Fluid Mechanics: Stream Function for Axisymmetric flow

I have problem in understanding the result of stream function in Axisymmetric 3D flow:

I know that the result is (for spherical coordinates):

$$u_r=\frac{1}{r^2sin\theta}\frac{\partial\psi}{\partial\theta},$$ and

$$u_\theta=-\frac{1}{rsin\theta}\frac{\partial\psi}{\partial r}.$$

But I cannot see how this comes from the continuity equation. What is derivation for this result?

• It doesn't come from the continuity equation, they are defined such that they exactly satisfy the continuity equation. There isn't only one definition of the stream functions, point in case you could chose to have the negative sign for $u_r$ instead of $u_{\theta}$ and still have valid stream functions (i.e. they satisfy continuity) – nluigi Apr 28 '16 at 9:36

1 Answer

The continuity equation is described by:

$$\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho{\vec u}) = 0$$

For incompressible and steady flow, it reduces to:

$$\nabla\cdot{\vec u} = 0$$

The incompressible continuity equation in spherical coordinates is:

$$\nabla\cdot{\vec u} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2q_r)+\frac{1}{r}\frac{\partial}{\partial \theta}(q_\theta \sin\theta) = 0$$

The condition a stream function must satisfy is that the mixed partial derivatives should be equal:

$$\frac{\partial^2 \psi}{\partial r \partial \theta} = \frac{\partial^2 \psi}{\partial \theta \partial r}$$

The stream function's relation to the radial and tangential components of the velocity as given in your question are:

$$u_r = \frac{1}{r^2 \sin\theta}\frac{\partial \psi}{\partial \theta}, \hspace{5 pt} u_\theta = -\frac{1}{r\sin\theta}\frac{\partial \psi}{\partial r}$$

Substituting these into the continuity equation gives: $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(\frac{r^2}{r^2 \sin\theta}\frac{\partial \psi}{\partial \theta}\right)+\frac{1}{r}\frac{\partial}{\partial \theta}\left(-\frac{\sin\theta}{r\sin\theta}\frac{\partial \psi}{\partial r}\right) = 0$$ $$\frac{1}{r^2\sin\theta}\frac{\partial}{\partial r}\left(\frac{\partial \psi}{\partial \theta}\right)-\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\frac{\partial \psi}{\partial r}\right) = 0$$ $$\frac{\partial^2 \psi}{\partial r \partial \theta} = \frac{\partial^2 \psi}{\partial \theta \partial r}$$

Which satisfies the required mathematical condition.

Note: I've just shown you how this satisfies the continuity equation. You should reason it out for yourself why the stream function satisfies the mixed partial derivatives condition.

Hint: You're attempting to solve a famous partial differential equation, which I suggest you try to solve in Cartesian coordinates first, and then transform into other coordinate systems. You'll get the corresponding result in cylindrical coordinates, which is axisymmetric 3D flow.

• Thank you very much! I will try solving the equality of mixed partials. – ValientProcess Apr 28 '16 at 10:12
• No problem! You should ideally look up complex variables and partial differential equations to see the connections I've stated here (incompressible conditions, also check out irrotational conditions). – GodotMisogi Apr 28 '16 at 10:41