Examples from Section 18.2
Math 421:03
Date: 6 November 2002
Proof.
[Solution]We will find the solution as
, where each
is
harmonic (just a fancy word for
). We will force
to be zero on the entire boundary except the ``North'' edge, where
for all
. Likewise,
will be zero all around the boundary except the ``South'' edge, where
as well.
The solution for is then
where
As for
,
where
So we have
See Figure
1 for the graph of
.
Figure 1:
from Example 1 ()

Let's see how we can use this to solve the heat equation with
inhomogenous boundary conditions. We will solve the boundary value problem
So our rectangular region has zeros all around it except at the top. We
know that we can solve the PDE part by writing
, where
,
. The boundary
conditions can be satisfied if

(1) 
and
That is, all the nonzero boundary stuff is given to . We can satisfy
the initial conditions if we require

(2) 
Now is easy; it's just from Example 1.

(3) 
where the are just the Fourier sine series of :
As for , we know that the general solution has to be
where to fit the inital conditions (2),
Because of the series expression for , we can simplify:
The integral should be familar; it is nonzero if and only if , in which case it is . So
So we are interested in the Fourier sine coefficients on of the
function
. They are
Thus
Putting this back into the general solution, we have
Putting it all together,
Example 2
Solve the heat equation with inhomogenous boundary conditions as those
given in Example 1.
Proof.
[Solution]We will have to
, and compute each
separately as in
the previous development. If we let
we have
As for
, the general solution is now of the form
where this time
So putting it all together we have
See Figure
2 for some graphs.
Figure 2:
Snapshots of the graph of from Example 2.
We have set
again. You can see that the homogenous part
approaches zero and the steady state solution is .

Matthew Leingang
20021108