Existence and multiplicity of solutions for critical elliptic equations with multi-polar potentials in symmetric domains

XYZ 75 5765 5789 wwwelseviercom/locate/xyz Existence and multiplicity of solutions for critical elliptic equations with multi-polar potentials in symmetric domains Qianqiao Guo, Junqiang Han, Pengcheng iu Department of Applied athematics, orthwestern Polytechnical University, Xi an, 79, China Received 9 October ; accepted 5 ay Communicated by S Ahmad Abstract In this paper, we consider the elliptic equations with critical Sobolev exponents and multi-polar potentials in bounded symmetric domains and prove the existence and multiplicity of symmetric positive solutions by using the Ekeland variational principle and the Lusternik Schnirelmann category theory c Elsevier Ltd All rights reserved Keywords: Elliptic equation; Sobolev exponent; ulti-polar potentials; Bounded symmetric domain; Positive solution Introduction The critical elliptic problem has been studied extensively since the initial work [] by Brézis and irenberg The critical elliptic problem with one Hardy-type potential has also attracted much attention in recent years We refer the interested readers to a partial list [ ] and the references therein Here, we are concerned with the critical elliptic problem with multi-polar Hardy-type potentials In [], Cao and Han considered the problem u u µ i x a i= i = Kx u u in, u = on, and proved the existence of positive and sign-changing solutions in bounded smooth domains, where := is the critical Sobolev exponent In [3], problem with Kx was investigated by Felli and Terracini, and the existence of positive solutions with the smallest energy was deduced both in R and in bounded smooth domains In [4], Felli and Terracini also studied problem with symmetric multi-polar potentials in R when Kx and showed the existence of symmetric positive solutions Other related results on critical elliptic problems with multiple Hardy-type potentials can be seen in [5 7] and the references therein This work was supported by the ational atural Science Foundation of China Grant o, atural Science Foundation Research Project of Shaanxi Province o J4 and PU Foundation for Fundamental Research PU-FFR-JC7 Corresponding author Tel: +86 3649765 E-mail addresses: gqianqiao@nwpueducn Q Guo, southhan@63com J Han, pengchengniu@nwpueducn P iu -XXXX/$ - see front matter c Elsevier Ltd All rights reserved doi:6/jxyz5

5766 Q Guo et al / XYZ 75 5765 5789 However, as far as we know, there are few results about the multiplicity of solutions for the critical elliptic problem with multi-polar potentials In this paper, motivated by [4,8], we consider the critical elliptic problem with symmetric multi-polar potentials in bounded symmetric domains and prove the existence and multiplicity of symmetric positive solutions ore accurately, we are interested in the problem P λ,k u u µ x l= u = i= u x a l i = λ u + Kx u u in, on, where R 4 is a Z k SO -invariant bounded smooth domain, k 3, µ, λ R, R, l =,,,m, and Kx is a Z k SO -invariant positive bounded function on Here the domain is said to be Z k SO -invariant if e π /k y, T z, x = y, z R R and Kx is said to be Z k SO -invariant on if Ky, z = Ke π /k y, T z, x = y, z R R, where T is any rotation of R ote that if is Z k SO -invariant, we can write = B, R R R, where is Z k -invariant in R that is, e π /k y, y and B, R is a ball in R centered at the origin with radius R The group Z k SO acts on H as uy, z ueπ /k y, T z Given m regular polygons with k sides, centered at the origin and lying on the plane R {} R, we assume that A ai l, i =,,,k, are the vertices of the lth polygon, l =,,,m, A B, R B, R with R > max{r l, l =,,,m}, where r l = a l = al = = al k It is easy to see that the above assumptions can be easily satisfied, such as, a domain = B, R B, R = {x, x, x 3,, x x + x < R, x 3 + + x < R } with ai l = {a l, i, } R {} R, a l, i = e i π /k {r l, }, < r < r < < r m < R, i =,,,k, and l =,,,m We will prove the existence and multiplicity of Z k SO -invariant positive solutions for the problem P λ,k Before that, we also consider the limiting case of the problem P λ,k, that is, u Pλ,K u µ x k δ Srl x l= u = λ u + Kx u u in B R, R, u = on B R, R, where S rl := {x, R R : x = r l }, the distribution δ Srl D R supported in S rl and defined by, as in [4], D R δ S rl, ϕ DR := ϕxdσx, ϕ DR πr l S rl with dσ the line element on S rl, DR the space of smooth functions with compact support in R, B R, R := B, R B, R with R > max{r l, l =,,,m}, 4 and Kx a SO SO -invariant positive bounded function on B R, R For simplicity of notation, we write B R instead of B R, R in the sequel Here Kx is said to be SO SO -invariant on B R if Ky, z = K y, z, x = y, z B R R R We will prove the existence of SO SO -invariant positive solutions for the problem Pλ,K The paper is organized as follows In Section, we give some preliminary results Section 3 is devoted to the existence of one SO SO -invariant positive solution for the problem Pλ,K, provided that Kx satisfies some growth condition at zero; for details see Theorem 3 In Section 4, we show the existence of one Z k SO - invariant positive solution for the problem P λ,k in Theorem 4 Some growth conditions on Kx are needed, of course In Section 5, the multiplicity of Z k SO -invariant positive solutions for the problem P λ,k

Q Guo et al / XYZ 75 5765 5789 5767 is obtained by the Ekeland variational principle and the Lusternik Schnirelmann category theory, respectively Here, besides the growth conditions on Kx, the parameters λ, µ,, l =,,,m, are requested to be close to zero; for details see Theorems 5 and 59 At last, a nonexistence result is proved in Appendix otations and preliminary results Throughout this paper, positive constants will be denoted by C Denote H circ B R := {uy, z H BR : uy, z = u y, z }, where y, z B R R R, and H k := {uy, z H : ueπ /k y, z = uy, z }, where y, z R R It is known that the nonzero critical points of the energy functional J circ u := u u µ B R x k u y l= B R πr l S rl x y dσx dy λ u Kx u B R B R defined on H circ B R, and the energy functional u µ u J k u := x Kx u u x ai l λ u defined on H k are equivalent to the nontrivial weak solutions for the problem P λ,k and P λ,k, respectively We show a Hardy-type inequality first, which is an improved version of Theorem in [4] Proposition Let, B R R 3, be bounded or not, and R > r > Then, for any u H, the map y uy dσx S r L and x y µ uy dσx πr x y dy uy dy, S r where the constant µ := is optimal and not attained Proof If is equal to B R, we can prove as Theorem in [4] that, = inf B R uy dy, u H BR\{} B R uy dσx πr S r dy x y and the constant is optimal and not attained Then for any domain, B R R, the results follow Remark If = R, then Proposition is reduced to Theorem in [4]

5768 Q Guo et al / XYZ 75 5765 5789 For any domain bounded or not with z, R R, if {x, R R : x z r}, then it is also easy to prove that µ uy dσx πr S r z x y dy uy dy, where S r z := {x, R R : x z = r}, the constant µ = is optimal and not attained Denote by D, R the closure space of C R with respect to the norm u D, R := R u The limiting problem u µ u x = u u in R, u > in R \ {}, u D, R, where µ < µ, admits a family of solutions Uµ := C µ x µ µ µ/ µ + x µ+ µ µ/ µ with > and C µ = 4µ µ 4 ; see [9 ] oreover, for µ < µ, all solutions of take the above form and these solutions minimize R u Sµ := µ u x R u, / and R min u D, R \{} U µ µ U µ x = U R µ = Sµ ote that S := S is the best Sobolev constant The following lemma is from [4] Lemma 3 Let 4, µ < µ If u D, R is a solution for problem, then there exist positive constants κ u and κ u such that ux = x ν µ [κ u + O x α ], as x, 3 ux = x +ν µ [κ u + O x α ], as x +, 4 for some α,, where ν µ = 4µ / And hence there exists a positive constant κu such that κu U µ ux κuu µ Denote, for any u D, R, u x := u x, for any > For any solution u µ D, R for problem, denote Vx = ϕx u µ x with ϕx satisfying ϕx C B, r, ϕx, ϕx if x B, r, ϕx C, 6 5

Q Guo et al / XYZ 75 5765 5789 5769 where < r < small enough ow we give the following lemma Lemma 4 Let 4, µ < µ Then there hold O if µ < µ, V = O ln if µ = µ, 7 B,r O µ µ if µ > µ, V = u µ O µ µ, 8 B,r R V µ V B,r x = u µ µ uµ R x { O ln if µ = µ, + O µ µ 9 if µ µ, and for any ξ R \ {}, B,r Vx ξ u µ + o if µ < µ, R x + ξ = κ uµ ln ξ + O if µ = µ, C, µ, u µ ξ µ µ + o µ µ if µ > µ Proof The proofs of 7 9 are essentially similarly to Lemmas A and A in [8] By using Lemma 3, V u µ x = u µ y dy B,r x <r y < r κ u Uµ y dy y < r = C µ κ u y < r = C µ κ uω r = C µ κ uω + r y µ µ µ/ µ + y µ+ µ µ/ µ t t µ µ µ/ µ + t µ+ µ µ/ µ t t µ µ µ/ µ + t µ+ µ µ/ µ t t µ µ µ/ µ + t µ+ µ µ/ µ Cµ κ uω t t dt + µ µ O if µ < µ, = O ln if µ = µ, O µ µ if µ > µ, dt r dy dt dt t t dt + µ µ where ω is the surface measure of the unit sphere in R So 7 is obtained

577 Q Guo et al / XYZ 75 5765 5789 and otice that V B,r y r u µ y dy y < r u µ y dy = u µ y dy R U κ u y r µ y dy = C µ κ u = C µ κ u ω y r C µ κ u ω y r u µ y dy, y µ µ µ/ µ + y µ+ µ µ/ µ dy r r t t µ µ µ/ µ + t µ+ µ µ/ µ dt t µ µ dt = O µ µ Then 8 follows ote that u µ D, R is a solution for problem, ie u µ µ uµ x = uµ u µ in R To prove 9, multiplying the above equation by ϕ u µ and integrating by parts, it holds ϕ y u µ y dy = ϕ y u µ y µ ϕ y u µ y y < r y < r y dy + ϕy ϕyu µ y u µ ydy Hence B,r V µ V x = = y < r y < r + + R + + Similarly to, u µ y dy = O µ µ, y r r y < r y < r y < r ϕ y u µ y µ ϕ y u µ y y ϕy ϕyu µ y u µ ydy ϕy u µ y dy u µ µ uµ r y < r y < r ϕ y u µ y dy = O µ µ x ϕ y u µ y dy ϕy u µ y dy dy u µ y dy y r

Hence 9 holds since ϕy u µ y dy C y < r = { O ln if µ = µ, O µ µ if µ µ, Q Guo et al / XYZ 75 5765 5789 577 r y < r u µ y dy where the last equality can be obtained similarly to ow we prove As in [], Vx = B,r x + ξ ξ u µ x + x <r x <r x + ξ u µ x ξ + ϕ u µ x x + ξ r < x <r := A + A + A 3 To continue we distinguish three cases: µ < µ, µ = µ and µ > µ For µ < µ, by 5 we have that R u µ x < Then as 34 in [], A = ξ u µ + o 3 R oticing 5, by using 35 and 37 in [], we have A κ u µ x <r x + ξ x ξ Uµ = o, 4 U x µ A 3 κ u µ r < x <r = O x + ξ µ µ 5 for r > small enough, where 5 holds for µ < µ Then combining 3 with 4, 5, it gives Vx = B,r x + ξ ξ u µ + o if µ < µ R For µ = µ, by using 35 in [4], A = ξ u µ y dy = ξ κ uµ ln T y < r +O = κ uµ ln ξ + O 6 By using 5, as 3 in [], A = O 7 Then 6, 7 and 5 imply that Vx x + ξ = κ uµ ln ξ + O if µ = µ B,r

577 Q Guo et al / XYZ 75 5765 5789 For µ > µ, by using 5, B,r On the other hand, B,r Vx x + ξ 4κ u µ ξ x <r u µ x y < r x + ξ U µ y dy r require r < ξ = 4C µ κ u µ ω t ξ t µ µ µ/ µ + t µ+ µ µ/ µ = 4C µ κ u µ ω t ξ t µ µ µ/ µ + t µ+ µ µ/ µ r t + t µ µ µ/ µ + t µ+ µ µ/ µ 4C µ κ u µ ω t ξ t dt + µ µ 4C µ κ u µ ω ξ µ µ µ µ + o µ µ dt r dt dt t t dt + µ µ = C, µ, uµ ξ µ µ + o µ µ 8 Vx x + ξ Then 8 and 9 imply that B,r 4 ξ κ u µ x < r u µ x y < r x + ξ U µ y dy require r < ξ = C, µ, uµ ξ µ µ + o µ µ 9 Vx x + ξ = C, µ, uµ ξ µ µ + o µ µ if µ > µ 3 Existence of one positive solution for P λ,k In this section, we show the existence of one positive solution for the problem P λ,k Let 4 and Kx be SO SO -invariant on B R in this section Denote D, circ R := {uy, z D, R : uy, z = u y, z }, where y, z R R For µ < µ, define S circ µ := inf u D, circ R \{} R u µ R u x R u /, 3

Q Guo et al / XYZ 75 5765 5789 5773 which is achieved in D, circ R see Lemma 6 in [4] It is also easy to know that S circ µ is independent of B R R in the sense that, for any B R R, S circ µ = inf u H circ B R\{} B R u µ B R u x B R u / 3 An assumption on Kx is as follows K There exists α > such that Kx = K + O x α as x and α > if µ < µ, α if µ = µ, α > µ µ if µ > µ > µ Set t + = max{t, } We state the main result in this section Theorem 3 Let 4, µ + + m l= kµ + l l= rl > L, < µ and K hold For given λ R, there exists L, such that if then the problem Pλ,K admits a SO SO -invariant positive solution A lemma is crucial Lemma 3 Let 4, µ + + m l= kµ l + < µ Assume that {u n } H circ B R is a Palais Smale PS in short sequence at level c for J circ restricted to H circ B R, that is If J circ u n c, J circ u n in the dual space H circ B R c < S circ µ, K then {u n } has a converging subsequence in H circ B R Proof The proof is omitted since it is standard and similarly to Theorem 4 in [3] Denote { circ := u H circ B R : Define k l= B R πr l B R S rl u µ u u y x y dσx x dy = λ π circ : H circ B R \ {} circ, B R u u µ λ x u m π circ u = l= B R Kx u k B R u + B R πr l S rl B R u y dσx x y Kx u dy 4 } u,

5774 Q Guo et al / XYZ 75 5765 5789 for all u H circ B R \ {} Then J circ π circ u = B R u u µ λ x u m k B R l= B R Kx u πr l S rl u y dσx x y dy, for all u H circ B R \ {} Denote m circ := inf circ J circ ow we estimate m circ Proposition 33 Let 4, µ + + m l= kµ + l then l= rl > L, < µ and K hold For given λ R, there exists L, such that if m circ < S circ µ K Proof Assume S circ µ is attained by some u µ D, circ R For simplicity we assume that R u µ = Therefore the function v µ = S circ µ / u µ is a nonnegative solution for Take U circ x = ϕx u µ x with ϕx satisfying 6 Then it follows from that, for some positive constants κ, κ, r U circ y B R πr l Srl u µ + o if µ < µ, l R x y dσx ln dy = κ r + O if µ = µ, l µ µ κ + o µ µ if µ > µ On the other hand, K and 8 imply that Kx U circ = K B R B R r l U circ + Kx K U circ B R = K + O µ µ + O α Combining the above two equalities and Lemma 4, for µ < µ, we have J circ π circ U circ = B R U circ Ucirc µ λ x Ucirc m k B R l= B R Kx U circ πr l S rl Ucirc y dσx x y dy = R u µ u µ µ x + O µ µ O k R u µ K + O µ µ + O α l= r l + o

Q Guo et al / XYZ 75 5765 5789 5775 Hence there exists L, such that if > L, then l= rl J circ π circ U circ < S circ µ S circ µ K Hence m circ < K When µ = µ and µ > µ, the proofs are similar Proof of Theorem 3 Let {u n } H circ B R be a minimizing sequence for J circ on circ By the Ekeland variational principle [4], we can assume {u n } is a PS sequence Proposition 33 gives m circ < S circ µ K Hence Lemma 3 implies that there exists u circ, then u circ, such that J circ u = J circ u = m circ Therefore we end the proof by the maximum principle 4 Existence of one positive solution for P λ,k In this section, we show the existence of one positive solution for the problem P λ,k Denote K := max x Kx Let 4 in this section and assume that Kx, are Z k SO -invariant in the sequel Denote D, k R := {uy, z D, R : ue π /k y, z = uy, z }, where y, z R R As in [4], for any µ < µ, define S k µ := min u D, k R \{} R u µ u R u 4 / x It is easy to see that S k µ is independent of the Z k SO -invariant domain R in the sense that, for any Z k SO -invariant domain, u µ u x S k µ = u 4 / inf u H k \{} Two assumptions are needed K There exist x \{, ai l, i =,,k, l =,,m} and α > such that Kx = K + O x x α as x x Denote x := x, x R R Since Kx is Z k SO -invariant, then for any x, j = e π j /k x, x, j =,,,k, there holds Kx = K + O x x, j α as x x, j K 3 There exists α 3 > such that Kx = Kai L + O x al i α 3 as x ai L, i =,,k, with α 3 > if µ L < µ, α 3 if µ L = µ, α 3 > µ µ L if µ L > µ, where L L m is a positive natural number such that µ L < µ and { } S µ L S = min, l =,,m 43 Kai L Kai l The main result in this section is as follows

5776 Q Guo et al / XYZ 75 5765 5789 Theorem 4 Let 4, µ + + m l= kµ l + < µ i When min k S, k S µ L, S k µ Kai L K = k S K K assume K holds, then for given λ R, there exists L, such that if m l= kj= P λ,k admits a Z k SO -invariant positive solution ii When min k S, k S µ L, S k µ Kai L K = k S µ L, Kai L K assume K 3 holds, then for given λ R, there exists L, such that if kj=, j, a l x, j > L, the problem ml=,l L kj= µ L a L al j > L, the problem P λ,k admits a Z k SO -invariant positive solution a l al j + iii For k large enough, assume K holds, then for given λ R, there exists L, such that if m k l= rl with r l = a l = al = = al k, the problem P λ,k admits a Z k SO -invariant positive solution The following lemma is important > L Lemma 4 Let 4, µ + + m l= kµ l + restricted to H k, that is < µ Assume that {u n } H k is a PS sequence at level c for J k If J k u n c, J k u n in the dual space H k c < cµ, µ L := min k S K, k S µ L Kai L then {u n } has a converging subsequence in H k, S k µ K, Proof We omit the proof here since it is standard and similarly to Theorem 4 in [3] Denote { k µ,, λ := u H k : u u µ x u x ai l = λ Define u + Kx u } π k : H k \ {} k µ,, λ, u u µ λ x u m π k u = Kx u u x ai l 4 u

Q Guo et al / XYZ 75 5765 5789 5777 Then J k π k u = u u µ x λ u m Kx u u x ai l, for all u H k \ {} Denote m k := inf k µ,,λ J k Proposition 43 Let 4, µ + + m l= kµ l + < µ I If the assumptions in i of Theorem 4 hold, then for given λ ml= kj= a l x, j > L, there holds R, there exists L, such that if m k < k S K II If the assumptions in ii of Theorem 4 hold, then for given λ R, there exists L, such that if m l=,l L a + k µ L l j=, j al j a > L, there holds L al j kj= m k < k S µ L Kai L III If the assumptions in iii of Theorem 4 hold, then for given λ R, there exists L, such that if > L, there holds ml= k r l m k < S k µ K Proof I For x given in K, take Ux = k j= ϕx U x x, j with ϕx satisfying ϕ, ϕ = if x k j= B x, j, r, ϕ = if x where r > small enough, x, j = e π j /k x, x, j =,,,k Then gives Ux x ai l = ai l x, j U R + o j= On the other hand, the condition K and 8 imply that Kx U = K ks + O + O α By summing the above two equalities and Lemma 4, we have J k π k U = U U µ λ x U m U x a i l Kx U k Bx, j, r, ϕ 4 r, j=

5778 Q Guo et al / XYZ 75 5765 5789 = ks + O O k R U l= j= K ks + O + O α a l x, j + o Therefore there exists L, such that if m l= kj= J k π k U < k S Hence m k < k S K K a l x, j > L, then II It is easy to see that µ L > in this case since S Sµ L if µ L Take U k x = k j= ϕx Uµ L x a L j with ϕx satisfying ϕ, ϕ = if x k j= B a L j, r, ϕ = if x k Ba L j, r, i= ϕ 4 r with r > small enough Then as in Lemma 6 in [3], by, for some positive constants κ, κ, κ 3, κ 4, it follows j= ai l U al j R µ L + o if µ L < µ, U k x ln x ai l = κ j= ai l + O if µ L = µ, for l L, al j µ µ L κ j= ai l + µ µl o if µ L > µ, al j U k x R U µl x x ai L = x + O j=, j i ai l U al j R µ L + o if µ L < µ, ln + κ 3 ai l + O if µ L = µ, al j j=, j i µ µ L κ 4 j=, j i ai l al j By using Lemma 4, for µ L < µ, U k U k µ x λ Uk + µ µl o if µ L > µ = ks µl + O µ µ L O k U R l=,l L j= a l + al j = ks µl k U R l=,l L j= U k x ai l j=, j µ L a l + al j a L + o al j j=, j µ L + O a L + o al j + O

On the other hand, the condition K 3 and 8 imply that Kx U k = Kai L ks µl + O µ µ L + O α 3 Hence there exists L, such that if m l=,l L kj= J k π k U k = Q Guo et al / XYZ 75 5765 5789 5779 U k Uk µ λ x Uk m = ks µ L k R U Kx U k a l al j + k j=, j µ L a L al j > L, then l=,l L j= Uk x a i l a + k l al j j=, j µ L a L al j Ka L i ks µ L + O µ µ L + O α 3 + o + O < S µ L Kai L, which gives m k < S µ L Kai L When µ L = µ and µ L > µ, the proofs are similar III For k large enough, we have { min k S, k S µ L, S } k µ = S k µ Kai L K K K Take U circ x = ϕx u µ x with ϕx satisfying 6 and argue as in Proposition 33, for µ < µ, J k π k U circ = U circ Ucirc µ λ x Ucirc = R u µ u µ µ m Kx U circ x Ucirc x a i l + O µ µ O R u µ kµ l + o r l= l K + O µ µ + O α < S circ µ K On the other hand, Theorem 73 in [4] indicates lim k + S k µ = S circ µ Then if k large enough, m k < S k µ K When µ = µ and µ > µ, the proofs are similar Proof of Theorem 4 Let {u n } H k be a minimizing sequence for J k on k µ,, λ Then by using Proposition 43 and Lemma 4, the results can be obtained as in the proof of Theorem 3

578 Q Guo et al / XYZ 75 5765 5789 5 Existence of multiple positive solutions for P λ,k This section is devoted to the multiplicity of positive solutions for the problem P λ,k The results here consist of two parts We state them as follows respectively We always assume 5 and 3 k < K K in this section 5 Part I Denote CK = {b Kb = max x Kx} We state some assumptions first K 4 Kx C, K = max x Kx > max{k, Kai l, i =,,k, l =,,m} K 5 The set CK is finite and b R {} for every b CK, say CK = {b i,s, i k, s k CardCK}, where b i,s = b i,s, = eπi /k b,s, R {} K 6 There exists α 4 > such that if b i,s CK, then Kx = Kb i,s + O x b i,s α 4 as x b i,s H The first eigenvalue of operator µ m ki= x l= x a is positive, that is, there exists i l λ > such that u u u µ x x ai l λ u, Let L be the positive natural number appeared in 43 u H Theorem 5 Let 5, µ + + m l= kµ + l < µ If K 4, K 5, K 6, H hold and for every s k CardCK, l= j= a l b j,s >, then there exist µ >, µl > l =,,m, λ >, such that for all < µ < µ, µ L < µl, < µl l =,,m, l L, λ < λ, the problem P λ,k admits k CardCK positive solutions which are Z k SO -invariant To prove the above theorem, we follow the arguments of [8] By using Lemma 4, we have immediately the following lemma Lemma 5 If µ + + m l= kµ + l < µ and K 4 hold, then there exist µ >, µ L > such that min{k µ L µ Kai L, µ µ } and K k K cµ, µ L = c := k S K for all < µ µ, µ L µ L Choose r > small enough such that Bb i,s, r Bb j,t, r = for all i j or s t, i, j k, s, t k CardCK Let δ = r 3 and denote, for any s k CardCK, T s u := ψs x u, where ψ s x = min{, x b i,s, i =,,k} u As in [8], if u and T s u δ, then r \ k u r u Bb i,s,r 3 i= Therefore we obtain immediately the following lemma

Lemma 53 If u H such that T s u δ, then u 3 \ k u Bb i,s,r i= Q Guo et al / XYZ 75 5765 5789 578 Corollary 54 If u H, u such that T s u δ, T t u δ, then s = t It is easy to prove that if H holds and λ < λ, µ+ + m l= kµ + l < µ, then there exists c > such that u H c, for any u kµ,, λ Definition 55 For any i k, s k CardCK, consider the set s µ,, λ := {u k µ,, λ : T s u < δ} and its boundary Define Γ s µ,, λ := {u k µ,, λ : T s u = δ} m s := inf{j k : u s µ,, λ }, η s := inf{j k : u Γ s µ,, λ } Lemma 56 Let 5, µ + + m l= kµ + l l= j= a l b j,s >, s k CardCK, < µ If K 4, K 5, K 6 hold and then s µ,, λ and there exist µ >, λ > such that m s < c for all < µ µ, λ λ 5 Proof Take V s x = k j= ϕx U x b j,s H k with ϕx, a radial function, satisfying ϕ, ϕ = if x where r > small enough It is obvious that Then π k V s = V s k j= µ V s B b j,s, r, ϕ = if x x := t s V s kµ,, λ λ V s m Kx V s T s π k V s = ψs x π k V s π kv s Bb j,s,r ψs x π k V s = j= j= Bb j,s,r π kv s k Bb j,s, r, ϕ 4 r, j= V s x a i l 4 V s

578 Q Guo et al / XYZ 75 5765 5789 k B, = r ψs b j,s + y ϕb j,s + y U y k B, r ϕb j,s + y U y ψ s b j,s =, as let x b j,s = y uniformly in µ,, λ Hence we get the existence of independent of µ,, λ such that if < <, then π k V s s µ,, λ By using Lemma 4, as in Proposition 43, V s = ks + O, Kx V s = K ks + O + O α 4, V s µ l x ai l = k l= j= a l b j,s U R + o, λ V s = O By 33 in [5], we also know that V s x c, as The above estimates imply that there exists t > such that t s t as small enough Hence { max J k tv s t t max t t V s t Kx V s t V s x ai l t λ V s t µ V s x Since 5, m l= kj= a l b j,s >, α 4 >, then there exist µ >, λ > such that 5 holds } Let R = k i= Ri with Ri := {x = y, z R R : y = y cos θ, sin θ, i π k,,,k θ < iπ k }, i = Lemma 57 Let 5, µ + + m l= kµ + l < µ Assume that K 4, K 5, K 6 and H hold, then there exist µ >, µ L >, µl > l =,,m, l L, λ > such that for all < µ < µ, µ L < µ L, < µl l =,,m, l L, λ < λ, it holds c < η s Proof By contradiction we assume that there exist µ n, µn l, λ n and u n Γ s µ,, λ such that J k,n u n := u n µ n λn u n c c = k S K u n x Kx u n µ n l u n x ai l

Q Guo et al / XYZ 75 5765 5789 5783 Then it is obvious that {u n } is bounded Denote u i n := u n R i ow let us consider u n in H R subsequence, there exists l > such that lim u n R n = lim Kx u n R n = l Up to a As in Lemma 3 in [8], we deduce that l = S lim K Kx u n R n =, which implies a contradiction K and then Lemma 58 Let 5, µ + + m l= kµ l + < µ Assume that K 4, K 5, K 6, H hold and < µ < min{µ, µ }, µ L < µ L, < µl l =,,m, l L, λ < min{λ, λ, λ } Then for all u s µ,, λ, there exist ρ u > and a differential function f : B, ρ u H k R such that f = and f wu w s µ,, λ, w B, ρ u oreover, for all v H k, uv f u v µ m uv µ x l x a λ i l, v = uv Kx u uv u u µ m u µ x l x a λ i l u Kx u Proof The proof is standard and we sketch it here For u s µ,, λ, define a function F : R H k R by F u t, w := J k u tu w, tu w = t w u w µ x t m u w x ai l t λ u w t Kx u w Then F u, = J k u, u = and d dt F u, = u u µ x By using the implicit function theorem the results follow ow we prove the main result in this part u x ai l λ u Kx u Proof of Theorem 5 Let < µ < µ := min{ µ, µ, µ }, µ L < µl := min{ µ L, µ L }, < µl l =,,m, l L, λ < λ := min{ λ, λ, λ }, where µ, µ, µ, µ L, µ L, µl l =,,m, l L, λ, λ are given in Lemmas 5, 56 and 57 Let {u n } H k be a minimizing sequence for J k in s µ,, λ, that is, J k u n m s as n We assume u n since J k u n = J k u n Then the Ekeland variational principle implies the existence of a subsequence of {u n }, denoted also by {u n }, such that J k u n m s + n, J kw J k u n n w u n, w s µ,, λ Choose < ρ < ρ n ρ un and f n f un, where ρ un, f un are given by Lemma 58 Set v ρ = ρv with v H k and v H =, then v ρ B, ρ n By using Lemma 58, we get w ρ = f n v ρ u n v ρ s µ,, λ As in Theorem 33 in [8], it follows J k u n as n Therefore {u n } is a PS sequence for J k Lemma 56 gives m s < c Then we end the proof by Lemmas 4 and 5

5784 Q Guo et al / XYZ 75 5765 5789 5 Part II ow we consider the existence of multiple solutions by using the Lusternik Schnirelmann category theory The ideas are borrowed from [8,6,7] For δ >, set C δ K := {x distx, CK δ} We need the following K 5 b R {} for every b CK ote that if b = b, CK, then b i := e πi /k b, CK for every i k K 6 There exists α 5 > such that if b CK, then Kx = Kb i + O x b i α 5 as x b i, for every i k K 7 There exist R and d > such that B, R and sup x, x >R Kx K d Set k µ,, λ := {u k µ,, λ : J k u < c}, where c is given in Lemma 5 Theorem 59 Let 5, δ >, µ + + m l= kµ + l < µ If K 4, K 5, K 6, K 7, H hold and a l b >, for every b CK, i then there exist µ >, > l =,,m, λ > such that for all < µ < µ, µ L < µ L, < l =,,m, l L, λ < λ, the problem P λ,k admits at least Cat Cδ K CK positive solutions which are Z k SO -invariant The proof of the above theorem depends on some lemmas Lemma 5 Let µ + + m l= kµ + l < µ, < µ µ, µ L µ L where µ, µ L are constants appearing in Lemma 5 and K 4 hold If {u n } k µ,, λ satisfies J k u n c < c, J k k µ,,λ u n, then {u n } has a converging subsequence in H k Proof By using Lemmas 4 and 5, the result follows as Lemma 4 in [8] Denote i := R i \ R i Lemma 5 Let 5, µ + + m l= kµ + l a l b >, for every b CK i < µ Assume that K 4, K 5, K 6 hold and Then there exist µ >, λ > such that if < µ < µ, λ < λ, it holds k µ,, λ oreover, for any µ n, µn l, λ n as n and {v n} k µ n, µn l, λn, there exist xi n := xi, n, x n i and r n R + such that xn i xi = xi,, CK, r n and S vn i 4 x i u n r in D, i, as n, i =,,,k, K r n where vn i = v n i, xn i = eπ /k xn i,, x n, x i = eπ /k x i,,, and x i u n C r,rn r = r n r + xi n r n

with C r,rn the normalizing constant such that u r xi n r n = Q Guo et al / XYZ 75 5765 5789 5785 Proof Using the arguments of Lemma 56, it is easy to get k µ,, λ ow we prove the second part Consider vn in H R and set as in Lemma 3 in [8] that lim v n R n = lim Kx v n R n = l Then l = S lim n K and hence R K Kx v n = Set w n x := v n vn, then w nx = and w n x S R Therefore w n x =, w n x S By using Corollary 4 in [8] and a similar proof to Lemma 4 in [8], there exist xn and r n R + such that xn x C δk, r n as n and x w n u n r in D,, as n r n Hence v n S K 4 x u n r in D, r n By recalling the symmetry of v n, we end the proof To continue, as in [8] we define { x if x < R, ξx := x R if x R x For any u H k, set Θu := ξx u u From the proof of Lemma 56, it is known that for any u H k, t µ,,λ uu k µ,, λ with u u µ t µ,,λ u = λ x u m Kx u u x a i l 4 For b = b, R {}, define Ψ k : H k as Ψ k bx = t µ,,λ U µ,,λ xu µ,,λ x := t µ,,λ U µ,,λ x,

5786 Q Guo et al / XYZ 75 5765 5789 where U µ,,λ x := k i= ϕx U µ,,λ x b i H k, b i = e πi /k b, and ϕx, a radial function, satisfying ϕ, ϕ = if x k i= B b i, r, ϕ = if x k Bb i, r, with r > small enough, and µ,, λ as µ, λ, The proof of the following lemma is almost the same as Lemma 56 Lemma 5 Let 5, b CK, µ + + m l= kµ + l < µ If K 4, K 5, K 6 hold and a l b i >, λ, then there exist µ, µ L such that J k Ψ k b = max t> J ktu µ,,λ < c for all < µ µ, µ L µ L i= ϕ 4 r By Lemma 5 we see that if < µ µ, µ L µ L, then Ψ k b k µ,, λ It is also obvious that J k Ψ k b = c + o as µ,, λ and there exist c >, c > such that c < t µ,,λ U µ,,λ x < c for all b CK Lemma 53 Let 5, µ + + m l= kµ + l < µ, b CK and K 4, K 5, K 6 hold For all b CK, it follows Ψ k b d µ = i= as µ,, λ S K δ bi, Ψ k b d ν = i= S K δ bi, Proof ote that Ψ k b is bounded in H k and Ψ k b = t µ,,λ ϕx U µ,,λ x b i, i= Bb i,r Kx Ψ k b = Kx t µ,,λ ϕx U µ,,λ x b i Bb i,r i= Assume µ n, µn l, λ n as n Up to a subsequence, we get the existence of l > such that lim t µ n n, n Bb i,r,λn ϕx Uµn,µn l,λn x b i = lim Kx t µ n n, n,λn ϕx Uµn,µn l,λn x b i = l Then l = S lim n K Bb i,r and K Kx t µ n, n Bb i,r,λn ϕx Uµn,µn l,λn x b i = Set w i x := t µ n,µn l,λn ϕx Uµn,µ l n,λn x b i t µ n,µ n l,λn ϕx Uµn,µn l,λn x b i w i x S Bb i,r, then w i x = and

Applying the arguments of Theorem 33 in [8], it holds w i d µ i = S δ bi, w i d ν i = δ bi, which ends the proof Q Guo et al / XYZ 75 5765 5789 5787 Lemma 54 Let 5, µ + + m l= kµ + l < µ and K 4, K 5, K 6, K 7 hold For µ,, λ, ΘΨ k b = b + o, uniformly for b B, R CK; sup{distθu, CK : u k µ,, λ } Proof Let b B, R CK By Lemma 53, we have ΘΨ k b = ξx Ψ kb Ψ kb = ξxd µ d µ + o = b + o, as µ,, λ Take µ n, µn l, λ n as n and {v n} k µ n, µn l, λn By Lemma 5, there exist xn i = xi, n, x n i and r n R + such that xn i xi = xi,, CK, r n and v i n S K 4 x i u n r in D, i, as n, i =,,,k r n Since Θu is continuous, then Θv n = ξx v n x x i i ξx u n r r n i v n = + o = ξx ur x x i n + o i r n oticing that x i B, R leads to ξx i = xi, the result desired is true Proof of Theorem 59 We follow the arguments of Theorem 45 in [8] and Theorem A in [7] Given δ >, by using Lemmas 5 and 54, there exist µ >, µ l > l =,,m, λ > such that for all < µ < µ, µ L < µ L, < µ l l =,,m, l L, λ < λ, we have Ψ k b k µ,, λ for any b CK and ΘΨ k b b < δ, b B, R CK, and Θu C δ K, u k µ,, λ Define Ht, x := x + tθψ k b b with t, x [, ] CK Then H is continuous, H[, ] CK C δ K and Θ H is homotopic to the inclusion CK C δ K By Lemma 5, it is enough to prove Cat k µ,, λ Cat Cδ K CK, which can be obtained as Theorem 45 in [8], and therefore the problem P λ,k admits at least Cat Cδ K CK solutions which are Z k SO -invariant It is also easy to show that any of these solutions has a fixed sign, as Theorem 45 in [8] and Theorem in [9] Remark 55 The problem considered here involves symmetry and multiple solutions are obtained, which are different from [,3,5], where the symmetry is not been considered and existence, but not multiplicity, of solutions is obtained In [4], Felli and Terracini considered with symmetric multi-polar potentials in R when Kx and proved the existence of positive solutions, while the domain we considered here is bounded and Kx is positive bounded In [3], Felli and Schneider considered the problem related to the Caffarelli Kohn irenberg type inequality and showed the symmetry-breaking phenomenon in the inequality, namely the existence of non-radial minimizers, which motivates us to study the symmetry-breaking phenomenon of problem in the near future In [3], Badiale and Rolando also proved the existence of positive radial solutions for the semilinear elliptic problem with singular potentials, where both sub-critical and super-critical nonlinearities, rather than critical nonlinearities, are considered 3 It can be seen that when λ and the domain is bounded, there exist nontrivial solutions for the problem P λ,k However, if we consider the problem in R, then λ may lead to the nonexistence of nontrivial solutions Please see Appendix for a nonexistence result

5788 Q Guo et al / XYZ 75 5765 5789 Acknowledgment The authors would like to thank the anonymous referee for very valuable suggestions and comments Appendix Consider the quasilinear elliptic problem p u µ u p u x p = λ u q u + Kx u p u in R, u W,p R, where p u := div u p u, < p <, λ R, µ < p p p, p := p p is the critical Sobolev exponent, < q < p, Kx C R satisfying K < When p =, Kx, the nonexistence of nontrivial solutions for problem A can be seen in [3] Here we state the nonexistence result for problem A motivated by [] Theorem A Assume one of the following three cases holds: λ >, x, K, for ae x R ; λ <, x, K, for ae x R ; 3 λ =, x, K >, for ae x R, or λ =, x, K <, for ae x R Then if u W,p R is a weak solution for problem A, there holds u Proof Denote f x, u = µ u p u x p + λ u q u + Kx u p u Then problem A can be rewritten as { p u = f x, u in R, u W,p R It is easy to see that, for any ω R \ {}, there exists Cω > such that f x, u Cω + u p, x ω, u R Then as Claim 55 in [], u C R \{} W, loc R \{} following from Lemmas and in [33], Corollary in [34] and Theorem, Proposition in [35] ow we prove u L q R Take a cut-off function see Claim 53 in [] h C R satisfying ht t, ht t, h Given > small enough, define η as: η x = h x / if x 3, η x = h/ x if x, and η x elsewhere Then it is obvious that η C R \ {} Since u is a weak solution for problem A, u p u p u η u µη R x p λη u p Kxη u p = R Then by the Hardy inequality, the Sobolev inequality and the Hölder inequality, λ η u p = Kxη u R R p + u p u u p η u µη R x p C u p + u p η u R R C, where the constant C > is independent of Letting, we have u L q R Therefore, by Claim 53 in [], λ p u p q R p u p x, K =, R which implies u When Kx, λ, Theorem A extends the result in [3] to the p-laplace case A A

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