ON A BIHARMONIC EQUATION INVOLVING NEARLY CRITICAL EXPONENT

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1 Available at: off IC/004/ United Nations Educational Scientific and Cultural Organization and International Atomic Energy Agency THE ADUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS ON A IHARMONIC EQUATION INVOLVING NEARLY CRITICAL EXPONENT Mohamed en Ayed Département de Mathématiques, Faculté des Sciences de Sfax, Route Soukra, Sfax, Tunisia and Khalil El Mehdi Faculté des Sciences et Techniques, Université de Nouakchott,.P. 506, Nouakchott, Mauritania and The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy. Abstract This paper is concerned with a biharmonic equation under the Navier boundary condition P : u = u n+4 n 4, u > 0 in and u = u = 0 on, where is a smooth bounded domain in R n, n 5, and > 0. We study the asymptotic behavior of solutions of P which are minimizing for the Sobolev quotient as goes to zero. We show that such solutions concentrate around a point x 0 as 0, moreover x 0 is a critical point of the Robin s function. Conversely, we show that for any nondegenerate critical point x 0 of the Robin s function, there exist solutions of P concentrating around x 0 as 0. Finally we prove that, in contrast with what happened in the subcritical equation P, the supercritical problem P + has no solutions which concentrate around a point of as 0. MIRAMARE TRIESTE January 004 Mohamed.enayed@fss.rnu.tn khalil@univ-nkc.mr, elmehdik@ictp.trieste.it

2 Introduction and Results In this paper, we are concerned with the following semilinear biharmonic equation under the Navier boundary condition P { u = u p, u > 0 in u = u = 0 on, where is a smooth bounded domain in R n, n 5, is a small positive parameter, and p+ = n/n 4 is the critical Sobolev exponent of the embedding H H 0 Lp+. When the biharmonic operator in P is replaced by the Laplacian operator, there are many works devoted to the study of the counterpart of P, see for example [], [], [3], [5], [9], [], [], [5], [6], [8], [9], [] and the references therein. When 0, p, the mountain pass lemma proves the existence of solution to P for any domain. When = 0, the situation is more complex, Van Der Vorst showed in [] that if is starshaped P 0 has no solution whereas Ebobisse and Ould Ahmedou proved in [3] that P 0 has a solution provided that some homology group of is nontrivial. This topological condition is sufficient, but not necessary, as examples of contractible domains on which a solution exists show [4]. In view of this qualitative change in the situation when = 0, it is interesting to study the asymptotic behavior of the subcritical solution u of P as 0. Chou and Geng [0] made a first study, when is a convex domain. The aim of the first result of this paper is to remove the convexity assumption on. To state this result, we need to introduce some notation. Let us define on the following Robin s function ϕx = Hx, x, with Hx, y = Gx, y, for x, y, x y n 4 where G is the Green s function of, that is, { x Gx,. = c n δ x Gx,. = Gx,. = 0 in on, where δ x denotes the Dirac mass at x and c n = n 4n S n. Let δ a,λ x = c 0 λ n 4 + λ x a n 4, c 0 = [n 4n nn + ] n 4/8, λ > 0, a R n. It is well known see [7] that δ a,λ are the only solutions of u = u n+4 n 4, u > 0 in R n, with u L p+ R n and u L R n and are also the only minimizers of the Sobolev inequality on the whole space, that is S = inf{ u L R n u, s.t. u L, u L n L n 4 n n 4, u 0}.. R n

3 We denote by P δ a,λ the projection of the δ a,λ s on H H0, defined by P δ a,λ = δ a,λ in and P δ a,λ = P δ a,λ = 0 on. Let / u = u, u H H0.3 u, v = u v, u, v H H0.4 u q = u L q..5 Now we state the first result of this paper. Theorem. Assume that n 6. Let u be a solution of P, and assume that H u u p+ S as 0, where S is the best Sobolev constant in R n defined by.. Then up to a subsequence there exist a, λ > 0, α > 0 and v such that u can be written as u = α P δ a,λ + v with α, v 0, a and λ da, + as 0. In addition, a converges to a critical point x 0 of ϕ and we have where c = c n/n 4 0 defined in.. R n lim u 0 L = c c 0 /c ϕx 0, dx, c + x n+4/ = n 4c n/n 4 0 log+ x R x dx and c n + x n+ 0 is Remark. It is important to point out that in the Laplacian case see [5], the method of moving planes has been used to show that blowup points are away from the boundary of domain. The process is standard if domains are convex. For nonconvex regions, the method of moving planes still works in the Laplacian case through the applications of Kelvin transformations [5]. For P, the method of moving planes also works for convex domains [0]. However, for nonconvex domains, a Kelvin transformation does not work for P because the Navier boundary condition is not invariant under the Kelvin transformation of biharmonic operator. Our method here is essential in overcoming the difficulty arising from the nonhomogeneity of Navier boundary condition under the Kelvin transformation. Our next result provides a kind of converse to Theorem.. Theorem.3 Assume that n 6, and x 0 is a nondegenerate critical point of ϕ. Then there exists an 0 > 0 such that for each 0, 0 ], P has a solution of the form u = α P δ a,λ + v with α, v 0, a x 0 and λ da, + as 0. 3

4 In view of the above results, a natural question arises: are equivalent results true for slightly supercritical exponent? The aim of the next result is to answer this question. Theorem.4 Let be any smooth bounded domain in R n, n 5. Then P + has no solution u of the form u = α P δ a,λ + v with v 0, α, a and λ da, + as 0. The proofs of our results are based on the same framework and methods of [0], [] and [5]. The next section will be devoted to prove Theorem., while Theorems.3 and.4 will be proved in sections 3 and 4 respectively. Proof of Theorem. efore starting the proof of Theorem., we need some preliminary results Proposition. [8] Let a and λ > 0 such that λda, is large enough. For θ a,λ = δ a,λ P δ a,λ, we have the following estimates where f a,λ satisfies f a,λ = O Ha,. a 0 θ a,λ δ a,λ, b θ a,λ = c 0 + f a,λ, λ n d n where d is the distance da,,, λ f a,λ λ = O λ n d n, λ n 4 f a,λ λ a = O λ n+ d n, c θ a,λ n = O θ a,λ L n 4, λ λd n 4 λ n = O L n 4, λd n 4 θ a,λ = O θ a,λ, λd n 4 λ a n = O L n 4. λd n Proposition. Let u be a solution of P which satisfies H. Then, we have a u S n/4, b u p+ S n/4. Proof. Since u is a solution of P, then we have u = u p+. Thus, using the assumption H, we derive that u u p+ = u p p+ = S + o. Therefore u = u p+ = S n/4 + o. The result follows. 4

5 Proposition.3 Let u be a solution of P which satisfies H, and let x such that u x = u L := M. Then, for small, we have a b M = + o. u can be written as with ṽ 0, where λ = c /4 n 0 M p /4. Proof. u = P δ x, λ + ṽ, First of all, we prove that M + as 0. To this end, arguing by contradiction, we suppose that M n remains bounded for a sequence n 0 as n +. Then, in view of elliptic regularity theory, we can extract a subsequence, still denoted by u n, which converges uniformly to a limit u o. y Proposition., u 0 0, hence by taking limit in H we find that u 0 achieves a best Sobolev constant S, a contradiction to the fact that S is never achieved on a bounded domain [3]. Now we define the rescaled functions ω satisfies ω y = M u x + M + p/4 y, y = M p /4 x,. { ω = ω p, 0 < ω in ω 0 =, ω = ω = 0 on.. Following the same argument as in Lemma.3 [6], we have M p /4 dx, + as 0. Then it follows from standard elliptic theory that there exists a positive function ω such that after passing to a subsequence ω ω in C 4 loc Rn, and ω satisfies { ω = ω p, 0 ω in R n ω0 =, ω0 = 0. It follows from [7] that ω can be written as ωy = δ 0,αn y, with α n = c /4 n 0. Observe that, for y = M p /4 x x, we have M δ 0,αn y = M c 0 α n 4/ n + α n M p / x x n 4/ = M n 4/8 δ x, λ.3 with λ = α n M p /4. Then, w y δ 0,αn y = M u x M n 4/8 δ x, λ. 5

6 Let us define we need to compute u x = u x M n 4/8 P δ x, λ x, On one hand, we have u, P δ x, λ = = Thus u = u + M n 4/4 P δ x, λ M n 4/8 u, P δ x, λ. u xδ n+4/n 4 x x, λ u x + M + p/4 u, P δ x, λ = = + M n 4/8 0,R 0,R yδ n+4/n 4 x, λ x + M + p/4 w yδ n+4/n 4 0,α n ydy M n 4/8 w yδ n+4/n 4 0,α n ydy M n 4/8 w yδ n+4/n 4 0,α n ydy, where R is a large positive constant such that R n 0,R δn/n 4 0,α n = o. Since w n/n 4 using Holder s inequality we derive that Now, since w δ 0,αn 0,R in C 4 loc Rn, we obtain On the other hand, one can easily verify that Thus = M n/4 u n/n 4 c, w δ n+4/n 4 0,α n = o. u, P δ x, λ = M n 4/8 S n/4 + o. P δ x, λ = S n/4 + o. ym n+ p/4 dy. u = u M n 4/4 S n/4 + o,.4 and, using the fact that u 0 and Proposition., we derive that ut, since M, we have M M n 4/4 + o. and therefore claim a follows. Now we are going to prove claim b. Observe that, using Proposition. and claim a,.4 becomes u = S n/4 + o S n/4 + o = o. Thus claim b follows. 6

7 Proposition.4 Let u be a solution of P which satisfies H. Then, there exist a, α > 0, λ > 0 and v such that u = α P δ a,λ + v with α, a x 0, λ da,, λ /λ and v 0. Furthermore, v satisfies V 0 v, P δ a,λ = v, P δ a,λ / λ = 0, v, P δ a,λ / a = 0. Proof. y Proposition.3, u can be written as u = P δ x, λ +ṽ with ṽ 0, λ dx, as 0. Thus, the following minimization problem min{ u αp δ a,λ, α > 0, a, λ > 0} has a unique solution α, a, λ. Then, for v = u α P δ a,λ, we have that v satisfies V 0. From the two forms of u, one can easily verify that P δ x, λ P δ a,λ = o. Therefore, we derive that a x = o and λ /λ = + o. The result follows. Next, we state a result in which its proof is similar to the proof of Lemma.3 of [5], so we will omit it. Lemma.5 λ = + o as goes to zero implies that δ c 0 λn 4/ = O Log + λ x a in. We are now able to study the v -part of u solution of P. Proposition.6 Let u be a solution of P which satisfies H. Proposition.4 satisfies v C + C λ d n 4 if n < + if n, λ d n+4 n 4 where C is a positive constant independent of. Proof. Thus v Multiplying P by v and integrating on, we obtain u. v u p v = 0. Using Lemma.5, we find Then v occurring in α P δ p + p α P δ p v δ p v χ v <δ + v p v = 0. Q v, v f v + o v = 0,.5 7

8 with Q v, v = v p α P δ p v and f v = α P δ p v. We observe that Q v, v = v p α P δ p v v = v pα p δ p δ p θ v + o v = v pαp δ p c 0 λn 4/ v δ c 0 λn 4/ Using Lemma.5 and the fact that α, we find Q v, v = Q 0 v, v + o v, v δp + o v with Q 0 v, v = v p δ p v. According to [4], Q 0 is coercive, that is, there exists some constant c > 0 independent of, for small enough, such that where V 0 is the condition defined in Proposition.4. We also observe that f v = α p δ p [ = α p c 0 λn 4/ Q 0 v, v c v v E a,λ,.6 E a,λ = {v E/v satisfies V 0 },.7 δ p θ v δ p v Log + λ x a δ p v + δ p θ v ]. The last equality follows from Lemma.5. Therefore we can write, with = a, d f v c v + δ p θ v + δ v p R n n+4 n+4 n p n n n n+4 c v + θ L δ + n 4 δ. R n We notice that R n δ n/n 4 = O λ d n 8.8

9 and θ L δ np n+4 n+4 n c λ d n+4 n 4 if n + c λ d n 4 if n <..9 Thus we obtain f v C v + λ d n 4 if n < + if n λ d n+4 n 4 Combining.5,.6 and.0, we obtain the desired estimate..0 Next we prove the following crucial result : Proposition.7 For u = α P δ a,λ + v solution of P with λ zero, we have the following estimate a c c Ha, a λ n 4 and for n 6, we also have b c 3 H λ n 3 a, a + o a + o λ d n 4 λ d n 3 = 0. = 0, = + o as goes to where c, c are the constants defined in Theorem., and where c 3 is a positive constant. Proof. We start by giving the proof of Claim a. Multiplying the equation P by λ P δ / λ and integrating on, we obtain 0 = P δ u λ λ u p P δ λ λ = α δλ p P δ λ [ α P δ p + p α P δ p v +O δ p v + v p ] P δ χ δ v λ λ.. We estimate each term on the right-hand side in.. First, using Proposition., we have δ p λ P δ c δ p+ = O c λ c λ d n δ p λ P δ λ = δ p λ δ λ + n 4c 0 λ n 4/ δ p H δ p λ f λ, with = a, d. Expanding Ha,. around a and using Proposition., we obtain δλ p P δ λ = O λ d n + n 4c 0 Ha, a δ p λ d n. λ n 4/ Therefore, estimating the integral, we obtain δ p λ P δ λ = n 4 Ha, a c λ n 4 9 λ d n.

10 with c = c n/n 4 0 R n Secondly, we compute dx + x n+4/. P δ p P δ λ λ = [ δ p = δ p δ λ λ p δ p δ p θ λ λ δ p θ θ λ λ + θ θ δp p θ δ p + and we have to estimate each term on the right hand-side of.3. δ Using the fact that λ λ = n 4 with c = n 4 δ p δ λ λ = n 4 n c n 4 0 c p+ 0 n 4 λ = n 4 λ n 4/ R n Log+ x + x n λ x a +λ x a δ, we derive that R n + x n 4 n c x x + dx > 0. For the other terms in.3, using Proposition., we have δ p θ λ λ = δ p c 0 H λ f λ and p = n 4 = n 4 = n 4 δ p δ θ λ λ λ n 4/ c p+ Ha, a 0 λ n 4/ Ha, a λ n 4 c Ha, a λ n 4 = p = c 0p λ n 4/ = p λ n 4 = n 4 c 0 λ n 4/ λ n 4/ Ha, a λ n 4 c λ n 4/ δ p + θ p ] P δ λ λ δ p θ λ δ λ θ p δ + x + x dx λ d n n 4 λ + λ x a 0,λd Ha, a c 0 λ n 4/ 0,λ d Ha, a λ n 4 c p+ 0 + x n+4 p n 4 n 4 n 4 n λ d λ d n n 4 λ d n 4 + λ d n δ Hλ λ δ p f.3.4 λ d n λ d n.5 δ p δ λ λ λ d n n 4 x + x n+6 n 4 λ d n λ d n 4 + λ d n..3,.4,.5 and additional integral estimates of the same type provide us with the expansion P δ p P δ λ λ = n 4 λ n 4/ +O [ c + c Ha, a λ n 4 λ d n + if n = 5 λ d ]..6 0

11 We note that P δ p P δ v λ λ = δ p δ p P δ θ v λ λ = δ p δ v λ λ δ p θ v λ λ δ p v θ = δ p n δ v λ λ v n+4/n p n+4 λ d n 4/ δ n n+4/n n+4 v..7 δ p c Using.8 and.9, we derive that P δ p P δ v λ λ = δ p δ v λ λ v λ d n+4 v + if n < O λ d n 4. n 4.8 We also have, using Lemma.5 δ p δ v λ λ = δ p δ c n λ v 0 λ λ + δ δ p δ c n v 0 λ λ = O Log + λ x a δ v p = O v. P δ Noticing that, in addition, λ λ = Oδ and δ p v = O v, δ< v v p δ = O v p+..9.,.6,...,.9, Proposition.6, Lemma.5 and the fact that λ = +OLogλ prove claim a of our Proposition. Now, since the proof of Claim b is similar to that of Claim a, we only point out some necessary changes in the proof. We multiply the equation P by /λ P δ / a and we integrate on, thus we obtain a similar equation as.. As., we have δ p P δ λ a = δ p δ λ a c 0 λ n / δ p H a λ d n = c H λ n 3 a a, a λ d n..0

12 Now, to obtain a similar result as.6, we need to estimate the following quantities δ p δ λ a = 0 δ p θ λ a = δ p c 0 H λ n / a f λ a H = λ n 4 λ n 3 a a, a c λ d n θ λ a δ p = p Dθa δ p δ λ a x a λ d n H = λ n 4 λ n 3 y a, a c λ d n. These estimates imply that P δ p P δ λ a = c λ n 4 λ n 3 + if n = 6O H a a, a λ d 4. λ d n 3 + λ d n. Now, it remains to prove a similar result as.7. Using the same arguments, we obtain P δ p P δ v λ a = O + λ d n+4 n 4 + if n < λ d n 4.. The proof of claim b follows. We are now able to prove Theorem.. Proof of Theorem. Let u be a solution of P which satisfies H. Then, using Proposition.4, u = α P δ a,λ + v with α, λ da,, v satisfies V 0 and v 0. y Propositions.3 and.4, we have λ. Now, using claim a of Proposition.7, we derive that Now claim b implies that = c Ha, a + o c λ d n 4 = O..3 λ d n 4 λ n 4 H a, a = o a d n 3..4 Using.4 and the fact that for a near the boundary H/ aa, a cd 3 n a,, we derive that a is away from the boundary and it converges to a critical point x 0 of ϕ. Finally, using.3, we obtain λ n 4 c c ϕx 0 as 0. Thus in order to complete the proof of our theorem, it only remains to show that M := u L c 0 λ n 4/ as 0..5 Using Propositions.3 and.4, we derive that c 0 u L λ n 4. Hence.5 follows. This concludes the proof of Theorem..

13 3 Proof of Theorem.3 Let x 0 be a nondegenerate critical point of ϕ. It is easy to see that da, > d 0 > 0 for a near x 0. We will take a function u = αp δ a,λ + v where α α 0 is very small, λ is large enough, v is very small, a is close to x 0 and α 0 = S n/8 and we will prove that we can choose the variables α, λ, a, v so that u is a critical point of J, where /p+ J u = u u p+ is the functional corresponding to problem P. Let M = {α, λ, a, v R + R + E/ α α 0 < ν 0, d a > d 0, λ > ν 0, Logλ < ν 0, v < ν 0 and v E a,λ }, where ν 0 and d 0 are two suitable positive constants and where d a = da,. Let us define the functional K : M R, K α, a, λ, v = J αp δ a,λ + v. Notice that α, λ, a, v is a critical point of K if and only if u = αp δ a,λ + v is a critical point of J on E. So this fact allows us to look for critical points of J by successive optimizations with respect to the different parameters on M. First, arguing as in Proposition 4 of [0] and using computations performed in the previous sections, we observe that the following problem min{j αp δ a,λ + v, v E a,λ and v < ν 0 } is achieved by a unique function v which satisfies the estimate of Proposition.6. This implies that there exist A, and C i s such that K v α, λ, a, v = J αp δ a,λ + v = AP δ a,λ + λ P δ a,λ + where a i is the i th component of a. n i= C i P δ a a,λ, 3. i Now, we need to estimate the constants A,, and C i s. For this purpose, we take the scalar product of J αp δ a,λ +v with P δ a,λ, P δ a,λ / λ and P δ a,λ / a i with i =,..., n. Thus, we get a quasi-diagonal system whose coefficients are given by 3

14 λ P δ a,λ, P δ a,λ = c, λn 4 P δ a,λ, P δ a a,λ = O i λ P δ a a,λ = O, i λn 3 P δ a,λ, n 4, λ P δ a,λ λ P δ a,λ a j P δ a,λ, a i P δ a,λ where c i s are positive constants and δ ij is the Kronecker symbol. = O λ n 3, = c λ, λn = c 3 λ δ ij λ n 5 The other hand-side of this quasi-diagonal system is given by J αp δ a,λ + v, ψ where ψ = P δ a,λ, P δ a,λ / λ, P δ a,λ / a i with i =,..., n. Observe that, for u = αp δ a,λ + v, J u, ψ = J u αp δ a,λ, ψ J u p+ / u p ψ. Expanding J, we obtain J αp δ a,λ + v = S + Logλ + λ n Now, using.0,.,.6,.8,.0,.,. and Proposition.6, we derive that, after taking the following change of variable: α = α 0 + β, J u, P δ = O Logλ + β + λ n 4 J u, P δ/ λ = O J u, P δ/ a j = O λ + λ n 3 λ + λ n 4, for each j =,..., n. The solution of the system in A,, and C i s shows that A = O Logλ + β + λ n 4, = O λ + λ n 5, C j = O λ + λ n. Now, to find critical points of K, we have to solve the following system K α + K v, v α = 0 K E λ + K v, v λ = 0 K = 0, for j =,..., n. K a j + v, v a j Taking the derivatives, with respect to the different parameters on M, of the following equalities V 0 v, P δ a,λ = v, P δ a,λ / λ i = 0, v, P δ a,λ / a i = 0 for i =,..., n and using 3., we see that system E is equivalent to K α = 0 K E λ = P δ, v + n λ i= C P δ i λ a i, v = P δ λ a j, v + n i= C P δ i a i a j, v, for each j =,..., n. K a j 4

15 The same computation as in the proof of Proposition.6 shows that K α = J αp δ + v, P δ = J u αs n/4 α p S n/n 4 Logλ + λ n 4. Furthermore, using the estimates provided in the proof of claim a of Proposition.7, we derive that λ K λ = J αp δ + v, λ P δ λ Ha, a αc λ n 4 = n 4J u +O Logλ + Logλ λ n 4 + λ n α p S n/n 4 + c S n n 4 α p. Following also the proof of claim b of Proposition.7, we obtain, for each j =,..., n, K = J αp δ + v, P δ a j a j = cα λ n 4 H a a, a α p S n/n 4 On the other hand, one can easily verify that i P δ λ = O λ, ii P δ = O, λ a i Now, we take the following change of variables: α = α 0 + β, a = x 0 + ξ, λ n 4 = λ + Logλ λ n 4 + λ n + if n = 6 λ 3. c c iii P δ a i a j = Oλ Hx 0, x 0 + ρ Then, using estimates 3.3, Proposition.6 and the fact that x 0 is a nondegenerate critical point of ϕ, the system E becomes β = O Log + β E 3 ρ = O /n 4 + β + ξ + ρ ξ = O /n 4 + β + ξ + ρ + if n = 6 /. Thus rower s fixed point theorem shows that the system E 3 has a solution β, ρ, ξ for small enough such that β = O Log, ρ = O /n 4 + if n = 6 /, ξ = O /n 4 + if n = 6 /. y construction, the corresponding u is a critical point of J that is w = J u n/8 u satisfies w = w 8/n 4 w in, w = w = 0 on. 3.4 with w L n/n 4 very small, where w As in Propostion 4. of [7], we prove that w = max0, w. = 0. Thus, since w is a non-negative function which satisfies 3.4, the strong maximum principle ensures that w > 0 on and then u is a solution of P, which blows-up at x 0 as goes to zero. This ends the proof of Theorem.3. 5

16 4 Proof of Theorem.4 First of all, we can easily show that for u satisfying the assumption of the theorem, there is a unique way to choose a, λ and v such that u = α P δ a,λ + v 4. with 4. α R, α a, λ R +, λ da, + v 0 in E := H H0, v E a,λ and for any a, λ R +, E a,λ denotes the subspace of E defined by.7. In the following, we always assume that u, satisfying the assumption of Theorem.4, is written as in 4.. To simplify the notations, we set δ a,λ = δ, P δ a,λ = P δ and θ a,λ = θ. Now we are going to estimate the v occurring in 4.. Lemma 4. Let u satisfy the assumption of Theorem.4. Then we have i u S n/4 ; ii u p++ S n/4 as 0, S denoting the Sobolev constant defined by.. Proof. We have u = α P δ + v = α P δ + v since v E a,λ. From the fact that δ satisfies δ = δ p in R n and is a minimizer for S, we deduce that R n δ = S n/4. On the other hand, an explicit computation provides us with δ a,λ = δ a,λ R n λda, n as λda, +. Using Proposition., claim i is a consequence of 4.. Claim ii follows from the fact that u solves P +. Lemma 4. Let u satisfy the assumption of Theorem.4. Then λ occurring in 4. satisfies λ as 0. 6

17 Proof. and Thus According to Lemma 4., we have u p++ = S n/4 + o as u p++ = α P δ + v p+ α P δ + = α p++ = α p++ = α p++ P δ p++ + P δ p++ P δ p++ u p++ We observe that P δ p++ = Using Proposition., we obtain P δ p++ u p+ v u v n 4 λ λ n 4/ P δ p+ v + P δ v p n 4 + λ v p+ P δ v p+ P δ + v v L p+ + λ n 4/ v p+ + v L p+ = α p++ P δ p++ + o λ n 4/ δ θ p++ = δ p++ = c p++ λ 0 R n + λ x a λ θ L + λ x a = c p++ 0 λ n 4/ R n δ p+ θ n+n 4/ p+n 4 n 4 + λ λ d n. n 4 dx + x n+n 4/ λ λ d n 4.. We note that c p++ 0 R n dx + x = cp+ n+n 4/ 0 R n = S n/4. dx + x n Therefore P δ p++ and 4.4 and 4.5 provide us with u p++ = λ n 4/ S n/4 + o. 4.5 = α p++ λ n 4/ A combination of 4.3 and 4.6 proves the lemma. S n/4 + o + o. 4.6 Next, as in Lemma.3 of [5], we can easily prove the following estimate : 7

18 Lemma 4.3 λ = + o as goes to zero implies that δ x c 0λ n 4/ = O Log + λ x a in. We are now able to study the v -part of u. Lemma 4.4 Let u satisfy the assumption of Theorem.4. Then v occurring in 4. satisfies v p++ = o as 0. Proof. We observe that u p++ = α P δ p++ + v p++ α P δ p+ v + v p+ α P δ. We are going to estimate each term on the right hand-side in the above equality. [ ] α P δ p++ = α p++ δ p++ δ p+ θ = α p++ δ p++ n 4 α P δ p+ v λ v = o + o n 4 v p+ α P δ λ v p+ = o λ n 4/ using Proposition., Holder inequality and Sobolev embedding theorem. From Lemma 4. we derive that As we also have S n/4 + o = + o δ p++ + v p δ p++ from Lemma 4.3 we deduce that δ p++ n 4 = λ c 0 S n/4 n 4 = λ c 0 δ p+ + δ p++ R n c n 4 0λ δ p+, δ p+ δ p+ Log + λ x a = + os n/4 λ d n. 4.8 Combining 4.7 and 4.8, we obtain the desired result. Lemma 4.5 Let u satisfy the assumption of Theorem.4. Then we have where v is defined in 4.. i u L = O ii v L = O, 8

19 Proof. We notice that Claim ii follows from Claim i and Lemma 4.. Then we only need to show that Claim i is true. We define the rescaled functions ω y = M u where x is such that ω satisfies x + M p /4 y, y = M p + 4 x, 4.9 M := u x = u L. 4.0 { ω = ω p+, 0 < ω in ω 0 =, ω = ω = 0 on. 4. Following the same argument as in Lemma.3 [6], we have M p +/4 dx, + as 0. Then it follows from standard elliptic theory that there exists a positive function ω such that after passing to a subsequence ω ω in C 4 loc Rn, and ω satisfies { ω = ω p, 0 ω in R n ω0 =, ω0 = 0. It follows from [7] that ω can be written as ωy = δ 0,αn y, with α n = c /4 n 0 Therefore n 4 4 M p x,m 4 u p++ xdx = ω p++ ydy c > 0 0, as We notice that, as in the proof of Lemma 4. p u p++ xdx = p α P δ + v p++ xdx x,m 4 x,m 4 = α p++ δ p++ a,λ xdx + o = α p++ p x,m 4 n 4 λ x, λ p + M 4 dy + y λ a x n 4 n+ + o. 4.3 Combining 4., 4.3 and Lemma 4., we obtain dy n 4 4 M p 0,λ M 4 + y λ a x n 4 n+ c > 0 as We also have M n 4 p x,m 4 u p+ xdx = ω p+ ydy c > 0 as 0. 0, 9

20 Consequently, we find in the same way as above M n 4 p 0,λ M 4 dy + y λ a x c > 0 as n One of the two following cases occurs : Case. λ M p 4 0 as 0. In this case, we can assume λ M p 4 c > 0, as The claim follows from 4.6 and Lemma 4.. Case. λ M p 4 0 as 0. Now we distinguish two subcases: Case.. λ a x +, as 0. We can assume that λ a x remains bounded when 0. Thus, using 4.5 we obtain n M n 4 λ M p 4 c > 0 as 0, which implies λ M /n 4 c > 0 as 0. Using Lemma 4., we derive a contradiction : this subcase cannot happen. Case.. λ a x + as 0. Using 4.4 and 4.5, we obtain λ n λ a x n+n 4 M n + n 4 C > 0 as and λ n λ a x n M n n 4 C > 0 as From 4.7 and 4.8, we deduce that M λ a x n 4 as and we also derive a contradiction. Consequently Case cannot occur, and the lemma is proved. Now, arguing as in the proof of Proposition.6, we can easily derive the following estimate Lemma 4.6 Let u satisfy the assumption of Theorem.4. Then v occurring in 4. satisfies v C + λ d n 4 if n < + Logλ d λ d 4 if n = + if n > λ d n+4 with C independent of. Now, using the same method as in the proof of Claim a of Proposition.7, we can easily obtain the following result : 0

21 Proposition 4.7 Let u satisfy the assumption of Theorem.4. Then there exist C > 0 and C > 0 such that Ha, a C λ n 4 + o + C + o = O where a, λ and d = da, are given in 4.. We are now able to prove Theorem.4. Proof of Theorem.4 Logλ d λ d n + if n = 5 λ d 4 Arguing by contradiction, let us suppose that P + has a solution u as stated in Theorem.4. From Proposition 4.7, we have Ha, a C λ n 4 + o + C + o = O with C > 0 and C > 0. Two cases may occur : Logλ d λ d n + if n = 5 λ d Case. d 0 as 0. Using 4.0 and the fact that Ha, a cd 4 n, we derive a contradiction. Case.d 0 as 0. We have Ha, a c > 0 as 0 and 4.0 also leads to a contradiction. This completes the proof of Theorem.4. Acknowledgments. This work was finished when the authors were visiting Mathematics Department of Roma University La Sapienza. They would like to thank the Mathematics Department for warm hospitality. The authors would also like to thank Professors M. Grossi and F. Pacella for their constant support. References [] J.G. Azorero and J.P. Alonso, On limits of solutions of elliptic problems with nearly critical exponent, Comm. Partial Diff. Equations 7 99, 3-6. [] F.V. Atkinson and L.A. Peletier, Elliptic equations with near critical growth, J. Diff. Equations , [3] A. ahri, Y.Y. Li and O. Rey, On a variational problem with lack of compactness : the topological effect of the critical points at infinity, Calc. Var. and Part. Diff. Equ , [4] M. en Ayed and K. El Mehdi, The Paneitz Curvature problem on lower dimensional spheres, Preprint the Abdus Salam ICTP, Trieste, Italy, IC/003/48. [5] M. en Ayed, K. El Mehdi, M. Grossi and O. Rey, A nonexistence result of single peaked solutions to a supercritical nonlinear problem, Comm. Contemp. Math ,

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