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
