Sur l ’ équation div u = f

The main result is the following. Let be a bounded Lipschitz domain in Rd , d 2. Then for every f ∈ Ld( ) with ∫ f = 0, there exists a solution u ∈ C0( ) ∩ W1,d ( ) of the equation divu= f in , satisfying in addition u= 0 on ∂ and the estimate ‖u‖L∞ + ‖u‖W1,d C‖f ‖Ld , where C depends only on . However one cannot choose u depending linearly on f . To cite this article: J. Bourgain, H. Brezis, C. R. Acad. Sci. Paris, Ser. I 334 (2002) 973–976.  2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS Abridged English version Consider the equation divu= f on the torus T , (1) i.e., on R with 2π -periodic functions in all variables. Clearly (1) is underdetermined and a standard way of tackling (1) is to look for a special solution u satisfying the condition curl u = 0, i.e., one looks for a Adresses e-mail : [email protected] (J. Bourgain); [email protected] (H. Brezis).  2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS. Tous droits réservés S1631-073X(02)02344-0/FLA 973 J. Bourgain, H. Brezis / C. R. Acad. Sci. Paris, Ser. I 334 (2002) 973–976 solution u of the form u= gradv. Eq. (1) then becomes v = f. (2) Consequently, for every f ∈ L, 1 <p <∞, with ∫ Td f = 0, Eq. (1) admits a solution u= grad( )−1f ∈ W1,p. Three limiting cases are of interest (a) p = 1. It is well known that when f ∈ L1, the solution v of (2) does not necessarily belong to W2,1. However one might still hope to have some solution u of (1) in W1,1. This is not true: for some f ’s in L1, Eq. (1) has no solution u ∈ W1,1 and not even in Ld/(d−1) (see [1]). (b) p =∞. It is well known that when f ∈ L∞ the solution v of (2) does not necessarily belong to W2,∞. However one might hope to have some solution u of (1) in W1,∞. This is not true (see [2] and [1]). (c) p = d . This is the heart of our work. For every f ∈ L with ∫ f = 0, (3) Eq. (2) admits a solution v ∈ W2,d and thus Eq. (1) admits a solution u= gradv ∈ W1,d . Since W1,d is not contained in L∞ (this is a limiting case for the Sobolev imbedding) we cannot assert that this u belongs to L∞. However one might still hope that, given any f ∈ L with (3), there is some u ∈ L∞ solving (1). This is indeed true. Set Ld# = { f ∈ Ld; ∫ f = 0 } . PROPOSITION 1. – Given any f ∈ L# there is some u ∈ L∞ solving (1) (in the sense of distributions) with ‖u‖L∞ C(d)‖f ‖Ld . (4) However there is no bounded linear operator K : L# → L∞ such that div(Kf )= f, ∀f ∈ L# (in the sense of distributions). The proof of the existence of a solution of (1) satisfying (4) is quite elementary. It follows easily by duality from the Sobolev–Nirenberg imbedding W1,1 ⊂ Ld/(d−1) and the corresponding estimate ∥∥∥v − ∫ / v ∥∥∥ Ld/(d−1) C‖gradv‖L1 ∀v ∈ W1,1. (5) Since it relies on Hahn–Banach, it is not constructive. In fact, one cannot choose u=Kf , solution of (1) depending linearly on f and satisfying (4). The argument is by contradiction. If such K exists, the averaged operator