# Asymptotic Behavior of Solutions of Delayed Difference Equations

Authors: J. Diblík; I. Hlavičková
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

This contribution is devoted to the investigation of the asymptotic behavior of delayed difference equations with an integer delay. We prove that under appropriate conditions there exists at least one solution with its graph staying in a prescribed domain. This is achieved by the application of a more general theorem which deals with systems of first-order difference equations. In the proof of this theorem we show that a good way is to connect two techniques—the so-called retract-type technique and Liapunov-type approach. In the end, we study a special class of delayed discrete equations and we show that there exists a positive and vanishing solution of such equations.


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## Body

## 1. Introduction

Throughout this paper, we use the following notation: for an integerq, we define(1.1)Zq∞:={q,q+1,…}.
We investigate the asymptotic behavior for n→∞ of the solutions of the discrete delayed equation of the (k+1)-th order(1.2)Δv(n)=f(n,v(n),v(n-1),…,v(n-k)),
where n is the independent variable assuming values from the set ℤa∞ with a fixed a∈ℕ. The number k∈ℕ, k≥1 is the fixed delay, Δv(n)=v(n+1)-v(n), and f:ℤa∞×ℝk+1→ℝ.A functionv:ℤa-k∞→ℝ is a solution of (1.2) if it satisfies (1.2) for every n∈ℤa∞.We will study (1.2) together with k+1 initial conditions(1.3)v(a+s-k)=va+s-k∈R,s=0,1,…,k.
Initial problem (1.2), (1.3) obviously has a unique solution, defined for every n∈ℤa-k∞. If the function f is continuous with respect to its last k+1 arguments, then the solution of (1.2) continuously depends on initial conditions (1.3).Now we give a general description of the problem solved in this paper.Problem 1.
Letb,c:ℤa-k∞→ℝ be functions such that b(n)<c(n) for every n∈ℤa-k∞. The problem under consideration is to find sufficient conditions for the right-hand side of (1.2) that will guarantee the existence of a solution v=v*(n) of initial problem (1.2), (1.3) such that
(1.4)b(n)<v*(n)<c(n),n∈Za-k∞.This problem can be solved with help of a result which is valid for systems of first-order difference equations and which will be presented in the next section. This is possible because the considered equation (1.2) can be rewritten as a system of k+1 first-order difference equations, similarly as a differential equation of a higher order can be transformed to a special system of first-order differential equations. Although the process of transforming a (k+1)-st order difference equation to a system of first order equations is simple and well-known (it is described in Section 3), the determination of the asymptotic properties of the solutions of the resulting system using either Liapunov approach or retract-type method is not trivial. These analogies of classical approaches, known from the qualitative theory of differential equations, were developed for difference systems in [1] (where an approach based on Liapunov method was formulated) and in [2–5] (where retract-type analysis was modified for discrete equations). It occurs that for the mentioned analysis of asymptotic problems of system (1.2), neither the ideas of Liapunov, nor the retract-type technique can be applied directly. However, in spite of the fact that each of the two mentioned methods fails when used independently, it appears that the combination of both these techniques works for this type of systems. Therefore, in Section 2 we prove the relevant result suitable for the asymptotic analysis of systems arising by transformation of (1.2) to a system of first-order differential equations (Theorem 2.1), where the assumptions put to the right-hand side of the system are of both types: those caused by the application of the Liapunov approach and those which are typical for the retract-type technique. Such an idea was applied in a particular case of investigation of asymptotic properties of solutions of the discrete analogue of the Emden-Fowler equation in [6, 7]. The approach is demonstrated in Section 3 where, moreover, its usefulness is illustrated on the problem of detecting the existence of positive solutions of linear equations with a single delay (in Section 3.4) and asymptotic estimation of solutions (in Section 3.3).Advantages of our approach can be summarized as follows. We give a general method of analysis which is different from the well-known comparison method (see, e.g., [8, 9]). Comparing our approach with the scheme of investigation in [10, 11] which is based on a result from [12], we can see that the presented method is more general because it unifies the investigation of systems of discrete equations and delayed discrete equations thanks to the Liapunov-retract-type technique.For related results concerning positive solutions and the asymptotics of solutions of discrete equations, the reader is referred also to [13–25].


## 2. The Result for Systems of First-Order Equations

Consider the system ofm difference equations(2.1)Δu(n)=F(n,u(n)),
where n∈ℤa∞, u=(u1,…,um), and F:ℤa∞×ℝm→ℝm, F=(F1,…,Fm). The solution of system (2.1) is defined as a vector function u:ℤa∞→ℝm such that for every n∈ℤa∞, (2.1) is fulfilled. Again, if we prescribe initial conditions(2.2)ui(a)=uia∈R,i=1,…,m
the initial problem (2.1), (2.2) has a unique solution. Let us define a set Ω⊂ℤa∞×ℝm as(2.3)Ω:=⋃n=a∞Ω(n),
where(2.4)Ω(n):={(n,u):n∈Za∞,ui∈R,bi(n)<ui<ci(n),i=1,…,m}
with bi,ci:ℤa∞→ℝ, i=1,…,m, being auxiliary functions such that bi(n)<ci(n) for each n∈ℤa∞. Such set Ω is called a polyfacial set.Our aim (in this part) is to solve, in correspondence with formulated Problem1, the following similar problem for systems of difference equations.Problem 2.
Derive sufficient conditions with respect to the right-hand sides of system (2.1) which guarantee the existence of at least one solution u(n)=(u1*(n),…,um*(n)), n∈ℤa∞, satisfying
(2.5)(n,u1*(n),…,um*(n))∈Ω(n)
for every n∈ℤa∞.As we mentioned above, in [1] the above described problem is solved via a Liapunov-type technique. Here we will combine this technique with the retract-type technique which was used in [2–5] so as the result can be applied easily to the system arising after transformation of (1.2). This brings a significant increase in the range of systems we are able to investigate. Before we start, we recall some basic notions that will be used.


### 2.1. Consequent Point

Define the mapping𝒞:ℤa∞×ℝm→ℤa∞×ℝm as(2.6)C:(n,u)⟼(n+1,u+F(n,u)).
For any point M=(n,u)∈ℤa∞×ℝm, the point 𝒞(M) is called the first consequent point of the point M. The geometrical meaning is that if a point M lies on the graph of some solution of system (2.1), then its first consequent point 𝒞(M) is the next point on this graph.


### 2.2. Liapunov-Type Polyfacial Set

We say that a polyfacial setΩ is Liapunov-type with respect to discrete system (2.1) if(2.7)bi(n+1)<ui+Fi(n,u)<ci(n+1)
for every i=1,…,m and every (n,u)∈Ω. The geometrical meaning of this property is this: if a point M=(n,u) lies inside the set Ω(n), then its first consequent point 𝒞(M) stays inside Ω(n+1).In this contribution we will deal with sets that need not be of Liapunov-type, but they will have, in a certain sense, a similar property. We say that a polyfacial setΩ is Liapunov-type with respect to thejth variable (j∈{1,…,m}) and to discrete system (2.1) if(2.8)(n,u)∈Ω⟹bj(n+1)<uj+Fj(n,u)<cj(n+1).
The geometrical meaning is that if M=(n,u)∈Ω(n), then the uj-coordinate of its first consequent point stays between bj(n+1) and cj(n+1), meanwhile the other coordinates of 𝒞(M) may be arbitrary.


### 2.3. Points of Strict Egress and Their Geometrical Sense

An important role in the application of the retract-type technique is played by the so called strict egress points. Before we define these points, let us describe the boundaries of the setsΩ(n), n∈ℤa∞, in detail. As one can easily see, (2.9)⋃n∈Za∞∂Ω(n)=(⋃j=1mΩBj)∪(⋃j=1mΩCj)
with(2.10)ΩBj:={(n,u):n∈Za∞,uj=bj(n),bi(n)≤ui≤ci(n),i=1,…,m,i≠j},ΩCj:={(n,u):n∈Za∞,uj=cj(n),bi(n)≤ui≤ci(n),i=1,…,m,i≠j}.
In accordance with [3, Lemmas 1 and 2], a point (n,u)∈∂Ω(n) is a point of the type of strict egress for the polyfacial set Ω with respect to discrete system (2.1) if and only if for some j∈{1,…,m}(2.11)uj=bj(n),Fj(n,u)<bj(n+1)-bj(n),
or(2.12)uj=cj(n),Fj(n,u)>cj(n+1)-cj(n).
Geometrically these inequalities mean the following: if a point M=(n,u)∈∂Ω(n) is a point of the type of strict egress, then the first consequent point 𝒞(M)∉Ω(n+1)¯.


### 2.4. Retract and Retraction

IfA⊂B are any two sets in a topological space and π:B→A is a continuous mapping from B onto A such that π(p)=p for every p∈A, then π is said to be a retraction of B onto A. If there exists a retraction of B onto A, then A is called a retract of B.


### 2.5. The Existence Theorem for the System of First-Order Equations (Solution of Problem2)

The following result, solving Problem2, gives sufficient conditions with respect to the right-hand sides of (2.1) which guarantee the existence of at least one solution satisfying (2.5) for every n∈ℤa∞.Theorem 2.1.
Letbi(n),ci(n),bi(n)<ci(n), i=1,…,m, be real functions defined on ℤa∞ and let Fi:ℤa∞×ℝm→ℝ, i=1,…,m, be continuous functions. Suppose that for one fixed j∈{1,…,m} all the points of the sets ΩBj, ΩCj are points of strict egress, that is, if (n,u)∈ΩBj, then
(2.13)Fj(n,u)<bj(n+1)-bj(n),
and if (n,u)∈ΩCj, then
(2.14)Fj(n,u)>cj(n+1)-cj(n).
Further suppose that the set Ω is of Liapunov-type with respect to the ith variable for every i∈{1,…,m}, i≠j, that is, that for every (n,u)∈Ω(2.15)bi(n+1)<ui+Fi(n,u)<ci(n+1).
Then there exists a solution u=(u1*(n),…,um*(n)) of system (2.1) satisfying the inequalities
(2.16)bi(n)<ui*(n)<ci(n),i=1,…,m,
for every n∈ℤa∞.Proof.
The proof will be by contradiction. We will suppose that there exists no solution satisfying inequalities (2.16) for every n∈ℤa∞. Under this supposition we prove that there exists a continuous mapping (a retraction) of a closed interval onto both its endpoints which is, by the intermediate value theorem of calculus, impossible.
Without the loss of generality we may suppose that the indexj in Theorem 2.1 is equal to 1, that is, all the points of the sets ΩB1 and ΩC1 are strict egress points. Each solution of system (2.1) is uniquely determined by the chosen initial condition(2.17)u(a)=(u1(a),…,um(a))=(u1a,…,uma)=ua.
For the following considerations, let uia with uia∈(bi(a),ci(a)), i=2,…,m, be chosen arbitrarily but fixed. Now the solution of (2.1) is given just by the choice of u1a, we can write
(2.18)u=u(n,u1a)=(u1(n,u1a),…,um(n,u1a)).
Define the closed interval I:=[b1(a),c1(a)]. Hereafter we show that, under the supposition that there exists no solution satisfying inequalities (2.16), there exists a retraction ℛ (which will be a composition of two auxiliary mappings ℛ1 and ℛ2 defined below) of the set B:=I onto the set A:=∂I={b1(a),c1(a)}. This contradiction will prove our result. To arrive at such a contradiction, we divide the remaining part of the proof into several steps.Construction of the Leaving Valuen*
Let a pointũ1∈I be fixed. The initial condition u1(a)=ũ1 defines a solution u=u(n,ũ1)=(u1(n,ũ1),…,um(n,ũ1)). According to our supposition, this solution does not satisfy inequalities (2.16) for every n∈ℤa∞. We will study the moment the solution leaves the domain Ω for the first time. The first value of n for which inequalities (2.16) are not valid will be denoted as s.
(I) First consider the caseũ1∈intI. Then there exists a value s>1 in ℤa+1∞ such that(2.19)(s,u(s,ũ1))∉Ω(s)
while
(2.20)(r,u(r,ũ1))∈Ω(r)fora≤r≤s-1.
As the set Ω is of the Liapunov-type with respect to all variables except the first one and (s-1,u(s-1,ũ1))∈Ω(s-1), then
(2.21)bi(s)<ui(s,ũ1)<ci(s),i=2,…,m.
Because j=1 was assumed, and Ω is of Liapunov-type for each variable ui, i≠j, then the validity of inequalities (2.16) has to be violated in the u1-coordinate. The geometrical meaning was explained in Section 2.2.
Now, two cases are possible: either(s,u(s,ũ1))∉Ω(s)¯ or (s,u(s,ũ1))∈∂Ω(s). In the first case u1(s,ũ1)<b1(s) or u1(s,ũ1)>c1(s). In the second case u1(s,ũ1)=b1(s) or u1(s,ũ1)=c1(s) and, due to (2.13) and (2.14), u1(s+1,ũ1)<b1(s+1) or u1(s+1,ũ1)>c1(s+1), respectively.
(II) Ifũ1∈∂I, then (a,u(a,ũ1))∉Ω(a). Thus, for this case, we could put s=a. Further, because of the strict egress property of ΩB1 and ΩC1, either u1(a+1,ũ1)<b1(a+1) (if ũ1=b1(a)) or u1(a+1,ũ1)>c1(a+1) (if ũ1=c1(a)) and thus (a+1,u(a+1,ũ1))∉Ω(a+1)¯.
Unfortunately, for the next consideration the values (the first value of the independent variable for which the graph of the solution is out of Ω) would be of little use. What we will need is the last value for which the graph of the solution stays in Ω¯. We will denote this value as n* and will call it the leaving value. We can define n* as(2.22)n*=s-1if(s,u(s,ũ1))∉Ω(s)¯,n*=sif(s,u(s,ũ1))∈∂Ω(s).
As the value of n* depends on the chosen initial point ũ1, we could write n*=n*(ũ1) but we will mostly omit the argument ũ1, unless it is necessary. From the above considerations it follows that
(2.23)b1(n*)≤u1(n*,ũ1)≤c1(n*),u1(n*+1,ũ1)<b1(n*+1)oru1(n*+1,ũ1)>c1(n*+1).Auxiliary Mappingℛ1
Now we construct the auxiliary mappingℛ1:I→ℝ×ℝ. First extend the discrete functions b1,c1 onto the whole interval [a,∞):
(2.24)b1(t):=b1(⌊t⌋)+(b1(⌊t⌋+1)-b1(⌊t⌋))(t-⌊t⌋),c1(t):=c1(⌊t⌋)+(c1(⌊t⌋+1)-c1(⌊t⌋))(t-⌊t⌋),⌊t⌋ being the integer part of t (the floor function). Note that b1,c1 are now piecewise linear continuous functions of a real variable t such that b1(t)<c1(t) for every t and that the original values of b1(n),c1(n) for n∈ℤa∞ are preserved. This means that the graphs of these functions connect the points (n,b1(n)) or (n,c1(n)) for n∈ℤa∞, respectively. Denote V the set
(2.25)V:={(t,u1):t∈[a,∞),b1(t)≤u1≤c1(t)}.
The boundary of V consists of three mutually disjoint parts Va, Vb, and Vc:
(2.26)∂V=Va∪Vb∪Vc,
where
(2.27)Va:={(a,u1):b1(a)<u1<c1(a)},Vb:={(t,u1):t∈[a,∞),u1=b1(t)},Vc:={(t,u1):t∈[a,∞),u1=c1(t)}.
Define the mapping ℛ1:I→Vb∪Vc as follows: let ℛ1(ũ1) be the point of intersection of the line segment defined by its end points (n*,u1(n*,ũ1)), (n*+1,u1(n*+1,ũ1)) with Vb∪Vc (see Figure 1). The mapping ℛ1 is obviously well defined on I and ℛ1(b1(a))=(a,b1(a)), ℛ1(c1(a))=(a,c1(a)).
Prove that the mappingℛ1 is continuous. The point ℛ1(ũ1)=(t(ũ1),u1(ũ1)) lies either on Vb or on Vc. Without the loss of generality, consider the second case (the first one is analogical). The relevant boundary line segment for t∈[n*,n*+1], which is a part of Vc, is described by (see (2.24))(2.28)u1=c(n*)+(c(n*+1)-c(n*))(t-n*),
and the line segment joining the points (n*,u1(n*,ũ1)), (n*+1,u1(n*+1,ũ1)) by the equation
(2.29)u1=u1(n*,ũ1)+(u1(n*+1,ũ1)-u1(n*,ũ1))(t-n*),t∈[n*,n*+1].
The coordinates of the point ℛ1(ũ1)=(t(ũ1),u1(ũ1)), which is the intersection of both these line segments, can be obtained as the solution of the system consisting of (2.28) and (2.29). Solving this system with respect to t and u1, we get
(2.30)t(ũ1)=n*+u1(n*,ũ1)-c1(n*)c1(n*+1)-u1(n*+1,ũ1)+u1(n*,ũ1)-c1(n*),(2.31)u1(ũ1)=c1(n*)+(u1(n*,ũ1)-c1(n*))(c1(n*+1)-c1(n*))c1(n*+1)-u1(n*+1,ũ1)+u1(n*,ũ1)-c1(n*).
Let {vk}k=1∞ be any sequence with vk∈I such that vk→ũ1. We will show that ℛ1(vk)→ℛ1(ũ1). Because of the continuity of the functions Fi, i=1,…,m,
(2.32)u1(n,vk)→u1(n,ũ1)foreveryfixedn∈Za∞.
We have to consider two cases: (I)
(n*,u(n*,ũ1))∈Ω(n*), that is, b1(n*)<u1(n*,ũ1)<c1(n*),(II)
(n*,u(n*,ũ1))∈∂Ω(n*), that is, u1(n*,ũ1)=c1(n*).
Recall that (due to our agreement) in both casesu1(n*+1,ũ1)>c1(n*+1).
(I)
In this case alsou1(n*,vk)<c1(n*) and u1(n*+1,vk)>c1(n*+1) for k sufficiently large. That means that the leaving value n*(vk) is the same as n* given by ũ1 and thus the point ℛ1(vk)=(t(vk),u1(vk)) is given by
(2.33)t(vk)=n*+u1(n*,vk)-c1(n*)c1(n*+1)-u1(n*+1,vk)+u1(n*,vk)-c1(n*),(2.34)u1(vk)=c1(n*)+(u1(n*,vk)-c1(n*))(c1(n*+1)-c1(n*))c1(n*+1)-u1(n*+1,vk)+u1(n*,vk)-c1(n*).
The desired convergence ℛ1(vk)→ℛ1(ũ1) is implied by equations (2.30) to (2.34).(II)
Supposen*=a. Then ũ1=c1(a), vk=u1(a,vk)<c1(a) for all k and as k→∞, u1(a+1,vk)>c1(a+1). A minor edit of the text in the case (I) proof provides the continuity proof.Supposen*>a. In this case there can be u1(n*,vk)≤c1(n*) for some members of the sequence {vk} and u1(n*,vk)>c1(n*) for the others. Without the loss of generality, we can suppose that {vk} splits into two infinite subsequences {vqk} and {vrk} such that(2.35)u1(n*,vqk)≤c1(n*),u1(n*+1,vqk)>c1(n*+1)u1(n*,vrk)>c1(n*).For the subsequence{vqk}, the text of the proof of (I) can be subjected to a minor edit to provide the proof of continuity. As for the subsequence {vrk}, the leaving value n*(vrk) is different from n* given by ũ1 because (n*,u1(n*,vrk)) is already out of Ω¯. For k sufficiently large,(2.36)n*(vrk)=n*-1becauseu1(n*-1,ũ1)<c1(n*-1) and thus, as k→∞, u1(n*-1,vrk)<c1(n*-1).Hence, the value of the mappingℛ1 for vrk is (in (2.33), (2.34) we replace n* by n*-1)(2.37)t(vrk)=n*-1+u1(n*-1,vrk)-c1(n*-1)c1(n*)-u1(n*,vrk)+u1(n*-1,vrk)-c1(n*-1),u1(vrk)=c(n*-1)+(u1(n*-1,vrk)-c1(n*-1))(c1(n*)-c1(n*-1))c1(n*)-u1(n*,vrk)+u1(n*-1,vrk)-c1(n*-1).Due to (2.32), u1(n*,vrk)→u1(n*,ũ1)=c1(n*) and thus(2.38)t(vrk)⟶n*-1+u1(n*-1,vrk)-c1(n*-1)u1(n*-1,vrk)-c1(n*-1)=n*,u1(vrk)⟶c(n*-1)+(u1(n*-1,vrk)-c1(n*-1))(c1(n*)-c1(n*-1))u1(n*-1,vrk)-c1(n*-1)=c1(n*),R1(vrk)=(t(vrk),u1(vrk))⟶(n*,c1(n*))=R1(ũ1).We have shown thatℛ1(vqk)→ℛ1(ũ1) and ℛ1(vrk)→ℛ1(ũ1) and thus ℛ1(vk)→ℛ1(ũ1).Auxiliary Mappingℛ2
Defineℛ2:Vb∪Vc→{b1(a),c1(a)} as
(2.39)R2(P)={b1(a)ifP∈Vb,c1(a)ifP∈Vc.
The mapping ℛ2 is obviously continuous.Resulting Mappingℛ and Its Properties
Defineℛ:=ℛ2∘ℛ1. Due to construction we have
(2.40)R(b1(a))=b1(a),R(c1(a))=c1(a),
and ℛ(I)=∂I. The mapping ℛ is continuous because of the continuity of the two mappings ℛ1 and ℛ2. Hence, it is the sought retraction of I onto ∂I. But such a retraction cannot exist and thus we get a contradiction and the proof is complete.Figure 1
Construction of the mappingℛ1.


## 2.1. Consequent Point

Define the mapping𝒞:ℤa∞×ℝm→ℤa∞×ℝm as(2.6)C:(n,u)⟼(n+1,u+F(n,u)).
For any point M=(n,u)∈ℤa∞×ℝm, the point 𝒞(M) is called the first consequent point of the point M. The geometrical meaning is that if a point M lies on the graph of some solution of system (2.1), then its first consequent point 𝒞(M) is the next point on this graph.


## 2.2. Liapunov-Type Polyfacial Set

We say that a polyfacial setΩ is Liapunov-type with respect to discrete system (2.1) if(2.7)bi(n+1)<ui+Fi(n,u)<ci(n+1)
for every i=1,…,m and every (n,u)∈Ω. The geometrical meaning of this property is this: if a point M=(n,u) lies inside the set Ω(n), then its first consequent point 𝒞(M) stays inside Ω(n+1).In this contribution we will deal with sets that need not be of Liapunov-type, but they will have, in a certain sense, a similar property. We say that a polyfacial setΩ is Liapunov-type with respect to thejth variable (j∈{1,…,m}) and to discrete system (2.1) if(2.8)(n,u)∈Ω⟹bj(n+1)<uj+Fj(n,u)<cj(n+1).
The geometrical meaning is that if M=(n,u)∈Ω(n), then the uj-coordinate of its first consequent point stays between bj(n+1) and cj(n+1), meanwhile the other coordinates of 𝒞(M) may be arbitrary.


## 2.3. Points of Strict Egress and Their Geometrical Sense

An important role in the application of the retract-type technique is played by the so called strict egress points. Before we define these points, let us describe the boundaries of the setsΩ(n), n∈ℤa∞, in detail. As one can easily see, (2.9)⋃n∈Za∞∂Ω(n)=(⋃j=1mΩBj)∪(⋃j=1mΩCj)
with(2.10)ΩBj:={(n,u):n∈Za∞,uj=bj(n),bi(n)≤ui≤ci(n),i=1,…,m,i≠j},ΩCj:={(n,u):n∈Za∞,uj=cj(n),bi(n)≤ui≤ci(n),i=1,…,m,i≠j}.
In accordance with [3, Lemmas 1 and 2], a point (n,u)∈∂Ω(n) is a point of the type of strict egress for the polyfacial set Ω with respect to discrete system (2.1) if and only if for some j∈{1,…,m}(2.11)uj=bj(n),Fj(n,u)<bj(n+1)-bj(n),
or(2.12)uj=cj(n),Fj(n,u)>cj(n+1)-cj(n).
Geometrically these inequalities mean the following: if a point M=(n,u)∈∂Ω(n) is a point of the type of strict egress, then the first consequent point 𝒞(M)∉Ω(n+1)¯.


## 2.4. Retract and Retraction

IfA⊂B are any two sets in a topological space and π:B→A is a continuous mapping from B onto A such that π(p)=p for every p∈A, then π is said to be a retraction of B onto A. If there exists a retraction of B onto A, then A is called a retract of B.


## 2.5. The Existence Theorem for the System of First-Order Equations (Solution of Problem2)

The following result, solving Problem2, gives sufficient conditions with respect to the right-hand sides of (2.1) which guarantee the existence of at least one solution satisfying (2.5) for every n∈ℤa∞.Theorem 2.1.
Letbi(n),ci(n),bi(n)<ci(n), i=1,…,m, be real functions defined on ℤa∞ and let Fi:ℤa∞×ℝm→ℝ, i=1,…,m, be continuous functions. Suppose that for one fixed j∈{1,…,m} all the points of the sets ΩBj, ΩCj are points of strict egress, that is, if (n,u)∈ΩBj, then
(2.13)Fj(n,u)<bj(n+1)-bj(n),
and if (n,u)∈ΩCj, then
(2.14)Fj(n,u)>cj(n+1)-cj(n).
Further suppose that the set Ω is of Liapunov-type with respect to the ith variable for every i∈{1,…,m}, i≠j, that is, that for every (n,u)∈Ω(2.15)bi(n+1)<ui+Fi(n,u)<ci(n+1).
Then there exists a solution u=(u1*(n),…,um*(n)) of system (2.1) satisfying the inequalities
(2.16)bi(n)<ui*(n)<ci(n),i=1,…,m,
for every n∈ℤa∞.Proof.
The proof will be by contradiction. We will suppose that there exists no solution satisfying inequalities (2.16) for every n∈ℤa∞. Under this supposition we prove that there exists a continuous mapping (a retraction) of a closed interval onto both its endpoints which is, by the intermediate value theorem of calculus, impossible.
Without the loss of generality we may suppose that the indexj in Theorem 2.1 is equal to 1, that is, all the points of the sets ΩB1 and ΩC1 are strict egress points. Each solution of system (2.1) is uniquely determined by the chosen initial condition(2.17)u(a)=(u1(a),…,um(a))=(u1a,…,uma)=ua.
For the following considerations, let uia with uia∈(bi(a),ci(a)), i=2,…,m, be chosen arbitrarily but fixed. Now the solution of (2.1) is given just by the choice of u1a, we can write
(2.18)u=u(n,u1a)=(u1(n,u1a),…,um(n,u1a)).
Define the closed interval I:=[b1(a),c1(a)]. Hereafter we show that, under the supposition that there exists no solution satisfying inequalities (2.16), there exists a retraction ℛ (which will be a composition of two auxiliary mappings ℛ1 and ℛ2 defined below) of the set B:=I onto the set A:=∂I={b1(a),c1(a)}. This contradiction will prove our result. To arrive at such a contradiction, we divide the remaining part of the proof into several steps.Construction of the Leaving Valuen*
Let a pointũ1∈I be fixed. The initial condition u1(a)=ũ1 defines a solution u=u(n,ũ1)=(u1(n,ũ1),…,um(n,ũ1)). According to our supposition, this solution does not satisfy inequalities (2.16) for every n∈ℤa∞. We will study the moment the solution leaves the domain Ω for the first time. The first value of n for which inequalities (2.16) are not valid will be denoted as s.
(I) First consider the caseũ1∈intI. Then there exists a value s>1 in ℤa+1∞ such that(2.19)(s,u(s,ũ1))∉Ω(s)
while
(2.20)(r,u(r,ũ1))∈Ω(r)fora≤r≤s-1.
As the set Ω is of the Liapunov-type with respect to all variables except the first one and (s-1,u(s-1,ũ1))∈Ω(s-1), then
(2.21)bi(s)<ui(s,ũ1)<ci(s),i=2,…,m.
Because j=1 was assumed, and Ω is of Liapunov-type for each variable ui, i≠j, then the validity of inequalities (2.16) has to be violated in the u1-coordinate. The geometrical meaning was explained in Section 2.2.
Now, two cases are possible: either(s,u(s,ũ1))∉Ω(s)¯ or (s,u(s,ũ1))∈∂Ω(s). In the first case u1(s,ũ1)<b1(s) or u1(s,ũ1)>c1(s). In the second case u1(s,ũ1)=b1(s) or u1(s,ũ1)=c1(s) and, due to (2.13) and (2.14), u1(s+1,ũ1)<b1(s+1) or u1(s+1,ũ1)>c1(s+1), respectively.
(II) Ifũ1∈∂I, then (a,u(a,ũ1))∉Ω(a). Thus, for this case, we could put s=a. Further, because of the strict egress property of ΩB1 and ΩC1, either u1(a+1,ũ1)<b1(a+1) (if ũ1=b1(a)) or u1(a+1,ũ1)>c1(a+1) (if ũ1=c1(a)) and thus (a+1,u(a+1,ũ1))∉Ω(a+1)¯.
Unfortunately, for the next consideration the values (the first value of the independent variable for which the graph of the solution is out of Ω) would be of little use. What we will need is the last value for which the graph of the solution stays in Ω¯. We will denote this value as n* and will call it the leaving value. We can define n* as(2.22)n*=s-1if(s,u(s,ũ1))∉Ω(s)¯,n*=sif(s,u(s,ũ1))∈∂Ω(s).
As the value of n* depends on the chosen initial point ũ1, we could write n*=n*(ũ1) but we will mostly omit the argument ũ1, unless it is necessary. From the above considerations it follows that
(2.23)b1(n*)≤u1(n*,ũ1)≤c1(n*),u1(n*+1,ũ1)<b1(n*+1)oru1(n*+1,ũ1)>c1(n*+1).Auxiliary Mappingℛ1
Now we construct the auxiliary mappingℛ1:I→ℝ×ℝ. First extend the discrete functions b1,c1 onto the whole interval [a,∞):
(2.24)b1(t):=b1(⌊t⌋)+(b1(⌊t⌋+1)-b1(⌊t⌋))(t-⌊t⌋),c1(t):=c1(⌊t⌋)+(c1(⌊t⌋+1)-c1(⌊t⌋))(t-⌊t⌋),⌊t⌋ being the integer part of t (the floor function). Note that b1,c1 are now piecewise linear continuous functions of a real variable t such that b1(t)<c1(t) for every t and that the original values of b1(n),c1(n) for n∈ℤa∞ are preserved. This means that the graphs of these functions connect the points (n,b1(n)) or (n,c1(n)) for n∈ℤa∞, respectively. Denote V the set
(2.25)V:={(t,u1):t∈[a,∞),b1(t)≤u1≤c1(t)}.
The boundary of V consists of three mutually disjoint parts Va, Vb, and Vc:
(2.26)∂V=Va∪Vb∪Vc,
where
(2.27)Va:={(a,u1):b1(a)<u1<c1(a)},Vb:={(t,u1):t∈[a,∞),u1=b1(t)},Vc:={(t,u1):t∈[a,∞),u1=c1(t)}.
Define the mapping ℛ1:I→Vb∪Vc as follows: let ℛ1(ũ1) be the point of intersection of the line segment defined by its end points (n*,u1(n*,ũ1)), (n*+1,u1(n*+1,ũ1)) with Vb∪Vc (see Figure 1). The mapping ℛ1 is obviously well defined on I and ℛ1(b1(a))=(a,b1(a)), ℛ1(c1(a))=(a,c1(a)).
Prove that the mappingℛ1 is continuous. The point ℛ1(ũ1)=(t(ũ1),u1(ũ1)) lies either on Vb or on Vc. Without the loss of generality, consider the second case (the first one is analogical). The relevant boundary line segment for t∈[n*,n*+1], which is a part of Vc, is described by (see (2.24))(2.28)u1=c(n*)+(c(n*+1)-c(n*))(t-n*),
and the line segment joining the points (n*,u1(n*,ũ1)), (n*+1,u1(n*+1,ũ1)) by the equation
(2.29)u1=u1(n*,ũ1)+(u1(n*+1,ũ1)-u1(n*,ũ1))(t-n*),t∈[n*,n*+1].
The coordinates of the point ℛ1(ũ1)=(t(ũ1),u1(ũ1)), which is the intersection of both these line segments, can be obtained as the solution of the system consisting of (2.28) and (2.29). Solving this system with respect to t and u1, we get
(2.30)t(ũ1)=n*+u1(n*,ũ1)-c1(n*)c1(n*+1)-u1(n*+1,ũ1)+u1(n*,ũ1)-c1(n*),(2.31)u1(ũ1)=c1(n*)+(u1(n*,ũ1)-c1(n*))(c1(n*+1)-c1(n*))c1(n*+1)-u1(n*+1,ũ1)+u1(n*,ũ1)-c1(n*).
Let {vk}k=1∞ be any sequence with vk∈I such that vk→ũ1. We will show that ℛ1(vk)→ℛ1(ũ1). Because of the continuity of the functions Fi, i=1,…,m,
(2.32)u1(n,vk)→u1(n,ũ1)foreveryfixedn∈Za∞.
We have to consider two cases: (I)
(n*,u(n*,ũ1))∈Ω(n*), that is, b1(n*)<u1(n*,ũ1)<c1(n*),(II)
(n*,u(n*,ũ1))∈∂Ω(n*), that is, u1(n*,ũ1)=c1(n*).
Recall that (due to our agreement) in both casesu1(n*+1,ũ1)>c1(n*+1).
(I)
In this case alsou1(n*,vk)<c1(n*) and u1(n*+1,vk)>c1(n*+1) for k sufficiently large. That means that the leaving value n*(vk) is the same as n* given by ũ1 and thus the point ℛ1(vk)=(t(vk),u1(vk)) is given by
(2.33)t(vk)=n*+u1(n*,vk)-c1(n*)c1(n*+1)-u1(n*+1,vk)+u1(n*,vk)-c1(n*),(2.34)u1(vk)=c1(n*)+(u1(n*,vk)-c1(n*))(c1(n*+1)-c1(n*))c1(n*+1)-u1(n*+1,vk)+u1(n*,vk)-c1(n*).
The desired convergence ℛ1(vk)→ℛ1(ũ1) is implied by equations (2.30) to (2.34).(II)
Supposen*=a. Then ũ1=c1(a), vk=u1(a,vk)<c1(a) for all k and as k→∞, u1(a+1,vk)>c1(a+1). A minor edit of the text in the case (I) proof provides the continuity proof.Supposen*>a. In this case there can be u1(n*,vk)≤c1(n*) for some members of the sequence {vk} and u1(n*,vk)>c1(n*) for the others. Without the loss of generality, we can suppose that {vk} splits into two infinite subsequences {vqk} and {vrk} such that(2.35)u1(n*,vqk)≤c1(n*),u1(n*+1,vqk)>c1(n*+1)u1(n*,vrk)>c1(n*).For the subsequence{vqk}, the text of the proof of (I) can be subjected to a minor edit to provide the proof of continuity. As for the subsequence {vrk}, the leaving value n*(vrk) is different from n* given by ũ1 because (n*,u1(n*,vrk)) is already out of Ω¯. For k sufficiently large,(2.36)n*(vrk)=n*-1becauseu1(n*-1,ũ1)<c1(n*-1) and thus, as k→∞, u1(n*-1,vrk)<c1(n*-1).Hence, the value of the mappingℛ1 for vrk is (in (2.33), (2.34) we replace n* by n*-1)(2.37)t(vrk)=n*-1+u1(n*-1,vrk)-c1(n*-1)c1(n*)-u1(n*,vrk)+u1(n*-1,vrk)-c1(n*-1),u1(vrk)=c(n*-1)+(u1(n*-1,vrk)-c1(n*-1))(c1(n*)-c1(n*-1))c1(n*)-u1(n*,vrk)+u1(n*-1,vrk)-c1(n*-1).Due to (2.32), u1(n*,vrk)→u1(n*,ũ1)=c1(n*) and thus(2.38)t(vrk)⟶n*-1+u1(n*-1,vrk)-c1(n*-1)u1(n*-1,vrk)-c1(n*-1)=n*,u1(vrk)⟶c(n*-1)+(u1(n*-1,vrk)-c1(n*-1))(c1(n*)-c1(n*-1))u1(n*-1,vrk)-c1(n*-1)=c1(n*),R1(vrk)=(t(vrk),u1(vrk))⟶(n*,c1(n*))=R1(ũ1).We have shown thatℛ1(vqk)→ℛ1(ũ1) and ℛ1(vrk)→ℛ1(ũ1) and thus ℛ1(vk)→ℛ1(ũ1).Auxiliary Mappingℛ2
Defineℛ2:Vb∪Vc→{b1(a),c1(a)} as
(2.39)R2(P)={b1(a)ifP∈Vb,c1(a)ifP∈Vc.
The mapping ℛ2 is obviously continuous.Resulting Mappingℛ and Its Properties
Defineℛ:=ℛ2∘ℛ1. Due to construction we have
(2.40)R(b1(a))=b1(a),R(c1(a))=c1(a),
and ℛ(I)=∂I. The mapping ℛ is continuous because of the continuity of the two mappings ℛ1 and ℛ2. Hence, it is the sought retraction of I onto ∂I. But such a retraction cannot exist and thus we get a contradiction and the proof is complete.Figure 1
Construction of the mappingℛ1.


## 3. Application of Theorem2.1 to the Delayed Discrete Equation

Now, let us return to the original delayed discrete equation (1.2), that is, (3.1)Δv(n)=f(n,v(n),v(n-1),…,v(n-k)).
As it was said in Section 1, this equation will be transformed to a system of k+1 first-order discrete equations. Then we will apply Theorem 2.1 to this system and prove that under certain conditions there exists a solution of delayed equation (1.2) that stays in the prescribed domain. In the end, we will study a special case of (1.2).


### 3.1. Transformation of (1.2) to the System of First-Order Equations

We will proceed in accordance with the well-known scheme similarly as when constructing the system of first-order differential equations from a differential equation of a higher order. Put(3.2)u1(n):=v(n),u2(n):=v(n-1),⋯uk+1(n):=v(n-k),
where u1,u2,…,uk+1 are new unknown functions. From (1.2) we get Δu1(n)=f(n,u1(n),u2(n),…,uk+1(n)). Obviously u2(n+1)=u1(n),…,uk+1(n+1)=uk(n). Rewriting these equalities in terms of differences, we have Δu2(n)=u1(n)-u2(n),…,Δuk+1(n)=uk(n)-uk+1(n). Altogether, we get the system(3.3)Δu1(n)=f(n,u1(n),…,uk+1(n)),Δu2(n)=u1(n)-u2(n),⋯Δuk+1(n)=uk(n)-uk+1(n)
which is equivalent to (1.2).


### 3.2. The Existence Theorem for the Delayed Equation (1.2)  (Solution of Problem 1)

The following theorem is a consequence of Theorem2.1. In fact, this theorem has been already proved in [12]. There, the proof is based upon a modification of the retract method for delayed equations. Our method (rearranging a delayed equation to a system of first-order equations) is, by its principle, more general than that used in [12].Theorem 3.1.
Letb(n), c(n), b(n)<c(n), be real functions defined on ℤa-k∞. Further, let f:ℤa∞×ℝk+1→ℝ be a continuous function and let the inequalities
(3.4)b(n)+f(n,b(n),v2,…,vk+1)<b(n+1),(3.5)c(n)+f(n,c(n),v2,…,vk+1)>c(n+1)
hold for every n∈ℤa∞ and every v2,…,vk+1 such that
(3.6)b(n-i+1)<vi<c(n-i+1),i=2,…,k+1.
Then there exists a solution v=v*(n) of (1.2) satisfying the inequalities
(3.7)b(n)<v*(n)<c(n)
for every n∈ℤa-k∞.Proof.
We have shown that (1.2) is equivalent to system (3.3) which can be seen as a special case of system (2.1) with m=k+1 and F=(F1,…Fk+1) where
(3.8)F1(n,u1,…,uk+1):=f(n,u1,…,uk+1),F2(n,u1,…,uk+1):=u1-u2,⋯Fk(n,u1,…,uk+1):=uk-1-uk,Fk+1(n,u1,…,uk+1):=uk-uk+1.
Define the polyfacial set Ω as
(3.9)Ω:={(n,u):n∈Za∞,bi(n)<ui<ci(n),i=1,…,k+1}
with
(3.10)bi(n):=b(n-i+1),ci(n):=c(n-i+1),i=1,…,k+1.
We will show that for system (3.3) and the set Ω, all the assumptions of Theorem 2.1 are satisfied.
As the functionf is supposed to be continuous, the mapping F is continuous, too. Put the index j from Theorem 2.1, characterizing the points of egress, equal to 1. We will verify that the set Ω is of Liapunov-type with respect to the ith variable for any i=2,…,k+1, that is, (see (2.8)) that for every (n,u)∈Ω(3.11)bi(n+1)<ui+Fi(n,u)<ci(n+1)fori=2,…,k+1.
First, we compute
(3.12)ui+Fi(n,u)=ui+ui-1-ui=ui-1fori=2,…,k+1.
Thus we have to show that for i=2,…,k+1(3.13)bi(n+1)<ui-1<ci(n+1).
Because (n,u)∈Ω, then bp(n)<up<cp(n) for any p∈{1,…,k+1}, and therefore
(3.14)bi-1(n)<ui-1<ci-1(n)fori=2,…,k+1.
But, by (3.10), we have
(3.15)bi-1(n)=b(n-i+1+1)=b(n-i+2),
meanwhile
(3.16)bi(n+1)=b(n+1-i+1)=b(n-i+2),
and thus bi-1(n)=bi(n+1). Analogously we get that ci-1(n)=ci(n+1). Thus inequalities (3.11) are fulfilled.
Further we will show that all the boundary pointsM∈ΩB1∪ΩC1 are points of strict egress for the set Ω with respect to system (3.3). According to (2.11), we have to show that if u1=b1(n) and bi(n)<ui<ci(n) for i=2,…,k+1, then(3.17)b1(n)+F1(n,u)<b1(n+1),
that is,
(3.18)b1(n)+f(n,b1(n),u2,…,uk+1)<b1(n+1).
Notice that the condition bi(n)<ui<ci(n) for i=2,…,k+1 is equivalent with condition b(n-i+1)<ui<c(n-i+1) (see (3.10)). Looking at the supposed inequality (3.4) and realizing that b1(n)=b(n) and b1(n+1)=b(n+1), we can see that inequality (3.18) is fulfilled.
Analogously, according to (2.12), we have to prove that for u1=c1(n) and bi(n)<ui<ci(n) for i=2,…,k+1 the inequality(3.19)c1(n)+F1(n,u)>c1(n+1),
that is,
(3.20)c1(n)+f(n,c1(n),u2,…,uk+1)>c1(n+1)
holds.
Again, considering (3.5) and the fact that c1(n)=c(n) and c1(n+1)=c(n+1), we can see that this inequality really holds.
Thus, by the assertion of Theorem2.1, there exists a solution u=u*(n) of system (3.3) such that for every n∈ℤa∞(3.21)bi(n)<ui*(n)<ci(n)fori=1,…,k+1.
In our case,v=v*(n)=u1*(n) is the solution of the original equation (1.2). Further, b1(n)=b(n) and c1(n)=c(n), and thus the existence of a solution of the delayed equation (1.2) such that inequalities (3.7) are satisfied is guaranteed.


### 3.3. Asymptotic Solution Estimates for Delayed Difference Equations

Let us suppose that two functionsu,w:ℤa-k∞→ℝ are given such that (3.22)u(n)<w(n),n∈Za-k∞,(3.23)Δu(n)≥f(n,u(n),u(n-1),…,u(n-k)),n∈Za∞,(3.24)Δw(n)≤f(n,w(n),w(n-1),…,w(n-k)),n∈Za∞.
Consider the problem of whether there exists a solution v=v*(n), n∈ℤa-k∞ of (1.2) such that (3.25)u(n)<v*(n)<w(n),n∈Za-k∞.
The following corollary of Theorem 3.1 presents sufficient conditions for the existence of a solution of this problem.Corollary 3.2.
Let functionsu,w:ℤa-k∞→ℝ satisfy inequalities (3.22)–(3.24). Let f:ℤa∞×ℝk+1→ℝ be a continuous function such that
(3.26)f(n,u(n),y2,…,yk+1)>f(n,u(n),z2,…,zk+1),(3.27)f(n,w(n),y2,…,yk+1)>f(n,w(n),z2,…,zk+1)
for every n∈ℤa∞ and every y2,…,yk+1,z2,…,zk+1∈ℝ such that
(3.28)yi<zi,i=2,…,k+1.
Then there exists a solution v=v*(n) of (1.2) satisfying inequalities (3.25) for every n∈ℤa-k∞.Proof.
This assertion is an easy consequence of Theorem3.1.
Putb(n):=u(n), c(n):=w(n). Considering inequalities (3.23) and (3.26), we can see that(3.29)Δu(n)>f(n,u(n),v2,…,vk+1)
for every n∈ℤa∞ and every v2,…,vk+1 such that
(3.30)b(n-i+1)<vi<c(n-i+1),i=2,…,k+1.
Similarly,
(3.31)Δw(n)<f(n,w(n),v2,…,vk+1)
for every n∈ℤa∞ and every b(n-i+1)<vi<c(n-i+1), i=2,…,k+1.
Obviously, inequalities (3.29) and (3.31) are equivalent with inequalities (3.4) and (3.5), respectively. Thus, all the assumptions of Theorem 3.1 are satisfied and there exists a solution v=v*(n) of (1.2) satisfying inequalities (3.25) for every n∈ℤa-k∞.Example 3.3.
Consider the equation(3.32)Δv(n)=v2(n)-v(n-1)
for n∈ℤ3∞ which is a second-order delayed discrete equation with delay k=1. We will show that there exists a solution v=v*(n) of (3.32) that satisfies the inequalities
(3.33)1<v*(n)<n
for n∈ℤ2∞.We will prove that for the functions(3.34)u(n):=1,w(n):=n,f(n,v1,v2):=v12-v2
all the assumptions of Corollary 3.2 are satisfied. Inequality (3.22) is obviously fulfilled for n∈ℤ2∞. Inequality (3.23) can be also proved very easily:(3.35)Δu(n)=0,f(n,u(n),u(n-1))=12-1=0,
and thus for every n∈ℤ3∞, Δu(n)≥f(n,u(n),u(n-1)).As for inequality (3.24), we get(3.36)Δw(n)=1,f(n,w(n),w(n-1))=n2-n+1
and thus Δw(n)≤f(n,w(n),w(n-1)) for n∈ℤ3∞.Finally, the functions(3.37)f(n,u(n),v2)=1-v2,f(n,w(n),v2)=n2-v2
are decreasing with respect to v2. Therefore, conditions (3.26) and (3.27) are satisfied, too. Hence, due to Corollary 3.2, there exists a solution of (3.32) satisfying (3.33).


### 3.4. Positive Solutions of a Linear Equation with a Single Delay

We will apply the result of Theorem3.1 to the investigation of a simple linear difference equation of the (k+1)-st order with only one delay, namely, the equation(3.38)Δv(n)=-p(n)v(n-k),
where, again, n∈ℤa∞ is the independent variable and k∈ℕ, k≥1, is the fixed delay. The function p:ℤa∞→ℝ is assumed to be positive. Our goal is to give sharp sufficient conditions for the existence of positive solutions. The existence of such solutions is very often substantial for a concrete model considered. For example, in biology, when a model of population dynamics is described by an equation, the positivity of a solution may mean that the studied biological species can survive in the supposed environment.For its simple form, (3.38) often serves for testing new results and is very frequently investigated. It was analyzed, for example, in papers [10, 11, 26]. A sharp result on existence of positive solutions given in [26] is proved by a comparison method [8, 9]. Here we will use Theorem 3.1 to generalize this result.For the purposes of this section, define the expressionlnqt, where q∈ℕ, as (3.39)ln⁡qt:=ln⁡(ln⁡q-1t)ln⁡0t:=t.
We will write only lnt instead of ln1t. Further, for a fixed integer ℓ≥0 define auxiliary functions(3.40)μl(n):=18n2+18(nln⁡n)2+⋯+18(nln⁡n⋯ln⁡ln)2,pl(n):=(kk+1)k⋅(1k+1+kμl(n)),(3.41)νl(n):=(kk+1)n⋅nln⁡nln⁡2n⋯ln⁡ln.
In [26], it was proved that if p(n) in (3.38) is a positive function bounded by pℓ(n) for some ℓ≥0, then there exists a positive solution of (3.38) bounded by the function νℓ(n) for n sufficiently large. Since limn→∞νℓ(n)=0, such solution will vanish for n→∞. Here we show that (3.38) has a positive solution bounded by νℓ(n) even if the coefficient p(n) satisfies a less restrictive inequality (see inequality (3.58) below). The proof of this statement will be based on the following four lemmas. The symbols “o” and “O” stand for the Landau order symbols and are used for n→∞.Lemma 3.4.
The formula(3.42)ln⁡(y-z)=ln⁡y-∑i=1∞ziiyi
holds for any numbers y,z∈ℝ such that y>0 and |z|<y.Proof.
The assertion is a simple consequence of the well-known Maclaurin expansion(3.43)ln⁡(1-x)=-∑i=1∞1ixifor-1≤x<1.
As ln(y-z)-lny=ln(1-z/y), substituting x=z/y we get
(3.44)ln⁡(y-z)-ln⁡y=-∑i=1∞ziiyifor-y≤z<y
and adding lny to both sides of this equality, we get (3.42).Lemma 3.5.
For fixedr∈ℝ∖{0} and fixed q∈ℕ, the asymptotic representation
(3.45)ln⁡q(n-r)=ln⁡qn-rnln⁡n⋯ln⁡q-1n-r22n2ln⁡n⋯ln⁡q-1n-r22(nln⁡n)2ln⁡2n⋯ln⁡q-1n-⋯-r22(nln⁡n⋯ln⁡q-1n)2-r3(1+o(1))3n3ln⁡n⋯ln⁡q-1n
holds for n→∞.Proof.
We will prove relation (3.45) by induction with respect to q. For q=1, (3.45) reduces to
(3.46)ln⁡(n-r)=ln⁡n-rn-r22n2-r3(1+o(1))3n3
which holds due to Lemma 3.4. Suppose that relation (3.45) holds for some q. We can write lnq(n-r)=y-z with y=lnqn and
(3.47)z=rnln⁡n⋯ln⁡q-1n+r22n2ln⁡n⋯ln⁡q-1n+r22(nln⁡n)2ln⁡2n⋯ln⁡q-1n+⋯+r22(nln⁡n⋯ln⁡q-1n)2+r3(1+o(1))3n3ln⁡n⋯ln⁡q-1n.
Now we will show that (3.45) holds for q+1. Notice that in our case, the condition |z|<y from Lemma 3.4 is fulfilled for n sufficiently large because z→0 for n→∞, meanwhile y→∞ for n→∞. Thus we are justified to use Lemma 3.4 and doing so, we get
(3.48)ln⁡q+1(n-r)=ln⁡(ln⁡q(n-r))=ln⁡(y-z)=ln⁡y-1yz-12y2z2-⋯=ln⁡(ln⁡qn)-1ln⁡qn⋅(rnln⁡n⋯ln⁡q-1n+r22n2ln⁡n⋯ln⁡q-1n+⋯+r22(nln⁡n⋯ln⁡q-1n)2+r3(1+o(1))3n3ln⁡n⋯ln⁡q-1n)-12(ln⁡qn)2⋅(r2(nln⁡n⋯ln⁡q-1n)2+O(1n3(ln⁡n⋯ln⁡q-1n)2))+O(1(nln⁡n⋯ln⁡qn)3)=ln⁡q+1n-rnln⁡n⋯ln⁡qn-r22n2ln⁡n⋯ln⁡qn-r22(nln⁡n)2ln⁡2n⋯ln⁡qn-⋯-r22(nln⁡n⋯ln⁡qn)2-r3(1+o(1))3n3ln⁡n⋯ln⁡qn.
Thus, formula (3.45) holds for q+1, too, which ends the proof.Lemma 3.6.
For fixedr∈ℝ∖{0} and fixed q∈ℕ, the asymptotic representations
(3.49)ln⁡q(n-r)ln⁡qn=1-r2nln⁡n⋯ln⁡qn-r24n2ln⁡n⋯ln⁡qn-r24(nln⁡n)2ln⁡2n⋯ln⁡qn-⋯-r24(nln⁡n⋯ln⁡q-1n)2ln⁡qn-r28(nln⁡n⋯ln⁡qn)2-r3(1+o(1))6n3ln⁡n⋯ln⁡qn,(3.50)n-rn=1-r2n-r28n2-r316n3+o(1n3)
hold for n→∞.Proof.
Both these relations are simple consequences of the asymptotic formula(3.51)1-x=1-12x-18x2-116x3+o(x3)forx⟶0
and of Lemma 3.5 (for formula (3.49)). In the case of relation (3.49), we put
(3.52)x=rnln⁡n⋯ln⁡qn+r22n2ln⁡n⋯ln⁡qn+⋯+r22(nln⁡n⋯ln⁡q-1n)2ln⁡qn+r3(1+o(1))3n3ln⁡n⋯ln⁡qn
and in the case of relation (3.50), we put x=r/n.Lemma 3.7.
For fixedr∈ℝ∖{0} and fixed q∈ℕ, the asymptotic representation
(3.53)(n-r)nln⁡(n-r)ln⁡n⋯ln⁡q(n-r)ln⁡qn=1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡qn)-r2μq(n)-r316n3+o(1n3)
holds for n→∞.Proof.
We will prove relation (3.53) by induction with respect to q. For q=1, (3.53) reduces to
(3.54)(n-r)nln⁡(n-r)ln⁡n=1-r(12n+12nln⁡n)-r2μ1(n)-r316n3+o(1n3)=1-r(12n+12nln⁡n)-r2(18n2+18(nln⁡n)2)-r316n3+o(1n3).
On the other hand, using Lemma 3.6, we get
(3.55)(n-r)nln⁡(n-r)ln⁡n=(1-r2n-r28n2-r316n3+o(1n3))×(1-r2nln⁡n-r24n2ln⁡n-r28(nln⁡n)2-r3(1+o(1))6n3ln⁡n)=1-r2nln⁡n-r24n2ln⁡n-r28(nln⁡n)2-r2n+r24n2ln⁡n-r28n2-r316n3+o(1n3)=1-r(12n+12nln⁡n)-r2(18n2+18(nln⁡n)2)-r316n3+o(1n3).
Thus, for q=1, relation (3.53) holds. Now suppose that (3.53) holds for some q and prove that it holds for q+1. In the following calculations, we use Lemma 3.6 and we skip some tedious expressions handling.
(3.56)(n-r)nln⁡(n-r)ln⁡n⋯ln⁡q+1(n-r)ln⁡q+1n=(n-r)nln⁡(n-r)ln⁡n⋯ln⁡q(n-r)ln⁡qn⋅ln⁡q+1(n-r)ln⁡q+1n=(1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡qn)-r2μq(n)-r316n3+o(1n3))×(1-r2nln⁡n⋯ln⁡q+1n-r24n2ln⁡n⋯ln⁡q+1n-⋯-r24(nln⁡n⋯ln⁡qn)2ln⁡q+1n-r28(nln⁡n⋯ln⁡q+1n)2+o(1n3))=1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡q+1n)-r2μq+1(n)-r316n3+o(1n3).
We can see that formula (3.53) holds for q+1, too, which ends the proof.Now we are ready to prove that there exists a bounded positive solution of (3.38). Remind that functions pℓ and νℓ were defined by (3.40) and (3.41), respectively.Theorem 3.8.
Letω:ℤa∞→ℝ satisfy the inequality
(3.57)|ω(n)|≤ɛ(kk+1)k⋅k(2k2+k-1)16n3(k+1),n∈Za∞,
for a fixed ɛ∈(0,1). Suppose that there exists an integer ℓ≥0 such that the function p satisfies the inequalities
(3.58)0<p(n)≤pl(n)+ω(n)
for every n∈ℤa∞. Then there exists a solution v=v*(n), n∈ℤa-k∞ of (3.38) such that for n sufficiently large the inequalities
(3.59)0<v*(n)<νl(n)
hold.Proof.
Show that all the assumptions of Theorem3.1 are fulfilled. For (3.38), f(n,v1,…,vk+1)=-p(n)vk+1. This is a continuous function. Put
(3.60)b(n):=0,c(n):=νl(n).
We have to prove that for every v2,…,vk+1 such that b(n-i+1)<vi<c(n-i+1), i=2,…,k+1, the inequalities (3.4) and (3.5) hold for n sufficiently large. Start with (3.4). That gives that for b(n-k)<vk+1<c(n-k), it has to be
(3.61)0-p(n)⋅vk+1<0.
This certainly holds, because the function p is positive and so is vk+1.
Next, according to (3.5), we have to prove that(3.62)νl(n)-p(n)vk+1>νl(n+1)
which is equivalent to the inequality
(3.63)-p(n)vk+1>νl(n+1)-νl(n).
Denote the left-hand side of (3.63) as L(3.63). As vk+1<c(n-k)=νℓ(n-k) and as by (3.40), (3.58), and (3.57)
(3.64)p(n)≤(kk+1)k⋅(1k+1+kμl(n))+ɛ(kk+1)k⋅k(2k2+k-1)16n3(k+1),
we have
(3.65)L(3.63)>-(kk+1)k(1k+1+kμl(n)+ɛ⋅k(2k2+k-1)16n3(k+1))×(kk+1)n-k(n-k)ln⁡(n-k)⋯ln⁡l(n-k)=-(kk+1)n(1k+1+kμl(n)+ɛ⋅k(2k2+k-1)16n3(k+1))⋅(n-k)ln⁡(n-k)⋯ln⁡l(n-k).
Further, we can easily see that
(3.66)νl(n+1)-νl(n)=(kk+1)nnln⁡n⋯ln⁡ln(kk+1(n+1)nln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡ln-1).
Thus, to prove (3.63), it suffices to show that for n sufficiently large,
(3.67)-(1k+1+kμl(n)+ε⋅k(2k2+k-1)16n3(k+1))(n-k)nln⁡(n-k)ln⁡n⋯ln⁡l(n-k)ln⁡ln>kk+1(n+1)nln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡ln-1.
Denote the left-hand side of inequality (3.67) as L(3.67) and the right-hand side as R(3.67). Using Lemma 3.7 with r=k and q=ℓ, we can write
(3.68)L(3.67)=-(1k+1+kμl(n)+ε⋅k(2k2+k-1)16n3(k+1))×(1-k(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-k2μl(n)-k316n3+o(1n3))=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)+k2k+1μl(n)+k316n3(k+1)-kμl(n)+k216n3-ε⋅k(2k2+k-1)16n3(k+1)+o(1n3)=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-kk+1μl(n)+2k3(1-ε)+k2(1-ε)+kε16n3(k+1)+o(1n3).
Using Lemma 3.7 with r=-1 and q=ℓ, we get for R(3.67)(3.69)R(3.67)=kk+1(1+12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-μl(n)+116n3+o(1n3))-1=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-kk+1⋅μl(n)+k16n3(k+1)+o(1n3).
It is easy to see that the inequality (3.67) reduces to
(3.70)2k3(1-ɛ)+k2(1-ɛ)+kɛ16n3(k+1)+o(1n3)>k16n3(k+1)+o(1n3).
This inequality is equivalent to
(3.71)k(2k2(1-ɛ)+k(1-ɛ)-(1-ɛ))16n3(k+1)+o(1n3)>0.
The last inequality holds for n sufficiently large because k≥1 and 1-ɛ∈(0,1). We have proved that all the assumptions of Theorem 3.1 are fulfilled and hence there exists a solution of (3.38) satisfying conditions (3.59).


## 3.1. Transformation of (1.2) to the System of First-Order Equations

We will proceed in accordance with the well-known scheme similarly as when constructing the system of first-order differential equations from a differential equation of a higher order. Put(3.2)u1(n):=v(n),u2(n):=v(n-1),⋯uk+1(n):=v(n-k),
where u1,u2,…,uk+1 are new unknown functions. From (1.2) we get Δu1(n)=f(n,u1(n),u2(n),…,uk+1(n)). Obviously u2(n+1)=u1(n),…,uk+1(n+1)=uk(n). Rewriting these equalities in terms of differences, we have Δu2(n)=u1(n)-u2(n),…,Δuk+1(n)=uk(n)-uk+1(n). Altogether, we get the system(3.3)Δu1(n)=f(n,u1(n),…,uk+1(n)),Δu2(n)=u1(n)-u2(n),⋯Δuk+1(n)=uk(n)-uk+1(n)
which is equivalent to (1.2).


## 3.2. The Existence Theorem for the Delayed Equation (1.2)  (Solution of Problem 1)

The following theorem is a consequence of Theorem2.1. In fact, this theorem has been already proved in [12]. There, the proof is based upon a modification of the retract method for delayed equations. Our method (rearranging a delayed equation to a system of first-order equations) is, by its principle, more general than that used in [12].Theorem 3.1.
Letb(n), c(n), b(n)<c(n), be real functions defined on ℤa-k∞. Further, let f:ℤa∞×ℝk+1→ℝ be a continuous function and let the inequalities
(3.4)b(n)+f(n,b(n),v2,…,vk+1)<b(n+1),(3.5)c(n)+f(n,c(n),v2,…,vk+1)>c(n+1)
hold for every n∈ℤa∞ and every v2,…,vk+1 such that
(3.6)b(n-i+1)<vi<c(n-i+1),i=2,…,k+1.
Then there exists a solution v=v*(n) of (1.2) satisfying the inequalities
(3.7)b(n)<v*(n)<c(n)
for every n∈ℤa-k∞.Proof.
We have shown that (1.2) is equivalent to system (3.3) which can be seen as a special case of system (2.1) with m=k+1 and F=(F1,…Fk+1) where
(3.8)F1(n,u1,…,uk+1):=f(n,u1,…,uk+1),F2(n,u1,…,uk+1):=u1-u2,⋯Fk(n,u1,…,uk+1):=uk-1-uk,Fk+1(n,u1,…,uk+1):=uk-uk+1.
Define the polyfacial set Ω as
(3.9)Ω:={(n,u):n∈Za∞,bi(n)<ui<ci(n),i=1,…,k+1}
with
(3.10)bi(n):=b(n-i+1),ci(n):=c(n-i+1),i=1,…,k+1.
We will show that for system (3.3) and the set Ω, all the assumptions of Theorem 2.1 are satisfied.
As the functionf is supposed to be continuous, the mapping F is continuous, too. Put the index j from Theorem 2.1, characterizing the points of egress, equal to 1. We will verify that the set Ω is of Liapunov-type with respect to the ith variable for any i=2,…,k+1, that is, (see (2.8)) that for every (n,u)∈Ω(3.11)bi(n+1)<ui+Fi(n,u)<ci(n+1)fori=2,…,k+1.
First, we compute
(3.12)ui+Fi(n,u)=ui+ui-1-ui=ui-1fori=2,…,k+1.
Thus we have to show that for i=2,…,k+1(3.13)bi(n+1)<ui-1<ci(n+1).
Because (n,u)∈Ω, then bp(n)<up<cp(n) for any p∈{1,…,k+1}, and therefore
(3.14)bi-1(n)<ui-1<ci-1(n)fori=2,…,k+1.
But, by (3.10), we have
(3.15)bi-1(n)=b(n-i+1+1)=b(n-i+2),
meanwhile
(3.16)bi(n+1)=b(n+1-i+1)=b(n-i+2),
and thus bi-1(n)=bi(n+1). Analogously we get that ci-1(n)=ci(n+1). Thus inequalities (3.11) are fulfilled.
Further we will show that all the boundary pointsM∈ΩB1∪ΩC1 are points of strict egress for the set Ω with respect to system (3.3). According to (2.11), we have to show that if u1=b1(n) and bi(n)<ui<ci(n) for i=2,…,k+1, then(3.17)b1(n)+F1(n,u)<b1(n+1),
that is,
(3.18)b1(n)+f(n,b1(n),u2,…,uk+1)<b1(n+1).
Notice that the condition bi(n)<ui<ci(n) for i=2,…,k+1 is equivalent with condition b(n-i+1)<ui<c(n-i+1) (see (3.10)). Looking at the supposed inequality (3.4) and realizing that b1(n)=b(n) and b1(n+1)=b(n+1), we can see that inequality (3.18) is fulfilled.
Analogously, according to (2.12), we have to prove that for u1=c1(n) and bi(n)<ui<ci(n) for i=2,…,k+1 the inequality(3.19)c1(n)+F1(n,u)>c1(n+1),
that is,
(3.20)c1(n)+f(n,c1(n),u2,…,uk+1)>c1(n+1)
holds.
Again, considering (3.5) and the fact that c1(n)=c(n) and c1(n+1)=c(n+1), we can see that this inequality really holds.
Thus, by the assertion of Theorem2.1, there exists a solution u=u*(n) of system (3.3) such that for every n∈ℤa∞(3.21)bi(n)<ui*(n)<ci(n)fori=1,…,k+1.
In our case,v=v*(n)=u1*(n) is the solution of the original equation (1.2). Further, b1(n)=b(n) and c1(n)=c(n), and thus the existence of a solution of the delayed equation (1.2) such that inequalities (3.7) are satisfied is guaranteed.


## 3.3. Asymptotic Solution Estimates for Delayed Difference Equations

Let us suppose that two functionsu,w:ℤa-k∞→ℝ are given such that (3.22)u(n)<w(n),n∈Za-k∞,(3.23)Δu(n)≥f(n,u(n),u(n-1),…,u(n-k)),n∈Za∞,(3.24)Δw(n)≤f(n,w(n),w(n-1),…,w(n-k)),n∈Za∞.
Consider the problem of whether there exists a solution v=v*(n), n∈ℤa-k∞ of (1.2) such that (3.25)u(n)<v*(n)<w(n),n∈Za-k∞.
The following corollary of Theorem 3.1 presents sufficient conditions for the existence of a solution of this problem.Corollary 3.2.
Let functionsu,w:ℤa-k∞→ℝ satisfy inequalities (3.22)–(3.24). Let f:ℤa∞×ℝk+1→ℝ be a continuous function such that
(3.26)f(n,u(n),y2,…,yk+1)>f(n,u(n),z2,…,zk+1),(3.27)f(n,w(n),y2,…,yk+1)>f(n,w(n),z2,…,zk+1)
for every n∈ℤa∞ and every y2,…,yk+1,z2,…,zk+1∈ℝ such that
(3.28)yi<zi,i=2,…,k+1.
Then there exists a solution v=v*(n) of (1.2) satisfying inequalities (3.25) for every n∈ℤa-k∞.Proof.
This assertion is an easy consequence of Theorem3.1.
Putb(n):=u(n), c(n):=w(n). Considering inequalities (3.23) and (3.26), we can see that(3.29)Δu(n)>f(n,u(n),v2,…,vk+1)
for every n∈ℤa∞ and every v2,…,vk+1 such that
(3.30)b(n-i+1)<vi<c(n-i+1),i=2,…,k+1.
Similarly,
(3.31)Δw(n)<f(n,w(n),v2,…,vk+1)
for every n∈ℤa∞ and every b(n-i+1)<vi<c(n-i+1), i=2,…,k+1.
Obviously, inequalities (3.29) and (3.31) are equivalent with inequalities (3.4) and (3.5), respectively. Thus, all the assumptions of Theorem 3.1 are satisfied and there exists a solution v=v*(n) of (1.2) satisfying inequalities (3.25) for every n∈ℤa-k∞.Example 3.3.
Consider the equation(3.32)Δv(n)=v2(n)-v(n-1)
for n∈ℤ3∞ which is a second-order delayed discrete equation with delay k=1. We will show that there exists a solution v=v*(n) of (3.32) that satisfies the inequalities
(3.33)1<v*(n)<n
for n∈ℤ2∞.We will prove that for the functions(3.34)u(n):=1,w(n):=n,f(n,v1,v2):=v12-v2
all the assumptions of Corollary 3.2 are satisfied. Inequality (3.22) is obviously fulfilled for n∈ℤ2∞. Inequality (3.23) can be also proved very easily:(3.35)Δu(n)=0,f(n,u(n),u(n-1))=12-1=0,
and thus for every n∈ℤ3∞, Δu(n)≥f(n,u(n),u(n-1)).As for inequality (3.24), we get(3.36)Δw(n)=1,f(n,w(n),w(n-1))=n2-n+1
and thus Δw(n)≤f(n,w(n),w(n-1)) for n∈ℤ3∞.Finally, the functions(3.37)f(n,u(n),v2)=1-v2,f(n,w(n),v2)=n2-v2
are decreasing with respect to v2. Therefore, conditions (3.26) and (3.27) are satisfied, too. Hence, due to Corollary 3.2, there exists a solution of (3.32) satisfying (3.33).


## 3.4. Positive Solutions of a Linear Equation with a Single Delay

We will apply the result of Theorem3.1 to the investigation of a simple linear difference equation of the (k+1)-st order with only one delay, namely, the equation(3.38)Δv(n)=-p(n)v(n-k),
where, again, n∈ℤa∞ is the independent variable and k∈ℕ, k≥1, is the fixed delay. The function p:ℤa∞→ℝ is assumed to be positive. Our goal is to give sharp sufficient conditions for the existence of positive solutions. The existence of such solutions is very often substantial for a concrete model considered. For example, in biology, when a model of population dynamics is described by an equation, the positivity of a solution may mean that the studied biological species can survive in the supposed environment.For its simple form, (3.38) often serves for testing new results and is very frequently investigated. It was analyzed, for example, in papers [10, 11, 26]. A sharp result on existence of positive solutions given in [26] is proved by a comparison method [8, 9]. Here we will use Theorem 3.1 to generalize this result.For the purposes of this section, define the expressionlnqt, where q∈ℕ, as (3.39)ln⁡qt:=ln⁡(ln⁡q-1t)ln⁡0t:=t.
We will write only lnt instead of ln1t. Further, for a fixed integer ℓ≥0 define auxiliary functions(3.40)μl(n):=18n2+18(nln⁡n)2+⋯+18(nln⁡n⋯ln⁡ln)2,pl(n):=(kk+1)k⋅(1k+1+kμl(n)),(3.41)νl(n):=(kk+1)n⋅nln⁡nln⁡2n⋯ln⁡ln.
In [26], it was proved that if p(n) in (3.38) is a positive function bounded by pℓ(n) for some ℓ≥0, then there exists a positive solution of (3.38) bounded by the function νℓ(n) for n sufficiently large. Since limn→∞νℓ(n)=0, such solution will vanish for n→∞. Here we show that (3.38) has a positive solution bounded by νℓ(n) even if the coefficient p(n) satisfies a less restrictive inequality (see inequality (3.58) below). The proof of this statement will be based on the following four lemmas. The symbols “o” and “O” stand for the Landau order symbols and are used for n→∞.Lemma 3.4.
The formula(3.42)ln⁡(y-z)=ln⁡y-∑i=1∞ziiyi
holds for any numbers y,z∈ℝ such that y>0 and |z|<y.Proof.
The assertion is a simple consequence of the well-known Maclaurin expansion(3.43)ln⁡(1-x)=-∑i=1∞1ixifor-1≤x<1.
As ln(y-z)-lny=ln(1-z/y), substituting x=z/y we get
(3.44)ln⁡(y-z)-ln⁡y=-∑i=1∞ziiyifor-y≤z<y
and adding lny to both sides of this equality, we get (3.42).Lemma 3.5.
For fixedr∈ℝ∖{0} and fixed q∈ℕ, the asymptotic representation
(3.45)ln⁡q(n-r)=ln⁡qn-rnln⁡n⋯ln⁡q-1n-r22n2ln⁡n⋯ln⁡q-1n-r22(nln⁡n)2ln⁡2n⋯ln⁡q-1n-⋯-r22(nln⁡n⋯ln⁡q-1n)2-r3(1+o(1))3n3ln⁡n⋯ln⁡q-1n
holds for n→∞.Proof.
We will prove relation (3.45) by induction with respect to q. For q=1, (3.45) reduces to
(3.46)ln⁡(n-r)=ln⁡n-rn-r22n2-r3(1+o(1))3n3
which holds due to Lemma 3.4. Suppose that relation (3.45) holds for some q. We can write lnq(n-r)=y-z with y=lnqn and
(3.47)z=rnln⁡n⋯ln⁡q-1n+r22n2ln⁡n⋯ln⁡q-1n+r22(nln⁡n)2ln⁡2n⋯ln⁡q-1n+⋯+r22(nln⁡n⋯ln⁡q-1n)2+r3(1+o(1))3n3ln⁡n⋯ln⁡q-1n.
Now we will show that (3.45) holds for q+1. Notice that in our case, the condition |z|<y from Lemma 3.4 is fulfilled for n sufficiently large because z→0 for n→∞, meanwhile y→∞ for n→∞. Thus we are justified to use Lemma 3.4 and doing so, we get
(3.48)ln⁡q+1(n-r)=ln⁡(ln⁡q(n-r))=ln⁡(y-z)=ln⁡y-1yz-12y2z2-⋯=ln⁡(ln⁡qn)-1ln⁡qn⋅(rnln⁡n⋯ln⁡q-1n+r22n2ln⁡n⋯ln⁡q-1n+⋯+r22(nln⁡n⋯ln⁡q-1n)2+r3(1+o(1))3n3ln⁡n⋯ln⁡q-1n)-12(ln⁡qn)2⋅(r2(nln⁡n⋯ln⁡q-1n)2+O(1n3(ln⁡n⋯ln⁡q-1n)2))+O(1(nln⁡n⋯ln⁡qn)3)=ln⁡q+1n-rnln⁡n⋯ln⁡qn-r22n2ln⁡n⋯ln⁡qn-r22(nln⁡n)2ln⁡2n⋯ln⁡qn-⋯-r22(nln⁡n⋯ln⁡qn)2-r3(1+o(1))3n3ln⁡n⋯ln⁡qn.
Thus, formula (3.45) holds for q+1, too, which ends the proof.Lemma 3.6.
For fixedr∈ℝ∖{0} and fixed q∈ℕ, the asymptotic representations
(3.49)ln⁡q(n-r)ln⁡qn=1-r2nln⁡n⋯ln⁡qn-r24n2ln⁡n⋯ln⁡qn-r24(nln⁡n)2ln⁡2n⋯ln⁡qn-⋯-r24(nln⁡n⋯ln⁡q-1n)2ln⁡qn-r28(nln⁡n⋯ln⁡qn)2-r3(1+o(1))6n3ln⁡n⋯ln⁡qn,(3.50)n-rn=1-r2n-r28n2-r316n3+o(1n3)
hold for n→∞.Proof.
Both these relations are simple consequences of the asymptotic formula(3.51)1-x=1-12x-18x2-116x3+o(x3)forx⟶0
and of Lemma 3.5 (for formula (3.49)). In the case of relation (3.49), we put
(3.52)x=rnln⁡n⋯ln⁡qn+r22n2ln⁡n⋯ln⁡qn+⋯+r22(nln⁡n⋯ln⁡q-1n)2ln⁡qn+r3(1+o(1))3n3ln⁡n⋯ln⁡qn
and in the case of relation (3.50), we put x=r/n.Lemma 3.7.
For fixedr∈ℝ∖{0} and fixed q∈ℕ, the asymptotic representation
(3.53)(n-r)nln⁡(n-r)ln⁡n⋯ln⁡q(n-r)ln⁡qn=1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡qn)-r2μq(n)-r316n3+o(1n3)
holds for n→∞.Proof.
We will prove relation (3.53) by induction with respect to q. For q=1, (3.53) reduces to
(3.54)(n-r)nln⁡(n-r)ln⁡n=1-r(12n+12nln⁡n)-r2μ1(n)-r316n3+o(1n3)=1-r(12n+12nln⁡n)-r2(18n2+18(nln⁡n)2)-r316n3+o(1n3).
On the other hand, using Lemma 3.6, we get
(3.55)(n-r)nln⁡(n-r)ln⁡n=(1-r2n-r28n2-r316n3+o(1n3))×(1-r2nln⁡n-r24n2ln⁡n-r28(nln⁡n)2-r3(1+o(1))6n3ln⁡n)=1-r2nln⁡n-r24n2ln⁡n-r28(nln⁡n)2-r2n+r24n2ln⁡n-r28n2-r316n3+o(1n3)=1-r(12n+12nln⁡n)-r2(18n2+18(nln⁡n)2)-r316n3+o(1n3).
Thus, for q=1, relation (3.53) holds. Now suppose that (3.53) holds for some q and prove that it holds for q+1. In the following calculations, we use Lemma 3.6 and we skip some tedious expressions handling.
(3.56)(n-r)nln⁡(n-r)ln⁡n⋯ln⁡q+1(n-r)ln⁡q+1n=(n-r)nln⁡(n-r)ln⁡n⋯ln⁡q(n-r)ln⁡qn⋅ln⁡q+1(n-r)ln⁡q+1n=(1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡qn)-r2μq(n)-r316n3+o(1n3))×(1-r2nln⁡n⋯ln⁡q+1n-r24n2ln⁡n⋯ln⁡q+1n-⋯-r24(nln⁡n⋯ln⁡qn)2ln⁡q+1n-r28(nln⁡n⋯ln⁡q+1n)2+o(1n3))=1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡q+1n)-r2μq+1(n)-r316n3+o(1n3).
We can see that formula (3.53) holds for q+1, too, which ends the proof.Now we are ready to prove that there exists a bounded positive solution of (3.38). Remind that functions pℓ and νℓ were defined by (3.40) and (3.41), respectively.Theorem 3.8.
Letω:ℤa∞→ℝ satisfy the inequality
(3.57)|ω(n)|≤ɛ(kk+1)k⋅k(2k2+k-1)16n3(k+1),n∈Za∞,
for a fixed ɛ∈(0,1). Suppose that there exists an integer ℓ≥0 such that the function p satisfies the inequalities
(3.58)0<p(n)≤pl(n)+ω(n)
for every n∈ℤa∞. Then there exists a solution v=v*(n), n∈ℤa-k∞ of (3.38) such that for n sufficiently large the inequalities
(3.59)0<v*(n)<νl(n)
hold.Proof.
Show that all the assumptions of Theorem3.1 are fulfilled. For (3.38), f(n,v1,…,vk+1)=-p(n)vk+1. This is a continuous function. Put
(3.60)b(n):=0,c(n):=νl(n).
We have to prove that for every v2,…,vk+1 such that b(n-i+1)<vi<c(n-i+1), i=2,…,k+1, the inequalities (3.4) and (3.5) hold for n sufficiently large. Start with (3.4). That gives that for b(n-k)<vk+1<c(n-k), it has to be
(3.61)0-p(n)⋅vk+1<0.
This certainly holds, because the function p is positive and so is vk+1.
Next, according to (3.5), we have to prove that(3.62)νl(n)-p(n)vk+1>νl(n+1)
which is equivalent to the inequality
(3.63)-p(n)vk+1>νl(n+1)-νl(n).
Denote the left-hand side of (3.63) as L(3.63). As vk+1<c(n-k)=νℓ(n-k) and as by (3.40), (3.58), and (3.57)
(3.64)p(n)≤(kk+1)k⋅(1k+1+kμl(n))+ɛ(kk+1)k⋅k(2k2+k-1)16n3(k+1),
we have
(3.65)L(3.63)>-(kk+1)k(1k+1+kμl(n)+ɛ⋅k(2k2+k-1)16n3(k+1))×(kk+1)n-k(n-k)ln⁡(n-k)⋯ln⁡l(n-k)=-(kk+1)n(1k+1+kμl(n)+ɛ⋅k(2k2+k-1)16n3(k+1))⋅(n-k)ln⁡(n-k)⋯ln⁡l(n-k).
Further, we can easily see that
(3.66)νl(n+1)-νl(n)=(kk+1)nnln⁡n⋯ln⁡ln(kk+1(n+1)nln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡ln-1).
Thus, to prove (3.63), it suffices to show that for n sufficiently large,
(3.67)-(1k+1+kμl(n)+ε⋅k(2k2+k-1)16n3(k+1))(n-k)nln⁡(n-k)ln⁡n⋯ln⁡l(n-k)ln⁡ln>kk+1(n+1)nln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡ln-1.
Denote the left-hand side of inequality (3.67) as L(3.67) and the right-hand side as R(3.67). Using Lemma 3.7 with r=k and q=ℓ, we can write
(3.68)L(3.67)=-(1k+1+kμl(n)+ε⋅k(2k2+k-1)16n3(k+1))×(1-k(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-k2μl(n)-k316n3+o(1n3))=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)+k2k+1μl(n)+k316n3(k+1)-kμl(n)+k216n3-ε⋅k(2k2+k-1)16n3(k+1)+o(1n3)=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-kk+1μl(n)+2k3(1-ε)+k2(1-ε)+kε16n3(k+1)+o(1n3).
Using Lemma 3.7 with r=-1 and q=ℓ, we get for R(3.67)(3.69)R(3.67)=kk+1(1+12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-μl(n)+116n3+o(1n3))-1=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-kk+1⋅μl(n)+k16n3(k+1)+o(1n3).
It is easy to see that the inequality (3.67) reduces to
(3.70)2k3(1-ɛ)+k2(1-ɛ)+kɛ16n3(k+1)+o(1n3)>k16n3(k+1)+o(1n3).
This inequality is equivalent to
(3.71)k(2k2(1-ɛ)+k(1-ɛ)-(1-ɛ))16n3(k+1)+o(1n3)>0.
The last inequality holds for n sufficiently large because k≥1 and 1-ɛ∈(0,1). We have proved that all the assumptions of Theorem 3.1 are fulfilled and hence there exists a solution of (3.38) satisfying conditions (3.59).



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*Source: 671967-2011-07-12.xml*