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An analytic box solution has exponential Galerkin p convergence

Example

Assume Countable Choice.

On the single reference square Ω=(−1,1)2 let u(x,y)=1/(4−x) and use a(v,w)=∫Ω(∇v⋅∇w+vw) on H1(Ω) with exact load F(w)=a(u,w). For every p≥1, the tensor-degree-p conforming Galerkin solution up satisfies

∥u−up∥H1(Ω)≤C 13(34)p

for a fixed constant C.

Facts & Assumptions

Given: Countable Choice, p≥1, u(x,y)=(4−x)−1 on [−1,1]2, and H1 with norm ∥v∥H12=∥v∥22+∥∇v∥22.

[F1]

Analytic data converge exponentially under fixed-mesh p refinement: on the fixed box, reference derivative bounds M∣ν∣!R−∣ν∣ with R−1<q<1 yield a trace-matching polynomial with H1 error at most CR,qMqp.

[F2]

Cea transfers the fixed-mesh analytic p rate to the energy error: the exact coercive problem transfers this best approximation rate to its conforming Galerkin solution.

[F3]

Cauchy–Schwarz: ∣⟨x,y⟩∣≤∥x∥ ∥y∥, with equality exactly for dependent pairs: the inner-product Cauchy–Schwarz inequality bounds ∣a(v,w)∣ by ∥v∥H1∥w∥H1.

Verification

1.1F1givenalgebra

For a multi-index ν=(n,j), Dνu=0 when j>0, while ∂xnu(x,y)=n!(4−x)−n−1. Since 4−x≥3 on the square, ∣Dνu∣≤13∣ν∣!2−∣ν∣ in every case. Thus the hypothesis of [F1] holds with M=1/3, R=2, and q=3/4.

2.1F2F3step 1.1given∎

The form a is the H1 inner product. By [F3], F(w)=a(u,w) is bounded, and u is the exact variational solution by its definition. Applying [F2] with the constants established in step 1.1 gives ∥u−up∥H1≤C(1/3)(3/4)p.

Depends on

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