Assume that $u = (u^1, u^2, u^3)$ solves the evolution equations of linear elasticity: $$u_{tt}-µ \Delta u − (λ + µ) D (\nabla\cdot u) = 0$$ in $\mathbf{R}^3 × (0, ∞)$. Show that $w := \nabla \cdot u $ and $w := \nabla \times u$ each solve wave equations, but with differing speeds of propagation.
This is problem 21 in chapter 2 of Evan's PDE.
I am able to do this problem when $ w := \nabla \times u$ to obtain $w_{tt} = \mu \Delta w$. For $ w:= \nabla \cdot u$, I am not recognizing how to proceed from
$$ w_{tt} = \mu \Delta w + (\lambda + \mu) \nabla (\nabla \cdot w) $$
to $$w_{tt} = \mu (\Delta w) + (\lambda + \mu)(\Delta w)$$
Since $ w = \nabla \cdot u$ is a scalar, I am not sure how the divergence of $w$ is defined here. The identity $\Delta w = \nabla(\nabla \cdot w) - \nabla \times \nabla \times w $ would be useful here but I can't see why the curl of curl of $w$ would be zero in this case or even defined when $w$ is a scalar. Maybe I am misunderstanding something in the statement of the problem? Any help will be appreciated.
Evans means that $\nabla\cdot u$ solves a scalar-valued wave equation and $\nabla\times u$ solves a vector-valued wave equation. Your issue is that you are overloading your derivative operators, in particular $\Delta$. There is a vector Laplacian and a scalar Laplacian in play here. The identity you stated for $\Delta w$ is for a vector Laplacian and only makes sense for a vector-valued function $w$. Since $w = \nabla\cdot u$ is a scalar function, one should use the definition of the scalar Laplacian $\Delta w = \nabla\cdot(\nabla w)$ instead.