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.