= Constant-Jacobian radial equilibrium for a stretch-sum energy
{title2=$\Phi=\mu(v_1+v_2+v_3)+h(v_1v_2v_3),\quad h''>0\implies r^3=DR^3+C$}
For a radial deformation $y=r(R)x/R$ with positive radial derivative, put $q=r/R$ and $\delta=r'q^2=\det Dy$. The principal nominal stresses are $\Phi_1=\mu+h'(\delta)q^2$ and $\Phi_2=\mu+h'(\delta)r'q$. Radial equilibrium reads $\Phi_1'=2(\Phi_2-\Phi_1)/R$. Since $q'=(r'-q)/R$, substitution cancels the terms proportional to $h'$, leaving $h''(\delta)q^2\delta'=0$. Strict convexity of $h$ and positivity of $q$ force a constant positive determinant $D$.
Integration of $r^2r'=DR^2$ gives $r^3=DR^3+C$. Two spherical boundary radii fix $D,C$ uniquely. A positive increasing cube-radius interpolation verifies positivity of $r$ and $r'$, so the algebraic solution is an admissible smooth radial equilibrium. The result is restricted to this energy family and radial class; strict convexity of its determinant term alone does not imply uniqueness among arbitrary three-dimensional deformations.
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