(-\frac{1}{12}+\frac{1}{3}+\frac{1}{2}-\frac{5}{3}+\frac{11}{12})}{h^2}\\ & +g^{(1)}(x_1)\frac{(-\frac{1}{12}(3h)+\frac{1}{3}(2h)+\frac{1}{2}(h)-\frac{5}{3}(0)+\frac{11}{12}(-h))}{h^2}\\ & +\frac{1}{2}{g}^{(2)}(x_1)\frac{(-\frac{1}{12}(3h{)}^2+\frac{1}{3}(2h{)}^2+\frac{1}{2}(h{)}^2-\frac{5}{3}(0{)}^2+\frac{11}{12}(-h{)}^2)}{h^2}\\ & +\frac{1}{6}{g}^{(3)}(x_1)\frac{(-\frac{1}{12}(3h{)}^3+\frac{1}{3}(2h{)}^3+\frac{1}{2}(h{)}^3-\frac{5}{3}(0{)}^3+\frac{11}{12}(-h{)}^3)}{h^2}\\ & +\frac{1}{24}{g}^{(4)}(x_1)\frac{(-\frac{1}{12}(3h{)}^4+\frac{1}{3}(2h{)}^4+\frac{1}{2}(h{)}^4-\frac{5}{3}(0{)}^4+\frac{11}{12}(-h{)}^4)}{h^2}\\ & +\frac{1}{120}\frac{(-\frac{1}{12}{g}^{(5)}(z_1)(3h{)}^5+\frac{1}{3}{g}^{(5)}(z_2)(2h{)}^5+\frac{1}{2}{g}^{(5)}(z_3)(h{)}^5-\frac{5}{3}(0{)}^5}{h^2}\\ & \frac{+\frac{11}{12}{g}^{(5)}(z_4)(-h{)}^5)}{h^2}\\ \end{array}$

$=g^{(2)}(x_1)+\frac{h^3}{120}(-\frac{81}{4}{g}^{(5)}(z_1)+\frac{32}{3}{g}^{(5)}(z_2)+\frac{1}{2}{g}^{(5)}(z_3)-\frac{11}{12}{g}^{(5)}(z_4))$

where

$\begin{array}{cc}& x_1<z_1<x_1+3h,x_1<z_2<x_1+2h,\\ & x_1<z_3<x_1+h,\text{and}{x}_1-h<z_4<x_1.\\ \end{array}$.

Since

$\frac{67}{6}\underset{0\le \varphi \le 2}{\min (g^{(5)}(x_1+\varphi h))}\le \frac{32}{3}{g}^{(5)}(z_2)+\frac{1}{2}{g}^{(5)}(z_3)\le \frac{67}{6}\underset{0\le \varphi \le 2}{\max (g^{(5)}(x_1+\varphi h))},$

$\frac{254}{12}\underset{-1\le \varphi \le 3}{\min (g^{(5)}(x_1+\varphi h))}\le \frac{81}{4}{g}^{(5)}(z_1)+\frac{11}{12}{g}^{(5)}(z_4)\le \frac{254}{12}\underset{-1\le \varphi \le 3}{\max (g^{(5)}(x_1+\varphi h))},$

from the intermediate value property

$\frac{32}{3}{g}^{(5)}(z_2)+\frac{1}{2}{g}^{(5)}(z_3)=\frac{67}{6}{g}^{(5)}(x_1+{\varphi }_{1}h)),\text{for}0<{\varphi }_1,<2,$

$\frac{81}{4}{g}^{(5)}(z_1)+\frac{11}{12}{g}^{(5)}(z_4)=\frac{254}{12}{g}^{(5)}(x_1+{\varphi }_{2}h),\text{for}-1<{\varphi }_2<3.$.

This gives

$\begin{array}{cc}& \frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_1,y_j)=g^{(2)}(x_1)+\frac{h^3}{120}(\frac{67}{6}{g}^{(5)}(x_1+{\varphi }_{1}h)-\frac{254}{12}{g}^{(5)}(x_1+{\varphi }_{2}h))\\ & =\frac{{\partial }^2}{\partial x^2}u(x_1,y_j)+\frac{h^3}{120}(\frac{67}{6}\frac{{\partial }^5}{\partial x^5}u(x_1+{\varphi }_{1}h,y_j)-\frac{254}{12}\frac{{\partial }^5}{\partial x^5}u(x_1+{\varphi }_{2}h,y_j))\\ \end{array}$

which is

${\tau }_x(x_1,y_j)=\frac{h^3}{120}{M}_x^5(x_1,y_j).$

For $i=2,3,\dots ,m-1$

$\begin{array}{cc}& \frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_i,y_j)=\\ & \frac{(-\frac{1}{12}g(x_i+2h)+\frac{4}{3}g(x_i+h)-\frac{5}{2}g(x_i)+\frac{4}{3}g(x_i-h)-\frac{1}{12}g(x_i-2h))}{h^2}\end{array}$

$\begin{array}{cc}& =g(x_i)\frac{(-\frac{1}{12}+\frac{4}{3}-\frac{5}{2}+\frac{4}{3}-\frac{1}{12})}{h^2}\\ & +g^{(1)}(x_i)\frac{(-\frac{1}{12}(2h)+\frac{4}{3}(h)-\frac{5}{2}(0)+\frac{4}{3}(-h)-\frac{1}{12}(-2h))}{h^2}\\ & +\frac{1}{2}{g}^{(2)}(x_i)\frac{(-\frac{1}{12}(2h{)}^2+\frac{4}{3}(h{)}^2-\frac{5}{2}(0{)}^2+\frac{4}{3}(-h{)}^2-\frac{1}{12}(-2h{)}^2)}{h^2}\\ & +\frac{1}{6}{g}^{(3)}(x_i)\frac{(-\frac{1}{12}(2h{)}^3+\frac{4}{3}(h{)}^3-\frac{5}{2}(0{)}^3+\frac{4}{3}(-h{)}^3-\frac{1}{12}(-2h{)}^3)}{h^2}\\ & +\frac{1}{24}{g}^{(4)}(x_i)\frac{(-\frac{1}{12}(2h{)}^4+\frac{4}{3}(h{)}^4-\frac{5}{2}(0{)}^4+\frac{4}{3}(-h{)}^4-\frac{1}{12}(-2h{)}^4)}{h^2}\\ & +\frac{1}{120}\frac{(-\frac{1}{12}{g}^{(5)}(z_1)(2h{)}^5+\frac{4}{3}{g}^{(5)}(z_2)(h{)}^5-\frac{5}{2}(0{)}^5+\frac{4}{3}{g}^{(5)}(z_3)(-h{)}^5}{h^2}\\ & \frac{-\frac{1}{12}{g}^{(5)}(z_4)(-2h{)}^5)}{h^2}\\ \end{array}$

$=g^{(2)}(x_i)+\frac{h^3}{120}(-\frac{8}{3}{g}^{(5)}(z_1)+\frac{4}{3}{g}^{(5)}(z_2)-\frac{4}{3}{g}^{(5)}(z_3)+\frac{8}{3}{g}^{(5)}(z_4))$

where

$\begin{array}{cc}& x_i<z_1<x_i+2h,x_i<z_2<x_i+h,\\ & x_i-h<z_3<x_i,\text{and}{x}_i-2h<z_4<x_i.\\ \end{array}$.

Reasoning as before, combining terms with like signs and using the intermediate value property,

$\frac{4}{3}{g}^{(5)}(z_2)+\frac{8}{3}{g}^{(5)}(z_4)=4g^{(5)}(x_i+{\varphi }_{1}h)),\text{for}-2<{\varphi }_1,<1,$

$\frac{8}{3}{g}^{(5)}(z_1)+\frac{4}{3}{g}^{(5)}(z_3)=4g^{(5)}(x_i+{\varphi }_{2}h),\text{for}-1<{\varphi }_2<2.$.

This gives

$\begin{array}{cc}& \frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_i,y_j)=g^{(2)}(x_i)+\frac{h^3}{120}(\frac{67}{6}{g}^{(5)}(x_i+{\varphi }_{1}h)-\frac{254}{12}{g}^{(5)}(x_i+{\varphi }_{2}h))\\ & =\frac{{\partial }^2}{\partial x^2}u(x_i,y_j)+\frac{h^3}{120}(4\frac{{\partial }^5}{\partial x^5}u(x_i+{\varphi }_{1}h,y_j)-4\frac{{\partial }^5}{\partial x^5}u(x_i+{\varphi }_{2}h,y_j))\\ \end{array}$

which is

${\tau }_x(x_i,y_j)=\frac{h^3}{120}{M}_x^5(x_i,y_j).$

Under the assumption that  $\frac{{\partial }^6}{\partial x^6}u(x,y)$  is continuous, in the preceding argument, the expression

$\begin{array}{cc}& +\frac{1}{120}\frac{(-\frac{1}{12}{g}^{(5)}(z_1)(2h{)}^5+\frac{4}{3}{g}^{(5)}(z_2)(h{)}^5-\frac{5}{2}(0{)}^5+\frac{4}{3}{g}^{(5)}(z_3)(-h{)}^5}{h^2}\\ & \frac{-\frac{1}{12}{g}^{(5)}(z_4)(-2h{)}^5)}{h^2}\\ \end{array}$

can be replaced by

$\begin{array}{cc}& +\frac{1}{120}{g}^{(5)}(x_i)\frac{(-\frac{1}{12}(2h{)}^5+\frac{4}{3}(h{)}^5-\frac{5}{2}(0{)}^5+\frac{4}{3}(-h{)}^5}{h^2}\\ & \frac{-\frac{1}{12}(-2h{)}^5)}{h^2}\\ & \\ & +\frac{1}{720}\frac{(-\frac{1}{12}{g}^{(6)}(z_1)(2h{)}^6+\frac{4}{3}{g}^{(6)}(z_2)(h{)}^6-\frac{5}{2}(0{)}^5+\frac{4}{3}{g}^{(6)}(z_3)(-h{)}^6}{h^2}\\ & \frac{-\frac{1}{12}{g}^{(6)}(z_4)(-2h{)}^6)}{h^2}\\ \end{array}$

This gives

$\begin{array}{cc}& \frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}u(x_i,y_j)=\\ & =\frac{{\partial }^2}{\partial x^2}u(x_i,y_j)+\frac{h^4}{720}(\frac{8}{3}\frac{{\partial }^6}{\partial x^6}u(x_i+{\varphi }_{1}h,y_j)-\frac{32}{3}\frac{{\partial }^6}{\partial x^6}u(x_i+{\varphi }_{2}h,y_j))\\ \end{array}$

with    $-1<{\varphi }_1<1,-2<{\varphi }_2<2$    which is

${\tau }_x(x_i,y_j)=\frac{h^4}{720}{M}_x^6(x_i,y_j).$

The remaining truncation error estimates are done in the same way.

end working

Let the error

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

be defined by

$e_{i,j}=U_{i,j}-u(x_i,y_j)$ .

$U$ is the solution of the finite difference scheme (xx) and $u(x_i,y_j)$ is the solution to (1.0).

Since

$\begin{array}{cc}-{\mathrm{\Delta }}_{i,j}e& =-{\mathrm{\Delta }}_{i,j}U+{\mathrm{\Delta }}_{h,k}u(x_i,y_j)\\ & =f(x_i,y_j)+(\mathrm{\Delta }u(x_i,y_j)+\tau (x_i,y_j))\\ & =\tau (x_i,y_j)\\ \end{array}$ 
we get that

$\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-{\mathrm{\Delta }}_{i,j}e)=\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}\tau (x_i,y_j)\le {\Vert e\Vert }_2{\Vert \tau \Vert }_2$ .

Next it will be shown that the operator  $-{\mathrm{\Delta }}_{i,j}$  is positive definite for  $e$,  in particular that

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

      $\ge \lambda (4a^{-2}{(m+1)}^2(something)+4b^{-2}{(n+1)}^2()){\Vert e\Vert }_2$ ,

with

$\lambda \ge 1-(1+\sqrt{1}0)/12$ .

Begin with  $\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-{\mathrm{\Delta }}_{i,j}e)=$
$\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e-(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}e)=$
$\sum _{j=1}^n\sum _{i=1}^m{e}_{i,j}(-(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e)+\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}e)$

The sum  $\sum _{i=1}^m{e}_{i,j}(-(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e)$  will be estimated first.

work

$-h^2(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{1,j}e=$.

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

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

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

$\begin{array}{cc}=& (\frac{1}{12}{e}_{3,j}-\frac{5}{4}{e}_{2,j}+\frac{5}{4}{e}_{1,j}-\frac{1}{12}{e}_{0,j})\\ -& (-\frac{1}{12}{e}_{4,j}+\frac{5}{12}{e}_{3,j}-\frac{3}{4}{e}_{2,j}-\frac{5}{12}{e}_{1,j}+\frac{5}{6}{e}_{0,j})\\ \end{array}$

$-h^2(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e=$.

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

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

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

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

$\begin{array}{ccccc}=& (\frac{1}{12}{e}_{i+2,j}& -\frac{5}{4}{e}_{i+1,j}& +\frac{5}{4}{e}_{i,j}& -\frac{1}{12}{e}_{i-1,j})\\ -& (\frac{1}{12}{e}_{i+1,j}& -\frac{5}{4}{e}_{i,j}& +\frac{5}{4}{e}_{i-1,j}& -\frac{1}{12}{e}_{i-2,j})\\ \end{array}$

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

$-h^2(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{m,j}e=$.

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

$\begin{array}{cc}=& (-\frac{5}{6}{e}_{m+1,j}+\frac{5}{12}{e}_{m,j}+\frac{3}{4}{e}_{m-1,j}+\frac{1}{4}{e}_{m-2,j}+\frac{1}{12}{e}_{m-3,j})\\ -& (\frac{1}{12}{e}_{m+1,j}-\frac{5}{4}{e}_{m,j}+\frac{5}{4}{e}_{m-1,j}-\frac{1}{12}{e}_{m-2,j})\\ \end{array}$

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

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

end work

The  summation by parts formula  is now stated so it can be used.

$\sum _{i=1}^m{w}_i(v_i-v_{i-1})=w_m{v}_m-w_0{v}_0-\sum _{i=0}^{m-1}{v}_i(w_{i+1}-w_i)$

$Now,-h^2(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e=v_{i,j}-v_{i-1,j},\text{with}$

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

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

$\begin{array}{cc}{v}_{i,j}& =(\frac{1}{12}{e}_{i+2,j}-\frac{5}{4}{e}_{i+1,j}+\frac{5}{4}{e}_{i,j}-\frac{1}{12}{e}_{i-1,j})\\ & =\frac{1}{12}(e_{i+2,j}-e_{i+1,j})-\frac{7}{6}(e_{i+1,j}-e_{i,j})+\frac{1}{12}(e_{i,j}-e_{i-1,j})\\ \end{array}$

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

$v_{m,j}=(-\frac{5}{6}{e}_{m+1,j}+\frac{5}{12}{e}_{m,j}+\frac{3}{4}{e}_{m-1,j}+\frac{1}{4}{e}_{m-2,j}+\frac{1}{12}{e}_{m-3,j})$

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

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

Taking into account that  $e_{m+1,j}=e_{0,j}=0$  it follows

$\sum _{i=1}^m{e}_{i,j}(-h^2(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e)=-\sum _{i=0}^m{v}_{i,j}(e_{i+1,j}-e_{i,j})$

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

$\begin{array}{cc}=& \frac{1}{12}(e_{4,j}-e_{3,j})(e_{1,j}-e_{0,j})-\frac{1}{3}(e_{3,j}-e_{2,j})(e_{1,j}-e_{0,j})\\ & +\frac{5}{12}(e_{2,j}-e_{1,j})(e_{1,j}-e_{0,j})+\frac{5}{6}(e_{1,j}-e_{0,j}{)}^2\\ & -\frac{1}{12}\sum _{i=1}^{m-1}(e_{i+2,j}-e_{i+1,j})(e_{i+1,j}-e_{i,j})+\frac{7}{6}\sum _{i=1}^{m-1}(e_{i+1,j}-e_{i,j}{)}^2\\ & -\frac{1}{12}\sum _{i=1}^{m-1}(e_{i,j}-e_{i-1,j})(e_{i+1,j}-e_{i,j})+\frac{5}{6}(e_{m+1,j}-e_{m,j}{)}^2\\ & +\frac{5}{12}(e_{m,j}-e_{m-1,j})(e_{m+1,j}-e_{m,j})-\frac{1}{3}(e_{m-1,j}-e_{m-2,j})(e_{m+1,j}-e_{m,j})\\ & +\frac{1}{12}(e_{m-2,j}-e_{m-3,j})(e_{m+1,j}-e_{m,j})\\ \end{array}$

working here

Collect like terms in the expression immediately above as follows.

$\frac{5}{6}(e_{1,j}-e_{0,j}{)}^2+\frac{7}{6}\sum _{i=1}^{m-1}(e_{i+1,j}-e_{i,j}{)}^2+\frac{5}{6}(e_{m+1,j}-e_{m,j}{)}^2$

$=\frac{7}{6}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{3}(e_{1,j}-e_{0,j}{)}^2-\frac{1}{3}(e_{m+1,j}-e_{m,j}{)}^2$

$-\frac{1}{12}\sum _{i=1}^{m-1}(e_{i+2,j}-e_{i+1,j})(e_{i+1,j}-e_{i,j})-\frac{1}{12}\sum _{i=1}^{m-1}(e_{i,j}-e_{i-1,j})(e_{i+1,j}-e_{i,j})$

$\begin{array}{cc}=& -\frac{1}{6}\sum _{i=2}^{m-1}(e_{i+1,j}-e_{i,j})(e_{i,j}-e_{i-1,j})-\frac{1}{12}(e_{2,j}-e_{1,j})(e_{1,j}-e_{0,j})\\ & -\frac{1}{12}(e_{m+1,j}-e_{m,j})(e_{m,j}-e_{m-1,j})\\ \end{array}$

Now, rewrite the expression after making cancellations.

$\begin{array}{cc}& \frac{7}{6}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{6}\sum _{i=2}^{m-1}(e_{i+1,j}-e_{i,j})(e_{i,j}-e_{i-1,j})-\frac{1}{3}(e_{1,j}-e_{0,j}{)}^2\\ & -\frac{1}{3}(e_{m+1,j}-e_{m,j}{)}^2+\frac{1}{12}(e_{4,j}-e_{3,j})(e_{1,j}-e_{0,j})\\ & -\frac{1}{3}(e_{3,j}-e_{2,j})(e_{1,j}-e_{0,j})+\frac{1}{3}(e_{2,j}-e_{1,j})(e_{1,j}-e_{0,j})\\ & +\frac{1}{3}(e_{m,j}-e_{m-1,j})(e_{m+1,j}-e_{m,j})-\frac{1}{3}(e_{m-1,j}-e_{m-2,j})(e_{m+1,j}-e_{m,j})\\ & +\frac{1}{12}(e_{m-2,j}-e_{m-3,j})(e_{m+1,j}-e_{m,j})\\ \end{array}$

The following simple inequality will be used to bound terms.

$\begin{array}{cc}(a-b{)}^2& =a^2-2ab+b^2\ge 0\\ 2ab& \le a^2+b^2\\ ab& \le \frac{1}{2}(a^2+b^2)\\ \end{array}$
and also

$ab\le \frac{1}{2}({\alpha }^2{a}^2+b^2/{\alpha }^2)$.

$\begin{array}{cc}& \sum _{i=2}^{m-1}|(e_{i+1,j}-e_{i,j})(e_{i,j}-e_{i-1,j})|=\sum _{i=3}^{m-2}|(e_{i+1,j}-e_{i,j})(e_{i,j}-e_{i-1,j})|\\ & +|(e_{3,j}-e_{2,j})(e_{2,j}-e_{1,j})|+|(e_{m,j}-e_{m-1,j})(e_{m-1,j}-e_{m-2,j})|\\ & \le \frac{1}{2}(\sum _{i=3}^{m-2}(e_{i+1,j}-e_{i,j}{)}^2+\sum _{i=3}^{m-2}(e_{i,j}-e_{i-1,j}{)}^2)\\ & +|(e_{3,j}-e_{2,j})(e_{2,j}-e_{1,j})|+|(e_{m,j}-e_{m-1,j})(e_{m-1,j}-e_{m-2,j})|\\ \end{array}$

$\begin{array}{cc}& =\sum _{i=3}^{m-3}(e_{i+1,j}-e_{i,j}{)}^2+\frac{1}{2}(e_{3,j}-e_{2,j}{)}^2+\frac{1}{2}(e_{m-1,j}-e_{m-2,j}{)}^2\\ & +|(e_{3,j}-e_{2,j})(e_{2,j}-e_{1,j})|+|(e_{m,j}-e_{m-1,j})(e_{m-1,j}-e_{m-2,j})|\\ \end{array}$

$\begin{array}{cc}& =\sum _{i=2}^{m-2}(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{2}(e_{3,j}-e_{2,j}{)}^2-\frac{1}{2}(e_{m-1,j}-e_{m-2,j}{)}^2\\ & +|(e_{3,j}-e_{2,j})(e_{2,j}-e_{1,j})|+|(e_{m,j}-e_{m-1,j})(e_{m-1,j}-e_{m-2,j})|\\ \end{array}$

$\begin{array}{cc}& \le \sum _{i=2}^{m-2}(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{2}(e_{3,j}-e_{2,j}{)}^2-\frac{1}{2}(e_{m-1,j}-e_{m-2,j}{)}^2\\ & +\frac{1}{2}({\alpha }_1^2(e_{3,j}-e_{2,j}{)}^2+(e_{2,j}-e_{1,j}{)}^2/{\alpha }_1^2)\\ & +\frac{1}{2}({\beta }_1^2(e_{m-1,j}-e_{m-2,j}{)}^2+(e_{m,j}-e_{m-1,j}{)}^2/{\beta }_1^2)\\ \end{array}$

$\begin{array}{cc}& |(e_{4,j}-e_{3,j})(e_{1,j}-e_{0,j})|\le \frac{1}{2}({\alpha }_2^2(e_{4,j}-e_{3,j}{)}^2+(e_{1,j}-e_{0,j}{)}^2/{\alpha }_2^2)\\ & |(e_{3,j}-e_{2,j})(e_{1,j}-e_{0,j})|\le \frac{1}{2}({\alpha }_3^2(e_{3,j}-e_{2,j}{)}^2+(e_{1,j}-e_{0,j}{)}^2/{\alpha }_3^2)\\ & |(e_{2,j}-e_{1,j})(e_{1,j}-e_{0,j})|\le \frac{1}{2}({\alpha }_4^2(e_{2,j}-e_{1,j}{)}^2+(e_{1,j}-e_{0,j}{)}^2/{\alpha }_4^2)\\ \end{array}$

$\begin{array}{cc}& |(e_{m,j}-e_{m-1,j})(e_{m+1,j}-e_{m,j})|\\ & \le \frac{1}{2}({\beta }_2^2(e_{m,j}-e_{m-1,j}{)}^2+(e_{m+1,j}-e_{m,j}{)}^2/{\beta }_2^2)\\ & |(e_{m-1,j}-e_{m-2,j})(e_{m+1,j}-e_{m,j})|\\ & \le \frac{1}{2}({\beta }_3^2(e_{m-1,j}-e_{m-2,j}{)}^2+(e_{m+1,j}-e_{m,j}{)}^2/{\beta }_3^2)\\ & |(e_{m-2,j}-e_{m-3,j})(e_{m+1,j}-e_{m,j})|\\ & \le \frac{1}{2}({\beta }_4^2(e_{m-2,j}-e_{m-3,j}{)}^2+(e_{m+1,j}-e_{m,j}{)}^2/{\beta }_4^2)\\ \end{array}$

Now, substitute all the inequalities into the expression.

$\begin{array}{cc}& \frac{7}{6}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{6}\sum _{i=2}^{m-1}(e_{i+1,j}-e_{i,j})(e_{i,j}-e_{i-1,j})-\frac{1}{3}(e_{1,j}-e_{0,j}{)}^2\\ & -\frac{1}{3}(e_{m+1,j}-e_{m,j}{)}^2+\frac{1}{12}(e_{4,j}-e_{3,j})(e_{1,j}-e_{0,j})\\ & -\frac{1}{3}(e_{3,j}-e_{2,j})(e_{1,j}-e_{0,j})+\frac{1}{3}(e_{2,j}-e_{1,j})(e_{1,j}-e_{0,j})\\ & +\frac{1}{3}(e_{m,j}-e_{m-1,j})(e_{m+1,j}-e_{m,j})-\frac{1}{3}(e_{m-1,j}-e_{m-2,j})(e_{m+1,j}-e_{m,j})\\ & +\frac{1}{12}(e_{m-2,j}-e_{m-3,j})(e_{m+1,j}-e_{m,j})\\ \end{array}$

$\begin{array}{cc}& \frac{7}{6}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{6}(\sum _{i=2}^{m-2}(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{2}(e_{3,j}-e_{2,j}{)}^2\\ & -\frac{1}{2}(e_{m-1,j}-e_{m-2,j}{)}^2+\frac{1}{2}({\alpha }_1^2(e_{3,j}-e_{2,j}{)}^2+(e_{2,j}-e_{1,j}{)}^2/{\alpha }_1^2)\\ & +\frac{1}{2}({\beta }_1^2(e_{m-1,j}-e_{m-2,j}{)}^2+(e_{m,j}-e_{m-1,j}{)}^2/{\beta }_1^2))\\ & -\frac{1}{3}(e_{1,j}-e_{0,j}{)}^2-\frac{1}{3}(e_{m+1,j}-e_{m,j}{)}^2\\ & -\frac{1}{12}(\frac{1}{2}({\alpha }_2^2(e_{4,j}-e_{3,j}{)}^2+(e_{1,j}-e_{0,j}{)}^2/{\alpha }_2^2))\\ & -\frac{1}{3}(\frac{1}{2}({\alpha }_3^2(e_{3,j}-e_{2,j}{)}^2+(e_{1,j}-e_{0,j}{)}^2/{\alpha }_3^2))\\ & -\frac{1}{3}(\frac{1}{2}({\alpha }_4^2(e_{2,j}-e_{1,j}{)}^2+(e_{1,j}-e_{0,j}{)}^2/{\alpha }_4^2))\\ & -\frac{1}{3}(\frac{1}{2}({\beta }_2^2(e_{m,j}-e_{m-1,j}{)}^2+(e_{m+1,j}-e_{m,j}{)}^2/{\beta }_2^2))\\ & -\frac{1}{3}(\frac{1}{2}({\beta }_3^2(e_{m-1,j}-e_{m-2,j}{)}^2+(e_{m+1,j}-e_{m,j}{)}^2/{\beta }_3^2))\\ & -\frac{1}{12}(\frac{1}{2}({\beta }_4^2(e_{m-2,j}-e_{m-3,j}{)}^2+(e_{m+1,j}-e_{m,j}{)}^2/{\beta }_4^2))\\ \end{array}$

$\begin{array}{cc}& =\frac{7}{6}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2-\frac{1}{6}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2\\ & -(\frac{1}{6}+(\frac{1}{24})/{\alpha }_2^2+(\frac{1}{6})/{\alpha }_3^2+(\frac{1}{6})/{\alpha }_4^2)(e_{1,j}-e_{0,j}{)}^2\\ & -(-\frac{1}{6}+(\frac{1}{12})/{\alpha }_1^2+\frac{1}{6}{\alpha }_4^2)(e_{2,j}-e_{1,j}{)}^2\\ & -(-\frac{1}{12}+\frac{1}{12}{\alpha }_1^2+\frac{1}{6}{\alpha }_3^2)(e_{3,j}-e_{2,j}{)}^2\\ & -\left(\frac{1}{24}{\alpha }_2^2\right)(e_{4,j}-e_{3,j}{)}^2\\ & -(\frac{1}{6}+(\frac{1}{6})/{\beta }_2^2+(\frac{1}{6})/{\beta }_3^2+(\frac{1}{24})/{\beta }_4^2)(e_{m+1,j}-e_{m,j}{)}^2\\ & -(-\frac{1}{6}+(\frac{1}{12})/{\beta }_1^2+\frac{1}{6}{\beta }_2^2)(e_{m,j}-e_{m-1,j}{)}^2\\ & -(-\frac{1}{12}+\frac{1}{12}{\beta }_1^2+\frac{1}{6}{\beta }_3^2)(e_{m-1,j}-e_{m-2,j}{)}^2\\ & -\left(\frac{1}{24}{\beta }_4^2\right)(e_{m-2,j}-e_{m-3,j}{)}^2\\ \end{array}$

The choice

$\begin{array}{cc}& {\alpha }_1^2={\beta }_1^2=(\sqrt{5}-1)/2,{\alpha }_2^2={\beta }_4^2=8,\\ & {\alpha }_3^2={\alpha }_4^2={\beta }_2^2={\beta }_3^2=(11-\sqrt{5})/4,\\ \end{array}$

bounds all of the coefficients in the  ${\alpha }_i\text{'s}$  and  ${\beta }_i\text{'s}$  by  $\frac{1}{3}$   and yields the long sought inequality

$\sum _{i=1}^m{e}_{i,j}(-h^2(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e)\ge \frac{2}{3}\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2$.

and

$\sum _{j=1}^n\sum _{i=1}^m{e}_{i,j}(-(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e)\ge \frac{2}{3}{h}^{-2}\sum _{j=1}^n\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2$.

Reasoning in the exact same manner for the dimension in  $y$

$\sum _{j=1}^n{e}_{i,j}(-k^2(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}e)\ge \frac{2}{3}\sum _{j=0}^n(e_{i,j+1}-e_{i,j}{)}^2$.

and

$\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}e)\ge \frac{2}{3}{k}^{-2}\sum _{i=1}^m\sum _{j=0}^n(e_{i,j+1}-e_{i,j}{)}^2$.

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

$\sum _{j=1}^n\sum _{i=1}^m{e}_{i,j}(-(\frac{\underset{h}{\overset{2}{\partial }}}{{\partial }_h{x}^2}{)}_{i,j}e)+\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-(\frac{\underset{k}{\overset{2}{\partial }}}{{\partial }_k{y}^2}{)}_{i,j}e)$

leads to the inequality   $\sum _{i=1}^m\sum _{j=1}^n{e}_{i,j}(-{\mathrm{\Delta }}_{i,j}e)$

$\ge \frac{2}{3}{h}^{-2}\sum _{j=1}^n\sum _{i=0}^m(e_{i+1,j}-e_{i,j}{)}^2+\frac{2}{3}{k}^{-2}\sum _{i=1}^m\sum _{j=0}^n(e_{i,j+1}-e_{i,j}{)}^2$.