Pressure on a compressible solid, dynamics of the relaxation I have a sphere made of a solid of bulk compressibility B, and of volumic mass $\rho_0$.
The pressure around that sphere is $P_0$.
At time $t=0$, we apply a pressure on the sphere $P=P_0+\Delta P$. The way we apply it is with some solid stuff (a vise). 
I would like to know how the sphere is going to get deformed, if we assume a spherical symmetry.
I assume we have to write (at least) the following equations :
$\left\{ \begin{split} 
& \partial_t(\rho v)+v\partial_r(\rho v)=-\partial_{r}p \\
& \partial_t \rho+\partial_r(\rho v)=0 \\
     \end{split} \right.$ 
but I don't know how to write $\partial_r p$, meaning how the stress depends on the radius knowing that we know it outside.
 A: Your system of differential equations also needs an equation for the pressure. To find this, you can use the definition for bulk compressibility
$\frac{1}{\rho} \frac{\partial \rho}{ \partial p} = B$.
For solids it is typical to assume that the density varies only slightly with pressure. $\rho_0$ is the density of the solid when the pressure is $p=P_0$. Thus, we can assume
$\rho- \rho_0 =B\rho_0(p - P_0)$
since $\frac{\partial \rho}{ \partial p}$ is a small quantity in comparison with $\rho_0$. In general, if the pressure difference $\Delta P$ is very high, the following holds (constant compressibility assumed):
$\rho = \rho_0 \exp(B(p-P_0))$.
A: The equations you have written so far are incorrect.  In spherical coordinates (for a spherically symmetric deformation), the continuity equation should read:
$$\frac{\partial \rho}{\partial t}+\frac{1}{r^2}\frac{\partial}{\partial r}\left(\rho r^2v\right)=0\tag{1}$$
And the equations of motion should read:
$$\rho\left(\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial r}\right)=\left[\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\sigma_{rr})-\frac{(\sigma_{\theta \theta}+\sigma_{\phi \phi})}{r}\right]$$and $$\sigma_{\theta \theta}=\sigma_{\phi \phi}\tag{2}$$where the $\sigma$'s are the components of the stress tensor.  These are related to the displacement in the radial direction by the 3D version of Hooke's law.
(@Chemomechanics answer gives the steady state solution to these equations, but not the transient solution.  The transient solution is anisotropic in stress and strain.)
COMPLETION OF THE FORMULATION
The principal directions of stress and strain in this symmetric problem are in the radial, latitudinal, and longitudinal directions.  The principal strains are as follows:
$$\epsilon _{rr}=\frac{\partial u}{\partial r}$$
$$\epsilon_{\theta \theta}=\epsilon_{\phi \phi}=\frac{u}{r}$$where u is the displacement in the radial direction.
From Hooke's 3D law for the stress-strain behavior of elastic solids, it then follows that the principal stresses are
$$\sigma_{rr}=\frac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\epsilon_{rr}+\nu(\epsilon_{\theta \theta}+\epsilon_{\phi \phi})]=\frac{E}{(1+\nu)(1-2\nu)}\left[(1-\nu)\frac{\partial u}{\partial r}+2\nu\frac{u}{r}\right]$$
$$\sigma_{\theta \theta}=\sigma_{\phi \phi}=\frac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\epsilon_{\theta \theta}+\nu(\epsilon_{rr}+\epsilon_{\phi \phi})]=\frac{E}{(1+\nu)(1-2\nu)}\left[\nu\frac{\partial u}{\partial r}+\frac{u}{r}\right]$$where E is Young's modulus and $\nu$ is Poisson's ratio.
If we substitute these relationships between the stresses and the displacements into Eqn. 2, the differential momentum balance equation, we obtain:
$$\rho_0\frac{\partial ^2 u}{\partial t^2}=\frac{E(1-\nu)}{(1+\nu)(1-2\nu)}\left[\frac{\partial^2 u}{\partial r^2}+\frac{2}{r}\frac{\partial u}{\partial r}-2\frac{u}{r^2}\right]$$where quadratic terms in the displacement have been neglected.  
Boundary and initial conditions on this problem are:
$\sigma_{rr}=\frac{E}{(1+\nu)(1-2\nu)}\left[(1-\nu)\frac{\partial u}{\partial r}+2\nu\frac{u}{r}\right]=-(P-P_0)$ at r = R, t > 0
and
$u=0$ at t=0, all r
