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Morse index of multiple blow-up solutions to the two-dimensional sinh-poisson equation

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l’Ingegneria

PhD thesis in Mathematical Models for

Engineering, Electromagnetism and Nanosciences

Ruggero Freddi

Morse Index of Multiple Blow-Up

Solutions to the Two-Dimensional

Sinh-Poisson Equation

Advisors:

Prof. Angela Pistoia

Prof. Massimo Grossi

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First, I would like to express my deepest gratitude to my advisors An-gela Pistoia and Massimo Grossi for the great support, both academic and human, they gave me over these years as PhD student, for the stimulating conversations we had, and for the many interesting ideas they shared with me.

I would also like to thank Federico Amadio Guidi for being a great friend, as well as for the precious help he gave me, especially in the writing process. Finally, I would like to thank Alessio Basti and Alice Ramassone for always being available whenever I needed.

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In this thesis we consider the Dirichlet problem    −∆u = ρ2(eu− e−u) in Ω u = 0 on ∂Ω, (1)

where ρ is a small parameter and Ω is a C2 bounded domain in R2. For each ρ sufficiently small, [1] proves the existence of a m-point blow-up solution uρ

jointly with its asymptotic behaviour. We compute the Morse index of uρ in

terms of the Morse index of the associated Hamilton function of this problem. In addition, we give an asymptotic estimate for the first 4m eigenvalues and eigenfunctions.

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Introduction and Main Results iii

1 Preliminaries and Notations 1

2 Eigenvalues from 1 to m 8

3 Eigenvalues from m + 1 to 3m 24

4 Morse Index Computations 62

Bibliography 76

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We are concerned with the study of the Morse index of the Dirichlet problem    −∆u = ρ2(eu− e−u ) in Ω u = 0 on ∂Ω, (2)

where ρ is a small parameter and Ω is a C2 bounded and connected domain in

R2 or a convex polygon with corner points {ζ1, . . . , ζn} ⊆ ∂Ω. This equation

has been widely studied as it is strictly related to the vortex-type configu-ration for 2D turbulent Euler flows (see [6], [7], and [24]). Its importance is due to the fact that, suitably adapted, it describes interesting phenomena in widely different areas like liquid helium, meteorology and oceanography; it highlights effects that are important in all those subjects. Moreover, its dynamics are isomorphic to those of the electrostatic guiding-center plasma, which have been widely extended to describe strongly magnetized plasmas (see for instance [34], [29], and [35]).

It has been known since Kirchhoff [21] that, if we let ξi ∈ Ω, i = 1, . . . , m,

be the centres of the vorticity blobs, then the ξi’s obey an approximate

Hamiltonian dynamic associated to the Hamiltonian function

F (ξ1, . . . , ξm) = 1 2 m X k=1 R(ξk) + 1 2 X 16k,j6m αkαjG(ξk, ξj), (3)

with αk ∈ {1, −1}, k = 1, . . . , m, depending on the sign of the corresponding

vorticity blob. Through two different approaches, Joyce [19] and Montgomery [28] proved at heuristic level that, if we let ω be the vorticity, ψ the flow’s

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steam function, β ∈ R the inverse of the temperature, and Z > 0 an appro-priate normalization constant, then we have ω(ψ) = Z2 sinh(−ψ) for a flow with total vorticity equal to zero, i.e. R

Ωω = 0, and ω(ψ) = 1 Ze

−βψ for a flow

with total vorticity equal to one, i.e. R

Ωω = 1. Setting u = −|β|ψ, the 2D

Euler equation in stationary form          w · ∇ω(ψ) = 0 in Ω −∆ψ = 0 in Ω w · ν = 0 on ∂Ω \ {ζ1, . . . , ζn}, (4)

where w is the velocity field, reduces to the sinh-Poisson equation (2), where ρ =

q

|β|

Z. More recently, in [5] and [4] was rigorously proved that for any

β > −9π, ω(ψ) = Z1e

−βψ is the mean field-limit vorticity for both the

micro-canonical and the micro-canonical equilibrium statistic distributions for the Hamil-tonian point-vortex model. The solutions to (4) with β < 0 of the type of those suggested in [19] (‘negative temperature’ states) have shown to rapre-sent very well the numerical experiment on the Navier-Stokes equations with high Reynolds number ([25], [26], and [30]). For further details and recent developments in the study of this problem see [4], [22], and [36]. Due to the just mentioned results much effort has been put into finding out explicit solutions for the Euler equations with Joy-Montgomery vorticity. Among the most relevants there are the Mallier-Maslowe [23] counter rotating vor-tices, and their generalization ([8] and [9]). The Mallier-Maslowe vortices are sign changing solutions to −∆u = ρ2sinh(u), with 1-periodic boundary conditions, one absolute maxima and minima and two nodal domains in each periodic cell, the resulting Euler flow is composed of symmetric and disjoint regions where the velocity fields are counter directed. This type of solutions are a suitable initial data for numerical computations ([36] and [22]). More-over, their explicit expression led to recent results which gave some insight of the properties of the non-linear dynamical stabilities of periodic array of vortices ([20], [18], and [10]).

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instance in Physics, where we can look at the Morse index of a solution to an equation as a measure of the stability of that solution, and in Mathematical Analysis, as it allows us to determine the number of critical points (see [27], [31] and [32]).

Different solutions to the equation (2) have been found. In [13] it has been built a positive solution whose concentration points converges to critical point of (3), as ρ goes to zero. More recently, in [1] and [3] it has been proved the existence of a sign changing solution whose velocity field converges to a sum of Dirac deltas centred at critical points of (3). Furthermore, in [17] it has been constructed a solution that converges to a sum of k Dirac deltas with alternate sign all centred at the same critical point of the Robin’s function.

In [1] it is proved that, under some assumptions on the points ξ1, . . . , ξm,

for each ρ sufficiently small there exists a solution uρ to the equation (2),

and its profile is given. Let µρ,j and vρ,j be respectively the j-th eigenvalue

and the j-th eigenfunction of the linearised equation around uρ. Moreover,

let ˜ vρ,n(i) = χB√ 8ε τiρ (0)(x)vρ,n  τiρx √ 8 + ξρ,i 

be the rescaled function of the n-th eigenfunction around ξρ,i. We begin by

proving some results about the asymptotic behaviour of the eigenvalues µρ,j,

the asymptotic behaviour of the eigenfunctions vρ,j away from the points

ξ1, . . . , ξm, and the asymptotic behaviour of the rescaled eigenfunctions ˜v (i) ρ,j.

We then conclude by giving an expression in term of the Morse index of the Hamiltonian function for the Morse index of uρ, for ρ suitably small.

Let us now state the main results of this work. We start by the first m eigenvalues and eigenfunctions

Theorem 0.1. For every j = 1, . . . , m there exists an k ∈ {1, . . . , m} and nonzero real constants Cj(k) ∈ R such that

1. µρ,j < −2 log(ρ)1 , for ρ small enough;

2. limρ→0˜v (k) ρ,j = C (k) j in Cloc2 (R 2 );

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3. limρ→0 vρ,j(x) µρ,j = 8π Pm k=1C (k) j G(x, ξk) in Cloc1 ( ¯Ω \ {ξ1, · · · , ξm}).

By the previous result we see that the j-th eigenfunction also concentrates at the points ξk, k = 1, .., m. If we consider higher eigenvalues, we again

have concentration at k points but a different behaviour occurs. In order to describe it let us denote by ηj , for j = 1, . . . , 2m, the eigenvalues of the

Hessian matrix of the Hamilton function F at the point (ξ1, . . . , ξm). We will

always work under the assumption that ηj 6= 0, for j = 1, . . . , 2m. Now we

are in position to state the result about the behaviour of µρ,j and vρ,j for

j = m + 1, .., 3m.

Theorem 0.2. For every j = m + 1, . . . , 3m there exists an k ∈ {1, . . . , m} such that

1. µρ,j = 1 − 3πρ2ηj+ o(ρ2), for ρ small enough;

2. limρ→0˜vρ,j(k)= s(k)1,jx1+s(k)2,jx2 8+|x|2 in C 2 loc(R 2 ) for some s = (s(k)1,j, s(k)2,j)) 6= (0, 0); 3. limρ→0 vρ,j ρ = 2π Pm k=1 τk √ 8  s(k)1,j ∂G ∂x1(x, ξk) + s (k) 2,j∂x∂G2(x, ξk)  in C1 loc( ¯Ω \ {ξ1, . . . , ξm}).

Our final results concerns the study of µρ,j and vρ,j for j = 3m, .., 4m.

This result, jointly with the previous theorems, is crucial for the computation of the Morse index of the solution.

Finally, we have

Theorem 0.3. For every j = 3m + 1, . . . , 4m there exists an k ∈ {1, . . . , m} such that

1. µρ,j = 1 − 32log(ρ)1 (1 + o(1)), for ρ small enough;

2. limρ→0˜vρ,j(k)(x) = t (k) j 8−|x|2 8+|x|2 in C 2 loc(R 2 ) for some t(k)j 6= 0; 3. log ρ vρ,j(x) → 2πPmk=1t (k) j G(x, ξk) in Cloc1 ( ¯Ω \ {ξ1, . . . , ξm}).

Now let us denote by M(uρ) the Morse index of the solution uρ. We

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Theorem 0.4. For a sufficiently small ρ we have

M(uρ) = 3m − M(Hess F ),

where by M(Hess F ) we denote the number of negative eigenvalues of the Hessian matrix of F at the point (ξ1, . . . , ξm).

From this, we also deduce that:

Corollary 0.5. For a sufficiently small ρ, we have that

m 6 M(uρ) 6 3m.

It’s interesting to notice that the above results are very similar to those obtained in [15] where the Morse index for the multiple blow-up solutions to the Gelfand’s problem is calculated and asymptotic estimates for the eigen-values and eigenfunctions of its associated linearised equation are found. The biggest difference between this work and [15] is that the solutions to the Gelfand’s problem can not have sign changing blow-up solutions, while this type of solutions are possible in our case.

On the other hand, confronting our results with those obtained in [11] we can see how different they are. In our case, the Morse index of a solution depends on the Morse index of the Hamiltonian function, while in [11] is proved that if u is the least energy sign-changing radial solution to

   −∆u = |u|p−1u in B u = 0 on ∂B, (5)

where B is the ball of radius one and centred at the origin and p > 1, if p is large enough, then its Morse index is twelve.

Finally, we want to mention the results obtained in [2] where the equation 

 

−∆u = f (u) in Ω u = 0 on ∂Ω,

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is studied. In [2] are given proper hypothesis on the regularity and the growth rate of the non-linear part f for the existence of sign changing solutions with Morse index at most equal to one and for sign changing solution with Morse index equal to two.

Before concluding this short introduction the author wants to acknowl-edge that many of the ideas and the techniques in this thesis have been inspired by those of [14], where the Morse index is computed for the Gelfand problem.

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Preliminaries and Notations

In the following work, we will denote by C a constant which may possibly change from step to step, with o(1) a function of x ∈ R2 and ρ > 0 whose norm in a suitable function space (which we will specify every time) goes to zero as ρ goes to zero, and with Ω a C2 bounded and connected domain in R2 or a convex polygon with corner points {ζ1, . . . , ζn} ⊆ ∂Ω. Moreover, we

will let ε > 0 be such that B2ε(ξρ,i) ⊆ Ω and B2ε(ξρ,i) ∩ B2ε(ξρ,j) = ∅ for any

i, j = 1, . . . , m and i 6= j, and for ρ > 0 sufficiently small.

We will denote by ψρ,ia function in C∞(R2) such that 0 6 ψρ,i(x) 6 1 and

ψρ,i(x) = 1 for each x in Bε(ξρ,i), and ψρ,i(x) = 0 for each x in R2\B2ε(ξρ,i).

Furthermore, we will denote by Uτ,ξ the function

Uτ,ξ = log  8τ2 (τ2ρ2+ |x − ξ|2)2  , solution to    −∆U (x) = ρ2eU (x) = 8τ2ρ2 (τ2ρ2+|x−ξ|2)2 x ∈ R 2 R R2e U < ∞. (1.1)

From now on, G(·, ·) and H(·, ·) will denote respectively the Laplacian Green’s function on Ω and its regular part.

G(x, y) = − 1

2πlog(|x − y|) + H(x, y).

Moreover, we will denote by R the Robin function associated to H, i.e. R(x) = H(x, x).

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We will consider the equation    −∆u = ρ2(eu− e−u) in Ω u = 0 on ∂Ω (1.2)

and the Hamilton function

F (x1, . . . , xm) = 1 2 m X k=1 R(xk) + 1 2 X 16k,j6m αkαjG(xk, xj), with αk ∈ {1, −1}.

We give the following definition.

Definition 1.1. For any integer k > 1 and any open set U ⊆ Rk, we let F : U → R be a C1 function, and K ⊆ U be a bounded set of critical points for F . We will say that K is C1-stable for F if for any F

n → F in C1(U ),

there exists at least one critical point yn ∈ U for Fn, and y ∈ K, such that

yn→ y as n → ∞.

Remark 1. It can be verified that a set K of critical points for F is C1-stable

if either one of the following conditions is satisfied:

(i) K is either a strict local maximum or a strict local minimum set for F ; (ii) the Brouwer degree deg(∇F, Bε(K), 0) is non-zero, for any ε > 0 small

enough, where Bε(K) = {x ∈ U : |x − K| 6 ε}.

The following result holds.

Theorem 1.1 ([1, Theorem 1.1]). Let (ξ1, . . . , ξm) be a C1-stable critical

point for the function F . Then, there exist ρ0 > 0 such that for any ρ ∈

(0, ρ0), the equation (1.2) has a solution uρ such that

uρ(x) → 8π m

X

k=1

αkG(x, ξk) as ρ → 0

in Cloc∞ (Ω \ {ξ1, . . . , ξm}) ∩ Cloc1,σ Ω \ {ξ¯ 1, . . . , ξm}, for some σ ∈ (0, 1) and

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In [1] it is also proved that the solution of the theorem above has the form uρ(x) = m X k=1 αkP Uτρ,k,ξρ,k(x) + ϕρ(x), (1.3)

where we denoted by P Uτρ,i,ξρ,i the projection of Uτρ,i,ξρ,i onto H

1

0(Ω). The

parameter point (ξρ,1, . . . , ξρ,m) ∈ Ωm converges to (ξ1, . . . , ξm) ∈ Ωm and

τρ,i = e4π[H(ξρ,i,ξρ,i)+ Pm k=1,k6=iαiαkG(ξρ,i,ξρ,k)] √ 8 → e4π[H(ξi,ξi)+ Pm k=1,k6=iαiαkG(ξi,ξk)] √ 8 = τi (1.4) as ρ goes to zero. For this reason, τρ,i will be called inappropriately τi. It’s

easy to prove that H(x, ξρ,i) and its derivatives of every order are bounded

in ¯Ω, uniformly in ρ. Let’s set Ui = Uτi,ξi.

We have:

Theorem 1.2 ([1, Theorem 5.2]). Let ϕρ be as in (1.3). Then

lim

ρ→0ϕρ= 0

in Cloc∞(Ω \ {ξ1, . . . , ξm}) ∩ C1,σ( ¯Ω \ {ξ1, . . . , ξm}) ∩ C0,α( ¯Ω) ∩ W2,q(Ω), for a

suitable σ ∈ (0, 1), any α ∈ (0, 1), and any q ∈ [1, 2). and

Theorem 1.3 ([1, Proposition 3.2]). For any p ∈ (1,43), there exist ρ0 > 0

and R > 0 such that, for any ρ ∈ (0, ρ0), we have kϕρkH1

0(Ω)6 Rρ 2−p

p | log(ρ)|.

In the following, we will consider β to be a fixed constant in (12, 1). Then, by the previous theorem, for ρ  1 we have kϕρkH1

0(Ω)6 Rρ

β| log(ρ)|.

We have:

Theorem 1.4 ([1, Theorem 5.1]). Let P Uk be as in (1.3). Then

P Uk(x) = Uk(x) + 8πH(x, ξρ,k) − log(8τk2) + O(ρ2) = log  1 (τ2 kρ2+ |x − ξρ,k|2)2  + 8πH(x, ξρ,k) + O(ρ2) in Cloc∞(Ω) ∩ C1,σ( ¯Ω).

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We will call ξρ,1, . . . , ξρ,m the blow-up points.

We are interested in calculating the Morse index of the solutions to (1.2) of the form (1.3). The Morse index of a solution u to a differential equation is the sum of the dimensions of the eigenspaces relative to the negative eigenvalues of the linearised equation around the solution u.

Let J : H1

0(Ω) → (H01(Ω)) ∗

be the operator defined as follows

J (u)(v) = Z Ω ∇u∇v − ρ2 Z Ω eu − e−u v.

Then uρ, as defined in (1.3), is a weak solution to (1.2) if J (uρ)(v) = 0

for all v ∈ H10(Ω).

We will denote by L (E, F ) the space of bounded linear operator between two Banach spaces E and F .

The Fr´echet derivative Ju0ρ ∈ L H1

0(Ω), (H01(Ω)) ∗

 of J in uρ is defined

as the operator that associates to v ∈ H01(Ω) the element Ju0ρ(v) ∈ (H01(Ω))∗ such that Ju0ρ(v)(w) = Z Ω ∇v∇w − ρ2 Z Ω euρ+ e−uρ vw, ∀w ∈ H1 0(Ω). (1.5) We will denote by (·, ·)H1

0(Ω) the standard inner product of H

1 0(Ω).

If the couple (¯µ, v) is a solution to

Ju0ρ(v)(·) = ¯µ(v, ·)H1 0(Ω), that is Z Ω ∇v∇w − ρ2 Z Ω euρ+ e−uρ vw = Ju0ρ(v)(w) = ¯µ(v, w)H1 0(Ω) = ¯µ Z Ω ∇v∇w, ∀w ∈ H01(Ω),

then ¯µ and v are respectively the eigenvalue and the eigenfunction of the linearised equation of (1.2) around the solution uρ.

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The above equation can be written as follows (1 − ¯µ) Z Ω ∇v∇w − ρ2 Z Ω euρ+ e−uρ vw = 0, ∀w ∈ H1 0(Ω),

which means that for any ¯µ 6= 1, v is a weak solution to 

 

−∆v = µρ2(e+ e−uρ)v in Ω

v = 0 on ∂Ω,

where µ = 1−¯1µ, (note that if ¯µ < 0 then µ < 1).

Therefore, the equation for the n-th eigenvalue and the n-th eigenfunction of the linearised equation of (1.2) around uρ of the form (1.3) is

         −∆vρ,n = µρ,nρ2(euρ+ e−uρ)vρ,n in Ω vρ,n= 0 on ∂Ω kvn,ρ(x)kL∞(Ω) = 1. (1.6)

Noting that the linearised of (1.2) around uρ is defined by (1.5), which

is a positive form, we can conclude that the eigenvalues are all positive and consequently the Morse index of uρ is the sum of the dimensions of the

eigenspaces of the solutions to (1.6) with µ ∈ [0, 1).

Let χA be the characteristic function of the set A. We will denote by vρ,i

and µρ,i respectively the i-th eigenfunction such that kvρ,ikC(Ω) = 1 and the

i-th eigenvalue of the linearised of the equation (1.2), and we will denote by

˜ vρ,n(i) = χB√ 8ε τiρ (0)(x)vρ,n  τiρx √ 8 + ξρ,i 

the rescaled function of the n-th eigenfunction around ξρ,i.

Remark 2. We notice that χB√ 8ε τiρ (0)(x) → 1 for all x ∈ R2. We have uρ(x) = m X i=1 αi −2 log τi2ρ 2+ |x − ξ

ρ,i|2 + 8πH(x, ξρ,i) + ϕρ(x) + o(1)

= −2αklog τk2ρ

2+ |x − ξ

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in Cloc∞(Ω) ∩ C1,σ( ¯Ω), where `ρ,k(x) = −2 m X i=1,i6=k αilog τi2ρ 2+ |x − ξ ρ,i|2 + 8π m X i=1 αiH(x, ξρ,i) + o(1) in Cloc∞(Ω) ∩ C1,σ( ¯Ω). Moreover, χB2√ 8ε τkρ (0)(x)`ρ,k  τkρx √ 8 + ξρ,k  = −4 m X i=1,i6=k αilog(|ξk− ξi|) + 8π m X i=1 αiH(ξk, ξi) + o(1) = `k+ o(1) in Cloc(R2) ∩ C1,σ(R2), where `k = −4 m X i=1,i6=k αilog(|ξk− ξi|) + 8π m X i=1 αiH(ξi, ξk) = 8παk H(ξk, ξk) + αk m X i=1,i6=k αi  − 1 2πlog(|ξk− ξi|) + H(ξi, ξk) ! = 8παk H(ξk, ξk) + m X i=1,i6=k αkαiG(ξi, ξk) ! , and χB2√ 8ε τkρ (0)(x)uρ  τkρx √ 8 + ξρ,k  = χB2√ 8ε τkρ (x) −4αklog(τkρ) − 2αklog  1 + |x| 2 8  + + `ρ,k  τkρx √ 8 + ξρ,k  + ϕρ  τkρx √ 8 + ξρ,k ! = −4αklog(τkρ) − 2αklog  1 + |x| 2 8  + `k(x) + o(1)

in Cloc0,α(R2). For the rescaled function of the derivative of uρ we have

χB2√ 8ε τkρ (0)(x) ∂uρ ∂xi  τkρx √ 8 + ξρ,k 

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= −√4αk 8τkρ xi  1 + |x|82 + ∂`ρ,k ∂xi  τkρx √ 8 + ξρ,k  + + ∂ϕρ ∂xi  τkρx √ 8 + ξρ,k  + o(1)

in Cloc(R2) ∩ C0,σ(R2), and for the gradient we have

χB2√ 8ε τkρ (0)(x)∇uρ  τkρx √ 8 + ξρ,k  = −√4αk 8τkρ x  1 + |x|82  + ∇`ρ,k  τkρx √ 8 + ξρ,k  + + ∇ϕρ  τkρx √ 8 + ξρ,k  + o(1) in Cloc(R2) ∩ C0,σ(R2 ). Finally, χB√ 8ε τkρ (0)(x)e ±uρ τkρx √ 8 +ξρ,k  = χB√ 8ε τkρ (0)(x)(τkρ)∓4αk  1 + |x| 2 8 ∓2αk e±  `ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k  = (τkρ)∓4αk  1 + |x| 2 8 ∓2αk  e4π(H(ξk,ξk)+ Pm i=1,i6=kαkαiG(ξk,ξi)) ±2αk eo(1) using 1.4 = τ 2 kρ4 8  1 + |x| 2 8 2!∓αk (1 + o(1)) in C0,α(R2).

In the following work, in order to apply Lebesgue dominated convergence theorem, we remark that for any k = 1, . . . , m the following conditions hold:

• χB2√ 8ε τkρ (x)`ρ,k  τkρx 8 + ξρ,k 

and its derivatives are bounded in R2 uni-formly in ρ; • χB2√ 8ε τkρ (x)ϕρ  τkρx 8 + ξρ,k 

is bounded in R2uniformly in ρ (see Theorem 1.2); • χB2√ 8ε τkρ (x)e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k  is bounded in R2 uniformly in ρ.

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Eigenvalues from 1 to m

In this chapter, we will study the behaviour of the first m eigenvalues and eigenfunctions. In particular, we will prove that the first m eigenvalues go to zero as ρ goes to zero, and hence the Morse index is greater or equal than m. In addition, we will provide an estimate for the asymptotic behaviour of the first m eigenfunctions, which will also be of fundamental importance in the next chapter.

We start by studying the asymptotic behaviour of the eigenfunctions rescaled around the blow-up points. We prove the following.

Proposition 2.1. The function ˜vρ,n(i) = χB ε

τiρ(0)(x)v  τiρx 8 + ξρ,i  , converges in C2 loc(R

2) to the function ˜v(i)

n , solution to the equation

−∆˜v(i)n (x) = µn 1  1 + |x|822 ˜ vn(i)(x), x ∈ R2, (2.1) where µn = limρ→0µρ,n. Furthermore, there exists at least an i ∈ {1, . . . , m}

such that ˜vn(i) 6= 0.

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Proof. For any x ∈ B ε τiρ(0) we have − ∆˜vρ,n(i)(x) = −∆  vn,ρ  τiρx √ 8 + ξρ,i  = τ 2 iρ2 8 (−∆vρ,n)  τiρx √ 8 + ξρ,i  = µρ,n τ2 iρ4 8  euρ  τiρx√ 8 +ξρ,i  + e−uρ  τiρx√ 8 +ξρ,i  vρ,n  τiρx √ 8 + ξρ,i  = µρ,neo(1)    1  1 + |x|82 2 + τ4 iρ8 64  1 + |x| 2 8 2   v˜ (i) ρ,n(x). (2.2) Note that τ4 iρ8 64  1 + |x| 2 8 2 6 Cρ8 1 ρ4 = o(1), ∀x ∈ Bτiρε (0), and k˜v(i)ρ,nk2 H1 0(R2) = Z B ε τiρ(0) ∇˜vρ,n(i)(x) 2 dx = − Z B ε τiρ(0) ˜ vρ,n(i)(x)∆˜vρ,n(i)(x)dx = µρ,neo(1) Z B ε τiρ(0)  ˜ v(i)ρ,n(x) 2  1 + |x|82 2dx+ + µρ,neo(1) τ4 iρ8 64 Z B ε τiρ(0)  1 + |x| 2 8 2 ˜ v(i)ρ,n(x)2dx 6 µρ,neo(1) Z B ε τiρ(0) 1  1 + |x|82 2dx + µρ,ne o(1)τ 4 iρ8 64 Z B ε τiρ(0)  1 + |x| 2 8 2 dx 6 8πµρ,n+ o(1) 6 C, (2.3)

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since τ4 iρ8 64 Z B ε τiρ(0)  1 + |x| 2 8 2 dx 6 Cρ8 1 ρ6.

Since the right hand side (2.2) is bounded in R2 uniformly in ρ, and since {∇vρ,n} is bounded in L2(R2), by the standard theory of elliptic equations,

we have that ˜v(i)ρ,n→ ˜v(i)n in Cloc2 (R2), where ˜vn(i) is as in (2.1).

For the proof of the second part of the theorem we refer to [15, Proposition 2.11].

We can now give an estimate for the first m eigenvalues.

Proposition 2.2. We have that

µρ,1 < −

1 2 log(ρ).

Proof. Using the formula for the Rayleigh quotient we have

µρ,1 = inf v∈H1 0(Ω),v6=0 R Ω|∇v| 2dx ρ2R Ω(euρ(x)+ e −uρ(x)) v2(x)dx 6 R Ω|∇(U1(x)ψρ,1(x))| 2dx ρ2R Ω(euρ(x)+ e −uρ(x)) U2 1(x)ψρ,12 (x)dx . (2.4)

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We start by estimating the numerator. Z Ω |∇(Ui(x)ψρ,i(x))|2dx = Z Ω |∇Ui(x)|2ψρ,i2 (x)dx + Z Ω Ui2(x) |∇ψρ,i(x)|2dx+ + 2 Z Ω

∇Ui(x)∇ψρ,i(x)Ui(x)ψρ,idx

= −2 Z

∇Ui(x)∇ψρ,i(x)Ui(x)ψρ,i(x)dx −

Z

∆Ui(x)Ui(x)ψρ,i2 (x)dx+

+ Z Ω Ui2(x)|∇ψρ,i(x)|2dx + 2 Z Ω

∇Ui(x)∇ψρ,i(x)Ui(x)ψρ,idx

= ρ2 Z Ω eUi(x)U i(x)ψρ,i2 (x)dx + Z Ω Ui2(x)|∇ψρ,i(x)|2dx 6 Z B2ε(ξρ,i) 8τ2 iρ2 (τ2 iρ2+ |x − ξρ,i|2) 2 log 8τ2 i (τ2 iρ2+ |x − ξρ,i|2) 2 ! ψρ,i2 (x) dx+ + C Z B2ε(ξρ,i)\Bε(ξρ,i) Ui2(x)dx = 8τ 4 iρ4 8τ4 iρ4 Z B2√ 8ε τiρ (0)  −4 log(ρ) − logτi2 8  − 2 log1 + |x|82ψ2 ρ,i  τiρx 8 + ξρ,i   1 + |x|82 2 dx + C

using Lebesgue theorem = 8π  −4 log(ρ) − log τ 2 i 8  − 2  + C + o(1) = −32π log(ρ)(1 + o(1)). (2.5)

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For the denominator we have ρ2 Z Ω euρ(x)+ e−uρ(x) U2 i(x)ψρ,i2 (x)dx > ρ2 Z Bε(ξi,ρ) euρ(x)+ e−uρ(x) U2 i(x)dx = ρ2τ 2 iρ2 8 Z B√ 8ε τiρ (0)  euρ τiρx √ 8 +ξρ,i  + e−uρ τiρx √ 8 +ξρ,i  Ui2 τ√iρx 8 + ξρ,i  dx = τ 2 iρ4 8 1 τi4ρ4 Z B√ 8ε τiρ (0) e`ρ,i  τiρx√ 8 +ξρ,i  +ϕρ  τiρx√ 8 +ξρ,i   1 + |x|82 2  −4 log(ρ) − log τ 2 i 8  − 2 log  1 + |x| 2 8 2 dx+ +τ 2 iρ4 8 τ 4 i ρ 4 Z B√ 8ε τiρ (0) e−`ρ,i  τiρx√ 8 +ξρ,i  −ϕρ  τiρx√ 8 +ξρ,i  1 + |x| 2 8 2  −4 log(ρ) − log τ 2 i 8  − 2 log  1 + |x| 2 8 2 dx using Lebesgue theorem

= 16 log2(ρ)    Z R2 dx  1 + |x|822 + o(1)    = 128π log2(ρ)(1 + o(1)), (2.6)

where we used the fact that τi6ρ8 8 Z B√ 8ε τiρ (0) e−`ρ,i  τiρx√ 8 +ξρ,i  −ϕρ  τiρx√ 8 +ξρ,i  1 + |x| 2 8 2  −4 log(ρ) − log τ 2 i 8  − 2 log  1 + |x| 2 8 2 dx 6 Cρ8 1 ρ6(log(ρ) + 1) 2 = o(1).

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Letting i = 1 in (2.5) and (2.6), and substituting in (2.4), we get µρ,1 6 −32π log(ρ)(1 + o(1)) 128π log2(ρ)(1 + o(1)) = − 1 + o(1) 4 log(ρ) < − 1 2 log(ρ). Moreover: Proposition 2.3. We have

1. µρ,j < −2 log(ρ)1 for every j = 1, . . . , m;

2. ˜vρ,j(k) → ˜vj(k)= Cj(k) in C2 loc(R

2) for every k, j = 1, . . . , m;

3. µρ,j 9 0 for every j > m.

Proof. We proceed by induction on m. By Proposition 2.2 we have that µρ,1 < −2 log(ρ)1 . Let us assume that µρ,j < −2 log(ρ)1 for every j = 1, . . . , m − 1.

By Proposition 2.1 we have that ˜v(k)ρ,j → ˜v(k)j in C2 loc(R 2 ), and that ˜vj(k) is a solution to −∆˜vj(k)= 0; x ∈ R2. Then ˜v(k)ρ,j → Cj(k) in C2 loc(R

2) for every j = 1, . . . , m − 1 and every k =

1, . . . , m. Furthermore, we know that there exists k ∈ {1, . . . , m} such that Cj(k)6= 0. Set Ψρ(x) = m X k=1 λkUk(x)ψρ,k(x) + m−1 X k=1 aρ,kvρ,k(x), where aρ,j = − (Pm k=1λkUkψρ,k, vρ,j)H1 0(Ω) (vρ,j, vρ,j)H1 0(Ω) , (2.7)

for λk ∈ R, and for each j = 1, . . . , m. It follows immediately that (Ψρ, vρ,j)H1 0(Ω) =

0 for every j = 1, . . . , m − 1.

Let us now show that for a suitable choice of λk, we have aρ,j = o (log(ρ))

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1 µρ,j m X k=1 λkUkψρ,k, vρ,j ! H1 0(Ω) = 1 µρ,j m X k=1 λk Z Ω ∇(Uk(x)ψρ,k(x))∇vρ,j(x)dx = m X k=1 λkρ2 Z B2ε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)Uk(x)ψρ,k(x)dx = m X k=1 λkρ2 τk2ρ2 8 Z B2√ 8ε τkρ (0)  euρ  τkρx√ 8 +ξρ,k  + e−uρ  τkρx√ 8 +ξρ,k  vρ,j  τkρx √ 8 + ξρ,k  Uk  τkρx √ 8 + ξρ,k  ψρ,k  τkρx √ 8 + ξρ,k  dx = m X k=1 λkρ2 τk2ρ2 8 1 τ4 kρ4 Z B2√ 8ε τkρ (0) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|82 2 v˜ (k) ρ,j(x) ψρ,k  τkρx √ 8 + ξρ,k   −4 log(ρ) − log τ 2 i 8  − 2 log  1 + |x| 2 8  dx+ + m X k=1 λkρ2 τk2ρ2 8 τ 4 kρ4 Z B2√ 8ε τkρ (0) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)ψρ,k  τkρx √ 8 + ξρ,k   −4 log(ρ) − log τ 2 i 8  − 2 log  1 + |x| 2 8  dx using Lebesgue theorem

= −4 log(ρ)    m X k=1 λkC (k) j Z R2 1  1 + |x|822 dx + o(1)    = −32π log(ρ) m X k=1 λkC (k) j + o(1) ! , (2.8) where we used the fact that

τ6ρ8 8 Z B2√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)

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ψρ,k  τkρx √ 8 + ξρ,k   −4 log(ρ) − log(τ 2 i 8) − 2 log  1 + |x| 2 8  dx 6 Cρ 8 ρ6(log(ρ) + 1) = o(1).

Hence, taking λk such that

Pm k=1λkC (k) j = 0 for j = 1, . . . , m − 1, we have 1 µρ,j m X k=1 λkUkψρ,k, vρ,j ! H1 0(Ω) = −32π log(ρ)o(1) = o (log(ρ)) . (2.9)

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For the denominator we have kvρ,jk2H1 0(Ω) = (vρ,j, vρ,j)H1 0(Ω) = Z Ω |∇vρ,j(x)|2dx = µρ,jρ2 Z Ω euρ(x)+ e−uρ(x) v2 ρ,j(x)dx = µρ,j m X k=1 ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v2 ρ,j(x)dx + o(1) ! = µρ,j m X k=1 ρ2τ 2 kρ2 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2  ˜ vρ,j(k)(x) 2 dx + o(1) ! + + µρ,j m X k=1 ρ2τ 2 kρ2 8 τ 4 kρ 4 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2  ˜ v(k)ρ,j(x) 2 dx + o(1) !

using Lebesgue theorem

= µj    m X k=1 (Cj(k))2 Z R2 dx  1 + |x|822 + o(1)    = 8πµj m X k=1 (Cj(k))2+ o(1) ! , (2.10) where we used the fact that

τk6ρ8 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2  ˜ v(k)ρ,j(x) 2 dx 6 Cρ8 1 ρ6

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= o(1).

Using (2.9) and (2.10) in (2.7) we get aρ,j = o (log(ρ)) 8πPm i=1  Cj(i) 2 + o(1) = o (log(ρ)) . (2.11)

Using the formula for the Rayleigh quotient, we have µρ,m = inf v∈H1 0(Ω),v6=0, v⊥vρ,1,...,v⊥vρ,m−1 R Ω|∇v| 2 dx ρ2R Ω(e uρ(x)+ e−uρ(x)) v2(x)dx 6 R Ω|∇Ψρ(x)| 2 dx ρ2R Ω(euρ(x)+ e −uρ(x)) Ψ2 ρ(x)dx = R Ω ∇ Pm k=1λkUk(x)ψρ,k(x) + Pm−1 k=1 aρ,kvρ,k(x)  2 dx ρ2R Ω(e uρ(x)+ e−uρ(x)) Pm k=1λkUk(x)ψρ,k(x) + Pm−1 k=1 aρ,kvρ,k(x) 2 dx (2.12) Let us start by studying the numerator.

Z Ω ∇ m X k=1 λkUk(x)ψρ,k(x) + m−1 X k=1 aρ,kvρ,k(x) ! 2 dx = Z Ω ∇ m X k=1 λkUk(x)ψρ,k(x) ! 2 dx + Z Ω ∇ m−1 X k=1 aρ,kvρ,k(x) ! 2 dx+ + 2 m−1 X j=1 aρ,j Z Ω ∇ m X k=1 λkUk(x)ψρ,k(x) ! ∇vρ,j(x)dx = m X k=1 λ2k Z Ω |∇ (Uk(x)ψρ,k(x))|2dx + m−1 X k=1 a2ρ,k Z Ω |∇vρ,k(x)|2dx+ + 2 m−1 X j=1 aρ,j m X k=1 λkUkψρ,k, vρ,j ! H1 0(Ω) = −32π log(ρ) m X k=1 λ2k(1 + o(1)). (2.13) Since by (2.5) we have Z Ω |∇ (Uk(x)ψρ,k(x))|2dx = −32π log(ρ)(1 + o(1)),

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by (2.10), (2.11), and since by hypothesis µρ,j < −2 log(ρ)1 , we have a2ρ,k Z Ω |∇vρ,k(x)|2dx = 8πµja2ρ,k m X k=1  Cj(k)2+ o(1) ! = o(log(ρ)),

and by (2.9), (2.11), and since by hypothesis µρ,j < −2 log(ρ)1 , we have

aρ,k m X k=1 λkUkψρ,k, vρ,j ! H1 0(Ω) = aρ,kµρ,jlog(ρ)o(1) = o(log(ρ)).

For the denominator we have

ρ2 Z Ω euρ(x)+ e−uρ(x) m X k=1 λkUk(x)ψρ,k(x) + m−1 X k=1 aρ,kvρ,k(x) !2 dx = ρ2 Z Ω euρ(x)+ e−uρ(x) m X k=1 λkUk(x)ψρ,k(x) !2 dx+ + ρ2 Z Ω euρ(x)+ e−uρ(x) m−1 X k=1 aρ,kvρ,k(x) !2 dx + 2 m−1 X j=1 aρ,jρ2 Z Ω euρ(x)+ e−uρ(x) v ρ,j(x) m X k=1 λkUk(x)ψρ,k(x)dx = m X k=1 λ2kρ2 Z Ω euρ(x)+ e−uρ(x) U2 k(x)ψ 2 ρ,k(x)dx+ + m X k=1 a2ρ,kρ2 Z Ω euρ(x)+ e−uρ(x) v2 ρ,k(x)dx + 2 m−1 X j=1 aρ,j 1 µρ,j m X k=1 λkUkψρ,k, vρ,j ! H1 0(Ω) = 128 log2(ρ) m X k=1 λ2k(1 + o(1)), (2.14) Since by (2.6) we have ρ2 Z Ω euρ(x)+ e−uρ(x) U k(x)2(x)ψρ,k2 (x)dx = 128 log 2(ρ)(1 + o(1)),

by (2.10), (2.11), and since by hypothesis µρ,j < −2 log(ρ)1 , we have

a2ρ,kρ2 Z

euρ(x)+ e−uρ(x) v2

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< −8π 1 2 log(ρ)log 2(ρ)o(1) m X j=1  Ck(j)2+ o(1) ! = o(log(ρ)),

and by (2.9) and (2.11), we have aρ,j µρ,j m X k=1 λkUkψρ,k, vρ,j ! H1 0(Ω) = o(log2(ρ)). Using (2.13) and (2.14) in (2.12) we get

µρ,m 6 −32 log(ρ)Pm k=1λ 2 k(1 + o(1)) 128 log2(ρ)Pm k=1λ2k(1 + o(1)) < − 1 2 log(ρ).

Since µρ,j → 0 for every j = 1, . . . , m, proceeding as in the beginning of

this proof we have that ˜vj(k)→ Cj(k) in C2 loc(R

2) per every k, j = 1, . . . , m.

Let us finally prove that µρ,i 9 0 for every i > m. Let i and j be such

that µρ,i → 0 and µρ,j → 0, and let Cj = (C (1)

j , . . . , C (n) j ).

By the orthogonality of the eigenfunctions we get 0 = (vρ,i, vρ,j)H1 0(Ω)= µρ,iρ 2 Z Ω euρ(x)+ e−uρ(x) v ρ,i(x)vρ,j(x)dx, so that 0 = ρ2 Z Ω euρ(x)+ e−uρ(x) v ρ,i(x)vρ,j(x)dx = m X k=1 ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,i(x)vρ,j(x)dx + o(1) = m X k=1 τ2 kρ2 8 ρ 2 Z B√ 8ε τkρ (0)  euρ  τkρx√ 8 +ξρ,k  + e−uρ  τkρx√ 8 +ξρ,k  ˜ vρ,i(k)(x)˜v(k)ρ,j(x)dx + o(1) = m X k=1 τ2 kρ4 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 v˜ (k) ρ,i(x)˜v (k) ρ,j(x)dx+ + m X k=1 τk2ρ4 8 τ 4 kρ 4

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Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,i(k)(x)˜v(k)ρ,j(x)dx + o(1) using Lebesgue theorem

= 8π m X k=1 Ci(k)Cj(k)+ o(1) = 8π(Ci, Cj)Rm + o(1),

where we used the fact that τk6ρ8 8 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ v(k)ρ,i(x)˜vρ,j(k)(x)dx 6 Cρ8 1 ρ6 = o(1). Therefore (Ci, Cj)Rm = 0. Since in R

m there are at most m orthogonal

vectors, then there are at most m eigenvalues which go to zero. Since in 1 we proved that the first m eigenvalues go to zero, we get the conclusion.

The next theorem gives an estimate for the asymptotic behaviour of the first m eigenfunctions away from the blow-up points.

Lemma 2.4. For every 1 6 j 6 m we have

lim ρ→0 vρ,j(x) µρ,j = 8π m X k=1 Cj(k)G(x, ξk) in C1 loc( ¯Ω \ {ξ1, . . . , ξm}).

Proof. By Proposition 2.3 we know that ˜vρ,j(k) → Cj(k) in Cloc2 (R2) for every j, k = 1, . . . , m. For each x ∈ ¯Ω \ {ξ1, . . . , ξm}, using Green’s representation

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formula we get vρ,j(x) µρ,j = ρ2 Z Ω

euρ(y)+ e−uρ(y) v

ρ,j(y)G(x, y)dy = m X k=1 ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)G(x, y)dy + o(1)

= m X k=1 G(x, ξk)ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)dy+ + m X k=1 ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy + o(1).

(2.15) Let us study the integrals in the sums separately. For the first integral we have

ρ2 Z

Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)dy = ρ2τ 2 kρ2 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕ  τkρx√ 8 +ξρ,k   1 + |x|82 2 v˜ (k) ρ,j(x)dx+ + ρ2τ 2 kρ2 8 τ 4 kρ 4 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕτkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ v(k)ρ,j(x)dx using Lebesgue theorem

= Cj(k) Z R2 dx  1 + |x|82 2 + o(1) = 8πCj(k)+ o(1), (2.16) where we used the fact that

τ6 kρ8 8 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕτkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ v(k)ρ,j(x)dx 6 Cρ8 1 ρ6

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= o(1).

Let us now move to the second integral. For every compact subset K ⊆ ¯

Ω \ {ξρ,1, . . . , ξρ,m} and for every k = 1, . . . , m, there exists ε0k such that, for

ρ small enough, we have

dist(K, ξρ,k) = inf x∈K|x − ξρ,k| = ε 0 ρ,k > ε 0 k.

Then, for each λ < min(ε, ε01, . . . , ε0m), if x ∈ K then x /∈ Bλ(ξρ,k). Therefore

ρ2 Z

Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy

= ρ2 Z

Bε(ξρ,k)\Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy+

+ ρ2 Z

Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy = ρ2 Z Bε(ξρ,k)\Bλ(ξρ,k) e`ρ,k(y)+ϕρ(y) (τ2 kρ2+ |y − ξρ,k|2) 2 + e −`ρ,k(y)−ϕρ(y) τ2 kρ2+ |y − ξρ,k|2 2 ! vρ,j(y)(G(x, y) − G(x, ξk))dy+ + ρ2 Z Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy.

For the first of these integrals we have

ρ2 Z B¯ε(ξρ,k)\Bλ(ξρ,k) e`ρ,k(y)+ϕρ(y) (τ2 kρ2+ |y − ξρ,k|2) 2 + e −`ρ,k(y)−ϕρ(y) τ2 kρ 2+ |y − ξ ρ,k|2 2 ! vρ,j(y)(G(x, y) − G(x, ξk))dy 6  Cρ2 (ρ2+ λ2)2 + Cρ 2  Z Bε¯(ξρ,k)\Bλ(ξρ,k) |G(x, y) − G(x, ξk)|dy since G(x, ·) ∈ L1(Ω) 6 C  ρ2 (ρ2+ λ2)2 + ρ 2  .

Choosing λ = ργ with γ < 12, this integral converges to zero, uniformly in x ∈ K.

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Let us now study the second integral. ρ2 Z Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy 6 sup y∈B(λ)(ξρ,k) |(G(x, y) − G(x, ξk))|ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) dy

using the uniform continuity of G in K × Bε0(ξρ,k) and (2.16)

= o(1).

Therefore

ρ2 Z

B¯ε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)(G(x, y) − G(x, ξk))dy = o(1). (2.17)

Using (2.16) and (2.17) in (2.15) we get vρ,j(x) µρ,j = 8π m X k=1 Cj(k)G(x, ξρ,k) + o(1) in C0( ¯Ω \ {ξ

1, . . . , ξm}). Using the fact that

∂vρ,j

∂xi

= ρ2 Z

euρ(y)+ e−uρ(y) v

ρ,j(y)

∂G ∂xi

(x, y)dy,

and proceeding as above, we get the estimate in C1( ¯Ω \ {ξ1, . . . , ξm}).

Proof of Theorem 0.1. Point 1 of Proposition 2.3 proves 1. Point 2 of Propo-sition 2.3 proves that

lim ρ→0v˜ (k) ρ,j = C (k) j ,

while Proposition 2.1 proves Cj(k) 6= 0 for some k ∈ {1, . . . , m}. Finally, Lemma 2.4 proves 3.

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Eigenvalues from m + 1 to 3m

In this chapter, we will consider the eigenvalues and the eigenfunctions from m + 1 to 4m. We will prove that such eigenvalues go to one as ρ approaches to zero. Being interested in the eigenvalues smaller than one, we will need to determine if they go to one from below or from above. For this purpose, we will study the asymptotic behaviour of the eigenfunctions under different conditions on the behaviour of the eigenfunctions themselves, rescaled around the blow-up points {ξ1, . . . , ξm}. Finally, we will prove that

the eigenvalues from 3m + 1 to 4m go to one from above. This gives that the Morse Index is smaller or equal than 3m.

We begin by studying the asymptotic behaviour of the eigenfunctions rescaled around the blow-up points, under the assumption that the relative eigenvalues are smaller than 1 + o(1). This condition will be proved to hold for every eigenvalue from m + 1 to 4m.

We prove the following:

Proposition 3.1. For every j > m, if µρ,j 6 1 + o(1) then µρ,j → 1, and

˜ vρ,j(k)(x) → ˜v(k)j (x) = s (k) 1,jx1+ s (k) 2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2, (3.1)

in Cloc2 (R2), and there exists k ∈ {1, . . . , m} such that (s(k)1,j, s(k)2,j, t(k)j ) 6= (0, 0, 0).

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Proof. In Proposition 2.1 we proved that µρ,j → µj, where µj is an eigenvalue

of (2.1). The proof in [12, Lemma 4.3], suitably modified, shows that the first eigenvalues of (2.1) are 0 and 1. By assumption µj 6 1, and by Proposition

2.3 we have that µj 6= 0 for every j > m, so that µj = 1. In [12, Lemma 4.3]

is proved that the eigenfunctions for the eigenvalue 1 of (2.1) are of the form (3.1).

Let us now prove the following lemma.

Lemma 3.2. For j such that µρ,j → 1, we have that vρ,j(x) → 0 in Cloc1 ( ¯Ω \

{ξ1, . . . , ξm}).

Proof. Since µj → 1, by Proposition 3.1 we have that

˜ v(k)ρ,j(x) → ˜vj(k)(x) = s (k) 1,jx1+ s (k) 2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 in C2 loc(R 2

). Let K ⊆ ¯Ω \ {ξ1, . . . , ξm} be compact. Then, by Green’s

repre-sentation formula, for each x ∈ K we have that vρ,j

µρ,j

(x) = ρ2 Z

euρ(y)+ e−uρ(y) v

ρ,j(y)G(x, y)dy = m X k=1 ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)G(x, y)dy + o(1)

= m X k=1 G(x, ξk)ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)dy+ + m X k=1 ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy + o(1)

= o(1)

in Cloc( ¯Ω \ {ξ1, . . . , ξm}).

In fact, for the fist integral we have ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dy = ρ2τ 2 kρ2 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 v˜ (k) ρ,j(x)dy+

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+ ρ2τ 2 kρ2 8 τ 4 kρ 4 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)dy using Lebesgue theorem

= Z R2 1  1 + |x|82 2 s(k)1,jx1+ s (k) 2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 ! dx + o(1) = o(1),

where we used the fact that τ6 kρ8 8 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ v(k)ρ,j(x)dy 6 Cρ8 1 ρ6 = o(1).

For the second integral, proceeding as in the proof of Lemma 2.4, for every λ < min{infρ,kdist(K, ξρ,k), ε} we have

ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy 6 ρ2 Z Bε(ξρ,k)\Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy + + ρ2 Z Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy 6 C  ρ2 (ρ2+ λ2)2 + ρ 2  Z Bε(ξρ,k)\Bλ(ξρ,k) |G(x, y) − G(x, ξk)|dy+ + sup y∈Bλ(ξρ,k) |G(x, y) − G(x, ξk)|ρ2 Z Bλ(ξρ,k)

euρ(y)+ e−uρ(y) dy

using the uniform continuity of G(x, ·) in K, and taking λ = ργ with γ < 12 = o(1)

in Cloc( ¯Ω \ {ξ1, . . . , ξm}). Using the fact that

∂vρ,j

∂xi

(x) = ρ2 Z

euρ(y)+ e−uρ(y) v

ρ,j(y)

∂G ∂xi

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and proceeding as above, we get the estimate in C1

loc( ¯Ω \ {ξ1, . . . , ξm}). We

remark that this also proves that for every j we have vρ,j(x) = µρ,j m X k=1 G(x, ξk)ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dx + o(1) (3.2) in C1 loc( ¯Ω \ {ξ1, . . . , ξm}).

Let us now study the behaviour of the eigenfunctions away from the blow-up points, under a condition that will be proved to select the eigenfunctions from 3m + 1 to 4m. Lemma 3.3. If µρ,j → 1 and tj = (t (1) j , . . . , t (m) j ) 6= (0, . . . , 0) then log(ρ)vρ,j(x) → 2π m X k=1 t(k)j G(x, ξk) in C1 loc( ¯Ω \ {ξ1, . . . , ξm}). Proof. We have ρ2 Z Bε(ξρ,k) euρ(x)− e−uρ(x) v ρ,j(x)dx + o(1) = ρ2 Z B2ε(ξρ,k) euρ(x)− e−uρ(x) ψ ρ,k(x)vρ,j(x)dx+ + Z B2ε(ξρ,k)\Bε(ξρ,k) (−∆ψρ,k)(x)uρ(x)vρ,j(x)dx+ + 2 Z B2ε(ξρ,k)\Bε(ξρ,k) ∇ψρ,k(x)∇uρ(x)vρ,j(x)dx = Z Ω (−∆uρ)(x)ψρ,k(x)vρ,j(x)dx + Z Ω (−∆ψρ,k)(x)uρ(x)vρ,j(x)dx+ + 2 Z Ω ∇ψρ,k(x)∇uρ(x)vρ,j(x)dx = Z Ω −∆(uρ(x)ψρ,k(x))vρ,j(x)dx = Z Ω −∆vρ,j(x)uρ(x)ψρ,k(x)dx = µρ,jρ2 Z Ω euρ(x)+ e−uρ(x) v ρ,j(x)uρ(x)ψρ,k(x)dx = µρ,jρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)uρ(x)dx + o(1), (3.3)

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where we used the fact that ψ and its derivatives are bounded in R2, kvρ,jkC(Ω)6

1, kuρkC1(B

2ε(ξρ,k)\Bε(ξρ,k)) 6 C, were C doesn’t depend on ρ, and that by

Lemma 3.2 we have vρ,j → 0 in Cloc1 ( ¯Ω \ {ξ1, . . . , ξm}).

Let us start by studying the first integral. By Proposition 3.1 we know that ˜ v(k)ρ,j(x) → ˜vj(k)(x) = s (k) 1,jx1+ s(k)2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 in C2 loc(R 2 ). Hence ρ2 Z Bε(ξρ,k) euρ(x)− e−uρ(x) v ρ,j(x)dx = αkρ2 τk2ρ2 8 1 τ4 kρ4 Z B2√ 8ε τkρ (0) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|82 2 v˜ (k) ρ,j(x)dx+ − αkρ2 τ2 kρ2 8 τ 4 kρ 4 Z B2√ 8ε τkρ (0) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)dx using Lebesgue theorem

= αk Z R2 1  1 + |x|82 2 s(k)1,jx1+ s (k) 2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 ! dx + o(1) = = o(1), (3.4)

where we used the fact that

τk6ρ8 8 Z B2√ 8ε τkρ (0) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 ˜ v(k)ρ,j(x)dx 6 Cρ8 1 ρ6 = o(1).

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Let us now study the second integral. µρ,jρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)uρ(x)dx = µρ,jρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)  −4αklog(τkρ) − 2αklog  1 + |x − ξρ,k| 2 τ2 iρ2  + `ρ,k(x) + ϕρ(x)  dx = −4αklog(τkρ)µρ,jρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dx+ + µρ,jρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)  −2αklog  1 + |x − ξρ,k| 2 τ2 iρ2  + `ρ,k(x) + ϕρ(x)  dx = −4αklog(τkρ)µρ,jρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dx+ + µρ,jρ2 τk2ρ2 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|822 ˜ v(k)ρ,j(x)  −2αklog  1 + |x| 2 8  + `ρ,k  τkρx √ 8 + ξρ,k  + ϕρ  τkρx √ 8 + ξρ,k  dx + µρ,jρ2 τk2ρ2 8 τ 4 kρ4 Z B√ 8ε τkρ (0) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)  −2αklog  1 + |x| 2 8  + `ρ,k  τkρx √ 8 + ξρ,k  + ϕρ  τkρx √ 8 + ξρ,k  dx using Lebesgue theorem

= −4αklog(ρ)(1 + o(1))ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dx+ + Z R2 1  1 + |x|82 2 s(k)1,jx1 + s (k) 2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 !  −2αklog  1 + |x| 2 8  + `k  dx + o(1) = −4αklog(ρ)(1 + o(1))ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dx + 8παkτ (k) j + o(1), (3.5)

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where we used the fact that µρ,j τk6ρ8 8 Z B√ 8ε τkρ (0) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)  −2αklog  1 + |x| 2 8  + `ρ,k  τkρx √ 8 + ξρ,k  + ϕρ  τkρx √ 8 + ξρ,k  dx 6 Cρ8 1 ρ6 (log(ρ) + 1) = o(1).

Using (3.4) and (3.5) in (3.3) we get

ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x)dx = 2πt(k)j + o(1) log(ρ)(1 + o(1)). (3.6) Let us now use Green’s representation formula to estimate vρ,j. Let K ⊆

¯

Ω \ {ξ1, . . . , ξm} be compact. Then, for each x ∈ K, we have

log(ρ)vρ,j(x)

= log(ρ)µρ,jρ2

Z

euρ(y)+ e−uρ(y) v

ρ,j(y)G(x, y)dy = log(ρ)µρ,j m X k=1 ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)G(x, y)dy + o(1)

= log(ρ)µρ,j m X k=1 G(x, ξk)ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y)dy+ + log(ρ)µρ,j m X k=1 ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy + o(1) using (3.6) = 2π m X k=1 t(k)j G(x, ξk) + o(1),

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every λ < min{infρ,kdist(K, ξρ,k), ε} we have log(ρ)ρ2 Z Bε(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy = log(ρ)ρ2 Z Bε(ξρ,k)\Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy+

+ log(ρ)ρ2 Z

Bλ(ξρ,k)

euρ(y)+ e−uρ(y) v

ρ,j(y) (G(x, y) − G(x, ξk)) dy 6 C| log(ρ)| ρ 2 (ρ2+ λ2)2 + ρ 2 + sup y∈Bλ(ξk) |G(x, y) − G(x, ξk)| !

using Taylor’s expansion in y = ξk, |G(x, y) − G(x, ξk)| = C|y − ξk| + o(|y − ξk|)

= C |log(ρ)|  ρ2 (ρ2+ λ2)2 + ρ 2+ Cλ + o(λ)  .

Taking λ = ργ with γ < 12, we have that the right hand side converges to zero. Using the fact that

∂vρ,j

∂xi

= ρ2 Z

euρ(y)+ e−uρ(y) v

ρ,j(y)

∂G ∂xi

(x, y)dy,

and proceeding as above, we get the estimate in C1( ¯Ω \ {ξ

1, . . . , ξm}).

We have:

Lemma 3.4 ([33, Proposition 5.5]). For each ξ, ∈ R2, R > 0 e f, g ∈ C2(BR(ξ)) we have Z BR(ξ) {[(x − ξ) · ∇f (x)] ∆g(x) + [(x − ξ) · ∇g(x)] ∆f (x)} dx = R Z ∂BR(ξ)  2∂f ∂ν(x) ∂g ∂ν(x) − ∇f (x) · ∇g(x)  dσ(x). (3.7)

In the following, we will denote by oR(f (R)) a function such that

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Lemma 3.5. Let R > 0 be such that BR(ξk)∩BR(ξj) = ∅ e BR(ξk)∩BR(ξi) = ∅. If k 6= j e k 6= i then we have Z ∂BR(ξk) ∇G(x, ξi) · ∇G(x, ξj)dσ(x) = oR(1) and Z ∂BR(ξk) ∂G ∂ν(x, ξi) ∂G ∂ν(x, ξj)dσ(x) = oR(1). If k = j 6= i or k = i 6= j then we have R Z ∂BR(ξk) ∇G(x, ξi) · ∇G(x, ξj)dσ(x) = oR(1) and R Z ∂BR(ξk) ∂G ∂ν(x, ξi) ∂G ∂ν(x, ξj)dσ(x) = oR(1). If k = j = i then we have R Z ∂BR(ξk) ∇G(x, ξi) · ∇G(x, ξj)dσ(x) = 1 2π + oR(1) and R Z ∂BR(ξk) ∂G ∂ν(x, ξi) ∂G ∂ν(x, ξj)dσ(x) = 1 2π + oR(1).

Proof. The case k 6= i and k 6= j is trivial. Let us study the case k = i. ∇G(x, ξk) = − 1 2π (x − ξk) |x − ξk|2 + ∇H(x, ξk), Then R Z ∂BR(ξk) ∇G(x, ξk) · ∇G(x, ξj)dσ(x) = R Z ∂BR(ξk)  − 1 2π (x − ξk) |x − ξk|2 + ∇H(x, ξk)  · ·  − 1 2π (x − ξj) |x − ξj|2 + ∇H(x, ξj)  dσ(x) = −R 2π Z ∂BR(ξk) (x − ξk) |x − ξk|2 · ∇H(x, ξj)dσ(x)+ + R 4π2 Z ∂BR(ξk) (x − ξk) · (x − ξj) |x − ξk|2|x − ξj|2 dσ(x) + oR(1).

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Where we used the fact that |∇H(x, ξk)| e 1 2π (x−ξj) |x−ξj|2 + ∇H(x, ξj) are bounded in BR(ξk) and hence are bounded in ∂BR(ξk), uniformly in R. Let us study

the two integrals separately. R 2π Z ∂BR(ξk) (x − ξk) |x − ξk|2 · ∇H(x, ξj)dσ(x) = R 3 2πR2 Z 2π 0 (cos(θ), sin(θ)) · ∇H(x, ξj)dθ = oR(1).

Where we used the fact that |∇H(x, ξj)| is bounded in ∂BR(ξk), uniformly

in R. For the second integral we have. R 4π2 Z ∂BR(ξk) (x − ξk) · (x − ξj) |x − ξk|2|x − ξj|2 dσ(x) = R 3 4π2R2 Z 2π 0

(cos(θ), sin(θ)) · (R(cos(θ), sin(θ) + (ξk− ξj))

|R(cos(θ), sin(θ)) + (ξk− ξj)|2 dθ. If j 6= k then we have R3 4π2R2 Z 2π 0

(cos(θ), sin(θ)) · (R(cos(θ), sin(θ) + (ξk− ξj))

|R(cos(θ), sin(θ) + (ξk− ξj)|2

dθ = oR(1),

where we used the fact that the function inside the integral is uniformly bounded in R. If j = k then we have

R3

4π2R2

Z 2π

0

(cos(θ), sin(θ)) · (R(cos(θ), sin(θ) + (ξk− ξj))

|R(cos(θ), sin(θ) + (ξk− ξj)|2 dθ = R 3 4π2R2 Z 2π 0

(cos(θ), sin(θ)) · (R(cos(θ), sin(θ)) |R(cos(θ), sin(θ))|2 dθ = R 4 4π2R42π = 1 2π.

Analogous computations give the other identities.

The next theorem shows that the hypotheses of Lemma 3.3 do not hold if the eigenvalues go to one too fast. Thanks to this theorem we will prove that

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the estimate provided by Lemma 3.3 does not holds for the eigenfunctions from m + 1 to 3m, but holds only for the eigenfunctions from 3m + 1 to 4m . Theorem 3.6. For every j > m such that µρ,j < 1 + Cρ2, in (3.1) we have

that tj =  t(1)j , . . . , t(m)j = (0, . . . , 0), that is ˜ vρ,j(k)→ s (k) 1,jx1+ s (k) 2,jx2 8 + |x|2

in Cloc2 (R2) for some s = (s(k)1,j, s(k)2,j)) 6= 0. Furthermore, if tj 6= (0, . . . , 0)

then µρ,j > 1.

Proof. By Proposition 3.1 we have that

˜ v(k)ρ,j → s (k) 1,jx1+ s(k)2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2

in Cloc2 (R2). If tj = 0 then s 6= 0, otherwise we would have that v (k)

ρ,j → 0. Let

us assume by contradiction that tj 6= 0,that is t (k)

j 6= 0 for some k = 1, . . . , m.

Using (3.7) with f = uρ and g = vρ,j we get

R Z ∂BR(ξk)  2∂uρ ∂ν (x) ∂vρ,j ∂ν (x) − ∇uρ(x) · ∇vρ,j(x)  dσ(x) = Z BR(ξk) {[(x − ξk) · ∇uρ(x)] ∆vρ,j(x) + [(x − ξk) · ∇vρ,j(x)] ∆uρ(x)} dx.

For the left hand side, by Lemma 3.3 we have

R Z ∂BR(ξk)  2∂uρ ∂ν (x) ∂vρ,j ∂ν (x) − ∇uρ(x) · ∇vρ,j(x)  dσ(x) = 2 1 log(ρ)R Z ∂BR(ξk) ∂ ∂ν 8π m X i=1 αiG(x, ξi) + o(1) ! ∂ ∂ν 2π m X n=1 t(n)j G(x, ξn) + o(1) ! dσ(x)+ − R 1 log(ρ) Z ∂BR(ξk) ∇ 8π m X i=1 αiG(x, ξi) + o(1) ! · ∇ 2π m X n=1 t(n)j G(x, ξn) + o(1) ! dσ(x)

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applying Lemma 3.5 = 8π

log(ρ)αkt

(k)

j + oR(1).

Let us now study the right hand side. Z BR(ξk) {[(x − ξk) · ∇uρ(x)] ∆vρ,j(x) + [(x − ξk) · ∇vρ,j(x)] ∆uρ(x)} dx = −µρ,jρ2 Z BR(ξk) [(x − ξk) · ∇uρ(x)] euρ(x)+ e−uρ(x) vρ,j(x)dx+ − ρ2 Z BR(ξk) [(x − ξk) · ∇vρ,j(x)] euρ(x)− e−uρ(x) dx = −ρ2 Z BR(ξk) (x − ξk) · ∇ euρ(x)− e−uρ(x) vρ,j(x) dx+ − (µρ,j − 1)ρ2 Z BR(ξk) [(x − ξk) · ∇uρ(x)] euρ(x)+ e−uρ(x) vρ,j(x)dx = −ρ2 Z ∂BR(ξk) (x − ξk) · ν euρ(x)− e−uρ(x) vρ,j(x)dσ(x)+ + 2ρ2 Z BR(ξk) euρ(x)− e−uρ(x) v ρ,j(x)dx+ + (1 − µρ,j)ρ2 Z BR(ξk) [(x − ξk) · ∇uρ(x)] euρ(x)+ e−uρ(x) vρ,j(x)dx.

Let us study these three integrals separately. For the first integral we have − ρ2 Z ∂BR(ξk) (x − ξk) · ν euρ(x)− e−uρ(x) vρ,j(x)dσ(x) = −Rρ2 Z ∂BR(ξk) euρ(x)− e−uρ(x) v ρ,j(x)dσ(x) = oR(1), since ρ2 Z ∂BR(ξk) euρ(x)− e−uρ(x) v ρ,j(x)dσ(x) 6 ρ2 Z BR(ξk) euρ(x)+ e−uρ(x) dx = ρ2τ 2 kρ2 8 1 τ4 kρ4 Z √ 8R τkρ(0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 dx+

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+ ρ2τ 2 kρ2 8 τ 4 kρ 4 Z √ 8R τkρ(0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 dx

using Lebesgue theorem = Z R2 dx  1 + |x|82 2 + o(1) = 8π + o(1),

where we used the fact that τk6ρ8 8 Z √ 8R τkρ(0) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 dx 6 Cρ8 1 ρ6 = o(1).

For the second integral we have 2ρ2 Z BR(ξk) euρ(x)− e−uρ(x) v ρ,j(x)dx = αk2ρ2 τ2 kρ2 8 1 τ4 kρ4 Z B√ 8R τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 ˜v (k) ρ,j(x)dx+ + αk2ρ2 τk2ρ2 8 τ 4 kρ 4 Z B√ 8R τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)dx using Lebesgue theorem

= 2αk Z R2 1  1 + |x|82 2 s(k)1,jx1+ s(k)2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 ! dx + o  1 log(ρ)  = o  1 log(ρ)  ,

where we used the fact that

τ6 kρ8 8 Z B√ 8R τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)dx 6 Cρ8 1 ρ6 = o  1 log(ρ)  .

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For the third integral we have ρ2 Z BR(ξk) [(x − ξk) · ∇uρ(x)] euρ(x) + e−uρ(x) vρ,j(x)dx = ρ2 Z BR(ξk) euρ(x)+ e−uρ(x) v ρ,j(x)(x − ξk)· ·  −4αk (x − ξk) τ2 kρ2+ |x − ξk|2 + ∇`ρ,k(x) + ∇ϕρ(x)  dx = ρ2τ 2 kρ2 8 1 τ4 kρ4 (1 + o(ρ2)) Z B√ 8R τkρ (ξk−ξρ,k) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|822 ˜ vρ,j(k)(x) −αk 2 |x|2 1 + |x|82 + τkρ √ 8x · ∇`ρ,k  τkρx √ 8 + ξρ,k  + τ√kρ 8x · ∇ϕρ  τkρx √ 8 + ξρ,k ! dx+ + ρ2τ 2 kρ2 8 τ 4 kρ 4 Z B√ 8R τkρ (ξk−ξρ,k) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x) −αk 2 |x|2 1 + |x|82 + τkρ √ 8x · ∇`ρ,k  τkρx √ 8 + ξρ,k  + τ√kρ 8x · ∇ϕρ  τkρx √ 8 + ξρ,k ! dx (1 + o (1))

using Lebesgue theorem

= −αk 2 Z R2 |x|2  1 + |x|82 3 s(k)1,jx1+ s (k) 2,jx2 8 + |x|2 + t (k) j 8 − |x|2 8 + |x|2 ! dx + o (1) = 16 3 παkt (k) j + o (1) ,

where we used the fact that τk6ρ8 8 Z B√ 8R τkρ (ξk−ξρ,k) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x) −αk 2 |x|2 1 + |x|82 + τkρ √ 8x · ∇`ρ,k  τkρx √ 8 + ξρ,k ! dx 6 Cρ8 1 ρ6 = o (1) ,

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and 1 8τ2 k Z B√ 8R τkρ (ξk−ξρ,k) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|822 ˜ vρ,j(k)(x)τ√kρ 8x· · ∇`ρ,k  τkρx √ 8 + ξρ,k  dx 6 Cρ Z R2 |x|  1 + |x|82 2dx = o (1) , and ρ8 Z B√ 8R τkρ (ξk−ξρ,k) e−`ρ,k τkρx √ 8 +ξρ,k  −ϕρ τkρx √ 8 +ξρ,k  1 + |x| 2 8 2 ˜ vρ,j(k)(x)τ√kρ 8x· · ∇ϕρ  τkρx √ 8 + ξρ,k  dx 6 Cρ9 Z B√ 8R τkρ (ξk−ξρ,k)  1 + |x| 2 8 2 |x| ∇ϕρ  τkρx √ 8 + ξρ,k  dx using Cauchy-Schwarz = Cρ9    Z B√ 8R τkρ (ξk−ξρ,k)  1 + |x| 2 8 2 |x| !2 dx    1 2    Z B√ 8R τkρ (ξk−ξρ,k) ∇ϕρ  τkρx √ 8 + ξρ,k  2 dx    1 2 6 Cρ9ρ−6ρ−1kϕρkH01(Ω) 6 Cρ2ρβ| log(ρ)| = o (1) , and that 1 8τ2 k Z B√ 8R τkρ (ξk−ξρ,k) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 ˜v (k) ρ,j(x) τkρ √ 8x·

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· ∇ϕρ  τkρx √ 8 + ξρ,k  dx 6 Cρ Z B√ 8R τkρ (ξk−ξρ,k) |x|  1 + |x|822 ∇ϕρ  τkρx √ 8 + ξρ,k  dx using Cauchy-Schwarz 6 Cρ    Z R2    |x|  1 + |x|82 2    2 dx    1 2    Z B√ 8R τkρ (ξk−ξρ,k) ∇ϕρ  τkρx √ 8 + ξρ,k  2 dx    1 2 6 CkϕρkH1 0(Ω) 6 Cρβ| log(ρ)| = o (1) . We then have (1−µρ,j) (1 + o (1)) 16 3 παkt (k) j +o  1 log(ρ)  +oR(1) = 8π 1 log(ρ)αkt (k) j +oR(1).

Since this relation holds true for every R, we have (1 − µρ,j) =  3 2 1 log(ρ) + o  1 log(ρ)  (1 + o(1)) < 2 log(ρ). (3.8) We supposed by assumption that

(1 − µρ,j) > −Cρ2,

so that

−Cρ2log(ρ) > 2,

which gives a contradiction as Cρ2log(ρ) = o(1). Furthermore, by (3.8) we

have that if t(k)j 6= 0 for some k = 1, . . . , m, then we have µρ,j > 1 −

2

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We prove the following lemma.

Lemma 3.7. For any domain Ω0 ⊆ Ω, and any eigenfunction vρ,n, the

fol-lowing integral identity holds.

(1 − µρ,n)ρ2 Z Ω0 euρ(x)+ e−uρ(x) ∂ uρ ∂xj (x)vρ,n(x)dx = Z ∂Ω0 ∂vρ,n ∂ν (x) ∂uρ ∂xj (x) − vρ,n(x) ∂2u ρ ∂ν∂xj (x)dσx, (3.9)

where we denoted by ν the extenal normal vector to ∂Ω0.

Proof. Differentiating (1.2) with respect to xj, for j = 1, 2, we get

−∆∂uρ ∂xj

(x) = ρ2 euρ(x)+ e−uρ(x) ∂ uρ

∂xj

(x), ∀x in Ω.

Multiplying this expression by vρ,n, and integrating both sides of the equation

we get Z Ω0 ∇ ∂uρ ∂xj  (x)∇vρ,n(x)dx − Z ∂Ω0 vρ,n(x) ∂2u ρ ∂ν∂xj (x)dσx = ρ2 Z Ω0 euρ(x)+ e−uρ(x) ∂ uρ ∂xj (x)vρ,n(x)dx.

On the other hand, multiplying (1.6) by ∂uρ

∂xj we get Z Ω0 ∇vρ,n∇  ∂uρ ∂xj  (x)dx − Z ∂Ω0 ∂vρ,n ∂ν (x) ∂uρ ∂xj (x)dσx = µρ,nρ2 Z Ω0 euρ(x)+ e−uρ(x) ∂ uρ ∂xj (x)vρ,n(x)dx.

Taking the difference of these two expressions we get the conclusion.

We can now prove that the hypothesis of Theorem 3.6 holds for the eigen-values from m + 1 to 3m, and therefore the estimate provided by Lemma 3.3 does not hold for the corresponding eigenfunctions. The next two theorems show that the eigenvalues from m + 1 to 3m go to one, give an estimate for the rate of convergence, and prove that the eigenvalues after 3m are bigger than one, thus providing an estimate from above for the Morse index.

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Lemma 3.8. We have that µρ,m+1 < 1 + Cρ2 and µρ,m+1→ 1. Proof. Set Ψρ(x) = ∂uρ ∂x1 (x)ψρ,k(x) + m X j=1 aρ,jvρ,j(x), where aρ,j = − ∂u ρ ∂x1ψρ,k, vρ,j  H1 0(Ω) (vρ,j, vρ,j)H1 0(Ω) , (3.10)

with k = 1. With this choice for the aρ,j’s we immediately get that Ψρ

is orthogonal to vρ,j for j = 1, . . . , m. By Proposition 2.3 we have that

µρ,k= o(1) for any k = 1, . . . , m. Let us prove that aρ,j = o(1). Let us start

by studying the numerator 1 µρ,j  ∂uρ ∂x1 ψρ,k, vρ,j  H1 0(Ω) = 1 µρ,j Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x)  ∇vρ,j(x)dx = ρ2 Z B2ε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x) ∂uρ ∂x1 (x)ψρ,k(x)dx = ρ2 Z B2ε(ξρ,k)\Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x) ∂uρ ∂x1 (x)ψρ,k(x)dx+ + ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x) ∂uρ ∂x1 (x)dx. (3.11)

Let us study these integral separately. For the first integral we have

ρ2 Z B2ε(ξρ,k)\Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x) ∂uρ ∂x1 (x)ψρ,k(x)dx = o(1). (3.12)

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By Lemma 3.7 and Lemma 2.4 we have that ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) v ρ,j(x) ∂uρ ∂x1 (x)dx = 1 1 − µρ,j Z ∂Bε(ξρ,k)  ∂vj,ρ ∂ν (x) ∂uρ ∂x1 (x) − vj,ρ(x) ∂2u ρ ∂ν∂x1 (x)  dσx = µρ,j 1 − µρ,j Z ∂Bε(ξρ,k)) " ∂ ∂ν 8π m X k=1 Cj(k)G(x, ξk) ! ∂ ∂x1 8π m X k=1 αkG(x, ξk) ! + − 8π m X k=1 Cj(k)G(x, ξk) ! ∂2 ∂ν∂x1 8π m X k=1 αkG(x, ξk) ! # dσx+ o(1) ! = o(1). (3.13)

Using (3.12) and (3.13) in (3.11) we get 1 µρ,j  ∂uρ ∂x1 ψρ,k, vρ,j  H1 0(Ω) = o(1). (3.14) By (2.10) we have that (vρ,j, vρ,j)H1 0(Ω) = 8πµρ,j m X k=1  Cj(k) 2 + o(1) ! . (3.15)

Using (3.14) and (3.15) in (3.10) we get that aρ,j = o(1). By the formula for

the Rayleigh quotient we have that µρ,m = inf v∈H1 0(Ω),v6=0, v⊥vρ,1,...,v⊥vρ,m R Ω|∇v| 2dx ρ2R Ω(e uρ(x)+ e−uρ(x)) v2(x)dx 6 R Ω|∇Ψρ(x)| 2dx ρ2R Ω(euρ(x)+ e −uρ(x)) Ψ2 ρ(x)dx = R Ω|∇ ∂u ρ ∂x1(x)ψρ,k(x) + Pm j=1aρ,jvρ,j(x)  |2dx ρ2R Ω(e uρ(x)+ e−uρ(x))  ∂uρ ∂x1(x)ψρ,k(x) + Pm j=1aρ,jvρ,j(x) 2 dx . (3.16) Let us start by studying the numerator.

Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x) + m X j=1 aρ,jvρ,j(x) ! 2 dx

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= Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x)  2 dx + Z Ω ∇ m X j=1 aρ,jvρ,j(x) ! 2 dx+ + 2 Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x)  ∇ m X j=1 aρ,jvρ,j(x) ! dx = Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x)  2 dx + m X j=1 a2ρ,j Z Ω |∇vρ,j(x)| 2 dx+ + 2 m X j=1 aρ,j  ∂uρ ∂x1 ψρ,k, vρ,j  H1 0(Ω) = Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x)  2 dx + o(1)

In fact, by (2.10), and since aρ,k = o(1), we have that

a2ρ,k Z Ω |∇vρ,k(x)| 2 dx = 8πa2ρ,kµj m X k=1  Cj(k) 2 + o(1) ! = o(1), (3.17)

and by (3.14), and since aρ,k= o(1), we have that

aρ,j  ∂uρ ∂x1 ψρ,k, vρ,j  H1 0(Ω) = aρ,jµρ,jo(1) = o(1). (3.18) We remark that −∆∂uρ ∂x1 (x) = − ∂ ∂x1 ∆uρ(x) = ∂ ∂x1 ρ2 euρ(x)− e−uρ(x) = ρ2 euρ(x)+ e−uρ(x) ∂ uρ ∂x1 (x), ∀x ∈ Ω.

Let us go back to the numerator. Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x)  2 dx = Z Ω ∇∂uρ ∂x1 (x) 2 ψ2ρ,k(x)dx + Z Ω ∂uρ ∂x1 2 (x) |∇ψρ,k(x)| 2 dx+ + 2 Z Ω ∂uρ ∂x1 (x)ψρ,k(x)∇ ∂uρ ∂x1 (x)∇ψρ,k(x)dx

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= −2 Z Ω ∂uρ ∂x1 (x)ψρ,k(x)∇ ∂uρ ∂x1 (x)∇ψρ,k(x)dx+ − Z Ω ∆ ∂uρ ∂x1  (x)∂uρ ∂x1 (x)ψ2ρ,k(x)dx+ + Z Ω ∂uρ ∂x1 2 (x) |∇ψρ,k(x)|2dx + 2 Z Ω ∂uρ ∂x1 (x)ψρ,k(x)∇ ∂uρ ∂x1 (x)∇ψρ,k(x)dx = ρ2 Z Ω euρ(x)+ e−uρ(x) ∂ uρ ∂x1 2 (x)ψρ,k2 (x)dx + Z Ω ∂uρ ∂x1 2 (x) |∇ψρ,k(x)|2dx, hence Z Ω ∇ ∂uρ ∂x1 (x)ψρ,k(x) + m X k=1 aρ,kvρ,k(x) ! 2 dx = ρ2 Z Ω euρ(x)+ e−uρ(x) ∂ uρ ∂x1 2 (x)ψρ,k2 (x)dx + Z Ω ∂uρ ∂x1 2 (x) |∇ψρ,k(x)| 2 dx + o(1). (3.19) Let us now study the denominator

ρ2 Z Ω euρ(x)+ e−uρ(x) ∂uρ ∂x1 (x)ψρ,k(x) + m X j=1 aρ,jvρ,j(x) !2 dx = ρ2 Z Ω euρ(x)+ e−uρ(x) ∂uρ ∂x1 (x)ψρ,k(x) 2 dx+ + ρ2 Z Ω euρ(x)+ e−uρ(x) m X j=1 aρ,jvρ,j(x) !2 dx+ + 2 m X j=1 aρ,jρ2 Z Ω euρ(x)+ e−uρ(x) ∂ uρ ∂x1 (x)ψρ,k(x)vρ,j(x)dx = ρ2 Z Ω euρ(x)+ e−uρ(x) ∂ uρ ∂x1 2 (x)ψ2ρ,k(x)dx+ + m X j=1 a2ρ,jρ2 Z Ω euρ(x)+ e−uρ(x) v2 ρ,j(x)dx + 2 m X j=1 aρ,j µρ,j  ∂uρ ∂x1 ψρ,k, vρ,j  H1 0(Ω) = ρ2 Z Ω euρ(x)+ e−uρ(x) ∂ uρ ∂x1 2 (x)ψ2ρ,k(x)dx + o(1). (3.20) In fact, by (2.10), and since aρ,k = o(1), we have

a2ρ,kρ2 Z

euρ(x)+ e−uρ(x) v2

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= a 2 ρ,k µρ,j kvρ,jk2H1 0(Ω) = 8πa2ρ,j m X k=1  Cj(k) 2 + o(1) ! = o(1),

and by (3.14), and since aρ,k= o(1), we have

aρ,j µρ,j  ∂uρ ∂x1 ψρ,k, vρ,j  H1 0(Ω) = o(1).

Using (3.19) and (3.20) in (3.16), we get

µρ,m 6 ρ2R Ω e uρ(x)+ e−uρ(x)∂uρ ∂x1 2 (x)ψ2 ρ,k(x)dx + R Ω ∂uρ ∂x1 2 (x)|∇ψρ,k(x)|2dx + o(1) ρ2R Ω(euρ(x)+ e −uρ(x))∂uρ ∂x1 2 (x)ψ2 ρ,k(x)dx + o(1) = 1 + R Ω ∂uρ ∂x1 2 (x)|∇ψρ,k(x)|2dx + o(1) ρ2R Ω(euρ(x)+ e −uρ(x))∂uρ ∂x1 2 (x)ψ2 ρ,k(x)dx + o(1) .

It remains to prove that

R Ω ∂uρ ∂x1 2 (x)|∇ψρ,k(x)|2dx + o(1) ρ2R Ω(euρ(x)+ e −uρ(x))∂uρ ∂x1 2 (x)ψ2 ρ,k(x)dx + o(1) 6 Cρ2. For the numerator we have

Z Ω ∂uρ ∂x1 2 (x)|∇ψρ,k(x)|2dx = Z B2ε(ξρ,k)\Bε(ξρ,k) ∂uρ ∂x1 2 (x)|∇ψρ,k(x)|2dx 6 C.

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For the denominator we have ρ2 Z Ω euρ(x)+ e−uρ(x) ∂ uρ ∂x1 2 (x)ψ2ρ,k(x)dx > ρ2 Z Bε(ξρ,k) euρ(x)+ e−uρ(x) ∂ uρ ∂x1 2 (x)dx = ρ2τ 2 kρ 2 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|822 −√4αk 8τkρ x1 1 + |x|82 + ∂`ρ,k ∂x1  τkρx √ 8 + ξρ,k  +∂ϕρ ∂x1  τkρx √ 8 + ξρ,k !2 dx+ + ρ2τ 2 kρ2 8 τ 4 kρ 4 Z B√ 8ε τkρ (0) e−`ρ,k  τkρx√ 8 +ξρ,k  −ϕρ  τkρx√ 8 +ξρ,k  1 + |x| 2 8 2 −√4αk 8τkρ x1 1 + |x|82 + ∂`ρ,k ∂x1  τkρx √ 8 + ξρ,k  +∂ϕρ ∂x1  τkρx √ 8 + ξρ,k !2 dx > ρ2τ 2 kρ 2 8 1 τ4 kρ4 Z B√ 8ε τkρ (0) e`ρ,k τkρx √ 8 +ξρ,k  +ϕρ τkρx √ 8 +ξρ,k   1 + |x|822 −√4αk 8τkρ x1 1 + |x|82 + ∂`ρ,k ∂x1  τkρx √ 8 + ξρ,k  +∂ϕρ ∂x1  τkρx √ 8 + ξρ,k !2 dx > 2 τ2 kρ2 1 8τ2 k Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 x2 1  1 + |x|82 2dx+ − √ 8αk τkρ 1 8τ2 k Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 x1 1 + |x|82 ∂`ρ,k ∂x1  τkρx √ 8 + ξρ,k  dx+ − √ 8αk τkρ 1 8τ2 k Z B√ 8ε τkρ (0) e`ρ,k  τkρx√ 8 +ξρ,k  +ϕρ  τkρx√ 8 +ξρ,k   1 + |x|82 2 x1 1 + |x|82 ∂ϕρ ∂x1  τkρx √ 8 + ξρ,k  dx

using Lebesgue theorem

= 2 τ2 kρ2    Z R2 x21  1 + |x|824 dx + o(1)    = 1 ρ2  32π 3τ2 k + o(1)  , (3.21)

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