# Uniqueness of Positive Solutions for a Perturbed Fractional Differential Equation

Authors: Chen Yang; Jieming Zhang
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We are concerned with the existence and uniqueness of positive solutions for the following nonlinear perturbed fractional two-point boundary value problem:D0+αu(t)+f(t,u,u',…,u(n-2))+g(t)=0, 0<t<1, n-1<α≤n, n≥2, u(0)=u'(0)=⋯=u(n-2)(0)=u(n-2)(1)=0, where D0+α is the standard Riemann-Liouville fractional derivative. Our analysis relies on a fixed-point theorem of generalized concave operators. An example is given to illustrate the main result.


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## Body

## 1. Introduction

In this paper, we are interested in the existence and uniqueness of positive solutions for the following nonlinear perturbed fractional two-point boundary value problem:(1.1)D0+αu(t)+f(t,u,u′,…,u(n-2))+g(t)=0,0<t<1,n-1<α≤n,n≥2,u(0)=u′(0)=⋯=u(n-2)(0)=u(n-2)(1)=0,
where D0+α is the standard Riemann-Liouville fractional derivative and g:[0,1]→[0,+∞) is continuous.Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, engineering, and biological sciences; see [1–15], for example. In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously. For a small sample of such work, we refer the reader to [16–26] and the references therein. On the other hand, the uniqueness of positive solutions for nonlinear fractional differential equation boundary value problems has been studied by some authors; see [19–22, 25], for example.Wheng(t)≡0, Yang and Chen [22] investigated the existence and uniqueness of positive solutions for the problem (1.1) by means of a fixed-point theorem for u0 concave operators. They present the following result.Theorem 1.1 (see [22]).
Assume that(H1)
f∈C([0,1]×[0,+∞)×Rn-2→[0,+∞)),f(t,y1,y2,…,yn-1) is increasing for yi≥0, i=1,2,…,n-1, and f is not identically vanishing;(H2)
for anyt∈[0,1], yi≥0, i=1,2,…,n-1, there exist constants m1, m2, m1≤0≤m2<1 such that
(1.2)cm2f(t,y1,y2,…,yn-1)≤f(t,cy1,cy2,…,cyn-1)≤cm1f(t,y1,y2,…,yn-1),∀0<c≤1;(H3)
m2<(2m1+1)/3.Then the problem (1.1) with g(t)≡0 has a unique positive solution.In this paper, we will remove the condition(H3) and improve on (H2). And we will use a fixed-point theorem of generalized concave operators to show the existence and uniqueness of positive solutions for the problem (1.1). Our main result is summarized in the following theorem.Theorem 1.2.
Assume that(H1) holds and(H4)
for anyc∈(0,1) and yi≥0, i=1,2,…,n-1, there exists a number φ(c)∈(c,1) such that
(1.3)f(t,cy1,cy2,…,cyn-1)≥φ(c)f(t,y1,y2,…,yn-1).Then the problem (1.1) has a unique positive solution u*=I0+n-2v*∈C([0,1],[0,+∞)) which satisfies λ(v*)tα-n+1(1-t)≤v*(t)≤μ(v*)tα-n+1(1-t), t∈[0,1], where μ(v*)≥λ(v*)>0.Remark 1.3.
Some examples ofφ(t) which satisfy the condition (H4) are (1)
φ(t)=tr,for allt∈(0,1), where r∈(0,1),(2)
φ(t)=t(1+η(t)) with 0<η(t)≤1/t-1, for all t∈(0,1).Remark 1.4.
It is easy to see that the condition(H4) is weaker than the condition (H2). Moreover, we do not need the condition (H3) in our main result.


## 2. Preliminaries and Previous Results

For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proof of our theorem.Definition 2.1 (see [4, Definition 2.1]).
The integral(2.1)I0+αf(x)=1Γ(α)∫0x‍f(t)(x-t)1-αdt,x>0,
is called the Riemann-Liouville fractional integral of order α, where α>0 and Γ(α) denotes the gamma function.Definition 2.2 (see [4, page 36-37]).
For a functionf(x) given in the interval [0,∞), the expression
(2.2)D0+αf(x)=1Γ(n-α)(ddx)n∫0x‍f(t)(x-t)α-n+1dt,
where n=[α]+1, [α] denotes the integer part of number α, is called the Riemann-Liouville fractional derivative of order α.
By using the same method in [21], the problem (1.1) can be transformed into the following boundary value problem:
(2.3)D0+α-n+2v(t)+f(t,I0+n-2v(t),I0+n-3v(t),…,I0+1v(t),v(t))+g(t)=0,0<t<1,v(0)=v(1)=0.
Moreover, from Lemma 2.5 and Lemma 2.7 in [21], we can easily obtain the following result.Lemma 2.3.
Ifv∈C([0,1],[0,+∞)) is a positive solution of the problem (2.3), then u(t)=I0+n-2v(t) is a positive solution of the problem (1.1). On the other hand, if v∈C([0,1],[0,+∞)) is a positive solution of the problem (2.3), then the solution is
(2.4)v(t)=∫01‍G(t,s)[f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))+g(s)]ds,t∈[0,1],
where
(2.5)G(t,s)=1Γ(α-n+2){(t(1-s))α-n+1-(t-s)α-n+1,0≤s≤t≤1,(t(1-s))α-n+1,0≤t≤s≤1.
Here G(t,s) is called the Green function of the problem (2.3). Evidently, G(t,s)≥0 for t,s∈[0,1].The following property of the Green function plays important roles in this paper.Lemma 2.4 (see [22]).
The Green functionG(t,s) in Lemma 2.3 has the following property:
(2.6)α-n+1Γ(α-n+2)s(1-s)α-n+1tα-n+1(1-t)≤G(t,s)≤1Γ(α-n+2)(1-s)α-ntα-n+1(1-t)fort,s∈[0,1].In the sequel, we present some basic concepts in ordered Banach spaces for completeness and a fixed-point theorem which we will be used later. For convenience of readers, we suggest that one refers to [27, 28] for details.Suppose that(E,∥·∥) is a real Banach space which is partially ordered by a cone P⊂E, that is,  x≤y if and only if y-x∈P. If x≤y and x≠y, then we denote x<y or y>x. By θ we denote the zero element of E. Recall that a nonempty closed convex set P⊂E is a cone if it satisfies (i) x∈P, λ≥0⇒λx∈P; (ii) x∈P, -x∈P⇒x=θ.P is called normal if there exists a constant M>0 such that, for all x,y∈E, θ≤x≤y implies ∥x∥≤M∥y∥; in this case M is called the normality constant of P. If x1,x2∈E, the set [x1,x2]={x∈E∣x1≤x≤x2} is called the order interval between x1 and x2.For allx,y∈E, the notation x~y means that there exist λ>0 and μ>0 such that λx≤y≤μx. Clearly, ~ is an equivalence relation. Given h>θ (i.e., h≥θ and h≠θ), we denote by Ph the set Ph={x∈E∣x~h}. It is easy to see that Ph⊂P is convex and λPh=Ph for all λ>0.In a recent paper [28], Zhai et al. considered the following operator equation:
(2.7)x=Ax+x0.
They established the existence and uniqueness of positive solutions for the above equation, and they present the following interesting result.Theorem 2.5 (see Theorem 2.1 in [28]).
Leth>θ and P be a normal cone. Assume that(D1)A:P→P is increasing;(D2)x0∈P satisfies Ah+x0∈Ph;(D3) for any x∈P and t∈(0,1), there exists φ(t)∈(t,1) such that A(tx)≥φ(t)Ax.Then the operator equation (2.7) has a unique solution in Ph.Remark 2.6.
An operatorA is said to be generalized concave if A satisfies condition (D3).


## 3. Proof of Theorem1.2

In this section, we apply Theorem2.5 to study the problem (1.1), and we obtain a new result on the existence and uniqueness of positive solutions. The method used here is new to the literature and so is the existence and uniqueness result to the fractional differential equations.In our considerations we will work in the Banach spaceC[0,1]={x:[0,1]→Ris  continuous} with the standard norm ∥x∥=sup{|x(t)|:t∈[0,1]}. Notice that this space can be equipped with a partial order given by
(3.1)x,y∈C[0,1],x≤y⟺x(t)≤y(t)fort∈[0,1].
Set P={x∈C[0,1]∣x(t)≥0,t∈[0,1]}, the standard cone. It is clear that P is a normal cone in C[0,1] and the normality constant is 1.Proof of Theorem1.2.
Leth(t)=tα-n+1(1-t), t∈[0,1]. Then
(3.2)Ph={x∈P∣andthere existλ(x),μ(x)>0such  thatλ(x)h≤x≤μ(x)h}.
For any v∈P, we define
(3.3)Av(t)=∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds,x0(t)=∫01‍G(t,s)g(s)ds,
where G(t,s) is given as in Lemma 2.3. Noting that I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s)≥0 and G(t,s)≥0, it follows from (H1) that A:P→P. In the sequel we check that A and x0 satisfy all assumptions of Theorem 2.5.
Firstly, we prove thatA:P→P is an increasing operator. In fact, for vi∈P, i=1,2 with v1≤v2, we know that v1(t)≤v2(t), t∈[0,1], by the monotonicity of Riemann-Liouville fractional integral I0+δ,δ>0 and (H1),
(3.4)Av1(t)=∫01‍G(t,s)f(s,I0+n-2v1(s),I0+n-3v1(s),…,I0+1v1(s),v1(s))ds≤∫01‍G(t,s)f(s,I0+n-2v2(s),I0+n-3v2(s),…,I0+1v2(s),v2(s))ds=Av2(t).
That is Av1≤Av2. Hence, the condition (D1) in Theorem 2.5 is satisfied.
Next we show that the condition(D3) holds. From (H4), for any γ∈(0,1) and v∈P, we obtain
(3.5)A(γv)(t)=∫01‍G(t,s)f(s,I0+n-2γv(s),I0+n-3γv(s),…,I0+1γv(s),γv(s))ds=∫01‍G(t,s)f(s,γI0+n-2v(s),γI0+n-3v(s),…,γI0+1v(s),γv(s))ds≥∫01‍G(t,s)φ(γ)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds=φ(γ)Av(t),t∈[0,1].
That is A(γv)≥φ(γ)Av,  for all v∈P, γ∈(0,1). So the condition (D3) in Theorem 2.5 is satisfied. Now we show that the condition (D2) is also satisfied. On one hand, it follows from (H1) and Lemma 2.4 that
(3.6)Ah(t)=∫01‍G(t,s)f(s,I0+n-2h(s),I0+n-3h(s),…,I0+1h(s),h(s))ds≥h(t)∫01‍α-n+1Γ(α-n+2)s(1-s)α-n+1f(s,I0+n-2h(s),I0+n-3h(s),…,I0+1h(s),h(s))ds=α-n+1Γ(α-n+2)h(t)×∫01‍s(1-s)α-n+1f(s,I0+n-2h(s),I0+n-3h(s),…,I0+1h(s),h(s))ds,t∈[0,1].
On the other hand, also from (H1) and Lemma 2.4, we obtain
(3.7)Ah(t)≤1Γ(α-n+2)h(t)∫01‍(1-s)α-nf(s,I0+n-2h(s),I0+n-3h(s),…,I0+1h(s),h(s))ds,t∈[0,1].
Let
(3.8)λ(h)=α-n+1Γ(α-n+2)∫01‍s(1-s)α-n+1f(s,I0+n-2h(s),I0+n-3h(s),…,I0+1h(s),h(s))ds,μ(h)=1Γ(α-n+2)∫01‍(1-s)α-nf(s,I0+n-2h(s),I0+n-3h(s),…,I0+1h(s),h(s))ds.
Since f is continuous and f≢0, we can get 0<λ(h)≤μ(h). Consequently,
(3.9)λ(h)h(t)≤Ah(t)≤μ(h)h(t).
Next we consider x0. If g(t)≡0, then x0(t)≡0; if g(t)≢0, let l=maxt∈[0,1]g(t), then l>0. It is easy to prove that
(3.10)0≤x0(t)≤lΓ(α-n+2)h(t)∫01‍(1-s)α-nds=l(α-n+1)Γ(α-n+2)h(t).
Hence,
(3.11)0≤x0≤l(α-n+1)Γ(α-n+2)h.
Further,
(3.12)λ(h)h≤x0+Ah≤[μ(h)+l(α-n+1)Γ(α-n+2)]h.
Hence x0+Ah∈Ph. Finally, using Theorem 2.5, v=Av+x0 has a unique solution v* in Ph. That is, v* is a unique positive solution of the problem (2.3) in Ph. So there are μ(v*), λ(v*) with μ(v*)≥λ(v*)>0 such that λ(v*)tα-n+1(1-t)≤v*(t)≤μ(v*)tα-n+1(1-t), t∈[0,1]. From Lemma 2.3, u*=I0+n-2v* is the solution of the problem (1.1). Evidently, u*∈C([0,1],[0,+∞)) is a unique positive solution of the problem (1.1).Remark 3.1.
Letf≡C>0. Then the conditions (H1), (H4) are satisfied and the problem (2.3) has a unique solution v(t)=∫01‍G(t,s)[C+g(t)]ds,  t∈[0,1]. From Lemma 2.4, the unique solution v is a positive solution and satisfies v∈Ph=Ptα-n+1(1-t). So u=I0+n-2v is a unique positive solution of the problem (1.1).
To illustrate how our main result can be used in practice we present an example.Example 3.2.
Consider the following problem:(3.13)D0+5/2u(t)+a(t){u1/3(t)+[u′(t)]1/4}+g(t)=0,0<t<1,u(0)=u′(0)=u′(1)=0,
where a,g:[0,1]→[0,+∞) are continuous with a≢0.
In this example, we haveα=5/2. Let f(t,x,y)=a(t)[x1/3+y1/4]. Evidently, f(t,x,y) is increasing in x for t∈[0,1], y≥0, and increasing in y for t∈[0,1], x≥0. Moreover, f≢0. Set φ(γ)=γ5/12,γ∈(0,1). Then
(3.14)f(t,γx,γy)=a(t)[γ1/3x1/3+γ1/4y1/4]≥φ(γ)f(t,x,y),t∈[0,1],x,y≥0.
Hence, all the conditions of Theorem 1.2 are satisfied. An application of Theorem 1.2 implies that the problem (3.13) has a unique positive solution.



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*Source: 672543-2012-11-27.xml*