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1 The Pacic Intitute for the Mathematical Science htt:// Multile Solution for Quai-linear PDE Involving the Critical Sobolev and Hardy Exonent N. Ghououb and C. Yuan Deartment of Mathematic The Univerity of Britih Columbia Vancouver, B. C. V6T 2, Canada Prerint number: PIMS-98-4 Received on October 22, 998.

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3 Multile Solution for Quai-linear PDE Involving the Critical Sobolev and Hardy Exonent N. Ghououb & C. Yuan Deartment of Mathematic The Univerity of Britih Columbia Vancouver, B. C. V6T 2, Canada October 22, 998 Abtract We ue variational method to tudy the exitence and multilicity of olution for the following quai-linear artial dierential equation: (?4 u = juj r?2 u + jujq?2 u in = : where and are two oitive arameter and i a mooth bounded domain in R n. The variational aroach require that < < n, q () n? n? and r () = n n? which we aume throughout. However, the ituation dier widely with q and r, and the intereting cae occur either at the critical Sobolev exonent (r = ) or in the Hardy-critical etting ( = = q) or in the more general Hardy- Sobolev etting when q = n? n?. In thee cae ome comactne can be retored by etablihing Palai-Smale tye condition around aroriately choen dual et. Many of the reult are new even in the cae = 2, eecially thoe correonding to ingularitie (i.e., when < ). Introduction Conider the following quai-linear artial dierential equation: (?4 u = juj r?2 u + jujq?2 u in = : (P ; ) where and are two oitive arameter and i a mooth bounded domain in R n. We hall aume throughout that < n. The tarting oint of the variational aroach to thee roblem i the following Sobolev- Hardy Inequality which i eentially due to Caarelli-Kohn-Nirenberg [8]. Aume that < < n and that q () n?, then there i a contant C > uch that n? juj q C( ) q dx jruj dx for all u 2 H ; ()

4 We ue ;q () to denote the bet Sobolev-Hardy contant, i.e. the larget contant C atifying the above inequality for all u 2 H ; (), that i: ;q () = u2h ; inf ();u6= R jruj dx : juj q ) q dx In the imortant cae where q = (), we hall imly denote ; () a. Note that i nothing but the bet contant in the Sobolev inequality while i the bet contant in the Hardy inequality, i.e., () = ( R R jruj dx inf R u2h ; juj ();u6= dx : We hall alway aume that r () = n a way that the functional E ; (u) = jruj dx? r n? juj r dx? q for the non-ingular term in uch juj q dx i then well dened on the Sobolev ace H ; (). The (weak) olution of the roblem (P ; ) are then the critical oint of the functional E ;. Another relevant arameter will be the rt \eigenvalue" of the -Lalacian? dened a () ; () = inff jrwj dx; w 2 H ; (); jwj dx = g: Here are the main reult of thi aer. Theorem. (Hardy-Sobolev ubcritical ingular and non-ingular term) Suoe < q < () and r <. Aume one of the following condition hold: () (High order ingular term) < q, r, > and >. (2) (Low order ingular term) = q, < r, > and ; > >. Then (P ; ) ha innitely many olution. Moreover, (P ; ) ha an everywhere oitive olution with leat energy and another one that i ign-changing. Theorem.2 (Hardy-critical ingular term) Suoe < = q = () (i.e., = ).. (Subcritical non-ingular term) If r <, then (P ; ) ha innitely many olution {at leat one of them being oitive{ for any > and any < <. 2. (Critical non-ingular term) If r = and i tar-haed, then (P ; ) ha no nontrivial olution for any > ; >. Theorem.3 (Hardy-Sobolev critical ingular term) Suoe < q = () (i.e., < ). 2

5 . (High order non-ingular term) Aume < r < and >, >. If n > (?)r+2 +(?)(r?) (?)r+ +(?)(r?) (in articular if n 2 ), then (P ; ) ha a oitive olution. If n > (in articular if n > 3? 2 + ), then (P ; ) ha alo a ign-changing olution. 2. (Low order non-ingular term) Aume = r < and < <, >. If n 2, then (P ; ) ha a oitive olution. If n > 3? 2 +, then (P ; ) ha alo a ign-changing olution. 3. (Soboelev-critical non-ingular term) Aume r = and i tar-haed, then (P ; ) ha no non-trivial olution for any > and any >. Theorem.4 (Sobolev-critical non-ingular term and ubcritical ingular term) Suoe < q < () and r =.. (High order ingular term) Aume that < q and > ; >. If n > (?)(q?)+2 +(?)(q?) olution. (in articular if n 2? (? )), then (P ; ) ha a oitive If n > (?)(q?)+ (in articular if n > (? )(? ) + ), then (P +(?)(q?) ; ) ha alo a ign-changing olution. 2. (Low order non-ingular term) Aume = q and >, ; > >. If n 2? (? ), then (P ; ) ha a oitive olution. If n > ((? )(? ) + ), then (P ; ) ha alo a ign-changing olution. The following table ummarize our reult. 2 A Pohozaev-tye Identity In thi ection, we tart by identifying the contraint to the roblem of exitence of olution for (P ; ). Here i the main reult Theorem 2. If i a tar-haed domain in R n, then roblem (P ; ) ha no olution in the doubly critical cae: That i, for r = and q = () = n? n?, the roblem (P ;) ha no non-trivial olution. Aume i a tar-haed domain, then if we let v denote the outward normal then hx; vi > We aume we have the neceary regularity in the following oeration, otherwie, we can ue an aroximation argument a in Guedda-Veron [2]. 3

6 Table : Sobolev-ubcritical non-ingular term Singular term Parameter Non-ingular term Dimenion # of olution (HS-ubcritical) Innite < q < () > ; > r < n > One oitive = q < () ; > > ; > < r < n > One oitive (H-critical) = q = () > ; > > < r < n > Innite (One oitive) (HS-critical) < q = () > ; > < r < n > (?)r+2 +(?)(r?) One oitive 2 < r < n > (?)r+ +(?)(r?) Two > > ; > = r < n 2 One oitive 2 = r < n > 3? 2 + Two Table 2: Sobolev-Critical non-ingular term Singular term Parameter Non-ingular term Dimenion # of olution = q < () ; > > r = n > 2? (? ) One oitive 2 = q < () and > n > ((? )(? ) + ) Two < q < () > ; > r = n > (?)(q?)+2 +(?)(q?) One oitive 2 < q < () n > (?)(q?)+ +(?)(q?) Two q = () >, > r = n > None Multily the equation (P ; ) by hx; rui on both ide and integrate by art, we jruj hx; vidx + n? jruj dx = n? q On the other hand, multily the equation by u and integrate, we get Put the two identitie together, we jruj hx; vid = ( n? q jruj juj q = dx + juj r dx:? n? juj q ) juj q dx + n juj r dx: r dx + (n r? n? So if r = n = n? and q = n?, the roblem ha no non-trivial olution. n? 4 ) juj r :

7 3 The Extremal Function in the Hardy-Sobolev Inequalitie In thi ection, we ummarize the needed reult concerning the Hardy-Sobolev inequalitie. We rt recall the Hardy inequality. Lemma 3. ([3]) Aume that < < n and u 2 H ; (R n ), then () u 2 L (R n ); (2) (Hardy Inequality) R R n juj dx C n; R (3) The contant C n; i otimal. R n jruj dx; where C n; = ( n? ) ; The following extenion of the Hardy and Sobolev inequalitie i eentially due to Caarelli-Kohn-Nirenberg[8]. Lemma 3.2 (Sobolev-Hardy Inequality) Aume that < < n and that q () :=, then: n? n? () There exit a contant C > uch that for any u 2 H (), juj q ( ) dx C( jruj ) q dx: (2) The embedding u! u x =q from H () into L q () i comact rovided q < (). Proof: () For = or =, thi i jut the Sobolev (re., the Hardy) inequality. Since (), we have. We can therefore only conider the cae where < <. By the Hardy, Sobolev and Holder inequalitie, we have Rn juj () Rn juj = ()? juj Rn ( j juj ) ( juj ( R n Rn = ( j juj C ( = C ( ) ( juj R n jruj ) ( R n R n jruj ) n? n? : ()?)?? ) )? jruj ) R n? Remark 3. If i the whole ace, one can how that the condition q = () := n? n? are alo neceary for the above inequality to hold. Indeed, a tandard caling argument 5

8 how that q mut be equal to (). On the other hand, if we inert into the inequality the following function ( and 2 S n? being the olar coordinate), and ince We get du(x) d for u(x) = f?n log for " < ; "?n log " for ": = By L'Hoital' rule, we have 8 >< >: R n jruj " (? n )? n log?? n " < : "? ( + ( n? ) log ) d: lim "! R "? ( + ( n? ) log ) d log + " = n? + ; and alo R n Thu from the inequality we get that q. juj q? log q "?n q n? = " logq log+q " : log + q " log+ The following i an extenion of what i well known in the cae = 2 and =. Theorem 3. Suoe < < n, < and q = (), then the following hold:. () i indeendent of and will henceforth be denoted by. 2. i attained when = R n by the function " ; y a (x) = (a (n? )( n?? )? ) n???n (?) (a +? )? for ome a >. Moreover the function y a are the only oitive radial olution of?div(jruj?2 ru) = u ()? in R n. Hence, Rn jy a j q ( ) Rn q = krya k jy a j q = = n?? : 6

9 Proof: We rove (2). We how the bet contant are attained at function u (x) = c( +?? )? n?? ( < ) where > i a contant. For any f, let f be it Schwarz ymmetrization (or rearrangement) [2], then we have jrf j jrfj R n R n and Rn jf j q t Rn jfj q t ; aume the above integral are well dened (refer to Lieb [2] [22]). By thee inequalitie, we may retrict our dicuion to radial ymmetric function. Thu we may conider the following variational roblem: Maximize I(g) = jg(r)j q r n?? dr; when J(g) = where C i a given contant. The Euler-Lagrange equation i jg (r)j r n? dr = C: (r n? ju (r)j?2 u (r)) r + kr n?? juj q? = : () It can be eaily veried that the function u (x) = ( +?? )? n?? ( < ) are olution of (), where >. To continue, we need the following lemma of Bli ([], [2]). Lemma 3.3 Let h(x) be a meaurable, real-valued function dened on uch that the integral J = R h (x)dx i nite and given. Set g(x) = R x h(t)dt, then I = R g q (x)x?q dx attain it maximum value at the function h(x) = (x + )? +, with and q two contant atifying q > > ; = q? and >, a real number. By thi lemma and with the change of variable x = r?n? we can deduce that I() attain it maximum at the function Note that if then u (x) = ( +?? )? n?? g(x) = h(x) = (x + )? + ; x ( < ): h(t)dt = ( + x? )? : And if q = n?, then n? = q? =?. The theorem i thu roved. n? Remark 3.2 A exected, the comactne of the embedding u! above, doe not hold when q = (). Indeed, let f k (x) = ( k ) n???n (?) ( +? )? ; k 7 u x =q in Lemma 3.2.(2)

10 and et krf k (x)k = A and R jf k (x)j () dx = C. Let h k (x) = f k (x)? ( k ) n??n (?) ( + )? k for o that h k (x) 2 H ; (B) and kh k k! A H ; (). Hence fhk g i bounded in H ; () and k h k k L ()! C (). Now, hk (x)! for 6= and i the only oible cluter oint of f () h k () g in L () () which i imoible ince C 6=. 4 The Comactne Lemma Thi ection deal with the comactne roertie of the functional E ; (u) = jruj dx? r We recall the following tandard denition. juj r dx? q juj q dx: Denition 4. Say that a C -functional E on Banach ace X verie the Palai-Smale condition at the level c (in hort (PS) c ), if for any equence (u n ) n atifying lim n E(u n ) = c and lim n ke (u n )k =, ha a convergent ubequence. Dene the following function: L(; ) = 8 >< >:? ( n? ) (n?) n?? if < + if = and < ; if = and : Theorem 4. Aume < n, q () and r. () If q < () and r <, then for any > and any >, the functional E ; atie (P S) c for all c. (2) If q = () and r <, then for any > and any >, the functional E ; atie (P S) c for all c < L(; ). (3) If q < () and r =, then for any > and any >, the functional E ; atie (P S) c for all c < L(; ) = ( n ) n n?. Note that tatement 2 above yield that E ; alo atie (P S) c for all c when = q = q () (i.e., when = ) a long a <. Since olve a roblem in [3] and [25] where only a certain ingular Palai-Smale condition i etablihed. On the other hand, when =, = 2 and =, we recover the {by now well known{retricted comactne roertie that aear in Yamabe-tye roblem[4]. We rt recall a few known reult. 8

11 Lemma 4. ([25]) Let x; y 2 R n and h; i e be the tandard calar roduct in R n. Then ( C jx? yj ; if 2 h?2 x? jyj?2 y; x? yi e jx?yj C 2 ; if < < 2: (+jyj) 2? The following reult of Brezi-Lieb ([4]) will be ueful in the equel. Lemma 4.2 Suoe f n! f a.e. and kf n k C < for all n and for ome < <. Then lim n! fkf nk? kf n? fk g = kfk : Lemma 4.3 Let (u n ) n be a bounded equence in H ; () and let (q n ) n be a equence uch that < q n (), q n! () a n!. There exit then a ubequence till denoted by (u n ) n {without lo of generality{ uch that () u n! u weakly in H ; (). (2) u n! u in L r () if < r < = n n?. (3) u n! u almot everywhere. (4) un x! u x weakly in L (). (5) For any f 2 H ; (), R ju nj qn?2 u n f! R juj ()?2 u n f. (6) If 2, then R ju nj qn R jujqn + R ju n? uj qn + o(). (7) R ju n?uj () = R ju nj ()? R juj () + o(). Proof: Thee are tandard alication of the Hardy-Sobolev embedding theorem and the reult of Brezi-Lieb. We jut give the roof of (5) and (6). Without lo of generality, we may aume that q n < (). For (5), it i clear that ju n j qn?2 u n qn? qn! juj ()?2 u ()? () a:e: and ju n j qn? j j (? () ) () ()? = ( ju n j () qn? ()? qn? ()? ju n j () ) qn? ()? ( qn? (? ()? ) ) ()?qn ()? i uniformly bounded in n. Since for any f 2 H ; (), follow. f () 2 L () (), and the concluion In order to rove (6), we need the following lemma (roved in 9

12 Calculu Lemma For every q 3, there exit a contant C (deending on q) uch that for ; 2 R we have ( Cjjjj j j + j q? jj q? jj q q? if jj jj;? q(jj q?2 + jj q?2 ) j Cjj q? jj if jj jj: For q 3, there exit a contant C ( deending on q ) uch that for ; 2 R we have j j + j q? jj q? jj q? q(jj q?2 + jj q?2 ) j C(jj q? jj q?2 ): >From thi inequality, we can actually deduce the following more convenient reult for any q. j j + j q? jj q? jj q? q(jj q?2 + jj q?2 ) j 2C(jj q? + jj q? ): Now, back to the roof of (6). let w n = u n? u, then w n! weakly in H ; (). By the above calculu lemma, ju n j qn = jw n + uj qn jw nj qn + jujqn + C jujjw n j qn? jw n jjuj qn? + C 2 : In view of (5), we only need to how that jw n jjuj qn? lim n! = lim n! jw n jjuj qn?2 (? () ) juj () = : For that, we check ( jw njjuj qn?2 (? ) () () ) ()? = jw n j () ()? ()? ju n j () ( ) juj qn?2 ()? () qn?2 ()? ()? ( Hence it i uniformly bounded in n and the claim follow. ()?qn ()? juj () ) qn?2 ()? ( ) ()?qn ()? : Lemma 4.4 Let E n (u) = R jruj dx? R juj qn q n dx? R r jujr dx; ( > ; > ), where q n atify the condition in the reviou Lemma and < r <. Aume the equence fu n g atie E n (u n )! c, En(u n )!, then, there exit a ubequence, till denoted by fu n g, uch that for ome u 2 H ; () () u n! u weakly in u 2 H ; (). (2) ru n! ru a.e. (3) R jru n? ruj = R jru n j? R jruj + o(): (4) jru m j?2 ru m! jruj?2 ru weakly in [L? ()] n.

13 Proof: Since and we have he n(u n ); u n i = lim n E n (u n ) = c and lim n E n(u n ) = ; jru n j ju n j qn?? ju n j r ; o()( + ku n k) + jcj E n (u n )? he n(u n ); u n i = 8 < : (? (? q n ) R q n ) R juj qn dx + (? ) R r ju n j r dx r > juj qn dx r = : Since i bounded, we have ju n j = ju n j qn qn dx M( ju n j qn ) qn ; and kru n k = E n (u n ) + q n ju n j qn dx + ju n j r dx; r we conclude that fu n g i a bounded equence in H ; (). We therefore can aume that fu n g atie all of the concluion in Lemma 4.3. Now we ue a technique initiated by Boccardo and Murat and already ued by Garcia and Peral in a related context. Dene the function ( if jj k k () = k jj if jj k: We may aume alo that k (u n? u)! weakly in H ; () for any xed oitive k, ince k (u n? u)! a.e. and it i bounded. Then from the aumtion we get o() = he n(u n )? (E ) (u); k (u n? u)i + o() = Since junjqn?2 have and hjru n j?2 ru n? jruj?2 ru; r k (u n? u)i e? ( ju nj qn?2 u n? juj ()?2 u) k (u n? u): ()?2 u n! juj u in the weak tar toology of H?; () (By Lemma 4.3), we j ( ju nj qn?2 u n? juj ()?2 u) k (u n? u)j Ck; lim u hjru n j?2 ru n? jruj?2 ru; r k (u n? u)i e dx Ck: n! Let e n (x) = hjru n j?2 ru n? jruj?2 ru; r k (u n? u)i e, then e n (x) by Lemma 3., and i uniformly bounded in L (). Take < < and lit into S k n = fx 2 j ju n? uj kg; G k n = fx 2 j ju n? uj > kg:

14 then e n dx = e n dx + ( S k n S k n G k n e n dx e n dx) js k nj? + ( G k n e n dx) jg k nj? : Now, xed k, then jg k nj! a n! and from the uniform boundedne in L we get lim u e ndx (Ck) jj? n and let k!, we get that e n! trongly in L. By Lemma 4., we have that ru n! ru in L q for < q <. By aing to a ubequence, we have ru n! ru a:e: Thu () hold. A for (2), jut aly Lemma 4.2. The roof of thi lemma i thu comlete. Proof of Theorem 4..(): If q < () and r <, it i tandard to how that the comactne of the Hardy-Sobolev embedding and of the Sobolev embedding imly that for any > and any >, the functional E ; atie (P S) c for all c. Proof of Theorem 4..(2): Recall that L(; ) = 8 >< >:? ( n? ) (n?) n?? if < + if = and < ; if = and : and aume that q = () and r <. We need to how that E ; (u) = jruj dx? juj ()? juj r dx () r atie the Palai-Smale condition at any energy level le than L(; ). For that, aume fu n g i a equence in H ; () atifying: E ; (u n )! c < L(; ) and E ;(u n )! : By Lemma 4.4, we may aume that fu n g atie the concluion of both Lemma 4.2 and Lemma 4.3. For any v 2 C (), he ;(u n ); vi = which converge a n! to = (hjru n j?2 ru n ; rvi? ju n j r?2 u n v? ju nj ()?2 u n v)dx; (hjruj?2 ru; rvi? juj r?2 uv? juj ()?2 u v)dx = he ;(u); vi: 2

15 Hence u 2 H ; () i a weak olution of (P ; ). Chooe v = u, we have and thu = he ;(u); ui = E ; (u) = (? r ) By Lemma 4.3 and 4.4, we have and (jruj? juj r? juj () )dx juj r + (? () ) juj () dx : E ; (u n ) = E ; (u) + E ; (u n? u) + o() o() = he ;(u n ); u n? ui = he ;(u n )? E ;(u); u n? ui If = = () and <, then o() = = (jru n? ruj? ju n? uj () ) + o(): (jru n? ruj? ju n? uj ) + o() (? ) (jru n? ruj + o(); that i u n! u trongly. If < (i.e., < ()), we have for large n, E ; (u n? u) = E ; (u n )? E ; (u) + o() E ; (u n ) + o() c < L(; ): Thu for uch n (? () )kru n? ruk c <? n? (n? )? ( ) n?? : By the Sobolev-Hardy inequality, we nally get o() = = ( C (jru n? ruj? ju n? uj () )dx jru n? ruj?? () ( jru n? ruj )[?? () ( jru n? ruj dx: So again u n! u in H ; () trongly. 3 jru n? ruj ) () jru n? ruj ) ()? ]

16 Proof of Theorem 4..(3): Suoe now q < () and r =, then we have comactne in the ingular term but we will be dealing with a non-ingular term involving the critical Sobolev exonent. We have again E (u n? u) = E ; (u n )? E ; (u) + o() E ; (u n ) + o() c < L(; ) = n n ( ) n? Thu, for uch n, (? )kru n? ruk c < n n ( ) n? ; o that thi time we get from the Sobolev inequality that: o() = he ;(u n ); u n? ui = he ;(u n )? E ;(u); u n? ui So again u n! u in H ; () trongly.. = = C (jru n? ruj? (jru n? ruj? [? )( jru n? ruj dx: ju n? uj ) + o() jru n? ruj )? ] + o() 5 Min-max rincile and dual et aociated to E ; For Banach ace X and Y, we ue C(X; Y ) to denote the ace of all continuou ma from X to Y. Denition 5. Let X be a Banach ace and B be a cloed ubet of X. We ay that a cla F of comact ubet of X i a homotoy-table family with boundary B rovided that () every et in F contain B. (2) for any et A in F and any 2 C([; ] X; X) atifying (t; x) = x for all (t; x) in (fg X) [ ([; ] B) we have that (fg A) 2 F. Say that the cla F i 2 -homotoy table if all et in F are ymmetric and if we only require tability under odd homotoie (i.e., (t;?x) =?(t; x)). Say that a cloed et M i dual to the family F if it verie the following M \ B = ; and M \ A 6= ; for all A 2 F: We hall need the following weakened verion of the Palai-Smale condition. Denition 5.2 Say that a C -functional E on Banach ace X verie the Palai-Smale condition at level c and around the et M (in hort (PS) M;c ), if for any equence (u n ) n atifying lim n E(u n ) = c, lim n ke (u n )k = and lim n dit(u n ; M) = ha a convergent ubequence. 4

17 The following theorem of Ghououb [7] will be frequently ued in the equel. Theorem 5. Let E be a C -functional on X and conider a homotoy table family F of comact ubet of X with a cloed boundary B. Let M be a dual et F uch that: inf E(x) = c := c(e; F) = inf max E(x): x2m A2F x2a If E atie (PS) M;c, then M \ K c 6=, where K c i the et of all critical oint of E at level c. If F i only 2 -homotoy table, then the reult till hold true a long a the functional E i even and the dual et M i ymmetric. Note that the above theorem include the claical min-max rincile which hold under the aumtion that u x2b E(x) < c. It i enough to notice that in that cae M = fx 2 X; E(x) cg i a dual et. Conider again the functional E ; (u) = jruj dx? r juj r dx? q juj q dx: Recall that we aume that < < n,, q () n? and that n? r n in uch a way that n? E i a C -functional on the Sobolev ace H ; (). A rt dual et: Dene the Mountain Pa cla to be F = f 2 C([; ]; H ; ()); () = ; () 6= and E(()) g which i clearly homotoy-table with boundary B = fe g. Let M = fu 2 H ; (); u 6= ; he (u); ui = g: The following Nehari-tye duality roerty i by now tandard. Theorem 5.2 Aume q () and either one of the following cae: () = r, < q and < < ; >. (2) < r, = q and < < ; ; >. (3) < r, < q and >, >. The et M i then cloed, i dual to F and atie inf E ; = c := c(e ; ; F ) M 5

18 Proof: By denition, he ;(u); ui = (jruj? juj r? jujq )dx: Note that B \ M = ; ince for every u 2 M, we have E ; (u) = (? r ) juj r dx + (? q ) juj q dx > ; under the aumtion that either r or q i dierent from. We alo how that under thi aumtion, we have the etimate c inf u2m E ; (u). Let u 6= in M, conider the traight ath (t) = tu, E ; (tu) = t jruj? tr r juj r? tq q juj q : Since lim t! E ; (tu) =?, we have that c u t< E ; (tu) = E (t u). From de ; (tu) dt = t? jruj? t r? juj r? t q? juj q and de ;(tu) (t dt ) =, we get that Since u 2 M, we hould have jruj = t r? juj r + t q? juj q : jruj = juj r juj q + : Thu, t mut be equal to a long a either r or q i ditinct from. Thi clearly how that in either one of the 3 condition above, we have c inf u2m E ; (u): For the ret, we have to ditinguih the 3 cae. Cae (): 2 = r, < q and < <. To rove that M i cloed, ue the Sobolev-Hardy inequality and the denition of to nd a contant c > uch that he ;(u); ui = (? )kuk H ; ()? ckukq H ; () = kuk H ; ()(?? ckuk q? H ; ()): Chooe ome > uch that if kuk <, then?? ckuk q? >. Thi mean that we H ; () can nd ome contant > uch that for any u 2 M, we have kuk. So M i cloed. 6

19 To rove the interection roerty, x 2 F joining to, where 6= and E ; (). Note that ince <, we have that he ;((t)); (t)i > for t cloe to (ame roof a for the cloedne of M ). On the other hand, ince 6=, we have that he ;(); i < E ; () : It follow from the intermediate value theorem that there exit t uch that (t ) 2 M. Thi rove the duality and conequently that c inffe ; (u) : u 2 M g. Cae (2): < < r, = q () and < < ;q. To rove that M i cloed, ue the Sobolev-Hardy inequality with it bet contant ; and the Sobolev inequality to get: he ;(u); ui = jruj? jruj? c( juj r juj q? jruj ) r? ;q = (? ;q )kuk H ; ()? ckukr H ; () = kuk H ; ()(?? ckuk r? ;q H ; ()): jruj Chooe ome > uch that if kuk <, then? ;q? ckuk r? >. Thi mean that H ; () we can nd ome contant > uch that for any u 2 M, we have kuk. So M i cloed. To rove the interection roerty, x 2 F joining to, where 6= and E ; (). Note that ince < ;q, we have that he;((t)); (t)i > for t cloe to (ame roof a for the cloedne of M ). On the other hand, ince 6=, we have that he;(); i < E ; (). It follow from the intermediate value theorem that there exit t uch that (t ) 2 M. Thi rove the duality and conequently c inffe ; (u) : u 2 M g. Cae (3): 2 < r and >. To rove that M i cloed, again ue the Sobolev-Hardy inequality and the Sobolev embedding to nd contant c > ; c > uch that he ;(u); ui kuk? c kuk r? c kuk q = kuk (? c kuk r?? c kuk q? ): Since both r and q are ditinct from, we may chooe > uch that for any u 2 H ; () o that kuk <, we have? c kuk r?? c 2 kuk q? >. Thi mean that for any u 2 M, kuk, hence M i cloed. For the interection roerty, conider any 2 F joining and v. Since < r, again the roof above of the cloedne of M yield that he ; ((t)); (t)i > for t cloe to. Alo ince 6=, we have that he ;(); vi < E ; () 6= : 7

20 Then again, by the intermediate value theorem, we conclude that there exit t uch that (t ) 2 M, thi rove the duality and the inequality c inffe ; (u); u 2 M g. Another Dual et: Denote by S the here S = fu 2 H ; (); kuk H ; = g and by H the et () H = fh : H ; ()! H ; () odd homeomorhimg: Let 2 denote the Kranoelkii genu dened for every cloed ymmetric ubet D of H ; () a 2 (D) = inffn; there exit an odd and continuou ma h : D! R n n fgg; and conider the cla F 2 = fa; A cloed ymmetric with 2 (h(a) \ S ) 2; 8h 2 Hg: It i eay to verify that F 2 i a 2 -homotoy table cla. Let c 2 = inf A2F2 u A E ;. We hall now conider an aroriate dual et to F 2. Firt, we recall a few fact about the following weighted eigenvalue roblem ( < < ): (?4 u = b(x)juj?2 u; u 2 H ; (); u 6= : () We will ay that 2 R i the eigenvalue and u 2 H ; (); u 6= i the correonding eigenfunction of the above roblem if jruj?2 rur'dx = b(x)juj?2 u'dx hold for any ' 2 H ; (). The following lemma i well known. Lemma 5. ([25] []) Aume that b(x), b(x) 2 L t (), and jfx 2 : b(x) > gj 6=, where t if > n, t > if = n and t > n > otherwie. Let = inff R jrvj ; R b(x)jvj = g, then () > i the rt eigenvalue of the eigenvalue roblem (); (2) i imle and there exit reciely one air of normalized eigenfunction correonding to which do not change ign in. Here, v being normalized mean that R b(x)jvj =. We ue the lemma to rove the following Lemma 5.2 For 2 r < ; q < (); > ; > and any u 2 H ; (); u 6=, there exit a unique v = v(u) 2 H ; () uch that (a) R (jujr? + jujq? )v =, ; 8

21 (b) krvk = inffkr!k : R(juj r? + jujq? )j!j = g. Furthermore, the ma u! v(u) i continuou from L r ()! H ; (). Remark 5. It i quite unfortunate that the above lemma i not alicable {unle = 2{ whenever r = or when q = (). Thi will create additional comlication in the earch of a econd olution for the critical roblem. Proof: Since n +(q?)n > n, chooe n < t < n +(q?)n then t?(q?)t < n. Becaue juj (q?)t ( juj ) t(q?) t ( t?(q?)t )?(q?)t ; we get that jujq? 2 L t (). The functional (u) = kuk = R jruj dx, i clearly weakly lower emi-continuou and coercive. Moreover, the contraint et C = f! 2 H : (juj r? + jujq? )j!j dx = g i weakly cloed in H ; () and () i bounded below on C. Therefore, by the direct method of calculu of variation (Struwe [26]. 4), the inmum in (b) i achieved and thi inmum i the rt eigenvalue of () and thu i imle. Any function where uch an inmum i achieved i the eigenfunction correonding to the rt eigenvalue of (). By Lemma 5., it cannot change ign in. Thi give the uniquene of v(u) and therefore it continuity for non-zero u. Note that ( (u); v(u)) correond to the rt eigenair of the (weighted) eigenvalue roblem (?4 v = (juj r? + jujq? )jvj?2 v in v = () Let now M 2 = M \ fu 2 H ; (); (juj r? + jujq? )v(u)? u = g: The following duality reult wa rt noticed by G. Tarantello [28] in the cae = and = 2. Theorem 5.3 Aume q < () and r <, then M 2 i a cloed et that i dual to F 2 and inf M 2 E ; = c 2 := c(e ; ; F 2 ) a long a we are in either one of the following cae: 9

22 () = r, < q and < < ; <. (2) < r, = q and < < ; ; <. (3) < r, < q and <, <. Proof: In the 3 cae, we get from Theorem 5.2 (and it roof) that M i cloed and that for any u 6=, there exit a unique t(u) > uch that t(u)u 2 M. Clearly, t(u) = t(juj) = t(?u) and E ; (t(u)u) = max E ; (tu): t The uniquene of t(u) and it roertie give that the ma u! t(u) i continuou on H ; () and that the ma u! t(u)u dene an odd homeomorhim between S and M which give that 2 (A \ M ) 2 for all A 2 F 2. On the other hand, the ma h : A \ M! R given by h(u) = (juj r? + jujq? )v(u)? udx dene an odd and continuou ma. Since 2 (h(a \ M )) 2, we get that 2 h(a \ M ) which mean that A \ M 2 6= ; and M 2 i dual to F 2. In articular, c 2 inf u2m2 E ; (u). To rove the revere inequality, take u 2 M 2 and let v(u) be uch that (juj r? + jujq? )v(u)? udx = : Let!(u) be a minimizer for the roblem: 2 = inff (!);! 2 H ; ; Since u 2 M, we obtain (juj r? + jujq? )v(u)?! = ; 2 kruk R (jujr + jujq ) = : Dene A = anfv(u);!(u)g 2 F 2. Then clearly, (juj r? + jujq? )j!j = g: 2 kr!k ; 8! 2 A;! 6= : R(juj r? + jujq? )j!j For! 2 A atifying E ; (! ) = u A E ; c 2, we have! 6= and! 2 M. From the above inequality, we derive Thi imlie (juj r? + jujq? )j! j kr! k : (juj r? + jujq? )(j! j? juj ) kr! k 2? kruk :

23 Alying the following inequality (valid for t and x; y 2 R) with t = r (re. t = q), we conclude that t (t? jyj t ) (? jyj )jyj t? : j! j r + j! j q r q? r juj r? q juj q kr! k? kruk ; that i, Thi nihe the roof of the theorem. E ; (u) E ; (! ) c 2 : 6 The Solution in the cae of an HS-Subcritical ingular term In thi ection, we conider the roblem (?4 u = juj r?2 u + jujq?2 u in = : (P ; ) where < n, in the reence of a ubcritical ingular term ( < q < ()) and a ubcritical non-ingular term ( < r < ). Theorem 6. (Hardy-Sobolev ubcritical ingular term) Suoe < q < () and r <. Alo aume one of the following condition: () < q, r and >, >. (2) = q, < r and >, ; > >. Then the equation (P ; ) ha innitely many olution. Moreover, it ha an everywhere oitive olution u with minimal energy and a ign-changing olution u 2 that atie (ju 2 j r? + ju 2j q? )v(u 2 )? u 2 = where v(u 2 ) i the rt eigenvector of the (weighted) eigenvalue roblem (?4 v = (ju 2 j r? + ju 2j q? )jvj?2 v in v = Proof: Now that under thee condition the functional E ; atie (P S) c for any c. It i now enough to aly Theorem 5. to F and it dual et M (re., to F 2 and it dual et M 2 ) to get a olution u (re. u 2 ) which minimize the energy functional on M (re. M 2 ). To obtain other olution, we need the following reult of Rabinowitz ([7]). 2

24 Lemma 6. Let E be an even C -functional atifying the Palai-Smale condition on a Banach ace X = Y with dim(y ) <. Aume E() = a well a the following condition: () There i > uch that inf S() E. (2) There exit an increaing equence fy n g n of nite dimenional ubace of X, all containing Y uch that lim n dim(y n ) = and for each n, u SRn (Y n) E for ome R n >. Then E ha an unbounded equence of critical value. We now how that the functional E(u) = jruj? q juj q? juj r r atie the hyothei of the lemma. Without lo of generality, we aume that = (; ) n. Let Y k be the k-dimenional ubace of X = H ; (), generated by the rt k function of the bai f(ink x ; ; ink n x n ); k i 2 N; i = ; ; ng Let k denote the comlement of Y k in X, that i, generated by the bae vector not in Y k. For any u 2 E c k?, the toological comlement of Y k?, kuk Ckruk =k n ( Peral [25] ). Claim : For k uciently large, there exit > uch that E(u) for all u 2 k? with kuk H ; = Proof of Claim in the cae where = q, < r and < ; : E(u) (? By the Gagliardo-Nirenberg inequality ) jruj? C juj r : ; ( juj r ) r C ( jruj ) a ( juj )?a with a = n (? r ). Hence, for u \ Y c k?, E(u) C 2 jruj? C ( jruj ) ra ( juj ) r(?a) = (C 2? C ra ( C ) r(?a)? ) k n = (C 2? C 3 r? ): kr(?a)=n 22

25 Chooing = ( C 2 2C 3 k r(?a) n ) r?, we get that E(u) C 2 2 = C 4 k r(?a) n(r?) enough. Thi comlete the roof of Claim., for k large Proof of Claim in the cae where < q: Since q < (), chooe " > uch that q < n??", then n? juj q C ( juj q n n??" n??" ) n : If q Thu If q n n??" n n??" r, then r, then n juj q n??" C ( juj r ) q juj q C 2( juj r ) q r : juj r C 3 ( juj q n n??" r : n n??" n??" ) n r q : Becaue of thee relationhi, we could combine the lat two term of the functional E together. In thi ene, we may aume that and the ret i a in cae (). E(u) jruj? C juj r : Let Y = Y k with the k choen in Claim, we now how the following Claim 2: In both cae, there exit for each nite dimenional ubace Y k H ; (), oitive contant C ; C 2 (deending on Y k ) uch that u E(u) C R? C 2 R r : u2@b R (Y k ) Indeed, for any u 2 H ; () and any R >, we have E(Ru) R Rr kuk? H ; () r kukr r : Since Y k i a nite dimenional ace, it i cloed and the two norm k k r and k k on H ; () Y k are equivalent. Thi imlie the Claim. Now we can aly Lemma 6. to conclude that E ; ha an unbounded equence of critical value. 7 The Solution in the cae of a Hardy-critical ingular term Theorem 7. (Hardy-critical ingular term) Suoe < = q = () (i.e., = ). 23

26 () If < r < (high order non-ingular term), then (P ; ) ha innitely many olution {at leat one of them being oitive{ for any > and any < <. (2) If r = (critical non-ingular term) and i tar-haed, then (P ; ) ha no nontrivial olution for any > ; >. Proof: If r <, then by Theorem 4..2, The functional E ; atie (P S) c for any c a long a <. Since < r, the roof i the ame a in Theorem 6..(2) while the econd cae of the theorem i covered in ection 2. Remark 7. The cae when = q = r i really an eigenvalue roblem. There are olution for (P ; ) a long a i an eigenvalue of the roblem? u? juj?2 u = juj?2 u in H ; (). If < one can how that there i an innite number of eigenvalue for the above roblem. Indeed, thee correond to the critical level of the retriction of the functional ~E(u) = jruj dx? juj dx to the ubmanifold fu; R juj dx = g. But a light variation of Theorem 4..(2) how that in thi cae, E ~ ha (P S)c for any c and therefore a tandard alication of Ljuternik- Schnirelman theory alied to the genu 2 will yield the reult. 8 A oitive Solution in the cae of a Hardy-Sobolev critical ingular term In thi ection, we conider the rt olution for the roblem (P ; ) with the critical Sobolev- Hardy exonent. Theorem 8. (Hardy-Sobolev critical ingular term) Suoe < < q = () (i.e., < ) in the equation: 8 < :?4 u = juj r?2 u + juj ()?2 u in = : (P ; ) If r <, then (P ; ) ha a olution that i trictly oitive everywhere on, under any one of the following condition: () = r < and n 2, < < and >. (2) < r <, i large enough and >. (3) < r < and n > (?)r+2, > ; >. +(?)(r?) If r = and i tar-haed, then (P ; ) ha no non-trivial olution for any > ; >. 24

27 Proof: Note that the lat cae (r = ) i covered in ection 2. Now if r <, then Theorem 5.2 aert that either one of the 3 condition yield that the et M = fu 2 H ; (); u 6= ; he ;(u); ui = g i cloed, that it i dual to the Mountain Pa cla and that F = f 2 C([; ]; H ; ()); () = ; () 6= and E ; (()) g inf M E ; = c := c(e ; ; F ): On the other hand, Theorem 4..(2) yield that E ; atie the (P S) c condition for any? n? (n?)? c <. Therefore, we hould be able to aly Theorem 5. and conclude a long a we rove the following Lemma 8. In any one of the above three cae, we have c <? ( n? ) (n?) n??. Proof: We may aume {without lo of generality{ that =. We rt conider the following: Cae (): < r and large. In order to etimate the energy level c, we conider the function g(t) = E ; (tv " ) = t jrv " j? t () ()? tr r jv " j r and g(t) = t jrv " j? t () () ; where v " i the extremal function dened in the aendix. Note that lim t! g(t) =? and g(t) > when t i cloe to o that u t g(t) i attained for ome t " >. From we have and therefore Thu = g (t " ) = t? " ( jrv " j? t ()? "? t r? " jv " j r ); jrv " j = t ()? " + t r? " jv " j r > t ()? " t " ( jrv " j ) ()? : jrv " j t ()? " + ( jrv " j ) r? ()? ( jv " j r ): Chooe " mall enough o that by () and (6) of Lemma., we have t ()? " 2. That i, we get a lower bound for t ", which i indeendent of ". 25

28 Now we etimate g(t " ). The function g(t) attain it maximum at t = ( R jrv " j ) and i increaing in the interval [; ( R jrv " j ) ()? ]. By Lemma., we have So for large enough, we have g(t " ) = g(t " )? r tr " jv " j r g(( jrv " j ) ()? )? r tr " jv " j r = (? () )( jrv " j ) () ()?? r tr " jv " j r? (n? ) n?? + O(" n?? )? r ( 2 ) r ()? jv " j r : g(t " ) <? n? (n? )? : ()? Cae (2): < r < and n > (?)r+2, >. +(?)(r?) Note rt that the above condition i equivalent to maxf;? g < r <? and >. For any >, the above etimate on g(t " ) and Lemma 6. yield g(t " ) o that if r i choen in uch a way that i.e. r >?? n? (n? )? + O(" n??? )? O("? n?? >? r(n? ) (n? );? then? g(t " ) <? n? (n? )? : r(n?) (n? ) ): Cae (3): = r, < < and n 2. We till ue the function g(t). Since <, we have g(t) > when t cloe to and lim t! g(t) =?. So again g(t) attain it maximum at ome t " >. From we get Thu g (t) = t? ( jrv " j? t ()?? jv " j ) = g(t " ) = (? () )( = 8 >< >: t " = ( jrv " j? n??? (n?)? n?? (n?) jrv " j? jv " j ) ()? : jv " j ) () ()? + O(" n?? )? O(" (?)? ) > (? ) + O(" n?? )? O(" n?? j log "j) = (? ): 26

29 In the cae where > (? n? ), we require that? > (?), but both are equivalent to? 2 < n. In the cae where = (? ), we have 2 = n and the roof the lemma i now comlete. Remark 8. () If 2 n, then?? (? ), o that r >? whenever? < r. In thi cae, r can take any value between and. (2) If 2 > n, then < (? ) <?? and then we require that?? < <. 9 A Sign Changing Solution in the Hardy-Sobolev Critical Cae In thi ection, we extend the argument of Tarantello [28] to etablih the following. Theorem 9. (Hardy-Sobolev critical ingular term) Suoe 2 < q = () and r < in the equation: 8 < :?4 u = juj r?2 u + juj ()?2 u in = : Aume any one of the following condition: () = r <, n > 3? 2 +, > and < <. (2) < r <, > and large enough. (3) < r <, n > (?)r+ +(?)(r?) and > ; >. Then (P ; ) ha alo a changing-ign olution u that verie (juj r? + juj ()? )v(u)? u = ; where v(u) i the rt eigenvector of the (weighted) eigenvalue roblem 8 < :?4 v = (juj r? + juj ()? )jvj?2 v in v = (P ; ) Proof: We aume without lo of generality that =. Theorem 5.2 aert that for any q < (), either one of the 3 condition yield that the cloed et M q 2 = M q \ fu 2 H ; (); (juj r? + jujq? )v(u)? u = g: i dual to the cla F 2 and that inf E ;q = c 2;q := c(e ;q ; F 2 ); M q 2 27

30 where for each q < q (), the et M q (re. M q 2 ) denote the dual et aociated to the functional E ;q := R jruj? R juj q q? R r jujr. E will denote E ; () and M (re. M 2 ) will denote M () (re. M () 2 ). Note that by Theorem 5.2, M i dual to F but the ame cannot be aid about M 2 and F 2 unle = 2. Therefore, to etablih the theorem above, we hall reort to a limiting argument a q! (). Lemma 9. Under any one of the 3 condition in Theorem 9., we have: () c i;q! c i (i=,2) a q! (). (2) There exit > and > uch that for < jq? ()j <, we have c 2;q c ;q + n S n?. Proof: () Firt we rove that (c ;q ) q and (c 2;q ) q are uniformly bounded in q. We hall only how it for c 2;q. For any u 2 M q 2, Thu jruj? E ;q (u) = (? r ) i.e. c 2;q inf M q 2 E ;q. Since now E ;q (u) = and for any u ; v 2 H ; (), juj q? juj r = : juj r + (? q ) jruj? q jruj? r juj q dx juj q? juj r r juj r E(u); lim E(u + v ) = lim ( jr(u + v )j? ju + v j r ) =?;!;!!;! r E(u + v ) attain it maximum at ome nite and. Thi mean c 2;q E( u + v ); which i indeendent of q. Thi imlie the exitence of contant C > and C 2 >, uch that C ju ;q j q C 2 Similar etimate alo hold for kru ;q k and ku ;q k r. Notice that for every u 6=, there exit unique t q (u) > and t(u) > uch that t(u)u 2 M and t q (u)u 2 M q : 28

31 Furthermore, t q (u)! t(u) a q! (). Set q = t(u ;q ) o that q u ;q 2 M. We have c E ( q u ;q ) Since u ;q 2 M q, we have Thu c E ( q u ;q ) = (? q ) = (? q ) = jr q u ;q j? j q u ;q j r? j q u ;q j () r () = (? () ) jr q u ;q j + ( ()? r ) j q u ;q j r : E ;q (u ;q ) = jru ;q j? ju ;q j r? r q = (? q ) jru ;q j + ( q? r ) ju ;q j q ju ;q j r : qjru ;q j + ( q? () ) qjru ;q j + ( ()? r ) j q u ;q j r jru ;q j + (? q )( q? ) jru ;q j +( q? () ) qjru ;q j + ( ()? r ) j q u ;q j r = E ;q (u ;q ) + (? q )( q? ) jru ;q j + ( q? () ) qjru ;q j ( q? r )(r q? ) ju ;q j r + ( ()? q ) r qju ;q j r : Note that q! a q! (), therefore c c ;q + o(): To obtain the revere inequality, et t q = t q (u ) >. Thu, t q u 2 M q ; t q! a q! () and c ;q E ;q (t q u ) = E (t q u ) + juj () () = E (u ) + o() = c + o():? q juj q Thi comlete Part () of the lemma. We rove Part (2) by etimating u A" E ;q where A " = anfu ; v " g 2 F 2 and u i the rt olution of the critical roblem. For that, we need the moothne of u but thi cannot be guaranteed unle = 2. However, an eay aroximation argument and the fact that u t E (tu)! u t E(tu ) a u! u trongly, allow u to aume that u ha the required moothne. 29

32 Therefore we may uoe that u ; ru 2 L (). We hall conider the cae where r > rt. Cae (): Aume < r < and >. For " > and q uciently cloe to () and by alying the calculu lemma, E ;q ( u + v " ) = + jr(u + v " )j? r jr(u )j? r + A [ + B [ + C [ + Therefore, for " uciently mall, jr(v " )j? r ju j r? q ju + v " j r? q jv " j r? q ju j q jv " j q jru jjrv " j? + jru j? jrv " j] ju jjv " j r? + ju j r? jv " j] ju j jv "j q? + ju j jv "j q? ] jr(u )j? r jr(v " )j? r ju j r? q jv " j r? q ju j q jv " j q + A 2 (jj + jj )" n? (?) + B2 (jj r + jj r )"?? + C 2 (jj q + jj q )" n? (?) : lim ;! E ;q(u + v " ) =?: So we may aume that ; are in a bounded et. A in the tudy of the rt olution, let u conider the function ju + v " j q (r?)(n?) (n? ) g(t) = E (tv " ) = t jrv " j? t () ()? tr jv " j r r again. A in the reviou ection, we have g(t " )? n? (n? )? If now r >?? >?, we have + C" n??? r ( 2 ) r ()? E ;q (u + v " ) E ;q (u ) + E (v " ) + jj () ()? jjq jv " j q q + A 3 " n? (?) + B3 "?? (r?)(n?) (n? ) 3 jv " j r :

33 c ;q +? n? (n? )? + A 3 " n? (n?) + B3 "?? c ;q +? n? (n? )? Chooe " mall enough uch that C" n?? + C" n?? (r?)(n?) (n? ) + C" n?? jj () + ()? jjq jv " j q q? r ( 2 ) r ()? jv " j r + A3 " n? (?) + B3 "??? C 4 "? r(n?) (n?? ) + jj () ()? jjq jv " j q q + A3 " n? (?) + B3 "?? (n? (r?)(n?) )? C 4 "?? for ome contant >. Now chooe > mall enough uch that (r?)(n?) (n? ) r(n?) (n? )?2 jj () ()? jjq jv " j q q < for < jq? ()j < : Thu the cae when r > i etablihed. Cae (2): r =, 3? 2 + < n and < <. The aumtion 3? 2 + < n imlie that 2 < n, > (? ) and? < (? ). We aume that and are in a bounded et and we etimate E (u + v " ). Again let g(t) = E (tv " ) = t jrv " j? t () ()? t jv " j then the maximum g(t " ) of g(t) verie g(t " ) =? n? (n? )? + O(" n?? )? O(" (?)? ): Thu we have E ;q ( u + v " ) = + jr(u + v " )j? jr(u )j? + A [ + B [ + C [ jr(v " )j? ju j? q ju + v " j? q jv " j? q ju j q jv " j q jru jjrv " j? + jru j? jrv " j] ju jjv " j? + ju j? jv " j] ju j jv "j q? + ju j jv "j q? ] jr(u )j? 3 ju j? q ju j q ju + v " j q

34 + jr(v " )j? jv " j? q jv " j q + A 2 (jj + jj )" n? (?) + B2 (jj + jj )" (n?)(?) (?) + C 2 (jj q + jj q )" n? (?) : E ;q (u ) + E (v " ) + jj () ()? jjq q + A 3 " n? (?) + B3 " (n?)(?) (?) c ;q +? n? (n? )? + C" n?? + A 3 " n? (n?) + B3 " (n?)(?) (?) c ;q +? n? (n? )?? C 4 " (?)? + C" n?? + jj () ()? jjq jv " j q q Since 3? 2 + < n, we may chooe " mall enough uch that C" n?? jv " j q jj () + ()? jjq jv " j q q + A3 " n? (?) + B3 " (n)(?) (?) + A3 " n? (?) + B3 " (n?)(?) (?)? C 4 " (?)??2 for ome contant >. Now chooe > mall enough uch that jj () ()? jjq jv " j q q < for < jq? ()j < : The roof of the lemma i now comlete. (?)? O("? ) Proof of Theorem 9.: In order to get the econd olution, we hall conider the econd olution u 2;q of the roblem correonding to q < () and we will nd a limit a q! (). The location of u 2;q on the dual et M q 2 will be crucial for the comactne. Since c 2;q i bounded uniformly in q, there i K >, uch that kru 2;q k K whenever < jq? ()j < : For x 2, dene (u 2;q ) + (x) = maxfu 2;q (x); g and (u 2;q )? (x) = maxf?u 2;q (x); g. Since u 2;q 2 M q 2, we have that both (u 2;q ) + and (u 2;q )? (x) 6= and belonging to H ; (). In addition, kr(u 2;q ) k K whenever < jq? ()j < : Thu, we can nd q n uch that q n! () a n! +; u + ; u? 2 H ; and (u 2;qn ) * u weakly in H ; a n! +: We claim that u + 6= and u? 6=. To horten notation, et u n = (u 2;qn ) ; c ;n = c ;qn ; E n = E ;qn and? n = M qn : Since u n i the olution of the correonding ub-critical roblem, we have that u n 2? n. In articular, E n (u n ) c ;n : 32

35 From Lemma 9., we alo know that E n (u + n ) + E n (u? n ) = E n (u n ) = c 2;qn c ;n +? n? (n? )?? for n large. Necearily, E n (u n )? n? (n? )?? for n large. From the fact that u n 2? n and c n! c >, we derive K ju n j qn K 2 with uitable oitive contant K and K 2. Arguing by contradiction, aume for examle, that u + =. From above and the fact that u n 2? n, we obtain kru+ n k? ju + n j qn? n? q n (n? )?? + o() and Conequently, kru + n k? ju + n j qn = o(): ( ju + n j () ) () kru + n k ju + n j qn = + o() = ( ju + n j qn ) qn () ( ) ()?qn () ju + n j qn qn () + o() qn (? () ) Since R ju + n j qn i bounded away from zero, we conclude that ( ju + n j qn ) qn? () ( ) qn? () () + o(): That i, Thu, we have n?? (n? )? +o()? (n? ) ju + n j qn ju + n j qn n?? + o(): = kru+ n k? q n ju + n j qn +o()? n? (n? )??+o(): Thi i a contradiction and we conclude that u + 6=. Similarly, one how that u? 6=. Set u = u +? u?, that i u change ign in and u n := u qn * u weakly in H ; (): 33

36 So, he(u); wi = for any w 2 H ; (), i.e. u i a weak olution of (P ). Now, we rove that a ubequence of fu n g converge to u trongly in H ; () and conclude that u i a olution of the (P ; ()) that i located on M 2. Since fe(u n )g i bounded and En(u n )!, we may aume that the concluion of Lemma 4.4 hold for the equence (u n ) n. Note that u 2 M, hence E(u) c. Set u n = u + w n, with w n * weakly in H ;. We have c ;n + n?? (n? )?? E n (u + w n ) kruk? juj qn q n? r kukr r + krw nk? q n c + krw nk? q n jw n j qn Since jc ;n? c j = o(), we derive Furthermore, i.e. + o(): krw nk? jw n j qn? n? q n (n? )? = he n(u n ); u n i = he (u); ui + krw n k? krw n k? jw n j qn = o():? + o(): jw n j qn jw n j qn + o(); + o() The lat two relation how that the equence krw n k cannot be bounded away from zero and therefore a ubequence of fw n g converge trongly to zero. Sobolev critical non-ingular term In thi ection, we rove Theorem.4. We reformulate it a follow. Theorem. Suoe < q < () and r = in the equation: (?4 u = juj?2 u + jujq?2 u in = : (P ; ) Then (P ; ) ha a oitive olution if any one of the following condition hold: () < q, n > (?)(q?)+2 +(?)(q?) and > ; >. (2) = q, n 2? (? ) and >, ; > >. If one of the following condition hold: 34

37 (') < q, n > (?)(q?)+ +(?)(q?) and > ; >. (2') = q, n > ((? )(? ) + ) and >, ; > >, then (P ; ) ha alo a ign-changing olution. Remark. The exitence of a oitive olution under condition (2) above ha already been noticed in [3] in the cae where = q. By caling, we can alway aume that =. The correonding functional i again E (u) = jruj? juj? q Recall that under any one of the above condition, the et i dual to the cla F dened a juj q : M = fu 2 H ; (); u 6= ; h(e (u); ui = g F = f 2 C([; ]; H ; ()); () = ; () 6= and E (()) g: Moreover, the energy level c = inf A2F u u2a E (u) i equal to inf u2m E (u). By Theorem 4..(3), E atie (P S) c for any c < n n. So, the exitence of the rt oitive olution follow immediately from the following etimate. Lemma. If r =, then c < n n in any of the following three cae: () q = ; < < ; and n 2? (? ): (2) < q < () and large enough. (3) < q < () and n > (?)(q?)+2 +(?)(q?). Proof: Take v " to be the function a in lemma.2 of the aendix, then a the roof of lemma 9., we conider: Cae q > : max E (tv " ) t< n n + O(" n? )? q ( ) q? jv " j q = n n + O(" n? (n?)(?) )? O(" 2 ( ()?q) ): where we require that q > n? (? ). The etimate then follow. n? Cae q = : max E (tv " ) = t< n ( jrv " j? jv "j ) n = 8 >< >: n n n n + O(" n? (n?)(?) )? O(" 2 ( ()?) ) > n? (? ) n? + O(" n? n? )? O(" j log "j) = n?(? ) 35 n?

38 Since (n?)(?) ( ()? ) = (?)(?), the concluion now follow immediately. 2 For the ign changing olution, we hall roceed a in the cae of the Hardy-Sobolev critical exonent. Firt we nd aroriate ign changing olution for the ub-critical roblem, i.e. when r < and then we a to the limit a r!. Write again E ;r (u) = jruj? juj r? juj q r q ; and F r = f 2 C([; ]; H ; ()); () = ; () 6= and E ;r (()) g; M r = fu 2 H ; (); u 6= ; h(e ;r ) (u); ui = g c ;r = inf u A2F u2a E ;r (u): Then a in reviou ection, we know that M r i dual to F r and c ;r = inf u2m E (u). Alo dene c 2;r = inf u A2F 2 u2a E ;r (u); where F 2 i dened in ection 5. We write c 2 (re. E ) for c 2; Lemma.2 Under either one of the following condition, () = q, < < ; and n > (? )(? ) +, (2) < q, > and n > (?)(q?)+ +(?)(q?), we have (i) c i;r! c i (i = ; 2) a r!, (re E ; ). (ii) c 2;r c ;r + n n? for ome > and r cloe to uciently. Proof: For the rt concluion, the roof i exactly the ame a in the lat ection. For the econd one, we can aume that the rt olution u i mooth and ru 2 L (). For " > and q uciently cloe to, aly the calculu lemma to obtain E ;r (u + v " ) E ;r (u ) + E ;r (v " ) + A [ + B [ jru jjrv " j? + jru j? jrv " j] ju jjv " j r? + ju j r? jv " j] + C [ ju j jv "j q? + ju j jv "j q? ] E ;r (u ) + E ;r (v " ) + A 2 (jj + jj )" n? 2 + B 2 (jj r + jj r )"? + C 2 (jj q + jj q )" n? 2 : 36 (r?)(n?) (n? )

39 Again, we note that for " uciently mall, lim ;! E ;r(u + v " ) =?: So we may aume that ; are in a bounded et. Cae : Aume < q, < and n > (?)(q?)+ +(?)(q?). Then, by the calculu lemma, E ;r (u + v " ) E ;r (u ) + E (v " ) + jj where, we require q > n? n? q > r?? n n? + A 3 " n? 2 + B 3 "? c ;r + n S n n? + C" + A 3 " n? 2 + B 3 "? c ;r + n S n n? + C". From n? 2 (r?)(n?) (n? ) + jj? jjr r? jjr r (r?)(n?) (n? )? C 4 " (n?)(?) 2 ( ()?q) + jj (? ). From? + A 3 " n? 2 + B 3 "?? jjr r (n? (r?)(n?) jv " j r jv " j r? q (S 2 ) q? (r?)(n?) (n? ) jv " j r ; jv " j q ) > (n?)(?) ( ()? q) we get 2 > (n?)(?) 2 ( ()? q), we get that q > ()?. Since? (i.e.? ()?? ()? =?? n? > r?? n, the hyothei n? q > ()? n > (?)(q?)+ ) i ucient to allow u to chooe " mall enough uch that +(?)(q?) C" n? + A 3 " n? 2 + B 3 "? (r?)(n?) (n? )? C 4 " (n?)(?) 2 ( ()?q)?2 for ome contant >. Now chooe > mall enough uch that jj? jjr r Thu we roved the cae for q >. jv " j r < for < jr? j < : Cae (2): q =, n > (? )(? ) + and < < ;. We aume that and are in a bounded et and we etimate E (u + v " ). E ;r ( u + v " ) E ;r (u ) + E ;r (v " ) + A [ + B [ + C [ jru jjrv " j? + jru j? jrv " j] ju j jv "j? + ju j? jv "j ] ju jjv " j r? + ju j r? jv " j] E ;r (u ) + E (v " ) + jj 37? jjr r jv " j r

40 + A 3 " n? 2 + B 3 "? c ;r + n S n n? + C" + A 3 " n? 2 + B 3 "? c ;r + n S n n? + C"? C 4 " (?)(?) + jj (r?)(n?) (n? ) + jj? jjr r (r?)(n?) (n? ) + A 3 " n? 2 + B 3 "?? jjr r jv " j r jv " j r? O(" (?)(?) ) (r?)(n?) (n? ) Note that we have required that > n? (? ). By the aumtion, we can chooe " mall n? enough uch that C" n? + A 3 " n? 2 + B 3 "? (r?)(n?) (n? )? C 4 " (?)(?)?2 for ome contant >. Now chooe > mall enough uch that jj? jjr r jv " j r < for < jq? j < : The roof of thi lemma i now comlete. The ret of the roof of the Theorem i now very imilar to Theorem 9.. The detail are left for the intereted reader. Aendix: Etimate on the Extremal Sobolev-Hardy Function Aume, without lo of generality, that 2, and let U " (x) = (" +?? )?n? : U " (x) i a function in H ; (R n ) where the bet contant in the Sobolev-Hardy inequality i attained. It i well known that they are, modulo the tranlation and dilation, the unique oitive one, where the bet contant i achieved. (See ection 2) Let (x) be a function in C () dened a (x) = ( if R if 2R where B 2R (). Set u " (x) = (x)u " (x). For "!, the behavior of u " ha to be the ame a U " but we need recie etimate of the error term. Lemma. Aume <, 2 and q = n? n?. By taking v u " = " R R jv "j q =, we have the following etimate: 38 ( ju"j q ) q o that