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The incompressible Stokes equations areTaking the curl gives ${\nabla }^2\omega =0$. For a planar stream function, $\omega =-{\nabla }^2\psi$, so the biharmonic stream function for planar Stokes flow satisfies

For $\psi =r^{2}f(\theta )$,The polar rate-of-strain components areHence

Sincebiharmonicity givesNo slip at $\theta =-\alpha$, no penetration at $\theta =0$, and the imposed tangential traction givePut $\Delta =\sin 2\alpha -2\alpha \cos 2\alpha$. Solving,The upper-surface velocity is radial and equals $r{f}^{\prime }(0)$. Therefore the similarity solution for tangentially forced Stokes wedge gives

Choose the stream function conventionwhich satisfies continuity identically. Introduceso thatTake the similarity solutionThenandA direct differentiation shows that the terms containing $\eta f^{\prime }{f}^{\prime \prime }$ cancel:Also,while the delta scaling givesThe definitions of $D$ and $U_0$ implyAfter cancelling the common dimensional factor, the similarity equation for a two-dimensional laminar plume isThus

The velocity induced by planar vorticity is a convolution with this kernel, whose sign depends on the stream function and vorticity conventions. Its magnitude is $(2\pi |z|{)}^{-1}$ and its derivative is bounded by $C|z{|}^{-2}$. Splitting the convolution at radius $R$ gives $\Vert K\ast \omega {\Vert }_{\infty }\le R\Vert \omega {\Vert }_{\infty }+\Vert \omega {\Vert }_1/(2\pi R)$. The same singularity yields a log-Lipschitz modulus, enough for a Yudovich characteristic flow.