# Positive Solutions for Multipoint Boundary Value Problems for Singular Fractional Differential Equations

Authors: Mohamed Jleli; Erdal Karapinar; Bessem Samet
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

A class of nonlinear multipoint  boundary value problems for singular fractional differential equations is considered. By means of a coupled fixed point theorem on ordered sets, some results on the existence and uniqueness of positive solutions are obtained.


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## Body

## 1. Introduction

Fractional calculus is used to formulate different phenomena in physics, biology, medicine, and so forth. For more details on the applications of fractional calculus, we refer the reader to [1–4]. On the other hand, some basic theory for the initial value problems of fractional differential equations involving Riemann-Liouville differential operator has been discussed by Lakshmikantham [5–7], Bai and Lü [8], El-Sayed et al. [9, 10], Bai [11, 12], Zhang [13], and so forth.Very recently, Liang and Zhang [14] considered the following m-point boundary value problem:
(1)D0+αu(t)+f(t,u(t))=0,0<t<1,2<α≤3,u(0)=u′(0)=0,u′(1)=∑i=1m-2‍βiu′(ξi),
where D0+α is the Riemann-Liouville fractional derivative, 0<ξ1<ξ2<⋯<ξm-2<1 satisfies 0<∑i=1m-2‍βiξiα-2<1, and f:[0,1]×[0,+∞)→[0,+∞) is continuous and nondecreasing with respect to the second variable. Under some hypotheses, using a fixed point theorem on ordered sets, the authors established the existence and uniqueness of solution to such problem.Motivated by the abovementioned work, in this paper, we deal with the following multipoint boundary value problem:(2)D0+αu(t)+f(t,u(t),u(t))=0,0<t<1,2<α≤3,(3)u(0)=u′(0)=0,u′(1)=∑i=1m-2‍βiu′(ξi),
where f:(0,1]×[0,+∞)×[0,+∞)→[0,+∞) is continuous and lim⁡t→0+f(t,·,·)=+∞ (f is singular at t=0).


## 2. Preliminaries

The following preliminaries will be useful later.Definition 1.
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function φ:(0,+∞)→ℝ is given by
(4)D0+αφ(t)=1Γ(n-α)(ddt)(n)∫0t‍φ(s)(t-s)α-n+1ds,
where n=[α]+1, [α] denotes the integer part of number α, provided that the right side is pointwise defined on (0,+∞). Here, Γ is the Euler gamma function defined by
(5)Γ(α)=∫0+∞‍tα-1e-tdt.Definition 2.
The Riemann-Liouville fractional integral of orderα>0 of a given function φ:(0,+∞)→ℝ is defined by
(6)I0+αφ(t)=1Γ(α)∫0t‍(t-s)α-1φ(s)ds,
provided that the right side is defined on (0,+∞).From the definition of the Riemann-Liouville derivative, we can obtain the following statement.Lemma 3 (see [15]).
Letα>0. If one assumes that u∈C(0,1)∩L(0,1), then the fractional differential equation
(7)D0+αu(t)=0
has u(t)=c1tα-1+c2tα-2+⋯+cNtα-N, ci∈ℝ, i=1,2,…,N as unique solutions, where N is the smallest integer greater than or equal to α.Lemma 4 (see [15]).
Assume thatu∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then
(8)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cNtα-N,
for some ci∈ℝ, i=1,2,…,N, where N is the smallest integer greater than or equal to α.Lemma 5 (see [14]).
Assume that∑i=1m-2‍βiξiα-2≠1. If h∈C([0,1]), then the boundary value problem
(9)D0+αu(t)+h(t)=0,0<t<1,2<α≤3,u(0)=u′(0)=0,u′(1)=∑i=1m-2‍βiu′(ξi),
has a unique solution
(10)u(t)=∫01‍G(t,s)h(s)ds+tα-1(α-1)(1-∑i=1m-2βiξiα-2)×∑i=1m-2‍βi∫01‍H(ξi,s)h(s)ds,
where
(11)G(t,s)={tα-1(1-s)α-2-(t-s)α-1Γ(α),if0≤s≤t≤1,tα-1(1-s)α-2Γ(α),if0≤t≤s≤1,H(t,s)=∂G(t,s)∂t.It is clear that, for all0≤t≤s≤1, the function H(t,s)≥0. For 0≤s≤t≤1, we have ts≤s; then tα-2(1-s)α-2-(t-s)α-2≥0. Hence for 0≤t≤s≤1 one has H(t,s)≥0. The following properties of G will be used later.Lemma 6 (see [14]).
The following properties hold.(i)
Gis a nonnegative continuous function on [0,1]×[0,1];(ii)
G(·,s) is strictly increasing for all s∈[0,1].Let(X,⪯) be a partially ordered set endowed with a metric d. Let F:X×X→X be a given mapping.Definition 7.
One says that(X,⪯) is directed if, for every (x,y)∈X×X, there exists z∈X such that x⪯z, y⪯z and there exists w∈X such that x≽w, y≽w.Definition 8.
We say that(X,⪯,d) is regular if the following conditions hold. (c1)
if{xn} is a nondecreasing sequence in X such that xn→x∈X, then xn⪯x for all n;(c2)
if{yn} is a decreasing sequence in X such that yn→y∈X, then yn≽y for all n.Example 9.
LetX=C([0,T]), T>0 be the set of real continuous functions on [0,T]. We endow X with the standard metric d given by
(12)d(u,v)=max⁡0≤t≤T|u(t)-v(t)|,u,v∈X.
We define the partial order ⪯ on X by
(13)u,v∈X,u⪯v⟺u(t)≤v(t)∀t∈[0,T].
Let x,y∈X. For z=max⁡{x,y}, that is, z(t)=max⁡{x(t),y(t)}, for all t∈[0,T], we have x⪯z and y⪯z. For w=min⁡{x,y}, that is, w(t)=min⁡{x(t),y(t)}, for all t∈[0,T], we have x≽w and y≽w. This implies that (X,⪯) is directed. Now, let {xn} be a nondecreasing sequence in X such that d(xn,x)→0 as n→∞, for some x∈X. Then, for all t∈[0,T], {xn(t)} is a nondecreasing sequence of real numbers converging to x(t). Thus, we have xn(t)≤x(t), for all n, that is, xn⪯x for all n. Similarly, if {yn} is a decreasing sequence in X such that d(yn,y)→0 as n→∞, for some y∈X, we get that yn≽y for all n. Then, we proved that (X,⪯,d) is regular.Definition 10 (see [16]).
An element(x,y)∈X×X is called a coupled fixed point of F if
(14)F(x,y)=x,F(y,x)=y.Definition 11 (see [16]).
One says thatF has the mixed monotone property if
(15)(x,y),(u,v)∈X×X,x⪯u,y≽v⟹F(x,y)⪯F(u,v).Denote byΦ the set of functions φ:[0,+∞)→[0,+∞) satisfying the following.(Φ1)
φ is continuous;(Φ2)
φ is nondecreasing;(Φ3)
φ-1({0})={0}.The following two lemmas are fundamental for the proof of our main result.Lemma 12 (see [17]).
Let(X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let F:X×X→X be a mapping having the mixed monotone property on X such that
(16)ψ(d(F(x,y),F(u,v)))≤ψ(max⁡{d(x,u),d(y,v)})-φ(max⁡{d(x,u),d(y,v)}),
for all x,y,u,v∈X with x≽u and y⪯v, where ψ,φ∈Φ. Suppose also that (X,⪯,d) is regular and there exist x0,y0∈X such that
(17)x0⪯F(x0,y0),y0≽F(y0,x0).
Then, F has a coupled fixed point (x*,y*)∈X×X. Moreover, if {xn} and {yn} are the sequences in X defined by
(18)xn+1=F(xn,yn),yn+1=F(yn,xn),n=0,1,…,
then
(19)lim⁡n→∞d(xn,x*)=lim⁡n→∞d(yn,y*)=0.Lemma 13 (see [17]).
Adding to the hypotheses of Lemma12 the condition (X,⪯) is directed; one obtains uniqueness of the coupled fixed point. Moreover, one has the equality x*=y*.


## 3. Main Result

Let Banach spaceE=C([0,1]) be endowed with the norm ∥u∥∞=max⁡0≤t≤1|u(t)|. We define the partial order ⪯ on E by
(20)u,v∈E,u⪯v⟺u(t)≤v(t)∀t∈[0,1].
In Example 9, we proved that (E,⪯) with the classic metric given by
(21)d(u,v)=∥u-v∥∞,u,v∈E
satisfies the following properties: (E,⪯) is directed and (E,⪯,d) is regular.Define the closed coneP⊂E by
(22)P={u∈E:u≽0},
where 0 denotes the zero function.Definition 14.
One says that(u-,u+)∈C([0,1])×C([0,1]) is a coupled lower and upper solution to (2)-(3) if, for all t∈[0,1], one has
(23)u-(t)≤∫01‍G(t,s)f(s,u-(s),u+(s))ds+tα-1(α-1)(1-∑i=1m-2βiξiα-2)×∑i=1m-2βi∫01H(ξi,s)f(s,u-(s),u+(s))ds,u+(t)≥∫01‍G(t,s)f(s,u+(s),u-(s))ds+tα-1(α-1)(1-∑i=1m-2βiξiα-2)×∑i=1m-2βi∫01H(ξi,s)f(s,u+(s),u-(s))ds.The main result of this paper is the following.Theorem 15.
Suppose that the following conditions hold.(i)
0<ξ1<ξ2<⋯<ξm-2<1 satisfies 0<∑i=1m-2‍βiξiα-2<1 with βi>0 for i=1,…,m-2;(ii)
f:(0,1]×[0,+∞)×[0,+∞)→[0,+∞) is continuous, lim⁡t→0+f(t,·,·)=+∞;(iii)
there existsσ∈(0,1) such that t↦tσf(t,x,y) is continuous on [0,1] for all x,y∈[0,+∞);(iv)
there exists(24)0<λ≤[(1-∑i=1m-2βiξiα-σ-11-∑i=1m-2βiξiα-2-α-1α-σ)Γ(1-σ)(α-1)Γ(α-σ)]-1
such that for all x,y,z,w∈[0,+∞) with x≥z, y≤w and t∈[0,1],
(25)0≤tσ(f(t,x,y)-f(t,z,w))≤λη(max⁡{(x-z),(w-y)}),
where η:[0,+∞)→[0,+∞) is nondecreasing, β:u↦u-η(u)∈Φ;(v)
equations (2)-(3) has a coupled lower and upper solution (u-,u+)∈P×P.
Then,(I)
the boundary value problem (2)-(3) has a unique positive solution u*∈C([0,1]);(II)
the sequences{un} and {vn} defined by
(26)vvu0=u-,un+1=∫01‍G(t,s)f(s,un(s),vn(s))dsvivv+tα-1(α-1)(1-∑i=1m-2βiξiα-2)vvvv×∑i=1m-2‍βi∫01‍H(ξi,s)f(s,un(s),vn(s))ds,vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvn=0,1,…,vvv0=u+,vn+1=∫01‍G(t,s)f(s,vn(s),un(s))dsvivv+tα-1(α-1)(1-∑i=1m-2βiξiα-2)vvvv×∑i=1m-2‍βi∫01‍H(ξi,s)f(s,vn(s),un(s))ds,vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvn=0,1,…,converge uniformly to u*.Proof.
Suppose thatu is a solution to the boundary value problem (2)-(3). Then, from Lemma 5, we have
(27)u(t)=∫01‍G(t,s)f(s,u(s),u(s))ds+tα-1(α-1)(1-∑i=1m-2βiξiα-2)×∑i=1m-2‍βi∫01‍H(ξi,s)f(s,u(s),u(s))ds,
for all t∈[0,1].
Consider the operatorF defined by
(28)F(u,v)(t)=∫01‍G(t,s)f(s,u(s),v(s))ds+tα-1(α-1)(1-∑i=1m-2βiξiα-2)×∑i=1m-2‍βi∫01‍H(ξi,s)f(s,u(s),v(s))ds,
for all t∈[0,1], for all u,v∈P. From (iii) and Lemma 6, we have that F(P×P)⊂P.
Let(x,y),(u,v)∈P×P such that x⪯u and y≽v. From (25), we have
(29)sσf(s,x(s),y(s))≤sσf(s,u(s),v(s)),∀s∈[0,1].
Since the operator F is linear and increasing with respect to the function f, we deduce that
(30)F(x,y)(t)≤F(u,v)(t),∀s∈[0,1].
This implies that F has the mixed monotone property with respect to the partial order ⪯ given by (20).
In the sequel, we denote(31)ξ=∑i=1m-2‍βiξiα-2.
Let (x,y),(u,v)∈P×P such that x≽u and y⪯v. For all t∈[0,1], using (25) and Lemma 6, we have
(32)|F(x,y)(t)-F(u,v)(t)|=∫01‍G(t,s)[f(s,x(s),y(s))-f(s,u(s),v(s))]ds+tα-1(α-1)(1-ξ)×∑i=1m-2‍βi∫01‍H(ξi,s)[f(s,x(s),y(s))vvvvvvvvvvvvvvvvvvv-f(s,u(s),v(s))]ds=∫01‍G(t,s)s-σsσ[f(s,x(s),y(s))-f(s,u(s),v(s))]ds+tα-1(α-1)(1-ξ)×∑i=1m-2‍βi∫01‍H(ξi,s)s-σsσ[f(s,x(s),y(s))vvvvvvvvvvvvvvvvvvvvvvvvvvv-f(s,u(s),v(s))]ds≤∫01‍G(t,s)s-σλη(max⁡{x(s)-u(s),v(s)-y(s)})ds+tα-1(α-1)(1-ξ)×∑i=1m-2‍βi∫01‍H(ξi,s)s-σλη×(max⁡{x(s)-u(s),v(s)-y(s)})ds≤λη(max⁡{d(x,u),d(y,v)})×max⁡z∈[0,1](∫01‍G(z,s)s-σds+zα-1(α-1)(1-ξ)vvvvvvvvivvvv×∑i=1m-2‍βi∫01‍H(ξi,s)s-σds).
Thus, for all (x,y),(u,v)∈P×P such that x≽u and y⪯v, we have
(33)d(F(x,y),F(u,v))≤λχη(max⁡{d(x,u),d(y,v)}),
where
(34)χ=max⁡z∈[0,1](∫01G(z,s)s-σds+zα-1(α-1)(1-ξ)×∑i=1m-2βi∫01‍H(ξi,s)s-σds)
Now, let z∈[0,1]. We have
(35)∫01‍G(z,s)s-σds=∫0z‍zα-1(1-s)α-2-(z-s)α-1Γ(α)s-σds+∫z1‍zα-1(1-s)α-2Γ(α)s-σds=1Γ(α)(zα-1∫01‍(1-s)α-2s-σdsvvvvvvvvvv-∫0z‍(z-s)α-1s-σds)=zα-1Γ(α)(∫01‍(1-s)α-2s-σdsvvvvvvvvvv-z1-σ∫01‍(1-s)α-1s-σds)=zα-1Γ(α)(B(1-σ,α-1)-z1-σB(1-σ,α)),
where B denotes the beta function. Recall that beta and gamma functions satisfy the following properties:
(36)B(a,b)=∫01‍(1-s)b-1sa-1ds=Γ(a)Γ(b)Γ(a+b),Γ(a+1)=aΓ(a)fora,b>0.
Thus we have
(37)∫01‍G(z,s)s-σds=zα-1Γ(α)(1-α-1α-σz1-σ)B(1-σ,α-1).
Using the same computation as above, we can show that, for all i=1,…,m-2, we have
(38)∫01‍H(ξi,s)s-σds=α-1Γ(α)ξiα-2(1-ξi1-σ)B(1-σ,α-1).
Now, (37) and (38) give us that
(39)∫01‍G(z,s)s-σds+zα-1(α-1)(1-ξ)∑i=1m-2‍βi∫01‍H(ξi,s)s-σds=zα-1Γ(α)(1-ξ~1-ξ-α-1α-σz1-σ)B(1-σ,α-1),
where
(40)ξ~=∑i=1m-2‍βiξiα-σ-1≤ξ.
But the maximum of the above function depending in z is attained at z=1. Then
(41)χ=(1-ξ~1-ξ-α-1α-σ)B(1-σ,α-1)Γ(α)=(1-ξ~1-ξ-α-1α-σ)Γ(1-σ)(α-1)Γ(α-σ).
Now, it follows from (33), (41), and condition (iv) that
(42)ψ(d(F(x,y),F(u,v)))≤ψ(max⁡{d(x,u),d(y,v)})-φ(max⁡{d(x,u),d(y,v)}),
with ψ(t)=t and φ≡β.
Finally, taking(x0,y0)=(u-,u+)∈P×P, we have from condition (v) that x0⪯F(x0,y0) and y0≽F(y0,x0).
Now, from Lemmas12 and 13, there exists a unique u*∈P such that u*=F(u*,u*); that is, u* is the unique positive solution to (2)-(3). The convergence of the sequences {un} and {vn} to u* follows immediately from (19). This makes end to the proof.Example 16.
Consider the fractional boundary value problem(43)D0+5/2u(t)+(t-1/2)22t(u(t)+1u(t)+1)=0,0<t<1,(44)u(0)=u′(0)=0,2u′(1)=u′(14).
In this case, we have
(45)2<α=52<3,β1=12,ξ1=14,ξ=14,ξ~=18,f(t,x,y)=(t-1/2)22t(x+1y+1),vvvvvvvvvvvvv∀t∈(0,1],x,y≥0.
Note that f is continuous on (0,1]×[0,+∞)×[0,+∞) and lim⁡t→0+f(t,·,·)=+∞. Let σ=λ=1/2 and η(t)=(1/2)t. For all x,y,z,w∈[0,+∞) with x≥z, y≤w and t∈[0,1], we have
(46)0≤t1/2(f(t,x,y)-f(t,z,w))=(t-1/2)22[(x-z)+(w-y)(y+1)(w+1)]≤(t-12)2max⁡{x-z,w-y}≤14max⁡{x-z,w-y}=λη(max⁡{x-z,w-y}).
On the other hand, we have
(47)0<λ=12<[(1-β1ξ1α-σ-11-β1ξ1α-2-α-1α-σ)Γ(1-σ)(α-1)Γ(α-σ)]-1=185π.
Now, sinceG, H, and f are nonnegative continuous functions, it is easy to prove that u-≡0 is a lower solution to (43)-(44). Moreover, for all c>0, we use (37), (38), and (41) to obtain
(48)∫01‍G(t,s)f(s,c,0)ds+tα-1(α-1)(1-ξ)β1∫01‍H(ξ1,s)f(s,c,0)ds≤c+18(∫01‍G(t,s)s-σdsvvvvvvvvv+tα-1(α-1)(1-ξ)β1∫01‍H(ξi,s)s-σds)≤c+18(1-ξ~1-ξ-α-1α-σ)Γ(1-σ)(α-1)Γ(α-σ)=5π144(c+1).
Then, an upper solution u+ to (43)-(44) will be any constant c>0 satisfying
(49)5π144(c+1)≤c.
Finally, from Theorem 15, Problems (43) and (44) have a unique positive solution.



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