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9.2 Homogeneous Mixed Conditions

The method of separation of variables is a useful technique for finding special solutions
of partial differential equations. We can combine these special solutions to solve certain
1
problems. Consider the temperature of a one-dimensional rod of length h . The left
end is held at zero temperature, the right end is insulated and the initial temperature
distribution is known at time t = 0. To find the temperature we solve the problem:

2
∂u ∂ u
= κ , 0 < x < h, t > 0
∂t ∂x 2
u(0, t) = u x (h, t) = 0
u(x, 0) = f(x)

We look for special solutions of the form, u(x, t) = X(x)T(t). Substituting this into the
partial differential equation yields

00
0
X(x)T (t) = κX (x)T(t)
0
00
T (t) X (x)
=
κT(t) X(x)
Since the left side is only dependent on t, the right side in only dependent on x, and the
relation is valid for all t and x, both sides of the equation must be constant.

T 0 X 00
= = −λ
κT X

Here −λ is an arbitrary constant. (You’ll see later that this form is convenient.) u(x, t) =
X(x)T(t) will satisfy the partial differential equation if X(x) and T(t) satisfy the ordinary
differential equations,
0
00
T = −κλT and X = −λX.
Now we see how lucky we are that this problem happens to have homogeneous boundary
2
conditions . If the left boundary condition had been u(0, t) = 1, this would imply
X(0)T(t) = 1 which tells us nothing very useful about either X or T. However the
boundary condition u(0, t) = X(0)T(t) = 0, tells us that either X(0) = 0 or T(t) = 0.
Since the latter case would give us the trivial solution, we must have X(0) = 0. Likewise
0
by looking at the right boundary condition we obtain X (h) = 0.
We have a regular Sturm-Liouville problem for X(x).

00
0
X + λX = 0, X(0) = X (h) = 0
The eigenvalues and orthonormal eigenfunctions are

r
2
(2n − 1)π 2 (2n − 1)π
+
λ n = , X n = sin x , n ∈ Z .
2h h 2h
1
Why h? Because l looks like 1 and we use L to denote linear operators
2
Actually luck has nothing to do with it. I planned it that way.

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