# Existence and Uniqueness of Positive Solution for a Fractional Dirichlet Problem with Combined Nonlinear Effects in Bounded Domains

Authors: Imed Bachar; Habib Mâagli
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We prove the existence and uniqueness of a positive continuous solution to the following singular semilinear fractional Dirichlet problem(-Δ)α/2u=a1(x)uσ1+a2(x)uσ2, in Dlimx→z∈∂D(δ(x))1-(α/2)u(x)=0, where 0<α<2,σ1,σ2∈(-1,1),D is a bounded C1,1-domain in ℝn,n≥2, and δ(x) denotes the Euclidian distance from x to the boundary of D. The nonnegative weight functions a1,a2 are required to satisfy certain hypotheses related to the Karamata class. We also investigate the global behavior of such solution.


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## Body

## 1. Introduction

In the last two decades, several studies have been performed for the so-called fractional Laplacian,(-Δ)α/2,  0<α<2, which can be defined by the integral representation
(1)(-Δ)α/2u(x)=cn,αlimε↘0∫(|x-y|>ε)‍u(x)-u(y)|x-y|n+αdy,
where cn,α=(α2α-1/πn/2)(Γ((n+α)/2)/Γ(1-(α/2))) is a normalization constant; see, for instance, [1, 2]. From a probabilistic point of view, the fractional Laplacian appears as the infinitesimal generator of the stable Lévy process [3, 4]; see also [5]. The fractional powers of the Laplacian arise in a numerous variety of equations in mathematical physics and related fields (see, for instance, [6–11] and the references therein). Motivation from mechanics appears in the Signorini problem (cf. [12, 13]). And there are applications in fluid mechanics, (cf. [14]). The systematic study of the corresponding PDE models is more recent and many of the results have arisen in the last decade. The linear or quasilinear elliptic theory has been actively studied recently in the works of Caffarelli and collaborators [15, 16], Kassmann [17], Silvestre [18], and many others. The standard linear evolution equation involving fractional diffusion is
(2)∂u∂t+(-Δ)α/2u=0.
This is a model of the so-called anomalous diffusion, a much studied topic in physics, probability, and finance (see [19–23] and their references). For more applications, we refer the reader to the survey papers [24, 25].Throughout this paper, we consider a boundedC1,1-domain D in ℝn, n≥2, and we denote by δ(x) the Euclidian distance from x to the boundary of D. For two nonnegative functions f and g defined on a set S, the notation f(x)≈g(x),  x∈S, means that there exists c>0 such that (1/c)f(x)≤g(x)≤cf(x), for all x∈S.Recently, in [26], the authors considered the following problem:
(3){(-Δ)α/2u=φ(·,u)inD(inthesenseofdistributions),u>0inD,limx→z∈∂D(δ(x))1-(α/2)u(x)=0,
where 0<α<2 and φ is a positive measurable function in D×(0,∞) satisfying the following:(A1)
the mapt→φ(x,t) is continuous and nonincreasing in (0,∞), for x∈D;(A2)
for eachc>0, the function x→(δ(x))1-(α/2)φ(x,c(δ(x))(α/2)-1) is in Kα(D) (see Definition 1 below).They have proved that problem (3) has a positive continuous solution u in D satisfying, for each x∈D,
(4)u(x)=∫D‍GDα(x,y)φ(y,u(y))dy,
where GDα(x,y) denotes the Green function of the fractional Laplacian (-Δ)α/2 in D. However they have not investigated the asymptotic behavior of such solution.As a typical example of functionφ satisfying (A1) and (A2), we quote φ(x,u)=a(x)uσ, where σ≤0 and a is a positive measurable function in D such that the function
(5)x⟶(δ(x))((α/2)-1)(σ-1)a(x)
belongs to the Kato class Kα(D) defined as follows.Definition 1 (see [26]).
A Borel measurable functionq in D belongs to the Kato class Kα(D) if
(6)limr→0(supx∈D∫(|x-y|≤r)∩D‍(δ(y)δ(x))α/2GDα(x,y)|q(y)|dy)=0.It has been proved in [26] that the function
(7)x⟶(δ(x))-λbelongstoKα(D)iffλ<α.
For more examples of functions belonging to Kα(D), we refer to [26]. Note that for the classical case (i.e., α=2) the class K2(D) was introduced and studied in [27].On the other hand, Chemmam et al. considered in [28] the following semilinear fractional Dirichlet problem:
(8){(-Δ)α/2u=a(x)uσinD(inthesenseofdistributions),u>0inD,limx→z∈∂D(δ(x))1-(α/2)u(x)=0,
where 0<α<2, σ<1, and a satisfies the following hypothesis:(
H
0
)  a∈Clocγ(D), 0<γ<1, satisfying D,
(9)a(x)≈(δ(x))-λL(δ(x)),
where λ<(α/2)(1+σ)+1-σ and L belongs to the Karamata class 𝒦 defined as follows.Definition 2.
The class𝒦 is the set of all the Karamata functions L defined on (0,η] by
(10)L(t)∶=cexp(∫tη‍z(s)sds),
where η>diam(D),  c>0, and z∈C([0,η]) such that z(0)=0.As a typical example of a function belonging to the class𝒦 (see [29–31]), we quote
(11)L(t)=∏k=1m(logk(ωt))-ξk,
where ξk are real numbers, logkx=log∘log∘⋯logx(ktimes), and ω is a sufficiently large positive real number such that L is defined and positive on (0,η].Using a fixed-point argument, the authors have proved in [28] the existence and uniqueness of a positive continuous solution u for (8) satisfying, for x∈D,
(12)u(x)≈(δ(x))min(α/2,(α-λ)/(1-σ))ΨL,λ,σ(δ(x)),
where the function ΨL,λ,σ is defined on (0,η) by
(13)ΨL,λ,σ(t)∶={(L(t))1/(1-σ),α2(1+σ)<λ<α2(1+σ)+1-σ,(∫tη‍L(s)sds)1/(1-σ),ifλ=α2(1+σ),1,ifλ<α2(1+σ).
In particular, they have extended the results of [32, 33].In the present paper, we aim at studying the following fractional nonlinear problem involving both singular and sublinear nonlinearities with the reformulated Dirichlet boundary condition:(14){(-Δ)α/2u=a1(x)uσ1+a2(x)uσ2inD(inthesenseofdistributions),u>0inD,limx→z∈∂D(δ(x))1-(α/2)u(x)=0,
where 0<α<2 and σ1, σ2∈(-1,1). We will address the question of existence, uniqueness, and global behavior of a positive continuous solution to problem (14).In the elliptic case (i.e.,α=2), problems related to (14) have been studied by several authors (see, e.g., [34–39] and references therein). Using the subsupersolution method, the authors in [36] have established the existence and uniqueness of a positive continuous solution to (14) for α=2,σ1,  σ2<1, where the functions a1, a2 are required to satisfy some adequate assumptions related to the Karamata class 𝒦.Here, our goal is to study problem (14) for 0<α<2. To this end, we assume that the potential functions a1, a2 satisfy the following hypothesis.(
H
) for i∈{1,2}, ai∈Clocγ(D), 0<γ<1, and satisfies, for x∈D,
(15)ai(x)≈(δ(x))-λiLi(δ(x)),
where λi<(α/2)(1+σi)+1-σi and Li∈𝒦 defined on (0,η] with η>diam(D).As it turns out, estimates (12) depend closely on min(α/2,(α-λ)/(1-σ)). Also, as it will be seen, the numbers
(16)β1∶=min(α2,α-λ11-σ1),β2:=min(α2,α-λ21-σ2)
play an important role in the combined effect of singular and superlinear nonlinearities in (14) and lead to a competition. It is not obvious which wins, essentially in the estimates of solution. From here on and without loss of generality, we may assume that (α-λ1)/(1-σ1)≤(α-λ2)/(1-σ2) and we introduce the function θ defined on (0,η) by
(17)θ(t)={tβ1ΨL1,λ1,σ1(t)ifβ1<β2,tβ1(ΨL1,λ1,σ1(t)+ΨL2,λ2,σ2(t))ifβ1=β2.
For an explicit form of the function θ, see (36).Throughout this paper, we define the potential kernelGDα by
(18)GDαf(x)∶=∫D‍GDα(x,y)f(y)dy,forx∈D,f∈B+(D),
where B+(D) denotes the set of the nonnegative Borel measurable functions in D.Our main results are the following.Theorem 3.
Letσ1, σ2∈(-1,1) and assume (H). Then one has, for x∈D,
(19)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈θ(δ(x)).Using Theorem 3 and the Schauder fixed-point theorem, we will prove the following.Theorem 4.
Letσ1, σ2∈(-1,1) and assume (H). Then problem (14) has a unique positive continuous solution u in D satisfying, for x∈D,
(20)u(x)≈θ(δ(x)).In particular, we generalize the result obtained in [36] to the fractional setting and we recover the result obtained in [28].The content of this paper is organized as follows. In Section2, we collect some properties of functions belonging to the Karamata class 𝒦 and the Kato class Kα(D), which are useful to establish our results. In Section 3, we prove our main results.As usual, we denote byC0(D) the set of continuous functions in D¯ vanishing continuously on ∂D. Note that C0(D) is a Banach space with respect to the uniform norm ∥u∥∞=supx∈D|u(x)|. As in the elliptic case, if f∈B+(D) satisfies ∫D‍(δ(y))α/2f(y)dy<∞, then the functions f and GDαf are in Lloc1(D) and we have in the distributional sense
(21)(-Δ)α/2GDαf=f,inD.


## 2. The Karamata Class𝒦 and the Kato Class Kα(D)

We collect in this paragraph some properties of the Karamata class𝒦 and the Kato class Kα(D). We recall that a function L defined on (0,η] belongs to the class 𝒦 if
(22)L(t)∶=cexp(∫tη‍z(s)sds),
where η>diam(D),  c>0, and z∈C([0,η]) such that z(0)=0.Proposition 5 (see [30, 31]).
(
i
) A function L is in 𝒦 if and only if L is a positive function in C1((0,η]) such that
(23)limt→0+tL′(t)L(t)=0.
(
ii
) Let L1, L2∈𝒦, p∈ℝ. Then one has
(24)L1+L2∈𝒦,L1L2∈𝒦,L1p∈𝒦.
(
iii
) Let L∈𝒦 and ε>0. Then one has
(25)limt→0+tεL(t)=0.Applying Karamata’s theorem (see [30, 31]), we get the following.Lemma 6.
Letμ∈ℝ and let L be a function in 𝒦. One has the following:(i)
ifμ<-1, then ∫0η‍sμL(s)ds diverges and ∫tη‍sμL(s)ds~t→0+(-t1+μL(t))/(μ+1);(ii)
ifμ>-1, then ∫0η‍sμL(s)ds converges and ∫0t‍sμL(s)ds~t→0+(t1+μL(t))/(μ+1).Lemma 7 (see [36]).
LetL be a function in 𝒦. Then one has
(26)limt→0+L(t)∫tη‍(L(s)/s)ds=0.
In particular
(27)t→∫tη‍L(s)sds∈𝒦.Proposition 8 (see [40, 41]).
For(x,y)∈D×D, one has
(28)GDα(x,y)≈|x-y|α-nmin(1,(δ(x)δ(y))α/2|x-y|α).Proposition 9 (see [26, Corollary 6]).
Letq be a nonnegative function in Kα(D); then the family of functions
(29)Λq={x⟶∫D‍(δ(y)δ(x))(α/2)-1GDα(x,y)f(y)dy,|f|≤q}
is uniformly bounded and equicontinuous in D¯. Consequently Λq is relatively compact in C0(D).


## 3. Proofs of the Main Results

In this section we aim at proving Theorems3 and 4. To this end, we need the following lemmas.


### 3.1. Technical Lemmas

Lemma 10.
Forr, s>0, one has
(30)2-max(1-σ1,1-σ2)(r+s)≤r1-σ1(r+s)σ1+s1-σ2(r+s)σ2≤2(r+s).Proof.
Letr, s>0 and put t=r/(r+s). Since 0≤t≤1, then we get obviously
(31)2-max(1-σ1,1-σ2)≤t1-σ1+(1-t)1-σ2≤2.Lemma11 provides sharp estimates on some Riesz potential functions.Lemma 11 (see [28, Proposition 3.1]).
Letμ≤(α/2)+1 and let L be a function in 𝒦 such that ∫0η‍t(α/2)-μL(t)dt<∞. Let q be a positive measurable function in D such that, for x∈D,
(32)q(x)≈(δ(x))-μL(δ(x)).
Then, for x∈D, one has
(33)GDαq(x)≈ψ(δ(x)),
where ψ is the function defined on (0,η) by
(34)ψ(t)∶={t(α/2)-1∫0t‍L(s)sds,ifμ=α2+1,tα-μL(t),ifα2<μ<α2+1,tα/2∫tη‍L(s)sds,ifμ=α2,tα/2,ifμ<α2.Lemma 12.
Assume(H). Let u be a continuous function in D such that, for x∈D, u(x)≈θ(δ(x)). Then u is a solution of problem (14) if and only if
(35)u(x)=∫D‍GDα(x,y)[a1(y)uσ1(y)+a2(y)uσ2(y)]dy,u(x)=∫D‍GDα(x,y)a1(y)uσ1(y)+a2(y)uσ2=x∈D.Proof.
Assume(H). First we will give an explicit form of the function θ. We recall that, for i∈{1,2},λi<(α/2)(1+σi)+1-σi and βi:=min(α/2,(α-λi)/(1-σi)). Since β1<β2 is equivalent to (α-λ1)/(1-σ1)<(α-λ2)/(1-σ2) and (α/2)(1+σ1)<λ1, we deduce that, for t∈(0,η), we have
(36)θ(t)={t(α-λ1)/(1-σ1)(L1(t))1/(1-σ1),ifα-λ11-σ1<α-λ21-σ2,α2(1+σ1)<λ1,t(α-λ1)/(1-σ1)L(t),ifα-λ11-σ1=α-λ21-σ2,α2(1+σ1)<λ1,tα/2M(t),ifλ1=α2(1+σ1),λ2=α2(1+σ2),tα/2(∫tη‍L1(s)sds)1/(1-σ1),ifλ1=α2(1+σ1),λ2<α2(1+σ2),tα/2ifλ1<α2(1+σ1),
where
(37)L(t)∶=(L1(t))1/(1-σ1)+(L2(t))1/(1-σ2),M(t)∶=(∫tη‍L1(s)sds)1/(1-σ1)+(∫tη‍L2(s)sds)1/(1-σ2).
Now using the fact that u(x)≈θ(δ(x)), we deduce by simple computation from hypothesis (H), (36), and Proposition 5 that
(38)a1(x)uσ1(x)+a2(x)uσ2(x)≈(δ(x))(α/2)-1h(δ(x)),
where h is defined in (0,η) by(39)h(t)∶={t1-(α/2)-((λ1-ασ1)/(1-σ1))(L1(t))1/(1-σ1),ifα-λ11-σ1<α-λ21-σ2,α2(1+σ1)<λ1,t1-(α/2)-((λ1-ασ1)/(1-σ1))(L1Lσ1+L2Lσ2)(t),ifα-λ11-σ1=α-λ21-σ2,α2(1+σ1)<λ1,t1-α(L1Mσ1+L2Mσ2)(t),ifλ1=α2(1+σ1),λ2=α2(1+σ2),t1-αL1(t)(∫tη‍L1(s)sds)σ1/(1-σ1),ifλ1=α2(1+σ1),λ2<α2(1+σ2),t1+(α/2)(σ1-1)-λ1L1(t)+t1+(α/2)(σ2-1)-λ2L2(t),ifλ1<α2(1+σ1).We point out that for each case, the function h(t) can be written as a sum of terms of the form t-μL~(t), where μ<α. By Proposition 5 and Lemma 7, we have L~∈𝒦. On the other hand, since by Proposition 5, the function x→(δ(x))(α-μ)/2L~(δ(x)) is positive and belongs to C0(D), then there exists c>0 such that for each x∈D(40)0<(δ(x))-μL~(δ(x))≤c(δ(x))-(α+μ)/2.
Hence we deduce from (7) that the function x→h(δ(x)) is in Kα(D).
Now using Proposition9, we obtain that x→(δ(x))1-(α/2)GDα[a1uσ1+a2uσ2](x) is in C0(D). In particular, we have
(41)limx→z∈∂D(δ(x))1-(α/2)GDα(a1uσ1+a2uσ2)(x)=0,(-Δ)α/2GDα(a1uσ1+a2uσ2)=a1(x)uσ1+a2(x)uσ2==========inD(inthesenseofdistributions).
Consequently, it follows by (41) that u is a weak continuous solution of problem (14) if and only if u satisfies
(42){(-Δ)α/2(u-GDα(a1uσ1+a2uσ2))=0inDlimx→z∈∂D(δ(x))1-(α/2)(u-GDα(a1uσ1+a2uσ2))=0.
We deduce by [26, Theorem 6] that u-GDα(a1uσ1+a2uσ2)=0 in D. The proof is complete.Lemma 13.
Fori∈{1,2}, let Li∈𝒦 defined on (0,η] with η>diam(D) and let M be the function given by (37). Then one has, for t∈(0,η),
(43)∫tη(L1Mσ1+L2Mσ2)(s)sds≈M(t).Proof.
The proof can be found in [36].Now we are ready to prove our main results.


### 3.2. Proof of Theorem3

Assume(H). For i∈{1,2}, let Li∈𝒦 defined on (0,η] with η>diam(D) and define the nonnegative functions bi in (0,η) by
(44)bi(t)=(∫tη‍Li(s)sds)1/(1-σi).
Let θ be the function given by (36). To prove Theorem 3, we distinguish the following cases.Case 1.
(
α
-
λ
1
)
/
(
1
-
σ
1
)
<
(
α
-
λ
2
)
/
(
1
-
σ
2
) and  (α/2)(1+σ1)<λ1<(α/2)(1+σ1)+1-σ1.
Sinceθ(t)=t(α-λ1)/(1-σ1)(L1(t))1/(1-σ1), then we have
(45)a1(x)θσ1(δ(x))≈(δ(x))(ασ1-λ1)/(1-σ1)(L1(δ(x)))1/(1-σ1),a2(x)θσ2(δ(x))≈(δ(x))((α-λ1)/(1-σ1))σ2-λ2(L2L1σ2/(1-σ1))(δ(x)).
Using the fact that (ασ1-λ1)/(1-σ1)<((α-λ1)/(1-σ1))σ2-λ2, we deduce by Proposition 5 that
(46)a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))(ασ1-λ1)/(1-σ1)(L1(δ(x)))1/(1-σ1).
Since, for μ=(λ1-ασ1)/(1-σ1)∈(α/2,(α/2)+1), we have ∫0η‍t(α/2)-μ(L1(t))1/(1-σ1)dt<∞, then applying Lemma 11, we deduce that
(47)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈GDα[(δ(·))(ασ1-λ1)/(1-σ1)(L1(δ(·)))1/(1-σ1)](x)≈(δ(x))(α-λ1)/(1-σ1)(L1(δ(x)))1/(1-σ1)=θ(δ(x)).Case 2.
(
α
-
λ
1
)
/
(
1
-
σ
1
)
=
(
α
-
λ
2
)
/
(
1
-
σ
2
) and  (α/2)(1+σ1)<λ1<(α/2)(1+σ1)+1-σ1.
In this caseθ(t)=t(α-λ1)/(1-σ1)L(t). Therefore
(48)a1(x)θσ1(δ(x))≈(δ(x))(ασ1-λ1)/(1-σ1)(L1Lσ1)(δ(x)).
So we obtain by Proposition 5 and Lemma 11 with μ=(λ1-ασ1)/(1-σ1)∈(α/2,(α/2)+1),
(49)GDα[a1θσ1(δ(·))](x)≈(δ(x))(α-λ1)/(1-σ1)(L1Lσ1)(δ(x)).
Similarly, since (α/2)(1+σ2)<λ2<(α/2)(1+σ2)+1-σ2, we obtain
(50)GDα[a2θσ2(δ(·))](x)≈(δ(x))(α-λ2)/(1-σ2)(L2Lσ2)(δ(x))≈(δ(x))(α-λ1)/(1-σ1)(L2Lσ2)(δ(x)).
Hence by using (30), we deduce that
(51)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈(δ(x))(α-λ1)/(1-σ1)(L1Lσ1+L2Lσ2)(δ(x))≈(δ(x))(α-λ1)/(1-σ1)L(δ(x))=θ(δ(x)).Case 3.
Ifλ1=(α/2)(1+σ1) and λ2=(α/2)(1+σ2) and since θ(t)=tα/2M(t), then we have
(52)a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))-α/2(L1Mσ1+L2Mσ2)(δ(x)).
So by Proposition 5, Lemma 11 with μ=α/2, and Lemma 13, we deduce that
(53)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈(δ(x))α/2∫δ(x)η‍(L1Mσ1+L2Mσ2)(s)sds≈(δ(x))α/2M(δ(x))=θ(δ(x)).Case 4.
λ
1
=
(
α
/
2
)
(
1
+
σ
1
) and λ2<(α/2)(1+σ2).
In this caseθ(t)=tα/2b1(t). Since λ2-(ασ2/2)<(α/2), we deduce by Proposition 5 that
(54)a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))-α/2(L1b1σ1)(δ(x))+(δ(x))(ασ2/2)-λ2(L2b1σ2)(δ(x))≈(δ(x))-α/2(L1b1σ1)(δ(x)).
Hence applying Lemma 11 with μ=α/2, we obtain
(55)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈GDα[(δ(·))-α/2(L1b1σ1)(δ(·))](x)≈(δ(x))α/2∫δ(x)η‍(L1b1σ1)(s)sds≈(δ(x))α/2b1(δ(x))=θ(δ(x)).Case 5.
λ
1
<
(
α
/
2
)
(
1
+
σ
1
).
We haveθ(t)=tα/2. So
(56)a1(x)θσ1(δ(x))≈(δ(x))-(λ1-(ασ1/2))L1(δ(x)).
Applying again Lemma 11 with μ=λ1-(ασ1/2)<(α/2), we obtain
(57)GDα[a1θσ1(δ(·))](x)≈(δ(x))α/2.
On the other hand, since (α/2)<(α-λ1)/(1-σ1)≤(α-λ2)/(1-σ2), then λ1<(α/2)(1+σ1) and therefore
(58)GDα[a2θσ2(δ(·))](x)≈(δ(x))α/2.
Hence
(59)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈(δ(x))α/2=θ(δ(x)).
The proof is complete.


### 3.3. Proof of Theorem4

Letσ1, σ2∈(-1,1), assume (H), and consider v∶=GαD[a1θσ1(δ(·))+a2θσ2(δ(·))]. Using Theorem 3, there exists m>1 such that
(60)1mv(x)≤θ(δ(x))≤mv(x).
Put σ:=max(|σ1|,|σ2|), c:=mσ/(1-σ) and consider the set
(61)Γ∶={ω∈C0(D):1c(δ(x))1-(α/2)v(x)≤ω(x)≤c(δ(x))1-(α/2)v(x),x∈D1c}.
Let h be the function given by (39). Since a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))(α/2)-1h(δ(x)) and the function x→h(δ(x)) is in Kα(D), it follows by Proposition 9 that x→(δ(x))1-(α/2)v(x) is in C0(D). So Γ is a nonempty, closed, bounded, and convex set in C0(D). Define the operator T on Γ by
(62)Tω(x)∶=(δ(x))1-(α/2)GαD(((δ(·))(α/2)-1ω)σ1a1===================+((δ(·))(α/2)-1ω)σ2a2)(x).
We will prove that T has a fixed point. Since there exists a constant c>0 such that for all ω∈Γ we have
(63)|a1(x)((δ(x))(α/2)-1ω(x))σ1+a2(x)((δ(x))(α/2)-1ω(x))σ2|≤c(δ(x))(α/2)-1h(δ(x)),
where the function x→h(δ(x)) is in Kα(D), it follows that T(Γ)⊂Λh(δ(·)), where Λh(δ(·)) is given by (29). Therefore by Proposition 9, the family of functions {x→Tω(x),ω∈Γ} is relatively compact in C0(D).Next, we will prove thatT maps Γ into itself.Indeed, by using (60) we have for all ω∈Γ(64)GαD(a1((δ(·))(α/2)-1ω)σ1+a2((δ(·))(α/2)-1ω)σ2)≤GαD(a1cσmσθσ1(δ(·))+a2cσmσθσ2(δ(·)))=cv.
On the other hand, we have
(65)GαD(a1((δ(·))(α/2)-1ω)σ1+a2((δ(·))(α/2)-1ω)σ2)≥GαD(a1c-σm-σθσ1(δ(·))+a2c-σm-σθσ2(δ(·)))=1cv.
This implies that T(Γ)⊂Γ.Now, we will prove the continuity of the operatorT in Γ in the supremum norm. Let (ωk)k∈ℕ be a sequence in Γ which converges uniformly to a function ω in Γ. Then, for each x∈D, we have
(66)|Tωk(x)-Tω(x)|≤(δ(x))1-(α/2)GαD(a1(δ(.))((α/2)-1)σ1|ωkσ1-ωσ1|================+a2(δ(·))((α/2)-1)σ2|ωkσ2-ωσ2|)(x).
On the other hand, by similar arguments to the previous ones, we have
(67)a1(x)(δ(x))((α/2)-1)σ1|ωkσ1-ωσ1|(x)+a2(x)(δ(x))((α/2)-1)σ2|ωkσ2-ωσ2|(x)≤c(δ(x))(α/2)-1h(δ(x)).
We conclude by Proposition 9 and the dominated convergence theorem that, for all x∈D,
(68)Tωk(x)⟶Tω(x)ask⟶+∞.
Consequently, as T(Γ) is relatively compact in C0(D), we deduce that the pointwise convergence implies the uniform convergence; namely,
(69)∥Tωk-Tω∥∞⟶0ask⟶+∞.
Therefore, T is a compact operator from Γ into itself. So the Schauder fixed-point theorem implies the existence of ω∈Γ such that
(70)ω(x)=(δ(x))1-(α/2)GαD(a1((δ(·))(α/2)-1ω)σ1+a2((δ(·))(α/2)-1ω)σ2)(x).
Put u(x)=(δ(x))(α/2)-1ω(x). Then u is continuous and satisfies
(71)u(x)=GαD(a1uσ1+a2uσ2)(x).
Hence by Lemma 12 and Theorem 3, u is a required solution.Next, we aim at proving the uniqueness in the cone(72)S:={u∈C(D):u(x)≈θ(δ(x))}.
Let u and v be two solutions of (14) in S. Then there exists a constant m>1 such that
(73)1m≤uv≤m.
This implies that the set
(74)J={t∈(1,∞),1tv≤u≤tv}
is not empty. Let c0:=infJ and put w=v-c0-σu with σ=max(|σ1|,|σ2|).We claim thatc0=1. Indeed, assume that c0>1; then by using Lemma 12, we deduce that
(75)w=GαD(a1(vσ1-c0-σuσ1)+a2(vσ2-c0-σuσ2))≥0inD,
which implies that
(76)v≥c0-σu.
By symmetry, we deduce that
(77)v≤c0σu.
So c0σ∈J. Since σ<1, then we have c0σ<c0. This is a contradiction to the fact that c0:=infJ. Hence c0=1 and so u=v. This completes the proof.Example 14.
Letσ1∈(-1,0),letσ2∈(0,1), and put d=diam(D). For i∈{1,2}, let ai∈Clocγ(D),0<γ<1, satisfying
(78)a1(x)≈(δ(x))-λ1(log(3dδ(x)))-1,a2(x)≈(δ(x))-λ2,
whereλi<(α/2)(1+σi)+1-σi, such that  (α-λ1)/(1-σ1)≤(α-λ2)/(1-σ2). Then using Theorem 4, problem (14) has a unique positive continuous solution u in D satisfying the following estimates:(i)
if(α-λ1)/(1-σ1)<(α-λ2)/(1-σ2) and (α/2)(1+σ1)<λ1, then, for x∈D,
(79)u(x)≈(δ(x))(α-λ1)/(1-σ1)(log(3dδ(x)))-1/(1-σ1);(ii)
if(α-λ1)/(1-σ1)=(α-λ2)/(1-σ2) and (α/2)(1+σ1)<λ1, then, for x∈D,
(80)u(x)≈(δ(x))(α-λ1)/(1-σ1);(iii)
ifλ1=(α/2)(1+σ1) and λ2=(α/2)(1+σ2), then, for x∈D,
(81)u(x)≈(δ(x))α/2(log(3dδ(x)))1/(1-σ2);(iv)
ifλ1=(α/2)(1+σ1) and λ2<(α/2)(1+σ2), then, for x∈D,
(82)u(x)≈(δ(x))α/2(log∘log(3dδ(x)))1/(1-σ1);(v)
ifλ1<(α/2)(1+σ1), then, for x∈D,
(83)u(x)≈(δ(x))(α/2).


## 3.1. Technical Lemmas

Lemma 10.
Forr, s>0, one has
(30)2-max(1-σ1,1-σ2)(r+s)≤r1-σ1(r+s)σ1+s1-σ2(r+s)σ2≤2(r+s).Proof.
Letr, s>0 and put t=r/(r+s). Since 0≤t≤1, then we get obviously
(31)2-max(1-σ1,1-σ2)≤t1-σ1+(1-t)1-σ2≤2.Lemma11 provides sharp estimates on some Riesz potential functions.Lemma 11 (see [28, Proposition 3.1]).
Letμ≤(α/2)+1 and let L be a function in 𝒦 such that ∫0η‍t(α/2)-μL(t)dt<∞. Let q be a positive measurable function in D such that, for x∈D,
(32)q(x)≈(δ(x))-μL(δ(x)).
Then, for x∈D, one has
(33)GDαq(x)≈ψ(δ(x)),
where ψ is the function defined on (0,η) by
(34)ψ(t)∶={t(α/2)-1∫0t‍L(s)sds,ifμ=α2+1,tα-μL(t),ifα2<μ<α2+1,tα/2∫tη‍L(s)sds,ifμ=α2,tα/2,ifμ<α2.Lemma 12.
Assume(H). Let u be a continuous function in D such that, for x∈D, u(x)≈θ(δ(x)). Then u is a solution of problem (14) if and only if
(35)u(x)=∫D‍GDα(x,y)[a1(y)uσ1(y)+a2(y)uσ2(y)]dy,u(x)=∫D‍GDα(x,y)a1(y)uσ1(y)+a2(y)uσ2=x∈D.Proof.
Assume(H). First we will give an explicit form of the function θ. We recall that, for i∈{1,2},λi<(α/2)(1+σi)+1-σi and βi:=min(α/2,(α-λi)/(1-σi)). Since β1<β2 is equivalent to (α-λ1)/(1-σ1)<(α-λ2)/(1-σ2) and (α/2)(1+σ1)<λ1, we deduce that, for t∈(0,η), we have
(36)θ(t)={t(α-λ1)/(1-σ1)(L1(t))1/(1-σ1),ifα-λ11-σ1<α-λ21-σ2,α2(1+σ1)<λ1,t(α-λ1)/(1-σ1)L(t),ifα-λ11-σ1=α-λ21-σ2,α2(1+σ1)<λ1,tα/2M(t),ifλ1=α2(1+σ1),λ2=α2(1+σ2),tα/2(∫tη‍L1(s)sds)1/(1-σ1),ifλ1=α2(1+σ1),λ2<α2(1+σ2),tα/2ifλ1<α2(1+σ1),
where
(37)L(t)∶=(L1(t))1/(1-σ1)+(L2(t))1/(1-σ2),M(t)∶=(∫tη‍L1(s)sds)1/(1-σ1)+(∫tη‍L2(s)sds)1/(1-σ2).
Now using the fact that u(x)≈θ(δ(x)), we deduce by simple computation from hypothesis (H), (36), and Proposition 5 that
(38)a1(x)uσ1(x)+a2(x)uσ2(x)≈(δ(x))(α/2)-1h(δ(x)),
where h is defined in (0,η) by(39)h(t)∶={t1-(α/2)-((λ1-ασ1)/(1-σ1))(L1(t))1/(1-σ1),ifα-λ11-σ1<α-λ21-σ2,α2(1+σ1)<λ1,t1-(α/2)-((λ1-ασ1)/(1-σ1))(L1Lσ1+L2Lσ2)(t),ifα-λ11-σ1=α-λ21-σ2,α2(1+σ1)<λ1,t1-α(L1Mσ1+L2Mσ2)(t),ifλ1=α2(1+σ1),λ2=α2(1+σ2),t1-αL1(t)(∫tη‍L1(s)sds)σ1/(1-σ1),ifλ1=α2(1+σ1),λ2<α2(1+σ2),t1+(α/2)(σ1-1)-λ1L1(t)+t1+(α/2)(σ2-1)-λ2L2(t),ifλ1<α2(1+σ1).We point out that for each case, the function h(t) can be written as a sum of terms of the form t-μL~(t), where μ<α. By Proposition 5 and Lemma 7, we have L~∈𝒦. On the other hand, since by Proposition 5, the function x→(δ(x))(α-μ)/2L~(δ(x)) is positive and belongs to C0(D), then there exists c>0 such that for each x∈D(40)0<(δ(x))-μL~(δ(x))≤c(δ(x))-(α+μ)/2.
Hence we deduce from (7) that the function x→h(δ(x)) is in Kα(D).
Now using Proposition9, we obtain that x→(δ(x))1-(α/2)GDα[a1uσ1+a2uσ2](x) is in C0(D). In particular, we have
(41)limx→z∈∂D(δ(x))1-(α/2)GDα(a1uσ1+a2uσ2)(x)=0,(-Δ)α/2GDα(a1uσ1+a2uσ2)=a1(x)uσ1+a2(x)uσ2==========inD(inthesenseofdistributions).
Consequently, it follows by (41) that u is a weak continuous solution of problem (14) if and only if u satisfies
(42){(-Δ)α/2(u-GDα(a1uσ1+a2uσ2))=0inDlimx→z∈∂D(δ(x))1-(α/2)(u-GDα(a1uσ1+a2uσ2))=0.
We deduce by [26, Theorem 6] that u-GDα(a1uσ1+a2uσ2)=0 in D. The proof is complete.Lemma 13.
Fori∈{1,2}, let Li∈𝒦 defined on (0,η] with η>diam(D) and let M be the function given by (37). Then one has, for t∈(0,η),
(43)∫tη(L1Mσ1+L2Mσ2)(s)sds≈M(t).Proof.
The proof can be found in [36].Now we are ready to prove our main results.


## 3.2. Proof of Theorem3

Assume(H). For i∈{1,2}, let Li∈𝒦 defined on (0,η] with η>diam(D) and define the nonnegative functions bi in (0,η) by
(44)bi(t)=(∫tη‍Li(s)sds)1/(1-σi).
Let θ be the function given by (36). To prove Theorem 3, we distinguish the following cases.Case 1.
(
α
-
λ
1
)
/
(
1
-
σ
1
)
<
(
α
-
λ
2
)
/
(
1
-
σ
2
) and  (α/2)(1+σ1)<λ1<(α/2)(1+σ1)+1-σ1.
Sinceθ(t)=t(α-λ1)/(1-σ1)(L1(t))1/(1-σ1), then we have
(45)a1(x)θσ1(δ(x))≈(δ(x))(ασ1-λ1)/(1-σ1)(L1(δ(x)))1/(1-σ1),a2(x)θσ2(δ(x))≈(δ(x))((α-λ1)/(1-σ1))σ2-λ2(L2L1σ2/(1-σ1))(δ(x)).
Using the fact that (ασ1-λ1)/(1-σ1)<((α-λ1)/(1-σ1))σ2-λ2, we deduce by Proposition 5 that
(46)a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))(ασ1-λ1)/(1-σ1)(L1(δ(x)))1/(1-σ1).
Since, for μ=(λ1-ασ1)/(1-σ1)∈(α/2,(α/2)+1), we have ∫0η‍t(α/2)-μ(L1(t))1/(1-σ1)dt<∞, then applying Lemma 11, we deduce that
(47)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈GDα[(δ(·))(ασ1-λ1)/(1-σ1)(L1(δ(·)))1/(1-σ1)](x)≈(δ(x))(α-λ1)/(1-σ1)(L1(δ(x)))1/(1-σ1)=θ(δ(x)).Case 2.
(
α
-
λ
1
)
/
(
1
-
σ
1
)
=
(
α
-
λ
2
)
/
(
1
-
σ
2
) and  (α/2)(1+σ1)<λ1<(α/2)(1+σ1)+1-σ1.
In this caseθ(t)=t(α-λ1)/(1-σ1)L(t). Therefore
(48)a1(x)θσ1(δ(x))≈(δ(x))(ασ1-λ1)/(1-σ1)(L1Lσ1)(δ(x)).
So we obtain by Proposition 5 and Lemma 11 with μ=(λ1-ασ1)/(1-σ1)∈(α/2,(α/2)+1),
(49)GDα[a1θσ1(δ(·))](x)≈(δ(x))(α-λ1)/(1-σ1)(L1Lσ1)(δ(x)).
Similarly, since (α/2)(1+σ2)<λ2<(α/2)(1+σ2)+1-σ2, we obtain
(50)GDα[a2θσ2(δ(·))](x)≈(δ(x))(α-λ2)/(1-σ2)(L2Lσ2)(δ(x))≈(δ(x))(α-λ1)/(1-σ1)(L2Lσ2)(δ(x)).
Hence by using (30), we deduce that
(51)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈(δ(x))(α-λ1)/(1-σ1)(L1Lσ1+L2Lσ2)(δ(x))≈(δ(x))(α-λ1)/(1-σ1)L(δ(x))=θ(δ(x)).Case 3.
Ifλ1=(α/2)(1+σ1) and λ2=(α/2)(1+σ2) and since θ(t)=tα/2M(t), then we have
(52)a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))-α/2(L1Mσ1+L2Mσ2)(δ(x)).
So by Proposition 5, Lemma 11 with μ=α/2, and Lemma 13, we deduce that
(53)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈(δ(x))α/2∫δ(x)η‍(L1Mσ1+L2Mσ2)(s)sds≈(δ(x))α/2M(δ(x))=θ(δ(x)).Case 4.
λ
1
=
(
α
/
2
)
(
1
+
σ
1
) and λ2<(α/2)(1+σ2).
In this caseθ(t)=tα/2b1(t). Since λ2-(ασ2/2)<(α/2), we deduce by Proposition 5 that
(54)a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))-α/2(L1b1σ1)(δ(x))+(δ(x))(ασ2/2)-λ2(L2b1σ2)(δ(x))≈(δ(x))-α/2(L1b1σ1)(δ(x)).
Hence applying Lemma 11 with μ=α/2, we obtain
(55)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈GDα[(δ(·))-α/2(L1b1σ1)(δ(·))](x)≈(δ(x))α/2∫δ(x)η‍(L1b1σ1)(s)sds≈(δ(x))α/2b1(δ(x))=θ(δ(x)).Case 5.
λ
1
<
(
α
/
2
)
(
1
+
σ
1
).
We haveθ(t)=tα/2. So
(56)a1(x)θσ1(δ(x))≈(δ(x))-(λ1-(ασ1/2))L1(δ(x)).
Applying again Lemma 11 with μ=λ1-(ασ1/2)<(α/2), we obtain
(57)GDα[a1θσ1(δ(·))](x)≈(δ(x))α/2.
On the other hand, since (α/2)<(α-λ1)/(1-σ1)≤(α-λ2)/(1-σ2), then λ1<(α/2)(1+σ1) and therefore
(58)GDα[a2θσ2(δ(·))](x)≈(δ(x))α/2.
Hence
(59)GDα[a1θσ1(δ(·))+a2θσ2(δ(·))](x)≈(δ(x))α/2=θ(δ(x)).
The proof is complete.


## 3.3. Proof of Theorem4

Letσ1, σ2∈(-1,1), assume (H), and consider v∶=GαD[a1θσ1(δ(·))+a2θσ2(δ(·))]. Using Theorem 3, there exists m>1 such that
(60)1mv(x)≤θ(δ(x))≤mv(x).
Put σ:=max(|σ1|,|σ2|), c:=mσ/(1-σ) and consider the set
(61)Γ∶={ω∈C0(D):1c(δ(x))1-(α/2)v(x)≤ω(x)≤c(δ(x))1-(α/2)v(x),x∈D1c}.
Let h be the function given by (39). Since a1(x)θσ1(δ(x))+a2(x)θσ2(δ(x))≈(δ(x))(α/2)-1h(δ(x)) and the function x→h(δ(x)) is in Kα(D), it follows by Proposition 9 that x→(δ(x))1-(α/2)v(x) is in C0(D). So Γ is a nonempty, closed, bounded, and convex set in C0(D). Define the operator T on Γ by
(62)Tω(x)∶=(δ(x))1-(α/2)GαD(((δ(·))(α/2)-1ω)σ1a1===================+((δ(·))(α/2)-1ω)σ2a2)(x).
We will prove that T has a fixed point. Since there exists a constant c>0 such that for all ω∈Γ we have
(63)|a1(x)((δ(x))(α/2)-1ω(x))σ1+a2(x)((δ(x))(α/2)-1ω(x))σ2|≤c(δ(x))(α/2)-1h(δ(x)),
where the function x→h(δ(x)) is in Kα(D), it follows that T(Γ)⊂Λh(δ(·)), where Λh(δ(·)) is given by (29). Therefore by Proposition 9, the family of functions {x→Tω(x),ω∈Γ} is relatively compact in C0(D).Next, we will prove thatT maps Γ into itself.Indeed, by using (60) we have for all ω∈Γ(64)GαD(a1((δ(·))(α/2)-1ω)σ1+a2((δ(·))(α/2)-1ω)σ2)≤GαD(a1cσmσθσ1(δ(·))+a2cσmσθσ2(δ(·)))=cv.
On the other hand, we have
(65)GαD(a1((δ(·))(α/2)-1ω)σ1+a2((δ(·))(α/2)-1ω)σ2)≥GαD(a1c-σm-σθσ1(δ(·))+a2c-σm-σθσ2(δ(·)))=1cv.
This implies that T(Γ)⊂Γ.Now, we will prove the continuity of the operatorT in Γ in the supremum norm. Let (ωk)k∈ℕ be a sequence in Γ which converges uniformly to a function ω in Γ. Then, for each x∈D, we have
(66)|Tωk(x)-Tω(x)|≤(δ(x))1-(α/2)GαD(a1(δ(.))((α/2)-1)σ1|ωkσ1-ωσ1|================+a2(δ(·))((α/2)-1)σ2|ωkσ2-ωσ2|)(x).
On the other hand, by similar arguments to the previous ones, we have
(67)a1(x)(δ(x))((α/2)-1)σ1|ωkσ1-ωσ1|(x)+a2(x)(δ(x))((α/2)-1)σ2|ωkσ2-ωσ2|(x)≤c(δ(x))(α/2)-1h(δ(x)).
We conclude by Proposition 9 and the dominated convergence theorem that, for all x∈D,
(68)Tωk(x)⟶Tω(x)ask⟶+∞.
Consequently, as T(Γ) is relatively compact in C0(D), we deduce that the pointwise convergence implies the uniform convergence; namely,
(69)∥Tωk-Tω∥∞⟶0ask⟶+∞.
Therefore, T is a compact operator from Γ into itself. So the Schauder fixed-point theorem implies the existence of ω∈Γ such that
(70)ω(x)=(δ(x))1-(α/2)GαD(a1((δ(·))(α/2)-1ω)σ1+a2((δ(·))(α/2)-1ω)σ2)(x).
Put u(x)=(δ(x))(α/2)-1ω(x). Then u is continuous and satisfies
(71)u(x)=GαD(a1uσ1+a2uσ2)(x).
Hence by Lemma 12 and Theorem 3, u is a required solution.Next, we aim at proving the uniqueness in the cone(72)S:={u∈C(D):u(x)≈θ(δ(x))}.
Let u and v be two solutions of (14) in S. Then there exists a constant m>1 such that
(73)1m≤uv≤m.
This implies that the set
(74)J={t∈(1,∞),1tv≤u≤tv}
is not empty. Let c0:=infJ and put w=v-c0-σu with σ=max(|σ1|,|σ2|).We claim thatc0=1. Indeed, assume that c0>1; then by using Lemma 12, we deduce that
(75)w=GαD(a1(vσ1-c0-σuσ1)+a2(vσ2-c0-σuσ2))≥0inD,
which implies that
(76)v≥c0-σu.
By symmetry, we deduce that
(77)v≤c0σu.
So c0σ∈J. Since σ<1, then we have c0σ<c0. This is a contradiction to the fact that c0:=infJ. Hence c0=1 and so u=v. This completes the proof.Example 14.
Letσ1∈(-1,0),letσ2∈(0,1), and put d=diam(D). For i∈{1,2}, let ai∈Clocγ(D),0<γ<1, satisfying
(78)a1(x)≈(δ(x))-λ1(log(3dδ(x)))-1,a2(x)≈(δ(x))-λ2,
whereλi<(α/2)(1+σi)+1-σi, such that  (α-λ1)/(1-σ1)≤(α-λ2)/(1-σ2). Then using Theorem 4, problem (14) has a unique positive continuous solution u in D satisfying the following estimates:(i)
if(α-λ1)/(1-σ1)<(α-λ2)/(1-σ2) and (α/2)(1+σ1)<λ1, then, for x∈D,
(79)u(x)≈(δ(x))(α-λ1)/(1-σ1)(log(3dδ(x)))-1/(1-σ1);(ii)
if(α-λ1)/(1-σ1)=(α-λ2)/(1-σ2) and (α/2)(1+σ1)<λ1, then, for x∈D,
(80)u(x)≈(δ(x))(α-λ1)/(1-σ1);(iii)
ifλ1=(α/2)(1+σ1) and λ2=(α/2)(1+σ2), then, for x∈D,
(81)u(x)≈(δ(x))α/2(log(3dδ(x)))1/(1-σ2);(iv)
ifλ1=(α/2)(1+σ1) and λ2<(α/2)(1+σ2), then, for x∈D,
(82)u(x)≈(δ(x))α/2(log∘log(3dδ(x)))1/(1-σ1);(v)
ifλ1<(α/2)(1+σ1), then, for x∈D,
(83)u(x)≈(δ(x))(α/2).



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*Source: 295480-2013-08-19.xml*