Multiple Solutions for an Elliptic Problem Related to Vortex Pairs

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1 Wright State University CORE Scholar Mathematics and Statistics Faculty Publications Mathematics and Statistics Multiple Solutions for an Elliptic Problem Related to Vortex Pairs Yi Li Wright State University - Main Campus, yili@csunedu Shuangjie Peng Follow this and additional works at: Part of the Applied Mathematics Commons, Applied Statistics Commons, and the Mathematics Commons Repository Citation Li, Y, & Peng, S 011 Multiple Solutions for an Elliptic Problem Related to Vortex Pairs Journal of Differential Equations, 50 8, This Article is brought to you for free and open access by the Mathematics and Statistics department at CORE Scholar It has been accepted for inclusion in Mathematics and Statistics Faculty Publications by an authorized administrator of CORE Scholar For more information, please contact corescholar@wwwlibrarieswrightedu

2 Multiple Solutions to an Elliptic Problem Related to Vortex Pairs Yi Li Department of Mathematics, the University of Iowa Iowa City, IA, 54 USA Shuangjie Peng School of Mathematics and Statistics Central China Normal University Wuhan, P R China sjpeng@mailccnueducn 1 Introduction We consider the problem 11 { u λu ϕ p 1, x, u 0, x, where is a bounded domain in R N N, λ R N, < p < N/N for N 3 and < p < for N, ϕ is a positive harmonic function in Problem?? is related to steady vortex pairs: a steady vortex ring corresponding mathematically to a Stokes stream function Ψ defined on a bounded domain, and an open set A R N called the cross-section of a steady vortex ring and unknown a priori The function Ψ C 1 C \ A and the open set A satisfy the following free-boundary problem in cylindrical coordinates see, for example, [?] { λfψ, x A, 1 LΨ 0, x \ A, 13 Ψ A 0, Ψ 1 B x N 1 κ < 0, where L 1/r N / rr N / r / x N, x x 1,, x N 1 The vorticity function ft is positive if t > 0 and is zero if t 0, while B > 0 and κ are prescribed constants In general, Problems?? and?? can be written as the following free boundary problem { λfψ, x A, 14 Ψ 0, x \ A, 1

3 15 Ψ A 0, Ψ ϕ 0 < 0, where ϕ 0 is a C 1 function defined on Let ϕ be the solution of 16 { ϕ 0, x, ϕ ϕ 0, x Then, ϕ > 0 achieves its maximum and minimum on Let u Ψ ϕ and A {x : u > ϕ}, then Problem?? and?? become 17 { u λfu ϕ, x, ϕ 0, x In this paper, we investigate Problem?? to obtain its solution pairs u λ, A λ for λ sufficiently large, where A λ {x : u λ > ϕ} There are many existence results for Problem?? under various assumptions In [?,?,?,?,?,?], the solutions were obtained by using the mountain pass lemma for various nonlinearities fx, u and any λ > 0 In [?,?,?,?,?], to find the solutions, the constrained variation methods were used, but the vorticity function f is unknown a priori Moreover, in [?,?,?], the solutions were obtained by regarding λ as eigenvalue, so λ is not arbitrary The asymptotic behavior of the solution pair u λ, A λ of Problem?? was investigated in [?,?,?,?,?,?,?] More precisely, it is verified that the cross-section A λ of a steady vortex ring shrinks to a point, and a vortex ring degenerates into a singular vortex circle as λ Moreover, the Stokes stream function Ψ λ of a vortex ring converges to the Stokes stream function of the filament, which is the Green s function of the operator in In [?], Ambrosetti and Yang discussed the existence of solutions by using the mountain pass lemma and investigated the asymptotic behavior of the solution pair by estimating the upper bound of the critical value of the functional corresponding to Problem?? Recently, Li, Yan and Yang[?] studied Problem?? in the case N More precisely, they proved that the vortex core A λ shrinks to point x 0 which is exactly the minimum point of ϕ 0 on the boundary as λ Furthermore, they verified that u λ λ tends to the Green s function of both in W 1,p and C 1,α fu λ ϕ loc \ x 0 as λ To obtain their results, Li, Yan and Yang gave very delicate estimates to the upper and lower bounds of the least energy solution In [?], the limiting behavior of the mountain pass solution to Problem?? with N 3 and ϕ 1 was given We want to point out that most of the solutions mentioned here are in some sense the least energy solutions and the vortex core shrinks to a single point In this paper, we want to find some high energy solutions whose vortex core consists of multiple components which shrink to distinct points on as λ In the paper, we assume that the harmonic function ϕ C 1 satisfies that H ϕ has k k 1 strictly local minimum points z 1,, z k

4 Our main result is Theorem 11 Let ϕ satisfy H Then exists λ 0 > 0 such that, for any λ [λ 0,, Problem?? has solution pair u λ, A λ satisfying that i the vortex core A λ has exactly k components A λ,j, j 1,, k which shrink to the points z 1,, z k respectively as λ Moreover, A λ,j is approximately a ball and diama λ,j φ 1 N ϕ z j φ p 1 ϕ z j p λ 1, N 3 λ 1 ln λ p, N, where φ is the function defined by??; ii u λ has exactly k local maximum points z 1 A λ,1,, z k A λ,k which satisfy z j z j Oλ N N 1, distzj, Oλ N N 1, N 3 ln ln λ 1 z j z j O, distz j, O, N, ln λ ln λ c where j 1,, k, c > 0 is a constant Let us outline the proof of the main result of this paper The solutions in [?] were obtained by finding a minimizer of the corresponding functional in a suitable function space These method is hard to obtain solutions whose vortex core has several components In the present paper, we will use a reduction argument to this kind of solutions To apply a reduction argument to prove Theorems??, we need to construct an approximate solution for?? In the case N 3, we can scale the solution to the corresponding limit problem But, in the case N, the corresponding limit problem in R has no bounded solution So, we will follow the method in [?] to construct an approximate solution We should point out here, as we can see later, in the both two cases, since the local maximum points of the solution approach the boundary, the regular part of the green s function tends to infinity, which causes a difficulty to us We must take this influence into careful consideration and give a very exact estimate on the distance between the maximum points of the solutions and the boundary when we construct the approximate solutions Compared with the known results, our result provides an exact descripion to the profile of the solution pair u λ, A λ This paper is organized as follows In section, we construct the approximate solution for?? We will carry out a reduction argument in section 3 and the main results will be proved in section 4 and section 5 3

5 Approximate solutions and main result It is convenient to change?? to an equivalent problem Let ε 1, that is, we consider λ the following problem { ε u u ϕ p 1, in, 1 u 0, on In this section, we will construct the approximate solution for?? Firstly, we consider the case N 3 For c > 0, the following equation u u c p 1, x R N, u0 max ux, y R N ux 0, as x, has a unique solution W c x W c x in D 1, R N, which can be written as R p c φ x [ ] c, x Rc φ p 1, Nc W c R c crc N x N, x R c, where φx φ x is the unique solution of 3 φ φ p 1, φ > 0, φ H 1 0B 1 0 From Flucher and Wei[?], we see that W c is also non-degenerate This is, the kernel of the operator Lv v p 1W c c p v, v D 1, R N is spanned by { Wc x 1,, Wc x N } Note the operator L is not non-degenerate in the space of bounded functions For any z, define W ε,z,c x W x z c ε Because Wε,z,c does not belong to H0, 1 we need to make a projection Let P W ε,z,c satisfies { ε v W ε,z,c c p 1, x, 4 v 0, x Then 5 P W ε,z,c W ε,z,c ε N crc N hx, z, where hx, z is the regular part of the Green s function in and satisfies { h 0, in, h x z N, on 4

6 Now, we consider the case N Since 1 has no solutions in this case, we use the approximate solution in [?] Let R > 0 be a large constant, such that for any x, B R x Consider the following problem { ε u u a p 1, in B R 0, 6 u 0, on B R 0, where a > 0 is a constant Then,?? has a unique solution U ε,a with the form a ε /p s /p ε φ x, x sε 7 U ε,a x s ε a ln x R / ln s ε R, s ε x R, where φ solves??, and s ε 0, R satisfies 8 ε /p s /p ε φ 1 which implies Then a lns ε /R, s ε φ ε ln ε 1 p /> 0, as ε 0 p / a For any z, define U ε,z,a x U ε,a x z Let P U ε,z,a be the solution of { ε w U ε,z,a a p 1, in, w 0, on 9 P U ε,z,a U ε,z,a a ln R s ε gx, z, where gx, z satisfies { g 0, in ; It is easy to see that g ln R, x z on gx, z ln R πhx, z, where hx, z is the regular part of the Green s function in We will construct solutions for?? of the form P W ε,zε,j,c ε,j ω ε, N 3 or P U ε,zε,j,a ε,j ω ε, N, 5

7 where ω ε is a perturbation term To obtain an appropriately approximate solution, we need to choose suitable c ε,j and a ε,j Denote Z z 1,, z k R Nk In the sequel, we always assume that z j j 1,, k satisfies 10 z j z j < δ, d j dz j, γε N /N 1 N 3, or d j dz j, 1, N, ln ε α where γ, δ > 0 are small constants and α > 0 is a large number For i 1,, k, set 11 c ε,i ϕz i ε N ϕz i R N ϕz i hz i, z i and let a ε,i Z be the solution of the following problem 1 a i ϕz i a ε,i gz ln R i, z i ε j i where Gx, z j ln x z j R 13 a ε,i Z 1 ln R ε a ε,j Gz i, z j, gy, z j We can solve?? and find ϕz i 1 ln R ε j i a ε,jzgz i, z j 1 gz i,z i ln R ε Moreover, c ε,i and a ε,i Z are smooth in z j, j 1,, k Define 14 P ε,i P W ε,zi,c ε,i and V ε,z,j P U ε,zj,a ε,j Z Set s ε,i to be the solution of then, we see ε /p s /p ε φ 1 a ε,i lns ε /R, 1 lnr/s ε,i 1 ln ln ε lnr/ε O ln ε Now, using the facts see [?] and [?] that for j 1,, k, C, N 3, hz j, z j d N j C ln d j, N, 6

8 hz j, z j z j,l C d N 1 j we find that for any fixed constant L > 0, P ε,i x ϕx, l 1,, N, N 3, C d j, l 1,, N, W ε,zi,c ε,i ϕz i ε N ϕz i R N ϕz i hz i, z i ϕx ϕz i ε N ϕz i R N ϕz i Dhx, z i, x z i Oε N/N 1 W ε,zi,c ε,i c ε,i Oε, for x B Lε z i, V ε,z,i x ϕx U ε,zi,a ε,i Zx ϕx a ε,i ln R s ε,i gx, z i U ε,zi,a ε,i Zx ϕz i a ε,i gz ln R i, z i Os ε,i ln ε α 1 Os ε,i s ε,i U ε,zi,a ε,i Zx ϕz i a ε,i gz ln R i, z i O ε O ε ln ε α p/ U ε,zi,a ε,i Zx ϕz i a ε,i gz ln R i, z i O ε and for j i, by?? ln ln ε gz ln ε i, z i ln ln ε, for x B ln ε Lsε,i z i, P ε,j x W ε,zj,c ε,j ε N ϕz j R N ϕz i hx, z j Oε N, for x B Lε z i, V ε,z,j x U ε,zj,a ε,j Zx a ε,j ln R s ε a ε,j gy, z j a ε,j ln R s ε Gy, z j So, we obtain a ln R ε,j Gz i, z j 1 a s ε ln R ε,j DGzi, z j, y z i O ε ln ε p 1 s ε 1 ln ln ε a ε,j Gz i, z j O, for x B ln ε Lsε,i z i ln R ε P ε,j x ϕx W ε,zi,c ε,i x c ε,i Oε, x B Lε z i, ln ln ε V ε,z,j x ϕx U ε,zi,a ε,i Zx a ε,i Z O, x B ln ε Lsε,i z i From the above discussion, we see to prove Theorem??, we just need to prove the following theorem 7

9 Theorem 1 Let ϕ satisfy H, then there exists ε 0 > 0, such that Problem?? has a solution u ε with the form u ε P ε,j ω ε N 3, or u ε V ε,zε,j ω ε N, satisfying for j 1,, k ω ε L Oε 1 θ, z ε,j z j C 1 ε N N 1, distzε,j, C ε N N 1, N 3, or ln ln ε ln ln ε ω ε L O, z ln ε ε,j z j C, distz ε,j, 1, N, ln ε ln ε α where C 1, C, C and α 1 are positive constants, θ > 0 is any small constant To complete this section, we give the following lemma which will be used in our estimates Lemma There is a large L > 0 such that P ε,j x ϕx < 0, x \ V ε,z,j x ϕx < 0, x \ k B Lε z j k B Lsε,j z j Proof The case N can be proved as in [?] For the case N 3, we just need to choose L > 0 such that z max ϕzr N ϕz L N 1 z min ϕz 3 The reduction We only consider the case N 3 in this section since for the case N, the argument is similar and some basic estimates are given in [?] We recall that Z z 1,, z N and z j satisfies 31 z j z j < δ, d j : dz j, γε N /N 1, where γ > 0 is sufficiently small 8

10 and Let F ε,z { u L : P ε,j u 0, j 1,, k, h 1,, N }, E ε,z { u W, H 1 0 : P ε,j u 0, j 1,, k, h 1,, N } We define Q ε to be the projection from L to F ε,z as follows: Q ε u u N h1 b j,h ε P ε,j Hence the constants b j,h, j 1,, k, h 1,, N, satisfy 3 h1 b j,h ε P ε,j P ε,i u P ε,i z i,l z i,l Define We have L ε u ε u p 1 P ε,j ϕ p u Lemma 31 There exist constants ρ 0 > 0 and ε 0 > 0, such that for any ε 0, ε 0 ], Z satisfying??, u E ε,z with Q ε L ε u 0 in \ k B Lε z j, then Q ε L ε u L ρ 0 u L Proof We will use to denote L We argue by contradiction Suppose that there are ε n 0, Z n satisfying??, and u n E εn,zn with Q εn L εn u n 0 in \ k B Lεn z j,n, such that Q εn L εn u n 1 n, and u n 1 Firstly, we estimate b j,h,n in the following formula 33 Q εn L εn u n L εn u n p 1 h1 N b j,h,n Wεn,z j,n,c εn,j c εn,j p W εn,z j,n,c εn,j c ε n,j 9

11 For each fixed i, multiplying?? by P εn,i,n z i,l, noting that Q εn L εn u n belongs to F εn,z, we obtain P εn,i P εn,i u n L εn L εn u n z i,l z i,l N p 1 b j,h,n Wεn,z j,n,c c p εn,j ε n,j By??, we see 34 h1 Wεn,z j,n,c εn,j c ε n,j p Wεn,z j,n,c εn,j c ε n,j p c 0 δ jihl Oε n ε N n, W εn,z j,n,c εn W εn,z j,n,c εn,j W εn,z j,n,c εn,j c ε n,j P εn,i z i,l P εn,i z i,l where δ jihl 1 if j i and h l, and δ jihl 0 if otherwise On the other hand, using?? and Lemma??, we obtain ε n P ε n,i p 1 P εn,j ϕ z i,l p 1 Wεn,z i,n,c c p ε,i,n ε n,i,n p 1 Oε N n, B Lεn z i,n B Lεn z i,n which, together with?? and??, implies p P εn,i u n z i,l W εn,z i,n,c εn,i,n z i,l Wεn,z i,n,c ε,i,n c ε n,i,n Oε p c ε n,j P εn,i z i,l Oε N 1 n P εn,i u n z i,l c ε n,i,n un z i,h Therefore, b i,h,n O ε n h1 p 1 O b j,h,n ε n P ε n,j h1 h1 b j,h,n W εn,zj,n,c εn,j c εn,j p Wεn,z j,n,c εn,j b j,h,n ε 1 n O ε n 10 c ε n,j

12 Thus, we obtain L εn u n Q εn L εn u n O 1 ε n O n ε n For any fixed i, define Let Then As a result, ũ i,n y u n ε n x z i,n L n u u p 1 P εn,jε n x z i,n ϕε n x z i,n L n ũ i,n L εn u n p u L εn ũ i,n O 1 n ε n, where n { y : ε n x z i,n } Since ũ i,n 1, by the regularity theory on elliptic equations, we may assume that It is easy to check that ũ i,n u i, in C 1 locr N p P εn,jε n x z i,n ϕε n x z i,n W εn,zi,n,c εn,i ε n x z i,n c εn,i Oε n W ϕz0 ϕz 0 p, p where z 0 Hence, u i stisfies the following equation 35 u p 1W ϕz0 ϕz 0 p u 0 Now following the proof of [?], we can prove that any solution v to equation?? must belong to span W ϕz 0 x 1,, W ϕz 0 Hence x N 36 u i k1 i W ϕz0 k i W ϕz0 N x 1 x N 11

13 Since P εn,i z i,h un 0, we have W ϕz0 ϕz 0 p W ϕz0 u i 0, R z N h which, together with??, gives u i 0 Thus, for any L > 0 From Lemma??, the assumption gives u n 0, in C 1 B Lεn z i,n, Q εn L εn u n 0, in \ k i1b Lεn z i,n u n 0, y \ k i1b Lεn z i,n However, u n 0 on and u n o1 on B Lεn z i,n, i 1,, k So we have This is a contradiction u n o1 Remark 3 Lemma?? and the Fredholm alternative imply that Q ε L ε is one to one and onto from E ε,z to F ε,z Now consider the equation 37 Q ε L ε ω Q ε l ε Q ε R ε ω, where 38 l ε k P ε,j ϕ p 1 Wε,zj,c c p 1 ε,j ε,j and 39 R ε ω k P ε,j ϕ ω p 1 k P ε,j ϕ p 1 p 1 k P ε,j ϕ p 1 ω By Remark??,?? can be rewritten as 310 ω G ε ω : Q ε L ε 1 Q ε lε R ε ω The next proposition enable us to reduce the problem of finding a solution for?? to a finite dimensional problem 1,

14 Proposition 33 There is an ε 0 > 0, such that for any ε 0, ε 0 ]?? has a unique solution ω ε with ω ε Oε 1 θ, where θ is any small positive number Proof Define where Ẽ ε,z { u L : M Ẽε,Z { ω ε 1 θ }, P ε,j u 0, j 1,, k, h 1,, N } Then M is complete under the L norm and G ε is a map from Ẽε,Z to Ẽε,Z We claim that G ε is a contraction map from M to M Indeed, we only need to prove the following two facts 1 G ε maps from M into M For any ω M, similar to Lemma??, we can prove that for large L > 0, k V ε,z,j ϕ ω 0 in \ k B Lε z j Note also that for any u L, Therefore, we find that for any ω M, Q ε u u in \ k B Lε z j Q ε l ε Q ε R ε ω l ε R ε ω 0, in \ k B Lε z j So, applying Proposition??, we see Q ε L ε 1 Q ε l ε Q ε R ε ω C Q ε l ε Q ε R ε ω Thus, for any ω M, we have 311 G ε ω Q ε L ε 1 Q ε lε R ε ω C Q ε lε R ε ω Using the argument in??, we deduce from?? that b j,h, corresponding to u L, satisfies b j,h Cε N P ε,i u i, l z i,h Since l ε R ε ω 0, in \ k B Lε z j, 13

15 we find As a result, b j,h Cε N i, l Cε l ε R ε ω B Lε z j P ε,i lε R ε ω z i,l Q ε l ε R ε ω l ε R ε ω C j, h b j,h W W ε,zj,c ε,j c ε,j p ε,zj,c ε,j c ε,j C l ε R ε ω It follows from?? and?? that l ε Oε W ε,zj,c ε,j c ε,j p Oε N 1 Oε For the estimate for R ε ω, noting the boundedness of W ε,zj,c ε,j c ε,j, we see As a consequence, R ε ω C ω 31 Gω Cε ω ε 1 θ Thus, G ε is a map from M to M G ε is a contraction In fact, for any ω i M, i 1,, we have Noting that G ε ω 1 G ε ω Q ε L ε 1 Q ε Rε ω 1 R ε ω we can deduce as the proof of Fact 1 that R ε ω 1 R ε ω 0, in \ k B Lsε,j z j, G ε ω 1 G ε ω C R ε ω 1 R ε ω C ω 1 ω ω 1 ω 1 Cε 1 θ ω 1 ω 1 < 1 ω 1 ω 1 Now we have proved that G ε is a contraction map from M to M By the contraction mapping theorem, there is an ω ε M, such that ω ε G ε ω ε Moreover, it follows from?? that ω ε ε 1 θ 14

16 4 Proof of Theorem?? in case N 3 In this section, we will choose Z, such that k P ε,j ω ε is a solution of??, where ω ε be the solution to?? obtained in Proposition?? Define Iu ε Du 1 u ϕ p p, and KZ I k P ε,j ω ε It is well known that if Z is a critical point of KZ, then k P ε,j ω ε, is a solution of?? Lemma 41 ω ε satisfies ε Proof ω ε satisfies??, so ω ε solves p 1 ε ω ε P ε,i ϕ ω ε Hence, by??, we see ε Dω ε Dω ε Cε N θ p 1 W ε,zi,cε,i c ε,i p 1 P ε,i ϕ ω ε N h1 p 1 W ε,zi,cε,i c ε,i ω ε b j,h ε P ε,j N b j,h ε P ε,j z j,h h1 p p 1 W ε,zi,cε,i c ε,i ωε Oε N1 ω ε Oε N3 3θ Oε N θ Lemma 4 Proof We have KZ ε D KZ I P ε,j O ε N θ P ε,j ε 15 DP ε,j Dω ε ε Dω ε

17 1 p I p P ε,j ϕ ω ε P ε,j ε Dω ε ε DP ε,j Dω ε P ε,j ϕ 1 p p P ε,j ϕ ω ε P ε,j ϕ p P ε,j ϕ p Employing Lemma?? and Proposition??, we see p p P ε,j ϕ ω ε P ε,j ϕ p P ε,j ϕ p 1 ω ε p p P ε,j ϕ ω ε P ε,j ϕ p P ε,j ϕ i1 B Lε z i O ω ε N O ε N θ It follows from Lemma?? and?? ε DP ε,j Dω ε P ε,j ϕ p 1 ω ε p 1 W p p 1 ε,zj,cε,j c ε,j ω ε P ε,j ϕ ω ε p 1 W p ε,zj,cε,j c ε,j ω ε P ε,j ϕ k i1 B Lεz i Oε ω ε ω ε Oε N θ B Lε z j B Lε z j p 1 p 1 W ε,zj,cε,j c ε,j Oε N ω p W ε,zj,cε,j c ε,j Oε N1 ω ω ε p 1 p 1 p 1 ω ε ω ε ω ε So we see KZ I P ε,j O ε N θ 16

18 Lemma 43 The function φ defined by?? satisfies φ p p B 10 N pn p φ 1, and B 1 0 where B 1 0 denotes the area of B 1 0 B 1 0 φ p 1 B 1 0 φ 1, Proof We can easily check the following Pohozaev identity N p N φ p 1 Dφ ν xds B 1 0 As a result, we see B 1 0 The other part is obvious, since φ p 1 B 1 0 Hence, we complete the proof Now we prove Theorem?? Proof of Theorem??: Define B 1 0 B 1 0 φ p p B 10 N pn p φ 1 φ φ B 1 0 ν ds B 10 φ 1 S {Z z 1,, z k k : z j z j δ, d j : dz j, τε N /N 1 }, where τ will be determined later Consider the problem min KZ Z S There exists a minimizer Z ε for KZ in S To complete the proof, it suffices to verify that Z ε is an interior point of S and hence is a critical point of KZ From Lemma??, we need to estimate I k P ε,j By??, c ε,i R N c ε,i ϕz i R N ϕz i OεN hz i, z i ϕz i R N ϕz i Oε We have ε D P ε,j i1 Wε,zj,c ε,j c ε,j p 1 P ε,i 17

19 i1 Wε,zj,c c p 1 ε,j ε,j Wε,zi,c c ε,i ε,i ϕz i Oε N hz i, z i Wε,zj,c c p ε,j ε,j ϕz j Wε,zj,c c p 1 ε,j ε,j OεN1 Wε,zj,c c p ε,j ε,j ϕz j Wε,zj,c c p 1 ε,j ε,j OεN1 B Rcε,j εz j Rc N p/p ε,j ε N φ p B 1 0 B Rcε,j εz j Rc N p 1/p ε,j ϕz j ε N φ p 1 Oε N1 B 1 0 Similarly, by Lemma?? and??, we see p P ε,j ϕ B Lε z j B Lε z j Wε,zj,c ε,j c ε,j Oε p R /p c ε,j x zj φ R cε,j ε Rc N p/p ε,j ε N φ p Oε N1 B 1 0 OεN p Oε Oε N Hence I k ε P ε,j D 1 1 p 1 P ε,j 1 p k Rc N p/p ε,j ε N φ p B 1 0 p P ε,j ϕ Rc N p 1/p ε,j ϕz j ε N φ p 1 Oε N1 B 1 0 Using R c [ ] φ p 1 Nc and??, we can check R N p/p c ε,j N φ 1 p p N/ϕzj p p N/ p N pn Ohz j, z j ε N N p p N/ϕzj p p N/ R N φ ϕz 1 j hz j, z j ε N 18

20 and R N p 1/p c ε,j ϕz j N p 1 p N/ϕzj p p N/ φ 1 p N pn Ohz j, z j ε N N p 1 p N/ϕzj p p N/ R N φ ϕz 1 j hz j, z j ε N Now combing Lemmas?? and??, we have then KZ I P ε,j O ε N θ N p 1 p N/ φ ε N 1 B φ 1 0 ϕz j p p N/ 1 N p pn 1 RN ϕz j hz j, z j ε N Oε N1 O ε N θ Let Zε zε,1,, zε,k S be such that As a consequence, KZ ε z ε,j z j d j ε N N 1, ϕz ε,j ϕ z j Oε N N 1, ε N hz ε,j, z ε,j Oε N N 1 ε N N p pn Oε N N 1 N φ 1 p N p 1 φ p N p 1 B 1 0 ϕ z j But, so, we deduce and Hence, we see KZ ε KZ ε, ϕz ε,j ϕ z j Cε N N 1, j 1,, k, hz ε,j, z ε,j Cε N N 1, j 1,, k z ε,j z j C 1 ε N N 1, distzε,j, C ε N N 1, j 1,, k, 19

21 which implies z ε,j z j C 1 ε N N 1, distzε,j, C ε N N 1, j 1,, k, where C 1 and C are independent of τ Therefore, Z ε is a interior point of S if we choose τ sufficiently small 5 Proof of Theorem?? in case N To investigate the case N, we use V ε,z,j, a ε,j Z and U ε,zj,a ε,j Z to replace P ε,j, c ε,j and W ε,zj,c ε,j in Section 3 respectively and proceeding as we have done in the case N 3 see also [?] for the details, we can prove Proposition 51 There is an ε 0 > 0, such that for any ε 0, ε 0 ]?? has a unique solution ω ε with ln ln ε ω ε O ln ε Moreover, Define Then we have ε ε Dω ln ln ε 4 O ln ε 4 MZ I k P ε,j ω ε Lemma 5 ε ln ln ε 4 MZ I V ε,z,j O ln ε 4 Proof By definition, MZ I V ε,z,j ε Dω ε ε DV ε,z,j Dω ε V ε,z,j ϕ 1 p p V ε,z,j ϕ ω ε V ε,z,j ϕ p V ε,z,j ϕ p Employing??, Lemma?? and Proposition??, we obtain p 1 p 1 ω ε ω ε 0

22 p p V ε,z,j ϕ ω ε V ε,z,j ϕ p V ε,z,j ϕ k i1 B Ls ε,i z i p 1 ω ε p p V ε,z,j ϕ ω ε V ε,z,j ϕ p V ε,z,j ϕ O ω ε ε ln ln ε 4 O, ln ε 4 p 1 ω ε and ε DV ε,z,j Dω ε V ε,z,j ϕ p 1 p 1 ω ε p p 1 U ε,zj,a ε,j Z a ε,j Z ω ε V ε,z,j ϕ ω ε p 1 p U ε,zj,a ε,j Z a ε,j Z ω ε V ε,z,j ϕ k i1 B Ls ε,i z i ln ln ε O ω ln ε ε ω ε ε ln ln ε 4 O ln ε 4 B Lsep,j z j B Lsε,j z j U ε,zj,a ε,j Z a ε,j Z U ε,zj,a ε,j Z a ε,j Z p p 1 p 1 ω ε So we see ε ln ln ε 4 MZ I V ε,z,j O ln ε 4 Proof of Theorem??: Define { D Z z 1,, z k k : z j z j δ, d j : dz j, where τ 1 and τ will be determined later Consider the problem min MZ Z D 1 1 ln ε, 1 τ ln ε τ 1 },

23 There exists a minimizer Z ε for MZ in D Now we verify that Z ε is an interior point of D and hence is a critical point of MZ Similarly to the case N 3, we have and ε D i1 V ε,z,j B sε,j z j i1 Uε,zj,a ε,jz a ε,j Z p 1 π aε,j Z p 1 aε,j Zs φ 1 p ε,j ln R/s ε,j O ε ln ε, So, noting that and p V ε,z,j ϕ B sε,j z j we obtain I k ε V ε,z,j As a result, MZ π Uε,zj,a ε,jz a ε,j Z p 1 V ε,z,i Uε,zi,c ε,i a ε,jz ln R/s ε,i gx, z i aε,j Z ps ln R/s ε,j gz j, z j ε,j W ε,zj,aε,j Z a ε,j Z O ln ln ε p O ε ε O ln ε ln ε ln ε a ε,j Z ϕz i 1 gz j, z j 1 ln R/ε O, ln ε ε /p s /p ε,j φ 1 π D V ε,z,j 1 p ϕ z j ε ln ε ϕ z j ε ln ε Let Z ε z ε,1,, z ε,k D be such that a ε,i lns ε,j /R, 1 p 1 gz j, z j ln R/ε 1 p 1 gz j, z j ln R/ε z ε,j z j d j 1 ln ε, p V ε,z,j ϕ O ε ln ln ε ln ε ε ln ln ε O ln ε

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