Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 50 / 1

Write $⟨q⟩=|\Omega {|}^{-1}{\int }_{\Omega }qdx$, $E=⟨\epsilon ⟩$ and $S=⟨\sigma ⟩$. Use the infinitesimal strain tensor, symmetric stress, zero body force, and the outward unit normal $n$. The divergence theorem applied separately to the two displacement derivatives givesFor static elastic equilibrium, ${\sigma }_{ik,k}=0$, and therefore ${\partial }_k({\sigma }_{ik}{x}_j)={\sigma }_{ij}$. With traction $t_i={\sigma }_{ik}{n}_k$, the divergence theorem now givesThe right-hand side is automatically symmetric because the volume stress tensor is symmetric; it can also be written as the integral of $(t_i{x}_j+t_j{x}_i)/2$.

To prove the work identity, put $v=u-Ex$. The constant symmetric tensor $S$ is equilibrated, so integration by parts givesHere the two cross terms each equal $|\Omega |S:E$, and the last constant term restores one of them. Division by the volume proves the required identity. In particular, affine boundary displacement makes $v=0$ on the boundary and yields the Hill-Mandel energy identity. The zero-body force hypothesis matters: otherwise the first integration by parts has an additional volume-force term.

Interpret derivatives on the space of symmetric strains, with inner product $A:B=A_{ij}{B}_{ij}$. Assume a differentiable strain energy density with local stress $\sigma =D_{\epsilon }W$, and a differentiable stable equilibrium branch under the prescribed affine boundary displacement. Thus $u_E(x)=Ex+v_E(x)$ with $v_E=0$ on $\partial \Omega$; part(a) ensures that the mean strain is exactly $E$.

Vary $E$ in any symmetric direction $H$. The induced displacement variation has the form $\dot{u}=Hx+\dot{v}$, where $\dot{v}$ vanishes on the boundary. Differentiating the effective elastic energy and using static elastic equilibrium givesThe last term is zero by integration by parts, since the body force is zero and $\dot{v}$ has zero boundary trace. As $H$ is arbitrary,This is a work-conjugacy statement on symmetric tensors; treating off-diagonal entries as independent scalar coordinates without their contraction weights would introduce erroneous factors of two. If the minimized effective elastic energy has a nondifferentiable point, the corresponding formulation uses a subgradient rather than a claimed classical derivative.

In the Voigt elastic approximation, every local strain is $E$. The corresponding local stress is ${\sigma }_V(x)=D_{\epsilon }W(E,x)$. Differentiating the averaged strain energy density under the integral givesThe uniform strain need not generate an equilibrated local stress; this calculation concerns the stated approximation, not an assertion that it solves the heterogeneous boundary problem.

For the Reuss elastic approximation, use the convex conjugateUnder differentiable strictly convex duality, the local strain produced by uniform stress $S$ is $e_R(x)=D_S{W}^{\ast }(S,x)$. Set $F(S)=⟨W^{\ast }(S,x)⟩$. Averaging the local strains gives $E=DF(S)$. At the maximizing stress in $W_R(E)=F^{\ast }(E)$, the stationarity condition is precisely $DF(S)=E$. ConsequentlyThus the derivative of each approximate energy is its own mean stress, includingFor positive linear elasticity, $W=e:C(x)e/2$ and $W^{\ast }=S:C(x{)}^{-1}S/2$, giving the useful check $C_V=⟨C⟩$ and $C_R=⟨C^{-1}{⟩}^{-1}$. More general convex duality gives the same conjugacy relation with subgradients when differentiability or strict convexity is absent.