= Solution
The perturbation preserves the prescribed <displacement> because $\mathbf w=0$ on the boundary. <Strain> varies as $e_{ij}(\mathbf u+\alpha\mathbf w)=e_{ij}(\mathbf u)+\alpha(w_{i,j}+w_{j,i})/2$. Taylor expansion of the <strain energy density> gives
$$
U[\mathbf u+\alpha\mathbf w]-U[\mathbf u]
=\alpha\int_B\left[\tau_{ij}\frac{w_{i,j}+w_{j,i}}2-\rho g_iw_i\right]dV+O(\alpha^2).
$$
Symmetry of the <stress> makes its contraction with the two symmetric terms equal, hence
$$
\boxed{U[\mathbf u+\alpha\mathbf w]-U[\mathbf u]=\alpha\int_B(\tau_{ij}w_{i,j}-\rho g_iw_i)\,dV+O(\alpha^2).}
$$
A <minimizer> must have zero <first variation> for both signs of $\alpha$. Integrating by parts and using the zero boundary variation yields $\int_B(-\partial_j\tau_{ij}-\rho g_i)w_i\,dV=0$. The <fundamental lemma of the calculus of variations>, applied to arbitrary compactly supported variations, gives
$$
\boxed{\partial_j\tau_{ij}+\rho g_i=0,\quad\text{or }\nabla\cdot\tau+\rho\mathbf g=0.}
$$
This is the static momentum equation. Fixed-<displacement> data remove the surface variation; prescribed <tractions> would instead appear through boundary work in the potential energy.
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