such that t→∞lim tA0(t) A(t) =α >1, g∈C1((0,∞),(0

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

EXACT ASYMPTOTIC BEHAVIOR OF THE POSITIVE SOLUTIONS FOR SOME SINGULAR DIRICHLET PROBLEMS

ON THE HALF LINE

HABIB M ˆAAGLI, RAMZI ALSAEDI, NOUREDDINE ZEDDINI

Abstract. In this article, we give an exact behavior at infinity of the unique solution to the following singular boundary value problem

−1

A(Au⁰)⁰=q(t)g(u), t∈(0,∞), u >0, lim

t→0A(t)u⁰(t) = 0, lim

t→∞u(t) = 0.

HereAis a nonnegative continuous function on [0,∞), positive and differen- tiable on (0,∞) such that

t→∞lim tA⁰(t)

A(t) =α >1, g∈C¹((0,∞),(0,∞)) is non-increasing on (0,∞) with limt→0g⁰(t)Rt

0 ds

g(s) = −Cg ≤ 0 and the functionqis a nonnegative continuous, satisfying

0< a1= lim inf

t→∞

q(t)

h(t)≤lim sup

t→∞

q(t)

h(t) =a2<∞, whereh(t) =ct^−λexp(Rt

1 y(s)

s ds),λ≥2,c >0 andyis continuous on [1,∞) such that limt→∞y(t) = 0.

1. Introduction

In this article, we give the exact asymptotic behavior at infinity of the unique positive solution to the singular problem

1

A(Au⁰)⁰ =−q(t)g(u), t∈(0,∞), u >0, in (0,∞)

lim

t→0⁺A(t)u⁰(t) = 0, lim

t→∞u(t) = 0,

(1.1)

where the functionsA,qandg satisfy the following assumptions.

(H1) A is a continuous function on [0,∞), positive and differentiable on (0,∞) such that

t→∞lim t A⁰(t)

A(t) =α >1.

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

Key words and phrases. Singular nonlinear boundary value problems; positive solution;

exact asymptotic behavior; Karamata regular variation theory.

2016 Texas State University.c

Submitted December 8, 2015. Published February 17, 2016.

1

(H2) qis a nonnegative continuous function on (0,∞) satisfying 0< a1= lim inf

t→∞

t^λq(t)

L(t) ≤lim sup

t→∞

t^λq(t)

L(t) =a2<∞, whereλ≥2 andL∈ K(see (1.3) below), such thatR∞

1 s^1−λL(s)ds <∞.

(H3) The function g : (0,∞)→ (0,∞) is nonincreasing, continuously differen- tiable such that

lim

t→0⁺g⁰(t) Z t

0

1

g(s)ds=−Cg withCg ≥0.

(H4) α+ 1−λ+ (λ−2)Cg>0.

Using thatg is non-increasing, fort >0, we obtain 0< g(t)

Z t

0

1

g(s)ds≤t.

This implies 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 ds=g(t) Z t

0

1

g(s)ds−t, we obtain

t→0lim g(t)

t Z t

0

1

g(s)ds= 1−Cg. (1.2)

This implies that 0 ≤ C_g ≤ 1. The functions t⁻¹log(1 +t), log(log(e+ ¹_t)), t^−νlog(1 +¹_t), exp{(log(1 +¹_t))^ν},ν ∈(0,1) satisfy the assumption (H3), as well as the function

t²e^1/t, if 0< t < 1 2, 1

4e², ift≥1 2.

Singular nonlinear boundary value problems appear in a variety of applications and often only positive solutions are important. WhenA(t) = 1, problems of type (1.1) with various boundary conditions arise in the study of boundary layer equa- tions for the class of pseudoplastic fluids and have been studied for both bounded and unbounded intervals ofR(see [6, 11, 16, 22, 23, 29]) and the references therein.

WhenA(t) =tⁿ⁻¹(n≥1), the operatoru→_A¹(Au⁰)⁰ appears as the radial part of the laplace operator ∆ (see [10, 30]). Other results of existence and uniqueness of positive solutions were obtained by Agarwal and O’Regan in [1] on the interval (0,1) and in the case whereAis continuous on [0,1], positive and differentiable on (0,1) and satisfying an integrability condition. In general the exact asymptotic behavior of the unique positive solution of (1.1) is extremely complex when the coefficients are in general continuous functions, even though upper and lower bounds for this solution are often given (see [1, 4, 10, 15]). Recent research (see [2, 8, 16]) show that these problems should be studied in the case of Karamata regularly varying functions. This approach was initiated by Avakumovic [3] and followed by Maric and Tomic (see [20, 21]). Our aim in this paper is to give a contribution to the qualitative analysis of problem (1.1) by giving the exact asymptotic behavior at infinity of the unique positive solution under the previous assumptions onA,qand g. We note that the existence and uniqueness of such a solution are established by Mˆaagli and Masmoudi in [17]. For related results, we refer to Barile and Salvatore

[5], Cencelj, Repovˇs, and Virk [7], Ghergu and R˘adulescu [13, 14, 15, 16], R˘adulescu and Repovˇs [24, 25], Repovˇs [26, 27].

To state our results, we denote byKthe set of Karamata functionsLdefined on [1,∞) by

L(t) :=cexpZ t 1

y(s) s ds

, (1.3)

wherec >0 andy∈C([1,∞)) such that lim_t→∞y(t) = 0.

Remark 1.1. It is clear that a function L is in K if and only if L is a positive function inC¹([1,∞)) such that

t→∞lim t L⁰(t)

L(t) = 0. (1.4)

Throughout this paper, we denote byψ_g the unique solution of the equation Z ψg(t)

0

ds

g(s) =t , fort∈[0,∞), (1.5) and we mention that

t→0limtg⁰(ψ_g(t)) =−C_g. (1.6) Theorem 1.2. Assume (H1)–(H4). Then problem (1.1) has a unique solution u∈C²((0,∞))∩C([0,∞))satisfying

(i) If λ >2, ξ1

λ−2 1−Cg

≤lim inf

t→∞

u(x)

ψg(t^2−λL(t)) ≤lim sup

t→∞

u(x)

ψg(t^2−λL(t)) ≤ ξ2

λ−2 1−Cg

, whereξ_i=α+1−λ+(λ−2)C^ai g fori∈ {1 2}.

(ii) If λ= 2, ξ11−Cg ≤lim inf

t→∞

u(t) ψ_g(R∞

t L(s)

s ds) ≤lim sup

t→∞

u(t) ψ_g(R∞

t L(s)

s ds) ≤ξ21−Cg

An immediate consequence of Theorem 1.2 is the following result.

Corollary 1.3. Letube the unique solution of (1.1). Then, we have the following exact asymptotic behavior:

(a) WhenCg= 1, we have (i) lim_t→∞ψ ^u(t)

g(t^2−λL(t)) = 1 ifλ >2;

(ii) lim_t→∞ ^u(t)

ψ_g(R∞ t

L(s)

s ds)= 1 ifλ= 2.

(b) When Cg<1 anda1=a2=a0, we have (i) lim_t→∞ψ ^u(t)

g(t^2−λL(t)) = ((λ−2)(α+1−λ+(λ−2)C^a0 _g))^1−Cg ifλ >2;

(ii) limt→∞ u(t) ψg(R∞

t L(s)

s ds)= (_α−1^a0 )^1−Cg ifλ= 2.

Remark 1.4. In the hypothesis (H3), we do not need the monotonicity of the functiong on (0,∞), but only the fact that gis non-increasing in a neighborhood of zero.

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

(t²e^1/t, if 0< t < ¹₂,

1

4e², ift≥¹₂. and letqbe a nonnegative function in (0,∞) such that

t→∞lim q(t)

h(t) =b₀∈(0,∞), whereh(t) =t^−λL(t),λ≥2 and L∈ Ksuch thatR∞

1 s^1−λL(s)ds <∞. Then, we have C_g = 1 and ψ_g(ξ) = _log(ξ)⁻¹ forξ ∈(0, e⁻²). Letu be the unique solution of (1.1), then we have the following exact behavior:

(i) limt→∞u(t) log _t2−λ¹L(t)

= 1 ifλ >2;

(ii) lim_t→∞u(t) log R∞ ¹ t

L(s) s ds

= 1 ifλ= 2.

To establish our second result, we consider the special case where g(t) = t^−γ with γ ≥ 0, and λ = α+ 1 +γ(α−1). Note that in this case C_g = _γ+1^γ and (α+ 1−λ) + (λ−2)C_g = 0. We assume that A and q satisfy the following hypotheses:

(H5) A is a continuous function on (0,∞) such thatA(t) =t^αB(t) with α > 1 and ^tν_B(t)^B0(t) is bounded fortlarge andν ∈(0,1).

(H6) qis a nonnegative continuous function in (0,∞) and satisfies 0< a₁= lim inf

t→∞

q(t)

t^{γ−1−α(γ+1)}L(t) ≤lim sup

t→∞

q(t)

t^{γ−1−α(γ+1)}L(t) =a₂<∞, whereL∈ KwithR∞

1 L(s)

s ds=∞.

Theorem 1.6. Assume (H5), (H6)are satisfied. Then the Dirichlet problem

−1

A(A u⁰)⁰=q(t)u^−γ, t∈(0,∞), lim

t→0⁺A(t)u⁰(t) = 0, lim

t→∞u(t) = 0,

(1.7)

has a unique solutionu∈C([0,∞))∩C²((0,∞)), satisfying (γ+ 1)a1

α−1 _1+γ¹

≤lim inf

t→∞

u(t) t^1−α Rt

1 L(s)

s ds_1+γ¹

≤lim sup

t→∞

u(t) t^1−α Rt

1 L(s)

s ds_1+γ¹

≤((γ+ 1)a2

α−1 )^1+γ1 , In particular ifa1=a2, then

t→∞lim

u(t) t^1−α Rt

1 L(s)

s ds_1+γ¹

=(γ+ 1)a1

α−1 _1+γ¹

.

2. On the Karamata class

To make the paper self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following result is due to [19, 28].

Lemma 2.1. (i) LetL∈ Kand ε >0, then

t→∞lim t^−εL(t) = 0.

(ii) LetL₁, L₂∈ Kandp∈R. ThenL₁+L₂∈ K,L₁L₂∈ KandL^p₁∈ K. Applying Karamata’s theorem (see [19, 28]), we get the following result.

Lemma 2.2. Let γ∈R,L be a function inK defined on [1,∞). We have (i) If γ <−1, thenR∞

1 s^γL(s)dsconverges. Moreover Z ∞

t

s^γL(s)ds∼_t→∞−t^1+γL(t) γ+ 1 . (ii) If γ >−1, thenR∞

1 s^γL(s)dsdiverges. Moreover Z t

1

s^γL(s)ds∼_t→∞ t^1+γL(t) γ+ 1 . Lemma 2.3 ([8, 18]). Let L∈ K be defined on[1,∞). Then

t→∞lim L(t) Rt

1 L(s)

s ds

= 0. (2.1)

If further R∞ 1

L(s)

s dsconverges, then

t→∞lim

L(t) R∞

t L(s)

s ds

= 0. (2.2)

Remark 2.4. LetL∈ K, then using Remark 1.1 and (2.1), we deduce that t→

Z t

1

L(s)

s ds∈ K.

If furtherR∞ 1

L(s)

s dsconverges, then t→R∞ t

L(s)

s ds∈ K.

Definition 2.5. A positive measurable function k is called normalized regularly varying at infinity with index ρ∈Rand we writek∈N RV Iρ ifk(s) =s^ρL(s) for s∈[1,∞) withL∈ K.

Using the definition of the classKand the above Lemmas we obtain the following lemma.

Lemma 2.6 ([2]). (i) If k∈ N RV I_ρ, then lim_t→∞^k(ξt)_k(t) =ξ^ρ, uniformly for ξ∈[c1, c2]⊂(0,∞).

(ii) A positive measurable functionkbelongs to the classN RV Iρ if and only if limt→∞tk⁰(t)

k(t) =ρ.

(iii) Let L ∈ K and assume that R∞

1 s^1−λL(s)ds < ∞. Then the function θ(t) =R∞

t s^1−λL(s)ds belongs toN RV I_(2−λ). (iv) The functionψ_g◦θ∈N RV I_{(2−λ)(1−Cg)}.

(v) Let m1, m2 be positive functions on (0,∞) such that limt→∞m1(t) = lim_t→∞m₂(t) = 0 andlim_t→∞^m_m^1(t)

2(t)= 1. Thenlim_t→∞^ψ_ψ^g(m1(t))

g(m2(t)) = 1.

3. Proofs of Theorems 1.2 and 1.6 In the sequel, we denote by

v0(t) = Z ∞

t

1

A(s)ds fort∈(0,∞).

Since the functionA satisfies (H1), then using Definition 2.5 and assertion (ii) of Lemma 2.6, we deduce that there exists L0 ∈ K such that A(t) = t^αL0(t), for t >1. Hence, using Lemma 2.1, we deduce that 1/Ais integrable near infinity. So the functionv0is well defined, and by Lemma 2.2 we have

v₀(t) = Z ∞

t

1

A(s)ds∼ t^1−α

(α−1)L0(t) ast→ ∞. (3.1) In the sequel, we denote also byLAu:= _A¹(Au⁰)⁰ =u⁰⁰+^A_A⁰u⁰ and we remark that LAv0= 0.

Proof of Theorem 1.2. Letε∈(0, a1/2). Put

ξ_i= a_i

(α+ 1−λ) + (λ−2)Cg

fori∈ {1,2}, τ1 = ξ1−ε_a^ξ1

1 and τ2 = ξ2+ε^ξ_a²

2. Clearly, we have ^ξ₂¹ < τ1 < τ2 < ³₂ξ2. Let θ(t) =R∞

t s^1−λL(s)dsand put ωi(t) =ψg

τi

Z ∞

t

s^1−λL(s)ds

=ψg(τiθ(t)), fort >0.

By a simple calculus, fori∈ {1,2}we obtain L_Aω_i(t) +q(t)g(ω_i(t))

=g(ωi(t))t^−λL(t))h

τi(τit^2−λL(t)g⁰(ωi(t)) + (λ−2)Cg)

−τi

t A⁰(t)

A(t) −α+t L⁰(t) L(t)

−τi((α+ 1−λ+ (λ−2)Cg) +ai

+ q(t) t^−λL(t)−ai

i .

So, for the fixedε >0, there existsM_ε>1 such that fort > M_εandi∈ {1,2}, we have

τi

t A⁰(t)

A(t) −α+t L⁰(t) L(t)

≤ 3

2ξ2

t A⁰(t) A(t) −α

+

t L⁰(t) L(t)

≤ ε 4, a1−ε

2 ≤ a(t)

t^−µL(t) ≤a2+ε 2

|τi(τit^2−λL(t)g⁰(ωi(t)) + (λ−2)Cg)| ≤ 3

2ξ2|τit^2−λL(t)g⁰(ωi(t)) + (λ−2)Cg| ≤ ε 4. Indeed, the last inequality follows from (1.6) and the fact that from Lemmas 2.2 and 2.3, we have

t→∞lim

t^2−λL(t) R∞

t s^1−λL(s)ds = 2−λ, for allλ≥2. This implies that for eacht > Mε, we have

L_Aω₁(t) +q(t)g(ω₁(t))≥g(ω₁(t))t^−λL(t)[−ε+a₁−τ₁((α+ 1−λ) + (λ−2)C_g)] = 0

and

LAω2(t) +q(t)g(ω2(t))≤g(ω2(t))t^−λL(t)[ε+a2−τ2((α+ 1−λ) + (λ−2)Cg)] = 0.

Let u∈ C²((0,∞))∩C([0,∞)) be the unique solution of (1.1) (see [17]). Then, there existsB >0 such that

ω1(Mε)−B v0(Mε)≤u(Mε)≤ω2(Mε) +B v0(Mε). (3.2) We claim that

ω1(t)−B v0(t)≤u(t)≤ω2(t) +B v0(t) for allt > Mε. (3.3) Assume for instance that the right inequality of (3.3) is not true. Then the function h(t) =u(t)−ω2(t)−B v0(t) for t > Mε is not negative. Consequently, there exists t1> Mεsuch thath(t1) = maxM_ε≤t<∞h(t)>0. Sincehis continuous on [Mε,∞), h(Mε)≤0 and limt→∞h(t) = 0, thenh⁰(t1) = 0 andh(t)>0 fort∈(t1−δ, t1+δ) for someδ >0, sufficiently small. Namelyu(t)> ω2(t)+B v0(t) fort∈(t1−δ, t1+δ).

Sinceg is non-increasing on (0,∞), then 1

A(t)(A(t)h⁰(t))⁰ =−q(t)g(u(t))− 1

A(t)(A(t)ω⁰₂(t))⁰ ≥q(t)(g(ω2(t))−g(u(t)))≥0, for t ∈ (t₁−δ, t₁+δ). Which implies h⁰(t) ≤h⁰(t₁) = 0 for t ∈ (t₁−δ, t₁) and h⁰(t)≥h⁰(t₁) = 0 fort∈(t₁, t₁+δ). This implies thathhas a local minimum at t₁. Which contradicts the fact that h a global maximum at t₁ on [M_ε,∞). This proves that

u(t)≤ω2(t) +B v0(t) for allt > Mε. Similarly, we show that

ω1(t)−B v0(t)≤u(t) for allt > Mε. This proves (3.3).

Now, since ψ_g◦θ ∈ N RV I_{(2−λ)(1−Cg)}, there exists ˆL ∈ K such that ψ_g◦θ = t^{(2−λ)(1−Cg)}L(t) forˆ t ∈[1,∞). Moreover since (α−1)−(λ−2)(1−Cg) >0, it follows by Lemma 2.1 that

t→∞lim

t^1−α

t^{(2−λ)(1−Cg)}L(t)ˆ = 0.

This implies that

t→∞lim

t^1−α ψg(τi

R∞

t s^1−λL(s)ds) = lim

t→∞

t^1−α

ψg(τiθ(t))= lim

t→∞

ψ_g(θ(t)) ψg(τiθ(t))

t^1−α ψg(θ(t))= 0 uniformly inτ_i∈[^ξ₂¹,³₂ξ₂]⊂(0,∞). This together with (3.1) implies

t→∞lim

v₀(t)

ψg(τ1θ(t))= lim

t→∞

v₀(t) ψg(τ2θ(t)) = 0.

So, we obtain

lim sup

t→∞

u(t)

ω2(t) ≤1≤lim inf

t→∞

u(t) ω1(t).

Using this fact and assertions (i) and (iv) of Lemma 2.6, we deduce that lim inf

t→∞

u(t)

ψg(θ(t))= lim inf

t→∞

u(t) ω1(t)

ω1(t)

ψg(θ(t))≥ lim

t→∞

ψg(τ1θ(t))

ψg(θ(t)) =τ11−Cg.

By lettingεapproach zero, we obtain ξ₁^1−Cg ≤lim inf

t→∞

u(t) ψg(θ(t)). Similarly, we obtain

lim sup

t→∞

u(t)

ψ_g(θ(t))≤ξ₂^1−Cg.

This proves in particular the exact behavior at infinity in the case λ = 2. Now, for λ >2, we have by Lemma 2.2 that θ(t) ∼t→∞ t^2−λ

λ−2L(t). Hence it follows by assertions (i), (iv) and (v) of Lemma 2.6 that forλ >2, we have

t→∞lim

ψg(θ(t))

ψ_g((t)^2−λL(t))= lim

t→∞

ψg(θ(t)) ψ_g((λ−2)θ(t))

ψg((λ−2)θ(t))

ψ_g((t)^2−λL(t)) = 1 (λ−2)^1−Cg.

This achieves the proof of the Theorem.

Proof of Theorem 1.6. We recall thatg(t) =t^−γ,λ=α+1+(α−1)γandC_g= _1+γ^γ . Let ε ∈ (0,^a₂¹) and put τ₁ = (γ+ 1)(a₁−ε) and τ₂ = (γ+ 1)(a₂+ε). Put k(t) =Rt

1 L(s)

s dsand ω_i(t) =

(1 +γ)τ_i Z ∞

t

s^1−λk(s)ds_1+γ¹

fori∈ {1,2},

whereLis the function given in hypothesis (H6). Then, by a simple computation, we have

L_Aω_i(t) +q(t)g(ω_i(t))

=g(ωi(t))t^−λL(t)h τi

k(t)

L(t)(τit^2−λk(t)g⁰(ωi) + (λ−1−α))− γ γ+ 1

−τ_ik(t) L(t)

t A⁰(t) A(t) −α

+ γ

γ+ 1τ_i−τ_i+a_i+ q(t)

t^−λL(t)−a_ii

=g(ω_i(t))t^−λL(t)h τ_ik(t)

L(t)(τ_it^2−λk(t)g⁰(ω_i) + (α−1)γ)− γ γ+ 1

−τi

k(t) L(t)

t B⁰(t) B(t) − τi

γ+ 1+ai+ q(t) t^−λL(t)−ai

i .

Sinceg(t) =t^−γ andλ=α+ 1 + (α−1)γ, integrating by parts, we obtain τit^2−λk(t)g⁰(ωi(t)) + (α−1)γ

=−γτit^2−λk(t)(ωi(t))^−(1+γ)+ (α−1)γ

=γ

(α−1)− t^2−αk(t) (γ+ 1)R∞

t s^1−λk(s)ds

=γ(α−1)(1 +γ)R∞

t s^1−λk(s)ds−t^2−λk(t) (1 +γ)R∞

t s^1−λk(s)ds

= γ

γ+ 1 R∞

t s^1−λL(s)ds R∞

t s^1−λk(s)ds. This gives

k(t)

L(t)(τ_it^2−λk(t)g⁰(ω_i(t)) + (α−1)γ)− γ γ+ 1

= γ γ+ 1

hR∞

t s^1−λL(s)ds t^2−λL(t)

t^2−λk(t) R∞

t s^1−λk(s)ds−1i .

This together with Lemma 2.2 and the fact thatkand Lare inK, implies

t→∞lim k(t)

L(t)(τ_it^2−λk(t)g⁰(ω_i(t)) + (α−1)γ)− γ γ+ 1 = 0.

Now since ^tν_B(t)^B0(t) is bounded for t large and by Lemma 2.1, we have _L^k ∈ K and lim_t→∞^t1−ν_L(t)^k(t) = 0, we deduce that

t→∞lim k(t) L(t)

t B⁰(t) B(t)

= lim

t→∞

t^1−νk(t) L(t)

t^νB⁰(t) B(t)

= 0.

So, for the fixedε >0, there existsM_ε>1 such that for t≥M_ε, we have LAω2(t) +q(t)g(ω2(t))≤g(ω2(t))t^−λL(t)hε

3+ε 3− τ2

γ+ 1 +a2+ε 3 i

= 0, L_Aω₁(t) +q(t)g(ω₁(t))≥g(ω₁(t))t^−µL(t)[−ε

3 −ε 3− τ₁

γ+ 1+a₁−ε 3

i= 0.

Letu∈C([0,∞))∩C²((0,∞)) be the unique solution of (1.7). As in the proof of Theorem 1.2, we chooseC >0 such that

ω1(t)−Cv0(t)≤u(t)≤ω2(t) +Cv0(t) fort≥Mε.

Moreover, thanks to (H6), we have lim_t→∞k(t) = ∞. So, using Lemma 2.2, we obtain

t→∞lim

1 t^α−1

(1 +γ)τ₁R∞

t s^1−λk(s)ds_1+γ¹ = lim

t→∞

1

t^α−1 t^(2−λ)τ1_α−1^k(t)_1+γ¹

= lim

t→∞

α−1 τ₁k(t)

_1+γ¹

= 0.

This and (3.1) gives lim_t→∞ω^v0(t)

1(t) = 0. Similarly, we obtain lim_t→∞ω^v0(t)

2(t) = 0. So we have

lim sup

t→∞

u(t)

ω2(t) ≤1≤lim inf

t→∞

u(t) ω1(t). This implies that

lim inf

t→∞

u(t) (1 +γ)R∞

t s^1−λk(s)ds_1+γ¹

≥τ

1 1+γ

1 . Now, asεtends to zero, we obtain

lim inf

t→∞

u(t)

(1 +γ)R∞

t s^1−λk(s)ds_1+γ¹ ≥((γ+ 1)a₁)^1+γ1 . Similarly, we obtain

lim sup

t→∞

u(t)

(1 +γ)R∞

t s^1−λk(s)ds_1+γ¹

≤((γ+ 1)a2)^1+γ1 .

Now, since (γ+1)R∞

t s^1−λk(s)ds∼t→∞ t^2−λk(t)

α−1 =^{t(1−α)(γ+1)}_α Rt 1

L(s)

s ds, we deduce that

(γ+ 1)a₁ α−1

_1+γ¹

≤lim inf

t→∞

u(t) t^1−α Rt

1 L(s)

s ds_1+γ¹

≤lim sup

t→∞

u(t) t^1−α Rt

1 L(s)

s ds_1+γ¹

≤((γ+ 1)a2

α )^1+γ1 . In particular, ifa1=a2, we obtain

t→∞lim

u(t) t^1−α Rt

1 L(s)

s ds_1+γ¹

=(γ+ 1)a1

α−1 _1+γ¹

.

4. Applications

Application 1. We consider the Dirichlet problem

−1

A(Au⁰)⁰+β

u(u⁰)²=q(t)g(u), t∈(0,∞), u >0, in (0,∞),

lim

t→0⁺A(t)u⁰(t) = 0 lim

t→∞u(t) = 0,

(4.1)

whereβ <1 and lim_t→∞t−λ^q(t)L(t) =a0>0 withλ≥2 andL∈ Kwith R∞

1 s^1−λL(s)ds <∞.

We assume thatAsatisfies (H1) andg satisfies the following hypotheses:

(A1) The functiont→t^−βg(t) is non-increasing from (0,∞) into (0,∞).

(A2) lim_t→0g⁰(t)Rt 0

1

g(s)ds=−Cg with max(0,_β−1^β )≤Cg≤1.

(A3) (α−1)−(λ−2)(1−β)(1−Cg)>0

Note that for γ >0 and −γ < β <1, the function g(t) =t^−γ satisfies (A₁) and (A2). Putu=v^1−β1 . Thenv satisfies

−1

A(Av⁰)⁰= (1−β)q(t)g(v^1−β1 )v^1−β−β , t∈(0,∞), v >0, in (0,∞),

lim

t→0⁺A(t)v⁰(t) = 0, lim

t→∞v(t) = 0,

(4.2)

The functionf(r) = (1−β)g(r^1−β1 )r^−1−ββ is non-increasing on (0,∞) and a simple computation shows thatψg= (ψf)^1−β1 and

r→0limf⁰(r) Z r

0

1

f(s)ds= (1−β)(1−Cg)−1 =:−Cf, with 0≤Cf ≤1.

Applying Corollary 1.3 to problem (4.2), we deduce that there exists a unique solutionvto (4.2) such that

(a) WhenC_f = 1, we have (i) lim_t→∞ψ ^v(t)

f(t^2−λL(t)) = 1 ifλ >2;

(ii) lim_t→∞ ^v(t)

ψf R∞ t

L(s)

s ds = 1 ifλ= 2;

(b) WhenC_f <1, we have:

(i) lim_t→∞ψ ^v(t)

f(t^2−λL(t)) = [α+1−λ+(λ−2)C^a0 f]^1−Cf if 2< λ <2 + _{(1−β)(1−c}^α−1

g); (ii) lim_t→∞ ^v(t)

ψ_f(R∞ t

L(s)

s ds) = [_α−1^a0 ]^1−Cf ifλ= 2.

This implies that problem (4.1) has a solutionu∈C([0,∞))∩C²((0,∞)) satisfying the following exact behavior

(a) WhenCg= 1, we have:

(i) ifλ >2, then

t→∞lim

u(t)

ψg(t^2−λL(t)) = 1;

(ii) ifλ= 2, then

t→∞lim

u(t) ψg(R∞

t L(s)

s ds)= 1;

(b) If max(0,_β−1^β )≤C_g<1, then:

(i) if 2< λ <2 +_{(1−β)(1−C}^α−1

g), then

t→∞lim

u(t)

ψg(t^2−λL(t)) = [ a₀

α−1−(λ−2)(1−β)(1−Cg)]^1−Cg (ii) ifλ= 2, then

lim

|x|→∞

u(t) ψg(R∞

t L(s)

s ds)

= [ a0

α−1]^1−Cg.

Application 2. In this subsection, we assume that the function A satisfy the following hypothesis

(A4) A is a continuous function on [0,∞), positive and differentiable on (0,∞) such that _A¹ is integrable near 0 and lim_t→∞^{t A}_A(t)^0(t)=σ∈R− {1}.

We are interested in the exact behavior at infinity of the unique positive solution of the problem

1

A(t)(A(t)u⁰(t))⁰=−p(t)u^−γ, t∈(0,∞), u >0, in (0,∞),

u(0) = 0, lim

t→∞

u(t) ρ(t) = 0,

(4.3)

where γ > 0 and ρ(t) = Rt 0

ds

A(s). Let u(t) = ρ(t)v(t) and B(t) = A(t)ρ²(t) for t ∈ [0,∞). Then u is a positive solution of (4.3) if and only if v is a positive solution of the problem

1

B(t)(B(t)v⁰(t))⁰ =− p(t)

(ρ(t))^γ+1v^−γ, t∈(0,∞), v >0, in (0,∞),

lim

t→0⁺B(t)v⁰(t) = 0, lim

t→∞v(t) = 0.

(4.4)

First, we claim that ifAsatisfies (A4), then

t→∞lim tB⁰(t)

B(t) = 1 +|σ−1|>1. (4.5) Since ^{t B}_B(t)^0(t) = ^{t A}_A(t)^0(t)+_A(t)ρ(t)^2t and by Definition 2.5 and assertion (ii) of Lemma 2.6, we have A(t) = t^σL0(t) for t ≥ a > 1 with L0 ∈ K, then we deduce from Lemma 2.2 that

•Forσ <1, we haveρ(∞) =∞and so ρ(t) =

Z a

0

ds A(s)+

Z t

a

1

s^σL0(s)ds∼ 1 1−σ

t^1−σ

L0(t) ast→ ∞.

So

2t

A(t)ρ(t) ∼(1−σ)L0(t) t^1−σ

2t

t^σL₀(t) = 2(1−σ) as t→ ∞.

Consequently in this case we have

t→∞lim t B⁰(t)

B(t) =σ+ 2(1−σ) = 2−σ= 1 +|σ−1|.

•Forσ >1, we haveρ(∞) =R∞ 0

ds

A(s)ds <∞. So 2t

A(t)ρ(t) ∼ 2t

t^σL₀(t)ρ(∞)→0 ast→ ∞.

In this case we have

t→∞lim t B⁰(t)

B(t) =σ= 1 +|σ−1|.

This proves (4.5). Taking into account this fact, we assume that the function p satisfies the following hypotheses

(A5) pis a nonnegative continuous function (0,∞) satisfying 0< a₀= lim

t→∞

t^λp(t)

L(t)(ρ(t))^γ+1 <∞, whereλ≥2 andL∈ Ksuch thatR∞

1 s^1−λL(s)ds <∞.

(A6) 2 +|σ−1| −λ+ (λ−2)_γ+1^γ >0.

Assume that A andpsatisfy (A4)–(A6) and let v be the unique positive solution of problem (4.4). Thenvhas the following exact behavior at infinity

(i) ifλ >2, then

t→∞lim

v(t)

[(γ+ 1)t^2−λL(t)]^1+γ1

=h a₀

(λ−2)(2 +|σ−1| −λ+ (λ−2)_γ+1^γ ) i_1+γ¹

. (ii) ifλ= 2, then

t→∞lim

v(t) [(γ+ 1)R∞

t L(s)

s ds]^1+γ1

= [ a₀

|σ−1|

i_1+γ¹

Consequently, the unique positive solution u of problem (4.3) has the following exact behavior at infinity

(i) ifλ >2, then

t→∞lim

u(t)

ρ(t)[(γ+ 1)t^2−λL(t)]^1+γ1

=h a0

(λ−2)(2 +|σ−1| −λ+ (λ−2)_γ+1^γ ) i_1+γ¹

.

ii) if λ= 2, then

t→∞lim

u(t) ρ(t)[(γ+ 1)R∞ t

L(s) s ds]^1+γ1

=h a0

|σ−1|

i_1+γ¹

Acknowledgements. This project was funded by the Deanship of Scientific Re- search (DSR), King Abdulaziz University, Jeddah, under grant No. (253-662-1436- G). The authors, therefore, acknowledge with thanks DSR technical and financial support.

The three authors have contributed equally for this article.

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Habib Mˆaagli

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

E-mail address:[email protected]

Ramzi Alsaedi

Department of Mathematics, Faculty of Sciences, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

E-mail address:[email protected]

Noureddine Zeddini

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

E-mail address:[email protected]