Statement of the Problem
edit
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.
discrete approximation
edit
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
.
finite difference operations defined
edit
For the grid vector
define the finite difference operations
by the following.
.
.
.
.
.
.
simulation of the problem
edit
To simulate the problem (1.0) let
Then solve the non-singular linear system
;
for the remaining
The error
,
satisfies
.
for
,
and
.
proof of truncation error estimates
edit
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
.