Past exam of the solid mechanics course of the University of Oxford / 2007 / 3 / Solution

For a hyperelastic material $T_R(F)=DW(F)$. Material frame indifference of this first Piola-Kirchhoff law meansThe spatial force components rotate under a superposed rigid rotation, whereas the reference normal does not.

If $W(QF)=W(F)$, differentiate with respect to $F$ in direction $H$ to obtain $DW(QF):(QH)=DW(F):H$, hence the displayed stress transformation. Conversely, assume that stress transformation. For fixed $Q$, the derivative of $W(QF)-W(F)$ with respect to $F$ vanishes. The positive-determinant matrix domain is connected, as follows from polar decomposition and the connectedness of $SO(3)$ and of the positive-definite cone. Thus $W(QF)-W(F)=c(Q)$, independent of $F$. Iterating givesThe assumed lower bound on $W$ forces $c(Q)=0$: either nonzero sign would tend to minus infinity for one direction of integer $n$. This proves the required equivalence and explains the use of the lower-bound hypothesis.

In the frame-indifferent setting, isotropy means invariance under reference rotations as well: $W(FQ)=W(F)$ for $Q\in SO(3)$. The distinction matters: left rotations encode changes of observer, and right rotations encode material symmetry.

An orientation-preserving singular value decomposition is $F=R\mathrm{diag}(v_1,v_2,v_3)S$, with $R,S\in SO(3)$ and $v_i>0$. Material frame indifference and isotropy remove $R,S$, leaving a function $\Phi$ of the stretches. Permuting the stretches using signed permutation rotations does not change the value, so $\Phi$ is symmetric. Conversely a symmetric stretch function is unchanged by both left and right rotations, in particular is isotropic.

The eigenvalues of $B=F{F}^T$ are $v_i^2$. Its three principal invariants areThey determine the unordered positive eigenvalues as the roots of ${\lambda }^3-I{\lambda }^2+II\lambda -III=0$. Therefore a symmetric stretch function defines a well-defined $h(I,II,III)$ on the admissible invariant triples. Conversely a function of these invariants is a symmetric stretch function, proving the equivalence of all three descriptions.

The printed domain of $\Phi$ must be $(0,\infty {)}^3$, since it takes three arguments. Also isotropy alone as right invariance would not imply the spectral representation without material frame indifference: $\mathrm{tr}(AF{F}^T)$ for a fixed nonscalar symmetric positive-definite $A$ is right invariant but not generally objective. The equivalence here uses the frame-indifferent constitutive setting.

For the specific energy, Newton identities for the three eigenvalues gives ${\sum }_i{v}_i^6=I^3-3III+3III$, while $v_1{v}_2{v}_3=J=\sqrt{III}$. ThusTo differentiate it, useThese follow from $I=|F{|}^2$, the displayed trace formula for $II$, and $III=J^2$. The three partial derivatives of $h$ are $h_I=\alpha +3\beta I^2-3\beta II$, $h_{II}=-3\beta I$ and $h_{III}=3\beta -\gamma II{I}^{-3/2}/2$. The chain rule therefore givesFinally $-6\beta II=-3\beta (\mathrm{tr}B{)}^2+3\beta \mathrm{tr}(B^2)$, and $BF=F{F}^{T}F$. This is precisely the requested expression, with the coefficient of $F$ separated from the $BF$ and cofactor terms. The positive sign of the $6\beta J$ term comes from the $3\beta III$ contribution in the energy.