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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)
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Linear vortex-sheet growing modes

Statement

For x,t∈R, assume the fields are independent of a third spatial coordinate. Let H+=[0,∞) and H−=(−∞,0]. The real-valued interface displacement is a C∞ map η:Rx×Rt→R, and the real-valued scalar potentials are maps ϕ±:Rx×H±×Rt→R. Each ϕ± is C∞ on the open half-domain; ϕ± and their first partial derivatives ∂xϕ±, ∂yϕ±, and ∂tϕ± admit continuous one-sided traces at y=0, locally in (x,t). On the open half-domains use ordinary pointwise Euclidean derivatives in (x,y) and set Δ=∂xx+∂yy. Let U+,U−∈R be constant horizontal base velocities, and let the two half-planes y>0 and y<0 have the same constant density ρ0>0. The potentials satisfy Δϕ±=0 and, for every fixed (x,t), ϕ±, ∂xϕ±, ∂yϕ±, and ∂tϕ± tend to zero as y→+∞ and y→−∞ respectively. The stipulated linear interface equations, interpreted by one-sided traces at y=0, are (∂t+U±∂x)η=∂yϕ± and continuity of the defined pressure perturbations p±′(x,t)=−ρ0[∂tϕ±+U±∂xϕ±]y=0±. Gravity and surface tension are absent. Complexify this real linear system for the following normal-mode calculation.

For every real k>0, every nonzero complex amplitude H, and every A+,A−,λ∈C, consider a mode of the form η=Heikx+λt, ϕ+=A+e−kyeikx+λt, ϕ−=A−ekyeikx+λt. Set U‾=(U++U−)/2 and ΔU=U+−U−. The mode equations require

λ=−ikU‾±k∣ΔU∣2.

If ΔU≠0, one root has positive real part k∣ΔU∣/2, and its real or imaginary part gives a real exponentially growing solution of the stated linear system.

For the 2π-periodic interface, require the fields to be periodic in x and set T2π=R/(2πZ). For a complex-valued trigonometric polynomial η on this torus, define η^j=(2π)−1∫02πη(x)e−ijx dx and, for m≥0, ∥η∥Hm2=∑j∈Z(1+j2)m∣η^j∣2. Let Hm(T2π;C) be the completion of the complex-valued trigonometric polynomials in this norm; real-valued fields are viewed as elements of its real subspace. If ΔU≠0, then for every fixed t>0 this exact mode family has bounded initial Hm norm but unbounded evolved Hm norm as its integer wavenumber k→∞. Thus no uniform estimate of the form ∥η(t)∥Hm≤Ct∥η(0)∥Hm holds for this family. This is a linear interface-model conclusion, not a theorem of nonlinear turbulent transition, viscous shear-layer behavior, or physical instability in a particular experiment.

Facts & Assumptions

Given: The equal-density, no-gravity/no-surface-tension linear interface equations and Fourier-mode form in [A1]. The formal derivative and incompressibility conventions are the planned interface [F1].

[A1]

ρ0>0, U+,U−∈R, k>0, H≠0, and the mode forms η=Heikx+λt, ϕ+=A+e−kyeikx+λt, ϕ−=A−ekyeikx+λt with complex A+,A−,λ. On y>0 and y<0, Δϕ±=0 and the potentials and their first derivatives decay at the corresponding infinity. At y=0, (∂t+U±∂x)η=∂yϕ± and p±′=−ρ0(∂t+U±∂x)ϕ± with p+′=p−′; gravity and surface tension are absent.

[F1]

Formal parameterized fluid systems and observables on fd−x01 defines Euclidean partial derivatives pointwise, defines the scalar Laplacian by Δf=div⁡∇f=∑j∂j2f, and applies vector Laplacians componentwise. Extending these potentials independently of a third coordinate makes this two-coordinate Laplacian the restriction of that defined three-coordinate Laplacian.

Proof

1.1givenA1F1algebra

Using the planned supplier's pointwise Euclidean derivative convention [F1], substitute the upper and lower mode forms into the Laplace equations. Since ∂xx multiplies each mode by −k2 and ∂yy multiplies the exponential profile by k2, both potentials are harmonic; the chosen signs make them decay in their respective half-planes.

2.1givenA1step 1.1algebra

Let D±=λ+ikU±. The kinematic conditions at y=0 give D+H=−kA+ and D−H=kA−. Since k>0, A+=−D+H/k and A−=D−H/k.

3.1givenA1step 2.1algebra

Pressure continuity gives −ρ0D+A+=−ρ0D−A−. Substitute step 2.1 and use ρ0>0, k>0, and H≠0 to obtain D+2+D−2=0.

4.1givenstep 3.1algebra

Put Λ=λ+ikU‾. Then D+=Λ+ikΔU/2 and D−=Λ−ikΔU/2, so step 3.1 becomes 2Λ2−k2(ΔU)2/2=0.

5.1givenA1step 4.1algebra

Solving the quadratic gives Λ=±k∣ΔU∣/2, hence λ=−ikU‾±k∣ΔU∣/2. When ΔU≠0, the plus branch has positive real part k∣ΔU∣/2. Because the interface equations have real coefficients, real and imaginary parts of a complex mode separately solve them.

6.1givenA1step 5.1algebra∎

On T2π take an integer k≥1 and the growing-branch interface mode ηk(x,t)=akek∣ΔU∣t/2cos⁡(k(x−U‾t)), where ak=(1+k2)−m/2. Its only nonzero Fourier coefficients at time t are at ±k, each of magnitude akek∣ΔU∣t/2/2. Therefore ∥ηk(0)∥Hm2=1/2 and ∥ηk(t)∥Hm2=ek∣ΔU∣t/2. For fixed t>0 and ΔU≠0, the first norm is uniformly bounded and the second diverges as k→∞, proving the stated failure of a uniform bound.

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