Equation of motion for flow in laminar boundary layer in cylindrical coordinates

Question

I know the 2D equation of motion for flow in a laminar boundary layer in cartesian co-ordinates: $$\frac{\partial u}{\partial t}+ u \frac{\partial u}{\partial x}+ v \frac{\partial u}{\partial y}=\frac{-1}{\rho} \frac{\partial p}{\partial x}+\nu \frac{\partial^2 u}{\partial y^2}$$

This is useful to analyse flow in a duct, but however, how would you convert this into cylindrical coordinates, to notably look at flow in a circular cross section pipe. Note, I am only required to analyse flow in fully developped boundary layer, so only the viscous term really is important.

The viscous term should become (in polar/cylindrical coordinate) $$\frac{1}{r} \frac{\partial}{\partial r} \bigg(r \frac{\partial u}{\partial r} \bigg).$$

How do you get this?

Answer 1

Using $x = r\cos\phi$ and $y = \sin\phi$, you can calculate $\partial x/\partial r$ and $\partial y/\partial r$ and then you can apply the multidimensional chain rule,

\begin{align} \frac{\partial }{\partial r} = \frac{\partial x}{\partial r}\frac{\partial}{\partial x} + \frac{\partial y}{\partial r}\frac{\partial }{\partial y}, \end{align}

to the $r$-derivatives in

\begin{align} \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right). \end{align}

First do the $r$-derivative in between the brackets, then make sure that everything in those brackets is expressed in $x$ and $y$ (i.e., no $r$ or $\phi$ appear there anymore), using again the relations $x = r\cos\phi$ and $y = \sin\phi$ (and in particular $r = \sqrt{x^2+y^2}$), and then do the other $r$-derivative, using the chain rule again, and also the product rule. In the end eliminate all appearing $r$ and $\phi$ by writing them in terms of $x$ and $y$, using the twice mentioned relations.