= Solution
Taking the divergence of the <elastic wave in an isotropic solid>[Cauchy momentum equation] and commuting constant-coefficient derivatives gives
$$
\rho_0\frac{\partial^2}{\partial t^2}(\nabla\cdot\mathbf u)
=(\lambda+2\mu)\nabla^2(\nabla\cdot\mathbf u).
$$
Thus the dilatation is a wave of speed
$$
\boxed{c_P=\sqrt{\frac{\lambda+2\mu}{\rho_0}}}.
$$
Taking the curl, and using $\nabla\times\nabla f=0$, gives
$$
\rho_0\frac{\partial^2}{\partial t^2}(\nabla\times\mathbf u)
=\mu\nabla^2(\nabla\times\mathbf u),
$$
so the rotation propagates at
$$
\boxed{c_S=\sqrt{\frac{\mu}{\rho_0}}}.
$$
These are respectively the speeds of <P wave>[P-waves] and <S wave>[S-waves].
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