WU YUE-XIANGAND HUO YAN-MEI

(1.College of Applied Mathematics,Shanxi University of Finance and Economics,Taiyuan,030006)

(2.College of Economics,Shanxi University of Finance and Economics,Taiyuan,030006)

Abstract:In this paper,a class of mixed monotone operators is studied.Some new fixed point theorems are presented by means of partial order theory,and the uniqueness and existence of fixed points are obtained without assuming the operator to be compact or continuous.Our conclusions extend the relevant results.Moreover,as an application of our result,the existence and uniqueness of positive solution for a class of fractional differential equation boundary value problem are proved.

Key words:cone and semiorder,mixed monotone operator, fixed point,fractional differential equation

1 Introduction and Preliminaries

In 2012,Borcut and Berinde[1]introduced the concept of tripled fixed point for nonlinear mappings in partially ordered complete metric spaces and obtained its existence.

In this paper,without assuming operators to be continuous or compact,we study tripled fixed point for a class of mixed monotone operator on ordered Banach spaces.This research done is important in comparison with others,as some times coupled fixed points are more perfect than other fixed points.Tripled fixed points are more practical than coupled fixed points and they are applicable for most differential equations which are not solve by the application of fixed points or coupled fixed points.As an application of our result,we prove the existence and uniqueness of a positive solution for a class of fractional differential equation boundary value problem which can not be solved by using previously available methods.

Let the real topological linear space E be partially ordered by a cone P of E,i.e.,x≤y(or alternatively denoted as y ≥ x)if and only if y?x ∈ P.We denote by θ the zero element of E.Recall that a non-empty closed convex set P?E is a cone if it satis fies

and

N is called the normal constant of P.We write x?y if and only if

De finition 1.1[2]Let D?E.Operator A:D×D→E is said to be mixed monotone if A(x,y)is nondecreasing in x and nonincreasing in y,i.e.,u1≤u2,v2≤v1,ui,vi∈D(i=1,2)implies A(u1,v1)≤ A(u2,v2).Point(x?,y?)∈ D×D,x?≤ y?,is called a coupled lower-upper fixed point of A if x?≤ A(x?,y?)and A(y?,x?)≤ y?.Point(x?,y?)∈ D × D is called a coupled fixed point of A if x?=A(x?,y?)and A(y?,x?)=y?.Element x?∈ D is called a fixed point of A if A(x?,x?)=x?.

For all x,y∈E,the notation x～y means that there exist 0＜λ≤μsuch that λx≤ y≤ μx.Clearly,“～”is an equivalence relation.Given h＞ θ(i.e.,h≥ θ andwe denote by Phthe set

Ph={x ∈ E,there exist λ(x),μ(x)＞ 0 such that λ(x)h ≤ x ≤ μ(x)h}.

It is easy to see that Ph?P.

De finition 1.2[1]Let(E,≤)be a partially ordered set and F:E3→ E.We say that F has the mixed monotone property if for any x,y,z∈E,

De finition 1.3[1]An element(x,y,z)∈E3is called a tripled fixed point of a mapping F:E3→E if F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.

More concepts and facts about mixed monotone operators and other related concepts the reader is referred to[3]–[13]and some of the references therein.

2 Main Result and Proof

Let P be a cone of a real topological linear space E,and A:P3→E be a mixed monotone operator.We assume that:

(B1)there exists an h ∈ P withsuch that A(h,h,h)∈ Ph;

Firstly,we need the following lemma.

Lemma 2.1Let P be a cone of a real topological linear space E,h＞θ,be a mixed monotone operator,and the conditions(B1)and(B2)be satis fied.Then there exist u0,v0,w0∈Phand r∈(0,1)such that

Proof. (i)From condition(B2),for any t∈(0,1)and x,y,z∈P,we have

For any u,v,w ∈ Ph,there existμ1,μ2,μ3∈ (0,1)such that

Letμ =min{μ1,μ2,μ3}.Then μ ∈ (0,1).And from the mixed monotone properties of operator A,A(h,h,h)∈Ph,we have A(u,v,w)∈Ph,i.e.,We can choose a sufficiently small number t0∈ (0,1)such that

Hence u0＜v0.

(ii)By the mixed monotone properties of A,we have

Furthermore,we have

On the other hand,we have

Thus we have

i.e.,

(iii)Similar to the proof of(ii),we can prove that A(w0,u0,w0)≥w0.

Theorem 2.1Let P be a normal cone of a real Banach spacebe a mixed monotone operator,and the conditions(B1)and(B2)be satis fied.Then A has exactly one fixed point x?in Ph.

Moreover,constructing successively the sequences

For any initial point x0,y0,z0∈Ph,we have

Proof. From Lemma 2.1,there exist u0,v0,w0∈Phand r∈(0,1)such that

Constructing successively the sequences

Evidently,

Notice that rv0≤u0≤w0,hence

Let tn=sup{t＞ 0:un≥ tvn},n=1,2,···.Then

It follows that

Thus tn≤ tn+1,i.e.,0 ＜ t1≤ ···≤ tn≤ ···＜ 1,so tnis nondecreasing in n.

For any natural number p,we have

Since the cone P is normal,we have

where M is the normality constant of P.So we can claim that un,vnand wnare Cauchy sequences.Because E is complete,there exist u?,v?,w?such that

It is easy to know that u?,v?,w?∈ Phand un≤ u?≤ w?≤ v?≤ vn.So

Futher,

Thus

Let x?=u?=v?=w?.Then

By letting n→∞,we get

that is,x?is a fixed point of A in Ph.

In the following,we prove that x?is the unique fixed point of A in Ph.

Then

It is easy to prove that e1=1,and we getTherefore A has a unique fixed point x?in Ph.

Finally,constructing successively the sequences

For any initial points x0,y0,z0∈Ph,we can choose t0,t1,t2such that

Let t?=min{t0,t1,t2}.Then t?∈ (0,1)and

Let

Similarly,it follows that there exists a y?∈ Phsuch that

By using the uniqueness of fixed point of operator A,we get x?=y?.By induction we can get

Taking into account that P is normal,we immediately conclude that

3 Example

As an application of our result,we give only an example.

Example 3.1Consider the following fractional differential equation

subject to the condition

Consider the Banach space of continuous functions on[0,1]× [0,1]× [1,2]with the maximum norm

and set

Then P is a normal cone.

The fractional differential equation(3.1)with the boundary value condition(3.2)has a solution u0if and only if u0is a solution of the following fractional integral equation

where g(t,ξ)is the corresponding green function,i.e.,

cu(s,r,t)+c?1v(s,r,t)+cw(s,r,t))≥ ?(t)(u(s,r,t)+v(s,r,t)+w(s,r,t)),

where

whenever g(t,s)＜0,and there exist C1,C2＞0,h∈P such that

Then the differential equation(3.1)with the condition(3.2)has a unique solution in Ph.

Moreover,constructing successively the sequences for any initial u0,v0,η0∈ Ph,

we have

Proof. De fine the operator A:P×P×P→P by

Then u is the solution for the problem if and only if u=A(u,u,u).

i.e.,the condition(B1)is satis fied.

Also

i.e.,A(h,h,h)∈Ph.Hence the condition(B2)is satis fied.Therefore,Conclusion 3.1 holds.