arxiv:1405.7734v1 [math.ap] 9 May 014 The effects of a discontinues weight for a problem with a critical nonlinearity Rejeb Hadiji and Habib Yazidi Abstract { We study the minimizing problem px) u dx, u H0), 1 u = 1 L ) where is a smooth bounded domain of IR, 3 and p a positive discontinuous function. We prove the existence of a minimizer under some assumptions. Keywords : Critical Sobolev exponent, Lack of compactness, Best Sobolev constant. 010 AMS subject classifications: 35J0, 35J5, 35H30, 35J60. 1 Introduction We consider the minimizing problem Sp) = u H 1 0 )\{0 px) u dx ), 1) u dx where be a smooth bounded domain of IR, 3, = and p a discontinuous function. This problem is closely related to the best constant in Sobolev inequality in IR. It posses many interesting properties, see Talenti [14], and arising in many areas of mathematics and in a geometric context namely for example in the Yamabe problem and the prescribed scalar curvature problem see Aubin [1]. It s invariance under dilations produces a lack of compactness. The phenomenon of lack of compactness and the failure of the Palais Smale condition of this type of problem has been the subject of several studies and it was analyzed in minute detail by Struwe [13]. In the case where p is a constant, it is well known that 1) is not achieved for a general domain. evertheless, Brezis and irenberg showed in [6] that 1) has minimizer under Université Paris-Est Créteil, Laboratoire d Analyse et de Mathématiques Appliquées, CRS UMR 8050, UFR des Sciences et Technologie, 61, Avenue du Général de Gaulle Bât. P3, 4e étage, 94010 Créteil Cedex, France. E-mail : hadiji@u-pec.fr Université de Tunis, École nationale supérieure d ingénieurs de Tunis, 5 Avenue Taha Hssine, Bab Mnar 1008 Tunis, Tunisie. E-mail : habib.yazidi@gmail.com
a linear perturbation. Bahri and Coron in [4] proved that the Euler equation associated to this problem is solvable when some homology group of the domain with coefficients in Z/Z is nontrivial, see also the work of Coron [7]. In the case where p is a smooth positive function, we proved that the study of problem 1) depends on the behavior of the weight p near its minima, see [9] see also [11]). One may ask whether the lack of compactness of the variational functional associated to 1) can be make up by the discontinuity of the weight. In this paper, we consider the discontinuous function p defined by { α if x 1 px) = ) β if x, where α and β are some positive constants such that α < β and 1, are two non empty, disjoints domains such that = 1 and Γ 1 = 1 is not empty. Let us remark that the general case of discontinuous function will be treated in [10]. Statements and proofs of results Let { = α u dx+β u dx, u H 1 IR ), u 0in IR ±, u L IR ) = 1. IR + R Set { S + = u dx, u H 1 IR + ), u 0in IR +, u L IR + ) = 1. IR + and S = { IR It is easy to verify that see for example [8]) u dx, u H 1 IR ), u 0in IR, u L IR ) = 1 S + = S = S, 3) where S is the best constant of the Sobolev embedding defined by u dx IR S = ) u H 1 IR. )\{0 u dx IR Our main results are the following.
Theorem 1 We have = α +β ) S. Theorem Let, 1, and p be as ). Assume that the following geometrical condition g.c.) on Γ 1 holds: there exists an x 0 in the interior of Γ 1 such that in a neighborhood of x 0, lies on one side of the tangent plane at x 0 and the mean curvature with respect to the unit inner normal of at x 0 is positive. then Sp) is attained by some u H 1 0 ). Remark 1 Let us give simple examples for which the condition g.c.) in Theorem is fulfilled or not. Let = B0,R), R > 1 and e 1 = 1,0...,0). Set = Be 1,R), 1 = \, Γ 1 = 1 and x 0 in the interior of Γ 1. We have condition g.c.) holds. For 1 = Be 1,R), = \ 1 and, Γ 1 = 1 and x 0 in the interior of Γ 1. We have condition g.c.) is not satisfied. More precisely, in any neighborhood of x 0, does not lie on one side of the tangent plane at x 0 and the mean curvature with respect to the unit inner normal of at x 0 is negative. Let 1 = {x 1,...,x ) s.t. x 1 > 0 and = {x 1,...,x ) s.t. x 1 < 0 and x 0 = 0. We have condition g.c.) hold, more precisely, in any neighborhood of 0, lies on one side of the tangent plane at 0 but the mean curvature with respect to the unit inner normal of at 0 is 0. If Γ 1 is flat, that is mean that mean curvature at any point of Γ 1 is zero, then we have the following non-existence result: Proposition 1 Let = B0,R) and Γ 1 = {x \x = 0. Then Sp) is not achieved. Indeed, If Sp) is achieved by some positive function u > 0, then there exists a Lagrange multiplier µ R such that u satisfies the Euler equation α u = µu 1 in 1, β u = µu 1 in, Γ 1 α u ν 1 +β u ν = 0 On Γ 1, u = 0 on, where ν 1 and ν are respectively the outward normal of 1 and. On one hand we multiply 4) by u x and we integrate, on the other hand we multiply 4) by u and we integrate, we obtain, after some computations, the Pohozaev identity [ αx ν 1 ) u +βx ν ) u ] ds x = αx ν) u ν 1 ν 1 ν +βx ν) u ν ds x, where ν is the outward of. Since B0,R) is star-shaped about 0 then x ν > 0 and then [ αx ν 1 ) u +βx ν ) u ] ds x > 0 ν 1 ν Γ 1 4)
which gives a contradiction since x ν 1 = x ν = 0 for every x in Γ 1. Therefore Sp) is not achieved. Proof of Theorem 1. On one hand, we claim that Indeed, we see that, for all t ]0, 1[ we have { = α Therefore IR + α + β u dx+β { { IR + R R α +β ) S. 5) u dx, u H 1 IR ), u L IR + ) = t, u L IR ) = 1 t u dx, u H 1 IR ), u = t, = 1 t L IR u + ) L IR ) u dx, u H 1 IR ), u L IR + ) = t, u L IR ) = 1 t At this stage, define { A t = u H 1 IR ), u = t, L IR u = 1 t, + ) L IR ) B t = { u H 1 IR +), u L IR +) = t and We have We rewrite 7) as Using 8), we see that C t = { u H 1 IR ), u L IR ) = 1 t. 6) 7). A t B t and A t C t. 8) α u dx+β u dx. A t IR A t + R α u dx+β u dx α u dx+β u dx. 9) A t IR A t + R B t IR C t + R Or, looking at 3), direct computations give that B t IR + u dx = t S.
and C t IR u dx = 1 t) S. Then, 9) becomes α t 1 = which gives 5). On the other hand, we claim that S +β 1 t) t [0, 1] α +β S [ αt +β1 t) ]S ) S α +β ) S. 10) In order to prove the previous claim, for every x IR we note x = x,x ) where x IR 1. Let {u + j be a minimizing sequence of S+. We define the sequence u j x, x ) = u + j x, x ) for all x IR 1 ]+, 0] and for all j. Easily we see that {u j is a minimizing sequence of S. There exists t 0 = α β ) 1+ α β ) such that αt +β1 t) )S t [0, 1] We define the following functions : = αt 0 +β1 t 0 ) )S = α +β ) S. v + j x, x ) = t 0 u + j x, t θ 0 x ) for all x IR 1 ]0, + ] v j x, x ) = t 0 u j x, 1 t 0 ) θ x ) for all x IR 1 ]+, 0], where θ = 1). ow, consider w j x, x ) = { v + j x, x ) for all x IR 1 and for all x 0 v j x, x ) for all x IR 1 and for all x 0 11)
An easy computation yields that, for large j, w j is a testing function for defined by 6). Therefore α v + j dx+β v j dx. IR + IR Using the definitions of v + j and v j, we obtain α t 0 S +β 1 t 0) S +o1). Then, using the definition of t 0 and letting j +, we obtain = α +β ) S, which gives 10). Finally, 5) and 10) give the conclusion of Theorem 1. The proof of Theorem follows from the following two Lemmas. Lemma 1 Following the hypothesis of Theorem, we have if Sp) < then the imum in 1) is achieved. Proof. Weadapt the arguments ofbrezis-irenberg [6], proofoflemma.1). Let {u j H0 1) be a minimizing sequence for 1) that is, px) u j dx = Sp)+o1), 1) u j L = 1. 13) Easily we see that {u j is bounded in H0 1 ) we may extract a subsequence still denoted by u j, such that u j u weakly in H 1 0 ), u j u strongly in L ), u j u a.e. on, with u L 1. Set v j = u j u, so that v j 0 weakly in H 1 0) v j 0 strongly in L ), v j 0 a.e. on.
Using 1) we write px) ux) dx+ px) v j x) dx = Sp)+o1), 14) since v j 0 weakly in H0 1 ). On the other hand, it follows from a result of [5] that u+v j L = u L + v j L +o1), which holds since v j is bounded in L and v j 0 a.e.). Thus, by 13), we have 1 = u L + v j L +o1) 15) and therefore 1 u L + v j L +o1). 16) Using the definition of, extending v j by 0 in IR still denoted by v j ) we obtain v j L 1 1 1 v j L 1 [ ] α v j x) dx+β v j x) dx IR +,x IR 0,x 0 [ ] α v j x) dx+β v j x) dx IR +,x IR 0,x 0 [ ] α v j x) dx+β v j x) dx px) v j x) dx. 17) where IR +,x 0 = {x = x,x ) IR, \ x IR 1, x > x 0 and IR,x 0 = {x = x,x ) IR, \ x IR 1, x < x 0 with x O is such that x 0 = x,x 0 ). We claim that u 0. Indeed, suppose that u 0. From 14) we obtain px) v j dx = Sp)+o1), then From 15) we see that Or 17) gives that lim px) v j dx = Sp). j + lim v j L = 1. j + v j L px) v j dx. Passing to limit in the previous inequality we obtain Sp). This contradicts the hypothesis Sp) <. Consequently u 0. ow, we deduce from 16) and 17) that Sp) Sp) u Sp) L + px) v j x) dx+o1). 18)
Combining 14) and 18) we obtain px) ux) + px) v j x) dx Sp) u Sp) L + px) v j x) dx+o1). Thus [ ] Sp) px) ux) dx Sp) u L + 1 px) v j x) dx+o1). Since Sp) <, we deduce Therefore this means that u is a minimum of Sp). px) ux) dx Sp) u L, 19) px) ux) dx = Sp) u L. Lemma Assume that there exists x 0 in the interior of Γ 1 such that the g.c.) holds. Then Sp) <. Proof. Let{λ i x 0 ) 1 i 1,denotetheprincipalcurvaturesandHx 0 ) = 1 λ i x 0 ) 1 i=1 the mean curvature at x 0 with respect to the unit normal. For the simplicity, we suppose that x 0 = 0. Therefore we note {λ i 1 i 1 the principal curvatures at 0 and H0) = 1 1 λ i. Let R > 0, such that 1 i=1 BR) 1 = {x, x ) BR); x > ρx ) BR) = {x, x ) BR); x < ρx ) BR) Γ 1 = {x, x ) BR); x = ρx ) where x = x 1,x,...,x 1 ) and ρx ) is defined by ρx ) = Σ 1 i=1 λ ix i +O x 3 ). 1 We note that the condition g.c.) implies that ρx ) 0. Let us define, for ε > 0 and for t ]0, 1[ the function ϕx) ε+ x u 0,ε,t x) = +t x ) ϕx) ε+ x +1 t) x ) If x > 0 If x < 0
where ϕ is a radial C -function such that { 1 if x R 4 ϕx) = 0 if x R. 4 There exists t 0 = α β ) 1+ α β ) such that αt +β1 t) )S t [0, 1] We note u 0,ε x) = u 0,ε,t0 x). Set, for i {1,, and = αt 0 +β1 t 0 ) )S u dx Q i u) = i u dx ) Qu) = αq 1 +βq u). = α +β ) S. In order to obtain the result of Lemma, we use u 0,ε as a test function for Sp). From [], page 13), direct computation gives t 0 S + SH0)A)ε lnε) +Oε 1 ) 1 if = 3 Q 1 u 0,ε ) = t 0 S + SH0)A)ε 1 +Oε lnε) ) if 4 and Q u 0,ε ) = 1 t 0 ) S 1 t 0 ) S SH0)A)ε 1 lnε) +Oε 1 ) if = 3 SH0)A)ε 1 +Oε lnε) ) if 4 where A) is a positive constant. Combining 0) and 1) we see that, αt 0 +β1 t 0) )S β α)sh0)a)ε lnε) +Oε 1 ) 1 if = 3 Qu 0,ε ) = αt 0 +β1 t 0) )S β α)sh0)a)ε 1 +Oε lnε) ) if 4. Therefore, using the definition of t 0, we obtain Qu 0,ε ) α +β ) S β α)sh0)a)ε 1 lnε) +Oε 1 ) if = 3 ) α +β 1 S β α)sh0)a)ε +Oε lnε) ) if 4. 0) 1) Finally, Since β > α and H0) > 0 then we obtain the desired result. Remark see []): By looking at the previous proof, it follows that we can relax the condition g.c.) by allowing some of the λ i s to be negative with mean curvature positive.
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