 Solution

The steady Euler equations for an inviscid fluid with no body force are $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho$. Use the convective acceleration identityto obtainFor the question's streamfunction convention, $\mathbf{u}=({\psi }_y,-{\psi }_x,0)$, so $\mathbf{u}\times \mathbf{\omega }=-\omega \nabla \psi$. Constant $\omega$ therefore gives the Bernoulli function for planar constant-vorticity flowon the connected fluid region.

For circular streamlines, write $\mathbf{u}=w(r)\hat{\mathbf{\varphi }}$. The cylindrical curl gives $(rw{)}^{\prime }/r=\omega$, so $w=\omega r/2+C_1/r$. The outer wall condition gives $C_1=-2\omega a^2$. Choose the positive azimuthal orientation so that $w(a)=V$, as implicit in the requested sign for $\omega$. ThenReversing the circulation reverses $\omega$ but leaves the speed and pressure difference unchanged. For this uniform-vorticity circular flow in an annulus, one convenient streamfunction is $\psi =V(r^2/2-4a^2\log (r/a))/(3a)+C_2$.

The radial component of the Euler equations for an inviscid fluid is $p_r=\rho w^2/r$. Thus pressure increases outwards, andTherefore