Stokes Stream Function - Zero Divergence

Zero Divergence

In cylindrical coordinates, the divergence of the velocity field u becomes:

$\begin{align} \nabla \cdot \mathbf{u} &= \frac{1}{\rho} \frac{\partial}{\partial \rho}\Bigl( \rho\, u_\rho \Bigr) + \frac{\partial u_z}{\partial z} \\ &= \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( - \frac{\partial \Psi}{\partial z} \right) + \frac{\partial}{\partial z} \left( \frac{1}{\rho} \frac{\partial \Psi}{\partial \rho} \right) = 0, \end{align}$

as expected for an incompressible flow.

And in spherical coordinates:

$\begin{align} \nabla \cdot \mathbf{u} &= \frac{1}{r\, \sin(\theta)} \frac{\partial}{\partial \theta}\Bigl( u_\theta\, \sin(\theta) \Bigr) + \frac{1}{r^2} \frac{\partial}{\partial r}\Bigl( r^2\, u_r \Bigr) \\ &= \frac{1}{r\, \sin(\theta)} \frac{\partial}{\partial \theta} \left( - \frac{1}{r} \frac{\partial \Psi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{1}{\sin(\theta)} \frac{\partial \Psi}{\partial \theta} \right) = 0. \end{align}$

Read more about this topic: Stokes Stream Function