2.4. WAVE EQUATION 65

Hence

(46) ¨(t)e(t) e ≥

(˙(t))2

e (0 ≤ t ≤ T ).

2. Now if e(t) = 0 for all 0 ≤ t ≤ T , we are done. Otherwise there exists

an interval [t1,t2] ⊂ [0,T ], with

(47) e(t) 0 for t1 ≤ t t2, e(t2) = 0.

3. Write

(48) f(t) := log e(t) (t1 ≤ t t2).

Then

¨(t)

f =

¨(t) e

e(t)

−

˙(t)2

e

e(t)2

≥ 0 by (46),

and so f is convex on the interval (t1,t2). Consequently if 0 τ 1,

t1 t t2, we have

f((1 − τ)t1 + τt) ≤ (1 − τ)f(t1) + τf(t).

Recalling (48), we deduce

e((1 − τ)t1 + τt) ≤

e(t1)1−τ e(t)τ

,

and so

0 ≤ e((1 − τ)t1 + τt2) ≤

e(t1)1−τ e(t2)τ

(0 τ 1).

But in view of (47) this inequality implies e(t) = 0 for all times t1 ≤ t ≤ t2,

a contradiction.

2.4. WAVE EQUATION

In this section we investigate the wave equation

(1) utt − Δu = 0

and the nonhomogeneous wave equation

(2) utt − Δu = f,

subject to appropriate initial and boundary conditions. Here t 0 and

x ∈ U, where U ⊂

Rn

is open. The unknown is u :

¯

U × [0, ∞) → R,

u = u(x, t), and the Laplacian Δ is taken with respect to the spatial variables