68 2. FOUR IMPORTANT LINEAR PDE
We lastly invoke the initial conditions in (3) to compute a and b. The
first initial condition in (3) gives
b(x) = g(x) (x ∈ R),
whereas the second initial condition and (5) imply
a(x) = v(x, 0) = ut(x, 0) − ux(x, 0) = h(x) − g (x) (x ∈ R).
Our substituting into (7) now yields
u(x, t) =
1
2
x+t
x−t
h(y) − g (y) dy + g(x + t).
Hence
(8) u(x, t) =
1
2
[g(x + t) + g(x − t)] +
1
2
x+t
x−t
h(y) dy (x ∈ R, t ≥ 0).
This is d’Alembert’s formula.
We have derived formula (8) assuming u is a (sufficiently smooth) solu-
tion of (3). We need to check that this really is a solution.
THEOREM 1 (Solution of wave equation, n = 1). Assume g ∈
C2(R),
h ∈
C1(R),
and define u by d’Alembert’s formula (8). Then
(i) u ∈
C2(R
× [0, ∞)),
(ii) utt − uxx = 0 in R × (0, ∞),
and
(iii) lim
(x,t)→(x0,0)
t0
u(x, t) =
g(x0),
lim
(x,t)→(x0,0)
t0
ut(x, t) =
h(x0)
for each point
x0
∈ R.
The proof is a straightforward calculation.
Remarks. (i) In view of (8), our solution u has the form
u(x, t) = F (x + t) + G(x − t)
for appropriate functions F and G. Conversely any function of this form
solves utt −uxx = 0. Hence the general solution of the one-dimensional wave
equation is a sum of the general solution of ut − ux = 0 and the general
solution of ut + ux = 0. This is a consequence of the factorization (4). See
Problem 19.
