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Odd reflection at a Dirichlet endpoint

Example

Assume the Axiom of Countable Choice. Let c>0, let u0∈Cc2(R) and u1∈Cc1(R) be odd — equivalently, data on the half-line x>0 extended oddly — and let U be the d'Alembert solution of the whole-line problem with data (u0,u1) (d'Alembert's formula and uniqueness in one dimension). Then:

(i) U(⋅,t) is odd for every t, so u:=U∣x>0 solves the Dirichlet half-line problem utt=c2uxx on x>0 with u(0,t)=0 and data u0∣x>0, u1∣x>0;

(ii) for data obtained by oddly extending u0=ϕ, u1=cϕ′ from x>0, where ϕ∈Cc2((0,∞)), the reflected part re-enters with reversed sign:

u(x,t)=ϕ(x+ct)−ϕ(ct−x)(0<x<ct);

(iii) the half-line energy E(0,∞)(t)=12∫0∞(ut2+c2ux2) dx equals half the whole-line energy of U and is constant in t (Ivrii's Dirichlet case of the half-line energy problem).

Facts & Assumptions

Given: ACω; c>0; odd compactly supported data u0∈Cc2(R), u1∈Cc1(R); the d'Alembert solution U of the whole-line problem, and for part (ii) a fixed ϕ∈Cc2((0,∞)) with u0=ϕ, u1=cϕ′ on x>0.

[F1]

D'Alembert's formula: for U0∈C2(R), U1∈C1(R) the whole-line solution is U(x,t)=12[U0(x+ct)+U0(x−ct)]+12c∫x−ctx+ctU1(s) ds, a C2 classical solution of utt=c2uxx. (d'Alembert's formula and uniqueness in one dimension)

[F2]

Conservation in case (a): a homogeneous solution whose spatial support is contained in a fixed compact set throughout a time interval has constant total energy on that interval. (Conservation of total wave energy in three admissible settings)

[F3]

The energy density is e=12(Ut2+c2Ux2). By [F1] and the support definition, the spatial support of U(⋅,t) is contained in (supp⁡U0∪supp⁡U1)+[−ct,ct]. (Wave energy density, energy flux and total energy, The support of a function on Rn and its compactly supported Riemann integral)

Verification

1.1givenF1algebra

Oddness is preserved and the half-line problem is solved: if U0=u0 and U1=u1 are odd, then each term of [F1] is odd in x: for the first term, replacing x by −x interchanges the two arguments of the odd function U0 and changes the sign, and for the integral term the substitution s↦−s together with oddness of U1 reverses the orientation of the interval and the sign of the integrand, leaving the integral odd in x; hence U(⋅,t) is odd for every t, so U(0,t)=0; therefore u:=U∣x>0 is a C2 solution of utt=c2uxx on x>0 with trace u(0,t)=0 and the prescribed initial data u0∣x>0, u1∣x>0, which is (i).

2.1givenstep 1.1F1F4algebra

Reflection with reversed sign: take u0=ϕ and u1=cϕ′ with ϕ compactly supported in (0,∞), extended oddly, and 0<x<ct; in [F1] the first term is 12[ϕ(x+ct)−ϕ(ct−x)] because x−ct<0<x+ct and U0(−ξ)=−ϕ(ξ) for ξ>0; the integral term is 12c[cϕ(x+ct)−cϕ(ct−x)] by [F4] applied to the odd extension of cϕ′ on the two subintervals cut by 0; adding, u(x,t)=ϕ(x+ct)−ϕ(ct−x) as claimed; for x>ct the same computation gives u(x,t)=ϕ(x+ct), the incoming left-moving profile, so the second term is precisely the reflection.

3.1givenstep 1.1F1F2F3algebraF5∎

Half-line energy: for each t the density e(U)(x,t)=12(Ut2+c2Ux2) is even in x, because U(⋅,t) odd makes Ut(⋅,t) odd and Ux(⋅,t) even; hence ∫Re dx=2∫0∞e dx, that is, E(0,∞)(t)=12ER(t); fix T0>0; by [F3] the support of U(⋅,t) is contained in the fixed compact set K:=(supp⁡u0∪supp⁡u1)+[−cT0,cT0] for every t∈[0,T0], so [F2] makes ER constant on (0,T0); moreover ER(t)=∫Ke(U)(x,t) dx there and at the endpoints, and uniform continuity of e(U) on K×[0,T0] together with the finite measure of K makes this energy continuous on [0,T0], so the constancy extends to both endpoints; hence E(0,∞)=12ER is constant on [0,T0], and since T0 was arbitrary it is constant on [0,∞), which is (iii).

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