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Z x+ct
1
+ g(ξ) dξ, x < ct
2c
Z x+ct
1 1
u(x, t) = (−f(−x + ct) + f(x + ct)) + g(ξ) dξ + h(t − x/c), x < ct
2 2c −x+ct
Finally, we collect the solutions in the two domains.

( R
1 (f(x − ct) + f(x + ct)) + 1 x+ct g(ξ) dξ, x > ct
u(x, t) = 2 2c x−ct
R
1 (−f(−x + ct) + f(x + ct)) + 1 x+ct g(ξ) dξ + h(t − x/c), x < ct
2 2c −x+ct
6.5 The Wave Equation for a Finite Domain

Consider the wave equation for the infinite domain.

2
u tt = c u xx , −∞ < x < ∞, t > 0
u(x, 0) = f(x), u t (x, 0) = g(x)

If f(x) and g(x) are odd about x = 0, (f(x) = −f(−x), g(x) = −g(−x)), then u(x, t)
is also odd about x = 0. We can demonstrate this with D’Alembert’s solution.
Z x+ct
1 1
u(x, t) = (f(x − ct) + f(x + ct)) + g(ξ) dξ
2 2c x−ct

Z −x+ct
1 1
−u(−x, t) = − (f(−x − ct) + f(−x + ct)) − g(ξ) dξ
2 2c −x−ct
Z x−ct
1 1
= (f(x + ct) + f(x − ct)) − g(−ξ) (−dξ)
2 2c x+ct
Z x+ct
1 1
= (f(x − ct) + f(x + ct)) + g(ξ) dξ
2 2c x−ct
= u(x, t)

Thus if the initial conditions f(x) and g(x) are odd about a point then the solution of
the wave equation u(x, t) is also odd about that point. The analogous result holds if
the initial conditions are even about a point. These results are useful in solving the wave
equation on a finite domain.
Consider a string of length L with fixed ends.

2
u tt = c u xx , 0 < x < L, t > 0
u(x, 0) = f(x), u t (x, 0) = g(x), u(0, t) = u(L, t) = 0

We extend the domain of the problem to x ∈ (−∞ . . . ∞). We form the odd periodic
˜
extensions f and ˜g which are odd about the points x = 0, L.

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