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A gradient load observed through Stokes pressure

Example

Assume Countable Choice. Let d∈{2,3}, let D⊂Rd be a nonempty bounded connected Lipschitz open set, and let ν>0. Use V=H01(D;Rd) with gradient norm and Q=L02(D). Define xˉ1:=∣D∣−1∫Dx1 dx and ϕ(x):=x1−xˉ1. The load Fϕ(v):=−∫Dϕ div⁡v dx is the gradient forcing ∇ϕ=e1 in the weak sense. The unique weak Stokes solution is (u,p)=(0,ϕ), and the bounded gauge-invariant sensor Sϕ(q):=∫Dϕq dx has reading Sϕ(p)=∫Dϕ2 dx>0.

Example references

Guermond, MATH 661, Chapter 53, §§53.1–53.2.2, homogeneous weak Stokes formulation, pressure gauge, and well-posedness.

Boffi–Brezzi–Fortin, Mixed Finite Element Methods and Applications, §§4.2.1–4.2.3 and §8.2.1, Example 4.2.2, pp. 225–229, 233–234, and 462–463, weak Stokes solvability and pressure modulo constants.

Facts & Assumptions

Given: Countable Choice; the stated domain and viscosity; the homogeneous weak Stokes spaces and equations; and the coordinate function x↦x1 on D.

[A1]

Countable Choice is required for the weak Stokes solvability input. Weak Stokes solvability on bounded connected Lipschitz domains

[F1]

Stokes velocity and pressure spaces and weak forms: V=H01(D;Rd) is the H1-closure of compactly supported smooth fields, Q=L02(D), and the pressure term is b(v,p)=−∫Dp div⁡v dx.

[F2]

Weak Stokes solvability on bounded connected Lipschitz domains: every bounded load in V′ has a unique weak Stokes pair with its pressure in Q.

[F3]

Lipschitz stability of bounded Stokes pressure observations: pF denotes that unique normalized pressure for load F.

[F4]

Bounded Stokes pressure sensors: a sensor is a bounded real-linear functional on Q.

[F5]

Gauge-invariant pressure sensor weights: a weight in Q defines a bounded gauge-invariant pressure sensor.

[F6]

Cauchy-Schwarz inequality for L2: the integral pairing of L2(D) functions is bounded by the product of their L2 norms.

Verification

1.1givenF1algebra

Since D is nonempty and open it contains a ball, so ∣D∣>0; boundedness gives ∣D∣<∞. Hence x1∈L2(D) and xˉ1 is finite. Thus ϕ∈L2(D), ∫Dϕ dx=0, and its weak gradient is ∇ϕ=e1.

2.1F1F6step 1.1algebra

For v∈Cc∞(D;Rd), weak integration by parts gives ∫D∇ϕ⋅v dx=−∫Dϕ div⁡v dx. By the H1-closure in [F1], approximate any v∈V by such fields; both integrals converge by Cauchy–Schwarz, so the identity holds for all v∈V. Pointwise ∣div⁡v∣=∣tr⁡∇v∣≤d ∣∇v∣F, whence ∣Fϕ(v)∣≤d ∥ϕ∥2∥∇v∥2 and Fϕ∈V′.

3.1givenA1F1F2F3step 2.1

For every v∈V, aν(0,v)+b(v,ϕ)=−∫Dϕ div⁡v dx=Fϕ(v) by step 2.1, while b(0,q)=0 for every q∈Q. Thus (0,ϕ) satisfies both weak Stokes equations. Uniqueness in [F2] identifies it with the solution, so the pressure notation in [F3] gives pFϕ=ϕ.

4.1F4F5F6step 1.1step 3.1∎

By [F6], Sϕ(q)=∫Dϕq dx is bounded on Q; by [F4], [F5], and ϕ∈Q, it is a bounded gauge-invariant sensor. Since a nonempty open D contains a ball and no ball is contained almost everywhere in a hyperplane x1=xˉ1, ϕ is not zero almost everywhere; therefore ∫Dϕ2 dx>0.

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