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An elastic deformation is recoverable when the deforming loads are removed. In a local elastic constitutive model the stress depends on the present deformation. When this response also derives from a stored elastic energy, the material is a hyperelastic material. In the small-deformation approximation, the relevant kinematic variable is the infinitesimal strain tensor.

For a homogeneous comparison elastic stiffness tensor $C^0$, the elastic strain Green operator maps an elastic polarization to minus the induced infinitesimal strain tensor. In a bounded domain the induced displacement field has zero boundary data; in an infinite medium it is the decaying perturbation. The defining static elastic equilibrium equation is $\nabla \cdot (C^0\mathrm{sym}\nabla u+\tau )=0$. The bounded-domain operator is self-adjoint, annihilates constant elastic polarizations, has zero mean output, and satisfies $\Gamma C^0\Gamma =\Gamma$.

The infinitesimal strain tensor is the symmetric part of a displacement gradient tensor. It is the linear approximation to a finite strain measure for small displacement gradients. An infinitesimal rigid rotation contributes no strain, so the elastic energy depends on the symmetric part rather than the entire gradient.

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.

A perfectly bonded elastic interface has continuous displacement field and continuous traction, with the same chosen normal on both sides. For piecewise affine displacement fields, the difference of their gradients annihilates every tangential vector, and hence has the form $a\otimes n$. Their infinitesimal strain tensors differ by $\mathrm{sym}(a\otimes n)$.