## speed of what (2)

Again this is about elastic wave.  Let me resume from this:

$\displaystyle \mu \Delta {\bf u} + (\lambda + \mu) \nabla \nabla \cdot {\bf u} = \rho \ddot{\bf u}.\ \ \ \ (1)$

The previous point is that the equation of motion above does not look very similar to simple wave equations we see for string vibration, plane wave of sound etc. in text books.

One of tricks is to take divergence for both sides, which leads to:

$\displaystyle \mu \nabla \cdot (\Delta {\bf u}) + (\lambda + \mu) \nabla \cdot (\nabla \nabla \cdot {\bf u}) = \rho \nabla \cdot \ddot{\bf u}.$

Changing the order of derivatives,

$\displaystyle \mu \Delta (\nabla \cdot {\bf u}) + (\lambda + \mu) \Delta (\nabla \cdot {\bf u}) = \rho \frac {\partial^2 (\nabla \cdot {\bf u})}{\partial t^2}.$

Then renaming the function $\nabla \cdot {\bf u}$ as ${\bf \phi}$, the equation above becomes:

$\displaystyle (\lambda + 2 \mu) \Delta {\bf \phi} = \rho \ddot {\bf \phi}.\ \ \ \ (2)$

Finally the equation (2) is surely a wave equation. And the function ${\bf \phi}$ is a wave.  Phase velocity $c$ of the wave is:

$\displaystyle c = {\sqrt \frac {\lambda + 2 \mu}{\rho}}.$

Okay I am a bit relieved now.  But what does this wave physically mean?

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