Talk:Prandtl–Batchelor theorem

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I got lost in the proof. I think using dω/dψ is odd since from ω=-Δψ we learn that it is the operator -Δ.

I propose a proof along the following line where a maximum principle is exploited. The vorticity equation can be written as div(ω u - 1/Re grad ω ) =0, where it is used that div u =0. Now for any domain, the Gauss divergence theorem gives ʃΓ (ω u - 1/Re grad ω )•n d Γ =0. Next, we take Γ to be a small circle. Since ω is at least a continuous function, then for a small enough circle, we can approximate the above to  ωʃΓ  u •n d Γ  - 1/Re ʃΓ  grad ω •n d Γ =0. The first integral is zero because of the incompressibility of the fluid. Thus the remaining integral must be zero, which says that the normal derivative of ω must change sign if it is nonzero somewhere on Γ. Hence, within Γ, no extremum can occur; at most ω can be constant. Since we can pick the small circle anywhere in the domain, ω will have no minimum nor maximum in the domain.

If there exists a closed contour on which ω is constant then that value also holds inside that contour, otherwise, it would have an extremum inside the contour. This situation occurs for a closed streamline and high Reynolds numbers. We know that the velocity is tangential to the streamline and hence, for large Re, the vorticity equation says that the gradient of the vorticity is perpendicular to the streamline. As a consequence, we have that the vorticity is constant along the streamline. FredWubs (talk) 15:09, 5 August 2020 (UTC)