# Vectors into Navier–Stokes equation for an incompressible fluid

If I have velocity u and vorticity $${\bf w}=\nabla \times {\bf u}$$

and the equation

$$({\bf u}\cdot \nabla){\bf u}=-{\bf u} \times {\bf w} +\nabla\frac{1}{2}|{\bf u}|^2 \tag{1}$$

How do I show that that the above equation can be cast into the form

$$\frac{\partial {\bf u}}{\partial t} - {\bf u}\times {\bf w}=-\nabla(\frac{p}{\rho} +\frac{1}{2}|{\bf u}|^2)-\nu\nabla \times w$$

where $p$=pressure and $\nu=\frac{\mu}{\rho}$

• It'd be useful if you could add the version of the NS equation you want to start from – caverac Oct 17 '17 at 14:50
• whoops I'm sorry its the Navier stokes equation for incompressible fluid, using the equation1) transform it into equation 2) Can be seen here on page 2 arxiv.org/pdf/1502.01206.pdf but I want to know how you derive it. – Rich Oct 17 '17 at 15:10

$\newcommand{\vect}[1]{{\bf #1}}$ Start with the Navier-Stokes equation

$$\frac{\partial \vect{u}}{\partial t} + (\vect{u} \cdot \nabla)\vect{u} - \nu\nabla^2\vect{u} = -\nabla \frac{p}{\rho_0} + \vect{g} \tag{1}$$

where $\rho_0$ is the (constant) density, $\vect{g}$ represent external forces and $\nu = \mu/\rho_0$ is the kinematic viscosity. I will assume you know how to show that

$$(\vect{u}\cdot \nabla)\vect{u} = \color{blue}{\nabla \frac{1}{2}\vect{u}^2 + (\nabla \times \vect{u})\times \vect{u}} \tag{3}$$

Evaluating this into Eq. (1) you get

\begin{eqnarray} \frac{\partial \vect{u}}{\partial t} + \color{blue}{\nabla \frac{1}{2}\vect{u}^2 + (\nabla \times \vect{u})\times \vect{u}} - \nu\nabla^2\vect{u} &=& -\nabla \frac{p}{\rho_0} + \vect{g} \\ \Rightarrow~~~ \frac{\partial \vect{u}}{\partial t} + (\nabla\times\vect{u})\times\vect{u} - \nu\nabla^2\vect{u} &=& -\nabla \frac{p}{\rho_0} - \nabla\frac{1}{2}\vect{u}^2 + \vect{g}\\ \Rightarrow~~~ \frac{\partial \vect{u}}{\partial t} + \vect{\omega}\times\vect{u} -\nu\nabla^2\vect{u} &=& -\nabla\left(\frac{p}{\rho_0} + \frac{1}{2}\vect{u}^2\right) + \vect{g} \tag{4} \end{eqnarray}

Finally I will use the identity

$$\nabla^2\vect{u} = \nabla(\underbrace{\nabla \cdot\vect{u}}_{=0}) -\nabla\times(\nabla\times\vect{u}) = \color{red}{-\nabla\times(\nabla\times\vect{u})} \tag{5}$$

Replacing this into Eq. (4)

$$\frac{\partial \vect{u}}{\partial t} + \vect{\omega}\times\vect{u} +\nu\color{red}{\nabla\times(\nabla\times\vect{u})} = -\nabla\left(\frac{p}{\rho_0} + \frac{1}{2}\vect{u}^2\right) + \vect{g} \tag{6}$$

Rearranging this will led you to

$$\frac{\partial \vect{u}}{\partial t} - \vect{u}\times\vect{\omega} = -\nabla\left(\frac{p}{\rho_0} + \frac{1}{2}\vect{u}^2\right) - \nu\nabla\times\vect{\omega} + \vect{g}$$

If $\vect{g} = -\nabla\phi$, that is, the force is conservative, you can also write this as

$$\frac{\partial \vect{u}}{\partial t} - \vect{u}\times\vect{\omega} = -\nabla\left(\frac{p}{\rho_0} + \frac{1}{2}\vect{u}^2 + \phi\right) - \nu\nabla\times\vect{\omega}$$

• you're amazing man thank you – Rich Oct 17 '17 at 18:12