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www.elsevier.com/locate/anihpc

Large solutions for the Laplacian with a power nonlinearity given by a variable exponent

Jorge García-Melián

a

, Julio D. Rossi

b,,1

, José C. Sabina de Lis

a

aDpto. de Análisis Matemático, Universidad de La Laguna, C/Astrofísico Francisco Sánchez s/n, 38271 La Laguna, Spain bIMDEA Matematicas, C-IX, Campus Cantoblanco UAM, Madrid, Spain

Received 23 November 2007; accepted 8 March 2008 Available online 27 April 2008

Abstract

In this paper we consider positive boundary blow-up solutions to the problemu=uq(x)in a smooth bounded domainΩ⊂Rn. The exponentq(x)is allowed to be a variable positive Hölder continuous function. The issues of existence, asymptotic behavior near the boundary and uniqueness of positive solutions are considered. Furthermore, sinceq(x)is also allowed to take values less than one, it is shown that the blow up of solutions on∂Ωis compatible with the occurrence of dead cores, i.e., nonempty interior regions where solutions vanish.

©2008 Elsevier Masson SAS. All rights reserved.

Keywords:Large solutions; Existence and uniqueness; Variable exponents

1. Introduction

Boundary blow-up problems for elliptic equations have been widely considered in the last few years. In general, they take the form

u(x)=f (x, u(x)) inΩ,

u(x)= +∞ on∂Ω, (1.1)

whereΩ is a smooth bounded domain of RN (say C2,η) and f (x, u) is a given function. By a solution of (1.1) we understand a functionuC2(Ω)verifying the equation in the classical sense andu(x)→ ∞asx∂Ω. The solutions to problem (1.1) are known as “large” solutions. We refer to the pioneering papers [4,17] and [25], and to [14,23] and [26] for a large list of references.

In most of the previous works, the dependence onxoff was not really significative. Three types of nonlinearities have been frequently treated:f =f (u),f (x, u)=a(x)g(u)orf (x, u)controlled in terms of a functiong(u)which does not depend onx.

* Corresponding author.

E-mail addresses:jjgarmel@ull.es (J. García-Melián), jrossi@dm.uba.ar (J.D. Rossi), josabina@ull.es (J.C. Sabina de Lis).

1 On leave from Departamento de Matemática, FCEyN, UBA Ciudad Universitaria, Pab 1 (1428), Buenos Aires, Argentina.

0294-1449/$ – see front matter ©2008 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2008.03.007

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For the particular case where f is increasing and does not depend on x, f =f (u), it is well known that the so-called Keller–Osserman condition is necessary and sufficient for existence of solutions to (1.1):

x0

ds

F (s)<+∞ (1.2)

for some x0∈R, where F (u)=u

0 f (s) ds is a primitive off (see [17] and [25]). Note that it has been recently shown that the monotonicity off is not necessary, even for largeuas shown in [13] (see also [10], where an existence result was obtained with a nonmonotonicf which is however increasing for largeu).

When the dependence off onx is of the formf (x, u)=a(x)g(u), and the weighta(x)is bounded, the Keller–

Osserman condition on g is also necessary and sufficient for existence so that the presence of a(x)is not really important. Whena(x)is not bounded on∂Ω the situation is slightly different: if the growth ofa near∂Ωis not too strong then solutions to (1.1) exist whengsatisfies (1.2) (see [5,6] and [28] for the caseg(t )=tp,p >1). However, solutions may exist with ag not satisfying (1.2), provideda is singular enough on∂Ω. We refer the reader to [5]

and [24].

Thus, at this point it is natural to ask what happens for a functionf (x, u)that depends onx in such a way that condition (1.2) (whereF (u)is replaced byF (x, u)=u

0 f (x, s) ds) is satisfied at some points ofΩ and not at other points. If we assumef (x, u)to be continuous inΩ×R(so the presence of an unbounded weight is ruled out), is it really needed that (1.2) is satisfied at all points ofΩ to obtain existence of a solution to (1.1)?

In this direction, the problem (1.1) withf (x, u)= −λu+a(x)uq(x) with a >0 in Ω andq >1 inΩ,q =1 on∂Ω, was considered in the pioneering paper [21], and the existence of a maximal and a minimal positive solution was obtained (see also [9] and [22] for works dealing with nonlinearities with a variable exponent and homogeneous Dirichlet boundary conditions). Notice that condition (1.2) (with F (x, u)=u

0 f (x, s) ds) holds at points where q(x) >1, while it ceases to be true whenq(x)1. In this respect, the results in [21] show that (1.2) may fail on the boundary∂Ω, and the existence of positive solutions is still possible.

In the present paper we are considering the problem:

u=uq(x) inΩ,

u= +∞ on∂Ω, (1.3)

where the exponentq(x)will be a positive Hölder continuous function. We note that a distinctive feature in this work with respect to the hypotheses in [21] is that q <1 is permitted at some points inΩ. In this respect, one of the contributions of the present work is to show that condition (1.2) is only needed in a neighborhood of the boundary in order to have a positive solution, while it may fail not only on the boundary, but also at interior points.

We mention in passing that the case whereqis constant is well understood, see [2,3,7,11,12,15,16,18,20,27], but, at the best of our knowledge, the only previous work where large solutions with nonlinearities with a variable exponent were considered is [21].

In addition to existence of positive solutions to (1.3), we also consider uniqueness and the determination of the blow-up rate of solutions near the boundary of the domain. Our techniques are mainly based on comparison, using as a reference problem (1.3) with aqconstant.

Now we state our results. We first show that positive solutions to (1.3) are only possible ifq1 on∂Ω. Whenq is constant, this is known to hold (see Theorem 2 in [19] and Theorem 2.2 in [8]).

Theorem 1.LetqCη(Ω)be a nonnegative function, and assume there existsx0∂Ω such thatq(x0) <1. Then problem(1.3)has no positive solutions. Moreover, the same conclusion holds ifq 1in a whole neighborhood of x0∂Ωrelative toΩ.

Remark 1. As kindly pointed out to the authors by the referee of this paper, a nonexistence result for the variable exponent related problem

u= −λu+a(x)uq(x) inΩ,

u= +∞ on∂Ω,

can be obtained by means of Theorem 7.1 in [21] provided the set{q(x) >1}is strictly contained inΩ,q=1 in a whole neighborhood of∂Ωandλis conveniently large.

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Thanks to Theorem 1 we always needq1 on∂Ωin order to have positive solutions. We will make the assumption thatq >1 in a neighborhood of∂Ω, althoughq may be 1 on∂Ω. We also remark thatq1 is permitted at interior points, and still we get a solution.

Theorem 2.LetqCη(Ω)be a positive function and assumeq >1in the stripUδ= {xΩ: dist(x, ∂Ω) < δ}for someδ >0. Then problem(1.3)admits at least a positive solution.

It is natural to ask under which conditions the solution provided by Theorem 2 is unique. It turns out thatq >1 on

∂Ω is sufficientas long asq1 in the wholeΩ.

Theorem 3.AssumeqCη(Ω)verifiesq1inΩ andq >1on∂Ω. Then problem(1.3)admits a unique positive solution.

The approach for proving Theorem 3 is to obtain the boundary behavior of all positive solutions. We remark that this is a local issue, and hence the obtained behavior is similar to that in the case whereq is constant, at least at points whereq >1. In the rest of the paper,d(x)will stand for the function dist(x, ∂Ω).

Theorem 4.AssumeqCη(Ω)and letx0∂Ω withq(x0) >1. Ifuis a positive solution to(1.3), then

xlimx0

d(x)α(x)u(x)= α(x0)

α(x0)+1 1

q(x0)−1, (1.4)

whereα(x)=2/(q(x)−1).

Remark 2.As a byproduct of the proof of Theorem 4, the exact rate of the normal derivative ofucan also be obtained.

More precisely we have

xlimx0

d(x)α(x)+1u(x)·ν(x)¯ =α(x0) α(x0)

α(x0)+1 1

q(x0)1,

whereνis the outward unit normal andx¯is the closest point tox lying on∂Ω.

Now, another natural question arises: is it essential thatq 1 in the whole Ω to have uniqueness? As we are showing next, the answer is no. Uniqueness of positive solutions to (1.3) also holds ifq <1 at interior points if we assumeqis large enough on∂Ω. As a technical hypothesis, we also needqto be smooth in a neighborhood of∂Ω. Theorem 5.AssumeqCη(Ω)C2(Uδ)for someδ >0whereUδ= {xΩ: dist(x, ∂Ω) < δ},q >0 inΩ and q >3on∂Ω. Then there exists a unique solution to(1.3), which in addition verifies

u(x)= α(x)

α(x)+1 1

q(x)1d(x)α(x)+O d(x)β

(1.5) for everyβ(0,qq03

01), whereα(x)=2/(q(x)−1)andq0=min∂Ωq.

It should be noticed that in the case q constant it was shown in [18] that u=Adα +O(1)as d →0, α as above,A=α(α+1)1/(q1), provided thatq3 (such feature was more precisely described in [16] where a two-term asymptotic expansion forunear∂Ω was obtained). Theorem 5 provides in particular a substantial extension of the previous results covering the case whereqis variable.

On the other hand and as a counterpart to the uniqueness question studied in Theorem 5, the fact thatq achieves values less than one inΩallows the existence of solutionsuof (1.3) that exhibit simultaneously a singular behavior on∂Ω together with the presence of a dead core, i.e., a nonempty interior regionO inΩ whereuvanishes. Our next result asserts that dead cores arise provided the subdomainQofΩ whereq <1 is large enough. We provide a statement with hypotheses that are not optimal for the sake of clarity.

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Theorem 6.Suppose thatqCη(Ω)is a positive functions satisfyingq >1on∂Ω whileQ:= {xΩ: q(x) <1}

constitutes a smooth subdomain ofΩ. Forλ >0letΩλ= {λx: xΩ}and letqλCηλ)be given byqλ(x)= q(x/λ). Then, there existsλ0>0such that forλλ0all positive solutionsuλto

u=uqλ(x) inΩλ,

u= +∞ on∂Ωλ, (1.6)

possess a nonempty dead coreOλ:= {xΩλ: uλ=0}. Moreover,Oλprogressively fillsQλasλ→ ∞.

Remark 3.It is possible to exhibit examples showing that dead cores are absent in problem (1.3) if the region{q <1}

does not exceed a critical size.

Finally, we briefly consider the issue of boundary behavior of positive solutions to (1.3) in the case whereq=1 somewhere on∂Ω. Since the problem becomes linear there it is to be expected that the exact rate of divergence of the solutionsucannot be obtained, as happens in [5]. It is possible however to obtain the exact behavior of the logarithm ofu, providedq−1 behaves like a nonnegative power of the distance.

Theorem 7.AssumeqCη(Ω)and letx0∂Ωbe a point withq(x0)=1. If there exist positive constantsγ andQ such that

xlimx0

q(x)−1 d(x)γ =Q

then for every positive solution to(1.3):

xlimx0

d(x)γlogu(x)

−logd(x) =2γ+2

Q . (1.7)

Of course it would be desirable to obtain uniqueness of solutions to (1.3), at least in the very special caseq(x)= 1+Qd(x)γ. According to (1.7), it would be natural to deal with the equation satisfied byv=logu, that is,

v+ |∇v|2=eQd(x)γv inΩ,

v= +∞ on∂Ω. (1.8)

However, the operator in the left-hand side of (1.8) does not have the right monotonicity, and it could even happen that uniqueness does not hold. We leave this question as an open problem.

The paper is organized as follows: in Section 2 we prove Theorems 1 and 2. Section 3 will be dedicated to prove the boundary estimates, Theorems 4 and 7. The uniqueness results, Theorems 3 and 5 will be collected in Section 3 while the issue of dead cores (Theorem 6) will be analyzed in Section 4.

2. Existence

In this section, we deal with the issues of existence and nonexistence of positive solutions to problem (1.3). We first show that there are no solutions ifq <1 somewhere on∂Ω(alternatively,q1 in a neighborhood of a boundary point). Throughout the paper, we denote byB(x, r)the ball of centerxand radiusr.

Proof of Theorem 1. Letr >0 such thatq <1 inB(x0,3r)∩Ω, and choose a smooth subdomainDofB(x0,3r)∩Ω such that∂D∂ΩcontainsB(x0,2r)∩∂Ω. Letψbe a smooth function supported on∂Dwhich verifies 0ψ1, ψ=1 onB(x0, r)∂Ω andψ=0 on∂D\(B(x0,2r)∩∂Ω). It can be checked that the problem

z=zq(x) inD, z= on∂D,

has a unique positive solutionzn for every positive integern(see the proof of Theorem 2 for a similar argument).

Moreover, if (1.3) has a positive solution, it follows by comparison that

uzn inD, (2.1)

sinceunψon∂Dfor everyn.

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On the other hand, we havezn=nwn, wherewnsolves w=nq(x)1wq(x) inD,

w=ψ on∂D.

Now 0wn1 andq(x) <1, so it is standard to conclude (for a subsequence if necessary) thatwnw0asn→ ∞, wherew0 is the harmonic function in D which equalsψ on ∂D. Since w0>0 inD, we obtain thatzn→ +∞

uniformly in compact subsets ofD(B(x0, r)∂Ω). But then (2.1) impliesu= +∞inD(B(x0, r)∂Ω), which is not possible. Hence no positive solution to (1.3) exists.

Finally, observe that the previous argument continues to be valid – with only minor changes – if q 1 in a neighborhood of a pointx0∂Ω. Thus the proof is concluded. 2

Now we prove our existence result. The approach is the standard one: we construct solutions with finite datum on

∂Ω and then show that they are locally uniformly bounded.

Proof of Theorem 2. Letnbe a positive integer. Then the problem u=uq(x) inΩ,

u=n on∂Ω, (2.2)

has a unique positive solution. Indeed,u=0 is a subsolution andu¯=nis a supersolution, and then, by a well-known approach (see [1]) the existence of a classical solutionuC2,η(Ω)follows. To prove uniqueness, letu,vbe positive solutions and consider the setΩ0= {xΩ: u < v}. IfΩ0is nonempty, sinceuvinΩ0andu=von∂Ω0, it follows from the maximum principle thatu > vinΩ0, which is impossible. Thusuvand the symmetric argument showsu=v, giving uniqueness. The solution to (2.2) will be denoted byun.

Thanks to uniqueness, the solutionsunis increasing inn. Indeed,un+1is a supersolution to (2.2) and by uniqueness un+1un.

Let us prove next thatun is bounded in compact subsets ofΩ. Taking δsmall, we can assumeun>1 in a strip Uδ= {x: dist(x, ∂Ω) < δ}for alln. Fixεwith 0< ε < δand a pointx0such thatd(x0)=ε/2. Sinceq >1 inUδ, we have thatqq0>1 inB(x0, ε/4)and thusunuqn0 inB(x0, ε/4). HenceunU, the unique solution to

U=Uq0 inB(x0, ε/4), U= +∞ on∂B(x0, ε/4).

This shows that un is uniformly bounded in B(x0, ε/8). A compactness argument proves that un is uniformly bounded in the set{xΩ: d(x)=ε/2}, and since everyunis subharmonic, we obtain uniform bounds in the whole {xΩ: d(x) > ε/2}. Sinceεwas arbitrarily small, the sequence{un}is locally uniformly bounded inΩ.

Finally, it is standard to obtain that {un}is precompact inCloc2 (Ω), and thus, passing to a subsequence, unu inCloc2 (Ω), whereu verifiesu=uq(x) inΩ. Notice that, since un is increasing inn, it also follows that the whole sequence converges tou. Moreover,u= +∞on∂Ω, and is thus a positive solution to (1.3). This finishes the proof. 2

3. Boundary estimates

This section is devoted to prove the assertions concerning the boundary behavior of the solutions to (1.3). To prove Theorem 4 we use ideas from [6]. To this aim, it is important to obtain first a rough estimate for the solutions. This is the content of the next lemma.

Lemma 8. Assumex0∂Ω is such that q(x0) >1, and let u be a positive solution to (1.3). Then there exist a neighborhoodVofx0(relative toΩ)and positive constantsC1, C2such that

C1d(x)α(x)u(x)C2d(x)α(x) inV, (3.1)

whereα(x)=2/(q(x)−1).

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Proof. Choose a neighborhoodVofx0such thatq >1 inV(we can take for instance a ball centered atx0intersected withΩ). By diminishing the radius ofV, we can select a smaller neighborhoodV such thatB(x, d(x)/2)Vfor xV. TakexVand define the scaled function

v(y)=d(x)α(x)u

x+d(x) 2 y

,

foryB:=B(0,1). It can be checked that the functionvsolves the equation v=1

4d(x)α(x){q(x)q(x+(d(x)/2)y)}vq(x+(d(x)/2)y) inB.

Sinceqisη-Hölder, there exists a constant such that q(x)q

x+d(x) 2 y

Cd(x)η, and hence

vCvq(x+(d(x)/2)y) inB

for some positive constantC (we are using throughout the paper the letter C to denote constants, that may change from one line to another but are independent of the relevant quantities). That is,v is a subsolution to the equation v=Cvq(x+(d(x)/2)y) inB. Now we will construct a supersolution to the same equation which blows up on the boundary ofB.

Letφbe the solution to−φ=1 inBwithφ=0 on∂B. For a large positiveA0and someβ >0 to be chosen, we definev¯=A0φβ. Thenv¯will be a supersolution tov=Cvq(x+(d(x)/2)y)inBprovided that

β(β+1)|∇φ|2+βφCA{0q(x+(d(x)/2)y)1}φ{β+2βq(x+(d(x)/2)y)} for allyB. This inequality can be obtained choosingβ large in order to have

β+2−βq

x+d(x) 2

<0,

and thenA0large enough. By comparison, we arrive atvv¯inB, and settingy=0 we obtain u(x)A0φ(0)βd(x)α(x)

forxV. This shows the upper inequality in (3.1).

To prove the lower inequality we take a pointxVand denote byx¯ the closest point tox on∂Ω. Modulus an extra reduction ofVif necessary it can be assumed thatd(x¯+d(x)ν(x))¯ =d(x)for everyxVwhereνstands for the outward unit normal andd(x)designates the distance fromxto∂Ω. Denoting byAthe annulus

A=

y∈RN: 1<|y|<2+τ ,

whereτ >0, we introduceAx= ¯x+d(x)ν(x)¯ +d(x)AandQx=AxΩ (observe thatxQx, while the annulus Axis tangent to∂Ω atx¯). We remark that the outer radius can be any fixed number greater than 2, but for its later use in the proof of Theorem 7 we let it depend on the parameterτ, which is of no importance in the present proof.

We can assume, by diminishing the radius ofV, thatQxVfor everyxV. Now define the normalized function w(y)=d(x)α(x)u

¯

x+d(x)ν(x)¯ +d(x)y ,

foryQx, whereQx=A∩ {y∈RN: x¯+d(x)ν(x)¯ +d(x)yΩ}. Thenwsatisfies w=d(x)α(x){q(x)q(x¯+d(x)ν(x)¯+d(x)y)}wq(x¯+d(x)ν(x)¯+d(x)y)

inQx. Thanks to the Hölder condition verified byqit follows as before that wCwq(x¯+d(x)ν(x)¯+d(x)y) inQx,

for a certain positive constantC.

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On the other hand, it can be seen as before that the problem

⎧⎨

z=Czq(x¯+d(x)ν(x)¯+d(x)y) inA,

z=1 on|y| =1,

z=0 on|y| =2+τ,

has a unique positive solutionz. Sincewzon∂Qx, it follows by comparison thatwzinQx. Settingy= −2ν(x),¯ we arrive at

u(x)z

−2ν(x)¯

d(x)α(x),

and the proof of (3.1) concludes by noticing that sincezis bounded from below in|y| =2 we obtainz(−2ν(x))¯ C >0, whereCis independent ofx. 2

Proof of Theorem 4. Choose an open neighborhood W of x0 such that ∂Ω admits C2,η local coordinatesξ = 1, . . . , ξN),ξ:W→RN, with xWΩ if and only ifξ1(x) >0. With no loss of generality we can assume ξ(x0)=0. Settingu(x)= ¯u(ξ(x)),q(x)= ¯q(ξ(x)), thenu¯verifies an equation

N i,j=1

aij(ξ ) 2u¯

∂ξi∂ξj + N i=1

bi(ξ )∂u¯

∂ξi = ¯uq(ξ )¯

inξ(WΩ)whose coefficientsaij,bi areCη functions andaij(0)=δij. We can assume further thatWΩV, whereV is the neighborhood given by Lemma 8.

Let{xn}be an arbitrary sequence converging tox0, and denote bytn the projection ofξ(xn)ontoξ(W∂Ω)(a subset of the hyperplaneξ1=0). We introduce the functions

vn(y)=dnαnu(tn+dny)

wheredn=d(xn),αn=α(xn). Thenvnverifies the equation N

i,j=1

aij(tn+dny) 2v

∂yi∂yj +dn

N i=1

bi(tn+dny)∂v

∂yi =dn{αn(q(ξ(x¯ n))− ¯q(tn+dny))}vq(t¯ n+dny).

We now use the estimates (3.1) provided by Lemma 8. They imply that for every compact set of the half-space D:= {y∈RN: y1>0}there exist positive constantsC1,C2such that

C1dn{αn− ¯α(tn+dny)}y− ¯1α(tn+dny)vn(y)C2dn{αn− ¯α(tn+dny)}y1− ¯α(tn+dny) (3.2) whereα(x)= ¯α(ξ(x)). As in the proof of Lemma 8, we use the Hölder condition onq to obtain that

dnαn− ¯α(tn+dny)→1

uniformly fory in compacts ofDasn→ ∞. Thus (3.2) gives bounds for the sequence{vn}, and it is now standard to obtain that for a subsequence we havevnvinCloc2 (D), wherevverifies

v=vq(x0)

C1y1α(x0)v(y)C2y1α(x0) inD. (3.3)

Theorem 3.4 and Remark 3.6 (b) in [6] imply that problem (3.3) has a unique solution, which can be checked to be v(y)=

α(x0)

α(x0)+1q(x1

0)1y1α(x0). Then (1.4) is proved just by settingy=e1. 2

Now we prove Theorem 7. The proof is based on that of Lemma 8, but taking into account that the exponents there may be variable, and the involved constants have to be precisely estimated.

Proof of Theorem 7. Letε >0 and choose a neighborhoodWofx0such thatq(y)1+(Qε)d(y)γ foryW.

Forx close tox0, and 0< τ <1, we haved(y)(1τ )d(x)ifyB(x, τ d(x))and hence q(y)1+(Qε)(1τ )γd(x)γ

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inB(x, τ d(x)), providedB(x, τ d(x))W, which is certainly true ifx is close enough tox0. Denote for simplicity σ=σε,τ,x=(Qε)(1τ )γd(x)γ.

Ifx is close enough tox0we may further assume thatu >1 inB(x, τ d(x)). Hence uu1+σ inB

x, τ d(x) . We now introduce the function

v(y)= τ d(x)σ2

u

x+τ d(x)y

yB=B(0,1), which satisfies

vv1+σ

inB. On the other hand, we may look for a supersolution to the equationv=v1+σ of the form v=β

whereβ=2/σ,A >0 andφis the solution to−φ=1 inB withφ=0 on∂B. Thenvis a supersolution provided that

2 σ

2 σ +1

|∇φ|2+ 2

σφAσ.

Sinceσ=(Qε)(1τ )γd(x)γ, it is enough to take A=

Cd(x) −2γ

(Qε)(1τ )γ d(x)γ,

for some positive, large enough constantC. By comparison, v(y)v(y)

ifyB. Setting in particulary=0 we obtain u(x)

Cd(x) −2γ

(Qε)(1τ )γ d(x)γ

τ d(x)φ(0) −2

(Qε)(1τ )γ d(x)γ. It follows from the last estimate that

lim sup

xx0

d(x)γlogu(x)

−logd(x) 2γ+2 (Qε)(1τ )γ. Lettingε→0 and thenτ→0, we obtain

lim sup

xx0

d(x)γlogu(x)

−logd(x) 2γ+2

Q . (3.4)

Next we prove the lower estimate. As in the first part of the proof, we may assume a neighborhoodWofx0has been chosen so that

q(y)1+(Q+ε)d(y)γ

foryW. Forxclose tox0, we consider the setsA,Ax,Qx,Qxintroduced in the proof of Lemma 8. InQxwe have q(y)1+(Q+ε)(1+τ )γd(x)γ

and then ifu >1 we have uu1+θ inQx,

where we now setθ=(Q+ε)(1+τ )γd(x)γ. Introduce the function w(y)=d(x)2θu

¯

x+d(x)ν(x)¯ +d(x)y

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foryQx. Thenww1+θinQx, and it follows by comparison thatwUinQx, whereUis the unique positive solution to

⎧⎨

U=U1+θ inA, U= ∞ on|y| =1, U=0 on|y| =2+τ.

Thus our next aim will be to estimate from below the solutionUwhenθ→0. SinceUis radial, it verifiesU=U (r), wherer= |y|and

⎧⎨

U+Nr1U=U1+θ, 1< r <2+τ, U (1)= ∞,

U (2+τ )=0.

We introduce the change of variables ρ=

1

N(1r1N), ifN3, logr, ifN=2, and denoteV (ρ)=U (r). ThenV verifies

⎧⎨

V=g(ρ)V1+θ, 0< ρ < L, V (0)= ∞,

V (L)=0,

withg(ρ)=r2(N1)andLis given byL=1/N (1−1/(2+τ )N)ifN3,L=log(2+τ )forN=2.

Notice thatV is convex, and hence thanks to the mean value theorem:

V (ρ)= −V(ξ )(Lρ)V(L)(Lρ) (3.5)

whereξ(ρ, L)and 0< ρ < L. This shows that it is enough to obtain a lower estimate forV(L).

SinceV<0 andg(ρ)(2+τ )2(N1)=:c, we get VVcV1+θV.

An integration in(ρ, L)gives

V(ρ)

V(L)2+(2c/(2+θ ))V (ρ)2+θ 1.

Integrating with respect toρin(0, L)and settingt=V (ρ), we obtain

0

V(L)2+ 2c 2+θt2+θ

1/2

dtL. (3.6)

We taket=((2+θ )V(L)2/2c)2+1θσ and denote I (θ )=

0

1+σ2+θ. Then, it follows from (3.6) that

V(L) 2+θ

2c 1θ

1 LI (θ )

2+θθ

. (3.7)

On the other hand, if we perform in the integral definingI the change of variable 1+σ2+θ=t1, we obtain

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I (θ )= 1 2+θ

1 0

(1t )2+1θ1t122+1θdt

= 1 2+θB

1 2+θ,1

2− 1 2+θ

= 1 2+θ

(2+1θ)(122+1θ) (12) ,

whereB andstand for Euler Beta and Gamma functions, respectively. Since(z)∼1/zasz→0, it follows that I (θ )∼2/θasθ→0, and henceI (θ )1/θfor smallθ. This implies, thanks to (3.7), that

V(L) 2+θ

2c 1

θ

1

2+θ

θ

,

and then (3.5) gives logV (ρ)1

θlog 2+θ

2c

+2+θ θ log

1

+log(L−ρ).

Going back to the original variables, we arrive at logU (y)1

θlog 2+θ

2c

+2+θ θ log

1

+H

|y| ,

whereH is a function which does not depend onθ. Taking into account thatw(y)U (y), and settingy= −2ν(x),¯ we get

log

d(x)θ2u(x)

1

θlog 2+θ

2c

+2+θ θ log

1

+H (2), and then, sinceθ=(Q+ε)(1+τ )γd(x), it follows that

lim inf

xx0

d(x)γlogu(x)

−logd(x) 2γ+2 (Q+ε)(1+τ )γ. Finally, lettingε→0 andτ→0 we have

lim inf

xx0

d(x)γlogu(x)

−logd(x) 2γ+2 Q which together with (3.4) proves (1.7). 2 4. Uniqueness

This section will be devoted to obtain the uniqueness results Theorems 3 and 5. We begin with the case in which q1 inΩandq >1 on∂Ω.

Proof of Theorem 3. Letu,vbe positive solutions to (1.3). Sinceq >1 on∂Ω, we have, thanks to Theorem 4, that

xlimx0

u(x) v(x) =1

for everyx0∂Ω. By the compactness ofΩ, this limit holds uniformly, and so for small enoughε >0 there exists δ >0 such that

(1ε)vu(1+ε)v

for allxΩ such thatd(x)δ. Consider the problem z=zq(x) inΩδ,

z=u on∂Ωδ, (4.1)

withΩδ= {xΩ: dist(x, ∂Ω) > δ}. Problem (4.1) has a unique positive solution, which is preciselyu. Now it can be checked that(1ε)vand(1+ε)vare a sub and a supersolution respectively to (4.1), sinceq1 inΩ. It follows

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from the uniqueness ofuthat(1ε)vu(1+ε)vinΩδ. Thus this inequality is valid throughoutΩ, and letting ε→0 we arrive atu=v, which shows the desired result. 2

Now we consider the case whereq may be less or equal than one somewhere inΩ, but it is strictly greater than 3 on∂Ωand smooth in a neighborhood of∂Ω.

Proof of Theorem 5. We first show that (1.5) holds for allβ(0,qq03

01). For this aim we construct sub and superso- lutions near the boundary. We claim that forβ(0,qq03

01)and large enoughB, the function

¯

u=A(x)d(x)α(x)+Bd(x)β

is a supersolution inUρ:= {xΩ: dist(x, ∂Ω) < δ}ifδ >0 is small enough, where A=

α(α+1) 1

q1

and

α=2/(q−1).

We chooseδ small to havedC2(Uδ)andq >3 inUδ. Notice that in the present situationα, AC2(Uδ). Thus a direct computation gives:

u¯=dαA−2αdα1Ad−2dαlogdAα−2Adα1αd+Aα(α+1)dα2 +2Aαdα1logdαdAαdα1dAdαlog+Adα(logd)2|∇α|2 +Bβ(β−1)dβ2+Bβdβ1d,

where the fact that the distanced(x)verifies|∇d| =1 inUδhas been used. Some further computations show thatu¯is a supersolution provided that the following inequality holds,

d2A−2αd∇Ad−2d2logdAα−2Ad∇αd+2AαdlogdαdAαddAd2log +Ad2(logd)2|∇α|2+Bβ(β−1)dα+β+Bβdα+β+1d

A+Bdα+βq(x)

Aq(x), where we have used thatAq(x)1=α(α+1).

On the other hand, we also have by convexity that(x+y)qxq+qxq1y for all real positive numbersx,y and thusu¯will be a supersolution if

d2A−2αd∇Ad−2d2logdAα−2Ad∇αd+2AαdlogdαdAαdd

Ad2log+Ad2(logd)2|∇α|2+Bβ(β−1)dα+β+Bβdα+β+1d

qBAq1dα+β. (4.2)

Now, since 0< β <qq03

01, we have 0< β <1−αon∂Ω, so that we can diminishδfurther to have this inequality inUδ. Thus (4.2) can be written as

Bβdα+β

(1β)dd +o

dα+β

qAq1Bdα+β (4.3)

inUδ, where theo-term does not depend onB. Notice that the first term in the left-hand side of (4.3) is positive for smallδ, and thus ifB >1 (4.3) is implied by the inequality

β

(1β)dd +o

dα+β

qAq1

inUδ, which can be achieved by takingδsmaller if necessary, sinceβ <1. The election ofδis thus independent ofB as long asB >1, and we have shown thatuis a supersolution inUδifB >1.

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Analogously it can be proved that w=A(x)d(x)α(x)Bd(x)β

is a subsolution in the subset ofUδwhere it is positive, and henceu=max{w,0}is a subsolution in the wholeΩ. Now letube any solution to (1.3). We chooseBlarge so thatuuu¯ind=δ, and then it follows by comparison and Theorem 3 thatuuu¯inUδ. This proves (1.5).

Finally, we show uniqueness. Letu,vbe solutions to (1.3). Then, according to (1.5):

u(x)v(x)=O d(x)β for everyβ(0,qq03

01). LetΩ0= {xΩ: u(x) < v(x)}. IfΩ0= ∅, we would haveu=von∂Ω0(sinceuv=0 on∂Ω) anduvinΩ0. The maximum principle would implyu > vinΩ0, which is impossible. ThusΩ0= ∅, that is,uv. The symmetric argument givesu=v, and uniqueness is shown. This concludes the proof. 2

5. Dead core formation

In this final section we analyze the existence of dead cores for problem (1.3).

Proof of Theorem 6. The proof rests on the construction of a suitable weak supersolution to (1.3). For this aim, considerΩ= {xΩ: d(x) > δ}for a fixed smallδ(we only require onδthatq >1 in a neighborhood of∂Ω). Set Ωλ=λΩ. We will construct a supersolutionu=uλC(Ωλ)Hloc1 λ)exhibiting the following features:uλ= ∞ on∂Ωλ,uλpossesses a dead coreOλ which uniformly fillsQλasλ→ ∞. Thus, onceuλhas been obtained, we will obtain by comparison thatuuλinΩλfor every positive solutionuto (1.3), sinceu <+∞on∂Ωλwhileuλ= +∞

on∂Ωλ, and the assertions of the theorem will follow.

The supersolutionuλwill be constructed separately in the setsQλandΩλ\Qλ. Let us proceed first inQλand for a fixed numberm0>0 letu= ˜uλC2,η(Qλ)C(Qλ)be the solution to

u=uqλ(x) inQλ, u=m0 on∂Qλ.

Then,v=vλ(x)C2,η(Q)C(Q)defined as vλ(x)= ˜uλ(λx),

solves

v=λ2vq(x) inQ, v=m0 on∂Q.

For large λ,vλ develops a dead coreOλ= {xQ: vλ(x)=0}such thatOλ⊃ {xQ: dist(x, ∂Q)d(λ)}with d(λ)→0 as λ→ ∞. In fact, choosed0>0 small and setQd0 = {xQ: dist(x, ∂Q) > d0}. Then∂Qd0 can be covered with a finite number of ballsBBQwith the same radius,d0/2. In each of such ballsB consider the auxiliary problem

w=λ2f (w), xB,

w=m0, x∂B, (5.1)

where foru0,f (u)=min{uq0, uq1}, 0< q0q1<1 being, respectively, the minimum and the maximum ofq extended to the region{xQ: dist(x, ∂Qd0)d0/2}.

Due to the fact thatvλis subharmonic inQwe can assert that vλ(x)wλ(x), xB,

wherewλstands for the unique positive solution to (5.1). We now claim the existence of a criticalλc, only depending on d0, such that forλλc there exists a positiver(λ) so that wλ=0 in the ball with the same center as B and radiusr(λ). From this fact and the subharmonicity ofvλ it follows thatvλ=0 inQd0 for allλλc. Therefore, the dead cores properties stated above hold true (both forvλandu˜λ).

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Let us now proceed with the construction ofuλinΩλ\Qλand to this goal setu= ˆuλC2,η((Ωλ\Qλ)∂Qλ) the unique solution to

⎧⎨

u=uqλ(x), xΩλ\Qλ, u= ∞, x∂Ωλ, u=m0, x∂Qλ.

(5.2) Define:

uλ(x)=

u˜λ(x), xQλ, ˆ

uλ(x), xΩλ\Qλ.

Thenuλis a weak supersolution to (1.3) providedm0>0 is large enough. In fact it is enough to show that

∂uˆλ

∂ν >0 (5.3)

on∂Qλ whereνstands for the outer unit normal toΩλ\Qλ on the component∂Qλof its boundary. To prove that fact, let

W=

xΩλ\Qλ: dist(x, ∂Qλ) < d1

for certain small positived1. Observe thatuˆλ→ ˆuinC2,ηλ\Qλ)as m0→ ∞ whereuˆ is the minimal solution to (5.2) with m0= ∞. This means thatuˆλ remains finite on ∂W\∂Qλ while it increases with m0 on ∂Qλ. By subharmonicity,uˆλ< m0inWprovidedm0is large and (5.3) follows from the maximum principle. This finishes the construction of the supersolutionuλwith the desired properties.

To complete the proof let us next show the claim concerning problem (5.1). Consider the normalized case of the ballB=B(0, R). In order to demonstrate the dead core features of (5.1) it is enough to handle the slight variation of the problem in the annulusA= {x: ε0<|x|< R} ⊂B, 0< ε0< R, consisting in settingw=0 on|x| =ε0. By radial symmetry such problem inAcan be written as

⎧⎨

w=λ2g(ρ)f (w), 0< ρ < L, w(0)=0,

w(L)=m0,

(5.4) where forε0< r < Rthe suitable modification of the change of variablesρ=ρ(r)introduced in the proof of Theo- rem 7 has been performed. As in that proof,g(ρ)=r2(N1)while now

L=log(R/ε0), and L= 1 N

1 ε0N − 1

RN

,

in the casesN=2 andN3, respectively.

In virtue of the uniqueness in the solvability of (5.4) it follows that its solutionwλsatisfies wλ(ρ)v˜λ(ρ)

0< ρ < L, wherev= ˜vλ(ρ)is the unique solution to

⎧⎪

⎪⎩

v=λ2ε2(N0 1)f (v), 0< ρ < L, v(0)=0,

w(L)=m0.

(5.5) Thus the proof of the claim reduces to perform a dead core analysis in problem (5.5). In such case, direct integration shows that forλgreater than some criticalλc,v˜λ(ρ)vanishes in the interval(0, ρ(λ))whereρ(λ)is expressed as

ρ(λ)=L− 1 εN01λ

m0

0

ds

F (s), (5.6)

whereF (u)=u

0 f (s) ds. More precisely,λcis given by the unique value ofλ >0 which makes zero the difference in (5.6). This concludes the proof of Theorem 6. 2

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