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?
Best Answer
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.