Existence of solutions for a coupled system of difference equations with causal operators

https://doi.org/10.1016/j.amc.2013.03.100Get rights and content

Abstract

In this paper, we discuss a coupled system of difference equations with causal operators. Sufficient conditions under which such problems have solutions are given, by using a monotone iterative method. Difference inequalities with causal operators are also investigated. An example illustrates the results.

Introduction

By N0, we denote the set of nonnegative integer numbers. Let Z[a,b]={a,a+1,,b},E=C(Z[a,b]×Z[a,b],R), and QC(E,E), where a,bN0 and a<b. We shall say that Q is a causal operator, or nonanticipative, if the following property holds: for each couple of elements of E such that ui(s)=vi(s),i=1,2 for ask,s,kN0, there results Q(u1,u2)(s)=Q(v1,v2)(s) for 0sk with k<b arbitrary; for details about causal operators, see [6].

Put E0=C(Z[0,T-1]×Z[0,T-1],R),E1=C(Z[1,T]×Z[1,T],R),E¯0=C(Z[0,T-1],R),E¯1=C(Z[1,T],R). In this paper, we investigate two-point boundary value problems for first order difference equations with causal operators Q1,Q2 of the form:Δy(k-1)=Q1(y,z)(k),kZ[1,T],Δz(k-1)=Q2(z,y)(k),kZ[1,T],h1(y,z)=0,h2(z,y)=0,where Δy(k-1)=y(k)-y(k-1),hi(y,z)=gi(y(0),y(T),z(0),z(T)),i=1,2 with assumption

H1: Q1,Q2C(E1,E1),g1,g2C(R4,R),Q1,Q2 are bounded, and the next type of equations:Δy(k)=Q1(y,z)(k),kZ[0,T-1],Δz(k)=Q2(z,y)(k),kZ[0,T-1],h1(y,z)=0,h2(z,y)=0,where Δy(k)=y(k+1)-y(k) with assumption

H2: Q1,Q2C(E0,E0),g1,g2C(R4,R),Q1,Q2 are bounded.

For example, causal operators Q1,Q2 may take forms:Q1(y,z)(k)=f1(k,y(k),y(α1(k)),,y(αr(k)),z(k),z(β1(k)),,z(βs(k))),Q2(z,y)(k)=f2(k,z(k),z(α1(k)),,z(αr(k)),y(k),y(β1(k)),,y(βs(k))),where f1,f2C(Z[1,T]×Rr+s+2,R),αi,βjC(Z[1,T], Z[0,T]),αi(k)k and βj(k)k,i=1,2,,r,j=1,2,,s.

Note that problems (1), (2) represent two discrete analogues of a coupled system of differential equations with causal operators Q1,Q2 of the formx(t)=Q1(x,y)(t),t[0,T],y(t)=Q2(y,x)(t),t[0,T],g1(x(0),x(T))=0,g2(y(0),y(T))=0.Let C=C(Z[0,T],R) denote the class of maps w continuous on Z[0,T] (discrete topology) with the normw=maxkZ[0,T]|w(k)|.Note that C is a Banach space. By a solution of (1), we mean a pair (w,v)C×C such that it satisfies problem (1). Similarly we define the solution of problem (2).

To find a solution of problem (1) or (2) we are going to apply a monotone iterative method combined with lower and upper solutions of discrete problems. For some results on lower and upper approximations of solutions of difference equations and their applications, usually in showing the existence of a specific solution, see, e.g. [1], [3], [4], [5], [12], [14], [15], [16], [18], [19] and the related references therein. The monotone iterative method is a well known one in case of continuous problems, see for example [11]. It was also applied to discrete problems, see for example papers [8], [9], [10], [17], [18]. The study of a coupled system of difference equations with causal operators is also significant because this kind of system may occur in applications. Note that difference equations with delayed arguments belong to difference equations with causal operators. Difference equations are also discussed, for example, in papers [2], [13].

This paper is organized as follows. In Section 2, we present some basic materials needed to prove the main result for problem (1). In Section 3, we prove the existence of solutions for system (1), by applying a theorem from [7]. An example is also included in this section to illustrate theoretical results. In the last section, we discuss problem of type (2) giving sufficient conditions which guarantee the existence of solutions.

Section snippets

Difference inequalities

We shall first concentrate our attention to difference inequalities with positive linear operators. We shall say that a linear operator LC(E1,E1) is a positive linear operator if (Lm)(k)0 provided that m(k)0,kZ[1,T]. LetSk=i=1k[1+M(i)],kZ[0,T]withi=10=1.Then we have the following result.

Lemma 1

see [8]

Assume that LC(E¯1,E¯1) is a positive linear operator and MC(Z[1,T],R+) with R+=[0,). Let yC(Z[0,T],R) andΔy(k-1)-M(k)y(k)-Ly(k),kZ[1,T],y(0)ry(T),0rST-11andρ1i=1TL1(i)Si-11with1(k)=1,kZ[1,T]

Existence solutions to problems of type (1)

First, let us consider the following linear problemΔy(k-1)=-M(k)y(k)-(Ly)(k)+h(k),kZ[1,T],y(0)=ry(T)+β,0rST-1<1,βR,where hC(Z[1,T],R) and bounded.

Theorem 1

see [8]

Assume that LC(E1,E1) is a positive linear operator and MC(Z[1,T],R+). Let hC(Z[1,T],R) and be bounded. Let condition (4) hold. Then problem (6) has a unique solution.

PutM(y,z)(k)=M1(k)y(k)+M2(k)z(k)+L1y(k)+L2z(k).

Lemma 3

Assume that linear operators L1,L2C(E¯1,E¯1) are positive. Let the operator L1-L2 be positive. Let h¯1,h¯2,M1,M2C(Z[1,T],R) and M1(

Existence solutions to problems of type (2)

Now, we will discuss problem (2). PutPk=i=0k-1[1-M(i)]-1,kZ[0,T].First, we formulate a result connected with a difference inequality.

Lemma 4

See [8]

Assume that LC(E¯0,E¯0) is a positive linear operator and MC(Z[0,T-1],[0,1)). Let yC(Z[0,T],R) andΔy(k)-M(k)y(k)-(Ly)(k),kZ[0,T-1],y(0)ry(T),0rPT-11,andρ2i=0T-1(L1)(i)Pi+11.Then y(k)0,kZ[0,T].

Remark 7

If M(k)=0,kZ[0,T-1], then Pk=1,kZ[0,T].

PutPk+=i=0k-1{1-[M1(i)+M2(i)]}-1,Pk-=i=0k-1{1-[M1(i)-M2(i)]}-1for kZ[0,T]. Now, we formulate a similar result to

Acknowledgement

The author is grateful to the reviewer for his/her suggestions which improved the paper.

References (19)

There are more references available in the full text version of this article.

Cited by (6)

View full text