Hill–Mandel condition

Statement

Let $V \subset \R^3$ be a bounded Lipschitz domain, the volume element, and write $\langle g \rangle = \frac{1}{\abs{V}} \int_V g \, dx$. Let $\sigma$ be a symmetric stress field on $V$ with $\operatorname{div} \sigma = 0$. Let $\varepsilon = \frac{1}{2}(\nabla u + \nabla u^T)$ be the strain of a displacement field $u$ on $V$. Assume that $\sigma$ and $u$ are smooth enough for the divergence theorem. Let $n$ be the outer unit normal on $\partial V$. Then (Hill's lemma) $$\langle \sigma : \varepsilon \rangle - \langle \sigma \rangle : \langle \varepsilon \rangle = \frac{1}{\abs{V}} \int_{\partial V} \big( u - \langle \varepsilon \rangle x \big) \cdot \big( \sigma n - \langle \sigma \rangle n \big) \, dS.$$ In particular the Hill–Mandel condition $\langle \sigma : \varepsilon \rangle = \langle \sigma \rangle : \langle \varepsilon \rangle$ holds in each of these three cases:

1. affine boundary displacements: $u = E x$ on $\partial V$ for a constant symmetric matrix $E$;
2. uniform boundary tractions: $\sigma n = \Sigma n$ on $\partial V$ for a constant symmetric matrix $\Sigma$;
3. $V$ is a box, $u - E x$ is periodic and $\sigma n$ is antiperiodic on opposite faces.