Wave equation from Navier Stokes

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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?

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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.