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Linear vortex-sheet growing modes — worked interpretation

Example

On the periodic interface x∈T2π, where T2π=R/(2πZ), with half-planes y>0 and y<0, take equal density ρ0=1, base velocities U+=1 and U−=−1, and no gravity or surface tension. For t≥0, define η(x,t)=etcos⁡x,ϕ+(x,y,t)=et−y(sin⁡x−cos⁡x),ϕ−(x,y,t)=et+y(cos⁡x+sin⁡x). The potentials are harmonic and decay at their respective infinities; at y=0 they satisfy (∂t+U±∂x)η=∂yϕ± and continuity of p±′=−(∂t+U±∂x)ϕ±. Thus these fields give an exact real growing mode of the stipulated linear interface equations.

Facts & Assumptions

Given: The periodic interface equations, equal density, base speeds, mode fields and pressure definitions in [A1].

[A1]

x∈T2π, y>0 above and y<0 below, t≥0, ρ0=1, U+=1, U−=−1, Δϕ±=0, decay of ϕ± and their first derivatives at the corresponding infinities, (∂t+U±∂x)η=∂yϕ± at y=0, and p±′=−(∂t+U±∂x)ϕ± with p+′=p−′; gravity and surface tension are absent.

[F1]

Linear vortex-sheet growing modes: for equal positive densities and the stated harmonic linear interface conditions without gravity or surface tension, a normal mode with k>0 has λ=−ikU‾±k∣ΔU∣/2; nonzero velocity jump gives the positive real branch.

Verification

1.1givenA1algebra

The x-dependence of each potential is a sine or cosine and its y-dependence is e−y above or ey below, so ∂xxϕ±=−ϕ± and ∂yyϕ±=ϕ±. Hence Δϕ±=0, and the exponential factors give the required decay.

1.2givenA1algebra

At y=0, (∂t+∂x)ϕ+=2etsin⁡x and (∂t−∂x)ϕ−=2etsin⁡x. Therefore p+′=p−′=−2etsin⁡x, so pressure continuity holds.

2.1givenA1step 1.1algebra

At y=0, (∂t+∂x)η=et(cos⁡x−sin⁡x)=∂yϕ+, while (∂t−∂x)η=et(cos⁡x+sin⁡x)=∂yϕ−. Thus both kinematic conditions hold.

3.1givenA1F1algebra∎

Here k=1, U‾=0, and ΔU=2. The growing branch in [F1] has λ=1, agreeing with the explicit factor et in η; this is one exact linear-model example only.

Depends on

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