#### Positive Solutions for Systems of Second-Order Difference Equations

Positive Solutions for Systems of Second-Order Difference Equations
Johnny Henderson1 and Rodica Luca2
1Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA
2Department of Mathematics, Gheorghe Asachi Technical University, 700506 Iasi, Romania
Received 17 March 2015; Accepted 30 August 2015
Academic Editor: Miguel Ángel López
Copyright © 2015 Johnny Henderson and Rodica Luca. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
We study the existence and nonexistence of positive solutions of some systems of nonlinear second-order difference equations subject to multipoint boundary conditions which contain some positive constants.
1. Introduction
The mathematical modeling of many nonlinear problems from computer science, economics, mechanical engineering, control systems, biological neural networks, and others leads to the consideration of nonlinear difference equations (see [1, 2]). In the last decades, many authors have investigated such problems by using various methods, such as fixed point theorems, the critical point theory, upper and lower solutions, the fixed point index theory, and the topological degree theory (see, e.g., [3–15]).
In this paper, we consider the system of nonlinear second-order difference equations: Δ 2 ? ? − 1 + ? ? ? ? ? = 0 , ? = Δ 1 , ? − 1 , 2 ? ? − 1 + ? ? ? ? ? = 0 , ? = 1 , ? − 1 , ( ? ) with the multipoint boundary conditions ? 0 = ? ? = 1 ? ? ? ? ? + ? 0 , ? ? = ? ? = 1 ? ? ? ? ? , ? 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? + ? 0 , ( ? ? ) where ? , ? , ? , ? , ? ∈ ℕ , ? ≥ 2 , Δ is the forward difference operator with stepsize 1 , Δ ? ? = ? ? + 1 − ? ? , Δ 2 ? ? − 1 = ? ? + 1 − 2 ? ? + ? ? − 1 , ? = ? , ? means that ? = ? , ? + 1 , … , ? for ? , ? ∈ ℕ , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , and ? 0 and ? 0 are positive constants.
Under some assumptions on the functions ? and ? , we will prove the existence of positive solutions of problem ( ? ) - ( ? ? ) . By a positive solution of ( ? ) - ( ? ? ) we mean a pair of sequences ( ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? ) satisfying ( ? ) and ( ? ? ) with ? ? , ? ? ≥ 0 for all ? = 0 , ? and ? ? > 0 for all ? = 0 , ? − 1 , ? ? > 0 for all ? = 1 , ? . We will also give sufficient conditions for the nonexistence of positive solutions for this problem. System ( ? ) with the multipoint boundary conditions ? ? 0 − ? Δ ? 0 = 0 , ? ? = ∑ ? − 2 ? = 1 ? ? ? ? ? + ? 0 , ? ? 0 − ? Δ ? 0 = 0 , and ? ? = ∑ ? − 2 ? = 1 ? ? ? ? ? + ? 0 ( ? 0 , ? 0 > 0 ) has been investigated in [16]. Some systems of difference equations with parameters, subject to multipoint boundary conditions, were studied in [17, 18], by using the Guo-Krasnosel’skii fixed point theorem. We also mention the paper [19], where the authors investigated the existence and multiplicity of positive solutions for the system Δ 2 ? ? − 1 + ? ( ? , ? ? ) = 0 , Δ 2 ? ? − 1 + ? ( ? , ? ? ) = 0 , ? = 1 , ? − 1 , with the multipoint boundary conditions ( ? ? ) with ? 0 = ? 0 = 0 , by using some theorems from the fixed point index theory.
In Section 2, we present some auxiliary results which investigate a second-order difference equation subject to multipoint boundary conditions. In Section 3, we will prove our main results, and in Section 4, we will present an example which illustrates the obtained theorems. Our main existence result is based on the Schauder fixed point theorem which we present now.
Theorem 1. Let ? be a Banach space and ? ⊂ ? a nonempty, bounded, convex, and closed subset. If the operator ? ∶ ? → ? is completely continuous (continuous and compact, i.e., mapping bounded sets into relatively compact sets), then ? has at least one fixed point.
2. Auxiliary Results
In this section, we present some auxiliary results from [17] related to the following second-order difference equation with the multipoint boundary conditions: Δ 2 ? ? − 1 + ? ? = 0 , ? = ? 1 , ? − 1 , 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? , ( 1 ) where ? , ? , ? ∈ ℕ , ? ≥ 2 , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , and 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 .
Lemma 2 (see [17]). If ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , Δ 1 = ∑ 1 − ? ? = 1 ? ? ∑ ? ? = 1 ? ? ? ? + ∑ 1 − ? ? = 1 ? ? ∑ ? − ? ? = 1 ? ? ? ? ≠ 0 , and ? ? ∈ ℝ for all ? = 1 , ? − 1 , then the solution of (1) is given by ? ? = ∑ ? − 1 ? = 1 ? 1 ( ? , ? ) ? ? for all ? = 0 , ? , where Green’s function ? 1 is defined by ? 1 ? , ? = ? 0 + 1 ? , ? Δ 1 ( ? − ? ) 1 − ? ? = 1 ? ? + ? ? = 1 ? ? ? − ? ? ⋅ ? ? = 1 ? ? ? 0 ? ? + 1 , ? Δ 1 ? 1 − ? ? = 1 ? ? + ? ? = 1 ? ? ? ? ⋅ ? ? = 1 ? ? ? 0 ? ? , ? , ? = 0 , ? , ? = ? 1 , ? − 1 , 0 = 1 ? , ? ? ? ? ( ? − ? ) , 1 ≤ ? ≤ ? ≤ ? , ( ? − ? ) , 0 ≤ ? ≤ ? ≤ ? − 1 . ( 2 )
Lemma 3 (see [17]). If ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ? 2 < ⋯ < ? ? ≤ ? − 1 , ? ? ∈ ℕ for all ? = 1 , ? , and 1 ≤ ? 1 < ? 2 < ⋯ < ? ? ≤ ? − 1 , then Green’s function ? 1 of problem (1) satisfies ? 1 ( ? , ? ) ≥ 0 for all ? = 0 , ? , ? = 1 , ? − 1 . Moreover, if ? ? ≥ 0 for all ? = 1 , ? − 1 , then the unique solution ? ? , ? = 0 , ? , of problem (1) satisfies ? ? ≥ 0 for all ? = 0 , ? .
Lemma 4 (see [17]). Assume that ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ? 2 < ⋯ < ? ? ≤ ? − 1 , ? ? ∈ ℕ for all ? = 1 , ? , and 1 ≤ ? 1 < ? 2 < ⋯ < ? ? ≤ ? − 1 . Then Green’s function ? 1 of problem (1) satisfies the following inequalities:(a) ? 1 ( ? , ? ) ≤ ? 1 ( ? ) , ∀ ? = 0 , ? , ? = 1 , ? − 1 , where ? 1 ? = ? 0 + 1 ? , ? Δ 1 ? − ? ? = 1 ? ? ? ? ? ? = 1 ? ? ? 0 ? ? + 1 , ? Δ 1 ? − ? ? = 1 ? ? ? − ? ? ? ? = 1 ? ? ? 0 ? ? . , ? ( 3 ) (b)For every ⟦ ⟧ } ? ∈ { 1 , … , ? / 2 , one has m i n ? = ? , ? − ? ? 1 ? , ? ≥ ? 1 ? 1 ? ≥ ? 1 ? 1 ? , , ? ∀ ? = 0 , ? , ? = 1 , ? − 1 , ( 4 ) where ? 1 ? = m i n , ? ∑ ? − 1 1 − ? ? = 1 ? ? + ∑ ? ? = 1 ? ? ? − ? ? ∑ ? − ? ? = 1 ? ? ? ? , ? ∑ 1 − ? ? = 1 ? ? + ∑ ? ? = 1 ? ? ? ? ∑ ? − ? ? = 1 ? ? ? − ? ? > 0 , ( 5 ) and ⟦ ⟧ ? / 2 is the largest integer not greater than ? / 2 .
Lemma 5 (see [17]). Assume that ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ? 2 < ⋯ < ? ? ≤ ? − 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ? 2 < ⋯ < ? ? ≤ ? − 1 , ⟦ ⟧ } ? ∈ { 1 , … , ? / 2 , and ? ? ≥ 0 for all ? = 1 , ? − 1 . Then the solution ? ? , ? = 0 , ? , of problem (1) satisfies the inequality ? ? ? ? = ? , ? − ? ? ? ≥ ? 1 ? ? ? ? = 0 , ? ? ? .
We can also formulate similar results as Lemmas 2–5 above for the discrete boundary value problem Δ 2 ? ? − 1 + ℎ ? = 0 , ? = ? 1 , ? − 1 , 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? , ( 6 ) where ? , ? , ? ∈ ℕ , ? ≥ 2 , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℝ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , and ℎ ? ∈ ℝ for all ? = 1 , ? − 1 . We denote by Δ 2 , ? 2 , ? 2 , and ? 2 the corresponding constants and functions for problem (6) defined in a similar manner as Δ 1 , ? 1 , ? 1 , and ? 1 , respectively.
3. Main Results
We present first the assumptions that we will use in the sequel:( ? 1 ) ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 , ? ? ∈ ℕ for all ? = 1 , ? , 1 ≤ ? 1 < ⋯ < ? ? ≤ ? − 1 , and ? ? ≥ 0 for all ? = 1 , ? , ∑ ? ? = 1 ? ? < 1 .( ? 2 )The constants ? ? , ? ? ≥ 0 for all ? = 1 , ? − 1 , and there exist ? 0 , ? 0 ∈ { 1 , … , ? − 1 } such that ? ? 0 > 0 , ? ? 0 > 0 .( ? 3 ) ? , ? ∶ [ 0 , ∞ ) → [ 0 , ∞ ) are continuous functions and there exists ? 0 > 0 such that ? ( ? ) < ? 0 / ? , ? ( ? ) < ? 0 / ? for all ? ∈ [ 0 , ? 0 ] , where ∑ ? = ? ? ? { ? − 1 ? = 1 ? ? ? 1 ∑ ( ? ) , ? − 1 ? = 1 ? ? ? 2 ( ? ) } and ? 1 , ? 2 are defined in Section 2.( ? 4 ) ? , ? ∶ [ 0 , ∞ ) → [ 0 , ∞ ) are continuous functions and satisfy the conditions ? ? ? ? → ∞ ( ? ( ? ) / ? ) = ∞ , ? ? ? ? → ∞ ( ? ( ? ) / ? ) = ∞ .
Our first theorem is the following existence result for problem ( ? ) - ( ? ? ) .
Theorem 6. Assume that assumptions ( ? 1 ) - - ( ? 3 ) hold. Then problem ( ? ) - ( ? ? ) has at least one positive solution for ? 0 > 0 and ? 0 > 0 sufficiently small.
Proof. We consider the problems Δ 2 ℎ ? − 1 = 0 , ? = ℎ 1 , ? − 1 , 0 = ? ? = 1 ? ? ℎ ? ? ℎ + 1 , ? = ? ? = 1 ? ? ℎ ? ? , ( 7 ) Δ 2 ? ? − 1 = 0 , ? = ? 1 , ? − 1 , 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? + 1 . ( 8 ) Problems (7) and (8) have the solutions ℎ ? = 1 Δ 1 − ? 1 − ? ? = 1 ? ? + ? − ? ? = 1 ? ? ? ? , ? = ? 0 , ? , ? = 1 Δ 2 ? 1 − ? ? = 1 ? ? + ? ? = 1 ? ? ? ? , ? = 0 , ? , ( 9 ) respectively, where Δ 1 and Δ 2 are defined in Section 2. By assumption ( ? 1 ) we obtain ℎ ? > 0 for all ? = 0 , ? − 1 and ? ? > 0 for all ? = 1 , ? .
We define the sequences ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? by ? ? = ? ? − ? 0 ℎ ? , ? ? = ? ? − ? 0 ? ? , ? = 0 , ? , ( 1 0 ) where ( ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? ) is a solution of ( ? ) - ( ? ? ) . Then ( ? ) - ( ? ? ) can be equivalently written as Δ 2 ? ? − 1 + ? ? ? ? ? + ? 0 ? ? = 0 , ? = Δ 1 , ? − 1 , 2 ? ? − 1 + ? ? ? ? ? + ? 0 ℎ ? = 0 , ? = 1 , ? − 1 , ( 1 1 ) with the boundary conditions ? 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? , ? 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? . ( 1 2 ) Using Green’s functions ? 1 and ? 2 from Section 2, a pair ( ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? ) is a solution of problem (11)-(12) if and only if it is a solution for the problem ? ? = ? − 1 ? = 1 ? 1 ( ? , ? ) ⋅ ? ? ? ? − 1 ? = 1 ? 2 ? ? , ? ? ? ? ? + ? 0 ℎ ? + ? 0 ? ? , ? = ? 0 , ? , ? = ? − 1 ? = 1 ? 2 ( ? , ? ) ? ? ? ? ? + ? 0 ℎ ? , ? = 0 , ? , ( 1 3 ) where ( ℎ ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? are given in (9).
We consider the Banach space ? = ℝ ? + 1 with the norm ‖ ? ‖ = ? ? ? ? = 0 , ? | ? ? | , ? = ( ? ? ) ? = 0 , ? , and we define the set ? = { ( ? ? ) ? = 0 , ? , 0 ≤ ? ? ≤ ? 0 , ∀ ? = 0 , ? } ⊂ ? .
We also define the operator ℰ ∶ ? → ? by ℰ ( ? ) = ? − 1 ? = 1 ? 1 ( ? , ? ) ⋅ ? ? ? ? − 1 ? = 1 ? 2 ? ? , ? ? ? ? ? + ? 0 ℎ ? + ? 0 ? ? ? = 0 , ? , ? ? = ? ? = 0 , ? ∈ ? . ( 1 4 ) For sufficiently small ? 0 > 0 and ? 0 > 0 , by ( ? 3 ) , we deduce ? ? ? + ? 0 ? ? ≤ ? 0 ? , ? ? ? + ? 0 ℎ ? ≤ ? 0 ? , ∀ ? = ? 0 , ? , ∀ ? ? , ? ? ? ∈ ? . ( 1 5 ) Then, by using Lemma 3, we obtain ℰ ( ? ) ? ≥ 0 for all ? = 0 , ? and ? = ( ? ? ) ? = 0 , ? ∈ ? . By Lemma 4, for all ? ∈ ? , we have ? − 1 ? = 1 ? 2 ? ? , ? ? ? ? ? + ? 0 ℎ ? ≤ ? − 1 ? = 1 ? 2 ? ? ? ? ? ? + ? 0 ℎ ? ≤ ? 0 ? ? − 1 ? = 1 ? ? ? 2 ? ≤ ? 0 , ∀ ? = 1 , ? − 1 , ℰ ( ? ) ? ≤ ? − 1 ? = 1 ? 1 ( ? ) ? ? ? ? − 1 ? = 1 ? 2 ? ? , ? ? ? ? ? + ? 0 ℎ ? + ? 0 ? ? ≤ ? 0 ? ? − 1 ? = 1 ? ? ? 1 ( ? ) ≤ ? 0 , ∀ ? = 0 , ? . ( 1 6 ) Therefore ℰ ( ? ) ⊂ ? .
Using standard arguments, we deduce that ℰ is completely continuous. By Theorem 1, we conclude that ℰ has a fixed point ? = ( ? ? ) ? = 0 , ? ∈ ? . This element together with ? = ( ? ? ) ? = 0 , ? given by (13) represents a solution for (11)-(12). This shows that our problem ( ? ) - ( ? ? ) has a positive solution ( ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? ) with ? ? = ? ? + ? 0 ℎ ? , ? ? = ? ? + ? 0 ? ? , ? = 0 , ? , ( ? ? > 0 for all ? = 0 , ? − 1 and ? ? > 0 for all ? = 1 , ? ) for sufficiently small ? 0 > 0 and ? 0 > 0 .
In what follows, we present sufficient conditions for the nonexistence of positive solutions of ( ? ) - ( ? ? ) .
Theorem 7. Assume that assumptions ( ? 1 ) , ( ? 2 ) , and ( ? 4 ) hold. Then problem ( ? ) - ( ? ? ) has no positive solution for ? 0 and ? 0 sufficiently large.
Proof. We suppose that ( ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? ) is a positive solution of ( ? ) - ( ? ? ) . Then ( ( ? ? ) ? = 0 , ? , ( ? ? ) ? = 0 , ? ) with ? ? = ? ? − ? 0 ℎ ? , ? ? = ? ? − ? 0 ? ? , ? = 0 , ? , is a solution for (11)-(12), where ( ℎ ? ) ? = 0 , ? and ( ? ? ) ? = 0 , ? are the solutions of problems (7) and (8), respectively (given by (9)). By ( ? 2 ) there exists ⟦ ⟧ } ? ∈ { 1 , 2 , … , ? / 2 such that ? 0 , ? 0 ∈ { ? , … , ? − ? } and then ∑ ? − ? ? = ? ? ? ? 1 ( ? ) > 0 and ∑ ? − ? ? = ? ? ? ? 2 ( ? ) > 0 . By using Lemma 3 we have ? ? ≥ 0 , ? ? ≥ 0 for all ? = 0 , ? , and by Lemma 5, we obtain ? ? ? ? = ? , ? − ? ? ? ≥ ? 1 ‖ ? ‖ and ? ? ? ? = ? , ? − ? ? ? ≥ ? 2 ‖ ? ‖ , where ? 1 and ? 2 are defined in Section 2.
Using now (9), we deduce that m i n ? = ? , ? − ? ℎ ? = ℎ ? − ? = ℎ ? − ? ℎ 0 ‖ ℎ ‖ , m i n ? = ? , ? − ? ? ? = ? ? = ? ? ? ? ‖ ? ‖ . ( 1 7 ) Therefore, we obtain m i n ? = ? , ? − ? ? ? + ? 0 ℎ ? ≥ ? 1 ? ‖ ? ‖ + 0 ℎ ? − ? ℎ 0 ‖ ℎ ‖ ≥ ? 1 ‖ ? ‖ + ? 0 ‖ ℎ ‖ ≥ ? 1 ‖ ‖ ? + ? 0 ℎ ‖ ‖ , m i n ? = ? , ? − ? ? ? + ? 0 ? ? ≥ ? 2 ‖ ‖ ? ‖ ‖ + ? 0 ? ? ? ? ‖ ? ‖ ≥ ? 2 ‖ ‖ ? ‖ ‖ + ? 0 ‖ ? ‖ ≥ ? 2 ‖ ‖ ? + ? 0 ? ‖ ‖ , ( 1 8 ) where ? 1 = ? ? ? { ? 1 , ℎ ? − ? / ℎ 0 } and ? 2 = ? ? ? { ? 2 , ? ? / ? ? } .
We now consider ? ? = ? ? ? 2 ? 1 ? − ? ? = ? ? ? ? 2 ( ? ) , ? 1 ? 2 ? − ? ? = ? ? ? ? 1 ( ? ) − 1 > 0 . ( 1 9 ) By using ( ? 4 ) , for ? defined above, we conclude that there exists ? 0 > 0 such that ? ( ? ) > 2 ? ? , ? ( ? ) > 2 ? ? for all ? ≥ ? 0 . We consider ? 0 > 0 and ? 0 > 0 sufficiently large such that m i n ? = ? , ? − ? ? ? + ? 0 ℎ ? ≥ ? 0 , m i n ? = ? , ? − ? ? ? + ? 0 ? ? ≥ ? 0 . ( 2 0 ) By ( ? 2 ) , (11), (12), and the above inequalities, we deduce that ‖ ? ‖ > 0 and ‖ ? ‖ > 0 .
Now, by using Lemma 4 and the above considerations, we have ? ? = ? − 1 ? = 1 ? 2 ( ? , ? ) ? ? ? ? ? + ? 0 ℎ ? ≥ ? 2 ? − 1 ? = 1 ? 2 ( ? ) ? ? ? ? ? + ? 0 ℎ ? ≥ ? 2 ? − ? ? = ? ? 2 ( ? ) ? ? ? ? ? + ? 0 ℎ ? ≥ 2 ? ? 2 ? − ? ? = ? ? 2 ( ? ) ? ? ? ? + ? 0 ℎ ? ≥ 2 ? ? 2 ? − ? ? = ? ? 2 ( ? ) ? ? m i n ? = ? , ? − ? ? ? + ? 0 ℎ ? ≥ 2 ? ? 2 ? 1 ? − ? ? = ? ? 2 ( ? ) ? ? ‖ ‖ ? + ? 0 ℎ ‖ ‖ ‖ ‖ ≥ 2 ? + ? 0 ℎ ‖ ‖ ≥ 2 ‖ ? ‖ . ( 2 1 ) Therefore, we obtain ? ‖ ? ‖ ≤ ? 2 ≤ ‖ ? ‖ 2 . ( 2 2 ) In a similar manner, we deduce ? ? = ? − 1 ? = 1 ? 1 ( ? , ? ) ? ? ? ? ? + ? 0 ? ? ≥ ? 1 ? − 1 ? = 1 ? 1 ( ? ) ? ? ? ? ? + ? 0 ? ? ≥ ? 1 ? − ? ? = ? ? 1 ( ? ) ? ? ? ? ? + ? 0 ? ? ≥ 2 ? ? 1 ? − ? ? = ? ? 1 ( ? ) ? ? ? ? + ? 0 ? ? ≥ 2 ? ? 1 ? − ? ? = ? ? 1 ( ? ) ? ? m i n ? = ? , ? − ? ? ? + ? 0 ? ? ≥ 2 ? ? 1 ? 2 ? − ? ? = ? ? 1 ( ? ) ? ? ‖ ‖ ? + ? 0 ? ‖ ‖ ‖ ‖ ≥ 2 ? + ? 0 ? ‖ ‖ ‖ ‖ ? ‖ ‖ . ≥ 2 ( 2 3 ) So, we obtain ‖ ‖ ? ‖ ‖ ≤ ? ? 2 ≤ ‖ ? ‖ 2 . ( 2 4 ) By (22) and (24), we obtain ‖ ? ‖ ≤ ‖ ? ‖ / 2 ≤ ‖ ? ‖ / 4 , which is a contradiction, because ‖ ? ‖ > 0 . Then, for ? 0 and ? 0 sufficiently large, problem ( ? ) - ( ? ? ) has no positive solution.
Similar results as Theorems 6 and 7 can be obtained if instead of boundary conditions ( ? ? ) we have ? 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? + ? 0 , ? 0 = ? ? = 1 ? ? ? ? ? + ? 0 , ? ? = ? ? = 1 ? ? ? ? ? ? ? 1 or ? 0 = ? ? = 1 ? ? ? ? ? + ? 0 , ? ? = ? ? = 1 ? ? ? ? ? , ? 0 = ? ? = 1 ? ? ? ? ? + ? 0 , ? ? = ? ? = 1 ? ? ? ? ? ? ? 2 or ? 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? + ? 0 , ? 0 = ? ? = 1 ? ? ? ? ? , ? ? = ? ? = 1 ? ? ? ? ? + ? 0 , ? ? 3 where ? 0 and ? 0 are positive constants.
For problem ( ? ) – ? ? 1 , instead of sequences ( ℎ ? ) ? = 0 , ? and ( ? ? ) ? = 0 , ? from the proof of Theorem 6, the solutions of problems Δ 2 ∼ ℎ ? − 1 = 0 , ? = 1 , ? − 1 , ∼ ℎ 0 = ? ? = 1 ? ? ∼ ℎ ? ? , ∼ ℎ ? = ? ? = 1 ? ? ∼ ℎ ? ? + 1 , ( 2 5 ) Δ 2 ∼ ? ? − 1 = 0 , ? = 1 , ? − 1 , ∼ ? 0 = ? ? = 1 ? ? ∼ ? ? ? + 1 , ∼ ? ? = ? ? = 1 ? ? ∼ ? ? ? ( 2 6 ) are ∼ ℎ ? = 1 Δ 1 ? 1 − ? ? = 1 ? ? + ? ? = 1 ? ? ? ? , ? = 0 , ? , ∼ ? ? = 1 Δ 2 − ? 1 − ? ? = 1 ? ? + ? − ? ? = 1 ? ? ? ? , ? = 0 , ? , ( 2 7 ) respectively. By assumption ( ? 1 ) we obtain ∼ ℎ ? > 0 , for all ? = 1 , ? , and ∼ ? ? > 0 for all ? = 0 , ? − 1 .
For problem ( ? ) – ? ? 2 , instead of sequences ( ℎ ? ) ? = 0 , ? and ( ? ? ) ? = 0 , ? from Theorem 6, the solutions of problems (7) and (26) are ( ℎ ? ) ? = 0 , ? and ( ∼ ? ? ) ? = 0 , ? , respectively, which satisfy ℎ ? > 0 , for all ? = 0 , ? − 1 , and ∼ ? ? > 0 for all ? = 0 , ? − 1 . For problem ( ? ) – ? ? 3 , instead of sequences ( ℎ ? ) ? = 0 , ? and ( ? ? ) ? = 0 , ? from Theorem 6, the solutions of problems (25) and (8) are ( ∼ ℎ ? ) ? = 0 , ? and ( ? ? ) ? = 0 , ? , respectively, which satisfy ∼ ℎ ? > 0 , for all ? = 1 , ? , and ? ? > 0 for all ? = 1 , ? .
Therefore we also obtain the following results.
Theorem 8. Assume that assumptions ( ? 1 ) - - ( ? 3 ) hold. Then problem ( ? ) – ? ? 1 has at least one positive solution ( ? ? > 0 , for all ? = 1 , ? , and ? ? > 0 for all ? = 0 , ? − 1 ) for ? 0 > 0 and ? 0 > 0 sufficiently small.
Theorem 9. Assume that assumptions ( ? 1 ) , ( ? 2 ) , and ( ? 4 ) hold. Then problem ( ? ) – ? ? 1 has no positive solution ( ? ? > 0 , for all ? = 1 , ? , and ? ? > 0 for all ? = 0 , ? − 1 ) for ? 0 and ? 0 sufficiently large.
Theorem 10. Assume that assumptions ( ? 1 ) - - ( ? 3 ) hold. Then problem ( ? ) – ? ? 2 has at least one positive solution ( ? ? > 0 , for all ? = 0 , ? − 1 , and ? ? > 0 for all ? = 0 , ? − 1 ) for ? 0 > 0 and ? 0 > 0 sufficiently small.
Theorem 11. Assume that assumptions ( ? 1 ) , ( ? 2 ) , and ( ? 4 ) hold. Then problem ( ? ) – ? ? 2 has no positive solution ( ? ? > 0 , for all ? = 0 , ? − 1 , and ? ? > 0 for all ? = 0 , ? − 1 ) for ? 0 and ? 0 sufficiently large.
Theorem 12. Assume that assumptions ( ? 1 ) - - ( ? 3 ) hold. Then problem ( ? ) – ? ? 3 has at least one positive solution ( ? ? > 0 , for all ? = 1 , ? , and ? ? > 0 for all ? = 1 , ? ) for ? 0 > 0 and ? 0 > 0 sufficiently small.
Theorem 13. Assume that assumptions ( ? 1 ) , ( ? 2 ) , and ( ? 4 ) hold. Then problem ( ? ) – ? ? 3 has no positive solution ( ? ? > 0 , for all ? = 1 , ? , and ? ? > 0 for all ? = 1 , ? ) for ? 0 and ? 0 sufficiently large.
4. An Example
We consider ? = 2 0 , ? ? = ̂ ? / ( ? + 1 ) , ? ? = ? / ? for all ? = 1 , 1 9 , ̂ ? > 0 , ? > 0 , ? = 2 , ? = 3 , ? = 1 , ? = 2 , ? 1 = 1 / 2 , ? 2 = 1 / 3 , ? 1 = 4 , ? 2 = 1 6 , ? 1 = 1 / 3 , ? 2 = 1 / 4 , ? 3 = 1 / 5 , ? 1 = 5 , ? 2 = 1 0 , ? 3 = 1 5 , ? 1 = 3 / 4 , ? 1 = 1 0 , ? 1 = 1 / 3 , ? 2 = 1 / 5 , ? 1 = 3 , and ? 2 = 1 8 . We also consider the functions ? , ? ∶ [ 0 , ∞ ) → [ 0 , ∞ ) , ? ( ? ) = ∼ ? ? ? 1 / ( 2 ? + 1 ) , and ? ( ? ) = ∼ ? ? ? 2 / ( 3 ? + 2 ) , for all ? ∈ [ 0 , ∞ ) , with ∼ ? , ∼ ? > 0 and ? 1 , ? 2 > 2 . We have ? ? ? ? → ∞ ? ( ? ) / ? = ? ? ? ? → ∞ ? ( ? ) / ? = ∞ .
Therefore, we consider the system of second-order difference equations Δ 2 ? ? − 1 + ̂ ? ∼ ? ? ? 1 ? ( ? + 1 ) 2 ? ? + 1 = 0 , ? = Δ 1 , 1 9 , 2 ? ? − 1 + ? ∼ ? ? ? 2 ? ? 3 ? ? + 2 = 0 , ? = 1 , 1 9 , ? 0 with the multipoint boundary conditions ? 0 = 1 2 ? 4 + 1 3 ? 1 6 + ? 0 , ? 2 0 = 1 3 ? 5 + 1 4 ? 1 0 + 1 5 ? 1 5 , ? 0 = 3 4 ? 1 0 , ? 2 0 = 1 3 ? 3 + 1 5 ? 1 8 + ? 0 , ? ? 0 where ? 0 and ? 0 are positive constants.
We have ∑ 2 ? = 1 ? ? = 5 / 6 < 1 , ∑ 3 ? = 1 ? ? = 4 7 / 6 0 < 1 , ∑ 1 ? = 1 ? ? = 3 / 4 < 1 , and ∑ 2 ? = 1 ? ? = 8 / 1 5 < 1 . The functions ? 1 and ? 2 are given by ? 1 ? = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 2 5 6 5 ? − ? 6 7 1 2 2 0 , 1 ≤ ? ≤ 4 , 7 8 2 0 + 6 7 1 7 0 ? − ? 6 1 2 2 0 , 5 ≤ ? ≤ 9 , 1 2 6 2 0 + 6 7 1 2 9 0 ? − ? 6 7 1 2 2 0 , 1 0 ≤ ? ≤ 1 5 , 3 0 7 0 0 − 6 7 1 8 6 4 ? − ? 6 7 1 2 ? 2 0 , 1 6 ≤ ? ≤ 1 9 . 2 ? = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 1 4 5 ? − ? 6 3 2 2 0 , 1 ≤ ? ≤ 2 , 2 5 0 + 1 4 7 8 5 ? − ? 4 9 2 2 0 , 3 ≤ ? ≤ 9 , 2 5 6 0 + 1 4 7 8 ? − ? 4 9 2 2 0 , 1 0 ≤ ? ≤ 1 7 , 3 4 6 0 − 1 4 7 2 6 ? − ? 1 4 7 2 2 0 , 1 8 ≤ ? ≤ 1 9 . ( 2 8 )
Hence, we deduce that assumptions ( ? 1 ) , ( ? 2 ) , and ( ? 4 ) are satisfied. In addition, by using the above functions ? 1 and ? 2 , we obtain ∼ ? ∑ ∶ = 1 9 ? = 1 ? 1 ( ? ) / ( ? + 1 ) ≈ 3 0 . 1 7 8 4 0 0 2 , ∼ ? ∑ ∶ = 1 9 ? = 1 ? 2 ( ? ) / ? ≈ 2 3 . 6 3 8 3 1 2 5 4 , and then ? = ? ? ? { ̂ ? ∼ ? , ? ∼ ? } . We choose ? 0 = 1 and if we select ∼ ? , ∼ ? satisfying the conditions ∼ ? < 3 / ? = 3 ? ? ? 1 / ( ̂ ? ∼ ? ? ) , 1 / ( ∼ ? ) , ∼ ? < 5 / ? = 5 ? ? ? 1 / ( ̂ ? ∼ ? ? ) , 1 / ( ∼ ? ) , then we conclude that ? ( ? ) ≤ ∼ ? / 3 < 1 / ? , ? ( ? ) ≤ ∼ ? / 5 < 1 / ? for all ? ∈ [ 0 , 1 ] . For example, if ̂ ? = 1 , ? = 2 , then for ∼ ? ≤ 0 . 0 6 3 and ∼ ? ≤ 0 . 1 0 5 the above conditions for ? and ? are satisfied. So, assumption ( ? 3 ) is also satisfied. By Theorems 6 and 7 we deduce that problem ? 0 - ? ? 0 has at least one positive solution (here ? ? > 0 and ? ? > 0 for all ? = 0 , 2 0 ) for sufficiently small ? 0 > 0 and ? 0 > 0 and no positive solution for sufficiently large ? 0 and ? 0 .
By the proofs of Theorems 6 and 7 we can find some intervals for ? 0 and ? 0 such that problem ? 0 - ? ? 0 has at least one positive solution, or it has no positive solution. We consider ∼ ? = 0 . 0 6 3 , ∼ ? = 0 . 1 0 5 , ̂ ? = 1 , ? = 2 , ? 0 = 1 , ? = ? ? ? { ∼ ? , 2 ∼ ? } = 2 ∼ ? (as above), ? 1 = 3 , and ? 2 = 4 . Then Δ 1 = 6 7 1 / 1 8 0 , Δ 2 = 1 4 7 / 2 0 , and the sequences ( ℎ ? ) ? = 0 , 2 0 and ( ? ? ) ? = 0 , 2 0 from (9) are ℎ ? = ( − 3 9 ? + 2 3 1 0 ) / 6 7 1 and ? ? = ( 5 ? + 1 5 0 ) / 1 4 7 for all ? = 0 , 2 0 . We also obtain ℎ ? ? ? = ℎ 0 = 2 1 0 / 6 1 and ? ? ? ? = ? 2 0 = 2 5 0 / 1 4 7 . If we choose ? 0 ≤ ( ? − 1 ( 1 / ? ) − 1 ) / ? ? ? ? and ? 0 ≤ ( ? − 1 ( 1 / ? ) − 1 ) / ℎ ? ? ? , then inequalities (15) are satisfied. Because ? − 1 ( 1 / ? ) ≈ 1 . 0 0 3 1 and ? − 1 ( 1 / ? ) ≈ 1 . 0 0 2 1 3 , for ? 0 ≤ 6 . 1 7 ⋅ 1 0 − 4 and ? 0 ≤ 1 8 . 2 1 ⋅ 1 0 − 4 problem ? 0 - ? ? 0 has at least one positive solution.
Now we choose ? = 5 (the constant from the beginning of the proof of Theorem 7), and then we obtain ? 1 = ? 2 = 5 / 1 9 , ℎ 1 5 = 1 7 2 5 / 6 7 1 , ? 5 = 1 7 5 / 1 4 7 , ? 1 = ? ? ? { ? 1 , ℎ 1 5 / ℎ 0 } = 5 / 1 9 , ? 2 = ? ? ? { ? 2 , ? 5 / ? 2 0 } = 5 / 1 9 , ∑ 1 5 ? = 5 ? ? ? 2 ( ? ) ≈ 2 7 . 8 8 3 2 1 3 9 1 , ∑ 1 5 ? = 5 ? ? ? 1 ( ? ) ≈ 1 8 . 4 1 0 8 3 2 3 1 , and ? ≈ 0 . 7 8 4 3 2 0 8 7 (given by (19)). For ∼ ? ∶ = 2 ? + 0 . 1 , the inequalities ? ( ? ) / ? ≥ ∼ ? and ? ( ? ) / ? ≥ ∼ ? are satisfied for ? ≥ ? 0 ≈ 5 3 . 4 6 8 1 2 1 2 3 1 and ? ≥ ? 0 ≈ 7 . 2 1 6 6 2 6 9 , respectively. We consider ? 0 = ? ? ? { ? 0 , ? 0 } = ? 0 , and then for ? 0 ≥ ? 0 / ℎ 1 5 and ? 0 ≥ ? 0 / ? 5 , inequalities (20) are satisfied. Therefore, if ? 0 ≥ 2 0 . 7 9 8 4 and ? 0 ≥ 4 4 . 9 1 3 3 , problem ? 0 - ? ? 0 has no positive solution.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors thank the referees for their valuable comments and suggestions. The work of R. Luca was supported by a grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, Project no. PN-II-ID-PCE-2011-3-0557.
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