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Therefore, the Fourier coefficients of f with respect to the system of eigenfunctions are
L Z L
R
0 f(x)dx 2
L 1dx L
a 0 = 2 R = f(x)dx, (8.35)
0 0
R
L cos(πmx/L)f(x)dx 2 Z L πmx
a m = 0 = cos dx, m = 1, 2, . . . (8.36)
R
L 2 L L
cos (πmx/L)dx 0
0
The Fourier coefficients ea n of g can be computed similarly. Substituting t = 0 into (8.30),
and assuming that the corresponding series converges uniformly, we obtain
∞ ∞
X πnx X πnx
A 0 a 0
u(x, 0) = + A n cos = f(x) = + a n cos .
2 L 2 L
n=1 n=1
Recall that the (generalized) Fourier coefficients are uniquely determined, and hence A n =
a n for all n ≥ 0. In order to compute B n , we differentiate (8.30) formally (term-by-term)
with respect to t and then substitute t = 0. We have

∞ ∞
X cπn πnx X πnx
B 0 ba 0
u t (x, 0) = + B n cos = g(x) + + ea n cos .
2 L L 2 L
n=1 n=1
Therefore, B n = ea n L/cπn for all n ≥ 1. Similarly, B 0 = ea 0 . Thus, the problem is
formally solved.
In (8.30) we have instead a (nondecaying) trigonometric factor. This is related to
the fact that hyperbolic equations preserve the singularities of the given data since the
rate of the decay of the generalized Fourier coefficients to zero usually depends on the
smoothness of the given function (under the assumption that this function satisfies the
prescribed boundary conditions). The precise decay rate of the Fourier coefficients is
provided for the classical Fourier system by the general theory of Fourier analysis.

Example 8.1 Solve the problem

u tt − 4u xx = 0, 0 < x < 1, t > 0, (8.37)
u x (0, t) = u x (1, t) = 0, t ≥ 0,
2
u(x, 0) = f(x) = cos πx, 0 ≤ x ≤ 1,
2
u t (x, 0) = g(x) = sin πx cos πx, 0 ≤ x ≤ 1.

The solution of (8.37) was shown to have the form

∞
A 0 + B 0 t X
u(x, t) = + (A n cos 2πnt + B n sin 2πnt) cos πnx. (8.38)
2
n=1
Substituting f into (8.38) implies
∞
X
A 0 2
u(x, 0) = + A n cos πnx = cos πx. (8.39)
2
n=1

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