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 $\nu=\frac{\mu}{\rho}$

  • $\begingroup$ It'd be useful if you could add the version of the NS equation you want to start from $\endgroup$ – caverac Oct 17 '17 at 14:50
  • $\begingroup$ 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. $\endgroup$ – 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

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

  • $\begingroup$ you're amazing man thank you $\endgroup$ – Rich Oct 17 '17 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.