Partial Differential Equations

1.4 The Wave Equation 25
Proof. Idea: Integrate the square of the difference of two solutions over U.
We set
Let u and ũ be solutions of the described type. Then, w = u − ũ solves the boundary value problem
{
ut − ∆u = 0 on U × ]0, T [,
u = 0 on (U × {0}) ∪ (∂U × [0, T [).
∫
e(t) :=
U
w(x, t) 2 dx ∀t ∈ [0, T ].
Since w(· , t) vanishes for every t ∈ ]0, T [ on ∂U, integration by parts as given in formula (1.7) yields
∫
∫
∫
e ′ (t) = 2 w(x, t) w t (x, t)dx = 2 w(x, t) ∆w(x, t)dx = −2 |Dw(x)| 2 dx ≤ 0.
U
U
Since e(t) ≥ 0 for all t ∈ [0, T ], and since e(0) = 0 by assumption, this gives e(t) = 0 for all t ∈ [0, T ],
and therefore, w = 0 on U× ]0, T [.
U
Exercise 1.3.10. Provide an explicit solution of the following initial value problem:
{
ut − ∆u + cu = f on R n × ]0, ∞[,
u = g on R n × {0}.
Here, c ∈ R is a constant.
Exercise 1.3.11. For g : [0, ∞[ → R with g(0) = 0, derive the solution formula
u(x, t) =
√ x ∫ t
4π
0
1
(t − s) 3 2
e −x2
4(t−s) g(s) ds
for the initial and the boundary value problem
⎧
⎨ u t − u xx = 0 on [0, ∞[ × ]0, ∞[,
u = 0 on [0, ∞[ ×{0},
⎩
u = g on {0} × [0, ∞[.
(Hint: Set v(x, t) = u(x, t) − g(t), and extend v as an odd function in x onto R.)
1.4 The Wave Equation
The wave equation is the partial differential equation
u tt − ∆u = 0,
where u: U× ]0, ∞[ → R for an open subset U ⊆ R n and ∆u = ∑ n
j=1 ∂2 u
∂x j
2 . Usually, one wants to have
solutions with boundary values for t = 0, i.e. one seeks functions u: U × [0, ∞[ → R. One also considers
the inhomogeneous variant
u tt − ∆u = f
where f : U × [0, ∞[ → R is a given function. A commonly used abbreviation of u tt − ∆u is ✷u. One calls
✷ the d’Alembert operator.
The wave equation describes vibrations of elastic materials like strings or membranes. Here, u(x, t) is
interpreted as a displacement (from some “ground state”). The acceleration of a small volume element
V is given by d2
dt 2 ∫V u(x, t) dx = ∫ V u tt(x, t) dx. The force acting on V attaches to the boundary ∂V of