Saint-Venant's compatibility condition

In the mathematical theory of elasticity the strain $varepsilon$ is related to a displacement field $u$ by:$varepsilon\_\{ij\}\; =\; frac\{1\}\{2\}\; left(\; frac\{partial\; u\_i\}\{partial\; x\_j\}\; +\; frac\{partial\; u\_j\}\{partial\; x\_i\}\; ight)$
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 [N.I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity. Leyden: Noordhoff Intern. Publ., 1975.] 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:$frac\{frac\{partial^2\; varepsilon\_\{22\{partial\; x\_3^2\}\; +\; frac\{partial^2\; varepsilon\_\{33\{partial\; x\_2^2\}\; -\; 2\; frac\{partial^2\; varepsilon\_\{23\{partial\; x\_2\; partial\; x\_3\{partial\; x\_1\}\; =frac\{frac\{partial^2\; varepsilon\_\{22\{partial\; x\_1\; partial\; x\_3\}\; -\; frac\{partial\}\{partial\; x\_2\}\; left\; (\; frac\{partial\; varepsilon\_\{23\{partial\; x\_1\}\; -\; frac\{partial\; varepsilon\_\{13\{partial\; x\_2\}\; +\; frac\{partial\; varepsilon\_\{12\{partial\; x\_3\}\; ight)\}\{partial\; x\_2\}\; +frac\{frac\{partial^2\; varepsilon\_\{33\{partial\; x\_1\; partial\; x\_2\}\; -\; frac\{partial\}\{partial\; x\_3\}\; left\; (\; frac\{partial\; varepsilon\_\{23\{partial\; x\_1\}\; +\; frac\{partial\; varepsilon\_\{13\{partial\; x\_2\}\; -\; frac\{partial\; varepsilon\_\{12\{partial\; x\_3\}\; ight)\}\{partial\; x\_3\}$and there are two further relations obtained by cyclic permutation. However, in practisethe six equations are preferred.In its simplest form of course the components of $varepsilon$ must be assumed twice continuously differentiable, but more recent work [C Amrouche, PG Ciarlet, L Gratie, S Kesavan, On Saint Venant's compatibility conditions and Poincaré's lemma, C. R. Acad. Sci. Paris, Ser. I, 342 (2006), 887-891. doi|10.1016/j.crma.2006.03.026] 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 indices following a semicolon indicate covariant differentiation. The vanishing of $W(T)$ 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. [V.A. Sharafutdinov, Integral Geometry of Tensor Fields, VSP 1994,ISBN 906764165X. Chapter 2. [http://www.math.nsc.ru/~sharafutdinov/files/book.pdf on-line version] ] 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:$V\_\{i\_1..i\_k\; j\_1...j\_k\}\; =\; sumlimits\_\{p=1\}^\{k\}\; (-1)^p\; \{m\; choose\; p\}\; U\_\{i\_1..i\_\{k-p\}j\_1...j\_p,j\_\{p+1\}...j\_k\; i\_\{k-p+1\}...i\_k\; \}$On a simply connected domain in Euclidean space $W=0$ implies that $U\; =\; dF$ for some rank k-1 symmetric tensor field $F$.

References