Exact boundary behavior of the unique positive solution for singular second-order

Exact boundary behavior of the unique positive solution for singular second-order

differential equations

Imed Bachar

and Habib Mâagli

1King Saud University, College of Science, Mathematics Department, P. O. Box 2455, Riyadh 11451, Saudi Arabia

2King Abdulaziz University, College of Sciences and Arts, Rabigh Campus, Department of Mathematics, P. O. Box 344, Rabigh 21911, Saudi Arabia

Received 5 February 2015, appeared 8 July 2015 Communicated by Paul Eloe

Abstract. In this paper, we give the exact asymptotic behavior of the unique positive solution to the following singular boundary value problem







−_A¹(Au⁰)⁰ =p(x)g(u), x∈(0, 1), u>0, in(0, 1),

limx→0⁺(Au⁰)(x) =0, u(1) =0,

whereAis a continuous function on[0, 1), positive and differentiable on(0, 1)such that

1

A is integrable in a neighborhood of 1,g∈C¹((_0,∞)_,(_0,∞))is nonincreasing on(_0,∞) with limt→0g⁰(t)Rt

0 1

g(s)ds = −Cg ≤ 0 and p is a nonnegative continuous function in (0, 1)satisfying

0<p₁=lim inf

x→1

p(x)

h(1−x) ≤lim sup

x→1

p(x)

h(1−x) =p₂<_∞, where h(_t) =_ct^−λ_exp(R_η

t z(s)

s ds)_,λ ≤ 2,c >_{0 andz}is continuous on[_0,η] _{for some} η>1 such thatz(0) =0.

Keywords: singular nonlinear boundary value problems, positive solution, exact asymptotic behavior, Karamata regular variation theory.

2010 Mathematics Subject Classification: 34B16, 34B18, 34D05.

BCorresponding author. Email: abachar@ksu.edu.sa

1 Introduction

In this paper, we give the exact asymptotic behavior near the boundary of the unique positive solution to the following singular problem

(−_A¹(Au⁰)⁰ = p(x)g(u), x∈(0, 1),

u>_{0, in} (_{0, 1})_, ^(1.1)

subject to the boundary conditions

xlim→0⁺(Au⁰)(x) =0, u(1) =0. (1.2) The functions A, pandgsatisfy the following assumptions.

(H₁) Ais a continuous function on [0, 1), positive and differentiable on (0, 1) such that _A¹ is integrable in a neighborhood of 1 and limx→1(x−1)A⁰(x)

A(x) =α<_1.

(H₂) pis a nonnegative continuous function in(0, 1)satisfying 0< p₁=lim inf

x→1

p(x)

h(1−x) ≤lim sup

x→1

p(x)

h(1−x) = p₂ <_∞, whereh(t) =t^−λL(t),λ≤2 such thatRη

0 s^1−λL(s)ds<_∞for some η>1 andLbelongs to the class of Karamata functionsK(see Definition1.1).

(_H3) The functiong: (_0,∞)→(_0,∞)is nonincreasing, continuously differentiable such that limt→0g⁰(t)

Z _t

0

1

g(s)^ds= −Cg withCg ≥0.

(H₄) λ+ (2−λ)Cg−α−1>0.

Observe thatCg ∈[0, 1]. Indeed, since the functiongis nonincreasing, we obtain fort>0 0<g(t)

Z _t

0

1

g(s)^ds≤t.

This implies that lim_t→0g(t)Rt 0 1

g(s)ds=0. Now, since fort>0 Z _t

0 g⁰(s)

Z _s

0

1

g(r)^dr=g(t)

Z _t

0

1

g(s)^ds−t, we get

limt→0

g(t) t

Z _t

0

1

g(s)^ds=1−C_g. HenceCg∈ [0, 1].

The functions t⁻¹ln(1+t), ln ln(e+ ¹_t), t^−νln 1+¹_t, exp{ ln(1+¹_t)^ν}, ν ∈ (0, 1) satisfy the assumption(H₃), as well as the function

(t²e^1t, if 0<t < ¹_2,

1

4e², ift ≥ ¹₂.

When A ≡ 1, problems of type (1.1) with various boundary conditions arise in the study of boundary layer equations for the class of pseudoplastic fluids and have been studied for both bounded and unbounded intervals ofR(see [4,5,23,27] and the references therein).

When A(t) = tⁿ⁻¹ (n ≥ 2), the operator u → _A¹(Au⁰)⁰ appears as the radial part of the Laplace operator ∆ (see [24]). Our setting includes the scalar curvature equation and the relativistic pendulum equation, which correspond to A(t) = (1+t²)⁻²¹, resp. A(t) = (1−t²)⁻²¹. For various existence, uniqueness and asymptotic behavior results of such problem, we refer the reader to [8–11,14,21,25,26] and the references therein. However, we emphasize that in problem (1.1) the function Acould be singular att =1.

On the other hand, the singular nonlinear problem

(−_A¹(Au⁰)⁰ = f(x,u), x∈ (0, 1),

u>0, in(0, 1), (1.3)

subject to different boundary conditions has been considered by many authors, where Ais a continuous function on[_{0, 1}), positive and differentiable on(_{0, 1})satisfying some appropriate conditions (see for example [1,2,13,16,17,19]). In [15, Theorem 5], Mâagli and Masmoudi investigated equation (1.3) with boundary value conditionsu⁰(0) =u(1) =0. They supposed that f is a nonnegative continuous function on(0, 1)×(0,∞)and nonincreasing with respect to the second variable. Under some appropriate conditions on the function A, they proved the existence of a unique positive solutionuinC([0, 1])∩C²((0, 1))to (1.3) and gave estimates on such a solution. In particular they extended some results of [1,2] and [19]. Our aim in this paper is to establish the exact boundary behavior of the unique solution to problem (1.1)–(1.2).

To state our results, we need some notations.

Definition 1.1. The class Kis the set of all Karamata functionsLdefined on(0,η]by L(t)_:=cexp

Z _η

t

z(s) s ds

,

for someη>1 and wherec>0 andz∈C([0,η])such thatz(0) =0.

Note that functions belonging to the class K are in particular slowly varying functions.

The theory of such functions was initiated by Karamata in a fundamental paper [12].

We also point out that the first use of the Karamata theory in the study of the growth rate of solutions near the boundary is done in the paper of Cîrstea and R˘adulescu [7].

Remark 1.2. A function Lis inKif and only if Lis a positive function inC¹((0,η]), for some η>1, such that lim_t→0⁺tL⁰(t)

L(t) =0.

Typical examples of functions belonging to the classK (see [3,18,22]) are:

L(t) =

∏

log_kω t

ξ_k

, L(t) =2+sin

log₂

e+ ¹

t

, L(t) =exp

( m k

∏

log_kω t

νk

) ,

where log_kx = log◦log◦ · · · ◦logx (k times),ξ_k ∈ _R,ν_k ∈(0, 1)andω is a sufficiently large positive real number such thatLis defined and positive on(0,η].

Throughout this paper, we denote byψ_gthe unique solution determined by Z _ψg(t)

0

1

g(s)^ds=t, t ∈[0,∞), (1.4) and we mention that

limt→0tg⁰(ψ_g(t)) =−C_g. (1.5) Our first result is the following.

Theorem 1.3. Assume that hypotheses(H₁)–(H₄)are fulfilled. Then problem(1.1)–(1.2)has a unique positive solution u∈C([0, 1])∩C²((0, 1))satisfying

(i) ifλ<_2,then ξ₁

2−λ 1−Cg

≤lim inf

x→1

u(x)

ψ_g((1−x)^2−λL(1−x))

≤lim sup

x→1

u(_x)

ψ_g((1−x)^2−λL(1−x)) ≤ ξ₂

2−λ 1−Cg

;

(ii) ifλ=2,then

ξ¹₁^−Cg ≤lim inf

x→1

u(_x) ψg

R1−x 0

L(t)

t dt ≤lim sup

x→1

u(x) ψg

R1−x 0

L(t)

t dt ≤ξ¹₂^−Cg, whereξ₁ = ^p1

λ+(2−λ)Cg−α−1 andξ₂ = ^p2

λ+(2−λ)Cg−α−1.

An immediate consequence of Theorem1.3is the following.

Corollary 1.4. Let u be the unique solution of problem(1.1)–(1.2). Then, we have the following exact boundary behavior.

(a) When C_g =1,then we have (i) _limx→1 ^u(x)

ψg((1−x)^2−λL(1−x)) =₁ifλ<_2;

(ii) lim_x→1 u(x) ψg(R1−x

0 L(_t)

t dt) =1ifλ=2.

(b) When C_g <1and p₁ = p₂ = p₀,then we have (i) lim_x→1 u(x)

ψg((1−x)^2−λL(1−x)) = ^p0

(2−λ)(λ+(2−λ)Cg−α−1) 1−Cg

ifλ<2;

(ii) lim_x→1 u(x) ψg(R1−x

0 L(t)

t dt) = ₁^p₋⁰

α

1−Cg

ifλ=2.

Example 1.5. Letg be the function defined by g(t) =

(t²e^1t, if 0<t < ¹_2,

1

4e², ift ≥ ¹₂,

and pbe a nonnegative continuous function in(0, 1)satisfying

xlim→1

p(x)

h(1−x) = p₀∈ (0,∞), where h(t) = t^−λL(t), λ≤ 2 and L∈ K such thatRη

0 s^1−λL(s)ds< ∞. Then, we haveC_g =1 andψ_g(ξ) = _log⁻₍¹

ξ) forξ ∈ (0,e⁻²). Letu be the unique solution of (1.1)–(1.2). Then, we have the following exact behavior.

(i) lim_x→1u(x)log

1 (1−x)^2−λL(1−x)

=1 ifλ<2;

(ii) lim_x→1u(x)log

1 R1−x

0 L(t)

t dt

=1 ifλ=2.

In order to establish our second result, we consider the special case whereg(t) =t^−γwith γ≥0 andλ= (α+1) + (α−1)γ. Note that in this caseC_g = ^γ

γ+1 andλ+ (2−λ)C_g−α−1= 0. We assume the following hypotheses.

(H₅) Ais a continuous function on[0, 1), positive and differentiable on(0, 1)such thatA(x) = (1−x)^αB(x) with α < 1 and ^(1−x_B⁾₍^ζ_x^B₎^0(x) is bounded in a neighborhood of 1 for some ζ ∈ (0, 1).

(H₆) pis a nonnegative continuous function in(0, 1)satisfying 0< p₁=lim inf

x→1

p(_x)

(1−x)^{γ−1−α(1+γ)}L(1−x)

≤lim sup

x→1

p(x)

(1−x)^{γ−1−α(1+γ)}L(1−x) = p₂ <_∞, whereγ≥0 andL ∈ KwithRη

0 L(s)

s ds=_∞.

Our second result is the following.

Theorem 1.6. Assume that hypotheses(H₅)and(H₆)are fulfilled.Then the problem











−_A¹(Au⁰)⁰ = p(x)u^−γ, x∈ (0, 1), u >0, in(0, 1),

lim_x→0⁺(Au⁰)(x) =0, u(1) =0,

(1.6)

has a unique positive solution u ∈C([0, 1])∩C²((0, 1))satisfying (b₁)^γ+11 ≤lim inf

x→1

u(x) (1−x)^1−αRη

1−x L(t)

t dt ¹

γ+1

≤lim sup

x→1

u(x) (1−x)^1−αRη

1−x L(t)

t dt ¹

γ+1

≤(b₂)^γ1+1,

where b₁ = ^(γ₁⁺₋^1)p1

α and b₂= ^(γ₁⁺₋^1)p2

α . In particular if p₁ = p2 = p0,then

limx→1

u(x) (1−x)^1−αRη

1−x L(t)

t dt ¹

γ+1

=

(γ+1)p₀ 1−α

_γ+¹₁ .

The content of this paper is organized as follows. In Section 2, we present some funda- mental properties of Karamata regular variation theory. In Section 3, exploiting the results of the previous section, we prove Theorems 1.3 and1.6 by constructing a convenient pair of subsolution and supersolution.

2 On the Karamata class K

In this section, we collect some properties of Karamata functions.

Proposition 2.1([18,22]).

(i) Let L₁,L₂∈ Kand q∈_R.Then the functions

L₁+L₂, L₁L₂ and L^q₁ belong to the classK. (ii) Let L be a function inKandε>0.

Then we have

tlim→0⁺t^εL(t) =0.

Lemma 2.2([18,22]). Letµ∈_Rand L be a function inKdefined on(0,η].Then the following hold.

(i) Ifµ<−_1,thenRη

0 s^µL(s)ds diverges andRη

t s^µL(s)ds ∼

t→0⁺ −^tµ+1L(t)

µ+1 . (ii) Ifµ>−1,thenRη

0 s^µL(s)ds converges andRt

0s^µL(s)ds ∼

t→0⁺

t^µ+1L(t) µ+1 . The proof of the next lemma can be found in [6].

Lemma 2.3. Let L be a function inKdefined on(_0,η]_.Then we have

tlim→0⁺

L(t) Rη

t L(s)

s ds =0.

If furtherRη 0

L(s)

s ds converges, then we have

tlim→0⁺

L(t) Rt

0 L(s)

s ds

=0.

Remark 2.4. LetLbe a function inKdefined on(0,η], then using Remark1.2and Lemma2.3, we deduce that

t →

Z _η

t

L(s)

s ds∈ K_.

Definition 2.5. A positive measurable function f is called normalized regularly varying at zero with indexρ∈_Rand we write f ∈NRVZρ if f(s) =s^ρL(s)fors∈(0,η)with L∈ K.

Using the definition of Karamata class and the previous lemmas, we obtain the following.

Lemma 2.6([25]).

(i) If f ∈NRVZρ,thenlimt→0 f(ξt)

f(t) =ξ^ρ,uniformly forξ ∈ [c₁,c₂]⊂(_0,∞). (ii) A positive measurable function f belongs toNRVZ_ρif and only if lim_t→0t f⁰(t)

f(t) =ρ.

(iii) Let L∈ Kandλ≤2such thatRη

0 s^1−λL(s)ds<∞. Then the functionθ(t):= Rt

0s^1−λL(s)ds belongs toNRVZ₂−λ.

(iv) The functionψ_g ∈NRVZ_(1−Cg).

(v) The functionψ_g◦θ ∈NRVZ_{(2−λ)(1−Cg)}.

(vi) Let f ∈NRVZ_ρand m₁,m₂be two positive functions on(0,∞)such that

tlim→0⁺m₁(t) = lim

t→0⁺m₂(t) =0 and lim

t→0⁺

m₁(t)

m₂(t) =1, then lim

t→0⁺

f(m₁(t)) f(m₂(t)) =1.

3 Proofs of Theorems 1.3 and 1.6

In the sequel, we denote by

v₀(x) =

Z ₁

x

t

A(t)^{dt, forx} ∈(_{0, 1})_,

and we let L_Au := _A¹(Au⁰)⁰ = u⁰⁰+ ^A_A⁰u⁰. Note that since the function A satisfies (H₁), the functionv0(x)is well defined and we haveL_Av0 =−_A¹.

The following result will play a crucial role in the proof of our main result.

Lemma 3.1. Assume(H₁),then there exists L0 ∈ Ksuch that v0(x) ∼

x→1

(1−x)^1−α

(1−α)L₀(1−x)^{. (3.1)}

Proof. It is clear that

v0(x) ∼

x→1

Z ₁

x

1

A(t)^{dt. (3.2)}

On the other hand, by(H₁), we have lim_x→1(1−x)A⁰(x)

A(x) =limt→0 tA⁰(1−t)

A(1−t) =−α> −1.

So by Lemma2.6, we deduce that the function f(t)_:= A(₁−t)belongs to NRVZα. There- fore, there exists L₀ ∈ Ksuch that

f(t):= A(1−t) =t^αL₀(t), fort ∈(0,δ).

Hence by using this fact, Proposition2.1(i) and Lemma2.2(ii), we deduce that Z ₁

x

1 A(t)^dt=

Z _1−x

0

1

A(1−t)^dt=

Z _1−x

0 t^−α 1

L₀(t)^dt ∼

x→1

(1−x)^1−α

(1−α)L₀(1−x)^{. (3.3)} Combining(3.2)and(3.3), we obtain the required result. This completes the proof.

Proof of Theorem1.3. Letε ∈ (0,^p₂¹)and putξ_i = ^pi

λ+(2−λ)Cg−α−1 fori= 1, 2, τ₁ = ξ₁−ε^ξ_p¹

1 and

τ₂= ξ₂+ε^ξ_p²

2. Clearly, we have ^ξ₂¹ <τ₁ <τ₂ < ³₂ξ₂. Letθ(t) =Rt

0s^1−λL(s)dsand define v_i(x) =ψ_g

τ_i

Z _1−x

0 s^1−λL(s)ds

=ψ_g(τ_iθ(1−x)), forx∈(0, 1)andi∈ {1, 2}. By a simple calculus, we obtain fori∈ {1, 2},

L_Av_i(x) +p(x)g(v_i(x)) =v_i⁰⁰(x) + ^A

0(_x) A(x)^v

0

i(x) +p(x)g(v_i(x))

=g(v_i(x))(1−x)^−λL(1−x)

×^hτ_i

τ_i(₁−x)^2−λ_L(₁−x)_g⁰(_vi(_x)) + (₂−λ)_Cg +τ_i

(x−1)A⁰(x)

A(x) −α+ (1−x)L⁰(1−x) L(1−x)

−τ_i λ+ (₂−λ)_Cg−α−1 +_pi +

p(x)

(1−x)^−λL(1−x)−p_i

.

So, for the fixedε>0, there existsδ_ε ∈(0, 1)such that for x∈(δ_ε, 1)andi∈ {1, 2}, we have

τ_i

(x−1)A⁰(x)

A(x) −α+ (1−x)L⁰(1−x) L(1−x)

≤ ³ 2ξ₂

(x−1)A⁰(x) A(x) −α

+

(1−x)L⁰(1−x) L(1−x)

≤ ^ε 4, p₁− ^ε

2 ≤ ^p(x)

(1−x)^−λL(1−x) ≤ p₂+ ^ε 2 and

τ_i

τ_i(1−x)^2−λL(1−x)g⁰(v_i(x)) + (2−λ)Cg

≤ ³ 2ξ₂

τ_i(₁−x)^2−λL(₁−x)g⁰(v_i(x)) + (₂−λ)C_g ≤ ^ε

4.

Indeed, the last inequality follows from (1.5) and the fact that from Lemmas 2.2 and2.3, we have lim_x→1(1−x)^2−λL(1−x)

θ(1−x) =2−λ. This implies that for eachx∈ (δε, 1), we have L_Av₁(x) +p(x)g(v₁(x))

≥g(v₁(x))(₁−x)^−λL(₁−x)−ε+p₁−τ₁ λ+ (₂−λ)C_g−α−₁=₀ and

L_Av₂(x) +p(x)g(v₂(x))

≤ g(v₂(x))(1−x)^−λL(1−x)ε+p₂−τ₂ λ+ (2−λ)C_g−α−1

=0.

Letu∈ C([0, 1])∩C²((0, 1))be the unique solution of (1.1)–(1.2) (see [15, Theorem 5]). Then, there existsM >0 such that

v₁(δ_ε)−Mv₀(δ_ε)≤u(δ_ε)≤v₂(δ_ε) +Mv₀(δ_ε)_{. (3.4)} We claim that

v₁(x)−Mv₀(x)≤u(x)≤v₂(x) +Mv₀(x) _{for each} x∈(δ_ε, 1)_{. (3.5)} Assume for instance that the left inequality of (3.5) is not true. Then, there existsx₀ ∈(δ_ε, 1) such that

v₁(x₀)−Mv₀(x₀)−u(x₀)>0.

By (3.4), the continuity of the functions v₁, v0 and u on [δ_ε, 1) and that limx→1v₁(x) = lim_x→1v₀(x) =lim_x→1u(x) =0, we deduce that there existsx₁∈(δε, 1)such that

0< v₁(x₁)−Mv₀(x₁)−u(x₁) = max

x∈[δε,1](v₁(x)−Mv₀(x)−u(x)). This implies thatv₁⁰(x₁)−Mv⁰₀(x₁)−u⁰(x₁) =0 and

L_A(v₁−Mv₀−u)(x₁) = (v₁−Mv₀−u)⁰⁰(x₁)≤_0.

On the other hand, using the fact that p is nonnegative and the monotonicity of g, we obtain

L_A(v₁−Mv₀−u)(x₁) =L_A(v₁)(x₁) + ^M

A(x₁)−L_Au(x₁)

≥ p(x₁) [g(u(x₁))−g(v₁(x₁))] + ^M A(x₁)

≥ p(x₁) [g(u(x₁) +Mv0(x₁))−g(v₁(x₁))] + ^M A(x₁)

>0.

This yields to a contradiction. In the same way, we prove the right inequality of (3.5).

Now, sinceψ_g◦θ ∈NRVZ_{(2−λ)(1−Cg)}, there existsbL∈ Ksuch thatψ_g(θ(t)) = t^{(2−λ)(1−Cg)}bL(t) for t ∈ (0,η). Moreover, since λ+ (2−λ)Cg−α−1 > 0, it follows by Proposition 2.1 that lim_t→0 t^1−α

t^{(2−λ)(1−Cg)}bL(t) =0. This implies that limt→0

t^1−α ψ_g(τ_iRt

0s^1−λL(s)ds) =lim

t→0

t^1−α

ψ_g(τ_iθ(t)) =lim

t→0

ψ_g(θ(t)) ψ_g(τ_i θ(t)

t^1−α ψ_g(θ(t)) =0 uniformly inτ_i ∈ [^ξ₂¹,^3ξ₂²]⊂(0,∞).

This together with (3.1) and Proposition2.1 implies that limx→1

v₀(x)

ψ_g(τ₁θ(1−x)) = lim

x→1

v₀(x)

ψ_g(τ₂θ(1−x)) =0.

So, we get by (3.5)

lim sup

x→1

u(x)

v₂(x) ≤1≤lim inf

x→1

u(x) v₁(x)^.

Using this fact and assertions (iv), (i) and (vi) of Lemma2.6, we deduce that lim inf

x→1

u(x)

ψ_g(θ(1−x)) ≥lim inf

x→1

u(x) v₁(x)

v₁(x)

ψ_g(θ(1−x)) ≥lim inf

x→1

ψ_g(τ₁θ(1−x))

ψ_g(θ(1−x)) =τ₁^1−Cg. By lettingεto zero, we obtain that ξ₁^1−Cg ≤lim infx→1 u(x)

ψg(θ(1−x)). Similarly, we obtain that lim sup_x→1ψ ^u(x)

g(θ(1−x)) ≤ξ¹₂^−Cg.

This proves in particular assertion (ii) of Theorem 1.3. Now, for λ < 2 we have by Lemma2.2

θ(1−x) ∼

x→1

(1−x)^2−λL(1−x)

2−λ .

Hence it follows by assertions (iv), (i) and (vi) of Lemma2.6that forλ<2, we have limx→1

ψ_g(θ(1−x))

ψg((1−x)^2−λL(1−x)) = lim

x→1

ψ_g(θ(1−x)) ψ_g((₂−λ)θ(₁−x))

ψ_g((2−λ)θ(1−x)) ψg((1−x)^2−λL(1−x))

= ¹

(2−λ)^1−Cg.

Proof of Theorem1.6. We recall thatg(t) =t^−γwithγ≥0 andλ=α+1+ (α−1)γ. In this case Cg= _γ^γ₊₁ andλ+ (2−λ)Cg−α−1=0. Let L∈ Kbe the function given in hypothesis(H6). Put

k(t) =

Z _η

t

L(s)

s ds and v_i(x) =

(1+γ)τ_i Z _1−x

0 s^{γ−α(1+γ)}k(s)ds

1/(γ+1)

fori∈ {1, 2}. Then we have

L_Av_i(x) +p(x)g(v_i(x)) =v⁰⁰_i(x) + ^A

0(x) A(x)^v

0

i(x) +p(x)g(v_i(x))

=g(v_i(x))(1−x)^{γ−α(1+γ)−1}L(1−x)

×

τ_ik(1−x) L(1−x)

τ_i(1−x)^{γ−α(1+γ)+1}k(1−x)g⁰(v_i(x)) +γ(1−α) +τ_ik(1−x)

L(1−x)

(x−1)A⁰(x) A(x) −α

−τ_i+p_i+

p(x)

(₁−x)^{γ−α(1+γ)−1}L(₁−x)−p_i

=g(v_i(x))(1−x)^{γ−α(1+γ)−1}L(1−x)

×

τ_i

k(1−x) L(1−x)

τ_i(1−x)^{γ−α(1+γ)+1}k(1−x)g⁰(v_i(x)) +γ(1−α)− ^γ γ+1

+τ_ik(₁−x) L(1−x)

(x−₁)A⁰(x) A(x) −α

+ ^γ

γ+1τ_i−τ_i+p_i+

p(x)

(1−x)^{γ−α(1+γ)−1}L(1−x)−p_i

=g(v_i(x))(1−x)^{γ−α(1+γ)−1}L(1−x)

×

τ_i

k(1−x) L(1−x)

τ_i(1−x)^{γ−α(1+γ)+1}k(1−x)g⁰(v_i(x)) +γ(1−α)− ^γ γ+1

+τ_ik(1−x) L(1−x)

(x−1)B⁰(x) B(x)

− ^τi

γ+1+p_i+

p(x)

(1−x)^{γ−α(1+γ)−1}L(1−x)−p_i

. Sinceg(t) =t^−γ, then, by integration by parts, we obtain

τ_i(1−x)^{γ−α(1+γ)+1}k(1−x)g⁰(v_i(x)) +γ(1−α)

=−γτ_i(₁−x)^{γ−α(1+γ)+1}_k(₁−x)(_vi(_x))^−(1+γ)+γ(₁−α)

=γ (1−α)− (₁−x)^{γ−α(1+γ)+1}k(₁−x) (1+γ)R1−x

0 s^{γ−α(1+γ)}k(s)ds

!

=γ

(₁−α)(₁+γ)R1−x

0 s^{γ−α(1+γ)}k(s)ds−(₁−x)^{γ−α(1+γ)+1}k(₁−x) (₁+γ)R1−x

0 s^{γ−α(1+γ)}k(_s)_ds

!

=γ

−R1−x

0 s^{γ−α(1+γ)+1}k⁰(s)ds (1+γ)R1−x

0 s^{γ−α(1+γ)}k(s)ds

!

= ^γ

γ+1 R1−x

0 s^{γ−α(1+γ)}L(s)ds R1−x

0 s^{γ−α(1+γ)}k(_s)_ds^.

On the other hand, by Remark2.4, we havekinK. This together with Lemma2.2, implies that

limx→1

k(1−x) L(1−x)

τ_i(1−x)^{γ−α(1+γ)+1}k(1−x)g⁰(v_i(x)) +γ(1−α)− ^γ

γ+1 =0.

Now since ^(1−x_B⁾₍^ζ_x^B₎^0(x) is bounded in a neighborhood of 1 and by Proposition2.1, we have

k

L ∈ Kand limx→1 (1−x)^1−ζk(1−x)

L(1−x) =0, we deduce that limx→1

k(1−x) L(1−x)

(1−x)B⁰(x) B(x)

= lim

x→1

(1−x)^1−ζk(1−x) L(1−x)

(1−x)^ζB⁰(x) B(x)

=0.

Letε∈(0, ^p₂¹)and putτ₁ = (γ+1)(p₁−ε)andτ₂ = (γ+1)(p₂+ε).

So, for the fixedε>0, there existsδ_ε ∈(0, 1)such that for x∈(δ_ε, 1), we have L_Av₁(x) +p(x)g(v₁(x))

≥ g(v₁(x))(1−x)^{γ−α(1+γ)−1}L(1−x)

−^ε 3− ^ε

3− ^τ1

γ+1 +p₁− ^ε 3

=0 and

L_Av₂(x) +p(x)g(v₂(x))

≤ g(v₂(x))(1−x)^{γ−α(1+γ)−1}L(1−x) ε

3 + ^ε

3− ^τ2

γ+1 +p₂+ ^ε 3

=0.

This implies thatv₁andv2are respectively a subsolution and a supersolution of the equa- tion−L_Au+p(x)u^−γ =0 in(δε, 1).

Letu ∈ C([0, 1])∩C²((0, 1)be the unique solution of (1.1)–(1.2) (see [15, Theorem 5]). As in the proof of Theorem1.3, we choose M>0 such that

v₁−Mv₀≤u≤v₂+Mv₀ in(δε, 1).

Moreover, thanks to assumption(H₆), we have lim_t→0k(t) =∞. So, using Lemma2.2, we obtain for i∈ {1, 2}

xlim→1

(₁−x)^1−α

(1+γ)τ_iR1−x

0 s^{γ−α(1+γ)}k(s)ds_γ+¹₁

= _lim

x→1

(₁−x)^1−α (1−x)^1−α ₁₋^τi

αk(1−x)^γ+11

=_0.

This together with the fact thatv₀(x) ∼

x→1

(1−x)^1−α

(1−α)B(1) gives that limx→1

v₀(x)

v₁(x) =₀= _lim

x→1

v₀(x) v2(x)^. So we have lim sup_x→1v^u(x)

2(x) ≤1≤lim inf_x→1 u(x)

v₁(x). This implies that lim inf

x→1

u(x)

(1+γ)R1−x

0 s^{γ−α(1+γ)}k(s)ds_γ+¹₁

≥τ₁

1 γ+1.

Sinceτ₁ = (γ+1)(p₁−ε), then by lettingεtends to zero, we get lim inf

x→1

u(x)

(1+γ)R1−x

0 s^{γ−α(1+γ)}k(s)ds_γ¹₊₁

≥((γ+1)p₁)^γ+11.