# Some New Existence and Uniqueness Results for an Integral Boundary Value Problem of Caputo Fractional Differential Equations

Authors: Haixing Feng; Chengbo Zhai
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

In this work, we consider an integral boundary value problem of Caputo fractional differential equations. Based on a fixed-point theorem of generalized concave operators, we obtain the existence and uniqueness of positive solutions. As applications of main results, we give two examples in the end.


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## Body

## 1. Introduction

The fractional differential equations appear naturally in fields of physics, chemistry, electrodynamics of complex medium, control of dynamical systems, and so on. So differential equations with fractional order have been studied by many researchers, and then the theory of fractional boundary value problems has been noticed more and more in recent years; see [1–26], for example.In [25], the author investigated the existence of positive solutions for an integral boundary value problem of fractional differential equations by using a fixed-point theorem on cones. In [21, 22], Yang gave the existence and nonexistence results for fractional differential equation integral boundary value problems. In these papers, we can find that there are no uniqueness results on positive solutions. Moreover, there are still very few results on the uniqueness of positive solutions for an integral boundary value problem of Caputo fractional differential equations. To fill this gap, we study the existence and uniqueness of positive solutions for the following integral boundary value problem for fractional differential equations (1)D0+αCxt+ft,xt=0,t∈0,1,ax0-bx′0=0,cx1+dx′1=0,x′′0+x′′′0=∫01x′′τdpτ,x′′1+x′′′1+∫01x′′τdqτ=0,where 3<α≤4,a,b,c,d≥0,ρ=ad+ac+bc>0.In [19], the authors studied the existence of positive solutions for (1) by the fixed-point theorems on cones. We also do not find the uniqueness results on positive solutions in [19]. In this paper, we will show the existence and uniqueness of positive solutions for problem (1). Our mail tool is a fixed-point theorem of generalized concave operators in ordered Banach spaces.The paper is organized as follows: in Section2, we will present some useful definitions, preliminaries, and lemmas. The existence and uniqueness results are proved in Section 3. In Section 4, we finish the paper with two examples.


## 2. Preliminaries

Now we list some conditions for convenience.(A1)
f(t,x)≥0,f(t,x) is increasing in x∈[0,+∞) for each t∈[0,1].(A2)
Forλ∈(0,1), there exists α(λ)∈(λ,1) such that f(t,λx)≥α(λ)f(t,x),t∈[0,1],x≥0.(A3)
p(t),q(t) are right continuous on [0,1), left continuous at t=0, and increasing on [0,1], with p(0)=q(0)=0;∫01x(τ)dp(τ),∫01x(τ)dq(τ) are the Riemann-Stieltjes integrals of x with respect to p,q; moreover, (2)0≤∫01dpτ<1,0≤∫01dqτ<1,2+∫01τdqτ1+∫01dqτ>1-∫01τdpτ1-∫01dpτ.(
A
4
)
f
(
t
,
x
) is continuous with ∫01(1-s)α-4f(s,x(s))ds existing.
(
A
1
′
)
f
(
t
,
x
) is decreasing in x∈[0,+∞) for each t∈[0,1] and f(t,x)≥0 for [0,1]×[0,+∞).
(
A
2
′
)
For anyλ∈(0,1), there exists β(λ)∈(0,1) such that (3)ft,λx≤λ-βλft,x,t∈0,1,x≥0.In the following, we present some definitions and lemmas.Definition 1 (see [12]).
The Riemann-Liouville fractional integral of orderα for function y is (4)I0+αyt=1Γα∫0tt-αα-1ysds,α>0,provided that the right side is point-wise defined on (0,+∞).Definition 2 (see [12]).
The Riemann-Liouville fractional derivative of orderα>0 for function y is (5)D0+αyt=1Γn-αddtn∫0tyst-sα-n+1ds,α>0,where n=[α]+1, provided that the right side is point-wise defined on (0,+∞).Definition 3 (see [23]).
Ify∈ACn[0,1], then the Caputo fractional derivatives of order α>0 exist almost everywhere on [0,1] and are represented by (6)D0+αCyt=1Γn-α∫0tynst-sα-n+1ds=I0+n-αDnyt.Next we recall some concepts. LetE be a real Banach space with norm ·, and let θ be the zero element of E. P is a cone in E. For all x,y∈E, the notation x~y means that there exist λ>0,μ>0 such that λx≤y≤μx. Clearly, ~ is an equivalence relation. Given h>θ, we denote by Ph the set Ph≔{x∈E∣x~y}. Clearly, Ph⊂P is convex and λPh=Ph for all λ>0.Definition 4.
A coneP∈E is said to be normal if and only if there exists a constant N>0 such that (7)θ≤x≤y⟹x≤Ny,x,y∈E,where N is called the normality constant of P.Definition 5.
An operatorA:E→E is increasing (decreasing) if x≤y implies Ax≤Ay(Ax≥Ay).Lemma 6 (see [19]).
Suppose that(A3),(A4) hold and 3<α≤4; then the integral boundary value problem (8)D0+αCxt+ft,xt=0,t∈0,1,ax0-bx′0=0,cx1+dx′1=0,x′′0+x′′′0=∫01x′′τdpτ,x′′1+x′′′1+∫01x′′τdqτ=0has a solution (9)xt=∫01G1t,ξ∫01G2ξ,sfs,xsdsdξ,where (10)G1t,s=1ρas+bc1-t+d,0≤s≤t≤1,1ρat+bc1-s+d,0≤t≤s≤1,(11)G2t,s=Gαt,s+Gpqt,s,t,s∈0,1,(12)Gαt,s=1-sα-31-t+t-sα-3Γα-2+1-sα-41-tΓα-3,s≤t,1-sα-31-tΓα-2+1-sα-41-tΓα-3,t≤s,Gpqt,s=1δ2+∫01tdqt-t1+∫01dqτ∫01Gατ,sdpτ+1δ1-∫01tdpt-t1-∫01dpτ∫01Gατ,sdqτ,t,s∈0,1,δ=1-∫01dpt1-∫01tdpt1+∫01dqt2+∫01tdqt>0.Lemma 7 (see [19]).
Let3<α≤4. Then Green’s function G2(t,s) has the following properties:(i)
(13)G2t,s∈C0,1×0,1,G2t,s>0,t,s∈0,1.(ii)
(14)G2t,s≥mtMpqMαs,max0≤t≤1G2t,s≤MpqMαs,t∈0,1,s∈0,1,where(15)Mpq=1+1δ2+∫01τdqτ∫01dpτ+∫01dqτ,(16)mt=1Mpq1-t2+1δ2+∫01τdqτ-t1+∫01dqτ∫011-τ2dpτ+1δ1-∫01τdpτ-t1-∫01dpτ∫011-τ2dqτ,t∈0,1,(17)Mαs=21-sα-3Γα-2+1-sα-4Γα-3.Lemma 8.
Letgt=∫01G1t,sds; then (18)gt=1ρ-t22bc+ac+ad+ac2+dt+bc2+d,t∈0,1 and g(t)≥g0,t∈[0,1], where g0=minb/ρ(c/2+d),d/ρ(a/2+b).We now present a fixed-point theorem of generalized concave operators which will be used in the later proofs.Theorem 9 (see [27]).
Leth>θ and P be a normal cone. Assume that(D1)
A:P→P is increasing and Ah∈Ph;(D2)
for anyx∈P and t∈(0,1), there exists α(t)∈(t,1) with respect to t such that (19)Atx≥αtAx.Then(i)
there areu0,v0∈Ph and r∈(0,1) such that (20)rv0≤u0<v0,u0≤Au0≤Av0≤v0;(ii)
operator equationx=Ax has a unique solution in Ph.Remark 10.
An operatorA is said to be generalized concave if A satisfies condition (D2).


## 3. Existence and Uniqueness of Positive Solutions for Problem (1)

In this section, we use Theorem9 to study problem (1) and we obtain some new results on the existence and uniqueness of positive solutions. This is also the main motivation for the study of (1) in the present work.SetJ=[0,1], C[J,R]={x∣x:J→Riscontinuous}. Evidently, C[J,R] is a Banach space with the norm xC=sup⁡xt:t∈J, P={x∈C[J,R]∣x(t)≥0,t∈J}, the standard cone, and it is normal. Our main result is summarized in the following theorem.Theorem 11.
Assume that(A1)–(A4) hold and f(t,bc+bd)>0,f(t,ad+bd)>0,t∈[0,1]. Then(i)
there existu0,v0∈Ph such that (21)u0t≤∫01G1t,ξ∫01G2ξ,sfs,u0sdsdξ,t∈J,v0t≥∫01G1t,ξ∫01G2ξ,sfs,v0sdsdξ,t∈J;(ii)
the integral boundary value problem (1) has a unique positive solution x∗ in Ph, where h(t)=(at+b)[c(1-t)+d],t∈J.Proof.
Define an operatorA:C[J,R]→C[J,R] by (22)Axt=∫01G1t,ξ∫01G2ξ,sfs,xsdsdξ.We know that x is a solution of problem (1) if and only if x is a fixed point of the operator A.
Firstly, we show thatA:P→P is increasing, generalized concave. From (A1), we know that f(t,x(t))≥0, and from (10) and  (13) we have G1(t,s)≥0,G2(t,s)≥0, so we have Ax(t)≥0 for x∈P. It follows from (A1) that A:P→P is increasing.
Now we prove thatA:P→P is generalized concave. For any x∈P and λ∈(0,1), from (A2) we have (23)Aλxt=∫01G1t,ξ∫01G2ξ,sfs,λxsdsdξ≥αλ∫01t,ξ∫01G2ξ,sfs,xsdsdξ=αλAxt. Thus, we have A(λx)≥α(λ)Ax,x∈P,λ∈(0,1).
Secondly, we prove thatAh∈Ph. Note that h(t)=(at+b)[c(1-t)+d], and we have (24)Aht=∫01G1t,ξ∫01G2ξ,sfs,hsdsdξ. From (10), it is easy to obtain that (25)hta+bc+dG1ξ,s≤G1t,s≤htρ,∀ξ∈0,1,t,s∈0,1.From (14), we have (26)mtMpqMαs≤G2t,s≤MpqMαs. We know by (16) that m(t) is decreasing, so when t=1,m(t) obtains the minimum, and m(1)>0, so we have (27)m1MpqMαs≤G2t,s≤MpqMαs,and from (17), we have (28)∫01Mαsds=∫0121-sα-3Γα-2+1-sα-4Γα-3ds=2α-2Γα-2+1α-3Γα-3.So from (25), (27), and (28), (22) has the following form: (29)Aht=∫01G1t,ξ∫01G2ξ,sfs,hsdsdξ≤htρ∫01G2ξ,sfs,hsds≤htρMpq∫01Mαsfs,hsds.After simple calculation, we get h′t0=0, where t0=ac+ad-bc/2ac. So we need to discuss several cases:
(
i
) When 0≤t0≤1, we have (30)ft,ht≤ft,ht0,t∈0,1. Let f01=max{f(t,h(t0)),t∈[0,1]}; then f01≥f(t,h(t0))≥f(t,h(0))=f(t,bc+bd)>0. From (28) and (29), we obtain that (31)Aht≤htρMpqf01∫01Mαsds=Mpqf01ρ2α-2Γα-2+1α-3Γα-3ht.
(
i
i
) When t0>1, we have (32)ft,ht≤ft,h1=ft,ad+bd,t∈0,1. Let f02=max{f(t,ad+bd),t∈[0,1]}; then f02≥f(t,h(1))=f(t,ad+bd)>0. From (28) and (29), we obtain that (33)Aht≤htρMpqf02∫01Mαsds=Mpqf02ρ2α-2Γα-2+1α-3Γα-3ht.
(
i
i
i
) When t0<0, we have (34)ft,ht≤ft,h0=ft,bc+bd,t∈0,1.Let f03=max{f(t,bc+bd),t∈[0,1]}; then f03≥f(t,h(0))=f(t,bc+bd)>0. From (28) and (29), we obtain that (35)Aht≤htρMpqf03∫01Mαsds=Mpqf03ρ2α-2Γα-2+1α-3Γα-3ht.
Let(36)μ1=Mpqρ2α-2Γα-2+1α-3Γα-3f01,t0∈0,1,f02,t0>1,f03,t0<0. Then from (31), (33), and (35), we have (37)Aht≤μ1ht,t∈0,1.On the other hand, from (25), (27), and (28), we have (38)Aht=∫01G1t,ξ∫01G2ξ,sfs,hsdsdξ≥hta+bc+d∫01G1t,ξ∫01G2ξ,sfs,hsdsdξ≥hta+bc+dm1Mpq∫01G1t,ξ∫01Mαsfs,hsdsdξ.Also, we need to discuss the following cases:
(
i
) When 0≤t0≤1, we have(39)ft,ht≥min⁡ft,h0,ft,h1,t∈0,1≔f04. We can see that f04=min⁡ft,h0,ft,h1,t∈0,1>0. From (28),  (38), and Lemma 8, we obtain that (40)Aht≥hta+bc+dm1Mpqf042α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf04g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.
(ii) Whent0>1, we have (41)ft,ht≥ft,h0=ft,bc+bd,t∈0,1.Let f05=min⁡{f(t,bc+bd),t∈[0,1]}, and then f05>0. From (28), (38), and Lemma 8, we obtain that (42)Aht≥hta+bc+dm1Mpqf052α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf05g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.
(iii) Whent0<0, we have (43)ft,ht≥ft,h1=ft,ad+bd,t∈0,1.Let f06=min⁡ft,ad+bd,t∈0,1, and then f06>0. From (28), (38), and Lemma 8, we obtain that (44)Aht≥hta+bc+dm1Mpqf062α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf06g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.Let (45)μ2=m1Mpqg0a+bc+d2α-2Γα-2+1α-3Γα-3f04,t0∈0,1,f05,t0>1,f06,t0<0.Then from (40), (42), and (44), we have (46)Aht≥μ2ht,t∈0,1.So, from (37) and (46) we have (47)μ2ht≤Aht≤μ1ht,that is, Ah∈Ph. Finally, an application of Theorem 9 implies that
(i) there existu0,v0∈Ph such that u0≤Au0,v0≥Av0, that is, (48)u0t≤∫01G1t,ξ∫01G2ξ,sfs,u0sdsdξ,t∈J,v0t≥∫01G1t,ξ∫01G2ξ,sfs,v0sdsdξ,t∈J; (ii) operator equation x=Ax has a unique solution x∗ in Ph. That is, x∗(t) is the unique positive solution for problem (1) in Ph.Theorem 12.
Assume that(A1′),(A2′),(A3),(A4) hold and f(t,x)>0 for x>0,t∈[0,1]. Then
(i) there existu0,v0∈Ph such that (49)u0t≤∫01G1t,ξ∫01G2ξ,sfs,u¯0sdsdξ,t∈J,v0t≥∫01G1t,ξ∫01G2ξ,sfs,v¯0sdsdξ,t∈J;where (50)u¯0t=∫01G1t,ξ∫01G2ξ,sfs,u0sdsdξ,t∈J,v¯0t=∫01G1t,ξ∫01G2ξ,sfs,v0sdsdξ,t∈J;
(ii) the integral boundary value problem (1) has a unique positive solution x∗ in Ph, where h(t)=(at+b)[c(1-t)+d],t∈J.Proof.
Similar to the proof of Theorem11, we consider the operator A:C[J,R]→C[J,R] by (51)Axt=∫01G1t,ξ∫01G2ξ,sfs,xsdsdξ.From (A1′), we know that Ax(t)≥0,t∈[0,1]. It follows from (A1′) that A:P→P is decreasing. For any x∈P,λ∈(0,1) and from (A2′), we have (52)Aλxt=∫01G1t,ξ∫01G2ξ,sfs,λxsdsdξ≤λ-βλ∫01t,ξ∫01G2ξ,sfs,xsdsdξ=λ-βλAxt. Thus, we have A(λx)≤λ-β(λ)Ax(t),x∈P,λ∈(0,1). Further, for λ∈(0,1),x∈P, we have (53)Ax=Aλ·1λx≤λ-βλA1λx.So we obtain A(1/λx)≥λβ(λ)Ax,x∈P,λ∈(0,1). Consequently, A2:P→P is increasing and for x∈P,λ∈(0,1), (54)A2λx=AAλx≥Aλ-βλAx=A1λβxAx≥λβλβλβλA2x≥λβλA2x. Let α(t)=tβ(t),t∈(0,1). Then t<α(t)<1 and A2(λx)≥α(λ)A2x,λ∈(0,1),x∈P. So the operator A2:P→P is generalized concave.
Next we prove thatA2h∈Ph. Note that h(t)=(at+b)[c(1-t)+d], and we can obtain (55)Aht=∫01G1t,ξ∫01G2ξ,sfs,hsdsdξ,A2ht=AAt=∫01G1t,ξ∫01G2ξ,sfs,Ahsdsdξ.Because h′(t0)=0, where t0=ac+ad-bc/2ac, we need to discuss several cases:
(i) When0≤t0≤1, we have (56)ft,ht≤min⁡ft,h0,ft,h1,t∈0,1≔f04, and we can see that f04=min⁡ft,h0,ft,h1,t∈0,1>0. From (28) and (29), we obtain that (57)Aht≤htρMpqf04∫01Mαsds=Mpqf04ρ2α-2Γα-2+1α-3Γα-3ht.
(ii) Whent0>1, we have (58)ft,ht≤ft,h0=ft,bc+bd,t∈0,1,Note that f03=max⁡ft,bc+bd,t∈0,1; then f03≥ft,h0=ft,bc+bd>0. From (28) and (29), we obtain that (59)Aht≤htρMpqf03∫01Mαsds=Mpqf03ρ2α-2Γα-2+1α-3Γα-3ht.
(iii) Whent0<0, we have (60)ft,ht≤ft,h1=ft,ad+bd,t∈0,1.Note that f02=max⁡ft,ad+bd,t∈0,1; then f02≥ft,h1=ft,ad+bd>0. From (28) and (29), we obtain that (61)Aht≤htρMpqf02∫01Mαsds=Mpqf02ρ2α-2Γα-2+1α-3Γα-3ht.Let (62)μ3=Mpqρ2α-2Γα-2+1α-3Γα-3f04,t0∈0,1,f03,t0>1,f02,t0<0. Then from (57), (59), and (61), we have (63)Aht≤μ3ht,t∈0,1.Further, we need to discuss the following cases:
(i) When0≤t0≤1, we have (64)ft,ht≥ft,ht0,t∈0,1. Let f07=min⁡ft,ht0,t∈0,1, and then f07>0. From (28), (38), and Lemma 8, we obtain that (65)Aht≥hta+bc+dm1Mpqf072α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf07g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.
(ii) Whent0>1, we have (66)ft,ht≥ft,h1=ft,ad+bd,t∈0,1. Note that f06=min⁡ft,ad+bd,t∈0,1>0 and from (28), (38), and Lemma 8, we obtain that (67)Aht≥hta+bc+dm1Mpqf062α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf06g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.
(iii) Whent0<0, we have (68)ft,ht≥ft,h0=ft,bc+bd,t∈0,1.Note that f05=min⁡ft,bc+bd,t∈0,1>0 and from (28), (38), and Lemma 8, we obtain that (69)Aht≥hta+bc+dm1Mpqf052α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf05g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.Let (70)μ4=m1Mpqg0a+bc+d2α-2Γα-2+1α-3Γα-3f07,t0∈0,1,f06,t0>1,f05,t0<0.Then from (65), (67), and (69), we have (71)Aht≥μ4ht,t∈0,1.So, from (63) and (71) we have (72)μ4ht≤Aht≤μ3ht.Hence, we have (73)A2ht≤∫01G1t,ξ∫01G2ξ,sfs,μ4hsdsdξ,(74)A2ht≥∫01G1t,ξ∫01G2ξ,sfs,μ3hsdsdξ.Because h′t0=0, where t0=ac+ad-bc/2ac, we need to discuss the following cases:
(i) When0≤t0≤1, we have (75)ft,μ4ht≤min⁡ft,μ4h0,ft,μ4h1,t∈0,1≔f08,and we can see that f08=min{ft,μ4h0,ft,μ4h1,t∈[0,1]}>0. From (28) and (73), we obtain that (76)A2ht≤htρMpqf08∫01Mαsds=Mpqf08ρ2α-2Γα-2+1α-3Γα-3ht.
(ii) Whent0>1, we have (77)ft,μ4ht≤ft,μ4h0=ft,μ4bc+bd,t∈0,1.Let f09=max{ft,μ4bc+bd,t∈[0,1]}; then f09≥ft,μ4h0=ft,μ4bc+bd>0. From (28) and (73), we obtain that (78)A2ht≤htρMpqf09∫01Mαsds=Mpqf09ρ2α-2Γα-2+1α-3Γα-3ht.
(iii) Whent0<0, we have (79)ft,μ4ht≤ft,μ4h1=ft,μ4ad+bd,t∈0,1. Let f10=max⁡ft,μ4ad+bd,t∈0,1; then f10≥ft,μ4h1=ft,μ4ad+bd>0. From (28) and (73), we obtain that (80)A2ht≤htρMpqf10∫01Mαsds=Mpqf10ρ2α-2Γα-2+1α-3Γα-3ht.Let (81)μ1′=Mpqρ2α-2Γα-2+1α-3Γα-3f08,t0∈0,1,f09,t0>1,f10,t0<0. Then from (76), (78), and (80), we have (82)A2ht≤μ1′ht,t∈0,1.Similarly, we need to discuss the following cases:
(i) When0≤t0≤1, we have (83)ft,μ3ht≥ft,μ3ht0,t∈0,1.Let f11=min⁡ft,μ3ht0,t∈0,1, and then f11>0. From (28), (74), and Lemma 8, we obtain that (84)A2ht≥hta+bc+dm1Mpqf112α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf11g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.
(ii) Whent0>1, we have (85)ft,μ3ht≥ft,μ3h1=ft,μ3ad+bd,t∈0,1.Let f12=min⁡ft,μ3ad+bd,t∈0,1, and then f12>0. From (28), (74), and Lemma 8, we obtain that (86)A2ht≥hta+bc+dm1Mpqf122α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf12g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.
(iii) Whent0<0, we have (87)ft,μ3ht≥ft,μ3h0=ft,μ3bc+bd,t∈0,1. Note that f13=min⁡ft,μ3bc+bd,t∈0,1>0 and from (28), (74), and Lemma 8, we obtain that (88)A2ht≥hta+bc+dm1Mpqf132α-2Γα-2+1α-3Γα-3∫01G1τ,ξdξ≥m1Mpqf13g0a+bc+d2α-2Γα-2+1α-3Γα-3ht.Let (89)μ2′=m1Mpqg0a+bc+d2α-2Γα-2+1α-3Γα-3f11,t0∈0,1,f12,t0>1,f13,t0<0.Then from (84), (86), and (88), we have (90)A2ht≥μ2′ht,t∈0,1.From (82) and (90), we can get (91)μ2′ht≤A2ht≤μ1′ht,that is, A2h∈Ph. Finally, an application of Theorem 9 implies the following: (i) there exist u0,v0∈Ph such that u0≤A2u0,v0≥A2v0; (ii) A2x=x has a unique solution x∗ in Ph. Let u¯0=Au0,v¯0=Av0, and then u0≤Au¯0,v0≥Av¯0. That is, (92)u0t≤∫01G1t,ξ∫01G2ξ,sfs,u¯0sdsdξ,t∈J,v0t≥∫01G1t,ξ∫01G2ξ,sfs,v¯0sdsdξ,t∈J.Moreover, A2x∗=x∗. Next, we show that x∗ is the unique fixed point of A in Ph. In view of A2(Ax∗)=A(A2x∗)=Ax∗, by the uniqueness of solutions for the operator equation x=A2x, we have that Ax∗=x∗. Suppose that y is another fixed point of A in Ph, so we have A2y=A(Ay)=Ay=y. Hence, we have x∗=y. So problem (1) has a unique positive solution x∗ in Ph.


## 4. Examples

Example 1.
Consider the following fractional boundary value problem of form (1): (93)D0+11/3Cxt+1-t1/3x1/3+1=0,t∈0,1,x0=x1=0,x′′0+x′′′0=∫01x′′τd12τ,x′′1+x′′′1+∫01x′′τd16τ=0.Conclusion 1.
Problem (93) has a unique positive solution in Ph, where h(t)=t(1-t),t∈[0,1].Proof.
From (93), we know α=11/3,f(t,x)=(1-t)1/3x1/3+1, and it is not difficult to find the following:(1)
f(t,bc+bd)=f(t,ad+bd)=f(t,0)=1>0,t∈[0,1] and f(t,x) is continuous, increasing in x∈[0,+∞).(2)
Letα(λ)=λ1/3, then α(λ)∈(λ,1), and we have (94)ft,λx=1-t1/3λx1/3+1≥λ1/31-t1/3x1/3+1=αλft,x.(3)
From (93), we know (95)pt=12t,qt=16t,∫01dpt=12,∫01dqt=16,so we can get(96)2+∫01τdqτ1+∫01dqτ=2+∫01τd1/6τ1+∫01d1/6τ=2514,1-∫01τdpτ1-∫01dpτ=1-∫01τd1/2τ1-∫01d1/2τ=2114.(4)
(97)∫011-s11/3-41-s1/3x1/3s+1ds=∫01x1/3sds+∫011-s-1/3dsexists.
So we claim that the conditions(A1)–(A4) hold. Hence, by Theorem 11, we can obtain that (93) has a unique positive solution in Ph, where h(t)=t(1-t),t∈[0,1].Example 2.
Consider the following fractional boundary value problem of form (1): (98)D0+11/3Cxt+1-t1/31+x1/3+1=0,t∈0,1,x0=x1=0,x′′0+x′′′0=∫01x′′τd12τ,x′′1+x′′′1+∫01x′′τd16τ=0.Conclusion 2.
Problem (98) has a unique positive solution in Ph, where h(t)=t(1-t),t∈[0,1].Proof.
From (98), we know α=11/3,f(t,x)=(1-t)1/3/1+x1/3+1, and it is not difficult to find the following:(1)
f(t,x)≥1>0,t∈[0,1],x∈[0,+∞) and f(t,x) is decreasing in x∈[0,+∞).(2)
Letβ(λ)=1/3, and then β(λ)∈(0,1), so we have (99)ft,λx=1-t1/31+λx1/3+1≤λ-1/31-t1/31+x1/3+1=λ-βλft,x.(3)
From (98), we know (100)pt=12t,qt=16t,∫01dpt=12,∫01dqt=16,so we can get(101)2+∫01τdqτ1+∫01dqτ=2+∫01τd1/6τ1+∫01d1/6τ=2514,1-∫01τdpτ1-∫01dpτ=1-∫01τd1/2τ1-∫01d1/2τ=2114.(4)
(102)∫011-s11/3-41-s1/31+x1/3s+1ds=∫0111+x1/3sds+∫011-s-1/3dsexists.
So the conditions(A1′),(A2′),(A3),(A4) hold. Hence, by Theorem 12, we can obtain that (98) has a unique positive solution in Ph, where h(t)=t(1-t),t∈[0,1].



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*Source: 4087403-2017-05-09.xml*