Radial deformation of an elastic solid ($Dy=(r/R)I+(r^{\prime }-r/R)\mathbf{\theta }\otimes \mathbf{\theta }$)

A spherical radial deformation is $y(x)=r(R)x/R$, with $R=|x|$ and $\mathbf{\theta }=x/R$. Differentiation gives the displayed symmetric gradient. Its radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $r/R$; when these are positive, they are the principal stretches. The determinant is $r^{\prime }(r/R{)}^2$. Positivity distinguishes an orientation-preserving radial elastic deformation from a map that collapses an interior sphere.

For a radial deformation $y=r(R)x/R$ with positive radial derivative, put $q=r/R$ and $\delta =r^{\prime }{q}^2=\det Dy$. The principal nominal stresses are ${\Phi }_1=\mu +h^{\prime }(\delta )q^2$ and ${\Phi }_2=\mu +h^{\prime }(\delta )r^{\prime }q$. Radial equilibrium reads ${\Phi }_1^{\prime }=2({\Phi }_2-{\Phi }_1)/R$. Since $q^{\prime }=(r^{\prime }-q)/R$, substitution cancels the terms proportional to $h^{\prime }$, leaving $h^{\prime \prime }(\delta )q^2{\delta }^{\prime }=0$. Strict convexity of $h$ and positivity of $q$ force a constant positive determinant $D$.

Integration of $r^2{r}^{\prime }=D{R}^2$ gives $r^3=D{R}^3+C$. Two spherical boundary radii fix $D,C$ uniquely. A positive increasing cube-radius interpolation verifies positivity of $r$ and $r^{\prime }$, 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.

Consider an orientation-preserving classical radial deformation of an elastic solid on a connected spherical annulus. If its nonnegative energy is zero everywhere and the material is isotropic with strict Baker-Ericksen inequalities, every gradient is a natural state, so $r^{\prime }=r/R$. Thus $(r/R{)}^{\prime }=0$ and $y=vx$.

A separate rigidity argument applies when the zero wells are $SO(3)\cup {\bigcup }_{i=1}^{N}SO(3)U_i$, with finitely many positive definite symmetric $U_i$. The radial gradient $F$ is positive symmetric; if $F=Q{U}_i$, then $F^2=U_i^2$, so uniqueness of the positive square root gives $F=U_i$. The rotation well similarly gives $F=I$. Hence the continuous ratio $r(R)/R$ takes values in a finite set of eigenvalues and must be constant on the connected radial interval. Differentiating $r=vR$ finishes the proof. This version needs neither material isotropy nor Baker-Ericksen inequalities and does not require an assumed continuity of $r^{\prime }$ beyond the radial-map regularity used for the ratio.