Using High Order Finite Differences/Third Order 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

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

is called a Poisson problem.

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

$$(x_i, y_j),\ 0 \le i \le m+1,\ 0 \le j \le n+1$$. 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:

$$\begin{align} \tau_x(x_i, y_j) & = ({\partial_h^2\over\partial_h x^2})\ u(x_i, y_j) - {\partial^2\over\partial x^2}\ u(x_i, y_j) \\ & = {h^3\over\ 120}\, M{_x^{(5)}}(x_i, y_j) \end{align}$$

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:

$$\begin{align} \tau_y(x_i, y_i) & = ({\partial_k^2\over\partial_k y^2})\ u(x_i, y_1) - {\partial^2\over\partial y^2}\ u(x_i, y_1) \\ & = {k^3\over\ 120}\, M{_y^{(5)}}(x_i, y_1) \end{align}$$

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

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

Then solve the non-singular linear system $$-\Delta_{i,\,j}\,U = f(x_i,\,y_j) \quad 1 \le i \le m, \ 1 \le j \le n$$ ; for the remaining $$U_{i,\,j}$$

error estimation
The error $$U - u = \{U_{i,\,j} - u(x_i,\,y_j)\}$$ , satisfies $${\lVert U - u \rVert}_2 \le {\rho\, a^2 b^2\over\ \pi^2(a^2+b^2)} {\lVert \tau \rVert}_2 (1 + O(h^2+k^2))$$. for $$\tau = \{\tau (x_i,\,y_j)\}, \quad 1 \le i \le m,\ 1 \le j \le n$$ , and $$\rho \ \le \ 1\, /\, (1 - (1 + \sqrt 10 )\,/\,12)$$.

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 $$x_i$$, $$\begin{align} g(x_i+\Delta x)\; & =\;g(x_i)+g^{(1)}(x_i)\Delta x +\frac{1}{2}\,g^{(2)}(x_i)({\Delta x})^2 +\frac{1}{6}\,g^{(3)}(x_i)({\Delta x})^3 \\ & +\frac{1}{24}\,g^{(4)}(x_i)({\Delta x})^4 +\frac{1}{120}\,g^{(5)}(z)({\Delta x})^5 \\ \end{align}$$. where $$z$$  is some number between  $$x_i$$ and $$x_i+\Delta x$$. Then $$\begin{align} & {\partial_h^2\over\partial_h x^2}\,u(x_1,\,y_j)\;= \\ & {\big ( -\frac{1}{12}g(x_1+3\,h)+\frac{1}{3}g(x_1+2\,h) +\frac{1}{2}g(x_1+h)-\frac{5}{3}g(x_1)+\frac{11}{12}g(x_1-h) \big ) \over\ h^2} \end{align}$$ $$\begin{align} & =\;g(x_1){(-\frac{1}{12}+\frac{1}{3}+\frac{1}{2}-\frac{5}{3}+\frac{11}{12})\over\ h^2} \\ & +\;g^{(1)}(x_1){(-\frac{1}{12}(3\,h)+\frac{1}{3}(2\,h)+\frac{1}{2}(h)-\frac{5}{3}(0)+\frac{11}{12}(-h))\over\ h^2} \\ & +\;\frac{1}{2}g^{(2)}(x_1){(-\frac{1}{12}(3\,h)^2+\frac{1}{3}(2\,h)^2+\frac{1}{2}(h)^2-\frac{5}{3}(0)^2+\frac{11}{12}(-h)^2)\over\ h^2} \\ & +\;\frac{1}{6}g^{(3)}(x_1){(-\frac{1}{12}(3\,h)^3+\frac{1}{3}(2\,h)^3+\frac{1}{2}(h)^3-\frac{5}{3}(0)^3+\frac{11}{12}(-h)^3)\over\ h^2} \\ & +\;\frac{1}{24}g^{(4)}(x_1){(-\frac{1}{12}(3\,h)^4+\frac{1}{3}(2\,h)^4+\frac{1}{2}(h)^4-\frac{5}{3}(0)^4+\frac{11}{12}(-h)^4)\over\ h^2} \\ & +\;\frac{1}{120}{(-\frac{1}{12}g^{(5)}(z_1)(3\,h)^5+\frac{1}{3}g^{(5)}(z_2)(2\,h)^5+\frac{1}{2}g^{(5)}(z_3)(h)^5-\frac{5}{3}(0)^5\over\ h^2} \\ & \quad \quad \quad \quad {+\frac{11}{12}g^{(5)}(z_4)(-h)^5)\over\ h^2} \\ \end{align}$$ $$=\;g^{(2)}(x_1) + {h^3\over\ 120}\big (-\frac{81}{4}g^{(5)}(z_1) +\frac{32}{3}g^{(5)}(z_2)+\frac{1}{2}g^{(5)}(z_3)-\frac{11}{12}g^{(5)}(z_4)\big ) $$ where $$\begin{align} & x_1\;<\;z_1\;<\;x_1+3\,h\,,\;x_1\;<\;z_2\;<\;x_1+2\,h\,,\; \\ & x_1\;<\;z_3\;<\;x_1+h\,,\;\;\text{and}\;\;x_1-h\;<\;z_4\;<\;x_1. \\ \end{align}$$. Since $$\frac{67}{6}\underset{0\; \le\; \phi \; \le\; 2}{\min(g^{(5)}(x_1\;+\;\phi\,h))}\; \le \;\frac{32}{3}g^{(5)}(z_2)+\frac{1}{2}g^{(5)}(z_3)\; \le \;\frac{67}{6}\underset{0\; \le\; \phi \; \le\; 2 }{\max(g^{(5)}(x_1\;+\;\phi\,h))},$$ $$\frac{254}{12}\underset{-1\; \le\; \phi \; \le\; 3}{\min(g^{(5)}(x_1\;+\;\phi\,h))}\; \le \;\frac{81}{4}g^{(5)}(z_1)+\frac{11}{12}g^{(5)}(z_4)\; \le \;\frac{254}{12}\underset{-1\; \le\; \phi \; \le\; 3 }{\max(g^{(5)}(x_1\;+\;\phi\,h))},$$ 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  $$\begin{align} & {\partial_h^2\over\partial_h x^2}\,u(x_1,\,y_j)\;=\; g^{(2)}(x_1) + {h^3\over\ 120}\big ( \frac{67}{6}\,g^{(5)}(x_1+\phi_1\,h) - \frac{254}{12}\,g^{(5)}(x_1+\phi_2\,h)\big ) \\ & \;=\; {\partial^2\over\partial x^2}\,u(x_1,\,y_j) + {h^3\over\ 120}\big ( \frac{67}{6}{\partial^5\over\partial x^5} u(x_1+\phi_1\,h,\;y_j) - \frac{254}{12}{\partial^5\over\partial x^5} u(x_1+\phi_2\,h,\;y_j)\big ) \\ \end{align}$$  which is  $$\tau_x(x_1,\,y_j)\;=\;{h^3\over\ 120}M_x^5(x_1,\,y_j).$$

For $$i\;=\;2,\;3,\;\ldots\,,\;m-1$$ $$\begin{align} & {\partial_h^2\over\partial_h x^2}\,u(x_i,\,y_j)\;= \\ & {\big ( -\frac{1}{12}g(x_i+2\,h)+\frac{4}{3}g(x_i+h) -\frac{5}{2}g(x_i)+\frac{4}{3}g(x_i-h)-\frac{1}{12}g(x_i-2\,h) \big ) \over\ h^2} \end{align}$$ $$\begin{align} & =\;g(x_i){(-\frac{1}{12}+\frac{4}{3}-\frac{5}{2}+\frac{4}{3}-\frac{1}{12})\over\ h^2} \\ & +\;g^{(1)}(x_i){(-\frac{1}{12}(2\,h)+\frac{4}{3}(h)-\frac{5}{2}(0)+\frac{4}{3}(-h)-\frac{1}{12}(-2\,h))\over\ h^2} \\ & +\;\frac{1}{2}g^{(2)}(x_i){(-\frac{1}{12}(2\,h)^2+\frac{4}{3}(h)^2-\frac{5}{2}(0)^2+\frac{4}{3}(-h)^2-\frac{1}{12}(-2\,h)^2)\over\ h^2} \\ & +\;\frac{1}{6}g^{(3)}(x_i){(-\frac{1}{12}(2\,h)^3+\frac{4}{3}(h)^3-\frac{5}{2}(0)^3+\frac{4}{3}(-h)^3-\frac{1}{12}(-2\,h)^3)\over\ h^2} \\ & +\;\frac{1}{24}g^{(4)}(x_i){(-\frac{1}{12}(2\,h)^4+\frac{4}{3}(h)^4-\frac{5}{2}(0)^4+\frac{4}{3}(-h)^4-\frac{1}{12}(-2\,h)^4)\over\ h^2} \\ & +\;\frac{1}{120}{(-\frac{1}{12}g^{(5)}(z_1)(2\,h)^5+\frac{4}{3}g^{(5)}(z_2)(h)^5-\frac{5}{2}(0)^5+\frac{4}{3}g^{(5)}(z_3)(-h)^5\over\ h^2} \\ & \quad \quad \quad \quad {-\frac{1}{12}g^{(5)}(z_4)(-2\,h)^5)\over\ h^2} \\ \end{align}$$ $$=\;g^{(2)}(x_i) + {h^3\over\ 120}\big (-\frac{8}{3}g^{(5)}(z_1) +\frac{4}{3}g^{(5)}(z_2)-\frac{4}{3}g^{(5)}(z_3)+\frac{8}{3}g^{(5)}(z_4)\big ) $$ where $$\begin{align} & x_i\;<\;z_1\;<\;x_i+2\,h\,,\;x_i\;<\;z_2\;<\;x_i+h\,,\; \\ & x_i-h\;<\;z_3\;<\;x_i\,,\;\;\text{and}\;\;x_i-2\,h\;<\;z_4\;<\;x_i. \\ \end{align}$$. 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  $$\begin{align} & {\partial_h^2\over\partial_h x^2}\,u(x_i,\,y_j)\;=\; g^{(2)}(x_i) + {h^3\over\ 120}\big ( \frac{67}{6}\,g^{(5)}(x_i+\phi_1\,h) - \frac{254}{12}\,g^{(5)}(x_i+\phi_2\,h)\big ) \\ & \;=\; {\partial^2\over\partial x^2}\,u(x_i,\,y_j) + {h^3\over\ 120}\big ( 4{\partial^5\over\partial x^5} u(x_i+\phi_1\,h,\;y_j) - 4{\partial^5\over\partial x^5} u(x_i+\phi_2\,h,\;y_j)\big ) \\ \end{align}$$  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 $$\begin{align} & +\;\frac{1}{120}{(-\frac{1}{12}g^{(5)}(z_1)(2\,h)^5+\frac{4}{3}g^{(5)}(z_2)(h)^5-\frac{5}{2}(0)^5+\frac{4}{3}g^{(5)}(z_3)(-h)^5\over\ h^2} \\ & \quad \quad \quad \quad {-\frac{1}{12}g^{(5)}(z_4)(-2\,h)^5)\over\ h^2} \\ \end{align}$$ can be replaced by $$\begin{align} & +\;\frac{1}{120}g^{(5)}(x_i){(-\frac{1}{12}(2\,h)^5+\frac{4}{3}(h)^5-\frac{5}{2}(0)^5+\frac{4}{3}(-h)^5\over\ h^2} \\ & \quad \quad \quad \quad {-\frac{1}{12}(-2\,h)^5)\over\ h^2} \\ & \; \\ & +\;\frac{1}{720}{(-\frac{1}{12}g^{(6)}(z_1)(2\,h)^6+\frac{4}{3}g^{(6)}(z_2)(h)^6-\frac{5}{2}(0)^5+\frac{4}{3}g^{(6)}(z_3)(-h)^6\over\ h^2} \\ & \quad \quad \quad \quad {-\frac{1}{12}g^{(6)}(z_4)(-2\,h)^6)\over\ h^2} \\ \end{align}$$

This gives $$\begin{align} & {\partial_h^2\over\partial_h x^2}\,u(x_i,\,y_j)\;=\; \\ & \;=\; {\partial^2\over\partial x^2}\,u(x_i,\,y_j) + {h^4\over\ 720}\big ( \frac{8}{3}{\partial^6\over\partial x^6} u(x_i+\phi_1\,h,\;y_j) - \frac{32}{3}{\partial^6\over\partial x^6} u(x_i+\phi_2\,h,\;y_j)\big ) \\ \end{align}$$ 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 $$e = \{e_{i,\,j}\}, \quad 1 \le i \le m, \ 1 \le j \le n,$$ 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 $$\begin{align} -\Delta_{i,\,j}\,e & = -\Delta_{i,\,j}\,U + \Delta_{h,\,k}\,u(x_i,\,y_j) \\ & = f(x_i,\,y_j) + (\Delta\,u(x_i,\,y_j) + \tau (x_i,\,y_j)) \\ & = \tau (x_i,\,y_j) \\ \end{align}$$ we get that $$ \sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e) \ = \ \sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,\tau (x_i,\,y_j) \ \le \ {\lVert e \rVert}_2\, {\lVert \tau \rVert}_2 $$ . Next it will be shown that the operator $$-\Delta_{i,\,j}$$ is positive definite for $$e$$, in particular that $$ \sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e) $$  $$ \ge \; \lambda (4 a^{-2}\,{(m+1)}^2\,(something)+4 b^{-2}\,{(n+1)}^2\,)\, {\lVert e \rVert}_2 $$ ,

with $$\lambda \ \ge \ 1 - (1+\sqrt 10)\,/\,12$$.

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
$$-h^2({\partial_h^2\over\partial_h x^2})_{1,\,j}\ e\ =$$. $$\frac{1}{12}\,e_{4,\,j}-\frac{1}{3}\,e_{3,\,j} -\frac{1}{2}\,e_{2,\,j}+\frac{5}{3}\,e_{1,\,j} -\frac{11}{12}\,e_{0,\,j}$$

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

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

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

$$-h^2({\partial_h^2\over\partial_h x^2})_{i,\,j}\ e\ =$$. $$\frac{1}{12}\,e_{i+2,\,j}-\frac{4}{3}\,e_{i+1,\,j} +\frac{5}{2}\,e_{i,\,j}-\frac{4}{3}\,e_{i-1,\,j} +\frac{1}{12}\,e_{i-2,\,j}$$

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

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

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

$$\begin{align} =\; & (\frac{1}{12}\,e_{i+2,\,j} & -\frac{5}{4}\,e_{i+1,\,j} & +\frac{5}{4}\,e_{i,\,j} & -\frac{1}{12}\,e_{i-1,\,j}) \\ -\; & (\frac{1}{12}\,e_{i+1,\,j} & -\frac{5}{4}\,e_{i,\,j} \quad & +\frac{5}{4}\,e_{i-1,\,j} & -\frac{1}{12}\,e_{i-2,\,j}) \\ \end{align}$$

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

$$-h^2({\partial_h^2\over\partial_h x^2})_{m,\,j}\ e\ =$$. $$-\frac{11}{12}\,e_{m+1,\,j}+\frac{5}{3}\,e_{m,\,j} -\frac{1}{2}\,e_{m-1,\,j}-\frac{1}{3}\,e_{m-2,\,j} +\frac{1}{12}\,e_{m-3,\,j}$$

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

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

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

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

$$ \sum_{i = 1}^m w_i\,(v_i - v_{i-1}) = w_m\,v_m - w_0\,v_0 - \sum_{i = 0}^{m-1} v_i\,(w_{i+1} - w_i) $$

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

$$\text{So the sum} \; \sum_{i = 1}^m e_{i,\,j}\,(-h^2({\partial_h^2\over\partial_h x^2})_{i,\,j}\,e)=\; \sum_{i = 1}^m e_{i,\,j}\,(v_{i,\,j} - v_{i-1,\,j})$$ $$ = e_{m,\,j}\,v_{m,\,j} - e_{0,\,j}\,v_{0,\,j} - \sum_{i = 0}^{m-1} v_{i,\,j}\,(e_{i+1,\,j} - e_{i,\,j}) $$ Taking into account that  $$e_{m+1,\,j}\, = \, e_{0,\,j}\, = \, 0$$ it follows $$ \sum_{i = 1}^m e_{i,\,j}\,(-h^2({\partial_h^2\over\partial_h x^2})_{i,\,j}\,e)=\; - \sum_{i = 0}^m v_{i,\,j}\,(e_{i+1,\,j} - e_{i,\,j}) $$ $$ =\; - v_{0,\,j}\,(e_{1,\,j} - e_{0,\,j}) - \sum_{i = 1}^{m-1} v_{i,\,j}\,(e_{i+1,\,j} - e_{i,\,j}) - v_{m,\,j}\,(e_{m+1,\,j} - e_{m,\,j}) $$

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

working here
Collect like terms in the expression immediately above as follows. $$\frac{5}{6}\,(e_{1,\,j} - e_{0,\,j})^2+\frac{7}{6}\sum_{i = 1}^{m-1}(e_{i+1,\,j} - e_{i,\,j})^2+\frac{5}{6}\,(e_{m+1,\,j} - e_{m,\,j})^2$$ $$=\, \frac{7}{6}\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2-\frac{1}{3}\,(e_{1,\,j} - e_{0,\,j})^2-\frac{1}{3}\,(e_{m+1,\,j} - e_{m,\,j})^2$$ $$-\frac{1}{12}\sum_{i = 1}^{m-1}(e_{i+2,\,j} - e_{i+1,\,j}) (e_{i+1,\,j} - e_{i,\,j})-\frac{1}{12}\sum_{i = 1}^{m-1}(e_{i,\,j} - e_{i-1,\,j})(e_{i+1,\,j} - e_{i,\,j})$$ $$\begin{align} =\, & -\frac{1}{6}\sum_{i = 2}^{m-1}(e_{i+1,\,j} - e_{i,\,j}) (e_{i,\,j} - e_{i-1,\,j})-\frac{1}{12}(e_{2,\,j} - e_{1,\,j}) (e_{1,\,j} - e_{0,\,j}) \\ \, & -\frac{1}{12}(e_{m+1,\,j} - e_{m,\,j})(e_{m,\,j} - e_{m-1,\,j}) \\ \end{align}$$

Now, rewrite the expression after making cancellations. $$\begin{align} & \frac{7}{6}\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2-\frac{1}{6}\sum_{i = 2}^{m-1}(e_{i+1,\,j} - e_{i,\,j})(e_{i,\,j} - e_{i-1,\,j})-\frac{1}{3}\,(e_{1,\,j} - e_{0,\,j})^2 \\ & -\frac{1}{3}\,(e_{m+1,\,j} - e_{m,\,j})^2+\frac{1}{12}\,(e_{4,\,j} - e_{3,\,j})\,(e_{1,\,j} - e_{0,\,j}) \\ & -\frac{1}{3}\,(e_{3,\,j} - e_{2,\,j})\,(e_{1,\,j} - e_{0,\,j})+\frac{1}{3}\,(e_{2,\,j} - e_{1,\,j})\,(e_{1,\,j} - e_{0,\,j}) \\ & +\frac{1}{3}\,(e_{m,\,j} - e_{m-1,\,j})\,(e_{m+1,\,j} - e_{m,\,j}) \; -\frac{1}{3}\,(e_{m-1,\,j} - e_{m-2,\,j})\,(e_{m+1,\,j} - e_{m,\,j}) \\ \; & +\frac{1}{12}\,(e_{m-2,\,j} - e_{m-3,\,j})\,(e_{m+1,\,j} - e_{m,\,j}) \\ \end{align}$$ The following simple inequality will be used to bound terms. $$\begin{align} (a-b)^2 & = a^2 - 2\,a\,b + b^2\, \ge \, 0 \\ 2\,a\,b \, & \le \, a^2 + b^2 \\ a\,b \, & \le \, \frac{1}{2}\,(a^2 + b^2) \\ \end{align}$$ and also $$ a\,b \, \le \, \frac{1}{2}\,(\alpha^2\,a^2 + b^2\,/\,\alpha^2) $$. $$\begin{align} & \sum_{i = 2}^{m-1}\left\vert (e_{i+1,\,j} - e_{i,\,j})(e_{i,\,j} - e_{i-1,\,j}) \right\vert\;=\;\sum_{i = 3}^{m-2}\left\vert (e_{i+1,\,j} - e_{i,\,j})(e_{i,\,j} - e_{i-1,\,j}) \right\vert \\ & +\;\left\vert (e_{3,\,j} - e_{2,\,j})(e_{2,\,j} - e_{1,\,j}) \right\vert\;+\;\left\vert (e_{m,\,j} - e_{m-1,\,j})(e_{m-1,\,j} - e_{m-2,\,j}) \right\vert \\ & \;\le\;\frac{1}{2}(\sum_{i = 3}^{m-2}(e_{i+1,\,j} - e_{i,\,j})^2\,+\,\sum_{i = 3}^{m-2}(e_{i,\,j} - e_{i-1,\,j})^2) \\ & +\;\left\vert (e_{3,\,j} - e_{2,\,j})(e_{2,\,j} - e_{1,\,j}) \right\vert\;+\;\left\vert (e_{m,\,j} - e_{m-1,\,j})(e_{m-1,\,j} - e_{m-2,\,j}) \right\vert \\ \end{align}$$

$$\begin{align} & =\;\sum_{i = 3}^{m-3}(e_{i+1,\,j} - e_{i,\,j})^2\;+\;\frac{1}{2}\,(e_{3,\,j} - e_{2,\,j})^2\;+\;\frac{1}{2}\,(e_{m-1,\,j} - e_{m-2,\,j})^2 \\ & +\;\left\vert (e_{3,\,j} - e_{2,\,j})(e_{2,\,j} - e_{1,\,j}) \right\vert\;+\;\left\vert (e_{m,\,j} - e_{m-1,\,j})(e_{m-1,\,j} - e_{m-2,\,j}) \right\vert \\ \end{align}$$ $$\begin{align} & =\;\sum_{i = 2}^{m-2}(e_{i+1,\,j} - e_{i,\,j})^2\;-\;\frac{1}{2}\,(e_{3,\,j} - e_{2,\,j})^2\;-\;\frac{1}{2}\,(e_{m-1,\,j} - e_{m-2,\,j})^2 \\ & +\;\left\vert (e_{3,\,j} - e_{2,\,j})(e_{2,\,j} - e_{1,\,j}) \right\vert\;+\;\left\vert (e_{m,\,j} - e_{m-1,\,j})(e_{m-1,\,j} - e_{m-2,\,j}) \right\vert \\ \end{align}$$ $$\begin{align} & \le\;\sum_{i = 2}^{m-2}(e_{i+1,\,j} - e_{i,\,j})^2\;-\;\frac{1}{2}\,(e_{3,\,j} - e_{2,\,j})^2\;-\;\frac{1}{2}\,(e_{m-1,\,j} - e_{m-2,\,j})^2 \\ & +\;\frac{1}{2}\,(\alpha_1^2\,(e_{3,\,j} - e_{2,\,j})^2+(e_{2,\,j} - e_{1,\,j})^2\,/\,\alpha_1^2) \\ & +\;\frac{1}{2}\,(\beta_1^2\,(e_{m-1,\,j} - e_{m-2,\,j})^2+(e_{m,\,j} - e_{m-1,\,j})^2\,/\,\beta_1^2) \\ \end{align}$$ $$\begin{align} & \;\left\vert (e_{4,\,j} - e_{3,\,j})(e_{1,\,j} - e_{0,\,j}) \right\vert \;\le \;\frac{1}{2}\,(\alpha_2^2\,(e_{4,\,j} - e_{3,\,j})^2+(e_{1,\,j} - e_{0,\,j})^2\,/\,\alpha_2^2) \\ & \;\left\vert (e_{3,\,j} - e_{2,\,j})(e_{1,\,j} - e_{0,\,j}) \right\vert \;\le \;\frac{1}{2}\,(\alpha_3^2\,(e_{3,\,j} - e_{2,\,j})^2+(e_{1,\,j} - e_{0,\,j})^2\,/\,\alpha_3^2) \\ & \;\left\vert (e_{2,\,j} - e_{1,\,j})(e_{1,\,j} - e_{0,\,j}) \right\vert \;\le \;\frac{1}{2}\,(\alpha_4^2\,(e_{2,\,j} - e_{1,\,j})^2+(e_{1,\,j} - e_{0,\,j})^2\,/\,\alpha_4^2) \\ \end{align}$$ $$\begin{align} & \;\left\vert (e_{m,\,j} - e_{m-1,\,j})(e_{m+1,\,j} - e_{m,\,j}) \right\vert \\ & \quad \quad \quad \le \;\frac{1}{2}\,(\beta_2^2\,(e_{m,\,j} - e_{m-1,\,j})^2+(e_{m+1,\,j} - e_{m,\,j})^2\,/\,\beta_2^2) \\ & \;\left\vert (e_{m-1,\,j} - e_{m-2,\,j})(e_{m+1,\,j} - e_{m,\,j}) \right\vert \\ & \quad \quad \quad \le \;\frac{1}{2}\,(\beta_3^2\,(e_{m-1,\,j} - e_{m-2,\,j})^2+(e_{m+1,\,j} - e_{m,\,j})^2\,/\,\beta_3^2) \\ & \;\left\vert (e_{m-2,\,j} - e_{m-3,\,j})(e_{m+1,\,j} - e_{m,\,j}) \right\vert \\ & \quad \quad \quad \le \;\frac{1}{2}\,(\beta_4^2\,(e_{m-2,\,j} - e_{m-3,\,j})^2+(e_{m+1,\,j} - e_{m,\,j})^2\,/\,\beta_4^2) \\ \end{align}$$

Now, substitute all the inequalities into the expression.

$$\begin{align} & \frac{7}{6}\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2-\frac{1}{6}\sum_{i = 2}^{m-1}(e_{i+1,\,j} - e_{i,\,j})(e_{i,\,j} - e_{i-1,\,j})-\frac{1}{3}\,(e_{1,\,j} - e_{0,\,j})^2 \\ & -\frac{1}{3}\,(e_{m+1,\,j} - e_{m,\,j})^2+\frac{1}{12}\,(e_{4,\,j} - e_{3,\,j})\,(e_{1,\,j} - e_{0,\,j}) \\ & -\frac{1}{3}\,(e_{3,\,j} - e_{2,\,j})\,(e_{1,\,j} - e_{0,\,j})+\frac{1}{3}\,(e_{2,\,j} - e_{1,\,j})\,(e_{1,\,j} - e_{0,\,j}) \\ & +\frac{1}{3}\,(e_{m,\,j} - e_{m-1,\,j})\,(e_{m+1,\,j} - e_{m,\,j}) \; -\frac{1}{3}\,(e_{m-1,\,j} - e_{m-2,\,j})\,(e_{m+1,\,j} - e_{m,\,j}) \\ \; & +\frac{1}{12}\,(e_{m-2,\,j} - e_{m-3,\,j})\,(e_{m+1,\,j} - e_{m,\,j}) \\ \end{align}$$

$$\begin{align} & \frac{7}{6}\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2-\frac{1}{6}\,\big ( \;\sum_{i = 2}^{m-2}(e_{i+1,\,j} - e_{i,\,j})^2\;-\;\frac{1}{2}\,(e_{3,\,j} - e_{2,\,j})^2\; \\ & -\;\frac{1}{2}\,(e_{m-1,\,j} - e_{m-2,\,j})^2 +\;\frac{1}{2}\,(\alpha_1^2\,(e_{3,\,j} - e_{2,\,j})^2+(e_{2,\,j} - e_{1,\,j})^2\,/\,\alpha_1^2) \\ & +\;\frac{1}{2}\,(\beta_1^2\,(e_{m-1,\,j} - e_{m-2,\,j})^2+(e_{m,\,j} - e_{m-1,\,j})^2\,/\,\beta_1^2)\; \big ) \\ & -\frac{1}{3}\,(e_{1,\,j} - e_{0,\,j})^2 -\frac{1}{3}\,(e_{m+1,\,j} - e_{m,\,j})^2 \\ & - \frac{1}{12}\,\big ( \frac{1}{2}\,(\alpha_2^2\,(e_{4,\,j} - e_{3,\,j})^2+(e_{1,\,j} - e_{0,\,j})^2\,/\,\alpha_2^2)\big ) \\ & - \frac{1}{3}\,\big ( \frac{1}{2}\,(\alpha_3^2\,(e_{3,\,j} - e_{2,\,j})^2+(e_{1,\,j} - e_{0,\,j})^2\,/\,\alpha_3^2)\big ) \\ & - \frac{1}{3}\,\big ( \frac{1}{2}\,(\alpha_4^2\,(e_{2,\,j} - e_{1,\,j})^2+(e_{1,\,j} - e_{0,\,j})^2\,/\,\alpha_4^2)\big ) \\ & - \frac{1}{3}\,\big ( \frac{1}{2}\,(\beta_2^2\,(e_{m,\,j} - e_{m-1,\,j})^2+(e_{m+1,\,j} - e_{m,\,j})^2\,/\,\beta_2^2)\big ) \\ & - \frac{1}{3}\,\big ( \frac{1}{2}\,(\beta_3^2\,(e_{m-1,\,j} - e_{m-2,\,j})^2+(e_{m+1,\,j} - e_{m,\,j})^2\,/\,\beta_3^2)\big ) \\ & - \frac{1}{12}\,\big ( \frac{1}{2}\,(\beta_4^2\,(e_{m-2,\,j} - e_{m-3,\,j})^2+(e_{m+1,\,j} - e_{m,\,j})^2\,/\,\beta_4^2)\big ) \\ \end{align}$$

$$\begin{align} & =\;\frac{7}{6}\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2-\frac{1}{6} \;\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2 \\ & \;-\;\big (\,\frac{1}{6}\,+\, (\frac{1}{24})\,/\,\alpha_2^2\,+\, (\frac{1}{6})\,/\,\alpha_3^2\,+\, (\frac{1}{6})\,/\,\alpha_4^2\,\big )\,(e_{1,\,j} - e_{0,\,j})^2 \\ & \;-\;\big (\,-\frac{1}{6}\,+\,(\frac{1}{12})\,/\,\alpha_1^2\,+\, \frac{1}{6}\,\alpha_4^2\,\big )\,(e_{2,\,j} - e_{1,\,j})^2 \\ & \;-\;\big (\,-\frac{1}{12}\,+\, \frac{1}{12}\,\alpha_1^2\,+\, \frac{1}{6}\,\alpha_3^2\,\big )\,(e_{3,\,j} - e_{2,\,j})^2 \\ & \;-\;\big (\,\frac{1}{24}\,\alpha_2^2\,\big )\,(e_{4,\,j} - e_{3,\,j})^2 \\ & \;-\;\big (\,\frac{1}{6}\,+\, (\frac{1}{6})\,/\,\beta_2^2\,+\, (\frac{1}{6})\,/\,\beta_3^2\,+\, (\frac{1}{24})\,/\,\beta_4^2\,\big )\,(e_{m+1,\,j} - e_{m,\,j})^2 \\ & \;-\;\big (\,-\frac{1}{6}\,+\,(\frac{1}{12})\,/\,\beta_1^2\,+\,\frac{1}{6}\,\beta_2^2\,\big )\,(e_{m,\,j} - e_{m-1,\,j})^2 \\ & \;-\;\big (-\frac{1}{12}\,+\,\frac{1}{12}\,\beta_1^2\,+\,\frac{1}{6}\,\beta_3^2\,\big )\,(e_{m-1,\,j} - e_{m-2,\,j})^2 \\ & \;-\;\big (\,\frac{1}{24}\,\beta_4^2\,\big )\,(e_{m-2,\,j} - e_{m-3,\,j})^2 \\ \end{align}$$ The choice $$\begin{align} & \alpha_1^2 = \beta_1^2 = (\sqrt 5-1)\,/\,2\,, \alpha_2^2 = \beta_4^2 = 8\,, \\ & \alpha_3^2 = \alpha_4^2 = \beta_2^2 = \beta_3^2 = (11-\sqrt 5)\,/\,4\,, \\ \end{align}$$ 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 $$\sum_{i = 1}^m e_{i,\,j}\,(-h^2\,({\partial_h^2\over\partial_h x^2})_{i,\,j}\,e)\;\ge\;\frac{2}{3}\sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2$$. and $$\sum_{j = 1}^n \sum_{i = 1}^m e_{i,\,j}\,(-\,({\partial_h^2\over\partial_h x^2})_{i,\,j}\,e)\;\ge\;\frac{2}{3}\,h^{-2}\,\sum_{j = 1}^n \sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2$$. Reasoning in the exact same manner for the dimension in $$y$$ $$\sum_{j = 1}^n e_{i,\,j}\,(-k^2\,({\partial_k^2\over\partial_k y^2})_{i,\,j}\,e)\;\ge\;\frac{2}{3}\sum_{j = 0}^n(e_{i,\,j+1} - e_{i,\,j})^2$$. and $$\sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\,({\partial_k^2\over\partial_k y^2})_{i,\,j}\,e)\;\ge\;\frac{2}{3}\,k^{-2}\,\sum_{i = 1}^m \sum_{j = 0}^n(e_{i,\,j+1} - e_{i,\,j})^2$$. Applying $$ \sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e) \; = $$ $$ \sum_{j = 1}^n \sum_{i = 1}^m e_{i,\,j}\,(-({\partial_h^2\over\partial_h x^2})_{i,\,j}\,e) \; + \; \sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-({\partial_k^2\over\partial_k y^2})_{i,\,j}\,e) $$ leads to the inequality $$ \sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e) $$ $$\ge\;\frac{2}{3}\,h^{-2}\,\sum_{j = 1}^n \sum_{i = 0}^m(e_{i+1,\,j} - e_{i,\,j})^2\;+\;\frac{2}{3}\,k^{-2}\,\sum_{i = 1}^m \sum_{j = 0}^n(e_{i,\,j+1} - e_{i,\,j})^2$$.