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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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Weak parabolic maximum principle

Statement

Let Q=Ω×(0,T] be a parabolic cylinder (Parabolic cylinder and parabolic boundary) with Ω bounded, and let u∈C2,1(Q‾) satisfy ut−Δu≤0 in Q. Then max⁡Q‾u=max⁡∂pQu.

Facts & Assumptions

Given: A parabolic cylinder Q=Ω×(0,T] with Ω bounded and u∈C2,1(Q‾) with ut−Δu≤0 in Q.

[F1]

Q‾ is compact, ∂pQ is a closed subset of it, and the class C2,1(Q‾) is the cylinder convention of Parabolic cylinder and parabolic boundary.

[F2]

For every ε>0 the function v=u+ε∣x∣2 is a strict subsolution, and the maximum of any strict subsolution is attained on ∂pQ (Strict-subsolution perturbation for the heat operator); the Euclidean squares ∣x∣2 are those of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn.

Proof

Given: A bounded parabolic cylinder Q and u∈C2,1(Q‾) with ut−Δu≤0 in Q.

1.1F1F3given

By [F1] and [F3] the maxima max⁡Q‾u and max⁡∂pQu exist, and R2:=sup⁡x∈Ω‾∣x∣2 is finite because Ω is bounded; also u≤max⁡Q‾u and, for every x∈Ω‾, ∣x∣2≤R2.

2.1step 1.1F2F3given

For every ε>0 put v:=u+ε∣x∣2; by [F2] v is a strict subsolution, so max⁡Q‾v=max⁡∂pQv, and max⁡Q‾u≤max⁡Q‾v=max⁡∂pQv≤max⁡∂pQu+εR2 by step 1.1; letting ε↓0 and using the Archimedean property [F3] gives max⁡Q‾u≤max⁡∂pQu.

3.1step 2.1given∎

Since ∂pQ⊆Q‾, the reverse inequality max⁡∂pQu≤max⁡Q‾u is immediate, so the two maxima coincide and the weak maximum principle holds.

Depends on

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Sources