Remarks on solutions of a fourth-order problem

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1 Applied Mathematics Letters ( ) Remarks on solutions of a fourth-order problem Anne Beaulieu a,rejeb Hadiji b, a A.B. Laboratoire d Analyse et de Mathématiques Appliquées, CNRS UMR 8050, Université demarne-la-vallée, 5, boulevard Descartes, Champs-sur-Marne, Marne la Vallée cedex 2, France b R.H. UFR des Sciences et Technologie, CNRS UMR 8050, Université Paris2 - Val-de-Marne, 6, avenue du Général de Gaulle, 9400 Créteil Cedex, France Received 8 June 2005; accepted 8 August 2005 Abstract In this paper, we study the two following minimization problems: S 0 (q,ϕ)= inf u 2 S θ (q,ϕ)= inf u 2. u H0 2(), u+ϕ q = u Hθ 2(), u+ϕ q = We prove that for aclass of maps ϕ,wehaves θ (q,ϕ)< S 0 (q,ϕ) for another class, we have S θ (q,ϕ)= S 0 (q,ϕ). c 2005 Elsevier Ltd. All rights reserved.. Introduction Let us consider the following two minimization problems: S θ (q,ϕ)= inf u 2 (I θ ) u Hθ 2(), u+ϕ q= S 0 (q,ϕ)= inf u 2, (I 0 ) u H0 2(), u+ϕ q = where is a domain in R N, N 3if q < q c = N 4 2N N 5ifq = q c.thefunction ϕ is given in C() L q () Hθ 2() = H 2 () H0 (). Recall that q c is the limiting Sobolev exponent in the imbedding H0 2() Lr (), r q c.note(see Van der Vorst [5][6]) that S θ (q c, 0) = S 0 (q c, 0) = S is thebest Sobolev constant S is not achieved. Corresponding author. Tel.: ; fax: address: hadiji@univ-paris2.fr (R. Hadiji) /$ - see front matter c 2005 Elsevier Ltd. All rights reserved. doi:0.06/j.aml

2 2 A.Beaulieu, R.Hadiji / Applied Mathematics Letters ( ) In [3]itisshown that if ϕ is not zero, then the infima S θ (q,ϕ)(resp. S 0 (q,ϕ))areachieved by u θ (resp. u 0 ), which satisfy respectively the following Euler Lagrange equations: { 2 u θ = Λ θ u θ + ϕ q 2 (u θ + ϕ) in, (E u θ = u θ = 0 on, θ ) 2 u 0 = Λ 0 u 0 + ϕ q 2 (u 0 + ϕ) in, u 0 ν = u (E 0 ) 0 = 0 on, where Λ θ (resp. Λ 0 )isthe Lagrange multiplier associated to u θ (resp. u 0 ). The interest in this type of equations comes from the fact that it resembles some geometrical equations involving the Paneitz operator, which is a fourth-order conformally covariant elliptic operator (see [4]). Since H 2 0 () H 2 () H 0 (), wehaves θ (q,ϕ) S 0 (q,ϕ).itisnatural to wonder if S θ (q,ϕ) < S 0 (q,ϕ) if the infimum on H 2 () H 0 () is achieved by a function of H 2 0 (). Remark. The signs of the Lagrange multipliers depend on ϕ.wehave(see[3]), if ϕ q < thenλ 0 > 0Λ θ > 0, if ϕ q > thenλ θ < 0Λ 0 < 0. Our main result is Theorem. Suppose that q [2, q c ] that ϕ is not identically 0. (i) If ϕ q < ϕ has a constant sign on, thenevery minimizer of (I θ ) is not in H0 2 () we have S θ (q,ϕ)< S 0 (q,ϕ). (ii) Let (H0 2()) be the orthogonal of H0 2() in the space H θ 2().If ϕ is in (H 0 2()),thenevery minimizer of (I θ ) is not in H0 2() we have S θ (q,ϕ)< S 0 (q,ϕ). (iii) For q 2,if ϕ q > ϕ is in H0 2(),thenS θ (q,ϕ)= S 0 (q,ϕ). We do not know if S θ (q,ϕ)= S 0 (q,ϕ)in the other cases. Proof of (i). Suppose that ϕ 0ϕ is not identically 0. We will adapt the argument of [6]toour situation. Let u θ be any minimizer of (I θ ).Weargue by contradiction. We suppose that u θ is in H 2 0 (). Let v be the solution of the following problem: { v = uθ in, v = 0 on. We have { (v uθ ) 0 in, v u θ = 0 on, { (v + uθ ) 0 in, (3) v + u θ = 0 on. We deduce from the maximum principle applied to (2) (3) that either v> u θ in or v = u θ or v = u θ.we use () to see that in both cases v = u θ v = u θ the function u θ has a constant sign. This fact together with u θ = u θ ν = 0on with the maximum principle lead to u θ = 0in, thatisfalse.thus we have v> u θ in. Using this inequality the fact that ϕ 0, we obtain u θ + ϕ<v+ ϕ in u θ ϕ<v+ ϕ in ; thus u θ + ϕ < v + ϕ in consequently we have v + ϕ q >. () (2)

3 A.Beaulieu, R.Hadiji / Applied Mathematics Letters ( ) 3 Now, let us consider the function f (t) = tv + ϕ q for t [0, ].Since f is continuous, f (0) < f () >, there is s [0, ] such that f (s) =. Then we have u θ 2 s 2 v 2, that contradicts the definition of u θ.thefirstpart of the theorem is proved. Proof of (ii). Let us distinguish two cases. Case. Let us suppose that ϕ q > thatϕ is in (H0 2()).Letu θ be any solution of (I θ ).Multiplying (E θ ) by u θ integrating by parts, we obtain u θ 2 + u θ ϕ = Λ θ. Since ϕ q >, we have Λ θ < 0, thus u θ ϕ < 0, which implies that u θ is not in H0 2 (), thenwe conclude that S θ (q,ϕ)< S 0 (q,ϕ). Case 2. If ϕ q <, let us suppose first that q = q c.fora, setu a,ε = ζu a,ε,whereζ is a smooth function, ε such that ζ is equal to near a U a,ε (x) = ( ) N 4 ε 2 + x a 2 2.Wehave,see[3], that u a,ε q = B + o(), u a,ε 2 A = A + o(),suchthat = S. B q 2 Since ϕ q <, there exists c ε > 0suchthat ϕ + c ε u a,ε q =. Using the Brezis-Lieb identity (see []) we obtain ] cε [ q = B ϕ q + o(), where o() tends to 0 as ε tends to 0. Direct computations give [ ] 2 S θ (q,ϕ) cε 2 u a,ε 2 = S ϕ q q + o(). As ε tends to 0, we find ( ) 2 S θ (q,ϕ) S ϕ q q. (4) On the other h, multiplying (E θ ) by (u θ + ϕ) integrating yields S θ (q,ϕ)= Λ θ u θ + ϕ q (u θ + ϕ)u θ, thereforethe Hölderinequality gives S θ (q,ϕ) Λ θ u θ q. (5) Using (5) the Sobolev inequality we find that ( ) S θ (q,ϕ) Λ θ u θ 2 2. (6) S Combining (4) (6) we see that S θ (q,ϕ) Λ θ S S 2 2 [ ] ϕ q q. Now, multiplying (E θ ) by (u θ + ϕ) integrating we obtain u θ. ϕ = Λ θ S θ (q,ϕ). (7) (8)

4 4 A.Beaulieu, R.Hadiji / Applied Mathematics Letters ( ) Finally, combining (7) (8) we are led to u θ. ϕ Λ θ [ ( ϕ q ) q ] > 0. This means that u θ is not in H0 2(). The case where q < q c can be obtained as this last case. Remark 2. The same argument as below shows that if ϕ is in H0 2(), theneverysolution of (I θ) is not in the orthogonal of H0 2 ().Thesecond part of the theorem is proved. Proof of (iii). Let ϕ be in H0 2().Weremark first that for ϕ q, we have S θ (q,ϕ)= inf u H θ 2() u+ϕ q u 2. (9) We have aconvexproblem. We are going to use a method of duality. We refer to [2]forthis proof. For all p L 2 (), let us define β θ = sup p u β 0 = sup p u. We have β θ = u H 2 θ () u+ϕ q sup v H 2 θ () v q p v p ϕ. u H 2 0 () u+ϕ q Let us prove that we have for every p L 2 () β θ = β 0. (0) First, we remark that β θ β 0 are finite. This follows from the Hölderinequality p v p 2 v 2,together with (9).Wededuce that the linear operator L : Hθ 2 () R v p v is continuous for the L q () topology. Thus there exists p L q q () such that for all v Hθ 2 () we have L(v) = pv. Wededuce that β θ = sup pv p ϕ; β 0 = sup pv p ϕ. () v H 2 θ () v q v H 2 θ () v q v H 2 0 () v q On the other h, it is easy to prove that, for all p L q q (),wehave sup pv = sup pv = sup v L q () v q 2 S θ (q,ϕ)= sup p L 2 () v H 2 0 () v q Thus we have proved (0).Now,inthecasewhere ϕ q, let us prove that 2 p 2 sup u H 2 θ () u+ϕ q pv = p q q. (2) p u (3)

5 A.Beaulieu, R.Hadiji / Applied Mathematics Letters ( ) 5 2 S 0(q,ϕ)= sup p L 2 () p 2 2 sup u H 2 0 () u+ϕ q p u. (4) Let us define, for p L 2 () for u Hθ 2(), L(u, p) = p 2 ( u)p. 2 We can see easily that sup L(u, p) = u 2. p L 2 () 2 By (9) (5) we have 2 S θ (q,ϕ)= inf sup L(u, p). u H θ 2() p L 2 () u+ϕ q Let us prove that (P)= (P ) where (P ) is the dual problem of (P),thatis sup L(u, p). inf p L 2 () u H θ 2() u+ϕ q Let us define A ={u Hθ 2 (); u + ϕ q }, let u θ be a minimizer that realizes S θ (q,ϕ) let p θ = u θ.by(5), wehave (5) (P) (P ) L(u θ, p) sup L(u θ, p) = p L 2 () 2 S θ (q,ϕ) for all p L 2 (). (6) Now, for all u A we have L(u, p θ ) p θ 2 sup ( u)p θ, 2 u A that gives, using u θ = 0on, L(u, p θ ) p θ 2 sup u( p θ ). 2 u A Thus we have, using (), L(u, p θ ) p θ 2 p θ q 2 q p θ ϕ. (7) But the Eulerequation (E θ ) for u θ gives 2 u θ q θ. q (8) On the other h, multiplying the Euler equation (E θ ) by u θ + ϕ,weobtain Λ θ = u θ 2 + u θ ϕ. (9) We know that Λ θ < 0, thus (8) (9) give p θ q + p θ ϕ = S θ (q,ϕ). q (20)

6 6 A.Beaulieu, R.Hadiji / Applied Mathematics Letters ( ) Now we obtain by (20) (7) that L(u, p θ ) 2 S θ(q,ϕ) for all u A. (2) It is classical (see [2]) that (2) (6) infer that (P)= (P ). The same proof remains valid for 2 S 0(q,ϕ)instead of 2 S θ (q,ϕ),thus we have proved (3) (4). Nowletus use (0) in order to conclude that S θ (q,ϕ)= S 0 (q,ϕ).thisends the proof of the theorem. References [] Brezis-Lieb, A relation between pointwise convergence of functions convergence of functionals, Proc. Amer. Math. Soc. 88 (983) [2] Ekl-Temam, Analyse convexe et problèmes variationnels, Dunod. [3] Guedda-Hadiji-Picard, A biharmonic problems with constraint involving critical Sobolev exponent, Proc. Roy. Soc. Edinburgh Sect. A 3A (200) [4] S. Paneitz, A quartic conformally covariant differential operator for arbitrary pseudo-riemannian manifolds, 983 (Preprint). [5] Van der Vorst, Fourth order elliptic equations with critical growth, C.R.A.S. 320 (I) (995) [6] Van der Vorst, Best constant for the embedding of the space H 2 H 0 () into L2N/(N 4) (),Differ. Integral Equ. 6 (993)