= Solution
Choose the homogeneous dilation $\boxed{r(R)=tR}$, which has the required endpoint values. For a radial field, put $J=r'r^2/R^2$. The radial energy density, after omitting its constant angular factor, is $R^2[r'^2+2r^2/R^2+(J-1)^2]$. Its Euler equation simplifies to
$$
R^2r''+2Rr'-2r+r^2J'=0.
$$
The chosen dilation has $r''=0$, $J=t^3$ constant, and satisfies it. More directly, the full first Piola-Kirchhoff response for the given energy is
$$
T(F)=2F+2(\det F-1)\operatorname{cof}F.
$$
At $F=tI$ 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 is
$$
\boxed{E_{\mathrm{hom}}(t)=\frac{4\pi}{3}[3t^2+(t^3-1)^2].}
$$
This grows like $(4\pi/3)t^6$, whereas the same-boundary cavitating competitor has energy asymptotic to $8\pi t^2$. Therefore \b[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>.
\Image[/past-exam-of-the-solid-mechanics-course-of-the-university-of-oxford/2010-cavitation-energy.png]
{title=Energy comparison in the finite-energy class: the homogeneous dilation grows like t to the sixth power while the cavitating competitor grows quadratically}
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