# Navier Stokes equation in cylindrical coordinates

A book on fluid mechanics gives the NS:
$$\dfrac{\partial\boldsymbol{u}}{\partial t}+(\boldsymbol{u}\cdot\boldsymbol{\nabla})\boldsymbol{u}=-\frac{1}{\rho}\nabla p+\nu\boldsymbol{\nabla}^2\boldsymbol{u}+\boldsymbol{g}$$
in cylindrical components:
$$\dfrac{\partial u_r}{\partial t}+(\boldsymbol{u}\cdot\nabla)u_r-\frac{u_{\theta}^2}{r}=-\frac{1}{\rho}\dfrac{\partial p}{\partial r}+\nu\left(\nabla^2u_r-\dfrac{u_r}{r^2}-\dfrac{2}{r^2}\dfrac{\partial u_{\theta}}{\partial\theta}\right)$$
$$\frac{\partial u_{\theta}}{\partial t}+(\boldsymbol{u}\cdot\nabla)u_{\theta}+\frac{u_{\theta}u_r}{r}=-\dfrac{1}{\rho r}\dfrac{\partial p}{\partial\theta}+\nu\left(\nabla^2u_{\theta}-\frac{u_{\theta}}{r^2}+\dfrac{2}{r^2}\dfrac{\partial u_r}{\partial\theta}\right)$$
and an obvious $$z$$-part.
I can see where the extra terms on the left come from (namely from $$\frac{u_{\theta}}{r}\frac{\partial}{\partial\theta}$$ applied to $$u_{\theta}\boldsymbol{e_{\theta}}$$ and $$u_r\boldsymbol{e_r}$$ respectively, where $$\boldsymbol{u}=u_r\boldsymbol{e_r}+u_{\theta}\boldsymbol{e_{\theta}}+u_z\boldsymbol{e_z}$$; I also see where the terms $$-\frac{u_r}{r^2}$$ and $$\frac{u_{\theta}}{r^2}$$ on the right come from, namely $$\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}$$ applied to $$u_r\boldsymbol{e_r}$$ and $$u_{\theta}\boldsymbol{e_{\theta}}$$ respectively. (Here
$$(\boldsymbol{u}\cdot\nabla)=u_r\frac{\partial}{\partial r}+\frac{u_{\theta}}{r}\frac{\partial}{\partial\theta}+u_z\frac{\partial}{\partial z}$$ and
$$\nabla^2=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial}{\partial r})+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}+\frac{\partial^2}{\partial z^2}$$.)
I do not see where $$-\frac{2}{r^2}\frac{\partial u_{\theta}}{\partial\theta}$$ and $$\frac{2}{r^2}\frac{\partial u_r}{\partial\theta}$$ come from though. By applying any of the Laplacian terms to the $$u_i$$ part of $$e_iu_i$$, the result should end up in the $$e_i$$ component of the NS because the $$e_i$$ wouldn't be switched. Where do these terms come from ?

$$\vec u$$ is a 'vector', not a scalar.

Therefore, we should use the identity ,$$\nabla^2 \vec u = \nabla(\nabla \cdot \vec u) - \nabla \times (\nabla \times \vec u)$$

Let's investigate $$r$$ component.

The first term of RHS

$$\nabla\cdot\vec u=\frac{1}{r}\frac{\partial(r\cdot u_r)}{\partial r}+\frac{1}{r}\frac{\partial u_\theta}{\partial\theta}+\frac{\partial u_z}{\partial z}$$

$$\nabla(\nabla\cdot\vec u)_r = \frac{\partial}{\partial r}\left((\frac{\partial u_r}{\partial r}+\frac{u_r}{r})+\frac{1}{r}\frac{\partial u_\theta}{\partial\theta}+\frac{\partial u_z}{\partial z})\right)$$

$$=\frac{\partial^2 u_r}{\partial r^2} + \frac{1}{r}\frac{\partial u_r}{\partial r}-\frac{u_r}{r^2} + \frac{1}{r} \frac{\partial ^2 u_\theta}{\partial r \partial \theta} - \frac{1}{r^2} \frac{\partial u_\theta}{\partial \theta} + \frac{\partial^2 u_z}{\partial r \partial z}$$

The second term of RHS

$$\nabla \times \vec u = \left(\frac{1}{r}\frac{\partial u_z}{\partial\theta}-\frac{\partial u_\theta}{\partial z}\right)\hat r+\left(\frac{\partial u_r}{\partial z}-\frac{\partial u_z}{\partial r}\right)\hat\theta+\frac{1}{r}\left(\frac{\partial(r u_\theta)}{\partial r}-\frac{\partial u_r}{\partial\theta}\right)\hat z$$

$$\nabla \times (\nabla \times \vec u)_r = \frac{1}{r} \frac{\partial}{\partial \theta}\left(\frac{1}{r}\left(\frac{\partial (r u_\theta)}{\partial r}- \frac{\partial u_r}{\partial \theta}\right)\right) - \frac{\partial}{\partial z} \left(\frac{\partial u_r}{\partial z} - \frac{\partial u_z}{\partial r}\right)$$

$$=\frac{1}{r}\left(\frac{\partial ^2 u_\theta}{\partial r \partial \theta} \right)+ \frac{1}{r^2}\frac{\partial u_\theta}{\partial \theta} - \frac{1}{r^2} \frac{\partial ^2 u_r}{\partial \theta^2} -\frac{\partial^2 u_r}{\partial z^2} + \frac{\partial^2 u_z}{\partial z \partial r}$$

Then, $$\nabla^2 \vec u = \nabla(\nabla \cdot \vec u) - \nabla \times (\nabla \times \vec u) = \frac{\partial^2 u_r}{\partial r^2} + \frac{1}{r} \frac{\partial u_r}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u_r}{\partial \theta^2} + \frac{\partial^2 u_r}{\partial z^2} - \frac{u_r}{r^2} - \frac{2}{r^2} \frac{\partial u_\theta}{\partial \theta}$$

$$= \nabla^2 u_r - \frac{u_r}{r^2} - \frac{2}{r^2} \frac{\partial u_\theta}{\partial \theta}$$

• Thanks a lot. I did manage to get the correct terms without the vector identity though (I wasn't applying the terms from the Laplacian step by step and that's why I was getting an incomplete and incorrect answer).
– chb
Commented Feb 10, 2023 at 23:32