If we substitute the equations for the stresses into the differential force balance of the previous post, we obtain $$\rho\frac{\partial ^2u}{\partial t^2}=(\lambda +2\mu)\frac{\partial}{\partial r}\left(\frac{1}{r^2}\frac{\partial(r^2u)}{\partial r}\right)$$or, equivalently, $$\rho\frac{\partial ^2\xi}{\partial t^2}=(\lambda +2\mu)r^2\frac{\partial}{\partial r}\left(\frac{1}{r^2}\frac{\partial\xi}{\partial r}\right)$$where ##\xi=r^2u##. At r = R, the boundary condition is $$\sigma_{rr}=\lambda\left(\frac{\partial u}{\partial r}+2\frac{u}{r}\right)+2\mu\frac{\partial u}{\partial r}=-P$$or, equivalently, $$(\lambda+2\mu)\frac{1}{r^2}\frac{\partial (r^2u)}{\partial r}-4\mu\frac{u}{r}=-P$$or$$(\lambda+2\mu)\frac{\partial \xi}{\partial r}-4\mu\frac{\xi}{R}=-PR^2$$
The boundary condition at large r is ##\xi=0##.
Initially the displacement is zero at all locations $$\xi(0,r)=0$$The initial velocity is also zero at all locations, except at the very boundary r = R (involving zero mass in the limit), where it is finite. From the solution for a sudden force applied to the end of a rod (at short times, the compression wave is very close to the boundary in our system, so that spherical curvature can be neglected at short times), we find that, for our case, $$\left(\frac{du}{dt}\right)_{t=0,r=R}=\frac{P}{\sqrt{\rho(\lambda+2\mu)}}$$or, equivalently, $$\left(\frac{d\xi}{dt}\right)_{t=0,r=R}=\frac{PR^2}{\sqrt{\rho(\lambda+2\mu)}}$$
This completes the formulation of the linearly elastic model equations for a suddenly pressurized spherical cavity.