ON THE BEST CONSTANT IN GAFFNEY INEQUALITY. Bernard DACOROGNA. EPFL - Lausanne - Suisse. avec G. CSATO et S. SIL
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1 ON THE BEST CONSTANT IN GAFFNEY INEQUALITY Bernard DACOROGNA EPFL - Lausanne - Suisse avec G. CSATO et S. SIL à paraître dans Journal of Functional Analysis (2018) En l honneur des 60 ans de Jérôme Pousin
2 GAFFNEY inequality for vector fields states that there exists a constant C = C (Ω) > 0 such that for every vector field u W 1,2 (Ω; R n ) u 2 L 2 C ( curl u 2 L 2 + div u 2 L 2 + u 2 L 2 ) where, on Ω, either ν u = 0 (i.e. u is parallel to ν and we write then u W 1,2 T ν u = 0 (Ω; Rn )) or (i.e. u is orthogonal to ν and we write then u W 1,2 N (Ω; Rn )).
3 GAFFNEY inequality Let 0 k n and Ω R n be a bounded open smooth set. Then there exists a constant C = C (Ω, k) > 0 such that ω 2 L 2 C ( dω 2 L 2 + δω 2 L 2 + ω 2 L 2 ) for every ω W 1,2 T ( Ω; Λ k ) W 1,2 N ( Ω; Λ k ).
4 Define ( stands for the L 2 norm) C T (Ω, k) = C N (Ω, k) = It is easy to see that sup ω W 1,2 T \{0} sup ω W 1,2 N \{0} { { ω 2 dω 2 + δω 2 + ω 2 ω 2 dω 2 + δω 2 + ω 2 C T (Ω, k), C N (Ω, k) 1. } } The cases k = 0 and k = n are trivial and C T = C N = 1. The discussion deals with the case 1 k n 1 (k = 1 is the case of vector fields).
5 Bibliography - Gaffney (1951) for manifolds without boundary - For manifolds with boundary: Friedrichs (1955), Morrey and Morrey-Eells (1956). - Bolik (in L p and in Hölder spaces), Csató-Dacorogna-Kneuss, Iwaniec-Martin, Iwaniec-Scott-Stroffolini (in L p ), Mitrea, Mitrea-Mitrea, Schwarz, Taylor, von Wahl. - For vector fields in dimension 2 and 3 : Amrouche-Bernardi-Dauge-Girault, Costabel, Dautray-Lions, Desvillettes-Villani.
6 A stronger version (in fact a regularity result) of Gaffney inequality is ω 2 C ( dω 2 + δω 2 + ω 2), ω W d,δ,2 ( T Ω; Λ k ) (and similarly with T replaced by N) where ω W d,δ,2 T ( Ω; Λ k ) = ω L 2 ( Ω; Λ k) : For smooth or convex Lipschitz domains dω L 2 ( Ω; Λ k+1) δω L 2 ( Ω; Λ k 1) ν ω = 0 on Ω. W d,δ,2 T ( Ω; Λ k ) = W 1,2 T For non-convex Lipschitz domains, in general, ( Ω; Λ k ). W 1,2 T = W d,δ,2 T.
7 Bibliography (for the stronger version) - Mitrea, Mitrea-Mitrea - For vector fields in dimension 2 and 3 : Amrouche-Bernardi-Dauge-Girault, Ben Belgacem-Bernardi-Costabel-Dauge, Ciarlet-Hazard-Lohrengel, Costabel, Costabel-Dauge, Girault-Raviart.
8 Plan of the talk I) The main theorem II) Some examples III) The case of polytopes IV) Idea of proof
9 I) The main theorem Definition Let Ω R n be an open smooth set and Σ = Ω be the associated (n 1) surface. Let γ 1,, γ n 1 be the principal curvatures of Σ. Let 1 k n 1. We say that Ω is k-convex if γ i1 + + γ ik 0 for every 1 i 1 < < i k n 1.
10 Remark (i) When k = 1 : Ω is convex if and only if Ω is 1-convex, i.e. γ i 0 for every 1 i n 1. The result is due to Hadamard and to Chern-Lashof. (ii) When k = n 1 : Ω is (n-1)-convex if and only if the mean curvature of Σ = Ω is non-negative, i.e. γ γ n 1 0.
11 Theorem Let Ω R n be a bounded open smooth set and 1 k n 1. Then the following statements are equivalent. (i) C T (Ω, k) = 1 (i.e. ω 2 dω 2 + δω 2 + ω 2 ) (ii) The sharper version of Gaffney inequality holds, namely ω 2 dω 2 + δω 2, ω W 1,2 ( T Ω; Λ k ). (iii) Ω is (n k) convex. (iv) The supremum is not attained. (v) C T is scale invariant, namely, for every t > 0 C T (t Ω, k) = C T (Ω, k).
12 Remark (i) Similar results for N, since C T (Ω, k) = C N (Ω, n k). (ii) When k = 1, we have the equivalence between - C T (Ω, 1) = 1 (i.e. ω 2 curl ω 2 + div ω 2 + ω 2 ) - The sharper version of Gaffney inequality ω 2 curl ω 2 + div ω 2, ω W 1,2 T (Ω; Rn ) - Ω is (n 1) convex i.e. the mean curvature of Σ = Ω is non-negative, meaning that γ γ n 1 0.
13 (iii) Still when k = 1, since C T (Ω, 1) = C N (Ω, n 1) - C N (Ω, 1) = 1 (i.e. ω 2 curl ω 2 + div ω 2 + ω 2 ) - The sharper version of Gaffney inequality - Ω is convex i.e. ω 2 curl ω 2 + div ω 2, ω W 1,2 N (Ω; Rn ) γ 1,, γ n 1 0.
14 II) Some examples Proposition Let 1 k n 1. Then there exists a set Ω k B (a fixed ball of R n ) such that C T (Ω k, k) is arbitrarily large (and similarly for C N ). Proof (k = 1) Choose Ω 1 an annulus with very small inner radius. choose ω (x) = x x n Then and observe that ν ω = 0 on Ω 1 and it is a harmonic field i.e. dω = 0 and δω = 0. Then [ ω 2 / ω 2] as the inner radius 0.
15 III) The case of Polytopes Definition Ω R n is said to be a generalized polytope, if there exist Ω 0, Ω 1,, Ω M bounded open polytopes such that, for every i, j = 1,, M with i j, Ω i Ω 0, Ω i Ω j = and Ω = Ω 0 \ In this case Ω i, i = 1,, M, are called the holes. ( ) M Ω i i=1 Theorem Let Ω R n be a generalized polytope. Then the following identity holds. ω 2 = dω 2 + δω 2, ω C 1 T ( Ω; Λ k ) C 1 N ( Ω; Λ k ).
16 Remark In particular if ω C 1 T C1 N is a harmonic field i.e. dω = 0 and δω = 0, then ω 0 independently of the topology!!!
17 IV) Idea of proof We discuss only the case k = 1 (i.e. d curl and δ div). Recall that (i) C T (Ω, 1) = 1 (i.e. ω 2 dω 2 + δω 2 + ω 2 ) (ii) The sharper version of Gaffney inequality holds, namely ω 2 dω 2 + δω 2, ω W 1,2 T (Ω; Rn ). (iii) Ω is (n 1) convex i.e. γ γ n 1 0. The implication (ii) (i) is trivial and we discuss only (iii) (ii).
18 For k = 1 we have curl ω 2 + div ω 2 ω 2 = 2 i<j ( ω i ω j ωj ω i x i x j x i x j ) Integrating the above we get, for every ω W 1,2 ( Ω; Λ k ), Ω ( curl ω 2 + div ω 2 ω 2 ) = Ω T ( ) ω; ν 2 γ1 + + γ n 1
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