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

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1 Available online at wwwsciencedirectcom XYZ 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, 709, China Received 9 October 0; accepted 5 ay 0 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 0 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 This work was supported by the ational atural Science Foundation of China Grant o 00, atural Science Foundation Research Project of Shaanxi Province o 0J04 and PU Foundation for Fundamental Research PU-FFR-JC007 Corresponding author Tel: addresses: gqianqiao@nwpueducn Q Guo, southhan@63com J Han, pengchengniu@nwpueducn P iu 0000-XXXX/$ - see front matter c 0 Elsevier Ltd All rights reserved doi:006/jxyz0050

2 5766 Q Guo et al / XYZ 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 = 0 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 multipolar 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 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 u m u µ P 0 λ0,k x u x a l = λ 0u + Kx u u in, i u = 0 on, where R 4 is a Z k SO -invariant bounded smooth domain, k 3, µ 0, λ 0 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, Tz, x = y,z R R and Kx is said to be Z k SO- invariant on if Ky, z = Ke π /k y,tz, x = y,z R R, where T is any rotation of R ote that if is Z k SO -invariant, we can write = B 0, R R R, where is Z k -invariant in R that is, e π /k y, y and B 0, R is a ball in R centered at the origin with radius R The group Z k SO acts on H 0 as uy, z ue π /k y,tz Given m regular polygons with k sides, centered at the origin and lying on the plane R {0} R, we assume that A a l i, i =,,,k, are the vertices of the lth polygon, l =,,,m, A B 0, R B 0, R with R > max{r l, l =,,,m}, where r l = a l = a l = = a l k It is easy to see that the above assumptions can be easily satisfied, such as, a domain = B 0, R B 0, R = {x, x, x 3,,x x + x < R, x x < R } with a l i = {al, i,0} R {0} R, a l, i = e i π /k {r l,0},0 < r < r < < r m < R, i =,,,k, and l =,,,m

3 Q Guo et al / XYZ We will prove the existence and multiplicity of Z k SO -invariant positive solutions for the problem P λ0,k Before that, we also consider the limiting case of the problem P λ0,k, that is, u Pλ u µ 0 0,K x u = 0 l= k δ Srl x u = λ 0 u + Kx u u in B R, R, on B R, R, where S rl := {x,0 R R : x = r l }, the distribution δ Srl D R supported in S rl and defined by, as in [4], D R δ Srl, 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 0, R B 0, 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 0,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 0,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 λ0,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 λ0,k is obtained by the Ekeland variational principle and the Lusternik Schnirelmann category theory, respectively Here, besides the growth conditions on Kx, the parameters λ 0, µ 0,, 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 0 circ B R := {uy, z H 0 B R : uy,z = u y, z }, where y, z B R R R, and H 0 k := {uy,z H 0 : ue π /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 0 x B R

4 5768 Q Guo et al / XYZ k u y l= B R πr l S rl x y dσx dy λ 0 u Kx u B R B R defined on H 0 circ B R, and the energy functional J k u := u µ 0 u x Kx u u x a l λ 0 u i defined on H 0 k are equivalent to the nontrivial weak solutions for the problem P λ 0,K and P λ 0,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 > 0 Then, for any u H0, the map y uy dσx S r x y L and µ uy πr S r dσx x y dy uy dy, 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 H0 BR\{0} B R uy dσx πr S r x y dy 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] For any domain bounded or not with z,0 R R, if {x,0 R R : x z r }, then it is also easy to prove that µ uy πr S rz dσx x y dy uy dy, where S r z := {x,0 R R : x z = r }, the constant µ = is optimal and not attained Denote by D, R the closure space of C 0 R with respect to the norm u D, R := u R

5 Q Guo et al / XYZ The limiting problem { u u µ x = u u in R, u > 0 in R \ {0}, 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 > 0 and C µ = 4µ µ 4 ; see [9 ] oreover, for 0 µ < µ, all solutions of take the above form and these solutions minimize Sµ := min u D, R \{0} R u µ u x R u /, and Uµ µ U µ R x = Uµ = Sµ R ote that S0 := 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 0 u and κ u such that ux = x ν µ [κ 0 u + O x α ], as x 0, 3 ux = x +ν µ [κ u + O x α ], as x +, 4 for some α 0,, where ν µ = 4µ / And hence there exists a positive constant κu such that κu U µ ux κuu µ 5 Denote, for any u D, R, u x := x u, for any > 0 For any solution u µ D, R for problem, denote V x = ϕx u µ x with ϕx satisfying ϕx C0 B0, r, 0 ϕx, ϕx if x B 0, r, ϕx 6 C, where 0 < r < small enough ow we give the following lemma

6 5770 Q Guo et al / XYZ Lemma 4 Let 4, µ < µ Then there hold O if µ < µ, V = O ln if µ = µ, 7 B0,r O µ µ if µ > µ, V = u µ O µ µ, 8 B0,r R V µ V x = u µ µ uµ x B0,r R { O ln if µ = µ, + O µ µ if µ µ, and for any ξ R \ {0}, ξ V x u µ + o if µ < µ, R B0,r x + ξ = κ u µ ln ξ + O if µ = µ, C, µ, u µ ξ µ µ + o µ µ if µ > µ 9 0 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 B0,r x <r y < r κ u Uµy dy y < r = Cµκ u y < r y µ µ µ/ µ + y µ+ µ µ/ µ dy r = Cµκ uω t t µ µ µ/ µ + t µ+ µ µ/ µ dt = C µκ uω + r 0 0 t t µ µ µ/ µ + t µ+ µ µ/ µ t t µ µ µ/ µ + t µ+ µ µ/ µ C µκ uω O if µ < µ, = O ln if µ = µ, O µ µ if µ > µ, 0 t t µ µ dt + dt r t t + µ µ dt dt

7 Q Guo et al / XYZ where ω is the surface measure of the unit sphere in R So 7 is obtained otice that V u µ y dy = u µ y dy u µ y dy, B0,r y < r R y r and y r u µ y dy u µ u κ = C κ = C κ y r µ u ω y r C µ κ u ω U µy 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 B0,r V µ V x = = y < r y < r + + 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 x ϕ y u µ y dy ϕy u µ y dy dy u µ y dy y r

8 577 Q Guo et al / XYZ Similarly to, u µ y dy = O µ µ, y r r y < r ϕ y u µ y dy = O µ µ Hence 9 holds since ϕy u µ y dy C y < r = { O ln if µ = µ, O µ µ if µ µ, r y < r u µ y dy where the last equality can be obtained similarly to ow we prove 0 As in [], V x B0,r x + ξ = ξ u µ x + x <r x <r x + ξ u µ x ξ + ϕ u µ x r < x <r x + ξ := 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 u µ x < Then as 34 in [], R 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 U µ = o, 4 U x A 3 κ u µ µ = µ µ O r < x <r x + ξ 5 for r > 0 small enough, where 5 holds for µ < µ Then combining 3 with 4, 5, it gives V x = B0,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 u µ ln T y < r

9 Q Guo et al / XYZ O = κ u µ ln ξ + O 6 By using 5, as 3 in [], A = O 7 Then 6, 7 and 5 imply that V x x + ξ = κ u µ ln ξ + O if µ = µ B0,r For µ > µ, by using 5, V x B0,r x + ξ 4κ u µ ξ Uµy dy y < r r x <r u µ x x + ξ require r < ξ = 4C µκ u µ ω t ξ 0 t µ µ µ/ µ + t µ+ µ µ/ µ dt = 4C µκ u µ ω t ξ 0 t µ µ µ/ µ + t µ+ µ µ/ µ dt + r t t µ µ µ/ µ + t µ+ µ µ/ µ 4C µκ u µ ω t ξ t dt + µ µ 4C µκ u µ ω ξ µ µ µ µ + o µ µ 0 dt r t t dt + µ µ = C, µ, uµ ξ µ µ + o µ µ 8 On the other hand, V x x + ξ B0,r 4 ξ κ u µ x < r u µ x x + ξ y < r U µy dy require r < ξ = C, µ, uµ ξ µ µ + o µ µ 9 Then 8 and 9 imply that V x x + ξ = C, µ, uµ ξ µ µ + o µ µ if µ > µ B0,r

10 5774 Q Guo et al / XYZ Existence of one positive solution for P λ 0,K In this section, we show the existence of one positive solution for the problem P λ 0,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 0 < µ, define S circ µ 0 := inf u D, circ R \{0} R u µ 0 R u x R u /, 3 which is achieved in D, circ R see Lemma 6 in [4] It is also easy to know that S circ µ 0 is independent of B R R in the sense that, for any B R R, S circ µ 0 = inf u H 0 circ B R\{0} B R u µ 0 B R u x BR u / 3 An assumption on Kx is as follows K There exists α > 0 such that Kx = K0 + O x α as x 0 and α > if µ 0 < µ, α if µ 0 = µ, α > µ µ 0 if µ > µ 0 > µ Set t + = max{t, 0} We state the main result in this section Theorem 3 Let exists L 0, such that if > L, r l= l 4, µ m l= kµ+ l < µ and K hold For given λ 0 R, there then the problem Pλ 0,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 } H0 circ B R is a Palais Smale PS in short sequence at level c for J circ restricted to H0 circ B R, that is If J circ u n c, J circu n 0 in the dual space H 0 circ B R c < S circ µ 0, K0 then {u n } has a converging subsequence in H0 circ B R Proof The proof is omitted since it is standard and similarly to Theorem 4 in [3]

11 Denote { circ := u H0 circ B R : Define Q Guo et al / XYZ k l= B R πr l = λ 0 u + B R B R B R S rl u u µ 0 x u y x y dσx dy } Kx u π circ : H0 circ BR \ {0} circ, u u µ B R 0 λ x 0u m π circu = l= B R Kx u k B R πr l Sr l u y dσx x y dy 4 u, for all u H 0 circ B R \ {0} Then J circ π circ u = B R u u µ 0 λ x 0 u m k B R l= BR Kx u πr l S rl u y dσx x y dy, for all u H 0 circ B R \ {0} Denote m circ := inf circ J circ ow we estimate m circ Proposition 33 Let 4, µ m l= kµ+ l < µ and K hold For given λ 0 R, there exists L 0, such that if r l= l > L, then m circ < S circ µ 0 K0 Proof Assume S circ µ 0 is attained by some u µ0 D, circ R For simplicity we assume that u µ0 = Therefore the function v µ0 = S R circ µ 0 / u µ0 is a nonnegative solution for Take U circ x = ϕx u µ0 x with ϕx satisfying 6 Then it follows

12 5776 Q Guo et al / XYZ from 0 that, for some positive constants κ, κ, r U circ y B R πr l Srl u µ0 + o if µ 0 < µ, l R x y dσx ln dy = κ r + O if µ 0 = µ, l µ µ0 κ + o µ µ 0 if µ 0 > µ On the other hand, K and 8 imply that Kx U circ = K0 U circ + B R B R r l B R = K0 + O µ µ 0 + O α Kx K0 U circ Combining the above two equalities and Lemma 4, for µ 0 < µ, we have J circ π circ U circ = B R U circ U µ circ 0 λ x 0U circ m k l= B R πr l BR Kx U circ Sr l U circ y x y dσx dy = R u µ 0 u µ µ O µ µ 0 O k x R u µ 0 µ l + o r l= l K0 + O µ µ 0 + O α Hence there exists L 0, such that if > L, then r l= l J circ π circ U circ < S circ µ 0 K0 Hence m circ < S circ µ0 K0 When µ 0 = µ and µ 0 > µ, the proofs are similar Proof of Theorem 3 Let {u n } H0 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 µ0 Hence Lemma 3 implies that there exists K0 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 λ0,k In this section, we show the existence of one positive solution for the problem P λ0,k

13 Q Guo et al / XYZ 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 \{0} inf u H 0 k \{0} R u µ u x R u / 4 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 = 4 / u Two assumptions are needed K There exist x 0 \ {0, a l i, i =,,k, l =,,m} and α > such that Kx = K + O x x 0 α as x x 0 Denote x 0 := x 0, x 0 R R Since Kx is Z k SO - invariant, then for any x 0,j = e πj /k x 0, x 0, j =,,,k, there holds Kx = K + O x x 0,j α as x x 0,j K 3 There exists α 3 > 0 such that Kx = Ka L i + O x al i α3 as x a L i, 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 Ka L i = min Ka l i, l =,,m 43 The main result in this section is as follows Theorem 4 Let i When min k S K 4, µ m l= kµ+ l < µ, k S µ L Ka L i, S k µ 0 K0 = k S K assume K holds, then for given λ 0 R, there exists L 0, such that if m k l= j= > L, the problem P a l λ0,k admits a Z k SO-invariant positive x0,j solution ii When min k S, k S µ L S k Ka L i, µ 0 K0 = k S µ L Ka L i, K,

14 5778 Q Guo et al / XYZ assume K 3 holds, then for given λ 0 R, there exists L 0, such that if m k l=,l L j= + k µ L a l al j j=,j > L, the problem P a L al j λ0,k admits a Z k SO -invariant positive solution iii For k large enough, assume K holds, then for given λ 0 R, there exists L 0, such that if m k l= > L with r rl l = a l = a l = = a l k, the problem P λ 0,K admits a Z k SO -invariant positive solution The following lemma is important Lemma 4 Let 4, µ m l= kµ+ l < µ Assume that {u n } H0 k is a PS sequence at level c for J k restricted to H0 k, that is If J k u n c, J ku n 0 in the dual space H 0 k c < cµ 0, µ L := min k S K, k S µ L Ka L i, then {u n } has a converging subsequence in H0 k S k µ 0 K0, Proof We omit the proof here since it is standard and similarly to Theorem 4 in [3] Denote { k µ 0,, λ 0 := u H0 k : Define u µ 0 u u x a l i = λ 0 π k : H0 k \ {0} k µ 0,, λ 0, u u µ 0 x λ 0 u π k u = Kx u x u + Kx u m u x a l i 4 u } Then J k π k u = u u µ 0 x λ 0 u Kx u m u x a l i, for all u H 0 k \ {0} Denote m k := inf J k k µ 0,,λ 0

15 Q Guo et al / XYZ Proposition 43 Let 4, µ m l= kµ+ l < µ I If the assumptions in i of Theorem 4 hold, then for given λ 0 R, there exists L 0, such that if m k l= j= > L, there holds a l x0,j m k < k S K II If the assumptions in ii of Theorem 4 hold, then for given λ 0 R, there exists L 0, such that if m k l=,l L j= + k µ L a l al j j=,j > L, there holds a L al j m k < k S µ L Ka L i III If the assumptions in iii of Theorem 4 hold, then for given λ 0 exists L 0, such that if m l= > L, there holds k r l R, there m k < S k µ 0 K0 Proof I For x 0 given in K, take Ux = k j= ϕx U 0x x 0,j with ϕx satisfying 0 ϕ, ϕ = if x ϕ 4 r, k j= B x 0,j, r, ϕ = 0 if x k Bx 0,j, r, where r > 0 small enough, x 0,j = e πj /k x 0, x 0, j =,,,k Then 0 gives Ux x a l = i a l i x U 0,j 0 + o R 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 0 λ x 0 U Kx U j= m U x a l i = ks + O O k R U0 µ l + o a l= j= l x 0,j K ks + O + O α

16 5780 Q Guo et al / XYZ Therefore there exists L J k π k U < k S Hence m k < k S K K 0, such that if m k l= j= > L, then a l x0,j II It is easy to see that µ L > 0 in this case since S Sµ L if µ L 0 Take U k x = k j= ϕx U µ L x a L j with ϕx satisfying 0 ϕ, ϕ = if x ϕ 4 r k j= B a L j, r, ϕ = 0 if x k Ba L j, r, with r > 0 small enough Then as in Lemma 6 in [3], by 0, for some positive constants κ, κ, κ 3, κ 4, it follows a l j= i U al j µ L + o if µ L < µ, R U k x ln x a l = κ i a l j= i + O if µ L = µ, for l L, al j µ µ L κ a l i + µ µl o if µ L > µ, al j U k x x a L = i j= R U µ L x + x + O j=,j i κ 3 κ 4 j=,j i j=,j i a l i al j By using Lemma 4, for µ L < µ, U k U k µ 0 x λ 0Uk i= R U µ L + o if µ L < µ, ln a l i al j + O if µ L = µ, µ µ L a l i + µ µl o if µ L > µ al j U k x a l i = ks µl + O µ µ L O k U0 R a l + al j + O l=,l L j= j=,j µ L a L + o al j

17 Q Guo et al / XYZ m = ks µl k U0 R + j=,j µ L a L + o al j l=,l L j= + O a l al j On the other hand, the condition K 3 and 8 imply that Kx U k = Ka L i ks µl + O µ µ L + O α3 then Hence there exists L J kπ ku k = = 0, such that if m l=,l L k j= U k U k µ 0 λ x 0U k m ks µ L k R U0 Kx Uk l=,l L j= + k a l al j j=,j U k x a l i a l + k al j j=,j µ L a L al j Ka L i ks µ L + O µ µ L + O α 3 + o µ L a L al j > L, + O < S µ L Ka L i K, which gives m k < S µ L Ka L i 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 Ka L i, S k µ 0 = S k µ 0 K0 K0 Take U circ x = ϕx u µ0 x with ϕx satisfying 6 and argue as in Proposition 33, for µ 0 < µ, J kπ ku circ = U circ U µ circ 0 λ x 0U circ m Kx Ucirc U circ x a l i = R u µ 0 u µ µ O µ µ 0 O x R u µ 0 m l= K0 + O µ µ 0 + O α < S circ µ0 K0 k r l + o

18 578 Q Guo et al / XYZ On the other hand, Theorem 73 in [4] indicates lim k + S k µ 0 = S circ µ 0 Then if k large enough, m k < S k µ 0 K0 When µ 0 = µ and µ 0 > µ, the proofs are similar Proof of Theorem 4 Let {u n } H0 k be a minimizing sequence for J k on k µ 0,, λ 0 Then by using Proposition 43 and Lemma 4, the results can be obtained as in the proof of Theorem 3 5 Existence of multiple positive solutions for P λ0,k This section is devoted to the multiplicity of positive solutions for the problem P λ0,k The results here consist of two parts We state them as follows respectively We always assume 5 and 3 k < K K0 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{k0, Ka l i, i =,,k, l =,,m} K 5 The set CK is finite and b R {0} for every b CK, say CK = {b i,s, i k, s k CardCK}, where b i,s = b i,s,0 = eπi /k b,s,0 R {0} K 6 There exists α 4 > such that if b i,s CK, then Kx = Kb i,s +O x b i,s α4 as x b i,s H The first eigenvalue of operator µ 0 that is, there exists λ 0 > 0 such that u µ 0 u x x m Let L be the positive natural number appeared in 43 l= k i= x a l i is positive, u x a l λ i 0 u, u H0 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 > 0, then there exist µ0 > 0, µl > 0 l =,,m, λ0 > 0, such that for all 0 < µ 0 < µ0,0 µ L < µl, < µl l =,,m, l L,0 λ 0 < λ0, the problem P λ0,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

19 k K Q Guo et al / XYZ Lemma 5 If µ m l= kµ+ l < µ and K 4 hold, then there exist 0 µ 0 > 0, 0 µ L > 0 such that min{k µ L µ, µ 0 µ } and Ka L i K0 cµ 0, µ L = c := k S K for all 0 < µ 0 0 µ 0,0 µ L 0 µ L Choose r 0 > 0 small enough such that Bb i,s, r 0 Bb j,t, r 0 = for all i j or s t, i, j k, s, t r0 kcardck Let δ = 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 0 and T s u δ, then r 0 u r 0 u \ k Bb i,s,r 0 3 i= Therefore we obtain immediately the following lemma Lemma 53 If u H0 such that T s u δ, then u 3 u \ k Bb i,s,r 0 i= Corollary 54 If u H 0, u 0 such that T s u δ, T t u δ, then s = t It is easy to prove that if H holds and 0 λ 0 < λ 0, µ m l= kµ+ l < µ, then there exists c > 0 such that u H 0 c, for any u k µ 0,, λ 0 Definition 55 For any i k, s kcardck, consider the set s µ 0,, λ 0 := {u k µ 0,, λ 0 : T s u < δ } and its boundary Define Γ s µ 0,, λ 0 := {u k µ 0,, λ 0 : T s u = δ } m s := inf {J k : u s µ 0,, λ 0 }, η s := inf {J k : u Γ s µ 0,, λ 0 } Lemma 56 Let l= j= 5, µ m l= kµ+ l < µ If K 4,K 5,K 6 hold and a l b j,s > 0, s k CardCK,

20 5784 Q Guo et al / XYZ then s µ 0,, λ 0 and there exist µ 0 > 0, 0 λ 0 > 0 such that m s < c for all 0 < µ 0 µ 0,0 λ 0 0 λ 0 5 Proof Take V s x = k j= ϕx U 0x b j,s H 0 k with ϕx, a radial function, satisfying 0 ϕ, ϕ = if x ϕ 4 r, k j= B b j,s, r, ϕ = 0 if x k Bb j,s, r, j= where r > 0 small enough It is obvious that π k V s = V s µ 0 V s x λ 0 V s Kx V s m V s x a l i 4 V s Then T s π k V s = := t s V s k µ 0,, λ 0 = ψs x π k V s π kv s Bb ψs j,s,r x π k V s j= Bb π j,s,r kv s j= = k B0, r ψs b j,s + y ϕb j,s + y U 0y k B0, r ϕb j,s + y U 0 y let x b j,s = y ψ s b j,s = 0, as 0 uniformly in µ 0,, λ 0 Hence we get the existence of 0 independent of µ 0,, λ 0 such that if 0 < < 0, then π k V s s µ 0,, λ 0 By using Lemma 4, as in Proposition 43, V s = ks + O, Kx V s = K ks + O + O α4, λ 0 V s x a l = m i V s = O l= j= k a l b U j,s 0 + o, R

21 By 33 in [5], we also know that V s x c, as 0 Q Guo et al / XYZ The above estimates imply that there exists t > 0 such that t s t as small enough Hence { t max J k tv s max t t t 0 V s t Kx V s t } V s x a l i t λ 0 V s t µ 0 V s x Since 5, m k l= j= > 0, α a l 4 >, then there exist bj,s µ 0 > 0, 0 λ 0 > 0 such that 5 holds Let R = k i= R i with R i := {x = y, z R R : y = y cos θ, sinθ, i π k θ < iπ k }, i =,,,k 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 0 > 0, µ L > 0, µl > 0 l =,,m, l L, λ 0 > 0 such that for all 0 < µ 0 < µ 0,0 µ L < µ L, < µl l =,,m, l L,0 λ 0 < λ 0, it holds c < η s Proof By contradiction we assume that there exist µ n 0 0, µ n l 0, λ n 0 0 and u n Γ s µ 0,, λ 0 such that J k,n u n := λn 0 u n µ n 0 u n c c = k S K u n x Kx u n µ n l u n x a l i Then it is obvious that {u n } is bounded Denote u i n := u n R i ow let us consider u n in H0R Up to a subsequence, there exists l > 0 such that lim u n R n = lim Kx u n R n = l As in Lemma 3 in [8], we deduce that l = S lim K Kx u n R n = 0, K and then

22 5786 Q Guo et al / XYZ which implies a contradiction Lemma 58 Let 5, µ m l= kµ+ l < µ Assume that K 4,K 5,K 6, H hold and 0 < µ 0 < min{ µ 0, µ 0 },0 µ L < µ L, < µl l =,,m, l L,0 λ 0 < min{ 0 λ 0, λ 0, λ 0 } Then for all u s µ 0,, λ 0, there exist ρ u > 0 and a differential function f : B0, ρ u H0 k R such that f0 = and fwu w s µ 0,, λ 0, w B0, ρ u oreover, for all v H0 k, f 0, v = uv u v µ 0 x u µ 0 u x m uv x a l i m u x a l i λ0 uv Kx u uv u Kx u λ0 Proof The proof is standard and we sketch it here For u s µ 0,, λ 0, define a function F : R H0 k R by F ut, w := J ktu w, tu w = t u w u w µ 0 t m Then F u,0 = J k u, u = 0 and d dt F u,0 = u u µ 0 x 0 λ 0 u u w x a l i t λ 0 Kx u w Kx u By using the implicit function theorem the results follow ow we prove the main result in this part x u w t u x a l i Proof of Theorem 5 Let 0 < µ 0 < µ0 := min{ 0 µ 0, µ 0, µ 0 },0 µ L < µl := min{ 0 µ L, µ L }, < µl l =,,m, l L,0 λ 0 < λ0 := min{ 0 λ 0, λ 0, λ 0 }, where 0 µ 0, µ 0, µ 0, 0 µ L, µ L, µl l =,,m, l L, 0 λ 0, λ 0 are given in Lemmas 5, 56 and 57 Let {u n } H 0 k be a minimizing sequence for J k in s µ 0,, λ 0, that is, J k u n m s as n We assume u n 0 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 0,, λ 0

23 Q Guo et al / XYZ Choose 0 < 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 H0 k and v H 0 =, then v ρ B0, ρ n By using Lemma 58, we get w ρ = f n v ρ u n v ρ s µ 0,, λ 0 As in Theorem 33 in [8], it follows J k u n 0 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 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 > 0, set C δ K := {x distx, CK δ } We need the following K 5 b R {0} for every b CK ote that if b = b,0 CK, then b i := e πi /k b,0 CK for every i k K 6 There exists α 5 > such that if b CK, then Kx = Kb i + O x b i α5 as x b i, for every i k K 7 There exist R 0 and d 0 > 0 such that B0, R 0 and sup x, x >R0 Kx K d 0 Set k µ 0,, λ 0 := {u k µ 0,, λ 0 : J k u < c}, where c is given in Lemma 5 Theorem 59 Let hold and 5, δ > 0, µ m l= kµ+ l < µ If K 4,K 5,K 6,K 7,H a l b > 0, for every b CK, i then there exist µ 0 > 0, > 0 l =,,m, λ 0 > 0 such that for all 0 < µ 0 < µ 0,0 µ L < µ L, < l =,,m, l L,0 λ 0 < λ 0, the problem P λ0,k admits at least Cat Cδ KCK positive solutions which are Z k SO -invariant The proof of the above theorem depends on some lemmas Lemma 50 Let µ m l= kµ+ l < µ,0 < µ 0 0 µ 0,0 µ L 0 µ L where 0 µ 0, 0 µ L are constants appearing in Lemma 5 and K 4 hold If {u n } k µ 0,, λ 0 satisfies J k u n c < c, J k k µ 0,,λ 0u n 0, then {u n } has a converging subsequence in H0 k Proof By using Lemmas 4 and 5, the result follows as Lemma 4 in [8] Denote i := R i \ R i

24 5788 Q Guo et al / XYZ Lemma 5 Let 5, µ m l= kµ+ l < µ Assume that K 4,K 5,K 6 hold and a l b > 0, for every b CK i Then there exist 0 µ 0 > 0, 0 λ 0 > 0 such that if 0 < µ 0 < 0 µ 0,0 λ 0 < 0 λ 0, it holds k µ 0,, λ 0 oreover, for any µ n 0 0, µ n l 0, λ n 0 0 as n and {v n } k µ n 0, µ n l, λn 0, there exist x i n := x i, n, x n i and r n R + such that x i n x i 0 = x i, 0,0 CK, r n 0 and S vn i 4 x i u n r 0 in D, i, as n, i =,,,k, K r n where vn i = v n i, x i n = e π /k x i, n, x x i u n C r,rn r = r n r + xi n r n n, x i 0 = e π /k x i, 0,0, and with C r,rn the normalizing constant such that u r xi n r n = Proof Using the arguments of Lemma 56, it is easy to get k µ 0,, λ 0 ow we prove the second part Consider vn in H0R and set as in Lemma 3 in [8] that lim n Then l = S lim n R v n = lim n K and hence R K Kx v n R Kx v n = 0 Set w n x := v n vn, then w nx = and w n x S R Therefore w n x =, w n x S = l By using Corollary 4 in [8] and a similar proof to Lemma 4 in [8], there exist x n and r n R + such that x n x 0 C δ K, r n 0 as n and x w n u n r 0 in D,, as n r n Hence S vn K 4 x u n r 0 in D, r n

25 Q Guo et al / XYZ By recalling the symmetry of v n, we end the proof To continue, as in [8] we define { x if x < R0, ξx := x R 0 if x R 0 x For any 0 u H0 k, set Θu := ξx u u From the proof of Lemma 56, it is known that for any 0 u H0 k, t µ0,,λ 0 uu k µ 0,, λ 0 with t µ0,,λ 0 u = u u µ 0 x λ 0 u Kx u m u x a l i 4 For b = b,0 R {0}, define Ψ k : H 0 k as Ψ k bx = t µ0,,λ 0 U µ0,,λ 0xU µ0,,λ 0x := t µ0,,λ 0 U µ0,,λ 0x, where U µ0,,λ 0x := k i= ϕx Uµ0,,λ 0 0 x b i H 0 k, b i = e πi /k b,0 and ϕx, a radial function, satisfying 0 ϕ, ϕ = if x ϕ 4 r k i= B b i, r, ϕ = 0 if x k Bb i, r, with r > 0 small enough, and µ 0,, λ 0 0 as µ 0 0, λ 0 0, 0 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 > 0, λ 0 0, then there exist µ 0, µ L such that J k Ψ k b = max t>0 J ktu µ0,,λ 0 < c for all 0 < µ 0 µ 0, 0 µ L µ L i= By Lemma 5 we see that if 0 < µ 0 µ 0,0 µ L µ L, then Ψ k b k µ 0,, λ 0 It is also obvious that J k Ψ k b = c + o as µ 0 0, 0, λ 0 0 and there exist c > 0, c > 0 such that c < t µ0,,λ 0 U µ0,,λ 0x < c for all b CK

26 5790 Q Guo et al / XYZ Lemma 53 Let all b CK, it follows Ψ k b d µ = 5, µ m l= kµ+ l < µ, b CK and K 4,K 5,K 6 hold For i= as µ 0 0, 0, λ 0 0 S K δ bi, Ψ k b d ν = i= S K Proof ote that Ψ k b is bounded in H0 k and Ψ k b = t µ0,,λ 0 ϕx U µ0,,λ 0 0 x b i, lim n i= Kx Ψ k b = Bb i,r Bb i,r i= Bb i,r δ bi, Kx t µ0,,λ 0 ϕx U µ0,,λ 0 0 x b i Assume µ n 0 0, µ n l 0, λ n 0 0 as n Up to a subsequence, we get the existence of l > 0 such that t µ n 0,µ n l,λn 0 ϕx Uµn 0,µn l,λn 0 0 x b i = lim n = l Then l = S lim n K Set w i x := Bb i,r Bb i,r Bb i,r and Kx t µ n 0,µ n l,λn 0 ϕx Uµn 0,µn l,λn 0 0 x b i K Kx t µ n 0,µ n l,λn 0 ϕx Uµn 0,µn l,λn 0 0 x b i = 0 t n µ n 0,µn l,λn 0 ϕx Uµ 0,µn l,λn 0 0 x b i t µ n 0,µ n l,λn 0 ϕx Uµn 0,µn l,λn 0 0 x b i w i x S 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, then w i x = and Lemma 54 Let 5, µ m l= kµ+ l < µ and K 4,K 5,K 6,K 7 hold For µ 0 0, 0, λ 0 0, ΘΨ k b = b + o, uniformly for b B0, R 0 CK; sup{distθu, CK : u k µ 0,, λ 0 } 0 Proof Let b B0, R 0 CK By Lemma 53, we have ΘΨ k b = ξx Ψ kb Ψ = ξxd µ kb d µ + o = b + o, as µ 0 0, 0, λ 0 0

27 Q Guo et al / XYZ Take µ n 0 0, µ n l 0, λ n 0 0 as n and {v n } k µ n 0, µ n l, λn 0 By Lemma 5, there exist x i n = x i, n, x n i and r n R + such that x i n x i 0 = x i, 0,0 CK, r n 0 and S vn i 4 x i u n r 0 in D, i, as n, i =,,,k K r n Since Θu is continuous, then Θv n = ξx v x x i n n i ξx u r r n i v = n ur x x i n + o = ξx0 + o i r n oticing that x i 0 B0, R 0 leads to ξx i 0 = x i 0, 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 > 0, by using Lemmas 5 and 54, there exist µ 0 > 0, > 0 l =,,m, λ 0 > 0 such that for all 0 < µ 0 < µ 0,0 µ L < µ L, < l =,,m, l L,0 λ 0 < λ 0, we have Ψ k b k µ 0,, λ 0 for any b CK and ΘΨ k b b < δ, b B0, R 0 CK, and Θu C δ K, u k µ 0,, λ 0 Define Ht, x := x + tθψ k b b with t, x [0,] CK Then H is continuous, H[0,] CK C δ K and Θ H is homotopic to the inclusion CK C δ K By Lemma 50, it is enough to prove Cat k µ 0,, λ 0 Cat Cδ KCK, which can be obtained as Theorem 45 in [8], and therefore the problem P λ0,k admits at least Cat Cδ KCK 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 [30], 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 0 0 and the domain is bounded, there exist nontrivial solutions for the problem P λ0,k However, if we consider the problem in R, then λ 0 0 may lead to the nonexistence of nontrivial solutions Please see Appendix for a nonexistence result

28 579 Q Guo et al / XYZ 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, A 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 > 0, x, K 0, for ae x R ; λ < 0, x, K 0, for ae x R ; 3 λ = 0, x, K > 0, for ae x R, or λ = 0, x, K < 0, for ae x R Then if u W,p R is a weak solution for problem A, there holds u 0 Proof Denote fx, u = µ u p u x + λ u q u + Kx u p u Then problem A can p be rewritten as { p u = fx, u in R, u W,p R A It is easy to see that, for any ω R \ {0}, there exists Cω > 0 such that fx, u Cω + u p, x ω, u R Then as Claim 55 in [], u C R \ {0} W, loc R \ {0} 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 0, ht t,0 h Given > 0 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 C0 R \ {0} 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 = 0 R Then by the Hardy inequality, the Sobolev inequality and the Hölder inequality, λ η u p = Kxη u p R R + u p u u p η u µη R x p

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