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The only way a function of x can be identically equal to a function of t is if both of the
functions are equal to the same constant; therefore similarly, as for the wave equation, we
can write
0
00
T (t) X (x)
= = −λ. (9.4)
2
c T(t) X(x)
The reason for using a negative constant is that it will give us a Sturm-Liouville
equation to solve for X(x) that looks like one we have already seen. The reason for calling
the constant λ is that λ turns out to be an eigenvalue and eigenvalues are conventionally
called λ.
Equation (9.4) can be written in the form of two ordinary differential equations:
0 2
T (t) + λc T(t) = 0
00
X (x) + λX(x) = 0

To obtain boundary conditions on X(x), we use the given boundary conditions (9.2) on
u(x, t), that is, since u(x, t) = X(x)T(t)

0 = u(0, t) ≡ X(0)T(t)
0 = u(l, t) ≡ X(l)T(t)

has to be true for all t > 0. The only way this can be true, without making T(t) identically
zero, is to require
X(0) = X(L) = 0.

This leads to a Sturm-Liouville problem for X of the following form:
00
X + λX = 0, X(0) = X(l) = 0.

This is one of the problems we solved in the previous class, and it was shown there that
there exists an infinite set of solutions of the form
πnx
X n (x) = A n sin (9.5)
l
corresponding to the eigenvalues

2 2
π n
λ n = , n = 1, 2, . . . .
l 2
To find the function T n (t) corresponding to X n (x), it is necessary to solve the first-order
ODE
2 2 2
π n c
0 2
T (t) = −λ n c T n (t) = − T n (t).
n
l 2
This is a separable first-order equation with general solution
2 2 2
π n c
T n (t) = a n e − l 2 t . (9.6)
We now have an infinite family of solutions of our heat equation; namely

2 2 2
πnx π n c
u n (x, t) = X n (x)T n (t) = b n sin e − l 2 t ,
l

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