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Projected stationary weak Stokes solvability

Statement

Assume full AC. Let $d \geq 2$ , let $R > 0$ , let $Ω \subset (- R, R)^{d}$ be nonempty bounded open, let $μ > 0$ , and let $f \in L^{2} (Ω; R^{d})$ . There is a unique $u \in V$ such that $μ \int_{Ω} \nabla u : \nabla v = \int_{Ω} f \cdot v (v \in V),$ where $V$ is the distributionally divergence-free subspace of $H_{0}^{1}$ defined above. It satisfies $∥ \nabla u ∥_{2} \leq 2 R ∥ f ∥_{2} / μ$ . This is a projected variational velocity result; it asserts no pressure, $L^{2}$ pressure regularity, or classical boundary regularity.

Statement references

Martin Vohralík, A posteriori error estimates for efficiency and error control in numerical simulations, Course NM497 lecture notes (2010–2011).

Theo Bühler and Dietmar Salamon, Functional Analysis.

Facts & Assumptions

Given: The hypotheses stated in the formulation, including the explicitly adopted choice principle where indicated.

[F1]

The Axiom of Choice: full AC is assumed; applying its choice function to a sequence of nonempty sets gives a selector for that sequence, hence Countable Choice.

[F2]

Projected no-slip Stokes Hilbert space: $V$ is the real Hilbert space of distributionally divergence-free $H_{0}^{1}$ vector fields.

[F3]

Directional Poincaré bound for zero trace: $∥ v ∥_{2} \leq 2 R ∥ \nabla v ∥_{2}$ for the specified domain.

[F4]

Riesz representation for Hilbert spaces: under its Countable Choice hypothesis, every bounded real linear functional on a real Hilbert space has a unique representing vector.

Proof

Given: Full AC, $μ > 0$ , $f \in L^{2}$ , and $V$ on $Ω \subset (- R, R)^{d}$ .

1.1givenF2F3F1

By [F2], $V$ is complete in its $H^{1}$ norm. The estimate in [F3] shows $∥ v ∥_{2} \leq 2 R ∥ \nabla v ∥_{2}$ , so the gradient norm is equivalent to the $H^{1}$ norm on $V$ and is complete. Thus $(v, w)_{V} = \int_{Ω} \nabla v : \nabla w$ is a real Hilbert inner product.

2.1givenF1F4step 1.1algebra

The functional $F (v) = \int f \cdot v$ is linear and satisfies $∣ F (v) ∣ \leq ∥ f ∥_{2} ∥ v ∥_{2} \leq 2 R ∥ f ∥_{2} ∥ v ∥_{V}$ . To verify the Countable Choice hypothesis of [F4], take any sequence $(X_{n})_{n \in N}$ of nonempty sets. Its range $F = {X_{n} : n \in N}$ is a family of nonempty sets, so full AC [F1] gives a choice function $g$ on $F$ ; then $n \mapsto g (X_{n})$ is a selector for the sequence. Riesz [F4] therefore applies to the real Hilbert space from step 1.1 and bounded functional $F$ , giving a unique $z \in V$ with $(z, v)_{V} = F (v)$ ; set $u = z / μ$ .

3.1givenF3step 2.1algebra∎

Taking $v = u$ in the variational identity gives $μ ∥ \nabla u ∥_{2}^{2} = \int f \cdot u \leq 2 R ∥ f ∥_{2} ∥ \nabla u ∥_{2}$ . If the gradient norm is zero, the desired bound is immediate; otherwise divide by $μ ∥ \nabla u ∥_{2}$ . If $u_{1}, u_{2}$ both solve, their difference tested against itself has zero gradient norm, hence is zero by Poincaré.

Projected stationary weak Stokes solvability

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources