Cylindrical coordinates in PDE

Question

I'm trying to write out the 3D incompressible Navier-Stokes equations in cylindrical coordinates but got stuck. The Navier-Stokes equations: $$u_t – \nu\Delta u + u\cdot \nabla u + \nabla p=0$$ $$\nabla \cdot u =0 $$
Here both $u$ denotes the velocity vector field and $p$ is the pressure.
We consider the cylindrical coordiantes: $x=r\cos \theta; y=r\sin \theta; z=z$. I know then $$\Delta=\frac{1}{r}\frac{\partial}{\partial_{r}}(r\frac{\partial}{\partial_{r}}) + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial z^2} $$. So the velocity components are $(u_{r}, u_{\theta},u_{z})$.
My question is: why is the following $r$ momentum equation true?
$$\frac{\partial u_r}{\partial t} + u\cdot \nabla u_{r}-\frac{1}{r}u_{\theta}^2+p_{r}=\nu(\Delta u_{r}-\frac{u_{r}}{r^2}-\frac{2}{r^2}\frac{\partial u_{\theta}}{\partial \theta})$$
I have no idea where the term $-\frac{1}{r}u_{\theta}^2$ comes from. And for the term on the right hand side, what are the two terms besides the laplacian operator?

Answer 1

Your confusion seems to stem from thinking $\mathbf{e}_r,\mathbf{e}_\theta$ are constants. They are not, and so when differentiating you need to take account of their change too.

Remember the operator $\nabla$ is $$ \nabla=\mathbf{e}_r\frac{\partial}{\partial r}+\mathbf{e}_\theta\frac{1}{r}\frac{\partial}{\partial\theta}+\mathbf{e}_z\frac{\partial}{\partial z} $$ and the vector laplacian is defined as $$ \nabla^2 \mathbf{A}=\nabla(\nabla\cdot\mathbf{A})-\nabla\times(\nabla\times\mathbf{A}). $$ In cylindrical coordinates, we have \begin{align*} (\mathbf{u}\cdot\nabla)\mathbf{u} &=\left(u_r\frac{\partial}{\partial r}+u_\theta\frac{1}{r}\frac{\partial}{\partial\theta}+u_z\frac{\partial}{\partial z}\right)(u_r\,\mathbf{e}_r+u_\theta\,\mathbf{e}_\theta+u_z\,\mathbf{e}_z)\\ \end{align*} and note that $\frac{\partial}{\partial\theta}\mathbf{e}_\theta=-\mathbf{e}_r$, giving the extra term $-\frac1ru_\theta^2$ that you see in the LHS.

Similarly, when you expand the vector laplacian, you get $$ (\nabla^2\mathbf{u})\cdot\mathbf{e}_r = \nabla^2 u_r -\frac2{r^2}\frac{\partial u_\theta}{\partial\theta}-\frac{u_r}{r^2}. $$