Mean field equation for equilibrium vortices with neutral orientation

Nonlinear Analysis 66 (2007) 509–526www.elsevier.com/locate/na

Mean field equation for equilibrium vortices withneutral orientation

Ken Sawadaa, Takashi Suzukia, Futoshi Takahashib,∗a Division of Mathematical Science, Department of System Innovation, Graduate School of Engineering Science,

Osaka University, Toyonaka, Japanb Department of Mathematics, Graduate School of Science/Faculty of Science, Osaka City University, Sumiyoshi-ku,

Osaka, Japan

Received 26 August 2005; accepted 29 November 2005

Abstract

This paper is concerned with the equilibrium mean field equation of many vortices of the perfect fluid

with neutral orientation, −�v = λ

(ev∫

Ω evdx− e−v∫

Ω e−vdx

)in Ω , v = 0 on ∂Ω , where Ω ⊂ R2 is a bounded

domain with smooth boundary ∂Ω , and λ ≥ 0 is a constant.Using the isoperimetric inequality of [T. Suzuki, Global analysis for a two-dimensional elliptic

eigenvalue problem with the exponential nonlinearity, Ann. Inst. H. Poincare 9 (1992) 367–398] and meanvalue theorem of [T. Suzuki, Semilinear Elliptic Equations, Gakkotosho, Tokyo, 1994], we prove the linearstability and a priori estimate of any solution under some assumptions on the domain and the parameterλ, which lead to the uniqueness theorem of the trivial solution on a simply connected domain and thecalculation of the Leray–Schauder degree on any domain for λ in a certain range.c© 2005 Elsevier Ltd. All rights reserved.

Keywords: Mean field equation; Spectrum analysis; Isoperimetric inequality; Mean value theorem

1. Introduction

This paper is concerned with the equilibrium mean field equation of many vortices of theperfect fluid with neutral orientation,

−�v = λ

(ev∫

Ω evdx− e−v∫

Ω e−vdx

)in Ω , v = 0 on ∂Ω , (1.1)

∗ Corresponding author.E-mail addresses: [email protected] (T. Suzuki), [email protected] (F. Takahashi).

0362-546X/$ - see front matter c© 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2005.11.044

http://www.elsevier.com/locate/na

mailto:[email protected]

mailto:[email protected]

http://dx.doi.org/10.1016/j.na.2005.11.044

510 K. Sawada et al. / Nonlinear Analysis 66 (2007) 509–526

where Ω ⊂ R2 is a bounded domain with smooth boundary ∂Ω , and λ ≥ 0 is a constant. It isobtained by Pointin and Lundgren [13], with v and λ derived from the stream function and theinverse temperature, respectively. If these vortices are provided with the same orientation, thenwe have

−�v = λev∫

Ω evdxin Ω , v = 0 on ∂Ω . (1.2)

See [3,18,19,5,6,9,10].Eq. (1.2) has been studied in detail, and for instance the following results are known. First, we

have the quantization of λ in 8πN for the non-compact solution sequence with the classificationof the singular limit using the Green’s function [11]. Second, there is a uniqueness of the solutionif λ ∈ [0, 8π) and Ω is simply connected [15]. Third, singular perturbation of the solution isvalid from the generic singular limit [2]. Finally, for each component of [0,+∞) \ 8πN we havethe topological degree calculation [8]. We refer the reader also to [16,7,17] and the referencestherein, for later developments and related backgrounds.

Similarly to (1.2), problem (1.1) is provided with a variational structure, and it is equivalentfor v to be a critical point of the variational functional

Jλ(v) = 1

2‖∇v‖2

2 −λ(

log∫Ω

evdx + log∫Ω

e−vdx

)(1.3)

defined for v ∈ H 10 (Ω). In more detail, by the standard argument using the Trudinger–Moser

inequality, any critical point v ∈ H 10 (Ω) is actually a classical solution to (1.1). In contrast with

the case for (1.2), on the other hand, v ≡ 0 is always a solution to (1.1) which we call the trivialsolution, and furthermore, if v is a solution to (1.1), then so is −v.

In the present paper, we show the following theorems concerning the solution set of (1.1).

Theorem 1. If Ω is simply connected and 0 ≤ λ ≤ 4π , then (1.1) possesses no non-trivialsolution.

Theorem 2. The Leray–Schauder degree d(λ) for (1.1) is well defined, and it holds that d(λ) =1 for λ ∈ [0, 8π).

These theorems are direct consequences of the following lemmas because only the trivialsolution v = 0 is admitted for λ = 0 in (1.1).

Lemma 1. If Ω is simply connected and 0 ≤ λ ≤ 4π , then any solution to (1.1) is linearlystable.

Lemma 2. Given ε > 0, we have C > 0 such that

‖v‖∞ ≤ C

for any solution v to (1.1) with 0 ≤ λ ≤ 8π − ε.

Here and henceforth, as usual we say that a critical point v of Jλ is linearly stable if thelinearized operator L around v, realized as a self-adjoint operator in L2(Ω) with the domainD(L) = (H 1

0 ∩ H 2)(Ω), is positive definite. In more detail, this L is defined by

K. Sawada et al. / Nonlinear Analysis 66 (2007) 509–526 511

〈Lφ, φ〉 = d2

ds2Jλ(v + sφ)

∣∣∣∣s=0

= (∇φ,∇φ)− λ

{∫Ωρ+φ2dx − (ρ+, φ)2 +

∫Ωρ−φ2dx − (ρ−, φ)2

}

for φ ∈ H 10 (Ω) and

ρ± = e±v∫Ω e±vdx

, (1.4)

with 〈 , 〉 and (, ) standing for the H 10 (Ω)

′−H 10 (Ω) duality and the L2 inner product, respectively,

and therefore, we have

Lφ = −�φ − λ {ρ+φ − (ρ+, φ)ρ+ + ρ−φ − (ρ−, φ)ρ−} (1.5)

with D(L) = (H 10 ∩ H 2)(Ω).

This paper is composed of seven sections. Taking preliminaries in Section 2, Lemmas 1 and 2are proven in Sections 3 and 4, respectively. Lemma 1 is proven using the transformationemployed by [18,15] and an isoperimetric inequality of [15], while Lemma 2 is proven by a meanvalue theorem of [16]. On the basis of these lemmas, Theorems 1 and 2 are proven in Section 5.

In Section 6, we show some results on the mixed vortex equation with a different ratio,

−�v = λev∫

Ω evdx− μ

e−v∫Ω e−vdx

in Ω , v = 0 on ∂Ω , (1.6)

where λ and μ are non-negative parameters. Obviously, if λ �= μ, then v ≡ 0 is not a solution to(1.6). Section 7 is an appendix, and some facts used in the preceding sections are proven.

2. Preliminaries

This section is concerned with the trivial solution to (1.1). We study its linearized stability andsee for what parameter λ it becomes unstable.

First, we recall the following lemma of [18,15] used in the study of (1.2), where the linearizedoperator around the solution v is given by

Lφ = −�φ − λρ {φ − (ρ, φ)} (2.1)

with D(L) = (H 2 ∩ H 10 )(Ω) and ρ = ev/

∫Ω evdx . Its proof is given in Section 7 for the reader’s

convenience.

Lemma 3. Let ρ = ρ(x) > 0 be a continuous function defined on Ω , and L be the self-adjointoperator in L2(Ω) defined by (2.1) with the domain D(L) = (H 1

0 ∩ H 2)(Ω). Then, counting the

multiplicity, the number of negative eigenvalues of L is equal to the number of eigenvalues μ inthe range 0 < μ < 1 of the eigenvalue problem

−�ψ = μλρψ in Ω , ψ = constant on ∂Ω ,∫∂Ω

∂ψ

∂νds = 0. (2.2)

The above lemma is applicable to the trivial solution v ≡ 0 to (1.1). In fact, we have

ρ+ = ρ− = ρ0 for ρ0 ≡ 1

|Ω | ,


and therefore, the linearized operator around this v ≡ 0 is given by

L0φ = −�φ − 2λρ0 {φ − (ρ0, φ)}with the domain D(L0) = (H 1

0 ∩ H 2)(Ω) by (1.5). Thus, by Lemma 3, the Morse index of thetrivial solution v ≡ 0 to (1.1) is equal to the number of eigenvaluesμ in the range 0 < μ < 1 for

−�ψ = 2μλρ0ψ in Ω , ψ = constant on ∂Ω ,∫∂Ω

∂ψ

∂νds = 0. (2.3)

This means

−�ψ = σψ in Ω , ψ = constant on ∂Ω ,∫∂Ω

∂ψ

∂νds = 0 (2.4)

for σ = 2μλρ0, and obviously, its first eigenvalue is 0 and the corresponding eigenfunction is aconstant.

The second eigenvalue, therefore, is equal to

σ2 = inf

{‖∇ψ‖2

2 | ψ ∈ H 1c (Ω), ‖ψ‖2 = 1,

∫Ωψdx = 0

}(2.5)

for

H 1c (Ω) =

{ψ ∈ H 1(Ω) | ψ = constant on ∂Ω

} ∼= H 10 (Ω)⊕ R.

Given

ψ ∈ H 1c (Ω), ‖ψ‖2 = 1,

∫Ωψdx = 0,

we have

φ = ψ − c ∈ H 10 (Ω)

for c = ψ|∂Ω . Then it holds that

1 = ‖ψ‖22 = ‖φ‖2

2 + 2c∫Ωφdx + c2 |Ω |

0 =∫Ωψdx =

∫Ωφdx + c |Ω | ,

and hence

‖φ‖22 = 1 + 1

|Ω |(∫

Ωφdx

)2

.

Thus, we obtain

σ2 = inf

{‖∇φ‖2

2 | φ ∈ H 10 (Ω), ‖φ‖2

2 = 1 + 1

|Ω |(∫

Ωφdx

)2}. (2.6)

This σ2 gives the second eigenvalue of (2.3), denoted by μ2 = σ2/(2λρ0), and if μ2 < 1, thatis,

λ >σ2

2|Ω | , (2.7)


Fig. 1. A bifurcation diagram.

then the trivial solution v = 0 is unstable, and therefore, a branch of non-trivial solutions emergesfrom (σ2 |Ω | /2, 0) in the λ–v plane.

When Ω is the unit disk B = {x ∈ R2 | |x | < 1

}, this σ2 is associated with a zero of a

Bessel function. More precisely, the solution ψ to (2.4) for Ω = B is given by the separation ofvariables, and it holds that

ψ(r, θ) = ψn,m(r, θ) = an,m Jn(√σn,mr)

{cos nθsin nθ

for some n = 0, 1, 2, . . . and m = 1, 2, . . . in polar coordinates, where an,m > 0 is a normalizingconstant, Jn(ρ) is the Bessel function of order n:

d2 J

dρ2+ 1

ρ

dJ

dρ+(

1 − n2

ρ2

)J = 0,

and σn,m > 0 is a constant defined by the boundary condition, i.e.,

√σn,m =

{m-th positive root of J ′

0(ρ) if n = 0m-th positive root of Jn(ρ) if n ≥ 1.

Since J ′0(ρ) = −J1(ρ), the second eigenvalue σ2 of (2.6) has the multiplicity 2, σ2 = σ0,1 =

σ1,1, and this value is approximately (3.832 · · ·)2. In other words, the trivial solution v ≡ 0 to(1.1) becomes unstable at

λ∗ = σ2

2|Ω | � 7.342π

for Ω = B , and a branch of non-trivial solutions emerges from (λ∗, 0) in the λ–v plane. Inparticular, the condition 0 ≤ λ ≤ 4π is not improved to 0 ≤ λ ≤ 8π in Theorem 1. See Fig. 1.

3. Spectrum of the linearized operator

This section is devoted to the proof of Lemma 1. Given a solution v to (1.1), let Q = Q( , )be the quadratic form on H 1

0 (Ω) associated with the linearized operator L defined by (1.5) with


the domain D(L) = (H 10 ∩ H 2)(Ω):

Q(φ, φ) = (∇φ,∇φ)− λ

{∫Ωρ+φ2dx − (ρ+, φ)2 +

∫Ωρ−φ2dx − (ρ−, φ)2

}.

Then, putting

φ± = φ − (ρ±, φ), (3.1)

we have

Q(φ, φ) = 1

2(∇φ+,∇φ+)− λ

∫Ωρ+φ2+dx + 1

2(∇φ−,∇φ−)− λ

∫Ωρ−φ2−dx,

with the right-hand side decomposed as

Q(h, h) ≡ A(h, h)− B(h, h)

for h = t (φ+, φ−) ∈ V , where

V ={

h =(φ+φ−

)| φ± ∈ H 1

c (Ω), φ+ − φ− = constant

}

A(h, h) = 1

2{(∇φ+,∇φ+)+ (∇φ−,∇φ−)}

B(h, h) = λ

{∫Ωρ + φ2+dx +

∫Ωρ−φ2−dx

}.

These A and B are extended as quadratic forms on V , and the above structure induces theeigenvalue problem; find h ∈ V and σ ∈ R satisfying

A(h, g) = σ B(h, g) for any g ∈ V . (3.2)

For this problem, we have a countably infinite discrete set of non-negative eigenvalues{σ j}∞

j=1

and a basis of V composed of eigenfunctions denoted by{h j =

(φ

j+φ

j−

)}j=1,2,...

,

such that

B(hi , h j ) = δi j .

See [1] for this well-known fact.Now, we prove the following.

Lemma 4. Let λ > 0 and X be the maximal subspace of V on which Q is negative definite.Then, the Morse index of the solution v to (1.1) is given by dim X − 2, which is also equal to thenumber of the eigenvalues of (3.2) in the range 0 < σ < 1 counting the multiplicity.

Proof. Since Q(h, h) < 0 is equivalent to A(h, h)/B(h, h) < 1 for h ∈ V , we confirm thatthe number of eigenvalues σ in the range [0, 1) is identical to the dimension of the maximalsubspace X in V on which Q is negative definite. The least eigenvalue of (3.2) is σ = 0 with the


multiplicity 2, and(10

),

(01

)

are corresponding eigenfunctions. Therefore, the number of eigenvalues of (3.2) in the range0 < σ < 1 is equal to dim X − 2.

Let

V0 = span

⟨(10

),

(01

)⟩⊂ V ,

and V1 be its orthogonal complement in V with respect to B . Thus, h = t (φ+, φ−) ∈ V belongs

to V1 if and only if

B

(h,

(10

))= B

(h,

(01

))= 0

which is equivalent to∫Ω(ρ+φ+ · 1 + ρ−φ− · 0)dx =

∫Ω(ρ+φ+ · 0 + ρ−φ− · 1)dx = 0.

In other words,

h =(φ+φ−

)∈ V1 ⇔

∫Ωρ±φ±dx = 0 for h ∈ V . (3.3)

Given h = t (φ+, φ−) ∈ V , on the other hand, we have φ+ − φ− = constant, and henceφ+ − φ+|∂Ω = φ− − φ−|∂Ω . This means well-definedness of

φ ≡ φ± − φ±|∂Ω ∈ H 10 (Ω). (3.4)

It is obvious, conversely, that φ± = φ+c± with φ ∈ H 10 (Ω) and c± ∈ R implies t (φ+, φ−) ∈ V ,

and thus, (3.4) induces an isomorphism,

V ∼= H 10 (Ω)⊕ R2.

From (3.3), h ∈ V , identified with (φ, c±) ∈ H 10 (Ω)⊕ R2, is in V1 if and only if

c± = −∫Ωρ±φdx .

This implies∫Ωρ±φ2±dx =

∫Ωρ±φ2dx + 2c±

∫Ωρ±φdx + c2±

=∫Ωρ±φ2dx − c2± =

∫Ωρ±φ2dx −

(∫Ωρ±φdx

)2

,

and therefore,

V1 ∼= H 10 (Ω) and Q(h, h) = Q(φ, φ) (3.5)

for h ∈ V1 identified with φ ∈ H 10 (Ω).


In particular, the maximal subspace Y in H 10 (Ω) on which Q is negative definite corresponds

isomorphically to the maximal subspace Y in V1 on which Q is negative definite. Since thedimension of Y is the Morse index of v and X = Y ⊕ V0, Lemma 4 is proven. �

To prove Lemma 1, we use the isoperimetric inequality of [15]. Originally, it is shown for∫Ω pdx < 8π , but the case

∫Ω pdx = 8π is also included [7].

Lemma 5. If Ω ⊂ R2 is a bounded simply connected domain and p ∈ C2(Ω) ∩ C0(Ω) is apositive function satisfying

−� log p ≤ p in Ω ,∫Ω

pdx ≤ 8π,

then it holds that

inf

{∫Ω

|∇φ|2 dx | φ ∈ H 1c (Ω),

∫Ω

pφ2dx = 1,∫Ω

pφdx = 0

}> 1. (3.6)

Proof of Lemma 1. Let v be a solution to (1.1) for 0 ≤ λ ≤ 4π and define ρ± by (1.4).Regarding (3.5), we show Q(h, h) > 0 for each h ∈ V1 \ {0}.

In fact, for p± = 2λρ± we have

−� log p± = ∓�v = ±λ(ρ+ − ρ−) ≤ 2λρ± = p± in Ω

and also∫Ω

p±dx = 2λ∫Ωρ±dx ≤ 8π.

On the other hand, it holds that

Q(h, h) = 1

2Q+(φ+, φ+)+ 1

2Q−(φ−, φ−)

for h = t (φ+, φ−) ∈ V , where

Q±(φ, φ) =∫Ω

|∇φ|2 dx −∫Ω

p±φ2dx,

and we have also

V1 ={(φ+φ−

)| φ± ∈ H 1

c (Ω), φ+ − φ− = constant,∫Ωρ±φ±dx = 0

}⊂ V +

1 × V −1

for

V ±1 =

{φ± ∈ H 1

c (Ω) |∫Ω

p±φ±dx = 0

}.

Finally, we obtain

Q±(φ±, φ±) =∫Ω

|∇φ±|2 dx − 2λ∫Ωρ±φ2±dx > 0

for each φ± ∈ V ±1 \ {0} by Lemma 5, and the proof is complete. �


4. A priori estimate of the solution

To prove Lemma 2, we use the following facts due to [16], which are some extensions ofthe mean value theorem for sub-harmonic functions. The proofs of these facts are provided inSection 7 for the reader’s convenience.

Lemma 6. Let Ω ⊂ R2 be an open set, and p = p(x) > 0 be a positive C2 function satisfying−� log p ≤ p in Ω . Then, we have

log p(x0) ≤ 1

|∂B|∫∂B

log p ds − 2 log

(1 − 1

8π

∫B

pdx

)+

log p(x0) ≤ 1

|B|∫

Blog p dx − 2 log

(1 − 1

8π

∫B

pdx

)+

(4.1)

for any x0 ∈ Ω and r > 0 sufficiently small such that B = Br (x0) ⊂⊂ Ω , where (·)+ =max(·, 0).

Lemma 7. Under the assumptions of the previous lemma, we have

maxω

p ≤(

1 − 1

8π

∫ω

pdx

)−2

+max∂ω

p

for each subdomain ω ⊂⊂ Ω with smooth boundary ∂ω.

Let {(λk, vk)} be a solution sequence to (1.1) with λk → λ0 for some λ0 ≥ 0 and‖vk‖∞ → +∞ as k → ∞. We introduce the blow-up set of the sequence (λk , vk), denotedby S, S = S+ ∪ S−, where

S± = {x0 ∈ Ω | there exists xk ∈ Ω such that xk → x0 and vk(xk) → ±∞}

.

First, we show the finiteness of the interior blow-up point.

Lemma 8. There holds

�(S ∩ Ω) ≤ lim infk→∞

[λk

4π

], (4.2)

where [γ ] denotes the largest integer not exceeding γ ∈ R.

Proof. Putting pk± = λke±vk/∫Ω e±vk , we have

−� log pk± ≤ pk± in Ω

similarly, and therefore, Lemma 6 is applicable. We obtain

±vk(x0) ≤ ± 1

|B|∫

Bvkdx − 2 log

(1 − 1

8πλk±(B)

)+

for

λk±(B) =∫

B λke±vk dx∫Ω e±vk dx

(4.3)


and B = Br (x0) ⊂⊂ Ω , and therefore,

1

|B|∫

Bvkdx + 2 log

(1 − 1

8πλk−(B)

)+

≤ vk(x0)

≤ 1

|B|∫

Bvkdx − 2 log

(1 − 1

8πλk+(B)

)+. (4.4)

Here, we have

‖−�vk‖1 ≤ 2λk

and the elliptic L1 estimate of [4] guarantees lim supk→∞ ‖vk‖W 1,q < ∞ for any 1 ≤ q < 2. Inparticular, we confirm∣∣∣∣

∫Bvkdx

∣∣∣∣ = O(1) as k → ∞. (4.5)

If

limk→∞

(λk+(B)+ λk−(B)

)< 8π (4.6)

for a subsequence, then we have lim supk→∞ λk±(B) < 8π , which implies

lim supk→∞

λk±(Br/2(x

′0))< 8π

for each x ′0 ∈ Br/2(x0). Applying (4.4) and (4.5) for x0 = x ′

0 and B = Br/2(x ′0), we conclude

that {vk} is uniformly bounded in Br/2(x0) along this subsequence, which means x0 �∈ S.In other words, we obtain

lim infk→∞

(λk+(B)+ λk−(B)

)≥ 8π for each x0 ∈ S ∩ Ω (4.7)

and B = Br (x0) with r > 0 sufficiently small. Then, the standard covering argument guarantees(4.2). �

Lemma 9. If {(λk, vk)} is a solution sequence for (1.1) satisfying lim supk→∞ λk < +∞, thenwe have a subsequence, denoted by the same symbol, and a constant δ > 0 such that∫

Ωe±vk dx ≥ δ (k = 1, 2, . . .).

Proof. Passing to a subsequence, we have

λke±vk∫Ω e±vk dx

dx ⇀ μ±(dx)

in the sense of measures on Ω from the assumption of the boundedness of {λk}. Then, the set

S = {x0 ∈ Ω | μ+({x0})+ μ−({x0}) ≥ 8π}is finite, and if x0 ∈ Ω \ S, we have

λk+(B)+ λk−(B) < 8π,

for B = Br (x0) with 0 < r � 1 and λk±(B) defined by (4.3). This implies supk ‖vk‖L∞(B ′) <+∞ for B ′ = Br/2(x0) from the proof of Lemma 8, and therefore, {vk} is locally uniformlybounded in Ω \ S.


In particular, given a domain ω ⊂ Ω \ S , we have ‖vk‖L∞(ω) = O(1), and therefore,∫Ω

e±vk dx ≥∫ω

e±vk dx ≥ δ

for some δ > 0 independent of k. �

Proof of Lemma 2. Let {(λk, vk)} be a solution to (1.1) satisfying λk → λ0 ∈ [0, 8π), and setpk± = λke±vk/

∫Ω e±vk dx . Then, we have −� log pk± ≤ pk± in Ω and

∫ω

pk±dx < 8π for anyω ⊂⊂ Ω , and therefore,

maxω

pk± ≤(

1 − 1

8π

∫ω

pk±dx

)−2

max∂ω

pk±

by Lemma 7. Letting ω ↑ Ω and taking account of the boundary condition, we obtain

∥∥∥pk±∥∥∥

L∞(Ω)≤(

1 − λk

8π

)−2

λk

(∫Ω

e±vk dx

)−1

= O(1) (4.8)

by Lemma 9.This means that any ε > 0 admits C > 0 such that 0 ≤ λ ≤ 8π − ε implies ‖p±‖L∞(Ω) ≤ C .

Then, the standard elliptic estimate applied to (1.1) guarantees Lemma 2. �

5. Proof of Theorems 1 and 2

Proof of Theorem 1. Let (λ0, v0) be a non-trivial solution to (1.1) with 0 < λ0 ≤ 4π . Then, thelinearized operator around v0 is positive by Lemma 1. Applying the implicit function theorem,we find a branch of non-trivial solutions C emerging from (λ0, v0) in the λ–v plane in the rangeλ0 − δ < λ < λ0 + δ for some δ > 0. Continuing this process, we confirm that this branchcontinues up to λ = 0 by Lemma 2 and the elliptic estimate. However, only the trivial solutionis admitted when λ = 0, and thus, the branch C should intersect the branch of trivial solutions at(λ, 0) for some 0 < λ < 4π . This (λ, 0) must be a bifurcation point, which is a contradiction tothe fact C is a non-bifurcating branch. This proves Theorem 1. �

Proof of Theorem 2. All solutions to (1.1) stay bounded when λ lies in a compact interval of0 ≤ λ < 8π by Lemma 2. Then, from the elliptic estimates we can define the Leray–Schauderdegree d(λ) for any 0 ≤ λ < 8π . By the homotopy invariance, d(λ) is constant for 0 ≤ λ < 8πand we know that only the trivial solution with Morse index 0 exists for λ = 0. This impliesd(λ) = d(0) = 1, and the proof is complete. �

6. Mixed vortices with different ratio

In this section, we consider a generalization of (1.1), that is, (1.6). Its variational functional is

Jλ,μ(v) = 1

2

∫Ω

|∇v|2 dx − λ log∫Ω

evdx − μ log∫Ω

e−vdx (6.1)

defined for ∈ H 10 (Ω). For Iλ(v) = Jλ,0(v), the Trudinger–Moser inequality guarantees

I8π (±v) = 1

2

∫Ω

|∇v|2 dx − 8π log

(∫Ω

e±vdx

)≥ −C


for v ∈ H 10 (Ω), where C is a constant. Thus, we obtain

Jλ,μ(v) = 1

2

(1 − λ

8π− μ

8π

)∫Ω

|∇v|2 dx + λ

8πI8π(v) + μ

8πI8π (−v) ≥ −C,

provided that

1 − λ

8π− μ

8π≥ 0

i.e., λ+μ ≤ 8π . In the case of λ+μ < 8π , therefore, Jλ,μ is obviously coercive on H 10 (Ω). We

have a non-trivial solution of (1.6), a global minimizer of Jλ,μ on H 10 (Ω). However, this range

is extended as (λ, μ) ∈ [0, 8π)× [0, 8π), and more precisely, J8π,8π is bounded from below inH 1

0 (Ω). In fact, the proof of the second case of Theorem 5 of [14] is valid for this J8π,8π , whilea different proof is given in [12].

We note that this inequality is optimal. In fact, by Jensen’s inequality we have

− log

(1

|Ω |∫Ω

e±vdx

)≤ ∓ 1

|Ω |∫Ωvdx,

and hence it holds that

Jλ,μ(v) ≤ Iλ(v) + μ

(1

|Ω |∫Ωvdx − log |Ω |

).

Since infv∈H10 (Ω)

Iλ(v) = −∞ for λ > 8π , Jλ,μ is unbounded from below if λ > 8π fromPoincare’s inequality. Similarly, we have infv∈H1

0 (Ω)Jλ,μ(v) = −∞ if μ > 8π .

In this section, first, we show the following.

Theorem 3. If λ,μ ≥ 0, λ �= μ, 4(λ+ μ) ≤ α(λ − μ)2, and Ω ⊂ R2 is a star-shaped domain,then we have no solution to (1.6), where α > 0 is a constant satisfying

(x · ν) ≥ α |∂Ω | (6.2)

for any x ∈ ∂Ω .

Proof. First, Pohozaev’s identity applied to (1.6) yields

1

2

∫∂Ω

(∂u

∂ν

)2

(x · ν)ds = 2λ

(1 − |Ω |∫

Ω eudx

)+ 2μ

(1 − |Ω |∫

Ω e−udx

). (6.3)

On the other hand, by integrating Eq. (1.6), we have

−∫∂Ω

∂u

∂νds =

∫Ω

−�udx = λ− μ, (6.4)

and Schwarz’s inequality(∫∂Ω

∂u

∂νds

)2

≤(∫

∂Ω

(∂u

∂ν

)2

(x · ν)ds

)(∫∂Ω

1

(x · ν)ds

)

and the assumption (6.2) lead to

α

(∫∂Ω

∂u

∂νds

)2

≤∫∂Ω

(∂u

∂ν

)2

(x · ν)ds. (6.5)


From (6.3), we have

2(λ+ μ) >1

2

∫∂Ω

(∂u

∂ν

)2

(x · ν)ds.

Then, (6.5) and (6.4) imply

2(λ+ μ) >α

2

(∫∂Ω

∂u

∂νds

)2

= α

2(λ− μ)2.

Thus, we obtain the conclusion. �

If Ω = BR(0), we have α = (2π)−1, and therefore, there exists no solution to (1.6) onΩ = BR(0) when

8π(λ+ μ) ≤ (λ− μ)2, λ, μ ≥ 0, λ �= μ.

This result may be compared with the following.

Theorem 4. If A = {x ∈ R2 | a < |x | < b

}and

H 10,rad(A) =

{v ∈ H 1

0 (A) | v : is radially symmetric},

then there is a minimizer of

inf{

Jλ,μ(v) | v ∈ H 10,rad(A)

}for any λ ≥ 0 and μ ≥ 0, which solves (1.6).

The proof of the above theorem is standard, and is omitted. See [16] for example. Here, wenote the following.

(1) When λ �= μ, v ≡ 0 is not a solution, and therefore, the minimizer in Theorem 4 is notidentically zero.

(2) When λ = μ, the minimizer in Theorem 4 may be the trivial solution to (1.6).

In the second case, we put Jλ = Jλ,λ, and take v0 ∈ H 10,rad(A) and β > 0. Then

Jλ(βv0) = β2

2

∫A

|∇v0|2 dx − λ

(log

∫A

eβv0dx + log∫

Ae−βv0dx

),

and we have

e±βv0 = 1 ± βv0 + 1

2β2v2

0 + o(β2),

log

(∫A

e±βv0

)= log |A| ± β

|A|∫

Av0dx + β2

2 |A|∫

Av2

0dx − β2

2 |A|2(∫

Av0dx

)2

+ o(β2)

for 0 < β � 1. Then, we obtain

Jλ(βv0) = β2

{1

2

∫A

|∇v0|2 dx − λ

|A|∫

Av2

0dx + λ

|A|2(∫

Av0dx

)2}

+Jλ(0)+ o(β2) (6.6)

with Jλ(0) = −2λ log |A|.


From (6.6), we conclude that if there exists a v0 ∈ H 10,rad(A) such that

1

2

∫A

|∇v0|2 dx − λ

|A|∫

Av2

0dx + λ

|A|2(∫

Av0dx

)2

< 0, (6.7)

then 0 ∈ H 10,rad(A) cannot be a global minimizer of Jλ, and therefore, the global minimizer in

Theorem 4 is a non-trivial solution to (1.6).

For example, if we set v0(r) = sin(

πb−a (r − a)

)for a ≤ r = |x | ≤ b, we see v0 ∈ H 1

0,rad(A)

and ∫A

|∇v0|2 dx = π3

2

(b + a

b − a

),∫

A|v0|2 dx = π

2(b2 − a2),

∫Av0dx = 2(b2 − a2),

then (6.7) is reduced to

π5

2(π2 − 8)

(b + a

b − a

)< λ.

7. Appendix

This section is devoted to the proof of Lemmas 7, 6, and 3.

Proof of Lemma 7. Given ω ⊂⊂ Ω , we take h satisfying

−�h = 0 in ω, h = log p on ∂ω,

and set q = pe−h . This q solves

−� log q ≤ qeh in ω, q = 1 on ∂ω. (7.1)

We define right continuous, strictly decreasing functions

K (t) =∫

{q>t}qehdx,

and

μ(t) =∫

{q>t}ehdx .

Then, the co-area formula implies

−K ′(t) =∫

{q=t}qeh

|∇q|ds =∫

{q=t}t

eh

|∇q|ds = −tμ′(t) (7.2)

for a.e. t . On the other hand, Green’s formula and (7.1) gives

K (t) ≥∫

{q>t}(−� log q)dx =

∫{q=t}

|∇q|q

ds = 1

t

∫{q=t}

|∇q|ds

for a.e. t > 1 since ∂{q > t} = {q = t} for t > 1 and the unit outer normal vector is −∇q/|∇q|on ∂{q > t}. Therefore, from the Nehari inequality which assures(∫

∂ω1

eh/2dx

)2

≥ 4π∫ω1

ehdx


for any subdomain ω1 ⊂ ω and the Schwarz inequality, we obtain

−K ′(t)K (t) ≥ t∫

{q=t}eh

|∇q|ds · 1

t

∫{q=t}

|∇q|ds ≥(∫

{q=t}eh/2dx

)2

≥ 4π∫

{q>t}ehdx = 4πμ(t)

for a.e. t > 1. In particular, we have

d

dt

{μ(t)t − K (t)+ 1

8πK 2(t)

}= μ(t)+ 1

4πK (t)K ′(t) ≤ 0 (7.3)

for a.e. t > 1. Here we have used (7.2).Next, we define the continuous function

j (t) = K (t)− μ(t)t =∫

{q>t}(q − t)ehdx .

By (7.3), we have for all t ≥ 1,[− j (τ )+ 1

8πK 2(τ )

]τ=∞

τ=t≤ 0.

That is, j (t) ≤ 18π K 2(t), or equivalently,

μ(t) ≥ K 2(t)

t

{1

K (t)− 1

8π

}, t ≥ 1. (7.4)

Setting

J (t) = μ(t)

K (t)− μ(t)

8π,

we have J (t + 0) = J (t) and

J (t − 0)− J (t) = μ(t − 0)

K (t − 0)− μ(t)

K (t)− 1

8π(μ(t − 0)− μ(t))

= (μ(t − 0)− μ(t)) K (t)+ μ(t) (K (t)− K (t − 0))

K (t − 0)K (t)− 1

8π(μ(t − 0)− μ(t))

= (μ(t − 0)− μ(t))

(K (t)− μ(t)t

K (t − 0)K (t)− 1

8π

),

since

μ(t − 0)− μ(t) =∫

{q=t}ehds ≥ 0, K (t − 0)− K (t) =

∫{q=t}

qehds ≥ 0,

which leads to

K (t − 0)− K (t) = t (μ(t − 0)− μ(t)) .

Thus

J (t − 0)− J (t) = (μ(t − 0)− μ(t))

{j (t)

K (t − 0)K (t)− 1

8π

}≤ 0, (7.5)


because

8π j (t) ≤ K 2(t) ≤ K (t)K (t − 0).

Furthermore, we have

J ′(t) = μ′(t)(

1

K (t)− 1

8π

)− μ(t)

K ′(t)K 2(t)

≥ tμ(t)μ′(t)K 2(t)

− μ(t)K ′(t)K 2(t)

= 0 a.e. t > 1

by (7.4) and (7.2).Set t0 = maxω q ≥ 1, then by (7.5) we have

limt↓t0

J (t) ≥ limt↑t0

J (t).

As K (t0 + 0) = μ(t0 + 0) = 0 by the definition of t0, l’Hospital’s theorem and (7.2) imply

limt↓t0

J (t) = limt↓t0

μ′(t)K ′(t)

= 1

t0.

For 1 ≤ t ≤ t0, on the other hand, it holds that

limt↑t0

J (t) ≥ J (t) = μ(t)

K (t)− μ(t)

8π≥ K 2(t)

t

{1

K (t)− 1

8π

}2

by (7.4) and (7.5). The right-hand side of the above inequality tends to (1 − K (1)8π )

2 as t ↓ 1, and

this is larger than (1 − 18π

∫ω pdx)

2+. Thus, we obtain

1

t0≥(

1 − 1

8π

∫ω

pdx

)2

+,

t0 = maxω

pe−h ≤(

1 − 1

8π

∫ω

pdx

)−2

+, (7.6)

so that

maxω

p ≤(

1 − 1

8π

∫ω

pdx

)−2

+maxω

eh ≤(

1 − 1

8π

∫ω

pdx

)−2

+max∂ω

p

from the maximum principle for the harmonic function h. �

Proof of Lemma 6. We use the same notation as in the preceding proof. Let ω = B = Br (x0) ⊂⊂ Ω and consider the inequality (7.6). Then we have

q(x0) = p(x0)e−h(x0) ≤ t0 = maxB

pe−h ≤(

1 − 1

8π

∫B

pdx

)−2

+.

By the mean value theorem of the harmonic function h, we know

h(x0) = 1

|∂B|∫∂B

log pds = 1

|B|∫

Blog pdx .

Then taking log of the both sides of the above inequality, we obtain Lemma 6. �


Proof of Lemma 3. The quadratic form associated with the operator L with the domain D(L) =(H 1

0 ∩ H 2)(Ω) is given by

Q(φ, φ) =∫Ω

|∇φ|2 dx − λ

∫Ωρφ2dx + λ(ρ, φ)2

defined for φ ∈ H 10 (Ω). We have∫

Ωρψdx = 0,

∫Ωρψ2dx =

∫Ωρφ2dx − (ρ, φ)2

for ψ = SW (φ) ≡ φ − (ρ, φ), and therefore, it holds that

Q(φ, φ) = Q(ψ,ψ),

where

Q(ψ,ψ) =∫Ω

|∇ψ|2 dx − λ

∫Ωρψ2dx .

Thus, the Morse index of v is equal to the dimension of the maximal subspace of SW (H 10 (Ω))

on which Q is negative definite. Now, we note the following.First, SW (φ) = 0 is equivalent to φ = 0 for φ ∈ H 1

0 (Ω), and therefore, SW : H 10 (Ω) →

SW (H 10 (Ω)) is an isomorphism. Next, we have{ψ ∈ H 1(Ω)

∣∣∣∣∫Ωρψdx = 0, ψ

∣∣∣∣∂Ω

= constant

}= SW (H 1

0 (Ω)). (7.7)

In fact, it is obvious that ψ = SW (φ) belongs to the set of the left-hand side of (7.7) forφ ∈ H 1

0 (Ω). For ψ ∈ H 1(Ω) satisfying∫Ω ρψdx = 0 and ψ|∂Ω = constant, on the other

hand, we have φ = ψ − ψ|∂Ω ∈ H 10 (Ω) and SW (φ) = ψ − (ρ,ψ) = ψ . This proves (7.7).

Thus, the maximal subspace of H 10 (Ω) on which Q is negative definite is isomorphic to the

maximal subspace of SW (H 10 (Ω)) on which Q is negative definite.

Since Q(ψ,ψ) < 0 is equivalent to∫Ω |∇ψ|2 dx/λ

∫Ω ρψ

2dx < 1, the latter subspace isassociated with the eigenvalue problem; find (ψ, σ) ∈ SW (H 1

0 (Ω))× R such that∫Ω

∇ψ · ∇φdx = σλ

∫Ωρψφdx

for any φ ∈ SW (H 10 (Ω)). More precisely, the maximal subspace of SW (H 1

0 (Ω)) on which Q isnegative definite is the union of the eigenspaces of

−�ψ = σλρψ in Ω , ψ = constant on ∂Ω ,∫Ωρψdx = 0 (7.8)

with the eigenvalue in the range of 0 ≤ σ < 1. But if σ = 0, then the only solution to (7.8) isψ ≡ 0. When σ > 0, on the other hand, the problem (7.8) is equivalent to (2.2). Thus, we obtainthe lemma. �

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