Flow inside and outside Cylinder

Question

Incompressible fluid with constant density ρ fills the three-dimensional domain below the free surface z = η(r) in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_θ$. Gravity acts upon the fluid.

Suppose the fluid in r < a rotates rigidly about the z-axis with angular velocity Ω, and the fluid in r 􏰌 a is irrotational. Show that the velocity in $r \ge$􏰌 a is given by $$u_θ = \frac{{Ωa^2}}{r}$$

I got stuck about how to show rigidly.

My idea:

For incompressible: $$\nabla\bullet u= \frac{1}{r}\frac{∂u_θ}{∂_θ}=0$$ For irrotational: $$\frac{∂u_θ}{∂_r}=0$$ For as $r \rightarrow$ a, $u_0$ approaches the tangential velocity Ωr.

the second part of the problem is suppose the free surface position satisfies η$\rightarrow$ 0 as r $\rightarrow \infty$. Show that the free surface position in $r\ge a$ is $$η= -\frac{Ω^2a^4}{2gr^2}$$

My idea is to use the Bernoulli's theorem which is expressed in the boundary condition of the free surface:

$\frac{∂\phi }{∂t}+ \frac{1}{2}|\nabla\phi|^2+gη=0$ at y =η

$\phi$ is the velocity potential, then it is $\frac{Ωa^2z}{r}$

$\frac{Ωa^2z}{r}$=-gη we can get η. But how to get z from the condition?

Answer 1

With the equation describing the incompressibility, not much can be taken:

$$\frac{1}{r}\frac{∂u_θ}{∂θ}=0$$

$$u_\theta(r)=f(r)+C$$

With the boundary conditions,

$$u_\theta(a)=\Omega a\implies f(a)+C=\Omega a$$

For an irrotational flux, the curl is zero and, with the restrictions, the only relevant term is (you miss the $ru_\theta$ thing into the partial):

$$\frac{1}{r}\frac{∂(ru_θ)}{∂r}=0$$

$$ru_\theta=K$$

With boundary condition $u_\theta(a)=\Omega a$

$$au_\theta(a)=K\implies K=\Omega a^2$$

$$u_\theta=\dfrac{\Omega a^2}{r}$$

For the added second question

In general, $\dfrac{∂\phi }{∂t}+ \dfrac{1}{2}|\nabla\phi|^2+\dfrac{p}{\rho}+gη=0$. For the free surface $p=0$ and because stationary $\dfrac{∂\phi }{∂t}=0$

We don't need the flow potential as we have the flow, so is, $\vert\nabla\phi\vert=u_\theta=\dfrac{\Omega a^2}{r}$, and $\eta=z$ because the potential energy depends on the heigh/depth, expressed by $z$.

$\dfrac{1}{2}|u_\theta|^2+gz=0$

$\dfrac{1}{2}\left(\dfrac{\Omega a^2}{r}\right)^2+gz=0\implies z=-\dfrac{\Omega^2a^4}{2gr^2}$