1.6. The law of reflection 31
Proposition 1.6.3. i. If u
satisfies u = 0 and is odd in
x1, then the restriction of u to {x1 0} is a smooth solution of u = 0
satisfying the Dirichlet boundary condition (1.6.1).
ii. Conversely, if u
0}) is a smooth solution of u = 0
satisfying (1.6.1), define ˜ u to be the odd extension of u to
Then ˜ u is
a smooth odd solution of ˜ u = 0.
Proof. i. Setting x1 = 0 in the identity u(t, x1,x ) = −u(t, −x1,x ) shows
that (1.6.1) is satisfied.
ii. First prove by induction on n that
(1.6.3) ∀n 0,
= 0 .
The case n = 0 is (1.6.1).
Since the derivatives ∂t and ∂j for j 1 are parallel to the boundary
along which u = 0, it follows that utt and ∂j
with j 1 vanish at x1 = 0.
The equation u = 0 implies

The right-hand side vanishes on {x1 = 0} proving the case n = 1.
If the case n 1 is known, consider v :=
∂1nu. 2
It satisfies the wave
equation in x1 0 and, by the inductive hypothesis, satisfies the Dirichlet
boundary condition at x1 = 0. The case n = 1 applied to v proves the case
n + 1. This completes the proof of (1.6.3).
It is not hard to prove using Taylor’s theorem that (1.6.3) is a necessary
and sufficient for the odd extension ˜ u to belong to
The equation
˜ u = 0 for x1 0 follows from the equation in x1 0 since ˜ u is odd.
Example 1.6.1. Suppose that d = 1 and that f C0
0[) so that for
0 t small u = f(x t) is a solution of the wave equation supported to the
left of and approaching the boundary x1 = 0. To describe the continuation
as a solution satisfying the Dirichlet condition, use the method of images
as follows. The solution in {x 0} is the restriction to x 0 of an odd
solution of the wave equation. For t 0 the odd extension is equal to the
given function in x 0 and to minus its reflection in {x 0},
˜ u = f(x t) f(−x t) .
The formula on the right is the unique odd solution of the wave equation
that is equal to u in {t 0} {x 0}. The solution u is the restriction of
˜ u to x 0.
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