# Positive Solutions of Nonlinear Eigenvalue Problems for a Nonlocal Fractional Differential Equation

Authors: Xiaoling Han; Hongliang Gao
Journal: Mathematical Problems in Engineering
 (2011)

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## Abstract

By using the fixed point theorem, positive solutions of nonlinear eigenvalue problems for a nonlocal fractional differential equationD0+αu(t)+λa(t)f(t,u(t))=0,0<t<1,u(0)=0,u(1)=Σi=1∞αiu(ξi) are considered, where 1<α≤2 is a real number, λ is a positive parameter, D0+α is the standard Riemann-Liouville differentiation, and ξi∈(0,1),αi∈[0,∞) with Σi=1∞αiξiα-1<1, a(t)∈C([0,1],[0,∞)),f(t,u)∈C([0,∞),[0,∞)).


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. This is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various sciences such as physics, mechanics, chemistry, and engineering. For details, see [1–6] and references therein.Recently, many results were obtained dealing with the existence and multiplicity of solutions of nonlinear fractional differential equations by the use of techniques of nonlinear analysis, see [7–21] and the reference therein. Bai and Lü [7] studied the existence of positive solutions of nonlinear fractional differential equation
(1.1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=0,
where 1<α≤2 is a real number, D0+α is the standard Riemann-Liouville differentiation, and f:[0,1]×[0,∞)→[0,∞) is continuous. They derived the corresponding Green function and obtained some properties as follows.Proposition 1.1.
Green’s functionG(t,s) satisfies the following conditions:(R1)
G(t,s)∈C([0,1]×[0,1]), and G(t,s)>0 for t,s∈(0,1);(R2)
there exists a positive functionγ∈C(0,1) such that(1.2)min⁡1/4≤t≤3/4G(t,s)≥γ(s)max⁡0≤t≤1G(t,s)≥γ(s)G(s,s),s∈(0,1),
where
(1.3)G(t,s)={[t(1-s)]α-1-(t-s)α-1Γ(α),0≤s≤t≤1,[t(1-s)]α-1Γ(α),0≤t≤s≤1.It is well known that the cone plays a very important role in applying Green’s function in research area. In [7], the authors cannot acquire a positive constant taken instead of the role of positive function γ(s) with 1<α<2 in (1.2). In [9], Jiang and Yuan obtained some new properties of the Green function and established a new cone. The results can be stated as follows.Proposition 1.2.
Green’s functionG(t,s) defined by (1.3) has the following properties: G(t,s)=G(1-s,1-t) and
(1.4)α-1Γ(α)tα-1(1-t)(1-s)α-1s≤G(t,s)≤1Γ(α)tα-1(1-t)(1-s)α-2,∀t,s∈(0,1).Proposition 1.3.
The functionG*(t,s):=t2-αG(t,s) has the following properties:
(1.5)q(t)Φ(s)≤G*(t,s)≤Φ(s),∀t,s∈[0,1],
where q(t)=(α-1)t(1-t),Φ(s)=(1/Γ(α))s(1-s)α-1.In this paper, we study the existence of positive solutions of nonlinear eigenvalue problems for a nonlocal fractional differential equation(1.6)D0+αu(t)+λa(t)f(t,u(t))=0,0<t<1,u(0)=0,u(1)=∑i=1∞αiu(ξi),
where 1<α≤2 is a real number, λ is a positive parameter, D0+α is the standard Riemann-Liouville differentiation, and ξi∈(0,1),αi∈[0,∞) with Σi=1∞αiξiα-1<1, a(t)∈C([0,1],[0,∞)),f(t,u)∈C([0,∞),[0,∞)).We assume the following conditions hold throughout the paper:(H1)
ξi∈(0,1),αi∈[0,∞) are both constants with Σi=1∞αiξiα-1<1;(H2)
a(t)∈C([0,1],[0,∞)), a(t)≢0;(H3)
f(t,u)∈C([0,1]×[0,∞),[0,∞)), and there exist g∈C([0,+∞),[0,+∞)),q1,q2∈C((0,1),(0,+∞)) such that(1.7)q1(t)g(y)≤f(t,tα-2y)≤q2(t)g(y),t∈(0,1),y∈[0,+∞),
where ∫01qi(s)ds<+∞,i=1,2.


## 2. The Preliminary Lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature.Definition 2.1.
The fractional integral of orderα>0 of a function y:(0,∞)→R is given by
(2.1)I0+αy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds
provided the right side is pointwise defined on (0,∞).Definition 2.2.
The fractional derivative of orderα>0 of a function y:(0,∞)→R is given by
(2.2)D0+αy(t)=1Γ(n-α)(ddt)n∫0ty(s)(t-s)α-n+1ds,
where n=[α]+1, provided the right side is pointwise defined on (0,∞).Lemma 2.3.
Letα>0. If one assumes u∈C(0,1)∩L(0,1), then the fractional differential equation
(2.3)D0+αu(t)=0
has u(t)=C1tα-1+C2tα-2+⋯+CNtα-N,Ci∈R,i=1,2,…,N, where N is the smallest integer greater than or equal to α, as unique solutions.Lemma 2.4.
Assume thatu∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to u∈C(0,1)∩L(0,1). Then
(2.4)I0+αD0+αu(t)=u(t)+C1tα-1+C2tα-2+⋯+CNtα-N
for some Ci∈R,i=1,2,…,N.Lemma 2.5 (see [7]).
Giveny∈C[0,1] and 1<α≤2, the unique solution of
(2.5)D0+αu(t)+y(t)=0,0<t<1,u(0)=u(1)=0
is
(2.6)u(t)=∫01G(t,s)y(s)ds,
where
(2.7)G(t,s)={[t(1-s)]α-1-(t-s)α-1Γ(α),0≤s≤t≤1,[t(1-s)]α-1Γ(α),0≤t≤s≤1.Lemma 2.6.
Suppose (H1) holds. Giveny∈C[0,1] and 1<α≤2, the unique solution of
(2.8)D0+αu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=∑i=1∞αiu(ξi)
is
(2.9)u(t)=∫01G(t,s)y(s)ds+A(y)tα-1,
where
(2.10)G(t,s)={[t(1-s)]α-1-(t-s)α-1Γ(α),0≤s≤t≤1,[t(1-s)]α-1Γ(α),0≤t≤s≤1,A(y)=Σi=1∞αi∫01G(ξi,s)y(s)ds1-Σi=1∞αiξiα-1.Proof.
By applying Lemmas2.4 and 2.5, we have
(2.11)u(t)=∫01G(t,s)y(s)ds+C1tα-1+C2tα-2.
Because
(2.12)∑i=1∞αi∫01G(ξi,s)ds=Σi=1∞αiξiα-1(1-ξi)αΓ(α),αiξiα-1(1-ξi)<αiξiα-1,
by (H1), Σi=1∞αiξiα-1(1-ξi) is convergent; therefore, Σi=1∞αi∫01G(ξi,s)ds is convergent. Note that y(t) is continuous function on [0,1], so Σi=1∞αi∫01G(ξi,s)y(s)ds is convergent.
Fromu(0)=0,u(1)=Σi=1∞αiu(ξi), we have C2=0,C1=Σi=1∞αi∫01G(ξi,s)y(s)ds/(1-Σi=1∞αiξiα-1). Therefore,
(2.13)u(t)=∫01G(t,s)y(s)ds+A(y)tα-1,A(y)=Σi=1∞αi∫01G(ξi,s)y(s)ds1-Σi=1∞αiξiα-1.Lemma 2.7 (see [7]).
LetE be a Banach space, P⊆E a cone, and Ω1,Ω2 two bounded open sets of E with 0∈Ω1⊂Ω̅1⊂Ω2. Suppose that T:P∩(Ω̅2∖Ω1)→P is a completely continuous operator such that either(i)
∥Tx∥≤∥x∥,x∈P∩∂Ω1, and ∥Tx∥≥∥x∥,x∈P∩∂Ω2, or(ii)
∥Tx∥≥∥x∥,x∈P∩∂Ω1, and ∥Tx∥≤∥x∥,x∈P∩∂Ω2,holds. Then T has a fixed point in P∩(Ω̅2∖Ω1).


## 3. The Main Results

Let(3.1)G0(t,s)=G(t,s)+tα-1Σi=1∞αiG(ξi,s)1-Σi=1∞αiξiα-1.
Then u(t) is the solution of BVP (1.6) if and only if  Tu(t)=u(t), where T is the operator defined by(3.2)Tu(t):=λ∫01G0(t,s)a(s)f(s,u(s))ds.
By similar arguments to Proposition 1.3, we obtain the following result.Lemma 3.1.
Suppose (H1) holds. The functionG¯(t,s):=t2-αG0(t,s) has the following properties:
(3.3)q(t)Ψ(s)≤G¯(t,s)≤Ψ(s),∀t,s∈[0,1],
where q(t)=(α-1)t(1-t),Ψ(s)=Φ(s)+Σi=1∞αiG(ξi,s)/(1-Σi=1∞αiξiα-1).LetE=C[0,1] be endowed with the ordering u≤v if u(t)≤v(t) for all t∈[0,1], and the maximum norm ∥u∥=max0≤t≤1|u(t)|. Define the cone P⊂E by P={u∈E∣u(t)≥0}, and(3.4)K={u∈P∣u(t)≥q(t)‖u‖},
where q(t) is defined by (3.3).It is easy to see thatP and K are cones in E. For any 0<r<R<+∞, let Kr={u∈K∥u∥<r},∂Kr={u∈K∥u∥=r},K¯r={u∈K∥u∥≤r}, and K¯R∖Kr={u∈K∣r≤∥u∥≤R}.For convenience, we introduce the following notations:(3.5)g0=lim⁡u→0+g(u)u,g∞=lim⁡u→+∞g(u)u.
By similar arguments to Lemma 4.1 of [9], we obtain the following result.Lemma 3.2.
Assume that (H1)–(H3) hold. LetT¯:K→E be the operator defined by
(3.6)T¯u(t):=λ∫01G¯(t,s)a(s)f(s,sα-2u(s))ds.
Then T¯:K→K is completely continuous.Theorem 3.3.
Assuming (H1)–(H3) hold,g0,g∞ exist. Then, for each λ satisfying
(3.7)1[((α-1)/16)2∫1/43/4Ψ(s)a(s)q1(s)ds]g0<λ<1(∫01Ψ(s)a(s)q2(s)ds)g∞,
there exists at least one positive solution of BVP (1.6) in P.Theorem 3.4.
Assuming  (H1)–(H3) hold,g0,g∞ exist. Then, for each λ satisfying
(3.8)1[((α-1)/16)2∫1/43/4Ψ(s)a(s)q1(s)ds]g∞<λ<1(∫01Ψ(s)a(s)q2(s)ds)g0,
there exists at least one positive solution of BVP (1.6) in P.Proof of Theorem3.3.
Letλ be given as in (3.7), and choose ε>0 such that
(3.9)1[((α-1)/16)2∫1/43/4Ψ(s)a(s)q1(s)ds](g0-ε)≤λ≤1(∫01Ψ(s)a(s)q2(s)ds)(g∞+ε).
Beginning with g0, there exists an H1>0 such that g(u)≥(g0-ε)u, for 0<u≤H1. So  u∈K and ∥u∥=H1. For t∈[1/4,3/4], we have
(3.10)T¯u(t)=λ∫01G¯(t,s)a(s)f(s,sα-2u(s))ds≥λ∫01q(t)Ψ(s)a(s)q1(s)g(u(s))ds≥λ∫01q(t)Ψ(s)a(s)q1(s)(g0-ε)u(s)ds≥λ(α-116)2∫1/43/4Ψ(s)a(s)q1(s)(g0-ε)ds‖u‖≥‖u‖=H1.
Thus, ∥T¯u∥≥∥u∥. So, if we let
(3.11)Ω1={u∈K∣‖u‖<H1},
then
(3.12)‖T¯u‖≥‖u‖,u∈∂Ω1.
It remains to considerg∞. There exists an H¯2 such that g(u)≤(g∞+ε)u, for all u≥H¯2. There are the two cases, (a), where  g is bounded, and (b), where g is unbounded.
Case a.
SupposeN>0 is such that g(u)≤N, for all 0<u<∞.
LetH2=max{2H1,Nλ∫01Ψ(s)a(s)q2(s)ds}. Then, for u∈K with ∥u∥=H2, we have
(3.13)T¯u(t)≤λ∫01Ψ(s)a(s)q2(s)g(u(s))ds≤λN∫01Ψ(s)a(s)q2(s)ds≤H2=‖u‖.
So, if we let
(3.14)Ω2={u∈K∣‖u‖<H2},
then
(3.15)‖T¯u‖≤‖u‖,u∈∂Ω2.
Case b.
LetH2>max{2H1,H¯2} be such that g(u)≤g(H2),0<u<H2. Choosing u∈K with ∥u∥=H2,
(3.16)T¯u(t)≤λ∫01Ψ(s)a(s)q2(s)g(H2)ds≤λ∫01Ψ(s)a(s)q2(s)(g∞+ε)dsH2≤H2=‖u‖,
and so ∥T¯u∥≤∥u∥. For this case, if we let
(3.17)Ω2={u∈K∣‖u‖<H2},
then
(3.18)‖T¯u‖≤‖u‖,u∈∂Ω2.
Therefore, by (ii) of Lemma 2.7, T¯ has a fixed point u such that H1≤∥u∥≤H2 and satisfies
(3.19)u(t)=∫01G¯(t,s)a(s)f(s,sα-2u(s))ds.
It is obvious that y(t)=tα-2u(t) is solution of (1.6) for t∈(0,1], and
(3.20)y(t)=∫01G0(t,s)a(s)f(s,y(s))ds,t∈(0,1].
Next, we will prove y(0)=0. From u∈C[0,1] and (H1)–(H3), we have
(3.21)lim⁡t→0+y(t)=lim⁡t→0+∫01G0(t,s)a(s)f(s,y(s))ds=lim⁡t→0+∫01G0(t,s)a(s)f(s,sα-2u(s))ds≤lim⁡t→0+∫01G0(t,s)a(s)q2(s)g(u(s))ds≤lim⁡t→0+∫01G0(t,s)a(s)q2(s)dsmax⁡‖u‖≤H2g(u)≤0.
Thus, y(0)=0. then y(t)=tα-2u(t) is solution of (1.6) for t∈[0,1].Proof of Theorem3.4.
Letλ be given as in (3.8), and choose ε>0 such that
(3.22)1[((α-1)/16)2∫1/43/4Ψ(s)a(s)q1(s)ds](g∞-ε)≤λ≤1(∫01Ψ(s)a(s)q2(s)ds)(g0+ε).
Beginning with g0, there exists an H1>0 such that g(u)≤(g0+ε)u, for 0<u≤H1. So, for u∈K and ∥u∥=H1, we have
(3.23)T¯u(t)=λ∫01G¯(t,s)a(s)f(s,sα-2u(s))ds≤λ∫01Ψ(s)a(s)q2(s)ds(g0+ε)‖u‖≤‖u‖.
Thus, ∥T¯u∥≤∥u∥. So, if we let
(3.24)Ω1={u∈K∣‖u‖<H1},
then
(3.25)‖T¯u‖≤‖u‖,u∈∂Ω1.
Next, consideringg∞, there exists an H¯2 such that g(u)≥(g∞-ε)u, for all u≥H¯2. Let H2=max{2H1,(16/(α-1))H¯2}. Then, u∈K∥u∥=H2. For t∈[1/4,3/4], we have
(3.26)T¯u(t)=λ∫01G¯(t,s)a(s)f(s,sα-2u(s))ds≥λ∫01q(t)Ψ(s)a(s)q1(s)g(u(s))ds≥λ∫01q(t)Ψ(s)a(s)q1(s)(g∞-ε)u(s)ds≥λ(α-116)2∫1/43/4Ψ(s)a(s)q1(s)(g∞-ε)ds‖u‖≥‖u‖=H2,
and so ∥T¯u∥≥∥u∥. For this case, if we let
(3.27)Ω2={u∈K∣‖u‖<H2},
then
(3.28)‖T¯u‖≥‖u‖,u∈∂Ω2.
Therefore, by (i) of Lemma2.7, T¯ has a fixed point u such that H1≤∥u∥≤H2 and satisfies
(3.29)u(t)=∫01G¯(t,s)a(s)f(s,sα-2u(s))ds.
By similar method to Theorem 3.3, we can get y(0)=0,  then y(t)=tα-2u(t) is solution of (1.6) for t∈[0,1]. We complete the proof.



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