 Solution

Put $n=x/R$ and $q=r(R)/R$. Differentiating $y=qx$ uses $\nabla R=n$ and $\nabla q=(r^{\prime }-q)n/R$. HenceIts radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $q$. They are positive by hypothesis, so they are also the principal stretches:For an isotropic stored energy the nominal stress has the same radial/tangential projectors,where symmetry of $\Phi$ gives equal tangential derivatives. Since $\mathrm{Div}(n\otimes n)=2n/R$ and the scalar coefficients depend only on $R$,Thus equilibrium is exactly $\boxed{d{\Phi }_1/dR=2({\Phi }_2-{\Phi }_1)/R}$.

For the particular energy define $\delta =r^{\prime }{q}^2=\det F>0$. Its two derivatives areAlso $q^{\prime }=(r^{\prime }-q)/R$. Differentiating the first formula and substituting the equilibrium equation givesThe last terms cancel. Strict positivity of $h^{\prime \prime }$ and $q$ yields $\boxed{{\delta }^{\prime }=0}$, so $r^2{r}^{\prime }=D{R}^2$ for one positive constant $D$. Integration gives $r^3=D{R}^3+C$.

The displacement data mean $r(1)={\lambda }_1$ and $r(2)=2{\lambda }_2$, not $r(2)={\lambda }_2$. Solving the two scalar equations givesThe given inequality makes $D>0$. Since $r^3$ increases from ${\lambda }_1^3>0$ to $8{\lambda }_2^3$, it is positive on the whole shell, and $r^{\prime }=D{R}^2/r^2>0$. Thus the formula is smooth and admissible and satisfies equilibrium. Conversely every admissible radial equilibrium had to have this constant determinant and these two integration constants, proving uniqueness. This is a constant-Jacobian radial equilibrium for a stretch-sum energy.