Let D be the rectangle
and let C be the boundary of D.
The operator
is the usual Laplacian. The problem, determine a function u(x, y)
such that
is called a Poisson problem.
To approximate u(x, y) numerically, use the grid
.
with

and

The second partial derivative

can be approximated on the grid by difference quotients
.
These difference quotients are given by
.
.
.
The second partial derivative

can be approximated on the grid by difference quotients
.
These difference quotients are given by
.
.
.
The difference quotients
are third order accurate with truncation errors:
with
,
for some
,
,
for some 
When
is
continuous, these estimates also hold.
with

for some

The case for
is
,
for some
.
The difference quotients
are third order accurate with truncation errors:
with
,
for some
,
,
for some 
When
is
continuous, these estimates also hold.
with

for some

The case for
is
,
for some
.
The Laplacian
then can be
approximated on the interior of the grid by
The truncation error
is given by
.
For the grid vector
define the finite difference operations
by the following.
.
.
.
.
.
.
To simulate the problem (1.0) let
Then solve the non-singular linear system
;
for the remaining
The error
,
satisfies
.
for
,
and
.
The truncation error estimates for
are done under the assumption that
is
sufficiently smooth so that
is continuous.
For notational convenience let

Expand
in it's Taylor expansion about
,
.
where
is some number between
and
. Then


where
.
Since


from the intermediate value property

.
This gives

which is

For 


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

.
This gives

which is

Under the assumption that
is continuous, in the preceding argument, the expression
can be replaced by
This gives

with
which is

The remaining truncation error estimates are done in the same way.
Let the error
be defined by
.
is the solution of the finite difference scheme
(xx) and
is the solution to (1.0).
Since
we get that
.
Next it will be shown that the operator
is positive definite for
,
in particular that
,
with
.
Begin with



The sum
will be estimated first.
.
.
.
The summation by parts formula is now stated
so it can be used.





.



Taking into account that
it follows



Collect like terms in the expression immediately above as follows.




Now, rewrite the expression after making cancellations.

The following simple inequality will be used to bound terms.

and also
.






Now, substitute all the inequalities into the expression.



The choice

bounds all of the coefficients in the
and
by
and yields the long sought inequality
.
and
.
Reasoning in the exact same manner for the dimension in
.
and
.
Applying


leads to the inequality

.