 Solution

This is a Rankine vortex. Write $r=\sqrt{x^2+y^2}$. Inside the core the velocity is $u=(\Gamma y/a^2,-\Gamma x/a^2,0)$, soThis is the core value obtained from the inner formula; outside, the given irrotationality makes the vorticity zero. For nonzero $\Gamma$, the sharp boundary $r=a$ has different one-sided velocity gradients, so a single classical vorticity is not defined exactly on that edge; the inner trace is the boxed value and the outer trace is zero.

Inside the core, direct differentiation gives $(u\cdot \nabla )u=-{\Gamma }^2(x,y,0)/a^4$. The steady Euler equations therefore give $p_r=\rho {\Gamma }^{2}r/a^4$ and $p_z=-\rho g$. Integrating,Outside, $|u{|}^2={\Gamma }^2/r^2$, so part (a) givesMatching pressure at the core edge $r=a$ yields $A=B-\rho {\Gamma }^2/a^2$. Hence the complete pressure field, up to one additive constant $B$, isBoth the velocity and pressure are continuous at $r=a$. The inside expression has been derived from the momentum equation because the core is rotational.