Saint-Venant's compatibility condition

In the mathematical theory of elasticity the strain $\varepsilon$ is related to a displacement field $\ u$ by

$\varepsilon _{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)$

Saint-Venant derived the compatibility condition for an arbitrary symmetric second rank tensor field to be of this form, this has now been generalized to higher rank symmetric tensor fields.

Rank 2 tensor fields

The integrability condition takes the form of the vanishing of the Saint-Venant's tensor defined by

$W_{ijkl}={\frac {\partial ^{2}\varepsilon _{ij}}{\partial x_{k}\partial x_{l}}}+{\frac {\partial ^{2}\varepsilon _{kl}}{\partial x_{i}\partial x_{j}}}-{\frac {\partial ^{2}\varepsilon _{il}}{\partial x_{j}\partial x_{k}}}-{\frac {\partial ^{2}\varepsilon _{jk}}{\partial x_{i}\partial x_{l}}}$

Due to the symmetry conditions $W_{ijkl}=W_{klij}=-W_{jikl}=W_{ijlk}$ there are only six (in the three dimensional case) distinct components of $W$. These six equations are not independent as verified by for example

${\frac {{\frac {\partial ^{2}\varepsilon _{22}}{\partial x_{3}^{2}}}+{\frac {\partial ^{2}\varepsilon _{33}}{\partial x_{2}^{2}}}-2{\frac {\partial ^{2}\varepsilon _{23}}{\partial x_{2}\partial x_{3}}}}{\partial x_{1}}}={\frac {{\frac {\partial ^{2}\varepsilon _{22}}{\partial x_{1}\partial x_{3}}}-{\frac {\partial }{\partial x_{2}}}\left({\frac {\partial \varepsilon _{23}}{\partial x_{1}}}-{\frac {\partial \varepsilon _{13}}{\partial x_{2}}}+{\frac {\partial \varepsilon _{12}}{\partial x_{3}}}\right)}{\partial x_{2}}}+{\frac {{\frac {\partial ^{2}\varepsilon _{33}}{\partial x_{1}\partial x_{2}}}-{\frac {\partial }{\partial x_{3}}}\left({\frac {\partial \varepsilon _{23}}{\partial x_{1}}}+{\frac {\partial \varepsilon _{13}}{\partial x_{2}}}-{\frac {\partial \varepsilon _{12}}{\partial x_{3}}}\right)}{\partial x_{3}}}$

and there are two further relations obtained by cyclic permutation. However, in practise the six equations are preferred. In its simplest form of course the components of $\varepsilon$ must be assumed twice continuously differentiable, but more recent work proves the result in a much more general case.

In differential geometry the symmetrised derivative of a vector field appears also as the Lie derivative of the metric tensor g with respect to the vector field.

$T_{ij}=({\mathcal {L}}_{U}g)_{ij}=U_{i;j}+U_{j;i}$

where indicies following a semicolon indicate covariant differentiation. The vanishing of $WT$ is thus the integrability condition for local existence of $U$ in the Euclidean case.

Generalization to higher rank tensors

Saint-Vanants compatibility condition can be thought of as an analogue, for symmetric tensor fields, of Poincare's lemma for skew-symmetric tensor fields (differential forms). The result can be generalized to higher rank symmetric tensor fields. Let F be a symmetric rank-k tensor field on an open set in n-dimensional Euclidean space, then the symmetric derivative is the rank k+1 tensor field defined by

$(dF)_{i_{1}...i_{k}i_{k+1}}=F_{(i_{1}...i_{k},i_{k+1})}$

where we use the classical notation that indices following a comma indicate differentiation and groups of indices enclosed in brackets indicate symmetrization over those indices. The Saint-Venant tensor $W$ of a symmetric rank-k tensor field $U$ is defined by

$W_{i_{1}..i_{k}j_{1}...j_{k}}=V_{(i_{1}..i_{k})(j_{1}...j_{k})}$

with

$V_{i_{1}..i_{k}j_{1}...j_{k}}=\sum \limits _{p=1}^{k}(-1)^{p}{m \choose p}U_{i_{1}..i_{k-p}j_{1}...j_{p},j_{p+1}...j_{k}i_{k-p+1}...i_{k}}$

On a simply connected domain in Euclidean space $W=0$ implies that $U=dF$ for some rank k-1 symmetric tensor field $F$.