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Energy uniqueness for the homogeneous heat equation

Statement

Assume ACω. Let n≥2, let Ω⊆Rn be a bounded C1 domain (or belong to the specified finite piecewise C1 class) of Bounded C1 domains and their outward normals, let T>0, and let u∈C2,1(Q‾) solve ut−Δu=0 in Q=Ω×(0,T] with u=0 on ∂Ω×[0,T] and u(⋅,0)=0 on Ω. Then u≡0 on Q‾; equivalently, the Dirichlet problem for the homogeneous heat equation is unique in this class by the energy method. No backward-in-time or terminal-data uniqueness is claimed here (the time-reversed function solves the backward heat equation, for which the energy is nondecreasing rather than nonincreasing).

Facts & Assumptions

Given: ACω, n≥2, a bounded C1 domain Ω in the Green-identity class, T>0, and u∈C2,1(Q‾) with ut−Δu=0 in Q, u=0 on ∂Ω×[0,T] and u(⋅,0)=0 on Ω.

[A1]

Countable Choice is the ambient hypothesis, carried by the Green identity supplier (The Axiom of Countable Choice (ACω)).

[F1]

Differentiation under the integral sign with an integrable majorant (Differentiation under the integral sign).

[F2]

Green's first identity: for real u∈C2(Ω‾) and v∈C1(Ω‾), ∫Ω(vΔu+Du⋅Dv) dx=∫∂Ωv∂νu dS (First Green identity).

[F4]

Δu=∑i∂i∂iu in the notation of The Laplacian of a C2 function and of a C2 vector field, and the domain class and boundary conventions are those of Bounded C1 domains and their outward normals.

[F6]

Dominated convergence (Dominated convergence).

[F7]

Every bounded open set has finite measure because it lies in a bounded box (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included). A nonnegative continuous function g with zero integral on an open set vanishes everywhere: if g(x0)>0, a small nondegenerate box B inside that set has g≥g(x0)/2 on B, so ∫g≥(g(x0)/2)λn(B)>0, a contradiction by the same box-volume formula.

Proof

Given: ACω, a bounded C1 domain Ω⊂Rn with n≥2, T>0, and u∈C2,1(Q‾) with ut−Δu=0 in Q, u=0 on ∂Ω×[0,T], u(⋅,0)=0 on Ω.

1.1A1F1F5F6given

Define E(t):=12∫Ωu(x,t)2 dx for t∈[0,T]. The maps u and ut are continuous on the compact cylinder [F5], hence bounded by constants M,M1<∞, so [F1] applies with the constant majorant 2MM1 and gives E′(t)=∫Ωu(x,t)ut(x,t) dx for every t∈(0,T); moreover E is continuous on [0,T], since for tk→t the integrands u(⋅,tk)2 converge pointwise to u(⋅,t)2 (continuity of u on Q‾) and are dominated by 4M2, so [F6] gives E(tk)→E(t).

2.1step 1.1F2F3F4given

Substituting ut=Δu into step 1.1 gives E′(t)=∫ΩuΔu dx; [F2] with v=u reads ∫Ω(uΔu+∣Du∣2) dx=∫∂Ωu ∂νu dS, and the boundary term vanishes because u=0 on ∂Ω×[0,T], so E′(t)=−∫Ω∣Du(x,t)∣2dx≤0; hence E is nonincreasing on (0,T) by [F3], and since E(0)=12∫Ωu(x,0)2dx=0 by the zero initial datum, continuity from step 1.1 gives E(t)≤0 and hence E≡0 on [0,T] because E≥0.

3.1step 2.1F7given

For every t, E(t)=0 is the integral of the nonnegative continuous function u(⋅,t)2, so [F7] gives u(x,t)=0 for every x∈Ω; by continuity of u on Q‾ this gives u≡0 on Q‾.

4.1step 3.1given∎

If u1,u2 are two solutions in this class with the same zero initial and lateral data, their difference w:=u1−u2 again satisfies wt−Δw=0 in Q, w=0 on ∂Ω×[0,T] and w(⋅,0)=0 on Ω, so step 3.1 gives w≡0; the Dirichlet problem for the homogeneous heat equation is therefore unique in this class. This is exactly the forward-in-time direction: the argument uses u(⋅,0)=0 and deduces vanishing for larger times, and no terminal data or backward uniqueness is used.

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