 Solution

Choose the homogeneous dilation $\boxed{r(R)=tR}$, which has the required endpoint values. For a radial field, put $J=r^{\prime }{r}^2/R^2$. The radial energy density, after omitting its constant angular factor, is $R^2[r^{\prime 2}+2r^2/R^2+(J-1{)}^2]$. Its Euler equation simplifies toThe chosen dilation has $r^{\prime \prime }=0$, $J=t^3$ constant, and satisfies it. More directly, the full first Piola-Kirchhoff response for the given energy isAt $F=tI$ this is $[2t+2(t^3-1)t^2]I$, constant throughout the body. Its divergence vanishes, so the deformation satisfies the full force-free equilibrium equations as well.

Its energy isThis grows like $(4\pi /3)t^6$, whereas the same-boundary cavitating competitor has energy asymptotic to $8\pi t^2$. Therefore for sufficiently large $t$, the homogeneous equilibrium is not a global energy minimum in the finite-energy class admitting the cavitating deformation.

The admissible class matters: the competitor from the preceding part is not continuous at the center. This comparison is in that Sobolev class, not a claim about minimization restricted to smooth cavity-free deformations.