Choose the homogeneous dilation $\boxed{r(R)=tR}$, which has the required endpoint values. For a radial field, put $J=r^{\prime }{r}^2/R^2$. The radial energy density, after omitting its constant angular factor, is $R^2[r^{\prime 2}+2r^2/R^2+(J-1{)}^2]$. Its Euler equation simplifies toThe chosen dilation has $r^{\prime \prime }=0$, $J=t^3$ constant, and satisfies it. More directly, the full first Piola-Kirchhoff response for the given energy isAt $F=tI$ 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 isThis grows like $(4\pi /3)t^6$, whereas the same-boundary cavitating competitor has energy asymptotic to $8\pi t^2$. Therefore 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.

The determinant condition is $r^{\prime }(r/R{)}^2=1$, or $r^2{r}^{\prime }=R^2$. Integration gives $r^3=R^3+C$ and the outer boundary value fixes $C=t^3-1$. ThusFor $t>1$, $r(R)$ tends to the positive cavity radius $(t^3-1{)}^{1/3}$ as $R\downarrow 0$. Its tangential stretches grow like that radius divided by $R$, so $|F|\to \infty$. The deformation is singular at the origin and is interpreted as a finite-energy Sobolev deformation, with a point value irrelevant to its energy.

Since $r^{\prime }=R^2/r^2$ and the determinant penalty is zero almost everywhere,For fixed $t>1$, the first integrand is $O(R^6)$ near zero and the second is bounded. Thus the energy is finite despite the unbounded gradient. In particular the gradient is square integrable, as required for this competitor.

For large $t$, $r(R)/t\to 1$ uniformly on $[0,1]$. The first integral is $O(t^{-4})$, while ${\int }_0^{1}r(R{)}^{2}dR\sim t^2$. ConsequentlyThe spherical volume factor $R^{2}dR$ is what makes the $R^{-1}$ tangential-gradient singularity integrable in three dimensions.

For $R>0$, differentiation gives ${\partial }_{j}R={\theta }_j$ and ${\partial }_j{\theta }_i=({\delta }_{ij}-{\theta }_i{\theta }_j)/R$. HenceThe two matrices are the orthogonal projections onto the tangent plane of the reference sphere and its radial line. The gradient multiplies the radial direction by $r^{\prime }$ and each tangent direction by $r/R$. With positive $r,r^{\prime }$, it is itself symmetric positive definite, so its eigenvalues are its principal stretches:

A deformation maps a material point in the reference body to its current position, $y:\Omega \to {\mathbb{R}}^3$, with deformation gradient $F=Dy$. A homogeneous deformation has constant gradient and is affine, $y(X)=FX+b$.

If $\det F>0$, the matrix $F^{T}F$ is symmetric positive definite. The matrix-square-root theorem says that any such matrix has a unique symmetric positive definite square root, obtained by taking positive square roots in its orthogonal spectral decomposition. SetThen $R^{T}R=U^{-1}{F}^{T}F{U}^{-1}=I$ and $\det R=\det F/\det U>0$, so $R\in \mathrm{SO}(3)$ and $F=RU$. If also $F=\tilde{R}\tilde{U}$, then $F^{T}F={\tilde{U}}^2$; uniqueness of the positive square root gives $\tilde{U}=U$ and then $\tilde{R}=R$. This proves existence and uniqueness of polar decomposition.

The principal stretches are the positive eigenvalues of $U$, equivalently the singular values of $F$ or square roots of the eigenvalues of $F^{T}F$.

As a function of the full deformation gradient, use $e=\mathrm{sym}(F-I)$. Let $Q=\mathrm{diag}(1,-1,-1)$, a proper rotation. Assumed full frame indifference would give $\tilde{W}(Q)=\tilde{W}(I)=0$. For $F_s=I+s(Q-I)$, its infinitesimal strain is $s\mathrm{sym}(Q-I)$. Quadratic homogeneity therefore givesChoose $s=1/4$. ThenThe homogeneous deformation $y(X)=F_{s}X$ changes volume but has zero energy. This illustrates why a nondegenerate small-strain quadratic energy cannot also be imposed as a fully frame-indifferent finite-deformation energy.

For small displacement gradients write $F=I+H$. Taylor expansion around the stress-free reference state givesThe linear term vanishes because $DW(I)=0$. Drop the irrelevant constant and retain the quadratic term:The major symmetry $\boxed{C_{i\alpha j\beta }=C_{j\beta i\alpha }}$ follows from equality of mixed second derivatives. For a skew matrix $K$, stress covariance along $e^{tK}$ givesDifferentiation at zero yields $D\hat{T}(I)[K]=C[K]=0$. Thus $C_{i\alpha j\beta }{K}_{j\beta }=0$ for every skew matrix, which forces the input minor symmetry $C_{i\alpha j\beta }=C_{i\alpha \beta j}$. Combining it with the major symmetry gives the output minor symmetry as well:These identify reference and spatial indices at $F=I$. Decompose $H=e+K$ into symmetric and skew parts. Since $C[K]=0$ and the major symmetry also eliminates the other cross term,Thus infinitesimal rotations cost no energy in the quadratic approximation.

A homogeneous hyperelastic material has a stored energy per reference volume $W(F)$ depending on the local deformation gradient but not explicitly on the material point. Its first Piola-Kirchhoff response isA superposed rigid rotation changes the observer's description without changing the energy of the deformation. Material frame indifference is thereforeThis left-rotation invariance is distinct from a material isotropy assumption on right rotations. Differentiate it in an arbitrary matrix direction $H$:The Frobenius inner product then gives $Q^T\hat{T}(QF)=\hat{T}(F)$, proving $\hat{T}(QF)=Q\hat{T}(F)$.

For the converse, that stress covariance makes the derivative in $F$ of $W(QF)-W(F)$ zero. The positive-determinant matrices form a connected set, as also follows from polar decomposition, so this difference is a constant $c(Q)$ independent of $F$. Composition gives $c(Q_1{Q}_2)=c(Q_1)+c(Q_2)$. For fixed $F$, continuity of $W$ and compactness of $\mathrm{SO}(3)$ make $c$ bounded. But $c(Q^n)=nc(Q)$ for every positive integer, forcing $c(Q)=0$. Thus stress covariance and frame indifference of the energy are equivalent. Compactness removes the possible additive energy constant in the converse.

The radial gradient $F=D\mathbf{y}$ is symmetric positive definite. If it belongs to a well $SO(3)U_i$, write $F=Q{U}_i$. Then $F^{T}F=U_i^2$, while positive symmetry makes $F$ itself the unique positive square root of an operator of $F^{T}F$. Hence $F=U_i$. Likewise a symmetric positive-definite matrix in $SO(3)$ must be $I$. ThereforeFor each radius, the tangential stretch $r(R)/R$ must belong to the finite set of eigenvalues of these matrices. The ratio is continuous on the connected radial interval, so it is one constant $v>0$. Thus $r(R)=vR$ and differentiation gives $r^{\prime }=v$ as well. HenceThis proof uses finiteness and positive definiteness of the wells, not isotropy or the Baker-Ericksen inequalities. It also covers $R_1=0$: the punctured three-dimensional ball remains connected, and no condition at the excluded center was used.

At every point the deformation gradient attains the differentiable energy's minimum, so $DW(D\mathbf{y})=0$ and it is a natural state. By the strict Baker-Ericksen argument just proved, all its principal stretches are equal. In the radial formula this givesThus $r(R)=vR$ for one constant $v>0$ on the connected radial interval, andThe constancy is global through the annulus, not a separately chosen dilation at each point.

Using $J=1$ and the preceding first Piola-Kirchhoff stress tensor,Its eigenvalues are $\mu (1+\sqrt{5})/2$, $\mu (1-\sqrt{5})/2$, and zero. Because it is symmetric, its largest traction magnitude on a unit spatial normal is the largest absolute eigenvalue. Put $\phi =(1+\sqrt{5})/2$. The positive eigenvalue has the strictly largest modulus, with normalThe question asks for the plane in the reference configuration, so this spatial normal must be pulled back. Nanson's formula gives $N$ parallel to $F^{T}n$, andTherefore the required plane and maximum areEither sign of its unit normal describes the same plane. Maximizing over reference normals is equivalent to maximizing over spatial normals because the map $N\mapsto F^{-T}N/|F^{-T}N|$ is a bijection of unit normals.

We first derive the stress and energy facts used in both parts. The Cauchy traction $s=Tn$ is force per unit area in the deformed configuration. The first Piola-Kirchhoff traction $s_R=T_{R}N$ is force per unit reference area. Their force vectors are measured in the same spatial coordinates, but the area measures and the normals used to specify the surface are different.

Parametrize a reference surface by $x=x(u,v)$. Its oriented area vector is $(x_u\times x_v)dudv$. Under deformation, the two tangents become $F{x}_u,F{x}_v$. The cross-product identitytherefore proves Nanson's formula $nda=J{F}^{-T}NdA$. Equality of the physical force on corresponding surface patches now givesSince this holds for every reference normal,The first Piola-Kirchhoff stress tensor is generally not symmetric, whereas the Cauchy stress tensor is symmetric in the absence of couple stresses.

For the stated energy on orientation-preserving gradients, differentiation givesA stress-free state satisfies $F{F}^T=J^{-1}I$. Taking determinants yields $J^2=J^{-3}$, so $J=1$ and $F{F}^T=I$: precisely $F\in SO(3)$. These states are also the global energy minimizers. If $v_i>0$ are the singular values, the arithmetic-geometric mean inequality gives $|F{|}^2\ge 3J^{2/3}$, with equality only when the singular values are equal. The function $3J^{2/3}/2+J^{-1}$ decreases up to $J=1$ and increases thereafter, with minimum $5/2$. Thus $W\ge 5\mu /2$, with equality precisely at rotations. A constant shift of the energy can set this minimum to zero without changing the stress.

For the given simple shear,For a unit normal $N$, $|T_{R}N{|}^2={\mu }^2(N_1^2+N_2^2)={\mu }^2(1-N_3^2)$. Hence every reference plane whose normal lies in the $e_1,e_2$ plane is maximizing:There is a family of maximizing planes, rather than one distinguished normal.