rewrite Euler equation to the next form I am trying to rewrite the Euler equation:
$\frac{Du}{Dt}=-\frac{1}{\rho}\nabla p +g$
to the form:
$\frac{\partial u}{\partial t}+(\nabla \times u) \times u = -\nabla(\frac{p}{\rho}+\frac{1}{2}u^2+\phi)$
that basically means to prove that vector identity given there.
I'm trying to use:


*

*$\frac{D}{Dt}=\frac{\partial}{\partial t}+(u\nabla)$

*$\nabla (u^2) = 2(u\nabla)u-2(\nabla \times u) \times u$
I think that it's easy, but I really don't know how to solve it!
Please help me.
 A: Looks pretty straight-forward to me. You start with
$$\frac{Du}{Dt}=-\frac{1}{\rho}\nabla p +g$$
Now use $\frac{D}{Dt}=\frac{\partial}{\partial t}+(u\nabla)$ on the left hand side to get
$$\frac{\partial u}{\partial t}+(u\nabla)u=-\frac{1}{\rho}\nabla p +g$$
Then use $\nabla (u^2) = 2(u\nabla)u-2(\nabla \times u) \times u$ on the second term of the left hand side (first solving for $u\nabla u$, of course) to get:
$$\frac{\partial u}{\partial t}+\frac12 \nabla(u^2) + (\nabla \times u) \times u=-\frac{1}{\rho}\nabla p +g$$
Assuming $\rho$ is constant and $g=-\nabla \phi$, you then get
$$\frac{\partial u}{\partial t}+  (\nabla \times u) \times u=-\nabla(\frac{1}{\rho}\nabla p + \frac12 u^2 +\phi),$$
where $\nabla(u^2)$ has been moved to the right hand side and the $\nabla$ pulled out using the aformentioned assumptions.
A: Remember that $g =-\nabla \phi$ (conservative 'force')
And then you have what you were looking for.
$$\frac{\partial u}{\partial t} + u\nabla u = -\nabla (\frac{p}{\rho}+\phi) $$
$$\Leftrightarrow \frac{\partial u}{\partial t} + (\nabla u)u +\frac{1}{2}\nabla u^2 = -\nabla (\frac{p}{\rho}+\phi)$$
and that's it.
