# Positive Solutions for Class of State Dependent Boundary Value Problems with Fractional Order Differential Operators

Authors: Dongyuan Liu; Zigen Ouyang; Huilan Wang
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We consider the following state dependent boundary-value problemD0+αy(t)-pD0+βg(t,y(σ(t)))+f(t,y(τ(t)))=0, 0<t<1; y(0)=0, ηy(σ(1))=y(1), where Dα is the standard Riemann-Liouville fractional derivative of order 1<α<2, 0<η<1, p≤0, 0<β<1, β+1-α≥0 the function g is defined as g(t,u):[0,1]×[0,∞)→[0,∞), and g(0,0)=0 the function f is defined as f(t,u):[0,1]×[0,∞)→[0,∞)σ(t), τ(t) are continuous on t and 0≤σ(t), τ(t)≤t. Using Banach contraction mapping principle and Leray-Schauder continuation principle, we obtain some sufficient conditions for the existence and uniqueness of the positive solutions for the above fractional order differential equations, which extend some references.


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## Body

## 1. Introduction

Fractional order differential equations has useful applications in many fields, such as physics, mechanics, chemistry, engineering, biology, and so on. There has been a significant development in fractional differential equations (e.g., [1–9]). In the previous papers, some authors investigated fractional order partial differential equations [10–15]. For example, Wu [15] used the wavelet operational method for solving fractional partial differential equations numerically. Since it is one of the important fields to be concerned with the boundary value problems for fractional order differential equations, some authors considered the existence of positive solutions for fractional differential equations or systems with boundary value conditions [16–25] and the stability [26].As early as 1994, Delbosco [27] investigated the nonlinear Dirichlet-type problem
(1)xα-1D0+αyx=fyx,0<x<1,1<α<2,y0=y′1=0,
where Dα is α order Riemann-Liouville derivative. The author had proved that if f is a Lipschitz function, then the problem has at least one solution y(x) in a certain subspace of C[0,1] in which the fractional derivative has a Hölder property.Later, using some fixed point theorems, Bai and Lü [20] obtained the existence of positive solutions of the following equation with boundary value conditions
(2)D0+αyt+ft,yt=0,0<t<1,1<α<2,y0=y1=0,ory0+y′0=y1+y′1=0.More recently, Bai [9] also considered the following boundary value problem
(3)D0+αyt+ft,yt=0,0<t<1,1<α≤2,y0=0,βyη=y1.
By constructing a Green’s function, and using contraction map principle, the author obtained some existence conditions of positive solutions for (3).Motivated by the above references, we consider a state dependent boundary value problem with fractional order differential operators(4)D0+αyt-pD0+βgt,yσt+ft,yτt=0,WWWWWWW000WWWWWWWW0<t<1,y0=0,ηyσ1=y1,
where Dα is the standard Riemann-Liouville fractional derivative of order 1<α<2, 0<β<1, 0<η<1, p≤0, β+1-α≥0, 1-ησα-1(1)>0; the function g is defined as g(t,u):[0,1]×[0,∞)→[0,∞), and g(0,0)=0; the function f is defined as f(t,u):[0,1]×[0,∞)→[0,∞) and σ(t), τ(t) are continuous on t and 0≤σ(t),τ(t)≤t.By using Banach contraction mapping principle and Leray-Schauder continuation principle, we obtain some sufficient conditions for the existence and uniqueness of the positive solutions for boundary value problem (4). Furthermore, we give an example to illustrate our results.


## 2. Preliminary

In this section, we introduce some definitions and preliminary facts which are used in this paper.Definition 1 (see [8, 16]).
The fractional integral of orderα with the lower limit t0 for a function f is defined as
(5)Iαft=1Γα∫t0tfst-s1-αds,t>t0,α>0,
provided the right-side is point-wise defined on [t0,∞], where Γ is the Gamma function.Definition 2 (see [8, 16]).
Riemann-Liouville derivative of orderα with the lower limit t0 for a function f:[0,∞)→R can be written as
(6)Dαft=1Γn-αdndtn∫t0tfst-sα+1-nds,WWWWWWWWWt>t0,n-1<α<n.Definition 3 (see [8, 16]).
Caputo’s derivative of orderα with the lower limit t0 for a function f:[0,∞)→R can be written as
(7)Dcαft=1Γn-α∫t0tfnst-sα+1-nds=In-αfnt,t>t0,n-1<α<n.It is well known that ifn-1≤α≤n, then Dαtα-k=0, k=1,2,…,n. Furthermore, if y(t)∈L1[0,T] and α>0, then for t∈[0,T], we have
(8)DαIαyt=yt,
which is with the semigroup property
(9)IδIα=Iδ+α=IαIδ,
for δ+α>0 and t∈[0,T].We also need to introduce some Lemmas as follows, which will be used in the proof of our main theorems.Lemma 4 (see [19, 20]).
Letα>0; then the fractional equation
(10)Dαht=0,
has solutions
(11)ht=c1tα-1+c2tα-2+⋯+cntα-n,0ci∈R,i=1,2,…,n,n=α+1.Lemma 5 (see [19, 20]).
Letα>0; then
(12)IαDαht=ht+c1tα-1+c2tα-2+⋯+cntα-n,
for some ci∈R, i=1,2,…,n, n=[α]+1.Lemma 6 (see [28], the Banach contraction mapping theorem).
LetT:M→M be a contraction mapping of a complete metric space M. Then T has one and only one fixed point.Lemma 7 (see [28–30], the Leray-Schauder continuation principle).
LetX be a Banach space with C⊂X being closed and convex. Assume that U is a relatively open subset of C with 0∈U and T:U¯→C is completely continuous. Then either(a)
Thas a fixed point in U¯, or(b)
there existsu∈∂U and λ∈(0,1) with u=λTu.Throughout this paper, we assume thatf,g∈C([0,1]×[0,∞),[0,∞,)) and we satisfy the following(H):
(i)f(t,u),g(t,u) is Lebesgue measurable with respect to t on [0,1];(ii)f(t,u),g(t,u) is continuous with respect to u on [0,∞).


## 3. Main Results

For convenience, we rewrite (4) as follows:
(13)D0+αyt=pD0+βgt,yσt-ft,yτt,WWWWWWWWWWWW0W00W0<t<1,(14)y0=0,ηyσ1=y1.Integrating both sides of (13) of α order with respect to t, it follows that
(15)ytW=p1Γα-β∫0tt-sα-β-1gs,yσsdsWW+c0tα-1+c1tα-2-1ΓαWW·∫0tt-sα-1fs,yτsds,0<t<1.
From (14) and (15), we have
(16)c1=0,(17)y1W=p1Γα-β∫011-sα-β-1gs,yσsdsWW+c0-1Γα∫011-sα-1fs,yτsds,WWWWWWWWWW0WWWWW0<t<1,yσ1W=p1Γα-β∫0σ1σ1-sα-β-1gs,yσsdsWW+c0σα-11-1ΓαWW·∫0σ1σ1-sα-1fs,yτsds,WWWWWWWWWiiiiWWW0<t<1.Combining (14) with (17), we obtain
(18)c0=1Γα1-ησα-11·∫011-sα-1fs,yτsds-ηΓα1-ησα-11·∫0σ1σ1-sα-1fs,yτsds-pΓα-β1-ησα-11·∫011-sα-β-1gs,yσsds+pηΓα-β1-ησα-11·∫0σ1σ1-sα-β-1gs,yσsds,WWWWWWWWWWWWW0<t<1.
According to (15) and (18), it follows that
(19)yt=pΓα-β∫0tt-sα-β-1gs,yσsdsW+1Γα1-ησα-11W·∫011-sα-1tα-1fs,yτsdsW-ηΓα1-ησα-11W·∫0σ1σ1-sα-1tα-1fs,yτsdsW-pΓα-β1-ησα-11W·∫011-sα-β-1tα-1gs,yσsdsW+pηΓα-β1-ησα-11W·∫0σ1σ1-sα-β-1tα-1gs,yσsdsW-1Γα∫0tt-sα-1fs,yτsds=pΓα-β∫0tt-sα-β-1gs,yσsdsW-pΓα-β1-ησα-11W·∫0t+∫tσ1+∫σ111-sα-β-1tα-1gs,yσsdsW-1Γα∫0tt-sα-1fs,yτsdsW+1Γα1-ησα-11W·∫0t+∫0σ1+∫σ111-sα-1tα-1fs,yτsdsW+pηΓα-β1-ησα-11W·∫0t+∫tσ1σ1-sα-β-1tα-1gs,yσsdsW-ηΓα1-ησα-11W·∫0t+∫tσ1σ1-sα-1tα-1fs,yτsds=p∫0t1Γα-β1-ησα-11WWW·1-ησα-11t-sα-β-1WWWW-1-sα-β-1tα-1WWWW+ησ1-sα-β-1tα-1WWWW·gs,yσs1-ησα-11t-sα-β-1dsW+p∫tσ1Γα-β1-ησα-11-1-1-sα-β-1tα-1WWWW00W+ησ1-sα-β-1tα-1WWWWWW·Γα-β1-ησα-11-1WWWWWW·gs,yσsdsW-p∫σ111-sα-β-1tα-1Γα-β1-ησα-11WWWWW·gs,yσsdsW-∫0tΓα1-ησα-11-11-ησα-11t-sα-1WWWW-1-sα-1tα-1+ησ1-sα-1tα-1WWWW·Γα1-ησα-11-1WWWW·fs,yτsdsW-∫tσ1-1-sα-1tα-1+ησ1-sα-1tα-1Γα1-ησα-11WWW·fs,yτsdsW+∫σ111-sα-1tα-1Γα1-ησα-11fs,yτsds,WWWWWWWW0W00WWWWW0<t<1.
Let
(20)Gt,s=1-sα-β-1tα-1-1-ησα-110·t-sα-β-1-ησ1-sα-β-1tα-1W·Γα-β1-ησα-11-1,WWWWW0≤s≤t≤1,s≤σ1;1-sα-β-1tα-10-1-ησα-11t-sα-β-1W·Γα-β1-ησα-11-1,WWWi00W0<σ1≤s≤t≤1;1-sα-β-1tα-1-ησ1-sα-β-1tα-1W·Γα-β1-ησα-11-1,WWWWW0<t≤s≤σ1≤1;1-sα-β-1tα-1W·Γα-β1-ησα-11-1,WW0W0≤t≤s≤1,σ1≤s,(21)G~t,s=1-sα-1tα-1-1-ησα-11t-sα-10-ησ1-sα-1tα-1W·Γα1-ησα-11-1,WWWWiW0≤s≤t≤1,s≤σ1;1-sα-1tα-10-1-ησα-11t-sα-1W·Γα1-ησα-11-1,WWlWW0<σ1≤s≤t≤1;1-sα-1tα-1-ησ1-sα-1tα-1W·Γα1-ησα-11-1,WWWWWWW0<t≤s≤σ1≤1;1-sα-1tα-1W·Γα1-ησα-11-1,WW0≤t≤s≤1,σ1≤s.
According to (19)–(21), it follows that
(22)yt=-p∫01Gt,sgs,yσsds+∫01G~t,sfs,yτsds,
which means that if y(t) satisfies (13)-(14), then it satisfies (22). It is easy to show that if y(t) satisfies (22), then it also satisfies (13)-(14). Thus, the boundary value problem (13)-(14) is actually equivalent to integral equation (22). Therefore, we have the following.Lemma 8.
Problem (13)-(14) is equivalent to (22).Lemma 9.
For any(t,s)∈(0,1)×(0,1), G(t,s),G~(t,s) are continuous and G(t,s)>0, G~(t,s)>0.Proof.
It is obvious thatG(t,s),G~(t,s) are continuous on [0,1]×[0,1]. We first prove that G(t,s)>0 on [0,1]×[0,1]. Let
(23)g1t,sW=1-sα-β-1tα-1-1-ησα-11t-sα-β-1WW-ησ1-sα-β-1tα-1,WWWWWWWWW0lW0≤s≤t≤1,s≤σ1;g2t,sW=1-sα-β-1tα-1-1-ησα-11t-sα-β-1,WWWWWWWWWWWW0<σ1≤s≤t≤1;g3t,sW=1-sα-β-1tα-1-ησ1-sα-β-1tα-1,WWWWWWWW0W0<t≤s≤σ1≤1;g4t,s=1-sα-β-1tα-1,0≤t≤s≤1,σ1≤s.
We first show that g1(t,s)>0, 0≤s≤t≤1, s≤σ(1). We rewrite g1(t,s) as follows:
(24)g1t,sW=tα-11-stα-β-11-sα-β-1-1-ησα-11WW00WW·1-stα-β-1t-β-ησ1-sα-β-1.
Let
(25)h1t,s=1-sα-β-1-1-ησα-111-stα-β-1t-β-ησ1-sα-β-1.
Since 0<ησα-1(1)<1, then
(26)h1s,s=1-sα-β-1-ησ1-sα-β-1>0.
Differentiating both sides of (25) with respect to t, it follows that
(27)∂h1t,s∂tW=1-ησα-111-stα-β-2t-β-2WW·β-α+1s+1-sttβW=1-ησα-111-stα-β-2t-β-2s-αs+tβ≥0,WWWWWWWWWWW0WWWWWW0<s<t≤1;
which means that h1(t,s) is nondecreasing with respect to t on [s,1]. Thus, for any t∈[s,1], h1(t,s)≥h1(s,s)>0, therefore, g1(t,s)=tα-1h1(t,s)>0.
Using the similar method, we can prove thatg2(t,s)>0, and it is obvious that g3(t,s)>0, g4(t,s)>0. Hence, combining (20) and (23), we obtain that G(t,s)>0.
Now, we prove thatG~(t,s)>0. Denote
(28)g~1t,sW=1-sα-1tα-1-1-ησα-11t-sα-1WW-ησ1-sα-1tα-1,WWWWWW00W0≤s≤t≤1,s≤σ1;g~2t,sW=1-sα-1tα-1-1-ησα-11t-sα-1,WWWWWWW0WWW0<σ1≤s≤t≤1;g~3t,sW=1-sα-1tα-1-ησ1-sα-1tα-1,WWWWWWWW0<t≤s≤σ1≤1;g~4t,s=1-sα-1tα-1,0≤t≤s≤1,σ1≤s.
Let
(29)h~1t,s=1-sα-1-1-ησα-111-stα-1-ησ1-sα-1.
Since 0<ησα-1(1)<1, then
(30)h~11,s≥1-sα-1-1-ησα-111-sα-1-ησ1-sα-1=ησα-111-sα-1-ησ1-sα-1=ησ1-sσ1α-1-ησ1-sα-1≥ησ1-sα-1-ησ1-sα-1=0.
Differentiating both sides of (29) with respect to t, it follows that
(31)∂h~1t,s∂t=-α-11-ησα-111-stα-2st-2≤0,0<s<t≤1;
which means that h~1(t,s) is nonincreasing with respect to t on [s,1]. Thus, for any t∈[s,1], h~1(t,s)≥h1(1,s)>0, therefore, g~1(t,s)=tα-1h~1(t,s)>0.
Using the similar method, we can prove thatg~2(t,s)>0, and the case that g~3(t,s)>0, g~4(t,s)>0 is obvious. Combining (21) and (28) and using the above argument, we obtain that G~(t,s)>0. The proof is complete.Lemma 10.
For any(t,s)∈(0,1)×(0,1), G(t,s),G~(t,s) are nondecreasing functions with respect to t∈(0,1); that is, for any t∈(0,1), G(t,s)>G(s,s),G~(t,s)>G~(s,s).Proof.
According to the proof of Lemma9, we notice that
(32)∂g1t,s∂t=α-1tα-2h1t,s+tα-1∂h1t,s∂t>0,WWWWWWWWWWWWW00W0<s<t≤1,∂g2t,s∂t>0,∂g3t,s∂t>0,∂g4t,s∂t>0.
At the same time, for s<t<1, we have
(33)∂g~1t,s∂tW=α-11-sα-1tα-2-1-ησα-11WW·t-sα-2-ησ1-sα-1tα-2W≥α-1tα-21-sα-1-ησ1-sα-1≥0,∂g~2t,s∂t>0,∂g~3t,s∂t>0,∂g~4t,s∂t>0.
The proof is complete.Now, we present our main results.Theorem 11.
Assume that (H) holds. Suppose that there are two functionsλ(t),μ(t)∈C([0,1],[0,∞)) such that
(34)ft,u-ft,v≤λtu-v,Whfort∈0,1,u,v∈0,∞,gt,u-gt,v≤μtu-v,Wfort∈0,1,u,v∈0,∞.
If
(35)-p∫01G1,sμsds+∫01G~1,sλs<1,
then the problem (13)-(14) has a unique positive solution.Proof.
Set(36)Ω=yt∈C0,1∣yt≥0,fort∈0,1,
with the maximum norm
(37)y=max⁡0≤t≤1yt.
It is easy to show that Ω is a complete metric space. We denote a operator T as follows:
(38)Tyt=-p∫01Gt,sgs,yσsds+∫01G~t,sfs,yτsds.
From Lemma 9 and the conditions p≤0, f>0, g>0, it follows that T maps Ω into itself and we only need to prove the contraction. In fact, according (H) and (34), for any u,v∈Ω, we have
(39)Tu-TvW=max⁡0≤t≤1-p∫01Gt,sgs,uσsdsWWWWW+p∫01Gt,sgs,vσsdsWWWWW+∫01G~t,sfs,uτsdsWWWWW-∫01G~t,sfs,vτsdsW≤-p∫01Gt,sμsuσs-vσsdsWW+∫01G~t,sλsuτs-vτsdsW≤-p∫01Gt,sμsds+∫01G~t,sλsu-vW≤-p∫01G1,sμsds+∫01G~1,sλsu-vW<u-v,
which means that
(40)Tu-Tv<u-v.
By the Banach contraction mapping principle (Lemma 6), we obtain that T has a unique fixed point y~(t) which is a positive solution of (13)-(14). The proof is complete.Remark 12.
Ifp=0,f(t,y(τ(t)))=f(y(t)), then problem (13)-(14) is problem (3).Theorem 13.
Assume that (H) holds. Suppose that there exists four nonnegative real-valued functionsm,n,l,q∈L1[0,1] such that
(41)ft,u≤nt+mtu,foralmosteveryt∈0,1andallu∈0,∞,(42)gt,u≤lt+qtu,foralmosteveryt∈0,1andallu∈0,∞.
If
(43)∫01G~1,sms-pG1,sqsds<1,
then the problem (13)-(14) has at least one positive solution.Proof.
We also consider the operatorT defined in (38). We divide the proof into four steps.
Step 1.  T:Ω→Ω is continuous.
Letyn(t) be a sequence in Ω such that yn(t)→y(t) as n→∞. Noticing that f(t,y),g(t,y) are continuous with respect to y, then for each t∈[0,1], we have
(44)lim⁡n→∞ft,ynτt=ft,yτt,lim⁡n→∞t,ynσt=gt,yσt;
thus,
(45)lim⁡n→∞sup⁡t∈0,1ft,ynτt-ft,yτt=0,lim⁡n→∞sup⁡t∈0,1gt,ynσt-gt,yσt=0,
which implies that
(46)Tynt-Tyt=-p∫01Gt,sgs,ynσsdsWW+p∫01Gt,sgs,yσsdsWW+∫01G~t,sfs,ynσsdsWW-∫01G~t,sfs,yσsds≤-p∫01Gt,sgs,ynσs-gs,yσsdsW+∫01G~t,sfs,ynσs-fs,yσsds≤-p∫01G1,sdssup⁡t∈0,1gs,ynσs-gs,yσsW+∫01G~1,sdssup⁡t∈0,1fs,ynσs-fs,yσs⟶0,asn⟶∞.Step 2. T maps bounded sets into bounded sets in Ω.
Indeed, it suffices to show that for anyγ>0, there exists a positive constant η⋆>0 such that for y∈Bγ={y∈Ω:y≤γ}, we have Ty≤η⋆. From (38)–(43), we have
(47)Ty=max⁡0≤t≤1-p∫01Gt,sgs,yσsdsW0WW+∫01G~t,sfs,yτsds≤-p∫01Gt,sls+qsyσsdsW+∫01G~t,sns+msyτsds≤∫01-pG1,sls+G~1,snsdsW+∫01-pG1,sqs+G~1,smsdsy≤∫01-pG1,sls+G~1,snsds+γ≔η⋆,
which means
(48)Ty≤η⋆.Step 3. T maps bounded sets into equicontinuous sets in Ω.
For anyt1,t2∈[0,1],t1<t2, and for each y(t)∈Bγ, we have
(49)Tyt1-Tyt2=-p∫01Gt1,s-Gt2,sgs,yσsdsW0+∫01G~t1,s-G~t2,sfs,yτsds≤-p∫01Gt1,s-Gt2,sls+qsyσsdsW+∫01G~t1,s-G~t2,snt+mtyτsds≤-p∫01Gt1,s-Gt2,sls+qsγdsW+∫01G~t1,s-G~t2,snt+mtγds.
Because G(t,s),G~(t,s) are continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1], which means that for any ε>0, there exists δ>0, when |t2-t1|<δ,s∈[0,1],
(50)Gt1,s-Gt2,s<ε2-p∫01ls+qsγds,G~t1,s-G~t2,s<ε2∫01ns+msγds.
Thus
(51)Tyt1-Tyt2<ε,
which means that {Ty:y∈Bγ} is equicontinuous.
Step 4. A priori bounds.
Let(52)U=y∈Ω:y<r,wherer=∫01G~1,sns-pG1,slsds1-∫01G~1,sms-pG1,sqsds.
Assume that there existsy∈U and 0<λ<1 such that y=λTy and we claim that y≠r. In fact,
(53)yt=λTyt=λ-p∫01Gt,sgs,yσsdsW0W+∫01G~t,sfs,yτsds,
which implies that
(54)yt=λ-p∫01Gt,sgs,yσsdsW0W+∫01G~t,sfs,yτsds<-p∫01Gt,sls+qsyσsdsW+∫01G~t,sns+msyτsds≤∫01-pG1,sls+G~1,snsdsW+∫01-pG1,sqs+G~1,smsdsy≤∫01-pG1,sls+G~1,snsdsW+r∫01-pG1,sqs+G~1,smsds=r,
which means that
(55)y<r.
That is, there is no y∈∂U, such that y=λTy for 0<λ<1.
From Step 1 to Step 3, it follows thatT is completely continuous. Along with Step 4 and Lemma 7, it follows that T has at least a fixed point in U. The proof is complete.


## 4. Examples

Example 1.
Consider(56)D3/2xt+12D3/4t2xt/2101+xt/2W=e-txt/29+et1+xt/2,t∈0,1,x0=0,x1=x12,
where
(57)n=2,p=-12,α=32,β=34,η=1,τt=σt=t2,g=t2xt/2101+xt/2,f=e-txt/29+et1+xt/2.
It is easy to show that
(58)λt=e-t9+et,μt=t210.
Let
(59)hx=x1+x,x>0;
then
(60)hx-hy=x1+x-y1+y=x-y1+x1+y<x-y,forx,y>0;
thus,
(61)ft,u-ft,v≤λtu-v,gt,u-gt,v≤μtu-v,
which satisfies (28), (29). Because
(62)G1,s=1-s-1/4-1-1/20·1-s-1/4-1/2-s-1/4W·Γ3/41-1/2-1,WWWW0WWW0≤s≤12;1-s-1/4-1-1/21-s-1/4W·Γ3/41-1/2-1,WWWWWWWWWWW12≤s≤1,G~1,s=1-s1/2-1-1/20·1-s1/2-1/2-s1/2W·Γ3/21-1/2-1,WWWWWW0≤s≤12;1-s1/2-1-1/21-s1/2W·Γ3/21-1/2-1,WWWWW00WWWW12≤s≤1,
thus
(63)-p∫01G1,sμsds+∫01G~1,sλsW=12∫01/21/21-s-1/4-1/2-s-1/4Γ3/41-1/2s210dsWW+12∫1/211/21-s-1/4Γ3/41-1/2s210dsWW+∫01/21/21-s1/2-1/2-s1/2Γ3/21-1/2e-s9+esdsWW+∫1/211/21-s1/2Γ3/21-1/2e-s9+esdsW<12∫01/21-s-1/4Γ3/42-1s210dsWW+12∫1/211-s-1/4Γ3/42-1s210dsWW+∫01/21-s1/2Γ3/22-1e-s9+esdsWW+∫1/211-s1/2Γ3/22-1e-s9+esdsW<120Γ3/42-1+19Γ3/22-1W≈0.4014<1;
which satisfies Theorem 11. Thus (56) has a unique positive solution on [0,1].



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*Source: 263748-2015-01-12.xml*