Past exam of the solid mechanics course of the University of Oxford / 2010 / 2

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 $F^{T}F$.

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:

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.

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.