34.2. HEAT AND WAVE EQUATIONS 671

In one dimension, this reduces to

ut = α2uxx

and this is the equation in what follows. There are other issues besides the equation toconsider.

You have a rod of length L. The heat equation for the temperature u in the rod is

ut = α2uxx

In addition to this, there are boundary conditions given on u at the ends of the rod. Forexample, you could have

u(0, t) = u(L, t) = 0

and there is also an initial temperature given

u(x,0) = f (x)

Then the idea is to find the unknown function u(t,x) . Here t is time and x is the coordinateof a point on the rod. The constant α2 varies from material to material. It is different foriron than for aluminum for example. Here you have x ∈ [0,L] and t > 0.

This is a rectangular shape and so it is reasonable to look for a nonzero solution to theabove partial differential equation and boundary condition in the form

u(x, t) = a(t)b (x)

Thena ′ (t)b (x) = α

2a (t)b ′′ (x)

One can separate the variables as follows.

a′ (t)α2a(t)

=b′′ (x)b(x)

(34.1)

Both sides must equal to some constant c since otherwise they could not be equal. One wayto see this is to differentiate both sides with respect to t. Then(

a′ (t)α2a(t)

)′= 0 and so

a′ (t)α2a(t)

= c,

a constant. Consider the side involving x.

b ′′ (x)− cb(x) = 0, b(0) = b(L) = 0

Of course you can’t have b(x) = 0 since if it were 0, you would have u(x, t) = 0. Therefore,from Example 33.2.1, −c = n2π2

L2 where n is a positive integer and

b(x) = sin(nπx

L

)Of course there is such a function for each n a positive integer. Having picked such apositive integer, 34.1 now forces a(t) to satisfy the equation

a′ (t)+n2π2α2

L2 a(t) = 0