# The Eigenvalue Problem for Caputo Type Fractional Differential Equation with Riemann-Stieltjes Integral Boundary Conditions

Authors: Wenjie Ma; Yujun Cui
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, we investigate the eigenvalue problem for Caputo fractional differential equation with Riemann-Stieltjes integral boundary conditionsDc0+θp(y)+μf(t,p(y))=0, y∈[0,1], p(0)=p′′(0)=0, p(1)=∫01p(y)dA(y), where Dc0+θ is Caputo fractional derivative, θ∈(2,3], and f:[0,1]×[0,+∞)→[0,+∞) is continuous. By using the Guo-Krasnoselskii’s fixed point theorem on cone and the properties of the Green’s function, some new results on the existence and nonexistence of positive solutions for the fractional differential equation are obtained.


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## Body

## 1. Introduction

The experience of the last few years has fully borne out the fact that the integer order calculus is not as widely used as fractional order calculus in some fields such as chemistry, control theory, and signal processing. On the remarkable survey of Agarwal, Benchohra, and Hamani [1] it is pointed out that fractional differential equations constitute a fundamental tool in the modeling of some phenomena (see also [2–4]). The use of fractional order is more accurate for the description of phenomena, so the study of fractional differential equations becomes the mainstream with the help of techniques of nonlinear analysis. We refer the reader to [5–36] for recent results. For example, in [9], the author studied the following fractional differential equation: (1)Dcαxy+fy,xy=0,0<y<1,with boundary conditions (2)x′0=x′′0=0,x1=μ∫01xsds,where α∈(2,3], μ∈[0,1), and Dcα is the Caputo derivative. They solved the above problem by means of classical fixed point theorems.In [5], the boundary value problem for the following nonlinear fractional differential equation was discussed: (3)D0+αxy+ayfy,xy,x′′y=0,0<y<1,with boundary conditions (4)x0=x′0=x′′0=x′′1=0,where D0+α is the Riemann-Liouville differentiation, α∈(3,4]. By using a fixed point theorem, a new result of the existence of three positive solutions is obtained.In [15], the authors investigated the following class of BVP: (5)Dc0+qxy=fy,xy,Dc0+σxy,0<y<1,with boundary conditions (6)x0=x′′0=0,x′1=αx′′1,where q∈(2,3), σ∈(1,2), f:[0,1]×R×R→R is a given function, and Dc0+q denotes the Caputo differentiation. The author investigated this problem by using Banach contraction principle, Leray-Schauder nonlinear alternative, properties of the Green’s function, and Guo-Krasnoselskii fixed point theorem on cone. Similar problems can be referred to in [25].In this paper, we investigate the eigenvalue problem for Caputo fractional boundary value problem with Riemann-Stieltjes integral boundary conditions(7)Dc0+θpy+μfy,py=0,y∈0,1,p0=p′′0=0,p1=∫01pydAy,where θ∈(2,3), f:[0,1]×[0,∞)→(0,∞) is continuous, μ>0 is a parameter, Dc0+θ is the Caputo fractional derivative, and A is a bounded variation function with positive measures with (8)B=∫01ydAy<1.Our proof is based on the properties of the Green’s function and the Guo-Krasnosel’skii fixed point theorem on cone.


## 2. Preliminaries

In order to solve problem (7), we provide the properties related to problem (7).Definition 1 (see [3]).
The Caputo’s fractional derivative of orderθ>0 for a function x∈Cn[0,+∞) is defined as (9)Dc0+θxy=1Γn-θ∫0yy-sn-θ-1xnsds,n-1<θ<n,where n is the smallest integer greater than or equal to θ.Lemma 2 (see [3]).
Letθ>0. If we assume x∈C(0,1)∩L(0,1), then the fractional differential equation (10)Dc0+θxy=0has the general solution x(y)=C0+C1y+⋯+Cn-1yn-1, Ck∈R, k=0,1…,n-1, where n is the smallest integer greater than or equal to θ.Lemma 3 (see [3]).
Given thatx∈C(0,1)∩L(0,1) with a fractional derivative of order θ that belongs to C(0,1)∩L(0,1). Then (11)I0+θDc0+θxy=xy+C0+C1y+⋯+Cn-1yn-1,forCk∈R,k=0,1…,n-1,where n is the smallest integer greater than or equal to θ.Firstly, we consider the following linear Caputo fractional differential equation:(12)Dc0+θpy+σy=0,y∈0,1,p0=p′′0=0,p1=∫01pydAy.Lemma 4.
Letθ∈(2,3] and assume that σ∈C[0,1]. Then p is the solution of the above boundary value problem (12), if and only if p satisfies the following integral equation: (13)py=∫01Gy,sσsds,where(14)
G
y
,
s
=
1
Γ
θ
y
1
-
B
1
-
s
θ
-
1
-
∫
s
1
y
-
s
θ
-
1
d
A
y
-
y
-
s
θ
-
1
,
0
≤
s
≤
y
≤
1
,
y
1
-
B
1
-
s
θ
-
1
-
∫
s
1
y
-
s
θ
-
1
d
A
y
,
0
≤
y
≤
s
≤
1
,and (15)B=∫01ydAy<1.Proof.
Applying the fractional integral of orderθ to both sides of (12) for y∈[0,1], we get the following formula: (16)I0+θDc0+θpy+I0+θσy=0.According to p(0)=p′′(0)=0 and Lemma 3, we obtain(17)py=cy-1Γθ∫0yy-sθ-1σsds,where c∈R. Since p(1)=∫01‍p(y)dA(y), we deduce that (18)c-1Γθ∫011-sθ-1σsds=∫01cy-1Γθ∫0yy-sθ-1σsdsdAy.Therefore, (19)c=1Γθ1-B∫011-sθ-1σsds-∫01∫0yy-sθ-1σsdsdAy.Substituting the above equality into (17), one has (20)py=yΓθ1-B∫011-sθ-1σsds-∫01∫0yy-sθ-1σsdsdAy-1Γθ∫0yy-sθ-1σsds=yΓθ1-B∫011-sθ-1σsds-∫01∫s1y-sθ-1dAyσsds-1Γθ∫0yy-sθ-1σsdsand (21)
G
y
,
s
=
1
Γ
θ
y
1
-
B
1
-
s
θ
-
1
-
∫
s
1
y
-
s
θ
-
1
d
A
y
-
y
-
s
θ
-
1
,
0
≤
s
≤
y
≤
1
,
y
1
-
B
1
-
s
θ
-
1
-
∫
s
1
y
-
s
θ
-
1
d
A
y
,
0
≤
y
≤
s
≤
1
.The proof is completed.Lemma 5.
The Green’s functionG(y,s) has the following properties:
(i)Γ(θ)G(y,s)≤1/(1-B)(1-s)θ-1, for y,s∈[0,1];
(ii)Γ(θ)G(y,s)≥N(1-s)θ-1, for y∈[1/4,3/4] and s∈[0,1], where (22)N=min⁡1-∫01yθ-1dAy41-B,miny∈1/4,3/4y1-yθ-2.Proof.
(i) Obviously, the inequalityΓ(θ)G(y,s)≤1/(1-B)(1-s)θ-1 holds from the representation of G(y,s).
(ii) In view ofB=∫01‍ydA(y)<1 and θ∈(2,3), we have 1-∫s1‍yθ-1dA(y)>0.
Fory≤s, we have (23)y1-B1-sθ-1-∫s1y-sθ-1dAy≥y1-B1-sθ-1-∫s1y-ysθ-1dAy≥y1-B1-sθ-11-∫s1yθ-1dAy≥y1-∫01yθ-1dAy1-B1-sθ-1.For s≤y, we have (24)
y
1
-
B
1
-
s
θ
-
1
-
∫
s
1
y
-
s
θ
-
1
d
A
y
-
y
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s
θ
-
1
≥
y
1
-
B
1
-
s
θ
-
1
-
∫
s
1
y
-
y
s
θ
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1
d
A
y
-
y
-
y
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θ
-
1
=
y
1
-
B
+
B
1
-
B
1
-
s
θ
-
1
-
y
1
-
B
∫
s
1
y
-
y
s
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1
d
A
y
-
y
θ
-
1
1
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-
1
=
y
1
-
s
θ
-
1
+
y
∫
0
1
y
d
A
y
1
-
B
1
-
s
θ
-
1
-
y
1
-
B
∫
s
1
y
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y
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y
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-
1
1
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=
1
-
s
θ
-
1
y
-
y
θ
-
1
+
y
1
-
B
1
-
s
θ
-
1
∫
0
1
y
d
A
y
-
∫
s
1
y
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y
s
θ
-
1
d
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y
=
1
-
s
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-
1
y
-
y
θ
-
1
+
y
1
-
B
1
-
s
θ
-
1
∫
0
1
y
d
A
y
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∫
0
1
y
θ
-
1
d
A
y
=
1
-
s
θ
-
1
y
-
y
θ
-
1
+
y
1
-
B
1
-
s
θ
-
1
∫
0
1
y
-
y
θ
-
1
d
A
y
≥
1
-
s
θ
-
1
y
-
y
θ
-
1
=
1
-
s
θ
-
1
y
1
-
y
θ
-
2
.Thus, the above two inequalities yield the inequality in (ii). The proof is completed.LetX=C[0,1], p=maxy∈[0,1]|p(y)|; then (X,·) is a Banach space. We define the cone P⊂X by(25)P=p∈X:py≥N1-Bp,y∈14,34.Let Aμ:X→X be the operator defined as(26)Aμpy=μ∫01Gy,sfs,psds.Thus, the fixed point of the above integral equation is equivalent to the solution of the BVP (7).Lemma 6.
A μ ( P ) ⊂ P and Aμ:P→P is a completely continuous operator, where Aμ is defined in (26).Proof.
By Lemma5, for ∀p∈P, we have (27)Aμpy=μ∫01Gy,sfs,psds≥μNΓθ∫011-sθ-1fs,psds≥μN1-Bmaxt∈0,1∫01Gt,sfs,psds=N1-BAμp,y∈14,34.Hence we have Aμ(P)⊂P. Let Ω⊂P be bounded. Then there exists a constant M>0 such that p≤M for ∀p∈Ω. Let L1=maxy∈[0,1],p∈[0,M](μf(y,p)+1). Then (28)Aμpy≤L1Γθ1-B∫011-sθ-1ds,y∈0,1.Thus, Aμ(Ω) is bounded. Put p∈Ω and y1,y2∈[0,1]. We deduce that (29)
A
μ
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1
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1
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θ
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θ
.Since y,y(y-1)θ,yθ+1,yθ are uniformly continuous on [0,1], Aμ(Ω) is equicontinuous, by using Arzela-Ascoli’s theorem, we can prove that Aμ:P→P is completely continuous. The proof is completed.The following Guo-Krasnoselskii’s fixed point theorem is used to prove the existence of positive solution of (7).Theorem 7 (see [37]).
LetP be a cone of real Banach space X and let Ω1 and Ω2 be two bounded open sets in X such that 0∈Ω1∈Ω¯1∈Ω2. Let operator A:P∩(Ω¯2∖Ω1)→P be completely continuous operator. If one of the following two conditions holds:
(1)Ap≤p for all p∈P∩∂Ω1, Ap≥p for all p∈P∩∂Ω2,
(2)Ap≥p for all p∈P∩∂Ω1, Ap≤p for all p∈P∩∂Ω2,
thenA has at least one fixed point in P∩(Ω¯2∖Ω1).


## 3. Existence of Positive Solutions

In this section, we investigate the existence of positive solutions for integral boundary value problems of fractional differential equation (7).For convenience, we denote them by(30)h0=limu→0+supy∈0,1fy,uu,h∞=limu→+∞supy∈0,1fy,uu,h0∗=limu→0+infy∈1/4,3/4fy,uu,h∞∗=limu→+∞infy∈1/4,3/4fy,uu,C=1Γθ+11-B,E=N21-BΓθ∫1/43/41-sθ-1ds.Theorem 8.
Suppose thatCh0<Eh∞∗ holds; then for μ∈(1/h∞∗E,1/Ch0), problem (7) has a positive solution. Here we impose h0-1=+∞ if h0=0 and [h∞∗]-1=0 if h∞∗=+∞.Proof.
Letμ∈(1/h∞∗E,1/Ch0) and ϵ>0 satisfy (31)1h∞∗-ϵE≤μ≤1Ch0+ε.According to the definition of h0, we know that there exists a constant E1>0 such that (32)fy,u≤h0+εu,foru∈0,E1,y∈0,1.Put Ω1={p∈P:p<E1}. Let p∈P∩∂Ω1. We have p=E1 and (33)Aμpy=μ∫01Gy,sfs,psds≤μΓθ1-B∫011-sθ-1h0+εpsds≤μh0+εΓθ1-B∫011-sθ-1ds·p=h0+εμC·p≤p,y∈0,1.Therefore, Aμp≤p for p∈P∩∂Ω1.
By the definition ofh∞∗, we know that there exists E2>0 such that (34)fy,u≥h∞∗-ϵu,foru≥E2andy∈14,34.
LetE3=max⁡2E1,E2/N(1-B), Ω2={p∈P:p<E3}. Then for p∈P∩∂Ω2, by (25) we have (35)miny∈1/4,3/4py≥N1-Bp≥E2,and thus (36)Aμpy=μ∫01Gy,sfs,psds≥μ∫1/43/4Gy,sfs,psds≥μNΓθ∫1/43/41-sθ-1fs,psds≥μNΓθ∫1/43/41-sθ-1h∞∗-ϵpsds≥μN2h∞∗-ϵ1-BΓθ∫1/43/41-sθ-1pds=μEh∞∗-ϵ·p≥p.Therefore, Aμp≥p for p∈P∩∂Ω2.
By Theorem7, if μ∈(1/h∞∗E,1/Ch0), we assert that Aμ has a fixed point in P∩(Ω2¯∖Ω1) and therefore problem (7) has at least one positive solution. The proof is completed.Theorem 9.
Assume thath∞C<Eh0∗ holds. Then for μ∈1/Eh0∗,1/h∞C, the problem (7) has at least a positive solution. Here we impose [h0∗]-1=0 if h0∗=+∞ and [h∞]-1=+∞ if h∞=0.Proof.
Letμ∈1/Eh0∗,1/h∞C and ϵ>0 such that (37)1h0∗-ϵE≤μ≤1Ch∞+ε.According to the definition of h0∗, there exists a constant E4>0 such that (38)fy,u≥h0∗-ϵu,foru∈0,E4andy∈14,34.Put Ω3={p∈P:p<E4}. Let p∈P∩∂Ω3; we have (39)Aμpy=μ∫01Gy,sfs,psds≥μ∫1/43/4Gy,sfs,psds≥μNΓθ∫1/43/41-sθ-1fs,psds≥μNΓθ∫1/43/41-sθ-1h0∗-ϵpsds≥μN2h0∗-ϵ1-BΓθ∫1/43/41-sθ-1pds=μEh0∗-ϵ·p≥p.Therefore, Aμp≥p for p∈P∩∂Ω3.
It follows from the definition ofh∞ that there exists a constant E5>0 such that (40)fy,u≤h∞+ε2u,foru≥E5andy∈0,1.This together with the continuity of f implies that (41)fy,u≤h∞+ε2u+M,foru∈Randy∈0,1holds for some M>0.
LetE6=max⁡{2E4,E5,2M/ε}, Ω4={p∈P:p<E6}. For ∀p∈P∩∂Ω4, we conclude that (42)Aμpy=μ∫01Gy,sfs,psds≤μΓθ1-B∫011-sθ-1h∞+ε2pds+MμΓθ1-B∫011-sθ-1ds=μCh∞+ε2p+M≤μCh∞+εp≤p.Therefore, Aμp≤p for ∀p∈P∩∂Ω4.
By Theorem7, if μ∈1/Eh0∗,1/h∞C, we conclude that Aμ has a fixed point in P∩(Ω4¯∖Ω3), and so problem (7) has one positive solution. The proof is completed.


## 4. Nonexistence of Positive Solutions

In this section, we present some sufficient conditions for nonexistence of positive solution to integral boundary value problems of fractional differential equation (7).Theorem 10.
Ifh0<+∞ and h∞<+∞, then there exists a μ∗>0 such that problem (7) has no positive solution for μ∈(0,μ∗).Proof.
Sinceh0<+∞, h∞<+∞, we have f(y,u)≤n1u for u∈[0,r1], and f(y,u)≤n2u for u∈[r2,+∞), where n1,n2,r1,r2 are positive numbers with r1<r2. Let n=max{n1,n2,maxr1≤u≤r2f(y,u)/u}; then we have f(y,u)≤nu for u∈[0,+∞). Suppose p0(y) is a positive solution of problem (7); then we are going to prove that this leads to a contradiction for 0<μ<μ∗≔1/Cn. Since (Aμp0)(y)=p0(y), for y∈[0,1], then (43)p0=Aμp0≤μΓθ1-B∫011-sθ-1fy,p0yds≤μCnp0<p0,which is a contradiction. Therefore this completes the proof.Theorem 11.
Ifh0∗>0, h∞∗>0, f(t,u)>0 for t∈1/4,3/4 and u>0, then there exists a μ∗>0 such that problem (7) has no positive solution for all μ>μ∗.Proof.
Sinceh0∗>+∞, h∞∗>+∞, we have f(y,u)≥m1u for u∈[0,r1], and f(y,u)≥m2u for u∈[r2,+∞), where m1,m2,r1,r2 are positive numbers and r1<r2. Let m=min{m1,m2,minr1≤u≤r2f(y,u)/u}>0; then we have f(y,u)≥mu for u∈[0,+∞). Suppose p1(y) is a positive solution of problem (7); then we are going to prove that this leads to a contradiction for μ>μ∗≔1/EmN(1-B). Since Aμp1=p1, then (44)p1=Aμp1≥μNΓθ∫1/43/41-sθ-1fs,psds≥μmNΓθ∫1/43/41-sθ-1psds≥μmNΓθ∫1/43/41-sθ-1N1-Bp1ds≥μEmN1-Bp1>p1,which is a contradiction. Therefore this completes the proof.


## 5. Example

Example 1.
We consider the following fractional equation:(45)Dc0+5/2py+μyp3y-py+epy-1=0,y∈0,1,p0=p′′0=0,p1=12∫01pydy,where θ=5/2, f(y,p)=yp3-p+ep-1, A(y)=(1/2)y, B=1/4<1. We obtain C=8/15Γ(5/2), N=3/64, E=27(93-1)/1310720Γ(5/2).
It is easy to see that, for allp>0, (46)supy∈0,1fy,pp=p2-1+ep-1pand (47)infy∈1/4,3/4fy,pp=14p2-1+ep-1p.Then h0=0, h∞∗=∞; from Theorem 8, problem (45) has a positive solution.Example 2.
We consider the following fractional equation:(48)Dc0+5/2py+μpy3+yln⁡py+3=0,y∈0,1,p0=p′′0=0,p1=12∫01pydy,where θ=5/2, fy,p=p3+yln⁡(p+3), A(y)=1/2y, and B=1/4<1. We obtain C=8/15Γ(5/2), N=3/64, and E=27(93-1)/1310720Γ(5/2).
One can easily see that, for allp>0, (49)supy∈0,1fy,pp=p3+ln⁡p+3pand (50)infy∈1/4,3/4fy,pp=p3+1/4ln⁡p+3p.Then h∞=0, h0∗=∞; from Theorem 9, problem (48) has a positive solution.Example 3.
We consider the following fractional equation:(51)Dc0+5/2py+μypy=0,y∈0,1,p0=p′′0=0,p1=12∫01pydy,where θ=5/2, f(y,p)=yp, A(y)=1/2y, and B=1/4<1. We obtain C=8/15Γ(5/2), N=3/64, and E=27(93-1)/1310720Γ(5/2).
By direct calculation, we obtain that(52)h0=limp→0+supy∈0,1fy,pp=1<+∞and (53)h∞=limp→+∞supy∈0,1fy,pp=1<+∞.Take n=1 and μ∗=45π/32>0. By Theorem 10, problem (51) has no positive solution for 0<μ<μ∗.Example 4.
We consider the following fractional equation:(54)Dc0+5/2py+μypy=0,y∈0,1,p0=p′′0=0,p1=12∫01pydy,where θ=5/2, f(y,p)=yp, A(y)=1/2y, and B=1/4<1. We obtain C=8/15Γ(5/2), N=3/64, and E=27(93-1)/1310720Γ(5/2).
Obviously, we can infer that(55)h0∗=limp→0+infy∈1/4,3/4fy,pp=14>0and(56)h∞∗=limp→+∞infy∈1/4,3/4fy,pp=14>0.
Takem=1/4 and μ∗=3.4π/81(93-1)×108. By Theorem 11, problem (54) has no positive solution for μ>μ∗.



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*Source: 2176809-2018-08-12.xml*