 Solution

A streamline is a curve whose tangent is parallel to the instantaneous fluid velocity field. In steady flow one may parametrize it by $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r})$. Dotting the steady Euler equations with $\mathbf{u}$ eliminates the cross-product term, since $\mathbf{u}\cdot ((\nabla \times \mathbf{u})\times \mathbf{u})=0$. Hencewhich is Bernoulli's principle along each streamline.

For the supplied field, $xu+yv=0$, so along a streamlineWriting $r^2=x^2+y^2$, the tangential speed is $v_{\theta }=-br+c/r$ and the radial speed is zero. Both $r$ and $|\mathbf{u}{|}^2=(-br+c/r{)}^2$ are constant around each nonstagnant circular streamline. The Bernoulli quantity is constant there, so the pressure is constant on every such circle.

If a circle is wholly stagnant, the same conclusion follows directly from the steady Euler equations: on it $\mathbf{u}=0$ and $\nabla |\mathbf{u}{|}^2=0$, hence $\nabla p=0$. Equivalently the radial momentum equation gives $\nabla p=\rho (-br+c/r{)}^2{\mathbf{e}}_r/r$ and no angular pressure variation, including at zero tangential speed. Thus $p$ is constant on $x^2+y^2=a^2$ for every nonzero $a$, with no exception for a stagnation circle.