Lucia M. a Blowing-up Branch of Solutions for a Mean Field Equation

Calc. Var. (2006) 26(3): 313–330DOI 10.1007/s00526-006-0007-3 Calculus of Variations

Marcello Lucia

A blowing-up branch of solutions for a meanfield equation

Received: 2 December 2004 / Revised: 14 September 2005 / Accepted: 10 October 2005 /Published online: 15 March 2006c© Springer-Verlag 2006

Abstract We consider the equation

−�u = λ

(eu∫�

eu− 1

|�|)

, u ∈ H10 (�).

If � is of class C2, we show that this problem has a non-trivial solution uλ foreach λ ∈ (8π, λ∗). The value λ∗ depends on the domain and is bounded frombelow by 2 j2

0 π , where j0 is the first zero of the Bessel function of the first kind oforder zero (λ∗ ≥ 2 j2

0 π > 8π). Moreover, the family of solution uλ blows-up asλ → 8π .

Keywords Mean field equations · Moser-Trudinger inequality · Mountain passtheorem · Faber-Krahn inequality

1 Introduction

This paper deals with the problem :

−�u = λ

(eu∫�

eu− 1

|�|)

, u ∈ H10 (�), (1.1)

where � is a bounded open set of IR2. Clearly u ≡ 0 is a solution of (1.1) and weare here interested in finding non-trivial solution for this problem.

Equations involving exponential nonlinearities are motivated by several prob-lems arising in geometry and physics. Already in 1853, in connection with the

M. Lucia (B)Dept. Math., National Center for Theoretical Sciences, Kuang Fu Rd, Hsinchu, TaiwanE-mail: [email protected]

314 M. Lucia

geometrical problem of finding a metric on � having a prescribed constant Gausscurvature, Liouville [17] studied the semilinear equation:

−�u = λeu . (1.2)

Beside geometrical interpretations, Eq. (1.2) is also related to several fields ofphysics. For example, it arises in some problems of combustion [4]. Also, in sta-tistical mechanics, Caglioti, et al. [6] and Kiessling [14] proved that the Gibbsmeasures defined by a system of N -particles having logarithmic interactions in adomain � of R

2 is described (when N → ∞) by the non-local equation

−�u = λeu∫�

eu, u ∈ H1

0 (�). (1.3)

They also proved that (1.3) has a solution for each λ < 8π , while uniqueness ofsolutions on simply connected domains was derived by Suzuki for each λ < 8πin [25] and by Chang et al. for λ = 8π in [7].

The PDE in (1.1) has been studied with several boundary conditions. In [24],in connection with the asymptotical behavior of some solutions derived in theframework of the Chern-Simons theory (see [26]), Struwe and Tarantello provedexistence of periodic solutions. In biology, PDE of the type (1.1) with Neumannboundary conditions are connected to some models of chemotaxis (see for ex-ample [27]). If we consider homogenous Dirichlet boundary conditions, Prob-lem (1.1) has also some very interesting applications in the theory of plasma orpoint-vortex gas. For example, in the study of the thermal equilibrium of a two-dimensional plasma confined by a magnetic field in a cylinder, Smith and O’Neil[23] analyzed the following equation:

−�u = λeu∫�

eu− µ, u ∈ H1

0 (�), (1.4)

where λ ∈ R, µ ≥ 0 are constants which are related to the inverse temperatureand the rotation frequency of the plasma. In [23] the problem is discussed on aball and their numerical simulations indicate that (1.4) has a unique solution ifλ < 8π , and may have multiple solutions when λ > 8π .

For the Problem (1.4), existence and uniqueness of solution on any boundeddomain are easy to prove when λ ≤ 0. The question becomes more subtle whenthis parameter is positive. In [20], we noted that Problem (1.4) has always a so-lution when λ < 8π . Moreover, we also noted that the method used already bySuzuki in [25] shows that this solution is unique when the domain is simply con-nected. On non-simply connected domain, we were able to handle the question ofuniqueness when µ = λ

|�| . In this case, we are reduced to the Problem (1.1) forwhich u ≡ 0 is a trivial solution. The main result of [20], shows that u ≡ 0 isactually the unique solution as far as λ ≤ 8π (for any bounded domain).

The goal of this paper is to construct non-trivial solution beyond this value8π . We will do this by using a variational method. Indeed, Problem (1.1) is theEuler-Lagrange equation of the functional :

I (λ, u) = 1

2

∫�

|∇u|2 − λ log

(1

|�|∫

�

eu)

+ λ

|�|∫

�

u, u ∈ H10 (�). (1.5)

A blowing-up branch of solutions for a mean field equation 315

As a consequence of the Moser-Trudinger inequality (see [22]), this functional iswell-defined and of class C∞. A straightforward calculation shows moreover thatits critical points are weak solution of Problem (1.1).

To state the main result of this paper, we introduce the space of H1(�)-functions of mean zero which are constant on the boundary :

H1c (�) :=

{ϕ ∈ H1(�) :

∫�

ϕ = 0, ϕ − α ∈ H10 (�) for some α ∈ R

}, (1.6)

(where α is a non-prescribed constant). In this space let us consider the least eigen-value of the Laplacian :

�(�) := inf

{ ∫�

|∇ϕ|2 : ϕ ∈ H1c (�),

∫�

ϕ2 = 1

}. (1.7)

If there is no risk of confusion, we will write � instead of �(�). By using Schwarzsymmetrization and calculating the value of � for a ball and the disjoint union oftwo balls, we will first prove that �|�| ≥ 2 j2

0 π , with j0 the first zero of theBessel function J0 (for the properties of J0, we refer to [28]). Let us emphasizethat since the functions in H1

c (�) have not a prescribed value on the boundary,such bound from below does not follow immediately from the classical Faber-Krahn inequality (see [3] or [8]) available for the first or second eigenvalue ofthe Laplacian with homogeneous Dirichlet boundary condition. As a byproduct ofthis bound on �, we see that the interval (8π, �|�|) is not empty and thereforethe following Theorem applies for any bounded open set :

Theorem 1.1 Let � ⊂⊂ R2 and set E := (8π, �|�|). Then,

(1) There exists D ⊂ E dense in E such that Problem (1.1) admits a nontrivialsolution (λ, uλ) for each λ ∈ D;

(2) If furthermore ∂� is of class C2, non-trivial solution exist for each λ ∈ E.

By setting S :={(λ, u) ∈ E × H10 (�) :(λ, u) solves (1.1), u �≡ 0}, we also

have

lim inf(λ,u)∈S,λ→8π

‖∇u‖L2 = ∞. (1.8)

This paper is organized in the following way. In Sect. 2, we state the two mainingredients that will be used to derive our existence result. The first is a deforma-tion Lemma that has been introduced in [19]. The second tool is a compactnessresult for sequences of solutions to (1.1). In Sect. 3, we prove a Faber-Krahn typeinequality for the value � and show that �|�| ≥ 2 j2

0 π > 8π . In Sect. 4, we showthat in the range (8π, �|�|) the functional exhibits a “Mountain Pass” geometry.This fact together with our Deformation Lemma allow to conclude in Sect. 4 theexistence of a non-trivial critical point for almost all λ ∈ (8π, �|�|). Using thensome a priori estimates derived from a “blow-up” analysis, we conclude the proofof Theorem 1.1.

316 M. Lucia

2 Deformation Lemma and a priori estimates

The main problem that we have to deal with the functional I (λ, ·) is the difficultyto verify the so-called “Palais-Smale condition”. Actually, in [18] we show thatwhen λ = 8π N (N ∈ N), the Palais-Smale condition is not satisfied for thefunctional

I (λ, u) = 1

2

∫�

|∇u|2 − λ log

(1

|�|∫

�

eu)

, u ∈ H10 (�). (2.1)

These same arguments show that the functional I (λ, ·) defined by (1.5) admitsunbounded Palais-Smale sequence when λ = 8π N . In [24] and [10], this problemis overcome by using a trick due to Struwe. They exploit some monotonicity ofthe “min-max” value c(λ) to construct bounded Palais-Smale sequence for almostall λ in some range of the parameter. Here, we use the approach presented in [18]and then in [19], where we proved a deformation Lemma for the functional (2.1).Instead of using the classical flow defined by the gradient of the functional I (λ, ·),we introduced a new flow to overcome the problem of verifying the “Palais-Smalecondition”.

We first introduce a definition:

Definition 2.1 Given two sets A ⊂ B ⊂ H10 (�), we say that A is a deformation

retract of B if there exists a map η : [0, 1] × H10 (�) → H1

0 (�) satisfying

(a) η is continuous;(b) η(t, u0) = u0 for all (t, u0) ∈ [0, 1] × A;(c) for t = 1, η(1, ·) maps B on A.

The deformation lemma we will need can be stated as follows :

Proposition 2.2 Let λ ∈ (0,∞) and 0 < a < b. Then,

(1) either there exists a sequence (λn, un) solution of Problem (1.1) satisfying:

a ≤ I (λn, un) ≤ b, λn ∈ (0, λ], λn → λ;(2) or, {u : I (λ, u) ≤ a} is a deformation retract of {u : I (λ, u) ≤ b}.Proof Let us set

J : H10 (�) → R, u → log

(1

|�|∫

�

eu)

− 1

|�|∫

�

u,

and

I : H10 (�) → R, u → 1

2

∫�

|∇u|2 − λJ (u).

Then, up to some minor modifications, we can follow the proof of the deformationLemma given in [18] (a more general version is given in [19]). ��

Above deformation lemma will allow to construct solution for a dense set ofvalue of the parameter belonging to some interval of R. For C2 domain, existencein the full set will be a consequence of the following a priori estimate.


Proposition 2.3 Let � ⊂⊂ R2 be of class C2. Assume (λn, un) is a sequence of

solution to Problem (1.1) with:

λn ≤ λ and λn → λ.

Then, there exists C > 0 such that

sup�

{eun∫�

eun

}≤ C ∀n ∈ N. (2.2)

Proof We first consider the a priori estimates near the boundary. Let hn be thesolution of

−�hn = λn

|�| , hn ∈ H10 (�). (2.3)

By setting

vn := un + hn and Kn(x) := λn∫�

eune−hn(x),

we rewrite Problem (1.1) as follows:

−�vn = Knevn , vn ∈ H10 (�). (2.4)

Since ∂� is of class C2, we have hn ∈ W 2,2(�) [Thm 8.12, [12]]. From Sobolevembedding [Thm. 7.26, [12]], we derive hn ∈ C1(�) and also that

|hn| + |∇hn| ≤ C0|λn|, (2.5)

where C0 := C0(�). From (2.5) and since Kn > 0, we get

|∇Kn|Kn

= |∇hn| ≤ C0|λn|. (2.6)

Since λn is bounded, the right hand-side of (2.6) is uniformly bounded and wecan therefore apply the arguments of [Prop.4, [21]] to Eq. (2.4) together with the[Thm. 2.1’, [11]] (note that vn > 0). In this way, we deduce the existence of aneighbourhood � ⊂ � of ∂� (depending on � and the uniform bound on |∇K |

K )in which

‖vn‖L∞(�) ≤ C ∀n ∈ N. (2.7)

Using (2.5), the a priori estimate (2.7) together with∫�

evn ≥ |�| (since vn ≥ 0),we get on the set � the following a priori estimate

eun(x)∫�

eun= e−hn(x)evn(x)∫

�e−hn evn

≤ C ∀x ∈ � ∀n ∈ N. (2.8)

We are left with the problem of obtaining interior a priori estimates. To do this,we consider again the function hn defined by (2.3) and we set

wn := un + hn − log

( ∫�

eun

)and Vn(x) := λne−hn . (2.9)

318 M. Lucia

Hence, if (λn, un) is a solution of Problem (1.1), the sequence wn satisfies :

−�wn = Vn(x)ewn on �,∫�

Vn(x)ewn = λn → λ.

Assume that sup�{wn} → ∞. We will show that this implies λ = 8π N .From (2.8) and (2.5), we see that

sup�

{wn} ≤ C ∀n ∈ N and sup�\�

{wn} → +∞ (as n → ∞).

Therefore, since 0 < C1 ≤ Vn ≤ C2, we may use the main result of Brezis-Merle [5] and find m disjoint balls Bi = B(ai , r) ⊂ � \ � (i = 1, . . . , m) suchthat

supBi

{wn} → +∞ and supK

{wn} → −∞,

where we have set K := �\ (�∪mi=1 B(ai ,

r2 )). Since Vn is furthermore relatively

compact in C0loc(�), the results of Li-Shafrir [16] give

m∑i=1

∫Bi

Vnewn → 8π N .

Therefore, we derive∫�

Vnewn =∫

∪mi=1 Bi

Vnewn +∫

KVnewn +

∫�

Vnewn ,

that isλn = 8π N + o(1) + O(|�|).

By taking n → ∞, we get λ = 8π N + O(|�|). Since |�| is arbitrarily small, weactually obtain λ = 8π N .

We can now say more by using the refined blow-up analysis that has been donein [9]. Indeed, let hn be defined by (2.3) and vn := un + hn . Then,

−�vn = λne−hn evn

e−hn evn, vn ∈ H1

0 (�), λn → 8π N . (2.10)

Since sup� wn → ∞ (wn defined in (2.9)), we have sup�evn∫� evn → ∞ and using

Eq. (2.10) we derive that sup� vn → ∞. Since

� log(e−hn ) = −�hn = λn

|�| > 0,

the results of [9] show that λn − 8π N > 0. In other words, blow-up can hap-pen only for λn approaching 8π N from the right. Since we assume λn ≤ λ, theproposition is proved. ��

We finally point out the following result which will be used in the last section.


Proposition 2.4 Assume (λn, un) ∈ R× H10 (�) is a sequence of solutions of (1.1)

satisfying :

(i)∫

�

|∇un|2 ≤ C or (ii)eun∫�

eun≤ C.

Then, up to a subsequence, (λn, un) → (λ, u) ∈ R × H10 (�) and (λ, u) is a

solution of Problem (1.1).

Proof If (i) holds, then as a consequence of the Moser-Trudinger equation (see[22] or [2]), we get

un ⇀ u in H10 (�) and

eun∫�

eun→ eu∫

�eu

in L2(�). (2.11)

Using (2.11) with the fact that∫

�

∇un∇ϕ = λn

∫�

(eun∫�

eun− 1

|�|)

ϕ, ∀ϕ ∈ H10 (�), (2.12)

we derive the conclusion. If (ii) holds, by using un as a test function in (2.12) wederive immediately that

∫�

|∇un|2 ≤ C∫

�

|un| ≤ C

{∫�

|∇un|2}1/2

,

which shows that∫�

|∇un|2 ≤ C . Hence the conclusion follows from (i). ��

3 A Faber-Krahn inequality

In order to prove in the following section that the functional exhibits some “moun-tain pass” geometry, we need to estimate the range of the parameter where thequadratic form D2 I(λ,0) is positive definite. By calculating explicitly D2 I(λ,0) (seeProp. 4.1), we are led to consider the value � defined by (1.7) and to find a boundfrom below for �. The goal of this section is to show :

�|�| ≥ 2 j20 π, (3.1)

where here and in the sequel jn denotes the first zero of the Bessel function Jnwhose definition and properties can be found in [28]. In particular, from (3.1), wesee that the interval (8π, �|�|) is never empty.

The difficulty here is that we deal with function in the space H1c (�) (defined

by (1.6)) for which the boundary value is not prescribed. If we were workingin the space H1

0 (�) then we would be able to apply directly results obtainedalready by Krahn [15]. More precisely, given any ω ⊂⊂ R

2, let us denote byλ1(ω) ≤ λ2(ω) ≤ . . . the eigenvalues of the Laplacian in H1

0 (ω). The Faber-Krahn inequality states that :

λ1(ω)|ω| ≥ j20 π, (3.2)

320 M. Lucia

(see for example [3] or [8]). By applying this inequality on the nodal domains ofthe second eigenfunction, Krahn [15] also gave a bound from below of the secondeigenvalue :

λ2(ω)|ω| ≥ 2 j20 π. (3.3)

Here, though we may not apply directly the results (3.2) and (3.3), we are never-theless able to derive a similar conclusion for the value �(�).

We start by noting that from classical results of variational theory the mini-mization problem associated with (1.7) admits always a solution. More precisely,there exists ψ ∈ H1

c (�) satisfying

�(�) =∫

�

|∇ψ |2 with∫

�

ψ2 = 1. (3.4)

We check then easily that (�(�), ψ) solves the eigenvalue problem :

−�ψ = �(�)ψ, ψ ∈ H1c (�), ψ �≡ 0. (3.5)

In the case of a ball B, the first eigenvalue �(B) of Problem (3.5) can be calculatedexplicitly and we have :

Proposition 3.1 Let B(0, R) be a ball of R2. Then,

�(B) = λ2(B) =(

j1R

)2

, (3.6)

where λ2(B) is the second eigenvalue of the Dirichlet-Laplacian on B and j1 isthe first zero of the Bessel function J1. In particular,

�(B)|B| = j21 π > 8π. (3.7)

Proof To simplify the notation we will write � instead of �(B). We first studyProblem (3.5) in the space of radial functions. If the eigenfunction ψ solving (3.5)is radial, it has to solve the ODE Problem:

− (rψ ′)′

r= �ψ(r), r ∈ (0, R),∫ R

0ψ(r)rdr = 0, ψ ′(0) = 0, ψ �≡ 0.

(3.8)

By integrating the left and right hand-side of above ODE on (0, R), we see thatProblem (3.8) can be reformulated as

− (rψ ′)′

r= �ψ(r), r ∈ (0, R),

ψ ′(0) = ψ ′(R) = 0, ψ �≡ 0.

(3.9)

Therefore, since � is the least eigenvalue of (3.5) it has to coincide with thefirst positive eigenvalue of the Neumann-Laplacian restricted to the class of ra-dial function. Denoting by j ′ the first zero of J

′0 and using well-known properties

of the Bessel functions, we deduce that:

ψ(r) = C J0

(j ′

Rr

)(C �= 0), � =

(j ′

R

)2

.


Since on the one hand j ′ coincides with the first zero j1 of J1 and on the otherhand (

j1R )2 = λ2(B), we actually have :

ψ(r) = C J0

(j1R

r

)and � =

(j1R

)2

= λ2(B).

This concludes the case when ψ is radial. Assume now that ψ is not radial. Dif-ferentiate now Eq. (3.5) with respect to the angle variable θ . We get :

−�

(∂ψ

∂θ

)= �

∂ψ

∂θon B,

∂ψ

∂θ= 0 on ∂ B.

Since ψ is not radial we have ∂ψ∂θ

�≡ 0 and therefore � coincides with an eigen-value of the Laplacian defined on the space H1

0 (B). By noting that ∂ψ∂θ

must changesign and since � is the least eigenvalue of (3.5) we obtain as in the radial case� = λ2(B). ��

We turn now our attention to open sets which are a union of two disjoint balls.

Proposition 3.2 Let B1 = B(a1, R1), B2 = B(a2, R2) be two disjoint balls withR2 ≤ R1. Then,

�(B1 ∪ B2)|B1 ∪ B2| ≥ 2 j20 π.

Proof To simplify the notation, we will write � instead of �(B1 ∪ B2). Considerψ ∈ H1

c (B1 ∪ B2) satisfying (3.4) and we remind that (�, ψ) solves (3.5). Wefirst deal with the “radial case” and then consider “non-radial function”. Moreprecisely, consider the subset of functions ϕ ∈ H1

c (B1 ∪ B2) satisfying :

B1 → R, x → ϕ(x − a1) is radial , (3.10)

B2 → R, x → ϕ(x − a2) is radial , (3.11)

and define

H1c,rad(B1 ∪ B2) := {ϕ ∈ H1

c (B1 ∪ B2) : ϕ satisfies (3.10) and (3.11)}.Case 1: ψ ∈ H1

c,rad(B1 ∪ B2) and ψ ≡ 0 on B2.In this case we have

−�ψ = �ψ on B1∫

B1

ψ = 0, ψ(· − a1) radial .

Hence, we are reduced to the first part of the proof of Prop. 3.1 and we deduce� = (

j1R1

)2. Therefore,

�|B1 ∪ B2| = j21 π

R21 + R2

2

R21

≥ j21 π. (3.12)

322 M. Lucia

Case 2: ψ ∈ H1c,rad(B1 ∪ B2), ψ �≡ 0 on B1 and ψ �≡ 0 on B2.

We see in this case that ψ is equal to{ψ(x − a1) = J0(

√�|x − a1|) ∀x ∈ B(a1, R1),

ψ(x − a2) = AJ0(√

�|x − a2|) ∀x ∈ B(a2, R2) (A �= 0),(3.13)

where (A, �) will be subjected to the conditions

J0(√

�R1) = AJ0(√

�R2),∫ R1

0J0(

√�r)r dr +

∫ R2

0AJ0(

√�r)r dr = 0.

(3.14)

From the system (3.14), we deduce that � must satisfy

J0(√

�R2)

∫ √�R1

0J0(t)t dt + J0(

√�R1)

∫ √�R2

0J0(t)t dt = 0. (3.15)

We check easily that (3.15) is equivalent to√

�R1 J0(√

�R2)J ′0(

√�R1) + √

�R2 J0(√

�R1)J ′0(

√�R2) = 0,

equation which can be written as

1√�R2

J0(√

�R2)

J ′0(

√�R2)

+ 1√�R1

J0(√

�R1)

J ′0(

√�R1)

= 0. (3.16)

Hence, we are led to study the zeros of the function

(0, j1) × (0, j1) → R, (x, y) → 1

x

J0(x)

J′0(x)

+ 1

y

J0(y)

J′0(y)

. (3.17)

In Prop. .2 of the appendix, we give a proof that any (x, y) zero of (3.17) has theproperty x2 + y2 ≥ 2 j2

0 . By applying this result to (3.16), we obtain :

�R21 + �R2

2 ≥ 2 j20 .

Therefore,�|B1 ∪ B2| = �

(R2

1 + R22

)π ≥ 2 j2

0 π. (3.18)

Case 3: ψ �∈ H1c,rad(B1 ∪ B2).

In this case up to translation we may assume that B1 is centered at the origin andthat ψ is not radial on B1. By considering the derivative of ψ with respect to theangular variable θ , we get :

−�

(∂ψ

∂θ

)= �

∂ψ

∂θ,

∂ψ

∂θ= 0 on ∂ B1,

∂ψ

∂θ�≡ 0 on B1.


Hence, � has to be an eigenvalue of Laplacian in H10 (B1). We then derive easily

that:

�|B1 ∪ B2| =(

j1R1

)2

π(R2

1 + R22

) ≥ j21 π. (3.19)

By comparing (3.12), (3.18) and (3.19), we conclude the proof of the theorem.��

By using Schwarz symmetrization together with Propositions 3.1 and 3.2, weare able to handle the case of a general open bounded set :

Proposition 3.3 Let � ⊂⊂ R2. Then, �|�| ≥ 2 j2

0 π .

Proof Let ψ be defined by (3.4). By setting c := ψ |∂�, we consider

�+ := {x ∈ � : ψ(x) ≥ c} and �− := {x ∈ � : ψ(x) < c}.Choose two disjoint balls B+, B− with

|B+| := |�+| and |B−| := |�−|.Consider on these two balls, the Schwarz symmetrization

ψ+ : B+ → R and ψ− : B− → R

of the functions ψ |�+ and (−ψ)|�− (for the definition and property of theSchwarz symmetrization, we refer to [13]). Let us finally set :

ψ∗ : B+ ∪ B− → R, x →{

ψ+(x) x ∈ B+;−ψ−(x) x ∈ B−.

(3.20)

The crucial point is that the sets {ψ |�+ > t} (respectively {−ψ |�− > t}) do nottouch the boundary of �+ (respectively �−). Hence, the same arguments used toderive the classical Faber-Krahn inequality give:

�(�) =∫�

|∇ψ |2∫�

ψ2=

∫�+ |∇ψ |2 + ∫

�− |∇ψ |2∫�+ ψ2 + ∫

�− ψ2

≥∫

B+ |∇ψ+|2 + ∫B− |∇ψ−|2∫

B+ |ψ+|2 + ∫B− |ψ−|2

=∫

B+∪B− |∇ψ∗|2∫B+∪B− |ψ∗|2

≥ �(B+ ∪ B−). (3.21)

In the case |�−| = 0, we have |B−| = 0 and ψ∗ ∈ H1c (B+). Hence, rela-

tion (3.21) together with Prop. 3.1 give

�(�)|�| ≥ �(B+)|�| = �(B+)|B+| = j21 π. (3.22)

We derive the same conclusion if |�+| = 0. If |�+| �= 0 and |�−| �= 0, fromrelation (3.21) and Prop. 3.2 we derive :

�(�)|�| ≥ �(B+ ∪ B−)|�| = �(B+ ∪ B−)|B+ ∪ B−| ≥ 2 j20 π. (3.23)

The estimates (3.22) and (3.23) give the conclusion since 2 j20 < j2

1 . ��

324 M. Lucia

To conclude this section, we emphasize that the value of �|�| may be verylarge. More precisely, given M > 0, we can always find a domain � that has theproperty �(�)|�| ≥ M . This can be justified easily by considering rectangles.

Proposition 3.4 Let R = (0, a) × (0, b) with a ≥ b ≥ 1. Then,

�(R)|R| = π2{

4

a2+ 1

b2

}ab. (3.24)

Proof We will set � := �(R). Let ψ be defined by (3.4) and differentiate (3.5)with respect to the variable y. We see that the function ϕ := ∂ψ

∂y satisfies thefollowing eigenvalue problem with mixed boundary conditions :

−�ϕ = �ϕ,

ϕ(0, y) = ϕ(a, y) = 0 ∀y ∈ (0, b),

∂ϕ

∂n(x, 0) = ∂ϕ

∂n(x, b) = 0 ∀x ∈ (0, a).

(3.25)

We check easily that ∂ψ∂y �≡ 0. Therefore, � is an eigenvalue of the Problem (3.25).

Hence, we necessarily have

∂ψ

∂y= Cm,n sin

(mπ

ax

)cos

(nπ

by

)m = 1, 2, 3, . . . n = 0, 1, 2 . . .

(3.26)Due to the boundary condition satisfied by ψ , we see that ∂ψ

∂y must change sign.Hence, we cannot have n = 0 in (3.26). Therefore, ψ has to be of the form :

ψ = Cm,n sin

(mπ

ax

)sin

(nπ

by

)+ f (y) m, n = 1, 2, 3, . . .

Using the boundary conditions, we deduce that f ≡ 0, which gives :

ψ = Cm,n sin

(mπ

ax

)sin

(nπ

by

)m, n = 1, 2, 3, . . . (3.27)

If (m, n) = (1, 1), we check easily that the function (3.27) cannot satisfy ourinitial problem (3.5). For (m, n) = (2, 1), the function (3.27) is of mean zero andsatisfies then the Problem (3.5). Hence, we deduce

ψ = Cm,n sin

(2π

ax

)sin

(π

by

)and � =

(2π

a

)2

+(

π

b

)2

. ��


4 Existence result

The following proposition shows that the functional (1.5) has a mountain passgeometry for each λ ∈ (8π, �|�|), where � is defined by (1.7).

Proposition 4.1 Let λ ∈ (8π, �|�|). Then, there exist ρ, α > 0 such that

I (λ, u) ≥ α ∀‖u‖ = ρ. (4.1)

Moreover, there exists u0 ∈ H10 (�) such that

‖∇u0‖L2 > ρ and I (λ, u0) < 0. (4.2)

Proof We check easily that

D2 I(0)(ξ, ξ) =∫

�

|∇ξ |2 − λ

|�|∫

�

{ξ2 − (

1

|�|∫

�

ξ)2}

=∫

�

|∇ξ |2 − λ

|�|∫

�

∣∣∣∣ξ − 1

|�|∫

�

ξ

∣∣∣∣2

≥(

1 − λ

�|�|)∫

�

|∇ξ |2.

Therefore, when λ < �|�|, the point u ≡ 0 is a strict local minimum of thefunctional (1.5). The first assertion of the Proposition follows.

Let us now prove the second statement. Given (a, µ) ∈ � × (0, ∞), considerthe function :

δ(x) = log8µ2

(1 + µ2|x − a|2)2x ∈ IR2, (4.3)

which is solution of the problem

−�u = eu on IR2,

∫IR2

eu < ∞. (4.4)

Fix then a ball B := B(a, R) ⊂ � and consider the “projection on H10 (B)” δ of δ

defined by�δ = �δ, δ ∈ H1

0 (B).

We see easily that δ − δ = log(8µ2

(1+µ2 R2)2 ), that is

δ(x) = δ(x) − log

(8µ2

(1 + µ2 R2)2

)∀x ∈ B. (4.5)

By using (4.5), a straightforward calculation shows then that∫B

|∇δ|2 = 16π log µ2 + O(1), (4.6)

log

( ∫B

eδ

)= log(µ2) + O(1), (4.7)

∫B

δ = O(1). (4.8)

326 M. Lucia

Define now the function

δ(x) :={

δ(x) if x ∈ B,

0 if x ∈ � \ B.(4.9)

Using (4.6), (4.7) and (4.8) we deduce that

I (λ, δ) = (8π − λ) log µ2 + O(1). (4.10)

Hence, on the one hand ‖∇δ‖L2 → ∞ (by 4.6) and on the other hand for eachλ > 8π , we have I (λ, δ) → −∞ as µ → ∞ (by 4.10). This proves the secondassertion of the proposition. ��

We are now able to prove that Problem (1.1) has a non-trivial solution whenthe parameter λ ∈ E := (8π, �|�|)Proof of Theorem 1.1 The proof is divided in three parts. Based on the Deforma-tion Lemma given in Sect. 2, we first prove existence of non-trivial critical pointsfor a dense set of value of λ ∈ E . In a second part, by assuming ∂� to be of classC2, existence is obtained for the full set E by using the a priori estimates given inProp. 2.3 (following the idea of [24]). Finally we prove that the set of solutions isunbounded in H1

0 when λ tends to 8π .

1) Existence for a dense set.Given λ ∈ (8π, �|�|), choose α, ρ > 0 and u0 ∈ H1

0 (�) given by Prop. 4.1. Letus consider the set of continuous curve in H1

0 (�) connecting u ≡ 0 to u0 :

� := {γ ∈ C([0, 1], H10 (�)) : γ (0) = 0, γ (1) = u0},

and consider also the minmax value :

c := infγ∈�

maxt∈[0,1]

I (λ, γ (t)). (4.11)

If the Palais-Smale condition would hold, the arguments of [1] would show thatthis minmax value c is a critical value for the functional I (λ, ·). Here, without thiscompactness condition, we prove a slightly different result. We first note that since‖u0‖L2 ≥ ρ and α > 0, we have

c ≥ α > 0. (4.12)

Assume there is no sequence (λn, un) such that:{∇ I (λn, un) = 0, λn ∈ [0, λ), λn → λ,

c − ε < I (λn, un) < c + ε, with ε = c

2.

(4.13)

Then, by Prop. 2.2, I c−ε is a deformation retract of I c+ε through a continuousmap η : [0, 1] × H1

0 (�) → H10 (�) where

I c+ε := {u : I (λ, u) < c + ε} and I c−ε := {u : I (λ, u) < c − ε}.Since u0, 0 ∈ I c−ε , for each γ ∈ � and t ∈ [0, 1], we note that the deformedcurve

γt : [0, 1] → H10 (�), s → η(t, γ (s)),


is still in the set �. Consider now any γ0 ∈ � having the property{γ0(s) ∈ I c+ε ∀s ∈ [0, 1],I (γ0(s)) ∈ (c − ε, c + ε) for some s ∈ [0, 1], (4.14)

(such curve exists by the definition of the minmax value c). Then, we have

infγ∈�

maxs∈[0,1]

I (γ (s)) ≤ maxs∈[0,1]

I (η(t, γ0(s))) ∀t ∈ [0, 1];but at t = 1, the right hand-side is less than c − ε due to (4.14) and the fact thatη is a deformation retract. A contradiction with the definition of c. Hence, theproperty (4.13) must hold.

2) Existence on (8π, �|�|) when ∂� is of class C2.Let λ ∈ (8π, �|�|) Consider a sequence (λn, un) satisfying (4.13). UsingProp. 2.3 and Prop. 2.4, we deduce that (up to a subsequence) (λn, un) convergesstrongly in R×H1

0 (�) to a (λ, u) solution of Problem (1.1). Moreover, from (4.13)we also derive

0 < c − ε < I (λ, u) < c + ε.

As a consequence, we deduce that u �≡ 0.

3) Proof of Property (1.8).If (1.8) does not hold, then there exists a sequence (λn, un) ∈ S such that

λn → 8π and ‖∇un‖2L2 ≤ C.

Using again Prop. 2.4, we see that, up to a subsequence, (λn, un) convergesstrongly in R × H1

0 (�) to (8π, u) solution of (1.1). Since � is of class C2, stan-dard regularity results show that u ∈ C2(�) ∩ C0(�). But, from [20], we knowthat u ≡ 0 is the unique solution in C2(�) ∩ C0(�) of Problem (1.1) at λ = 8π .Therefore, we deduce that u ≡ 0, namely (8π, 0) is a bifurcation point for Prob-lem (1.1). We claim that this is impossible. Indeed, as a consequence of the Moser-Trudinger inequality ([2, 22]), we derive that the mapping

F : H10 (�) → L2(�), u → eu∫

�eu

,

is of class C1 with

DF(0)(ξ) = 1

|�|(

ξ − 1

|�|∫

�

ξ

). (4.15)

Using (4.15) and (λn, ‖∇un‖L2) → (8π, 0), we deduce easily that un‖∇un‖L2con-

verges weakly in H10 (�) to some u �≡ 0 solving :

−�u = 8π

|�|(

u − 1

|�|∫

�

u

), u ∈ H1

0 (�).

Therefore, recalling the definition of �(�) in (1.7) we obtain

8π

|�| ≥ �(�). (4.16)

But (4.16) is a contradiction since the Faber-Krahn inequality proved in Proposi-tion 3.3 says that actually �(�)|�| > 8π . This shows that (1.8) must hold. ��

328 M. Lucia

Acknowledgements The author is grateful to Prof. A. Bahri, Prof. M. Kiessling, Prof. Y.Y. Lifor very fruitful discussions and to Prof. C.S. Lin for bringing [9] to his attention. He alsothanks Prof. Cao Daomin for his warm hospitality and his support at the Institute of AppliedMathematics (CAS, Beijing) where this paper was accomplished.

Appendix A: Zeros of 1x

J0(x)

J ′0(x)

+ 1y

J0( y)J ′

0( y)

Let Jn be the Bessel function of the first kind of order n and denote by jn the first zero of Jn . Wenote for the sequel that j1 coincides with the first zero of J ′

0 (see [28]). The goal of this appendixis to locate the zeros of the problem:

1

x

J0(x)

J ′0(x)

+ 1

y

J0(y)

J ′0(y)

= 0, (x, y) ∈ (0, j1) × (0, j1). (A.1)

With this aim, let us define the function

� : (0, j1) → R, s → 1

s

J0(s)

J ′0(s)

. (A.2)

Rewrite Eq. (A.1) as follows :

�(x) + �(y) = 0, (x, y) ∈ (0, j1) × (0, j1), (A.3)

and defineZ := {(x, y) ∈ (0, j1) × (0, j1) : �(x) + �(y) = 0}.

Using the well-known properties of the Bessel functions [28], we check easily

�( j0) = 0, (A.4)

lims→0

�(s) = −∞, lims→ j1

�(s) = +∞, (A.5)

�′(s) = 1

s+ s|�(s)|2 > 0 ∀s ∈ (0, j1). (A.6)

From (A.4), we have ( j0, j0) ∈ Z and the relation (A.6) shows that the function � is strictlymonotone. In particular, we deduce :

y → �(x) + �(y) strictly monotone ∀x ∈ (0, j1), (A.7)

x → �(x) + �(y) strictly monotone ∀y ∈ (0, j1). (A.8)

We now prove the following result on the location of the set Z :

Proposition .2 Let (x, y) ∈ Z . Then, x2 + y2 ≥ 2 j20 .

Proof We will show that

Z ∩ {(x, y) : x2 + y2 = 2 j20 } = {( j0, j0)}. (A.9)

This fact together with the monotonicity properties (A.7) and (A.8) will then imply that

Z ∩ {(x, y) ∈ (0, ∞) × (0, ∞) : x2 + y2 ≤ 2 j20 } = {( j0, j0)},

which will conclude the proposition.To prove (A.9), let us consider the function

� : (0,√

2 j0) → R, x → �(x) + �(√

2 j20 − x2

),


and note that (A.9) is equivalent to the fact that j0 is the unique zero of �. Using (A.6), thederivative of the function � is given by :

�′(x) = 1

x+ x�2(x) − x√

2 j20 − x2

{1√

2 j20 − x2

+√

2 j20 − x2�2(√2 j2

0 − x2)}

= 2j20 − x2

x(2 j20 − x2)

+ x{�(x) − �

(√2 j2

0 − x2)}

�(x). (A.10)

Moreover at the point j0, we have

�( j0) = 0, �′( j0) = 0, (A.11)

and from (A.10) a straightforward calculation shows that

�′′( j0) = − 4

j20

< 0. (A.12)

Assume now that the function � has some zero on the open interval ( j0, j1). Consider the firstzero x0 > j0 on this interval. By the properties (A.11) and (A.12) we deduce that �′(x0) ≥ 0.But on the other hand, by evaluating (A.10) at x0, we deduce that �′(x0) < 0. A contradiction.Hence, � has no zero on the interval ( j0, j1) and by symmetry, the same property holds on theinterval (0, j0). Therefore, j0 is the unique zero of � which establishes the statement (A.9). ��

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Lucia M. a Blowing-up Branch of Solutions for a Mean Field Equation