Wave equation from Navier Stokes I have the following equations, similar to the Navier Stokes equations for an incompressible fluid
$$
\frac{1}{c_s^2}P_t + \nabla \cdot u = 0 \\
u_t + u\cdot \nabla u = -\nabla P + \nu \nabla^2 u
$$
According to my book, one can derive a damped wave equation from these two equations for the velocity $u$. However, I have differentiated the second equation wrt. $t$ and inserted the first equation. 
This does not give me anything useful at all. Can someone reproduce this damped wave equation from these two expressions?
 A: Actually, you've basically gotten there already. I'm pretty sure that the claim is only true under irrotational assumptions; at least, all other previous times I've seen a wave equation derived from Navier-Stokes/Euler the irrotational assumption is enforced. So you need to use that. To make use of that assumption you also need to use a vector identity. 
I'll for now throw away the nonlinear terms just for ease of typing, and use $\triangle$ for the Laplacian:
$$ u_{tt} + \nabla P_t - \nu \triangle u_t = \text{nonlinear} $$
$$ u_{tt} - \nabla (c_s^2 \nabla\cdot u) - \nu^2 \triangle\triangle u + \triangle \nabla P = \text{nonlinear} $$
And for the second term you use the vector identity
$$ \nabla \times (\nabla \times u) = \nabla(\nabla\cdot u) - \triangle u $$
so you have
$$ u_{tt} - c_s^2 \triangle u - \nu^2\triangle^2 u + \triangle \nabla P - c_s^2 \nabla\times(\nabla\times u) = \text{nonlinear} $$
In the irrotational case you have $\nabla \times u \equiv 0$ and the equation reduces to damped nonlinear wave equation. 
