# A Contraction Fixed Point Theorem in  Partially Ordered Metric Spaces and Application to Fractional Differential Equations

Authors: Xiangbing Zhou; Wenquan Wu; Hongjiang Ma
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We generalize a fixed point theorem in partially ordered complete metric spaces in the study of A. Amini-Harandi and H. Emami (2010). We also give an application on the existence and uniqueness of the positive solution of a multipoint boundary value problem with fractional derivatives.


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## Body

## 1. Introduction

Let𝒮 denote the class of those functions β:[0,+∞)→[0,1) which satisfies the condition(1.1)β(tn)→1impliestn→0.
The following generalization of Banach's contraction principle is due to Geraghty [1].Theorem 1.1.
Let(M,d) be a complete metric space and let f:M→M be a map. Suppose there exists β∈𝒮 such that for each x,y∈M,(1.2)d(f(x),f(y))≤β(d(x,y))d(x,y).
Then f has a unique fixed point z∈M, and {fn(x)} converges to z, for each x∈M.And then, Amini-Harandi and  Emami [2] proved a version of Theorem 1.1 in the context of partially ordered complete metric spaces.Theorem 1.2.
Let(M,⪯) be a partially ordered set and suppose that there exists a metric d in M such that (M,d) is a complete metric space. Let f:M→M be a nondecreasing mapping, and there exists an element x0∈M with x0⪯f(x0). Suppose that there exists β∈𝒮 such that
(1.3)d(f(x),f(y))≤β(d(x,y))d(x,y),foreachx,y∈Mwithx≽y.
Assume that either f is continuous or M is such that
(1.4)ifanincreasingsequence{xn}→xinM,thenxn⪯x,∀n.
Besides, if
(1.5)foranyx,y∈M,thereexistsz∈Mwhichiscomparabletoxandy,
Then f has a unique fixed point.Theorem2.2 was also applied to obtain  the existence and uniqueness  of the solution of a periodic boundary value problem by Amini-Harandi and Emami [2] and a singular fractional three-point boundary value problem by  Cabrera et al. [3].


## 2. Main Results

In this section, we firstly define a class of functions𝒜 by β:[0,+∞)→[0,+∞), and there exists a constant K>0 such that
(2.1)supt∈[0,+∞){β(t)}≤K.Remark 2.1.
Function classes𝒜 include all bounded functions on [0,+∞) with upper bound K which are more extensive than those of 𝒮. For example, β(t)=Ksint∈𝒜 but not in 𝒮.Now, we give an extended version of Theorem1.2 in the context of partially ordered complete metric spaces.Theorem 2.2.
Let(M,⪯) be a partially ordered set and suppose that there exists a metric d in M such that (M,d) is a complete metric space. Let f:M→M be a nondecreasing mapping, and there exists s an element x0∈M with x0⪯f(x0). Suppose that there exists a constant θ∈(0,1/K) and β∈𝒮 such that
(2.2)d(f(x),f(y))≤θβ(θd(x,y))d(x,y),foreachx,y∈Mwithx≽y.
Assume that either f is continuous or
(2.3)ifanincreasingsequence{xn}→xinM,thenxn⪯x,∀n.
Besides, if
(2.4)foranyx,y∈M,thereexistsz∈Mwhichiscomparabletoxandy,
then f has a unique fixed point.Proof.
We first show thatf has a fixed point. Since x0⪯f(x0) and f is an increasing function, we obtain by induction that
(2.5)x0⪯f(x0)⪯f2(x0)⪯f3(x0)⪯⋯⪯fn(x0)⪯⋯.
Put xn+1=fn(x0), n=1,2,… For each integer n≥1, from (2.5), we have xn⪯xn+1, then by (2.2)
(2.6)d(xn+1,xn)=d(fn(x0),fn-1(x0))≤θβ(θd(xn,xn-1))d(xn,xn-1)≤Kθd(xn,xn-1).
If there exists n0∈ℕ such that d(xn0,xn0-1)=0, then xn0=fn0-1(x0)=f(xn0-1)=xn0-1 and xn0-1 is a fixed point; in this case, the proof is finished. Otherwise, for any n∈ℕ, d(xn,xn-1)≠0. Then by (2.6), we have
(2.7)d(xn+1,xn)≤Kθd(xn,xn-1)≤⋯≤(Kθ)n+1d(xn0,xn0-1)→0,asn→+∞,
that is,
(2.8)d(xn+1,xn)→0,asn→+∞.
Now, we show thatxn is a Cauchy sequence. By the triangle inequality and (2.2), we have
(2.9)d(xn,xm)≤d(xn,xn+1)+d(xn+1,xm+1)+d(xm+1,xm)≤d(xn,xn+1)+θβ(θd(xn,xm))d(xn,xm)+d(xm+1,xm)≤d(xn,xn+1)+θKd(xn,xm)+d(xm+1,xm),
and then
(2.10)d(xn,xm)≤(1-Kθ)-1[d(xn,xn+1)+d(xm+1,xm)]→0,asm,n→+∞,
which implies that xn is a Cauchy sequence in M. Since (M,d) is a complete metric space, then there exists a z∈M such that limn→+∞xn=z. To prove that z is a fixed point of f, if f is continuous, then
(2.11)z=limn→+∞xn=limn→+∞fn(x0)=limn→+∞fn+1(x0)=f(limn→+∞fn(x0))=f(z),
hence z=f(z). If case (2.3) holds, then we claim that f(z)=z still holds. In fact,
(2.12)d(f(z),z)≤d(f(z),f(xn))+d(f(xn),z)≤θKd(z,xn)+d(xn+1,z),
since d(z,xn)→0, taking limit as n→+∞, then d(f(z),z)≤0, this proves that d(f(z),z)=0; consequently, f(z)=z.
Lety be another fixed point of f. From (2.4) there exists x∈M which is comparable to y and z. Monotonicity implies that fn(x) is comparable to fn(y)=y and fn(z)=z for n=0,1,2… Moreover,
(2.13)d(z,fn(x))=d(fn(z),fn(x))≤Kθd(z,fn-1(x)).
Taking limit, and then limn→+∞d(z,fn(x))=0. Similar to [2], we have d(z,y)=0. The proof of the uniqueness of the fixed point is completed.


## 3. Application to Fractional Differential Equations

In this section, we consider the unique positive solution for a general higher order fractional differential equation by using the generalized fixed point theorem(3.1)-Dαx(t)=f(t,x(t),Dμ1x(t),Dμ2x(t),…,Dμn-1x(t)),Dμix(0)=0,1≤i≤n-1,Dμn-1+1x(0)=0,Dμn-1x(1)=∑j=1m-2‍ajDμn-1x(ξj),
where n-1<α≤n,  n∈ℕ and n≥3 with 0<μ1<μ2<⋯<μn-2<μn-1 and n-3<μn-1<α-2, aj∈ℝ, 0<ξ1<ξ2<⋯<ξm-2<1 satisfying 0<∑j=1m-2‍ajξjα-μn-1-1<1 and  Dα is the standard Riemann-Liouville derivative, f∈C([0,1]×ℝn,[0,+∞)). Recently, there has been a significant development in the study of fractional differential equations; for more details we refer the reader to [4–12] and the references cited therein.For the convenience of the reader, we present some notations and lemmas which will be used in the proof of our results.Definition 3.1 (see [13, 14]).
The Riemann-Liouville fractional integral of orderα>0 of a function x:(0,+∞)→ℝ is given by
(3.2)Iαx(t)=1Γ(α)∫0t‍(t-s)α-1x(s)ds,
provided that the right-hand side is pointwise defined on (0,+∞).Definition 3.2 (see [13, 14]).
The Riemann-Liouville fractional derivative of orderα>0 of a function x:(0,+∞)→ℝ is given by
(3.3)Dtαx(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1x(s)ds,
where n=[α]+1, [α] denotes the integer part of number α, provided that the right-hand side is pointwise defined on (0,+∞).Proposition 3.3 (see [13, 14]).
Consider the following.(1)
Ifx∈L1(0,1), ν>σ>0, then
(3.4)IνIσx(t)=Iν+σx(t),DtσIνx(t)=Iν-σx(t),DtσIσx(t)=x(t).(2)
Ifν>0,  σ>0, then
(3.5)Dtνtσ-1=Γ(σ)Γ(σ-ν)tσ-ν-1.Proposition 3.4 (see [13, 14]).
Letα>0, and f(x) is integrable, then
(3.6)IαDtαx(t)=f(x)+c1xα-1+c2xα-2+⋯+cnxα-n,
where ci∈ℝ  (i=1,2,…,n), n is the smallest integer greater than or equal to α.Lemma 3.5 (see [15]).
Letx(t)=Iμn-1u(t), then BVP (3.1) is equivalent to the following BVP:
(3.7)-Dα-μn-1u(t)=f(t,Iμn-1u(t),Iμn-1-μ1u(t),…,Iμn-1-μn-2u(t),u(t)),u(0)=u′(0)=0u(1)=∑j=1m-2‍aju(ξj).
Moreover, if v∈C([0,1],[0,+∞)) is a solution of problem (3.7), then the function x(t)=Iμn-1u(t) is a positive solution of problem (3.1).Let(3.8)k(t,s)={(t(1-s))α-μn-1-1-(t-s)α-μn-1-1Γ(α-μn-1),0≤s≤t≤1,(t(1-s))α-μn-1-1Γ(α-μn-1),0≤t≤s≤1,
from [15], the Green function of (3.7) is
(3.9)H(t,s)=k(t,s)+tα-μn-1-11-∑j=1m-2ajξjα-μn-1-1∑j=1m-2‍ajk(ξj,s),
and with property
(3.10)0≤H(t,s)≤1Γ(α-μn-1)(1+∑j=1m-2aj1-∑j=1m-2ajξjα-μn-1-1)=η.In our considerations, we will work in the Banach spaceE=C([0,1];ℝ) with the classical metric given by d(x,y)=sup0≤t≤1|x(t)-y(t)|. Notice that this space can be equipped with a partial order given by
(3.11)x,y∈C[0,1],x≤y⟺x(t)≤y(t)foranyt∈[0,1].
In [2], it is proved that (C[0,1],≤) satisfies condition (2.3) of Theorem 2.2. Moreover, for x,y∈C[0,1], as the function max{x,y}∈C[0,1],  (C[0,1],≤) satisfies condition [15]. Consider the cone
(3.12)P={u∈C[0,1]:u(t)≥0}.
Note that as P is a closed set of C[0,1], P is a complete metric space.It is well known that the BVP (3.7) is equivalent to the integral equation
(3.13)u(t)=∫01‍H(t,s)f(s,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds.
Now, for u∈P we define the operator T by
(3.14)(Tu)(t)=∫01‍H(t,s)f(s,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds.
Then from the assumption on f and (3.10), we have
(3.15)T(P)⊂P.We also introduce the following class of nondecreasing functionsℬ by ϕ:[0,+∞)→[0,+∞) satisfying the following:
(3.16)ϕ(x)≤Kx,foranyx>0.
Clearly, if ϕ∈ℬ, then ϕ(x)/x∈𝒜. The standard functions ϕ∈ℬ, for example, ϕ(x)=Kx, ϕ(x)=(2K/π)xarctanx, and ϕ(x)=Kx2/(1+x).Theorem 3.6.
Supposef(t,x1,x2,…,xn) is nondecreasing in xi on [0,+∞); moreover, there exist n positive constants ρi, i=1,2,…,n that satisfy
(3.17)max{ρ1,ρ2,…,ρn}≤(nη)-1,
and there exist a function ϕ∈ℬ and constants 0<θ1<Γ(μn-1)/K, 0<θn<1/K, 0<θi<Γ(μn-1-μi)/K, i=1,2,…,n-2 such that
(3.18)|f(t,x1,x2,…,xn)-f(t,y1,y2,…,yn)|≤∑i=1n‍ρiϕ(θi(xi-yi)),
for xi,yi∈[0,+∞), i=1,2,…,n with xi≥yi and t∈[0,1]. Then problem (3.1) has a unique nonnegative solution.Proof.
We check that the hypotheses in Theorem2.2 are satisfied.
Firstly, the operatorT is nondecreasing by the hypothesis. Then for any u≥v, we have
(3.19)(Tu)(t)=∫01‍H(t,s)f(s,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds≥∫01‍H(t,s)f(s,Iμn-1v(s),Iμn-1-μ1v(s),…,Iμn-1-μn-2v(s),v(s))ds=(Tv)(t).
Noticing that
(3.20)u(s)-v(s)≤d(u,v),Iμn-1u(s)-Iμn-1v(s)≤∫0t‍(t-s)μn-1-1|u(s)-v(s)|Γ(μn-1)ds≤d(u,v)Γ(μn-1),Iμn-1-μiu(s)-Iμn-1-μiv(s)≤∫0t‍(t-s)μn-1-μi-1|u(s)-v(s)|Γ(μn-1-μi)ds≤d(u,v)Γ(μn-1-μi),i=1,2,…,n-2,
and take
(3.21)θ=max{θ1Γ(μn-1),θ2Γ(μn-1-μ1),…,θn-1Γ(μn-1-μn-2),θn}.
Thus, for anyu≥v, we have
(3.22)d(Tu,Tv)=maxt∈[0,1]|Tu(t)-Tv(t)|≤η∫01‍|f(s,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds-f(s,Iμn-1v(s),Iμn-1-μ1v(s),…,Iμn-1-μn-2v(s),v(s))|≤η∫01‍[ρ1ϕ(θ1(Iμn-1u(s)-Iμn-1v(s)))+ρ2ϕ(θ2(Iμn-1-μ1u(s)-Iμn-1-μ1v(s)))+⋯+ρn-1ϕ(θn-1(Iμn-1-μn-2u(s)-Iμn-1-μn-2v(s)))+ρnϕ(θn(u(s)-v(s)))]ds≤nηmax{ρ1,ρ2,…,ρn}ϕ(θd(u,v))≤ϕ(θd(u,v)).
For u≠v, we have
(3.23)d(Tu,Tv)≤ϕ(θd(u,v))θd(u,v)θd(u,v)=β(θd(u,v))θd(u,v),
and this inequality is obviously satisfied for u=v. Thus, we have
(3.24)d(Tu,Tv)≤ϕ(θd(u,v))θd(u,v)θd(u,v)=β(θd(u,v))θd(u,v),foranyu,v∈Pwithu≥v.
Finally, since the zero function satisfies0≤T0, Theorem 2.2 tells us that the operator T has a unique fixed point in P, or, equivalently, the BVP (3.1) has a unique nonnegative solution x in C[0,1].Theorem 3.7.
If the assumptions of Theorem3.6 are satisfied, and there exists t0∈[0,1] such that f(t0,0,…,0)≠0, then the unique solution of (3.1) is positive (a positive solution means a solution satisfying x(t)>0 for t∈(0,1)).Proof.
By Theorem3.6, the problem (3.1) has a unique nonnegative solution. We prove the nonnegative solution is also positive.
Otherwise, there exists0<t*<1 such that x(t*)=0, and
(3.25)x(t*)=∫01‍H(t*,s)f(s,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds=0.
Then
(3.26)0=x(t*)=∫01‍H(t*,s)f(s,Iμn-1u(s),Iμn-1-μ1u(s),…,Iμn-1-μn-2u(s),u(s))ds≥∫01‍H(t*,s)f(s,0,…,0)ds≥0.
Consequently,
(3.27)∫01‍H(t*,s)f(s,0,…,0)ds=0,
this yields
(3.28)H(t*,s)f(s,0,…,0)=0,a.e.s∈[0,1].
Note that H(t*,s)>0,s∈(0,1), then we have
(3.29)f(s,0,…,0)=0,a.e.s∈[0,1].
But on the other hand, sincef(t0,0,…,0)≠0, t0∈[0,1], we have f(t0,0,…,0)>0, by the continuity of f, we can find a set Ω⊂[0,1] satisfying t0∈Ω and the Lebesgue measure μ(Ω)>0 such that f(t,0,…,0)>0 for any t∈Ω. This contradicts to (3.29). Therefore, x(t)>0, that is, x(t) is positive solution of (3.1).Example 3.8.
Consider the following fractional boundary value problem:(3.30)-D5/2x(t)=et+110x(t)+12sin2(D1/8x(t))+13cos2(D1/4x(t)),0<t<1,D1/4x(0)=D5/4x(0)=0,D1/4x(1)=14D1/4x(14)+12D1/4x(34).
Then the BVP (3.30) has a unique positive solution.Proof.
Since(3.31)∑j=1m-2‍ajξjα-μn-1-1=14(14)5/4+12(34)5/4=0.39316<1,η=1Γ(α-μn-1)(1+∑j=1m-2aj1-∑j=1m-2ajξjα-μn-1-1)=0.10964.
Thus,
(3.32)(nη)-1=3.0405.
LetK=15 and ϕ(x)=15x, and take
(3.33)f(t,x1,x2,x3)=et+110x1+12sin2x2+13cos2x3,(t,x1,x2,x3)∈[0,1]×[0,+∞)3.
Then, for any x1≥y1, x2≥y2, x3≥y3,
(3.34)|f(t,x1,x2,x3)-f(t,y1,y2,y3)|=|x1-y110+sin2x2-sin2y22+cos2x3-cos2y33|≤x1-y110+x2-y22+x3-y33=275×15×14(x1-y1)+110×15×13(x2-y3)+49×15×120(x3-y3)=275ϕ(14(x1-y1))+110ϕ(13(x2-y3))+49ϕ(120(x3-y3))=∑i=13‍ρiϕ(θi(xi-yi)),
where
(3.35)ρ1=275,ρ2=110,ρ3=49,θ1=14,θ2=13,θ3=120.
Thus, ϕ∈ℬ and all of the conditions of Theorem 3.6 are satisfied.
On the other hand,f(0,0,…,0)=4/3≠0, by Theorems 3.6 and 3.7, the BVP (19) has a unique positive solution.Remark 3.9.
In Example3.8, β(x)=ϕ(x)/x=K=15 which does not possess property β:[0,+∞)→[0,1) and
(3.36)β(tn)→1impliestn→0.
Thus, the unique positive solution of BVP (3.30) cannot be obtained via Theorem 1.2, but we obtain the unique positive solution of BVP (3.30) by using the generalized fixed point Theorem 2.2, which implies that Theorem 2.2 is an essential promotion of Theorem 1.2.



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