On the boundedness of the solutions in nonlinear discrete Volterra difference equations

Abstract

In this article, we investigate the boundedness property of the solutions of linear and nonlinear discrete Volterra equations in both convolution and non-convolution case. Strong interest in these kind of discrete equations is motivated as because they represent a discrete analogue of some integral equations. The most important result of this article is a simple new criterion, which unifies and extends several earlier results in both discrete and continuous cases. Examples are also given to illustrate our main theorem.

1 Introduction

We consider the nonlinear system of Volterra difference equations

$$x\left(n+1\right)=\sum _{j=0}^{n}f\left(n,j,x\left(j\right)\right)+h\left(n\right),n\ge 0,$$

(1.1)

with the initial condition

$$x\left(0\right)=x_0,$$

(1.2)

where

(A) The function f(n,j,.) : ℝ^d→ ℝ^dis a mapping for any fixed 0 ≤ j ≤ n, x₀ ∈ ℝ^dand h(n) ∈ ℝ^d, n ≥ 0.

(B) For any 0 ≤ j ≤ n, there exists an a(n, j) ∈ ℝ₊, such that

$$\left|\right|f\left(n,j,x\right)\left|\right|\le a\left(n,j\right)\varphi \left(\left|\right|x\left|\right|\right),0\le j\le n,x\in {ℝ}^d,$$

and

$$\underset{n\ge 0}{\text{sup}}\left|\right|h\left(n\right)\left|\right|<\mathrm{\infty }$$

hold. Here ϕ : ℝ₊ → ℝ₊ is a monotone non-decreasing mapping such that ϕ (v) > 0, v > 0, and ϕ(0) = 0 where ||.|| is any fixed norm on ℝ^d.

In recent years, there has been an increasing interest in the study of the asymptotic behavior of the solutions of both convolution and non-convolution-type linear and nonlinear Volterra difference equations (see [1–17] and references therein). Appleby et al. [2], under appropriate assumptions, have proved that the solutions of the discrete linear Volterra equation converge to a finite limit, which in general is non-trivial. The main result on the boundedness of solutions of a linear Volterra difference system in [2] was improved by Györi and Horváth [8]. In terms of the kernel of a linear system Györi and Reynolds [10] found necessary conditions for the solutions to be bounded. Also Györi and Reynolds [9] studied some connections between results obtained in [2, 8]. Elaydi et al. [6] have shown that under certain conditions there is a one-to-one correspondence between bounded solutions of linear Volterra difference equations with infinite delay and its perturbation. Also Cuevas and Pinto [4] have shown that under certain conditions there is a one to one correspondence between weighted bounded solutions of a linear Volterra difference equation with unbounded delay and its perturbation. In most of our references linear and perturbed linear equations are investigated, moreover the boundedness and estimation of the solutions are founded by using the resolvent of the equations.

This article studies the boundedness of the solution of (1.1) under initial condition (1.2). As an illustration, we formulate the following statement which is an interesting consequence of our Corollary 5.7.

Consider the linear convolution-type Volterra equation

$$x\left(n+1\right)=\sum _{j=0}^{n}A\left(n-j\right)x\left(j\right)+h\left(n\right),n\ge 0,$$

(1.3)

with the initial condition

$$x\left(0\right)=x_0.$$

(1.4)

Here A(n) ∈ ℝ^{d × d}are given matrices, h(n) ∈ ℝ^dare given vectors and x₀ ∈ ℝ^d.

Proposition 1.1. Assume that one of the following conditions is satisfied:

$$\begin{array}{c}\left(\alpha \right)\sum _{i=0}^{\mathrm{\infty }}\left|\right|A\left(i\right)\left|\right|<1and\underset{n\ge 0}{\text{sup}}\left|\right|h\left(n\right)\left|\right|<\mathrm{\infty };\\ \left(\beta \right)\sum _{i=0}^{\mathrm{\infty }}\left|\right|A\left(i\right)\left|\right|=1and\sum _{i=0}^n\left|\right|A\left(i\right)\left|\right|<1,n\ge 0,moreover\\ \underset{n\ge 0}{\text{sup}}\frac{\left|\right|h\left(n\right)\left|\right|}{\sum _{i=n+1}^{\mathrm{\infty }}\left|\right|A\left(i\right)\left|\right|}<\mathrm{\infty }.\end{array}$$

Then the solution of (1.3) and (1.4) is bounded for any x₀ ∈ ℝ^d.

We remark that the above proposition gives sufficient conditions for the boundedness, but they are not necessary in general, see Remark 6.5 below.

To the best of our knowledge, this is the first article dealing with the boundedness property of the solutions of a linear inhomogeneous Volterra difference system with the critical case ${\sum }_{i=0}^{\mathrm{\infty }}\left|\right|A\left(i\right)\left|\right|=1$. For some recent literature on the boundedness of the solutions of linear Volterra difference equations, we refer the readers to [10–12]. We give some applications of our main result for sub-linear, linear, and super-linear Volterra difference equations. We study the boundedness of solutions of convolution cases and we get a result parallel to the corresponding result of Lipovan [18] for integral equation. Also we give some examples to illustrate our main results.

The rest of the article is organized as follows. In Section 2, we briefly explain some notation and two definitions which are used to state and to prove our results. In Section 3, we sate our main result with its proof. In Section 4, we give three applications based on our main result. In Section 5, some corollaries with convolution estimations and boundedness of convolution infinite delay equation are given. Examples are also given to illustrate our main theorem in Section 6.

2 Preliminaries

In this section, we give some notation and some definitions which are used in this article.

Let ℝ be the set of real numbers, ℝ₊ the set of non-negative real numbers, ℤ is the set of integer numbers, and ℤ₊ = {n ∈ ℤ : n ≥ 0}. Let d be a positive integer, ℝ^dis the space of d-dimensional real column vectors with convenient norm ||.||. Let ℝ^{d × d}be the space of all d × d real matrices. By the norm of a matrix A ∈ ℝ^{d × d}, we mean its induced norm ||A|| = sup{||Ax|| |x ∈ ℝ^d, ||x|| = 1}. The zero matrix in ℝ^{d × d}is denoted by 0 and the identity matrix by I. The vector x and the matrix A are non-negative if x _i ≥ 0 and A _ij ≥ 0,1 ≤ i,j ≤ d, respectively. Sequence (x(n))_{n ≥ 0}in ℝ^dis denoted by x :ℤ₊ → ℝ^d.

The following definitions will be useful to prove the main results.

Definition 2.1. Let the function ϕ and the sequence a(n,j) ∈ ℝ₊, 0 ≤ j ≤ n, be given in condition (B). We say that the non-negative constant u has property (P _N ) with an integer N ≥ 0 if there is v >0, such that

$$\sum _{j=0}^{N}a\left(N,j\right)\varphi \left(u\right)+\left|\right|h\left(N\right)\left|\right|\le v,$$

(2.1)

and

$$\sum _{j=0}^{N}a\left(n,j\right)\varphi \left(u\right)+\sum _{j=N+1}^{n}a\left(n,j\right)\varphi \left(v\right)+\left|\right|h\left(n\right)\left|\right|\le v,n\ge N+1$$

(2.2)

hold.

Definition 2.2. We say that the vector x₀ ∈ ℝ^dbelongs to the set S if there exist a non-negative constant u and an integer N ≥ 0 such that u has property (P _N ) and the solution x(n;x₀), n ≥ 0, of (1.1) and (1.2) satisfies

$$\parallel x\left(n;x_0\right)\parallel \le u,0\le n\le N.$$

(2.3)

Remark 2.3. x₀ ∈ ℝ^dbelongs to the set S if

$$a\left(0,0\right)\varphi \left(\parallel x_0\parallel \right)+\parallel h\left(0\right)\parallel \le v,$$

(2.4)

and

$$a\left(n,0\right)\varphi \left(\parallel x_0\parallel \right)+\sum _{j=1}^{n}a\left(n,j\right)\varphi \left(v\right)+\parallel h\left(n\right)\parallel \le v,n\ge 1$$

(2.5)

hold. In this case N = 0 and u = ||x₀|| have property (P₀).

Remark 2.4. If there exists an N ≥ 0 and two positive constants u and v such that (2.2) holds, then

$${\alpha }_N:=\underset{n\ge N+1}{\text{sup}}\sum _{j=0}^{N}a\left(n,j\right)<\mathrm{\infty },$$

(2.6)

$${\beta }_N:=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)<\mathrm{\infty },$$

(2.7)

$$\gamma :=\underset{n\ge 0}{\text{sup}}\parallel h\left(n\right)\parallel <\mathrm{\infty }.$$

(2.8)

Conditions (2.6) and (2.7) are equivalent to

$$\underset{n\ge 0}{\text{sup}}\sum _{j=0}^{n}a\left(n,j\right)<\mathrm{\infty }.$$

(2.9)

3 Main result

Our main goal in this section is to establish the following result with the proof.

Theorem 3.1. Let (A) and (B) be satisfied and assume that the initial vector x₀belongs to the set S. Then the solution x(n;x₀), n ≥ 0, of (1.1) and (1.2) is bounded. More exactly the solution satisfies (2.3) with suitable u and N, such that

$$\parallel x\left(n;x_0\right)\parallel \le v,n\ge N+1,$$

(3.1)

where v is defined in (2.1) and (2.2).

Proof. Let x₀ ∈ S and consider the solution x(n) = x(n; x₀), n ≥ 0, of (1.1) with the condition (1.2) and let u and N be defined in (2.3). Then

$$\begin{array}{ll}\hfill \parallel x\left(N+1\right)\parallel & \le \sum _{j=0}^N\parallel f\left(N,j,x\left(j\right)\right)\parallel +\parallel h\left(N\right)\parallel \\ \le \sum _{j=0}^{N}a\left(N,j\right)\varphi \left(\parallel x\left(j\right)\parallel \right)+\parallel h\left(N\right)\parallel \\ \le \sum _{j=0}^{N}a\left(N,j\right)\varphi \left(u\right)+\parallel h\left(N\right)\parallel \le v,\end{array}$$

where we used the monotonicity of ϕ, and the definition of v. Thus (3.1) holds for n = N + 1.

Now we show that (3.1) holds for any n ≥ N + 1. Assume, for the sake of contradiction, that (3.1) is not satisfied for all n ≥ N + 1. Then there exists n₀ ≥ N + 1 such that

$$\parallel x\left(n_0+1\right)\parallel =\parallel x\left(n_0+1;x_0\right)\parallel >v,$$

(3.2)

and

$$\parallel x\left(n\right)\parallel =\parallel x\left(n;x_0\right)\parallel \le v,N+1\le n\le n_0.$$

(3.3)

Hence, from Equation (1.1), we get

$$\begin{array}{ll}\hfill \parallel x\left(n_0+1\right)\parallel & \le \sum _{j=0}^N\parallel f\left(n_0,j,x\left(j\right)\right)\parallel +\sum _{j=N+1}^{n_0}\parallel f\left(n_0,j,x\left(j\right)\right)\parallel +\parallel h\left(n_0\right)\parallel \\ \le \sum _{j=0}^{N}a\left(n_0,j\right)\varphi \left(\parallel x\left(j\right)\parallel \right)+\sum _{j=N+1}^{n_0}a\left(n_0,j\right)\varphi \left(\parallel x\left(j\right)\parallel \right)+\parallel h\left(n_0\right)\parallel .\end{array}$$

Since ϕ is a monotone non-decreasing mapping, (2.3) and (3.3) yield

$$\parallel x\left(n_0+1\right)\parallel \le \sum _{j=0}^{N}a\left(n_0,j\right)\varphi \left(u\right)+\sum _{j=N+1}^{n_0}a\left(n_0,j\right)\varphi \left(v\right)+\parallel h\left(n_0\right)\parallel .$$

But x₀ ∈ S and u has property (P _N ), and hence ||x(n₀ + 1)|| ≤ v. This contradicts the hypothesis that (3.1) does not hold for n₀ ≥ N + 1. So inequality (3.1) holds.

4 Applications

In this section, we give some applications of our main result. Throughout this section we take ϕ (t) = t^p, t > 0 with p > 0. There are three cases:

• 1. Sub-linear case when 0 < p < 1;

• 2. Linear case when p = 1;

• 3. Super-linear case when p > 1.

4.1 Sub-linear case

Our aim in this section is to establish a sufficient, as well as a necessary and sufficient, condition for the boundedness of all solutions of (1.1) and the scalar case of (1.1), respectively.

The next result provides a sufficient condition for the boundedness of solutions of (1.1).

Theorem 4.1. Let (A), (B) be satisfied and ϕ (t) = t^p, t > 0, with fixed p ∈ (0,1). If (2.8) and (2.9) hold, then for any x₀ ∈ ℝ^dthe solution x(n; x₀), n ≥ 0 of (1.1) and (1.2) is bounded.

The next Lemma provides a necessary and sufficient condition for the condition (2.2) be satisfied, and will be useful in the proof of Theorem 4.1.

Lemma 4.2. Assume ϕ (t) = t^p, t > 0 and 0 < p < 1. Any positive constant u has property (P₀) if and only if (2.8) and (2.9) are satisfied.

Proof. Let the non-negative constant u have property (P₀) (N = 0 in Definition 2.1). Then the condition (2.2) is satisfied for some positive v and for all n ≥ 1, so

$${\alpha }_0:=\underset{n\ge 1}{\text{sup}}a\left(n,0\right)<\mathrm{\infty },{\beta }_0:=\underset{n\ge 1}{\text{sup}}\sum _{j=1}^{n}a\left(n,j\right)<\mathrm{\infty },$$

(4.1)

$$\mathsf{\text{and}}{\gamma }_0:=\underset{n\ge 1}{\text{sup}}\parallel h\left(n\right)\parallel <\mathrm{\infty },$$

(4.2)

this imply that conditions (2.8) and (2.9) are satisfied.

Conversely, we assume (2.8), (2.9) and we prove that any positive constant u has property (P₀). Clearly, (2.9) is equivalent to α₀ < ∞, β₀ < ∞ and (2.8) implies γ₀ < ∞.

Since p ∈ (0,1), it is clear that for an arbitrarily fixed u > 0, (2.1) and

$$\frac{{\alpha }_0{u}^p}{v}+{\beta }_0{v}^{p-1}+\frac{{\gamma }_0}{v}\le 1,$$

(4.3)

are satisfied for any v large enough. From (4.3) we get

$${\alpha }_0{u}^p+{\beta }_0{v}^p+{\gamma }_0\le v,$$

that (2.5) is satisfied for ||x₀|| = u and all n ≥ 1. Then by Definition 2.1, u has property (P₀).

Now we prove Theorem 4.1.

Proof. Let (A) and (B) be satisfied. By Lemma 4.2, we have that for any x₀ ∈ ℝ^d, u = ||x₀|| has property (P₀) (see Remark 2.3). Thus, the conditions of Theorem 3.1 hold, and the initial vector x₀ belongs to S, and hence the solution x(n; x₀) of (1.1) and (1.2) is bounded.

We consider the scalar case of Volterra difference equation

$$x\left(n+1\right)=\sum _{j=0}^{n}a\left(n,j\right)x^p\left(j\right)+h\left(n\right),n\ge 0,$$

(4.4)

$$x\left(0\right)=x_0,$$

(4.5)

where x₀ ∈ ℝ₊, a(n,j) ∈ ℝ₊, h(n) ∈ ℝ₊, 0 ≤ j ≤ n and p ∈ (0,1).

The following result provides a necessary and sufficient condition for the boundedness of the solution of (4.4) and (4.5). The necessary part of the next theorem was motivated by a similar result of Lipovan [18] proved for convolution-type integral equation.

Theorem 4.3. Assume

$$\underset{n\to \mathrm{\infty }}{\text{lim}\mathsf{\text{inf}}}\left(\sum _{j=0}^{n}a\left(n,j\right)\right)>0,$$

(4.6)

moreover for any n ≥ 0, there exists an index j _n such that

$$0\le j_n\le nanda\left(n,j_n\right)+h\left(n\right)>0.$$

(4.7)

For any x₀ ∈ (0, ∞), the solution of (4.4) is bounded, if and only if (2.8) and (2.9) are satisfied.

Proof. Assume (2.8) and (2.9) are satisfied. Clearly, by Theorem 4.1 the solution of (4.4) is bounded. Conversely, let the solution x(n) = x(n;x₀) of (4.4) be bounded on ℝ₊, with x₀ > 0. Under condition (4.7), by mathematical induction we show that x(n) > 0, n ≥ 0. For n = 0 this is clear. Suppose that required inequality is not satisfied for all n ≥ 0. Then there exists index ℓ ≥ 0 such that x(0) > 0, ..., x(ℓ) > 0 and x(ℓ+1) ≤ 0. But by condition (4.7), we get

$$\begin{array}{ll}\hfill x\left(\ell +1\right)& =\sum _{j=0}^{\ell }a\left(\ell ,j\right)x^p\left(j\right)+h\left(\ell \right)\\ \ge a\left(\ell ,j_{\ell }\right)x^p\left(j_{\ell }\right)+h\left(\ell \right)>0,0\le j_{\ell }\le \ell ,\end{array}$$

which is a contradiction. So x(n) > 0 for all n ≥ 0. On the other hand, for any n ≥ N* ≥ 1

$$x\left(n+1\right)\ge \sum _{j=0}^{N^{*}-1}a\left(n,j\right)x^p\left(j\right)\ge \sum _{j=0}^{N^{*}-1}a\left(n,j\right)\underset{0\le j\le N^{*}}{\text{min}}{x}^p\left(j\right),$$

hence

$$\underset{n\ge 0}{\text{sup}}\sum _{j=0}^{N^{*}-1}a\left(n,j\right)<\mathrm{\infty }.$$

(4.8)

Since x(n + 1) ≥ h(n) for all n ≥ 0, clearly sup_n≥0h(n) is finite.

Define now

$$m=\underset{n\to \mathrm{\infty }}{\text{lim}\mathsf{\text{inf}}}x\left(n\right),$$

which is finite. First we show that m > 0.

Assume for the sake of contradiction that m = 0. In this case we can find a strictly increasing sequence (N _k )_k≥1, such that

$$x\left(N_k\right)=\underset{0\le n\le N_k}{\text{min}}x\left(n\right)>0,\mathsf{\text{and }}x\left(N_k\right)\to 0\mathsf{\text{ as }}k\to \mathrm{\infty }\mathsf{\text{.}}$$

From (4.4) if n = N _k - 1, we deduce

$$\begin{array}{ll}\hfill x\left(N_k\right)& =\sum _{j=0}^{N_k-1}a\left(N_k-1,j\right)x^p\left(j\right)+h\left(N_k-1\right)\\ \ge \sum _{j=0}^{N_k-1}a\left(N_k-1,j\right)\underset{0\le j\le N_k}{\text{min}}{x}^p\left(j\right)\\ =x^p\left(N_k\right)\sum _{j=0}^{N_k-1}a\left(N_k-1,j\right).\end{array}$$

Since x(N _k ) > 0, we have that

$$x^{1-p}\left(N_k\right)\ge \sum _{j=0}^{N_k-1}a\left(N_k-1,j\right),k\ge 1.$$

Since p ∈ (0,1) and x(N _k ) → 0, an k → ∞, we get

$$\underset{k\to \mathrm{\infty }}{\text{lim}\mathsf{\text{inf}}}\left(\sum _{j=0}^{N_k-1}a\left(N_k-1,j\right)\right)=0,$$

which contradicts (4.6). So m > 0 and for $\frac{1}{2}m$, there exists N* ≥ 0 such that

$$x\left(n\right)\ge \frac{1}{2}m,n\ge N^{*}.$$

Hence,

$$\begin{array}{ll}\hfill x\left(n+1\right)& =\sum _{j=0}^{n}a\left(n,j\right)x^p\left(j\right)+h\left(n\right)\\ \ge \sum _{j=N^{*}}^{n}a\left(n,j\right)x^p\left(j\right)\ge \frac{1}{2^p}{m}^p\sum _{j=N^{*}}^{n}a\left(n,j\right),n\ge N^{*}.\end{array}$$

But the solution x(n) is a bounded sequence, and hence

$$\underset{n\ge N^{*}}{\text{sup}}\sum _{j=N^{*}}^{n}a\left(n,j\right)<\mathrm{\infty }.$$

This and (4.8) imply condition (2.9).

Remark 4.4. In general, without condition (4.7) the necessary part of Theorem 4.3 is not true. In fact if a(0, 0) = 0, and h(0) = 0, that is (4.7) does not hold for n = 0, then for any x₀ ∈ (0, ∞) the solution of (4.4) satisfies x(1;x₀) = 0, and hence

$$x\left(n+1\right)=\sum _{\underset{j\ne 1}{j=0}}^{n}a\left(n,j\right)x^p\left(j\right)+h\left(n\right),n\ge 1.$$

Thus, the solution x(n; x₀) does not depend on the choice of the sequence (a(n, 1))_n≥1. This shows that the boundedness of the solutions does not imply (2.9), in general.

4.2 Linear case

Our aim in this section is to obtain sufficient condition for the boundedness of the solution of (1.1) under the initial condition (1.2), but in the linear case of Volterra difference equation.

The following result gives a sufficient condition for the boundedness.

Theorem 4.5. Assume (A), (B) are satisfied and ϕ (t) = t, t ≥ 0. Then the solution x(n;x₀), x₀ ∈ S, n ≥ 0 of (1.1) and (1.2) is bounded, if (2.9) is satisfied and there exists an N ≥ 0 such that one of the following conditions holds:

• (i) condition (2.8) holds and

  $${\beta }_N=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)<1;$$

  (4.9)

• (ii) ${\beta }_N=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)=1,$

and for any$n\in {\Gamma }_N^{\left(1\right)}$,

$$\sum _{j=0}^{N}a\left(n,j\right)=0,h\left(n\right)=0$$

(4.10)

hold, moreover

$$\underset{n\in {\Gamma }_N^{\left(2\right)}}{\text{sup}}{\left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{-1}\sum _{j=0}^{N}a\left(n,j\right)<\mathrm{\infty },$$

(4.11)

and

$$\underset{n\in {\Gamma }_N^{\left(2\right)}}{\text{sup}}{\left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{-1}\parallel h\left(n\right)\parallel <\mathrm{\infty },$$

(4.12)

where

$${\Gamma }_N^{\left(1\right)}=\left\{n\ge N+1:\sum _{j=N+1}^{n}a\left(n,j\right)=1\right\},$$

and

$${\Gamma }_N^{\left(2\right)}=\left\{n\ge N+1:\sum _{j=N+1}^{n}a\left(n,j\right)<1\right\}.$$

For the proof of Theorem 4.4, we need the following lemma.

Lemma 4.6. Assume ϕ (t) = t, t ≥ 0. A positive constant u has property (P _N ) with an integer N ≥ 0 if and only if the condition (2.9) and either (i) or (ii) are satisfied.

Proof. Necessity. We show that (P _N ) implies (2.9) and either (i) or (ii). Suppose a positive constant u has property (P _N ) with an integer N ≥ 0, hence (2.1) and (2.2) are satisfied for v > 0 and for any n ≥ N + 1.

From (2.1) and (2.2), it is clear that (2.8), (2.9) are satisfied and

$$\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)v,n\ge N+1.$$

(4.13)

Therefore

$$1-\sum _{j=N+1}^{n}a\left(n,j\right)\ge 0,n\ge N+1,$$

and hence

$${\beta }_N=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)\le 1.$$

The latest inequality implies two cases with respect the value of β _N .

• The first case β _N < 1. In this case the condition (i) is satisfied.

• Consider now the second case where β _N = 1. Clearly if ${\sum }_{j=N+1}^{n}a\left(n,j\right)=1$, then from (4.13), we get

  $$\sum _{j=0}^{N}a\left(n,j\right)u+h\left(n\right)=0,n\in {\Gamma }_N^{\left(1\right)},$$

or equivalently (4.10).

But if ${\sum }_{j=N+1}^{n}a\left(n,j\right)<1$, then

$$1-\sum _{j=N+1}^{n}a\left(n,j\right)>0,n\in {\Gamma }_N^{\left(2\right)},$$

and (4.13) yields

$${\left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{-1}\left(\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \right)\le v,n\in {\Gamma }_N^{\left(2\right)}$$

and hence (4.11) and (4.12) are satisfied. Then condition (ii) holds.

Sufficiency. We show that if (2.9) and one of the conditions (i) and (ii) is satisfied with some u ≥ 0 and N ≥ 0, then u has property (P _N ). It is easy to observe that (2.9) yields

$$\underset{n\ge N+1}{\text{sup}}\sum _{j=0}^{N}a\left(n,j\right)<\mathrm{\infty },\mathsf{\text{and}}\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)<\mathrm{\infty }.$$

Let (i) of Theorem 4.5 be satisfied, that is β _N < 1. Then for u ≥ 0 and n ≥ N + 1, N ≥ 0, there exists v > 0 such that

$$\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1-{\beta }_N\right)v,n\ge N+1.$$

It implies

$$\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)v,$$

i.e. (2.2) is satisfied and (2.1) also is satisfied for all v large enough, hence u has property P _N .

Now suppose β _N = 1, and (4.10) holds. Then, clearly (2.1) and (2.2) are satisfied for any v ≥ 0 and $n\in {\Gamma }_N^{\left(1\right)}$.

If $n\in {\Gamma }_N^{\left(2\right)}$ and (4.11) and (4.12) are satisfied, then for u ≥ 0, we have

$$\underset{n\in {\Gamma }_N^{\left(2\right)}}{\text{sup}}{\left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{-1}\left[\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \right]<\mathrm{\infty }.$$

Then there exists v > 0 large enough such that

$${\left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{-1}\left[\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \right]\le v,n\in {\Gamma }_N^{\left(2\right)},$$

since $1-{\sum }_{j=N+1}^{n}a\left(n,j\right)>0$, for all $n\in {\Gamma }_N^{\left(2\right)}$.

Therefore

$$\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1-\sum _{j=N+1}^{n}a\left(n,j\right)\right)v,$$

i.e. for all v large enough the conditions (2.2) and (2.1) are satisfied. Hence, u has property (P _N ).

The following lemma is extracted from [2] (Lemma 5.3) and will be needed in this section.

Lemma 4.7. Assume (A), (B) are satisfied and ϕ (t) = t, t ≥ 0. For every integer N > 0, there exists a non-negative constant K₁(N) independent of the sequence (h(n))_n≥0and x₀, such that the solution (x(n))_n≥0of (1.1) and (1.2), satisfies

$$\parallel x\left(n\right)\parallel \le K_1\left(N\right)\left(\underset{0\le m\le N}{\text{max}}\parallel h\left(m\right)\parallel +\parallel x_0\parallel \right),0\le n\le N.$$

(4.14)

Now we give the proof of Theorem 4.5.

Proof. Let (A), (B), (2.9) and either (i) or (ii) in Theorem 4.5 be satisfied. By Lemma 4.7 the solution of (1.1) and (1.2) satisfies (4.14) for all 0 ≤ n ≤ N. This means that, there exists a non-negative constant u such that

$$\parallel x\left(n\right)\parallel \le u,0\le n\le N.$$

By Lemma 4.6 we have u has property (P _N ). Then the conditions of Theorem 3.1 hold for the initial vector x₀ belonging to S, and hence the solution x(n; x₀) of (1.1) and (1.2) is bounded.

4.3 Super-linear case

Our aim in this section is to obtain sufficient condition for the boundedness in the super-linear case.

Theorem 4.8. Assume that conditions (A) and (B) are satisfied with the function ϕ(t) = t^p, t > 0, where p > 1. Suppose also (2.8) and (2.9) hold. Then the solution x(n;x₀) of (1.1) and (1.2) is bounded for some x₀ ∈ ℝ^dif there exists$v\in \left[{\left(\frac{1}{p{\beta }_0}\right)}^{\frac{1}{p-1}},{\left(\frac{1}{{\beta }_0}\right)}^{\frac{1}{p-1}}\right]$such that

$$a\left(0,0\right)\parallel x_0{\parallel }^p+\parallel h\left(0\right)\parallel \le v,$$

(4.15)

and

$${\alpha }_0\parallel x_0{\parallel }^p+{\gamma }_0\le v-{\beta }_0{v}^p,$$

(4.16)

where α₀, β₀and γ₀are defined in (4.1) and (4.2).

Proof. Assume (2.8), (2.9), (4.15), and (4.16) are satisfied. The case β₀ = 0 is clear. So we assume that β₀ > 0. In this case, clearly, $v-{\beta }_0{v}^p\ge 0$ if

$$v\in \left[0,{\left(\frac{1}{{\beta }_0}\right)}^{\frac{1}{p-1}}\right],$$

and the maximum value of the function $g\left(v\right)=v-{\beta }_0{v}^p\mathsf{\text{ is }}{\left(\frac{1}{p{\beta }_0}\right)}^{1/\left(p-1\right)}$.

Then there exists v such that

$$v\in \left[{\left(\frac{1}{p{\beta }_0}\right)}^{\frac{1}{p-1}},{\left(\frac{1}{{\beta }_0}\right)}^{\frac{1}{p-1}}\right],$$

and the conditions (2.4) and (2.5) hold. By Remark 2.3 we get that under conditions (4.15) and (4.16), u = ||x₀|| has property (P₀) and x₀ belongs to S. Then the conditions of Theorem 3.1 hold, so the solution of (1.1) with the initial condition (1.2) is bounded.

5 Some corollaries with convolution estimations

In this section we give some corollaries on the boundedness of the solutions of (1.1) and (1.2) but in the convolution-type. Through out in this section we take a(n, i) = α(n - i), n ≥ 0, and 0 ≤ i ≤ n, and the following condition

• (C) For any n ≥ 0, there exists an α(n) ∈ ℝ₊, such that

  $$\parallel f\left(n,j,x\right)\parallel \le \alpha \left(n-j\right)\varphi \left(\parallel x\parallel \right),$$

with a monotone non-decreasing mapping ϕ : ℝ₊ → ℝ₊ and ||.|| is any norm on ℝ^d.

Remark 5.1. If a(n,j) = α(n - j), α(n) ∈ ℝ₊, n ≥ 0, and ϕ(t) > 0, t > 0, then the non-negative constant u has property (P _N ) with an integer N ≥ 0 if and only if

$$\sum _{j=0}^N\alpha \left(j\right)\varphi \left(u\right)+\parallel h\left(N\right)\parallel \le v,$$

(5.1)

and

$$\sum _{j=n-N}^n\alpha \left(j\right)\varphi \left(u\right)+\sum _{j=0}^{n-N-1}\alpha \left(j\right)\varphi \left(v\right)+\parallel h\left(n\right)\parallel \le v,n\ge N+1$$

(5.2)

are satisfied.

Remark 5.2. x₀ ∈ ℝ^dbelongs to the set S if

$$\alpha \left(0\right)\varphi \left(\parallel x_0\parallel \right)+\parallel h\left(0\right)\parallel \le v,$$

(5.3)

and

$$\alpha \left(n\right)\varphi \left(\parallel x_0\parallel \right)+\sum _{j=0}^{n-1}\alpha \left(j\right)\varphi \left(v\right)+\parallel h\left(n\right)\parallel \le v,n\ge 1$$

(5.4)

hold. In this case N = 0 and u = ||x₀|| has property (P₀).

By our main result, we have the following corollary.

Corollary 5.3. Let (A), (C) and sup_n≥0||h(n)|| < ∞ be satisfied, and assume that the initial vector x₀belongs to the set S. Then the solution x(n;x₀), n ≥ 0, of (1.1) and (1.2) is bounded.

Proof. Assume (A), (C) are satisfied. By Theorem 3.1 and Remark 5.1, it is easy to prove that the solution of (1.1) and (1.2) is bounded.

The following two corollaries are immediate consequence of Theorems 4.1 and 4.3 of the sub-linear convolution case, respectively.

Corollary 5.4. Assume (A), (C) are satisfied and ϕ(t) = t^p, t > 0, with fixed p ∈ (0,1). If

$$\underset{n\ge 0}{\text{sup}}\parallel h\left(n\right)\parallel <\mathrm{\infty },and\sum _{j=0}^{\mathrm{\infty }}\alpha \left(j\right)<\mathrm{\infty },$$

(5.5)

then for any x₀ ∈ ℝ^dthe solution x(n; x₀), n ≥ 0 of (1.1) and (1.2) is bounded.

Corollary 5.5. Consider Equation (4.4) with p ∈ (0, 1) and non-negative coefficients. Assume

$$\sum _{n=0}^{\mathrm{\infty }}\alpha \left(n\right)>0,$$

and for any n ≥ 0 one has h(n) > 0 if α(j) = 0, 0 ≤ j ≤ n. Then the solution of (4.4) and (4.5) is bounded if and only if the condition (5.5) is satisfied.

Proof. The proof is immediate consequence from proof of Theorem 4.3 with Remark 5.1.

Remark 5.6. The Corollary 5.5 is analogous to the corresponding result of Lipovan (Theorem 3.1, [18]) for integral equation.

In the next corollary we assume that

$$\sum _{n=0}^{\mathrm{\infty }}\alpha \left(n\right)\le 1.$$

(5.6)

Under condition (5.6) we have three cases

Case 1. $\sum _{n=0}^{\mathrm{\infty }}\alpha \left(n\right)<1$;

Case 2. $\sum _{j=0}^{\mathrm{\infty }}\alpha \left(j\right)=1$ and for any n ≥ 1,

$$\sum _{j=0}^{n-1}\alpha \left(j\right)<1,\mathsf{\text{or equivalently}}\sum _{j=n}^{\mathrm{\infty }}\alpha \left(n\right)>0;$$

Case 3. There exists an index M ≥ 0 such that $\sum _{n=0}^M\alpha \left(n\right)=1$, moreover α(M) ≠ 0 and α(n) = 0, n ≥ M + 1.

Corollary 5.7. Assume (A), (C) and (5.6), and let ϕ(t) = t, t ≥ 0. Then the solution of (1.1) and (1.2) is bounded for any x₀ ∈ ℝ, if one of the following conditions holds

• (a) Case 1. holds and sup_n≥0||h(n)|| < ∞.

• (b) Case 2. holds and

  $$\underset{n\ge 1}{\text{sup}}\frac{\parallel h\left(n\right)\parallel }{\sum _{j=n}^{\mathrm{\infty }}\alpha \left(j\right)}<\mathrm{\infty }.$$

• (c) Case 3. holds and h(n) = 0, n ≥ M + 1.

Proof. (a) This part is an immediate consequence of (i) from Theorem 4.5 with Remark 5.1.

• (b) Assume that (b) is satisfied. For a fixed N ≥ 0 and u ∈ ℝ₊, there exists a positive constant v such that

  $$\frac{\sum _{j=n-N}^n\alpha \left(j\right)}{\sum _{j=n-N}^{\mathrm{\infty }}\alpha \left(j\right)}u+\frac{\parallel h\left(n\right)\parallel }{\sum _{j=n-N}^{\mathrm{\infty }}\alpha \left(j\right)}\le u+\frac{\parallel h\left(n\right)\parallel }{\sum _{j=n}^{\mathrm{\infty }}\alpha \left(j\right)}\le v,n\ge N+1,$$

and

$$\sum _{j=0}^N\alpha \left(j\right)u+\parallel h\left(N\right)\parallel \le v$$

hold. Thus

$$\sum _{j=n-N}^n\alpha \left(j\right)u+\parallel h\left(n\right)\parallel \le \sum _{j=n-N}^{\mathrm{\infty }}\alpha \left(j\right)v\le \left(1-\sum _{j=0}^{n-N-1}\alpha \left(j\right)\right)v,$$

i.e.

$$\sum _{j=n-N}^n\alpha \left(j\right)u+\sum _{j=0}^{n-N-1}\alpha \left(j\right)v+\parallel h\left(n\right)\parallel \le v,n\ge N+1.$$

(5.7)

By Remark 5.1, u has property (P _N ). For the initial value x₀ ∈ ℝ^dapplying Corollary 5.3, we get the boundedness of the solution of (1.1) and (1.2).

• (c) Assume the condition (c). Clearly, for all v ≥ 0 the conditions (2.1) and (2.2) are satisfied and u has property (P _N ). Then for any x ₀ ∈ S, the solution of (1.1) is bounded according to Corollary 5.3.

The proof of the following corollary is an immediate consequence of Theorem 4.8 and Remark 5.1 and it is therefore omitted.

Corollary 5.8. Assume that conditions (A) and (C) are satisfied with the function ϕ (t) = t^p, t > 0,p > 1, and (5.5) holds. Then for an x₀ ∈ ℝ^dthe solution x(n;x₀) of (1.1) and (1.2) is bounded, if there exists$v\in \left[{\left(\frac{1}{p{\beta }_c}\right)}^{\frac{1}{p-1}},{\left(\frac{1}{{\beta }_c}\right)}^{\frac{1}{p-1}}\right]$such that

$$\alpha \left(0\right)\parallel x_0{\parallel }^p+\parallel h\left(0\right)\parallel \le v,$$

and

$${\alpha }_c\parallel x_0{\parallel }^p+{\gamma }_0\le v-{\beta }_c{v}^p,$$

where

$${\alpha }_c=\underset{n\ge 1}{\sup }\alpha (n)<\mathrm{\infty },{\beta }_c={\displaystyle \sum _{j=0}^{\mathrm{\infty }}\alpha (j)<\mathrm{\infty },}and{\gamma }_0=\underset{n\ge 1}{\sup }\parallel h(n)\parallel \mathrm{\infty }.$$

Now we show the application of Corollary 5.7 to the linear convolution Volterra difference equation with infinite delay

$$x\left(n+1\right)=\sum _{i=-\mathrm{\infty }}^{n}Q\left(n-i\right)x\left(i\right),n\ge 0,$$

(5.8)

with initial condition

$$x\left(n\right)=\phi \left(n\right),n\le 0,$$

(5.9)

where Q(n) ∈ ℝ^{d × d}, n ≥ 0 and φ(m) ∈ ℝ^d, m ≤ 0.

From (5.8), we have

$$x\left(n+1\right)=\sum _{i=0}^{n}Q\left(n-i\right)x\left(i\right)+\sum _{i=-\mathrm{\infty }}^{-1}Q\left(n-i\right)\phi \left(i\right)n\ge 0.$$

If we compare the latest equation with Equation (1.1), we have f(n, i, x) = Q(n - i)x and

$$h\left(n\right)=\sum _{i=-\mathrm{\infty }}^{-1}Q\left(n-i\right)\phi \left(i\right)n\ge 0.$$

If $M_{\phi }=\underset{m\le -1}{\text{sup}}\parallel \phi \left(m\right)\parallel$, then we get

$$\parallel h\left(n\right)\parallel \le \sum _{j=n+1}^{\mathrm{\infty }}\parallel Q\left(j\right)\parallel M_{\phi }.$$

Corollary 5.9. Assume Q(n) ∈ ℝ^{d × d}, n ≥ 0, and

$$\sum _{n=0}^{\mathrm{\infty }}\parallel Q\left(n\right)\parallel \le 1.$$

(5.10)

Then the solution of (5.8) with the initial condition (5.9) is bounded, if

$$M_{\phi }=\underset{m\le -1}{\text{sup}}\parallel \phi \left(m\right)\parallel <\mathrm{\infty }.$$

(5.11)

Proof. To prove that the solution of (5.8) with the initial condition (5.9) is bounded, it is enough to show that one of the hypotheses (a), (b), (c) of Corollary 5.7 holds. Let (5.10) and (5.11) be satisfied. There are three cases.

First consider the case when

$$\sum _{n=0}^{\mathrm{\infty }}\parallel Q\left(n\right)\parallel <1,$$

and (5.11) are satisfied. This implies (a) of Corollary 5.7.

Second consider the case when

$$\sum _{j=0}^{\mathrm{\infty }}\parallel Q\left(j\right)\parallel =1,\mathsf{\text{and}}\sum _{j=n}^{\mathrm{\infty }}\parallel Q\left(j\right)\parallel >0,n\ge 1.$$

This yields

$$\underset{n\ge 1}{\text{sup}}\frac{\parallel h\left(n\right)\parallel }{\sum _{j=n}^{\mathrm{\infty }}\parallel Q\left(j\right)\parallel }\le \underset{n\ge 1}{\text{sup}}\frac{\sum _{j=n+1}^{\mathrm{\infty }}\parallel Q\left(j\right)\parallel M_{\phi }}{\sum _{j=n}^{\mathrm{\infty }}\parallel Q\left(j\right)\parallel }\le M_{\phi }<\mathrm{\infty },$$

and hence condition (b) of Corollary 5.7 is satisfied.

The last case is when there exists N ≥ 0, such that

$$\sum _{j=0}^N\parallel Q\left(j\right)\parallel =1,\mathsf{\text{and}}\parallel Q\left(n\right)\parallel =0,\mathsf{\text{for all}}n\ge N+1.$$

In this case, we get ||h(n)|| = 0 for all n ≥ N + 1, hence (c) of Corollary 5.7 holds.

6 Examples

In this section we give some examples to illustrate our results.

Example 6.1. Let us consider the case when

$$a\left(n,j\right)=\frac{2n+1}{\left(n+j+1\right)\left(n+j+2\right)}\mathsf{\text{for all}}0\le j\le n.$$

We have $a\left(0,0\right)=\frac{1}{2},a\left(n,0\right)=\frac{2n+1}{\left(n+1\right)\left(n+2\right)},{\text{sup}}_{n\ge 1}a\left(n,0\right)=\frac{1}{2}$,

$$\sum _{j=1}^n\frac{2n+1}{\left(n+j+1\right)\left(n+j+2\right)}=\frac{n\left(2n+1\right)}{2\left(n+1\right)\left(n+2\right)},$$

and

$$\underset{n\ge 0}{\text{sup}}\sum _{j=0}^n\frac{2n+1}{\left(n+j+1\right)\left(n+j+2\right)}=1.$$

(6.1)

One can easily see that conditions (2.4) and (2.5) are equivalent to the inequalities

$$\frac{1}{2}\varphi \left(\parallel x_0\parallel \right)+\parallel h\left(0\right)\parallel \le v,$$

(6.2)

and

$$\frac{2n+1}{\left(n+1\right)\left(n+2\right)}\varphi \left(\parallel x_0\parallel \right)+\frac{n\left(2n+1\right)}{2\left(n+1\right)\left(n+2\right)}\varphi \left(v\right)+\parallel h\left(n\right)\parallel \le v,n\ge 1.$$

(6.3)

Consider the scalar equation

$$x\left(n+1\right)=\sum _{j=0}^n\frac{\left(2n+1\right)}{\left(n+j+1\right)\left(n+j+2\right)}{x}^p\left(j\right)+h\left(n\right),n\ge 0,$$

(6.4)

with the initial condition

$$x\left(0\right)=x_0$$

(6.5)

where x₀ ∈ ℝ₊, h(n) ∈ ℝ₊, n ≥ 0 and p > 0. In fact there are three cases with respect to the value of p.

(a1) p ∈ (0,1) and ϕ (t) = t^p, t > 0. If sup_n≥0h(n) < ∞ and (6.1) are satisfied, then for any x₀ ≥ 0, the solution of (6.4) and (6.5) is bounded by Theorem 4.1.

(a2) p = 1 and ϕ (t) = t, t >0. It is not difficult to see that for any v > 0 large enough the inequalities (6.2) and (6.3) are equivalent to the inequalities

$$\frac{1}{2}{x}_0+h\left(0\right)\le v,$$

and

$$\frac{2\left(2n+1\right)}{5n+4}{x}_0+\frac{2\left(n+1\right)\left(n+2\right)}{n\left(5n+4\right)}nh\left(n\right)\le v,n\ge 1.$$

(6.6)

Let k = sup_n≥1nh(n) < ∞, it is easily to see that the inequality (6.6) is satisfied if

$$\frac{4}{5}{x}_0+\frac{4}{3}k\le v.$$

Hence (2.4) and (2.5) are satisfied, by Lemma 4.6, x₀ has property (P₀). It is worth to note that in this case our Theorem 4.5 is applicable for any x₀ ∈ ℝ₊, but the results in [2, 8–12, 14] are not applicable in this case.

(a3) p > 1 and ϕ(t) = t^p, t > 0. Assume k = sup_n≥1h(n), and for x₀ ≥ 0, there exists $v\in \left[{\left(\frac{1}{p}\right)}^{\frac{1}{p-1}},1\right]$, such that

$$\frac{1}{2}{x}_0^p+h\left(0\right)\le v$$

and

$$\frac{1}{2}{x}_0^p+k\le v-v^p$$

hold. Hence x₀ has property (P₀), and by Theorem 4.8, for small x₀, the solution of (6.4) and (6.5) is bounded.

Summarizing the observations and applying Theorem 4.5, we get the next new result.

Proposition 6.2. If Equation (6.4) is linear, that is p = 1, and sup_{n ≥ 1}n h(n) < ∞, then every positive solution of (6.4) with initial condition (6.5) is bounded.

Proposition 6.3. Assume Equation (6.4) is super-linear, that is p > 1. If for some x₀ ≥ 0 there exists$v\in \left[{\left(\frac{1}{p}\right)}^{\frac{1}{p-1}},1\right]$, such that

$$\frac{1}{2}{x}_0^p+h\left(0\right)\le v,$$

and

$$\frac{1}{2}{x}_0^p+k\le v-v^p$$

hold, then the positive solution of (6.4) with initial condition (6.5) is bounded.

The following example shows that applicability of our result in the critical case.

Example 6.4. Consider the equation

$$x\left(n+1\right)=\sum _{i=0}^{n}c{q}^{n-i}x\left(i\right)+h\left(n\right),n\ge 0,$$

(6.7)

$$x\left(0\right)=x_0,$$

(6.8)

where x₀ ∈ ℝ₊, q ∈ (0,1), c ∈ (0,1) and h(n) ∈ ℝ₊, n ≥ 0.

If q + c < 1 and sup_n≥0h(n) < ∞, then our condition (a) in Corollary 5.7 holds. Relation q + c = 1 implies

$$\sum _{n=0}^{\mathrm{\infty }}c{q}^n=1.$$

Note that the results in [2, 8–12] are not applicable. But our Corollary 5.7 is applicable under condition

$$\underset{n\ge 1}{\text{sup}}\frac{h\left(n\right)}{q^n}<\mathrm{\infty },$$

since it implies condition (b) in Corollary 5.7:

$$\underset{n\ge 1}{\text{sup}}\frac{h\left(n\right)}{\sum _{j=n}^{\mathrm{\infty }}c{q}^j}=\underset{n\ge 1}{\text{sup}}\frac{\left(1-q\right)h\left(n\right)}{c{q}^n}<\mathrm{\infty }.$$

Remark 6.5. By mathematical induction, it is easy to see the solution of (6.7) and (6.8) is given in the form

$$x\left(n\right)=c{\left(q+c\right)}^{n-1}{x}_0+\sum _{j=0}^{n-2}c{\left(q+c\right)}^{n-j-2}h\left(j\right)+h\left(n-1\right),n\ge 2,$$

where x(0) = x₀and x(1) = cx₀ + h(0). Let${\sum }_{j=0}^{\mathrm{\infty }}c{q}^j=1$or equivalently q + c = 1 moreover h(n) = kⁿ⁺¹, 0 < q < k < 1. In this case the above solution is bounded for any x₀ ∈ ℝ₊. At the same time condition (b) in Corollary 5.7 does not hold, and hence condition (β) in Proposition 1.1 is not necessary.

Therefore by statements (a) and (b) of Corollary 5.7 we get the following.

Proposition 6.6. The solution of (6.7) and (6.8) is bounded if either q + c < 1 and sup_n≥0h(n) < ∞, or q + c = 1, and

$$\underset{n\ge 1}{\text{sup}}\frac{h\left(n\right)}{q^n}<\mathrm{\infty }.$$

The next example shows the sharpness of our Corollary 5.8.

Example 6.7. Consider the equation

$$x\left(n+1\right)=\sum _{i=0}^n{q}^{n-i}{x}^p\left(i\right),n\ge 0,$$

(6.9)

where x(0) = x₀ > 0, p > 1, q ∈ (0,1).

Since x₀ > 0, implies x(n) > 0 for all n ≥ 0, therefore from (6.9), we have x(n + 1) ≥ x^p(n), n ≥ 0. For fixed p > 1, inequality $x_0^p>x_0$ holds if and only if x₀ > 1. In the latest, by mathematical induction it is easy to prove that the sequence (x(n))_n≥0is strictly increasing. Now for x₀ > 1, we prove that x(n) → ∞ as n → ∞. Assume, for the sake of contradiction, that the sequence (x(n))_n≥0is bounded. Since it is strictly increasing, x* = lim_n→∞x(n) is finite and x* > x₀. On the other hand x(n+ 1) ≥ x^p(n), and hence we get x* ≥ (x*) p > x*, which is a contradiction. Hence for all x₀ > 1 the solution of (6.9) is unbounded. Now applying Corollary 5.8 to (6.9), there exists

$$v\in \left[{\left(\frac{1-q}{p}\right)}^{\frac{1}{p-1}},{\left(1-q\right)}^{\frac{1}{p-1}}\right],$$

such that $x_0^p\le v$ and

$$q{x}_0^p\le v-\frac{1}{1-q}{v}^p.$$

Hence

$$v=\frac{1}{q}\left(v-\frac{1}{1-q}{v}^p\right)\mathsf{\text{implies}}v={\left(1-q\right)}^{\frac{2}{p-1}},$$

i.e.

$$x_0\le {\left(1-q\right)}^{\frac{2}{p\left(p-1\right)}}.$$

Then the solution of (6.9) with initial value $x_0\le {\left(1-q\right)}^{\frac{2}{p\left(p-1\right)}}$ is bounded, but the solution of (6.9) with initial value x₀ > 1 is unbounded. Our results do not give any information about the boundedness of the solutions, whenever $x_0\in \left({\left(1-q\right)}^{\frac{2}{p\left(p-1\right)}},1\right]$ but this gap tends to zero if either p is large enough or q is very close to zero. Hence, our results for the super-linear case are sharp in some sense. As a special case let p = 2. In this case, the solution of (6.9) with initial condition x(0) = x₀ is bounded whenever x₀ ∈ [0,1 - q] and it is unbounded whenever x₀ > 1.

Based on our results we state the following conjecture as an open problem.

Conjecture 6.8. Let p > 1 and 0 < q <1. Then there exists a constant κ > 1 such that the solution of (6.9) with initial condition x(0) = x₀is bounded whenever x₀ ∈ [0, κ) and it is unbounded whenever x₀ > κ.