# Existence and Global Asymptotic Behavior of Positive Solutions for Nonlinear Fractional Dirichlet Problems on the Half-Line

Authors: Imed Bachar; Habib Mâagli
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We are interested in the following fractional boundary value problem:Dαu(t)+atuσ=0, t∈(0,∞), limt→0⁡t2-αu(t)=0, limt→∞⁡t1-αu(t)=0, where 1<α<2, σ∈(-1,1), Dα is the standard Riemann-Liouville fractional derivative, and a is a nonnegative continuous function on (0,∞) satisfying some appropriate assumptions related to Karamata regular variation theory. Using the Schauder fixed point theorem, we prove the existence and the uniqueness of a positive solution. We also give a global behavior of such solution.


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## Body

## 1. Introduction

Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetic. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see [1–3]). In consequence, the subject of fractional differential equations has been gaining much importance and attention. Most of the related results focused on developing the global existence and uniqueness of solutions on finite intervals (see [4–12]) and the references therein). However, to the best of our knowledge, there exist few articles dealing with the existence of solutions to fractional differential equations on the half-line; see, for instance, [13–21]. In [17], by using the recent Leggett-Williams norm-type theorem due to O’Regan and Zima, the author established the existence of positive solutions for fractional boundary value problems of resonance on infinite intervals. On the other hand, in [20], Su and Zhang studied the following fractional differential problem on the half-line by using Schauder’s fixed point theorem:
(1)Dαu(t)=f(t,u,Dα-1u),t∈(0,∞),1<α≤2,u(0)=0,lim⁡t→∞Dα-1u(t)=u∞,u∞∈R,
where Dα is the standard Riemann-Liouville fractional derivative (see Definition 8 below).In [21], by means of the Leray-Schauder alternative theorem, Zhao and Ge proved the existence of solutions to the following boundary value problem:
(2)Dαu(t)+f(t,u)=0,t∈(0,∞),1<α<2,u(0)=0,lim⁡t→∞Dα-1u(t)=βu(ξ),
where β∈R and 0<ξ<∞.In this paper, we aim at studying the existence, uniqueness, and the exact asymptotic behavior of a positive solution to the following fractional boundary value problem:(3)Dαu(t)+a(t)uσ=0,t∈(0,∞),lim⁡t→0t2-αu(t)=0,lim⁡t→∞t1-αu(t)=0,
where 1<α<2, σ∈(-1,1), and a is a nonnegative continuous function on (0,∞) that may be singular at 0.To state our result, we need some notations. We first introduce the following Karamata classes.Definition 1.
The classK is the set of all Karamata functions L defined on (0,η] by
(4)L(t)≔cexp⁡(∫tηz(s)sds),
for some η>1 and where c>0 and z∈C([0,η]) such that z(0)=0.Definition 2.
The classK∞ is the set of all Karamata functions L defined on [1,∞) by
(5)L(t)≔cexp⁡(∫1tz(s)sds),
where c>0 and z∈C([1,∞)) such that lim⁡t→∞z(t)=0.It is easy to verify the following.Remark 3.
(i) A functionL is in K if and only if L is a positive function in C1((0,η]), for some η>1, such that lim⁡t→0+(tL′(t)/L(t))=0.
(ii) A functionL is in K∞ if and only if L is a positive function in C1([1,∞)) such that lim⁡t→∞(tL′(t)/L(t))=0.Remark 4 (see [22]).
LetL be a function in K∞, and then there exists m≥0 such that for every β>0 and t≥1 we have
(6)(1+β)-mL(t)≤L(β+t)≤(1+β)mL(t).As a typical example of function belonging to the classK, we quote
(7)L(t)=Πk=1m(log⁡k(ωt))ξk,
where ξk are real numbers, log⁡kx=log⁡∘log⁡∘⋯log⁡x (k times), and ω is a sufficiently large positive real number such that L is defined and positive on (0,η], for some η>1.In the sequel, we denote byB+((0,∞)) the set of nonnegative Borel measurable functions in (0,∞) and by C2-α([0,∞)) the set of all functions f such that t→t2-αf(t) is continuous on [0,∞).We also denote byC0([0,∞)) the set of continuous functions v on [0,∞) such that lim⁡t→∞v(t)=0. It is easy to see that C0([0,∞)) is a Banach space with the uniform norm ∥v∥∞=sup⁡t>0|v(t)|.For two nonnegative functionsf and g defined on a set S, the notation f(t)≈g(t), t∈S means that there exists c>0 such that (1/c)f(t)≤g(t)≤cf(t), for all t∈S.Finally, forλ∈R, we put λ+=max⁡(λ,0).Throughout this paper we assume that the functiona is nonnegative on (0,∞) and satisfies the following condition:(H)a∈C((0,∞)) such that
(8)a(t)≈t-λ(1+t)λ-μL1(min⁡(t,1))L2(max⁡(t,1)),t>0,
where λ≤2+(α-2)σ, μ≥1+(α-1)σ, L1∈K defined on (0,η], for some η>1 and L2∈K∞ satisfying
(9)∫0ηL1(s)sλ-(α-2)σ-1ds<∞,∫1∞L2(s)sμ-(α-1)σds<∞.
In what follows, we put
(10)ν=min⁡(1,2-λ+(α-2)σ1-σ),ζ=max⁡(0,2-μ+(α-2)σ1-σ),
and we define the function θ on (0,∞) by
(11)θ(t)=tν(1+t)ζ-ν(L~1(min⁡(t,1)))1/(1-σ)×(L~2(max⁡(t,1)))1/(1-σ),
where, for t∈(0,η),
(12)L~1(t)={∫0tL1(s)sdsifλ=2+(α-2)σ,L1(t)if1+(α-1)σ<λ<2+(α-2)σ,∫tηL1(s)sdsifλ=1+(α-1)σ,1ifλ<1+(α-1)σ
and, for t≥1,
(13)L~2(t)={∫t∞L2(s)sdsifμ=1+(α-1)σ,L2(t)if1+(α-1)σ<μ<2+(α-2)σ,∫1t+1L2(s)sdsifμ=2+(α-2)σ,1ifμ>2+(α-2)σ.
Our main result is the following.Theorem 5.
Let1<α<2, σ∈(-1,1) and assume (H). Then problem (3) has a unique positive solution u∈C2-α([0,∞)) satisfying, for t∈(0,∞),
(14)u(t)≈tα-2θ(t).Remark 6.
The conclusion of Theorem5 remains valid for the case α=2 and σ<1 (see [23]).The content of this paper is organized as follows. In Section2, we present some properties of the Green function Gα(t,s) of the operator u→-Dαu on (0,∞) with Dirichlet conditions lim⁡t→0t2-αu(t)=0 and lim⁡t→∞t1-αu(t)=0. Next, we give some fundamental properties of the two Karamata classes K and K∞ and we establish sharp estimates on some potential functions. In Section 3, exploiting the results of the previous section and using the Schauder fixed point theorem, we prove Theorem 5.


## 2. Preliminaries

### 2.1. Fractional Calculus and Green Function

For the convenience of the reader, we recall in this section some basic definitions on fractional calculus (see [2, 24, 25]) and we give some properties of the Green function Gα(t,s).Definition 7.
The Riemann-Liouville fractional integral of orderβ>0 of a function h:(0,∞)→R is given by
(15)Iβh(t)=1Γ(β)∫0t(t-s)β-1h(s)ds,t>0,
provided that the right-hand side is pointwise defined on (0,∞).Definition 8.
The Riemann-Liouville fractional derivative of orderβ>0 of a function h:(0,∞)→R is given by
(16)Dβh(t)=1Γ(n-β)(ddt)n∫0t(t-s)n-β-1h(s)ds,t>0,
where n=[β]+1 provided that the right-hand side is pointwise defined on (0,∞). Here [β] means the integer part of the number β.So we have the following properties (see [2]).Proposition 9.
(1) Letβ>0 and let h∈L1(0,∞), and then one has
(17)DβIβh(t)=h(t),fort>0.
(2) Letβ>0, and then
(18)Dβh(t)=0iffh(t)=∑‍mj=1cjtβ-j,
where m is the smallest integer greater than or equal to β and (c1,c2,…,cm)∈Rm.Corollary 10.
Letβ>0 and assume that Dβh∈L1(0,∞). Then,
(19)IβDβh(t)=h(t)+c1tβ-1+c2tβ-2+⋯+cmtβ-m,t>0,
where m is the smallest integer greater than or equal to β and (c1,c2,…,cm)∈Rm.Lemma 11.
Let1<α<2 and h∈L1(0,∞). The unique solution of
(20)Dαu(t)+h(t)=0,t>0lim⁡t→0t2-αu(t)=0,lim⁡t→∞t1-αu(t)=0
is given by
(21)u(t)=∫0∞Gα(t,s)h(s)ds,
where
(22)Gα(t,s)=1Γ(α)[tα-1-((t-s)+)α-1]
is Green’s function for the boundary value problem (20).Proof.
We may apply Corollary10 and Proposition 9 to reduce equation Dαu(t)+h(t)=0 to an equivalent integral equation
(23)u(t)=-Iαh(t)+c1tα-1+c2tα-2,
where (c1,c2)∈R2. Hence the general solution of Dαu(t)+h(t)=0 is
(24)u(t)=c1tα-1+c2tα-2-1Γ(α)∫0t(t-s)α-1h(s)ds.
By using lim⁡t→0t2-αu(t)=0 and lim⁡t→∞t1-αu(t)=0, we get
(25)c2=0,c1=1Γ(α)∫0∞h(s)ds.
Therefore, the unique solution of problem (20) is
(26)u(t)=tα-1Γ(α)∫0∞h(s)ds-1Γ(α)∫0t(t-s)α-1h(s)ds=∫0∞Gα(t,s)h(s)ds.
The proof is complete.Next we give sharp estimates on the Green functionGα(t,s). To this end, we need the following lemma.Lemma 12.
(i) Forλ,μ∈(0,∞) and x∈[0,1], one has
(27)min⁡(1,μλ)(1-xλ)≤1-xμ≤max⁡(1,μλ)(1-xλ).
(ii) For (t,s)∈(0,∞)×(0,∞) one has
(28)min⁡(1,t)min⁡(1,s)≤min⁡(t,s)≤max⁡(1,t)min⁡(1,s).Proposition 13.
The Green functionGα(t,s) defined by (22) satisfies
(29)Gα(t,s)≈tα-2min⁡(t,s),for(t,s)∈(0,∞)×(0,∞).Proof.
For(t,s)∈(0,∞)×(0,∞) we have
(30)Gα(t,s)=tα-1Γ(α)[1-((t-s)+t)α-1].
Since ((t-s)+/t)∈[0,1) for (t,s)∈(0,∞)×(0,∞), then by applying Lemma 12 (i) with μ=α-1 and λ=1, we obtain
(31)Gα(t,s)≈tα-1[1-((t-s)+t)]=tα-2min⁡(t,s).From here on, we define the potential kernelGα on B+((0,∞)) by
(32)Gαh(t)≔∫0∞Gα(t,s)h(s)ds,fort>0.Using Proposition13 and Lemma 12 (ii), we deduce the following.Corollary 14.
Let1<α<2 and h∈B+((0,∞)), and then the function t→Gαh(t) belongs to C2-α([0,∞)) if and only if the integral ∫0∞min⁡(1,s)h(s)ds converges.Proposition 15.
Let1<α<2 and f be a function such that the map s→min⁡(1,s)f(s) is continuous and integrable on (0,∞). Then Gαf is the unique solution in C2-α([0,∞)) of the boundary value problem
(33)Dαu(t)+f(t)=0,t>0lim⁡t→0t2-αu(t)=0,lim⁡t→∞t1-αu(t)=0.Proof.
From Corollary14, the function Gαf is well defined in (0,∞). Using Proposition 13 and Lemma 12 (ii), we get
(34)Gα|f|(t)≤c(tα-2(t+1)∫0∞min⁡(1,s)|f(s)|ds).
This implies that I2-α(Gα|f|) is finite on (0,∞). So by using Fubini’s theorem, we obtain
(35)I2-α(Gαf)(t)=1Γ(2-α)∫0t(t-s)1-αGαf(s)ds=1Γ(2-α)∫0∞(∫0t(t-s)1-αGα(s,r)ds)f(r)dr.
Observe that by considering the substitution s=r+(t-r)θ, we obtain
(36)∫rt(t-s)1-α(s-r)α-1ds=Γ(α)Γ(2-α)(t-r).
Using this fact and (22) we deduce that
(37)1Γ(2-α)∫0t(t-s)1-αGα(s,r)ds=1Γ(2-α)Γ(α)×[∫0t(t-s)1-αsα-1ds-∫0t(t-s)1-α((s-r)+)α-1ds]=t-1Γ(2-α)Γ(α)∫0t(t-s)1-α((s-r)+)α-1ds.
Now, assume that r≤t, and then by (36) we have
(38)∫0t(t-s)1-α((s-r)+)α-1ds=∫rt(t-s)1-α(s-r)α-1ds=Γ(α)Γ(2-α)(t-r).
On the other hand, if t≤r and s∈(0,t), we have
(39)∫0t(t-s)1-α((s-r)+)α-1ds=0.
So combining (38) and (39), we obtain
(40)1Γ(2-α)∫0t(t-s)1-αGα(s,r)ds=t-(t-r)+=min⁡(t,r).
This implies that
(41)I2-α(Gαf)(t)=∫0∞min⁡(t,r)f(r)dr=∫0trf(r)dr+t∫t∞f(r)dr,
and Dα(Gαf)(t)=(d2/dt2)(I2-α(Gαf))(t)=-f(t), for t>0.
Moreover, using Proposition13 and the dominated convergence theorem, we deduce that
(42)lim⁡t→0t2-αGαf(t)=0,lim⁡t→∞t1-αGαf(t)=0.
Finally, we need to prove the uniqueness. Let u,v∈C2-α([0,∞)) be two solutions of (33) and put ω=u-v. Then ω∈C2-α([0,∞)) and Dαω=0. Hence, it follows from Corollary 10 that ω(t)=c1tα-1+c2tα-2. Using the fact that lim⁡t→0t2-αω(t)=lim⁡t→∞t1-αω(t)=0, we deduce that ω=0 and therefore u=v. The proof is complete.


### 2.2. Sharp Estimates on the Potential of Some Karamata Functions

We collect in this paragraph some properties of functions belonging to the Karamata classK (resp., K∞) and we give estimates on some potential functions.Proposition 16 (see [26, 27]).
(i) LetL1,L2∈K (resp, K∞) and p∈R. Then the functions
(43)L1+L2,L1L2andL1pbelongtotheclassK(resp,K∞).
(ii) Let L be a function in K (resp., K∞) and ε>0. Then one has
(44)lim⁡t→0+tεL(t)=0,(resp.lim⁡t→∞t-εL(t)=0).Theorem 17 (see [26, 27]).
(a) Letγ∈R and L be a function in K defined on (0,η]. One has the following.(i)
Ifγ<-1, then ∫0ηsγL(s)ds diverges and ∫tηsγL(s)ds~t→0+-tγ+1L(t)/(γ+1).(ii)
Ifγ>-1, then ∫0ηsγL(s)ds converges and ∫0tsγL(s)ds~t→0+tγ+1L(t)/(γ+1).
(b) Letγ∈R and L be a function in K∞. One has the following.(i)
Ifγ>-1, then ∫1∞sγL(s)ds diverges and ∫1tsγL(s)ds~t→∞tγ+1L(t)/(γ+1).(ii)
Ifγ<-1, then ∫1∞sγL(s)ds converges and ∫t∞sγL(s)ds~t→∞-tγ+1L(t)/(γ+1).The proof of the next lemma can be found in [11].Lemma 18.
LetL be a function in K defined on (0,η]. Then one has
(45)lim⁡t→0+L(t)∫tη(L(s)/s)ds=0.
In particular,
(46)t⟶∫tηL(s)sds∈K.
If further ∫0η(L(s)/s)ds converges, then one has lim⁡t→0+(L(t)/∫0t(L(s)/s)ds)=0.
In particular,(47)t⟶∫0tL(s)sds∈K.In the next lemma, we have the following properties related to the classK∞. For the proof we refer to [22].Lemma 19.
LetL be a function in K∞. Then one has
(48)lim⁡t→∞L(t)∫1t(L(s)/s)ds=0.
In particular,
(49)t⟶∫1t+1L(s)sds∈K∞.
If further ∫1∞(L(s)/s)ds converges, then one has
(50)lim⁡t→∞L(t)∫t∞(L(s)/s)ds=0.
In particular,
(51)t⟶∫t∞L(s)sds∈K∞.Now, we put(52)b(t)=t-β(1+t)β-γL3(min⁡(t,1))L4(max⁡(t,1)),t>0,
where L3∈K and L4∈K∞. We aim at giving sharp estimates on the potential function Gαb(t).Proposition 20.
Assume thatL3∈K defined on (0,η], for some η>1 and L4∈K∞. Let β≤2 and γ≥1 such that
(53)∫0ηs1-βL3(s)ds<∞,∫1∞s-γL4(s)ds<∞.
Then for t>0(54)Gαb(t)≈tα-2ψβ(min⁡(t,1))ϕγ(max⁡(t,1)),
where, for t∈(0,1],
(55)ψβ(t)={∫0tL3(s)sdsifβ=2,t2-βL3(t)if1<β<2,t∫tηL3(s)sdsifβ=1,tifβ<1
and, for t≥1,
(56)ϕγ(t)={t∫t∞L4(s)sdsifγ=1,t2-γL4(t)if1<γ<2,∫1t+1L4(s)sdsifγ=2,1ifγ>2.Proof.
Using Proposition13 and Remark 4, we have
(57)t2-αGαb(t)≈∫0ηmin⁡(t,s)s-βL3(s)ds+∫η∞min⁡⁡(t,s)s-γL4(s)ds=I(t)+J(t).
Case  1. Assume that 0<t≤1.
By using (53), we deduce that
(58)J(t)≈t.
On the other hand,
(59)I(t)=∫0ts1-βL3(s)ds+t∫tηs-βL3(s)ds=I1(t)+I2(t).
Using Theorem 17 and hypothesis (53), we deduce that
(60)I1(t)≈{t2-βL3(t)ifβ<2,∫0tL3(s)sdsifβ=2,I2(t)≈{t2-βL3(t)if1<β≤2,t∫tηs-βL3(s)dsifβ≤1.
Hence, it follows by Lemma 18, Proposition 16, and hypothesis (53) that
(61)I(t)≈{∫0tL3(s)sdsifβ=2,t2-βL3(t)if1<β<2,t∫tηL3(s)sdsifβ=1,tifβ<1.
Combining (58) and (61) and using Proposition 16 and hypothesis (53), we deduce that, for 0<t≤1,
(62)t2-αGαb(t)≈ψβ(t).Case2. Assume that t>η+1.
By using (53), we deduce that
(63)I(t)≈1.
On the other hand,
(64)J(t)=∫tηs1-γL4(s)ds+t∫t∞s-γL4(s)ds=J1(t)+J2(t).
Using again Theorem 17 and hypothesis (53), we deduce that
(65)J1(t)≈{t2-γL4(t)if1≤γ<2,∫ηts1-γL4(s)dsifγ≥2,J2(t)≈{t2-γL4(t)ifγ>1,t∫t∞L4(s)sdsifγ=1.
Hence, it follows from Lemma 19 and hypothesis (53) that
(66)J(t)≈{t∫t∞L4(s)sdsifγ=1,t2-γL4(t)if1<γ<2,∫ηtL4(s)sdsifγ=2,1ifγ>2.
Combining (63) and (66) and using Proposition 16, hypothesis (53), and Remark 4, we deduce that, for t>η+1,
(67)t2-αGαb(t)≈{t∫t∞L4(s)sdsifγ=1,t2-γL4(t)if1<γ<2,∫ηtL4(s)sdsifγ=2,1ifγ>2.≈ϕγ(t).
Now since the functions t→t2-αGαb(t) and t→ϕγ(t) are positive and continuous on [1,η+1], we deduce that, for t∈[1,η+1],
(68)t2-αGαb(t)≈ϕγ(t).
Finally, using (62), (67), and (68), we obtain the required result.


## 2.1. Fractional Calculus and Green Function

For the convenience of the reader, we recall in this section some basic definitions on fractional calculus (see [2, 24, 25]) and we give some properties of the Green function Gα(t,s).Definition 7.
The Riemann-Liouville fractional integral of orderβ>0 of a function h:(0,∞)→R is given by
(15)Iβh(t)=1Γ(β)∫0t(t-s)β-1h(s)ds,t>0,
provided that the right-hand side is pointwise defined on (0,∞).Definition 8.
The Riemann-Liouville fractional derivative of orderβ>0 of a function h:(0,∞)→R is given by
(16)Dβh(t)=1Γ(n-β)(ddt)n∫0t(t-s)n-β-1h(s)ds,t>0,
where n=[β]+1 provided that the right-hand side is pointwise defined on (0,∞). Here [β] means the integer part of the number β.So we have the following properties (see [2]).Proposition 9.
(1) Letβ>0 and let h∈L1(0,∞), and then one has
(17)DβIβh(t)=h(t),fort>0.
(2) Letβ>0, and then
(18)Dβh(t)=0iffh(t)=∑‍mj=1cjtβ-j,
where m is the smallest integer greater than or equal to β and (c1,c2,…,cm)∈Rm.Corollary 10.
Letβ>0 and assume that Dβh∈L1(0,∞). Then,
(19)IβDβh(t)=h(t)+c1tβ-1+c2tβ-2+⋯+cmtβ-m,t>0,
where m is the smallest integer greater than or equal to β and (c1,c2,…,cm)∈Rm.Lemma 11.
Let1<α<2 and h∈L1(0,∞). The unique solution of
(20)Dαu(t)+h(t)=0,t>0lim⁡t→0t2-αu(t)=0,lim⁡t→∞t1-αu(t)=0
is given by
(21)u(t)=∫0∞Gα(t,s)h(s)ds,
where
(22)Gα(t,s)=1Γ(α)[tα-1-((t-s)+)α-1]
is Green’s function for the boundary value problem (20).Proof.
We may apply Corollary10 and Proposition 9 to reduce equation Dαu(t)+h(t)=0 to an equivalent integral equation
(23)u(t)=-Iαh(t)+c1tα-1+c2tα-2,
where (c1,c2)∈R2. Hence the general solution of Dαu(t)+h(t)=0 is
(24)u(t)=c1tα-1+c2tα-2-1Γ(α)∫0t(t-s)α-1h(s)ds.
By using lim⁡t→0t2-αu(t)=0 and lim⁡t→∞t1-αu(t)=0, we get
(25)c2=0,c1=1Γ(α)∫0∞h(s)ds.
Therefore, the unique solution of problem (20) is
(26)u(t)=tα-1Γ(α)∫0∞h(s)ds-1Γ(α)∫0t(t-s)α-1h(s)ds=∫0∞Gα(t,s)h(s)ds.
The proof is complete.Next we give sharp estimates on the Green functionGα(t,s). To this end, we need the following lemma.Lemma 12.
(i) Forλ,μ∈(0,∞) and x∈[0,1], one has
(27)min⁡(1,μλ)(1-xλ)≤1-xμ≤max⁡(1,μλ)(1-xλ).
(ii) For (t,s)∈(0,∞)×(0,∞) one has
(28)min⁡(1,t)min⁡(1,s)≤min⁡(t,s)≤max⁡(1,t)min⁡(1,s).Proposition 13.
The Green functionGα(t,s) defined by (22) satisfies
(29)Gα(t,s)≈tα-2min⁡(t,s),for(t,s)∈(0,∞)×(0,∞).Proof.
For(t,s)∈(0,∞)×(0,∞) we have
(30)Gα(t,s)=tα-1Γ(α)[1-((t-s)+t)α-1].
Since ((t-s)+/t)∈[0,1) for (t,s)∈(0,∞)×(0,∞), then by applying Lemma 12 (i) with μ=α-1 and λ=1, we obtain
(31)Gα(t,s)≈tα-1[1-((t-s)+t)]=tα-2min⁡(t,s).From here on, we define the potential kernelGα on B+((0,∞)) by
(32)Gαh(t)≔∫0∞Gα(t,s)h(s)ds,fort>0.Using Proposition13 and Lemma 12 (ii), we deduce the following.Corollary 14.
Let1<α<2 and h∈B+((0,∞)), and then the function t→Gαh(t) belongs to C2-α([0,∞)) if and only if the integral ∫0∞min⁡(1,s)h(s)ds converges.Proposition 15.
Let1<α<2 and f be a function such that the map s→min⁡(1,s)f(s) is continuous and integrable on (0,∞). Then Gαf is the unique solution in C2-α([0,∞)) of the boundary value problem
(33)Dαu(t)+f(t)=0,t>0lim⁡t→0t2-αu(t)=0,lim⁡t→∞t1-αu(t)=0.Proof.
From Corollary14, the function Gαf is well defined in (0,∞). Using Proposition 13 and Lemma 12 (ii), we get
(34)Gα|f|(t)≤c(tα-2(t+1)∫0∞min⁡(1,s)|f(s)|ds).
This implies that I2-α(Gα|f|) is finite on (0,∞). So by using Fubini’s theorem, we obtain
(35)I2-α(Gαf)(t)=1Γ(2-α)∫0t(t-s)1-αGαf(s)ds=1Γ(2-α)∫0∞(∫0t(t-s)1-αGα(s,r)ds)f(r)dr.
Observe that by considering the substitution s=r+(t-r)θ, we obtain
(36)∫rt(t-s)1-α(s-r)α-1ds=Γ(α)Γ(2-α)(t-r).
Using this fact and (22) we deduce that
(37)1Γ(2-α)∫0t(t-s)1-αGα(s,r)ds=1Γ(2-α)Γ(α)×[∫0t(t-s)1-αsα-1ds-∫0t(t-s)1-α((s-r)+)α-1ds]=t-1Γ(2-α)Γ(α)∫0t(t-s)1-α((s-r)+)α-1ds.
Now, assume that r≤t, and then by (36) we have
(38)∫0t(t-s)1-α((s-r)+)α-1ds=∫rt(t-s)1-α(s-r)α-1ds=Γ(α)Γ(2-α)(t-r).
On the other hand, if t≤r and s∈(0,t), we have
(39)∫0t(t-s)1-α((s-r)+)α-1ds=0.
So combining (38) and (39), we obtain
(40)1Γ(2-α)∫0t(t-s)1-αGα(s,r)ds=t-(t-r)+=min⁡(t,r).
This implies that
(41)I2-α(Gαf)(t)=∫0∞min⁡(t,r)f(r)dr=∫0trf(r)dr+t∫t∞f(r)dr,
and Dα(Gαf)(t)=(d2/dt2)(I2-α(Gαf))(t)=-f(t), for t>0.
Moreover, using Proposition13 and the dominated convergence theorem, we deduce that
(42)lim⁡t→0t2-αGαf(t)=0,lim⁡t→∞t1-αGαf(t)=0.
Finally, we need to prove the uniqueness. Let u,v∈C2-α([0,∞)) be two solutions of (33) and put ω=u-v. Then ω∈C2-α([0,∞)) and Dαω=0. Hence, it follows from Corollary 10 that ω(t)=c1tα-1+c2tα-2. Using the fact that lim⁡t→0t2-αω(t)=lim⁡t→∞t1-αω(t)=0, we deduce that ω=0 and therefore u=v. The proof is complete.


## 2.2. Sharp Estimates on the Potential of Some Karamata Functions

We collect in this paragraph some properties of functions belonging to the Karamata classK (resp., K∞) and we give estimates on some potential functions.Proposition 16 (see [26, 27]).
(i) LetL1,L2∈K (resp, K∞) and p∈R. Then the functions
(43)L1+L2,L1L2andL1pbelongtotheclassK(resp,K∞).
(ii) Let L be a function in K (resp., K∞) and ε>0. Then one has
(44)lim⁡t→0+tεL(t)=0,(resp.lim⁡t→∞t-εL(t)=0).Theorem 17 (see [26, 27]).
(a) Letγ∈R and L be a function in K defined on (0,η]. One has the following.(i)
Ifγ<-1, then ∫0ηsγL(s)ds diverges and ∫tηsγL(s)ds~t→0+-tγ+1L(t)/(γ+1).(ii)
Ifγ>-1, then ∫0ηsγL(s)ds converges and ∫0tsγL(s)ds~t→0+tγ+1L(t)/(γ+1).
(b) Letγ∈R and L be a function in K∞. One has the following.(i)
Ifγ>-1, then ∫1∞sγL(s)ds diverges and ∫1tsγL(s)ds~t→∞tγ+1L(t)/(γ+1).(ii)
Ifγ<-1, then ∫1∞sγL(s)ds converges and ∫t∞sγL(s)ds~t→∞-tγ+1L(t)/(γ+1).The proof of the next lemma can be found in [11].Lemma 18.
LetL be a function in K defined on (0,η]. Then one has
(45)lim⁡t→0+L(t)∫tη(L(s)/s)ds=0.
In particular,
(46)t⟶∫tηL(s)sds∈K.
If further ∫0η(L(s)/s)ds converges, then one has lim⁡t→0+(L(t)/∫0t(L(s)/s)ds)=0.
In particular,(47)t⟶∫0tL(s)sds∈K.In the next lemma, we have the following properties related to the classK∞. For the proof we refer to [22].Lemma 19.
LetL be a function in K∞. Then one has
(48)lim⁡t→∞L(t)∫1t(L(s)/s)ds=0.
In particular,
(49)t⟶∫1t+1L(s)sds∈K∞.
If further ∫1∞(L(s)/s)ds converges, then one has
(50)lim⁡t→∞L(t)∫t∞(L(s)/s)ds=0.
In particular,
(51)t⟶∫t∞L(s)sds∈K∞.Now, we put(52)b(t)=t-β(1+t)β-γL3(min⁡(t,1))L4(max⁡(t,1)),t>0,
where L3∈K and L4∈K∞. We aim at giving sharp estimates on the potential function Gαb(t).Proposition 20.
Assume thatL3∈K defined on (0,η], for some η>1 and L4∈K∞. Let β≤2 and γ≥1 such that
(53)∫0ηs1-βL3(s)ds<∞,∫1∞s-γL4(s)ds<∞.
Then for t>0(54)Gαb(t)≈tα-2ψβ(min⁡(t,1))ϕγ(max⁡(t,1)),
where, for t∈(0,1],
(55)ψβ(t)={∫0tL3(s)sdsifβ=2,t2-βL3(t)if1<β<2,t∫tηL3(s)sdsifβ=1,tifβ<1
and, for t≥1,
(56)ϕγ(t)={t∫t∞L4(s)sdsifγ=1,t2-γL4(t)if1<γ<2,∫1t+1L4(s)sdsifγ=2,1ifγ>2.Proof.
Using Proposition13 and Remark 4, we have
(57)t2-αGαb(t)≈∫0ηmin⁡(t,s)s-βL3(s)ds+∫η∞min⁡⁡(t,s)s-γL4(s)ds=I(t)+J(t).
Case  1. Assume that 0<t≤1.
By using (53), we deduce that
(58)J(t)≈t.
On the other hand,
(59)I(t)=∫0ts1-βL3(s)ds+t∫tηs-βL3(s)ds=I1(t)+I2(t).
Using Theorem 17 and hypothesis (53), we deduce that
(60)I1(t)≈{t2-βL3(t)ifβ<2,∫0tL3(s)sdsifβ=2,I2(t)≈{t2-βL3(t)if1<β≤2,t∫tηs-βL3(s)dsifβ≤1.
Hence, it follows by Lemma 18, Proposition 16, and hypothesis (53) that
(61)I(t)≈{∫0tL3(s)sdsifβ=2,t2-βL3(t)if1<β<2,t∫tηL3(s)sdsifβ=1,tifβ<1.
Combining (58) and (61) and using Proposition 16 and hypothesis (53), we deduce that, for 0<t≤1,
(62)t2-αGαb(t)≈ψβ(t).Case2. Assume that t>η+1.
By using (53), we deduce that
(63)I(t)≈1.
On the other hand,
(64)J(t)=∫tηs1-γL4(s)ds+t∫t∞s-γL4(s)ds=J1(t)+J2(t).
Using again Theorem 17 and hypothesis (53), we deduce that
(65)J1(t)≈{t2-γL4(t)if1≤γ<2,∫ηts1-γL4(s)dsifγ≥2,J2(t)≈{t2-γL4(t)ifγ>1,t∫t∞L4(s)sdsifγ=1.
Hence, it follows from Lemma 19 and hypothesis (53) that
(66)J(t)≈{t∫t∞L4(s)sdsifγ=1,t2-γL4(t)if1<γ<2,∫ηtL4(s)sdsifγ=2,1ifγ>2.
Combining (63) and (66) and using Proposition 16, hypothesis (53), and Remark 4, we deduce that, for t>η+1,
(67)t2-αGαb(t)≈{t∫t∞L4(s)sdsifγ=1,t2-γL4(t)if1<γ<2,∫ηtL4(s)sdsifγ=2,1ifγ>2.≈ϕγ(t).
Now since the functions t→t2-αGαb(t) and t→ϕγ(t) are positive and continuous on [1,η+1], we deduce that, for t∈[1,η+1],
(68)t2-αGαb(t)≈ϕγ(t).
Finally, using (62), (67), and (68), we obtain the required result.


## 3. Proof of the Main Result

The next lemma will play a crucial role in the proof of Theorem5Lemma 21.
Assume that the functiona satisfies (H) and put ω(t)=a(t)t(α-2)σθσ(t) for t>0. Then one has for t∈(0,∞)(69)t2-αGαω(t)≈θ(t).Proof.
We recall that(70)ν=min⁡(1,2-λ+(α-2)σ1-σ),ζ=max⁡(0,2-μ+(α-2)σ1-σ),θ(t)=tν(1+t)ζ-ν(L~1(min⁡(t,1)))1/(1-σ)×(L~2(max⁡(t,1)))1/(1-σ),
where, for t∈(0,1],
(71)L~1(t)={∫0tL1(s)sdsifλ=2+(α-2)σ,L1(t)if1+(α-1)σ<λ<2+(α-2)σ,∫tηL1(s)sdsifλ=1+(α-1)σ,1ifλ<1+(α-1)σ
and, for t≥1,
(72)L~2(t)={∫t∞L2(s)sdsifμ=1+(α-1)σ,L2(t)if1+(α-1)σ<μ<2+(α-2)σ,∫1t+1L2(s)sdsifμ=2+(α-2)σ,1ifμ>2+(α-2)σ.
For t>0, we have
(73)ω(t)≈t-λ+νσ+(α-2)σ(1+t)λ-μ+(ζ-ν)σ×L1(min⁡(t,1))(L~1(min⁡(t,1)))σ/(1-σ)×L2(max⁡(t,1))(L~2(max⁡(t,1)))σ/(1-σ).
Using Proposition 20 with β=λ-νσ-(α-2)σ and γ=μ-ζσ-(α-2)σ,L3(t)=L1(t)(L~1(t))σ/(1-σ) and L4(t)=L2(t)(L~2(t))σ/(1-σ), we obtain for t∈(0,1](74)t2-αGαω(t)≈{∫0tL1(s)s(∫0sL1(r)rdr)σ/(1-σ)dsifλ=2+(α-2)σ,t(2-λ+(α-2)σ)/(1-σ)L1(t)(L1(t))σ/(1-σ)if1+(α-1)σ<λ<2+(α-2)σ,t∫tηL1(s)s(∫sηL1(r)rdr)σ/(1-σ)dsifλ=1+(α-1)σ,tifλ<1+(α-1)σ,≈{∫0t(L1(s)sds)1/(1-σ)ifλ=2+(α-2)σ,t(2-λ+(α-2)σ)/(1-σ)(L1(t))1/(1-σ)if1+(α-1)σ<λ<2+(α-2)σ,t(∫tηL1(s)sds)1/(1-σ)ifλ=1+(α-1)σ,tifλ<1+(α-1)σ,≈θ(t).On the other hand, using again Proposition 20 and Remark 4, we get, for t≥1,(75)t2-αGαω(t)≈{t∫t∞L2(s)s(∫s∞L2(r)rdr)σ/(1-σ)dsifμ=1+(α-1)σ,tζL2(t)(L2(t))σ/(1-σ)if1+(α-1)σ<μ<2+(α-2)σ,∫1t+1L2(s)s(∫1s+1L2(r)rdr)σ/(1-σ)dsifμ=2+(α-2)σ,1ifμ>2+(α-2)σ.≈{t(∫t∞L2(s)sds)1/(1-σ)ifμ=1+(α-1)σ,tζ(L2(t))1/(1-σ)if1+(α-1)σ<μ<2+(α-2)σ,(∫1t+1L2(s)sds)1/(1-σ)ifμ=2+(α-2)σ,1ifμ>2+(α-2)σ.≈θ(t).This completes the proof.Proof of Theorem5.
From Lemma21, there exists M>1 such that for each t>0(76)1Mθ(t)≤t2-αGαω(t)≤Mθ(t),
where ω(t)=a(t)t(α-2)σθσ(t).
Putc0=M1/(1-|σ|) and let
(77)Λ={v∈C0([0,∞)):θ(t)c0(1+t)≤v(t)≤c0θ(t)1+t,t>0}.
In order to use a fixed point theorem, we define the operator T on Λ by
(78)Tv(t)=t2-α1+t∫0∞Gα(t,s)a(s)s(α-2)σ(1+s)σvσ(s)ds.
For this choice of c0 and using (76), we easily prove that for all v∈Λ and t>0(79)Tv(t)≤c0θ(t)1+t,Tv(t)≥θ(t)c0(1+t).
On the other hand, using Proposition 13 and Lemma 12 (ii), there exists c>0 such that for all t,s>0, we have
(80)t2-αGα(t,s)1+t≤cmin⁡(1,s).
This implies that there exists c>0 such that, for each v∈Λ and t>0,
(81)|Tv(t)|≤c∫0∞min⁡(1,s)ω(s)ds.
Now by hypothesis (H) and Theorem 17, the function t→min⁡(1,t)ω(t) is in L1(0,∞), which implies that the family {Tv(t),v∈Λ} is uniformly bounded.
Using (80) and the fact that, for each s>0, the function t→(t2-αGα(t,s))/(1+t) is in C0([0,∞)), we deduce that the family {Tv(t),v∈Λ} is equicontinuous in [0,∞].
Hence, it follows by Ascoli’s theorem thatT(Λ) is relatively compact in C0([0,∞)) and therefore T(Λ)⊂Λ.
Next, we will prove the continuity ofT in the supremum norm. Let (vk)k be a sequence in Λ which converges to v in Λ. Using again (80) and Lebesgue’s theorem, we deduce that Tvk(t)→Tv(t) as k→∞, for t>0.
SinceT(Λ) is relatively compact in C0([0,∞)), then the pointwise convergence implies the uniform convergence. Thus we have proved that T is a compact mapping from Λ to itself.
Now, the Schauder fixed point theorem implies the existence ofv∈Λ such that
(82)v(t)=t2-α1+t∫0∞Gα(t,s)a(s)s(α-2)σ(1+s)σvσ(s)ds.
Put u(t)=tα-2(1+t)v(t). Then u∈C2-α([0,∞)) and u satisfies the equation
(83)u(t)=Gα(auσ)(t).
Since the function s→min⁡(1,s)a(s)uσ(s) is continuous and integrable on (0,∞), then by Proposition 15, the function u is a positive solution in C2-α([0,∞)) of problem (3).
Finally, it remains to prove thatu is the unique positive solution in C2-α([0,∞)) satisfying (14). To this end, assume that problem (3) has two positive solutions u,v∈C2-α([0,∞)) satisfying (14). Then there exists a constant m>1 such that
(84)1m≤uv≤m.
This implies that the set
(85)J={m≥1:1m≤uv≤m}
is not empty. Let c=inf⁡J. Then c≥1 and we have (1/c)v≤u≤cv. It follows that uσ≤c|σ|vσ and consequently
(86)-Dα(c|σ|v-u)=a(c|σ|vσ-uσ)≥0,(87)lim⁡t→0t2-α(c|σ|v-u)(t)=0,lim⁡t→∞t1-α(c|σ|v-u)(t)=0.
This implies by Proposition 15 that c|σ|v-u=Gα(a(c|σ|vσ-uσ))≥0. By symmetry, we obtain also v≤c|σ|u. Hence c|σ|∈J and c≤c|σ|. Since |σ|<1, then c=1 and therefore u=v.Example 22.
Let1<α<2, σ∈(-1,1), and a be a positive continuous function on (0,∞) such that
(88)a(t)≈t-λ(1+t)λ-μlog⁡(2min⁡⁡(t,1)),t>0,
where λ<2+(α-2)σ and μ>1+(α-1)σ. Then by Theorem 5, problem (3) has a unique positive solution u∈C2-α([0,∞)) satisfying, for t>0,
(89)u(t)≈tα-2+ν(1+t)ζ-ν(L~1(min⁡(t,1)))1/(1-σ)×(L~2(max⁡(t,1)))1/(1-σ),
where ν=min⁡(1,(2-λ+(α-2)σ)/(1-σ)), ζ=max⁡(0,(2-μ+(α-2)σ)/(1-σ)),
(90)L~1(t)={log⁡(2t)if1+(α-1)σ<λ<2+(α-2)σ,(log⁡(2t))2ifλ=1+(α-1)σ,1ifλ<1+(α-1)σ,L~2(t)={log⁡(1+t)ifμ=2+(α-2)σ,1ifμ≠2+(α-2)σ.



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*Source: 537971-2014-07-16.xml*