1.6. The law of reflection 31 Proposition 1.6.3. i. If u C∞(R1+d) 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 C∞({x1 0}) is a smooth solution of u = 0 satisfying (1.6.1), define ˜ to be the odd extension of u to R1+d. Then ˜ is a smooth odd solution of ˜ = 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, ∂2nu ∂2nx1 x1=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 ∂2u j with j 1 vanish at x1 = 0. The equation u = 0 implies ∂2u ∂x2 1 = ∂2u ∂t2 d j=2 ∂2u ∂x2 j . The right-hand side vanishes on {x1 = 0} proving the case n = 1. If the case n 1 is known, consider v := ∂2nu. 1 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 ˜ to belong to C∞(R1+d). The equation ˜ = 0 for x1 0 follows from the equation in x1 0 since ˜ is odd. Example 1.6.1. Suppose that d = 1 and that f C∞(]−∞, 0 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}, ˜ = 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 ˜ to x 0.
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