34.3. NONHOMOGENEOUS PROBLEMS 679

and it is assumed that one can obtain a valid Fourier series expansion in terms of theseeigenfunctions of all the functions of interest. Note how, for the sake of simplicity, it isassumed that ∫ L

0y2

n (x)dx = 1

You multiply by an appropriate constant to make it this way. Thus, if the eigenfunctionsare multiples of sin

( nπ

L x), you choose the multiple to satisfy the above equation. Let

f (x, t) =∞

∑n=0

fn (t)yn (x)

Thus it is desired to have

∞

∑n=0

b′n (t)yn (x) =−a∞

∑n=0

λ nbn (t)yn (x)+∞

∑n=0

fn (t)yn (x)

and this is achieved ifb′n (t) =−aλ nbn (t)+ fn (t)

which is a familiar equation, the solution being

bn (t) = e−aλ ntbn (0)+∫ t

0e−aλ n(t−s) fn (s)ds

Then the solution is

u(x, t) =∞

∑n=0

(e−aλ ntbn (0)+

∫ t

0e−aλ n(t−s) fn (s)ds

)yn (x)

where bn (0) needs to be chosen to satisfy the initial condition. Thus it is required that

bn (0) =∫ L

0g(u)yn (u)du

In what was done earlier, yn was typically something like (2/L)1/2 sin( nπx

L

). Then the

solution is

u(x, t) =∞

∑n=0

(e−aλ nt

(∫ L

0g(u)yn (u)du

)+∫ t

0e−aλ n(t−s) fn (s)ds

)yn (x)

Example 34.3.1 Find the solution to

ut (x, t) = uxx (x, t)+ f (x, t)

u(0, t) = 0 = u(2, t)u(x,0) = x

where f (x, t) = xt.

First find the eigenfunctions and eigenvalues for the equation

y′′+λy = 0,y(0) = 0 = y(2)