# Positive Solutions for Nonlinear Fractional Differential Equations with Boundary Conditions Involving Riemann-Stieltjes Integrals

Authors: Jiqiang Jiang; Lishan Liu; Yonghong Wu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider the existence of positive solutions for a class of nonlinear integral boundary value problems for fractional differential equations. By using some fixed point theorems, the existence and multiplicity results of positive solutions are obtained. The results obtained in this paper improve and generalize some well-known results.


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## Body

## 1. Introduction

This paper is concerned with the existence of positive solutions to the following boundary value problem (BVP) for fractional differential equation:(1.1)D0+αu(t)+f(t,u(t),u′(t),…,u(n-2)(t))=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(n-2)(1)=β[u(n-2)],
where β[v]=∫01‍v(t)dA(t) is a linear functional on C[0,1] given by a Riemann-Stieltjes integral with A representing a suitable function of bounded variation, D0+α is the Riemann-Liouville fractional derivative of order n-1<α≤n, n≥2, f:[0,1]×ℝn-1→ℝ+ satisfies the Carathéodory type conditions, ℝ=(-∞,+∞) and ℝ+=[0,+∞).Fractional differential equations arise in the modeling and control of many real-world systems and processes particularly in the fields of physics, chemistry, aerodynamics, electrodynamics of complex media, and polymer rheology. Fractional differential equations also serve as an excellent tool for the description of hereditary properties of various materials and processes. Hence, intensive research has been carried out worldwide to study the existence of solutions of nonlinear fractional differential equations (see [1–25]). For example, by means of a mixed monotone method, Zhang [11] studied a unique positive solution for the singular boundary value problem
(1.2)D0+αu(t)+q(t)f(t,u(t),u′(t),…,u(n-2)(t))=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=u(n-2)(1)=0,
where α∈(n-1,n], n≥2, D0+α is the standard Riemann-Liouville derivative, f=g+h is nonlinear, and g and h have different monotone properties.Recently, nonlocal boundary value problems for fractional differential equations were investigated intensively [13–23]. In [14], Bai concerned the existence and uniqueness of a positive solution for the following nonlocal problem:
(1.3)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=0,βu(η)=u(1),
where 1<α≤2, 0<βηα-1<1, 0<η<1, D0+α is the standard Riemann-Liouville differentiation. The function f is continuous on [0,1]×ℝ+.In [20], El-Shahed and Nieto investigated the existence of nontrivial solutions for the following nonlinear m-point boundary value problem of fractional type:
(1.4)RD0+αu(t)+f(t,u(t))=0,t∈[0,1],α∈(n-1,n],n∈ℕ,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=∑i=1m-2‍aiu(ηi),
where n≥2, ai>0(i=1,2,…,m-2), 0<η1<η2<⋯<ηm-2<1, f∈C([0,1]×ℝ,ℝ). Also the authors considered the analogous problem using the Caputo fractional derivative:
(1.5)CD0+αu(t)+f(t,u(t))=0,t∈[0,1],α∈(n-1,n],n∈ℕ,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=∑i=1m-2‍aiu(ηi).
Under certain growth conditions on the nonlinearity, several sufficient conditions for the existence of nontrivial solution are obtained by using the Leray-Schauder nonlinear alternative.Inspired by the work of the above papers, the aim of this paper is to establish the existence and multiplicity of positive solutions of the BVP (1.1). We discuss the boundary value problem with the Riemann-Stieltjes integral boundary conditions, that is, the BVP (1.1), which includes fractional order two-point, three-point, multipoint, and nonlocal boundary value problems as special cases. Moreover, the β[·] in (1.1) is a linear function on C[0,1] denoting the Riemann-Stieltjes integral; the A in the Riemann-Stieltjes integral is of bounded variation, namely, dA can be a signed measure. By using the Krasnosel’skii fixed point theorem, the Leray-Schauder nonlinear alternative and the Leggett-Williams fixed point theorem, some existence and multiplicity results of positive solutions are obtained.The rest of this paper is organized as follows. In Section2, we present some lemmas that are used to prove our main results. In Section 3, the existence and multiplicity of positive solutions of the BVP (1.1) are established by using some fixed point theorems. In Section 4, we give four examples to demonstrate the application of our theoretical results.


## 2. Basic Definitions and Preliminaries

We begin this section with some preliminaries of fractional calculus. Letα>0 and n=[α]+1, where [α] is the largest integer smaller than or equal to α. For a function f:(0,+∞)→R, we define the fractional integral of order α of f as
(2.1)I0+αf(t)=1Γ(α)∫0t‍(t-s)α-1f(s)ds,
provided the integral exists. The fractional derivative of order α>0 of a continuous function f is defined by
(2.2)D0+αf(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1f(s)ds,
provided the right-hand side is pointwise defined on (0,+∞). We recall the following properties [26, 27] which are useful for the sequel. For α>0, β>0, we have
(2.3)I0+αI0+βf(t)=I0+α+βf(t),D0+αI0+αf(t)=f(t).
As an example, we can choose a function f such that f,D0+αf∈C(0,+∞)∩Lloc1(0,+∞).Forα>0, the general solution of the fractional differential equation D0+αu(t)=0 with u∈C(0,1)∩L(0,1) is given by
(2.4)u(t)=c1tα-1+c2tα-2···+cntα-n,
where ci∈R(i=1,2,…,n). Hence for u∈C(0,1)∩L(0,1), we have
(2.5)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2···+cntα-n.
Set
(2.6)G0(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.Lemma 2.1 (see [11]).
Lety∈Cr[0,1](Cr[0,1]={y∈C[0,1],try∈C[0,1],0≤r<1}). Then the boundary value problem,
(2.7)D0+α-n+2v(t)=y(t),0<t<1,n-1<α≤n,n≥2,v(0)=0,v(1)=0,
has a unique solution
(2.8)v(t)=∫01‍G0(t,s)y(s)ds.Lemma 2.2 (see [11]).
The functionG0(t,s) defined by (2.6) satisfies the following properties: (i)
G0(t,s)≥0, G0(t,s)≤G0(s,s) for all t,s∈[0,1];(ii)
there exist a positive functionρ∈C(0,1) and 0<ξ<η<1 such that
(2.9)mint∈[ξ,η]G0(t,s)≥ρ(s)G0(s,s),s∈(0,1),where
(2.10)ρ(s)={[η(1-s)]α-n+1-(η-s)α-n+1[s(1-s)]α-n+1,s∈(0,r],(ξs)α-n+1,s∈[r,1).By (2.4), the unique solution of the problem
(2.11)D0+α-n+2v(t)=0,0<t<1,n-1<α≤n,n≥2,v(0)=0,v(1)=β[v]
is γ(t)=tα-n+1, with β[v] replaced by 1. As in [21], the Green’s function for boundary value problem (2.11) is given by
(2.12)G(t,s)=γ(t)1-β[γ]𝒢(s)+G0(t,s),
where 𝒢(s):=∫01‍G0(t,s)dA(t).Lemma 2.3.
Let0≤β[γ]<1 and 𝒢(s)≥0 for s∈[0,1], the Green function G(t,s) defined by (2.12) has the following properties: (i)
G(t,s)≥0,G(t,s)≤(1+β[1]/(1-β[γ]))G0(s,s) for all t,s∈[0,1];(ii)
mint∈[ξ,η]G(t,s)≥mint∈[ξ,η]G0(t,s)≥ρ(s)G0(s,s), s∈(0,1).Proof.
By Lemma2.2, it is easy to prove this lemma, so we omit it.LetX=C[0,1]. It follows that (X,∥·∥) is a Banach space, where ∥·∥ is defined by the supernorm ∥x∥=supt∈[0,1]|x(t)|. P={x∈X:x(t)≥0,t∈[0,1]}. Clearly P is a cone of X. Now, in the following, we give the assumptions to be used throughout the rest of this paper.(H1)
A is a function of bounded variation, 𝒢(s)≥0 for s∈[0,1] and 0≤β[γ]<1.(H2)
f:[0,1]×ℝn-1→ℝ+ satisfies the following conditions of Carathéodory type:
(i)
f(·,x) is Lebesgue measurable for each fixed x∈ℝn-1;(ii)
f(t,·) is continuous for a.e. t∈[0,1].In order to overcome the difficulty due to the dependance off on derivatives, we consider the following modified problem:
(2.13)D0+α-n+2v(t)+f(t,I0+n-2v(t),I0+n-3v(t),…,I0+1v(t),v(t))=0,0<t<1,v(0)=0,v(1)=β[v],
where n-1<α≤n, n≥2.Lemma 2.4.
The nonlocal fractional order boundary value problem (1.1) has a positive solution if and only if the nonlinear fractional integrodifferential equation (2.13) has a positive solution.Proof.
Ifu is a positive solution of the fractional order boundary value problem (1.1), let v(t)=D0+n-2u(t). Then from the boundary value conditions of (1.1) and the definition of the Riemann-Liouville fractional integral and derivative, we have(2.14)v(t)=D0+n-2u(t)=u(n-2)(t),I0+1v(t)=I0+1u(n-2)(t)=1Γ(1)∫0t‍u(n-2)(s)ds=u(n-3)(t),I0+2v(t)=I0+2u(n-2)(t)=1Γ(2)∫0t‍(t-s)u(n-2)(s)ds=u(n-4)(t),⋮I0+n-2v(t)=I0+n-2u(n-2)(t)=1Γ(n-2)∫0t‍(t-s)n-3u(n-2)(s)ds=u(t),(2.15)D0+α-n+2v(t)=1Γ(2n-α-2)(ddt)n∫0t‍(t-s)2n-α-3u(n-2)(s)ds=1Γ(2n-α-3)(ddt)n∫0t‍(t-s)2n-α-4u(n-3)(s)ds=⋯=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1u(s)ds=D0+αu(t),
which imply that v(0)=u(n-2)(0)=0, v(1)=u(n-2)(1)=β[u(n-2)]=β[v]. Thus v(t) is a positive solution of the nonlinear fractional integrodifferential equation (2.13).
On the other hand, ifv is a positive solution of the nonlinear fractional integrodifferential equation (2.13), let u(t)=I0+n-2v(t), then by (2.3) and the definition of the Riemann-Liouville fractional derivative, we have
(2.16)u′(t)=D0+1u(t)=D0+1I0+n-2v(t)=D0+1I0+1I0+n-3v(t)=I0+n-3v(t),u′′(t)=D0+2u(t)=D0+2I0+n-2v(t)=D0+2I0+2I0+n-4v(t)=I0+n-4v(t),⋮u(n-3)(t)=D0+n-3u(t)=D0+n-3I0+n-2v(t)=D0+n-3I0+n-3I0+1v(t)=I0+1v(t),u(n-2)(t)=D0+n-2u(t)=D0+n-2I0+n-2v(t)=v(t),D0+αu(t)=dndtnI0+n-αu(t)=dndtnI0+n-αI0+n-2v(t)=dndtnI0+2n-α-2v(t)=D0+α-n+2v(t)D0+αu(t)=-f(t,I0+n-2v(t),I0+n-3v(t),…,I0+1v(t),v(t))D0+αu(t)=-f(t,u(t),u′(t),…,u(n-3)(t),u(n-2)(t)),0<t<1,
which imply that u(0)=u'(0)=⋯=u(n-3)(0)=0,   u(n-2)(0)=v(0)=0,   u(n-2)(1)=v(1)=β[v]=β[u(n-2)]. Moreover, it follows from the monotonicity and property of I0+n-2 that I0+n-2v(t)∈C([0,1],ℝ+). Consequently, u(t)=I0+n-2v(t) is a positive solution of the fractional order boundary value problem (1.1).By Lemma2.4, we will concentrate our study on (2.13). We here define an operator T:P→P by
(2.17)Tv(t)=∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds,t∈[0,1].
Clearly, v is a fixed point of T in P, and so v is a positive solution of BVP (2.13).In order to prove our main results, we need the following lemmas.Lemma 2.5 (see [28]).
LetX be a real Banach space, Ω be a bounded open subset of X, where θ∈Ω, T:Ω¯→X is a completely continuous operator. Then, either there exist x∈∂Ω, μ∈(0,1) such that μT(x)=x, or there exists a fixed point x*∈Ω¯.Lemma 2.6 (see [29]).
LetX be a real Banach space, P be a cone in X. Assume that Ω1 and Ω2 are two bounded open sets of X with θ∈Ω1 and Ω¯1⊂Ω2. Let T:P∩(Ω¯2∖Ω1)→P be 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.Then T has a fixed point in P∩(Ω¯2∖Ω1).Lemma 2.7 (see [30, 31]).
LetP be a cone in a real Banach space X, Pc={x∈P:∥x∥<c},  φ be a nonnegative continuous concave functional on P such that φ(x)≤∥x∥ for all x∈P¯c, and P(φ,b,d)={x∈P:b≤φ(x),∥x∥≤d}. Suppose that T:P¯c→P¯c is completely continuous and there exist positive constants 0<a<b<d≤c such that(C1)
{x∈P(φ,b,d):φ(x)>b}≠ϕ and φ(Tx)>b for x∈P(φ,b,d),(C2)
∥Tx∥<a for x∈P¯a,(C3)
φ(Tx)>b for x∈P(φ,b,c) with ∥Tx∥>d.Then T has at least three fixed points x1,x2, and x3 satisfying
(2.18)∥x1∥<a,b<φ(x2),a<∥x3∥withφ(x3)<b.Remark 2.8.
Ifd=c, then condition (C1) of Lemma 2.7 implies condition (C3) of Lemma 2.7.
For notational convenience, we introduce the following constants:(2.19)L1=∫ξη‍ρ(s)G0(s,s)ds,L2=(1+β[1]1-β[γ])∫01‍G0(s,s)ds,
and a nonnegative continuous concave functional φ on the cone P defined by
(2.20)φ(v)=minξ≤t≤η|v(t)|.


## 3. Main Results

In this section, we present and prove our main results.Theorem 3.1.
Assume that (H1) and (H2) hold and there exist nonnegative functions h1,h2,…,hn-1∈L[0,1] such that
(3.1)|f(t,x1,x2,…,xn-1)-f(t,y1,y2,…,yn-1)|≤∑i=1n-1‍hi(t)|xi-yi|,
for  almost  every t∈[0,1] and  all (x1,x2,…,xn-1), (y1,y2,…,yn-1)∈Rn-1.
If(3.2)0<∫01‍G0(s,s)∑i=1n-1‍hi(s)ds<(1+β[1]1-β[γ])-1,
then BVP (1.1) has a unique positive solution.Proof.
We will show thatT is a contraction mapping. For any v1,v2∈P and 1≤i≤n-2, by the definition of fractional integral, we obtain
(3.3)|I0+iv1(t)-I0+iv2(t)|=|1(i-1)!∫0t‍(t-s)i-1(v1(s)-v2(s))ds|≤1(i-1)!∫0t‍(t-s)i-1|v1(s)-v2(s)|ds≤1(i-1)!∫0t‍(t-s)i-1ds∥v1-v2∥=1i!ti∥v1-v2∥≤∥v1-v2∥.
So, for any v1,v2∈P, by (3.3) and Lemma 2.3, we have
(3.4)|Tv1(t)-Tv2(t)|=|∫01‍G(t,s)[f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))-f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))]ds{∫01‍G(t,s)[f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))}|≤∫01‍G(t,s)|f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))-f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))|ds≤∫01‍(1+β[1]1-β[γ])G0(s,s)∑i=1n-1‍hi(s)|I0+n-1-iv1(s)-I0+n-1-iv2(s)|ds≤(1+β[1]1-β[γ])∫01‍G0(s,s)∑i=1n-1‍hi(s)ds∥v1-v2∥.
This implies that
(3.5)∥Tv1-Tv2∥≤κ∥v1-v2∥,
where κ=(1+β[1]/(1-β[γ]))∫01‍G0(s,s)∑i=1n-1‍hi(s)ds∈(0,1). By the Banach contraction mapping principle, we deduce that T has a unique fixed point v*. Thus, by Lemma 2.4, u*(t)=I0+n-2v*(t) is a unique positive solution of BVP (1.1).Lemma 3.2.
Assume that (H1) and (H2) hold and the following conditions are satisfied.(H3)
There exist nonnegative real-valued functionsq,p1,p2,…,pn-1∈L[0,1] such that
(3.6)f(t,x1,x2,…,xn-1)≤q(t)+∑i=1n-1‍pi(t)|xi|,for  almost  every t∈[0,1] and  all (x1,x2,…,xn-1)∈Rn-1.
ThenT:P→P is a completely continuous operator.Proof.
For anyv∈P, as G(t,s)≥0 for all t,s∈[0,1], we have Tv(t)≥0, so T(P)⊂P. Let D⊂P be any bounded set. Then there exists a constant L>0 such that ∥v∥≤L for any v∈D. Moreover for any v∈D, s∈[0,1], v(s)≤∥v∥≤L. Proceeding as for (3.3), we obtain
(3.7)|I0+n-1-iv(s)|=I0+n-1-iv(s)≤∥v∥≤L,i=1,2,…,n-1.
Thus,
(3.8)|Tv(t)|=∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≤∫01‍(1+β[1]1-β[γ])G0(s,s)[q(s)+∑i=1n-1‍pi(s)I0+n-1-iv(s)]ds≤(1+β[1]1-β[γ])∫01‍G0(s,s)[q(s)+∑i=1n-1‍pi(s)∥v∥]ds≤(1+β[1]1-β[γ])(L+1)∫01‍G0(s,s)[q(s)+∑i=1n-1‍pi(s)]ds<+∞.
Therefore, T(D) is uniformly bounded.
Now we show thatT(D) is equicontinuous on [0,1]. Since G(t,s) is continuous on [0,1]×[0,1], G(t,s) is uniformly continuous on [0,1]×[0,1]. Hence, for any ε>0, there exists a constant δ0>0 such that for any s∈[0,1], t,t'∈[0,1], when |t-t'|<δ0, it holds
(3.9)|G(t,s)-G(t′,s)|<[1+(L+1)∫01‍(q(s)+∑i=1n-1‍pi(s))ds]-1ε.
Consequently, for any t,t'∈[0,1] and |t-t'|<δ0, we have
(3.10)|Tv(t)-Tv(t′)|≤∫01‍|G(t,s)-G(t′,s)|f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≤∫01‍|G(t,s)-G(t′,s)|[q(s)+∑i=1n-1‍pi(s)∥v∥]ds≤(L+1)∫01‍|G(t,s)-G(t′,s)|[q(s)+∑i=1n-1‍pi(s)]ds<ε.
This implies that T(D) is equicontinuous. Thus according to the Ascoli-Arzela Theorem, T(D) is a relatively compact set.
In the end, we show thatT:P→P is continuous. Assume that vm,v0∈P(m=1,2,…), vm→v0(m→+∞), then
(3.11)|vm(t)-v0(t)|≤∥vm-v0∥→0,
and ∥vm∥≤L (m=0,1,2,…), where L is a positive constant. Keeping in mind that f satisfies Carathéodory conditions on [0,1]×ℝn-1, we have
(3.12)limm→+∞f(t,I0+n-2vm(t),I0+n-3vm(t),…,I0+1vm(t),vm(t))=f(t,I0+n-2v0(t),I0+n-3v0(t),…,I0+1v0(t),v0(t)),fora.e.t∈[0,1].
Proceeding as for (3.3), for m∈ℕ we obtain
(3.13)|I0+n-1-ivm(s)|=I0+n-1-ivm(s)≤∥vm∥≤Li=1,2,…,n-1.
This together with (3.6),
(3.14)0≤f(t,I0+n-2vm(t),I0+n-3vm(t),…,I0+1vm(t),vm(t))≤q(t)+L∑i=1n-1‍pi(t).
The Lebesgue dominated convergence theorem gives
(3.15)limm→+∞∫01‍|f(s,I0+n-2vm(s),I0+n-3vm(s),…,I0+1vm(s),vm(s))-f(s,I0+n-2v0(s),I0+n-3v0(s),…,I0+1v0(s),v0(s))|ds=0.
Now we deduce from (3.15), Lemma 2.3(3.16)|Tvm(t)-Tv0(t)|=|∫01‍G(t,s)[f(s,I0+n-2vm(s),I0+n-3vm(s),…,I0+1vm(s),vm(s))-f(s,I0+n-2v0(s),I0+n-3v0(s),…,I0+1v0(s),v0(s))]ds{|∫01‍G(t,s)[f(s,I0+n-2vm(s),I0+n-3vm(s),…,I0+1vm(s),vm(s))}|≤(1+β[1]1-β[γ])∫01‍G0(s,s)|f(s,I0+n-2vm(s),I0+n-3vm(s),…,I0+1vm(s),vm(s))-f(s,I0+n-2v0(s),I0+n-3v0(s),…,I0+1v0(s),v0(s))|ds≤(1+β[1]1-β[γ])maxs∈[0,1]G0(s,s)∫01|f(s,I0+n-2vm(s),I0+n-3vm(s),…,I0+1vm(s),vm(s))-f(s,I0+n-2v0(s),I0+n-3v0(s),…,I0+1v0(s),v0(s))|ds
that ∥Tvm-Tv0∥→0, as m→+∞. So T:P→P is continuous. Therefore T:P→P is completely continuous.Remark 3.3.
Iff:[0,1]×ℝn-1 is continuous, by similar argument as above, we can show that T is completely continuous.Theorem 3.4.
Assume that (H1)–(H3) hold. If
(3.17)∫01‍G0(s,s)∑i=1n-1‍pi(s)ds<(1+β[1]1-β[γ])-1,
then BVP (1.1) has at least one positive solution.Proof.
Let(3.18)Ω={v∈P:∥v∥<r},wherer=(1+β[1]/(1-β[γ]))∫01‍G0(s,s)q(s)ds1-(1+β[1]/(1-β[γ]))∫01‍G0(s,s)∑i=1n-1pi(s)ds,
we have Ω⊂P. From Lemma 3.2, we know that T:Ω→P is completely continuous. If there exists v∈∂Ω,μ∈(0,1) such that
(3.19)v=μTv,
then by (H3) and (3.19), we have
(3.20)v(t)=μTv(t)=μ∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≤μ∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≤μ(1+β[1]1-β[γ])[∫01‍G0(s,s)q(s)ds+∫01‍G0(s,s)∑i=1n-1‍pi(s)ds∥v∥],
which implies that
(3.21)∥v∥≤μ(1+β[1]1-β[γ])[∫01‍G0(s,s)q(s)ds+r∫01‍G0(s,s)∑i=1n-1‍pi(s)ds]<(1+β[1]1-β[γ])[∫01‍G0(s,s)q(s)ds+r∫01‍G0(s,s)∑i=1n-1‍pi(s)ds]=r.
This means that v∉∂Ω. By Lemma 2.5, T has a fixed point v^∈Ω¯. By Lemma 2.4, BVP (1.1) has at least one positive solution u^(t)=I0+n-2v^(t).Theorem 3.5.
Assume that (H1)–(H3) hold. If there exist two positive constants r1<r2 such that (i)
f(t,x1,x2,…,xn-1)≤L2-1r2, for (t,x1,x2,…,xn-1)∈[0,1]×[0,r2]×⋯×[0,r2],(ii)
f(t,x1,x2,…,xn-1)≥L1-1r1, for (t,x1,x2,…,xn-1)∈[0,1]×[0,r1]×⋯×[0,r1],where L1, L2 are defined by (2.19), then BVP (1.1) has at least one positive solution.Proof.
LetΩ2={v∈P:∥v∥<r2}. For any v∈∂Ω2, we have ∥v∥=r2 and 0≤v(t)≤r2 for every t∈[0,1]. Similar to (3.7), for 0≤v(s)≤r2, we have
(3.22)0≤|I0+n-1-iv(s)|=I0+n-1-iv(s)≤∥v∥≤r2,i=1,2,…,n-1.
It follows from condition (i) that
(3.23)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))≤L2-1r2,for(s,v)∈[0,1]×[0,r2].
Thus, for any v∈∂Ω2, by (3.23) and Lemma 2.3, we have
(3.24)|Tv(t)|=∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≤∫01‍G(t,s)L2-1r2ds≤(1+β[1]/(1-β[γ]))L2-1r2∫01‍G0(s,s)ds=r2=∥v∥,t∈[0,1],
which means that
(3.25)∥Tv∥≤∥v∥,v∈∂Ω2.
On the other hand, letΩ1={v∈P:∥v∥<r1}. For any v∈∂Ω1, we have ∥v∥=r1 and 0≤v(t)≤r1 for every t∈[0,1]. Similar to (3.23), from condition (ii), we can get
(3.26)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))≥L1-1r1,for(s,v)∈[0,1]×[0,r1].
Hence for any t∈[ξ,η], v∈∂Ω1, by (3.26) and Lemma 2.3 we have
(3.27)|Tv(t)|=∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≥∫01‍G(t,s)L1-1r1ds≥L1-1r1∫ξη‍G(t,s)ds≥L1-1r1∫ξη‍ρ(s)G0(s,s)ds=r1=∥v∥.
Thus we get
(3.28)∥Tv∥≥∥v∥,v∈∂Ω1.
By (3.25), (3.28), and Lemma 2.6, T has a fixed point v~∈Ω¯2∖Ω1 such that r1≤∥v~∥≤r2. By Lemma 2.4, BVP (1.1) has at least one positive solution u~(t)=I0+n-2v~(t).Theorem 3.6.
Assume that (H1)–(H3) hold. If there exist constants 0<a<b<c such that (I)
f(t,x1,x2,…,xn-1)<L2-1a, for (t,x1,x2,…,xn-1)∈[0,1]×[0,a]×⋯×[0,a],(II)
f(t,x1,x2,…,xn-1)≤L2-1c, for (t,x1,x2,…,xn-1)∈[0,1]×[0,c]×⋯×[0,c],(III)
f(t,x1,x2,…,xn-1)≥L1-1b, for (t,x1,x2,…,xn-1)∈[ξ,η]×[(b/(n-2)!)ξn-2,c]×[(b/(n-3)!)ξn-3,c]×⋯×[bξ,c]×[b,c],where L1, L2 are defined by (2.19), then BVP (1.1) has at least three positive solutions u1, u2, and u3 satisfying
(3.29)∥D0+n-2u1∥<a,b<φ(D0+n-2u2)<∥D0+n-2u2∥≤c,a<∥D0+n-2u3∥,φ(D0+n-2u3)<b.Proof.
We will show that all conditions of Lemma2.7 are satisfied.
First, ifv∈P¯c, then ∥v∥≤c. So we have 0≤v(t)≤c, t∈[0,1]. Similar to (3.23), it follows from condition (II) that
(3.30)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))≤L2-1c,for(s,v)∈[0,1]×[0,c].
Thus, for any v∈P¯c, by (3.30), we have
(3.31)|Tv(t)|=∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≤∫01‍G(t,s)L2-1cds≤(1+β[1]1-β[γ])L2-1c∫01‍G0(s,s)ds=c,
which means that ∥Tv∥≤c, v∈P¯c. Therefore, T:P¯c→P¯c. By Lemma 3.2, we know that T:P¯c→P¯c is completely continuous.
Next, similar to (3.30) and (3.31), it follows from condition (I) that if v∈P¯a then ∥Tv∥<a. So the condition (C2) of Lemma 2.7 holds.
Now, we takev(t)=(b+c)/2, t∈[0,1]. It is easy to see that v(t)=(b+c)/2∈P(φ,b,c), and so
(3.32)φ(v)=minξ≤t≤η|v(t)|=b+c2>b,
where φ(v) is defined by (2.20). This proves that {v∈P(φ,b,c):φ(v)>b}≠ϕ.
On the other hand, ifv∈P(φ,b,c), then b≤v(t)≤c, t∈[ξ,η]. By the definition of fractional integral, for any t∈[ξ,η], 1≤i≤n-2, we obtain
(3.33)bi!ξi≤b(i-1)!∫0t‍(t-s)i-1ds≤I0+iv(t)=1(i-1)!∫0t‍(t-s)i-1v(s)ds≤c(i-1)!∫0t‍(t-s)i-1ds≤ci!ηi≤c.
It follows from (3.33) and condition (III) that
(3.34)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))≥L1-1b,fors∈[ξ,η],v∈P(φ,b,c).
Hence we have
(3.35)φ(Tv)=minξ≤t≤η|Tv(t)|=minξ≤t≤η∫01‍G(t,s)f(s,I0+n-2v(s),I0+n-3v(s),…,I0+1v(s),v(s))ds≥L1-1bminξ≤t≤η∫01‍G(t,s)ds>L1-1b∫ξη‍ρ(s)G0(s,s)ds=b,
which implies that φ(Tv)>b, for v∈P(φ,b,c). This shows that condition (C1) of Lemma 2.7 is also satisfied.
By Lemma2.7 and Remark 2.8, BVP (2.13) has at least three positive solutions v1,v2, and v3 such that ∥v1∥<a, b<φ(v2)<∥v2∥≤c, and a<∥v3∥, φ(v3)<b. By Lemma 2.4, BVP (1.1) has at least three positive solutions ui(t)=I0+n-2vi(t), (i=1,2,3). By (2.3), we have D0+n-2ui(t)=D0+n-2I0+n-2vi(t)=vi, i=1,2,3. So u1,u2,u3 are three positive solutions of BVP (1.1) satisfying
(3.36)∥D0+n-2u1∥<a,b<φ(D0+n-2u2)<∥D0+n-2u2∥≤c,a<∥D0+n-2u3∥,φ(D0+n-2u3)<b.
The proof of Theorem 3.6 is completed.


## 4. Examples

Example 4.1.
Consider the following problem:(4.1)D0+7/2u(t)+(1-t)3etu(1+et)(1+u)+12t2sin2u'+14tu′′=0,0<t<1,u(0)=u′(0)=u′′(0)=0,u′′(1)=β[u′′].
Let β[u′′]=(1/2)u′′(1/2). Then
(4.2)G0(t,s)=1Γ(3/2){[t(1-s)]1/2-(t-s)1/2,0≤s≤t≤1,[t(1-s)]1/2,0≤t≤s≤1.(4.3)𝒢(s)=12G0(12,s)≥0,β[1]=∫01‍dA(t)=12,β[γ]=∫01‍t1/2dA(t)=24<1.
Let
(4.4)f(t,x,y,z)=(1-t)3etx(1+et)(1+x)+12t2sin2y+14tz,h1(t)=(1-t)3et1+et,h2(t)=12t2,h3(t)=14t.
Then f is a nonnegative continuous function on [0,1]×(ℝ+)3 and, for any (t,x1,y1,z1) and (t,x2,y2,z2)∈[0,1]×(ℝ+)3, satisfies
(4.5)|f(t,x1,y1,z1)-f(t,x2,y2,z2)|≤h1(t)|x1-x2|+h2(t)|y1-y2|+h3(t)|z1-z2|.
So we have
(4.6)∫01‍G0(s,s)∑i=13‍hi(s)ds≤1Γ(3/2)∫01‍(s(1-s))1/2((1-s)3+s2+s)ds=B(3/2,9/2)+B(7/2,3/2)+B(5/2,3/2)Γ(3/2)=33128π≈0.4569608<(1+β[1]1-β[γ])-1=4-26-2≈0.5638901,
where B(·,·) denotes a Beta function. So all conditions of Theorem 3.1 are satisfied. Thus, by Theorem 3.1, BVP (4.1) has at least one positive solution.Example 4.2.
Consider the following problem:(4.7)D0+5/2u(t)+12(t-t2)ln(1+u)+12t2u'+t3+sint=0,0<t<1,u(0)=u′(0)=0,u′(1)=β[u′],
where β[u']=∫01‍u'(s)dA(s) with
(4.8)A(s)={0,s∈[0,14),2,s∈[14,916),1,s∈[916,1].
Set
(4.9)f(t,x,y)=12(t-t2)ln(1+x)+12t2y+t3+sint,p1(t)=12(t-t2),p2(t)=12t2,q(t)=t3+1.
Then f:[0,1]×[0,+∞)×[0,+∞)→[0,+∞) is continuous and
(4.10)f(t,x,y)≤p1(t)x+p2(t)y+q(t).
As in [21], β[γ]=∫01‍γ(t)dA(t)=21/4+(-1)×9/16=1/4<1, 𝒢(s)=∫01‍G0(t,s)dA(t)≥0,
(4.11)∫01‍G0(s,s)(p1(s)+p2(s))ds≤1Γ(3/2)∫01‍(s(1-s))1/2((s-s2)+s2)ds=B(5/2,5/2)+B(7/2,3/2)Γ(3/2)=18π≈0.22155673<(1+β[1]1-β[γ])-1=37≈0.42857143,
where B(·,·) denotes a Beta function and G0(t,s) is defined by (4.2). So all conditions of Theorem 3.4 are satisfied. Thus, by Theorem 3.4, BVP (4.7) has at least one positive solution.Example 4.3.
Consider the following problem:(4.12)D0+5/2u(t)+t25ln(1+u)+tetu'10+10et+sint20+12=0,0<t<1,u(0)=u′(0)=0,u′(1)=β[u′],
where β[u']=∫01‍u'(s)dA(s) with A(s) as given by (4.8). Set
(4.13)f(t,x,y)=t25ln(1+x)+tety10+10et+sint20+12,p1(t)=t25,p2(t)=tet10+10et,q(t)=sint20+12.
Then f:[0,1]×[0,+∞)×[0,+∞)→[0,+∞) is continuous and
(4.14)f(t,x,y)≤p1(t)x+p2(t)y+q(t).
By Example 4.2, β[γ]=∫01‍γ(t)dA(t)=1/4<1, 𝒢(s)=∫01‍G0(t,s)dA(t)≥0, where G0(t,s) is defined by (4.2). As in [1, 3], we also take ξ=1/4,η=3/4, then
(4.15)L1-1=(∫1/43/4‍ρ(s)G0(s,s)ds)-1≈13.6649,L2-1=(1+β[1]1-β[γ])-1(∫01‍G0(s,s)ds)-1=127π≈0.967182.
Choosing r1=1/30,r2=1, we have
(4.16)f(t,x,y)≤0.85≤L2-1r2,for(t,x,y)∈[0,1]×[0,1]×[0,1],f(t,x,y)≥0.5≥L1-1r1,for(t,x,y)∈[0,1]×[0,130]×[0,130].
So all conditions of Theorem 3.5 are satisfied. Thus, by Theorem 3.5, BVP (4.12) has at least one positive solution.Example 4.4.
Consider the following problem:(4.17)D0+5/2u(t)+f(t,u(t),u′(t))=0,0<t<1,u(0)=u′(0)=0,u′(1)=β[u′],
where β[u']=∫01‍u'(s)dA(s) with A(s) as given by (4.8). Set
(4.18)f(t,x,y)={t2100ln(1+x)+15y2+t1000,(t,x,y)∈[0,1]×[0,1]×[0,1],t2100ln(1+x)+292+12y+t1000,(t,x,y)∈[0,1]×((ℝ+)2∖[0,1]2),p1(t)=t2100,p2(t)=15,q(t)=t1000+292.
Then
(4.19)f(t,x,y)≤p1(t)x+p2(t)y+q(t).
By Example 4.3, β[γ]=∫01‍γ(t)dA(t)=1/4<1, 𝒢(s)=∫01‍G0(t,s)dA(t)≥0,
(4.20)L1-1=(∫1/43/4‍ρ(s)G0(s,s)ds)-1≈13.6649,L2-1=(1+β[1]1-β[γ])-1(∫01‍G0(s,s)ds)-1=127π≈0.967182.
Choosing a=1/20, b=1, c=100, we have
(4.21)f(t,x,y)≤0.044845<L2-1a≈0.048359,for(t,x,y)∈[0,1]×[0,120]×[0,120],f(t,x,y)≥14.5025≥L1-1b≈13.6649,for(t,x,y)∈[14,34]×[14,100]×[1,100],f(t,x,y)≤65.501≤L2-1c≈96.7182,for(t,x,y)∈[0,1]×[0,100]×[0,100].
So all conditions of Theorem 3.6 are satisfied. Thus, by Theorem 3.6, BVP (4.17) has at least three positive solutions u1,u2, and u3, satisfying
(4.22)max0≤t≤1|u1′(t)|<120,1<min1/4≤t≤3/4|u2′(t)|<max0≤t≤1|u2′(t)|≤100,120≤max0≤t≤1|u3′(t)|≤100withmin1/4≤t≤3/4|u3′(t)|<1.



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*Source: 708192-2012-09-27.xml*