Abstract: We prove the existence of bubbling solutions for the following Chern-Simons-Higgs equation:
�u +1
ε2 eu(1 − eu) = 4π
N∑
j=1
δp j , in �,
where � is a torus. We show that if N > 4 and p1 �= p j , j = 2, . . . , N , then for smallε > 0, the above problem has a solution uε, and as ε → 0, uε blows up at the vertexpoint p1, and satisfies
1
ε2 eu(1 − eu) → 4π Nδp1 .
This is the first result for the existence of a solution which blows up at a vertex point.
1. Introduction
The Chern-Simons theories were developed to explain certain condensed matter phe-nomena, anyon physics, superconductivity, quantum mechanics and so on. In this paper,we will study the (2 + 1)-dimensional relativistic Abelian Chern-Simons-Higgs modelon a torus �. This theory was proposed in [9,10] in an attempt to explain superconduc-tivity of type 2. A self-dual system for stationary solutions is derived and can be furtherreduced to an elliptic partial differential equation with exponential nonlinearity:
�u +1
ε2 eu(1 − eu) = 4π
N∑
j=1
δp j , in �, (1.1)
734 C.-S. Lin, S. Yan
where δp j (x) is the Dirac measure at p j ∈ �, and ε is the Chern-Simons constant. See[9,10,17,21] and the references therein. Note that pi and p j may coincide for somei �= j .
For a torus �, it is well-known (see for example [18]) that there exists a constantε∗ > 0, such that if ε ≥ ε∗, (1.1) has no solution, while if ε < ε∗, there are at leasttwo solutions for (1.1), one of which is the maximal solution, and the other one can beobtained through the mini-max variational method. As ε → 0, the maximal solutiontends to 0 uniformly in any compact subset of �\{p1, . . . , pN }. But the second solutionhas a different asymptotic behavior.
In [4], it is proved that for any sequence of solution un with εn > 0, one of thefollowing holds true:
(a) un → 0 uniformly in any compact subset of �\{p1, . . . , pN };(b) un + ln 1
ε2n
is bounded from above;
(c) there is a finite set S = {q1, . . . , qL } ⊂ � and x1,n, . . . , xL ,n ∈ �, such that asn → +∞, x j,n → q j ,
un(x j,n) + ln1
ε2n
→ +∞, ∀ j = 1, . . . , L ,
and
un(x) + ln1
ε2n
→ −∞, uniformly on any compact subset of �\S.
Moreover,
1
ε2n
eun (1 − eun ) →L∑
j=1
M jδq j , M j ≥ 8π,
in the sense of measure.
In case (c), q j is called a blow up point for the solution un . Let
u0(x) = −4π
N∑
j=1
G(x, p j ), (1.2)
where G(x, p j ) is the Green function of −� in � with singularity at p j . That is, G(x, p j )
satisfies
−�G(x, p j ) = δp j − 1
|�| ,
where |�| is the measure of �. For the second solution, Choe [3] proved the followingtheorem:
Theorem A. Assume that N > 2 and let uε be a sequence of solutions of (1.1), which isobtained from the mountain pass lemma in the variational theory. As ε → 0, uε blowsup at exactly one point q ∈ �\{p1, . . . , pN }, satisfying u0(q) = maxx∈� u0(x) and
1
ε2 euε (1 − euε ) → 4π Nδq ,
in the sense of measure.
Bubbling Solutions for Abelian Chern-Simons Model 735
When N = 1, 2, it was shown in [19] that the conclusion of Theorem A does notnecessarily hold. Theorem A is the first result on the existence of bubbling solutions forthe Chern-Simons-Higgs equation on a torus, but it only tells us that (1.1) has solutionswhich blow up at a regular point q of the function u0. A question raised in [17] is whether(1.1) has a solution which blows up at one of the vortex points p j . This kind of solutionis physically meaningful, because the order parameter eu is always zero at any vortex.However, the existence of such solutions implies even in a tiny neighborhood of a vertex,the order parameter may have a positive lower bound, which makes superconductivitypossible.
In this paper, we will give a positive result to the problem raised by Tarantello. Themain result of this paper is the following:
Theorem 1.1. Let � ⊂ R2 be a torus and N > 4. If p1 �= p j , ∀ j �= 1, there is anε0 > 0, such that for any ε ∈ (0, ε0), (1.1) has a solution uε, satisfying that as ε → 0,
1
ε2 euε(1 − euε
) → 4π Nδp1 .
The solution uε in Theorem 1.1 satisfies a stronger inequality:
maxx∈�
uε(x) ≥ c > −∞,
although uε + ln 1ε2 → −∞ uniformly in any compact subset of �\{p1}. As a result,
p1 is a blow-up point of uε. As far as we know, Theorem 1.1 is the first result on theexistence of solutions for (1.1), which blow up at a vortex point.
It is also natural to ask whether there is a sequence of bubbling solutions with
maxx∈�
uε(x) → −∞,
as ε → 0. For the case N = 2, the readers can refer to [15,19] for the existence of suchsolutions. For the general case, we will discuss it in a future work.
It seems hard to find a min-max theorem as in [3,16,19] to prove the existence of asolution blowing up at a vortex point. In this paper, we will use the contraction mappingtheorem to prove Theorem 1.1. The crucial step in this procedure is to construct a goodapproximation solution ϕε for (1.1), such that the linear operator Lε of (1.1) at ϕε isinvertible, although L−1
ε has a large norm.Let us point out the question of the existence of a solution which blows up at several
vortex points is still open.In this paper, we will also study the system of equations corresponding to the rela-
tivistic Abelian Chern-Simons model involving two Higgs scalar fields and two gaugefields on a torus �:
⎧⎪⎪⎪⎨
⎪⎪⎪⎩
�u = 1ε2 ev(eu − 1) + 4π
N1∑j=1
δp j
�v = 1ε2 eu(ev − 1) + 4π
N2∑j=1
δq j .
(1.3)
The readers can refer to [7,8,12,13] and references therein for the background physics.The mathematical analysis of the system (1.3) has been recently initiated in [13]
where the problem (1.3) is studied on the plane. More recently, the paper [6] studied theuniqueness of topological solutions and existence of non-topological solutions for (1.3)
736 C.-S. Lin, S. Yan
in a plane. In [14], it is shown that there is a constant ε∗ > 0, such that if ε < ε∗, (1.3)has at least two solutions, while if ε ≥ ε∗, (1.3) has no solution.
It is interesting to study this problem on a torus since the solutions of the ChernSimons equations depend on the topology of underlying space. For example, the relativ-istic Abelian Chern-Simons theory (scalar equation) ([1]), non-Abelian SU(3) Chern-Simons theory ([20]), Maxwell-Chern-Simons theory ([2,17]) models on a torus arewell studied. In spite of similarity with the scalar Abelian Chern Simons equation andAbelian Higgs vortex equations (see [11] for details), the system (1.3) is difficult notonly since it involves two equations, but also due to the “mixed” nature of the nonlinearterms.
Although it is difficult to obtain a blow-up solution for the system by using the moun-tain pass lemma as in [3,16,19], it turns out that the method we use to prove Theorem 1.1applies to (1.3) in some special case. Since system (1.3) is symmetric in (u, v) exceptthe terms involving the Dirac measure, as in [13], we make the following the change ofvariable
u → u + v, v → u − v.
Then the system (1.3) can be transformed into an equivalent problem⎧⎪⎪⎪⎨
⎪⎪⎪⎩
�u = 2ε2 eu − 1
ε2 eu+v
2 − 1ε2 e
u−v2 + 4π
N1∑j=1
δp j + 4πN2∑j=1
δq j
�v = 1ε2 e
u+v2 − 1
ε2 eu−v
2 + 4πN1∑j=1
δp j − 4πN2∑j=1
δq j .
(1.4)
Define
u0(x) = −4π
N1∑
j=1
G(x, p j ) − 4π
N2∑
j=1
G(x, q j ),
and
v0(x) = −4π
N1∑
j=1
G(x, p j ) + 4π
N2∑
j=1
G(x, q j ).
Making a further change of variable
u → u0 + u, v → v0 + v,
we can reduce system (1.4) to⎧⎨
⎩�u = 2
ε2 eu0+u − 1ε2 e
u0+v02 + u+v
2 − 1ε2 e
u0−v02 + u−v
2 + 4π(N1+N2)|�| , in �;
�v = 1ε2 e
u0+v02 + u+v
2 − 1ε2 e
u0−v02 + u−v
2 + 4π(N1−N2)|�| , in �.(1.5)
Note that the second equation in (1.5) is uniquely solvable for any given u. Of course,v depends on u. The main observation for (1.5) is that if N1 = N2, then the second equa-tion of (1.5) becomes
�v = 1
ε2 eu0+u
2
(e
v+v02 − e− v+v0
2
). (1.6)
Bubbling Solutions for Abelian Chern-Simons Model 737
If we further assume that p1 = q1, then the function v0 is regular near p1. As a result,the constant −v0(p1) can be used as an approximate solution of (1.6) near p1 for anygiven u. Note that v + v0 ≈ 0 near p1, we find that near p1, the first equation in (1.5) isa perturbation of the following equation:
�u = 2
ε2
(eu0+u − e
u0+u2
)+
8π N1
|�| .
So, using the method as in the proof of Theorem 1.1, we can obtain
Theorem 1.2. Let � ⊂ R2 be a torus. Suppose that N1 = N2 = N > 4, p1 = q1 andp1 �= p j , q j for j > 1. Then, there is an ε0 > 0, such that for any ε ∈ (0, ε0), (1.5)has a solution (uε, vε), satisfying
vε → 4π
N∑
j=2
G(p1, p j ) − 4π
N∑
j=2
G(q1, q j ), uniformly in �;
and
1
ε2 eu0+uε
2
(1 − e
u0+uε2
)→ 4π Nδp1 ,
as ε → 0.
Theorem 1.2 roughly states that if the two terms in (1.3) involving the Dirac measuresare equal at p1 and the total weights of the two equations are equal, then (1.3) has asolution (uε, vε), such that uε and vε have the same bubbling behavior at p1. That is,uε − vε has no bubble at p1.
We need the condition N1 = N2 and p1 = q1 to show that (1.5) has a solution(uε, vε), such that vε is nearly a constant. Let us point out that if N1 �= N2, or p1 �= qi ,i = 1, . . . , N2, it is still an open problem whether (1.3) has a solution concentratingat p1.
We will use that contraction mapping theorem to prove Theorem 1.1. The first cru-cial step in this paper is to construct a good approximation solution ϕε, which will bepresented in Sect. 2. The most difficult part of this paper is the analysis of the linearoperator Lε corresponding to ϕε, which has the form Lεu = �u + 1
ε2 fε(x)u, wherefε(x) has compact support and fε(x) may not have fixed sign. Here, we will encountera typical difficulty for an elliptic problem in R2: the L2 norm of the gradient of u can notcontrol any L p norm of u. Thus, the operator Lε will not generate naturally a space inwhich we can carry out our analysis. Instead, we have to work on some weighted spaceswhich are not related to Lε in a natural way. The estimate of the norm of Lε involves lotsof complicated analysis, so we put this part in the appendixes. In Sect. 3, we will proveTheorem 1.1. Since the proof of Theorem 1.2 is just a bit more technically complicatedthan that of Theorem 1.1, we sketch it in Sect. 4.
2. The Approximate Solutions
Recall that
u0(x) = −4π
N∑
j=1
G(x, p j ).
738 C.-S. Lin, S. Yan
We consider the following equivalent problem of (1.1):
�u +1
ε2 eu0+u (1 − eu0+u) = 4π N
|�| , in �. (2.1)
In this section, we will construct an approximate solution for (2.1). Suppose thatN > 4. Then by Theorem 2.1 of [5], we know that the following problem has a uniqueradial solution V (|x |):
{�V + |x |2eV (1 − |x |2eV ) = 0, in R2;−tV ′(t) → 2N , as t → +∞.
(2.2)
Let γ (x, p) be the regular part of G(x, p). That is,
γ (x, p) = G(x, p) − 1
2πln
1
|x − p| .
We define the approximate solution ϕε for (2.1) as follows:
ϕε(x) = V
( |x − p1|ε
)+ 4π N (γ (x, p1) − γ (p1, p1))
(1 − ε2N−4−θ
)+ M,
x ∈ Bdε (p1), (2.3)
while
ϕε(x) = V
(dε
ε
)+ 4π N
(G(x, p1) +
1
2πln dε − γ (p1, p1)
)(1 − ε2N−4−θ
)+ M,
x ∈ �\Bdε (p1), (2.4)
where V (|x |) is the solution of (2.2), θ > 0 is a fixed small constant, dε is the constantto make ϕε ∈ C1, and
M = 2 ln1
ε+ 4πγ (p1, p1) + 4π
N∑
j=2
G(p1, p j ). (2.5)
We will prove that the constant dε has a following estimate:
dε = (a + o(1)) εθ
2N−4 , for some constant a > 0. (2.6)
In the rest of this section, we explain how to find this approximate solution ϕε. We candecompose (2.1) into two problems:
�u1 = 4π N
|�| , in �, (2.7)
and
�u2 +1
ε2 eu0+u2(1 − eu0+u2
) = 0, in �. (2.8)
For any constant M ,
u1 = 4π N (γ (x, p1) − γ (p1, p1)) + M. (2.9)
is a solution of (2.7).
Bubbling Solutions for Abelian Chern-Simons Model 739
On the other hand, if we blow up (2.8) at p1, we then obtain the following limitproblem:
�V + |x |2eV (1 − |x |2eV ) = 0, in R2.
By Theorem 2.1 of [5], if β > 8, the following problem has a unique radial solution:{
�V + |x |2eV (1 − |x |2eV ) = 0, in R2;−tV ′(t) → β, as t → +∞.
(2.10)
Let V be the solution of (2.10), where the constant β is to be determined later. So, near
p1, u2(x) ≈ V( |x−p1|
ε
).
For a constant dε > 0 small with dε >> ε, we define
ϕε(x) = V
( |x − p1|ε
)+ u1(x), x ∈ Bdε (p1),
where u1(x) is defined in (2.9). Then in Bdε (p1),
−�ϕε = 1
ε2
|x − p1|2ε2 e
V( |x−p1|
ε
) (1 − |x − p1|2
ε2 eV
( |x−p1|ε
))− 4π N
|�|
= 1
ε2 eu0(x)+ϕε(x)−u0(x)−u1(x)+ln |x−p1|2
ε2
(1 − e
u0(x)+ϕε(x)−u0(x)−u1(x)+ln |x−p1|2ε2
)
−4π N
|�| .
Since
−u0(x) − u1(x) + ln|x − p1|2
ε2
= 4πγ (x, p1) + 4π
N∑
j=2
G(x, p j ) + ln1
ε2 − 4π N (γ (x, p1) − γ (p1.p1)) − M,
if we choose
M = 2 ln1
ε+ 4πγ (p1, p1) + 4π
N∑
j=2
G(p1, p j ), (2.11)
then, near p1,
−�ϕε ≈ 1
ε2 eu0(x)+ϕε(x)(
1 − eu0(x)+ϕε(x))
− 4π N
|�| .
So, we find
ϕε(x) = V
( |x − p1|ε
)+ 4π N (γ (x, p1) − γ (p1, p1)) + M, x ∈ Bdε (p1),
(2.12)
where M is given by (2.11).
740 C.-S. Lin, S. Yan
We only define ϕε near p1. But we need to define ϕε in �. Noting that V(
dε
ε
)→ −∞,
we expect that in �\Bdε (p1), �φε ≈ 0. So we can define
ϕε(x)=V
(dε
ε
)+4π N
(G(x, p1)+
1
2πln dε−γ (p1, p1)
)+M, x ∈ �\Bdε (p1).
(2.13)
For any V , which is the solution of (2.10), the function ϕε is continuous. But asan approximate solution of a second order equation, we need to choose a suitable β in(2.10) to make ϕε ∈ C1. If ϕε ∈ C1(�), then
1
εV ′
(dε
ε
)= −2N
1
dε
holds.
If we let tε = dε
ε→ +∞, then
− tεV ′(tε) = 2N . (2.14)
Thus, we see that β must satisfy
β = 2N .
Unfortunately, (2.14) has no solution. To solve this problem, we modify ϕε as follows:
ϕε(x)=V
( |x − p1|ε
)+4π N (γ (x, p1)−γ (p1, p1))
(1−εL
)+M, x ∈ Bdε (p1),
(2.15)
while
ϕε(x) = V
(dε
ε
)+ 4π N
(G(x, p1) +
1
2πln dε − γ (p1, p1)
)(1 − εL
)+ M,
x ∈ �\Bdε (p1), (2.16)
where L > 0 is a constant to be determined, M is given by (2.11), and dε > 0 satisfies
1
εV ′
(dε
ε
)= −2N
(1 − εL
) 1
dε
.
Let tε = dε
ε→ +∞, then
− tεV ′(tε) = 2N(
1 − εL)
. (2.17)
It follows from Lemma 2.6 of [5] that as t → +∞,
V (t) = −2N ln t + I − a
t2N−4 + O
(1
t2N−2
),
and
V ′(t) = −2N
t+
a(2N − 4)
t2N−3 + O
(1
t2N−1
),
Bubbling Solutions for Abelian Chern-Simons Model 741
where a > 0 and I are some constant. Thus, tε satisfying
−a(2N − 4)
t2N−4ε
+ O
(1
t2N−2ε
)= −2NεL ,
which implies that there is a constant a > 0,
tε = (a + o(1)) ε− L2N−4 . (2.18)
So, we find
dε = (a + o(1)) ε1− L2N−4 . (2.19)
In order to keep dε << 1, we need to choose L , such that 1− L2N−4 > 0. So, we take
L = 2N − 4 − θ, (2.20)
where θ > 0 is a fixed small constant. With such L , we see that (2.6) comes from (2.19).
3. Solutions Concentrating at a Vortex Point
In this section, we will construct a solution for (2.1) with
uε = ϕε + ω,
where the approximate solution ϕε is defined in (2.3) and (2.4). So, ω should solve
Lε,1ω = gε,1(x, ω), (3.1)
where
Lε,1ω = �ω + fε,1(x)ω, (3.2)
fε,1(x) =
⎧⎪⎨
⎪⎩− 1
ε2 eV
( |x−p1|ε
)+2 ln |x−p1|
ε
(2e
V( |x−p1|
ε
)+2 ln |x−p1|
ε − 1
), x ∈ B2d (p1);
0, x ∈ �\B2d (p1),
(3.3)
d > 0 is a fixed constant, and
gε,1(x, s) = fε,1(x)s +1
ε2 eu0+ϕε+s (eu0+ϕε+s − 1
)+
4Nπ
|�| − �ϕε. (3.4)
By (2.3) and (2.4), we have
−�ϕε +4Nπ
|�|
=⎧⎨
⎩− 1
ε2 eV (|x−p1 |
ε)+2 ln |x−p1 |
ε
(eV (
|x−p1 |ε
)+2 ln |x−p1 |ε − 1
)− 4π N
|�| ε2N−4−θ , x ∈ Bdε (p1);− 4π N
|�| ε2N−4−θ , x ∈ �\Bdε (p1).
(3.5)
742 C.-S. Lin, S. Yan
Let us introduce two function spaces Xα,ε and Yα,ε, such that for any ξ ∈ Xα,ε, wenot only control the L2 norm of ξ and �ξ , but also take into account the concentrationat p1. For this purpose, we fix a small α > 0 and define
ρ(x) = (1 + |x |)1+ α2 , ρ(x) = 1
(1 + |x |) (ln(2 + |x |))1+ α2.
We say a function ξ is in Xα,ε if
‖ξ‖2Xα,ε
= ε4‖(�ξ)ρ‖2L2(B2d/ε)
+ ‖ξ ρ‖2L2(B2d/ε)
+ ‖�ξ‖2L2(�\L2(Bd (p1))
+ ‖ξ‖2L2(�\Bd (p1))
< +∞, (3.6)
where ξ (y) = ξ(p1 + εy), Bt = Bt (0). On the other hand, we say ξ ∈ Yα,ε if
‖ξ‖2Yα,ε
= ε4‖ξρ‖2L2(B2d/ε)
+ ‖ξ‖2L2(�\L2(Bd (p1))
< +∞. (3.7)
By using Theorem B.1 in Appendix B, we are now ready to prove the main result ofthis paper.
Proof of Theorem 1.1. It follows from Theorem B.1 that Lε,1 is invertible from Xα,ε toYα,ε. Rewrite (3.1) as
ω = Bε,1ω =: L−1ε,1gε,1(x, ω), ω ∈ Xα,ε.
Fix a small constant σ > 0. Define
Sε,1 ={ω : ω ∈ Xα,ε, ‖ω‖L∞ + ‖ω‖Xα,ε ≤ ε1−σ
}.
We will prove that Bε,1 is a contraction map from Sε,1 to Sε,1. So, by the contractionmapping theorem, there is an ωε ∈ Sε,1, such that ωε = Bε,1ωε.
First, we prove that Bε,1 maps Sε,1 to Sε,1. It follows from Theorem B.1 that
‖Bε,1ω‖L∞(�) + ‖Bε,1ω‖Xα,ε ≤ C ln1
ε‖gε,1(x, ω)‖Yα,ε .
To estimate ‖gε,1(x, ω)‖Yα,ε , by (3.4) and (3.5), we first note that for x ∈ Bdε (p1),
gε,1(x, ω) = fε,1(x)ω +1
ε2 eu0+ϕε+ω(eu0+ϕε+ω − 1
)
− 1
ε2 eV (|x−p1|
ε)+2 ln |x−p1|
ε
(eV (
|x−p1|ε
)+2 ln |x−p1|ε − 1
)+ O
(ε2N−4−θ
).
(3.8)
On other hand, by (2.5), for x ∈ Bdε (p1),
u0 + ϕε = −4π
N∑
j=1
G(x, p j ) + V
( |x − p1|ε
)
+ 4π N (γ (x, p1) − γ (p1, p1))(
1 − ε2N−4−θ)
+ 2 ln1
ε+ 4πγ (p1, p1) + 4π
N∑
j=2
G(p1, p j )
= V
( |x − p1|ε
)+ 2 ln
|x − p1|ε
+ O(|x − p1|), (3.9)
Bubbling Solutions for Abelian Chern-Simons Model 743
which, together with (3.8), gives
gε,1(x, ω) = O
(1
ε2 eV (|x−p1|
ε)+2 ln |x−p1|
ε
(|ω|2 + |x − p1|
))+ O
(ε2N−4−θ
).
(3.10)
As a result,
ε4‖gε,1(εx + p1, ω)ρ(x)‖2L2(Bdε/ε)
≤ C‖eV (x)+2 ln |x | (|ω(εy + p1)|2 + ε|x |)
ρ(x)‖2L2(Bdε/ε)
+ O(ε4
)
≤ C(ε4(1−σ) + ε2
) ∫
R2
1
(1 + |x |)2(2N−4)−αdx + O
(ε4
)≤ Cε2,
since |ω| ≤ ε1−σ , |x |ρ(x) ≤ C(1 + |x |)2+ α2 and 2(2N − 4) − α > 2 if α > 0 is small.
On the other hand, in �\Bdε (p1),
| fε,1(x)| ≤ C
ε2 eV (|x−p1|
ε)+2 ln |x−p1|
ε ≤ C
ε4
(ε
dε
)2N
, (3.11)
and∣∣∣∣
1
ε2 eu0+ϕε
∣∣∣∣ ≤ C
ε2 eV (dεε
)+2 ln |x−p1|ε ≤ C
ε4
(ε
dε
)2N
. (3.12)
Combining (3.4), (3.5), (3.11) and (3.12), we obtain
gε,1(x, ω) = O(ε2N−4−θ
), x ∈ �\Bdε (p1). (3.13)
So, from ρ(x) = (1 + |x |)1+ α2 ≤ Cε−1− α
2 in B2d/ε\Bdε/ε,
ε4‖gε,1(εx + p1, ω)ρ(x)‖2L2(B2d/ε\Bdε/ε)
≤ Cε2(2N−2−θ)ε−4−α ≤ Cε2,
and
‖gε,1(x, ω)‖L2(�\Bd (p1))≤ Cε2N−4−θ ≤ Cε2.
So, we have proved
‖gε,1(x, ω)‖Yα,ε ≤ Cε, (3.14)
which gives
‖Bε,1ω‖L∞(�) + ‖Bε,1ω‖Xα,ε ≤ C ln1
ε‖gε,1(x, ω)‖Yα,ε ≤ ε1−σ .
So, Bε,1 maps Sε,1 to Sε,1.Next, we show that Bε,1 is a contraction map. For any ω1, ω2 ∈ Sε,1, we have
So, we have proved that Bε,1 is a contraction map. ��
Bubbling Solutions for Abelian Chern-Simons Model 745
4. Bubbling Solutions for the System
In this section, we will discuss the system under the assumption that p1 = q1 andN1 = N2. Since the proof of Theorem 1.2 is similar to the proof of Theorem 1.1, wejust sketch it.
Let ϕε be the function defined in (2.3) and (2.4), where in the definition of the constantM in (2.5), we need to replace u0 by u0. Denote
v∗ = 4π
N1∑
j=2
G(p1, p j ) − 4π
N2∑
j=2
G(p1, q j ).
Our objective is to prove that (1.5) has a solution of the form
uε = 2ϕε + ωε, v = v∗ + ηε,
where (ωε, ηε) is a perturbation term.For any ω, we consider the second equation of (1.5):
�η = 1
ε2 eϕε+u0+ω
2
(e
v∗+v0+η
2 − e− v∗+v0+η
2
). (4.1)
Rewrite (4.1) as follows:
Lε,2η = gε,2(x, ω, η), (4.2)
where
Lε,2η = �η + fε,2(x)η, (4.3)
fε,2(x) ={
− 1ε2 e
V( |x−p1|
ε
)+2 ln |x−p1|
ε , x ∈ B2d(p1);0, x ∈ �\B2d(p1),
(4.4)
gε,2(x, s, t) is defined as
gε,2(x, s, t) = fε,2(x)t +1
ε2 e2ϕε+s+u0
2
(e
v∗+v0+t2 − e− v∗+v0+t
2
). (4.5)
It follows from Theorem B.1 that Lε,2 is an isomorphism from Xα,ε to Yα,ε. Note thatunder the condition p1 = q1, v0 has no singularity at p1. Using the contraction mappingtheorem as in Sect. 3, we can prove
Proposition 4.1. Fix τ1 ∈ (0, 1). Then for any ω ∈ Xα,ε with ‖ω‖L∞(�)+‖ω‖Xα,ε ≤ ετ1 ,there is an ηε ∈ Xα,ε, ηε = ηε(ω), satisfying (4.2). Moreover,
‖ηε‖L∞(�) + ‖ηε‖Xα,ε ≤ Cε ln1
ε, (4.6)
and
‖ηε(ω1) − ηε(ω2)‖L∞ + ‖ηε(ω1) − ηε(ω2)‖Xα,ε ≤ Cε
(ln
1
ε
)2
‖ω1 − ω2‖Xα,ε .
(4.7)
746 C.-S. Lin, S. Yan
Proof. The proof of the existence part and (4.6) is similar to the proof of Theorem 1.1.We thus omit it.
To prove (4.7), we let ηε =: η1 − η2, ηi = ηε(ωi ), i = 1, 2. Then
Lε,2ηε = gε,2(x, ω1, η1) − gε,2(x, ω2, η2).
By Theorem B.1, we have
‖ηε‖L∞ + ‖ηε‖Xα,ε ≤ C ln1
ε‖gε,2(x, ω1, η1) − gε,2(x, ω2, η2)‖Yα,ε
≤ C ln1
ε
(‖gε,2(x, ω1, η1) − gε,2(x, ω2, η1)‖Yα,ε
+ ‖gε,2(x, ω2, η1) − gε,2(x, ω2, η2)‖Yα,ε
)
≤ C ln1
ε
(Cε ln
1
ε‖ω1 − ω2‖Yα,ε + Cε ln
1
ε‖ηε‖Xα,ε
). (4.8)
So we obtain (4.7). ��Proof Theorem 1.2. For any ω ∈ Xα,ε with ‖ω‖L∞ + ‖ω‖Xα,ε ≤ ετ1 , let ηε(ω) be themap obtained in Proposition 4.1. We substitute vε = v∗ + ηε(ω) into the first equationof (1.5). Then, ω satisfies
Lε,1ω = gε,1(x, ω, ηε(ω)), (4.9)
where Lε,1 is the linear operator defined in (3.2) and (3.3), and
gε,1(x, s, t) = fε,1(x)s +2
ε2 eu0+ϕε+s − 1
ε2 eu0+ϕε+s
2 + v0+v∗+t2
− 1
ε2 eu0+ϕε+s
2 − v0+v∗+t2 +
8π N1
|�| − �ϕε.
By (4.6), we find that gε,1(x, ω, ηε(ω)) ≈ gε,1(x, ω), where gε,1(x, ω) is defined in(3.4). So we can prove
Appendix A. The Invertibility of Linear Operators in R2
In this section, we consider the invertiblity of the following linear operator:
Lw = �w + a(x)w, in R2, (A.1)
Bubbling Solutions for Abelian Chern-Simons Model 747
where a(x) is a continuous function satisfying
|a(x)| ≤ C
(1 + |x |)β , β > 2.
Here, we do not assume that a is negative. So∫
R2
(|Du|2 + au2)
may not be a norm. Weneed to discuss the invertiblity of Lw in a suitable weighted space.
Define
ρ(x) = (1 + |x |)1+ α2 , ρ(x) = 1
(1 + |x |) (ln(2 + |x |))1+ α2,
where α > 0 is a fixed small constant.Let Xα be the closure of C∞
0 (R2) under the norm
‖ξ‖Xα =(∫
R2|�ξ |2ρ2 +
∫
R2ξ2ρ2
) 12
,
while Yα is the closure of C∞0 (R2) under the norm
‖ξ‖Yα =(∫
R2ξ2ρ2
) 12
.
Then we have,
Theorem A.1. Suppose that L has trivial kernel in L∞(R2). If w satisfies Lw = h inR2, and w ∈ Xα and h ∈ Yα , then
‖w‖Xα + ‖w‖L∞(R2) ≤ C‖h‖Yα ,
where C > 0 is a constant, independent of w and h.
Proof. The proof of this theorem is exactly the same as that of Theorem 4.1 in [5]. ��By Theorem 2.1 in [5], we know that if N > 4, then
�u + eu(1 − eu) = 4πδ0, in R2, (A.2)
has a unique radial solution W (|x |), satisfying −tW ′(t) → 2N as t → +∞.In the following, we consider the invertibility of the linear operator:
L1w = �w + eW (r)(1 − 2eW (r))w, in R2.
For this purpose and the discussion of the system, we also introduce
L2w = �w − eW (r)w, in R2.
Lemma A.2. Suppose that w is a bounded solution of Liw = 0, i = 1, 2. Then w = 0.
748 C.-S. Lin, S. Yan
Proof. The result for i = 1 follows from the first part of Corollary 2.8 in [5].Suppose that w is bounded in R2 satisfying L2w = 0. It is easy to check that for any
φ ∈ C∞0 (R2) with φ ≥ 0,
∫
R2w+�φ ≥
∫
w>0φ�w =
∫
R2eW (|x |)w+φ ≥ 0.
Thus, w+ is a bounded sub-harmonic function in R2. So w+ must be a constant.Similarly, w− is a bounded super-harmonic function in R2. So w− must be a constant.
As a result, w = 0. ��One of the main results in this section is
Theorem A.3. Li is an isomorphism from Xα to Yα . Moreover, if w satisfies Liw = hin R2, and w ∈ Xα and h ∈ Yα , then
‖w‖Xα + ‖w‖L∞(R2) ≤ C‖h‖Yα ,
where C > 0 is a constant, independent of w and h.
Proof. It follows from Theorem A.1, Lemma A.2 that we only need to prove that Li isa onto map from Xα to Yα .
We consider L2 first. By the definition of Yα , we only need to show that for anyh ∈ C∞
0 (R2), there is a w ∈ Xα , such that L2w = h. Suppose that spt h ⊂ BR(0).Consider
�w − eW w = h, in Bn(0), w ∈ H10 (Bn(0)). (A.3)
By the Riesz representation theorem, (A.3) has a solution wn , satisfying∫
R2(|Dwn|2 + eW w2
n) ≤ C. (A.4)
We claim that
|wn| ≤ C. (A.5)
To prove this, for any n > R and x ∈ BR(0),
supB1(x)
|wn| ≤ C‖wn‖L2(B2(x)) + C‖h‖L2(B2(x))
≤ C‖e12 W wn‖L2(B2(x)) + C‖h‖L2(B2(x)) ≤ C.
Moreover, we have
�wn − eW wn = 0, in Bn(0)\BR(0),
from which we find
|wn(y)| ≤ max|x |=R|wn(x)| ≤ C, ∀ y ∈ Bn(0)\BR(0).
So we have prove (A.5).
Bubbling Solutions for Abelian Chern-Simons Model 749
From (A.4) and (A.5), we know that there is w, satisfying
�w − eW w = h, in R2, w ∈ L∞(R2),
∫
R2(|Dw|2 + eW w2) ≤ C. (A.6)
We claim that w ∈ Xα . In fact, noting that ρ2eW ≤ C , we obtain∫
R2|�w|2ρ2 ≤
∫
R2h2ρ2 +
∫
R2e2W w2ρ2
≤ C + C∫
R2eW w2 < ∞,
which, together with w ∈ L∞(R2), implies w ∈ Xα .Next, we consider L1. Write
L1w = L2w + 2eW (1 − eW )w.
Then L1w = h, h ∈ Yα is equivalent to
w = L−12 (2eW (1 − eW )w) + L−1
2 h, w ∈ Xα. (A.7)
It is easy to check that for any w ∈ Xα , 2eW (1 − eW )w ∈ Yα by using the fact thate2W ρ2ρ−2 ≤ C . Thus, L−1
2 (2eW (1−eW )w) is defined. On the other hand, L−12 (2eW (1−
eW )w) is a compact operator in Xα . To check this, we just need to use
(i) e2W ρ2ρ−2 → 0 as |x | → ∞;(ii) any bounded sequence in Xα has a subsequence which is strongly convergent for
any R > 0 in
Yα,R ={
ξ :(∫
BR(0)
ξ2ρ2) 1
2
< ∞}
.
By the Fredholm alternative, (A.7) has solution if w = L−12 (2eW (1 − eW )w) just has
zero solution. By Lemma A.2, we obtain the result. ��
Appendix B. Invertibility of Some Linear Operators
In this section, we will discuss the invertibility of the following linear operators
Lε,1w = �w +1
ε2 fε,1(x)w,
and
Lε,2w = �w +1
ε2 fε,2(x)w,
where
fε,1(x) =⎧⎨
⎩−e
W( |x−p1|
ε
) (2e
W( |x−p1|
ε
)
− 1
), x ∈ B2d(p1);
0, x ∈ �\B2d(p1),
750 C.-S. Lin, S. Yan
fε,2(x) ={
−eW
( |x−p1|ε
)
, x ∈ B2d(p1);0, x ∈ �\B2d(p1),
d > 0 is a fixed small constant, and W (|x |) is the solution of
�u + eu(1 − eu) = 4πδ0, in R2, (B.1)
satisfying −tW ′(t) → 2N as t → +∞.First, let us recall that ξ ∈ Xα,ε if ‖ξ‖2
Xα,ε< ∞, where ‖ξ‖2
Xα,εis defined in (3.6);
while ξ ∈ Yα,ε if ‖ξ‖2Yα,ε
< ∞, where ‖ξ‖2Yα,ε
is defined in (3.7).The main result of this section is the following:
Theorem B.1. Lε,i is an isomorphism from Xα,ε to Yα,ε. Moreover, if wε ∈ Xα,ε andhε ∈ Yα,ε satisfy
Lε,iwε = hε,
then there is a constant C > 0, independent of ε, such that
‖wε‖L∞(�) + ‖wε‖Xα,ε ≤ C
(ln
1
ε
)‖hε‖Yα,ε .
When hε has compact support in B2d(p1), after blowing-up at p1, the problemLε,iw = hε is a perturbation problem of Liw = hε(εx + p1) discussed in Appendix A.So we just need to concentrate on the case that hε = 0 in B2d(p1). We will prove
Lemma B.2. Assume hε = 0 in B2d(p). Suppose that wε ∈ Xα,ε and hε ∈ Yα,ε satisfy
�wε +1
ε2 fε,i (x)wε = hε.
Then there is a constant C > 0, independent of ε, such that
‖wε‖L∞(�) + ‖wε‖Xα,ε ≤ C
(ln
1
ε
)‖hε‖Yα,ε .
The proof of Lemma B.2 is quite long. We leave it to the end of this section.
Proof of Theorem B.1. Let
Lε,1w = �w +1
ε2 eW (|x−p1|
ε)(
1 − 2eW (|x−p1|
ε))
w,
and
Lε,2w = �w − 1
ε2 eW (|x−p1|
ε)w.
Then, we may regard Lε,i as a cut-off of Lε,i .Suppose that wε ∈ Xα,ε and hε ∈ Yα,ε satisfy
Lε,iwε = hε.
Bubbling Solutions for Abelian Chern-Simons Model 751
Let h∗ε = hε in B2d(p), and h∗
ε = 0 in �\B2d(p). By Theorem A.3, there is w∗ε ∈ Xα ,
such that
Lε,iw∗ε = h∗
ε , in R2,
and(∫
R2
(|�w∗|2ρ2 + |w∗|2ρ2
)) 12
+ ‖w∗‖L∞ ≤ C
(∫
R2|h∗|2ρ2
) 12
, (B.2)
where w∗(y) = w∗(εy + p1) and h∗(y) = h∗(εy + p1).Let ξ ∈ C∞
0 (B2d+δ(p1)), with ξ = 1 in B2d(p1), 0 ≤ ξ ≤ 1, where δ > 0 is a smallnumber. Then
Lε,i (wε − ξw∗ε ) = hε,
where
hε = hε − ξh∗ε − 2Dξ Dw∗
ε − �ξw∗ε +
(Lε,i (ξw∗
ε ) − Lε,i (ξw∗ε )
).
Note that hε = 0 in B2d(p). By Lemma B.2,
‖wε − ξw∗ε‖Xα,ε + ‖wε − ξw∗
ε‖L∞ ≤ C ln1
ε‖hε‖Yα,ε ≤ C ln
1
ε‖hε‖Yα,ε . (B.3)
In the last inequality, we have used (B.2). So, we obtain
‖wε‖L∞(�) + ‖wε‖Xα,ε
≤ ‖wε − ξw∗ε‖Xα,ε + ‖wε − ξw∗
ε‖L∞ + ‖ξw∗ε‖Xα,ε + ‖ξw∗
ε‖L∞
≤ C
(ln
1
ε
)‖hε‖Yα,ε . (B.4)
To show that Lε,i is an onto map, we can rewrite
Lε,iw = h,
as
w = −(� − I )−1(
1
ε2 fε,i (x)w + w
)+ (� − I )−1h, w ∈ Xα,ε.
Since � is bounded, it is easy to check that −(� − I )−1( 1ε2 fε,i (x)w + w) is compact.
Thus the result follows from the Fredholm alternative. ��We now go back to the proof of Lemma B.2. In Lemma B.2, we assume hε = 0 in
B2d(p1). So we need to study the solution of the following problem:
�wε +1
ε2 fε,i (x)wε = 0, x ∈ B2d(p1). (B.5)
Define f1(t) = eW (t)(1 − 2eW (t)) and f2(t) = −eW (t). It is easy to show that thefollowing problem:
�ξ + fi (t)ξ = 0, ξ(0) = 1, (B.6)
752 C.-S. Lin, S. Yan
has a radial solution ξi , satisfying
ξi (t) = ξi ln t + Ii + O
(1
tτi
), (B.7)
for some constants ξi �= 0, Ii and τi > 0.The major step in the proof of Lemma B.2 is to show that the solution of (B.5) is a
perturbation of the radial solution ξi
( |x−p1|ε
). The following decomposition lemma is
crucial to the proof of Lemma B.2.
Lemma B.3. Let wε,i be a solution of (B.5). Then
wε,i (x) = wε,i (p1)ξi
( |x − p1|ε
)+ φε, in B2d(p1),
where ξi is the solution of (B.6), φε satisfies{
�φε + 1ε2 fε,i (x)φε = 0, x ∈ B2d(p1),
φε(p1) = 0.(B.8)
Moreover, the maximum point xε of |φε| in B2d(p1) satisfies
|xε − p1| → 2d, as ε → 0,
and (up to a subsequence)
φε
με
→ φ, in C2loc(B2d\{p1}),
where με = maxx∈B2d (p1) |φε(x)|, and φ is a harmonic function in B2d(p1) withφ(p1) = 0.
Proof. Step 1. Let xε ∈ B2d(p1) be a maximum point of |φε| in B2d(p1). We claim
|xε − p1| ≥ r0 > 0.
We argue by contradiction. Suppose that xε → 0. First, we claim that
|xε − p1|ε
→ +∞. (B.9)
Otherwise, define φε(y) = φε(εy+p1)φε(xε)
. Then |φε| ≤ 1 and φε(yε) = 1 with yε = xε−p1ε
→y0 ∈ R2. Using |φε| ≤ 1 , we may assume that φε → φ, and φ is a bounded solution of
�φ + fi (|x |)φ = 0.
By Lemma A.2, φ = 0, which contradicts φε(yε) = 1. So, we have proved the claim.Let rε = |xε − p1|. Define
φε(y) = φε(rε y + p1)
φε(xε).
Bubbling Solutions for Abelian Chern-Simons Model 753
Then,
�φε +1
η2ε
fε,i (rε y + p1)φε = 0,
|φε| ≤ 1, and φε
(xε−p1
rε
)= 1, where ηε = ε
rε. Using (B.9), we see that φε con-
verges uniformly to a bounded harmonic function φ in any compact subset of R2\{0}.By Liouville theorem, φ = 1 in R2. In particular, for any large constant C > 0, andC−1 ≤ |x−p1|
rε≤ C , we have
φε(x)
φε(xε)≥ 1
2. (B.10)
Without loss of generality, we may assume φε(xε) > 0. Define
φ∗ε (r) =
∫ 2π
0φε(r, θ) dθ, r = |x − p1|.
Then, φ∗ε satisfies
{�φ∗
ε + 1ε2 fε,i (x)φ∗
ε = 0, in B2d(0)
φ∗ε (0) = 0.
(B.11)
Thus, by the uniqueness of the initial value problem for the ordinary differential equa-tions, φ∗
ε = 0. On the other hand, by (B.10),
φ∗ε (r) > 0, C−1rε ≤ r ≤ Crε.
This is a contradiction. So, we have proved |xε − p1| ≥ r0 > 0.
Step 2. Since fi
( |x−p1|ε
)→ 0 uniformly in any compact subset of B2d(p1)\{p1}, we
may assume that φε
μεconverge to a harmonic function φ in C1
loc(B2d(p1)\{p1}). We claimφ(p1) = 0 if φ �= 0. Suppose not. We may assume that φ(p1) > 0. Then, we can findr0 > 0, such that φε(x) > c′με, |x − p1| = r0, for some constant c′ > 0. Define
φ∗ε (r) =
∫ 2π
0φε(r, θ) dθ.
Then φ∗ε (r0) > 0. So we can obtain a contradiction as in Step 2. Therefore, φ(p1) = 0.
Finally, if φ �= 0, the maximum point of |φ| must be on the boundary of B2d(p1).Thus, |xε − p1| → 2d. If φ = 0, then φε(x) = o(1)με in B2d−θ (p1) for any θ > 0. Sowe also have |xε − p1| → 2d. ��Proof of Lemma B.2. We argue by contradiction. Suppose that there are εn → 0,wn ∈ Xα,εn , hn ∈ Yα,εn , hn = 0 in B2d(p1), satisfying
�wn +1
ε2 fε,i (x)wn = hn, in �,
754 C.-S. Lin, S. Yan
such that
‖wn‖Xα,εn+ ‖wn‖L∞(�) = 1, ‖hn‖Yα,εn
= o(1)
(ln
1
ε
)−1
. (B.12)
By Lemma B.3, wn has the following decomposition:
wn = mnξi
( |x − p1|ε
)+ φn, in B2d(p1),
with mn = wn(p1). We may assume that mn ξi ≥ 0. Thus,
mnξ0
( |x − p1|ε
)≥ 0, ∀|x − p1| ≥ d.
Recall μn = maxx∈B2d (p1) |φn(x)|. Now we define
μn = maxx∈B 7
4 d(p1)
|φn(x)|. (B.13)
The crucial step to obtain a contradiction is to show that φn is a perturbation term of
mnξi
( |x−p1|ε
). We will prove
limn→+∞
μn
mn ln 1εn
< +∞, limn→+∞
μn
mn ln 1εn
= 0. (B.14)
Assume (B.14) at the moment. We are now ready to obtain a contradiction. By (B.14),we find
This is a contradiction. Thus, we prove (B.25).Next, we show
μn ≤ C |mn| ln1
εn. (B.27)
Suppose we have
|mn| ln 1εn
μn→ 0. (B.28)
Then, from Lemma B.3 and (B.28), we find
τn = maxx∈B2d (p1)
(φεn (x) + o(με)) = (1 + o(1))μn . (B.29)
Let wn = wnτn
. Then |wn| ≤ 1 in B2d(p1), and wn satisfies
�wn +1
ε2n
fεn ,i (x))wn = hn
τn. (B.30)
Using (B.25) and the L p estimate, we can deduce from (B.30) that
‖wn‖L∞(�\B2d (p1)) ≤ C +C‖hn‖L2(�\B2d (p1))
τn≤ C.
Bubbling Solutions for Abelian Chern-Simons Model 757
Thus, wn is a bounded function in �. So, wn converges in Cloc(�\{p1}) to a boundedharmonic function ω1. Noting that hn = 0 in B2d(p1), similar to the proof of Lemma B.3,we can prove that the maximum point xn of wn satisfies |xn − p1| ≥ r0 > 0. Thus,ω1 = 1. On the other hand, from Lemma B.3, (B.28) and (B.29),
wn = wn
μn
μn
τn→ φ, in B2d(p1).
So, φ = ω1 = 1 in B2d(p1). But φ(p1) = 0. This is a contradiction and (B.27) isproved.
Using Lemma B.3 and (B.27), we obtain
τn ≤ C |mn| ln1
εn+ μn ≤ C |mn| ln
1
εn. (B.31)
Define wn = wn
|mn | ln 1εn
. Then wn is bounded in B2d(p1) and satisfies
�wn +1
ε2n
fεn ,i (x)wn = hn
|mn| ln 1εn
. (B.32)
From (B.25) and (B.31), we find
‖hn‖Yα,εn
|mn| ln 1εn
≤ ‖hn‖Yα,εn
τn
τn
|mn| ln 1εn
→ 0.
As a result, we can deduce as above that wn → C in Cloc(�\{p1}). But
wn = mnξi
|mn| ln 1εn
+μn
|mn| ln 1εn
φn
μn.
Taking a limit in the above relation, we obtain
C = C1 + limn→+∞
μn
|mn| ln 1εn
φ, in B2d(p1)\{p1},
where C1 is a constant. If φ �= 0, then φ is not a constant, since φ(p1) = 0. Thus
limn→+∞
μn
|mn| ln 1εn
= 0.
So, we prove (B.14). If φ = 0, then μn = o(1)μn . So, in this case, we obtain the resultfrom (B.27). ��
758 C.-S. Lin, S. Yan
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