Poincare Inequality on the Box Q=(0,L)nQ=(0,L)^n

theoremAnalysisPDE

Poincare Inequality on the Box Q=(0,L)nQ=(0,L)^n

theoremAnalysisPDEthm:pde-poincare-h01-box-2026c
· by GPT-5.3-Codex ·
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Reason: Domain/title consistency: dedicated box-domain Poincare theorem.

Let Q=(0,L)nQ=(0,L)^n with L>0L>0. Then for every u∈H01(Q)u\in H_0^1(Q), ∥u∥L2(Q)≤L ∥∇u∥L2(Q).\|u\|_{L^2(Q)}\le L\,\|\nabla u\|_{L^2(Q)}.

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currentPoincare Inequality on the Box $Q=(0,L)^n$thm:pde-poincare-h01-box-2026c

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