A Note on Superlinear Ambrosetti-Prodi Type Problem in a Ball
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1 A Noe on Superlinear Ambrosei-Prodi Type Problem in a Ball by P. N. Srikanh 1, Sanjiban Sanra 2 Absrac Using a careful analysis of he Morse Indices of he soluions obained by using he Mounain Pass Theorem applied o he associaed Euler Lagrange funcional acing boh on he full space H(Ω) 1 and on is subspace H,r(Ω) 1 of radially symmeric funcions we prove he exisence of non-radially symmeric soluions of a problem of Ambrosei Prodi ype in a Ball. Keywords: Mounain Pass Theorem, Non-radial soluions, Morse Index, Concenraion. AMS Subjec Classificaions (2): 35J65, 35J2 1 Inroducion We consider he following semi-linear ellipic problem (P ) { u = (u + ) p ϕ 1 in Ω u = on Ω where Ω = {x IR n : x < 1}. ϕ 1 is he eigenfuncion corresponding o he firs eigenvalue λ 1 of ( ) wih zero Dirichle daa on he boundary and normalized by aking ϕ 1 () = 1. is a real parameer. Due o requiremens of he variaional mehod used here, we have o resric he values of he dimension n o 1 < n < 6. In [2],i has been proved ha (S ) wih g(u) = u 2 admis a non-radially symmeric soluion for large > even if Ω is a ball.i is also easy o see ha he resul in [2] follows if g(u) = u p 1 TIFR Cenre, P.B. 1234, IISc Campus, Bangalore , India, srikanh@mah.ifrbng.res.in 2 Deparmen of Mahemaics, Indian Insiue of Science, Bangalore , India, sanjiban@mah.iisc.erne.in 1
2 where p (1, n+2 ) if n 3 and p (1, ) if n = 2.Moivaed by he resuls in [2] i has n 2 been proved in [1] ha (S ) has a non-symmeric soluion in Ω where Ω is a smooh bounded domain in R n where p (1, n+2 ) if n 3 and p (1, ) if n = 2. n 2 (S ) { u = g(u) ϕ1 in Ω u = on Ω where g(u) = u p.in [1] he mehod works for nonlineariy of he form g(u) = a u + p +b u p for a >,b >. I is no clear in wheher he mehod [1] works for a nonlineariy of he form g(u) = (u + ) p because of he esimae one needs o conrol he negaive soluion.in his noe we conclude ha he resul in [2] holds for g(u) = (u + ) p. So our main resuls are he following Theorem 1. For > sufficienly large,all he non-negaive radial soluions of (P ) have Morse Index larger han one. As a consequence of his heorem, we have Corollary. For > sufficienly large a soluion of (P ) in H 1,r(Ω) wih Morse Index 1 has o change sign. Theorem 2. For > sufficienly large he radial soluion of (P ) has Morse Index one in he space H 1,r(Ω) has Morse Index a leas wo on he whole space H 1 (Ω). Theorem 3. For > sufficienly large (P ) has soluions which are no radially symmeric. 2 Proof of he Theorems. Since we are in he case n 5, regulariy resuls imply ha he H(Ω)-soluions 1 obained here are indeed classical soluions, and his fac will be used hroughou he paper.also noe ha if u is a posiive soluion of (P ) hen u + = u. Remark 1.To sar wih, using his fac we obain he following properies of he radial 2
3 soluions of (P ). We recall ha a radial soluion u of (P ) saisfies he following equaion: u rr (n 1)u r r and boundary condiions u r () = u(1) = = ( u + ) p ϕ 1, 1) Each non-negaive radial soluion of (P ) has a mos a finie number of poins of maxima and minima. 2) A a poin of maximum a of a non-negaive radial soluion u we have u p (a ) ϕ 1 (a ). Also, here is an open inerval (a, d ) where u p (r) ϕ 1 (r) >. 3) A a poin of minimum m of a non-negaive radial soluion u we have u p (m ) ϕ 1 (m ). Also, here is an open inerval (e, m ) where u p (r) ϕ 1 (r) <. 4)The proofs of Theorem 1 and Corollary goes in he same way as in [2] so we will no prove his in our noe. 5)If u is a negaive soluion o (P ) hen we have { u = ϕ1 in Ω u = on Ω Hence he soluion o his problem is u = ϕ 1 λ 1 which is unique. 6) When he nonlineariy is u 2, hen one can easily show ha u + L C which is a consequence of he negaive par saisfying u L C. However in he conex of he problem we are discussing he negaive par behaves like u L which calls for a subsle argumen o obain he main resul of he Paper.Infac his seem o be he main reason why he resuls of [1] do no seem o work for he nonlineariies like (u + ) p. 3
4 The sraegy used o prove Theorem 1 consiss in showing ha any posiive soluion u of (P ) has Morse index a leas wo, for large. This can be done by jus considering hree ypes of posiive soluions: Type I: u has a unique maximum a. Type II: u has a leas wo maxima. Type III: u has a unique maximum a a poin a (, 1). Lemma 1: Suppose u is a posiive radial soluion of (P ) having unique maximum a he origin, hen for large such a soluion will have a high Morse index. Proof. Similar as in [2]. Noe, by definiion, if u is a soluion of (P ), is Morse Index is given by he dimension of he space where he form below is negaive definie, ha is w p u p 1 w, w <, w H 1 (Ω). Since he linearized problem is given by ( ) = (p u p 1 )( ) i follows ha, if he weigh ges very large on a inerval of fixed lengh, hen he Morse Index also ges large. Remark 2. We have made no assumpion ( u p ϕ 1 )(). All we wan in Lemma is ha is he unique maximum. In he case ( u p ϕ 1 )() =, u decreases near can be seen by differeniaing equaion u rr r (n 1) u r = ( u p ϕ 1 )r o obain u rrrr () < and u rrr () =. Noe u rr () = follows from ( u p ϕ 1 )() =. This implies u p ϕ 1 > in an inerval (, d ). Lemma 2. If u is a Radial soluion of (P ) for some > wih more han one maximum, hen is Morse Index is a leas wo on he Radial space. Proof. There are wo possibiliies. A) The soluion has precisely wo maxima: and a > ; B) The soluion has wo maxima < a 1 < a 2 < 1, and evenually more. We assume ha hose are he wo larges ones. 4
5 The proof in eiher case follows in he same way as in [2]. The basic fac is o use he funcion F (r) := u p ϕ 1, and noing ha F (r) = (p u p 1 )F (r) p(p 1) u 2 r u p 2 λ 1 ϕ 1 and observing also ha p(p 1) u p 1 u 2 r + λ 1 ϕ 1 >, r [, 1). Lemma 3. Suppose ha for all large here exiss a radial posiive soluion u in (, 1) of (P ), wih Morse Index one on he Radial space H 1,r(Ω) wih ( u p ϕ 1 )(), hen a 1 as, where a is he unique maximum in (, 1). Proof of Lemma 3. From he fac ha here is a unique maximum, we have u r > in (, a ) and u r in (a, 1). We denoe by b he unique poin in (, a ) where ( u p ϕ 1 ) =. i.e ( u p ϕ 1 )(b ) =. The uniqueness of b follows from he fac u is increasing in ([, a ) where as ϕ 1 is decreasing in (, 1). The proof of his lemma is essenially conained in Lemma 1, and similar o [2]. Remark 3. Le b be he poin in (, a ) as specified in he proof of Lemma 3. If b =, large, we see easily ha he Morse index canno be one. I is also clear from Lemma 3, since a 1, one should have ha b 1 as well in order o mainain Morse Index one in he conex of he posiive soluions we are discussing. We will now prove ha b 1 will lead o a conradicion, hus compleing he proof of Theorem 1. Lemma 4. Suppose ha for all large here exiss a radial posiive soluion u in (, 1) of (P ), wih Morse Index one on he Radial space wih ( u p ϕ 1 )() <. Then b canno converge o 1 as, where b is he unique poin in (, a ) such ha ( u p ϕ 1 )(b ) =, a being he unique maximum in (, 1). Proof of Lemma 4.See [2]. Proof of Theorem 1. Theorem 1 now follows direcly from he above lemmas. Proof of Theorem 2.By Theorem 1, we know ha he soluion obained by Mounain Pass in H 1,r(Ω) has o change sign. Noe ha here canno be more han one posiive par if Morse Index is one.noe ha each posiive par conribues o he Morse Index by one in 5
6 he radial space follows from, < u (pu p 1 bi )u, u > H 1,r (a i,b i )= (u p + ϕ 1 )u < a i in each (a i, b i ), where (a i, b i ) are disjoin inervals of posiiviy of u. u (ai,b i ) H 1,r(a i, b i ). In his conex we are in, due o he above facs we need o look a wo ypes of soluions. Type IV: he radial soluion u has a (unique) posiive maximum a and a unique negaive minimum a a (, 1). Type V: he radial soluion u has a (unique) negaive minimum a and a unique posiive maximum a a (, 1). All oher possibiliies are covered by reasonings similar o he ones used o handle Types I hrough V. We proceed now wih he proof of Theorem 2 by showing Lemmas 5, Proposiion 1 and Lemma 6. Lemma 5. Suppose u is a radial soluion of (P ) wih u() > and wih a negaive par hen u has Morse Index a leas wo on he space H 1 (Ω). Proof of Lemma 5. Noe ha, under he assumpions of his lemma, a ypical soluion is of Type 4. Dropping he prefix for he soluion we have ha i saisfies he equaion i.e. u = (u + ) p ϕ 1 in Ω u = on Ω (1) In course of our proof, we will discuss he case Ω IR 2 (See Remark 4 for he general case), jus o make he idea more explici. Denoing he coordinaes by (x, y) = (r cos θ, r sin θ), and differeniaing (1) wih respec o x and wriing w = u r cos θ, we see w saisfies w = (pu p 1 )w ϕ 1,r cos θ in B(, d ) w = ϕ 1,r cos θ in B(, a ) B(, d ) w = on B(, a ) Noe ha u r in (, a ), where a is he firs minima of u and u(d ) =. Since ϕ 1r, we have w (pu p 1 )w, w < H 1 (Ω) where w is defined by w in B(, a ) and zero elsewhere. Noe ha u + conribues o he Morse Index and w also conribues o he Morse Index and w and u + are orhogonal. In 6
7 fac he Morse Index of u is a leas wo. In fac he Morse Index of u is a leas hree since w 1 = u r sin θ will play he same role as w. Hence he lemma. Remark 4. I is clear u xi (i = 1,..., n) will work exacly like u r cos θ if we are in higher dimensions, leading o Morse index being (n + 1). Remark 5. In view of Theorem 1 and Lemma 5 i is clear ha Theorem 2 will follow if we can show ha Radial soluions wih u() < and having a unique zero in (, 1) has Morse Index a leas wo on he whole space for large. A ypical soluion we need o consider is of Type V. Noe ha if u() < and has wo nodal zeros, hen argumens as he ones used in Lemma 5 would lead o a high Morse Index on he whole space. Le d denoes he unique zero of u in (, 1), a he poin of maximum and b is he unique poin in (d, a ) such ha u p (b ) ϕ 1 (b ) =. Remark 6. If u denoes a soluion as discussed in Remark 5, we have ha u L if u r in [, a ], where u = max( u, ). I is easy o show ha u L.We now proceed o show ha u + L C 2 p+1 for C independen of, for all soluions under consideraion which is a major cause of difficuly. Proposiion 1. Suppose u is a radial soluion of (P ) for > wih u() < and wih u changing sign a a unique poin say d in (, 1) wih u > in (d, 1), having a unique maximum a and u r in (, a ). Then here exiss C > independen of such ha u + L C 2 p+1. (2) Remark 7. In he conex of Theorem 2, which is our main goal, i is clear, from earlier discussions, ha now we need jus o consider soluions saisfying hypohesis of Proposiion 1. Tha is, soluions saring wih u() < and having a unique posiive maximum a some poin a wih u r in [, a ]. Noe ha argumens similar o Lemma 5 would imply higher Morse Index if u r < in some subinerval of [, a ]. Also as observed earlier wo posiive maxima would make he soluions o have higher Morse Index on Radial space 7
8 iself. Also u() > has already been aken care hrough Lemma 5. Proof of Proposiion 1. Similar o [2]. Remark 8. In he conex of soluions discussed in Proposiion 1 i is clear ha argumens of Lemma 3 are applicable and ha a 1 as as long as we assume u has Morse Index one on he Radial space. Lemma 6. Le u be a soluion of (P ) wih Morse Index one on he Radial space and saisfying he hypohesis of Proposiion 1, hen d 1 as. Proof. Similar as [2]. Proof of Theorem 2 (compleed).in order o finish he proof of Theorem 2, we have jus o use he previous lemmas o show ha a radial soluion of Type V, has Morse index higher han 1 in he full space H 1 (Ω). This is done nex. We will discuss he proof Ω IR 2, and Ω IR n, n 3 separaely. Recall we are working in a siuaion where he soluion is like in Type V.Le us firs discuss he case when Ω IR n, when n 3 Hence Le w = u(r)(x 1 + x x n ) hen w = (x 1 + x x n ) u 2 n u xi i=1 w = (x 1 + x x n )( u p ϕ 1 ) 2 Then w saisfies on (d, 1) he equaion n u xi i=1 w = p( u p 1 ) w (p 1) u p (x 1 + x x n ) (x 1 + x x n )ϕ 1 2 n i=1 u xi in Ω d w = on Ω d where Ω d = {r : d < r < 1}. Thus w p( u p 1 )w, w = (p 1) (x 1 + x x n ) 2 u p+1 dx (x 1 + x x n ) 2 uϕ 1 dx Ω Ω n 2 (x 1 + x x n ) u u xi dx Ω i=1 8
9 i.e. w p( u p 1 ) w, w = (p 1) r 2 u p+1 r 2 uϕ 1 + n Ω Ω Ω u 2 Now ha u(r) > in Ω d.in order o prove ha he soluion has Morse Index higher han one on he whole space,i is enough o show ha he funcion F (r) = (p 1) u p r 2 ϕ 1 r 2 + n u(r) is negaive in (d, 1).We will show his o be he case for large. By Lemma 6,we know ha d 1.Hence aking > large enough we can assume r 2 3 in he region of ineres.observe ha 4 F (r) < if u p 1 (r) 4n.Hence 3(p 1) F (r) can occur only if < u p 1 (r) < 4n.Now noe ha 3(p 1) F (d ) <. and F (1) =. Firs we claim ha λ 1 u r (r) ϕ 1,r (r) r (d, a ) (3) Proof of he claim:we have ( u r r n 1 ) r = (( u + ) p ϕ 1 )r n 1 and inegraion from o s, where s a,we have Hence we have, s s s ( u r r n 1 ) r dr = ( u + ) p r n 1 dr ϕ 1 r n 1 dr. Thus we have, s n 1 u r (s) = s n 1 u r (s) = s d u p r n 1 dr s ϕ 1 r n 1 dr s n 1 u r (s) λ 1 s n 1 ϕ 1,r (s) s s ϕ 1 r n 1 dr d u p r n 1 dr λ 1 u r (r) ϕ 1,r (r) as u r (r) in (d, a ). 9
10 Noe ha in his case p < n+2 n 2 and λ 1 > nπ2 4. For r (d, a ) and < u p 1 (r) < 4n 3(p 1) we have F r (r) F r (r) = p(p 1) u r (r) u p 1 (r)r 2 2 u p (r)r ϕ 1,r r 2 2ϕ 1 r + n u r (r) = p(p 1) u r (r) u p 1 (r)r 2 + ϕ 1,r (r) r 2 + n u r (r) 2 u p (r)r 2ϕ 1 r (4) Now le us look o wha happens o F r (1). We have F r (1) = n u r (1) + ϕ 1r (1). Hence we have F r (1) = { n ur(1) + ϕ 1r (1) }. Dropping he prefix for he ime being, we have in (, a ). 1 2 (r n 1 u r (r)) r = ((u + ) p ϕ 1 )r n 1 Muliplying boh sides by r n 1 u r and inegraing beween o a we have a (r n 1 u r (r)) r (r n 1 a u r (r))dr = u p u r r 2n 2 a dr u r ϕ 1 r 2n 2 dr d a (r n 1 u r ) r (r n 1 u r )dr = 1 a (u p+1 ) p + 1 r r 2n 2 a dr u r ϕ 1 r 2n 2 dr d a ((r n 1 u r ) 2 ) r dr = 1 p + 1 up+1 (a )a 2n 2 2n 2 a u p+1 r 2n 3 a dr u r ϕ 1 r 2n 2 dr p + 1 d = 1 p + 1 up+1 (a )a 2n 2 + 2n 2 a u p+1 r 2n 3 a dr + u r ϕ 1 r 2n 2 dr (5) p + 1 d Again we have in (a, 1) (r n 1 u r (r)) r = (u p ϕ 1 )r n 1 Muliplying boh sides by r n 1 u r and inegraing beween a o 1 we have 1 (r n 1 u r ) r (r n 1 u r )dr = (u p+1 ) r r 2n 2 dr u r ϕ 1 r 2n 2 dr a p + 1 a a 1
11 1 2 1 ((r n 1 u r ) 2 ) r dr = 1 a p + 1 up+1 (a )a 2n 2 2n 2 p a u p+1 r 2n 3 dr 1 a u r ϕ 1 r 2n 2 dr u2 r(1) 2 = 1 p + 1 up+1 (a )a 2n 2 2n u p+1 r 2n 3 dr u r ϕ 1 r 2n 2 dr p + 1 a a u2 r(1) 2 = 1 p + 1 up+1 (a )a 2n 2 + 2n u p+1 r 2n 3 dr + u r ϕ 1 r 2n 2 dr (6) p + 1 a a Adding (5) and (6) we have u 2 r(1) 2 = 2n 2 p d u p+1 r 2n 3 dr + 1 a u r ϕ 1 r 2n 2 dr + a u r ϕ 1 r 2n 2 dr Hence we have, u 2 r(1) 2 2n 2 p d u p+1 r 2n 3 dr + a u r ϕ 1 r 2n 2 dr (7) Noe ha we have r n 1 ϕ 1r (r) is an increasing funcion and 1 rn 1 ϕ 1 dr = ϕ 1r(1) λ 1. Hence for large, (7) yeilds u 2 r(1) 2 2 ϕ 1r(1) λ 1 a r n 1 ϕ 1 dr λ 1 u r (1) 2 ϕ 1r(1) π 2 n 4 u r (1) 2 ϕ 1r(1) Thus we have u r (1) n < ϕ 1r(1) So we have F r (1) >.Hence F (r) < in a neighborhood of 1. Now we are required o prove ha F r (r) > for whenever < u p 1 (r) and r (a, 1). 11 4n 3(p 1)
12 Dropping he prefix for he ime being (for he sake of simpliciy). In (a, 1) we have, (r n 1 u r ) r = (u p ϕ 1 )r n 1 have Muliplying he above equaion by r n 1 u r and inegraing from a o r,(r > a ) we r (s n 1 u s ) s (s n 1 u s )ds = 1 r r (u p+1 ) s s 2n 2 ds u s ϕ 1 s 2n 2 ds a p + 1 a a 1 2 r ((s n 1 u s ) 2 ) s ds = a 1 p + 1 up+1 (r)r 2n 2 1 p + 1 up+1 (a )a 2n 2 2n 2 p + 1 r a u s ϕ 1 s 2n 2 ds r a u p+1 s 2n 3 ds 1 2 (rn 1 u r ) 2 (r) = 1 p + 1 up+1 (r)r 2n 2 1 p + 1 up+1 (a )a 2n 2 2n 2 p + 1 r a u s ϕ 1 s 2n 2 ds r a u p+1 s 2n 3 ds 1 2 r2n 2 u 2 r(r) = 1 p + 1 up+1 (r)r 2n p + 1 up+1 (a )a 2n 2 + 2n 2 r r u p+1 s 2n 3 ds + u s ϕ 1 s 2n 2 ds (8) p + 1 a a Adding (5) and (8) we have Hence we have i.e. for large we have 1 2 r2n 2 u 2 r(r) = 1 p + 1 up+1 (r)r 2n 2 + 2n 2 p r a u s ϕ 1 s 2n 2 ds + a r u s ϕ 1 s 2n 2 ds d u p+1 s 2n 3 ds 1 2 r2n 2 u 2 r(r) 2n 2 r u p+1 s 2n 3 a ds + u s ϕ 1 s 2n 2 ds p + 1 d 1 2 r2n 2 u 2 a r(r) u s ϕ 1 s 2n 2 ds 12
13 1 2 r2n 2 u 2 r(r) 2 r a n 1 ϕ 1 (a ) s n 1 ϕ 1 ds λ r2n 2 u 2 r(r) 2 r 2n 2 ϕ λ 2 1r (r) 2 1 λ 1 u r (r) 2 ϕ 1r(r) 4n 3 u r (r) < ϕ 1r(r) Now we have for r (a, 1) F r (r) = p(p 1) u r (r) u p 1 (r)r 2 p u p (r)r ϕ 1,r r 2 2ϕ 1 r + n u r (r) For large we have F r (r) u r (r) p(p 1) u p 1 (r)r 2 > Hence F r (r) >. Hence F (r) < r (a, 1) for large. Now we have o show ha F (r) < r (d, a ). Noe ha F r (d ) > and F r (1) > which implies F (r) is increasing in a neighborhood of d and 1. Hence if F has a sric zero in (d, 1), hen i mus have a leas wo zeros in (d, 1).We claim ha his canno happen. If possible, le here exis r (d, a ) such ha F (r ) =. Then we claim ha (1 r ) C. Since < u p 1 (r) < 4n 3(p 1) we have (p 1) u p (r )r 2 ϕ 1 (r )r 2 + u(r ) = (p 1) u p (r )r 2 + ϕ 1 (r )r 2 = u(r ) 13
14 ϕ 1 (r ) C Hence we have by mean value heorem ϕ 1 (1) ϕ 1 (r ) = ϕ 1,r (ξ )(1 r ) for some ξ (r, 1, ) i.e we have (1 r ) C. Now we claim ha ur(r) If r (d, a ) we have Inegraing beween from r o a we have can be made as small as we wish. (r n 1 u r ) r = ( u p ϕ 1 )r n 1 a (r n 1 a u r ) r dr = u p r n 1 a dr ϕ 1 r n 1 dr r r r r n 1 a u r (r ) = u p r n 1 a dr ϕ 1 r n 1 dr r r u r (r ) as. Hence we have F r (r ) > for large which is a conradicion. Hence F (r) <, r (d, 1). Now we discuss he case when Ω IR 2. Le w(x, y) be defined by Then w saisfies on (d, 1) he equaion w(x, y) = u(r) cos θ. { w = (p u p 1 ) w (p 1) u p cos θ ϕ 1 cos θ + u cos θ in Ω r 2 d w = on Ω d where Ω d = {r : d < r < 1} he annular domain. 14
15 Noe ha u(r) > in Ω d. In order o prove ha he soluion has a Morse Index higher han one on he whole space, i is enough o prove ha he funcion F (r) defined by F (r) = (p 1) u p (r)r 2 ϕ 1 (r)r 2 + u(r) is negaive in (d, 1). We will show his o be he case for large. Similar as above F (r) can only occur if < u(r) <, r (d, 1). 4n. We claim ha 3(p 1) F (r) < Noe ha from (4), F r (d ) > and F r (1) <.Hence F (r) is increasing in a neighborhood of d and 1,i now follows ha if F has a sric zero in (d, 1),hen i mus have a leas wo zeros in (d, 1). If possible, le here exis r (d, 1) such ha F (r ) =. Then we claim ha (1 r ) C. Since < u p 1 (r) < 4n 3(p 1) we have (p 1) u p (r )r 2 ϕ 1 (r )r 2 + u(r ) = (p 1) u p (r )r 2 + ϕ 1 (r )r 2 = u(r ) Hence we have by mean value heorem ϕ 1 (r ) C ϕ 1 (1) ϕ 1 (r ) = ϕ 1r (ξ )(1 r ) for some ξ (r, 1) i.e we have (1 r ) C. Now we claim ha ur(r) If r (d, 1) we have can be made as small as we wish. (r u r ) r = ( u p ϕ 1 )r Inegraing beween from r o a we have a a (r u r ) r dr = u p a rdr ϕ 1 rdr r r r 15
16 a r u r (r ) = u p a rdr ϕ 1 rdr r r u r (r ) as. Hence we have F r (r ) > for large, which is a conradicion. Similar is he case when r (a, 1). Hence Theorem 2 follows. Proof of Theorem 3. The funcional associaed o (P ) is I(u) = 1 2 Ω u 2 1 (u + ) p+1 + ϕ 1 u p + 1 Ω Ω on H 1 (Ω).From Remark 1, i follows ha he negaive soluion u is unique and is a sric local minimum for he funcional I.Also noe ha he funional is unbounded below and hence i saisfies all he condiions of Mounain Pass Theorem and if I(u ) = c and hence here exis a criical level c > c as a consequence of Mounain Pass heorem.also i is clear ha I saisfies all he hypohesis of Theorem 1.2 of [5](See page 222 and Theorem 5.1 of [3]) and hence a a level c here exis a soluion u of (P ) wih Morse Index less han or equal o one.if his soluion is non-radial hen we are done.on he oher hand if his soluion is radial,since I(u ) = c > c, u u and u canno be negaive. Hence i has a posiive par.this implies u has Morse Index one on he radial space and hence he soluion we are analysizing has o mainain Morse Index one on he whole space.however, Theorem 2 implies such a soluion has Morse Index a leas wo on he whole space.then i implies ha u has o be non-radial.hence he Theorem. Acknowledgemen. The second auhor acknowledges he parial suppor received from CSIR, India during he course of his work. References [1] Dancer E.N. and Yan Shusen, Mulipliciy and Profile of he changing sign soluions for an Ellipic Problem of Ambrosei-Prodi ype, (Pre-Prin). 16
17 [2] defigueiredo, D.G.,Srikanh P.N.,Sanra Sanjiban,Non-radially Symmeric Soluions for a Superlinear Ambrosei-Prodi Type Problem in a Ball, Communicaions in Conemporarary Mahemaics,(o appear). [3] Ekeland,I.,Ghoussoub,N.,Seleced new aspecs of he calculus of variaion in he large,bull.of AMS,39(22) [4] Fang,G.,Ghoussoub,N.,Morse-Type Informaion on Palais-Smale sequences obained by Min-Max Principles,Communicaions in Pure and Applied Mahemaics,47 (1994) [5] Ghoussoub,N.,Dualiy and Perurbaion Mehods in Criical Poin Theory, Cambridge Tracs in Mahemaics,17,Cambridge Universiy Press. 17
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