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Conditional nonlinear Couette energy stability — formal worked field

Example

Assume ACω. Define T2π=R/(2πZ) and let Q=T2π×(0,1)×T2π, with Lebesgue integration over one fundamental cell. Take ρ0=μ=ν=1, S=1, base field Ub=(y,0,0), and pressure p=0. For any a∈R, set u(x,y,z,t)=(y+ae−π2tsin⁡(πy),0,0),t≥0. This field is smooth, periodic in x,z, divergence-free, and has wall values 0 at y=0 and (1,0,0) at y=1. Its perturbation w=u−Ub satisfies ∥w(t)∥L2(Q)2=e−2π2t∥w(0)∥L2(Q)2≤e−t∥w(0)∥L2(Q)2. The sufficient parameter condition in the supplier is ∣S∣h2/ν=1<2; this example verifies that conditional bound for one exact field.

Facts & Assumptions

Given: The periodic channel, ρ0=μ=ν=1, S=1, h=1, p=0, the field u, and ACω as stated.

[A1]

On Q=T2π×(0,1)×T2π, ρ0=μ=ν=1, u=(y+ae−π2tsin⁡(πy),0,0) for t≥0, with periodic x,z, wall values u∣y=0=0, u∣y=1=(1,0,0), and pressure p=0.

[F1]

Conditional nonlinear Couette energy stability: for an already-existing smooth incompressible Navier-Stokes solution on a periodic channel of height h with Couette shear S, if ∣S∣h2/ν<2 and the theorem's stated dependencies hold, its perturbation energy satisfies E(t)≤e−(2ν/h2−∣S∣)tE(0) on the assumed smooth interval.

Verification

1.1givenA1algebra

The only nonzero component of u is independent of x, so ∇⋅u=0 and (u⋅∇)u=0. Direct differentiation gives ut=−aπ2e−π2tsin⁡(πy)(1,0,0) and Δu=−aπ2e−π2tsin⁡(πy)(1,0,0); hence ut+(u⋅∇)u+ρ0−1∇p−νΔu=0. Since sin⁡(0)=sin⁡(π)=0, the stated wall values hold.

2.1givenA1step 1.1algebra

The perturbation is w=ae−π2tsin⁡(πy)(1,0,0). Integrating sin⁡2(πy) over 0<y<1 and over both periods gives E(t)=∥w(t)∥L2(Q)2=2π2a2e−2π2t=e−2π2tE(0).

3.1givenA1F1step 2.1algebra∎

Here ∣S∣h2/ν=1 and 2ν/h2−∣S∣=1. Applying [F1] gives E(t)≤e−tE(0), while step 2.1 verifies the sharper exact decay because 2π2≥1. The estimate is conditional on the stipulated supplier and assumptions.

Depends on

Dependency tree · 2 levels