# Combined Effects in Singular Elliptic Problems in Punctured Domain

Authors: Imed Bachar; Habib Mâagli; Hassan Eltayeb
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The paper deals with nonlinear elliptic differential equations subject to some boundary value conditions in a regular bounded punctured domain. By means of properties of slowly regularly varying functions at zero and the Schauder fixed-point theorem, we establish the existence of a positive continuous solution for the suggested problem. Global estimates on such solution, which could blow up at the origin, are also obtained.


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## Body

## 1. Introduction and Statement of Main Results

Consider the problem(1)−Δvz=p1zvα1+p2zvα2inΩ\0,v>0inΩ\0,limz⟶0zN−2vz=0,v=0on∂Ω,where Ω be a bounded C1,1-domain in ℝNN≥3 containing 0,α1,α2<1 and p1,p2∈C+Ω\0 satisfying some adequate conditions. In this paper, we are interested in the study of existence and global asymptotic behavior of positive solutions for problem (1). In particular, as it will be seen, the solution may blow up at the origin. The main feature of this paper consists in the presence of the combined effects of singular and sublinear terms in the nonlinearity. Our approach combines properties of Karamata class and Kato class (see [1–4]) with the Schauder fixed-point theorem.In literature, many researches have studied similar problems for both bounded and unbounded domains (see, for example, [5–13] and the references therein).In [12], Shi and Yao studied the problem
(2)−Δv=λvr+bzv−αinD,v>0inD,v=0on∂D,where D⊂ℝNN≥2 is a bounded domain in ℝNN≥2,r,α∈0,1, and λ is a real parameter. The function b is required to be in C2,γD¯,γ∈0,1, with bz≠0 in D¯. Under various assumptions on bz, they have proved some existence results for λ belonging to a certain range.Ifb≡1, Colcite and Palmieri [14] showed that (2) has at least one solution provided that λ≥0 and r∈0,1. However, if r≥1, they have also proved that there exists λ∗>0 such that (1.2) has a solution for λ∈0,λ∗, and no solutions exist if λ>λ∗.In [11], by combining monotonicity arguments with variational techniques, Ra˘dulescu and Repovš studied the competition between convex and concave nonlinearity and variable potentials. They have proved that problem (2), with −1<α<0 and r>1, has a solution, provided that λ>0 is small enough.In [7], Cîrstea et al. investigated the following bifurcation problem:
(3)−Δv=λfv+bzgvinD,v>0inD,v=0on∂D,where D⊂ℝNN≥2 is a bounded domain, λ∈ℝ,b is a nonnegative Hölder function, and f is positive, continuous, and nondecreasing function on 0,∞ such that the mapping fs/s is nonincreasing in 0,∞. The nonnegative, continuous nonlinearity g is assumed to fulfill the hypotheses:g is nonincreasing on 0,∞ with lims⟶0gs=+∞.There existC0,η0>0 and α∈0,1 so that gs≤C0s−α,∀s∈0,η0.They have proved the existence of a unique solution to this problem.Ifλ=0, problem (3) becomes
(4)−Δv=bzgvinD,v>0inD,v=0on∂D.Many physical phenomena can be described by this kind of problems, known as the Lane–Emden–Fowler equation, see, for example, [7–9, 11, 15] and the references therein.The study of such problems has been attracted by many researches (see, for instance, [2, 16–25] and the references therein).In [18], Ghergu et al. considered the semilinear elliptic equation
(5)−Δv=vαlogvβinB\0,where B is the the unit open ball in ℝN, with N≥3,N/N−2<α<N+2/N−2, and −∞<β<∞. They have proved that the nonnegative solution v∈C2B\0 of the above equation either has a removable singularity at the origin or it behaves like
(6)vz=A1+o1z−2α−1log1z−βα−1asz⟶0,with A=2/α−11−βN−2−2/α−11/α−1.In [25], Zhao et al. investigated the problem
(7)Δv+12vΔv2−b−vr−2v+b+vs−2v=0inD,v=0on∂D,where D⊂ℝN,N≥3, is a bounded smooth domain, r>4,4<s<4N/N−2, and b±∈C+D¯.Using variational techniques, they have proved the existence of infinitely many solutions.Quoirin and Umezu [20] dealt with the following concave-convex problems under Neumann boundary conditions
(8)−Δv=vr−2vinD,∂v∂n=λbzvs−2von∂D,where D is a bounded and smooth domain of ℝNN≥2,1<s<2<r,λ>0, and r∈C1+α∂D for some α∈0,1.They have proved that∫∂Db<0 is a necessary and sufficient condition for the existence of nontrivial nonnegative solutions of this problem.In our analysis, we shall extensively use the class of slowly regularly varying functions at zero introduced by Karamata in [26] as follows.Definition 1.
A measurable functionM:0,η⟶0,∞, η>0, is said to be slowly varying at zero if Ms is represented in the form
(9)Ms≔cexp∫sηwrrdr,where c>0 and w∈C0,η with w0=0.The set of slowly varying at zero (called also Karamata class) is denoted byK. It is clear that M belongs to the class K if and only if M is a positive function in C10,η, for some η>0, with
(10)lims⟶0+sM′sMs=0.Typical examples of slowly varying functions at zero, also used as weight functions (see, [27, 28]), are
(11)Πj=1nlogjρsζj,where ρ is a positive real number, ζk∈ℝ and logjs=log∘log∘⋯∘logs (j-times). For more examples, we refer the reader to [1, 3, 4].The Karamata class has been frequently used in describing the asymptotic analysis of solutions (see, for examples, [29–33] and the subsequent papers [5, 6, 15, 34–44]).To simplify our statements in this paper, we need some notations.(i)
d≔diamΩ and δΩz=dz,∂Ω denote the Euclidean distance from z to ∂Ω(ii)
η denotes a positive real number such that η>d(iii)
Forα<1,ν≤N+2−Nα,λ≤2, and L,L∈K defined on 0,η, such that(12)∫0ηrN+2−Nα−ν−1Lrdr<∞and∫0ηr1−λLrdr<∞define the functions ΦL,ν,αs and ΨL,λ,αs, for s∈0,η by
(13)ΦL,ν,αs=1,ifν<2,∫sηLrrdr11−α,ifν=2,Ls11−α,if2<ν<N+2−Nα,∫0sLrrdr11−α,ifν=N+2−Nα,ΨL,λ,αs=1,ifλ<1+α,∫sηLrrdr11−α,ifλ=1+α,Ls11−α,if1+α<λ<2,∫0sLrrdr11−α,ifλ=2.(iv)
BΩ (resp., B+Ω) denotes the collection of all (resp., nonnegative) Borel measurable functions in Ω(v)
Forf,g∈B+Ω, we say f≈g in Ω, if there exists c>0 such that 1/cfz≤gz≤cfz, for all z∈Ω(14)C0Ω¯=v∈CΩ¯:limz⟶ξ∈∂Ωvz=0(vi)
GΩ denotes Green’s function of the Laplace operator in Ω with Dirichlet conditions(vii)
Forf∈B+Ω, we set(15)Vfz=∫ΩGΩz,yfydy(viii)
From ([45], Lemma 9), we know that for any function f∈B+Ω such that f∈Lloc1Ω and Vf∈Lloc1Ω, we have(16)−ΔVf=f,inΩinthedistributionalsense(ix)
The letterc will denote a positive constant which may vary from line to lineIn [46], the authors investigated the problem
(17)−Δvz=bzvαz,z∈Ω\0,v>0,inΩ\0,limz⟶0zN−2vz=0,vz=0,z∈∂Ω,where α<1 and b are a positive continuous function in Ω\0 satisfying
(18)bz≈z−νLzδΩz−λLδΩz,forz∈Ω\0,where α<1,ν≤N+2−Nα,λ≤2, and L,L∈K defined on 0,η such that (12) is fulfilled.They have proved that problem (17) has at least one positive continuous solution v on Ω¯\0 satisfying
(19)vz≈zmin0,2−ν1−αΦL,ν,αzδΩzmin1,2−λ1−αΨL,λ,αδΩz.In this paper, we aim at generalizing the results obtained in [46] to problem (1).To this end, we make the following hypothesis:H For i∈1,2,pi is a positive continuous functions in Ω\0 satisfying
(20)piz≈z−νiLizδΩz−λiLiδΩz,where αi<1,νi≤N+2−Nαi,λi≤2, and Li,Li∈K defined on 0,η such that
(21)∫0ηrN+2−Nαi−νi−1Lirdr<∞and∫0ηr1−λiLirdr<∞.We may assume that(22)2−ν11−α1≤2−ν21−α2and2−λ11−α1≤2−λ21−α2.Fori∈1,2, we denote by
(23)σi≔min0,2−νi1−αiandβi≔min1,2−λi1−αi.Define the functionθ on Ω¯\0 as follows.Ifσ1<σ2 and β1<β2, then
(24)θz=zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1δΩz.Ifσ1<σ2 and β1=β2, then
(25)θz=zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.Ifσ1=σ2 and β1<β2, then
(26)θz=zσ1ΦL1,ν1,α1z+ΦL2,ν2,α2zδΩzβ1ΨL1,λ1,α1δΩz.Ifσ1=σ2 and β1=β2, then
(27)θz=zσ1ΦL1,ν1,α1z+ΦL2,ν2,α2z×δΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.By using properties of slowly regularly varying functions at zero, we prove the following results.Theorem 2.
Under hypothesisH, we have for z∈Ω¯\0,
(28)Vpz≈θz,where pz≔p1zθα1z+p2zθα2z.Theorem 3.
Assume that hypothesisH is satisfied. Then, problem (1) admits a solution v on C+Ω¯\0 satisfying
(29)vz≈θz.Remark 4.
(i)
Theorem3 generalizes the main result in [46].(ii)
Forν1>2, we have limz⟶0vz=∞(iii)
limz⟶0zN−2θz=0, where θ is the function given by (24), (25), (26), and (27).The paper is organized as follows. In Section2, we recall some fundamental properties of functions belonging to the Karamata class, and we prove Theorem 2. In Section 3, we prove Theorem 3 by means of the Schauder fixed-point theorem.


## 2. Karamata Class and Proof of Theorem2

### 2.1. Basic Properties of Karamata Class

Lemma 5 (See [1, 3, 4]).
IfM1,M2∈K, then
(i)
For everyc≥0 and r∈ℝ,(30)M1+cM2,M1M2,andM1rarealsoinK.(ii)
For everyε>0,(31)lims⟶0+sεM1s=0andlims⟶0+s−εM1s=∞.Lemma 6 (Asymptotic behavior, see [1, 3, 4]).
IfM∈K, then
(i)
For everyγ>−1,∫0ηrγMrdr converges and(32)∫0srγMrdr~s⟶0+s1+γMs1+γ.(ii)
For everyγ<−1,∫0ηrγMrdr diverges and(33)∫sηrγMrdr~s⟶0+−s1+γMs1+γ.Lemma 7 (See [4, 6]).
LetM∈K, then
(i)
s↦∫sηMr/rdr belongs to K and(34)lims⟶0+Ms∫sηMr/rdr=0.(ii)
If∫0ηMr/rdr<∞, then s↦∫0sMr/rdr belongs to K and(35)lims⟶0+Ms∫0sMr/rdr=0.


### 2.2. Proof of Theorem2

The next Proposition, which follows from ([46], Proposition 2.12), will be used.Proposition 8.
Let≤N,ζ≤2, and M1, M2∈K such that
(36)∫0ηrN−γ−1M1rdr<∞and∫0ηr1−ζM2rdr<∞.
Then forz∈Ω\0,(37)Vhz≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζΨM2,ζ,0δΩz,where hz≔z−γM1zδΩz−ζM2δΩz.We recall that fori∈1,2,pi∈C+Ω\0 satisfying
(38)piz≈z−νiLizδΩz−λiLiδΩz,where αi<1,νi≤N+2−Nαi,λi≤2, and Li,Li∈K defined on 0,η such that (21) is fulfilled.We aim at proving that onΩ\0,(39)Vpz≈θz,where pz≔p1zθα1z+p2zθα2z and the function θ given by (24), (25), (26), and (27).Throughout the proof, we will apply Lemma5 and Lemma 7 to verify that some functions are in K.We distinguish the following cases:Case 9.
Ifσ1<σ2 and β1<β2, then for z∈Ω\0,(40)θz=zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1δΩz.
Therefore,(41)pz≈z−ν1+σ1α1L1ΦL1,ν1,α1α1zδΩz−λ1+β1α1L1ΨL1,λ1,α1α1δΩz+z−ν2+σ1α2L2ΦL1,ν1,α1α2zδΩz−λ2+β1α2L2ΨL1,λ1,α1α2δΩz.
Sinceσ1<σ2 and β1<β2, we deduce by Lemma 5 that
(42)pz≈z−ν1+σ1α1L1ΦL1,ν1,α1α1zδΩz−λ1+β1α1L1ΨL1,λ1,α1α1δΩz.
Using Lemma6 (i), (21), and Proposition 8 with γ=ν1−σ1α1,ζ=λ1−β1α1,M1=L1ΦL1,ν1,α1α1, and M2=L1ΨL1,λ1,α1α1, we obtain for z∈Ω\0,(43)Vpz≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζΨM2,ζ,0δΩz.
Sincemin0,2−γ=σ1 and min1,2−ζ=β1, we deduce that
(44)Vpz≈zσ1ΦM1,γ,0zδΩzβ1ΨM2,ζ,0δΩz.
Now, by using a simple computation, we obtain forz∈Ω\0,(45)ΦM1,γ,0z≈ΦL1,ν1,α1zandΨM2,ζ,0δΩz≈ΨL1,λ1,α1δΩz.
Combining this fact with (44), we obtain for z∈Ω\0,(46)Vpz≈θz.Case 10.
Ifσ1<σ2 and β1=β2, then for z∈Ω\0,(47)θz=zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
In this case,(48)pz≈z−ν1+σ1α1L1ΦL1,λ1,α1α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1δΩz+z−ν2+σ1α2L2ΦL1,λ1,α1α2z×δΩz−λ2+β2α2L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2δΩz.
Sinceσ1<σ2 and β1=β2, we deduce that
(49)pz≈z−ν1+σ1α1L1ΦL1,λ1,α1α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1δΩz+z−ν1+σ1α1L1ΦL1,λ1,α1α1z×δΩz−λ2+β2α2L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2δΩz=b1z+b2z.
Applying again Proposition8 with γ=ν1−σ1α1,ζ1=λ1−β1α1,M1=L1ΦL1,ν1,α1α1 and M2=L1×ΨL1,λ1,α1+ΨL2,λ2,α2α1, we obtain for z∈Ω\0,(50)Vb1z≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ1ΨM2,ζ1,0δΩz.
Similarly, by using Proposition8 with γ=ν1−σ1α1,ζ2=λ2−β2α2,M1=L1ΦL1,ν1,α1α1, and M2=L2×ΨL1,λ1,α1+ΨL2,λ2,α2α2, we obtain for z∈Ω\0,(51)Vb2z≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ2ΨM2,ζ2,0δΩz.
Hence,(52)Vpz≈Vb1z+Vb2z≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ1ΨM2,ζ1,0δΩz+zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ2ΨM2,ζ2,0δΩz.
Using the fact thatmin0,2−γ=σ1 and min1,2−ζ1=β1=β2=min1,2−ζ2, we deduce that
(53)Vpz≈zσ1ΦM1,γ,0zδΩzβ1ΨM2,ζ1,0δΩz+zσ1ΦM1,γ,0zδΩzβ1ΨM2,ζ2,0δΩz.
Using Lemma7, ([6], Lemmas 8 and 9) and a simple computation, we deduce that
(54)ΦM1,γ,0z≈ΦL1,ν1,α1zandΨM2,ζ1,0+ΨM2,ζ2,0δΩz≈ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
Hence,(55)Vpz≈zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz=θz.Case 11.
Ifσ1=σ2 and β1<β2, then for z∈Ω\0,(56)θz=zσ1ΦL1,ν1,α1z+ΦL2,ν2,α2zδΩzβ1ΨL1,λ1,α1δΩz.
Therefore,(57)pz≈z−ν1+σ1α1L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1α1δΩz+z−ν2+σ1α2L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2z×δΩz−λ2+β1α2L2.ΨL1,λ1,α1α2δΩz.
Sinceσ1=σ2 and β1<β2, we deduce that
(58)pz≈z−ν1+σ1α1L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1α1δΩz+z−ν2+σ2α2L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1α1δΩz=b1z+b2z.
Using Proposition8 with γ1=ν1−σ1α1,ζ=λ1−β1α1,M1=L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1, and M2=L1.ΨL1,λ1,α1α1, we obtain for z∈Ω\0,(59)Vb1z≈zmin0,2−γ1ΦM1,γ1,0zδΩzmin1,2−ζ1ΨM2,ζ,0δΩz.
Similarly, by using Proposition8 with γ2=ν2−σ2α2,ζ=λ1−β1α1,M1=L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2, and M2=L1.ΨL1,λ1,α1α1, we obtain for z∈Ω\0,(60)Vb2z≈zmin0,2−γ2ΦM1,γ2,0zδΩzmin1,2−ζΨM2,ζ,0δΩz.
Sincemin0,2−γ1=σ1=σ2=min0,2−γ2 and min1,2−ζ=β1, we deduce that
(61)Vpz≈Vb1z+Vb2z≈zσ1ΦM1,γ1,0z+ΦM1,γ2,0zδΩzβ1ΨM2,ζ,0δΩz.
By using similar arguments as in the proof of Case10, we deduce that
(62)ΦM1,γ1,0+ΦM1,γ2,0z≈ΦL1,ν1,α1+ΦL2,ν2,α2zandΨM2,ζ,0δΩz≈ΨL1,λ1,α1δΩz.
Hence,(63)Vpz≈θz.Case 12.
Ifσ1=σ2 and β1=β2, then
(64)θz=zσ1ΦL1,ν1,α1z+ΦL2,ν2,α2z×δΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
In this case,(65)pz≈z−ν1+σ1α1L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1δΩz+z−ν2+σ2α2L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2z×δΩz−λ2+β2α2L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2δΩz=b1z+b2z.
Now, by using Proposition8 with γ1=ν1−σ1α1,ζ1=λ1−β1α1,M1=L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1, and M2=L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1, we obtain for z∈Ω\0,(66)Vb1z≈zmin0,2−γ1ΦM1,γ1,0zδΩzmin1,2−ζ1ΨM2,ζ1,0δΩz.
By applying again Proposition8 with γ2=ν2−σ2α2,ζ2=λ2−β2α2,M1=L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2, and M2=L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2, we obtain for z∈Ω\0,(67)Vb2z≈zmin0,2−γ2ΦM1,γ2,0zδΩzmin1,2−ζ2ΨM2,ζ2,0δΩz.
Sincemin0,2−γ1=σ1=σ2=min0,2−γ2 and min1,2−ζ1=β1=β2=min1,2−ζ2, we deduce that
(68)Vpz≈zσ1δΩzβ1ΦM1,γ1,0zΨM2,ζ1,0δΩz+ΦM1,γ2,0zΨM2,ζ2,0δΩz≈zσ1δΩzβ1ΦM1,γ1,0+ΦM1,γ2,0zΨM2,ζ1,0+ΨM2,ζ2,0δΩz.
As in the proof of Case11 and Case 10, we deduce that
(69)ΦM1,γ1,0+ΦM1,γ2,0z≈ΦL1,ν1,α1+ΦL2,ν2,α2zandΨM2,ζ1,0+ΨM2,ζ2,0δΩz≈ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
That is(70)Vpz≈θz.
The proof is completed.


## 2.1. Basic Properties of Karamata Class

Lemma 5 (See [1, 3, 4]).
IfM1,M2∈K, then
(i)
For everyc≥0 and r∈ℝ,(30)M1+cM2,M1M2,andM1rarealsoinK.(ii)
For everyε>0,(31)lims⟶0+sεM1s=0andlims⟶0+s−εM1s=∞.Lemma 6 (Asymptotic behavior, see [1, 3, 4]).
IfM∈K, then
(i)
For everyγ>−1,∫0ηrγMrdr converges and(32)∫0srγMrdr~s⟶0+s1+γMs1+γ.(ii)
For everyγ<−1,∫0ηrγMrdr diverges and(33)∫sηrγMrdr~s⟶0+−s1+γMs1+γ.Lemma 7 (See [4, 6]).
LetM∈K, then
(i)
s↦∫sηMr/rdr belongs to K and(34)lims⟶0+Ms∫sηMr/rdr=0.(ii)
If∫0ηMr/rdr<∞, then s↦∫0sMr/rdr belongs to K and(35)lims⟶0+Ms∫0sMr/rdr=0.


## 2.2. Proof of Theorem2

The next Proposition, which follows from ([46], Proposition 2.12), will be used.Proposition 8.
Let≤N,ζ≤2, and M1, M2∈K such that
(36)∫0ηrN−γ−1M1rdr<∞and∫0ηr1−ζM2rdr<∞.
Then forz∈Ω\0,(37)Vhz≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζΨM2,ζ,0δΩz,where hz≔z−γM1zδΩz−ζM2δΩz.We recall that fori∈1,2,pi∈C+Ω\0 satisfying
(38)piz≈z−νiLizδΩz−λiLiδΩz,where αi<1,νi≤N+2−Nαi,λi≤2, and Li,Li∈K defined on 0,η such that (21) is fulfilled.We aim at proving that onΩ\0,(39)Vpz≈θz,where pz≔p1zθα1z+p2zθα2z and the function θ given by (24), (25), (26), and (27).Throughout the proof, we will apply Lemma5 and Lemma 7 to verify that some functions are in K.We distinguish the following cases:Case 9.
Ifσ1<σ2 and β1<β2, then for z∈Ω\0,(40)θz=zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1δΩz.
Therefore,(41)pz≈z−ν1+σ1α1L1ΦL1,ν1,α1α1zδΩz−λ1+β1α1L1ΨL1,λ1,α1α1δΩz+z−ν2+σ1α2L2ΦL1,ν1,α1α2zδΩz−λ2+β1α2L2ΨL1,λ1,α1α2δΩz.
Sinceσ1<σ2 and β1<β2, we deduce by Lemma 5 that
(42)pz≈z−ν1+σ1α1L1ΦL1,ν1,α1α1zδΩz−λ1+β1α1L1ΨL1,λ1,α1α1δΩz.
Using Lemma6 (i), (21), and Proposition 8 with γ=ν1−σ1α1,ζ=λ1−β1α1,M1=L1ΦL1,ν1,α1α1, and M2=L1ΨL1,λ1,α1α1, we obtain for z∈Ω\0,(43)Vpz≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζΨM2,ζ,0δΩz.
Sincemin0,2−γ=σ1 and min1,2−ζ=β1, we deduce that
(44)Vpz≈zσ1ΦM1,γ,0zδΩzβ1ΨM2,ζ,0δΩz.
Now, by using a simple computation, we obtain forz∈Ω\0,(45)ΦM1,γ,0z≈ΦL1,ν1,α1zandΨM2,ζ,0δΩz≈ΨL1,λ1,α1δΩz.
Combining this fact with (44), we obtain for z∈Ω\0,(46)Vpz≈θz.Case 10.
Ifσ1<σ2 and β1=β2, then for z∈Ω\0,(47)θz=zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
In this case,(48)pz≈z−ν1+σ1α1L1ΦL1,λ1,α1α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1δΩz+z−ν2+σ1α2L2ΦL1,λ1,α1α2z×δΩz−λ2+β2α2L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2δΩz.
Sinceσ1<σ2 and β1=β2, we deduce that
(49)pz≈z−ν1+σ1α1L1ΦL1,λ1,α1α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1δΩz+z−ν1+σ1α1L1ΦL1,λ1,α1α1z×δΩz−λ2+β2α2L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2δΩz=b1z+b2z.
Applying again Proposition8 with γ=ν1−σ1α1,ζ1=λ1−β1α1,M1=L1ΦL1,ν1,α1α1 and M2=L1×ΨL1,λ1,α1+ΨL2,λ2,α2α1, we obtain for z∈Ω\0,(50)Vb1z≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ1ΨM2,ζ1,0δΩz.
Similarly, by using Proposition8 with γ=ν1−σ1α1,ζ2=λ2−β2α2,M1=L1ΦL1,ν1,α1α1, and M2=L2×ΨL1,λ1,α1+ΨL2,λ2,α2α2, we obtain for z∈Ω\0,(51)Vb2z≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ2ΨM2,ζ2,0δΩz.
Hence,(52)Vpz≈Vb1z+Vb2z≈zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ1ΨM2,ζ1,0δΩz+zmin0,2−γΦM1,γ,0zδΩzmin1,2−ζ2ΨM2,ζ2,0δΩz.
Using the fact thatmin0,2−γ=σ1 and min1,2−ζ1=β1=β2=min1,2−ζ2, we deduce that
(53)Vpz≈zσ1ΦM1,γ,0zδΩzβ1ΨM2,ζ1,0δΩz+zσ1ΦM1,γ,0zδΩzβ1ΨM2,ζ2,0δΩz.
Using Lemma7, ([6], Lemmas 8 and 9) and a simple computation, we deduce that
(54)ΦM1,γ,0z≈ΦL1,ν1,α1zandΨM2,ζ1,0+ΨM2,ζ2,0δΩz≈ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
Hence,(55)Vpz≈zσ1ΦL1,ν1,α1zδΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz=θz.Case 11.
Ifσ1=σ2 and β1<β2, then for z∈Ω\0,(56)θz=zσ1ΦL1,ν1,α1z+ΦL2,ν2,α2zδΩzβ1ΨL1,λ1,α1δΩz.
Therefore,(57)pz≈z−ν1+σ1α1L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1α1δΩz+z−ν2+σ1α2L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2z×δΩz−λ2+β1α2L2.ΨL1,λ1,α1α2δΩz.
Sinceσ1=σ2 and β1<β2, we deduce that
(58)pz≈z−ν1+σ1α1L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1α1δΩz+z−ν2+σ2α2L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1α1δΩz=b1z+b2z.
Using Proposition8 with γ1=ν1−σ1α1,ζ=λ1−β1α1,M1=L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1, and M2=L1.ΨL1,λ1,α1α1, we obtain for z∈Ω\0,(59)Vb1z≈zmin0,2−γ1ΦM1,γ1,0zδΩzmin1,2−ζ1ΨM2,ζ,0δΩz.
Similarly, by using Proposition8 with γ2=ν2−σ2α2,ζ=λ1−β1α1,M1=L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2, and M2=L1.ΨL1,λ1,α1α1, we obtain for z∈Ω\0,(60)Vb2z≈zmin0,2−γ2ΦM1,γ2,0zδΩzmin1,2−ζΨM2,ζ,0δΩz.
Sincemin0,2−γ1=σ1=σ2=min0,2−γ2 and min1,2−ζ=β1, we deduce that
(61)Vpz≈Vb1z+Vb2z≈zσ1ΦM1,γ1,0z+ΦM1,γ2,0zδΩzβ1ΨM2,ζ,0δΩz.
By using similar arguments as in the proof of Case10, we deduce that
(62)ΦM1,γ1,0+ΦM1,γ2,0z≈ΦL1,ν1,α1+ΦL2,ν2,α2zandΨM2,ζ,0δΩz≈ΨL1,λ1,α1δΩz.
Hence,(63)Vpz≈θz.Case 12.
Ifσ1=σ2 and β1=β2, then
(64)θz=zσ1ΦL1,ν1,α1z+ΦL2,ν2,α2z×δΩzβ1ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
In this case,(65)pz≈z−ν1+σ1α1L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1z×δΩz−λ1+β1α1L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1δΩz+z−ν2+σ2α2L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2z×δΩz−λ2+β2α2L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2δΩz=b1z+b2z.
Now, by using Proposition8 with γ1=ν1−σ1α1,ζ1=λ1−β1α1,M1=L1.ΦL1,ν1,α1+ΦL2,ν2,α2α1, and M2=L1.ΨL1,λ1,α1+ΨL2,λ2,α2α1, we obtain for z∈Ω\0,(66)Vb1z≈zmin0,2−γ1ΦM1,γ1,0zδΩzmin1,2−ζ1ΨM2,ζ1,0δΩz.
By applying again Proposition8 with γ2=ν2−σ2α2,ζ2=λ2−β2α2,M1=L2.ΦL1,ν1,α1+ΦL2,ν2,α2α2, and M2=L2.ΨL1,λ1,α1+ΨL2,λ2,α2α2, we obtain for z∈Ω\0,(67)Vb2z≈zmin0,2−γ2ΦM1,γ2,0zδΩzmin1,2−ζ2ΨM2,ζ2,0δΩz.
Sincemin0,2−γ1=σ1=σ2=min0,2−γ2 and min1,2−ζ1=β1=β2=min1,2−ζ2, we deduce that
(68)Vpz≈zσ1δΩzβ1ΦM1,γ1,0zΨM2,ζ1,0δΩz+ΦM1,γ2,0zΨM2,ζ2,0δΩz≈zσ1δΩzβ1ΦM1,γ1,0+ΦM1,γ2,0zΨM2,ζ1,0+ΨM2,ζ2,0δΩz.
As in the proof of Case11 and Case 10, we deduce that
(69)ΦM1,γ1,0+ΦM1,γ2,0z≈ΦL1,ν1,α1+ΦL2,ν2,α2zandΨM2,ζ1,0+ΨM2,ζ2,0δΩz≈ΨL1,λ1,α1+ΨL2,λ2,α2δΩz.
That is(70)Vpz≈θz.
The proof is completed.


## 3. Kato Class and Proof of Theorem3

### 3.1. Kato ClassKΩ

From [24], we recall that Green’s function GΩz,y satisfied
(71)GΩz,y≈1z−yN−2min1,δΩzδΩyz−y2,z,y∈Ω.Definition 13.
A Borel measurable functionq in Ω is in the Kato class KΩ if
(72)lims⟶0supz∈Ω∫Ω∩Bz,sδΩyδΩzGΩz,y∣qy∣dy.
Note that this classKΩ was introduced in [2] and properly contains the usual Kato class KNΩ defined (see, [24, 47]) as
(73)KNΩ=q∈BΩ,lims⟶0supz∈Ω∫Ω∩Bz,s1z−yN−2∣qy∣dy=0.Proposition 14 (See [46], Proposition 2.8).
Fori∈1,2, let Mi∈K and λi∈ℝ. The following properties are equivalent:
(i)
The functionz↦z−λ1M1zδΩz−λ2M2δΩz is in KΩ(ii)
∫0ηr1−λiMirdr<∞, for i∈1,2(iii)
λi<2 or λi=2 with ∫0ηMir/rdr<∞, for i∈1,2As a consequence of the above Proposition, hypothesisH, Lemma 5, and Lemma 6 (i), we obtain the following.Corollary 15.
Assume that hypothesisH is satisfied and let
(74)pz≔p1zθα1z+p2zθα2z,where the function θ is given by (24), (25), (26), and (27).
Then, the functionz⟶zN−2pz∈KΩ.Proposition 16 (See [46], Proposition 2.5).
Letq∈KΩ with q≥0. Then, the family
(75)Λq=z⟶zN−2∫ΩGΩz,yy2−Nfydy,f≤qis uniformly bounded and equicontinuous in Ω¯. Consequently, Λq is relatively compact in C0Ω¯.


### 3.2. Proof of Theorem3

Assume that hypothesisH is satisfied and let θ be the function given by (24), (25), (26), and (27).By Theorem2, there exists M>1 such that for each Ω\0,
(76)1MVpz≤θz≤MVpz,where pz is given by (74).Putα=maxα1,α2,c0=Mα/1−α, and consider the set
(77)S=w∈C0Ω¯:1c0zN−2Vpz≤wz≤c0zN−2Vpz.By Corollary15 and Proposition 16, it follows that z⟶zN−2Vpz∈C0Ω¯. So, S≠∅, closed, bounded, and convex set in C0Ω¯.Forw∈S, set
(78)Fwz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy.By using (76), for all w∈S, we have
(79)p1zz2−Nα1wα1z+p2zz2−Nα2wα2z≤c0z2−Nqz,where qz=zN−2pz, which belongs to KΩ by Corollary 15. Hence, by Proposition 16, the family of functions z⟶Fwz,w∈S is relatively compact in C0Ω¯.Next, we aim at proving thatFS⊂S.Indeed, by using (76), for all w∈S, we have
(80)Fwz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy≤zN−2∫ΩGΩz,yp1yc0αMαθα1y+p2yc0αMαθα2ydy=c0zN−2Vpz.On the other hand,(81)Fwz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy≥zN−2∫ΩGΩz,yp1yc0−αM−αθα1y+p2yc0−αM−αθα2ydy=1c0zN−2Vpz.Hence,FS⊂S.Next, we shall prove the continuity ofF in the supremum norm. Let wkk be a sequence in S which converges to w in S. For each z∈Ω, we have
(82)Fwkz−Fwz≤∫ΩGΩz,yp1yy2−Nα1wkα1y−wα1y+p2yy2−Nα2wkα2y−wα2ydy.Since from (79), we have
(83)p1yy2−Nα1wkα1y−wα1y+p2yy2−Nα2wkα2y−wα2y≤c0y2−Nqy,and we deduce by Proposition 16 and the dominated convergence theorem that for all z∈Ω,limk⟶∞Fwkz=Fwz.SinceFS is relatively compact in C0Ω¯, then limk⟶∞Fwkz=Fwz uniformly. Thus, we have proved that F is a compact mapping from S to itself.So, by the Schauder fixed-point theorem, there exists a functionw∈S such that
(84)wz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy.Putvz=z2−Nwz. Then, v∈CΩ¯\0 and satisfies the equation
(85)vz=Vp1vα1+p2vα2z,forz∈Ω¯\0.Since the functiony↦ψy≔p1yvα1y+p2yvα2y∈Lloc1Ω\0 and from (85) the function z↦Vψz∈Lloc1Ω\0, we deduce by (16) that
(86)−Δvz=p1zvα1z+p2zvα2z,z∈Ω\0.By (76), we deduce that
(87)M−11−αθz≤vz≤M11−αθz.Using (87), Remark 4 (iii), and the fact that w∈S, we obtain
(88)limz⟶0zN−2vz=0andlimz⟶∂Ωvz=0.So,v is a solution of problem (1) satisfying (29).Example 17.
Letα1<0<α2<1. Consider p1,p2∈CΩ\0 such that
(89)p1z≈z−ν1log4dz−αδΩz−2log4dδΩz−2,(90)p2z≈z−ν2log4dz−αδΩz−λ2log4dδΩz−ζ,where νi<N+2−Nαi with 2−ν1/1−α1≤2−ν2/1−α2,λ2<2,α∈0,1, and ζ∈ℝ. Then, by Theorem 3, problem (1) has at least one positive solution v∈CΩ¯\0 satisfying the following estimates:
(i)
If2<ν1<N+2−Nα1, then for z∈Ω\0,(91)vz≈z2−ν11−α1log4dz−α1−α1log4dδΩz−11−α1(ii)
Ifν1=ν2=2, then for z∈Ω\0,(92)vz≈log4dz1−α1−α2log4dδΩz−11−α1(iii)
Ifν1=2 and ν2<2, then for z∈Ω\0,(93)vz≈log4dz1−α1−α1log4dδΩz−11−α(iv)
Ifν1<1, then for z∈Ω\0,(94)vz≈log4dδΩz−11−α


## 3.1. Kato ClassKΩ

From [24], we recall that Green’s function GΩz,y satisfied
(71)GΩz,y≈1z−yN−2min1,δΩzδΩyz−y2,z,y∈Ω.Definition 13.
A Borel measurable functionq in Ω is in the Kato class KΩ if
(72)lims⟶0supz∈Ω∫Ω∩Bz,sδΩyδΩzGΩz,y∣qy∣dy.
Note that this classKΩ was introduced in [2] and properly contains the usual Kato class KNΩ defined (see, [24, 47]) as
(73)KNΩ=q∈BΩ,lims⟶0supz∈Ω∫Ω∩Bz,s1z−yN−2∣qy∣dy=0.Proposition 14 (See [46], Proposition 2.8).
Fori∈1,2, let Mi∈K and λi∈ℝ. The following properties are equivalent:
(i)
The functionz↦z−λ1M1zδΩz−λ2M2δΩz is in KΩ(ii)
∫0ηr1−λiMirdr<∞, for i∈1,2(iii)
λi<2 or λi=2 with ∫0ηMir/rdr<∞, for i∈1,2As a consequence of the above Proposition, hypothesisH, Lemma 5, and Lemma 6 (i), we obtain the following.Corollary 15.
Assume that hypothesisH is satisfied and let
(74)pz≔p1zθα1z+p2zθα2z,where the function θ is given by (24), (25), (26), and (27).
Then, the functionz⟶zN−2pz∈KΩ.Proposition 16 (See [46], Proposition 2.5).
Letq∈KΩ with q≥0. Then, the family
(75)Λq=z⟶zN−2∫ΩGΩz,yy2−Nfydy,f≤qis uniformly bounded and equicontinuous in Ω¯. Consequently, Λq is relatively compact in C0Ω¯.


## 3.2. Proof of Theorem3

Assume that hypothesisH is satisfied and let θ be the function given by (24), (25), (26), and (27).By Theorem2, there exists M>1 such that for each Ω\0,
(76)1MVpz≤θz≤MVpz,where pz is given by (74).Putα=maxα1,α2,c0=Mα/1−α, and consider the set
(77)S=w∈C0Ω¯:1c0zN−2Vpz≤wz≤c0zN−2Vpz.By Corollary15 and Proposition 16, it follows that z⟶zN−2Vpz∈C0Ω¯. So, S≠∅, closed, bounded, and convex set in C0Ω¯.Forw∈S, set
(78)Fwz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy.By using (76), for all w∈S, we have
(79)p1zz2−Nα1wα1z+p2zz2−Nα2wα2z≤c0z2−Nqz,where qz=zN−2pz, which belongs to KΩ by Corollary 15. Hence, by Proposition 16, the family of functions z⟶Fwz,w∈S is relatively compact in C0Ω¯.Next, we aim at proving thatFS⊂S.Indeed, by using (76), for all w∈S, we have
(80)Fwz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy≤zN−2∫ΩGΩz,yp1yc0αMαθα1y+p2yc0αMαθα2ydy=c0zN−2Vpz.On the other hand,(81)Fwz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy≥zN−2∫ΩGΩz,yp1yc0−αM−αθα1y+p2yc0−αM−αθα2ydy=1c0zN−2Vpz.Hence,FS⊂S.Next, we shall prove the continuity ofF in the supremum norm. Let wkk be a sequence in S which converges to w in S. For each z∈Ω, we have
(82)Fwkz−Fwz≤∫ΩGΩz,yp1yy2−Nα1wkα1y−wα1y+p2yy2−Nα2wkα2y−wα2ydy.Since from (79), we have
(83)p1yy2−Nα1wkα1y−wα1y+p2yy2−Nα2wkα2y−wα2y≤c0y2−Nqy,and we deduce by Proposition 16 and the dominated convergence theorem that for all z∈Ω,limk⟶∞Fwkz=Fwz.SinceFS is relatively compact in C0Ω¯, then limk⟶∞Fwkz=Fwz uniformly. Thus, we have proved that F is a compact mapping from S to itself.So, by the Schauder fixed-point theorem, there exists a functionw∈S such that
(84)wz=zN−2∫ΩGΩz,yp1yy2−Nα1wα1y+p2yy2−Nα2wα2ydy.Putvz=z2−Nwz. Then, v∈CΩ¯\0 and satisfies the equation
(85)vz=Vp1vα1+p2vα2z,forz∈Ω¯\0.Since the functiony↦ψy≔p1yvα1y+p2yvα2y∈Lloc1Ω\0 and from (85) the function z↦Vψz∈Lloc1Ω\0, we deduce by (16) that
(86)−Δvz=p1zvα1z+p2zvα2z,z∈Ω\0.By (76), we deduce that
(87)M−11−αθz≤vz≤M11−αθz.Using (87), Remark 4 (iii), and the fact that w∈S, we obtain
(88)limz⟶0zN−2vz=0andlimz⟶∂Ωvz=0.So,v is a solution of problem (1) satisfying (29).Example 17.
Letα1<0<α2<1. Consider p1,p2∈CΩ\0 such that
(89)p1z≈z−ν1log4dz−αδΩz−2log4dδΩz−2,(90)p2z≈z−ν2log4dz−αδΩz−λ2log4dδΩz−ζ,where νi<N+2−Nαi with 2−ν1/1−α1≤2−ν2/1−α2,λ2<2,α∈0,1, and ζ∈ℝ. Then, by Theorem 3, problem (1) has at least one positive solution v∈CΩ¯\0 satisfying the following estimates:
(i)
If2<ν1<N+2−Nα1, then for z∈Ω\0,(91)vz≈z2−ν11−α1log4dz−α1−α1log4dδΩz−11−α1(ii)
Ifν1=ν2=2, then for z∈Ω\0,(92)vz≈log4dz1−α1−α2log4dδΩz−11−α1(iii)
Ifν1=2 and ν2<2, then for z∈Ω\0,(93)vz≈log4dz1−α1−α1log4dδΩz−11−α(iv)
Ifν1<1, then for z∈Ω\0,(94)vz≈log4dδΩz−11−α



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