# Positive Solutions for Three-Point Boundary Value Problem of Fractional Differential Equation withp-Laplacian Operator

Authors: Shang-lin Yao; Guo-hui Wang; Zhi-ping Li; Li-jun Yu
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We investigate the existence of multiple positive solutions for three-point boundary value problem of fractional differential equation withp-Laplacian operator -𝒟tβ(φp(𝒟tαx))(t)=h(t)f(t,x(t)),t∈(0,1),x(0)=0,𝒟tγx(1)=a𝒟tγx(ξ),𝒟tαx(0)=0, where 𝒟tβ,𝒟tα,𝒟tγ are the standard Riemann-Liouville derivatives with 1<α≤2,0<β≤1,0<γ≤1,0≤α−γ−1, ξ∈(0,1) and the constant a is a positive number satisfying aξα-γ-2≤1-γ; p-Laplacian operator is defined as φp(s)=|s|p-2s,p>1. By applying monotone iterative technique, some sufficient conditions for the existence of multiple positive solutions are established; moreover iterative schemes for approximating these solutions are also obtained, which start off a known simple linear function. In the end, an example is worked out to illustrate our main results.


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## Body

## 1. Introduction

In this paper, we study the existence of multiple positive solutions for the following three-point boundary value problem of fractional differential equation withp-Laplacian operator
(1)-𝒟tβ(φp(𝒟tαx))(t)=h(t)f(t,x(t)),t∈(0,1),x(0)=0,𝒟tγx(1)=a𝒟tγx(ξ),𝒟tαx(0)=0,
where 𝒟tβ,𝒟tα,𝒟tγ are the standard Riemann-Liouville derivatives with 1<α≤2,0<β≤1,0<γ≤1,0≤α-γ-1,ξ∈(0,1) and the constant a is a positive number satisfying aξα-γ-2≤1-γ; p-Laplacian operator is defined as φp(s)=|s|p-2s,p>1.Fractional differential equations have gained considerable importance due to their application in various sciences, such as physics, mechanics, chemistry, and engineering. In the recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives; see the monographs of Kilbas et al. [1], Miller and Ross [2], Podlubny [3], and the papers [4–14] and the references therein.In [15], Li et al. were concerned with the nonlinear differential equation of fractional order
(2)𝒟tαx(t)+f(t,x(t))=0,t∈(0,1),1<α≤2
subject to the boundary conditions
(3)x(0)=0,𝒟tβx(1)=a𝒟tβx(ξ).
By using some fixed point theorems, the existence and multiplicity results of positive solutions were established.On the other hand, the differential equations withp-Laplacian have also been widely studied owing to the fact that p-Laplacian boundary value problems have important application in theory and application of mathematics and physics. For example, in [16], by using the fixed point index, Yang and Yan investigated the existence of positive solution for the third-order Sturm-Liouville boundary value problems with p-Laplacian operator:
(4)(φp(x′′(t)))′+f(t,x(t))=0,t∈(0,1),ax(0)-bx′(0)=0cx(1)+dx′(1)=0,x′′(0)=0.
However, there are few articles dealing with the existence of solutions to boundary value problems for fractional differential equation with p-Laplacian operator. In [17], the authors investigated the nonlinear nonlocal problem
(5)𝒟tβ(φp(𝒟tαx))(t)+f(t,x(t))=0,t∈(0,1),x(0)=0,x(1)=ax(ξ),𝒟tαx(0)=0,
where 0<α≤2,0<β≤1,0≤a≤1,0<ξ<1. By using Krasnoselskii’s fixed point theorem and Leggett-Williams theorem, some sufficient conditions for the existence of positive solutions to the above BVP are obtained. In [18], by using upper and lower solutions method, under suitable monotone conditions, Wang et al. investigated the existence of positive solutions to the following nonlocal problem:
(6)𝒟tβ&(φp(𝒟tαx))(t)+f(t,x(t))=0,t∈(0,1)x(0)=0,x(1)=ax(ξ),𝒟tαx(0)=0,𝒟tαx(1)=b𝒟tαx(η),
where 1<α,β≤2,0≤a,b≤1,0<ξ,η<1. Recently, Chai [19] investigated the two-point boundary value problem of fractional differential equation with p-Laplacian operator:
(7)𝒟tβ(φp(𝒟tαx))(t)+f(t,x(t))=0,t∈(0,1),x(0)=0,x(1)+a𝒟tγx(1)=0,𝒟tαx(0)=0.
By means of the fixed point theorem on cones, some existence and multiplicity results of positive solutions are obtained.Motivated by the above mentioned works, in this paper, we consider the multiplicity results of positive solutions for the three point boundary value problem of fractional differential equation withp-Laplacian operator. Difference to [15–19], by using monotone iterative technique, we not only establish the existence of multiple positive solutions but also obtain the iterative sequences of these positive solutions.


## 2. Preliminaries and Lemmas

In this section, we introduce some preliminary facts which are used throughout this paper.Definition 1 (see [1–3]).
The Riemann-Liouville fractional integral of orderα>0 of a function x:(0,+∞)→ℝ is given by
(8)Iαx(t)=1Γ(α)∫0t(t-s)α-1x(s)ds
provided that the right-hand side is pointwise defined on (0,+∞).Definition 2 (see [1–3]).
The Riemann-Liouville fractional derivative of orderα>0 of a function x:(0,+∞)→ℝ is given by
(9)𝒟tαx(t)=1Γ(n-α)(ddt)n×∫0t(t-s)n-α-1x(s)ds,
where n=[α]+1, [α] denotes the integer part of number α, provided that the right-hand side is pointwise defined on (0,+∞).Proposition 3 (see [1–3]).
(1) Ifx∈L1(0,1),ν>σ>0, then
(10)IνIσx(t)=Iν+σx(t),𝒟tσIνx(t)=Iν-σx(t),𝒟tσIσx(t)=x(t).
(2) Ifν>0,σ>0, then
(11)𝒟tνtσ-1=Γ(σ)Γ(σ-ν)tσ-ν-1.Proposition 4 (see [1–3]).
Letα>0, and f(x) is integrable, then
(12)Iα𝒟tαx(t)=f(x)+c1xα-1+c2xα-2+⋯+cnxα-n,
where ci∈ℝ(i=1,2,…,n), n is the smallest integer greater than or equal to α.Definition 5.
A functionu∈C(I,R) is called a nonnegative solution of BVP (1), if u≥0 on [0,1] and satisfies (1). Moreover, if u(t)>0,t∈(0,1), then u is said to be a positive solution of BVP (1).
For forthcoming analysis, we first consider the following fractional differential equation:(13)𝒟tαx(t)+g(t)=0,t∈(0,1),x(0)=0,𝒟tγx(1)=a𝒟tγx(ξ).Lemma 6 (see [15]).
If1<α≤2 and g∈L1[0,1], then the boundary value problem (13) has the unique solution
(14)x(t)=∫01‍G(t,s)g(s)ds,
where
(15)G(t,s)={dtα-1(1-s)α-γ-1-adtα-1(ξ-s)α-γ-1-(t-s)α-1Γ(α),0≤s≤min{t,ξ}≤1,dtα-1(1-s)α-γ-1-(t-s)α-1Γ(α),0<ξ≤s≤t≤1,dtα-1(1-s)α-γ-1-adtα-1(ξ-s)α-γ-1Γ(α),0≤t≤s≤ξ<1,dtα-1(1-s)α-γ-1Γ(α),max{t,ξ}≤s≤1,
where d=(1-aξα-γ-1)-1.Lemma 7 (see [15]).
The Green functionG(t,s) in Lemma 6 has the following properties:(i)
G(t,s) is continuous on [0,1]×[0,1],(ii)
G(t,s)>0 for any s,t∈(0,1).And ifaξα-γ-2≤1-γ, the Green function G(t,s) also satisfies(iii)
G(t,s)≤G(s,s) for any s,t∈[0,1],(iv)
there exists a positive functionσ(s)∈C(0,1) such that
(16)minξ≤t≤1G(t,s)≥σ(s)G(s,s),where
(17)G(s,s)={dsα-1(1-s)α-γ-1-adsα-1(ξ-s)α-γ-1Γ(α),0≤s≤ξ,dsα-1(1-s)α-γ-1Γ(α),ξ≤s≤1.Letℕ be the set of positive integers, let ℝ be the set of real numbers, and let ℝ+ be the set of nonnegative real numbers. Let I=[0,1]. Denote by C(I,ℝ) the Banach space of all continuous functions from I into ℝ with the norm
(18)∥x∥=max{x(t):t∈I}.Define the coneP in C(I,ℝ+) as
(19)P={∑u∈C(I,ℝ+):minξ≤t≤1x(t)≥σ(t)∥x∥,t∈I}.
Let q>1 satisfy the relation (1/q)+(1/p)=1, where p is given by (1).To study BVP (1), we first consider the associated linear BVP:
(20)𝒟tβ(φp(𝒟tαx))(t)+g(t)=0,t∈(0,1)x(0)=0,𝒟tγx(1)=a𝒟tγx(ξ),𝒟tαx(0)=0,
for g∈L1[0,1] and g≥0.Letw=𝒟tαx,v=φp(w). By Proposition 4, the solution of initial value problem
(21)𝒟tβv(t)+g(t)=0,t∈(0,1),v(0)=0,
is given by v(t)=C1tβ-1-Iβg(t),t∈[0,1]. From the relations v(0)=0,0<β≤1, it follows that C1=0, and so
(22)v(t)=-Iβg(t),t∈[0,1].
Noting that 𝒟tαx=w,w=φp-1(v), from (22), we know that the solution of (20) satisfies
(23)𝒟tαx(t)=φp-1(-Iβg(t)),t∈(0,1)x(0)=0,𝒟tγx(1)=a𝒟tγx(ξ).
By Lemma 6, the solution of (23) can be written as
(24)x(t)=-∫01G(t,s)φp-1(-Iβg(s))ds,t∈I.
Since h(s)≥0, s∈I, we have φp-1(-Iβg(s))=-(Iβg(s))q-1, s∈I, and so
(25)x(t)=∫01G(t,s)(Iβg(s))q-1ds=(Γ(β))1-q∫01G(t,s)(∫0s(s-τ)β-1g(τ)dτ)q-1ds,t∈I,
which implies that the solution of (23) is given by
(26)x(t)=∫01G(t,s)(Iβg(s))q-1ds=(Γ(β))1-q∫01G(t,s)(∫0s‍(s-τ)β-1g(τ)dτ)q-1ds.
For the convenience, we make the following assumptions.(H1)
f:[0,1]×[0,+∞)→(0,+∞) is continuous and nondecreasing, and there exists a constant λ>0 such that, for any t∈[0,1],x∈[0,+∞),
(27)f(t,cx)≥cλf(t,x),∀0<c≤1.(H2)
h(t)∈L1(0,1) is nonnegative on (0, 1), and
(28)0<∫01G(s,s)(∫0s(s-τ)β-1h(τ)dτ)q-1ds<+∞.Remark 8.
By (27), for any c≥1,(t,x)∈[0,1]×[0,+∞), clearly,
(29)f(t,cx)≤cλf(t,x).
Now, for any x∈P, define one operator T as follows:
(30)(Tx)(t)=(Γ(β))1-q∫01G(t,s)×(∫0s‍(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1ds,t∈I.
Then by (20) and (23), the BVP (1) is equivalent to the fixed point problems of the operators T.Lemma 9.
Assume that (H1) and (H2) hold. ThenT:P→P are continuous, compact, and nondecreasing.Proof.
In fact, for anyx∈P,
(31)∥Tx∥=maxt∈I{∫0s‍(Γ(β))1-q∫01‍G(t,s)×(∫0s‍(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1ds}≤(Γ(β))1-q×∫01G(s,s)(∫0s(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1ds.
On the other hand, by Lemma 7,
(32)minξ≤t≤1Tx(t)=minξ≤t≤1{(Γ(β))1-q∫01G(t,s)×(∫0s‍(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1ds}≥σ(t)(Γ(β))1-q∫01G(s,s)×(∫0s(s-τ)β-1×h(τ)f(τ,x(τ))dτ∫0s(s-τ)β-1)q-1ds=σ(t)∥Tx∥.
So T(P)⊂P.
Next, supposingD⊂P is a bounded set, then for any x∈D, there exists a constant M>0 such that ||x||≤M. Thus for any x∈D, we have
(33)∥Tx∥=maxt∈I{∫0s(s-τ)β-1(Γ(β))1-q×∫01G(t,s)(∫0s(s-τ)β-1×h(τ)f(τ,x(τ))dτ∫0s(s-τ)β-1)q-1ds∫0s(s-τ)β-1}≤(Γ(β))1-q×∫01G(s,s)(∫0s(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1ds≤(Γ(β))1-q∫01G(s,s)(∫0s(s-τ)β-1h(τ)dτ)q-1ds×(sup[0,1]×[0,M]f(t,x))q-1<+∞,
which implies T(D) is bounded. On the other hand, according to the Arzela-Ascoli theorem and Lebesgue dominated convergence theorem, we easily see T:P→P is completely continuous. In the end, noticing the monotonicity of f on x and the definition of T, we also have that the operator T is nondecreasing.


## 3. Main Results

Define two constants(34)B=(Γ(β))1-q∫01G(s,s)(∫0s(s-τ)β-1h(τ)dτ)q-1ds,l=maxt∈If(t,1).Theorem 10.
Suppose conditions (H1) and (H2) hold. If there exists a positive constantb>1 such that
(35)lq-1B≤b1-λ(q-1),
where l and B are defined by (34), then the BVP (1) has the maximal and minimal solutions x* and y*, which are positive, and there exist two positive constants m1≤m2 such that
(36)m2σ(t)≤x*(t)≤b,m1σ(t)≤y*(t)≤b,t∈I.
Moreover for initial values x0=b,y0=0, define the iterative sequences xn,yn by
(37)xn(t)=(Txn-1)(t)=Tnx0(t),yn(t)=(Tyn-1)(t)=Tny0(t).
Then
(38)limn→+∞xn=x*,limn→+∞yn=y*
for t∈I uniformly, respectively.Proof.
LetP[0,b]={x∈P:0≤||x||≤b}; we firstly prove T(P[0,b])⊂P[0,b]. In fact, for any x∈P[0,b], we have
(39)0≤x(t)≤maxt∈Ix(t)=∥x∥≤b.
By the assumption (H1), we have
(40)0≤f(t,x(t))≤f(t,b)≤bλf(t,1)≤bλmaxt∈If(t,1)=lbλ.
It follows from Lemma 9 that T:P→P is completely continuous operator; thus by (35) and (40), we have
(41)∥Tx∥=maxt∈I{∫01G(t,s)(∫0s(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1(Γ(β))1-q×∫01G(t,s)(∫0s(s-τ)β-1×h(τ)f(τ,x(τ))dt∫)q-1ds}≤(Γ(β))1-q×∫01G(s,s)(∫0s(s-τ)β-1h(τ)f(τ,x(τ))dτ)q-1ds≤(Γ(β))1-q×∫01G(s,s)(∫0s(s-τ)β-1h(τ)dτ)q-1ds×(lbλ)q-1=B(lbλ)q-1≤b,
which implies that T(P[0,b])⊂P[0,b].
Lety0(t)=0,t∈[0,1]; then y0(t)∈P[0,b]. Leting y1(t)=(Ty0)(t), we have y1∈P[0,b]. Denote
(42)yn+1=Tyn=Tn+1y0,n=1,2,….
It follows from T(P[0,b])⊂P[0,b] that yn∈P[0,b]. Since T is compact, we obtain that {yn} is a sequentially compact set.
Sincey1=Ty0=T0∈P[0,b], we have
(43)y1(t)=(Ty0)(t)=(T0)(t)≥0=y0(t),t∈[0,1].
By the induction, we get
(44)yn+1≥yn,n=0,1,2,….
Consequently, there exists y*∈P[0,b] such that yn→y*. Letting n→+∞, from the continuity of T and Tyn=yn-1, we obtain Ty*=y*, which implies that y* is a nonnegative solution of boundary value problem (1). Since f:[0,1]×[0,+∞)→(0,+∞), we know the zero function is not the solution of boundary value problem (1), thus max0≤t≤1|y*(t)|>0; by y*∈P, we have
(45)y*(t)≥∥y*∥σ(t)>0,t∈(0,1),
that is, y* is a positive solution of boundary value problem (1).
On the other hand, letx0(t)=b,t∈[0,1]; then x0(t)∈P[0,b]. Leting x1=Tx0, from the previous expressions, we have x1∈P[0,b]. Thus let us denote
(46)xn+1=Txn=Tn+1x0,n=1,2,….
It follows from T(P[0,b])⊂P[0,b] that
(47)xn∈P[0,b]n=0,1,2,….
Since T is compact by Lemma 9, we can assert that {xn} is a sequentially compact set.
Now, sincex1∈P[0,b], we have
(48)0≤x1(t)≤∥x1∥≤b=x0(t).
It follows from Lemma 9 that T:P→P is nondecreasing, so
(49)x2=Tx1≤Tx0=x1.
By the induction, we have
(50)xn+1≤xn,n=0,1,2,….
Consequently, there exists x*∈P[0,b] such that xn→x*. Letting n→+∞, from the continuity of T and Txn=xn-1, we obtain Tx*=x*, which implies that x* is a nonnegative solution of boundary value problem (1).
Next, notingx0(t)=b≥y0(t)=0,t∈[0,1], thus it follows from monotonicity of T that Tx0≥Ty0; by the induction, we have xn≥yn,n=0,1,2,…, which implies that x*≥y*. Thus by (45) we have
(51)x*(t)≥y*(t)≥∥x*∥σ(t)>0,t∈(0,1).
This means that x* is also a positive solution of boundary value problem (1).
In the end, letu* be any fixed point of T in P[0,b], then
(52)y0=0≤u*≤b=x0,
and then
(53)y1=Ty0≤Tu*=u*≤Tb=x1.
By induction, we have
(54)yn≤u*≤xn,n=1,2,3,….
Taking the limit, we have
(55)y*≤u*≤x*.
This implies that x* and y* are maximal and minimal solutions of the BVP (1).Let m1=||y*||,m2=||x*||, then we have
(56)m2σ(t)≤x*(t)≤b,m1σ(t)≤y*(t)≤b,t∈I.
The proof is completed.Remark 11.
Ifh(t)≡1, then (H2) holds naturally, and in this case we take
(57)B*=(Γ(β))1-q∫01dΓ(α)(∫0s(s-τ)β-1dτ)q-1ds=d(βΓ(β))1-qΓ(α)(β(q-1)+1).
Thus we have the following Corollary 12.Corollary 12.
Suppose condition (H1) holds andh(t)≡1. If there exists a positive constant b>1 such that
(58)lq-1B*≤b1-λ(q-1),
where l is defined by (34), then the BVP (1) has the maximal and minimal solutions x* and y*, which are positive, and there exist two positive constants m1≤m2 such that
(59)m2σ(t)≤x*(t)≤b,m1σ(t)≤y*(t)≤b,t∈I.
Moreover for initial values x0=b,y0=0, define the iterative sequences xn,yn by
(60)xn(t)=(Txn-1)(t)=Tnx0(t),yn(t)=(Tyn-1)(t)=Tny0(t).
Then
(61)limn→+∞xn=x*,limn→+∞yn=y*
for t∈I uniformly, respectively.Corollary 13.
Suppose conditions(H1)and(H2) hold. If
(62)λ<p-1.
Then there exists a constant b>1 such that the BVP (1) has the maximal and minimal solutions x* and y*, which are positive, and there exist two positive constants m1≤m2 such that
(63)m2σ(t)≤x*(t)≤b,m1σ(t)≤y*(t)≤b,t∈I.
Moreover for initial values x0=b,y0=0, define the iterative sequences xn,yn by
(64)xn(t)=(Txn-1)(t)=Tnx0(t),yn(t)=(Tyn-1)(t)=Tny0(t).
Then
(65)limn→+∞xn=x*,limn→+∞yn=y*
for t∈I uniformly, respectively.Proof .
It follows fromλ<p-1 that
(66)limu→+∞uλup-1=0,
which implies that there exists b>1 large enough such that
(67)bλbp-1<1B*p-1l.
Notice that (1/p)+(1/q)=1; (67) is equivalent to
(68)lq-1B*≤b1-λ(q-1).
By Theorem 10, the conclusion of Corollary 13 holds.Remark 14.
In Corollary13, we obtain that the BVP (1) has the maximal and minimal solutions x* and y* only by comparing p-1 to λ. But note that p and λ are irrelative, so (62) is easy to be satisfied; this implies that Corollary 13 is very interesting.Example 15.
Consider the following boundary value problem:(69)-𝒟t1/2(φ6(𝒟t3/2x))(t)=sintx4(t)+t2x3/4(t)+2t3+5t+2,t∈(0,1),x(0)=0,𝒟t1/6x(1)=16163𝒟t1/6x(14),𝒟t3/2x(0)=0.
Letα=3/2,β=1/2,γ=1/6,p=6, and f(t,x)=sintx4+t2x3/4+2t3+5t+2,h(t)≡1, then
(70)aξα-γ-2=16163(14)-2/3=16<1-γ=56.
For any 0<c≤1 and x∈[0,+∞), we have
(71)f(t,cx)=sint(cx)4+t2(cx)3/4+2t3+5t+2≥sint(cx)4+t2c4x3/4+c4(2t3+5t+2)=c4f(t,x).
Taking λ=4, then
(72)λ=4<p-1=5,
which implies that (62) holds. By Corollary 13, we know the BVP (69) has at least two positive solutions.



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