Beltrami–Michell compatibility equations

Remarks

The proof writes the strain in terms of the stress by the inverted Hooke's law $\varepsilon = \frac{1 + \nu}{E} \sigma - \frac{\nu}{E} (\tr \sigma) I$ and substitutes it into the Saint-Venant compatibility condition. Contracting the result and using the equilibrium equations to replace divergences of $\sigma$ by derivatives of $b$ gives first $\Delta (\tr \sigma) = -\frac{1 + \nu}{1 - \nu} \partial_k b_k$, and then the six equations above. Conversely, if $\Omega$ is simply connected, a symmetric stress field that satisfies the equilibrium equations and these six equations comes from a displacement field. With zero body force the equations reduce to Beltrami's form $\Delta \sigma_{ij} + \frac{1}{1 + \nu} \partial_i \partial_j (\tr \sigma) = 0$; then $\tr \sigma$ is harmonic and each $\sigma_{ij}$ is biharmonic. Example: a constant stress field with $b = 0$ satisfies them, since every term is a derivative of a constant; it is the stress of an affine displacement.

Going deeper

For a student

The stress formulation of linear elasticity solves for $\sigma$ from the equilibrium equations and the Beltrami–Michell equations, together with traction conditions on the boundary. It is convenient when all boundary data are tractions. In two dimensions the same idea, combined with an Airy stress function, leads to a single biharmonic equation. The equations depend only on Poisson's ratio, so in this setting the stress field does not depend on the shear modulus.