Some results for the Gelfand’s problem

Francesca Gladiali and Massimo Grossi∗

Abstract

Under suitable assumptions on Ω ⊂ IR2 we prove nondegeneracy,uniqueness and star-shapedness of the level sets of solutions to the prob-lem

−∆u = λeu in Ωu = 0 on ∂Ω,

(0.1)

where λ∫Ω

euλ→8π as λ→0.

A. M. S. subject classification 2000 35J60

Keywords: Blowing up solutions, Green’s function.

1 Introduction

In this paper we study qualitative properties of solutions of the Gelfand′s prob-lem

−∆u = λeu in Ωu = 0 on ∂Ω,

(1.1)

where Ω is an smooth bounded domain in IR2 and λ > 0.Problem (1.1) has been studied by several authors because of its wide appli-

cations. It arise in the classical field theory defined on a (2+1)-Minkowsky spacewhere the Lagrangian couples the scalar Maxwell (or Yang-Mills) field with theChern-Simons terms coming from a gauge field (see [12] and [29]). Moreover thisproblems has been derived in [10] from Onsanger’s vortex model for turbulentEuler flows.

Let us recall some results to (1.1). By standard regularity result we havethat a weak solution to (1.1) is a classical solution

Moreover it is known that there exists λ∗ = λ∗(Ω) > 0 s. t.

- If λ > λ∗ there is no solution of (1.1), even in distributional sense;

- if 0 ≤ λ < λ∗ problem (1.1) admits at least a minimal solution uλ;

∗Dipartimento di Matematica, Universita di Roma ”La Sapienza” P.le Aldo Moro, 2 - 00185Roma. e-mail: gladiali@mat.uniroma1.it, grossi@mat.uniroma1.it

Supported by M.U.R.S.T., project “Variational methods and nonlinear differential equa-tions”.

1

- if λ = λ∗ there exists a unique solution u∗.

Moreover the minimal solution uλ goes to 0 uniformly as λ → 0.Using standard variational methods (see [9] for example) it is possible to

show the existence of a second solution uλ of mountain pass type which blowsup at an interior point as λ → 0. The asymptotic behavior of this solution uλ

in Ω as λ → 0 is a well understood problem. In [24],[25],[31] it was proved that,denoting by G(x, y) the Green function of −∆ in Ω,

uλ(x) → G(x, x0) in Ω \ x0.

Moreover the blow up point x0 satisfies

∇R(x0) = 0

where R(x) is the Robin function defined as R(x) = H(x, x) with

H(x, y) = G(x, y)− 12π

log1

|x− y|.

In [28] Suzuki proved that any nondegenerate critical point of the Robinfunction generates at least one solution to (1.1) as λ→0. This result was gener-alized by Baraket and Pacard in [2] to the case of multiple blow-up solutions.

In this paper we study some qualitative properties of the mountain passsolution uλ as λ → 0. Our first result concerns the nondegeneracy of thissolution. Let us recall that a solution u to (1.1) is nondegenerate if the problem

−∆v = λeuv in Ωv = 0 on ∂Ω,

(1.2)

admits only the trivial solution v ≡ 0.In [27] Suzuki proved that if Ω is simply connected and

λ

∫Ω

euλ < 8π (1.3)

then the solution of (1.1) is nondegenerate. In this paper we analyze the non-degeneracy of any solution uλ blowing up at one point for λ small in a moregeneral setting.

Theorem 1 Let us consider a domain Ω ⊂ IR2 and let uλ be a blowing upsolution to (1.1)satisfying

λ

∫Ω

euλ→8π as λ→0 (1.4)

Then if x0 is a nondegenerate critical point of the Robin function we get thatthere exists λ > 0 such that for any 0 < λ < λ the solution uλ is nondegenerate.

2

The proof of Theorem 1 is given in Section 3.We point out that we do not need to assume that λ

∫Ω

euλ < 8π. This allowsto consider a wider class of domains than Suzuki’s one. For example, in [10]and [11], it was proved that if Ω is a rectangle of sides with length a and 1, fora sufficiently large we have

λ

∫Ω

euλ > 8π

Although there exist in literature similar statements to Theorem 1, mainly inelliptic problems involving the Sobolev critical exponent, the proof of Theorem(1) requires new techniques. For example, in the problem −∆u = n(n− 2)u

n+2n−2−ε in Ω

u > 0 in Ω,u = 0 on ∂Ω,

(1.5)

where Ω ⊂ IRn, n ≥ 3, the analogous of Theorem 1 is proved by splitting thesolution uε like

uε = αεPUλε,xε + vε (1.6)

where αε, λε and xε satisfy suitable condition and PUλ,x : IRN 7→ IR is theprojection of Uλ,x(y) = λ

N−22 (1+λ2|y−x|2)−N−2

2 onto H10 (Ω). In this approach

the computation of some estimates involving vε plays a crucial role (see [1] andthe references therein).

In our case we do not have a decomposition of the solution uλ as in [1] butwe use a contradiction argument which leads to study the linearized equationat infinity, namely

−∆v =1

(1 + |x−y|28 )2

v in IRN . (1.7)

Some integral identities involving the Robin function allow then to conclude theproof of Theorem 1.

Next results concerns the uniqueness of the solution in domains for whichGidas, Ni, Nirenberg’s Theorem holds.

Theorem 2 Let Ω be a bounded domain which is convex in the direction x1, x2

and symmetric with respect to the axis xi = 0 for i = 1, 2. Then there exists λ,with 0 < λ < λ∗ such that for all 0 < λ < λ there is only one solution of (1.1)which blows up at the origin.

Note that in Theorem 2 we do not make any assumption on the integralλ∫Ω

eu. Indeed an important step in the proof of Theorem 2 (which is a conse-quence of the Gidas, Ni, Nirenberg’s Theorem) is to derive that any blowing upsolution uλ of (1.1) satisfies

λ

∫Ω

euλ→8π.

3

¿From this fact we derive the uniqueness as in the case of the nondegeneracy.We quote that the a local uniqueness result for solutions making a blow-up

point in a non-degenerate setting was proved in a different way by Mizoguchiand Suzuki in [23]. Concerning again Theorem 2 we recall the results by Kieloferin [18] and[19] where it is proved that the set of solutions forms a curve whoseend points are the trivial solution and the singular limit.

Then (quoting a result in [16]) and recalling known results on the minimalsolution uλ we have the following

Corollary 1 Let us consider a domain Ω ⊂ IR2 which is convex in the directionx1, x2 and symmetric with respect to the planes x1 = 0 and x2 = 0. Then thereexists λ > 0 such that for any 0 < λ < λ problem (1.1) has exactly two solutionswhich are nondegenerate.

We again recall that in the case where λ∫Ω

euλ < 8π Theorem 2 can be deducedby some results in [27].

In Section 5 we study the shape of the level sets in convex domains. Let usdenote by xλ a point where uλ(x) achieves its maximum. The first result is thefollowing

Theorem 3 Let uλ be a solution of (1.1) satisfying (1.3). Then if Ω is convex,there exists λ > 0 such that for all 0 < λ < λ,

(x− xλ) · ∇uλ(x) < 0

A consequence of the previous theorem is that uλ has strictly starshaped levelsets and then xλ is the only critical point of uλ in Ω. Under some additionalassumption on the boundary of Ω we obtain the strict convexity of the level set.

Theorem 4 Let uλ be a solution of (1.1) satisfying (2.2). If ∂Ω has strictlypositive curvature at any point, there exists λ > 0 such that for all 0 < λ < λthe level sets of uλ have strictly positive curvature at any point different fromthe maximum one. In particular the level sets are strictly convex.

We stress that in the proof of the Theorem 3 and 4 an important role isplayed by the Green function in a convex domain.

The paper is organized as follows: in Section 2 we recall some known factson the problem (1.1) and we prove some asymptotic estimates. In Section 3 weprove Theorem 1. In Section 4 we prove Theorem 2 and Corollary 1. In Section5 we obtain results on the shape of the level sets of a solution to (1.1) satisfying(1.4). Finally in the last section we come back to study the minimal solutionuλ and we prove, for λ small enough the strictly convexity of the level sets indomains having positive curvature on the boundary.

Acknowledgement. We would like to thank the anonymous referee for theseveral references.

4

2 Preliminaries and asymptotic estimates

Let uλ be a solution of −∆u = λeu in Ωu = 0 on ∂Ω,

(2.1)

satisfying

λ

∫Ω

euλ→8π. (2.2)

Theorem 5 Every solution U ∈ C2(IR2) of the problem−∆u = eu in IR2,∫IR2 eu < ∞,

(2.3)

is given by

Uδ,y(x) = log8δ

(δ + |x− y|2)2(2.4)

for any (δ, y) ∈ IR+ × IR2.

Proof. See Chen and Li ([8]). utIn the following we set U(x) = U8,0.

We will denote in the sequel || · ||L∞(Ω) by || · ||. Let us denote by R(x) theRobin function of a domain Ω ⊂ R2, i.e.

R(x) = H(x, x), (2.5)

where H(x, y) is the regular part of the Green function G(x, y), i.e.,

H(x, y) = G(x, y) +12π

log1|x|

. (2.6)

Let xλ ∈ Ω be the point where uλ achieves its maximum. We recall someknown results about the asymptotic behavior of the solution of (1.1).

Theorem 6 Let uλ be a solution of (1.1) satisfying (2.2). Then

xλ→x0 ∈ Ω and ∇R(x0) = 0. (2.7)

Moreoveruλ(x)→8πG(x, x0) in C1(Ω \ x0), (2.8)

and ∣∣∣∣∣uλ(x)− logeuλ(xλ)(

1 + 18λeuλ(xλ)|x− xλ|2

)2∣∣∣∣∣ ≤ C. (2.9)

5

Proof See [25] for the proof of (2.7) and (2.8).Then we prove estimate (2.9).Setting vλ(x) = uλ(x) + log λ we derive that vλ(x) satisfies the equation

−∆vλ = evλ in Ωvλ = + log λ on ∂Ω,

(2.10)

We claim that vλ(xλ)→+∞ as λ→0.By contradiction let us suppose that there exist a sequence λn→0 and a

sequence of points xn ∈ Ω such that either

i) vλn(xn)→−∞

orii) vλn

(xn)→C ∈ R

By (2.2) we get∫Ω

evλn→8π and since i) implies that evλn→0 in Ω we get acontradiction. On the other hand if ii) occurs by a Brezis-Merle result ([3]) wededuce that vλn

is bounded L∞loc(Ω). However (2.8) implies that vλn(x)→−∞

in Ω \ x0, again a contradiction.So vλ(xλ)→+∞ as λ→0. Hence we are in the same setting of Theorem 0.3

in [21] which gives∣∣∣∣∣vλ(x)− logevλ(xλ)(

1 + 18evλ(xλ)|x− xλ|2

)2∣∣∣∣∣ ≤ C.

Recalling the definiton of vλ we deduce (2.9). ut

We have the following result:

Lemma 1 Let us consider the equation−∆v = 1

(1+|x|28 )2

v in IR2

v ∈ L∞(IR2)(2.11)

Then

v(x) =2∑

i=1

aixi

8 + |x|2+ b

8− |x|2

(8 + |x|2)(2.12)

for some ai, b ∈ IR.

Proof See [9] or also [13] for a more detailed proof.

Lemma 2 Let u be a solution of the following problem−∆u = f in Ωu = 0 in ∂Ω (2.13)

and let ω be a neighborhood of ∂Ω.

6

Then, for α ∈ (0, 1) and any neighborhood ω′ ⊂⊂ ω with ∂Ω ⊂ ∂ω, thereexists C > 0, not depending on u such that

||∇u||C0,α(ω) ≤ C(||f ||L1(Ω) + ||f ||L∞(ω)), (2.14)

Proof See [4] and [17].

Remark 1 As a consequence of the previous lemma we have that if fλ isbounded in L1(Ω) ∩ L∞(ω), then the corresponding solutions uλ has a com-pact closure in L2(Ω), whilst ∇fλ, restricted to ∂Ω, has a compact closure inL2(∂Ω).

Lemma 3 Let Ω be a smooth bounded domain of IR2. For any y ∈ Ω we have∫∂Ω

(x− y) · ν(x)(∂G(x, y)

∂νx

)2

dSx =12π

(2.15)

and ∫∂Ω

νi(x)(∂G(x, y)

∂νx

)2

dSx = −∂R(y)∂yi

(2.16)

Proof Let us consider a solution uλ of (1.1). By Pohozaev identity ([26]) wehave

2λ

∫Ω

(euλ − 1) =12

∫∂Ω

(x− y) · ν(x)(∂uλ

∂νx

)2

dSx (2.17)

Passing to the limit in (2.17) we get, by (2.2) and (2.8)

16π = 32π2

∫∂Ω

(x− y) · ν(x)(∂G(x, y)

∂νx

)2

dSx (2.18)

and (2.20) follows.On the other hand (2.16) follows by direct computation (see [17] for the

analogous formula where Ω ⊂ IRn, n ≥ 3). utLet us consider the linearized equation associated to (1.1), i.e.

−∆v = λeuv in Ωv = 0 on ∂Ω,

(2.19)

Lemma 4 Let vλ be a solution of (2.19). The following identities hold

∫∂Ω

(x− y) · ν(x)∂uλ

∂ν

∂vλ

∂νdSx = −2

∫∂Ω

∂vλ

∂νdSx for any y ∈ IR2 (2.20)

and ∫∂Ω

∂uλ

∂xi

∂vλ

∂νdSx = 0 for any i = 1, 2. (2.21)

7

Proof Let us consider the function

wλ(x) = (x− y) · ∇uλ + 2 (2.22)

which satisfies

−∆wλ = λeuλwλ (2.23)

Multiplying (2.23) by vλ and (2.19) by wλ we get∫∂Ω

wλ∂vλ

∂νdSx = 0 (2.24)

Since uλ = 0 on ∂Ω we obtain (2.20).To prove (2.21) we differentiate (1.1) with respect to xi:

−∆∂uλ

∂xi= λeuλ

∂uλ

∂xi(2.25)

Multiplying (2.25) by vλ and (2.19) by ∂uλ

∂xiwe get (2.21). ut

Lemma 5 Let uλ be a solution of (1.1) and set δ2λ = 1

λe||uλ||. Then δλ→0 as

λ→0 and the function zλ : Ωλ = Ω−xλ

δλ7→ IR,

zλ(x) = uλ(δλx + xλ)− ||uλ|| (2.26)

verifies

zλ(x)→ log1(

1 + |x|28

)2 in C2loc(IR

2) (2.27)

Proof First we show that δλ→0 as λ→0. Formula (2.9) implies that, for anyP 6= x0, ∣∣∣∣∣uλ(P )− ||uλ||+ log

(1 +

18δ2

λ

|P − xλ|2)2∣∣∣∣∣ ≤ C. (2.28)

If δλ ≥ C > 0, recalling (2.8) and passing to the limit in (2.28) we reach acontradiction.

Let us write down the equation satisfied by zλ.−∆zλ = ezλ in Ωλ

zλ(0) = 0 zλ(x) ≤ 0 ∀x ∈ Ωλ∫Ωλ

ezλ→8π.(2.29)

Since zλ(0) = 0, by a Brezis-Merle’s result ([3]) we deduce that zλ is boundedin L∞loc(IR

2). Standard regularity theory implies that zλ(x)→z(x) in C2loc(IR

2)where z is a solution of −∆z = ez in IR2

z(0) = 0 z(x) ≤ 0 ∀x ∈ IR2∫IR2 ez < +∞.

(2.30)

8

By Theorem 5 we get that z(x) = log 1(1+

|x|28

)2 and this finishes the proof. ut

Theorem 7 Let uλ be a solution of (1.1) satisfying (2.2). Then

||uλ||∞ = −2 log λ + C0 − 8πR(x0) + o(1) as λ→0 (2.31)

where C0 = − 12π

∫IR2 log 1

|x|1(

1+|x|28

)2 dx = 2 log 8

Proof Using Green’s representation formula we get

||uλ||∞ = uλ(xλ) = λ

∫Ω

G(xλ, y)euλ(y)dy =

− λ

2π

∫Ω

log |xλ − y|euλ(y)dy + λ

∫Ω

H(xλ, y)euλ(y)dy = I1 + I2.(2.32)

Using the notations of the proof of Lemma 5 we have

I1 = − λ

2π

∫Ω

log |xλ − y|euλ(y)dy = −λδ2e||uλ||

2π

∫Ωλ

log(δλ|y|)ezλ(y)dy =

− log δλ

2π

∫Ωλ

ezλ(y)dy − 12π

∫Ωλ

log |y|ezλ(y)dy. (2.33)

By (2.9) and since we have that∫Ωλ

ezλ(y)dy = 8π + O(λ) (see Theorem 1.3 in[7]) we get

I1 =(

14π

log λ +14π||uλ||∞

)∫Ωλ

ezλ(y)dy − 12π

∫IR2

log |y| 1(1 + |y|2

8

)2 dy + o(1) =

(2 log λ + 2||uλ||∞) (1 + O(λ))− 12π

∫IR2

log |y| 1(1 + |y|2

8

)2 dy + o(1) (2.34)

since δ2λ = 1

λe||uλ||→0 by Lemma 5. Moreover

I2 = λ

∫Ω

H(xλ, y)euλ(y)dy = λδ2e||uλ||∞∫

Ωλ

H(xλ, xλ + δλy)ezλ(y)dy =

8πH(x0, x0) + o(1) (2.35)

again by (2.9). Finally

||uλ||∞ = I1 + I2 = (2 log λ + 2||uλ||∞) (1 + O(λ))− 12π

∫IR2

log |y| 1(1 + |y|2

8

)2 dy

+8πH(x0, x0) + o(1). (2.36)

Then

||uλ||∞(1 + O(λ)) = −2 log λ(1 + O(λ)) +12π

∫IR2

log |y| 1(1 + |y|2

8

)2 dy

−8πH(x0, x0) + o(1)

9

and so

||uλ||∞ = −2 log λ +C0 − 8πR(x0)

1 + O(λ)+ o(1) = −2 log λ + C0 − 8πR(x0) + o(1)

(2.37)and then (2.31) follows (here we have that 1

2π

∫IR2 log |y| 1(

1+|y|28

)2 dy = 2 log 8 by

straightforward computation). ut

Lemma 6 Let f ∈ C1(IR2) be a function satisfying f(x) = O( 1|x|4 ) at infinity.

Set x = (x1, x2) and

w(x1, x2) = (2.38)

−+∞∫

a1a21+a2

2x1+

a2a21+a2

2x2

f

(a1t +

a22

a21 + a2

2

x1 −a1a2

a21 + a2

2

x2, a2t−a1a2

a21 + a2

2

x1 +a21

a21 + a2

2

x2

)dt

for (a1, a2) ∈ IR2 \ (0, 0). Then w satisfies

a1∂w

∂x1(x1, x2) + a2

∂w

∂x2(x1, x2) = f(x1, x2) (2.39)

Moreover, setting

ζ(x1, x2) = log

1x2

1 + x22

√x21+x2

2∫−∞

tf

(t

x1√x2

1 + x22

, tx2√

x21 + x2

2

)dt

(2.40)

we have that ζ satisfies

(x1

∂ζ

∂x1(x1, x2) + x2

∂ζ

∂x2(x1, x2) + 2

)eζ(x1,x2) = f(x1, x2) (2.41)

Proof It is a straightforward computation. q.e.d.

3 The nondegeneracy result.

In this section we prove Theorem 1. We will use the following lemma.

Lemma 7 It holds

∂2R(y)∂yi∂yj

= −2∫

∂Ω

∂G(x, y)∂xi

∂

∂yj

(∂G(x, y)∂νx

)dSx. (3.1)

10

Proof By (2.16) and the definition of the Robin function we get

∂R(y)∂yi

= −∫

∂Ω

νi(x)(∂G(x, y)

∂νx

)2

dSx. (3.2)

Differentiating (3.2) with respect to yj we get

∂2R(y)∂yi∂yj

= −2∫

∂Ω

νi(x)∂G(x, y)

∂νx

∂

∂yj

(∂G(x, y)∂νx

)dSx. (3.3)

Since the Green function G(x, y) is zero on the boundary ∂Ω and G(x, y) =G(y, x) we have νi(x)∂G(x,y)

∂νx= ∂G(x,y)

∂xiand it proves the lemma. q.e.d.

Proof of Theorem 1We argue by contradiction and let us suppose that there exists vλ 6≡ 0 which

solves (2.19). Hence vλ which satisfies −∆vλ = λeuλvλ in Ω||vλ||∞ = 1vλ = 0 on ∂Ω,

(3.4)

where uλ satisfies (1.1) and (2.2). Let us set for δ2λλe||uλ||∞ = 1,

uλ(x) = uλ(δλx + xλ)− ||uλ||∞ (3.5)

andvλ(x) = vλ(δλx + xλ). (3.6)

We have that vλ verifies the equation −∆vλ = euλ vλ in Ωλ

||vλ||∞ = 1vλ = 0 on ∂Ωλ,

(3.7)

where Ωλ = Ω−xλ

δλ. We remark that, since |vλ| ≤ 1 we obtain by (3.7) that∫

Ωλ|∇vλ|2 ≤

∫Ωλ

euλ ≤ 9π for λ large.Then, using the standard regularity theory it is possible to show that vλ

converges to v0 in C2loc(IR

2). Passing to the limit in (2.16) and using Theorem5 we deduce that v0 satisfies

−∆v0 =1

(1 + |x|28 )2

v0 (3.8)

Since |v0| ≤ 1, by Lemma 1 we get

v0(x) =2∑

i=1

aixi

8 + |x|2+ b

8− |x|2

8 + |x|2(3.9)

for some a1, a2, b ∈ IR.

Step 1: ai = 0 for i = 1, 2 in (3.9).

11

Here we use the assumption that x0 is a nondegenerate critical point of theRobin function. By contradiction let us suppose that (a1, a2) 6= (0, 0).

Set fλ(y) = euλ(y)vλ(y) − 64b 8−|y|2(8+|y|2)3 . Using the Green’s rapresentation

formula we derive

vλ(x) =∫

Ω

G(x, y)λeuλ(y)vλ(y)dy =∫

Ωλ

G(x, xλ + δλy)euλ(y)vλ(y)dy =∫Ωλ

G(x, xλ + δλy)(

fλ(y) + 64b8− |y|2

(8 + |y|2)3

)dy = I1,λ + I2,λ (3.10)

Since vλ converges to v0 pointwise in IR2 we have that fλ(y)→64∑2

i=1aiyi

(8+|y|2)3

pointwise in IR2. Moreover from Theorem 6 we get that fλ(y) = O( 1|y|4 ) at

infinity. Hence Lemma 6 applies and then there exists wλ(y) which solves (2.39)with f = fλ. Moreover wλ(y)→− 16

(8+|y|2)2 . Then

I1,λ =∫

Ωλ

G(x, xλ + δλy)fλ(y)dy = (3.11)∫Ωλ

G(x, xλ + δλy)2∑

i=1

ai∂wλ

∂yidy = −δλ

2∑i=1

ai

∫Ωλ

∂G(x, xλ + δλy)∂yi

wλdy =

δλ

(2∑

i=1

ai∂G(x, x0)

∂yi

∫IR2

16(8 + |y|2)2

dy + o(1)

)=

δλ

(2π

2∑i=1

ai∂G(x, x0)

∂yi+ o(1)

)

On the other hand we have

I2,λ = 64b

∫Ωλ

G(x, xλ + δλy)8− |y|2

(8 + |y|2)3dy = (3.12)

b

2

∫Ωλ

G(x, xλ + δλy)2∑

i=1

∂(yie

U(y))

∂yidy =

− b

2δλ

2∑i=1

∫Ωλ

∂G(x, xλ + δλy)∂yi

yieU(y)dy =

− b

2δλ

(2∑

i=1

∂G(x, x0)∂yi

∫IR2

yieU(y)dy + o(1)

)= o(δλ).

Finally by (3.10)-(3.12) we easily derive

vλ(x)δλ

→2π2∑

i=1

ai∂G(x, x0)

∂yipointwise in (Ω \ x0) (3.13)

12

By Lemma 2 we deduce that

vλ(x)δλ

→2π2∑

i=1

ai∂G(x, x0)

∂yiin C1(ω \ x0) (3.14)

where ω is a neighborhood of the boundary ∂Ω Hence we can pass to the limitin the identity (2.21) and we get

2∑j=1

aj

∫∂Ω

∂G(x, x0)∂xi

∂

∂νx

(∂G(x, x0)

∂yj

)dSx = 0 for any i = 1, 2. (3.15)

By Lemma 7 we deduce

2∑i=1

ai∂2R

∂yi∂yj(x0) = 0 (3.16)

Since x0 is a nondegenerate critical point of the Robin function we get thatthe matrix ∂2R

∂xi∂xj(x0) is invertible. So ai = 0 for i = 1, 2, a contradiction. This

finishes the proof of Step 1.

Step 2: b = 0 in (3.9)By contradiction let us suppose that b 6= 0. By the previous step we have

vλ(x)→b8− |x|2

8 + |x|2in C2

loc(IR2) (3.17)

We multiply by uλ in (3.4) and by vλ in (1.1). We get

λ

∫Ω

euλvλ = λ

∫Ω

euλuλvλ (3.18)

Then

λ

∫Ω

euλuλvλ = λ

∫Ω

euλ(uλ − ||uλ||∞)vλ + ||uλ||∞λ

∫Ω

euλvλ =∫Ωλ

euλ(uλ)vλ + ||uλ||∞λ

∫Ω

euλvλ = (3.19)∫IR2

1

(1 + |x|28 )2

log1

(1 + |x|28 )2

(b8− |x|2

8 + |x|2

)dx + ||uλ||∞λ

∫Ω

euλvλ + o(1) =

128b

∫IR2

|x|2

(8 + |x|2)2+ ||uλ||∞λ

∫Ω

euλvλ + o(1) =

8πb + ||uλ||∞λ

∫Ω

euλvλ + o(1) =

Let us recall that−∫

∂Ω

∂vλ

∂ν= λ

∫Ω

euλvλ (3.20)

13

Let us show that∂vλ

∂xi= o(δλ) (3.21)

To prove (3.21) we use the Green’s representation formula,

∂vλ(x)∂xi

= λ

∫Ω

∂G(x, y)∂xi

euλ(y)vλ(y) =∫Ωλ

∂G(x, xλ + δλy)∂xi

euλ(y)vλ(y)dy (3.22)

We use (2.40) and (2.41) of Lemma 6 with fλ(y) = euλ(y)vλ(y). Since euλ(y)vλ(y) →

64b 8−|y|2(8+|y|2)3 = b

2

2∑i=1

∂∂yi

(yi

(8+|y|2)2

)it is not difficult to see that, with this choice

of fλ, the function ζ = ζλ in Lemma 6 verifies eζλ(y)→ b2(8+|y|2)2 . Then (3.22)

becomes, for x 6= x0,

∂vλ(x)∂xi

=∫

Ωλ

∂G(x, xλ + δλz)∂xi

(y1

∂ζλ

∂y1(y1, y2) + x2

∂ζλ

∂y2(y1, y2) + 2

)eζλ(y1,y2)dy =

2∑i=1

∫Ωλ

∂G(x, xλ + δλz)∂xi

∂(yieζλ(y1,y2))∂yi

dy = −δλ

2∑i=1

∫Ωλ

∂2G(x, xλ + δλz)∂xi∂yi

yieζλ(y1,y2)dy =

−δλ

2∑i=1

∂2G(x, x0)∂xi∂yi

(∫IR2

yieU(y1,y2)dy + o(1)

)= o(δλ) (3.23)

Hence, by (3.20) and 2.31) we derive

||uλ||∞λ

∫Ω

euλvλ = o (||uλ||∞δλ) = o(1) (3.24)

By (3.18) we get

λ

∫Ω

euλuλvλ = o(1) (3.25)

which implies b = 0 in (3.19). This finishes the proof of Step 2.

Step 3: the contradiction.We have proved that v0 ≡ 0 in IR2. Let us denote by xλ the point where

vλ achieves its maximum, so that vλ(xλ) = 1. Since vλ → 0 in C2loc(IR

2) wenecessarily have |xλ| → ∞. Let us define the functions

uλ(x) = uλ(x

|x|2), vλ(x) = vλ(

x

|x|2), x ∈ IR2 \ 0.

Clearly, we have vλ( xλ

|xλ|2 ) = 1. It is easily seen that vλ satisfies the followingequation:

−∆vλ =1|x|4

euλ(x)vλ(x) (3.26)

14

We observe that 1|x|4 euλ(x) ≤ C by (2.9) of Theorem 6. Moreover |vλ(x)| ≤ 1

and vλ(x) → 0 in C2loc(IR

2 \ 0). Hence vλ → 0 in L2(B1(0)).Since the capacity of one point is zero, we can apply the regularity theory to

vλ (see Theorem 8.17 [15]) and we get that ||vλ||L∞(B 12(0)) ≤ C||vλ||L2(B1(0)) →

0. This gives a contradiction since ||vλ||L∞(B 12(0)) = vλ( xλ

|xλ|2 ) = 1. This provesthe theorem. ut

4 A uniqueness result

In this section we consider a bounded domain Ω which is convex in the directionx1, x2 and symmetric with respect to the axis xi = 0 for i = 1, 2. We know by theGidas, Ni, Nirenberg’s results (see [14]) that any solution u of (1.1) is symmetricwith respect to the axis xi = 0 for i = 1, 2 and ∂u

∂xi< 0 for xi > 0, i = 1, 2.

Moreover u has a unique maximum point in 0. A consequence of this result isthe following proposition

Proposition 1 Let us consider a solution uλ of (1.1) where Ω is a boundeddomain which is convex in the direction x1, x2 and symmetric with respect tothe axis xi = 0 for i = 1, 2. Then we have either

limλ→0

λ

∫Ω

euλ = 0 (4.27)

orlimλ→0

λ

∫Ω

euλ = 8π (4.28)

Proof First of all we show that

λ

∫Ω

euλ ≤ C (4.29)

for some positive constant C depending only on Ω. By Pohozaev’s identity weget

2λ

∫Ω

(euλ − 1) =12

∫∂Ω

(x · ν)(

∂uλ

∂ν

)2

(4.30)

Since Ω is strictly starshaped with respect to the origin, by 1.1 we get∫∂Ω

(x · ν)(

∂uλ

∂ν

)2

≥ γ

|Ω|

(∫∂Ω

∂uλ

∂ν

)2

=γ

|Ω|

(∫Ω

euλ

)2

(4.31)

¿From (4.30) and (4.31) we get (refz3). Setting vλ = uλ + log λ and arguingas in Theorem 6 we derive that

∫Ω

evλ ≤ C(Ω) and vλ(0)→∞.By Theorem 3 in Brezis and Merle ([3]) we have that there exist finitely many

blow up points x1, . . . , xl such that vλ tends to −∞ uniformly on compactsubsets of Ω \ x1, . . . , xl and

evλ l∑

i=1

αiδxi (4.32)

15

with αi ≥ 4π. Here δxi is the Dirac mass at xi.We recall that y is called a blow up point if there exist yλ→y such that

vλ(yλ)→∞ as λ→0.

By (4.32) we get λeuλ l∑

i=1

αiδxi and Gidas-Ni-Nirenberg’s theorem implies

that the set x1, . . . , xl is the singleton O. We have proved that

λeuλ αδO. (4.33)

Let us show that either α = 0 or α = 8π.By (4.33) we derive that uλ→G(x,O) in C1(ω) where ω is a neighborhood

of the boundary ∂Ω. Again by the Pohozaev identity (4.30), passing to the limitas λ→0

2α =α2

2

∫∂Ω

(x · ν)(

∂G(x,O)∂ν

)2

(4.34)

and by (2.15) of Lemma 3 we get that either α = 0 or α = 8π. This proves thelemma. ut

We can prove Theorem 2

Proof of Theorem 2 We argue by contradiction. Let us suppose thereexists a sequence λn → 0 and function un, vn ∈ C∞(Ω) which solves (1.1) and(2.2) with λ replaced by λn.

By the previous proposition we have that

limn→∞

λn

∫Ω

eun = 8π, limn→∞

λn

∫Ω

evn = 8π

Then we can apply the asymptotic estimates of Section 2 to both un and vn.Let wn(x) = un(δnx)−vn(δnx)

‖un(δnx)−vn(δnx)‖∞ where δ2n = 1

λne‖un‖ . Then wn(x) : Ωn → IRsolves −∆wn = cn(x)w(x) in Ωn

‖wn‖ = 1wn = 0 on ∂Ωn

(4.35)

where

cn(x) =∫ 1

0

etun(δnx)+(1−t)vn(δnx)−‖un‖dt =∫ 1

0

etun(x)+(1−t)vn(x)+(1−t)(‖vn‖−‖un‖)dt.

As in the proof of Theorem 1 we call un = un(δnx)− ‖un‖ and vn = vn(δnx)−‖vn‖. We had already proved that un(x) → log 1

(1+|x|28 )2

in C2loc(IR

2). In a

similar manner we can see that vn(x) → log 1

(1+|x|28 )2

in C2loc(IR

2). In fact vn

solves

16

−∆vn = Cevn in Ωn

vn(0) = 0 in Ωn

vn ≤ 0 in Ωn

(4.36)

where C = e‖vn‖−‖un‖ = 1 + o(1) from Theorem 7. By a Brezis-Merle resultwe can deduce that vn is bounded in L∞loc(Ω) and then by standard regularitytheory vn → v in C2

loc(IR2), where v solves

−∆v = ev in IR2

vn(0) = 0 in IR2 (4.37)

and so v(x) = log 1

(1+|x|28 )2

. Then we obtain that cn(x) → 1

(1+|x|28 )2

. From equa-

tion (4.35) and standard regularity results we have that wn → w in C2loc(IR

2)where w solves

−∆w = 1

(1+|x|28 )2

w in IR2

‖w‖∞ = 1(4.38)

¿From Lemma 1 we have w =∑ aixi

8+|x|2 + b 8−|x|28+|x|2 for some ai, b ∈ IR. Since wn

is symmetric with respect to the axis x1 = 0 and x2 = 0 we also get that w issymmetric with respect to the axis x1 = 0 and x2 = 0. This implies ai = 0 fori = 1, 2 and

w = b8− |x|2

8 + |x|2.

Now let wn(x) = un(x)−vn(x)‖un−vn‖∞ . It solves

−∆wn = cn(x)w(x) in Ωwn = 0 on ∂Ω (4.39)

where cn(x) = λn

∫ 1

0etun(x)+(1−t)vn(x)dt. Using un as a test function in (4.39),

and wn in (1.1), we get

λn

∫Ω

eunwndx =∫

Ω

cnwnundx (4.40)

for all n.We want to show that b = 0. By contradiction let us suppose that b 6= 0.

From equation (4.40) we have

λn

∫Ω

eunwndx =∫

Ω

cnwn(un − ‖un‖)dx + ‖un‖∫

Ω

cnwndx = (4.41)∫Ωn

cnwnundx + ‖un‖∫

Ωn

cnwndx = 8πb + ‖un‖∫

Ωn

cnwndx + o(1).

An easy computation shows that

λn

∫Ω

eunwndx =∫

Ωn

eunwndx→b

∫IR2

b64

(8 + |x|2)28− |x|2

8 + |x|2= 0 (4.42)

17

If we also prove that ‖un‖∫Ωn

cnwndx → 0 we reach a contradiction in(4.41). We proceed as in the proof of Theorem (1), Step 2.

Let us consider the equation (4.39). Integrating over Ω we get −∫

∂Ω∂wn

∂µ =∫Ω

cnwn, and as in the proof of Theorem (1) we want to show that

∂wn

∂xi= o(δn). (4.43)

This proves that ‖un‖∫Ω

cnwn = o(‖un‖δn) → 0, by Theorem 7.Let us prove (4.43). Using the Green’s representation formula we have

∂wn(x)∂xi

=∫

Ω

∂G(x, y)∂xi

cn(y)wn(y)dy =∫

Ωn

∂G(x, δny)∂xi

cn(y)wn(y).

Let fn(y) = cn(y)wn(y). We can use Lemma 6 and we have for x 6= 0

∂wn

∂xi=∫

Ωn

∂G(x, δny)∂xi

(y1

∂ξn

∂y1(y1, y2) + y2

∂ξn

∂y2(y1, y2) + 2

)eξn(y1,y2)dy =

2∑i=1

∫Ωn

∂G(x, δny)∂xi

∂yieξn(y1,y2)

∂yidy = −δn

2∑i=1

∫Ωn

∂2G(x, δny)∂xi∂yi

yieξn(y1,y2)dy =

−δn

2∑i=1

∂2G(x, 0)∂xi∂yi

(∫IR2

yieb2 U(y1,y2)dy + o(1)

)= o(δn).

Then 4.43) holds and it implies that b = 0.Hence we have shown that w = 0 in IR2. Let us denote by xn as the

maximum point of wn, so that wn(xn) = 1. Since wn → 0 in C2loc(IR

2) we havethat xn →∞. Let us set

un(x) = un(x

|x|2), vn(x) = vn(

x

|x|2), wn(x) = wn(

x

|x|2), x ∈ IR2 \ 0.

We have wn( xn

|xn|2 ) = 1, and wn satisfies the following equation

−∆wn =1|x|4

cn(x)wn(x)

where cn(x) =∫ 1

0etun(x)+(1−t)vn(x)+(1−t)(‖vn‖−‖un‖)dt. Using the estimate (2.9)

we have that 1|x|4 cn(x) ≤ C. Furthermore |wn(x)| ≤ 1 and wn → 0 in C2

loc(IR2 \

0). Hence wn → 0 in L2(B1(0)). We can apply the regularity theory to wn

and get ‖wn‖L∞(B 12(0)) ≤ ‖wn‖L2(B1(0)) → 0. This gives a contradiction to

‖wn‖L∞(B 12(0)) = wn( xn

|xn|2 ) = 1 and it proves the Theorem. ut

We now prove Corollary 1.Proof of Corollary 1 By Proposition 1 we have that any solution to (1.1)

satisfies either (4.27) or 4.28). If (4.27) holds, by classical results we know that inthis case the corresponding solution uλ is the minimal solution, which is unique.On the other hand if (4.28) holds, by Theorem 2 we have the uniqueness of thesolution blowing up at one point. This proves the corollary. ut

18

5 Starshapedness and convexity results.

In this section we prove some results about the level sets of the solutions. Letus start with the proof of Theorem 3.

Proof of Theorem 3 By contradiction let us suppose that there exist asequence λn → 0 and points yn ∈ Ω, yn 6= xλn

= xn, such that

(yn − xn) · ∇un(yn) ≥ 0 (5.1)

for all n ∈ N (we set ∇uλn≡ ∇un). Up to a subsequence yn → y ∈ Ω. Let

δ2n = 1

λne||un||∞ . We consider three different cases:

i) y 6= x0;

ii) y = x0 and yn ∈ BδnR(xn), for some real number R > 0 ;

iii) y = x0 but yn /∈ BδnR(xn), for any real number R.

Step 1: Consider the case i).We have y ∈ Ω \Bδ(x0) for some δ > 0. From Theorem 6 we have un(x) →

8πG(x, x0) in C1(Ω\x0) and so (yn−xn)·∇un(yn) → 8π(y−x0)·∇G(y, x0) ≥ 0and this is not possible. In fact consider the function (x − x0) · ∇G(x, x0); itis harmonic in Ω \ Bδ(x0) and from the Hopf’s boundary lemma (x − x0) ·∇G(x, x0) = (x − x0) · n∂G

∂n ≤ 0 on ∂Ω since Ω is convex. Moreover (x −x0) · ∇G(x, x0) = − 1

2π + (x − x0) · ∇H(x, x0) and if δ is small enough it isstrictly negative on ∂Bδ(x0). We can then apply the maximum principle getting(x− x0) · ∇G(x, x0) < 0 in Ω \Bδ(x0). This leads to a contradiction.

Step 2: We are in the case ii).Consider the function un(x) as in Lemma 5. Let yn = yn−xn

δn. Then |yn| <

δnRδn

= R and so up to a subsequence yn → y ∈ BR(0). From (5.1) we haveyn ·∇un(yn) = (yn−xn) ·∇un(yn) ≥ 0 and passing to the limit yn ·∇un(yn) →− 1

2|y|2(

1+|y|28

) . If y 6= 0 we reach a contradiction. On the other hand if y = 0 let us

consider the function φn(t) = un(tyn). This function has a maximum at t = 0,since 0 is a maximum for un, and another critical point in [0, 1]. So there existsa value t ∈ [0, 1] such that φ′′n(t) = 0 and then D2un(tyn) = 0. Letting n →∞we derive that 0 is a degenerate maximum point for the function log 1(

1+|x|28

)2which is impossible.

Step 3: Finally we consider the case iii).¿From equation (1.1) and the Green’s representation formula we have

un(x) = λn

∫Ω

G(x, y)eun(y)dy (5.2)

for all x ∈ Ω, where G(x, y) = 12π log 1

|x−y| + H(x, y). Let rn = |yn − xn|.Then rn → 0 but 1 ≥ rn ≥ Rδn for ant R > 0. Let vn(x) = un(rnx + xn) −

19

12π log 1

rnλn

∫Ω

eun(y)dy. From equation (5.2) we have

vn(x) = λn

∫Ω

G(rnx + xn, y)eun(y)dy − 12π

log1rn

λn

∫Ω

eun(y)dy

for all x ∈ Ωn = Ω−xn

rn. Letting y = δnt + xn, we get

vn(x) = λn

∫Ωn

G(rnx+xn, δnt+xn)eun(δnt+xn)δ2ndt− 1

2πlog

1rn

λn

∫Ωn

eun(δnt+xn)δ2ndt

where Ωn = Ω−xn

δn. Then

vn(x) =∫

Ωn

G(rnx + xn, δnt + xn)eun(t)dt− 12π

log1rn

∫Ωn

eun(t)dt (5.3)

=∫

Ωn

[12π

log1

|rnx− δnt|+ H(rnx + xn, δnt + xn)

]eun(t)dt− 1

2πlog

1rn

∫Ωn

eun(t)dt

=12π

∫Ωn

log1

|x− δn

rnt|

eun(t)dt +∫

Ωn

H(rnx + xn, δnt + xn)eun(t)dt.

¿From equation (2.2) we know that∫Ωn

eun(t)dt → 8π. Moreover from (2.9)∫Ωn

H(rnx + xn, δnt + xn)eun(t)dt → 8πH(x0, x0).

We observe that δn

rn< 1

R for any R > 0 ∈ IR and so δn

rn→ 0. Again from (2.9)

we derive that for any x ∈ Ωn, and |t ≥ 1| it holds

log1

|x− δn

rnt|

eun(t) ≤ C|x|+ |t||t|4

for n large enough. We can pass to the limit in (5.3) getting

vn(x) → 4 log1|x|

+ 8πH(x0, x0) in C(IR2 \ 0).

We can do the same for the derivatives of un. In fact

∇vn(x) = λ

∫Ω

∇xG(rnx + xn, y)eun(y)dy (5.4)

=∫

Ωn

[− 1

2π

rn(rnx− δnt)|rnx− δnt|2

+ rn∇xH(rnx + xn, δnt + xn)]

eun(t)dt(5.5)

The function ∇xH(rnx+xn, δnt+xn) is bounded in a neighborhood of (x0, x0)and so

rn

∫Ωn

∇xH(rnx + xn, δnt + xn)eun(t)dt → 0

20

as n →∞. Passing to the limit in 5.4) we get

∇vn(x) → − 4x

|x|2in C(IR2 \ 0).

Now let yn = yn−xn

rn. Then yn → y, where |y| = 1. From (5.1) and the C1

convergence of vn to 4 log 1|x| + 8πH(x0, x0) we get

yn · ∇vn(yn) = (yn − xn) · ∇un(yn) ≥ 0 ∀n

and passing to the limit y · − 4y|y|2 = −4 ≥ 0 which is impossible. This finishes

the proof of this Theorem. ut

We now prove Theorem 4.

Proof of Theorem 4 We point out that from the previous theorem we havethat ∇uλ(x) 6= 0 for any x 6= xλ. Then the level set are regular curves.

We argue by contradiction. Let us suppose there exists a sequence λn → 0and points yn ∈ Ω, yn 6= xn such that

Kun(yn) ≤ 0 ∀n (5.6)

where Kun(yn) denotes the curvature of the level set x ∈ Ω s.t. un(x) = un(yn)at the point yn. It is easily seen that Ku(y) = − 1

‖∇u(y)‖vtHu(y)v‖v‖2 where Hu(y) is

the Hessian matrix of u at the point y and v is any vector v 6= 0, perpendicularto∇u(y) (see [30] for definition). Up to a subsequence and using our assumptionon ∂Ω we get yn → y ∈ Ω. Then as in the proof of Theorem 3 we distinguishthree cases

i) y 6= x0;

ii) y = x0 and yn ∈ BδnR(xn);

iii) y = x0 but yn /∈ BδnR(xn).

Step 1: Let us consider case i). Since y 6= x0 we have y ∈ Ω \Bδ(x0) for someδ > 0, and from Theorem 6 un(x) → 8πG(x, x0) in C2(Ω \ Bδ(x0)). If y ∈ Ωpassing to the limit in (5.6) we get

KG(y,x0)(y) ≤ 0

which is impossible from a Theorem of Lewis, see [22].

Step 2: Now let y = x0 and yn ∈ BδnR(xn). We consider the function un.Note that from equation (5.6) we have

Kun(yn) = δnKun(yn) ≤ 0 (5.7)

where yn = yn−xn

δn. Up to a subsequence yn → y ∈ BR(0) and if y 6= 0 passing

to the limit in (5.7) we getKU (y) ≤ 0

21

which is impossible since U is a radial function. So let y = 0. From (5.7) andthe definition of curvature we know that for each vector τn = α(−∂un

∂x2, ∂un

∂x1) 6= 0

Kun(yn) = − 1

‖∇un(yn)‖τt

nHun (yn)τn

‖τn‖2 ≤ 0. In particular we choose ‖τn‖ = 1for each n. Hence we have τ t

nHun(yn)τn ≥ 0 for all n. Passing to the limit

τn → τ with ‖τ‖ = 1 and τ tHU (0)τ ≥ 0. But this is impossible since 0 is anondegenerate maximum point for U and its Hessian matrix is negative defined.

Step 3: Finally we consider case iii). Let rn = |yn − xn| and let vn(x) =un(rnx + xn) − 1

2π log 1rn

λn

∫Ω

eun(y)dy. As in Theorem 3 we can proove thatvn → W = 4 log 1

|x| + H(x0, x0) in C1(IR2 \ 0) and in the same manner wecan see the convergence is C2(IR2 \ 0). Letting yn = yn−xn

rnthen yn → y such

that |y| = 1 and from (5.6) we have

Kvn(yn) = rnKun(yn) ≤ 0 ∀n.

Passing to the limit we get KW (y) ≤ 0 and this is not possible since W is aradial function. ut

6 Convexity results for the minimal solutions.

In this section we study the minimal solution of problem (1.1). We will considerthe more general problem: −∆u = λf(u) in Ω

u > 0 in Ωu = 0 on ∂Ω

(6.1)

where Ω is a bounded smooth convex domain in IRN , N ≥ 2 with positive Gausscurvature at any point p ∈ ∂Ω, and f : [0,+∞) → (0,+∞) is a continuous,increasing and convex function such that,

f(0) > 0, lims→∞

f(s)s

= +∞.

Under these conditions it is known that there exists a real number λ∗ =λ∗(Ω), with 0 < λ∗ < ∞ such that for any 0 < λ < λ∗ equation (1.1) hasa minimal positive classical solution uλ which is the unique stable solution of(1.1), in the sense that λ1 (−∆− λeuλ) > 0. Moreover

Theorem 8 The minimal solution uλ is strictly increasing in λ for all x ∈ Ωand

limλ→0

uλ(x) = 0 in C2,α(Ω).

In an interesting paper (see [5]) Cabre and Chanillo have considered the minimalsolution of problem (6.1) in Ω ⊂ IR2 and have shown the following

22

Theorem 9 Let Ω be a smooth, bounded, strictly convex domain of IR2 whoseboundary has positive curvature. If uλ is the minimal solution of problem (6.1)in Ω for 0 < λ < λ∗, then uλ has a unique critical point in x0 ∈ Ω. Moreoverx0 is the maximum of uλ and it is nondegenerate in the sense that the Hessianmatrix of uλ at x0 is negative defined.

Here we want to study the convexity of the level sets of the minimal solutionuλ for λ small and Ω ⊂ IRN , N ≥ 3. We will need the following lemma

Lemma 8 Let us consider the function w(x) = −v12 (x) where v0 is a solution

of −∆v = 1 in Ωv > 0 in Ωv = 0 on ∂Ω.

(6.2)

Then the hessian matrix of w(x) is definite positive in a neighborhood of theboundary of Ω

Proof We repeat some standard computations (see [6]). Take a point y ∈ Ωwith small distance d to ∂Ω, and let z ∈ ∂Ω such that d(y, ∂Ω) = d = |z − y|.Let τ1, . . . , τN−1, ν be an orthonormal system of coordinates at z. Here ν is theunit normal vector to ∂Ω at z and τ1, . . . , τN−1 is a basis of the tangent spaceTN−1(z) to ∂Ω at z. We compute the second derivatives of wt with respect tothe directions τ1, . . . , τN−1, ν at y:

wτiτj =14v−

32 vτivτj −

12v−

12 vτiτj

wτiν =14v−

32 vτi

vν −12v−

12 vτiν

wνν =14v−

32 v2

ν −12v−

12 vνν .

¿From the boundary conditions we get that v(z) = 0 and vτi(z) = 0 ∀i =

1, . . . , N − 1 and consequently from Hopf’s Lemma we have v(y) ∼ αd withα > 0 and vτi

(y) = O(d) ∀i = 1, . . . , N − 1. Then we get

wτiτj= O(

1d

12)

wτiν = O(1d

12)

wνν ∼c1

d32

where c1 > 0. Now let ξ ∈ IRN be any vector different from zero. Thenξ =

∑N−1i=1 a1τi + aNν where ai ∈ IR for i = 1, . . . , N and let us denote by

Hw(x) the Hessian matrix of the function w at the point x.

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Then

ξtHw(y)ξ = a2Nwνν +

N−1∑i,j=1

aiajwτiτj + 2N−1∑i=1

aiaNwτiν

∼ a2Nc1

d32

+ O(1d

12) > 0 (6.3)

if d is small enough and aN 6= 0.Let us show that we have the same claim even if aN = 0. In this case we

have that ξ is a tangent vector to ∂Ω at z. Hence from the hypothesis on thecurvature of ∂Ω we have:

ku,z(ξ) = − 1|∇u(z)|

ξtHu(z)ξ > 0

where ku,z(ξ) is the normal curvature of ∂Ω at z in the direction ξ, and it isstrictly greater than zero. Then ξtHu(z)ξ < 0 and ξtHu(y)ξ < 0 if d is smallenough. Let us calculate

ξtHw(y)ξ = −12

1u

12ξtHu(y)ξ +

14

1u

32

N−1∑i=1

aiτi · ∇uN−1∑j=1

ajτj · ∇u =

−12

1u

12ξtHu(y)ξ +

14

1u

32(ξ · ∇u)2 > 0.

which proves the lemma utWa can prove the main result of this section

Theorem 10 Let Ω be a smooth, bounded, convex domain of IRN , N ≥ 2, withpositive Gauss curvature at any point p ∈ ∂Ω. If uλ is the minimal solution ofproblem (6.1) in Ω then there exists λ > 0, 0 < λ < λ∗, such that uλ has strictlyconvex level surfaces for all 0 < λ < λ. Moreover it has a unique nondegeneratecritical point, which is the maximum one.

Proof Let us consider the function vλ(x) = 1λf(0)uλ(x). It solves

−∆vλ = 1f(0)f(uλ) in Ω

vλ > 0 in Ωvλ = 0 on ∂Ω

(6.4)

Since ‖uλ‖∞ → 0 as λ → 0, by classical regularity results vλ → v in C2(Ω),where v is a solution to (6.2).

Now we can use some known results about the convexity of v, see [20].First of all we can assume that the unit ball B1(0) is contained in Ω and

let us consider the minimal solution v0 of the problem (6.2) in B1(0). An easycomputaction shows that v0(x) = 1−|x|2

2N .

24

Then we consider the function

w0(x) = −v120 (x).

We have Hw0(x) ≥ c > 0 in B1(0), where Hf (x) denotes the Hessian matrix ofthe function f at the point x. Now we deform continuously B1(0) in Ω by afamily Ωt, 0 ≤ t < 1, of bounded and strictly convex domains Ωt (with positivecurvature on ∂Ωt) such that Ω0 = B1, Ω1 = Ω, Ωt ⊂ Ωs for t < s, ∂Ωt varies

smoothly in the C2,α sense in t for 0 ≤ t < 1, and Ωt ↑ Ω if t ↑ 1. Let wt = −v12t

where vt is the solution of (6.2) in Ωt. It is straightforward to see that wt

satisfies ∆wt = −

12+|∇wt|2

wt> 0 in Ωt

wt < 0 in Ωt

wt = 0 on ∂Ωt.

(6.5)

¿From the C2,α(Ω) convergence of wt to w0 we have Hwt(x) > 0 in Ωt for some

t > 0. Let t = supt : Hwt(x) > 0 in Ωt. Then Hwt(x) ≥ 0 in Ωt. We want to

show thatHwt

(x) ≥ c > 0. (6.6)

By the previous lemma this is true for x belonging to a neighborhood of theboundary of Ω. On the other hand if there exists a point x such that Hi,j(x)ξiξj =0 by a Korevaar and Lewis results (see Theorem 1 in [20]) we get (for some vectorη) Hi,j(x)ηiηj = 0 in Ωt. But this is impossible by the previous lemma.

¿From (6.6) we deduce that Hwt+t′ (x) > 0 in Ωt+t′ for some t′ > 0, contra-dicting the definition of t. Hence t = 1 and Hw(x) ≥ 0 in Ω. Again using theresult by Korevaar and Lewis and the previous lemma we get that Hw(x) > 0in Ω.

Finally we consider the function wλ = −v12λ (x) in Ω. From the C2 con-

vergence of vλ → v we also have the C2 convergence of wλ → w in Ω andHwλ

(x) → Hw(x) in C0(Ω). Hence if λ is small enough Hwλ(x) > 0 in Ω and

then vλ has strictly convex level surfaces in Ω. The same is true for the functionuλ.

We observe that being wλ a strictly convex function it has a unique crit-ical point in x0 ∈ Ω, which is the minimum of wλ. Moreover ∇uλ(x) =λf(0)∇vλ(x) = λf(0)

2wλ∇wλ and hence uλ has a unique critical point in x0 which

is a the minimum and it is nondegenerate since Huλ(x0) = λf(0)Hvλ

(x0) =λf(0)2wλ

Hwλ(x0) < 0. ut

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