Nonlinear problems with solutions exhibiting a free boundary on the boundary

Ann. I. H. Poincaré – AN 22 (2005) 303–330 www.elsevier.com/locate/anihpc

Nonlinear problems with solutions exhibiting a free boundary on the boundary Problèmes nonlinéaires avec frontière libre sur le bord du domaine Juan Dávila a,∗ , Marcelo Montenegro b a Departamento de Ingeniería Matemática, CMM (UMR CNRS), Universidad de Chile, Casilla 170/3, Correo 3, Santiago, Chile b Universidade Estadual de Campinas, IMECC, Departamento de Matemática, Caixa Postal 6065, CEP 13083-970, Campinas, SP, Brasil

Received 14 June 2004; accepted 15 July 2004 Available online 7 April 2005

Abstract We prove existence of nonnegative solutions to −
u + u = 0 on a smooth bounded domain Ω subject to the singular −β + λf (x, u) on ∂Ω ∩ {u > 0} with 0 < β < 1. There is a constant λ∗ such that for boundary derivative condition ∂u ∂ν = −u ∗ 0 < λ < λ every nonnegative solution vanishes on a subset of the boundary with positive surface measure. For λ > λ∗ we show the existence of a maximal positive solution. We analyze its linearized stability and its regularity. Minimizers of the energy functional related to the problem are shown to be regular and satisfy the equation together with the boundary condition.  2005 Elsevier SAS. All rights reserved. Résumé Nous démontrons l’existence de solutions u
0 de l’équation −
u + u = 0 sur un domaine borné régulier Ω avec la −β + λf (x, u) sur ∂Ω ∩ {u > 0} oú 0 < β < 1. Il existe une constante condition de Neumann singulière suivante : ∂u ∂ν = −u λ∗ telle que pour 0 < λ < λ∗ , toute solution u
0 s’annule sur une partie du bord, de mesure (surfacique) strictement positive. Pour λ > λ∗ , nous démontrons l’existence d’une solution maximale positive. Nous analysons ses propriétés de stabilité linéaire et de régularité. On démontre que les minimiseurs de la fonctionnelle d’énergie associée sont réguliers et vérifient l’équation ainsi que la condition de bord.  2005 Elsevier SAS. All rights reserved. MSC: 35R35; 35J65; 35J25

* Corresponding author.

E-mail addresses: [email protected] (J. Dávila), [email protected] (M. Montenegro). 0294-1449/$ – see front matter  2005 Elsevier SAS. All rights reserved. doi:10.1016/j.anihpc.2004.07.006

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1. Introduction We study the existence and regularity of solutions of the following nonlinear boundary value problem  −
u + u = 0 in Ω,   u
0 in Ω, ∂u   = −u−β + f (x, u) on ∂Ω ∩ {u > 0}, ∂ν

(1)

where Ω ⊂ Rn , n
2, is a bounded domain with smooth boundary, 0 < β < 1 and ν is the exterior unit normal vector to ∂Ω. We assume that f : ∂Ω × R → R

is C 1 and f
0.

By a solution of (1) we mean a function u ∈ H 1 (Ω) ∩ C(Ω) satisfying    −β     −u + f (x, u) ϕ, ∇u · ∇ϕ + uϕ = ∀ϕ ∈ C01 Ω ∪ ∂Ω ∩ {u > 0} . Ω

(2)

(3)

∂Ω∩{u>0}

An equivalent way to write problem (1) is  −
u + u = 0 in Ω,  ∂u −β = −u + f (x, u) on Γ+ (u),  ∂ν u=0 on Γ0 (u),

(4)

where  Γ+ (u) = x ∈ ∂Ω: u(x) > 0 ,

 Γ0 (u) = x ∈ ∂Ω: u(x) = 0 .

The last boundary condition in (4) is trivial by the definition of Γ0 (u) itself. This notation emphasizes the fact that u satisfies a boundary condition of mixed type: a nonlinear Neumann condition on Γ+ (u) and Dirichlet on Γ0 (u). Observe that Γ+ (u) and Γ0 (u) form a partition of the boundary that depends on the solution u. In this sense u solves a free boundary problem on the boundary. In principle one may try to find solutions of (1) which are positive on ∂Ω, but it turns out that there are situations where no such a solution exists, and nonetheless there are nontrivial solutions of (1), see Theorem 1.8. There are at least two approaches to tackle the question of existence of a solution: one is to work with a regularization of problem (1) and the second one is a variational formulation for (1), see (7) below. In our first approach we consider the following regularization of (1)

−
u + u = 0 in Ω, ∂u (5) u = − (u+ε)1+β + f (x, u) on ∂Ω, ∂ν where ε > 0 is a parameter tending to zero. The solutions of (5) have the following convergence property. Theorem 1.1. Suppose f satisfies (2) and f (x, u) → 0 for u → ∞ u

uniformly in x.

(6)

and u = limε→0 u¯ ε exists. The convergence is Then Eq. (5) possesses a maximal solution u¯ ε which is positive in Ω uniform in Ω and u is a solution to the free boundary problem (1).

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Another approach to find solutions to (1) is to consider the functional φ : H 1 (Ω) → R given by    (u+ )1−β 1  |∇u|2 + u2 + φ(u) = − F (x, u+ ), 2 1−β Ω

where F (x, u) =

u 0

305

(7)

∂Ω

f (x, s) ds and u+ = max{u, 0}.

Theorem 1.2. Suppose that f satisfies (2) and (6). Then φ attains its minimum in H 1 (Ω) and any minimizer u of φ solves the free boundary problem (1). Next we deal with the regularity of the maximal solution u¯ ε to (5) or a minimizer u of φ. Theorem 1.3. Suppose f satisfies (2) and (6). Then there exists a constant C independent of ε such that the maximal solution u¯ ε to (5) satisfies |∇ u¯ ε |  C(u¯ ε )−β

in Ω.

As a consequence we have u¯ ε C 1/(1+β) (Ω)

 C. Remark 1.4. A consequence of the previous theorem is that the convergence u¯ ε → u of the maximal solution to 1

for all 0 < µ < 1 . And hence u ∈ C 1+β (Ω).

(5) in Theorem 1.1 is in the norm of C µ (Ω) 1+β

Theorem 1.5. Suppose f satisfies (2) and (6). Let u denote a minimizer of φ (cf. (7)). Then there exists a constant C such that |∇u|  Cu−β

in Ω

and hence u ∈ C 1/(1+β) (Ω). Remark 1.6. A prototype function describing the behavior of the solutions of (1) near a free boundary point is given by u(r, θ ) = cr α sin(αθ ), expressed in polar coordinates (x1 = r cos θ , x2 = r sin θ ), where α=

1 . 1+β

The function u is harmonic in the upper half-plane R2+ = {(x1 , x2 ) ∈ R2 | x2 > 0} and satisfies u = 0 on {(x1 , 0): x1
0} and  ∂u = −u−β on (x1 , 0): x1 < 0 , ∂ν for a suitable choice of the constant c > 0. This example indicates that the regularity stated in Theorems 1.3 and 1.5 is optimal with respect to the Hölder exponent. A modification of this harmonic function u will be useful later in the proof of the regularity theorems. For the proof of Theorem 1.5 we will use a Hardy type inequality.

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Proposition 1.7. Let G ⊂ Rn be a smooth domain (not necessarily bounded) and let Γ ⊂ ∂G be a bounded, relatively open subset with smooth boundary ∂Γ (the boundary is taken relative to ∂G). Let us define dΓ c (x) = dist(x, Γ c ),

Γ c = ∂G \ Γ.

There exists a constant Ch such that   ψ2  Ch |∇ψ|2 , ∀ψ ∈ C0∞ (G ∪ Γ ), dΓ c Γ

(8)

G

where Ch depends on Γ and G. Finally we address the question of whether there are in fact situations where the solutions that we construct in Theorems 1.1 and 1.2 are positive in Ω or whether there are nontrivial solutions which are zero at some subset of ∂Ω. For this purpose we consider (1) with f replaced by λf where λ > 0 is a parameter  −
u + u = 0 in Ω,   u
0 in Ω, (9) ∂u   = −u−β + λf (x, u) on ∂Ω ∩ {u > 0} . ∂ν Theorem 1.8. Assume f satisfies (2), (6) and f (x, u) is increasing in u and there is a ξ > 0 such that f (x, ξ ) ≡ 0.

(10)

Then for any λ > 0 Eq. (9) has a maximal solution u¯ λ and the map λ ∈ (0, ∞) → u¯ λ is nondecreasing. Moreover, (a) There exists λ∗ > 0 such that for λ > λ∗ , u¯ λ > 0 in Ω. (b) For 0 < λ < λ∗ all solutions must vanish in a nontrivial subset of ∂Ω, that is, the surface measure of {x ∈ ∂Ω: u(x) = 0} is positive. (c) The extremal solution u¯ λ∗ is positive a.e. on ∂Ω. (d) For λ > λ∗ , u¯ λ is stable in the sense that  Λ(u¯ λ ) =

inf ϕ∈C 1 (Ω)

Ω

|∇ϕ|2 + ϕ 2 −



−1−β (β u¯ λ ∂Ω  2 ∂Ω ϕ

+ λfu (x, u¯ λ ))ϕ 2

> 0.

(11)

Remark 1.9. If in addition to (2) and (6) we assume that f is concave, then actually the stability condition (11) characterizes the maximal solution in a similar way as in [4]. Depending on the dimension, the maximal solution u¯ λ∗ could be positive on ∂Ω, not only a.e. The relation between β and n is (3β + 1 + 2 β 2 + β)/(β + 1) > n − 1, which can only hold for some 0 < β < 1 in dimensions n = 2, 3, 4. The proof of this assertion is related to the stability of uλ . Proposition 1.10. The extremal solution satisfies u¯ λ∗
c > 0 on ∂Ω if (3β + 1 + 2 β 2 + β)/(β + 1) > n − 1, where c is a constant. There are a few works dealing with a singular derivative boundary condition. For example in [5,6] the authors study an evolution equation in one space dimension with a Neumann condition involving the singular term −u−β . In higher dimensions a similar evolution problem was addressed in [8] with a positive unbounded nonlinearity such as 1/(1 − u) and with a time interval [0, T ) where 0  u(t) < 1. One of the main contributions of this paper is

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307

that we deal with the possibility that the solution of the stationary problem in dimension n
2 vanishes on a large subset of ∂Ω. In this situation the solution develops a free boundary, a phenomenon that can occur only when Ω is at least a two-dimensional domain. Elliptic equations involving a nonlinear Neumann boundary condition have been studied elsewhere in the literature, see for instance [3,11,13] and [14]. The plan of the paper is the following. In Section 2 we present additional results and examples concerning the behavior of the maximal solution of (9) as λ varies. The proof of these assertions is postponed until Section 9. The Hardy-type inequality of Proposition 1.7 is proved in Section 3. Section 4 is devoted to introduce notations for later purposes. In Section 5 we construct a local subsolution [4,12] which is then used in Sections 6 and 7 for the proof of the regularity Theorems 1.3 and 1.5 respectively. In the proof of these theorems we employ frequently auxiliary results for linear equations that for convenience we have collected in the Appendix. Theorem 1.1 is proved in Section 6 and the proof of Theorem 1.2 is given in Section 7. Finally in Section 8 we prove Theorem 1.8 and Proposition 1.10.

2. Examples In this section we give some examples illustrating the exact vanishing properties of the maximal solution of (9) when λ varies. Proposition 2.1. Let Ω be a ball in Rn and assume that f satisfies (2) and (6) (the requirements of Theorem 1.8) and depends only on u. Then u¯ λ = 0 for 0 < λ < λ∗ . Proposition 2.2. For any smooth domain Ω and any function f satisfying the hypotheses of Theorem 1.8 we have that u¯ λ ≡ 0 for λ sufficiently small. Example 2.3. Let Ω be a ball in Rn . We construct a function f = f (x) (depending only on x) such that u¯ λ ≡ 0 for 0 < λ < λ¯ and u¯ λ ≡ 0 for λ¯  λ  λ∗ where 0 < λ¯ < λ∗ . The construction of this example is presented in Section 9.

3. Hardy inequalities In order to achieve our regularity results we need to establish the Hardy type inequality (8). For this aim we begin proving a Hardy inequality in a half space. Consider the upper half plane  R2+ = (x1 , x2 ) ∈ R2 | x2 > 0 and

 Γ = (x1 , x2 ) ∈ R2 | x1 < 0, x2 = 0 .

We use the standard notation r, θ for polar coordinates. Proposition 3.1. inf

ψ∈C0∞ (R2+ ∪Γ )

 R2

+ Γ

|∇ψ|2 ψ 2 /r

=

1 . π

(12)

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Proof. Let ψ ∈ C0∞ (R2+ ). Then π ψ(r, π) =

∂ψ (r, θ ) dθ  π 1/2 ∂θ

π 

0

Hence 

ψ2 π r

Γ

∂ψ ∂θ

1/2

2 (r, θ ) dθ

.

0

  π∞  1 ∂ψ 2 dr dθ  π |∇ψ|2 . r ∂θ R2+

0 0

For the opposite inequality we consider functions of the special form ψ(r, θ ) = ϕ(r)θ, where ϕ ∈ C0∞ (0, ∞). Observe that 

ψ(r)2 dr = π 2 r

∞

Γ

ϕ(r)2 dr r

0

and 

1 |∇ψ| = π 3 3

∞

R2+

Hence



∞

ϕ (r) r dr + π

2

2

0



ϕ(r)2 dr. r

0

∞  2 ϕ (r) r dr π 1  =  ∞0 + . 2 2 )/r) dr 3 π (ψ(r) )/r dr ((ϕ(r) Γ 0 R2+

|∇ψ|2

But

∞ inf ∞

ϕ∈C0 (0,∞)

 ∞0 0

ϕ  (r)2 r dr

(ϕ(r)2 /r) dr

= 0,

which can be seen by taking  log ε  − r if 0  r  ε, ε ϕε (r) = if ε  r  1 ,   − log r 0 if r
1. In fact ∞ 1 ϕε (r)2 r dr = log2 ε − log ε, 2

(13)

0

and ∞

1 ϕε (r)2 1 dr = log2 ε − log3 ε. r 2 3

0

Thus

∞  ∞0 0

ϕε (r)2 r dr

(ϕε (r)2 /r) dr

→0

as ε → 0+ .

2

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309

Remark 3.2. The fact that the functions ψ in (12) are required to vanish on a half of ∂R2+ is important for the infimum to be positive, since  2 R2+ |∇ψ|  = 0. inf ¯ 2 \{0}) R (ψ 2 /r) ψ∈C0∞ (R + This can be seen by taking ψ = ϕε as defined in (13). We prove now the Hardy inequality in a domain. Proof of Proposition 1.7. First we claim that it suffices to prove (8) for functions in C0∞ (G ∪ Γ ) which have support near ∂Γ . Indeed, let σ > 0 and consider  (∂Γ )σ = x ∈ G: dist(x, ∂Γ ) < σ .

be such that η ≡ 0 in G \ (∂Γ )σ and η ≡ 1 in (∂Γ )σ/2 . Suppose that (8) is true for functions in Let η ∈ C 1 (G) ∞ C0 (G ∪ Γ ) with support in (∂Γ )σ . Then for any ψ ∈ C0∞ (G ∪ Γ )    2  ψ2 (ηψ)2   C ∇(ηψ) dΓ c dΓ c Γ

(∂Γ )σ/2

G

and therefore 



ψ2 C dΓ c

(∂Γ )σ/2

|∇ψ|2 + ψ 2 .

G∩supp(η)

Since ψ vanishes on ∂G \ Γ , by Poincaré’s inequality   ψ 2  C |∇ψ|2 G

G∩supp(η)

and thus 

ψ2 C dΓ c

 |∇ψ|2 .

(14)

G

(∂Γ )σ/2

On the other hand by the trace theorem we have   2 ψ  C |∇ψ|2 . Γ

G

Since 1/dΓ c is bounded away from ∂Γ and ψ = 0 on ∂G \ Γ we obtain   ψ2  C |∇ψ|2 . dΓ c ∂G\(∂Γ )σ/2

(15)

G

Combining (14) and (15) we see that (8) holds. Using a partition of unity and the same argument as before we see that it is sufficient to consider the case of ψ ∈ C0∞ (G ∪ Γ ) with support in a small ball Bσ (x0 ) centered at x0 ∈ ∂Γ . In this situation choose an open set W ⊃ Bσ (x0 ) and a change of variables ϕ : W → Bσ (0) which flattens the boundary of G, that is, ϕ(W ∩ G) = Bσ (0) ∩ H where  H = (x  , xn ): x  ∈ Rn−1 , xn > 0 ,

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= Bσ (0) \ H

. We can also assume that W ∩ Γ is mapped into and ϕ(W ∩ ∂G) = Bσ (0) ∩ ∂H , ϕ(W \ G)  (x1 , x2 , . . . , xn ): x1 > 0, xn = 0 ∩ Bσ (0). Then, applying Proposition 3.1 to ψ ◦ ϕ in the half plane {(x1 , x2 , . . . , xn ): x1 > 0, x2 ∈ R, x3 = · · · = xn = 0} and integrating in the variables x3 , . . . , xn we see that (8) is valid. 2 4. Notations Let us choose and fix τ0 > 0 small enough so that for any x0 ∈ ∂Ω there exists an open set W containing the ball Bτ0 (x0 ) and a smooth diffeomorphism ϕ: W ⊂ Rn → Bτ0 (0) which flattens the boundary of Ω, that is ϕ(W ∩ Ω) = Bτ0 (0) ∩ H, ϕ(W ∩ ∂Ω) = Bτ0 (0) ∩ ∂H,

,

= Bτ0 (0) \ H ϕ(W \ Ω) where

 H = (x  , xn ): x  ∈ Rn−1 , xn > 0 .

We can also assume that ϕ(x0 ) = 0, ∇ϕ(x0 ) = I and ϕ preserves the normal direction on the surface W ∩ ∂Ω. For 0 < τ < τ0 and x0 ∈ ∂Ω let us adopt the notation Bτ+ = Bτ (x0 ) ∩ Ω,

(16)

and let us decompose its boundary as Γ = ∂Bτ (x0 ) ∩ Ω, i

Γe

=Γ1

∂Bτ+

=Γe

∪Γi

(the external and internal boundaries of

Γ = Bτ (x0 ) ∩ ∂Ω. e

∪Γ2

We also decompose   Γ 1 = ϕ −1 Bτ/2 (0) ∩ ∂Ω,

Bτ+ ) (17)

with Γ 2 = Γ e \ Γ 1.

(18) Γ1

In Fig. 1. we show the above defined sets. For convenience we have flattened and portions of the boundary ∂Ω. Let us introduce the rescaled variables x˜ and y˜ which allow us to work in the unit ball: 1 1 y˜ = ϕτ (x) ˜ = ϕ(τ x˜ + x0 ) = ϕ(x). x = τ x˜ + x0 , τ τ Henceforth we use the notation: 1 + = 1 (Bτ+ − x0 ) = B1 (0) ∩ 1 (Ω − x0 ),  = (Ω − x0 ), B Ω τ τ τ 1 1 1 Γi = (Γ i − x0 ), Γe = (Γ e − x0 ), Γk = (Γ k − x0 ), k = 1, 2. τ τ τ

Fig. 1.

Γ 2,

but in fact they are

(19)

(20)

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311

5. Construction of a local subsolution + . This construction is inspired by the Given x0 ∈ ∂Ω and 0 < τ < τ0 we construct a special function v˜ in B explicit solution of Remark 1.6. + ), a
0. For a parameter s > 0 consider the linear equation Let a ∈ L∞ (B  + , −
v˜ + a(x) ˜ v˜ = 0 in B     ∂ v˜ (x) ˜ = − dist(x, ˜ Γ2 )−β/(1+β) x˜ ∈ Γ1 , (21) ∂ν   v( 2 , ˜ x) ˜ = 0 x ˜ ∈ Γ    v( ˜ x) ˜ = s dist(x, ˜ ∂ Ω) x˜ ∈ Γi . Remark 5.1. The above problem (21) has a solution. By Hardy’s inequality of Lemma 1.7 the solution belongs to + ) and by potential theory methods it can be shown that the solution is in C 0 (B + ). H 1 (B The main goal of this section is to prove: + ), a
0. There exist τ0 > 0 (even smaller than that one in (16)) and s0 > 0 such that Lemma 5.2. Let a ∈ L∞ (B + and satisfies if 0 < τ < τ0 and s
s0 the solution of (21) is positive in B v( ˜ x) ˜
cs dist(x, ˜ Γ2 )1/(1+β) ,

∀x˜ ∈ Γ1 ,

(22)

where c > 0 is independent of x0 , τ and s (c depends only on Ω, n, β and aL∞ (B+ ) ). Remark 5.3. Note that in particular, for large s the solution v˜ to (21) satisfies ∂ v˜ (23)  −v˜ −β on Γ1 . ∂ν For this reason we call v˜ a local subsolution. We will take s0 larger if necessary so that (23) holds for s
s0 . For the construction of the subsolution consider the function   1 u(y) , ˜ = ρ α sin(γ θ ) + bθ 2 , α = 1+β

(24)

where γ , b > 0 are constants to be chosen later and ρ, θ, ω are toroidal coordinates:   1 n−2 + ρ cos θ ω + ρ sin θ en . (ρ, θ, ω) ∈ (0, ∞) × (0, 2π) × S → y˜ = 2 Here w = (w1 , . . . , wn−1 , 0) ∈ S n−2 which is the unit sphere of n − 2 dimensions, and en = (0, . . . , 0, 1). Abusing notation and using (19) we write   u(x) ˜ = u ϕτ (x) ˜ .

(25)

+ ). For σ > 0 define Lemma 5.4. Let a ∈ L∞ (B  + : dist(x, ˜ ∂ Γ1 ) < σ . Vσ = x˜ ∈ B There exist 12 < γ < α (recall that 12 < α < 1) and b > 0, σ > 0, τ0 > 0 small so that for all 0 < τ < τ0 the function u(x) ˜ defined by (24) and (25) satisfies ˜  0 in Vσ . −
x˜ u + a(x)u

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Moreover −C dist(x, ˜ Γ2 )−β/(1+β) 

∂u 1 (x) ˜  − dist(x, ˜ Γ2 )−β/(1+β) ∂νx˜ C

∀x˜ ∈ Γ1 ,

(26)

where C > 0. The constants γ , b, σ , τ0 , C depend only on Ω, n, β and aL∞ (B+ ) . Proof. A calculation shows that the Laplacian of u with respect to y˜ is given by     1 ρ α−2 + ρ cos(θ ) (α 2 − γ 2 ) sin(γ θ ) + bθ 2 + 2b
y˜ u = 1/2 + ρ cos(θ ) 2       + ρ α cos(θ ) bθ 2 + sin(γ θ ) − sin(θ ) 2bθ + γ cos(γ θ )   = ρ α−2 (α 2 − γ 2 ) sin(γ θ ) + bθ 2 + 2b + O(ρ) ,

(27)

where O(ρ) stands for a function bounded by a constant times ρ. We fix γ such that 12 < γ < α. Observe that    ∂u 1 ∂u  = = ρ α−1 γ cos(γ π) + 2bπ .  ∂νy˜ ρ ∂θ θ=π We choose now b > 0 small enough so that γ cos(γ π) + 2bπ < 0. This ensures the validity of (26). A computation shows that
x˜ u = Aij ∂y˜i y˜j u + Bi ∂y˜i u,

(28)

where we have adopted the convention of summation over repeated indices. The functions Aij and Bi are given by Aij =

∂ y˜i ∂ y˜j , ∂ x˜k ∂ x˜k

Bi =

 ∂ 2 y˜i k

∂ x˜k2

.

Since y˜ = τ1 ϕ(τ x˜ + x0 ), ϕ(x0 ) = 0 and ∇ϕ(x0 ) = I we have 1 1 ϕ(τ x˜ + x0 ) = x˜ + τ D 2 ϕ(0)x˜ 2 + O(τ 2 ). τ 2 Consequently, Aij = δij + O(τ ),

Bi = O(τ ).

Thus from (28)
x˜ u =
y˜ u + O(τ )Dy2˜ u + O(τ )Dy˜ u. On the other hand observe that Dy2˜ u = O(ρ α−2 ),

Dy˜ u = O(ρ α−1 ).

Hence (27) implies

 
x˜ u − a(x)u ˜
ρ α−2 2b + O(ρ) + O(τ ) + O(τρ) − O(ρ 2 ) ,

and thus, since b > 0 we can choose τ , σ small enough so that
x˜ u − a(x)u ˜
0

in Vσ .

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313

Proof of Lemma 5.2. Let us write v˜ = v˜1 + v˜2 where v˜1 is the solution of the following problem  + , ˜ v˜1 = 0 in B   −
v˜1 + a(x) ∂ v˜1 (x) ˜ = − dist(x, ˜ Γ2 )−β/(1+β) x˜ ∈ Γ1 ,   ∂ν v˜1 (x) ˜ =0 x˜ ∈ Γ2 ∪ Γi , and v˜2 satisfies  −
v˜2 + a(x) ˜ v˜2 = 0     ∂ v˜2 (x) ˜ =0 ∂ν   ˜ =0   v˜2 (x)  ˜ = s dist(x, ˜ ∂ Ω) v˜2 (x)

+ , in B x˜ ∈ Γ1 , x˜ ∈ Γ2 , x˜ ∈ Γi .

+ . Observe that v˜1  0 while v˜2
0 in B Let σ > 0 be small as in Lemma 5.4 and recall:  + : dist(x, Vσ = x˜ ∈ B ˜ ∂ Γ1 ) < σ . + let P (x)  denote the closest point in ∂ Ω  closest to x. ˜ ∈ ∂Ω ˜ If τ is small enough this projection is well For x˜ ∈ B +  defined and smooth on B . Hence we fix τ0 even smaller than that one in (16), so that this property holds for 0 < τ < τ0 . + define For x˜ ∈ B   if P (x) ˜ ∂ Ω) ˜ ∈ Γ2 , g(x) ˜ = dist(x, (29) 1 if P (x) ˜ ∈ Γ1 . By the strong maximum principle and Hopf’s lemma applied to v˜2 /s we have v˜2 (x) ˜
csg(x), ˜

+ \ Vσ , ∀x˜ ∈ B

where c > 0 depends only on Ω and σ . It follows that for s large v( ˜ x) ˜
csg(x), ˜

+ \ Vσ , ∀x˜ ∈ B

(30)

+ \ Vσ ). for a new constant c > 0 (here we use only the fact that v˜1 is bounded from above in B ˜ be the function of Lemma 5.4. We are going to show that for large s there holds Let u(x) su  C v˜

in Vσ ,

 for some where C is a constant. Indeed, first recall that
u − a(x)u ˜
0 in Vσ . Also one has u  Cg on ∂Vσ ∩ Ω constant C. By (30) with s sufficiently large one obtains su  C v˜

 on ∂Vσ ∩ Ω.

(31)

Estimate (26) implies ∂u s (x) ˜  − dist(x, ˜ Γ2 )−β/(1+β) ∀x˜ ∈ ∂Vσ ∩ Γ1 . ∂ν C Hence choosing s > 0 large enough we have s

∂u ∂ v˜ C ∂ν ∂ν On the other hand s

on ∂Vσ ∩ Γ1 .

u = v˜ = 0 on ∂Vσ ∩ Γ2 .

(32)

(33)

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Hence, by the maximum principle for s large su  C v˜

in Vσ .

(34)

+ and We fix s0 sufficiently large such that (30)–(34) are valid for s
s0 . This shows that v˜ is positive in B satisfies (22). 2 6. Existence, regularity and convergence of u¯ ε In this section we prove Theorems 1.1 and 1.3. First we remark that there exists a maximal solution u¯ ε to (5) because 0 is a subsolution and a large constant is a supersolution by (10). It is convenient to introduce some notation. Let M > 0 be such that   M
sup f x, u¯ ε (x) , x∈∂Ω

for all 0 < ε < 1. We need a Harnack inequality which, for completeness, we prove in the Appendix.  u
0 satisfies Lemma 6.1. Suppose that u ∈ H 1 (B3 ∩ Ω),

 −
u + a(x)u ˜ = 0 in B3 ∩ Ω, ∂u N on Γe , ∂ν where N is a constant. Then there is a constant ck > 0 such that   + and ∀x˜1 ∈ B1/2 ∩ B + . ˜ Γe ) ck u(x˜1 ) − N , ∀x˜ ∈ B u(x) ˜
ck dist(x, The constant ck can be chosen independent of x0 ∈ ∂Ω and of 0 < τ < τ0 (τ0 > 0 was introduced in Section 4).  > 0, Let τ0 and s0 be the constants in the statement of Lemma 5.2 and Remark 5.3. Let us fix a large constant C independent of 0 < ε < 1, such that 1  s0 < ck2 C, 2 ε 1+β 1+β , u¯ L∞ (Ω) < τ0 C β

u¯ ε L∞ (Ω) <

ck

(35) (36)

1+β C

. 2M Next we fix C0 large enough such that  1+β C0
6.  C

(37)

(38)

For the sake of notation, from this point on we write u = u¯ ε . Given a point x0 ∈ ∂Ω and 0 < τ < τ0 we define u( ˜ x) ˜ = τ −1/(1+β) u(τ x˜ + x0 ), which satisfies

−
u˜ + τ 2 u˜ = 0 ∂ u˜ = gτε (x, ˜ u) ˜ ∂ ν˜

 in Ω,  on ∂ Ω,

= x˜ ∈ Ω

1 (Ω − x0 ), τ

(39)

(40)

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315

 and gτε is given by where ν˜ is the exterior unit normal vector to ∂ Ω ˜ u) ˜ = τ β/(1+β) g ε (τ x˜ + x0 , τ 1/(1+β) u), ˜ gτε (x, and g ε (x, u) = −

u + f (x, u). (u + ε)1+β

Lemma 6.2. Let x1 ∈ Ω and assume that u(x1 )
C0 δ(x1 )1/(1+β) . Fix

 τ=

u(x1 )  C

(41)

1+β ,

(42)

and let x0 ∈ ∂Ω be such that dist(x1 , ∂Ω) = |x0 − x1 |.

(43)

Then τ < τ0 and u˜ (defined in (39)) verifies  ˜ ∂ Ω), u( ˜ x) ˜
s0 dist(x,

∀x˜ ∈ Γi .

Proof. By (36) we have τ < τ0 . Let x˜1 denote the point τ1 (x1 − x0 ) which satisfies 1 6 by (38), (41)–(43). Observe that by the choice of τ we have |x˜1 | 

(44)

 u( ˜ x˜1 ) = C.

(45)

Using Harnack’s Lemma 6.1 and (45) we obtain   ∂ u˜ + .   , ∀x˜ ∈ B ˜ ∂ Ω) ck C − sup u( ˜ x) ˜
ck dist(x, Γe ∂ν

(46)

From the boundary condition in (40) and the definition of M sup Γe

∂ u˜  τ β/(1+β) M. ∂ν

Notice that from (37) it follows that u(x1 )β 

1+β ck C 2M

and therefore τ β/(1+β) M = M



u(x1 )  C

β

1   ck C. 2

Inserting this in (46) and recalling (35) we find 1   
s0 dist(x, ˜ ∂ Ω), u( ˜ x) ˜
ck2 C dist(x, ˜ ∂ Ω) 2

∀x˜ ∈ Γi .

2

Proof of Theorem 1.3. Let x1 be a point in Ω. We distinguish two cases.

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Case 1. Assume u(x1 )
C0 δ(x1 )1/(1+β) .

(47)

Let τ be given by (42) and x0 ∈ ∂Ω be such that dist(x1 , ∂Ω) = |x0 − x1 |. Let v˜ be the solution of problem (21) ˜ = τ 2 . We know by Lemma 5.2 that v˜ satisfies with s = s0 and a(x) v( ˜ x) ˜
cs0 dist(x, ˜ Γ2 )1/(1+β) ∀x˜ ∈ Γ1 . (48) By Lemma 6.2 u˜
v˜

on Γi .

(49) + . B

We claim that u˜
v˜ on To see this, define   u(x) ˜  (x)  for x˜ ∈ Ω \ B1 (0), U ˜ = + .  ∩ B1 (0) = B max u(x), ˜ v(x) ˜ for x˜ ∈ Ω Then the function

   1 (x − x0 ) U (x) = τ 1/(1+β) U τ

is a subsolution of Eq. (5) (cf. (23)) and since u is the maximal solution we have U  u in Ω and as consequence + . v˜  u˜ in B This fact in combination with (48) yields the estimate u( ˜ x) ˜
cs0 dist(x, ˜ Γ2 )1/(1+β) , ∀x˜ ∈ Γ1 .

(50)

From the boundary condition in (40) we deduce that    ∂ u˜     C dist(x, ˜ Γ2 )−β/(1+β) + τ β/(1+β) M on Γ1 ,  ∂ ν˜ 

(51)

and therefore, on a smaller set we obtain an estimate    ∂ u˜     C on B1/3 ∩ ∂ Ω,   ∂ ν˜ 

(52)

with a constant C independent of ε. We will deduce from this an estimate of the form   ∇ u( ˜ x˜1 )  C,

(53)

with C independent of ε. From the definition of u˜ it will follow that   ∇u(x1 )  Cu(x1 )−β . Let us prove (53). For this purpose choose p > n and take n < r <     ∂ u˜    u ˜ W 1,r (B1/4 ∩Ω) +  u ˜ 1  C   L (B1/3 ∩Ω) , ∂ ν˜ Lp (B1/3 ∩∂ Ω) 

np n−1 .

By Lemma 9.3

and by the embedding W 1,r ⊂ C µ we have for some 0 < µ < 1     ∂ u˜    + u ˜ L1 (B1/3 ∩Ω) u ˜ C µ (B1/4 ∩Ω)  C   . ∂ ν˜  p  L (B1/3 ∩∂ Ω)

By the assumption (2) and the lower bound (50) we see that the right-hand side of the boundary condition in (40) satisfies     ε   ∂ u˜   g (x,   ˜ C µ (B ∩∂ Ω) + u ˜ L1 (B1/3 ∩Ω)  .  C  τ ˜ u) 1/4 ∂ ν˜  p  L (B1/3 ∩∂ Ω)

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317

Using Schauder estimates (see e.g. [9]) we deduce     ∂ u˜    u ˜ C 1,µ (B1/5 ∩Ω) + u ˜ L1 (B1/3 ∩Ω)  C   . ∂ ν˜  p  L (B1/3 ∩∂ Ω)

Recalling that |x˜1 |  by (44) we obtain     ∂ u˜     ∇ u( ˜ x˜1 )  C  +  u ˜  . L1 (B1/3 ∩Ω)  ∂ ν˜  p  L (B1/3 ∩∂ Ω) 1 6

By (52) we can assert that    ∂ u˜    C  ∂ ν˜  p  L (B1/3 ∩∂ Ω)

with C independent of ε. It suffices then to find an estimate for u ˜ L1 (B1/3 ∩Ω)  . Using (51) we see that    ∂ u˜     C on B5/12 ∩ ∂ Ω   ∂ ν˜  and therefore, using Lemma 9.5 we find     u˜  C u( ˜ x) ˜ + 1 , ∀x˜ ∈ B1/2 ∩ Ω.  B1/3 ∩Ω

Putting x˜ = x˜1 in the latter estimate and recalling (44) and (45) we obtain the desired conclusion. Case 2. Assume u(x1 )  C0 δ(x1 )1/(1+β) . Define u( ˜ x) ˜ = τ −1/(1+β) u(τ x˜ + x1 ), where τ = 12 δ(x1 ). Then −
u˜ + τ 2 u˜ = 0 in B1 (0), u˜
0 in B1 (0) and u(0) ˜  21/(1+β) C0 . Since u˜
0, by elliptic estimates we have |∇ u(0)| ˜  C, where C depends only on n, β, C0 , τ0 . This implies |∇u(x1 )|  Cτ −β/(1+β)  Cu(x1 )−β . 2 Proof of Theorem 1.1. Note that u¯ ε decreases as ε decreases to 0 and by the uniform estimate of Theorem 1.3

such that u¯ ε → u in C µ (Ω)

for 0 < µ < 1 . Let ϕ ∈ C 1 (Ω ∪ (∂Ω ∩ {u > 0})). It is there exists u ∈ C 1/(1+β) (Ω) 0 1+β easy to pass to the limit in     u¯ ε ε − ε ∇ u¯ ε · ∇ϕ + u¯ ε ϕ = + f (x, u ¯ ) ϕ (54) (u¯ + ε)1+β Ω

∂Ω

 for all terms, except possibly for Ω ∇ u¯ ε · ∇ϕ. But u¯ ε is bounded in H 1 (Ω) since, by taking ϕ = u¯ ε in (54) we get   ε 2 ε 2 |∇ u¯ | + (u¯ )  f (x, u¯ ε )u¯ ε  C. Ω

∂Ω

εk 1 Thus for a sequence εk → 0, we conclude that  weakly in H (Ω) to a function which necessarily  u¯ ε converges coincides with u. Then by weak convergence Ω ∇ u¯ ·∇ϕ → Ω ∇u·∇ϕ. This shows that u is a solution of (1). 2

7. Regularity for minimizers of φ Let us recall the functional corresponding to problem (1):    1  (u+ )1−β 2 2 φ(u) = |∇u| + u + − F (x, u+ ). 2 1−β Ω

∂Ω

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Proof of Theorem 1.5. Suppose that u ∈ H 1 (Ω) is a minimizer of φ. First observe that u
0. Indeed u+ is also a minimizer and φ(u+ )  φ(u). But if u+ ≡ u, then φ(u+ ) < φ(u), which is an absurd. Next remark that u satisfies −
u + u = 0 in Ω. This follows by observing that for any ϕ ∈ C0∞ (Ω) the function s → φ(u + sϕ) is differentiable and attains its minimum value at s = 0. Thus u is smooth in Ω and the objective now is to show that   ∇u(x1 )  Cu(x1 )−β , ∀x1 ∈ Ω, (55) for some constant C. The argument follows the same scheme as in the proof of Theorem 1.3: given x1 ∈ Ω we distinguish two cases: Case 1. u(x1 )
C0 δ(x1 )1/(1+β) , and Case 2. u(x1 )  C0 δ(x1 )1/(1+β) . In Case 2, the argument is exactly the same as in the proof of Theorem 1.3. Suppose that Case 1 occurs. Let τ be given by (42) and x0 ∈ ∂Ω be such that dist(x1 , ∂Ω) = |x0 − x1 |. Let v˜ be the solution of problem (21) with s = s0 . By Lemma 6.2 we deduce (49), that is u˜
v˜

on Γi .

We claim that u˜
v˜

+ . in B

(56)

˜ τ1 (x − x0 )) and define To prove this let v(x) = τ 1/(1+β) v(  u(x)  for x ∈ Ω \ Bτ (x0 ), U (x) = max u(x), v(x) for x ∈ Bτ (x0 ) ∩ Ω. Then U satisfies

−
U + U  0 in Ω, ∂U −β  −U + f (x, U ) on ∂Ω ∩ {U > 0}. ∂ν We shall prove that if τ is small enough then φ(U ) < φ(u) unless u ≡ U . First note that   2  1  ∇(U − u) + (U − u)2 + ∇U · ∇(U − u) + U (U − u) φ(U ) − φ(u) = − 2 Ω Ω  1 (U 1−β − u1−β ) − F (x, U ) + F (x, u). + 1−β

(57)

(58)

∂Ω

Next we multiply (57) by U − u
0 and integrate by parts to obtain      ∂U (U − u)  −U −β + f (x, U ) (U − u). ∇U · ∇(U − u) + U (U − u)  ∂ν Ω

∂Ω

∂Ω∩{U >0}

Combining (58) and (59) φ(U ) − φ(u)  −

1 2

 Ω

  ∇(U − u)2 + 

− ∂Ω∩{U >0}



∂Ω∩{U >0}

U 1−β u1−β − − U −β (U − u) 1−β 1−β

F (x, U ) − F (x, u) − f (x, U )(U − u).

(59)

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319

We claim that

and

U 1−β u1−β − − U −β (U − u)  CU −1−β (U − u)2 1−β 1−β

(60)

  F (x, U ) − F (x, u) − f (x, U )(U − u)  CU −1−β (U − u)2 .

(61)

To verify (60) we consider first the case U  2u. By the mean value theorem there is a ξ
0 such that u  ξ  U and u1−β 1 β U 1−β − − U −β (U − u) = βξ −1−β (U − u)2  u−1−β (U − u)2  CU −1−β (U − u)2 . 1−β 1−β 2 2 In the second case, U
2u, we have U  2(U − u) and therefore u1−β U 1−β 1 1 4 − − U −β (U − u)  U 1−β = U −1−β U 2  U −1−β (U − u)2 . 1−β 1−β 1−β 1−β 1−β To prove (61) observe that 1 F (x, U ) − F (x, u) − f (x, U )(U − u) = fu (x, ξ )(U − u)2 , 2 for some u  ξ  U and (61) follows because fu (x, ξ ) is bounded on ∂Ω × [0, max∂Ω U ]. Hence   2 1  ∇(U − u) + C φ(U ) − φ(u)  − U −1−β (U − u)2 . 2 Ω

(62)

∂Ω∩{U >0}

(x)  ≡ u˜ in Ω + we can rewrite (62) as \B Define U ˜ = τ −1/(1+β) U (τ x˜ + x0 ). Using that U     2 1    − u) −1−β (U ∇(U − u) ˜  +C ˜ 2 . φ(U ) − φ(u)  τ n−2β/(1+β) − U 2 + B

>0} Γe ∩{U

Using the explicit lower bound (22) we obtain     2 1   C n−2β/(1+β) 2 −1  2   ∇(U − u) ˜ + 1+β dist(x, − , φ(U ) − φ(u)  τ ˜ Γ ) (U − u) ˜ 2 s0 + B

Γ1

(x) where we have also used the fact that if v( ˜ x) ˜ = 0 then U ˜ − u( ˜ x) ˜ = 0, which allows us to restrict the integral 1 to Γ . By Hardy’s inequality (cf. (8))    2 1 CCh n−2β/(1+β) ∇(U  − u) − + 1+β ˜  . φ(U ) − φ(u)  τ 2 s 0

+ B

(1+β)

) < 0. Thus, we see that φ(U ) < φ(u) We can choose s0 larger if necessary in order to make (−1/2 + CCh /s0  ≡ u, unless U ˜ which implies our claim (56). The rest of the argument continues in exactly the same manner as in the proof of Theorem 1.3. 2 Proof of Theorem 1.2. Since f is sublinear and 0 < β < 1 the functional φ attains its minimum in H 1 (Ω). Let u be a minimizer of φ. We have shown at the beginning of the proof of Theorem 1.5 that u is smooth in Ω and solves −
u + u = 0

in Ω.

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and hence ∂Ω ∩ {u > 0} is an open subset of ∂Ω. To verify By Theorem 1.5 we know also that u ∈ C 1/(1+β) (Ω) the boundary condition in (1) observe that    d φ(u + sϕ)|s=0 = 0, ∀ϕ ∈ C01 Ω ∪ ∂Ω ∩ {u > 0} ds which is equivalent to (3). 2

8. Proof of Theorem 1.8 and Proposition 1.10 Proof of Theorem 1.8. By Theorem 1.1 for every λ > 0 there is a solution of (1). The solution u¯ ελ of (21) is unique, since f is sublinear. Furthermore u¯ λ = limε→0 u¯ ελ is the maximal solution of (1). Indeed, if v is a solution of (1), then it is a subsolution to (21), so v  u¯ ελ . As ε → 0 one obtains v  u¯ λ . This solution u¯ λ is also nondecreasing with respect to λ because the same is true for (21), that is, λ1  λ2 implies that u¯ ελ1  u¯ ελ2 , since f (x, u) is increasing in u. For λ > 0 small enough there is no nontrivial solution, see the proof of Proposition 2.2 in Section 9. For λ > 0 large enough we will see that there exists a positive solution. To prove this we will follow the method

such that the surface of sub-super solutions. By a subsolution of (9) we mean a function u ∈ H 1 (Ω) ∩ C(Ω) measure of {x ∈ ∂Ω: u(x) = 0} is zero and that verifies

in Ω, −
u + u  0 ∂u (63) −β  −u + λf (x, u) on ∂Ω, ∂ν

the surface measure of {x ∈ ∂Ω: u(x) = 0} is zero satisfying the A supersolution is a function u¯ ∈ H 1 (Ω) ∩ C(Ω) above (63) with the inequality signs reversed. Our aim is to find a subsolution u
c > 0 for some constant c and a ¯ implying the existence of a solution u such that u  u  u, ¯ see [1]. supersolution u¯ such that u  u, Construction of the subsolution for sufficiently large λ: Let Y be the solution of

−
Y + Y = 0 in Ω, ∂Y =1 on ∂Ω. ∂ν Let ξ0 be as in the assumption (10), that is, such that f (x, ξ0 ) ≡ 0 and solve

−
v + v = 0 in Ω, ∂v = f (x, ξ0 ) on ∂Ω. ∂ν

Let us fix ε > 0 such that By the maximum principle and Hopf’s lemma v is positive in Ω. b := inf(v − εY ) > 0.

Ω

Define u = k(v − εY ), where we choose k large enough in such a way that kb
ξ0

and b−β  εk 1+β .

Subsequently we choose λ
k. This results in kf (x, ξ0 )  λf (x, kb) and k −β (v − εY )−β  kε, implying the normal derivative inequality in (63).

(64)

(65)

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321

Construction of the ordered supersolution: Let Y be the solution to (64) and define u¯ = AY , where A > 0 is a large constant such that A
λf (x, A) for every x ∈ ∂Ω. This makes u¯ a supersolution of (9). We may take A even ¯ 2 larger in order to have u  u. We now proceed with the proof of the remaining items of Theorem 1.8. Proof of Theorem 1.8(a) and (b). Define 

. λ∗ = inf λ > 0: u¯ λ > 0 in Ω Observe that 0 < λ∗ < ∞. Let 0 < λ < λ∗ and suppose that u¯ λ > 0 a.e. on ∂Ω. Fix λ such that λ < λ < λ∗ . Let ζ be the solution of

−
ζ + ζ = 0 in Ω, ∂ζ (66) = f (x, u¯ λ ) on ∂Ω. ∂ν Clearly ζ > 0 in Ω since f (x, u¯ λ )
0 and f (x, u¯ λ ) ≡ 0 on ∂Ω. Let 0 < ε  λ − λ . Then the function w =

and satisfies u¯ λ + εζ is positive in Ω ∂w −β = −u¯ λ + λ f (x, u¯ λ ) + εf (x, u¯ λ )  −w −β + λf (x, w). ∂ν Hence w is a subsolution of (9) corresponding to λ, thus u¯ λ
w in Ω. Since λ < λ∗ this contradicts the definition of λ∗ . 2 Proof of Theorem 1.8(c). We have   −β u¯ λ  λ f (x, u¯ λ ). ∂Ω



−β

∂Ω

u¯ λ  C as λ decreases to λ∗ . In fact, integrating Eq. (9) in Ω we find

∂Ω −β

−β

But u¯ λ decreases to u¯ λ∗ and hence u¯ λ increases to u¯ λ∗ as λ decreases to λ∗ , and by monotone convergence we  −β deduce that ∂Ω u¯ λ∗  C. In particular u¯ λ∗ > 0 a.e. on ∂Ω. 2 Proof of Theorem 1.8(d). Fix λ > λ∗ . From now on we drop the dependence on λ and write u = u¯ λ , we also write uε = u¯ ελ . Our aim is to prove that Λ(u) > 0, see the definition in (11). First we prove that for λ > λ∗ the maximal solution u is weakly stable [2,4] in the sense that Λ(u)
0.

(67)

In fact, we know that the maximal solution uε of (5) is weakly stable, that is     βuε − ε 2

+ λfu (x, uε ) ϕ  |∇ϕ|2 + ϕ 2 , ∀ϕ ∈ C 1 (Ω). (uε + ε)2+β ∂Ω

(68)

Ω

Since uε
c for some c > 0 independent of ε one can let ε → 0 in (68) and obtain (67). Let us show that Λ(u) > 0. We introduce a new parameter θ  0 and consider the family of problems  −
u + u = 0 in Ω,   u
0 in Ω, ∂u   = −u−β + λf (x, u) + θ on ∂Ω ∩ {u > 0}. ∂ν The main observations to conclude are: there exists θ0 < 0 such that for θ > θ0 (69) has a positive maximal solution u¯ θ , and

(69)

(70)

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for θ > θ0 we have Λ(u¯ θ )
0 and Λ(u¯ θ ) is strictly increasing with θ .

(71)

Indeed, assuming (70) and (71) we have a maximal solution u¯ θ to (69) for some θ < 0. But then 0  Λ(u¯ θ ) < Λ(u¯ 0 ) and finally note that u¯ 0 is just the maximal solution of (9).

2

Proof of (70). Fix λ such that λ∗ < λ < λ and let 0 < ε  λ − λ . Let ζ be the solution of (66) and Y be the solution of (64). Take θ0 < 0 with |θ0 | small enough so that 1 |θ0 |Y  εζ. 2 For θ > θ0 set w = u¯ λ + εζ + θ Y, and observe that w > u¯ λ . We compute ∂w −β = −u¯ λ + λ f (x, u¯ λ ) + εf (x, u¯ λ ) + θ  −w −β + λf (x, w) + θ. ∂ν Thus w is a positive subsolution, and by the method of sub and supersolutions we can find a maximal solution u¯ θ . 2 Proof of (71). Let θ0 < θ1 < θ2 and let u¯ θ1 , u¯ θ2 denote the maximal solution of (69) with parameters θ1 and θ2 . Note that u¯ θ1 < u¯ θ2 .

(72)

Let ψ1 and ψ2 denote the first eigenfunctions, i.e.,

−
ψi + ψi = 0 in Ω, ∂ψi −β−1 = βu ψi + λfu (x, u)ψi + Λ(u¯ θi )ψi on ∂Ω, ∂ν i = 1, 2, normalized so that ψi L2 (∂Ω) = 1. Then     β(u¯ θ1 )−β−1 + λfu (x, u¯ θ1 ) ψ12 Λ(u¯ θ1 ) = |∇ψ1 |2 + ψ12 − Ω

∂Ω







|∇ψ2 |2 + ψ22 − Ω

  β(u¯ θ1 )−β−1 + λfu (x, u¯ θ1 ) ψ22

∂Ω





|∇ψ2 |

2

<

(73)

+ ψ22

−

Ω

  β(u¯ θ2 )−β−1 + λfu (x, u¯ θ2 ) ψ22

∂Ω

= Λ(u¯ θ2 ), where the last inequality is strict because ψ2 > 0 and (72).

2

Proof of Proposition 1.10. We shall prove that there exists a constant c > 0 independent of λ such that u¯ λ
c

on ∂Ω, ∀λ > λ∗ .

The conclusion follows by letting λ  λ∗ .

(74)

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323

and integrate in Ω: Multiply the equation by ϕ ∈ C 1 (Ω)    ∂ u¯ λ ϕ + ∇ u¯ λ · ∇ϕ + u¯ λ ϕ = 0. − ∂ν Ω

Ω

(75)

Ω

−γ = u¯ λ

for λ > λ∗ . Take ϕ where γ > 1 will be specified later. Notice that ϕ is well defined because u¯ λ > 0 in Ω −γ −1 We have ∇ϕ = −γ u¯ λ ∇u and using (75)     −β−γ −γ −γ −1 1−γ 2 u¯ λ − λ f (x, u¯ λ )u¯ λ − γ u¯ λ |∇ u¯ λ | + u¯ λ = 0. ∂Ω

Ω

∂Ω

Ω

Using the hypothesis f (x, u)
0 we find    4γ (1−γ )/2 2 −β−γ 1−γ |∇ u ¯ |  u ¯ + u¯ λ . λ λ (1 − γ )2 Ω

(76)

Ω

∂Ω

(1−γ )/2

We use the stability condition (67) with ϕ = u¯ λ and the fact that fu is bounded to obtain     −β−γ (1−γ )/2 2 1−γ 1−γ β u¯ λ  |∇ u¯ λ | + u¯ λ + C u¯ λ . Ω

∂Ω

Ω

(77)

∂Ω

Let ε > 0 be a small constant to be fixed later. Then from (76) and (77) we obtain         4γ (1 − ε) 4γ ε −β−γ 1−γ 1−γ (1−γ )/2 2 −β−γ 1−γ β u¯ λ + − u¯ λ − C u¯ λ |∇ u¯ λ |  u¯ λ + u¯ λ (1 − γ )2 (1 − γ )2 Ω

∂Ω

Ω

∂Ω

∂Ω

which is equivalent to      4γβ(1 − ε) 4γ ε −β−γ (1−γ )/2 2 1−γ 1−γ − 1 u ¯ + |∇ u ¯ |  C u ¯ + C u¯ λ . λ λ λ (1 − γ )2 (1 − γ )2 Ω

∂Ω

Ω

Ω

(78)

∂Ω

We need the following versions of Sobolev’s inequality and trace inequality. Lemma 8.1. For any µ > 0 and d1 > 0, d2 > 0 there exists C = C(Ω, µ, d1 , d2 ) such that 2/d1   

ϕ 2  µ |∇ϕ|2 + C |ϕ|d1 , ∀ϕ ∈ C 1 (Ω), and ∂Ω

Ω



 ϕ µ

Ω

∂Ω

 |∇ϕ| + C

2

(79)

|ϕ|

d2

2

Ω

2/d2 ,

∀ϕ ∈ C 1 (Ω).

(80)

Ω (1−γ )/2

and d1 > 0, d2 > 0 to be fixed Proof of Proposition 1.10 continued. We use (79) and (80) with ϕ = u¯ λ shortly, and combine with (78)      4γβ(1 − ε) 2γ ε −β−γ 1−γ 1−γ − 1 u ¯ + u ¯ + u ¯ λ λ λ (1 − γ )2 µ(1 − γ )2 Ω ∂Ω ∂Ω      2γ εC 1−γ 1−γ d1 (1−γ )/2 d2 (1−γ )/2 .  C u¯ λ + C u¯ λ + u¯ λ + u¯ λ µ(1 − γ )2 Ω

∂Ω

∂Ω

Ω

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We fix µ small enough such that 2γ ε/(µ(1 − γ )2 ) = C and deduce that     4γβ(1 − ε) −β−γ d1 (1−γ )/2 d (1−γ )/2 −1 u¯ λ  C u¯ λ + C u¯ λ2 . (1 − γ )2 ∂Ω

Ω

∂Ω

Observe that by taking ϕ ≡ 1 in (11) we deduce  −β−1 u¯ λ C ∂Ω

with a constant C independent of λ, for λ in a bounded interval, say, λ∗ < λ  λ0 . For this reason we take d1 > 0 so that d1 1−γ 2 = −β − 1 and conclude    4γβ(1 − ε) −β−γ d (1−γ )/2 − 1 u ¯  C + C u¯ λ2 , (81) λ (1 − γ )2 Ω

∂Ω

λ∗

where C independent of λ for < λ  λ0 .  d (1−γ )/2 To proceed further we have to bound the integral Ω u¯ λ2 by a constant independent of λ, λ∗ < λ  λ0 . By Lemma 9.1,  u¯ λ (x)
c dist(x, ∂Ω) u¯ λ , ∂Ω

where c depends only on Ω. Since u¯ λ
u¯ λ∗ > 0 a.e. we see that there is a constant C independent of λ > λ∗ such that   d (1−γ )/2 u¯ λ2  C dist(x, ∂Ω)d2 (1−γ )/2 dx < ∞, (82) Ω

Ω

if we fix d2 > 0 small so that d2 1−γ 2 > −1. Finally, from (81) and (82) we deduce   4γβ(1 − ε) −β−γ −1 u¯ λ  C. 2 (1 − γ ) ∂Ω

2 2 The latter estimate is useful if 4γβ(1 − ε)/(1 − γ ) − 1 > 0 for some ε > 0, that is, if 4γβ/(1 − γ ) > 1. This is 2 β . the case for 1 < γ < 1 + 2β + 2 β + In summary, if 0 < p < 1 + 3β + 2 β + β 2 then there is a constant C independent of λ∗ < λ  λ0 such that  −p u¯ λ  C. ∂Ω

Let vλ = 1/u¯ λ . Then vλ satisfies

−
vλ + vλ  0 in Ω, ∂vλ 2+β  vλ on ∂Ω. ∂ν The proof of (74) will be completed with the aid of the following lemma.

satisfies Lemma 8.2. Suppose v ∈ C 2 (Ω)

−
v + v  0 in Ω, ∂v  vq on ∂Ω, ∂ν

J. Dávila, M. Montenegro / Ann. I. H. Poincaré – AN 22 (2005) 303–330

325

where q > 1. If  vp < K ∂Ω

for some p > (n − 1)(q − 1) then vL∞ (∂Ω)  C, where C depends only on Ω, p, q and K. Proof of Lemma 8.1. We only prove (79) the other inequality being analogous. By Hölder’s inequality we can assume that d < 2(n−1) n−2 , the Sobolev exponent for the trace inequality. For the sake of contradiction suppose that

such that there exists a sequence ϕn ∈ C 1 (Ω) 2/d    |ϕn |d . 1 = ϕn2 > µ |∇ϕn |2 + n ∂Ω

Ω

∂Ω

H 1 (Ω)

Then ϕn is bounded in up to a subsequence ϕn  ϕ weakly in H 1 (Ω). By the compact embedding  and 1 2 2 H (Ω) ⊂ L (∂Ω) we have ∂Ω ϕ = 1. But the embedding H 1 (Ω) ⊂ Ld (∂Ω) is also compact and therefore we  also have ∂Ω |ϕ|d = 0, a contradiction. 2 Proof of Lemma 8.2. We have v q ∈ Lp/q (∂Ω). By Lp theory v ∈ W 1,r (Ω) for all 1  r < pn/(q(n − 1)). By the n 1 trace inequality v ∈ W 1−1/r,r (∂Ω) and by Sobolev’s embedding v ∈ Lt (∂Ω) for 1t = 1r n−1 − n−1 . It follows that q 1 1 t ∗ ∗ v ∈ L (∂Ω) for t < t with t ∗ = p − n−1 . But t > p. Repeating this process a finite number of times (bootstrap) we obtain the conclusion. 2

9. Proofs for Section 2 Proof of Proposition 2.1. If the domain is a ball and f = f (u), the maximal solutions u¯ ε to (5) are radial and hence u¯ λ is radial. This means that u¯ λ is a constant on ∂Ω and this constant is either positive or zero. 2 Proof of Proposition 2.2. Assume u¯ λ ≡ 0 for a sufficiently small λ > 0. Since
u¯ λ = u¯ λ
0, then u¯ λ attains u¯ λ (y)
0, implies λ1  u¯ λ (y)β f (y, u¯ λ (y)). But u¯ λ (y)β f (y, u¯ λ (y))  K it maximum at a point y ∈ ∂Ω. Thus ∂∂ν where K > 0 is a constant independent of λ for λ small, because the map λ → u¯ λ is nondecreasing. This is an absurd. 2 Construction of Example 2.3. Let Ω = BR (0) and pick a point x0 ∈ ∂BR . We take f smooth such that 0  f  1, f ≡ 1 in Br0 (x0 ) and f ≡ 0 in Rn \ B2r0 (x0 ) for a fixed r0 > 0 and consider the problem

−
u + u = 0 in BR , ∂u (83) = −u−β + λf (x) on ∂BR ∩ {u > 0}. ∂ν We will proceed in two steps. First we will prove that for large R and λ the maximal solution of (83) is nontrivial. Then we fix such a large λ = λ¯ and take R even larger in order to prove that the maximal solution vanishes on a subset of ∂BR (0) with positive surface measure. Claim 1. If λ is large enough then for any R large the maximal solution of (83) is nontrivial.

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In fact, we will construct a nontrivial subsolution of (83) for λ large. Let D = B3r0 (x0 ) ∩ BR (0) and write its boundary as ∂D = Γ ∪ Υ where Γ = Br0 (x0 ) ∩ ∂BR (0) and Υ = ∂D \ Γ . Let w = w1 + w2 where w1 solves  in D,  −
w1 + w1 = 0 ∂w1 −β/(1+β) on Γ , (x) = − dist(x, Υ )  ∂ν w1 = 0 on Υ , and w2 satisfies   −
w2 + w2 = 0 in D, ∂w2 (x) = λ on Γ ,  ∂ν on Υ . w2 = 0 For σ > 0 define Wσ = {x ∈ D: dist(x, ∂Γ ) < σ }. A similar calculation as in Lemma 5.4 implies that for R large enough there is a function u defined in Wσ that satisfies  in Wσ ,   −
u + u  0 ∂u 1 −β/(1+β) (x)  − C dist(x, Υ ) on Γ ∩ ∂Wσ ,   ∂ν u=0 on Υ ∩ ∂Wσ . (Taking R large here corresponds to work with small τ in Lemma 5.4.) Moreover, as in (22) u
c dist(x, Υ )1/(1+β)

on Γ ∩ ∂Wσ ,

where c > 0. Here C and c are positive constants that are independent of R and λ. Following the argument of Lemma 5.4 the following assertions hold: (1) w
cλu in ∂Wσ ∩ D for λ large enough, (2) w = cλu = 0 on ∂Wσ ∩ Υ , ∂u (3) cλ ∂ν (x)  −λ dist(x, Υ )−β/(1+β)  ∂w ∂ν (x) for every x ∈ ∂Wσ ∩ Γ and λ large, see (26). By the maximum principle cλu  w

in Wσ .

Therefore w(x)
cλ dist(x, Υ )1/(1+β)

on Γ.

Hence on Γ ∂w (x) = − dist(x, Υ )−β/(1+β) + λ  −cβ λβ w −β + λ. ∂ν Thus, if λ is sufficiently large, w is a subsolution in D. Next, we extend w by zero to BR (0) \ D and this is a nontrivial subsolution of (83). Claim 2. We fix a sufficiently large λ = λ¯ and prove that for R large enough the maximal solution vanishes on a subset of ∂BR with positive surface measure. Define v(y) = u(Ry) and y = Rx for y ∈ B1 then v satisfies

−R −2
v + v = 0 in B1 , ∂v −β = −v R + λRf (Ry) on ∂B1 ∩ {v > 0}. ∂ν

J. Dávila, M. Montenegro / Ann. I. H. Poincaré – AN 22 (2005) 303–330

327

Integration over B1 gives   ∂v −2 = v R ∂n ∂B1

B1

and using the boundary condition     −β  −v + λf (Ry) + v = R −1 B1

But

∂v ∂n



∂B1 ∩{v>0}



∂B1 ∩{v=0}

 ∂v . ∂n

 0 in the set ∂B1 ∩ {v = 0} and therefore   v + R −1 v −β  λR −1 f (Ry)  CλR −n .

B1

∂B1 ∩{v>0}

(84)

∂B1

1 2 r + Rn , which satisfies To proceed further we need to estimate v on ∂B1 . Define Y (r) = 2R

−
Y + Y
0 in BR , ∂Y  = 1 on ∂BR . Y (R) = ∂ν The function U = MY is a supersolution of (83) if one takes M large enough (independently of λ). Therefore u  CY on ∂BR . Notice that Y (R) = R2 + Rn and hence v  CR on ∂B1 . Using this estimate in (84) we deduce    v −β  λCR −n+1 . CR −β ∂B1 ∩ {v > 0}  ∂B1 ∩{v>0}

We conclude that for R large enough   ∂B1 ∩ {v > 0}  λCR −n+1+β < |∂B1 |. This implies that u vanishes on subset of ∂BR of positive surface measure.

2

Acknowledgements J.D. was supported by FONDECYT No. 1020815. M.M. was supported by CNPq-478896/2003-4 and FAPESP 04/03273-9.

Appendix. Auxiliary results In this section we collect a number of auxiliary results for linear equations that we use in the paper. Lemma 9.1. Let D ⊂ Rn be a bounded, smooth domain, a ∈ L∞ (D) and a
0. Suppose u ∈ H 1 (D), u
0 satisfies −
u + a(x)u
0 in D. Then there is a constant C > 0 independent of u such that   u dist(x, ∂D), ∀x ∈ D. u(x)
C ∂D

(85)

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J. Dávila, M. Montenegro / Ann. I. H. Poincaré – AN 22 (2005) 303–330

Proof. Let D  ⊂⊂ D be a nonempty open set, with smooth boundary. By a weak version of Harnack’s inequality (see [7, Theorem 8.18])  u(x)
C u ∀x ∈ D  , D

where C > 0. We proceed to verify that   u  C u. ∂D

D

Let w be the solution of  −
w + a(x)w = χD  in D, w=0 on ∂D. Then w
0 and       ∂w u+ u = (−
w + a(x)w)u = − ∇w · ∇u + a(x)wu ∂ν D D D D     ∂w ∂w ∂u =− u+ w + (−
u + a(x)u)w
− u. ∂ν ∂ν ∂ν D



D

D



∂D

 −C, then D  u
C ∂D u. But From here we deduce that (85) holds for x ∈ D  . If x ∈ D \ D  we argue as follows. Let z solve 

 ,  −
z + a(x)z = 0 in D \ D z = 0 on ∂D,  z=1 on ∂D  .  By Hopf’s lemma z(x)
c dist(x, ∂D) for all x ∈ D \ D  , where c > 0. Then v(x) = u(x)/(C ∂D u) satisfies v
z on ∂(D \ D  ). By the maximum principle v(x)
z(x) ∀x ∈ D \ D  . So     u(x)
C u z(x)
C u dist(x, ∂D) ∀x ∈ D \ D  . 2 ∂w ∂ν

∂D

∂D

The next estimate follows from [10]. Lemma 9.2. Let D ⊂ Rn be a bounded, smooth domain, a ∈ L∞ (D) and a
0. Suppose u ∈ H 1 (D) satisfies  −
u + a(x)u = 0 in D, u=g on ∂D, where g ∈ Lp (∂Ω) and p
1. Let 1  r <

np n−1

Then there exists C independent of g and u such that

uLr (D)  CgLp (∂D) . We state the next results in this section in a form suitable for the proof of the regularity results. Therefore we recall the notation introduced in Section 4: 1  = (Ω − x0 ) where x0 ∈ ∂Ω, 0 < τ < τ0 , Ω τ + = Ω  ∩ B1 (0), Γe = ∂ Ω  ∩ B1 (0). B (86) The constants that appear in the next lemmas can be chosen independently of x0 ∈ ∂Ω and 0 < τ < τ0 . The following estimate follows from [10].

J. Dávila, M. Montenegro / Ann. I. H. Poincaré – AN 22 (2005) 303–330

329

+ ). Suppose u ∈ H 1 (B + ) satisfies Lemma 9.3. Let a ∈ L∞ (B

+ , −
u + a(x)u = 0 in B ∂u =g on Γe , ∂ν np where g ∈ Lp (∂Ω) and p
1. Let 1  r < n−1 . Then there exists C independent of g and u such that   uW 1,r (B3/4 ∩Ω)   C gLp (Γe ) + uL1 (B + ) . + ) and suppose that u ∈ H 1 (B + ) satisfies Lemma 9.4. Let a ∈ L∞ (B

+ , −
u + a(x)u  0 in B ∂u N on Γe , ∂ν where N is a constant. If p > 1 then there is a constant C > 0 independent of u, N such that   + . u(x)  C u+  p + + N , ∀x ∈ B1/2 ∩ B L (B3/4 ∩B )

Proof. Use Moser’s iteration scheme, see e.g. [7, p. 195].

2

+ such that (B2 ∩ Ω)  ⊂D⊂ Proof of Lemma 6.1. By (86) we may assume that there is a smooth domain D ⊂ B  Denote x = x˜ and x1 = x˜1 . By Lemma 9.4 (B3 ∩ Ω). u(x1 )  CuLr (B3/4 ∩B+ ) + C N,

(87)

n . On the other hand, by Lemma 9.1 where we fix 1 < r < n−1   u dist(x, ∂D) ∀x ∈ D. u(x)
c ∂D

By Lp estimates (Lemma 9.2)  uLr (B3/4 ∩B+ )  C u, ∂D

+ , we find and by (88) and the fact that dist(x, ∂D) = dist(x, Γe ) ∀x ∈ B + . e ) ∀x ∈ B u(x)
cu r + dist(x, Γ L (B3/4 ∩B )

Using (87) we obtain

  u(x)
c dist(x, Γe ) u(x1 ) − C N .

2

+ ) and suppose that u ∈ H 1 (B + ), u
0 satisfies Lemma 9.5. Let a ∈ L∞ (B

+ , −
u + a(x)u
0 in B ∂u
−N on Γe , ∂ν where N is a constant. Then there is a constant C > 0 independent of u, N such that    + . u  C u(x) + N ∀x ∈ B1/2 ∩ B + B3/4 ∩B

Proof. Use Moser’s iterations, see e.g. [7, p. 195].

2

(88)

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J. Dávila, M. Montenegro / Ann. I. H. Poincaré – AN 22 (2005) 303–330

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