$($Pressurized hole$)$ For spherical symmetric problems in elasticity, such as a pressurized spherical hole in an infinite medium, there is only one non$-$zero displacement

$u=u_r(r),u_{}=u_{}=0$

In the spherical coordinates, the strains associated with the above displacements are

${}_{rr}=\frac{du}{dr},{}_{}={}_{}=\frac{u}{r}$

The equilibrium equation in the absence of body force is

$\frac{d{}_{rr}}{dr}+\frac{2}{r}({}_{rr}-{}_{})=0$

a$)$ Use Hooke's law and equilibrium equation, show that the displacement satisfies

$\frac{d}{dr}[\frac{1}{r^2}\frac{d}{dr}(r^{2}u)]=0$

And, therefore, the general solution under spherical symmetry is

$u=C_{1}r+C_2{r}^{-2}$

b$)$ Apply the general solution to a spherical hole subjected to internal pressure $p.$ Show that the solution is

$u=\frac{1+v}{2E}\frac{p{a}^3}{r^2},{}_{rr}=-\frac{p{a}^3}{r^3}$

c$)$ Derive the expression for ${}_{}$ and ${}_{+}.$