 Solution

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.