Mixed Monotone Model Operator Yue-Xiang Wu , Yan-Mei

2016 International Conference on Mathematical, Computational and Statistical Sciences and Engineering (MCSSE 2016) ISBN: 978-1-60595-396-0 Fixed Point for t−φ(t,u,v) Mixed Monotone Model Operator 1* 2 Yue-xiang Wu , Yan-mei Huo 1Department of Applied Mathematics, Shanxi University of Finance and Economics, Taiyuan, 030006, China. 2College of Economics, Shanxi University of Finance and Economics, Taiyuan, 030006, China Abstract. This paper studies the uniqueness and existence of fixed point of t−φ(t,u,v) mixed monotone model operator in the partial order Banach space. Our conclusions essentially improve the relevant results. Moreover, as an application of our results, we prove the existence and uniqueness of a positive solution for a class of integral equations which can not be solved by using previously available methods. Keywords. Cone and semiorder; t−φ(t,u,v) mixed monotone model operator; Fixed point 1 Introduction and Preliminaries In this paper, we will study the existence and uniqueness of fixed point of a kind of mixed monotone operator. The resulting conclusion essentially improves the relevant results obtained by [3,8]. As an application of our results, we prove the existence and uniqueness of a positive solution for a class of integral equation which can not be solved by using previously available methods. Let the real Banach space E (or, more general case, topological linear space) 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. By Θ we denote the zero element of E. Recall that a non-empty closed convex set P⊂E is a cone if it satisfies x∈P, λ≥0⇒λx∈P and x, −x∈P⇒x=Θ. We denote by P the interior set of P. A cone P is said to be solid cone if P≠φ. P is said to be normal if there exists a positive constant N such that Θ≤x≤y implies |x|≤N|y|. N is called the normal constant of P. We write x≪y if and only if y−x∈P. Definition 1.1 (see [2,6,8]) 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). 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≤μ such that λx≤y≤μx. Clearly, ∼ is an equivalence relation. Given h>Θ(i.e.h≥Θ and h≠Θ), we denote by Ph the 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. Definition 1.2 Let P be a cone of a real topological linear space E, and A:P×P→E a mixed monotone operator. Assume that for all 0<t<1,u,v∈P, there exists a single function φ: t<φ=φ(t,u,v)<1, such that 1 A(tu, tv)≥φ(t,u,v)A(u,v).(1.1) Then A is called t−φ(t,u,v) mixed monotone model operator. For more facts about mixed monotone operators and other related concepts the reader is referred to [1,4,5,7,9-11] and some of the references therein. 321 2. t−φ(t,u,v) mixed monotone model operator We give the following Theorem on t−φ(t,u,v) mixed monotone model operator in ordered Banach spaces. Theorem 2.1 Let P be a normal cone of a real Banach space E, h>Θ. A:Ph×Ph→Ph is a t−φ(t,u,v) mixed monotone model operator. Suppose (i) φ(t,u,v) is nonincreasing in u; (ii) φ(t,u,v) is nondecreasing in v; (iii) lim φ(t,t−1h,th)=l>0 . t→0+ * * * * Then there exists unique fixed point x ∈Ph such that A(x ,x )=x . Moreover, for any initial point x0, y0∈Ph , constructing successively the sequences * xn=A(xn−1, yn−1),yn=A(yn−1, xn−1), n=1,2,⋅⋅⋅ , we have |xn−x |→0 and * |yn−x |→0 as n→∞. Proof Firstly, choose t0∈(0,1) such that 1 l t h≤A(h,h)≤ h.(2.1) 0 t0 2 According to condition (iii), we can also adjust above 0<t0<1 sufficiently small such that −1 1 l φ(t0,t0 h,t0h) h≤ h.(2.2) t0 2 t0 Thus we have −1 φ(t0,t0 h,t0h) t h≤A(h,h)≤ h.(2.3) 0 t0 Since φ(t0,h,h)>t0 , hence, when k→+∞, φ(t0,h,h) ( )k→ +∞.(2.4) t0 Select sufficiently large positive integer k such that φ(t0,h,h) 1 ( )k≥ .(2.5) t0 t0 Let k −k u0=t0h, v0=t0 h,(2.6) and constructing the iterative sequences un=A(un−1,vn−1), vn=A(vn−1,un−1), n=1,2,⋅⋅⋅.(2.7) It is clear that u0, v0∈Ph and u0<v0, and u1=A(u0, v0)≤A(v0,u0)=v1. Likewise, u2=A(u1, v1)≤A(v1, v1)≤A(v1, u1)=v2. In general, un≤vn, n=1,2,⋅⋅⋅.(2.8) Furthermore, by using formulas (2.1, 2.4, 2.6–2.8) and conditions (i) and (ii), we have 322 k −k k−1 −k+1 k−1 −k+1 u1=A(u0,v0)=A(t0h,t0 h)≥φ(t0,t0 h,t0 h)A(t0 h,t0 h) k−1 −k+1 k−2 −k+2 k−2 −k+2 ≥φ(t0,t0 h,t0 h)φ(t0,t0 h,t0 h)A(t0 h,t0 h) k−1 −k+1 −1 −1 ≥⋅⋅⋅≥φ(t0,t0 h,t0 h)⋅⋅⋅φ(t0,t0h,t0 h)A(t0h,t0 h) k−1 −1 k−1 ≥φ(t0,h,h) A(t0h,t0 h)≥φ(t0,h,h) φ(t0,h,h)A(h,h) k k k−1 k =φ(t0,h,h) A(h,h)≥φ(t0,h,h) t0h≥t0 t0h=t0h=u0. i.e. u1≥u0.(2.9) On the other hand, since 1 1 u,tv)≤φ(t, u,tv)−1A(u,v),(2.10) A( t t similarly, we can get −k k −1 −k+1 k−1 −k k −1 −k+1 k−1 v1=A(t0 h,t0h)=A(t0 t0 h,t0t0 h)≤φ(t0,t0 h,t0h) A(t0 h,t0 h) −k k −1 −k+1 k−1 −1 −k+2 k−2 ≤φ(t0,t0 h,t0h) φ(t0,t0 h,t0 h) A(t0 h,t0 h)≤⋅⋅⋅ −k k −1 −k+1 k−1 −1 −1 −1 ≤φ(t0,t0 h,t0h) φ(t0,t0 h,t0 h) ⋅⋅⋅φ(t0,t0 h,t0h) A(h,h) 1 A(h,h) 1 1 −k ≤ ≤ h=t h=v , −k k k−1 −1 k−1 t0 0 0 φ(t0,t0 h,t0h) φ(t0,t0 h,t0h) t0 i.e. v1≤v0.(2.11) Hence we have u0≤u1≤v1≤v0.(2.12) By using (2.8),(2.12) and the mixed monotone properties of operator A, we easily have u0≤u1≤⋅⋅⋅≤un≤⋅⋅⋅≤vn≤⋅⋅⋅≤v1≤v0.(2.13) Let tn=sup{t>0:un≥tvn}, n=0,1,2,⋅⋅⋅.(2.14) Then un≥tnvn.(2.15) * It is easy to obtain that tn is nondecreasing in n and there exists t such that lim * n→∞tn=t .(2.16) We prove that t*=1. In fact, suppose 0<t*<1, Consider two cases: * (i) there exists integer N such that tN=t . * * In this case, we have tn=t , and un≥t vn for all n≥N holds. Hence * * −1 * * un+1=A(un,vn)≥A(t vn,(t ) un)≥φ(t ,vn,un)A(vn,un)=φ(t ,vn,un)vn+1, n≥N. 323 * * Therefore by the definition of tn+1 , we have tn+1≥φ(t ,vn,un)>t , n≥N. This is a contradiction. * (ii) For all integer n, tn<t . In this case, t * t n * t * −1 n * * −1 * * −1 u =A(u ,vn)≥A( t vn, (t ) un)≥φ( ,t vn,(t ) un)A(t vn,(t ) un) n+1 n t* tn t* * * tn tn φ(t ,vn,un) φ(t ,vn,un) ≥φ( ,t*v ,(t*)−1u )×φ(t*,v ,u )A(v ,u )> ×t*[ ]v =t [ ]v . t* n n n n n n t* t* n+1 n t* n+1 Thus, by the definition of tn+1 and conditions (i) and (ii), we have * * φ(t ,vn,un) φ(t ,v0,u0) t ≥t [ ]≥tn[ ]. n+1 n t* t* * φ(t ,v0,u0) Let n→∞, we have t*≥t*[ ]>t*, this is a contradiction. t* * lim Hence t =1. i.e. n→∞tn=1. * Furthermore, similar the proof of Theorem 2.1, there exist x ∈Ph such that lim lim * n→∞un=n→∞vn=x , and x* is the fixed point of operator A. Next, the fixed point of operator A is uniqueness. In fact, suppose u*∈Ph is other fixed point of operator A. Let * −1 t*=sup{0<t≤1:tu*≤x ≤t u*}.(2.17) * −1 Clearly, 0<t*≤1 and t*u*≤x ≤t* u*. If t*<1, then * * * −1 x =A(x ,x )≥A(t*u*,t* u*)≥φ(t*,u*,u*)A(u*,u*)=φ(t*u*,u*)u*, and u u * * * −1 * −1 * −1 x =A(x ,x )≤A(t u*,t u )≤φ(t , ,t u ) A(u ,u )=φ(t , ,t u ) u .

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