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Bounded-domain Stokes residual estimate — formal worked field

Example

Assume ACω. Let Lx,Lz,h,μ>0, define TL=R/(LZ) for L>0, and set Q=TLx×(0,h)×TLz. Regard fields on Q as periodic lifts to R×(0,h)×R with periods Lx,Lz in x,z, and integrate with Lebesgue measure over one fundamental cell in each periodic coordinate. Set w(x,y,z)=(sin⁡(πy/h),0,0), q=0, and f=μ(π/h)2w. Then w is a smooth divergence-free periodic field with zero trace on both y walls and satisfies −μΔw+∇q=f. Its norms satisfy ∥w∥L2(Q)2=LxLzh/2 and ∥∇w∥L2(Q)2=(π/h)2∥w∥L2(Q)2, so ∥w∥L2(Q)=(h/π)∥∇w∥L2(Q)≤h∥∇w∥L2(Q). This is a formal worked field for the periodic-channel Poincare clause; the general bounded C1 Euclidean residual theorem is not applied to this periodic cell.

Facts & Assumptions

Given: The field, periodic channel, positive parameters, residual equation, and ACω assumption in [A1].

[A1]

Lx,Lz,h,μ>0, Q=TLx×(0,h)×TLz with Lebesgue measure on one fundamental cell, w=(sin⁡(πy/h),0,0), q=0, f=μ(π/h)2w, periodicity in x,z, and zero trace at y=0,h; assume ACω.

[F1]

Bounded-domain Stokes residual estimate: on this periodic channel, every C1 field v periodic in the torus variables and zero on y=0,h satisfies ∥v∥L2(Q)≤h∥∇v∥L2(Q). The theorem's global statement carries ACω.

Verification

1.1givenA1algebra

The field is periodic in x,z and vanishes at y=0,h because sin⁡(0)=sin⁡(π)=0. Its only nonzero component is independent of x, so ∇⋅w=0, and Δw=−(π/h)2w. Hence −μΔw+∇q=μ(π/h)2w=f.

1.2givenA1algebra

Integrating sin⁡2(πy/h) over 0<y<h gives h/2, so ∥w∥L2(Q)2=LxLzh/2. The only nonzero derivative is ∂yw1=(π/h)cos⁡(πy/h), whose squared integral gives ∥∇w∥L2(Q)2=LxLz(π/h)2(h/2)=(π/h)2∥w∥L2(Q)2.

2.1givenA1F1step 1.2algebra∎

Since π>1, step 1.2 gives ∥w∥L2(Q)=(h/π)∥∇w∥L2(Q)≤h∥∇w∥L2(Q). This explicitly verifies the channel inequality in [F1]; the field is only a worked channel residual, not an application of the general-domain residual estimate.

Depends on

Dependency tree · 2 levels