# 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
$\Delta u(x,y)={\partial ^{2} \over \partial x^{2}}u(x,y)+{\partial ^{2} \over \partial y^{2}}u(x,y)$
is the usual Laplacian. The problem, determine a function u(x, y) such that
${\begin{cases}{\begin{aligned}-\Delta u(x,y)&=f(x,y),\ {\text{for}}\ (x,y)\in D\\\\u(x,y)&=g(x,y),\ {\text{for}}\ (x,y)\in C,\\\end{aligned}}\end{cases}}\quad _{(1.0)}$
is called a Poisson problem.

### discrete approximation

To approximate u(x, y) numerically, use the grid
$(x_{i},y_{j}),\ 0\leq i\leq m+1,\ 0\leq j\leq n+1$. with $x_{i}=i\,h,\,y_{i}=j\ k$ and $h=a\,/\,(m+1),k=b\,/\,(n+1)$
The second partial derivative ${\partial ^{2} \over \partial x^{2}}\ u(x,y)$ can be approximated on the grid by difference quotients $({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})$.
These difference quotients are given by
$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{1},y_{j})\ =$.
${-{\frac {1}{12}}\,u(x_{1}+3\,h,y_{j})+{\frac {1}{3}}\,u(x_{1}+2\,h,y_{j})+{\frac {1}{2}}\,u(x_{1}+h,y_{j})-{\frac {5}{3}}\,u(x_{1},y_{j})+{\frac {11}{12}}\,u(x_{1}-h,y_{j}) \over \ h^{2}}$
$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})\ =$.
${-{\frac {1}{12}}\,u(x_{i}+2\,h,y_{j})+{\frac {4}{3}}\,u(x_{i}+h,y_{j})-{\frac {5}{2}}\,u(x_{i},y_{j})+{\frac {4}{3}}\,u(x_{i}-h,y_{j})-{\frac {1}{12}}\,u(x_{i}-2\,h,y_{j}) \over \ h^{2}}$
${\text{for}}\ i\,=\,2,\ 3,\ \ldots \,m-1\quad {\text{and}}$
$({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{m},y_{j})\ =$.
${{\frac {11}{12}}\,u(x_{m}+h,y_{j})-{\frac {5}{3}}\,u(x_{m},y_{j})+{\frac {1}{2}}\,u(x_{m}-h,y_{j})+{\frac {1}{3}}\,u(x_{m}-2\,h,y_{j})-{\frac {1}{12}}\,u(x_{m}-3\,h,y_{j}) \over \ h^{2}}$
The second partial derivative ${\partial ^{2} \over \partial y^{2}}\ u(x,y)$ can be approximated on the grid by difference quotients $({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})$.
These difference quotients are given by
$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{1})\ =$.
${-{\frac {1}{12}}\,u(x_{i},y_{1}+3\,k)+{\frac {1}{3}}\,u(x_{i},y_{1}+2\,k)+{\frac {1}{2}}\,u(x_{i},y_{1}+k)-{\frac {5}{3}}\,u(x_{i},y_{1})+{\frac {11}{12}}\,u(x_{i},y_{1}-k) \over \ k^{2}}$
$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})\ =$.
${-{\frac {1}{12}}\,u(x_{i},y_{j}+2\,k)+{\frac {4}{3}}\,u(x_{i},y_{j}+k)-{\frac {5}{2}}\,u(x_{i},y_{j})+{\frac {4}{3}}\,u(x_{i},y_{j}-k)-{\frac {1}{12}}\,u(x_{i},y_{j}-2\,k) \over \ k^{2}}$
${\text{for}}\ j\,=\,2,\ 3,\ \ldots \,n-1\quad {\text{and}}$
$({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{n})\ =$.
${{\frac {11}{12}}\,u(x_{i},y_{n}+k)-{\frac {5}{3}}\,u(x_{i},y_{n})+{\frac {1}{2}}\,u(x_{i},y_{n}-k)+{\frac {1}{3}}\,u(x_{i},y_{n}-2\,k)-{\frac {1}{12}}\,u(x_{i},y_{n}-3\,k) \over \ k^{2}}$

### truncation errors

The difference quotients  $({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})$  are third order accurate with truncation errors:
${\begin{aligned}\tau _{x}(x_{i},y_{j})&=({\partial _{h}^{2} \over \partial _{h}x^{2}})\ u(x_{i},y_{j})-{\partial ^{2} \over \partial x^{2}}\ u(x_{i},y_{j})\\&={h^{3} \over \ 120}\,M{_{x}^{(5)}}(x_{i},y_{j})\end{aligned}}$
with
$M{_{x}^{(5)}}(x_{1},y_{j})={\frac {67}{6}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{1}+\phi {_{1}^{(1,\,j)}}\,h,\ y_{j})-{\frac {254}{12}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{1}+\phi {_{2}^{(1,\,j)}}\,h,\ y_{j})$ ,
for some $0<\phi {_{1}^{(1,\,j)}}<2\,,\quad -1<\phi {_{2}^{(1,\,j)}}<3$ ,
$M{_{x}^{(5)}}(x_{i},y_{j})=4\,{\partial ^{5} \over \partial x^{5}}\,u(x_{i}+\phi {_{1}^{(i,\,j)}}\,h,\ y_{j})-4\,{\partial ^{5} \over \partial x^{5}}\,u(x_{i}+\phi {_{2}^{(i,\,j)}}\,h,\ y_{j})$ ,
for some $-2<\phi {_{1}^{(i,\,j)}}<1\,,\quad -1<\phi {_{2}^{(i,\,j)}}<2\,,$
${\text{for}}\ \ i\ =\ 2,\ 3,\ \ldots \,,\ m-1.$
When ${\partial ^{6} \over \partial x^{6}}\,u(x,\,y)$ is continuous, these estimates also hold.
$\tau _{x}(x_{i},\,y_{j})\;=\;{h^{4} \over \ 720}M_{x}^{6}(x_{i},\,y_{j}),$  with
$M_{x}^{6}(x_{i},\,y_{j})\;=\;{\frac {8}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi {_{1}^{(i,\,j)}}\,h,\;y_{j})-{\frac {32}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi {_{2}^{(i,\,j)}}\,h,\;y_{j})$
for some $-1\;<\;\phi {_{1}^{(i,\,j)}}\;<\;1\,,\quad -2\;<\;\phi {_{2}^{(i,\,j)}}\;<\;2,$  ${\text{for}}\ \ i\ =\ 2,\ 3,\ \ldots \,,\ m-1.$
The case for $i\;=\;m$ is
$M{_{x}^{(5)}}(x_{m},y_{j})={\frac {254}{12}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{m}+\phi {_{1}^{(m,\,j)}}\,h,\ y_{j})-{\frac {67}{6}}\,{\partial ^{5} \over \partial x^{5}}\,u(x_{m}+\phi {_{2}^{(m,\,j)}}\,h,\ y_{j})$ ,
for some $-3<\phi {_{1}^{(m,\,j)}}<1\,,\quad -2<\phi {_{2}^{(m,\,j)}}<0$ .
The difference quotients  $({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{j})$  are third order accurate with truncation errors:
${\begin{aligned}\tau _{y}(x_{i},y_{i})&=({\partial _{k}^{2} \over \partial _{k}y^{2}})\ u(x_{i},y_{1})-{\partial ^{2} \over \partial y^{2}}\ u(x_{i},y_{1})\\&={k^{3} \over \ 120}\,M{_{y}^{(5)}}(x_{i},y_{1})\end{aligned}}$
with
$M{_{y}^{(5)}}(x_{i},y_{1})={\frac {67}{6}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{1}+\psi {_{1}^{(i,\,1)}}\,k)-{\frac {254}{12}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{1}+\psi {_{2}^{(i,\,1)}}\,k)$ ,
for some $0<\psi {_{1}^{(i,\,1)}}<2\,,\quad -1<\psi {_{2}^{(i,\,1)}}<3$ ,
$M{_{y}^{(5)}}(x_{i},y_{j})=4\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{j}+\psi {_{1}^{(i,\,j)}}\,k)-4\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{j}+\psi {_{2}^{(i,\,j)}}\,k)$ ,
for some $-2<\psi {_{1}^{(i,\,j)}}<1\,,\quad -1<\psi {_{2}^{(i,\,j)}}<2\,,$
${\text{for}}\ \ j\ =\ 2,\ 3,\ \ldots \,,\ n-1.$
When ${\partial ^{6} \over \partial y^{6}}\,u(x,\,y)$ is continuous, these estimates also hold.
$\tau _{y}(x_{i},\,y_{j})\;=\;{k^{4} \over \ 720}M_{y}^{6}(x_{i},\,y_{j}),$  with
$M_{y}^{6}(x_{i},\,y_{j})\;=\;{\frac {8}{3}}{\partial ^{6} \over \partial y^{6}}u(x_{i}+\psi {_{1}^{(i,\,j)}}\,k,\;y_{j})-{\frac {32}{3}}{\partial ^{6} \over \partial y^{6}}u(x_{i}+\psi {_{2}^{(i,\,j)}}\,k,\;y_{j})$
for some $-1\;<\;\psi {_{1}^{(i,\,j)}}\;<\;1\,,\quad -2\;<\;\psi {_{2}^{(i,\,j)}}\;<\;2,$  ${\text{for}}\ \ j\ =\ 2,\ 3,\ \ldots \,,\ n-1.$
The case for $j\;=\;n$ is
$M{_{y}^{(5)}}(x_{i},y_{n})={\frac {254}{12}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{n}+\psi {_{1}^{(i,\,n)}}\,k)-{\frac {67}{6}}\,{\partial ^{5} \over \partial y^{5}}\,u(x_{i},\ y_{n}+\psi {_{2}^{(i,\,n)}}\,k)$ ,
for some $-3<\psi {_{1}^{(i,\,n)}}<1\,,\quad -2<\psi {_{2}^{(i,\,n)}}<0$ .
The Laplacian $\Delta u(x,y)$ then can be approximated on the interior of the grid by
$\Delta _{h,\,k}u(x_{i},y_{j})={\partial _{h}^{2} \over \partial _{h}x^{2}}u(x_{i},y_{j})+{\partial _{k}^{2} \over \partial _{k}y^{2}}u(x_{i},y_{j})$
The truncation error
$\tau (x_{i},y_{j})=\Delta _{h,\,k}u(x_{i},y_{j})-\Delta u(x_{i},y_{j})$ 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
$U=\{U_{i,\,j}\}\quad 0\leq i\leq m+1,\ 0\leq j\leq n+1$ define the finite difference operations
$\Delta _{i,\,j}=({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,U+({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,U$ by the following.
$({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1,\,j}\ U\ =$.
${-{\frac {1}{12}}\,U_{4,\,j}+{\frac {1}{3}}\,U_{3,\,j}+{\frac {1}{2}}\,U_{2,\,j}-{\frac {5}{3}}\,U_{1,\,j}+{\frac {11}{12}}\,U_{0,\,j} \over \ h^{2}}$
$({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\ U\ =$.
${-{\frac {1}{12}}\,U_{i+2,\,j}+{\frac {4}{3}}\,U_{i+1,\,j}-{\frac {5}{2}}\,U_{i,\,j}+{\frac {4}{3}}\,U_{i-1,\,j}-{\frac {1}{12}}\,U_{i-2,\,j} \over \ h^{2}}$
${\text{for}}\ i\,=\,2,\ 3,\ \ldots \,m-1\quad {\text{and}}$
$({\partial _{h}^{2} \over \partial _{h}x^{2}})_{m,\,j}\ U\ =$.
${{\frac {11}{12}}\,U_{m+1,\,j}-{\frac {5}{3}}\,U_{m,\,j}+{\frac {1}{2}}\,U_{m-1,\,j}+{\frac {1}{3}}\,U_{m-2,\,j}-{\frac {1}{12}}\,U_{m-3,\,j} \over \ h^{2}}$
$({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,1}\ U\ =$.
${-{\frac {1}{12}}\,U_{i,\,4}+{\frac {1}{3}}\,U_{i,\,3}+{\frac {1}{2}}\,U_{i,\,2}-{\frac {5}{3}}\,U_{i,\,1}+{\frac {11}{12}}\,U_{i,\,0} \over \ k^{2}}$
$({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\ U\ =$.
${-{\frac {1}{12}}\,U_{i,\,j+2}+{\frac {4}{3}}\,U_{i,\,j+1}-{\frac {5}{2}}\,U_{i,\,j}+{\frac {4}{3}}\,U_{i,\,j-1}-{\frac {1}{12}}\,U_{i,\,j-2} \over \ k^{2}}$
${\text{for}}\ j\,=\,2,\ 3,\ \ldots \,n-1\quad {\text{and}}$
$({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,n}\ U\ =$.
${{\frac {11}{12}}\,U_{i,\,n+1}-{\frac {5}{3}}\,U_{i,\,n}+{\frac {1}{2}}\,U_{i,\,n-1}+{\frac {1}{3}}\,U_{i,\,n-2}-{\frac {1}{12}}\,U_{i,\,n-3} \over \ k^{2}}$

### simulation of the problem

To simulate the problem (1.0) let
${\begin{aligned}U_{0,\,j}\ \ &=&g(x_{0},\ y_{j})\,,&\quad 0\leq j\leq n+1\\U_{m+1,\,j}\ \ &=&g(x_{m+1},\ y_{j})\,,&\quad 0\leq j\leq n+1\\U_{i,\,0}\ \ &=&g(x_{i},\ y_{0})\,,&\quad 0\leq i\leq m+1\\U_{i,\,n+1}\ \ &=&g(x_{i},\ y_{n+1)}\,,&\quad 0\leq i\leq m+1\\\end{aligned}}$
Then solve the non-singular linear system
$-\Delta _{i,\,j}\,U=f(x_{i},\,y_{j})\quad 1\leq i\leq m,\ 1\leq j\leq n$ ; for the remaining $U_{i,\,j}$

### error estimation

The error
$U-u=\{U_{i,\,j}-u(x_{i},\,y_{j})\}$ , satisfies
${\lVert U-u\rVert }_{2}\leq {\rho \,a^{2}b^{2} \over \ \pi ^{2}(a^{2}+b^{2})}{\lVert \tau \rVert }_{2}(1+O(h^{2}+k^{2}))$ . for
$\tau =\{\tau (x_{i},\,y_{j})\},\quad 1\leq i\leq m,\ 1\leq j\leq n$  , and
$\rho \ \leq \ 1\,/\,(1-(1+{\sqrt {1}}0)\,/\,12)$ .

### proof of truncation error estimates

The truncation error estimates for  ${\big{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j}){\big )}$  are done under the assumption that $u(x,\;y)$ is sufficiently smooth so that  ${\partial ^{5} \over \partial x^{5}}\,u(x,\;y)$ is continuous. For notational convenience let $g(x)\;=\;u(x,\;y_{j}).$
Expand $g(x)$ in it's Taylor expansion about  $x_{i}$,
${\begin{aligned}g(x_{i}+\Delta x)\;&=\;g(x_{i})+g^{(1)}(x_{i})\Delta x+{\frac {1}{2}}\,g^{(2)}(x_{i})({\Delta x})^{2}+{\frac {1}{6}}\,g^{(3)}(x_{i})({\Delta x})^{3}\\&+{\frac {1}{24}}\,g^{(4)}(x_{i})({\Delta x})^{4}+{\frac {1}{120}}\,g^{(5)}(z)({\Delta x})^{5}\\\end{aligned}}$.
where $z$ is some number between $x_{i}$  and $x_{i}+\Delta x$. Then
${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{1},\,y_{j})\;=\\&{{\big-{\frac {1}{12}}g(x_{1}+3\,h)+{\frac {1}{3}}g(x_{1}+2\,h)+{\frac {1}{2}}g(x_{1}+h)-{\frac {5}{3}}g(x_{1})+{\frac {11}{12}}g(x_{1}-h){\big )} \over \ h^{2}}\end{aligned}}$
${\begin{aligned}&=\;g(x_{1}){(-{\frac {1}{12}}+{\frac {1}{3}}+{\frac {1}{2}}-{\frac {5}{3}}+{\frac {11}{12}}) \over \ h^{2}}\\&+\;g^{(1)}(x_{1}){(-{\frac {1}{12}}(3\,h)+{\frac {1}{3}}(2\,h)+{\frac {1}{2}}(h)-{\frac {5}{3}}(0)+{\frac {11}{12}}(-h)) \over \ h^{2}}\\&+\;{\frac {1}{2}}g^{(2)}(x_{1}){(-{\frac {1}{12}}(3\,h)^{2}+{\frac {1}{3}}(2\,h)^{2}+{\frac {1}{2}}(h)^{2}-{\frac {5}{3}}(0)^{2}+{\frac {11}{12}}(-h)^{2}) \over \ h^{2}}\\&+\;{\frac {1}{6}}g^{(3)}(x_{1}){(-{\frac {1}{12}}(3\,h)^{3}+{\frac {1}{3}}(2\,h)^{3}+{\frac {1}{2}}(h)^{3}-{\frac {5}{3}}(0)^{3}+{\frac {11}{12}}(-h)^{3}) \over \ h^{2}}\\&+\;{\frac {1}{24}}g^{(4)}(x_{1}){(-{\frac {1}{12}}(3\,h)^{4}+{\frac {1}{3}}(2\,h)^{4}+{\frac {1}{2}}(h)^{4}-{\frac {5}{3}}(0)^{4}+{\frac {11}{12}}(-h)^{4}) \over \ h^{2}}\\&+\;{\frac {1}{120}}{(-{\frac {1}{12}}g^{(5)}(z_{1})(3\,h)^{5}+{\frac {1}{3}}g^{(5)}(z_{2})(2\,h)^{5}+{\frac {1}{2}}g^{(5)}(z_{3})(h)^{5}-{\frac {5}{3}}(0)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {+{\frac {11}{12}}g^{(5)}(z_{4})(-h)^{5}) \over \ h^{2}}\\\end{aligned}}$
$=\;g^{(2)}(x_{1})+{h^{3} \over \ 120}{\big-{\frac {81}{4}}g^{(5)}(z_{1})+{\frac {32}{3}}g^{(5)}(z_{2})+{\frac {1}{2}}g^{(5)}(z_{3})-{\frac {11}{12}}g^{(5)}(z_{4}){\big )}$ where
${\begin{aligned}&x_{1}\;<\;z_{1}\;<\;x_{1}+3\,h\,,\;x_{1}\;<\;z_{2}\;<\;x_{1}+2\,h\,,\;\\&x_{1}\;<\;z_{3}\;<\;x_{1}+h\,,\;\;{\text{and}}\;\;x_{1}-h\;<\;z_{4}\;<\;x_{1}.\\\end{aligned}}$.
Since
${\frac {67}{6}}{\underset {0\;\leq \;\phi \;\leq \;2}{\min(g^{(5)}(x_{1}\;+\;\phi \,h))}}\;\leq \;{\frac {32}{3}}g^{(5)}(z_{2})+{\frac {1}{2}}g^{(5)}(z_{3})\;\leq \;{\frac {67}{6}}{\underset {0\;\leq \;\phi \;\leq \;2}{\max(g^{(5)}(x_{1}\;+\;\phi \,h))}},$
${\frac {254}{12}}{\underset {-1\;\leq \;\phi \;\leq \;3}{\min(g^{(5)}(x_{1}\;+\;\phi \,h))}}\;\leq \;{\frac {81}{4}}g^{(5)}(z_{1})+{\frac {11}{12}}g^{(5)}(z_{4})\;\leq \;{\frac {254}{12}}{\underset {-1\;\leq \;\phi \;\leq \;3}{\max(g^{(5)}(x_{1}\;+\;\phi \,h))}},$
from the intermediate value property
${\frac {32}{3}}g^{(5)}(z_{2})+{\frac {1}{2}}g^{(5)}(z_{3})\;=\;{\frac {67}{6}}\,g^{(5)}(x_{1}\;+\;\phi _{1}\,h)),\quad {\text{for}}\;0\;<\;\phi _{1}\,,\;<\;2\,,$
${\frac {81}{4}}g^{(5)}(z_{1})+{\frac {11}{12}}g^{(5)}(z_{4})\;=\;{\frac {254}{12}}\,g^{(5)}(x_{1}\;+\;\phi _{2}\,h),\quad {\text{for}}\;-1\;<\;\phi _{2}\;<\;3\,.$.
This gives
${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{1},\,y_{j})\;=\;g^{(2)}(x_{1})+{h^{3} \over \ 120}{\big{\frac {67}{6}}\,g^{(5)}(x_{1}+\phi _{1}\,h)-{\frac {254}{12}}\,g^{(5)}(x_{1}+\phi _{2}\,h){\big )}\\&\;=\;{\partial ^{2} \over \partial x^{2}}\,u(x_{1},\,y_{j})+{h^{3} \over \ 120}{\big{\frac {67}{6}}{\partial ^{5} \over \partial x^{5}}u(x_{1}+\phi _{1}\,h,\;y_{j})-{\frac {254}{12}}{\partial ^{5} \over \partial x^{5}}u(x_{1}+\phi _{2}\,h,\;y_{j}){\big )}\\\end{aligned}}$
which is $\tau _{x}(x_{1},\,y_{j})\;=\;{h^{3} \over \ 120}M_{x}^{5}(x_{1},\,y_{j}).$
For $i\;=\;2,\;3,\;\ldots \,,\;m-1$
${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j})\;=\\&{{\big-{\frac {1}{12}}g(x_{i}+2\,h)+{\frac {4}{3}}g(x_{i}+h)-{\frac {5}{2}}g(x_{i})+{\frac {4}{3}}g(x_{i}-h)-{\frac {1}{12}}g(x_{i}-2\,h){\big )} \over \ h^{2}}\end{aligned}}$
${\begin{aligned}&=\;g(x_{i}){(-{\frac {1}{12}}+{\frac {4}{3}}-{\frac {5}{2}}+{\frac {4}{3}}-{\frac {1}{12}}) \over \ h^{2}}\\&+\;g^{(1)}(x_{i}){(-{\frac {1}{12}}(2\,h)+{\frac {4}{3}}(h)-{\frac {5}{2}}(0)+{\frac {4}{3}}(-h)-{\frac {1}{12}}(-2\,h)) \over \ h^{2}}\\&+\;{\frac {1}{2}}g^{(2)}(x_{i}){(-{\frac {1}{12}}(2\,h)^{2}+{\frac {4}{3}}(h)^{2}-{\frac {5}{2}}(0)^{2}+{\frac {4}{3}}(-h)^{2}-{\frac {1}{12}}(-2\,h)^{2}) \over \ h^{2}}\\&+\;{\frac {1}{6}}g^{(3)}(x_{i}){(-{\frac {1}{12}}(2\,h)^{3}+{\frac {4}{3}}(h)^{3}-{\frac {5}{2}}(0)^{3}+{\frac {4}{3}}(-h)^{3}-{\frac {1}{12}}(-2\,h)^{3}) \over \ h^{2}}\\&+\;{\frac {1}{24}}g^{(4)}(x_{i}){(-{\frac {1}{12}}(2\,h)^{4}+{\frac {4}{3}}(h)^{4}-{\frac {5}{2}}(0)^{4}+{\frac {4}{3}}(-h)^{4}-{\frac {1}{12}}(-2\,h)^{4}) \over \ h^{2}}\\&+\;{\frac {1}{120}}{(-{\frac {1}{12}}g^{(5)}(z_{1})(2\,h)^{5}+{\frac {4}{3}}g^{(5)}(z_{2})(h)^{5}-{\frac {5}{2}}(0)^{5}+{\frac {4}{3}}g^{(5)}(z_{3})(-h)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {-{\frac {1}{12}}g^{(5)}(z_{4})(-2\,h)^{5}) \over \ h^{2}}\\\end{aligned}}$
$=\;g^{(2)}(x_{i})+{h^{3} \over \ 120}{\big-{\frac {8}{3}}g^{(5)}(z_{1})+{\frac {4}{3}}g^{(5)}(z_{2})-{\frac {4}{3}}g^{(5)}(z_{3})+{\frac {8}{3}}g^{(5)}(z_{4}){\big )}$ where
${\begin{aligned}&x_{i}\;<\;z_{1}\;<\;x_{i}+2\,h\,,\;x_{i}\;<\;z_{2}\;<\;x_{i}+h\,,\;\\&x_{i}-h\;<\;z_{3}\;<\;x_{i}\,,\;\;{\text{and}}\;\;x_{i}-2\,h\;<\;z_{4}\;<\;x_{i}.\\\end{aligned}}$.
Reasoning as before, combining terms with like signs and using the intermediate value property,
${\frac {4}{3}}g^{(5)}(z_{2})+{\frac {8}{3}}g^{(5)}(z_{4})\;=\;4\,g^{(5)}(x_{i}\;+\;\phi _{1}\,h)),\quad {\text{for}}\;-2\;<\;\phi _{1}\,,\;<\;1\,,$
${\frac {8}{3}}g^{(5)}(z_{1})+{\frac {4}{3}}g^{(5)}(z_{3})\;=\;4\,g^{(5)}(x_{i}\;+\;\phi _{2}\,h),\quad {\text{for}}\;-1\;<\;\phi _{2}\;<\;2\,.$.
This gives
${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j})\;=\;g^{(2)}(x_{i})+{h^{3} \over \ 120}{\big{\frac {67}{6}}\,g^{(5)}(x_{i}+\phi _{1}\,h)-{\frac {254}{12}}\,g^{(5)}(x_{i}+\phi _{2}\,h){\big )}\\&\;=\;{\partial ^{2} \over \partial x^{2}}\,u(x_{i},\,y_{j})+{h^{3} \over \ 120}{\big4{\partial ^{5} \over \partial x^{5}}u(x_{i}+\phi _{1}\,h,\;y_{j})-4{\partial ^{5} \over \partial x^{5}}u(x_{i}+\phi _{2}\,h,\;y_{j}){\big )}\\\end{aligned}}$
which is $\tau _{x}(x_{i},\,y_{j})\;=\;{h^{3} \over \ 120}M_{x}^{5}(x_{i},\,y_{j}).$
Under the assumption that  ${\partial ^{6} \over \partial x^{6}}\,u(x,\;y)$ is continuous, in the preceding argument, the expression
${\begin{aligned}&+\;{\frac {1}{120}}{(-{\frac {1}{12}}g^{(5)}(z_{1})(2\,h)^{5}+{\frac {4}{3}}g^{(5)}(z_{2})(h)^{5}-{\frac {5}{2}}(0)^{5}+{\frac {4}{3}}g^{(5)}(z_{3})(-h)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {-{\frac {1}{12}}g^{(5)}(z_{4})(-2\,h)^{5}) \over \ h^{2}}\\\end{aligned}}$ can be replaced by
${\begin{aligned}&+\;{\frac {1}{120}}g^{(5)}(x_{i}){(-{\frac {1}{12}}(2\,h)^{5}+{\frac {4}{3}}(h)^{5}-{\frac {5}{2}}(0)^{5}+{\frac {4}{3}}(-h)^{5} \over \ h^{2}}\\&\quad \quad \quad \quad {-{\frac {1}{12}}(-2\,h)^{5}) \over \ h^{2}}\\&\;\\&+\;{\frac {1}{720}}{(-{\frac {1}{12}}g^{(6)}(z_{1})(2\,h)^{6}+{\frac {4}{3}}g^{(6)}(z_{2})(h)^{6}-{\frac {5}{2}}(0)^{5}+{\frac {4}{3}}g^{(6)}(z_{3})(-h)^{6} \over \ h^{2}}\\&\quad \quad \quad \quad {-{\frac {1}{12}}g^{(6)}(z_{4})(-2\,h)^{6}) \over \ h^{2}}\\\end{aligned}}$
This gives
${\begin{aligned}&{\partial _{h}^{2} \over \partial _{h}x^{2}}\,u(x_{i},\,y_{j})\;=\;\\&\;=\;{\partial ^{2} \over \partial x^{2}}\,u(x_{i},\,y_{j})+{h^{4} \over \ 720}{\big{\frac {8}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi _{1}\,h,\;y_{j})-{\frac {32}{3}}{\partial ^{6} \over \partial x^{6}}u(x_{i}+\phi _{2}\,h,\;y_{j}){\big )}\\\end{aligned}}$
with $-1\;<\;\phi _{1}\;<\;1\,,\quad -2\;<\;\phi _{2}\;<\;2$ which is
$\tau _{x}(x_{i},\,y_{j})\;=\;{h^{4} \over \ 720}M_{x}^{6}(x_{i},\,y_{j}).$
The remaining truncation error estimates are done in the same way.

### end working

Let the error
$e=\{e_{i,\,j}\},\quad 1\leq i\leq m,\ 1\leq j\leq n,$ 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
${\begin{aligned}-\Delta _{i,\,j}\,e&=-\Delta _{i,\,j}\,U+\Delta _{h,\,k}\,u(x_{i},\,y_{j})\\&=f(x_{i},\,y_{j})+(\Delta \,u(x_{i},\,y_{j})+\tau (x_{i},\,y_{j}))\\&=\tau (x_{i},\,y_{j})\\\end{aligned}}$  we get that
$\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)\ =\ \sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,\tau (x_{i},\,y_{j})\ \leq \ {\lVert e\rVert }_{2}\,{\lVert \tau \rVert }_{2}$ .
Next it will be shown that the operator  $-\Delta _{i,\,j}$  is positive definite for $e$,  in particular that
$\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)$  $\geq \;\lambda (4a^{-2}\,{(m+1)}^{2}\,(something)+4b^{-2}\,{(n+1)}^{2}\,)\,{\lVert e\rVert }_{2}$ ,
with $\lambda \ \geq \ 1-(1+{\sqrt {1}}0)\,/\,12$ .
Begin with  $\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)\;=$ $\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e-({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)\;=$ $\sum _{j=1}^{n}\sum _{i=1}^{m}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)\;+\;\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)$
The sum  $\sum _{i=1}^{m}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)$ will be estimated first.

### work

$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{1,\,j}\ e\ =$.
${\frac {1}{12}}\,e_{4,\,j}-{\frac {1}{3}}\,e_{3,\,j}-{\frac {1}{2}}\,e_{2,\,j}+{\frac {5}{3}}\,e_{1,\,j}-{\frac {11}{12}}\,e_{0,\,j}$
$={\frac {7}{6}}(-{\frac {1}{1}}\,e_{2,\,j}+{\frac {2}{1}}\,e_{1,\,j}-{\frac {1}{1}}\,e_{0,\,j})$
${\begin{aligned}+\;&({\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})-{\frac {1}{4}}\,(e_{3,\,j}-e_{2,\,j})+{\frac {5}{12}}\,(e_{2,\,j}-e_{1,\,j})-{\frac {1}{4}}\,(e_{1,\,j}-e_{0,\,j}))\\\end{aligned}}$
${\begin{aligned}=\;&({\frac {1}{12}}\,e_{3,\,j}-{\frac {5}{4}}\,e_{2,\,j}+{\frac {5}{4}}\,e_{1,\,j}-{\frac {1}{12}}\,e_{0,\,j})\\-\;&(-{\frac {1}{12}}\,e_{4,\,j}+{\frac {5}{12}}\,e_{3,\,j}-{\frac {3}{4}}\,e_{2,\,j}-{\frac {5}{12}}\,e_{1,\,j}+{\frac {5}{6}}\,e_{0,\,j})\\\end{aligned}}$
$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\ e\ =$.
${\frac {1}{12}}\,e_{i+2,\,j}-{\frac {4}{3}}\,e_{i+1,\,j}+{\frac {5}{2}}\,e_{i,\,j}-{\frac {4}{3}}\,e_{i-1,\,j}+{\frac {1}{12}}\,e_{i-2,\,j}$
$={\frac {7}{6}}\,(-{\frac {1}{1}}\,e_{i+1,\,j}+{\frac {2}{1}}\,e_{i,\,j}-{\frac {1}{1}}\,e_{i-1,\,j})$
$+{\frac {1}{12}}\,(e_{i+2,\,j}-e_{i+1,\,j})-{\frac {1}{12}}\,(e_{i+1,\,j}-e_{i,\,j})$
$+{\frac {1}{12}}\,(e_{i,\,j}-e_{i-1,\,j})-{\frac {1}{12}}\,(e_{i-1,\,j}-e_{i-2,\,j})$
${\begin{aligned}=\;&({\frac {1}{12}}\,e_{i+2,\,j}&-{\frac {5}{4}}\,e_{i+1,\,j}&+{\frac {5}{4}}\,e_{i,\,j}&-{\frac {1}{12}}\,e_{i-1,\,j})\\-\;&({\frac {1}{12}}\,e_{i+1,\,j}&-{\frac {5}{4}}\,e_{i,\,j}\quad &+{\frac {5}{4}}\,e_{i-1,\,j}&-{\frac {1}{12}}\,e_{i-2,\,j})\\\end{aligned}}$
${\text{for}}\ i\,=\,2,\ 3,\ \ldots \,m-1\quad {\text{and}}$
$-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{m,\,j}\ e\ =$.
$-{\frac {11}{12}}\,e_{m+1,\,j}+{\frac {5}{3}}\,e_{m,\,j}-{\frac {1}{2}}\,e_{m-1,\,j}-{\frac {1}{3}}\,e_{m-2,\,j}+{\frac {1}{12}}\,e_{m-3,\,j}$
${\begin{aligned}=\;&(-{\frac {5}{6}}\,e_{m+1,\,j}+{\frac {5}{12}}\,e_{m,\,j}+{\frac {3}{4}}\,e_{m-1,\,j}+{\frac {1}{4}}\,e_{m-2,\,j}+{\frac {1}{12}}\,e_{m-3,\,j})\\-\;&({\frac {1}{12}}\,e_{m+1,\,j}-{\frac {5}{4}}\,e_{m,\,j}+{\frac {5}{4}}\,e_{m-1,\,j}-{\frac {1}{12}}\,e_{m-2,\,j})\\\end{aligned}}$
$={\frac {7}{6}}(-{\frac {1}{1}}\,e_{m+1,\,j}+{\frac {2}{1}}\,e_{m,\,j}-{\frac {1}{1}}\,e_{m-1,\,j})$
${\begin{aligned}+\;&(-{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})+{\frac {1}{4}}\,(e_{m-1,\,j}-e_{m-2,\,j})-{\frac {5}{12}}\,(e_{m,\,j}-e_{m-1,\,j})+{\frac {1}{4}}\,(e_{m+1,\,j}-e_{m,\,j}))\\\end{aligned}}$

### end work

The summation by parts formula is now stated so it can be used.
$\sum _{i=1}^{m}w_{i}\,(v_{i}-v_{i-1})=w_{m}\,v_{m}-w_{0}\,v_{0}-\sum _{i=0}^{m-1}v_{i}\,(w_{i+1}-w_{i})$
$Now,\quad -h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\ e\ =v_{i,\,j}-v_{i-1,\,j},\quad {\text{with}}$
$v_{0,\,j}\;=\;(-{\frac {1}{12}}\,e_{4,\,j}+{\frac {5}{12}}\,e_{3,\,j}-{\frac {3}{4}}\,e_{2,\,j}-{\frac {5}{12}}\,e_{1,\,j}+{\frac {5}{6}}\,e_{0,\,j})$
$=\;-{\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})+{\frac {1}{3}}\,(e_{3,\,j}-e_{2,\,j})-{\frac {5}{12}}\,(e_{2,\,j}-e_{1,\,j})-{\frac {5}{6}}\,(e_{1,\,j}-e_{0,\,j})$
${\begin{aligned}v_{i,\,j}\;&=\;({\frac {1}{12}}\,e_{i+2,\,j}-{\frac {5}{4}}\,e_{i+1,\,j}+{\frac {5}{4}}\,e_{i,\,j}-{\frac {1}{12}}\,e_{i-1,\,j})\\\;&=\;{\frac {1}{12}}\,(e_{i+2,\,j}-e_{i+1,\,j})-{\frac {7}{6}}\,(e_{i+1,\,j}-e_{i,\,j})+{\frac {1}{12}}\,(e_{i,\,j}-e_{i-1,\,j})\\\end{aligned}}$
${\text{for}}\ i\,=\,1,\ 2,\ \ldots \,m-1\quad {\text{and}}$.
$v_{m,\,j}\;=\;(-{\frac {5}{6}}\,e_{m+1,\,j}+{\frac {5}{12}}\,e_{m,\,j}+{\frac {3}{4}}\,e_{m-1,\,j}+{\frac {1}{4}}\,e_{m-2,\,j}+{\frac {1}{12}}\,e_{m-3,\,j})$
$=\;-{\frac {5}{6}}\,(e_{m+1,\,j}-e_{m,\,j})-{\frac {5}{12}}\,(e_{m,\,j}-e_{m-1,\,j})+{\frac {1}{3}}\,(e_{m-1,\,j}-e_{m-2,\,j})-{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})$
${\text{So the sum}}\;\sum _{i=1}^{m}e_{i,\,j}\,(-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)=\;\sum _{i=1}^{m}e_{i,\,j}\,(v_{i,\,j}-v_{i-1,\,j})$ $=e_{m,\,j}\,v_{m,\,j}-e_{0,\,j}\,v_{0,\,j}-\sum _{i=0}^{m-1}v_{i,\,j}\,(e_{i+1,\,j}-e_{i,\,j})$
Taking into account that $e_{m+1,\,j}\,=\,e_{0,\,j}\,=\,0$  it follows
$\sum _{i=1}^{m}e_{i,\,j}\,(-h^{2}({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)=\;-\sum _{i=0}^{m}v_{i,\,j}\,(e_{i+1,\,j}-e_{i,\,j})$
$=\;-v_{0,\,j}\,(e_{1,\,j}-e_{0,\,j})-\sum _{i=1}^{m-1}v_{i,\,j}\,(e_{i+1,\,j}-e_{i,\,j})-v_{m,\,j}\,(e_{m+1,\,j}-e_{m,\,j})$
${\begin{aligned}=\;&{\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})\,(e_{1,\,j}-e_{0,\,j})-{\frac {1}{3}}\,(e_{3,\,j}-e_{2,\,j})\,(e_{1,\,j}-e_{0,\,j})\\&+{\frac {5}{12}}\,(e_{2,\,j}-e_{1,\,j})\,(e_{1,\,j}-e_{0,\,j})+{\frac {5}{6}}\,(e_{1,\,j}-e_{0,\,j})^{2}\\\;&-{\frac {1}{12}}\sum _{i=1}^{m-1}(e_{i+2,\,j}-e_{i+1,\,j})(e_{i+1,\,j}-e_{i,\,j})+{\frac {7}{6}}\sum _{i=1}^{m-1}(e_{i+1,\,j}-e_{i,\,j})^{2}\\\;&-{\frac {1}{12}}\sum _{i=1}^{m-1}(e_{i,\,j}-e_{i-1,\,j})(e_{i+1,\,j}-e_{i,\,j})+{\frac {5}{6}}\,(e_{m+1,\,j}-e_{m,\,j})^{2}\\\;&+{\frac {5}{12}}\,(e_{m,\,j}-e_{m-1,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\;-{\frac {1}{3}}\,(e_{m-1,\,j}-e_{m-2,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\\\;&+{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\\\end{aligned}}$

### working here

Collect like terms in the expression immediately above as follows.
${\frac {5}{6}}\,(e_{1,\,j}-e_{0,\,j})^{2}+{\frac {7}{6}}\sum _{i=1}^{m-1}(e_{i+1,\,j}-e_{i,\,j})^{2}+{\frac {5}{6}}\,(e_{m+1,\,j}-e_{m,\,j})^{2}$
$=\,{\frac {7}{6}}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}-{\frac {1}{3}}\,(e_{1,\,j}-e_{0,\,j})^{2}-{\frac {1}{3}}\,(e_{m+1,\,j}-e_{m,\,j})^{2}$
$-{\frac {1}{12}}\sum _{i=1}^{m-1}(e_{i+2,\,j}-e_{i+1,\,j})(e_{i+1,\,j}-e_{i,\,j})-{\frac {1}{12}}\sum _{i=1}^{m-1}(e_{i,\,j}-e_{i-1,\,j})(e_{i+1,\,j}-e_{i,\,j})$
${\begin{aligned}=\,&-{\frac {1}{6}}\sum _{i=2}^{m-1}(e_{i+1,\,j}-e_{i,\,j})(e_{i,\,j}-e_{i-1,\,j})-{\frac {1}{12}}(e_{2,\,j}-e_{1,\,j})(e_{1,\,j}-e_{0,\,j})\\\,&-{\frac {1}{12}}(e_{m+1,\,j}-e_{m,\,j})(e_{m,\,j}-e_{m-1,\,j})\\\end{aligned}}$
Now, rewrite the expression after making cancellations.
${\begin{aligned}&{\frac {7}{6}}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}-{\frac {1}{6}}\sum _{i=2}^{m-1}(e_{i+1,\,j}-e_{i,\,j})(e_{i,\,j}-e_{i-1,\,j})-{\frac {1}{3}}\,(e_{1,\,j}-e_{0,\,j})^{2}\\&-{\frac {1}{3}}\,(e_{m+1,\,j}-e_{m,\,j})^{2}+{\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})\,(e_{1,\,j}-e_{0,\,j})\\&-{\frac {1}{3}}\,(e_{3,\,j}-e_{2,\,j})\,(e_{1,\,j}-e_{0,\,j})+{\frac {1}{3}}\,(e_{2,\,j}-e_{1,\,j})\,(e_{1,\,j}-e_{0,\,j})\\&+{\frac {1}{3}}\,(e_{m,\,j}-e_{m-1,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\;-{\frac {1}{3}}\,(e_{m-1,\,j}-e_{m-2,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\\\;&+{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\\\end{aligned}}$
The following simple inequality will be used to bound terms.
${\begin{aligned}(a-b)^{2}&=a^{2}-2\,a\,b+b^{2}\,\geq \,0\\2\,a\,b\,&\leq \,a^{2}+b^{2}\\a\,b\,&\leq \,{\frac {1}{2}}\,(a^{2}+b^{2})\\\end{aligned}}$ and also
$a\,b\,\leq \,{\frac {1}{2}}\,(\alpha ^{2}\,a^{2}+b^{2}\,/\,\alpha ^{2})$.
${\begin{aligned}&\sum _{i=2}^{m-1}\left\vert (e_{i+1,\,j}-e_{i,\,j})(e_{i,\,j}-e_{i-1,\,j})\right\vert \;=\;\sum _{i=3}^{m-2}\left\vert (e_{i+1,\,j}-e_{i,\,j})(e_{i,\,j}-e_{i-1,\,j})\right\vert \\&+\;\left\vert (e_{3,\,j}-e_{2,\,j})(e_{2,\,j}-e_{1,\,j})\right\vert \;+\;\left\vert (e_{m,\,j}-e_{m-1,\,j})(e_{m-1,\,j}-e_{m-2,\,j})\right\vert \\&\;\leq \;{\frac {1}{2}}(\sum _{i=3}^{m-2}(e_{i+1,\,j}-e_{i,\,j})^{2}\,+\,\sum _{i=3}^{m-2}(e_{i,\,j}-e_{i-1,\,j})^{2})\\&+\;\left\vert (e_{3,\,j}-e_{2,\,j})(e_{2,\,j}-e_{1,\,j})\right\vert \;+\;\left\vert (e_{m,\,j}-e_{m-1,\,j})(e_{m-1,\,j}-e_{m-2,\,j})\right\vert \\\end{aligned}}$
${\begin{aligned}&=\;\sum _{i=3}^{m-3}(e_{i+1,\,j}-e_{i,\,j})^{2}\;+\;{\frac {1}{2}}\,(e_{3,\,j}-e_{2,\,j})^{2}\;+\;{\frac {1}{2}}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}\\&+\;\left\vert (e_{3,\,j}-e_{2,\,j})(e_{2,\,j}-e_{1,\,j})\right\vert \;+\;\left\vert (e_{m,\,j}-e_{m-1,\,j})(e_{m-1,\,j}-e_{m-2,\,j})\right\vert \\\end{aligned}}$
${\begin{aligned}&=\;\sum _{i=2}^{m-2}(e_{i+1,\,j}-e_{i,\,j})^{2}\;-\;{\frac {1}{2}}\,(e_{3,\,j}-e_{2,\,j})^{2}\;-\;{\frac {1}{2}}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}\\&+\;\left\vert (e_{3,\,j}-e_{2,\,j})(e_{2,\,j}-e_{1,\,j})\right\vert \;+\;\left\vert (e_{m,\,j}-e_{m-1,\,j})(e_{m-1,\,j}-e_{m-2,\,j})\right\vert \\\end{aligned}}$
${\begin{aligned}&\leq \;\sum _{i=2}^{m-2}(e_{i+1,\,j}-e_{i,\,j})^{2}\;-\;{\frac {1}{2}}\,(e_{3,\,j}-e_{2,\,j})^{2}\;-\;{\frac {1}{2}}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}\\&+\;{\frac {1}{2}}\,(\alpha _{1}^{2}\,(e_{3,\,j}-e_{2,\,j})^{2}+(e_{2,\,j}-e_{1,\,j})^{2}\,/\,\alpha _{1}^{2})\\&+\;{\frac {1}{2}}\,(\beta _{1}^{2}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}+(e_{m,\,j}-e_{m-1,\,j})^{2}\,/\,\beta _{1}^{2})\\\end{aligned}}$
${\begin{aligned}&\;\left\vert (e_{4,\,j}-e_{3,\,j})(e_{1,\,j}-e_{0,\,j})\right\vert \;\leq \;{\frac {1}{2}}\,(\alpha _{2}^{2}\,(e_{4,\,j}-e_{3,\,j})^{2}+(e_{1,\,j}-e_{0,\,j})^{2}\,/\,\alpha _{2}^{2})\\&\;\left\vert (e_{3,\,j}-e_{2,\,j})(e_{1,\,j}-e_{0,\,j})\right\vert \;\leq \;{\frac {1}{2}}\,(\alpha _{3}^{2}\,(e_{3,\,j}-e_{2,\,j})^{2}+(e_{1,\,j}-e_{0,\,j})^{2}\,/\,\alpha _{3}^{2})\\&\;\left\vert (e_{2,\,j}-e_{1,\,j})(e_{1,\,j}-e_{0,\,j})\right\vert \;\leq \;{\frac {1}{2}}\,(\alpha _{4}^{2}\,(e_{2,\,j}-e_{1,\,j})^{2}+(e_{1,\,j}-e_{0,\,j})^{2}\,/\,\alpha _{4}^{2})\\\end{aligned}}$
${\begin{aligned}&\;\left\vert (e_{m,\,j}-e_{m-1,\,j})(e_{m+1,\,j}-e_{m,\,j})\right\vert \\&\quad \quad \quad \leq \;{\frac {1}{2}}\,(\beta _{2}^{2}\,(e_{m,\,j}-e_{m-1,\,j})^{2}+(e_{m+1,\,j}-e_{m,\,j})^{2}\,/\,\beta _{2}^{2})\\&\;\left\vert (e_{m-1,\,j}-e_{m-2,\,j})(e_{m+1,\,j}-e_{m,\,j})\right\vert \\&\quad \quad \quad \leq \;{\frac {1}{2}}\,(\beta _{3}^{2}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}+(e_{m+1,\,j}-e_{m,\,j})^{2}\,/\,\beta _{3}^{2})\\&\;\left\vert (e_{m-2,\,j}-e_{m-3,\,j})(e_{m+1,\,j}-e_{m,\,j})\right\vert \\&\quad \quad \quad \leq \;{\frac {1}{2}}\,(\beta _{4}^{2}\,(e_{m-2,\,j}-e_{m-3,\,j})^{2}+(e_{m+1,\,j}-e_{m,\,j})^{2}\,/\,\beta _{4}^{2})\\\end{aligned}}$
Now, substitute all the inequalities into the expression.
${\begin{aligned}&{\frac {7}{6}}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}-{\frac {1}{6}}\sum _{i=2}^{m-1}(e_{i+1,\,j}-e_{i,\,j})(e_{i,\,j}-e_{i-1,\,j})-{\frac {1}{3}}\,(e_{1,\,j}-e_{0,\,j})^{2}\\&-{\frac {1}{3}}\,(e_{m+1,\,j}-e_{m,\,j})^{2}+{\frac {1}{12}}\,(e_{4,\,j}-e_{3,\,j})\,(e_{1,\,j}-e_{0,\,j})\\&-{\frac {1}{3}}\,(e_{3,\,j}-e_{2,\,j})\,(e_{1,\,j}-e_{0,\,j})+{\frac {1}{3}}\,(e_{2,\,j}-e_{1,\,j})\,(e_{1,\,j}-e_{0,\,j})\\&+{\frac {1}{3}}\,(e_{m,\,j}-e_{m-1,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\;-{\frac {1}{3}}\,(e_{m-1,\,j}-e_{m-2,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\\\;&+{\frac {1}{12}}\,(e_{m-2,\,j}-e_{m-3,\,j})\,(e_{m+1,\,j}-e_{m,\,j})\\\end{aligned}}$
${\begin{aligned}&{\frac {7}{6}}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}-{\frac {1}{6}}\,{\big\;\sum _{i=2}^{m-2}(e_{i+1,\,j}-e_{i,\,j})^{2}\;-\;{\frac {1}{2}}\,(e_{3,\,j}-e_{2,\,j})^{2}\;\\&-\;{\frac {1}{2}}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}+\;{\frac {1}{2}}\,(\alpha _{1}^{2}\,(e_{3,\,j}-e_{2,\,j})^{2}+(e_{2,\,j}-e_{1,\,j})^{2}\,/\,\alpha _{1}^{2})\\&+\;{\frac {1}{2}}\,(\beta _{1}^{2}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}+(e_{m,\,j}-e_{m-1,\,j})^{2}\,/\,\beta _{1}^{2})\;{\big )}\\&-{\frac {1}{3}}\,(e_{1,\,j}-e_{0,\,j})^{2}-{\frac {1}{3}}\,(e_{m+1,\,j}-e_{m,\,j})^{2}\\&-{\frac {1}{12}}\,{\big{\frac {1}{2}}\,(\alpha _{2}^{2}\,(e_{4,\,j}-e_{3,\,j})^{2}+(e_{1,\,j}-e_{0,\,j})^{2}\,/\,\alpha _{2}^{2}){\big )}\\&-{\frac {1}{3}}\,{\big{\frac {1}{2}}\,(\alpha _{3}^{2}\,(e_{3,\,j}-e_{2,\,j})^{2}+(e_{1,\,j}-e_{0,\,j})^{2}\,/\,\alpha _{3}^{2}){\big )}\\&-{\frac {1}{3}}\,{\big{\frac {1}{2}}\,(\alpha _{4}^{2}\,(e_{2,\,j}-e_{1,\,j})^{2}+(e_{1,\,j}-e_{0,\,j})^{2}\,/\,\alpha _{4}^{2}){\big )}\\&-{\frac {1}{3}}\,{\big{\frac {1}{2}}\,(\beta _{2}^{2}\,(e_{m,\,j}-e_{m-1,\,j})^{2}+(e_{m+1,\,j}-e_{m,\,j})^{2}\,/\,\beta _{2}^{2}){\big )}\\&-{\frac {1}{3}}\,{\big{\frac {1}{2}}\,(\beta _{3}^{2}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}+(e_{m+1,\,j}-e_{m,\,j})^{2}\,/\,\beta _{3}^{2}){\big )}\\&-{\frac {1}{12}}\,{\big{\frac {1}{2}}\,(\beta _{4}^{2}\,(e_{m-2,\,j}-e_{m-3,\,j})^{2}+(e_{m+1,\,j}-e_{m,\,j})^{2}\,/\,\beta _{4}^{2}){\big )}\\\end{aligned}}$
${\begin{aligned}&=\;{\frac {7}{6}}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}-{\frac {1}{6}}\;\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}\\&\;-\;{\big\,{\frac {1}{6}}\,+\,({\frac {1}{24}})\,/\,\alpha _{2}^{2}\,+\,({\frac {1}{6}})\,/\,\alpha _{3}^{2}\,+\,({\frac {1}{6}})\,/\,\alpha _{4}^{2}\,{\big )}\,(e_{1,\,j}-e_{0,\,j})^{2}\\&\;-\;{\big\,-{\frac {1}{6}}\,+\,({\frac {1}{12}})\,/\,\alpha _{1}^{2}\,+\,{\frac {1}{6}}\,\alpha _{4}^{2}\,{\big )}\,(e_{2,\,j}-e_{1,\,j})^{2}\\&\;-\;{\big\,-{\frac {1}{12}}\,+\,{\frac {1}{12}}\,\alpha _{1}^{2}\,+\,{\frac {1}{6}}\,\alpha _{3}^{2}\,{\big )}\,(e_{3,\,j}-e_{2,\,j})^{2}\\&\;-\;{\big\,{\frac {1}{24}}\,\alpha _{2}^{2}\,{\big )}\,(e_{4,\,j}-e_{3,\,j})^{2}\\&\;-\;{\big\,{\frac {1}{6}}\,+\,({\frac {1}{6}})\,/\,\beta _{2}^{2}\,+\,({\frac {1}{6}})\,/\,\beta _{3}^{2}\,+\,({\frac {1}{24}})\,/\,\beta _{4}^{2}\,{\big )}\,(e_{m+1,\,j}-e_{m,\,j})^{2}\\&\;-\;{\big\,-{\frac {1}{6}}\,+\,({\frac {1}{12}})\,/\,\beta _{1}^{2}\,+\,{\frac {1}{6}}\,\beta _{2}^{2}\,{\big )}\,(e_{m,\,j}-e_{m-1,\,j})^{2}\\&\;-\;{\big-{\frac {1}{12}}\,+\,{\frac {1}{12}}\,\beta _{1}^{2}\,+\,{\frac {1}{6}}\,\beta _{3}^{2}\,{\big )}\,(e_{m-1,\,j}-e_{m-2,\,j})^{2}\\&\;-\;{\big\,{\frac {1}{24}}\,\beta _{4}^{2}\,{\big )}\,(e_{m-2,\,j}-e_{m-3,\,j})^{2}\\\end{aligned}}$
The choice
${\begin{aligned}&\alpha _{1}^{2}=\beta _{1}^{2}=({\sqrt {5}}-1)\,/\,2\,,\alpha _{2}^{2}=\beta _{4}^{2}=8\,,\\&\alpha _{3}^{2}=\alpha _{4}^{2}=\beta _{2}^{2}=\beta _{3}^{2}=(11-{\sqrt {5}})\,/\,4\,,\\\end{aligned}}$
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
$\sum _{i=1}^{m}e_{i,\,j}\,(-h^{2}\,({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)\;\geq \;{\frac {2}{3}}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}$. and
$\sum _{j=1}^{n}\sum _{i=1}^{m}e_{i,\,j}\,(-\,({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)\;\geq \;{\frac {2}{3}}\,h^{-2}\,\sum _{j=1}^{n}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}$. Reasoning in the exact same manner for the dimension in $y$
$\sum _{j=1}^{n}e_{i,\,j}\,(-k^{2}\,({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)\;\geq \;{\frac {2}{3}}\sum _{j=0}^{n}(e_{i,\,j+1}-e_{i,\,j})^{2}$. and
$\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\,({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)\;\geq \;{\frac {2}{3}}\,k^{-2}\,\sum _{i=1}^{m}\sum _{j=0}^{n}(e_{i,\,j+1}-e_{i,\,j})^{2}$. Applying  $\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)\;=$
$\sum _{j=1}^{n}\sum _{i=1}^{m}e_{i,\,j}\,(-({\partial _{h}^{2} \over \partial _{h}x^{2}})_{i,\,j}\,e)\;+\;\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-({\partial _{k}^{2} \over \partial _{k}y^{2}})_{i,\,j}\,e)$ leads to the inequality  $\sum _{i=1}^{m}\sum _{j=1}^{n}e_{i,\,j}\,(-\Delta _{i,\,j}\,e)$
$\geq \;{\frac {2}{3}}\,h^{-2}\,\sum _{j=1}^{n}\sum _{i=0}^{m}(e_{i+1,\,j}-e_{i,\,j})^{2}\;+\;{\frac {2}{3}}\,k^{-2}\,\sum _{i=1}^{m}\sum _{j=0}^{n}(e_{i,\,j+1}-e_{i,\,j})^{2}$.