Example 1 Find
a solution to the following partial differential equation.
Solution
One of the main differences here that we’re going to have
to deal with is the fact that we’ve now got two initial conditions. That is not something we’ve seen to this
point, but will not be all that difficult to deal with when the time rolls
around.
We’ve already done the separation of variables for this
problem, but let’s go ahead and redo it here so we can say we’ve got another
problem almost completely worked out.
So, let’s start off with the product solution.
Plugging this
into the two boundary conditions gives,
Plugging the
product solution into the differential equation, separating and introducing a
separation constant gives,
We moved the to the left side for convenience and chose for the separation constant so the
differential equation for would match a known (and solved) case.
The two
ordinary differential equations we get from separation of variables are then,
We solved the
boundary value problem above in Example 1 of the Solving
the Heat Equation section of this chapter and so the eigenvalues and
eigenfunctions for this problem are,
The first
ordinary differential equation is now,
and because the
coefficient of the h is clearly
positive the solution to this is,
Because there
is no reason to think that either of the coefficients above are zero we then
get two product solutions,
The solution is
then,
Now, in order
to apply the second initial condition we’ll need to differentiate this with
respect to t so,
If we now apply
the initial conditions we get,
Both of these
are Fourier sine series. The first is
for on while the second is for on with a slightly messy coefficient. As in the last few sections we’re faced
with the choice of either using the orthogonality of the sines to derive
formulas for and or we could reuse formula from previous
work.
It’s easier to
reuse formulas so using the formulas form the Fourier
sine series section we get,
Upon solving
the second one we get,
