 Finite-energy radial cavitation versus homogeneous dilation

For the unit ball and energy $W(F)=|F{|}^2+(\det F-1{)}^2$, boundary dilation $y=tx$ admits a singular incompressible radial competitor $r(R)=(R^3+t^3-1{)}^{1/3}$. Its radial stretch is $R^2/r^2$ and its double tangential stretch is $r/R$. Although the latter diverges at zero, the three-dimensional volume element makes its squared gradient integrable. Its energy is $4\pi {\int }_0^1[R^6/r^4+2r^2],dR\sim 8\pi t^2$.

The smooth homogeneous equilibrium $y=tx$ has constant stress and energy $(4\pi /3)[3t^2+(t^3-1{)}^2]$, of sixth-order growth. It is therefore beaten for large $t$ in the finite-energy Sobolev space class admitting the cavity. This conclusion depends on that class: the singular map is not continuous at the center and cannot simply be substituted as a smooth cavity-free competitor.