# Existence Results and the Monotone Iterative Technique for Nonlinear Fractional Differential Systems with Coupled Four-Point Boundary Value Problems

Authors: Yujun Cui; Yumei Zou
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

By establishing a comparison result and using the monotone iterative technique combined with the method of upper and lower solutions, we investigate the
existence of solutions for nonlinear fractional differential systems with coupled four-point boundary value problems.


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## Body

## 1. Introduction

This paper discusses the coupled four-point boundary value problems(1)Dpx(t)+f(t,x(t),y(t))=0,t∈(0,1),1<p≤2,Dqy(t)+g(t,x(t),y(t))=0,t∈(0,1),1<q≤2,x(0)=y(0)=0,x(1)=ay(ξ),y(1)=bx(η),
where f and g:(0,1)×R×R→R are continuous, ξ,η∈(0,1), a,b>0 with ab<1, and Dpx denotes the Caputo fractional derivative of x with 1<p≤2 defined by
(2)Dpx(t)=I2-px′′(t)=1Γ(p)∫0t‍(t-s)1-px′′(s)ds.I2-p is the Riemann-Liouville fractional integral of order 2-p; see [1–4].It is well known that(3)I2-p(D2-px(t))=x(t)-∑k=01‍x(k)(0+)k!tk,D2-p(I2-px(t))=x(t).Fractional differential equation’s modeling capabilities in physics, chemistry, economics, and other fields, over the last few decades, have resulted in the rapid development of the theory of fractional differential equations; we refer the reader to the books [1–4]. On the other hand, the study of systems involving coupled boundary value problems is also important as such systems occur in the study of reaction-diffusion equations and Sturm-Liouville problems, for example, [5–16]. In [17–25], using the upper and lower solutions method and the monotone iterative method, the authors considered the existence of solutions of initial value problems and boundary value problems for fractional differential equations. But, as far as we know, there have been few papers which have considered the existence of solutions of (1) by means of the monotone iterative method.Motivated by the above papers, in this paper, we will investigate the existence of a solution of problem (1) by means of the upper and lower solutions method and the monotone iterative method. The novelty of this paper is that Caputo-type fractional differential systems involve two different fractional derivatives Dp and Dq and that the nonlinear terms f, g in the systems (1) involve unknown functions x(t) and y(t).In the following, we denote(4)E=C2([0,1],R),E1=C([0,1],(0,+∞)).


## 2. Preliminaries and Lemmas

In this section, we introduce the definition of the lower and upper solutions and present some existence and uniqueness results for linear problems together with comparison results for differential systems (1) which will be needed in the next section.Throughout this paper, we always assume that the following condition is satisfied:(
H
1
)   0<ab<1.Definition 1.
(
u
0
,
v
0
)
∈
E
×
E is called a lower system of solutions of differential system (1) if
(5)Dpu0(t)+f(t,u0(t),v0(t))≥0,t∈(0,1),Dqv0(t)+g(t,u0(t),v0(t))≥0,t∈(0,1),u0(0)≤0,v0(0)≤0,u0(1)≤av0(ξ),v0(1)≤bu0(η).
Analogously, (α0,β0)∈E×E is called an upper system of solutions if it satisfies the reversed inequalities.Ifu0(t)≤α0(t) and v0(t)≤β0(t), t∈[0,1], we say that (u0,v0) and (α0,β0) are ordered lower and upper system of solutions of (1). In what follows, we assume that (u0,v0) and (α0,β0) are ordered lower and upper system of solutions of (1) and define the sector
(6)Ω={≤(x(t),y(t))≤(α0(t),β0(t)),t∈[0,1](x,y)∈E×E:(u0(t),v0(t))iiiiiiiiiiiiiiiiiiiiiii≤(x(t),y(t))iiiiiiiiiiiiiiiiiiiiiii≤(α0(t),β0(t)),t∈[0,1]},
where the vectorial inequalities mean that the same inequalities hold between their corresponding components.Lemma 2 (see [17]).
Letz(t)∈E and r(t)∈E1. If z(t) satisfies the inequality
(7)-Dpz(t)≤-r(t)z(t),p∈(1,2],t∈(0,1),z(0)≤0,z(1)≤0,
then z(t)≤0, ∀t∈[0,1].We have the following important result.Lemma 3 (comparison theorem).
LetM(t),N(t)∈E1 be given. Assume that x(t),y(t) satisfy
(8)-Dpx(t)≤-M(t)x(t),t∈(0,1),-Dqy(t)≤-N(t)y(t),t∈(0,1),x(0)≤0,y(0)≤0,x(1)≤ay(ξ),y(1)≤bx(η).
Then x(t)≤0, y(t)≤0,  ∀t∈[0,1].Proof.
Suppose the contrary. By Lemma2, We consider the following three possible cases.
Case  1. Consider x(1)≤0 and y(1)>0. By Lemma 2, x(t)≤0, ∀t∈[0,1]. Then y(1)≤bx(η)≤0 which contradicts y(1)>0.
Case  2. Consider y(1)≤0 and x(1)>0. By Lemma 2, y(t)≤0, ∀t∈[0,1]. Then x(1)≤ay(ξ)≤0 which contradicts x(1)>0.
Case  3. Consider x(1)>0 and y(1)>0. By Lemma 2, we have x(1)=max⁡t∈[0,1]⁡x(t)>0 and y(1)=max⁡t∈[0,1]⁡y(t)>0. We only prove that x(1)=max⁡t∈[0,1]⁡x(t)>0. If not, x(t) has a local positive maximum at some t0∈(0,1) such that x(t0)=max⁡t∈[0,1]⁡x(t)>0. Then, by Theorem 2.1 in [21], we have the fact that the Caputo derivative of the function x is nonpositive at the point t0. Thus, (9)0≤-Dpx(t0)≤-M(t0)x(t0)<0,
which is a contradiction. Furthermore, considering the boundary condition y(1)≤bx(η), there exists t1∈[0,η) such that
(10)x(t)≤0,t∈[0,t1];x(t)≥0,t∈[t1,1].
A similar proof, for y(t), gives us that there exists t2∈[0,ξ) such that
(11)y(t)≤0,t∈[0,t2];y(t)≥0,t∈[t2,1].
It follows from (10) and (11) that
(12)x(1)≤ay(ξ)≤ay(1)≤abx(η)≤abx(1),
which implies that ab≥1, a contradiction. Hence x(t)≤0,  y(t)≤0,  ∀t∈[0,1].Corollary 4.
LetM(t),N(t)∈E1 be given. Assume that x(t),y(t) satisfy
(13)-Dpx(t)=-M(t)x(t),t∈(0,1),-Dqy(t)=-N(t)y(t),t∈(0,1),x(0)=0,y(0)=0,x(1)=ay(ξ),y(1)=bx(η).
Then x(t)=y(t)=0, ∀t∈[0,1].Lemma 5.
Letρ, σ∈C[0,1], then the linear differential system with coupled four-point boundary value problem
(14)Dpx(t)+ρ(t)=0,t∈(0,1),1<p≤2,Dqy(t)+σ(t)=0,t∈(0,1),1<q≤2,x(0)=y(0)=0,x(1)=ay(ξ),y(1)=bx(η)
has integral representation
(15)x(t)=∫01‍G1(t,s)ρ(s)ds+∫01‍H1(t,s)σ(s)ds,y(t)=∫01‍G2(t,s)σ(s)ds+∫01‍H2(t,s)ρ(s)ds,
where
(16)G1(t,s)=Gp(t,s)+abξt1-abξηGp(η,s),H1(t,s)=at1-abξηGq(ξ,s),G2(t,s)=Gq(t,s)+abηt1-abξηGq(ξ,s),H2(t,s)=bt1-abξηGp(η,s),Gp(t,s)={t(1-s)p-1-(t-s)p-1Γ(p),0≤s≤t≤1,t(1-s)p-1Γ(p),0≤t≤s≤1.Proof.
It follows from [21] that (14) is equivalent to the system of integral equations
(17)x(t)=x(1)t+∫01‍Gp(t,s)ρ(s)ds,t∈[0,1],y(t)=y(1)t+∫01‍Gq(t,s)σ(s)ds,t∈[0,1].
By coupled four-point boundary value conditions of problem (14), we have
(18)y(1)=bx(η)=bηx(1)+b∫01‍Gp(η,s)ρ(s)ds,(19)x(1)=ay(ξ)=aξy(1)+a∫01‍Gq(ξ,s)σ(s)ds.
After simple computation, we get
(20)x(1)=abξ1-abξη∫01‍Gp(η,s)ρ(s)ds+a1-abξη∫01‍Gq(ξ,s)σ(s)ds,(21)y(1)=abη1-abξη∫01‍Gq(ξ,s)σ(s)ds+b1-abξη∫01‍Gp(η,s)ρ(s)ds.
Substituting (20) into (18) and (21) into (19), respectively, we obtain the desired results.Now we enunciate the following existence and uniqueness results for differential system:(22)Dpx(t)-M(t)x(t)+ρ(t)=0,t∈(0,1),1<p≤2,Dqy(t)-N(t)y(t)+σ(t)=0,t∈(0,1),1<q≤2,x(0)=y(0)=0,x(1)=ay(ξ),y(1)=bx(η),
where M, N∈E1.Lemma 6.
LetM, N∈E1. Then differential system (22) has a unique solution.Proof.
Indeed, by Lemma5, differential system (22) is equivalent to the operator equation
(23)(x,y)=T(x,y)+(ρ~,σ~),
where
(24)T(x,y)(t)=(-∫01‍G1(t,s)M(s)x(s)ds-∫01‍H1(t,s)N(s)y(s)ds,-∫01‍G2(t,s)N(s)y(s)(s)ds-∫01‍H2(t,s)M(s)x(s)ds),ρ~(t)=∫01‍G1(t,s)ρ(s)ds+∫01‍H1(t,s)σ(s)ds,σ~(t)=∫01‍G2(t,s)σ(s)ds+∫01‍H2(t,s)ρ(s)ds.
We apply the Fredholm theorem to find the unique solution of differential system (22). By using standard arguments, we can easily show that T:C[0,1]×C[0,1]→C[0,1]×C[0,1] is linear completely continuous. Also, by Corollary 4, the operator equation (x,y)=T(x,y) has only the zero solution. Thus, for given (ρ~,σ~)∈C[0,1]×C[0,1], operator equation (23) has a unique solution in C[0,1]×C[0,1], by the Fredholm theorem. This ends the proof.


## 3. Main Results

In this section, we prove the existence of extremal solutions of differential system (1).Theorem 7.
Assume thatf∈C([0,1]×R×R,R), g∈C([0,1]×R×R,R). Let (u0,v0) and (α0,β0) be ordered lower and upper system of solutions of (1). In addition, we assume that
(
H
2
)  f(t,x,y) is nondecreasing in y and there exists M(t)∈E1 such that
(25)f(t,x1,y)-f(t,x2,y)≥-M(t)(x1-x2),
where u0(t)≤x2≤x1≤α0(t),  v0(t)≤y≤β0(t);
(
H
3
)  g(t,x,y) is nondecreasing in x and there exists N(t)∈E1 such that
(26)g(t,x,y1)-g(t,x,y2)≥-N(t)(y1-y2),
where v0(t)≤y2≤y1≤β0(t), u0(t)≤x≤α0(t).Then differential system (1) has extremal solutions in the section Ω.Proof.
Let us define two sequences{(un,vn),(αn,βn)} by relations
(27)Dpun+1(t)-M(t)un+1(t)+f(t,un(t),vn(t))+M(t)un(t)=0,t∈(0,1),Dqvn+1(t)-N(t)vn+1(t)+g(t,un(t),vn(t))+N(t)vn(t)=0,t∈(0,1),un+1(0)=vn+1(0)=0,un+1(1)=avn+1(ξ),vn+1(1)=bun+1(η),Dpαn+1(t)-M(t)αn+1(t)+f(t,αn(t),βn(t))+M(t)αn(t)=0,t∈(0,1),Dqβn+1(t)-N(t)βn+1(t)+g(t,αn(t),βn(t))+N(t)βn(t)=0,t∈(0,1),αn+1(0)=βn+1(0)=0,αn+1(1)=aβn+1(ξ),βn+1(1)=bαn+1(η),
for n=1,2,…. Note that {(u1,v1),(α1,β1)} are well defined, by Lemma 6. First, we show that
(28)(u0,v0)≤(u1,v1)≤(α1,β1)≤(α0,β0).
Let z=u0-u1, w=v0-v1. This and the assumption that (u0,v0) is a lower system of solutions of (1) yield
(29)-Dpz(t)≤-M(t)z(t),t∈(0,1),-Dqw(t)≤-N(t)w(t),t∈(0,1),z(0)≤0,w(0)≤0,z(1)≤aw(ξ),w(1)≤bz(η).
Hence, u0(t)≤u1(t) and v0(t)≤v1(t), t∈[0,1], by Lemma 3. By a similar way, we can show that α1(t)≤α0(t) and β1(t)≤β0(t), t∈[0,1]. Now we put z=u1-α1, w=v1-β1. Hence, in view of assumptions (H2), (H3), we have
(30)-Dpz(t)=-M(t)z(t)-f(t,α0(t),β0(t))-M(t)α0(t)+f(t,u0(t),v0(t))+M(t)u0(t)≤-M(t)z(t)-f(t,α0(t),β0(t))-M(t)α0(t)+f(t,u0(t),β0(t))+M(t)u0(t)≤-M(t)z(t),t∈(0,1),-Dqw(t)≤-N(t)w(t),t∈(0,1),z(0)=0,w(0)=0,z(1)=aw(ξ),w(1)=bz(η).
This and Lemma 3 prove that (u1,v1)≤(α1,β1), t∈[0,1], so, relation (28) holds.
Now we show that(u1,v1) is a lower system of solution of problem (1). Note that
(31)Dpu1(t)+f(t,u1(t),v1(t))≥M(t)u1(t)-f(t,u0(t),v0(t))-M(t)u0(t)+f(t,u1(t),v0(t))≥0,t∈(0,1),Dqv1(t)+g(t,u1(t),v1(t))≥0t∈(0,1),u1(0)=v1(0)=0,u1(1)=av1(ξ),v1(1)=bu1(η),
by assumptions (H2), (H3). It proves that (u1,v1) is a lower system of solution of (1). Similarly, we can prove that (α1,β1) is an upper system of solution of problem (1).
By mathematical induction we can show that(32)(u0,v0)≤(u1,v1)≤⋯≤(un,vn)≤(αn,βn)≤⋯≤(α1,β1)≤(α0,β0)
for t∈[0,1] and n=1,2,…. Employing standard arguments we see that the sequences {(un,vn),(αn,βn)} converge to their limit functions (u*,v*), (α*,β*), respectively. Indeed, (u*,v*) and (α*,β*) are solutions of problem (1) and (u0(t),v0(t))≤(u*,v*)≤(α*,β*)≤(α0,β0) on [0,1].
We need to show now that(u*,v*) and (α*,β*) are extremal solutions of problem (1) in the segment Ω. To prove it, we assume that (x,y) is another solution of problem (1) and (un,vn)≤(x(t),y(t))≤(αn(t),βn(t)), t∈[0,1] for some positive integer n. Put z=un+1-x, w=vn+1-y. Hence, in view of assumptions (H2), (H3), we have
(33)-Dpz(t)=-M(t)z(t)-f(t,x(t),y(t))-M(t)x(t)+f(t,un(t),vn(t))+M(t)un(t)≤-M(t)z(t)-f(t,x(t),y(t))-M(t)x(t)+f(t,un(t),y(t))+M(t)un(t)≤-M(t)z(t),t∈(0,1),-Dqw(t)≤-N(t)w(t),t∈(0,1),z(0)=0,w(0)=0,z(1)=aw(ξ),w(1)=bz(η).
Hence, (un+1(t),vn+1(t))≤(x(t),y(t)), t∈[0,1], by Lemma 3. By a similar way, we can show that (x(t),y(t))≤(αn+1(t),βn+1(t)), t∈[0,1]. So by induction, (un(t),vn(t))≤(x(t),y(t))≤(αn(t),βn(t)), t∈[0,1] on [0,1] for all n. Taking the limit as n→+∞, we conclude (u*(t),v*(t))≤(x(t),y(t))≤(α*(t),β*(t)), t∈[0,1]. That is, (u*(t),v*(t)) and (α*(t),β*(t)) are extremal systems of solutions of (1) in Ω.


## 4. Example

Consider the following problems:(34)D5/4x(t)+sint-2x(t)+18y3(t)t3=0,t∈(0,1),D7/4y(t)-y3(t)+x2(t)+1=0,t∈(0,1),x(0)=y(0)=0,x(1)=14y(12),y(1)=2x(34).
Obviously,
(35)f(t,x,y)=sint-2x+18y3t3,g(t,x,y)=-y3+x2+1.
Take (u0(t),v0(t))=(0,0), (α0(t),β0(t))=(t,2); then
(36)D5/4u0(t)+sint-2u0(t)+18v03(t)t3=sint≥0,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiit∈(0,1),D7/4v0(t)-v03(t)+u02(t)+1=1≥0,t∈(0,1),u0(0)=v0(0)=0,u0(1)=14v0(12),v0(1)=2u0(34),D5/4α0(t)+sint-2α0(t)+18β03(t)t3=sint-2t+t3≤0,t∈(0,1),D7/4β0(t)-β03(t)+α02(t)+1=-7+t2≤0,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiit∈(0,1),α0(0)=0,β0(0)≥0,α0(1)≥14β0(12),β0(1)≥2α0(34).
It shows that (u0(t),v0(t)) and (α0(t),β0(t)) are lower and upper systems of solutions of (34).On the other hand, it is easy to verify that conditions(H2), (H3) hold for M(t)=2 and N(t)=12.By Theorem7, problem (34) has an extremal system of solutions (u*(t),v*(t)) and (α*(t),β*(t)), which can be obtained by taking limits from some iterative sequences.



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