A GLOBAL COMPACTNESS RESULT FOR SINGULAR ELLIPTIC PROBLEMS INVOLVING CRITICAL SOBOLEV EXPONENT

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 131, Number 6, Pages 1857 1866 S 000-9939(0)0679-1 Article electronically published on October 1, 00 A GLOBAL COMPACTNESS RESULT FOR SINGULAR ELLIPTIC PROBLEMS INVOLVING CRITICAL SOBOLEV EXPONENT DAOMIN CAO AND SHUANGJIE PENG (Communicated by David S. Tartakoff) Abstract. Let R N be a bounded domain such that 0,N 3, = N N, R, ɛ R. Let {un} H1 0 () be a (P.S.) sequence of the functional E,ɛ (u) = 1 ( u u x ɛu ) 1 u. We study the limit behaviour of u n and obtain a global compactness result. 1. Introduction Let be a bounded domain in R N such that 0,N 3, = N N, R, ɛ R. In recent years, much attention has been paid to the existence of nontrivial solutions to the following problem: { u = (P,ɛ ) x u + u u + ɛu, x, u =0, x, where 0 < =( N ),ɛ R. For instance, in [11], by using a variational approach, E. Jannelli proved that if 1, then problem (P,ɛ ) has at least one solution u H0 1 () when 0 <ɛ<ɛ 1 (). If 1 <<, then problem (P,ɛ ) has at least one solution u H0 1() when ɛ () <ɛ<ɛ 1 (), where ɛ 1 (), ɛ () are some positive constants depending on. Let (1.1) E,ɛ (u) = 1 ( u u x ɛu ) 1 u, for u H0 1 (). The crucial step in [11] was to overcome the lack of compactness for E,ɛ (u). Indeed, the invariance of H0 1norm, norm and with respect to rescaling u L u r = r N u(r( )) and the existence of non-trivial entire solution of the limiting u x Received by the editors December, 001 and, in revised form, January 31, 00. 000 Mathematics Subject Classification. Primary 35J60; Secondary 35B33. Key words and phrases. Palais-Smale sequence, compactness, Sobolev and Hardy critical exponents. The first author was supported by Special Funds For Major States Basic Research Projects of China (G1999075107) and Knowledge Innovation Funds of CAS in China. The second author was supported by Knowledge Innovation Funds of CAS in China. 1857 c 00 American Mathematical Society

1858 DAOMIN CAO AND SHUANGJIE PENG problem (see [4], [10], [1] and [14]) { u = (1.) x u + u u, x R N, u 0, x, result in the failure of the classical Palais-Smale (P.S. for short) condition for E,ɛ on H0 1(). However a local (P.S.) condition can be established. Indeed, let u p p = u p for p (1, ) and RN (1.3) S := inf{ ( u u x ) u H1 0 (RN ), u =1}. Suppose {u m } H 1 0 () is a sequence such that E,ɛ(u m ) c< 1 N S N,DE,ɛ (u m ) 0 strongly in H 1 () = (H 1 0 ()) ;then{u m } contains a strongly convergent subsequence. Using this local (P.S.) condition, E.Jannelli was able to obtain the existence of one nontrivial solution in [11]. For earlier work, see [3]. As indicated above it is always very important to understand the convergence of the (P.S.) sequence in using variational methods. When = 0, M.Struwe[14] gives a complete description of all energy levels c of E,ɛ on which the (P.S.) c sequence is not compact, and he obtains the well known global compactness theorem, with which J.M.Coron [6] and W.-Y. Ding [7] obtained the positive critical points of E 0,0 with some non-starshaped domains. In [16] S.Yan generalizes this global compactness result to the case of p-laplacian successfully. Very recently, Adimurthi and M.Struwe in [1] have studied the convergence of (P.S.) sequences of the energy functional associated with a semilinear elliptic problem on a bounded domain in R with critical nonlinearity f(s) growing like exp(4πs )ass. As can easily be seen, problem (P,ɛ ) is the general case of problem (P 0,ɛ ). So it is very interesting and important to see whether a global compactness result similartothatofproblem(p 0,ɛ ) still exists or not. In this paper we investigate this problem and obtain an affirmative answer. Our proof is based on rescaling arguments. Such methods have been repeatedly used to extract convergent subsequence from families of solutions or minimizing sequence to nonlinear variational problems; see [13], [14], and [16]). In the following, let D 1, (R N ) be the completion of C 0 (RN ) with respect to the inner product (u, v) = R N u v, S 0 =inf{ R N u u H 1 0 (R N ), u =1}. Let B(x, r) denote the ball centered at x with radius r. For simplicity, we use the same notation m and ṽ m in different situations. Remark 1.1. (i) By Proposition 8.1 in [4], the solutions of (1.) are in one-to-one correspondence to solutions of the following type of problem: div( x a w) = x b w 1,w 0inR N, where a and b are constants depending on. From Theorem B in [5], any solution of the above equations is radially symmetric with respect to the origin, that is, u(x) depends on x only. So according to the theory of ordinary differential equations, the solution of (1.) is unique (up to a dilation). Using the results in [4], we can deduce that when 0 <, the solutions of (1.3) are of the following: u(x) =ε N U(εx), ε > 0, U(x) =C(, ) x + (1 + x 4 N ),

A GLOBAL COMPACTNESS RESULT 1859 where C(, ) is a constant depending only on and. SoS can be attained in R N when 0 <. (ii) S <S 0 when 0 <<. While <0, we can prove that S = S 0 and S cannot be attained, even though in R N.. Global compactness To state the main result, it is convenient to introduce the problem at infinity. The first one is (P0 ) v = v, v D 1, (R N ). v Moreover, for 0 <, the problem at infinity is given by (P v ) v = x + v, v D 1, (R N ). v To unify the notations we shall refer to the solutions of problems at infinity as critical points of the following family of functionals: F (u) = 1 RN ( u u x ) 1 R N u and in the cases of (P 0 )and(p ), the respective energy functionals are F 0 (u) and F (u). Theorem.1. Let N 3, 0 <, ɛ R, and suppose that {u m } H0 1() satisfies E,ɛ (u m ) c, DE,ɛ (u m ) 0 strongly in H 1 () as m. Then there exist a critical point u 0 of E,ɛ, k sequences of positive numbers {rm j } m (1 j k), l sequences of positive numbers {km j } m (1 j l) and l sequences of points {x j m } m (1 j l) in which converge to x j 0, such that up to a subsequence, k l (i) u m = u 0 + (rm j ) N v j 1 (rj m x)+ (km j ) N v j 0 (kj m (x xj m )) + ω m, where ω m H 1 0 () 0,r j m,k j m, v j 0 solves (P 0 ), andv j 1 solves (P ). (ii) E,ɛ (u m ) E,ɛ (u 0 )+ l F 0 k (vj 0 )+ F (vj 1 ). To prove this theorem, let us first recall some known results and establish some preliminary lemmas. The following lemma can be found in [9]. Lemma. (Hardy inequality). If u H0 1 (), then u u x 1 u, moreover is optimal. x L () and Lemma.3. Let {u m } H0 1(),u m u weakly in H0 1 (). Then (i) u m = u m u + + o(1), u (ii) u m = (u m u) + u + o(1), (iii) u m x = u mu x + u x + o(1).

1860 DAOMIN CAO AND SHUANGJIE PENG Proof. (i) and (ii) are the well known results of Brezis-Lieb in []. (iii) can be easily proved by Vitali s theorem and we omit the detail. Lemma.4. Let {v m } H0 1() be a (P.S.) sequence for E,0 at level β, that is, E,0 (v m ) β as m, and assume that v m converges weakly but not strongly to zero in H0 1 (). Then either (i) there exists a sequence of positive numbers {r m }, such that up to a subsequence, w m (x) =v m (x) r N m V 1 (r m x)+o(1) (x ) is a (P.S.) sequence for E,0 in H0 1() at level β F (V 1); moreover, w m 0 weakly in H0 1 (), or (ii) there exists a sequence of positive numbers {k m } and a sequence of points {y m } satisfying y m x 0, such that up to a subsequence, w m (x) =v m (x) k N m V 0 (k m (x y m )) + o(1) (x ) is a (P.S.) sequence for E,0 in H0 1 () at level β F0 (V 0 ). Moreover, w m 0 weakly in H0 1(), k mdist(x m, ),wherev 0 and V 1 solve (P0 ) and (P ) respectively, o(1) 0 in D 1, (R N ) as m. Proof. If E,0 (v m ) β< 1 N S N, then sequence {v m } is strongly relatively compact and hence v m 0,β = 0. Therefore we may assume that E,0 (v m ) β 1 N S N. Moreover, from DE,0 (v m ) 0, we also have 1 (.1) ( v m v m N x )=E,0(v m ) 1 DE,0(v m ),v m β 1 N S N By Hardy inequality, we have for m large enough (.) S N Nβ v m Nβ 1 and hence there are positive constants c i (i =1, ) such that (.3) c 1 v m c, m. Let δ>0 small (will be determined later) such that (.4) lim sup v m >δ. m 1 Up to a subsequence, choose minimal r m > 0 such that B(0, 1 rm ) v m = δ. Define ṽ m := r N m v m ( x r m ); then B(0,1) ṽ m = δ. Let us point out that, thanks to (.3) and (.4), the sequence {r m } is bounded away from zero. Denote m := {x R N x r m }; thenṽ m H0 1( m ) D 1, (R N ). Moreover, ṽ m D 1, (R N ) = v m D 1, (R N ) Nβ <. Up to a subsequence there exists V 1 D 1, (R N ) satisfying ṽ m V 1 weakly in D 1, (R N ), ṽ m V 1 a.e. in R N. We have either V 1 0orV 1 0..

A GLOBAL COMPACTNESS RESULT 1861 (I): Assume V 1 0. Since v m 0weaklyinH0 1(), we have r m, m R N. For this case we claim that V 1 satisfies (P ) and the sequence w m(x) := ṽ m (x) r N m V 1 (r m x) is a (P.S.) sequence for E,0 at level β F (V 1). Indeed, let us fix a ball B(0,r) and a test function ψ C0 (B(0,r)) and remark that for sufficiently large m, B(0,r) m.so DF (V 1 ),ψ = V 1 ψ B(0,r) = ṽ m ψ m m = v m ψ m B(0,r) ṽ m ψ x v m ψm x V 1 ψ x V 1 V 1 ψ B(0,r) ṽ m ṽ m ψ + o(1) m v m v m ψm + o(1) = DE,0 (v m ), ψ m + o(1) (by the (P.S.) condition) where ψ m (x) =r N m ψ(r m x). Thus V 1 solves (P ). Let ϕ C0 (RN ) satisfying 0 ϕ 1, ϕ inr N,ϕ 1, in B(0, 1),ϕ 0 outside B(0, ), and let w m (x) =v m (x) r N m V 1 (r m x)ϕ( r m x) H0 1 () where the sequence { r m } is chosen such that r m := rm r m while r m dist(0, m ) as m,thatis, w m (x) =ṽ m (x) V 1 (x)ϕ( x r m ). Set ϕ m (x) =ϕ( x r m ). Note that V 1 L (R N ), so as m, V 1 (ϕ m 1) R N V 1 (ϕ m 1) + V 1 (ϕ m 1) R N R N V 1 +8 r m V 1 R N \B(0,r m) B(0, r m)\b(0, r m) V 1 +8( V 1 ) = o(1). R N \B(0,r m) B(0, r m)\b(0, r m) Thus we have w m =ṽ m V 1 +o(1), where o(1) 0inD 1, (R N ). So by Lemma.3 and the invariance of dilation, we have for m large E,0 (w m )=E,0 (ṽ m ) F (V 1)+o(1) = β F (V 1)+o(1), DE,0 (w m ) H 1 () = o(1). Also, it is obvious that w m 0weaklyinH 1 0 ().

186 DAOMIN CAO AND SHUANGJIE PENG (II): Assume V 1 0. Let h C0 (B(0, 1)); we have (ṽ m h) R N = ṽ m (h ṽ m )+ ṽm h R N R N = ṽ m (h ṽ m )+o(1) R N = DE,0 (ṽ m ),h h ṽ m ṽ m + R x + ṽ m h + o(1) N R N 4 (N ) (ṽ m h) + S0 1 ( ṽ m ) N (ṽ m h) + o(1). R N B(0,1) R N Choose δ suitably small. From the above inequality and the fact 0 < (N) 4, we can find an a (0, 1) such that (.5) ṽ m 0. B(0,a) For simplicity, substitute ṽ m by z m and m by (because of dilation invariance), and without loss of generality, we still suppose that z m satisfies (.3) and (.4). Denote Q m (r) =sup z m x B(x,r) the concentration function of z m,choosex m andscale (.6) z m z m (x) :=R N m z m ( x + x m ) R m such that (.7) Qm (1) = sup z m = z m = 1 x Rm +xm,x RN B(x,1) B(0,1) L S N, where L denotes the least number of balls with radius 1 in R N that are needed to cover a ball of radius. Note that {R m } is obviously bounded away from zero. Setting m := {x R N x R m + x m }, we may regard z m D 1, (R N ), moreover, { z m } is bounded uniformly in D 1, (R N ). Thus up to a subsequence, (.8) z m V 0 weakly in D 1, (R N ). We are going to prove that the convergence actually holds in the strong Hloc 1 (RN ) sense. To do this, following the analogous argument in [14], for any x R N,we can find ρ [1, ] such that the solution φ m of the Dirichlet problems { φ =0, x B(x, 3) \ B(x, ρ), φ B(x,ρ) = z m V 0, φ B(x,3) =0, satisfies (.9) φ m 0inH 1 (B(x, 3) \ B(x, ρ)). Let { zm V (.10) ϕ m = 0, x B(x, ρ), φ m, x B(x, 3) \ B(x, ρ).

A GLOBAL COMPACTNESS RESULT 1863 It follows that (.11) ϕ m = ϕ m + o(1) H0 1 ( m )+D 1, (R N ), where ϕ m H0 1( m )ando(1) 0inD 1, (R N )asm (see for instance [14]). Hence, scaling back the ϕ m s, ϕ m (x) :=R N m ϕ m (R m (x x m )), we have, taking into account of (.9), (.10) and (.11), (.1) ϕ m L () = ϕ m D 1, (R N ) + o(1) = z m V 0 D 1, (B(x,ρ)) + o(1). By scale invariance, (.11) and the fact {z m } is a (P.S.) sequence for E,0, DE m ( z m ),ϕ m = DE,0 (z m ), ϕ m + o(1) = o(1) where E m ( z m ):= 1 ( z m z m m x + R m x m ) 1 z m. m Therefore, from the definitions of E m, ϕ m and (.9), we infer by using arguments similar to those in [14] z m ( z m V 0 ) o(1) = z m ( z m V 0 ) m B(x,ρ) m B(x,ρ) x + R m x m z m z m ( z m V 0 ) m B(x,ρ) = ( z m V 0 ) m B(x,ρ) m B(x,ρ) ( z m V 0 ) x + R m x m z m V 0 + o(1) m B(x,ρ) = ϕ m ϕ m m m x + R m x m ϕ m + o(1). m Moreover, by scale invariance, Sobolev inequality and Hardy inequality, (.13) o(1) = ( ϕ m c ϕ m ϕ m x ϕ m ϕ m x )(1 ϕ m ϕ m (1 z m D 1, (B(x,ρ)) S S ), L () where (.1) and the convergence of z m to V 0 have been used to get the last inequality. Recalling (.7) we have z m L (B(x,ρ)) L z m L (B(0,1)) 1 S N,sothat (.13) yields ϕ m L () 0. From (.1) we obtain z m V 0 D 1, (B(x,ρ)) 0. In particular, B(0,1) V 0 = 1 L S N > 0andV 0 0. By z m 0weakly,wealso have R m as m. Now using the result of case (I), we have z m (x) =R N m ṽ m ( x + x m )=(R m r m ) N x v m ( + x m ). R m R m r m r m )

1864 DAOMIN CAO AND SHUANGJIE PENG Define k m = R m r m,y m = xm r m ;theny m y 0, k m y m = R m x m. By (.5) we have x m >a>0, so k m y m +. Also, by the fact that {r m } is bounded away from zero, k m + (m ). Denote m = {x R N x k m + y m }, ṽ m = k N m v m ( x k m + y m ). Now we distinguish two cases: (1) k m dist(y m, ) c<, uniformly. Then after an orthogonal transformation, m = R N + = {x =(x 1,...,x N ),x 1 > 0}, () k m dist(y m, ),inthiscase m = R N. In each case we have for large m and any given ϕ C0 ( ), ṽ mϕ m x+k my m = o(1). Hence there holds, as m, ( V 0 ϕ V 0 V 0 ϕ) ṽ m ϕ = ( ṽ m ϕ m x + k m y m ṽ ṽ m m ϕ)+o(1) (.14) = ( v m ϕ v mϕ x v m v m ϕ)+o(1) = o(1). If = R N +, (.14) implies that V 0 H 1 0 ( ) is a weakly solution of the equation u = u 1, u > 0, x R N + ; u =0,x RN +. Thus, V 0 0, which is impossible. So case () is true, that is, V 0 solves (P0 ). Let w m (x) =v m (x) k N m V 0 (k m (x y m ))ϕ( k m (x y m )) H0 1 (), where k m,ϕ are defined similarly to case (I). We can also get w m =ṽ m V 0 + o(1), where o(1) 0inD 1, (R N ). Using the fact that k m y m +, the invariance of scaling and Lemma.3 we obtain that w m (x) =v m (x)k N m V 0 (k m (xx m ))+o(1) is still a (P.S.) sequence for E,0 at level β F0 (V 0) and of course it converges weakly to zero. This concludes the proof of Lemma.4. Proof of Theorem.1. By E,ɛ (u m ) c, DE,ɛ (u m ) 0 strongly in H 1 (), we have u m u 0 weakly in H0 1() and u0 solves problem (P,ɛ ). Moreover, setting v m = u m u 0,wethenhavev m 0 strongly in L () and by Lemma.3 we get u m = u m u + u + o(1), u m = (u m u) + u + o(1), u m x = u m u x + u x + o(1).

A GLOBAL COMPACTNESS RESULT 1865 Hence, E,ɛ (u m )=E,ɛ (u 0 )+E,0 (v m )+o(1), DE,ɛ (u m )=DE,ɛ (u 0 )+DE,0 (v m )+o(1), DE,0 (v m )=o(1). By applying Lemma.4 to {v m } recursively, the iteration must stop after finite steps; moreover the last (P.S.) sequence must strongly converge to zero. Applying Theorem.1, we can prove the following: Corollary.5. Suppose 1 N S N <c< N S N and c 1 N S N 0,thenfunctionalE,0 satisfies the (P.S.) c condition. Proof. Suppose {u m } is a (P.S.) c sequence for E,0 (u). Notice that for any u D 1, (R N )\{0} satisfying (P )(0 < ), F (u) 1 N S N. Moreover, if u changes sign, then F (u) N S N. By Theorem.1, c + o(1) = E,0 (u m ) E,0 (u 0 )+ l F 0 k (v j 0 )+ F (v j 1 ). Suppose that u 0 0; then c = l F 0 (vj 0 )+ k F (vj 1 ). If there exist some l or k such that v0 l or vk 1 changes sign, then by Remark 1.1 (ii) we get c N S N, which is a contradiction, so we can assume v0 i 0,vj 1 0(1 i l, 1 j k). By the uniqueness of solutions of problem (P0 )and(p ) (see Remark 1.1 (i)), we have c = l N S N 0 + k N S N, which is impossible. So u 0 0, and we can infer E,0 (u 0 ) 1 N S N, which follows that l =0,k = 0. Hence u m u 0 in H0 1(). Remark.6. (i) Let σ = {E,ɛ (u) u H 1 0 () solves (P )}, σ 0 = {F 0 (u) u H1 0 (RN )solves(p 0 )}, σ = {F (u) u H1 0 (RN )solves(p )}, be the spectral of (P ), (P0 )and(p ), respectively. Then in particular, Theorem.1 implies that any sequence {u m } satisfying E,ɛ (u m ) β,de,ɛ (u m ) 0 in H 1 ()(m ) is strongly relatively compact in H0 1 (), provided β { β + k β0 i + i=1 l β i β σ \ β,β0 i σ 0 \ 0,β i σ \ 0}. i=1 (ii) Assume u ɛ H0 1() is the ground state solution of (P,ɛ), 0 << 1. Then u ɛ u ɛ x S δ 0 in the sense of measure as ɛ 0, where δ x denotes the Dirac mass at x. Acknowledgement The authors are grateful to Dr. Didier Smets for helpful discussion.

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