Deriving the Poisson equation for pressure

Question

I am trying to derive the Poisson equation for pressure, specifically that $$bigtriangledown^2(p + frac{1}{2}|vec{u}|^2) = bigtriangledown cdot(vec{u}times vec{w}) + varepsilon kappa^2psicdotbigtriangledownphi $$ where the vorticity field is given by $$w = bigtriangledowntimes vec{u}$$
I am unsure how to manipulate the left-hand side to reach the right-hand side.I know that I can write $$|vec{u}|^2 = |bigtriangledownphi|^2 = |bigtriangledownpsi|^2$$ But I am unsure what else I can do.This problem is in reference to a particular paper I am reading,so some of the notation on the RHS may not be known without context,however,I just need help with the manipulation of the LHS.

Thanks, any help is greatly appreciated.

talrefae · Answer

Given the Euler equation (the inviscid Navier-Stokes equation, so $mu = 0$),
$$ partial_t mathbf{u} + mathbf{u} cdot nabla mathbf{u} = -nabla p + mathbf{F}$$
we can exploit vector identities to rewrite the advective acceleration term as the gradient of the kinetic energy (density), minus a vorticity-velocity coupling 
$$mathbf{u} cdot nabla mathbf{u} = nablaleft(frac{1}{2} mathbf{u}^2right) - mathbf{u} times mathbf{omega}$$
(where $mathbf{omega} = nabla times mathbf{u}$, like you define.) So the momentum equation becomes
$$ partial_t mathbf{u} + nablaleft(frac{1}{2} mathbf{u}^2right) - mathbf{u} times mathbf{omega} = -nabla p + mathbf{F}$$
now take the divergence of both sides (noting that incompressibility lets us set $nabla cdot partial_t mathbf{u} = partial_t nabla cdot mathbf{u} = 0$)

$$ nabla^2 left( p + frac{1}{2} mathbf{u}^2 right) = nabla cdot (mathbf{u} times mathbf{omega}) + nabla cdot mathbf{F}$$

where, for you, apparently,
$$nabla cdot mathbf{F} = varepsilon kappa^2 psi cdot nabla phi$$

but you have not specified the exact form of $mathbf{F}$. I'm sure it will be easy to retrace, though.

Deriving the Poisson equation for pressure

One Answer

Add your own answers!

Ask a Question