# Existence of Positive Solutions for Fractional Differential Equation with Nonlocal Boundary Condition

Authors: Hongliang Gao; Xiaoling Han
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

By using the fixed point theorem, existence of positive solutions for fractional differential equation with nonlocal boundary conditionD0+αu(t)+a(t)f(t,u(t))=0, 0<t<1, u(0)=0, u(1)=∑i=1∞αiu(ξi) is considered, where 1<α≤2 is a real number, D0+α is the standard Riemann-Liouville differentiation, and ξi∈(0,1),αi∈[0,∞) with ∑i=1∞αiξiα-1<1, a(t)∈C([0,1],[0,∞)),f(t,u)∈C([0,1]×[0,∞),[0,∞)).


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. It 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.It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations in terms of special functions [6–8]. Recently, there are some papers that deal with the existence and multiplicity of solution (or positive solution) of nonlinear initial fractional differential equation by the use of techniques of nonlinear analysis (fixed-point theorems, Leray-Shauder theory, etc.); see [9–17].Recently, Bai and Lü [15] 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.In this paper, we study the existence of positive solutions for fractional differential equation with nonlocal boundary condition(1.2)D0+αu(t)+a(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, D0+α is the standard Riemann-Liouville differentiation, and ξi∈(0,1),αi∈[0,∞) with ∑i=1∞αiξiα-1<1, a(t)∈C([0,1],[0,∞)),f(t,u)∈C([0,1]×[0,∞),[0,∞)).We assume the following conditions hold throughout the paper:(H1)
ξi∈(0,1),αi∈[0,∞) is both constants with ∑i=1∞αiξiα-1<1,(H2)
a(t)∈C([0,1],[0,∞)), a(t)≢0 on [a,b]⊂(0,1),(H3)
f(t,u)∈C([0,1]×[0,∞),[0,∞)).Remark 1.1.
To our knowledge, there are no results about the existence of positive solutions for problem (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,∞).Definition 2.3.
The mapθ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E, provided that θ:P→[0,∞) is continuous and
(2.3)θ(tx+(1-t)y)≥tθ(x)+(1-t)θ(y),
for all x,y∈P and 0≤t≤1.Remark 2.4.
As a basic example, we quote forλ>-1,
(2.4)D0+αtλ=Γ(λ+1)Γ(λ-α+1)tλ-α,
giving in particular D0+αtα-m=0,m=1,2,…,N, where N is the smallest integer greater than or equal to α.
From Definition2.2 and Remark 2.4, we then obtain the following.Lemma 2.5.
Letα>0. If one assumes u∈C(0,1)∩L(0,1), then the fractional differential equation
(2.5)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 2.6.
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,
(2.6)I0+αD0+αu(t)=u(t)+C1tα-1+C2tα-2+⋯+CNtα-N,
for some Ci∈R,i=1,2,…,N.Lemma 2.7 (see [15]).
Giveny∈C[0,1] and 1<α≤2, the unique solution of
(2.7)D0+αu(t)+y(t)=0,0<t<1,u(0)=u(1)=0
is
(2.8)u(t)=∫01G(t,s)y(s)ds,
where
(2.9)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 2.8.
Suppose(H1) holds. Given y∈C[0,1] and 1<α≤2, the unique solution of
(2.10)D0+αu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=∑i=1∞αiu(ξi)
is
(2.11)u(t)=∫01G(t,s)y(s)ds+B(y)tα-1,
where
(2.12)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,B(y)=∑i=1∞αi∫01G(ξi,s)y(s)ds1-∑i=1∞αiξiα-1.Proof.
By applying Lemmas2.6 and 2.7, we have
(2.13)u(t)=∫01G(t,s)y(s)ds+C1tα-1+C2tα-2.
Because
(2.14)∑i=1∞αi∫01G(ξi,s)ds=∑i=1∞αiξiα-1(1-ξi)αΓ(α),αiξiα-1(1-ξi)<αiξiα-1,
by (H1), ∑i=1∞αiξiα-1(1-ξi) is converge. Therefore, ∑i=1∞αi∫01G(ξi,s)ds is converge. y(t) is continuous function on [0,1], so ∑i=1∞αi∫01G(ξi,s)y(s)ds is converge.
Byu(0)=0, u(1)=∑i=1∞αiu(ξi), there are C2=0,C1=(∑i=1∞αi∫01G(ξi,s)y(s)ds)/(1-∑i=1∞αiξiα-1). Therefore,
(2.15)u(t)=∫01G(t,s)y(s)ds+B(y)tα-1,B(y)=∑i=1∞αi∫01G(ξi,s)y(s)ds1-∑i=1∞αiξiα-1.Lemma 2.9 (see [15]).
The functionG(t,s) defined by (2.9) satisfies the following conditions:(1)
G(t,s)>0, for t,s∈(0,1),(2)
there exists a positive functionγ∈C(0,1) such that(2.16)min⁡(1/4)≤t≤(3/4)⁡G(t,s)≥γ(s)max⁡0≤t≤1⁡G(t,s)=γ(s)G(s,s),0<s<1.Lemma 2.10 (see [18]).
Let E be a Banach space,P⊆E a cone and Ω1,Ω2 two bounded open sets of E with 0∈Ω1⊂Ω¯1⊂Ω2. Suppose that A:P∩(Ω¯2∖Ω1)→P is a completely continuous operator such that either(i)
∥Ax∥≤∥x∥,x∈P∩∂Ω1 and ∥Ax∥≥∥x∥,x∈P∩∂Ω2, or(ii)
∥Ax∥≥∥x∥,x∈P∩∂Ω1 and ∥Ax∥≤∥x∥,x∈P∩∂Ω2holds. Then, A has a fixed point in P∩(Ω¯2∖Ω1).Lemma 2.11 (see [19]).
LetP be a cone in real Banach space E,Pc={x∈P∣∥x∥≤c},θ a nonnegative continuous concave functional on P such that θ(x)≤∥x∥, for all x∈P¯c, and P(θ,b,d)={x∈P∣b≤θ(x),∥x∥≤d}. Suppose that A:P¯c→P¯c is completely continuous, and there exist constants 0<a<b<d≤c such that(C1)
{x∈P(θ,b,d)∣θ(x)>b}≠∅, and θ(Ax)>b,x∈P(θ,b,d),(C2)
∥Ax∥≤a, for x≤a,(C3)
θ(Ax)>b for x∈P(θ,b,c) with ∥Ax∥>d.Then, A has at least three fixed points x1,x2,x3 with
(2.17)‖x1‖<a,b<θ(x2),a<‖x3‖,θ(x3)<b.Remark 2.12.
If there holdsd=c, then condition (C1) of Lemma 2.11 implies condition (C3) of Lemma 2.11.


## 3. The Main Results

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}.Let the nonnegative continuous concave functionalθ on the cone P be defined by θ(u)=min(1/4)≤t≤(3/4)|u(t)|.Lemma 3.1 (see [15]).
LetT:P→E be the operator defined by Tu(t):=∫01G(t,s)f(s,u(s))ds, then T:P→P is completely continuous.Lemma 3.2.
LetA:P→E be the operator defined by
(3.1)Au(t):=∫01G(t,s)a(s)f(s,u(s))ds+B(a(⋅)f(⋅,u(⋅)))tα-1,
then A:P→P is completely continuous.Proof.
The proof is similar to Lemma3.1, so we omit.
Denote(*)M=(∫01G(s,s)a(s)ds+∑i=1∞αi∫01G(ξi,s)a(s)ds1-∑i=1∞αiξiα-1)-1,N=(∫1/43/4γ(s)G(s,s)a(s)ds)-1.Theorem 3.3.
Assume (H1)–(H3) hold, and there exist two positive constantsr2>r1>0 such that(1)
f(t,u)≤Mr2,forallt∈[0,1],u∈[0,r2],(2)
f(t,u)≥Nr1,forallt∈[0,1],u∈[0,r1], where M,N is defined in (*),then problem (1.2) has at least one positive solution u such that r1≤∥u∥≤r2.Proof.
By Lemmas2.8 and 3.2, we know A:P→P is completely continuous, and problem (1.2) has a solution u=u(t) if and only if u solves the operator equation u=Au. In order to apply Lemma 2.10, we separate the proof into the following two steps.
Step 1.
LetΩ2={u∈P∣∥u∥≤r2}. For u∈∂Ω2, we have 0≤u(t)≤r2 for all t∈[0,1]. It follows from (1) that for t∈[0,1],
(3.2)‖Au‖≤∫01G(s,s)a(s)f(s,u(s))ds+∑i=1∞αi∫01G(ξi,s)a(s)f(s,u(s))ds1-∑i=1∞αiξiα-1≤Mr2[∫01G(s,s)a(s)ds+∑i=1∞αi∫01G(ξi,s)a(s)ds1-∑i=1∞αiξiα-1]=r2=‖u‖.
Therefore,
(3.3)‖Au‖≤‖u‖,u∈P∩∂Ω2.Step 2.
LetΩ1={u∈P∣∥u∥≤r1}. For u∈∂Ω1, we have 0≤u(t)≤r1 for all t∈[0,1]. By assumption (2), for t∈[1/4,3/4], there is
(3.4)(Au)(t)=∫01G(t,s)a(s)f(s,u(s))ds+tα-1∑i=1∞αi∫01G(ξi,s)a(s)f(s,u(s))ds1-∑i=1∞αiξiα-1≥∫01γ(s)G(s,s)a(s)f(s,u(s))ds≥Nr∫1/43/4γ(s)G(s,s)a(s)ds=r1=‖u‖.
So,
(3.5)‖Au‖≥‖u‖,u∈P∩∂Ω1.
Therefore, by (ii) of Lemma 2.10, we complete the proof.Example 3.4.
Consider the problem(3.6)D0+3/2u(t)+u2+sin⁡⁡t4+15=0,0<t<1,u(0)=0,u(1)=∑i=1∞αiu(ξi),
where ∑i=1∞αiξi1/2=1/5.
A simple computation showedM≥1.4,N≈13.6649. Choosing r1=(1/70),r2=(9/10), we have
(3.7)f(t,u)=u2+sin⁡⁡t4+15≤1.2207≤Mr2,(t,u)∈[0,1]×[0,910],f(t,u)=u2+sin⁡⁡t4+15≥15≥Nr1,(t,u)∈[0,1]×[0,170].
With the use of Theorem3.3, problem (3.6) has at least one positive solutions u such that (1/70)≤∥u∥≤(9/10).Theorem 3.5.
Assume (H1)–(H3) hold, and there exist constants0<a<b<c such that the following assumptions hold:(A1)
f(t,u)<Ma for (t,u)∈[0,1]×[0,a],(A2)
f(t,u)≥Nb for (t,u)∈[1/4,3/4]×[b,c],(A3)
f(t,u)≤Mc for (t,u)∈[0,1]×[0,c], where M,N is defined in (*).Then, the boundary value problem (1.2) has at least three positive solutions u1,u2,u3 with
(3.8)‖u1‖<a,b<min⁡(1/4)≤t≤(3/4)⁡|u2|<‖u2‖≤c,a<‖u3‖≤c,min⁡(1/4)≤t≤(3/4)⁡|u3|<b.Proof.
We show that all the conditions of Lemma2.9 are satisfied.
Ifu∈P¯c, then ∥u∥≤c. Assumption (A3) implies f(t,u(t))≤Mc for 0≤t≤1. Consequently,
(3.9)‖Au‖=max⁡0≤t≤1⁡|∫01G(t,s)a(s)f(s,u(s))ds+tα-1∑i=1∞αi∫01G(ξi,s)a(s)f(s,u(s))ds1-∑i=1∞αiξiα-1|≤∫01G(s,s)a(s)f(s,u(s))ds+tα-1∑i=1∞αi∫01G(ξi,s)a(s)f(s,u(s))ds1-∑i=1∞αiξiα-1≤[∫01G(s,s)a(s)ds+∑i=1∞αi∫01G(ξi,s)a(s)ds1-∑i=1∞αiξiα-1]‖u‖≤‖u‖.
Hence, A:P¯c→P¯c. In the same way, if u∈P¯a, then assumption (A1) yields f(t,u(t))<Ma,0≤t≤1. Therefore, condition (C2) of Lemma 2.11 is satisfied.
To check condition (C1) of Lemma2.11, we choose u(t)=(b+c)/2,0≤t≤1. It is easy to see that u(t)=(b+c)/2∈P(θ,b,c),θ(u)=(θ(b+c))/2>b, and consequently, {u∈P(θ,b,d)∣θ(u)>b}≠∅ Hence, if u∈P(θ,b,c), then b≤u(t)≤c for (1/4)≤t≤(3/4). From assumption (A2), we have f(t,u(t))≥Nb for (1/4)≤t≤(3/4). So,
(3.10)θ(Au)=min⁡(1/4)≤t≤(3/4)⁡|(Au)(t)|≥∫01γ(s)G(s,s)a(s)f(s,u(s))ds>Nb∫1/43/4γ(s)G(s,s)a(s)ds=b=‖u‖θ(Au)>b, for all u∈P(θ,b,c).
This shows that condition (C1) of Lemma2.11 is also satisfied.
By Lemma2.11 and Remark 2.12, the boundary value problem (1.2) has at least three positive solutions u1,u2,andu3 with
(3.11)‖u1‖<a,b<min⁡(1/4)≤t≤(3/4)⁡|u2|,a<‖u3‖,min⁡(1/4)≤t≤(3/4)⁡|u3|<b.
The proof is complete.Example 3.6.
Consider the problem(3.12)D0+3/2u(t)+f(t,u)=0,0<t<1,u(0)=0,u(1)=∑i=1∞αiu(ξi),
where ∑i=1∞αiξi1/2=(1/5),
(3.13)f(t,u)={(t40)+14u2,u≤1,13+(t40)+u,u>1.
We haveM≥1.4,N≈13.6649. Choosing a=(1/14),b=1,c=36, there hold
(3.14)f(t,u)=t40+14u2≤0.097≤Ma,(t,u)∈[0,1]×[0,114],f(t,u)=13+t40+u≥14.025≥Nb≈13.7,(t,u)∈[14,34]×[1,36],f(t,u)≤13+t40+u≤48.025≤Mc≈50.4,(t,u)∈[0,1]×[0,36].
With the use of Theorem 3.5, problem (3.12) has at least three positive solutions u1,u2, and u3 with
(3.15)max⁡0≤t≤1⁡|u1(t)|<114,1<min⁡(1/4)≤t≤(3/4)⁡|u2(t)|<max⁡0≤t≤1⁡|u2(t)|≤36,114<max⁡0≤t≤1⁡|u3(t)|≤36,min⁡(1/4)≤t≤(3/4)⁡|u3(t)|<1.



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