# Asymptotic Upper and Lower Estimates of a Class of Positive Solutions of a Discrete Linear Equation with a Single Delay

Authors: J. Diblík; I. Hlavičková
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study a frequently investigated class of linear difference equationsΔv(n)=−p(n)v(n−k)
with a positive coefficient p(n) and a single delay k. Recently, it was proved that if the function p(n) is bounded above by a certain function, then there exists a positive vanishing solution of the considered equation, and the upper bound was found. Here we improve this result by finding even the lower bound for the positive solution, supposing the function p(n) is bounded above and below by certain functions.


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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 as n→∞ of the solutions of the discrete delayed equation of the (k+1)-th order
(1.2)Δv(n)=-p(n)v(n-k),
where n is the independent variable assuming values from the set ℤa∞ with a fixed a∈ℕ={0,1,2,…}. The number k∈ℕ,k≥1 is the fixed delay, Δv(n)=v(n+1)-v(n), and p:ℤa∞→ℝ+=(0,∞).Along with (1.2), we consider 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∞. Moreover, the solution of (1.2) continuously depends on initial conditions (1.3).Equation (1.2) is investigated very frequently. It was analyzed, for example, in papers [1–3] (where the comparison method [4, 5] was used) and [6]. Similar problems for differential and dynamic equations are studied, for example, in [7–10].In a recent work of the authors [6], it is proved that if the function p(n) is bounded above by a certain function, then there exists a positive vanishing (i.e., tending to 0 as n→∞) solution of the considered equation. Moreover, its upper bound was found. Our aim is to improve this result and to show that if the coefficient p(n) is between two functions pℓ(n)-ψ(n) and pℓ(n)+ω(n) (see (2.3), (2.6), and (2.7) below) then (1.2) has a positive vanishing solution which is bounded from below by the function αℓ(n) (see (2.5)) and from above by the function νℓ(n) (see (2.4)). Due to the linearity of equation considered it becomes clear that a similar result holds for a one-parametric family of positive vanishing solutions of (1.2).To prove this, we will use Theorem1.1 which is one of the main results of [6]. This theorem is valid for any delayed difference equation of the form:
(1.4)Δv(n)=f(n,v(n),v(n-1),…,v(n-k)).Theorem 1.1.
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:
(1.5)b(n)+f(n,b(n),v2,…,vk+1)<b(n+1),(1.6)c(n)+f(n,c(n),v2,…,vk+1)>c(n+1),
hold for every n∈ℤa∞ and every v2,…,vk+1 such that
(1.7)b(n-i+1)<vi<c(n-i+1),i=2,…,k+1.
Then there exists a solution v=v*(n) of (1.4) satisfying the inequalities
(1.8)b(n)<v*(n)<c(n),
for everyn∈ℤa-k∞.For related comparison theorems for solutions of difference equations as well as related methods and their applications, see, for example, [1, 11–21] and the related references therein. Investigation of positive solutions (and connected problems of oscillating solutions) attracted recently large attention. Except the references given above, one refers as well to [11, 22–33], and to the references therein. Existence of positive solutions of some classes of difference equations has been also studied in papers [12–16]. The existence of unbounded solutions by some comparison methods can be found, for example, in [17, 18].


## 2. Auxiliary Functions and Lemmas

Define the expressionlnqn, q∈ℕ∖{0}, as
(2.1)ln⁡qn:=ln⁡(ln⁡q-1n),where ln0n:=n. We will write only lnn instead of ln1n. Further, for a fixed integer ℓ≥0 define auxiliary functions:
(2.2)μl(n):=18n2+18(nln⁡n)2+⋯+18(nln⁡n⋯ln⁡ln)2,(2.3)pl(n):=(kk+1)k⋅(1k+1+kμl(n)),(2.4)νl(n):=(kk+1)n⋅nln⁡nln⁡2n⋯ln⁡ln,(2.5)αl(n):=(kk+1)n⋅nln⁡nln⁡2n⋯ln⁡lnln⁡l+1-σn,
where σ∈ℝ,σ>0, is a constant. Notice that if a is sufficiently large, all these functions are well defined for n∈ℤa∞.Finally, let functionsψ,ω:ℤa∞→ℝ+ satisfy for n∈ℤa∞ the inequalities:
(2.6)ψ(n)≤(kk+1)k⋅δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn,(2.7)ω(n)≤ε(kk+1)k⋅k(2k-1)16n3,
for fixed δ>0,β>2 and ε∈(0,1).In [3], it was proved that if p(n) in (1.2) is a positive function bounded above by pℓ(n) for some ℓ≥0, then there exists a positive solution of (1.2) bounded above by the function νℓ(n) for n sufficiently large. Since limn→∞νℓ(n)=0, such solution will vanish as n→∞. This result was further improved in [6], where it was shown that (1.2) has a positive solution bounded above by νℓ(n) even if the coefficient p(n) satisfies a less restrictive inequality, namely, p(n)<pℓ(n)+ω(n). Here we will prove that function αℓ provides the lower estimate of the solution, supposing pℓ(n)-ψ(n)≤p(n)≤pℓ(n)+ω(n). 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 as n→∞.Lemma 2.1.
For fixedr∈ℝ∖{0} and fixed  q∈ℕ, the asymptotic representation:
(2.8)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+o(1n3),
holds as n→∞.Proof.
Relation (2.8) can be proved by induction with respect to q, for details, see [6].Lemma 2.2.
For fixedr,s∈ℝ∖{0} and fixed  q∈ℕ, the asymptotic representations:
(2.9)(ln⁡q(n-r)ln⁡qn)s=1-rsnln⁡n⋯ln⁡qn-r2s2n2ln⁡n⋯ln⁡qn-r2s2(nln⁡n)2ln⁡2n⋯ln⁡qn-⋯-r2s2(nln⁡n⋯ln⁡q-1n)2ln⁡qn+r2s(s-1)2(nln⁡n⋯ln⁡qn)2+o(1n3),(2.10)n-rn=1-r2n-r28n2-r316n3+o(1n3),
hold as n→∞.Proof.
Both these relations are simple consequences of the asymptotic formula:(2.11)(1-x)s=1-sx+s(s-1)2x2-s(s-1)(s-2)6x3+o(x3)asx⟶0.
and of Lemma 2.1 (for formula (2.9)).
In the case of relation (2.10), we put x=r/n and s=1/2.
To prove relation (2.9), first notice that dividing (2.8) by lnqn, we get(2.12)ln⁡q(n-r)ln⁡qn=1-rnln⁡n⋯ln⁡q-1nln⁡qn-r22n2ln⁡n⋯ln⁡q-1nln⁡qn-r22(nln⁡n)2ln⁡2n⋯ln⁡q-1nln⁡qn-⋯-r22(nln⁡n⋯ln⁡q-1n)2ln⁡qn+o(1n3).
Thus, putting
(2.13)x=1-ln⁡q(n-r)ln⁡qn=rnln⁡n⋯ln⁡qn+r22n2ln⁡n⋯ln⁡qn+⋯+r22(nln⁡n⋯ln⁡q-1n)2ln⁡qn+o(1n3)
and using (2.11), we get (2.9).The following lemma is proved in [6].Lemma 2.3.
For fixedr∈ℝ∖{0} and fixed  q∈ℕ, the asymptotic representation:
(2.14)(n-r)n⋅ln⁡(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 as n→∞.Lemma 2.4.
For fixedr∈ℝ∖{0},  σ∈ℝ+ and  q∈ℕ, the asymptotic representation:
(2.15)(n-r)n⋅ln⁡(n-r)ln⁡n⋯ln⁡q(n-r)ln⁡qn⋅ln⁡q+1-σ(n-r)ln⁡q+1-σn=1-r(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡qn-σ2nln⁡n⋯ln⁡qnln⁡q+1n)-r2μq(n)+σ(σ+2)8⋅r2(nln⁡n⋯ln⁡q+1n)2-r316n3+o(1n3),
holds as n→∞.Proof.
Using Lemma2.2 with s=-σ/2 and q+1 instead of q, we get for n→∞(2.16)ln⁡q+1-σ(n-r)ln⁡q+1-σn=1+rσ2nln⁡n⋯ln⁡q+1n+r2σ4n2ln⁡n⋯ln⁡q+1n+r2σ4(nln⁡n)2ln⁡2n⋯ln⁡q+1n+⋯+r2σ4(nln⁡n⋯ln⁡qn)2ln⁡q+1n+r2σ(σ+2)8(nln⁡n⋯ln⁡q+1n)2+o(1n3).
Multiplying the asymptotic representations (2.14) and (2.16), we get (2.15).


## 3. Main Result

Now we are ready to prove that there exists a positive solution of (1.2) which is bounded below and above. Remind the functions pℓ, νℓ, αℓ, ψ, and ω were defined by (2.3)–(2.6) and (2.7), respectively.Theorem 3.1.
Suppose that there exist numbersa,ℓ∈ℕ, and  σ>0, such that the function  p in (1.2) satisfies the inequalities
(3.1)0<pl(n)-ψ(n)≤p(n)≤pl(n)+ω(n),
for every n∈ℤa∞. Then there exists a solution v=v*(n),n∈ℤa-k∞ of (1.2) such that for n sufficiently large the inequalities:
(3.2)αl(n)<v*(n)<νl(n),
hold.Proof.
Show that all the assumptions of Theorem1.1 are fulfilled. For (1.2),
(3.3)f(n,v1,…,vk+1)=-p(n)vk+1.
This is a continuous function. Put
(3.4)b(n)≔αl(n),c(n):=νl(n).
We have to prove that for every v2,…,vk+1 such that
(3.5)b(n-i+1)<vi<c(n-i+1),i=2,…,k+1,
the inequalities (1.5) and (1.6) hold for n sufficiently large. Start with (1.5). That gives that for
(3.6)b(n-k)<vk+1<c(n-k),
it has to be
(3.7)αl(n)-p(n)⋅vk+1<αl(n+1),
which is equivalent to the inequality
(3.8)-p(n)vk+1<αl(n+1)-αl(n).
Denote the left-hand side of (3.8) as L(3.8). As vk+1>b(n-k)=αℓ(n-k) and as by (2.3), (2.6), and (3.1)
(3.9)p(n)≥(kk+1)k⋅(1k+1+kμl(n))-(kk+1)k⋅δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn,
we have
(3.10)L(3.8)<-(kk+1)k(1k+1+kμl(n)-δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn)×(kk+1)n-k(n-k)ln⁡(n-k)⋯ln⁡l(n-k)ln⁡l+1-σ(n-k)=-(kk+1)n(1k+1+kμl(n)-δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn)×(n-k)ln⁡(n-k)⋯ln⁡l(n-k)ln⁡l+1-σ(n-k).
Further, we can easily see that
(3.11)αl(n+1)-αl(n)=(kk+1)nnln⁡n⋯ln⁡lnln⁡l+1-σn×(kk+1(n+1)nln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡lnln⁡l+1-σ(n+1)ln⁡l+1-σn-1).
Thus, to prove (3.8), it suffices to show that for n sufficiently large, the following inequality holds:
(3.12)-(1k+1+kμl(n)-δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn)(n-k)nln⁡(n-k)ln⁡n⋯ln⁡l(n-k)ln⁡lnln⁡l+1-σ(n-k)ln⁡l+1-σn<kk+1(n+1)nln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡lnln⁡l+1-σ(n+1)ln⁡l+1-σn-1.
Denote the left-hand side of inequality (3.12) as L(3.12) and the right-hand side as R(3.12). In the following computation we will use the fact that β>2 and
(3.13)μl(n)=18n2+O(1n2ln⁡2n),
and we will omit all the terms which are of order o(1/n3). Applying Lemma 2.4 with r=k and q=ℓ, we can write
(3.14)L(3.12)=-(1k+1+kμl(n)-δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn)×(σ(σ+2)8⋅k2(nln⁡n⋯ln⁡l+1n)2-k316n3+o(1n3)1-k(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-σ2nln⁡n⋯ln⁡lnln⁡l+1n)-k2μl(n)+σ(σ+2)8⋅k2(nln⁡n⋯ln⁡l+1n)2-k316n3+o(1n3))=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-σ2nln⁡n⋯ln⁡lnln⁡l+1n)+k2k+1μl(n)-σ(σ+2)8(k+1)⋅k2(nln⁡n⋯ln⁡l+1n)2+k316n3(k+1)-kμl(n)+k216n3+δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn+o(1n3)=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-σ2nln⁡n⋯ln⁡lnln⁡l+1n)-kk+1μl(n)-σ(σ+2)8(k+1)⋅k2(nln⁡n⋯ln⁡l+1n)2+2k3+k216n3(k+1)+δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn+o(1n3).
Using Lemma 2.4 with r=-1 and q=ℓ, we get for R(3.12)(3.15)R(3.12)=kk+1(1+12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-σ2nln⁡n⋯ln⁡lnln⁡l+1n-μl(n)+σ(σ+2)8⋅1(nln⁡n⋯ln⁡l+1n)2+116n3+o(1n3))-1=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln-σ2nln⁡n⋯ln⁡lnln⁡l+1n)-kk+1⋅μl(n)+σ(σ+2)k8(k+1)⋅1(nln⁡n⋯ln⁡l+1n)2+k16n3(k+1)+o(1n3).
It is easy to see that the inequality (3.12) reduces to
(3.16)-σ(σ+2)k28(k+1)⋅1(nln⁡n⋯ln⁡l+1n)2+2k3+k216n3(k+1)+δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn+o(1n3)<σ(σ+2)k8(k+1)⋅1(nln⁡n⋯ln⁡l+1n)2+k16n3(k+1)+o(1n3).
This inequality is equivalent to
(3.17)-σ(σ+2)k8⋅1(nln⁡n⋯ln⁡l+1n)2+k(2k-1)16n3+δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn+o(1n3)<0.
The last inequality holds for n sufficiently large because k≥1, σ>0, β>2, and as n→∞,
(3.18)δ(nln⁡n⋯ln⁡ln)2ln⁡l+1βn,1n3
tend to zero faster than
(3.19)1(nln⁡n⋯ln⁡l+1n)2
does.
Thus, we have proved that inequality (1.5) holds.
Next, according to (1.6), we have to prove that(3.20)νl(n)-p(n)vk+1>νl(n+1),
which is equivalent to the inequality:
(3.21)-p(n)vk+1>νl(n+1)-νl(n).
Denote the left-hand side of (3.21) as L(3.21). As vk+1<c(n-k)=νℓ(n-k) and as by (2.3), (3.1), and (2.7)
(3.22)p(n)≤(kk+1)k⋅(1k+1+kμl(n))+ε(kk+1)k⋅k(2k-1)16n3,
we have
(3.23)L(3.21)>-(kk+1)k(1k+1+kμl(n)+ε⋅k(2k-1)16n3)×(kk+1)n-k(n-k)ln⁡(n-k)⋯ln⁡l(n-k)=-(kk+1)n(1k+1+kμl(n)+ε⋅k(2k-1)16n3)⋅(n-k)ln⁡(n-k)⋯ln⁡l(n-k).
Further, we can easily see that
(3.24)νl(n+1)-νl(n)=(kk+1)nnln⁡n⋯ln⁡ln(kk+1(n+1)n⋅ln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡ln-1).
Thus, to prove (3.21), it suffices to show that for n sufficiently large, the following inequality holds:
(3.25)-(1k+1+kμl(n)+ε⋅k(2k-1)16n3)(n-k)n⋅ln⁡(n-k)ln⁡n⋯ln⁡l(n-k)ln⁡ln>kk+1(n+1)n⋅ln⁡(n+1)ln⁡n⋯ln⁡l(n+1)ln⁡ln-1.
Denote the left-hand side of inequality (3.25) as L(3.25) and the right-hand side as R(3.25). Using Lemma 2.3 with r=k and q=ℓ, we can write
(3.26)L(3.25)=-(1k+1+kμl(n)+ε⋅k(2k-1)16n3)×(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(2k-1)16n3+o(1n3)=-1k+1+kk+1(12n+12nln⁡n+⋯+12nln⁡n⋯ln⁡ln)-kk+1μl(n)+k316n3(k+1)+k(k-ε(2k-1))16n3+o(1n3).
Using Lemma 2.3 with r=-1 and q=ℓ, we get for R(3.25)(3.27)R(3.25)=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 (3.25) reduces to
(3.28)k316n3(k+1)+k(k-ε(2k-1))16n3+o(1n3)>k16n3(k+1)+o(1n3).
This inequality is equivalent to
(3.29)k(2k-1)(1-ε)16n3+o(1n3)>0.
The last inequality holds for n sufficiently large because k≥1 and 1-ε∈(0,1). We have proved that all the assumptions of Theorem 1.1 are fulfilled and hence there exists a solution of (1.2) satisfying conditions (1.8), that is, in our case, conditions (3.2).



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