= Solution
Divide the <Navier-Cauchy equation> by <mass density>. Since $\mu/\rho=c_T^2$ and $(\lambda+\mu)/\rho=c_L^2-c_T^2$, it becomes
$$
\boxed{u_{tt}=c_T^2\Delta u+(c_L^2-c_T^2)\nabla\operatorname{div}u.}
$$
Equivalently, using the <vector calculus> identity $\Delta u=\nabla\operatorname{div}u-\nabla\times(\nabla\times u)$,
$$
\boxed{u_{tt}=c_L^2\nabla\operatorname{div}u-c_T^2\nabla\times(\nabla\times u).}
$$
The <divergence> and <curl> consequently propagate at $c_L$ and $c_T$, respectively.
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