Some nonlinear elliptic equations in R N

Nonlinear Analysis 39 000) 837 860 www.elsevier.nl/locate/na Some nonlinear elliptic equations in Monica Musso, Donato Passaseo Dipartimento di Matematica, Universita di Pisa, Via Buonarroti,, 5617 Pisa, Italy Received 1 April 1998; accepted 15 May 1998 Keywords: Superlinear elliptic equations; Lack of compactness; Unbounded domains; Multiplicity of positive solutions 1. Introduction Let us consider the following problem: u + u + ax)u = u p u in ; P) u 0 in ; ux)=0; x where ax) is a nonnegative function in L N= ), p p N=N ) if N 3. Problems of this kind arise in several contexts: for example, in the study of the sting waves solutions of nonlinear Schrodinger equations, or equations of Klein Gordon type, or also in reaction diusion equations. For example, if we look for sting waves solutions of the Schrodinger equation i @ @t = m + V x) p ; 1.1) i.e. solutions of the form x; t) = exp iet ) vx); Corresponding author. Tel.: 0039 50 84439; fax: 0039 50 8444. E-mail address: passaseo@dm.unipi.it D. Passaseo) 036-546X/00/$ - see front matter? 000 Elsevier Science Ltd. All rights reserved. PII: S036-546X98)0051-X

838 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 then one can easily verify that satises Eq. 1.1) if only if the function vx) solves the elliptic equation in m v V x) E)v + v p v =0: 1.) In [16] Floer Weinstein considered the case N = 1 p = 3 with a potential V which is required to be a smooth function globally bounded in ; for a given nondegenerate critical point of V they proved that, if 0 E inf V, then for small enough there exists a solution of Eq. 1.), which concentrates around the critical point as 0. Their method based on a Lyapunov Schmidt reductio was extended in several directions in order to obtain similar results for N 1 p ]; N=N )[ or also to nd solutions with multiple peaks, which concentrate around any prescribed nite set of nondegenerate critical points of V see, for instance, [3 5]). In [6] Rabinowitz used a global variational method to nd solutions with minimal energy under the assumption inf x V x) inf V. Other related results have been recently stated in [1,, 5, 14]. Notice that a simple change of variables shows that Eq. 1.) is obviously equivalent to with u + u + ax)u= u p u = m inf V E ax)= Vx) inf V inf V E 0: In this paper we are concerned with problem P) in the case that ax) has the form ax)= k j j j x x j )) 1.3) j=1 where j R +, x j j L N= ), with j 0 j 0. Our aim is to give sucient conditions on the numbers 1 ;:::; k the points x 1 ;:::;x k in order to guarantee existence multiplicity of solutions for P). We shall prove that, if x i x j is large enough for all i j i; j =1;:::;k), then there exist at least k 1 solutions of P); if, in addition, 1 ;:::; k are small or large enough, then we have at least k 1 solutions see Theorems.1 4.1 also Remark 4.). Let us denote by H 1; ) the closure of C0 RN ) with respect to the norm u = [ Du + u ]dx ) 1= by N= the usual norm in L N= ).

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 839 The solutions of problem P) correspond to the positive functions which are critical points for the functional f a : H 1; ) R dened by f a u)= [ Du +1+ax))u ]dx; 1.4) constrained on the manifold } M = u H 1; ): u p dx =1 : 1.5) Since the embedding H 1; ), L p ) is not compact, the Palais Smale compactness condition for f a constrained on M is not satised. Therefore, the classical variational methods cannot be applied in the usual way to nd critical points. In particular, the inmum inf M f a is not achieved if a N= 0 see Proposition.). Similar diculties also occur in the study of elliptic problems in other unbounded domains exterior domains, for example) which have been investigated in several recent papers see [3, 4, 6, 8 1, 15, 18, 19]). Some methods elaborated in these papers apply in our problem too. In particular, using the concentration-compactness principle see [1]), it is possible to analyse the obstructions to the compactness: in fact, it can be shown that every Palais Smale sequence for f a constrained on M either converges strongly to its weak it or diers from it by one or more sequences, which, after suitable translations, converge to a solution of the it problem u + u = u p u in ; u H 1; 1.6) ): Taking also into account the uniqueness result of [0], this property allows us to nd an energy range where the Palais Smale condition is satised see Proposition.3). Notice that in this paper we only require ax) L N= ) exploit the parameters 1 ;:::; k in order to nd critical values for f a on M in the range where the Palais Smale condition holds. However let us mention that, arguing as in [4], it is possible to nd critical values, in the same energy range, under a suitable fast decay condition on ax) as x see []); under this condition, in [] it is proved the existence of k 1 solutions for problem P) with ax) of the form ax)= k j x x j ); j=1 without introducing the parameters j. Indeed there exist k 1 lower energy solutions u i i =1;:::;k 1), which up to translations) converge to a positive solution of Eq. 1.6) as x i+1 x i ; moreover, there exist k highest energy solutions u j j =1;:::;k) which, as j + a.e. in, tend to split as sum of two positive solutions of Eq. 1.6), sliding to innity in opposite directions, while they converge up to translations) to a positive solution of Eq. 1.6) as j 0inL N= ).

840 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 Finally, let us remark that unlike [1,, 14, 16, 3 5], etc., our methods are not local in nature: we use global variational methods, inspired by [13, 4], which relate the number of critical points of f a on M to the topological properties of its sublevels. The paper is organized as follows: the main multiplicity results are stated in Theorems.1 4.1; in Section we describe some preinary properties of the functional f a constrained on M; in Section 3 we obtain some asymptotic estimates which give informations on the topological properties of the sublevels of the functional f a ;in Section 4, we prove the main theorems compare them with analogous multiplicity results one could obtain by Morse theory see Remark 4.).. Statement of the main theorem preinary results We shall prove the following theorem. Theorem.1. Let p p N=N ) if N 3. Let 1 ;:::; k be given nonnegative functions belonging to L N= ) such that j N= 0 for all j =1;:::;k. Then, there exist 1 0, = 1 ) 0, 3 = 3 1 ; ) 0;:::; k = k 1 ;:::; k 1 ) 0 % 1 = % 1 1 ;:::; k ; x 1 ) 0, % = % 1 ;:::; k ; x 1 ; x ) 0;:::;% k 1 =% k 1 1 ; :::; k ; x 1 ;:::; x k 1 ) 0 such that if i i or i 1 i for any i =1;:::;k x j % j 1 for any j =;:::;k; then problem P), with ax) of form 1:3), has at least k 1 distinct solutions. The proof is reported in Section 4. Let us recall some known facts which will be useful in the sequel. Let us consider the functional f : M R see Eq. 1.5)) dened by fu)= [ Du + u ]dx the following minimization problem = inf fu): u M}:.1).) It has been shown see, for instance, [8, 8, 0, 17]) that is achieved by a positive function U which is unique modulo translation radially symmetric with respect to 0, decreasing when the radial coordinate increases such that x + Ux) x N 1)= e x = 1 0;.3) x + DUx) x N 1)= e x = 0:.4)

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 841 Moreover, it is well known see [1]) that any minimizing sequence for in H 1; ) has the form w n x)+ux y n );.5) where w n ) n 1 is a sequence of functions converging to 0 in H 1; ), y n ) n 1 is a sequence of points in U is the positive function, spherically symmetric with respect to 0, that realizes. Proposition.. Let a L N= ) be a nonnegative function such that a N= 0. Then inf M f a =.6) the inmum is not achieved see 1:4); 1:5); :)). Proof. Put m a = inf M f a. Clearly m a. Let us consider the sequence u n ) n 1 in H 1; ) dened by u n x)=ux y n ), where U M is the positive function, spherically symmetric with respect to 0, that realizes see Eq..)) y n ) n 1 is a sequence of points in such that y n =+. In order to obtain = m a, it suces to show that ax)unx)dx=0:.7) Indeed, for any % 0, we have ax)unx)dx= ax)unx)dx+ ax)unx)dx \B0;%) B0;%) ) =N ) N )=N a N= x)dx \B0;%) Ux) N=N ) dx + ax)u x y n )dx: B0;%) Now, since y n, by the Lebesgue convergence theorem we obtain ax)u x y n )dx=0 % 0: B0;%) Moreover, we have % + \B0;%) a N= x)dx=0 because a L N= ). Hence Eq..7) is proved.

84 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 Now, let us argue by contradiction assume that the inmum m a is achieved by a function u. Without any loss of generality, we can also assume u 0. Thus, we have [ Du + u ]dx [ Du +1+ax))u ]dx=; which implies that ux)=ux y), for a suitable y, with Ux y) 0 for all x. Hence, we deduce 0= ax)u x)dx= ax)u x y)dx 0; which is impossible. be a se- Proposition.3. Let a L N= ) be a nonnegative function. Let u n ) n 1 quence in M that satises f au n )=c ]; 1 =p [; f a M u n ) 0 in H 1; ) i.e. u n ) n 1 is a Palais Smale sequence for the functional f a constrained on M at level c). Then u n ) n 1 is relatively compact. For the proof it suces to argue as in [6]. Proposition.4. Let a L N= ) be a nonnegative function. If a function u is a critical point for f a constrained on M, such that f a u) 1 =p, then u has a constant sign. Proof. Let us suppose, by contradiction, that u = u + u u + x) = max ux); 0), u x)= min ux); 0)), with u + 0 u 0. Taking into account Eq..6), we have u ± p [ Du ± +1+ax)) u ± ]dx:.8) Moreover, [ Du ± +1+ax)) u ± ]dx=f a u) u ± p p.9) because u is a critical point for f a on M. Comparing Eqs..8).9), we get u ± p p =f a u) so f a u) 1 =p), contradicting our assumption.

3. Some asymptotic estimates M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 843 In this section we will provide some estimates in order to get information on the topological properties of the sublevels of f a constrained on M. Lemma 3.1. Let be a nonnegative function in L N= ) with N= 0. Then we have } a) sup x)u x y)dx: y =0; 0 R N } b) sup x)u x y)dx: y =0; 3.1) + R N } c) sup x)u x y)dx: 0; y = R =0; R + where U is the positive function, spherically symmetric with respect to 0, that realizes see Eq..)). Proof. In order to prove Eq. 3.1)a) we argue by contradiction: suppose there exist a sequence y n ) n 1 of points in a sequence n ) n 1 of positive numbers such that n = 0 n n x)u x y n )dx 0: 3.) For all n 1, Holder inequality implies that n n x)u x y n )dx n x)u x y n )dx = n + n n By n;1= \By n;1= n x)u x y n )dx ) =N By N= n x)dx n;1= By n; 1= Ux y n ) N=N ) dx ) =N + n \By n; 1= N= n x)dx \By n; 1= Ux y n ) N=N ) dx ) N )=N ) N )=N

844 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 ) =N ) N )=N B N= x)dx Ux) N=N ) dx ny n; ) =N N )=N + N= x)dx \B0; 1= Ux) dx) N=N ) : Since L N= ) n 0as, we have N= x)dx=0; B ny n; moreover, since U H 1; ), we have that Ux) N=N ) dx =0: \B0; 1= The previous computations imply that n n x)u x y n )dx=0; which is a contradiction with Eq. 3.). Let us now prove 3.1b). By contradiction, let us suppose that there exist a sequence y n ) n 1 of points in a sequence n ) n 1 of positive numbers such that n =+ Eq. 3.) holds. For all n 1, by using Holder inequality, we get n n x)u x y n )dx n x)u x y n )dx = n + n n B0; 1= \B0; 1= n x)u x y n )dx ) =N B0; 1= N= n x)dx B0; 1= Ux y n ) N=N ) dx ) =N +n \B0; 1= N= n x)dx \B0; 1= Ux y n ) N=N ) dx ) N )=N ) N )=N

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 845 ) =N B0; N= x)dx B0; 1= Ux y n ) N=N ) dx ) =N ) N )=N + \B0; N= x)dx Ux) N=N ) dx : Since L N= ), we have B0; N= x)dx= N= x)dx + \B0; N= x)dx=0: Moreover, since U H 1; ), Ux y n ) N=N ) dx = Hence, we get n B0; 1= n x)u x y n )dx=0; By n; 1= ) N )=N Ux) N=N ) dx =0: that is a contradiction with Eq. 3.). Finally, let us argue again by contradiction to prove 3.1c): suppose that there exist a sequence n ) n 1 of positive numbers a sequence y n ) n 1 of points in such that y n =+ Eq. 3.) holds. From 3.1 a) b) we get 0 inf n sup n + ; hence, it is not restrictive to assume that n = ]0; + [. By Holder inequality we can write, for any n % 0, n n x)u x y n )dx n x)u x y n )dx = n + n n B0;%) \B0;%) B0;%) n x)u x y n )dx n x)u x y n )dx ) =N ) N )=N + N= x)dx Ux) N=N ) dx : \B0;%

846 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 Since n = ]0; + [ y n =+, we have a n x)u x y n )dx=0 % 0: B0;%) It follows that =N sup n n x)u x y n )dx C x)dx) N= % 0; \B0;%) with C 0. Now let us remark that N= x)dx=0 % + \B0;%) because 0 L N= ). Hence we infer that [ n n x)u x y n )dx= 0], contradicting Eq. 3.). Let us dene :, : M, for any j =1;:::;k, j :M in the following way: x)= u)= x 1+ x ; 3.3) x 1+ x ux) p dx 3.4) j u)= x x j 1+ x x j ux) p dx 3.5) where x j is the point of which appears in Eq. 1.3). Moreover, for any j =1;:::;k, we dene f j : H 1; ) R f j : H 1; ) R to be, respectively, f j u)= [ Du + u + j j j x x j ))u ]dx 3.6) f j u)= [ Du + u + j j j x)u ]dx 3.7) where j, j x j are dened as in Eq. 1.3).

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 847 Proposition 3.. Assume that j N= 0 for any j =1;:::;k. Then inf f j u): u M; u)=0} 3.8) see Eqs. 3.7), 1.3), 3.4)). Proof. Let us notice that Proposition. implies that inf f j u): u M; u)=0} : 3.9) By contradiction, let us suppose that equality holds in Eq. 3.9). Hence there exists a sequence u n ) n 1 of functions belonging to M such that u n ) =0 n 1 f j u n )=: Then see [19] Eq..5)) there exist a sequence y n ) n 1 of points in a sequence w n ) n 1 in H 1; ), with w n 0inH 1; ), such that u n x)=ux y n )+w n x): First, let us remark that y n ) n 1 must be a bounded sequence in : otherwise, we should have up to a subsequence) y n +, which implies u y n 1+ y n =0, as a consequence, u n ) = 1. But this is impossible since u n )=0 n 1. Now let us prove that y n 0: in fact assume, by contradiction, that up to a subsequence) y n y 0; then, u n ) U y)), with U y)) 0 if y 0 as follows easily taking into account Eq. 3.4) the radial symmetry of U with respect to zero); but u n )=0 n 1, so we must have y n 0. It follows that j j j x)[ux y n )+w n x)] dx = j j j x)u x)dx 0; where the inequality holds because Ux) 0 for all x, j 0in j 0. Therefore, we have = f j u n ) = [ D[Ux y n )+w n x)] + Ux y n )+w n x) ]dx +j j j x) Ux y n )+w n x) dx =+j j j x)u x)dx ; which is impossible.

848 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 Remark 3.3. Taking into account the radial symmetry of U with respect to zero the denition of see Eq. 3.4)), it is easy to verify that U)=0 U y)) = y )y y 3.10) where : R + R is a continuous function satisfying %) 0 for all % 0. Hence, for any R 0, the map y U y)) is homotopically equivalent in \0} to the identity map of @B0;R). For any h =;:::;k, let f 1; :::; h : H 1; ) R be the functional dened by h f 1; :::; h u)= Du + u + j j j x x j ))u dx: 3.11) j=1 Now, for all z, let us dene z : z x)=x z x z)z; where z : R R is the following function: 1 if t z ; t z t)= z if t z ; 1 if t z ; by 3.1) 3.13) moreover, for any j =;:::;k, let j 1;j :M be the map dened by j 1;j u)= xj x j 1)=) x x ) j 1 +x j ux) p dx 3.14) R N see Eq. 3.3)). Finally, for j =;:::;k,ifx j x j 1, let us put } xj x j 1 S j 1;j = t 1): t [0; 1] 3.15) x j x j 1 T j 1;j = x : x x ) } j 1 +x j x j x j 1 )=0 : 3.16) The following proposition can be proved arguing as in [19]. Proposition 3.4. Assume j N= 0 j 0 for j =1;:::;k. Then we have see Eqs. 3.11) 3.16)): a) for any h =;:::;k j =;:::;h, inf f 1; :::; h u): u M; j 1;j u)= ± x } j x j 1 x j x j 1 ;x j;x j 1 ; x j x j 1 ; 3.17)

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 849 b) for any h =;:::;k j =;:::;h, if x j x j 1, inf f 1; :::; h u): u M; j 1;j u) S j 1;j } ; 3.18) c) for any j =;:::;k, sup R + z @B0; 1) j 1;j U x )) j 1+x j Rz z =0: 3.19) Remark 3.5. Taking into account the radial symmetry of U with respect to zero Eqs. 3.1) 3.16), it is easy to verify that, for j =;:::;k, j 1;j U x )) j 1+x j =0 j 1;j U z)) = z x ) j 1 + x j z x ) j 1 + x j z T j 1;j ; where %) 0 % 0. 4. Proof of the main result Proof of Theorem.1. The idea is to choose the parameters 1 ;:::; k, x 1 ;:::;x k in order to obtain suitable inequalities which describe the topological properties of the sublevels of f a constrained on M give rise to k 1 distinct critical values. More precisely, we choose consecutively 1 ;:::; k, x x 1 ; x 3 x ;:::; x k x k 1 in such a way that every choice does not modify the inequalities previously stated produces new inequalities. The proof consists of four steps. Step 1: Choice of the parameters 1 ;:::; k. Lemma 3:1 implies that there exists 1 0 such that, if 1 1 or 1 1= 1, then sup f 1 U y)): y B0;R)} 1 =p R 0: 4.1) Taking into account Proposition 3:, it follows that we can choose a positive number R 1 suciently large such that sup f 1 U y)): y @B0;R 1 )} inf f 1 u): u M; u)=0} sup f 1 U y)): y B0;R 1 )} 1 =p : 4.)

850 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 Now, using again Lemma 3:1, we can x = 1 ) 0 such that, if or 1=, then sup f U y)): y B0;R)} inf f 1 u): u M; u)=0} R 0: 4.3) From Proposition 3:, we infer that there exists R 0 suciently large such that sup f U y)): y @B0;R )} inf f u): u M; u)=0} sup f U y)): y B0;R )} inf f 1 u): u M; u)=0}: 4.4) Iterating this procedure, we can choose 3 ;:::; k ; in particular, we have that there exist positive numbers 1, = 1 ) 0;:::; k = k 1 ;:::; k 1 ) 0 R 1 ;:::;R k such that, if j j or j 1= j, then sup f j U y)): y @B0;R j )} inf f j u): u M; u)=0} sup f j U y)): y B0;R j )} 1 =p for any j =1;:::;k 4.5) sup f j U y)): y B0;R j )} inf f j 1 u): u M; u)=0} for any j =;:::;k: 4.6) We shall consider 1 ;:::; k xed as before. Step. Behaviour of the functional with respect to the parameters x 1 ;:::;x k. Let us rst observe that } sup i i i x x i ))U x y)dx: y T 1; = 0 for i =1;; x x 1 since x x 1 distx 1;T 1; ) = x x 1 distx ;T 1; )=+ : 4.7) Therefore, Eqs. 4.7) 3.8) imply that, if x x 1 is suciently large, then supf 1; U y)): y T 1; } inf f k u): u M; u)=0}: 4.8) Moreover, from Eqs. 4.7), 3.17), 3.18) Remark 3:5 it follows that, if x x 1 is large enough, then inf f 1; u): u M; 1; u) S 1; } supf 1; U y)): y T 1; } inf f 1; u): u M; 1; u)= ± x } x 1 : 4.9) x x 1

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 851 Arguing as in the proof of Lemma 3.1c) one can show that, if r 1 is suciently large, then )} x1 + x sup f 1; U y)): y @B ;r 1 inf f 1; u): u M; 1; u) S 1; }: 4.10) Finally, Eq. 3.19) Remark 3:5 imply that, if r 1 is large enough, the map y 1; U x )) 1+x r 1 y is homotopically equivalent in ± x } x 1 to the identity map on @B0; 1) T 1; x ) 1 + x : x x 1 4.11) Analogous properties hold when x j x j 1 is large enough for j =;:::;k. Step 3. Choice of the points x 1 ;:::;x k in. Let x 1 be a xed point in. Arguing as in the proof of Lemma 3.1c), one can show that for i =1;, supf 1; U y)): y @Bx i ;R i )} x = sup f i U y)): y @B0;R i )} 4.1) supf 1; U y)): y Bx i ;R i )} x = sup f i U y)): y B0;R i )}; 4.13) where R 1 R are the positive numbers xed in Step 1. Moreover, it is clear that for i =1;, inf f i u): u M; u)=0} = inf f i u): u M; i u)=0} inf f 1; u): u M; i u)=0}: 4.14) Consequently, from Eqs. 4.5), 4.6), 4.1) 4.14) we infer that, if x is suciently large, supf 1; U y)): y @Bx ;R )} inf f 1; u): u M; u)=0} supf 1; U y)): y Bx ;R )} inf f 1; u): u M; 1 u)=0} 4.15) supf 1; U y)): y @Bx 1 ;R 1 )} inf f 1; u): u M; 1 u)=0} supf 1; U y)): y Bx 1 ;R 1 )}: 4.16)

85 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 We can now conclude that there exists % 1 = % 1 1 ;:::; k ; x 1 ) such that, if x % 1, Eqs. 4.15), 4.16) all the properties stated in Step hold. Now x x as before let x 3. One can easily verify that supf 1; ; 3U y)): y @Bx 3 ;R 3 )} x 3 = sup f 3 U y)): y @B0;R 3 )}; 4.17) supf 1; ; 3U y)): y Bx 3 ;R 3 )} x 3 = sup f 3 U y)): y B0;R 3 )} 4.18), for i =1;, supf 1; ; 3U y)): y @Bx i ;R i )} x 3 = supf 1; U y)): y @Bx i ;R i )} 4.19) supf 1; ; 3U y)): y Bx i ;R i )} x 3 = supf 1; U y)): y Bx i ;R i )}; 4.0) where R 1, R R 3 are the positive numbers xed in Step 1. It is obvious that inf f 3 u): u M; u)=0} = inf f 3 u): u M; 3 u)=0}, for i =1;, inf f 1; ; 3 u): u M; 3 u)=0} 4.1) inf f 1; u): u M; i u)=0} inf f 1; ; 3 u): u M; i u)=0}: 4.) Taking into account Eqs. 4.5), 4.6), 4.17) 4.), we obtain supf 1; ; 3 U y)): y @Bx i ;R i )} inf f 1; ; 3 u): u M; i u)=0} supf 1; ; 3 U y)): y Bx i ;R i )} for i =1;;3 4.3) supf 1; ; 3 U y)): y Bx i ;R i )} inf f 1; ; 3 u): u M; i 1 u)=0} for i =;3: 4.4)

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 853 Let r 1 0 be xed in such a way that Eqs. 4.10) 4.11) hold; one can easily verify that )} sup x1 + x f 1; ; 3 U y)): y @B ;r 1 x 3 )} x1 + x = sup f 1; U y)): y @B ;r 1 4.5) )} sup x1 + x f 1; ; 3 U y)): y T 1; B ;r 1 x 3 x1 + x = sup f 1; U y)): y T 1; B ;r 1 )}: 4.6) Taking into account Eqs. 4.8) 4.10), it follows that, for x 3 suciently large, )} x1 + x sup f 1; ; 3 U y)): y T 1; B ;r 1 inf f k u): u M; u)=0} 4.7) )} x1 + x sup f 1; ; 3 U y)): y @B ;r 1 inf f 1; ; 3 u): u M; 1; u) S 1; } )} x1 + x sup f 1; ; 3 U y)): y T 1; B ;r 1 inf f 1; ; 3 u): u M; 1; u)= ± x } x 1 : 4.8) x x 1 Now, arguing as in Step, one can verify that there exists % = % 1 ;:::; k ; x 1 ; x ) 0 such that, if x 3 %, then Eqs. 4.3), 4.4), 4.7) 4.8) hold, supf 1; ; 3 U y)): y T ; 3 } inf f 1; ; 3 u): u M; 1; u) S 1; } 4.9), for r suciently large, )} x + x 3 sup f 1; ; 3 U y)): y @B ;r inf f 1; ; 3 u): u M; ; 3 u) S ; 3 } supf 1; ; 3 U y)): y T ; 3 } inf f 1; ; 3 u): u M; ; 3 u)= ± x } 3 x : 4.30) x 3 x

854 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 Furthermore, r can be chosen large enough such that the map y ; 3 U x )) +x 3 yr is homotopically equivalent in \±x 3 x )= x 3 x )} to the identity map on @B0; 1) T ; 3 x + x 3 )=). Iterating this procedure we obtain that there exist r i 0 % i = % i 1 ;:::; k ; x 1 ;:::; x i ) 0, for i =1;:::;k 1, such that, if x i+1 % i, then the functional f a = f 1; :::; k satises inf f a u): u M; i; i+1 u) S i; i+1 } )} xi + x i+1 sup f a U y)): y T i; i+1 B ;r i inf f a u): u M; i 1;i u) S i 1;i } )} xi 1 +x i sup f a U y)): y T i 1;i B ;r i 1 inf f a u): u M; k u)=0} for i =;:::;k 1 4.31), for i =1;:::;k 1, )} xi + x i+1 sup f a U y)): y @B ;r i inf f a u): u M; i; i+1 u) S i; i+1 } )} xi + x i+1 sup f a U y)): y T i; i+1 B ;r i inf f a u): u M; i; i+1 u)= ± where r i is large enough, such that the map y i; i+1 U x i +x i+1 x i+1 x i x i+1 x i )) r i y ± x } i+1 x i to the identity map on @B0; 1) x i+1 x i Furthermore, we have )} x1 + x sup f a U y)): y T 1; B ;r 1 inf f a u): u M; i+1 u)=0} } 1 =p) ; 4.3) is homotopically equivalent in T i; i+1 x i + x i+1 ) ) : 4.33)

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 855 supf a U y)): y Bx i+1 ;R i+1 )} inf f a u): u M; i u)=0} supf a U y)): y Bx i ;R i )} 1 =p) for i =1;:::;k 1; 4.34) supf a U y)): y @Bx j ;R j )} inf f a u): u M; j u)=0} supf a U y)): y Bx j ;R j )} 1 =p) for j =1;:::;k 4.35) see Remark 3.3) the map y j U x j R j y)) is homotopically equivalent in \0} to the identity map on @B0; 1): 4.36) Step 4. Our aim is now to prove the existence of k 1 critical points for f a constrained on M. Let us dene, for i =1;:::;k 1; b i = inf f a u): u M; i; i+1 u) S i; i+1 }; )} xi + x i+1 d i = sup f a U y)): y T i; i+1 B ;r i, for j =1;:::;k, e j = inf f a u): u M; j u)=0}; g j = supf a U y)): y Bx j ;R j )}: Using the inequalities stated in the previous steps, we can now show the existence of a critical value in [b i ;d i ], for any i =1;:::;k 1, the existence of a critical value in [e j ;g j ], for any j =1;:::;k. Let us observe that, since b k 1 d k 1 b k d k b 1 d 1 e k g k e k 1 g k 1 e 1 g 1 1 =p) ; the critical values we shall nd are pairwise distinct; so they correspond to k 1 distinct critical points for f a constrained on M; moreover, these critical points are nonnegative functions, because of Proposition :4. Let us now x i 1;:::;k 1} let us show the existence of a critical value in [b i ;d i ]. Arguing by contradiction, assume that [b i ;d i ] does not contain any critical value. Since b i d i 1 =p) the Palais Smale condition holds in ]; 1 =p) [, it follows that there exists 0 such that see, for instance, [7]) the sublevel fa bi is a deformation retract of the sublevel fa di as usual, we set fa c = u M: f a u) c}

856 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 c R). This means, in particular, that there exists a continuous map i :[0;1] fa di fa di such that Let i 0;u)=u i 1;u) f bi a u f di a ; u f di a : i :[0;1] @B0; 1) T i; i+1 x )) i + x i+1 ± x } i+1 x i x i+1 x i be a homotopy such that see 3:19), Remark 3:5 Eq. 4.33)) i0;y)=y i1;y)= i; i+1 U x )) i +x i+1 r i y for all y @B0; 1) T i; i+1 x i + x i+1 )=). Now dene i :[0;1] @B0; 1) T i; i+1 x ) ) i + x i+1 B0; 1) by it; y) if t [0; 1 ]; it; y)= i; i+1 i t 1;U x ))) i +x i+1 r i y if t [ 1 ; 1]: The function i is well dened because of Eq. 4.3)), it is continuous, for all y @B0; 1) T i; i+1 x i + x i+1 )=) B0; 1), it satises i0;y)=y; it; y) ± x i+1 x i t [0; 1] x i+1 x i as we infer from Eq. 4.3) the properties of i i ), moreover, since Thus i1;y) = S i; i+1 i 1;U x )) i +x i+1 r i y fa bi : i is a continuous deformation in ± x } i+1 x i from x i+1 x i @B0; 1) T i; i+1 x ) ) i + x i+1 B0; 1) to a set which does not intersect S i; i+1, which is impossible. Therefore [b i ;d i ] must contain a critical value c i. On the whole we get k 1 distinct critical points for f a constrained on M, say v 1 ;:::;v k 1, such that b i f a v i ) d i e k 1 =p) for i =1;:::;k 1.

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 857 Let us now x j 1;:::;k} prove that there exists a critical value c j [e j ;g j ]. Assume, by contradiction, that [e j ;g j ] does not contain any critical value for f a constrained on M. Since the Palais Smale condition holds in ]; 1 =p) [, taking into account Eq. 4.35), it follows that there exists 0 such that the sublevel fa ej is a deformation retract of fa gj supf a U y)): y @Bx j ;R j )} e j : 4.37) Thus, in particular, there exists a continuous function j : f gj a f ej a such that j u)=u Moreover, we have u f ej a : 4.38) U y): y @Bx j ;R j )} f ej a 4.39) because of Eq. 4.37)) U y): y Bx j ;R j )} f gj a ; 4.40) as follows from the denition of g j. Hence, we can consider the map j :[0;1] @B0; 1) @B0; 1), by dened, for all z 1 t)z +tj U x j R j z)) for t [0; 1 j t; z)= ]; j j U x j 1 t)r j z)) for t [ 1 ; 1]: The function j is well dened because of Eq. 4.40)), it is continuous because of Eqs. 4.38) 4.39)) it satises j 0;z)=z j 1;z)= j j U x j )) z @B0; 1): 4.41) Moreover, taking into account Remark 3:3, Eq. 4.35), the denition of e j properties of j, we infer that the j t; z) 0 t [0; 1]; z @B0; 1): 4.4) It is clear that Eqs. 4.41) 4.4) give a contradiction, since @B0; 1) is not contractible in \0}. Hence [e j ;g j ] must contain a critical value c j. Since j 1;:::;k}, we have k distinct critical points for f a constrained on M, say v 1 ;:::;v k, such that d k 1 e j f a v j ) g j 1 =p). Summarizing, if we set u i x)=[f a v i )] 1=p ) v i x) for i =1;:::;k 1 u i x)=[f a v j )] 1=p ) v j x) for j =1;:::;k; we have on the whole k 1 distinct solutions of problem P).

858 M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 The proof of Theorem :1 suggests that a weaker multiplicity result can be stated even when no assumption is required on the positive numbers 1 ;:::; k, which appears in Eq. 1.3). In fact, missing Step 1 arguing as in the other steps of the proof of Theorem :1, one can prove the following: Theorem 4.1. Let p p N=N ) if N 3. Let 1 ;::: k be given nonnegative functions belonging to L N= ) such that j N= 0 for all j =1;:::;k. Then there exist % 1 = % 1 x 1 ) 0;:::;% k 1 =% k 1 x 1 ;:::; x k 1 ) 0 such that, if x j % j 1 for j =;:::;k, problem P) with ax) of the form ax)= k j x x j ) 4.43) j=1 has at least k 1 distinct solutions. Remark 4.. Let us notice that the multiplicity results stated in this paper show some possible way to choose the positive numbers 1 ;:::; k the points x 1 ;:::;x k in order to obtain distinct solutions of problem P) indeed distinct critical values of the corresponding functional). On the other h, using Morse theory see, for example, [7]), it is possible to obtain the same number of solutions, choosing 1 ;:::; k large or small enough x i x j suciently large for i j i; j =1;:::;k), without any other relation between them. But let us point out that the solutions one could obtain by means of Morse theory are not really distinct: they are counted with their own multiplicity dened in a suitable way). For example, results like the following ones could be proved by means of Morse theory. Let p p N=N ) if N 3. Let 1 ;:::; k be given nonnegative functions belonging to L N= ), such that j N= 0 for all j =1;:::;k. Then: a) there exists % 0 such that, if x i x j % for i j i; j =1;:::;k); 4.44) then problem P), with ax) of the form Eq. 4.43), has at least k 1 solutions, which are counted with their multiplicity; b) there exist 0 % = % 1 ;:::; k ) such that, if i or i 1 for all i =1;:::;k; Eq. 4.44) holds, then problem P) with ax) of the form Eq. 1.3) has at least k 1 solutions, which are counted with their own multiplicity.

M. Musso, D. Passaseo / Nonlinear Analysis 39 000) 837 860 859 Finally, let us remark that the proof of Theorem :1 gives some information about the behaviour of the solutions u 1 ;:::;u k 1, u 1 ;:::;u k. In fact, one can infer that ) ui f a as x i+1 x i for i =1;:::;k 1 u i p ) uj f a as j 0or j for j =1;:::;k: u j p Therefore, taking into account [1], we obtain: a) if x i+1 x i, there exist z i w i 0inH 1; ) such that u i x)= 1=p ) [Ux z i )+w i x)]; b) if j 0or j, there exist z j w j 0inH 1; ) such that u j x)= 1=p ) [Ux z i )+w i x)]: References [1] A. Ambrosetti, M. Badiale, Homoclinics: Poincare Melnikov type results via a variational approach, C.R.A.S. Paris, T. 33, S.1 1996) 753 758 Annales I.H.P., Anal. Nonlineaire, to appear. [] A. Ambrosetti, M. Badiale, S. Cingolani, Semiclassical states of nonlinear Schrodinger equations, Arch. Rational Mech. Analysis, to appear. [3] A. Bahri, Y.Y. Li, On a min-max procedure for the existence of a positive solution for certain scalar eld equations in, Rev. Mat. Iberoamericana 6 1/ 1990) 1 15. [4] A. Bahri, P.L. Lions, On the existence of a positive solution of semilinear elliptic equations in unbounded domains, to appear. [5] T. Bartsch, Z.Q. Wang, Existence multiplicity results for some superlinear elliptic problems on, Commun. Partial Dierential Equations 0 1995) 175 1741. [6] V. Benci, G. Cerami, Positive solutions of some nonlinear elliptic problems in exterior domains, Arch. Rational Mech. Anal. 99 1987) 83 300. [7] V. Benci, G. Cerami, Multiple positive solutions of some elliptic problems via the Morse theory the domain topology, Calculus of Variations 1994) 9 48. [8] H. Berestycki, P.L. Lions, Nonlinear scalar eld equations I: existence of ground state, Arch. Rational Mech. Anal. 8 1983) 313 346. [9] D.M. Cao, Existence of positive solutions of semilinear elliptic equations in, Dierential Integral Equations 6 1993) 655 661. [10] G. Cerami, D. Passaseo, Existence multiplicity results of positive solution for nonlinear elliptic problems in exterior domains with rich topology, Nonlinear Analysis T.M.A. 18 199) 109 119. [11] G. Cerami, D. Passaseo, Existence multiplicity results for semilinear elliptic Dirichlet problems in exterior domains, Nonlinear Anal. T.M.A. 4 11) 1995) 1533 1547. [1] C.V. Coman, M.M. Marcus, Existence theorems for superlinear elliptic Dirichlet problems in exterior domains, to appear. [13] J.M. Coron, Topologie et cas ite des injections de Sobolev, C.R. Acad. Sci. Paris, Ser. I Math. 99 1984) 09 1. [14] M. del Pino, P. Felmer, Local mountain passes for semilinear elliptic problems in unbounded domains, Calculus Variations 4 1996) 11 137. [15] M.J. Esteban, P.L. Lions, Existence nonexistence for semilinear elliptic problems in unbounded domains, Proc. Royal Soc. Edinburgh 93A 198) 1 14.

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