Hodge decomposition for symmetric matrix fields and the elasticity complex in Lipschitz domains

. In 1999 M. Eastwood has used the general construction known as the Bernstein-Gelfand-Gelfand (BGG) resolution to prove, at least in smooth situation, the equivalence of the linear elasticity complex and of the de Rham complex in R 3 . The main objective of this paper is to study the linear elasticity complex for general Lipschitz domains in R 3 and deduce a complete Hodge orthogonal decomposition for symmetric matrix ﬁelds in L 2 , counterpart of the Hodge decomposition for vector ﬁelds. As a byproduct one obtains that the ﬁnite dimensional terms of this Hodge decomposition can be interpreted in homological terms as the corresponding terms for the de Rham complex if one takes the homology with value in rig ∼ = R 6 as in the (BGG) resolution.


Introduction
Let Ω be an open, connected and bounded domain in R 3 with a smooth boundary ∂Ω. Relative to an orthonormal cartesian basis {e i }, (i=1,2,3), the coordinates of a generic point will be denoted by {x 1 , x 2 , x 3 }, the components of a vector field v by v i and the components of a square matrix field S of order three by S ij . M 3 (resp. M 3 sym ) denotes the vector space of second-order (resp. symmetric secondorder) matrices. Latin indices range in the set {1, 2, 3}. The summation convention with respect to the repeated indices is used. The scalar product of the matrices E and S is denoted by E:S := E ij S ij .
Let v a smooth vector field defined on Ω; then the corresponding strain field ∇ s (v) = 1 2 (∇v T + ∇v) is a symmetric matrix field. The characterization of the smooth symmetric matrix fields E that are strain fields, i.e. that can be written as E = ∇ s (v) for some v, goes 1 back to the second half of the Nineteenth Century. Indeed, it was discovered by A. J.C. B. de Saint Venant (1864) the following result: Theorem 1.1 (Saint Venant's necessary compatibility theorem). The strain field E corresponding to a class C ∞ (Ω; R 3 ) displacement vector field v satisfies the compatibility equations.
The components of the matrix CU RL E are: (CU RL E) ij = ipk E jk,p where the commas stand for partial differentiations with respect to x and ipk denotes the alternator: , (i,p,k) is an even permutation of (1,2,3); −1, (i,p,k) is an odd permutation of (1,2,3); 0, (i,p,k) is not a permutation of (1,2,3).
The first 1 rigorous proof of sufficiency was given by E. Beltrami (1886) in the following form. When Ω is simply connected, thanks to the theorem 1.2, the Donati's theorem gives an orthogonal (in the L 2 -sense) decomposition for symmetric matrix fields analogous to the Helmoltz decomposition of vector fields. The Donati's theorem has been extended in various directions: see e.g. [11], [3] and their bibliography and theorems 2.1 and 2.4.
Beltrami (1892) has remarked that for a smooth symmetric matrix field E one has Div CU RL CU RL E = 0 and, moreover, that with the representation (called Beltrami's solution): it is possible to recover, with a suitable choice of E, the Airy's, the Maxwell's and the Morera's solution, see Gurtin [16], Sect.17. There exists stress fields that do not admit a Beltrami representation. The characterization of the smooth stress fields S admitting the representation (1.5) for some symmetric matrix field E is given by the following theorem of Gurtin [15]: Let us remark that (1.3) implies that this condition is automatically satisfied when the boundary of Ω consists in a single closed surface.
The previous results can also be presented in the framework of differential forms and exterior calculus as a complex of linear operators (called the linear elasticity complex, see [8]): where the kernel of each linear operator contains the image of the previous one and: For later use we denote by e i and p i = − ijk x k e j , i = 1, 2, 3, a basis of rig. The Beltrami's sufficiency compatibility theorem and the Gurtin's theorem give sufficient conditions in order that the sequence is exact, i.e. that each operator provides the integrability conditions for the one which precedes it. Volterra (1906) has given a characterization of ker(CU RL CU RL) for a general non simply connected domain Ω, inspired by the results of Poincaré on the complex (also called the De Rham-Poincaré complex): The analogy between the two complexes can be further developed. Indeed M. Eastwood [8] has proved that, with a general construction known as the Bernstein-Gelfand-Gelfand (BGG) resolution, the two sequences are equivalent for smooth functions. He has also conjectured that "anything which is true of the de Rham complex should have a counterpart for linear elasticity". For example, a smooth matrix field S has a generalized Beltrami representation of the form (see [9]): where A can be chosen such that DivA = 0; this representation is the counterpart to the Helmholtz decomposition of vector fields, 1 2 (∇v T + ∇v) being the analogous of the irrotational term and CU RL CU RLA the analogous of the solenoidal term 2 .
The aim of this paper is to prove "a further possible instance of this": a general Hodge decomposition for symmetric matrix fields analogous to the classical Hodge decomposition for vector fields (see e.g. [4]); for this we study the linear elasticity complex for general Lipschitz domains in various Sobolev spaces settings. Let us define {E ∈ L 2 Ω; M 3 sym ; CU RL CU RL E ∈ L 2 Ω; M 3 sym } where the operators Div and CU RL CU RL are to be taken in the distribution sense. We will consider the following situations: i: Beltrami completeness condition in L 2 setting: ii: Saint Venant compatibility condition in L 2 setting: (1.14) rig Some of the results presented concerning the Beltrami completeness condition has been given with different proofs in [12] and some on the Saint Venant compatibility condition in L 2 setting have been announced in [6].

Beltrami completeness condition in L 2 setting
Let Ω be a general bounded Lipschitz connected but eventually multiply connected domain in R 3 with boundary ∂Ω. Let be γ 0 the exterior boundary of Ω, i. e. the boundary of the unbounded connected component of R 3 \Ω, and γ q , q = 1, . . . , Q the others connected components of ∂Ω. Let denote with B an open ball such that Ω is contained in B and, for q = 1, . . . , Q, let Ω q be the connected component of B\Ω with boundary γ q . At last let Ω 0 be the connected component of B\Ω with boundary γ 0 ∂B. For future use let remark that like every multiply-connected domain, Ω can be reduced to be a simplyconnected one, Ω * , with a finite number N of planar, non-intersecting cuts, C α , α = 1, . . . , N , linking the γ q , q = 0, . . . , Q, i.e. such that the boundary of C α is contained in ∂Ω. Moreover the cuts are such that the simply-connected domain Ω * = Ω\ N α=1 C α verifies the cone condition.
Theorem 2.1. The following decomposition of L 2 Ω; M 3 sym in mutually orthogonal closed subspaces holds true: In [3] it has been proved that also ∇ s (H 1 (Ω; R 3 )) is a closed subspace of L 2 Ω; M 3 sym . Proposition 2.2. One has the following orthogonal decomposition: where: Proof. The orthogonality of the decomposition follows from the Green's formula (2.1). Let be given u ∈ H 1 (Ω; R 3 ) and set f : It is worth note that when u| ∂Ω =û| ∂Ω then S = ∇ s w = ∇ sŵ .
Let introduce the spaces: sym ; divS = 0 in Ω and Γ n (S) = 0 on ∂Ω The next theorem provides an answer to a remark of [3]; a slightly different result has been given in a general L p setting in [11].
Proof. Let L(S) be a linear and continuous functional on Σ ad (Ω); then there exist S ∈ Σ ad (Ω) such that : The second extension of the Donati's theorem (for a proof in a general L p setting see [11]) provides a complement to the Theorem 4.3 of [3].
It also follows from this theorem, theorem 2.1 and (2.3) that: The image of the operator CU RL CU RL, linear and continuous from is a closed subspace of G, from theorem 2.4 one deduces the orthogonal decomposition: where: The extension of the Gurtin's theorem 1.4 is the object of the following: (2.14) CU RL CU RL A = S and Div A = Div CU RL A = 0 The proof given here is inspired by a proof of a similar result for solenoidal vector fields given in , Chapter I, Theorem 3.4, see also [2]). A different proof largely inspired to the proof of Gurtin [16], Sect. 17 and that also uses the results of [14] has been given in [12] (However, let us explicitly remark that the statement (ii) of Theorem 2.2 there is not correct). The case of the Airy's stress function in Lipschitz domain has been considered in [10].
sym . Since Div S = 0 in order to prove that S ∈ G one has only to verify the conditions (b) and (c) or else that for q = 0, . . . , Q and for every v ∈ rig one has Γ q n (S) , v γq = 0. For this let χ q ∈ D Ω with χ q = 1 in a neighborhood of γ q and χ q = 0 in a neighborhood of every γ r with r = q. Then one has: (ii) Let S ∈ G and for every q = 0, . . . , Q let v q ∈ H 1 (Ω q ; R 3 ) the solution (unique up to a rigid displacement) of the problem: From the symmetry of S it follows that also A is symmetric; from (2.15) it follows that and, using once more (2.15), At last one also has: If one defines A = B(S) as the restriction to Ω of the inverse Fourier transform of (2.16) it is immediately seen from (2.17), (2.18) and (2.19) that the conditions of (2.14) are all satisfied. In order to conclude it is enough to prove that A ∈ H 2 (Ω; M 3 sym ). One can at first remark that from (2.16) it follows that ξ r ξ s A ij ∈ L 2 (R 3 ) and hence the second order derivatives of A ij are in L 2 (Ω). It is enough to prove that also A ij ∈ L 2 (Ω). For this let ω(ξ) ∈ D(R 3 ) with ω(ξ) = 1 for |ξ| ≤ 1. Since ω(ξ) A ij has compact support its inverse Fourier transform is analytic and hence its restriction to Ω is in L 2 (Ω). Since (1 − ω(ξ)) A ij ∈ L 2 (R 3 ) it follows that A ij ∈ L 2 (Ω) and the proof is complete. Proposition 2.6. Let be: Y := {T ∈ Ker(Div; L 2 ); T = ∇ s (u), u ∈ H 1 (Ω; R 3 ), u |γ 0 = 0 and u |γq ∈ rig, q = 1, . . . , Q}. Y has dimension 6Q and: with orthogonality in L 2 Ω; M 3 sym . Proof. Since the orthogonality of the decomposition follows from the Green's formula, one has only to prove that the sum in (2.20) is direct. For this let be S ∈ Ker(Div; L 2 ) and let us remark that there exists a unique u ∈ Y := {v ∈ H 1 (Ω; R 3 ), v |γ 0 = 0 and v |γ q ∈ rig, q = 1, . . . , Q} such that for all v ∈ Y: Remark 2.7. One can easily give a variational characterization of a basis of Y.
From this proposition and from theorem 2.1 one deduces the following decompositions of L 2 Ω; M 3 sym in mutually orthogonal subspaces: (2.21) and from (2.12) 3) and theorem 2.4 one also has: Let us explicitly remark that these decompositions of L 2 Ω; M 3 sym in mutually orthogonal subspaces are the counterpart of the decompositions of Helmoltz type for vector fields in L 2 (Ω; R 3 ). In order to obtain a complete decomposition of Hodge type we need the results of the next section. In order to completely characterize ker(CU RL CU RL ; L 2 ) it suffices, thanks to theorem 2.4, to study its intersection with Σ ad (Ω). From a direct inspection it appears that this space is:

CU RL CU RL is a linear and continuous operator from
and Div S = 0 in Ω, Γ n (S) = 0 on ∂Ω} .
When Ω is simply connected then K = 0.
In [5] the proof that ∇ s (H 1 (Ω; R 3 )) = ker(CU RL CU RL ; L 2 ) when Ω is simply connected is reduced to a weak version of the classical Theorem of Poincaré. We give here a different proof.
Proof. One remarks at first that the following identities hold true: Hence when S ∈ K one has :  [3]) implies that v ∈ H 1 (Ω; R 3 ) and the conditions Div S = 0 in Ω and Γ n (S) = 0 on ∂Ω imply that v ∈ rig and hence S = 0.
One can now prove the following result correcting the statement (ii) of Theorem 2.2 of [12]. As it has been remarked for the first time by V. Volterra in 1906 [18], K = 0 for a general non simply-connected domain. In order to study this general situation, let us define the following space of Volterra's dislocations: ∈ rig, for α = 1, . . . , N , where [[v]] C α is the jump of v across the cut C α and Ω * = Ω\ N α=1 C α . Let explicitly remark that when all these jumps vanish then v ∈ H 1 (Ω; R 3 ).
Let us write the jump of v across the cut C α in the form Let us also define on VD the functionals: Proposition 3.3. For every α = 1, . . . , N and i = 1, 2, 3 there exist u α i ∈ VD and r α i ∈ VD such that: Moreover, each vector field u α i and r α i is uniquely determined modulo a global infinitesimal rigid displacement on Ω.
The proof is obvious. For later use we write the corresponding Euler equations: Since meas (Ω) = meas (Ω * ), there is a canonical isomorphism of L 2 (Ω * ; M 3 sym ) with L 2 (Ω; M 3 sym ). Hence for a given v ∈ VD, one can associate with ∇ s v ∈ L 2 (Ω * ; M 3 sym ) the corresponding element in Hence, in the distribution sense, Taking v ∈ H 1 (Ω; R 3 ) in (3.10) one then finds from (2.1): Hence we conclude that ∇ s u α i belongs to Σ ad (Ω). The same arguments can be repeated to obtain that ∇ s r α i belongs to Σ ad (Ω). With suitable ] C α = p j for j = 1, 2, 3, etc.) one further finds that: It remains to prove that CU RL CU RL ( ∇ s u α i ) = 0 in the distribution sense. This result is a consequence of the relation: C α e i · (CU RL CU RL S) n dC Since the cuts C α are planar and each u α i is unique modulo a global rigid displacement of Ω, one can take for simplicity C α contained in the plane spanned by e 1 and e 2 and so n = e 3 and dC = dx 1 dx 2 . A simple computation then gives: and hence ( ∇ s u α i ) ∈ K. The proof that ( ∇ s r α i ) ∈ K is analogous. One obtains : From (3.16) one has Integrating by parts every term, one finds since k = 3 and S ij | Cα ∈ D(C α ) : Hence still integrating by parts one obtains, thanks to the symmetry of S: This concludes the proof for ∇ s r α i . From (3.14),(3.15), (3.10) and (3.11) it follows: . For later use one needs another form of the Green's formula. For this let remark at first that when W ∈ Σ ad (Ω), then the map W → {( Wn)| C α } α=N α=1 well defined for W ∈ W, can be extended to a linear and continuous map from Σ ad (Ω) into α=N α=1 H −1/2 (C α ; R 3 ). Indeed, for any fixed α and any g ∈ H 1/2 (C α ; R 3 ) one can find u ∈ H 1 (Ω * ; R 3 ) such that [[u]] C α = g and [[u]] C β = 0 when β = α and such that the map g → u is linear and continuous. The Green's formula for W ∈ W and u ∈ H 1 (Ω * ; R 3 ) reads: The density of W in Σ ad (Ω) (Theorem 2.3) allows to extend the map W → ( Wn)| Cα to a linear and continuous map from Σ ad (Ω) into H −1/2 (C α ; R 3 ). Moreover the following extended Green's formula holds true for W ∈ Σ ad (Ω) and u ∈ H 1 (Ω * ; R 3 ): (3.17) where ., . C α denotes the duality pairing between H −1/2 (C α ; R 3 ) and H 1/2 (C α ; R 3 ). We can now give the announced characterization.
Proof. Given W ∈ K, let V ∈ K be defined by: Let be V * , W * ∈ L 2 (Ω * ; M 3 sym ) the restriction of V and W to Ω * . From an inspection of (3.18) it appears that CU RL CU RL ( V * ) = 0.
Since the matrix fields ( ∇ s u α i ) and ( ∇ s r α i ) are linearly independent in L 2 (Ω; M 3 sym ), we also have Corollary 3.7. The space K is of dimension 6N . Proof. In order to prove this, let change the first cut C 1 into another one C 1 equivalent in the sense that (i) its boundary is contained in ∂Ω, (ii) the set of planar non -intersecting cuts { C 1 , C 2 , . . . , C N } is such that Ω * = Ω\{ C 1 N α=2 C α } is simply connected and satisfies the cone condition. Let O the subset of Ω whose boundary is C 1 , C 1 and the part of ∂Ω connecting the boundaries of C 1 , C 1 . Then the Green's formula in O implies that for all v ∈ rig and all S ∈ Σ ad (Ω) one has: Remark 3.10. From the previous proof it appears that the assumption that the cuts are planar is not necessary; it is enough that the nonintersecting cuts be Lipschitz. Proof. It is enough to remark that ker(CU RL CU RL ; L 2 ) = K ⊥ ⊕ ∇ s (H 1 (Ω; R 3 )).

The complete Hodge decomposition
Collecting the different results we get the following general Hodge orthogonal decompositions of L 2 (Ω; M 3 sym ): Let us stress that this orthogonal decomposition is the analogous of the following Hodge decomposition of L 2 (Ω; R 3 ) (see e.g. [2], [4]): R) ∼ = R Q In our situation it follows from proposition 2.6 that K ∼ = R 6N and from corollary 3.7 that Y ∼ = R 6Q . This corresponds to homology with value in rig ∼ = R 6 and so it is coherent with the Bernstein-Gelfand-Gelfand resolution as suggested by Eastwood [8].