Multipli ity results for the two-vortexChern-Simons Higgs model on the two-sphereWeiyue Ding � J�urgen Jost y Jiayu Li z Guofang Wang xAbstra tWe onsider a Ginzburg-Landau type fun tional on S2 with a 6thorder potential and the orresponding selfduality equations. We studythe limiting behavior in the two vortex ase when a oupling parametertends to zero. This two vortex ase is a limiting ase for the Moserinequality, and we orrespondingly dete t a ri h and varied asymptoti behavior depending on the position of the vorti es. We exploit analo-gies with the Nirenberg problem for the pres ribed Gauss urvatureequation on S2.keywords: Ginzburg-Landau fun tional, �6 theory, Moser-Trudinger inequality,Nirenberg problem, phase transition, Chern-Simons Higgs theory1 Introdu tionFun tionals that exhibit a selfduality phenomenon in the sense that theabsolute minimizers satisfy a set of �rst order partial di�erential equationsare important in various areas of geometry and physi s.In the present paper, we investigate a spe ial lass of su h fun tionals,namely Ginzburg-Landau type fun tionals with a 6th order potential. Su hfun tionals arise in Chern-Simons Higgs theories, as will be explained in x2.We onsider a line bundle L over a ompa t Riemann surfa e �, and theLagrangian densityL(A;�) = jrA�j2 + k24 jF j2j�j2 + 1k2 j�j2(1� j�j2)2:Here, � is a se tion of L, and A is a unitary onne tion on L with urvatureF . k is a oupling parameter, and we are parti ularly interested in the�Institute of Mathemati s, A ademia Sini a, Beijing 100080, P. R. China.yMax-Plan k-Institute for Mathemati s in the S ien es, Inselstr. 22-26, 04103 Leipzig,Germany.zInstitute of Mathemati s, A ademia Sini a, Beijing 100080, P. R. China.xInstitute of Mathemati s, A ademia Sini a, Beijing 100080, P. R. China.1

limit analysis as k tends to 0. This limit analysis reveals a geometri allyinteresting phase transition that may also be relevant in super ondu tivity.The selfduality be omes manifest by rewritingL(A;�) = Z� L(A;�) = Z��j��A�j2 + ( kj�jF + 2k j�j(j�j2 � 1))2�+ Z� F:If � is ompa t, or, more generally, if one requires ertain de ay onditionsat in�nity, Z� F = 2�Nis a topologi ally quantity, where N is an integer, the so- alled vortex num-ber that �xes the number of zeroes of �, the vorti es. For N � 0, absoluteminimizers of L then have to satisfy the selfduality equations��A� = 0F = 2k2 j�j2(1� j�j2):As k ! 0, one expe ts that the minima of the potentialV (�) = j�j2(1� j�j2)2at j�j = 1 and � = 0 dominate the behavior of minimizers of L, ex ept thatthe topologi al onstraint Z� F = 2�N�xes the number of zeroes of � as well as the integral of F . One thus expe tsa solution � with j�j lose to one ex ept in the vi inity of N vorti es. In the ase were � is a torus, su h a solution has been onstru ted by Ca�arelli-Yang [CaY℄. One also expe ts a solution that approa hes 0. Su h a solutionwas re ently obtained in an interesting paper of Tarantello [T℄ in ase N = 1,again for a torus. While the methods employed in the proofs of those resultsextend to the ase of an arbitrary ompa t Riemann surfa e �, the method ofTarantello only works for N = 1, be ause it depends on the Moser inequality.(She does obtain a se ond solution for arbitrary N , but as we shall see inthe present paper, the limiting behavior will depend on N in general.) Herewe onsider the ase N = 2 on the sphere S2 . This ase is a limiting ase for the Moser inequality, and onsequently the analysis and the resultsbe ome more subtle than for N = 1. In fa t, one may rewrite the selfdualityequations by putting u(x) = log j�(x)j22

to obtain �u = 4k2 eu(eu � 1) + 4� NXj=1 Æpjwhere Æp is the Dira distribution on entrated at p, and p1, ..., pN are thepres ribed zeroes of �, not ne essarily all distin t. As will be explained inse tion 3, in our ase N = 2 this equation an be related to the pres ribedGauss urvature equation �u = �2Keu + 2:Thus, one expe ts that the methods developed for the Nirenberg problem,i.e. the existen e problem for that equation, be ome relevant (see se tion3for referen es). That is indeed the ase, and in the present paper we shallobtain two families of solutions depending on the oupling parameter k witha pre ise asymptoti behavior di�erent from the one of the Ca�arelli-Yangsolution. The only ex eption is the ase of a single vortex with multipli itytwo where the Kazdan-Warner equation prevents the existen e of a solutionof the limiting equation and where we only �nd one additional family. Su ha ase distin tion is not untypi al for limit ases of embedding theorems. Onthe other hand, if the two vorti es are antipodal, then an easy symmetryargument produ es one-parameter solution spa es, i.e. in�nitely many solu-tions for ea h suÆ iently small value of k. The ase of the torus has been in-vestigated in our ompanion paper [DJLW℄ and by Nolas o-Tarantello [NT℄.By their results, it may be possible that a solution with a blow-up of the urvature at a non-vortex point also exists for ertain onformal lasses oftori.In on lusion, the asymptoti analysis of the Chern-Simons Higgs fun -tional onsidered here is mu h ri her than the orresponding one for theGinzburg-Landau fun tional with a 4th order potential (j�j2 � 1)2. There,it was shown in [HJS℄ that asymptoti ally, as k tends to 0, � be omes a ovariantly onstant se tion of L with j�j=1, and the onne tion A be omes at, ex ept near the vorti es where all the topology on entrates. Solutionsof the type found by Tarantello and in the present paper do not o ur inthat model. This is somewhat similar to the situation in the Seiberg-Wittenfun tional that again has a 4th order nonlinearity where the limiting analy-sis was arried out by Taubes [T3℄. We expe t that a Seiberg-Witten typefun tional with a 6th order potential will exhibit very interesting features,partly analogous to the ones found in the present paper. We hope to beable to study this more losely. In fa t, we onsider the present analysis asa model study for that problem.A knowledgement: This paper was written at the Max-Plan k-Institutefor Mathemati s in the S ien es in Leipzig. W.Y.D thanks the Max-Plan k-institute for its hospitality and good working onditions. J.Y.L. was sup-3

ported by a fellowship of the Humboldt Foundation. G.F.W. was supportedby the DFG through the Leibniz award of J.J.2 The Chern-Simons Higgs modelLet S2 be the standard sphere in R3 with the standard metri g0, andM = R�S2 with the Lorentzian metri g = dx20� g0. Consider the (trivial)prin iple bundle M � U(1) ! M . Let A = �i A� dx�, A�(x) 2 R, x =(x0; x1; x2) 2M be a onne tion on this prin iple bundle. The urvature ofA is given by FA = �i2 F�;� dx� ^ dx�with F�;� = ��A� � ��A�; �; � = 0; 1; 2. The ve tor bundle asso iated toM � U(1) is M � C , where C is the omplex plane. Let �(x) be a se tionof the ve tor bundle M � C , i.e. �(x) is a Higgs �eld, in physi al notation.Let DA� denote D��dx� with D�� = ��� � iA��. In this paper, we areinterested in the following Chern-Simons-Higgs Lagrangian a tion densityL(A;�) = D��D��+ 14k"�� F��A � V (�)(2.1)where k > 0 is the oupling onstant whi h determines the strength of theChern-Simons term "�� F��A , V (�) is the potential and the Levi-Civitatensor "�� ; �; �; = 0; 1; 2 is �xed by "012 = 1. This Lagrange density was�rst introdu ed by Hong-Kim-Pa in [HKP℄ and Ja kiw-Weinberg in [JW℄.The Euler-Lagrange equations for (2.1) are( 12k"�� F�� = j = i(�D �� ��D �);�� = ��V (�)�� ;(2.2)where j is the onserved matter urrent density. We are interested in stati solutions of (2.2) with V (�) = 1k2 j�j2(1� j�j2).The energy density orresponding to the Lagrange density (2.1) isE = jD0�j2 + jD1�j2 + jD2�j2 + 1k2 j�j2(1� j�j2)(2.3)supplemented by the Gauss lawF12 = 2kJ0 = �2ik (�D0�� �D0�):(2.4)Let ��A� = D1 + iD2. We havejD1�j2 + jD2�j2 = j��A�j2 + F12j�j2 � 12"ik�ijk;4

Therefore, the energy density (2.3) may be written asE = 14 � kj�jF12 + 2k j�j(j�j2 � 1)�2 + j���j2 + F12 + Im f�j"jk ��Dk�g;where "jk = �"kj; j; k = 1; 2 and "12 = 1. Thus we obtain the followingenergy fun tionalE(A;�) = ZS2 E = ZS2 14 �� kj�jF12 + 2k j�j(j�j2 � 1)j2 + ZS2 j��A�j2 + ZS2 F12:(2.5)The absolute minimizers of E under the homotopi ally invariant onstraint12� ZS2 F12 = N(2.6)satisfy the Bogomolny type self-dual equations� ��A� = 0;F12 + 2k2 j�j2(j�j2 � 1) = 0;(2.7)with the Gauss law kF12 + 2A0j�j2 = 0 (see (2.4)). Here N is an integer.One an easily he k that a solution of (2.7) with the Gauss law satis�es(2.2). In this paper, we are interested in �nding su h spe ial solutions of(2.2).As in [CaY℄ and [T℄, one an �rst obtain a maximal solution as followsTheorem 2.1 ([CaY℄) Let p1; :::; pm be given points (or vorti es) on S2and n1; :::; nm positive integers su h that Pmj=1 nj = N � 0. There exists ak 2 (0; 12pjS2j=�N) su h that (2.7) admits a solution (Ak; �k) for whi hp1; :::; pm are the zeroes of � with multipli ity n1; :::nm if and only if 0 <k � k . Moreover(i) The energy, magneti ux and ele tri hange of (Ak; �k) are respe -tively given by E = 2�N;� = 2�N;Q = 2�kN(2.8)(ii) The solution (Ak; �k) is maximal in the sense that if (A0; �0) is anothersolution of (2.7) with the same vorti es as (A;�), then j�0j < j�j.(iii) j�kj < 1 in S2 and j�kj ! 1 as k ! 0 a.e. in S2 and in H1;q(S2); 1 <q < 2. F (k)12 ! 2� NXj=1 Æpj in the sense of measures as k ! 0:(2.9)where ea h Dira distribution Æpj o urs with multipli ity nj; j = 1; :::;m.5

One an also obtain another solution by using the mountain pass Lemmaas [T℄. Here we are interested in solutions of (2.7) with a di�erent asymptoti behavior when k ! 0. Motivated by Ca�arelli-Yang's variational method,when N = 1, Tarantello obtained in [T℄,Theorem 2.2 There exists a solution ( ~Ak; ~�k) of (2.7) for small k > 0 su hthat (2.8) holds and k~�kkCq(S2) ! 0 as k ! 0 for any q � 0.Although they did not onsider (2.7) on S2, the methods of Ca�arelli-Yang and Tarantello extend to this ase.Tarantello used the Moser inequality [M1,2℄ to study this problem. Herewe onsider the ase N = 2. As we already mentioned in the introdu tion,this ase is a limiting ase (a riti al ase). The main diÆ ulty to �ndsolutions of (2.7) is the la k of a oer ivity ondition. A ru ial observation isthat our problem an be seen as a perturbation of the well-known Nirenbergproblem. Hen e, methods developed in the Nirenberg problem may be usedto study (2.7). First we obtain a solution of (2.7) whi h has a new asymptoti behavior as k ! 0.Theorem 2.3 Let N = 2 and P� two vorti es. For small k > 0, thereexists a solution (A2k; �2k) su h that(i) (2.8) holds,(ii) j�2kj ! 0 in C0 uniformly,(iii) F12(A2k)! 4�ÆQ,where Q 6= P� is determined by P� (see se tion 3.). Moreover, if P+ = �P�there exists a family of solutions (A2k(#); �2k(#)) su h that (i), (ii) and (iii)hold with T#Q, where T# is the rotation with angle # about the axis from P+to P�.This is a new interesting situation. We guess that su h a solution exists inthe general ase.Theorem 2.4 Let N = 2 and P� two vorti es. If P+ 6= P�, then for smallk there exists another solution (A3k; �3k) of (2.7) su h that(i) (2.8) holds,(ii) �3k ! 0 in Cq, as k ! 0, for any q � 0.The potential j�j2(j�j2 � 1)2 has a minimum at j�j = 1 and at � = 0.The solution of Theorem 2.1 orresponds to the minimum at j�j = 1, theone of Theorem 2.3 to the one at � = 0, while the solution of Theorem 2.4 isa saddle point solution for an asso iated fun tional. Of ourse, the vorti esprevent that j�j = 1 or � � 0 are exa t solutions, but in the limit k ! 0,the obstru tions on entrate at isolated points. A ording to the theorems,for N = 2, we have 3 di�erent ases for small k.6

(1) If P+ = P�, (2.7) admits at least two solutions,(2) If P+ = �P�, (2.7) admits in�nitely many solutions,(3) If P+ 6= �P�, (2.7) admits at least three solutions.It is lear that ase (3) is the generi ase. Before we start to prove thetheorems, we �rst redu e (2.7) to a semilinear equation. Su h a redu tionwas �rst used by Taubes in [T1℄, [T2℄.It is lear that the �rst equation of (2.7) may be written as2���� iA� = 0;(2.10)where A = A1 + iA2 and �� = 12(�1 + i�2) is the usual Cau hy-Riemannoperator. Hen e � an be onsidered as a holomorphi se tion of a linebundle, and it therefore admits a �nite number of zeroes in S2 with integermultipli ities. Outside the zero set of �;Z(�) we haveA = �2i�� log �:(2.11)Set u(x) = log j�(x)j2. From (2.7) and (2.11) u satis�es�u = 4k2 eu(eu � 1) in S2 n Z(�)(2.12)and u(z) = nk log jz � Pkj2 as z ! Pk:(2.13)On the other hand, if we have a solution u of (2.12)-(2.13), set�(z) = exp 0�12u(z) + i NXj=1 arg(z � Pj)1Aand A = �2i�� log�, then one an he k that (A;�) satis�es (2.7). Therefore,we only have to onsider (2.12) and (2.13). Clearly (2.12)-(2.13) is equivalentto �u = 4k2 eu(eu � 1) + 4� NXj=1 ÆPj ;(2.14)where ÆP is the Dira distribution on entrated at P .7

3 Proof of Theorem 2.3Let P+ and P� be two vorti es on S2. Let u0 be the unique solution of� �u0 = �2 + 4�(ÆP+ + ÆP�); in S2RS2 u0 = 0:(3.1)Let � = 4=k2 and K = eu0 . (2.12) is equivalent to�u = �Keu(Keu � 1) + 2 in S2:(3.2)We �rst summarize some simple properties of K in three di�erent ases.Lemma 3.1 ase (i): P+ = �P�. After a hange of oordinates, we mayassume that P+ is the north pole. Then K is axially symmetri , i.e.invariant under rotations about the axis between north and south pole,i.e. between P+ and P�, as well as invariant under re e tions aboutthe equator of S2, i.e. K(x) = K(�x) for all x 2 S2. K a hieves itsmaximum for any point on the equator. ase (ii): P+ = P�, i.e. P+ is a double vortex. K is again axially sym-metri about the line between P+ and �P+. It a hieves its uniquemaximum at �P+. ase (iii): P+ 6= �P� ( the generi ase). K has a unique maximum point�Q 6= P+; P� and a unique saddle point Q = � �Q. �Equation (3.2) is the Euler-Lagrange equation of the following fun tionalI�(u) = Z 12 jruj2 + 2u+ �2 (Keu � 1)2;where R u is the average of u over S2, i.e. R u = 14� RS2 u. As in [T℄,motivated by the variational method used in [CaY℄, we onsider the followingfun tional J�(u) = Z (12 jruj2 + 2u+ �2 (Ke(u+�(u)) � 1)2)(3.3) +2�(u)� �2 + 2 log �in A� = (u 2 H1;2(S2)�� Z eu = 1 & (Z Keu)2 � 8� Z K2e2u � 0) ;(3.4) 8

where �(u) = log0BB�R Keu �r(R Keu)2 � 8� R K2e2uR K2e2u 1CCA :The term ��2 +log � ensures that J� has a uniform lower bound (see Lemma3.4 below). This value of �(u) is needed to satisfy the onstraint that omesfrom integrating (3.2). Alternatively, this value of �(u) is determined byminimizing J� among fun tions of the form u+ � w.r.t. � for given u satis-fying R eu = 1.Set H = fu 2 H1;2(S2)j R eu = 1g.Lemma 3.2 If u 2 ÆA�, the interior of A�, is a riti al point of J�, thenv = u+ �(u) is a solution of (3.2).Proof. The proof is straightforward ( f. [T℄). �A ru ial observation is that we may rewrite J� in a suitable form as follows.By the de�nition of �, we have�e�(u) Z Keu = 4 R KeuR Keu +r(R Keu)2 � 8� R K2e2u :(3.5)Consequently, 2 < �e�(u) Z Keu � 4; for any u 2 A�:(3.6)Set B�(u) = �e�(u) R Keu. Again, by de�nition, we have�Z K2e2(u+�(u)) � �Z Keu+�(u) = �2:Thus, we an rewrite J� as follows (deleting an irrelevant additive onstant)J�(u) = Z (12 jruj2 + 2u)� 2 log Z Keu + 2 logB�(u)� 12B�(u)Set �� = infu2A� J�(u). In this se tion, we shall prove that �� is a hievedby some u� 2 ÆA�. For simpli ity of notation, let f(t) = 2 log t � 12 t andf� = f Æ B�. Then J� is written asJ� = J + f�;9

where J(u) = Z (12 jruj2 + 2u)� 2 log Z Keu:The orresponding Euler-Lagrange equation of J is given by�u = �2Keu + 2;(3.7)whi h is the so- alled pres ribed Gauss urvature equation. The orrespond-ing problem of existen e of solutions of (3.7) is alled the Nirenberg problem.This problem has been studied by many mathemati ians. (See [M2℄, [A℄, [H℄,[CD℄, [CY1,2℄ and [CkL℄ and referen es therein.) J� an be onsidered as aperturbation of J for large �. So it is natural to apply methods developedfor the Nirenberg problem in our problem.Now let us �rst introdu e the de�nition of the enter of mass of a fun tionu 2 H1;2(S2) whi h was �rst used in [CD℄ in the Nirenberg problem. Foru 2 H1;2(S2), the enter of mass is de�ned asP (u) = RS2xeuRS2eu :Given q 2 S2, we hoose oordinates x = (x1; x2; x3) 2 S2 su h that q =(0; 0; 1). The stereographi proje tion � : S2 ! C = C [ f1g with respe tto q is de�ned by (x1; x2; x3) 7! z = x1 + ix21� x3 :For t > 0, let mt : C ! C be the usual multipli ation by t, i.e mt(z) = tzfor any z 2 C . For any u 2 H1;2(S2), there is (q; t) 2 S2 � [1;+1) andw = u'q;t := u Æ 'q;t + q;t:su h that P (w) = 0, where 'q;t = ��1 Æmt Æ � and q;t = log det(d'q;t).(Note that our notation di�ers slightly from the one in [CY1℄ and [CkL℄.)In [CkL℄, (see also [O℄, [CD℄ and [CY1℄) the authors provedLemma 3.3 H is di�eomorphi to H0 � B3 by sending u 2 H to (w =u'q;t ; q; 1 � t�2 log t), where H = fu 2 H1;2(S2) : R eu = 1g and H0 = fu 2HjP (u) = 0g.Now we an rewrite J� by this de omposition. First, let S(u) = R 12 jruj2+2u. It is important that S is invariant under onformal transformations,namely, S(u) = S(u'q;t)10

for any onformal transformations 'q;t of S2. Let u = (w; q; t) 2 H. Wewrite J� asJ� = J�(w; q; t) = S(w) � 2 log Z K Æ 'q;tew + f�(w; q);where f�(u) = f ÆB�(w; q; t)(3.8)and B�(w; q; t) = 40B�1 +vuuut1� 8� R K2 Æ 'q;te2w(det(d'q;t))�1(R K Æ 'q;tew)2 1CA�1(3.9)For simpli ity of notation, let b(u) = b(w; q; t) = R K2Æ'q;te2w(det(d'q;t)�1)=(R K Æ 'q;tew)2. We needLemma 3.4 ([CY1℄) If u 2 H0, then R jruj2 � 2(1 � a0)�1S(w) for a onstant a0 < 1.The following asymptoti behavior for large t is ru ial for the proofs of theTheorems.Lemma 3.5 For any b0 > 0, we have for all w with S(w) � b0 uniformlyin t as t!1(i) R K Æ 'q;tew = K(q) +O(t�1 log 1=2t);(ii) Z K2 Æ 'q;te2w(det(d'q;t))�1 = t2K2(q)Z e2w �1 + x32 �2 +O(t);(iii) Z K4 Æ 'q;te4w(det(d'q;t))�2 = t4K4(q)Z e4w(1 + x32 )4 +O(t3):Proof. (i) was proved in [CY1℄.(ii) We use the plane oordinates indu ed from the stereographi proje tion11

with respe t to q (see above). By the Taylor expansion of K around q =(0; 0; 1), we have K(x) = K(q) + a1x1 + a2x2 +O(jxj)in fx 2 C ��jxj � Mg for a �xed large M > 0. By a dire t omputation, wehave det(d'q;t)(z) = t2� 1 + jzj21 + t2jzj2�2Let Rt = fz 2 C ��jzj �M=tg and R t = C nRt. We de ompose the left handside of (ii) as follows14� ZRt +ZR t!K2(tz) e2w(z) t�2 (1 + t2jzj2)2(1 + jzj2)2 dA(z)where dA(z) = djzj2(1+jzj2)2 . A dire t omputation shows thatZR t K2(tz) e2w(z) t�2 (1 + t2jzj2)2(1 + jzj2)2 dA(z) � O(t�2)and ZRt x1(tz) e2w(z) t�2 (1 + t2jzj2)2(1 + jzj2)2 dA(z)� 2ZRt tjzj1 + t2jzj2 e2w(z) t�2 (1 + t2jzj2)2(1 + jzj2)2 dA(z)= 2ZRt 1 + t2jzj21 + jzj2 t�1 jzj e2w(z) dA(z)= 4Zjzj�Mt 1 + t2jzj21 + jzj2 t�1 jzj e2w(z) jzj(1 + jzj2)2 djzj� 8tZjzj�Mt jzj4(1 + jzj2)3 djzj� O(t):Similar ZRt x2(tz) e2w(z) t�2 (1 + t2jzj2)2(1 + jzj2)2 dA(z) = O(t):12

We also have ZRt K2(tz) e2w(z) t�2 (1 + t2jzj2)2(1 + jzj2)2 dA(z)= t2 ZRtK2(q) e2w(z) jzj4(1 + jzj2)2 dA(z) +O(t)= t2 Z K2(q) e2w(z) �1 + x32 �2 dA(z) +O(t):for z = (x1 + ix2)=(1 � x3). The pre eding estimates prove (ii).(iii) The proof is similar to that of (ii). �Lemma 3.6 ([H℄,[CY2℄) Infu2HJ(u) = � log(maxx2S2K(x)), and J doesnot a hieve its in�mum, if K is not onstant. �Now we haveProposition 3.7 For all suÆ iently large �, there exists u� 2 ÆA� with J�(u�) =��.To prove the proposition, we need one more Lemma. We re all �� =infu2A� J�(u).Lemma 3.8 (i) lim�!+1 �� = � log(max K)� 1 + 2 log 2(ii) infu2�A� J�(u) � � log(maxK)� 2 + 2 log 4.Proof. By Lemma 3.6, for any " > 0 there exists u0 2 H withJ(u0) � � log(maxK) + "=2On the other hand, B� := �u 2 A���b(u) � log � onverges to H as � !+1. Hen e we an hoose �0 su h that u0 2 B� for any � � �0. If we hoose �0 large enough, we haveJ�(u0) = J(u0) + f�(u0)� � log(maxK) + "=2 + f( 11 +q1� 8� log �)� � log(maxK)� 1 + 2 log 2 + ":13

Clearly, for any u 2 A�, J�(u) � � log(maxK) � 1 + 2 log 2. Hen elim�!1 �� = � log(maxK)� 1 + 2 log 2.(ii) If u 2 �A�, by de�nition, f Æ B�(u) = �2 + 2 log 4. Hen e J�(u) �� log(maxK)� 2 + 2 log 4. �Remark. By Lemma 3.8, lim�!1 �� < infu2�A� J�(u) for any �.Proof of Proposition 3.7. Consider a minimizing sequen e fuig 2 A� of J�.A ording to Lemma 3.3, we rewrite it as ui = (wi; qi; ti). First we laimthat S(wi) is bounded. This is easy to prove, for f�(ui) is bounded and� log R K Æ 'qi;ti eui � � log(maxK). By Lemma 3.4, the boundedness ofS(wi) implies that Z jrwij2 is bounded.Se ond, we laim that ftig is also bounded. Assume by ontradi tion thatftig is unbounded. There are two possibilities(1) qi ! q and q is one of the vorti es,(2) qi ! q and q is not equal to P+ or P�.Case (1). Sin e S(wi) is bounded, from (i) of Lemma 3.5 we have� log Z K Æ 'qi;ti ewi ! +1as i ! +1. Sin e S(wi) � 0 and f� is bounded, it follows that J�(ui) !+1, a ontradi tion.Case (2). Re alling (3.8), (3.9) and the boundedness of f�(ui), Lemma 3.5implies that t2i Z e2wi (1 + x3)2 � �:Re all that we havet2i Z e2wi = 1 and Z ewi x3 = 0:This implies thatZ e2wi(1 + x3)2 � Z ewi(1 + x3) = 1:14

So t2i � �is bounded.Now it is lear that the boundedness of ftig implies that of fkuikH1;2g.Hen e we may assume that there exists u� 2 H1;2(S2) su h that ui onvergesto u� weakly inH1;2, and strongly in Lq for any q > 1 and almost everywherein S2. It follows that Z Keui ! Z Keu�and Z K2e2ui ! Z K2e2u�as i ! +1. Therefore J�(u�) � �� and u� 2 A�. Now, in view of Lemma3.8, u� 2 ÆA�.Remark. We an prove the proposition in a di�erent way whi h does notuse onformal transformations. A tually, we an prove proposition 3.7 forany ompa t surfa e in [DJLW℄.By Lemma 3.2, we know that u� + �(u�) is a solution of (3.2).Now we onsider the behavior of u� as � ! +1. First, it is lear thatu� annot onverge in H1;2. Otherwise, we an obtain a minimum of J inH, whi h ontradi ts Lemma 3.6.Proposition 3.9 If we write u� as (w�; q�; t�) then w� ! 0 strongly inH1;2; t� ! 1 and q� ! Q as � ! +1, where Q is one of the maximumpoints of K. Moreover, w� ! 0 strongly in C1 as �! +1.Proof. Sin e lim�!1 �� = � log(maxK) � 1 + 2 log 2; fS(w�)g, hen efkw�kH1;2g is bounded. From the above dis ussion, we know that ft�g isunbounded. Assume q� ! Q and w� ! w0 weakly in H1;2 as �!1. By adire t omputation we havelim�!1�� = lim�!1J�(u�) � S(w0)� logK(Q)� 1 + 2 log 2Consequently, S(w0) = 0, hen e w0 � 0. K(Q) = maxK and w� onvergesto w0 = 0 strongly in H1;2. Clearly, w� satis�es a suitable equation similarto (3.7), from whi h we an show that w� ! 0 strongly in C1 as � ! +1by ellipti estimates. �15

Corollary 3.10 When P+ = �P�, there are in�nitely many solutions of(3.2).Proof. In this ase, K is axially symmetri along the axis rossing P+and P�. Denote by T# the rotation along this axis with angle #. It is learthat T#u� is also a riti al point of J� for any # 2 (0; 2�℄. From the previousproposition, we know P (u�)! Q as �!1. Hen e, for large �, P (u�) is notthe origin of R3 . On the other hand, it is lear that T#(P (u�)) = P (T# u�)and T# has no �xed points ex ept the origin. Hen e P (T#u�) 6= P (u�) forany # 2 (0; 2�℄, whi h impliesu� 6= T#u� for any # 2 (0; 2�℄:Hen e (3.2) admits in�nitely many solutions. �Now we an prove Theorem 2.3.Proof of Theorem 2.3 From propositions 3.8 and 3.9, all properties ex ept(ii) are easy to he k. Now we prove (ii). Re all that j��j = eu0+u�+�(u�) =Keu�+�(u�). We laim t2�� ! 0 as �!1;where u� = (w�; q�; t�). If the laim is true, by (3.5) and Lemma 3.5 wehave �e�(u�)! 2K(Q) as �! +1:By proposition 3.9, we an show thatmax eu� � t2�for some onstant > 0. In fa t, we haveeu�Æ'q;t = ew�(det d'q;t)�1and w� ! 0 strongly in C1. Hen e, we havej��j = Keu�+�(u�)� ��1t2 ! 0again by the laim. Therefore, we only have to prove the laim.Assume t2�� ! a0 as �!1 with a0 2 (0;1℄. By Lemma 3.5b(w�; q; t�)! a0whi h implies thatJ�(u�)! � log(maxK) + f(a0) > � log(maxK)� 1 + 2 log 2;a ontradi tion. This ompletes the proof of the Theorem.16

4 Proof of Theorem 2.4We �rst onsider a simple aseProposition 4.1 . If P+ = �P�, then for large � > 0, there exists asolution v� of (3.2) with v�(x) = v�(�x) (8x 2 S2) su h that v� � R ev� onverges to u0 2 H1;2(�) strongly for � ! 1, where u0 is the solution of(3.7) obtained by Moser [M2℄.Proof. The proof of the existen e of a solution is very similar to the onein [M2℄ (see also [T℄). If P+ = �P�, Lemma 3.1 says that K(x) = K(�x)for ea h x 2 S2. Therefore, we onsider a spe ial subspa e Hs = fu 2H1;2 �� u(x) = u(�x);8x 2 S2g. For ea h u 2 Hs, there is the improvedMoser inequality log ZS2 eu � 18 ZS2 jruj2 + ZS2 u(4.1)for some onstant > 0. From this inequality, it is easy to show that J�satis�es the Palais-Smale ondition and the oer ivity inHs\A�. The latteris J�(u) � 1 Z jruj2 � 2 for u 2 Hs \A�(4.2)for some positive onstant 1; 2. A tually, by (4.1) one an hoose = 1=8.Set �s� = infu2Hs\A� J�(u). As in se tion 3, we havelim�!1�s� = �s0 � 1 + 2 log 2(4.3)and infu2�A�\Hs J�(u) � �s0 � 2 + 2 log 4;(4.4)where �s0 = infu2Hs J(u) was studied by Moser in [M2℄. By a standardmethod, we show that �s� is a hieved by us� 2 ÆA� \Hs. The \symmetri variational prin iple" [P℄ implies that us� is a riti al point of J� in ÆA�.Hen e v� = us� + �(us�) is a solution of (3.2) by Lemma 3.2.Moreover, (4.2) and (4.3) imply thatZ jrus�j2 � for some onstant , provided that � is large enough. In view of the normal-ization R eu = 1. This implies that us� is bounded in H1;2.17

Assume us� ! us� 2 H1;2 weakly in H1;2 and strongly in Lp(S2) for any1 < p < +1. As in se tion 3, we have R Keus� ! R Keus� and R K2e2us� !R K2e2us� . Thus (4.3) implies that us� onverges to us� strongly in H1;2.Clearly, us� satis�es (3.7) and was obtained in [M2℄. �Proposition 4.2 If P+ = P�, there is no solution v� of (3.2) for large �su h that v� � R ev� onverges strongly in H1;2.Proof. Assume that u� = v� � R ev� onverges to ~u strongly in H1;2. Itis easy to he k that ~u satis�es (3.7). However, in this ase, i.e. P+ = P�the equation (3.7) admits no solution by the Kazdan-Warner identityZ hrK;rxiieu = 0that has to hold for any solution of (3.7), see [KW℄.Now we onsider the general and more diÆ ult ase P+ 6= �P�. In this ase, by Lemma 3.1, we know that K has a unique saddle point Q and aunique maximum point �Q(= �Q). Moreover minx2�K(x) = K(Q), where� is the great ir le rossing Q and ( �Q). This � satis�es the ondition (5.1)in [CY2℄, hen e we an de�ne a minimax value of J� as in [CY2℄ (see also[CkL℄ and [CD℄).Let : �D ! � be a parametrization of �, where D is the unit dis inR2 with boundary �D.De�nition 4.3 ([CY2℄) D(�) = fh : D ! H is a ontinuous map with thefollowing asymptoti behavior for all z0 2 �D :limz!z0 S(h(z)) = 0(4.5) limz!z0 P (h(z)) = (z0) 2 S2g:(4.6)(Here, P is the enter of mass de�ned in se tion 3, and S(h) = R 12(jrhj2+2h)).Set �0 = infh2Dmaxz2D J(h(z)).Lemma 4.4 ([CY2℄, [CkL℄) �0 > � logK(Q) + 0 for a onstant 0 > 0.For our problem, we need to modify the de�nition of D.18

De�nition 4.5 D0 = fh : D ! H is a ontinuous map satisfying the fol-lowing asymptoti onditions for all z0 2 �Dlimz!z0 S(h(z)) � Æ(4.7) limz!z0 P (h(z)) 2 BÆ( (z0))(4.8)for a �xed small Æ > 0g.We show that for small Æ;D0 and D are essentially the same in the fol-lowing Lemma.Lemma 4.6 There exists Æ0 > 0 su h that for any Æ < Æ0,(i) maxz2D J(h(z)) is a hieved in the interior of D for any h 2 D0,(ii) �00 := infh2D0 maxz2D J(h(t)) = �0:Proof. For ea h h 2 D0, we �rst onstru t an ~h 2 D su h that(1) S(~h(t)) � Æ and P (~h(t)) 2 BÆ( ( zjzj)), if 1=2 � jzj � 1(2) ~h(z) = h(2z), if jzj � 1=2.As in se tion 3, we de ompose h(z) as (wz; tz; qz) for z 2 D, whereqz = P (h(z))jP (h(z))j , 1 � t�2z log tz = jP (h(z))j and wz = h(z) Æ 'qz;tz + qz;tz . Bythe de�nition of D0, q(z0) 2 BÆ( (z0)) and S(wz) = S(h(z)) < Æ for anyz0 2 �D. We extend h to D2 = fz 2 C ��jzj � 2g byh0(z) = ( (wz; qz; tz) if jzj � 1;((2 � jzj)w zjzj ; ~qz; ~tz) if 1 � jzj � 2;where ~qz and ~tz (1 � jzj � 2) are de�ned by~qz = QzjQzj and 1� ~tz�2 log ~tz = Qzand Qz = (2� jzj)P (h( zjzj )) + (jzj � 1) zjzj :Sin e P (h( zjzj)) 2 BÆ( ( zjzj)), h0 is well-de�ned. Now, let ~h(z) = h0(2z) forz 2 D. Clearly, ~h 2 D. By Lemma 4.4. we havemaxz2D J(~h(z)) � �0 > � logK(Q) + 0On the other hand, it is lear that for small Æ > 0 S(u) � Æ and P (u) 2BÆ( (z0)) for some z0 2 �D imply thatJ(u) � � logK(Q) + "019

for "0 < 0=2. Hen e, we have maxz2D J(~h(z)) = maxz2D J(h(z)). Now itis lear that (i) and (ii) follow. �Now we return to onsider our fun tional J�. LetD� = fh 2 D0��h(D) � B�g.Re all that B� = fu 2 H��b(u) � log �g. For a �xed Æ0 (for example Æ0 as inLemma 4.6) it is lear that D� 6� ;, if � is suÆ iently large. Set�� = infh2D�maxz2D J�(h(z))Lemma 4.7 lim�!1 �� = �0 � 1 + 2 log 2.Proof. In view of Lemma 4.6, for any " > 0 there exists h0 2 D0 su hthat jmaxz2D J(h0(z)) � �0j < "=2:Sin e B� ! H as �! +1, we an hoose �0 > 0 su h that h0(D) � B� forany � > �0. Hen emaxz2D J�(h0(z)) � maxJ(h0(z)) + f( 11 +q1� 8� log �)� �0 + "� 1 + 2 log 2provided that �0 is large enough. On the other hand, for ea h h 2 D�, wehave maxz2D J(h(t)) � �0 � 1 + 2 log 2:This proves the Lemma. �Lemma 4.8 J� satis�es the Palais-Smale ondition in A�.In fa t, this was proved in the argument of proposition 3.7. �Now we state our main result in this se tion.Proposition 4.9 �� is a hieved by �u� 2 ÆB�, provided that � is large enough.Proof. We divide the proof into several steps.Step 1. We have(i) For "0 > 0 and large T1, there exist M0 > 0 and 0 > 0 su h that anyu = (w; q; t) with J�(u) � �0� 1+2 log 2+ "0 and t � T1 satis�es thatS(u) < M0 and u = (w; q; t) with q 62 B 0(P�).20

(ii) There exist small Æ > 0 and large T2 > 0 su h that, if u = (w; q; t) withS(u) � Æ, q 2 BÆ( ) and t � T2, then J�(u) � �0 � 1 + 2 log 2� 0=2:Note that (ii) was used in the proof of Lemma 4.6.Step 2. From [CkL, p51℄ there exist b0; e0; e1; T3 and N0; N1 > 0 su h thatif u = (w; q; t) with q 62 B 0(P�) and S(w) �M0, then(i) k�wJk � e0 �N0t�1 log1=2 t, if S(w) � b0 and t > T3,(ii) h�wJ; ~vi � e1S1=2(w) � N1t�1 log1=2 t, if S(w) � b0 and for any ~v 2TwH0.Here �wJ(u) is the derivative with respe t to w. Let T0 = maxfT; T1; T2; T3g,where T is determined in Lemma 3.5.Note that in steps 1 and 2, all onstants are independent of �.Step 3. There exists �0 > 0 su h that for any � > �0X := fu = (w; q; t)jt � T0g � B�:Step 4. If u = (w; q; t) 2 B� with S(w) �M0, q 62 B 0(P�) and t � T0, thenk�wf�(w; q; t)k � 1 t� � 1te� 2tk�tf�(w; q; t)k � 1te� 2t2and k�qf�(w; q; t)k � 1 t� � 1te� 2t2for some positive onstant 1 and 2.It is enough to he k that there exists a onstant su h thatjh�wd(u); vij � tkvkfor any v 2 TwH0.h�wd(u)vi = 2 R K2 Æ 'q;t e2w � v(det(d'q;t)�1)(R K Æ 'q;t ew)2� 2 R K2 Æ 'q;t e2w(det(d'q;t)�1) R K Æ 'q;t ew � v(R K Æ 'q;t ew)3� 2fK4 Æ 'q;t e4w(det(d'q;t)�2g1=2(R jvj2)1=2(R K Æ 'q;t ew)2+ R K2 Æ 'q;t e2w(det(d'g;t)�1)(R K2 Æ 'q;t e2w)1=2(R jvj2)(R K Æ 'q;t ew)2� t2(Z jvj2)1=2, by Lemma 3.5.21

Sin e u 2 B�, we have, by Lemma 3.5 t2 � log �:This is equivalent to ��1 � e� t2 . Step 4 follows.Step 5. Extend J�jX to Y := ft � T0g [ fu 2 (w; q; t) �� S(w) � M0 q 62B 0(P�)g. Let u 2 Y de�ne a fun tional H : Y ! R as in [CkL℄ byH(w; q; t) = S(w)� logK(q)� 2�K(q)K(q) t�2 log t:One an he k thatjJ �Hj � 8<: N0t�1 log1=2 t if S(w) � b t > T0N1 (jrK(q)j t�1 log1=2 t S(w)+t�2 + S2(w)) if S(w) � b t > T0see [CkL, p51℄. Extending J as in [CkL℄, we obtain ~J in Y . On the otherhand, by step 4 we an extend f�jX smoothly to ~f� = Y ! R su h that(i) ~f�jX = f�jX ,(ii) ~f satis�es Step 4,(iii) ~f� = �1 + 2 log 2, when t is suÆ ient large.Now we obtain a new fun tional ~J� = ~J + ~f� de�ned in Y satisfying(a) � ~J�(w) 6� 0 if u = (w; q; t) with t � T0(b) ~J� satis�es the Palais-Smale onditions on (���"0; ��+"0) for a �xedsmall onstant "0 > 0( ) ~J�jX = J�jX :(d) ~J� satis�es Steps 1-4.(a) We an follow [CkL℄ to prove (a). Here we give a sket h.If S(w) � t�1 log t, then k�w ~Jk � N0t�1 log1=2 t (see [CkL℄). (ii) andStep 4 implies that if t � T0 k�w ~fk2 � te� t2 :(4.9)Hen e k�w ~J�k � N02 t�1 log1=2 t, provided that T0 is suÆ ient large.If S(w) � t�1 log t, it was shown in [CkL℄ that near Q or �Q���t ~J(u)� 2�K(q)K(q) ��(4.10) 22

is ontrolled by log�1 t if t � T0 and away from Q and �Q���q ~J(u) + �qK(q)K(q) ��(4.11)is ontrolled by t�1+� log t, if t � T0. Here 0 < � < 1. By (ii) and Step 4it is lear that (4.10) and (4.11) hold also for ~J�. Thus there are no riti alpoints of ~J� if t � T0.(b) If fuig � Y is a Palais-Smale sequen e for ~J�, the argument in (i) im-plies that we may assume that fuig � X. Hen e (b) follows from Lemma 4.8.( ) and (d) are lear.Step 6. Now we set �?� = infh2D?� supz2D ~J�(h(z)), where D?� = fh 2D0��h(D) � Y g. Using the above argument, we haveD?� 6� ;and lim�!1�?� = �0 � 1 + 2 log 2Thus, for large � > 0, �?� 2 (�0 � 1 + 2 log 2 � "0; � � 1 + 2 log 2 + "0).Sin e ~J� satis�es the Palais-Smale ondition on (�0 � 1 + 2 log 2 � "0; � �1+2 log 2+ "0), if �?� is not a riti al value of ~J�, we an �nd a deformationT (�; t) : Y � [0; 1℄! Y su h that(i) T (u; 0) = u,(ii) T (u; 1) � ~J�?��"1 , if u 2 ~J�?�+"1(iii) T (u; t) = u, if u 2 ~J�?��2"1 [ (Y n ~J�?�+2"1);where ~Jb := fu 2 Y j ~J�(u) � bg and �1 < �0=4. The onstru tion of su h adeformation is standard. We refer to [Ck℄. We laimT (u; t) = uif u 2 �Y [ fu = (w; q; t)jS(w) � Æ; P (u) 2 BÆ( )g.If u = (w; q; t) with S(w) � Æ and P (u) 2 BÆ( ), then by Step 1 J�(u) ��0 � 1 + 2 log 2 � 02 . By (iii) in the onstru tion of the deformation T ,T (u; t) = u. If u 2 �Y , then u = (w; q; t) satis�es either S(w) = M0,q 2 �B 0(P�) and t � T0 or S � M0, q 2 B 0(P�) and t = T0. Again byStep 1, we have J�(u) > �0�1+2 log 2+�0, hen e T (u; t) = u by (iii) above.Hen e TÆh 2 D?� for any h 2 D?�. Now it is lear that the existen e of su ha deformation ontradi ts the de�nition of �?�. Therefore �?� is a riti al valueand there is u� 2 Y su h that u� is a riti al point of ~J� and ~J�(u�) = �?�.By the onstru tion of ~J�, we know u� 2 X and ~J�(u�) = J�(u) = ��. This ompletes the proof of the proposition. �23

Proposition 4.10 By taking a subsequen e, u� onverges to u0 strongly inCq for any q > 1, where u0 is a solution of (3.7) obtained in [CkL℄ and[CY℄.Proof. From the argument in the proof of the previous proposition, wehave u� 2 X. Therefore u� is bounded in H1;2. Assume u� onverges to u0weakly in H1;2, strongly in Lp for p > 1 and almost everywhere. As before,by Lebesgue's theorem, we haveZ Keu� ! Z Keu0 and Z K2e2u� ! Z K2e2u0 :Hen e, �e�(u�) ! 2R Keu0as �! +1.Sin e v� = u� + �(u�) satis�es (3.2), i.e.Z hrv� � r'i+ �Z Kev�(Kev� � 1)' + 2Z ' = 0;(4.12)we have Z ru0 � r'� 2Z Keuo'+ 2Z ' = 0;(4.13)whi h implies that u0 is a solution of (3.7). Choosing ' = u� � u0 in (4.12)and (4.13), we on ludeZ jru� �ru0j2 = ��Z Ke�(u�)eu�(Ke�(u�)eu� � 1)(u� � u0)+2Z Keu0(u� � uo)= ��Z Ke�(u�)eu�(u� � u0)+2Z Keu0(u� � uo) + o(1)= o(1) as �! +1It follows that u� onverges to u0 in H1;2. Now it is easy to on lude thatu� ! u0 in Cq(S2) for any q � 1 by the ellipti estimates. �Proof of Theorem 2.4. It follows from proposition 4.9 and 4.10. �Remark. FA2k ! 4� eu0+u0R eu0+u0 as k ! 0, where u0 is a solution of (3.2) andu0 is a solution of (3.7) with K = eu0 .24

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