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BY-NC-ND 3.0 license Open Access Published by De Gruyter August 20, 2015

Existence and global asymptotic behavior of positive solutions for combined second-order differential equations on the half-line

  • Imed Bachar EMAIL logo and Habib Mâagli

Abstract

We are concerned with the existence, uniqueness and global asymptotic behavior of positive continuous solutions to the second-order boundary value problem

1A(Au)+a1(t)uσ1+a2(t)uσ2=0,t(0,),

subject to the boundary conditions limt0+u(t)=0, limtu(t)/ρ(t)=0, where σ1,σ2<1 and A is a continuous function on [0,) which is positive and differentiable on (0,) such that 011/A(t)𝑑t< and 01/A(t)𝑑t=. Here, ρ(t)=0t1/A(s)𝑑s for t>0 and a1,a2 are nonnegative continuous functions on (0,) that may be singular at t=0 and satisfying some appropriate assumptions related to the Karamata regular variation theory. Our approach is based on the sub-supersolution method.

MSC 2010: 34B15; 34B18; 34B27

1 Introduction

Problems of the type

(1.1){1A(Au)+φ(,u)=0on (0,),u>0on (0,),limt0+u(t)=0,

where A is a positive and differentiable function on (0,) and φ is a measurable function on (0,)×(0,) that satisfies some further hypotheses, have been extensively studied in the literature (see, for example, [1, 2, 3, 18, 19, 26, 28] and the references therein).

In [28], Zhao considered (1.1) with A(t)1, where φ is a measurable function on (0,)×(0,) dominated by a convex positive function. Then, he showed that there exists b>0 such that for each μ(0,b], there exists a positive continuous solution u of (1.1) satisfying limtu(t)/t=μ.

In [18], Mâagli and Masmoudi generalized the result of Zhao to the more general boundary value problem

(1.2){1A(Au)+f(,u,Au)=0on (0,),u>0on (0,),limt0+u(t)=0,

where A is a positive and differentiable function on (0,) and f is a measurable function on (0,)×(0,)×(0,) which may change sign and is dominated by a regular function. Then, they proved the existence of a constant b>0 such that for each μ(0,b], (1.2) has a continuous solution u satisfying

limtu(t)ρ(t)=μ,

where

ρ(t)=0t1A(s)𝑑s,t0.

On the other hand, in [2], the first author studied (1.1) with A satisfying some appropriate conditions and where φ:(0,)×(0,)[0,) is a continuous function, nonincreasing with respect to the second variable such that for each c>0, φ(,c)>0 and

0A(s)min(1,ρ(s))φ(s,c)𝑑s<.

He proved the existence of a unique positive solution u in C([0,))C2((0,)) to (1.1) satisfying

limtu(t)ρ(t)=0.

Recently, in [3], we have proved the existence, uniqueness and the global asymptotic behavior for a positive solution to the boundary value problem

(1.3){1A(Au)+a(t)uσ=0for t(0,),u>0on (0,),limt0+u(t)=0,limtu(t)ρ(t)=0,

where σ<1 and where A is a continuous function on [0,) which is positive and differentiable on (0,) such that

011A(t)𝑑t<,01A(t)𝑑t=

with

ρ(t)=0t1A(s)𝑑s.

The nonnegative continuous function a is required to satisfy some appropriate assumptions related to the Karamata regular variation theory.

In this paper, we deal with the boundary value problem

(1.4){1A(Au)+a1(t)uσ1+a2(t)uσ2=0for t(0,),u>0on (0,),limt0+u(t)=0,limtu(t)ρ(t)=0,

where σ1,σ2<1 and A is a continuous function on [0,) which is positive and differentiable on (0,) such that

011A(t)𝑑t<,01A(t)𝑑t=

with

ρ(t)=0t1A(s)𝑑s,t0.

Our goal is to study (1.4) especially when the nonlinearity is the sum of a singular and a sublinear term. Under appropriate assumptions on a1 and a2 related to the Karamata classes 𝒦 and 𝒦 (see Definitions 1.1 and 1.2), we prove the existence, uniqueness and the global asymptotic behavior of a positive continuous solution to (1.4). In particular, we extend the results obtained in [3].

Observe that in the case A(t)=tn-1, n2, the operator

u1A(Au)

is the radially symmetric Laplacian Δu in n, where u is a function that depends only on t=|x|, xn (see [26]). For various existence, uniqueness and asymptotic behavior results for such a problem, we refer the reader to [4, 6, 7, 8, 9, 10, 11, 15, 14, 13, 12, 17, 21, 22, 23, 24, 26, 27] and the references therein. However, we emphasize that in (1.4), the function A(t)=tn-1, n2, is not allowed and the potential functions a1 and a2 may be singular at t=0.

Throughout this paper and without loss of generality, we assume that

011A(t)𝑑t=1.

To state our result, we need some definitions and notation.

Definition 1.1

The class 𝒦 is the set of all Karamata functions L defined on (0,η] by

L(t):=cexp(tηz(s)s𝑑s),η>1,

where c>0 and zC([0,η]) such that z(0)=0.

Definition 1.2

The class 𝒦 is the set of all Karamata functions L~ defined on [1,) by

L~(t):=cexp(1tz(s)s𝑑s),

where c>0 and zC([1,)) such that

limtz(t)=0.

Note that the functions belonging to the classes 𝒦 and 𝒦 are in particular slowly varying functions. The theory of such functions was initiated by Karamata in the fundamental paper [16]. 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 Cirstea and Rădulescu [10].

It is easy to verify the following result.

Proposition 1.3

  1. A function L is in 𝒦 if and only if L is a positive function in C1((0,η]) for some η>1 such that

    limt0+tL(t)L(t)=0.
  2. A function L~ is in 𝒦 if and only if L~ is a positive function in C1([1,)) such that

    limttL~(t)L~(t)=0.
  3. Let L~ be a function in 𝒦. Then, there exists m0 such that for every α>0 and t1 we have

    (1+α)-mL~(t)L~(α+t)(1+α)mL~(t).

As a typical example of a function belonging to the classes 𝒦 and 𝒦 (see [5, 20, 25]), we quote

L(t)=k=1m(logk(ωt))ξk,L~(t)=2+sin(log2(ωt)),L~(t)=exp(k=1m(logk(ωt))τk),

where

logkx=logloglogk timesx

and ξk, τk(0,1) and where ω is a sufficiently large positive real number such that L (resp. L~) is defined and positive on (0,η] for η>1 (resp. on [1,)).

In the sequel, we will denote by B+((0,)) the set of nonnegative Borel measurable functions in (0,) and by C0([0,)) the set of continuous functions v on [0,) such that limtv(t)=0. It is known that C0([0,)) is a Banach space with the uniform norm v=supt>0|v(t)|.

For s,t, we write min(s,t)=st and max(s,t)=st. For two nonnegative functions f and g defined on a set S, the notation f(t)g(t), tS, means that there exists c>0 such that

1cf(t)g(t)cf(t)

for all tS. Furthermore, we denote by G(t,s)=A(s)ρ(ts) the Green’s function of the operator

u-1A(Au)

on (0,) with the Dirichlet conditions

limt0+u(t)=0,limtu(t)ρ(t)=0.

For fB+((0,)), we put

Vf(t)=0G(t,s)f(s)𝑑t,t>0.

We point out that if the map

sA(s)ρ(1s)f(s)

is continuous and integrable on (0,), then Vf is the nonnegative continuous solution of the boundary value problem

(1.5){-1A(Au)=fin (0,),limt0+u(t)=0,limtu(t)ρ(t)=0.

To simplify our statement, we introduce the following. Let λ2, σ<1 and μ1+σ. Let also L𝒦 be defined on (0,η] for η>1 and L~𝒦 such that

0ηL(s)sλ-1𝑑s<,1L~(s)sμ-σ𝑑s<.

Define the function ΨL,λ,σ(t) for t(0,η) by

(1.6)ΨL,λ,σ(t)={(0tL(s)s𝑑s)11-σif λ=2,(L(t))11-σif  1+σ<λ<2,(tηL(s)s𝑑s)11-σif λ=1+σ,1if λ<1+σ,

and Ψ~L~,μ,σ(t) for t[1,) by

(1.7)Ψ~L~,μ,σ(t)={(tL~(s)s𝑑s)11-σif μ=1+σ,(L~(t))11-σif  1+σ<μ<2,(1t+1L~(s)s𝑑s)11-σif μ=2,1if μ>2.

Throughout this paper we will make the following assumption.

  1. For i{1,2}, let σi<1 and let ai be a nonnegative continuous function on (0,) such that

    (1.8)ai(t)1(A(t))2(ρ(t))-λi(1+ρ(t))λi-μiLi(ρ(1t))L~i(ρ(1t)),t>0,

    where λi2, μi1+σi and with Li𝒦, L~i𝒦 satisfying

    (1.9)0ηLi(s)sλi-1𝑑s<,1L~i(s)sμi-σi𝑑s<.

As it will be seen, for i{1,2}, the numbers

(1.10)νi=min(1,2-λi1-σi),ζi=max(0,2-μi1-σi)

will play an important role in the study of the asymptotic behavior of the solution. From here on and without loss of generality, we may assume that

2-λ11-σ12-λ21-σ2,2-μ11-σ12-μ21-σ2

and we introduce the function θ defined on (0,) by

θ(t)={(ρ(t))ν1(1+ρ(t))ζ2-ν1ΨL1,λ1,σ1(ρ(1t))Ψ~L~2,μ2,σ2(ρ(1t))if ν1<ν2,ζ1<ζ2,(ρ(t))ν1(1+ρ(t))ζ2-ν1ΨL1,λ1,σ1(ρ(1t))(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)(ρ(1t))if ν1<ν2,ζ1=ζ2,(ρ(t))ν1(1+ρ(t))ζ2-ν1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)(ρ(1t))Ψ~L~2,μ2,σ2(ρ(1t))if ν1=ν2,ζ1<ζ2,(ρ(t))ν1(1+ρ(t))ζ2-ν1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)(ρ(1t))(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)(ρ(1t))if ν1=ν2,ζ1=ζ2.

For t(0,1], observe that

(1.11)θ(t){(ρ(t))ν1ΨL1,λ1,σ1(ρ(t))if ν1<ν2,(ρ(t))ν1(ΨL1,λ1,σ1(ρ(t))+ΨL2,λ2,σ2(ρ(t)))if ν1=ν2,

and for t1,

(1.12)θ(t){(ρ(t))ζ2Ψ~L~2,μ2,σ2(ρ(t))if ζ1<ζ2,(ρ(t))ζ2(Ψ~L~1,μ1,σ1(ρ(t))+Ψ~L~2,μ2,σ2(ρ(t))))if ζ1=ζ2.

Our main results are the following two theorems.

Theorem 1.4

Let σ1,σ2<1 and assume that (H) is satisfied. Then,

(1.13)V(a1θσ1+a2θσ2)(t)θ(t)

for t(0,).

Theorem 1.5

Let σ1,σ2<1 and assume that (H) is satisfied. Then, (1.4) has a positive continuous solution u satisfying

(1.14)u(t)θ(t)

for t(0,).

The content of this paper is organized as follows. In Section 2, we present some fundamental properties of the Karamata classes 𝒦 and 𝒦. In particular, we recall some sharp estimates on some convenient potential functions. In Section 3, exploiting the results of the previous section, we first prove Theorem 1.4, which allows us to prove Theorem 1.5 by means of the sub-supersolution method. We conclude this section by studying the uniqueness of solutions and giving an example to illustrate our existence result.

2 Properties of the Karamata classes 𝒦 and 𝒦

We collect in this paragraph some properties of functions belonging to the Karamata classes 𝒦 and 𝒦.

Proposition 2.1

Proposition 2.1 (see [20, 25])

  1. Let L1,L2𝒦 (resp. 𝒦) and p. Then, the functions L1+L2, L1L2 and L1p belong to the class 𝒦 (resp. 𝒦).

  2. Let L be a function in 𝒦 (resp. L~𝒦) and ε>0. Then, we have

    limt0+tεL(t)=0(resp.limtt-εL~(t)=0).

Lemma 2.2

Lemma 2.2 (see [20, 25])

  1. Let γ and L be a function in 𝒦 defined on (0,η]. Then, the following hold.

    1. If γ<-1, then the integral

      0ηsγL(s)𝑑s

      diverges and

      tηsγL(s)𝑑st0+-tγ+1L(t)γ+1.
    2. If γ>-1, then the integral

      0ηsγL(s)𝑑s

      converges and

      0tsγL(s)𝑑st0+tγ+1L(t)γ+1.

  2. Let γ and L~ be a function in 𝒦. Then, the following hold.

    1. If γ>-1, then the integral

      1sγL~(s)𝑑s

      diverges and

      1tsγL~(s)𝑑sttγ+1L~(t)γ+1.
    2. If γ<-1, then the integral

      1sγL~(s)𝑑s

      converges and

      tsγL~(s)𝑑st-tγ+1L~(t)γ+1.

The proof of the next lemma can be found in [9].

Lemma 2.3

Let L be a function in K defined on (0,η]. Then, we have

limt0+L(t)(tηL(s)s𝑑s)-1=0.

In particular, we have

ttηL(s)s𝑑s𝒦.

If further the integral

0ηL(s)s𝑑s

converges, then we have

limt0+L(t)(0tL(s)s𝑑s)-1=0.

In particular, we have

t0tL(s)s𝑑s𝒦.

In the next lemma, we have similar properties related to the class 𝒦. For the proof we refer to [8].

Lemma 2.4

Let L~ be a function in K. Then, we have

limtL~(t)(1tL~(s)s𝑑s)-1=0.

In particular, we have

t1t+1L~(s)s𝑑s𝒦.

If further the integral

1L~(s)s𝑑s

converges, then we have

limtL~(t)(tL~(s)s𝑑s)-1=0.

In particular, we have

ttL~(s)s𝑑s𝒦.

Next, we recall the following key sharp estimates.

Proposition 2.5

Proposition 2.5 (see [3])

Let L0K be defined on (0,η] for η>1 and L~0K. Let β2 and γ1 such that

0ηs1-γL0(s)𝑑s<,1s-γL~0(s)𝑑s<.

Put

b(t)=1(A(t))2(ρ(t))-β(1+ρ(t))β-γL0(ρ(1t))L~0(ρ(1t)),t>0.

Then, for t>0,

Vb(t)ψβ(ρ(1t))ϕγ(ρ(1t)),

where for r(0,1],

ψβ(r)={0rL0(s)s𝑑sif β=2,r2-βL0(r)if  1<β<2,rrηL0(s)s𝑑sif β=1,rif β<2,

and for r1,

ϕγ(r)={rrL~0(s)s𝑑sif γ=1,r2-γL~0(r)if  1<γ<2,1r+1L~0(s)s𝑑sif γ=2,1if γ>2.

3 Proof of the main results

We recall that for i{1,2}, σi<1 and ai is a nonnegative continuous function on (0,) such that

ai(t)1(A(t))2(ρ(t))-λi(1+ρ(t))λi-μiLi(ρ(1t))L~i(ρ(1t)),t>0,

where λi2, μi1+σi, Li𝒦 defined on (0,η] and L~i𝒦 satisfying

0ηLi(s)sλi-1𝑑s<,1L~i(s)sμi-σi𝑑s<.

First, we will give sharp estimates of the function θ on (0,1] and on [1,), respectively. To this end, let L, M and N be the nonnegative functions defined on (0,η) by

L(t):=(L1(t))11-σ1+(L2(t))11-σ2,
M(t):=(tηL1(s)s𝑑s)11-σ1+(tηL2(s)s𝑑s)11-σ2,
N(t):=(0tL1(s)s𝑑s)11-σ1+(0tL2(s)s𝑑s)11-σ2  if 0ηLi(s)s𝑑s<,

and L~, M~ and N~ the nonnegative functions defined on [1,) by

L~(t):=(L~1(t))11-σ1+(L~2(t))11-σ2,
M~(t):=(1t+1L~1(s)s𝑑s)11-σ1+(1t+1L~2(s)s𝑑s)11-σ2,
N~(t):=(tL~1(s)s𝑑s)11-σ1+(tL~2(s)s𝑑s)11-σ2  if 1L~i(s)s𝑑s<.

Remark 3.1

Let L0𝒦 defined on (0,η] for η>1 and L~0𝒦. Then, we have

  1. 1+tηL0(s)s𝑑stηL0(s)s𝑑s  on  (0,η],

  2. 1+11+tL~0(s)s𝑑s11+tL~0(s)s𝑑s  on  [1,).

We recall that for i{1,2}, we have λi2, μi1+σi and

νi=min(1,2-λi1-σi),ζi=max(0,2-μi1-σi).

Since ν1<ν2 is equivalent to

2-λ11-σ1<2-λ21-σ2

and

1+σ1<λ1,

we deduce from (1.11) and Remark 3.1 (i) that for t(0,1],

(3.1)θ(t){(0ρ(t)L1(s)s𝑑s)11-σ1if λ1=2,λ2<2,(ρ(t))2-λ11-σ1(L1(ρ(t)))11-σ1if 2-λ11-σ1<2-λ21-σ2, 1+σ1<λ1<2,(ρ(t))2-λ11-σ1L(ρ(t))if 2-λ11-σ1=2-λ21-σ2, 1+σ1<λ1<2,ρ(t)M(ρ(t))if λ1=1+σ1,λ2=1+σ2,ρ(t)(ρ(t)ηL1(s)s𝑑s)11-σ1if λ1=1+σ1,λ2<1+σ2,ρ(t)if λ1<1+σ1,N(ρ(t))if λ1=λ2=2.

On the other hand, since ζ1<ζ2 is equivalent to

2-μ11-σ1<2-μ21-σ2

and

1+σ2μ2<2,

we deduce from (1.12) and Remark 3.1 (ii) that for t1,

(3.2)θ(t){(1ρ(t)+1L~2(s)s𝑑s)11-σ2if μ2=2,μ1>2,(ρ(t))2-μ21-σ2(L~2(ρ(t)))11-σ2if 2-μ11-σ1<2-μ21-σ2, 1+σ2<μ2<2,(ρ(t))2-μ21-σ2L~(ρ(t))if 2-μ11-σ1=2-μ21-σ2, 1+σ2<μ2<2,ρ(t)N~(ρ(t))if μ2=1+σ2,μ1=1+σ1,ρ(t)(ρ(t)L~2(s)s𝑑s)11-σ2if μ2=1+σ2,μ1>1+σ1,1if μ2>2,M~(ρ(t))if μ1=μ2=2.

3.1 Proof of Theorem 1.4

The next lemma will be used in the following.

Lemma 3.2

For r,s>0, we have

2-max(1-σ1,1-σ2)(r+s)r1-σ1(r+s)σ1+s1-σ2(r+s)σ22(r+s).

Proof.

Let r,s>0 and put

t=rr+s.

Since 0t1, we have

2-max(1-σ1,1-σ2)t1-σ1+(1-t)1-σ22.

This implies the result. ∎

The following lemmas will be useful in the proof of Theorem 1.4.

Lemma 3.3

Lemma 3.3 (see [9])

  1. On (0,η), we have

    tη(Mσ1L1+Mσ2L2)(s)s𝑑sM(t).
  2. Assume that

    0ηLi(s)s𝑑s<.

    Then, for t(0,η), we have

    0t(Nσ1L1+Nσ2L2)(s)s𝑑sN(t).

Lemma 3.4

Lemma 3.4 (see [7])

  1. On (1,), we have

    1t+1(M~σ1L~1+M~σ2L~2)(s)s𝑑sM~(t).
  2. Assume that

    1L~i(s)s𝑑s<.

    Then, for t>1, we have

    t(N~σ1L~1+N~σ2L~2)(s)s𝑑sN~(t).

We are ready to prove Theorem 1.4. For t>0, we let

ω(t)=a1(t)θσ1(t)+a2(t)θσ2(t)

and we will prove that

V(ω)(t)θ(t).

We recall that for i{1,2},

νi=min(1,2-λi1-σi),ζi=max(0,2-μi1-σi)

and

ai(t)1(A(t))2(ρ(t))-λi(1+ρ(t))λi-μiLi(ρ(1t))L~i(ρ(1t)),t>0.

Throughout the proof, we will use Proposition 2.1, Lemma 2.3 and Lemma 2.4 to verify that some functions are in 𝒦 and in 𝒦.

We distinguish the following cases.

Case 1. If ν1<ν2 and ζ1<ζ2, then

θ(t)=(ρ(t))ν1(1+ρ(t))ζ2-ν1ΨL1,λ1,σ1(ρ(1t))Ψ~L~2,μ2,σ2(ρ(1t)).

Therefore,

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-μ1+(ζ2-ν1)σ1(L1ΨL1,λ1,σ1σ1)(ρ(1t))(L~1Ψ~L~2,μ2,σ2σ1)(ρ(1t))
+1(A(t))2(ρ(t))-λ2+ν1σ2(1+ρ(t))λ2-μ2+(ζ2-ν1)σ2(L2ΨL1,λ1,σ1σ2)(ρ(1t))(L~2Ψ~L~2,μ2,σ2σ2)(ρ(1t)).

Since ν1<ν2 and ζ1<ζ2, we deduce by Proposition 2.1 that

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(L1ΨL1,λ1,σ1σ1)(ρ(1t))(1+ρ(t))λ1-ν1σ1-(μ2-ζ2σ2)(L~2Ψ~L~2,μ2,σ2σ2)(ρ(1t)).

Furthermore, by using (1.9) and Lemma 2.2, we have

0ηs1-λ1+ν1σ1(L1ΨL1,λ1,σ1σ1)(s)𝑑s<,1s-μ2+ζ2σ2(L~2Ψ~L~2,μ2,σ2σ2)(s)𝑑s<.

So, by applying Proposition 2.5 with β=λ1-ν1σ1, γ=μ2-ζ2σ2 and

L0=L1ΨL1,λ1,σ1σ1𝒦,L~0=L~2Ψ~L~2,μ2,σ2σ2𝒦

for t>0, we obtain

Vω(t)ψβ(ρ(1t))ϕγ(ρ(1t)),

where for r(0,1],

ψβ(r)={0rL(s)s(0sL(t)t𝑑t)σ11-σ1𝑑sif λ1=2,λ2<2,r2-βL1(r)(L1(r))σ11-σ1if 2-λ11-σ1<2-λ21-σ2, 1+σ1<λ1<2,
={(0rL(s)s𝑑s)11-σ1if λ1=2,λ2<2,r2-λ11-σ1(L1(r))11-σ1if 2-λ11-σ1<2-λ21-σ2, 1+σ1<λ1<2,

and for r1,

ϕγ(r)={rrL~2(s)s(sL~2(t)t𝑑t)σ21-σ2𝑑sif μ2=1+σ2,μ1>1+σ1,r2-γL~2(r)(L~2(r))σ21-σ2if 2-μ11-σ1<2-μ21-σ2, 1+σ2<μ2<2,
={r(rL~2(s)s𝑑s)11-σ2if μ2=1+σ2,μ1>1+σ1,r2-μ21-σ2(L~2(r))11-σ2if 2-μ11-σ1<2-μ21-σ2, 1+σ2<μ2<2.

Hence, by using the fact that

ρ(1t)ρ(t)1+ρ(t),ρ(1t)1+ρ(t),

Proposition 1.3 and (3.1)–(3.2) give that

Vω(t)θ(t).

Case 2. If ν1<ν2 and ζ1=ζ2, then, for t>0,

θ(t)=(ρ(t))ν1(1+ρ(t))ζ2-ν1ΨL1,λ1,σ1(ρ(1t))(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)(ρ(1t)).

In this case,

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-μ1+(ζ2-ν1)σ1×
×(L1ΨL1,λ1,σ1σ1)(ρ(1t))(L~1(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ1)(ρ(1t))
  +1(A(t))2(ρ(t))-λ2+ν1σ2(1+ρ(t))λ2-μ2+(ζ2-ν1)σ2×
×(L2ΨL1,λ1,σ1σ2)(ρ(1t))(L~2(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ2)(ρ(1t)).

Since ν1<ν2 and ζ1=ζ2, we deduce that

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-μ1+(ζ1-ν1)σ1
×(L1ΨL1,λ1,σ1σ1)(ρ(1t))(L~1(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ1)(ρ(1t))
  +1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-ν1σ1-(μ2-ζ2σ2)
×(L1ΨL1,λ1,σ1σ1)(ρ(1t)))L~2(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ2)(ρ(1t))
=ω1(t)+ω2(t).

So, by applying Proposition 2.5 with β=λ1-ν1σ1 and γi=μi-ζiσi, i{1,2}, to estimate Vω1(t) and Vω2(t), we obtain

Vω(t)Vω1(t)+Vω2(t)ψβ(ρ(1t))(ϕγ1(ρ(1t))+ϕγ2(ρ(1t))),

where, for r(0,1],

ψβ(r)={(0rL(s)s𝑑s)11-σ1if λ1=2,λ2<2,r2-λ11-σ1(L1(r))11-σ1if 2-λ11-σ1<2-λ21-σ2, 1+σ1<λ1<2,

and for r1,

(3.3)ϕγ1(r)+ϕγ2(r)={1+1r+1L~2(s)s(1+(1s+1L~2(t)t𝑑t)11-σ2)σ2𝑑sif μ2=2,μ1>2,r2-μ21-σ2(L~1(r)(L~(r))σ1+L~2(r)(L~(r))σ2)if 2-μ11-σ1=2-μ21-σ2, 1+σ2<μ2<2,rr(L~1N~σ1(s)+L~2N~σ2)(s)s𝑑sif μ2=1+σ2,μ1=1+σ1,2if μ2>2,1r+1(M~σ1L~1+M~σ2L~2)(s)s𝑑sif μ1=μ2=2.

So, by using Remark 3.1, Lemma 3.2 and Lemma 3.4, we deduce that

ϕγ1(r)+ϕγ2(r){(1r+1L~2(s)s𝑑s)11-σ2if μ2=2,μ1>2,r2-μ21-σ2L~(r)if 2-μ11-σ1=2-μ21-σ2, 1+σ2<μ2<2,rN~(r)if μ2=1+σ2,μ1=1+σ1,1if μ2>2,M~(r)if μ1=μ2=2.

So, by using again the fact that

ρ(1t)ρ(t)1+ρ(t),ρ(1t)1+ρ(t),

Proposition 1.3 and (3.1)–(3.2), we deduce that

Vω(t)θ(t).

Case 3. If ν1=ν2 and ζ1<ζ2, then, for t>0,

θ(t)=(ρ(t))ν1(1+ρ(t))ζ2-ν1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)(ρ(1t))Ψ~L~2,μ2,σ2(ρ(1t)).

So, we obtain

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-μ1+(ζ2-ν1)σ1
×(L1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ1)(ρ(1t))(L~1Ψ~L~2,μ2,σ2σ1)(ρ(1t))
  +1(A(t))2(ρ(t))-λ2+ν1σ2(1+ρ(t))λ2-μ2+(ζ2-ν1)σ2
×(L2(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ2)(ρ(1t))(L~2Ψ~L~2,μ2,σ2σ2)(ρ(1t)).

Since ν1=ν2 and ζ1<ζ2, we deduce that

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-ν1σ1-(μ2-ζ2σ2)
×(L1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ1)(ρ(1t))(L~2Ψ~L~2,μ2,σ2σ2)(ρ(1t))
  +1(A(t))2(ρ(t))-λ2+ν2σ2(1+ρ(t))λ2-μ2+(ζ2-ν2)σ2
×(L2(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ2)(ρ(1t))(L~2Ψ~L~2,μ2,σ2σ2)(ρ(1t))
=ω1(t)+ω2(t).

So, using again Proposition 2.5 with βi=λi-νiσi, i{1,2}, and γ=μ2-ζ2σ2 to estimate Vω1(t) and Vω2(t), we obtain

Vω(t)Vω1(t)+Vω2(t)(ψβ1(ρ(1t))+ψβ2(ρ(1t))ϕγ(ρ(1t)),

where, for r1 (see Case 1),

ϕγ(r)={r(rL~2(s)s𝑑s)11-σ2if μ2=1+σ2,μ1>1+σ1,r2-μ21-σ2(L~2(r))11-σ2if 2-μ11-σ1<2-μ21-σ2, 1+σ2<μ2<2,

and for r(0,1],

ψβ1(r)+ψβ2(r)={r2-λ11-σ1(L1(r)(L(r))σ1+L2(r)(L(r))σ2)if 2-λ11-σ1=2-λ21-σ2, 1+σ1<λ1<2,rrη(L1Mσ1+L2Mσ2)(s)s𝑑sif λ1=1+σ1,λ2=1+σ2,rrηL1(s)s(1+(sηL1(t)t𝑑t)11-σ1)σ1+rif λ1=1+σ1,λ2<1+σ2,2rif λ1<1+σ1,0r(L1Nσ1+L2Nσ2)(s)s𝑑sif λ1=λ2=2.

Hence, by using Lemma 3.2, Remark 3.1 and Lemma 3.3, we deduce that

(3.4)ψβ1(r)+ψβ2(r){r2-λ11-σ1L(r)if 2-λ11-σ1=2-λ21-σ2, 1+σ1<λ1<2,rM(r)if λ1=1+σ1,λ2=1+σ2,r(rηL1(s)s𝑑s)11-σ1if λ1=1+σ1,λ2<1+σ2,2rif λ1<1+σ1,N(r)if λ1=λ2=2.

Using the fact that

ρ(1t)ρ(t)1+ρ(t),ρ(1t)1+ρ(t),

Proposition 1.3 and (3.1)–(3.2), we deduce that

Vω(t)θ(t).

Case 4. If ν1=ν2 and ζ1=ζ2, then, for t>0,

θ(t)=(ρ(t))ν1(1+ρ(t))ζ2-ν1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)(ρ(1t))(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)(ρ(1t)).

In this case, we have

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-μ1+(ζ2-ν1)σ1
×(L1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ1)(ρ(1t))(L~1(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ1)(ρ(1t))
  +1(A(t))2(ρ(t))-λ2+ν1σ2(1+ρ(t))λ2-μ2+(ζ2-ν1)σ2
×(L2(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ2)(ρ(1t))(L~2(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ2)(ρ(1t)).

Since ν1=ν2 and ζ1=ζ2, we deduce that

ω(t)1(A(t))2(ρ(t))-λ1+ν1σ1(1+ρ(t))λ1-μ1+(ζ1-ν1)σ1
×(L1(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ1)(ρ(1t))(L~1(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ1)(ρ(1t))
  +1(A(t))2(ρ(t))-λ2+ν2σ2(1+ρ(t))λ2-μ2+(ζ2-ν2)σ2
×(L2(ΨL1,λ1,σ1+ΨL2,λ2,σ2)σ2)(ρ(1t))(L~2(Ψ~L~1,μ1,σ1+Ψ~L~2,μ2,σ2)σ2)(ρ(1t))
=ω1(t)+ω2(t).

Applying Proposition 2.5 with βi=λi-νiσi and γi=μi-ζiσi, i{1,2}, to estimate Vω1(t) and Vω2(t), we obtain

Vω(t)Vω1(t)+Vω2(t)(ψβ1(ρ(1t))+ψβ2(ρ(1t)))(ϕγ1(ρ(1t))+ϕγ2(ρ(1t))),

where sharp estimates for ψβ1(ρ(1t))+ψβ2(ρ(1t)) (resp. ϕγ1(ρ(1t))+ϕγ2(ρ(1t))) are given in (3.4) (resp. (3.3)).

The required result follows by similar arguments as in the previous cases.

3.2 Proof of Theorem 1.5

Let σ1,σ2<1 and assume that hypothesis (H) is satisfied. Since, by Theorem 1.4,

(3.5)0A(s)min(1,ρ(s))(a1θσ1+a2θσ2)(s)𝑑sθ(1)<,

we deduce that

v(t):=V(a1θσ1+a2θσ2)(t)

is a positive continuous solution of

(3.6){-1A(Av)(t)=(a1θσ1+a2θσ2)(t)in (0,),limt0+v(t)=0,limtv(t)ρ(t)=0.

Now, from Theorem 1.4, there exists M>1 such that for each t>0,

(3.7)1Mθ(t)v(t)Mθ(t).

Put

c=max(M|σ1|1-σ1,M|σ2|1-σ2).

Using (3.6) and (3.7), we deduce that the functions

u¯=1cv,u¯=cv

are, respectively, a sub-solution and a super-solution of (1.4) with

u¯u¯.

So, thanks to the sub-supersolution method (see [22, Theorem 5.1]), there exists a positive continuous solution u to (1.4) such that

1cvucv.

From (3.7), we conclude that u also satisfies (1.14).

3.3 A uniqueness result

Let σ1,σ2<1 and assume that hypothesis (H) is satisfied. Since for σ1,σ20 the function

φ(x,t):=a1(x)tσ1+a2(x)tσ2

is continuous on (0,)×(0,) and nonincreasing with respect to the second variable, then the uniqueness of solutions to (1.4) follows from [2, Lemma 1]. In the following, we will be interested in the case where σ1,σ2(-1,1).

The next lemma will be used in the following.

Lemma 3.5

Lemma 3.5 (see [2])

Let uC1((0,)) be a function satisfying

{-1A(Au)0in (0,),limta+u(t)=0,limtu(t)ρ(t)=0.

Then, u is a nondecreasing and nonnegative function on (0,).

Proposition 3.6

Let σ1,σ2(-1,1) and assume hypothesis (H) is satisfied. Then,(1.4) has a unique positive continuous solution u satisfying

u(t)θ(t)

for t(0,).

Proof.

The existence follows from Theorem 1.5. For the uniqueness, assume that (1.4) has two positive continuous solutions u,v satisfying u(t)θ(t)v(t). Then, there exists a constant m>1 such that

1muvm.

Therefore, the set

J={m1:1muvm}

is not empty. Let σ:=max(|σ1|,|σ2|) and put c0:=infJ. Then, c01 and we have

1c0vuc0v.

It follows that for i{1,2}, uσic0σvσi and that the function w:=c0σv-u satisfies

{-1A(Aw)=a1(c0σvσ1-uσ1)+a2(c0σvσ2-uσ2)0,limt0+w(t)=0,limtw(t)ρ(t)=0.

By Lemma 3.5, this implies that the function w=c0σv-u is nonnegative. By symmetry, we also have vc0σu. Hence, c0σJ and c0c0σ. Since 0σ<1, then c0=1 and therefore u=v. ∎

Example 3.7

Let σ1(-1,0), σ2(0,1), λ1<1+σ1 with

2-λ11-σ12-λ21-σ2

and μ1>1+σ1, μ2>1+σ2 such that

2-μ11-σ12-μ21-σ2.

Let a1,a2 be positive continuous functions on (0,) such that

ai(t)1(A(t))2(ρ(t))-λi(1+ρ(t))λi-μi

for i{1,2}. Then, by Proposition 3.6, (1.4) has a unique positive continuous solution u satisfying

u(t){ρ(t)(1+ρ(t))-1(log(ρ(t)+2))11-σ2if μ2=2,ρ(t)(1+ρ(t))2-μ21-σ2-1if 1+σ2<μ2<2,ρ(t)(1+ρ(t))-1if μ2>2,

for t>0.

Award Identifier / Grant number: 13-MAT2137-02

Funding statement: This project was funded by the National Plan for Science, Technology and Innovation (MAARIFAH), King Abdulaziz City for Science and Technology, Kingdom of Saudi Arabia, Award No. 13-MAT2137-02.

The authors are thankful to the referees for their careful reading of the paper.

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Received: 2015-6-26
Revised: 2015-7-10
Accepted: 2015-7-14
Published Online: 2015-8-20
Published in Print: 2016-8-1

© 2016 by De Gruyter

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