Lemme — — — - Weak vector and scalar potentials . Applications to Poincaré ’ s theorem and Korn ’ s inequality in Sobolev spaces with negative exponents .

In this paper, we present several results concerning vector potentials and scalar potentials with data in Sobolev spaces with negative exponents, in a not necessarily simply-connected, three-dimensional domain. We then apply these results to Poincaré’s theorem and to Korn’s inequality. 1 Weak versions of a classical theorem of Poincaré In this work, (the results of which were announced in [2]), Ω is a bounded open connected subset of R with a Lipschitz-continuous boundary Γ. The notation X′〈, 〉X denotes the duality pairing between a topological space X and its dual X ′. The letter C denotes a constant that is not necessarily the same at its various occurrences. We begin with a weak version of a well-known theorem of Poincaré. Here as elsewhere in this paper, “weak” means that the result to which it is attached holds as well in Sobolev spaces with negative exponents. Theorem 1.1. Let f ∈ H−m(Ω)3 for some integer m ≥ 0. Then the following properties are equivalent: (i) H−m(Ω)3〈f , φ〉Hm 0 (Ω)3 = 0 for all φ ∈ Vm = {φ ∈ H m 0 (Ω) ; div φ = 0}, (ii) H−m(Ω)3〈f , φ〉Hm 0 (Ω)3 = 0 for all φ ∈ V = {φ ∈ D(Ω) ; div φ = 0}, (iii) There exists a distribution χ ∈ H−m+1(Ω), unique up to an additive constant, such that f = grad χ in Ω. If in addition Ω is simply-connected, the above properties are equivalent to: (iv) curl f = 0 in Ω. Proof. For the equivalence between (i), (ii) and (iii), we refer to [4]. Since the implication (iii) =⇒ (iv) clearly holds, it remains to prove that (iv) =⇒ (iii). Email addresses: [email protected] (Chérif Amrouche), [email protected] (Philippe G. Ciarlet), [email protected] (Patrick Ciarlet, Jr.). Preprint submitted to Elsevier October 29, 2009 ha l-0 04 44 18 5, v er si on 1 6 Ja n 20 10 To begin with, let f ∈ H−m(Ω)3 be such that curl f = 0 in Ω. We then use the same argument as in [8]: We know that there exist a unique u ∈ H 0 (Ω) and a unique p ∈ H−m+1(Ω)/R (see [5]) such that ∆u + grad p = f and div u = 0 in Ω. (1) Hence ∆curl u = 0 in Ω so that the hypoellipticity (see [10]) of the polyharmonic operator ∆ implies that curl u ∈ C∞(Ω)3. Since div u = 0, we deduce that ∆u = curl curl u ∈ C∞(Ω)3. This also implies that ∆u belongs to C∞(Ω)3 and is an irrotational vector field. By the classical Poincaré theorem, there exists q ∈ C∞(Ω)3 such that ∆u = grad q. Thus, f = grad (p + q) and, thanks to [4] proposition 2.10, the function p + q belongs to the space H−m+1(Ω). We can give another proof of the implication (iv) =⇒ (iii) by using the following theorem: Theorem 1.2. Assume that both Ω and R \ Ω are simply-connected. Let u ∈ H 0 (Ω), m ≥ 0, be a vector field that satisfies div u = 0 in Ω. Then there exists a vector potential ψ in H 0 (Ω) 3 such that u = curl ψ, div ∆ψ = 0 in Ω, and ‖ψ‖Hm+1(Ω)3 ≤ C‖u ‖Hm(Ω)3 . (2) Proof. Let u ∈ H 0 (Ω) be such that div u = 0 in Ω and let ũ denote the extension of u by 0 in R \Ω. Thus ũ ∈ H 0 (R), div ũ = 0 in R, and there exist an open ball B containing Ω and a vector field w ∈ H 0 (B) such that ũ = curl w , div ∆w = 0 in B, and ‖w ‖Hm+1(B)3 ≤ C‖u ‖Hm(B)3 . The open set Ω′ := B \Ω is bounded, has a Lipschitz-continuous boundary and is simply-connected. Furthermore, the vector field w ′ := w |Ω′ belongs to Hm+1(Ω′)3 and satisfies curl w ′ = 0 in Ω′. Therefore there exists a function χ′ ∈ H1(Ω′) such that w ′ = grad χ′ in Ω′. Hence in fact χ′ ∈ Hm+2(Ω′) and the estimate ‖χ‖Hm+2(Ω′) ≤ C‖w ‖Hm+1(Ω′)3 holds. Since the function χ′ ∈ Hm+2(Ω′) can be extended to a function χ̃ in H(R), with ‖χ̃‖Hm+2(R3) ≤ C‖χ‖Hm+2(Ω′) ≤ C‖w ‖Hm+1(Ω′)3 , the vector field φ̃ := w −grad χ̃ belongs to the space H(B) and satisfies φ̃|Ω′ = 0. Then the restriction φ := φ̃|Ω belongs to the space H 0 (Ω), satisfies the estimate (2), and curl φ̃ = curl w = ũ in B. Thus u = curl φ 2 ha l-0 04 44 18 5, v er si on 1 6 Ja n 20 10 in Ω. Let now p denote the unique solution in the space H 0 (Ω) of ∆ p = div ∆φ, so that the estimate ‖p‖Hm+2(Ω) ≤ C‖φ‖Hm+1(Ω)3 holds. Then the function ψ = φ− grad p satisfies (2). We can give yet another proof of the above implication (iv) =⇒ (iii): Consider again the solution u ∈ H 0 (Ω) to (1) and let v ∈ H 0 (Ω) denote the vector potential of u as given by theorem 1.2. We then have ∆curl u = 0. If m = 2k, for some integer k ≥ 1, then H−m−1(Ω)3〈∆curl u , v 〉Hm+1 0 (Ω)3 = H−1(Ω)3〈∆ curl u , ∆v 〉H1 0 (Ω)3 = ∫ Ω ∆u ·∆curl v dx = ‖∆u ‖L2(Ω)3 . This implies that ∆u = 0 in Ω and thus u = 0 since u ∈ H 0 (Ω). The case m = 2k + 1 follows by a similar argument.