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Weak Stokes solvability on bounded connected Lipschitz domains

Statement

Assume Countable Choice. Let D⊂Rd be nonempty, bounded, connected and Lipschitz, d=2 or 3, and let ν>0. For every bounded linear functional F on V=H01(D;Rd) equipped with ∥v∥V=∥∇v∥2, there is a unique (u,p)∈V×Q, Q=L02(D), satisfying 2ν∫Dε(u):ε(v)−∫Dpdiv⁡v=F(v),−∫Dqdiv⁡u=0 for every v∈V,q∈Q. The pressure is normalized to have mean zero. One stability bound is ∥u∥V≤ν−1∥F∥V′ and ∥p∥2≤3Cdiv(D)∥F∥V′.

Facts & Assumptions

Given: Countable Choice; a bounded connected Lipschitz D; the real Sobolev spaces; ν>0 and F∈V′.

[A1]

Countable Choice is assumed by the Hilbert-space completeness and Lax–Milgram branch used below; no full Axiom of Choice is used by the local Korn or Poincare arguments. (The Axiom of Countable Choice (ACω))

[F1]

H1(D) is Hilbert under Countable Choice, H01(D) is its closed subspace, and bounded-domain Poincare makes the gradient norm equivalent to the H1 norm. (The Sobolev space H1 is a Hilbert space, Zero-boundary Sobolev space as a norm closure, A bounded-domain Poincare inequality for H1-zero fields, The Axiom of Countable Choice (ACω))

[F2]

Real L2(D) is Hilbert under Countable Choice, and the integral functional is bounded by ∣D∣1/2∥q∥2, so its zero-mean kernel Q is closed and Hilbert. (L2 with the integral pairing is a Hilbert space, Holder's inequality for integrals, including the endpoint cases, The Axiom of Countable Choice (ACω))

[F3]

2∥ε(v)∥22=∥∇v∥22+∥div⁡v∥22 for every zero-boundary field. (The homogeneous Dirichlet strain identity)

[F4]

The divergence form is bounded and has continuous inf-sup constant at least Cdiv(D)−1. (Continuous Stokes divergence inf-sup on a Lipschitz domain)

[F5]

On real Hilbert spaces, bounded a,b, coercivity on the constraint kernel and positive inf-sup imply unique mixed solvability with the stated norm bounds. (Babuska-Brezzi solvability by kernel coercivity and inf-sup)

[F6]

Every zero-boundary velocity has zero-mean divergence. (Zero mean of the divergence of a zero-boundary field)

[F7]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1A1F1F2

By [F1], V is complete in the H1 norm. If (vn) is Cauchy in ∥∇⋅∥2, [F1] and Poincare make it Cauchy in H1; its H1 limit is in the closed subspace H01 and its gradients converge. Thus V is Hilbert in the gradient norm. By [F2], Q is Hilbert.

2.1F3F7step 1.1

Cauchy–Schwarz and ∣ε(v)∣≤∣∇v∣F give ∣aν(u,v)∣≤2ν∥u∥V∥v∥V. By [F3], aν(v,v)=2ν∥ε(v)∥22≥ν∥v∥V2 on all of V, hence on ker⁡B; take Ma=2ν and α=ν.

3.1A1F4F5step 1.1step 2.1

The form b has Mb≤d and inf-sup β≥Cdiv(D)−1 by [F4]. Apply [F5] with G=0. It yields a unique velocity and a unique mean-zero pressure satisfying the two displayed equations. The bound on velocity is ∥u∥V≤α−1∥F∥=ν−1∥F∥.

4.1A1F4F5F6step 3.1∎

By [F6], div⁡u∈Q. The constraint equation tested with q=div⁡u therefore gives ∥div⁡u∥22=0, so the weak constraint is exact incompressibility. Also [F5] gives ∥p∥2≤β−1(∥F∥+Ma∥u∥V). Use β−1≤Cdiv(D), Ma=2ν and the velocity estimate of step 3.1 to obtain ∥p∥2≤3Cdiv(D)∥F∥. Mean-zero gauge is part of Q, so the pressure is unique in that normalization.

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