View source for Using High Order Finite Differences/Third Order Method

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==A Third Order Accurate Method==

===Statement of the Problem===

Let <b><i>D</i></b> be the <i>rectangle</i>

<math>D = \{(x, y): 0 < x < a,\ 0 < y < b\}</math>

and let <b><i>C</i></b> be the <i>boundary</i> of <b><i>D</i></b>.<br>

The <i>operator</i>

<math>\Delta u(x, y) = {\partial^2\over\partial x^2}u(x, y) + {\partial^2\over\partial y^2}u(x, y)</math>

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

<math> \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)}</math>

is called a <big><b><i>Poisson</i></b></big> problem.<br><br>

===discrete approximation===

To approximate u(x, y) numerically, use the <i>grid</i>

<math>(x_i, y_j),\ 0 \le i \le m+1,\ 0 \le j \le n+1</math>.<br>
with<br><math>x_i = i\,h,\,y_i = j\ k</math><br>
and<br><math>h = a\,/\,(m+1), k = b\,/\,(n+1)</math><br><br>

The <i>second partial derivative</i><br>
<math>{\partial^2\over\partial x^2}\ u(x, y)</math><br>
can be approximated on the grid by <i>difference quotients</i><br>
<math>({\partial_h^2\over\partial_h x^2})\ u(x_i, y_j)</math>.<br><br>
These <i>difference quotients</i> are given by<br>

<math>({\partial_h^2\over\partial_h x^2})\ u(x_1, y_j)\ =</math>.<br><br>
<math>{-\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}</math>

<math>({\partial_h^2\over\partial_h x^2})\ u(x_i, y_j)\ =</math>.<br><br>
<math>{-\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}</math>

<math>\text{for}\ i\,=\,2,\ 3,\ \ldots\, m-1  \quad \text{and}</math>

<math>({\partial_h^2\over\partial_h x^2})\ u(x_m, y_j)\ =</math>.<br><br>
<math>{\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}</math>


The <i>second partial derivative</i><br>
<math>{\partial^2\over\partial y^2}\ u(x, y)</math><br>
can be approximated on the grid by <i>difference quotients</i><br>
<math>({\partial_k^2\over\partial_k y^2})\ u(x_i, y_j)</math>.<br><br>
These <i>difference quotients</i> are given by<br>

<math>({\partial_k^2\over\partial_k y^2})\ u(x_i, y_1)\ =</math>.<br><br>
<math>{-\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}</math>

<math>({\partial_k^2\over\partial_k y^2})\ u(x_i, y_j)\ =</math>.<br><br>
<math>{-\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}</math>

<math>\text{for}\ j\,=\,2,\ 3,\ \ldots\, n-1  \quad \text{and}</math>

<math>({\partial_k^2\over\partial_k y^2})\ u(x_i, y_n)\ =</math>.<br><br>
<math>{\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}</math>

===truncation errors===

<br>The <i>difference quotients</i>&nbsp;&nbsp;
<math>({\partial_h^2\over\partial_h x^2})\ u(x_i, y_j)</math>
&nbsp;&nbsp;are <i>third order</i> accurate with <i>truncation</i> errors:<br>

<math>\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}</math>

with<br>

<math>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)</math>&nbsp;,

for some&nbsp;&nbsp;&nbsp; <math>0 < \phi{_1^{(1,\,j)}} < 2\,, \quad
-1 < \phi{_2^{(1,\,j)}} < 3</math>&nbsp;,<br>

<math>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)</math>&nbsp;,

for some&nbsp;&nbsp;&nbsp; <math>-2 < \phi{_1^{(i,\,j)}} < 1\,, \quad
-1 < \phi{_2^{(i,\,j)}} < 2\,,</math><br><br>
<math>\text{for}\ \ i\ =\ 2,\ 3,\ \ldots\,,\ m-1.</math>
&nbsp;&nbsp;&nbsp;&nbsp;<br>

When&nbsp; <math>{\partial^6\over\partial x^6}\,u(x,\,y)</math> &nbsp; is
continuous, these estimates also hold.<br><br>
<math>\tau_x(x_i,\,y_j)\;=\;{h^4\over\ 720}M_x^6(x_i,\,y_j),</math>
&nbsp;&nbsp;with<br><br>
<math>
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)
</math><br><br>
for some&nbsp;&nbsp;&nbsp; <math>-1\;<\;\phi{_1^{(i,\,j)}}\;<\;1\,,\quad -2\;<\;\phi{_2^{(i,\,j)}}\;<\;2,</math> &nbsp;&nbsp;&nbsp;<br><br>
<math>\text{for}\ \ i\ =\ 2,\ 3,\ \ldots\,,\ m-1.</math><br><br>
The case for&nbsp; <math>i\;=\;m</math> &nbsp;is<br><br>

<math>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)</math>&nbsp;,

for some&nbsp;&nbsp;&nbsp; <math>-3 < \phi{_1^{(m,\,j)}} < 1\,, \quad
-2 < \phi{_2^{(m,\,j)}} < 0</math>&nbsp;.<br>

<br>The <i>difference quotients</i>&nbsp;&nbsp;
<math>({\partial_k^2\over\partial_k y^2})\ u(x_i, y_j)</math>
&nbsp;&nbsp;are <i>third order</i> accurate with <i>truncation</i> errors:<br>

<math>\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}</math>

with<br>

<math>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)</math>&nbsp;,

for some&nbsp;&nbsp;&nbsp; <math>0 < \psi{_1^{(i,\,1)}} < 2\,, \quad
-1 < \psi{_2^{(i,\,1)}} < 3</math>&nbsp;,<br>

<math>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)</math>&nbsp;,

for some&nbsp;&nbsp;&nbsp; <math>-2 < \psi{_1^{(i,\,j)}} < 1\,, \quad
-1 < \psi{_2^{(i,\,j)}} < 2\,,</math><br><br>
<math>\text{for}\ \ j\ =\ 2,\ 3,\ \ldots\,,\ n-1.</math>
&nbsp;&nbsp;&nbsp;&nbsp;<br>

When&nbsp; <math>{\partial^6\over\partial y^6}\,u(x,\,y)</math> &nbsp; is
continuous, these estimates also hold.<br><br>
<math>\tau_y(x_i,\,y_j)\;=\;{k^4\over\ 720}M_y^6(x_i,\,y_j),</math>
&nbsp;&nbsp;with<br><br>
<math>
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)
</math><br><br>
for some&nbsp;&nbsp;&nbsp; <math>-1\;<\;\psi{_1^{(i,\,j)}}\;<\;1\,,\quad -2\;<\;\psi{_2^{(i,\,j)}}\;<\;2,</math> &nbsp;&nbsp;&nbsp;<br><br>
<math>\text{for}\ \ j\ =\ 2,\ 3,\ \ldots\,,\ n-1.</math><br><br>
The case for&nbsp; <math>j\;=\;n</math> &nbsp;is<br><br>

<math>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)</math>&nbsp;,

for some&nbsp;&nbsp;&nbsp; <math>-3 < \psi{_1^{(i,\,n)}} < 1\,, \quad
-2 < \psi{_2^{(i,\,n)}} < 0</math>&nbsp;.<br>

The <i>Laplacian</i> &nbsp;<math>\Delta u(x, y)</math>&nbsp; then can be
<i>approximated</i> on the <i>interior</i> of the <i>grid</i> by

<math>\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)</math>

The <i>truncation error</i><br><br>
<math>\tau (x_i, y_j) = \Delta_{h,\,k} u(x_i, y_j)-\Delta u(x_i, y_j)</math>
<br><br>is given by<br><br>
<math>\tau (x_i, y_j) = \tau_x (x_i, y_j)+\tau_y (x_i, y_j)</math>.

===finite difference operations defined===

<br>For the <i>grid vector</i><br><br>
<math>U = \{U_{i,\,j}\} \quad 0 \le i \le m+1, \ 0 \le j \le n+1</math>
<br><br>define the <i>finite difference operations</i><br><br>
<math>\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</math>
<br><br>by the following.<br><br>

<math>({\partial_h^2\over\partial_h x^2})_{1,\,j}\ U\ =</math>.<br><br>
<math>{-\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}</math>

<math>({\partial_h^2\over\partial_h x^2})_{i,\,j}\ U\ =</math>.<br><br>
<math>{-\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}</math>

<math>\text{for}\ i\,=\,2,\ 3,\ \ldots\, m-1  \quad \text{and}</math>

<math>({\partial_h^2\over\partial_h x^2})_{m,\,j}\ U\ =</math>.<br><br>
<math>{\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}</math>

<math>({\partial_k^2\over\partial_k y^2})_{i,\,1}\ U\ =</math>.<br><br>
<math>{-\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}</math>

<math>({\partial_k^2\over\partial_k y^2})_{i,\,j}\ U\ =</math>.<br><br>
<math>{-\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}</math>

<math>\text{for}\ j\,=\,2,\ 3,\ \ldots\, n-1  \quad \text{and}</math>

<math>({\partial_k^2\over\partial_k y^2})_{i,\,n}\ U\ =</math>.<br><br>
<math>{\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}</math>

===simulation of the problem===

<br>To <i>simulate</i> the problem&nbsp; <b>(1.0)</b> &nbsp; let<br>

<math>\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}</math>

<br>Then solve the <i>non-singular</i> <b>linear</b> system<br><br>
<math>-\Delta_{i,\,j}\,U = f(x_i,\,y_j) \quad 1 \le i \le m,
\ 1 \le j \le n</math>&nbsp; ;
<br><br>for the remaining&nbsp; <math>U_{i,\,j}</math>&nbsp;

===error estimation===

<br>The <i>error</i><br><br>
<math>U - u = \{U_{i,\,j} - u(x_i,\,y_j)\}</math>&nbsp;,<br>
<br>satisfies<br><br>
<math>{\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))</math>&nbsp;.<br>
<br>for<br><br>
<math>\tau = \{\tau (x_i,\,y_j)\}, \quad 1 \le i \le m,\ 1 \le j \le n</math>
&nbsp;,<br><br>and<br><br>
<math>\rho \ \le \ 1\, /\, (1 - (1 + \sqrt 10 )\,/\,12)</math>&nbsp;.<br>

===proof of truncation error estimates===

The <i>truncation error</i> estimates for&nbsp;
<math>\big ({\partial_h^2\over\partial_h x^2}\,u(x_i,\,y_j)\big )</math>
&nbsp;are done under the assumption that&nbsp; <math>u(x,\;y)</math> &nbsp;is
sufficiently smooth so that&nbsp;
<math>{\partial^5\over\partial x^5}\,u(x,\;y)</math> &nbsp;is continuous.
For notational convenience let<br><br><math>g(x)\;=\;u(x,\;y_j).</math><br><br>
Expand&nbsp; <math>g(x)</math> &nbsp;in it's <i>Taylor expansion</i> about&nbsp;
<math>x_i</math>,<br><br>
<math>\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}</math>.<br><br>
where&nbsp; <math>z</math> &nbsp;is some number between&nbsp; <math>x_i</math>
&nbsp;and&nbsp; <math>x_i+\Delta x</math>. Then<br><br>
<math>\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}</math><br><br>
<math>\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}</math><br><br>
<math>=\;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 )
</math>
<br><br>where<br><br>
<math>\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}</math>.<br><br>
Since<br><br>
<math>\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))},</math><br><br>
<math>\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))},</math><br><br>
from the <i>intermediate value property</i><br><br>
<math>\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\,,</math><br><br>
<math>\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\,.</math>.<br><br>
This gives<br><br>
<math>\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}</math><br><br>
which is<br><br><math>\tau_x(x_1,\,y_j)\;=\;{h^3\over\ 120}M_x^5(x_1,\,y_j).</math><br><br><br>



For <math>i\;=\;2,\;3,\;\ldots\,,\;m-1</math><br><br>
<math>\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}</math><br><br>
<math>\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}</math><br><br>
<math>=\;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 )
</math>
<br><br>where<br><br>
<math>\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}</math>.<br><br>
Reasoning as before, combining terms with like signs and using the <i>intermediate value property</i>,<br><br>
<math>\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\,,</math><br><br>
<math>\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\,.</math>.<br><br>
This gives<br><br>
<math>\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}</math><br><br>
which is<br><br><math>\tau_x(x_i,\,y_j)\;=\;{h^3\over\ 120}M_x^5(x_i,\,y_j).</math><br><br>

Under the assumption that&nbsp;
<math>{\partial^6\over\partial x^6}\,u(x,\;y)</math> &nbsp;is continuous, in the preceding argument, the expression<br><br>
<math>\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}</math>
<br><br>can be replaced by<br><br>
<math>\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}</math>

This gives<br><br>
<math>\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}</math><br><br>
with&nbsp;&nbsp;&nbsp; <math>-1\;<\;\phi_1\;<\;1\,,\quad -2\;<\;\phi_2\;<\;2</math> &nbsp;&nbsp;&nbsp;which is<br><br>
<math>\tau_x(x_i,\,y_j)\;=\;{h^4\over\ 720}M_x^6(x_i,\,y_j).</math><br><br>
The remaining <i>truncation error</i> estimates are done in the same way.<br><br>

===end working===
Let the <i>error</i><br><br>
<math>e = \{e_{i,\,j}\}, \quad 1 \le i \le m, \ 1 \le j \le n,</math>
<br><br>be defined by<br><br>
<math>e_{i,\,j} = U_{i,\,j} - u(x_i, y_j)</math>&nbsp;.<br><br>
<math>U</math> is the solution of the <i>finite difference scheme</i>
<b>(xx)</b> and <math>u(x_i,\,y_j)</math> is the solution to <b>(1.0)</b>.
<br><br>Since<br><br>
<math>\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}</math>&nbsp;<br>
we get that<br><br>
<math>
\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
</math>&nbsp;.<br><br>
Next it will be shown that the <i>operator</i>&nbsp;
<math>-\Delta_{i,\,j}</math>&nbsp;
is <b><i>positive definite</i></b> for&nbsp; <math>e</math>,&nbsp;
in particular that<br><br>
<math>
\sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e)
</math>&nbsp;<br><br style="line-spacing;50%;">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>
 \ge \;
 \lambda (4 a^{-2}\,{(m+1)}^2\,(something)+4 b^{-2}\,{(n+1)}^2\,())\,
{\lVert e \rVert}_2
</math>&nbsp;,<br><br>

with<br><br><math>\lambda \ \ge \ 1 - (1+\sqrt 10)\,/\,12</math>&nbsp;.<br>

<br>Begin with&nbsp;
<math>
\sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e) \; =
</math><br>
<math>
\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) \; =
</math><br>
<math>
\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)
</math><br><br>
The sum&nbsp;
<math>
\sum_{i = 1}^m e_{i,\,j}\,(-({\partial_h^2\over\partial_h x^2})_{i,\,j}\,e)</math>&nbsp; will be estimated first.<br>

===work===
<math>-h^2({\partial_h^2\over\partial_h x^2})_{1,\,j}\ e\ =</math>.<br><br>
<math>\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}</math>

<br><br>

<math>=\frac{7}{6}(-\frac{1}{1}\,e_{2,\,j}
+\frac{2}{1}\,e_{1,\,j}-\frac{1}{1}\,e_{0,\,j})
</math>
<br><br>

<math>\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}</math>


<br><br>

<math>\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}</math>

<math>-h^2({\partial_h^2\over\partial_h x^2})_{i,\,j}\ e\ =</math>.<br><br>
<math>\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}</math>
<br><br>


<math>=\frac{7}{6}\,(-\frac{1}{1}\,e_{i+1,\,j}
+\frac{2}{1}\,e_{i,\,j}-\frac{1}{1}\,e_{i-1,\,j})
</math>
<br><br>

<math>+\frac{1}{12}\,(e_{i+2,\,j}-e_{i+1,\,j})
-\frac{1}{12}\,(e_{i+1,\,j}
-e_{i,\,j})</math>

<math>+\frac{1}{12}\,(e_{i,\,j}-e_{i-1,\,j})
-\frac{1}{12}\,(e_{i-1,\,j}
-e_{i-2,\,j})</math>

<br><br><br><br>

<math>\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}</math>

<math>\text{for}\ i\,=\,2,\ 3,\ \ldots\, m-1  \quad \text{and}</math>

<math>-h^2({\partial_h^2\over\partial_h x^2})_{m,\,j}\ e\ =</math>.<br><br>
<math>-\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}</math>

<math>\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}</math>

<br><br>

<math>=\frac{7}{6}(-\frac{1}{1}\,e_{m+1,\,j}
+\frac{2}{1}\,e_{m,\,j}-\frac{1}{1}\,e_{m-1,\,j})
</math>
<br><br>

<math>\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}</math>

<br><br>

===end work===
<br>The &nbsp;<b><i>summation by parts formula</i></b>&nbsp; is now stated
so it can be used.<br><br>



<math>
\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)
</math><br>

<math>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}</math><br><br>
<math>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})
</math><br><br>
<math>
 =\; -\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})
</math><br><br>
<math>\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}</math><br><br>
<math>\text{for}\ i\,=\,1,\ 2,\ \ldots\, m-1 \quad \text{and}</math>.<br><br>
<math>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})</math><br><br>
<math>
 =\; -\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})
</math><br><br>

<math>\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})</math>
<math>
= 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})
</math><br><br>
Taking into account that&nbsp; <math>e_{m+1,\,j}\, = \, e_{0,\,j}\, = \, 0</math>
&nbsp;it follows<br><br>
<math>
\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})
</math><br><br>
<math>
=\; - 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})
</math><br><br>

<math>\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}</math><br><br>

===working here===

Collect like terms in the expression immediately above as follows.<br><br>
<math>\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</math><br><br>
<math>=\, \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</math><br><br>
<math>-\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})</math><br><br>
<math>\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}</math><br><br>

Now, rewrite the expression after making cancellations.<br><br>
<math>\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}</math><br><br>
The following simple inequality will be used to bound terms.<br><br>
<math>\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}</math><br>
and also<br><br>
<math>
a\,b \, \le \, \frac{1}{2}\,(\alpha^2\,a^2 + b^2\,/\,\alpha^2)
</math>.<br><br>
<math>\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}</math><br><br>

<math>\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}</math><br><br>
<math>\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}</math><br><br>
<math>\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}</math><br><br>
<math>\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}</math><br><br>
<math>\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}</math><br><br>

Now, substitute all the inequalities into the expression.<br><br>

<math>\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}</math><br><br>


<math>\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}</math><br><br>

<math>\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}</math><br><br>
The choice<br><br>
<math>\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}</math><br><br>
bounds all of the coefficients in the&nbsp; <math>\alpha_i \text{'s}</math>
&nbsp;and&nbsp; <math>\beta_i \text{'s}</math> &nbsp;by&nbsp; <math>\frac{1}{3}</math> &nbsp; and yields the long sought inequality<br><br>
<math>\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</math>.
<br><br>and<br><br>
<math>\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</math>.
<br><br>Reasoning in the exact same manner for the dimension in&nbsp; <math>y</math>
<br><br>
<math>\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</math>.
<br><br>and<br><br>
<math>\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</math>.
<br><br>Applying&nbsp;&nbsp;
<math>
\sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e) \; =
</math><br><br>
<math>
\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)
</math><br><br>leads to the inequality&nbsp;&nbsp;
<math>
\sum_{i = 1}^m \sum_{j = 1}^n e_{i,\,j}\,(-\Delta_{i,\,j}\,e)
</math><br><br>
<math>\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</math>.<br><br>

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