fluid dynamics in polar coordinates

Question

On page 12 of Malham's fluid dynamics notes the following flow field is considered: $\boldsymbol u= (u,v) = (kx, -ky)$. It's easy to see in these Cartesian coordinates that this is solenoidal: $\nabla\cdot \boldsymbol u = k-k=0$, and he derives that the stream function is $\psi=kxy$.

Now he moves to polar coordinates of the same flow and denotes $\boldsymbol u=(u_r,u_\theta)=(kr\cos 2\theta, -kr\sin 2\theta)$. I don't see why this parametrisation is acceptable - why the factor of 2? I can kind of derive this if I parametrise the flow as $(u_r,u_\theta)=(kr\cos \alpha\theta, -kr\sin \alpha\theta)$ and then enforce $\nabla\cdot \boldsymbol u=0$ for the polar divergence operator, which yields that $\alpha=2$, but I still don't understand why this is the same flow as the original. Why am I allowed to deviate from $(u_r,u_\theta)=(kr\cos \theta, -kr\sin \theta)$?

Answer 1

The flow in polar coordinates must satisfy the appropriate dynamics. For example in the $r$ coordinate we need $dr/dt = u_r(r,\theta,t)$.

Now, $r=r(x(t),y(t))=\sqrt{x^2+y^2}$, so by the chain rule, \begin{align} \frac {dr}{dt} &= \frac {\partial r}{\partial x} \frac {dx}{dt} + \frac {\partial r}{\partial y}\frac {dy}{dt} \\ &= \frac xr u + \frac yr v \\ &= \frac kr(x^2 - y^2) \\ &= kr(\cos^2\theta - \sin^2\theta) \\ &= kr\cos 2\theta \end{align}

Similarly we can derive that $\frac {d\theta}{dt}=-k\sin 2\theta$, and take careful note that $u_\theta = rd\theta/dt$ since the function $\boldsymbol u=(u_r,u_\theta)$ is w.r.t. unit basis vectors, whilst $d\theta/dt$ is w.r.t. the raw parameters. That is $\boldsymbol u = u_r \hat r + u_\theta \hat \theta$ where these are unit vectors.