Boundary Value Problems for Second-Order Finite Difference Equations and Systems 9783111040370, 9783111039312

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Johnny Henderson, Rodica Luca Boundary Value Problems for Second-Order Finite Difference Equations and Systems

De Gruyter Studies in Mathematics

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Edited by Carsten Carstensen, Berlin, Germany Gavril Farkas, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Waco, Texas, USA Niels Jacob, Swansea, United Kingdom Zenghu Li, Beijing, China Guozhen Lu, Storrs, USA Karl-Hermann Neeb, Erlangen, Germany René L. Schilling, Dresden, Germany Volkmar Welker, Marburg, Germany

Volume 91

Johnny Henderson, Rodica Luca

Boundary Value Problems for Second-Order Finite Difference Equations and Systems

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Mathematics Subject Classification 2020 Primary: 39-02, 39A27; Secondary: 39A06, 39A12, 39A99 Authors Prof. Dr. Johnny Henderson Baylor University Department of Mathematics 97328 One Bear Place Waco TX 76798-7328 USA [email protected]

Prof. Dr. Rodica Luca “Gheorghe Asachi” Technical University of Iasi Department of Mathematics Blvd. Carol I 11 700506 Iasi Romania [email protected]

ISBN 978-3-11-103931-2 e-ISBN (PDF) 978-3-11-104037-0 e-ISBN (EPUB) 978-3-11-104045-5 ISSN 0179-0986 Library of Congress Control Number: 2022949628 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2023 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

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Johnny Henderson’s dedication is to his longtime friends and collaborators Rodica Luca Tudorache, John R. Graef, Mouffak Benchohra, Abdelghani Ouahab, Richard I. Avery, Douglas R. Anderson, Samira Hamani, Paul W. Eloe, and Sotiris K. Ntouyas Rodica Luca’s dedication is to her family

Preface The mathematical modeling of many problems from economics, computer science, control systems, mechanical engineering, biological neural networks, and others leads to the consideration of nonlinear difference equations. In the last decades, many authors have studied such problems by using various methods, such as fixed point theory, fixed point index theory, variational methods, critical point theory, different transformations, extensions of Perron’s second theorem, and diversified criteria for the stability of solutions (see, e. g., the monographs [2, 16, 38, 48, 51] and the papers [1, 6, 11, 13, 17, 23, 26–28, 60, 62, 63]). In this brief monograph we study the existence of positive solutions for some classes of second-order nonlinear finite difference equations and systems of second-order nonlinear finite difference equations, subject to various multi-point boundary conditions. In the case of systems, these boundary conditions may be uncoupled or coupled. We also investigate a class of nonlinear νth-order Atıcı–Eloe fractional difference equations supplemented with varied boundary conditions. The book draws together our results that have been obtained in the last years. Chapter 1 deals with the existence of positive solutions for two second-order finite difference equations which contain a linear term and a sign-changing nonlinearity, with or without parameters, subject to multi-point boundary conditions. Chapter 2 is focused on the existence and multiplicity of positive solutions for two systems of nonlinear second-order difference equations with uncoupled multi-point boundary conditions. The nonlinearities from the systems are nonnegative functions and satisfy some assumptions containing concave functions, or they are signchanging functions. In Chapter 3 we study the existence and nonexistence of positive solutions for two systems of nonlinear second-order difference equations supplemented with coupled multi-point boundary conditions, with positive parameters in the systems or in the boundary conditions. The nonlinearities of the systems are nonnegative functions and satisfy various assumptions. Chapter 4 is concerned with the existence and multiplicity of positive solutions for two systems of nonlinear second-order difference equations subject to coupled multi-point boundary conditions, without parameters. The nonlinearities of the systems are nonnegative functions and satisfy various assumptions. Chapter 5 is devoted to the existence of positive solutions for a system of nonlinear second-order difference equations with parameters and sign-changing nonlinearities, supplemented with multi-point coupled boundary conditions. Finally, Chapter 6 deals with the existence of nontrivial solutions, nonnegative solutions, and positive solutions for a class of nonlinear νth-order Atıcı–Eloe fractional difference equations with left focal boundary conditions or Dirichlet boundary conditions. In each chapter, various examples are presented which support the main results. The methods used in the proof of our theorems include results from fixed point theory and fixed point index theory. The book complements the existing literature on finite difference equations. This brief monograph can serve as a good resource for mathematical and scientific rehttps://doi.org/10.1515/9783111040370-201

VIII � Preface searchers and for the graduate students in mathematics and science interested in the existence of positive solutions for finite difference equations and systems. We would like to express our warm thanks to the De Gruyter Publisher, especially to J. Scott Bentley, Consulting Editor, Steven Elliot, Senior Editor Mathematics, Nadja Schedensack, Content Editor - Books, and Vilma Vaičeliūnienė, Project Manager at VTeX Book Production for their support and work on our book. Johnny Henderson Rodica Luca

Contents Preface � VII 1 1.1 1.1.1 1.1.2 1.1.3 1.2 1.2.1 1.2.2 2 2.1 2.1.1 2.1.2 2.2 2.2.1 2.2.2 3

3.1 3.1.1 3.1.2 3.1.3 3.1.4 3.2 3.2.1 3.2.2 4

4.1 4.1.1

Second-order finite difference equations with multi-point boundary conditions � 1 A second-order difference equation with a linear term and a sign-changing nonlinearity � 1 Preliminary results � 1 Existence of positive solutions � 13 An example � 19 A second-order difference equation with a linear term, a sign-changing nonlinearity, and a positive parameter � 21 Existence of positive solutions � 22 Examples � 28 Systems of second-order finite difference equations with uncoupled multi-point boundary conditions � 30 Nonnegative nonlinearities � 30 Auxiliary results � 32 Existence and multiplicity of positive solutions � 35 Sign-changing nonlinearities and positive parameters � 47 Main result � 48 An example � 52 Systems of second-order finite difference equations with nonnegative nonlinearities, coupled multi-point boundary conditions, and positive parameters � 54 Positive parameters in the system of difference equations � 54 Preliminary results � 54 Existence of positive solutions � 63 Nonexistence of positive solutions � 73 Examples � 78 Positive parameters in the boundary conditions � 80 Main results � 80 An example � 89 Systems of second-order finite difference equations with nonnegative nonlinearities, coupled multi-point boundary conditions, without parameters � 91 Some particular nonlinearities � 91 Main results � 92

X � Contents 4.1.2 4.2 4.2.1 4.2.2 5 5.1 5.1.1 5.1.2 6 6.1 6.1.1 6.1.2 6.2 6.2.1 6.2.2

An example � 100 General nonlinearities � 101 Auxiliary results � 101 Existence of positive solutions � 106 Systems of second-order finite difference equations with sign-changing nonlinearities and coupled multi-point boundary conditions � 118 Positive parameters in the system of difference equations � 118 Existence of positive solutions � 118 Examples � 138 Nonlinear fractional difference equations with nonlocal boundary conditions � 141 Nonlinear νth-order Atıcı–Eloe fractional difference equations with left focal boundary conditions � 141 Some preliminaries and the Green function � 142 Existence results � 144 Nonlinear νth-order Atıcı–Eloe fractional difference equations with Dirichlet boundary conditions � 147 Preliminary results � 147 Existence results � 148

Bibliography � 151 Index � 155

1 Second-order finite difference equations with multi-point boundary conditions In this chapter we investigate the existence of positive solutions for two second-order finite difference equations which contain a linear term and a sign-changing nonlinearity, with or without parameters, subject to multi-point boundary conditions.

1.1 A second-order difference equation with a linear term and a sign-changing nonlinearity In this section we consider the nonlinear second-order difference equation Δ2 un−1 − Lun + f (n, un ) = 0,

n = 1, N − 1,

(1.1)

with the multi-point boundary conditions p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi , i=1

(1.2)

where N ∈ ℕ, N > 2, p, q ∈ ℕ, Δ is the forward difference operator with step size 1, Δun = un+1 − un , Δ2 un−1 = un+1 − 2un + un−1 , n = k, m means that n = k, k + 1, . . . , m for k, m ∈ ℕ, ξi ∈ ℕ for all i = 1, p, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, L is a positive constant, and f is a sign-changing nonlinearity. Under some assumptions on the function f we will investigate the existence of at least one or two positive solutions for problem (1.1), (1.2). In the proof of the main results we use the Guo–Krasnosel’skii fixed point theorem. Equation (1.1) with L = 0, where the nonlinearity f may be unbounded below or nonpositive, subject to the boundary conditions u0 = u1 and uN = uN−1 , which is a resonant problem, has been investigated in the paper [12] by transforming it into a nonresonant problem.

1.1.1 Preliminary results We study in this section the second-order difference equation Δ2 un−1 − Lun + yn = 0,

n = 1, N − 1,

with the multi-point boundary conditions (1.2), where yn ∈ ℝ for all n = 1, N − 1. 2

(1.3)

We denote by A = L+2+ 2L +4L the biggest solution of the characteristic equation 2 r − (L + 2)r + 1 = 0 associated with equation (1.3). The other solution is 1/A. We also √

https://doi.org/10.1515/9783111040370-001

2 � 1 Second-order difference equations denote p

d0 = (∑ ai Aξi − 1)( i=1

p

+ (1 − ∑ ai i=1

q

1 1 − ∑b ) AN i=1 i Aηi q

1 )(AN − ∑ bi Aηi ). Aξi i=1

(1.4)

Lemma 1.1.1. If d0 ≠ 0, then the unique solution of problem (1.3), (1.2) is given by un = −

A n−1 n−j ∑ (A − Aj−n )yj A2 − 1 j=1 p

q

+

ξ −1

i An+1 1 1 [( ) a ( − (Aξi −j − Aj−ξi )yj ) b ∑ ∑ ∑ i i AN i=1 Aηi i=1 d0 (A2 − 1) j=1

p

+ (1 − ∑ ai i=1 p

− (1 − ∑ ai i=1

1 N−1 N−j ) ∑ (A − Aj−N )yj Aξi j=1 η −1

q

i 1 ) ∑ bi ( ∑ (Aηi −j − Aj−ηi )yj )] ξ A i i=1 j=1

p

+

N−1 1 ξi [( a A − 1) ∑ ∑ (AN−j − Aj−N )yj i An−1 d0 (A2 − 1) i=1 j=1 p

ηi −1

q

− (∑ ai Aξi − 1) ∑ bi ( ∑ (Aηi −j − Aj−ηi )yj ) i=1

i=1

j=1

q

p

ξi −1

i=1

i=1

j=1

− (AN − ∑ bi Aηi ) ∑ ai ( ∑ (Aξi −j − Aj−ξi )yj )],

n = 0, N.

(1.5)

0 We consider here that ∑−1 j=1 zj = 0 and ∑j=1 zj = 0.

Proof. The general solution for the homogeneous equation associated with (1.3), that is, the equation Δ2 un−1 − Lun = 0, n = 1, N − 1, is unh = C1 An + C2

1 , An

n = 0, N,

with C1 , C2 ∈ ℝ. We will determine a particular solution for the nonhomogeneous equation (1.3), by using the variation of constants method. Namely, we will look for a solution of (1.3) of the form ũn = Pn An + Qn

1 , An

n = 0, N,

1.1 A difference equation with a sign-changing nonlinearity

� 3

where Pn , Qn ∈ ℝ for all n = 0, N. For the sequences (Pn )n=0,N and (Qn )n=0,N we obtain the system 1 { (P − Pn−1 )An + (Qn − Qn−1 ) n = 0, { { n A { { {(P − P )(An − An−1 ) + (Q − Q )( 1 − 1 ) = −y n n−1 n n−1 An An−1 { n (we consider that y0 = 0 and yN = 0). By solving the above system in the unknowns Pn − Pn−1 and Qn − Qn−1 , we deduce Pn − Pn−1 = −

1 yn , An−1 (A2 − 1)

Qn − Qn−1 =

An+1 yn , A2 − 1

from which we conclude n

Pn = − ∑ j=1

1 yj , Aj−1 (A2 − 1)

n

Qn = ∑ j=1

Aj+1 yj , A2 − 1

n = 0, N.

Then we obtain for ũn the expression ũn = −

A n−1 n−j ∑ (A − Aj−n )yj , A2 − 1 j=1

n = 0, N.

Therefore, the general solution of equation (1.3) is un = unh + ũn = C1 An + C2

1 A n−1 n−j − 2 ∑ (A − Aj−n )yj , n A A − 1 j=1

n = 0, N.

(1.6)

Now we impose to sequence (un )n=0,N (given by (1.6)) the boundary conditions (1.2), p q namely, u0 = ∑i=1 ai uξi and uN = ∑i=1 bi uηi . We obtain for the constants C1 and C2 the system p

ξ −1

i { 1 A ξi { { C + C = a [C A + C − (Aξi −j − Aj−ξi )yj ], ∑ ∑ { 1 2 i 1 2 2−1 ξi { A A { { i=1 j=1 { { { { N−1 { { 1 A C1 AN + C2 N − 2 ∑ (AN−j − Aj−N )yj { { A A − 1 j=1 { { { { { ηi −1 { q { A 1 { ηi { { ∑ (Aηi −j − Aj−ηi )yj ], { = ∑ bi [C1 A + C2 Aηi − 2 A − 1 j=1 i=1 {

4 � 1 Second-order difference equations or equivalently p

p

1 { { { C1 (∑ ai Aξi − 1) − C2 (1 − ∑ ai ξ ) { { i A { { i=1 i=1 { { { { ξi −1 p { { A { { a ( = ∑ ∑ (Aξi −j − Aj−ξi )yj ), { i 2−1 { A { { i=1 j=1 { { { { q q { { 1 1 N ηi − bi η ) C (A − b A ) + C ( ∑ ∑ i 2 { N i { 1 A A { i=1 i=1 { { { { { { A N−1 N−j { { ∑ (A − Aj−N )yj { = 2 { { A − 1 { j=1 { { { { { ηi −1 q { { A ηi −j { { − Aj−ηi )yj ). − b ( ∑ { i ∑ (A A2 − 1 i=1 j=1 { The determinant of the above system in the unknowns C1 and C2 is d0 , which by assumption of this lemma is different from zero. So, the above system has a unique solution, given by C1 =

q

p

ξ −1

i A 1 1 {( − b ) a ( (Aξi −j − Aj−ξi )yj ) ∑ ∑ ∑ i i η AN i=1 A i i=1 d0 (A2 − 1) j=1

p

+ (1 − ∑ ai i=1

q

ηi −1

i=1

j=1

N−1 1 )[ ∑ (AN−j − Aj−N )yj ξ Ai j=1

− ∑ bi ( ∑ (Aηi −j − Aj−ηi )yj )]}, C2 =

p

A {(∑ ai Aξi − 1) d0 (A2 − 1) i=1 N−1

q

ηi −1

i=1

j=1

× [ ∑ (AN−j − Aj−N )yj − ∑ bi ( ∑ (Aηi −j − Aj−ηi )yj )] j=1

q

p

ξi −1

i=1

i=1

j=1

− (AN − ∑ bi Aηi ) ∑ ai ( ∑ (Aξi −j − Aj−ξi )yj )}.

(1.7)

By replacing C1 and C2 in (1.6) with the expressions from (1.7) we obtain the solution of problem (1.3), (1.2) given by (1.5). To express the solution of problem (1.3), (1.2) by using the associated Green function, we will firstly investigate the solution of problem (1.3), (1.2) with ai = 0 for all i = 1, p and bj = 0 for all j = 1, q, and discover the corresponding Green function.

1.1 A difference equation with a sign-changing nonlinearity

� 5

Lemma 1.1.2. The unique solution of equation (1.3) with the boundary conditions u0 = 0 and uN = 0 is un0 = ∑N−1 i=1 g(n, i)yi , n = 0, N, where the Green function g is given by g(n, i) =

(A2 ×{

A − 1)(AN − A−N )

(Ai − A−i )(AN−n − An−N ), (An − A−n )(AN−i − Ai−N ),

1 ≤ i < n ≤ N, 0 ≤ n ≤ i ≤ N − 1.

(1.8)

Proof. By using Lemma 1.1.1 and (1.5) we deduce that the solution of equation (1.3) with the boundary conditions u0 = 0 and uN = 0 is un0 = − + −

A n−1 n−i ∑ (A − Ai−n )yi A2 − 1 i=1

N−1 AN+n+1 ∑ (AN−i − Ai−N )yi (A2 − 1)(A2N − 1) i=1

N−1 AN−n+1 ∑ (AN−i − Ai−N )yi , (A2 − 1)(A2N − 1) i=1

n = 0, N.

Therefore, by (1.9) we obtain un0 = −

A n−1 n−i ∑ (A − Ai−n )yi A2 − 1 i=1

+( n−1

N−1 AN+n+1 AN−n+1 − ) ∑ (AN−i − Ai−N )yi (A2 − 1)(A2N − 1) (A2 − 1)(A2N − 1) i=1

= ∑[ i=1

AN+n+1 − AN−n+1 N−i A (A − Ai−N ) − 2 (An−i − Ai−n )]yi 2 2N (A − 1)(A − 1) A −1

N−1

+∑

i=n

n−1

=∑ i=1

+ =

AN+1 (An − A−n ) N−i (A − Ai−N )yi (A2 − 1)(A2N − 1)

AN+1 (An − A−n )(AN−i − Ai−N ) − A(A2N − 1)(An−i − Ai−n ) yi (A2 − 1)(A2N − 1)

N−1 AN+1 ∑ (An − A−n )(AN−i − Ai−N )yi (A2 − 1)(A2N − 1) i=n

n−1 AN+1 ∑ [(An − A−n )(AN−i − Ai−N ) (A2 − 1)(A2N − 1) i=1

− (AN − A−N )(An−i − Ai−n )]yi +

(A2

N−1 AN+1 ∑ (An − A−n )(AN−i − Ai−N )yi 2N − 1)(A − 1) i=n

(1.9)

6 � 1 Second-order difference equations

=

n−1 AN+1 ∑ (Ai − A−i )(AN−n − An−N )yi (A2 − 1)(A2N − 1) i=1

+

N−1 AN+1 ∑ (An − A−n )(AN−i − Ai−N )yi , (A2 − 1)(A2N − 1) i=n

n = 0, N.

Then we deduce un0 = ∑N−1 i=1 g(n, i)yi for all n = 0, N, where g is given by (1.8). Lemma 1.1.3. If d0 ≠ 0, then the solution of problem (1.3), (1.2) given by (1.5) can be expressed as un = ∑N−1 j=1 G(n, j)yj , n = 0, N, where the Green function G is given by G(n, j) = g(n, j) q

q

p

+

1 1 1 1 [An (∑ bi η − N ) + n (AN − ∑ bi Aηi )] ∑ ai g(ξi , j) i d0 A A A i=1 i=1 i=1

+

1 1 1 [An (1 − ∑ ai ξ ) + n (∑ ai Aξi − 1)] ∑ bi g(ηi , j), i d0 A i=1 A i=1 i=1

p

p

q

for all n = 0, N and j = 1, N − 1, with g given by (1.8). Proof. By (1.5) and using the function g given by (1.8), we obtain un = −

1 A n−1 (An−j − Aj−n )(AN − A−N ) yj + C1 An + C2 n ∑ N −N 2 A A −A A − 1 j=1

=

n−1 A ∑ [−(An − A−n )(AN−j − Aj−N ) N −N − 1)(A − A ) j=1

(A2

+ (Aj − A−j )(AN−n − An−N )]yj + C1 An + C2 =

1 An

n−1 A [ ∑ (Aj − A−j )(AN−n − An−N )yj (A2 − 1)(AN − A−N ) j=1 N−1

+ ∑ (An − A−n )(AN−j − Aj−N )yj j=n

N−1

− ∑ (An − A−n )(AN−j − Aj−N )yj ] + C1 An + C2 j=1

n−1

= ∑ g(n, j)yj − j=1

1 An

A (A2 − 1)(AN − A−N )

N−1

× ∑ (An − A−n )(AN−j − Aj−N )yj + C1 An + C2 j=1

where C1 and C2 are defined by (1.7). Then we deduce

1 , An

∀ n = 0, N,

(1.10)

1.1 A difference equation with a sign-changing nonlinearity n−1

un = ∑ g(n, j)yj + j=1

(A2

A − 1)(AN − A−N )d0

p

× {[(∑ ai Aξi − 1)( i=1 p

+ (1 − ∑ ai i=1

q

1 1 ) − ∑b AN i=1 i Aηi q

1 )(AN − ∑ bi Aηi )] Aξi i=1

N−1

× [− ∑ (An − A−n )(AN−j − Aj−N )yj ] j=1

+ An (

q

1 1 − ∑ bi η )(AN − A−N ) N i A A i=1

p

ξi −1

i=1

j=1

× [∑ ai ( ∑ (Aξi −j − Aj−ξi )yj )] p

+ An (1 − ∑ ai i=1 p

− An (1 − ∑ ai i=1

N−1 1 N −N )(A − A ) ∑ (AN−j − Aj−N )yj Aξi j=1 q

η −1

i 1 N −N )(A − A )[ b ( (Aηi −j − Aj−ηi )yj )] ∑ ∑ i Aξi i=1 j=1

p

+

N−1 1 (∑ ai Aξi − 1)(AN − A−N ) ∑ (AN−j − Aj−N )yj n A i=1 j=1

−

i 1 ξi N −N ( a A − 1)(A − A )[ b ( (Aηi −j − Aj−ηi )yj )] ∑ ∑ ∑ i i An i=1 i=1 j=1

−

1 (AN − ∑ bi Aηi )(AN − A−N ) An i=1

p

q

q

p

ξi −1

i=1

j=1

× [∑ ai ( ∑ (Aξi −j − Aj−ξi )yj )]} n−1

= ∑ g(n, j)yj + j=1

A (A2 − 1)(AN − A−N )d0

p

× {−(∑ ai Aξi − 1)( i=1

N−1

q

1 1 − ∑b ) AN i=1 i Aηi

× (An − A−n ) ∑ (AN−j − Aj−N )yj j=1

p

− (1 − ∑ ai i=1

q

1 )(AN − ∑ bi Aηi ) Aξi i=1

η −1

� 7

8 � 1 Second-order difference equations N−1

× (An − A−n ) ∑ (AN−j − Aj−N )yj j=1

p

+ An (1 − ∑ ai i=1

N−1 1 N −N )(A − A ) ∑ (AN−j − Aj−N )yj Aξi j=1

p

+

N−1 1 ξi N −N ( a A − 1)(A − A ) ∑ ∑ (AN−j − Aj−N )yj An i=1 i j=1

+ [An (

q

q

1 1 1 − ∑ bi η ) − n (AN − ∑ bi Aηi )] N i A A A i=1 i=1

p

ξi −1

i=1

j=1

× [∑ ai ( ∑ (AN − A−N )(Aξi −j − Aj−ξi )yj )] p

+ [−An (1 − ∑ ai i=1

q

ηi −1

i=1

j=1

p

1 1 ) − n (∑ ai Aξi − 1)] ξ i A i=1 A

× [∑ bi ( ∑ (AN − A−N )(Aηi −j − Aj−ηi )yj )]} n−1

= ∑ g(n, j)yj + j=1

A (A2 − 1)(AN − A−N )d0

p

× {−(∑ ai Aξi − 1)( i=1

N−1

q

1 1 − ∑b ) AN i=1 i Aηi

× (An − A−n ) ∑ (AN−j − Aj−N )yj j=1

p

− (1 − ∑ ai i=1

q

1 )(AN − ∑ bi Aηi ) Aξi i=1 N−1

× (An − A−n ) ∑ (AN−j − Aj−N )yj j=1

p

+ An (1 − ∑ ai i=1

N−1 1 N −N )(A − A ) ∑ (AN−j − Aj−N )yj Aξi j=1

p

+

N−1 1 (∑ ai Aξi − 1)(AN − A−N ) ∑ (AN−j − Aj−N )yj n A i=1 j=1

+ [An (

q

q

1 1 1 − ∑b ) − n (AN − ∑ bi Aηi )] A AN i=1 i Aηi i=1

p

ξi −1

i=1

j=1

× [∑ ai [ ∑ [(Aξi − A−ξi )(AN−j − Aj−N )

1.1 A difference equation with a sign-changing nonlinearity

− (Aj − A−j )(AN−ξi − Aξi −N )]]yj ] p

+ [−An (1 − ∑ ai i=1

q

ηi −1

i=1

j=1

p

1 1 ) − n (∑ ai Aξi − 1)] A i=1 Aξi

× [∑ bi [ ∑ [(Aηi − A−ηi )(AN−j − Aj−N ) − (Aj − A−j )(AN−ηi − Aηi −N )]]yj ]}, for all n = 0, N. Therefore, we obtain n−1

un = ∑ g(n, j)yj + j=1

(A2

p

A − 1)(AN − A−N )d0

× {−(∑ ai Aξi − 1)( i=1

N−1

q

1 1 − ∑b ) AN i=1 i Aηi

× (An − A−n ) ∑ (AN−j − Aj−N )yj j=1

p

− (1 − ∑ ai i=1

q

1 )(AN − ∑ bi Aηi ) Aξi i=1 N−1

× (An − A−n ) ∑ (AN−j − Aj−N )yj j=1

p

+ An (1 − ∑ ai i=1

N−1 1 N −N )(A − A ) ∑ (AN−j − Aj−N )yj Aξi j=1

p

+

N−1 1 ξi N −N ( a A − 1)(A − A ) ∑ ∑ (AN−j − Aj−N )yj i An i=1 j=1

+ [An (

q

q

1 1 1 − ∑b ) − n (AN − ∑ bi Aηi )] A AN i=1 i Aηi i=1

p

ξi −1

i=1

j=1

× {∑ ai [ ∑ (−(Aj − A−j )(AN−ξi − Aξi −N ))yj N−1

− ∑ (Aξi − A−ξi )(AN−j − Aj−N )yj ] j=ξi p

N−1

i=1

j=1

+ ∑ ai ( ∑ (Aξi − A−ξi )(AN−j − Aj−N )yj )}

� 9

10 � 1 Second-order difference equations p

+ [−An (1 − ∑ ai i=1

q

ηi −1

i=1

j=1

p

1 1 ) − n (∑ ai Aξi − 1)] ξ A i=1 Ai

× {∑ bi [ ∑ (−(Aj − A−j )(AN−ηi − Aηi −N ))yj N−1

− ∑ (Aηi − A−ηi )(AN−j − Aj−N )yj ] j=ηi q

N−1

i=1

j=1

+ ∑ bi ( ∑ (Aηi − A−ηi )(AN−j − Aj−N )yj )}} N−1

= ∑ g(n, j)yj + j=1

N−1

p

j=1

i=1

q

q

1 1 1 1 [−An ( N − ∑ bi η ) + n (AN − ∑ bi Aηi )] i d0 A A A i=1 i=1

× [ ∑ (∑ ai g(ξi , j))yj ] p

+

p

1 1 1 [An (1 − ∑ ai ξ ) + n (∑ ai Aξi − 1)] i d0 A i=1 A i=1 N−1

q

j=1

i=1

× [ ∑ (∑ bi g(ηi , j))yj ] +

A (A2 − 1)(AN − A−N )d0 p

× {−(∑ ai Aξi − 1)( i=1

q

1 1 − ∑b ) AN i=1 i Aηi

N−1

× (An − A−n )( ∑ (AN−j − Aj−N )yj ) j=1

p

− (1 − ∑ ai i=1

q

1 )(AN − ∑ bi Aηi ) Aξi i=1 N−1

× (An − A−n )( ∑ (AN−j − Aj−N )yj ) j=1

p

+ An (1 − ∑ ai i=1

N−1 1 )(AN − A−N )( ∑ (AN−j − Aj−N )yj ) ξ Ai j=1

p

+

N−1 1 ξi N −N ( a A − 1)(A − A )( ∑ ∑ (AN−j − Aj−N )yj ) An i=1 i j=1

+ An (

q

p

1 1 − ∑b )(∑ ai (Aξi − A−ξi )) AN i=1 i Aηi i=1

1.1 A difference equation with a sign-changing nonlinearity

� 11

N−1

× ( ∑ (AN−j − Aj−N )yj ) j=1

q

−

p

1 (AN − ∑ bi Aηi )(∑ ai (Aξi − A−ξi )) An i=1 i=1 N−1

× ( ∑ (AN−j − Aj−N )yj ) j=1

p

− An (1 − ∑ ai i=1

q

1 )(∑ bi (Aηi − A−ηi )) Aξi i=1

N−1

× ( ∑ (AN−j − Aj−N )yj ) j=1

p

−

q

1 (∑ a Aξi − 1)(∑ bi (Aηi − A−ηi )) An i=1 i i=1 N−1

× ( ∑ (AN−j − Aj−N )yj )} j=1

N−1

= ∑ {g(n, j) + j=1

p

q

× (∑ ai g(ξi , j)) + i=1

+

p

q

1 1 1 1 [An (∑ bi η − N ) + n (AN − ∑ bi Aηi )] i d0 A A A i=1 i=1 p

1 1 [An (1 − ∑ ai ξ ) i d0 A i=1 q

1 (∑ a Aξi − 1)](∑ bi g(ηi , j))}yj An i=1 i i=1

N−1

= ∑ G(n, j)yj , j=1

∀ n = 0, N,

where the Green function G is given by (1.10). Lemma 1.1.4. The Green function g given by (1.8) satisfies the inequalities k(n)h(j) ≤ g(n, j) ≤ h(j), where h(j) = g(j, j) = k(n) =

A (Aj (A2 −1)(AN −A−N )

∀ n = 0, N, j = 1, N − 1,

− A−j )(AN−j − Aj−N ) for all j = 1, N − 1 and

1 min{An − A−n , AN−n − An−N }, AN−1 − A1−N

∀ n = 0, N.

Proof. Because the function φ(x) = Ax − A−x , x ≥ 0, is strictly increasing, with φ(0) = 0, we deduce that g(n, j) ≤ g(j, j) = h(j) for all n = 0, N, j = 1, N − 1.

12 � 1 Second-order difference equations For 1 ≤ j < n ≤ N we have g(n, j) AN−n − An−N AN−n − An−N ≥ , = N−j g(j, j) AN−1 − A1−N A − Aj−N and for 0 ≤ n ≤ j ≤ N − 1 we have g(n, j) An − A−n An − A−n = j ≥ . N−1 g(j, j) A − A1−N A − A−j Then we obtain g(n, j) ≥ k(n)h(j) for all n = 0, N, j = 1, N − 1, where k(n) = min{An − A−n , AN−n − An−N } for all n = 0, N.

1 AN−1 −A1−N

×

Remark 1.1.1. Because N > 2 we have k(n) < 1 for all n = 0, N, and then minn=1,N−1 k(n) ∈ (0, 1). p

Lemma 1.1.5. We assume that ai ≥ 0 for all i = 1, p, bj ≥ 0 for all j = 1, q, ∑i=1 ai Aξi ≥ 1, q q p ∑i=1 ai A1ξi ≤ 1, ∑i=1 bi A1ηi ≥ A1N , ∑i=1 bi Aηi ≤ AN , and d0 > 0. Then the Green function G given by (1.10) satisfies the inequalities k(n)h(j) ≤ G(n, j) ≤ Λh(j),

∀ n = 0, N, j = 1, N − 1,

where Λ=1+

q

q

p

+

p

1 1 1 [AN (∑ bi η − N ) + AN − ∑ bi Aηi ](∑ ai ) i d0 A A i=1 i=1 i=1 p

q

1 1 [AN (1 − ∑ ai ξ ) + ∑ ai Aξi − 1](∑ bi ). i d0 i=1 A i=1 i=1

(1.11)

Proof. Under the assumptions of this lemma, by using Lemma 1.1.4, we have G(n, j) ≥ g(n, j) ≥ k(n)h(j) for all n = 0, N, j = 1, N − 1. By using again Lemma 1.1.4, we obtain G(n, j) ≤ h(j) +

q

p

+

q

p

1 1 1 [AN (∑ bi η − N ) + AN − ∑ bi Aηi ](∑ ai h(j)) i d0 A A i=1 i=1 i=1 p

q

1 1 [AN (1 − ∑ ai ξ ) + ∑ ai Aξi − 1](∑ bi h(j)) i d0 A i=1 i=1 i=1

= Λh(j),

∀ n = 0, N, j = 1, N − 1.

Lemma 1.1.6. Under the assumptions of Lemma 1.1.5, the solution un , n = 0, N, of problem (1.3), (1.2) satisfies the inequality un ≥ Λ1 k(n)um for all n, m = 0, N.

1.1 A difference equation with a sign-changing nonlinearity

� 13

Proof. By using Lemma 1.1.3 and Lemma 1.1.5, we obtain N−1

N−1

N−1

j=1

j=1

un = ∑ G(n, j)yj ≥ ∑ k(n)h(j)yj = k(n) ∑ h(j)yj j=1

N−1

≥

k(n) k(n) u , ∑ G(m, j)yj = Λ j=1 Λ m

∀ n, m = 0, N.

In the proof of our main results we will use the Guo–Krasnosel’skii fixed point theorem presented below (see [29]). Theorem 1.1.1. Let X be a Banach space and let C ⊂ X be a cone in X. Assume Ω1 and Ω2 are bounded open subsets of X with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 and let 𝒜 : C ∩ (Ω2 \ Ω1 ) → C be a completely continuous operator (continuous and compact, that is, it maps bounded sets into relatively compact sets) such that either (i) ‖𝒜u‖ ≤ ‖u‖, u ∈ C ∩ 𝜕Ω1 , and ‖𝒜u‖ ≥ ‖u‖, u ∈ C ∩ 𝜕Ω2 , or (ii) ‖𝒜u‖ ≥ ‖u‖, u ∈ C ∩ 𝜕Ω1 , and ‖𝒜u‖ ≤ ‖u‖, u ∈ C ∩ 𝜕Ω2 . Then 𝒜 has a fixed point in C ∩ (Ω2 \ Ω1 ). 1.1.2 Existence of positive solutions In this section, we will investigate the existence of at least one or two positive solutions for problem (1.1), (1.2). We present now the basic assumptions that we will use in the sequel. p p (H1) We have ai ≥ 0 for all i = 1, p, bj ≥ 0 for all j = 1, q, ∑i=1 ai Aξi ≥ 1, ∑i=1 ai A1ξi ≤ 1, q

q

∑i=1 bi A1ηi ≥ A1N , ∑i=1 bi Aηi ≤ AN , d0 > 0, and L > 0. (H2) The function f : {1, . . . , N − 1} × ℝ+ → ℝ is continuous, and there exist cn ≥ 0, n = 1, N − 1, with ∑N−1 i=1 ci > 0 such that f (n, u) ≥ −cn for all n = 1, N − 1, u ∈ ℝ+ (ℝ+ = [0, ∞)).

We remark that under assumption (H2), the nonlinearity in equation (1.1), namely, −Lu+ f (n, u), may be unbounded below. We denote by (rn )n=0,N the solution of problem (1.3), (1.2) with yn = cn for all n = 1, N − 1, namely, the solution of problem Δ2 un−1 − Lun + cn = 0, n = 1, N − 1, { { { p q { { {u0 = ∑ ai uξi , uN = ∑ bi uηi , i=1 i=1 { where cn , n = 0, N, are given in (H2). So, by using the Green function G and Lemma 1.1.3, we have rn = ∑N−1 j=1 G(n, j)cj for all n = 0, N.

14 � 1 Second-order difference equations We consider now the difference equation Δ2 vn−1 − Lvn + f (n, (vn − rn )∗ ) + cn = 0,

n = 1, N − 1,

(1.12)

with the multi-point boundary conditions p

v0 = ∑ ai vξi , i=1

q

vN = ∑ bi vηi ,

(1.13)

i=1

where z∗ = z if z ≥ 0 and z∗ = 0 if z < 0. We obtain easily the following lemma. Lemma 1.1.7. The sequence (un )n=0,N is a positive solution of problem (1.1), (1.2) (un > 0 for all n = 0, N) if and only if (vn )n=0,N , vn = un + rn , n = 0, N, is a solution of the boundary value problem (1.12), (1.13) with vn > rn for all n = 0, N. By using Lemma 1.1.3, we also obtain the following result. Lemma 1.1.8. The sequence (vn )n=0,N is a solution of problem (1.12), (1.13) if and only if (vn )n=0,N is a solution of the problem N−1

vn = ∑ G(n, j)(f (j, (vj − rj )∗ ) + cj ), j=1

n = 0, N.

(1.14)

We consider the Banach space X = ℝN+1 = {v = (vn )n=0,N , vn ∈ ℝ, n = 0, N}, endowed with the maximum norm ‖v‖ = maxn=0,N |vn |, and we define the operator Q : X → X, Q(v) = (Qn (v))n=0,N , where N−1

Qn (v) = ∑ G(n, j)(f (j, (vj − rj )∗ ) + cj ), j=1

n = 0, N,

v = (vn )n=0,N .

By (H2), the operator Q is completely continuous. We also define the cone P = {v ∈ X, v = (vn )n=0,N , vn ≥

k(n) ‖v‖, ∀ n = 0, N}. Λ

By using Lemma 1.1.6, we deduce that Q(P) ⊂ P. In addition, we have the following lemma. Lemma 1.1.9. The sequence (vn )n=0,N is a solution of problem (1.14) if and only if (vn )n=0,N is a fixed point of operator Q. So, the existence of positive solutions of problem (1.1), (1.2) is reduced in three steps (Lemmas 1.1.7–1.1.9) to the fixed point problem of operator Q in the cone P.

� 15

1.1 A difference equation with a sign-changing nonlinearity

Let k0 = min{k(n), n = 1, N − 1}. By Remark 1.1.1 we have k0 ∈ (0, 1). We define the functions Φ(r) = max{f (n, u) + cn , n = 1, N − 1, u ∈ [ Ψ(r) = min{f (n, u) + cn , n = 1, N − 1, u ∈ [ for

r>

N−1 k0 r − Λ ∑ h(j)cj , r]}, Λ j=1

N−1 k0 r − Λ ∑ h(j)cj , r]}, Λ j=1

Λ2 N−1 ∑ h(j)cj . k0 j=1

Theorem 1.1.2. We assume that (H1) and (H2) hold. Additionally, we make the following assumption. 2 (H3) There exist r, R > 0 such that Λk ∑N−1 j=1 h(j)cj < r < R and 0

Φ(r) ≤

r

maxn=0,N ∑N−1 j=1 G(n, j)

and Ψ(R) >

R

maxn=1,N−1 ∑N−1 j=1 G(n, j)

.

(1.15)

Then problem (1.1), (1.2) has at least one positive solution. Now we also assume the following. (H4) We have f (n, u) ≤ 0 for all n = 1, N − 1 and u > 0 sufficiently large. Then problem (1.1), (1.2) has at least two positive solutions. Proof. We assume that (H1)–(H3) hold. We define the sets Ω1 = {v ∈ X, ‖v‖ < r} and Ω2 = {v ∈ X, ‖v‖ < R}, where r and R are given in assumption (H3). For v ∈ P ∩ 𝜕Ω1 , by Lemma 1.1.5, we obtain r ≥ vj − rj ≥ ≥

N−1 k(j) k(j) ‖v‖ − rj ≥ ‖v‖ − Λ ∑ h(i)ci Λ Λ i=1

N−1 k0 r − Λ ∑ h(i)ci > 0, Λ i=1

∀ j = 1, N − 1,

and then f (j, vj − rj ) + cj ≤ Φ(r),

∀ j = 1, N − 1.

Therefore, by Lemma 1.1.5 we deduce that N−1

Qn (v) ≤ ∑ G(n, j)Φ(r), j=1

and then

∀ n = 0, N,

16 � 1 Second-order difference equations 󵄩 󵄨 󵄩󵄩 󵄨 󵄩󵄩Q(v)󵄩󵄩󵄩 = max 󵄨󵄨󵄨Qn (v)󵄨󵄨󵄨 ≤ r = ‖v‖, n=0,N

∀ v ∈ P ∩ 𝜕Ω1 .

(1.16)

For v ∈ P ∩ 𝜕Ω2 , by Lemma 1.1.5, for j = 1, N − 1, we obtain R ≥ vj − rj ≥

N−1 N−1 k k(j) ‖v‖ − Λ ∑ h(i)ci ≥ 0 R − Λ ∑ h(i)ci , Λ Λ i=1 i=1

and then f (j, vj − rj ) + cj ≥ Ψ(R),

∀ j = 1, N − 1.

Then by (H3) we have for all v ∈ P ∩ 𝜕Ω2 and n = 1, N − 1, N−1

Qn (v) ≥ ∑ G(n, j)Ψ(R), j=1

so 󵄨 󵄨 󵄩󵄩 󵄩 󵄩󵄩Q(v)󵄩󵄩󵄩 = max 󵄨󵄨󵄨Qn (v)󵄨󵄨󵄨 n=0,N

󵄨 󵄨 ≥ max 󵄨󵄨󵄨Qn (v)󵄨󵄨󵄨 > R = ‖v‖, n=1,N−1

∀ v ∈ P ∩ 𝜕Ω2 .

(1.17)

Therefore, by (1.16), (1.17), and Theorem 1.1.1, we deduce that operator Q has a fixed point v1 = (v1n )n=0,N ∈ P satisfying r ≤ ‖v1 ‖ < R. By Lemma 1.1.5 we obtain un1 = v1n − rn ≥ ≥

N−1 k(n) 󵄩󵄩 1 󵄩󵄩 󵄩󵄩v 󵄩󵄩 − Λ ∑ h(j)cj Λ j=1

N−1 k0 r − Λ ∑ h(j)cj > 0, Λ j=1 p

u01 = v10 − r0 = ∑ ai uξ1i > 0, i=1

∀ n = 1, N − 1, q

uN1 = v1N − rN = ∑ bi uη1 i > 0. i=1

Hence, by using Lemmas 1.1.7–1.1.9, we conclude that u1 = (un1 )n=0,N is a positive solution of problem (1.1), (1.2). Now we assume in addition that (H4) holds. We prove that problem (1.1), (1.2) has a distinct second positive solution u2 . By (H4) we deduce that there exists M > 0 such that f (n, u) + cn ≤ cn for all n = 1, N − 1 and u > M. We choose R1 > max{R,

N−1 Λ (M + Λ ∑ h(j)cj )}, k0 j=1

N−1

R1 ≥ cΛ ∑ h(j), j=1

1.1 A difference equation with a sign-changing nonlinearity

� 17

where c = max{cn , n = 1, N − 1} > 0. Let Ω3 = {v ∈ X, ‖v‖ < R1 }. For v ∈ P ∩ 𝜕Ω3 , we obtain vn − rn ≥

N−1 k0 R1 − Λ ∑ h(j)cj > M, Λ j=1

∀ n = 1, N − 1,

which implies that f (n, (vn − rn )∗ ) + cn ≤ cn ≤ c for all n = 1, N − 1. Then for all v ∈ P ∩ 𝜕Ω3 we conclude N−1

N−1

j=1

j=1

Qn (v) ≤ Λ ∑ h(j)(f (j, (vj − rj )∗ ) + cj ) ≤ cΛ ∑ h(j) ≤ R1 , so 󵄩󵄩 󵄩 󵄩󵄩Q(v)󵄩󵄩󵄩 ≤ ‖v‖,

∀ v ∈ P ∩ 𝜕Ω3 .

(1.18)

Therefore, by (1.17), (1.18), and Theorem 1.1.1, the operator Q has a fixed point v2 = ∈ P such that R < ‖v2 ‖ ≤ R1 . By similar arguments used for v1 , we have v2n > rn for all n = 0, N, so by Lemmas 1.1.7–1.1.9, we obtain u2 = v2 − ̃r , where ̃r = (rn )n=0,N is a second positive solution of problem (1.1), (1.2). (v2n )n=0,N

Theorem 1.1.3. We assume that (H1), (H2), and (H3) hold. If limu→∞ f (n,u) = 0 uniformly u for n = 1, N − 1, then problem (1.1), (1.2) has at least two positive solutions. Proof. We consider the sets Ω1 and Ω2 defined at the beginning of the proof of Theorem 1.1.2. By the proof of Theorem 1.1.2, we know that operator Q has a fixed point v1 = (v1n )n=0,N ∈ P such that r ≤ ‖v1 ‖ < R with v1n > rn for all n = 0, N. Then u1 = (un1 )n=0,N , un1 = v1n − rn , n = 0, N, is a positive solution of problem (1.1), (1.2). We will prove that Q has a second fixed point ṽ ∈ P. = 0 uniformly for n = 1, N − 1, we deduce that From the condition limu→∞ f (n,u) u f (n,u)+c

−1 n limu→∞ = 0 uniformly with respect to n = 1, N −1. Then for ε ∈ (0, (Λ ∑N−1 j=1 h(j)) ], u there exists M1 > 0 such that f (n, u) + cn ≤ εu for all u > M1 and n = 1, N − 1. We choose R2 > max{R, kΛ (M1 + Λ ∑N−1 j=1 h(j)cj )} and Ω3 = {v ∈ X, ‖v‖ < R2 }. For v ∈ P ∩ 𝜕Ω3 we 0 obtain

vn − rn ≥

N−1 k0 R2 − Λ ∑ h(j)cj > M1 , Λ j=1

∀ n = 1, N − 1,

and hence f (n, vn − rn ) + cn ≤ ε(vn − rn ) ≤ εvn ≤ εR2 , Thus, for v ∈ P ∩ 𝜕Ω3 , we deduce

∀ n = 1, N − 1.

18 � 1 Second-order difference equations N−1

Qn (v) ≤ εΛR2 ∑ h(j) ≤ R2 , j=1

∀ n = 0, N,

and then 󵄩 󵄩󵄩 󵄩󵄩Q(v)󵄩󵄩󵄩 ≤ ‖v‖,

∀ v ∈ P ∩ 𝜕Ω3 .

Therefore, by Theorem 1.1.1, we conclude that operator Q has a fixed point ṽ = (ṽn )n=0,N ∈ P such that R < ‖ṽ‖ ≤ R2 and hence ũ = (ũn )n=0,N , where ũn = ṽn − ̃rn , n = 0, N, is a second positive solution of problem (1.1), (1.2). Theorem 1.1.4. We assume that (H1) and (H2) hold. Additionally, we make the following assumption. 2 (H5) There exist r, R > 0 such that Λk ∑N−1 j=1 h(j)cj < r < R and 0

Φ(R)
Λ(k02 ∑N−1 j=1 h(j)) such that f (n, u) ≥ C0 u for all n = 1, N − 1 and u > 0 sufficiently large. Then problem (1.1), (1.2) has at least two positive solutions. Proof. We suppose that assumptions (H1), (H2), and (H5) hold. We consider the sets Ω1 and Ω2 defined at the beginning of the proof of Theorem 1.1.2. In a similar manner as in the proof of Theorem 1.1.2, we obtain ‖Q(v)‖ ≥ ‖v‖ for all v ∈ P ∩ 𝜕Ω1 and ‖Q(v)‖ < ‖v‖ for all v ∈ P ∩ 𝜕Ω2 . Thus, by Theorem 1.1.1, operator Q has a fixed point v1 = (v1n )n=0,N ∈ P with r ≤ ‖v1 ‖ < R satisfying v1n − rn > 0 for all n = 0, N. Therefore, by Lemmas 1.1.7–1.1.9, (un1 )n=0,N , where un1 = v1n − rn , n = 0, N, is a positive solution of problem (1.1), (1.2). Now, we assume (H6) holds. Then there exists M2 > 0 such that for u > M2 we have f (n, u) + cn ≥ C0 u + cn ≥ C0 u,

∀ n = 1, N − 1,

u > M2 .

We choose R3 > max{R, kΛ (M2 + Λ ∑N−1 j=1 h(j)cj )}, 0

−1

k 2 C N−1 R3 ≥ k0 C0 Λ( ∑ h(j))( ∑ h(j)cj )( 0 0 ∑ h(j) − 1) . Λ j=1 j=1 j=1 N−1

N−1

Let Ω3 = {v ∈ X, ‖v‖ < R3 }. Then for v ∈ P ∩ 𝜕Ω3 we obtain vn − rn ≥

n−1 k0 R3 − Λ ∑ h(j)cj > M2 , Λ j=1

∀ n = 1, N − 1,

1.1 A difference equation with a sign-changing nonlinearity

� 19

which implies ∀ n = 1, N − 1.

f (n, (vn − rn )∗ ) + cn ≥ C0 (vn − rn ),

Thus, for all v ∈ P ∩ 𝜕Ω3 and n = 1, N − 1, we deduce N−1

Qn (v) ≥ C0 k0 ∑ h(j)(vj − rj ) j=1

N−1

≥ C0 k0 ( ∑ hj )( j=1

N−1 k0 R3 − Λ ∑ h(i)ci ) ≥ R3 , Λ i=1

and then 󵄩󵄩 󵄩 󵄩󵄩Q(v)󵄩󵄩󵄩 ≥ ‖v‖,

∀ v ∈ P ∩ 𝜕Ω3 .

Therefore, operator Q has another fixed point v2 ∈ P such that R < ‖v2 ‖ ≤ R3 , so u2 = v2 −̃r is a second positive solution of problem (1.1), (1.2). The following theorem results from Theorem 1.1.4. Theorem 1.1.5. We assume that (H1), (H2), and (H5) hold. If limu→∞ f (n,u) = ∞ uniformly u for n = 1, N − 1, then problem (1.1), (1.2) has at least two positive solutions. We also obtain existence of at least one positive solution of problem (1.1), (1.2) if we combine the first condition of (1.15) with condition limu→∞ f (n, u)/u = ∞ uniformly for n = 1, N − 1. So, we have the following theorem. Theorem 1.1.6. We assume that (H1) and (H2) hold. Additionally, we make the following assumption. 2 r ̃ There exists r > Λ ∑N−1 h(j)cj such that Φ(r) ≤ (H3) . N−1 j=1 k 0

maxn=0,N ∑j=1 G(n,j)

= ∞ uniformly for n = 1, N − 1. Then problem (1.1), (1.2) has at Moreover, limu→∞ f (n,u) u least one positive solution. Proof. The proof of this theorem follows from the first part of the proof of Theorem 1.1.2 (with the set Ω1 , for which ‖Q(v)‖ ≤ ‖v‖ for all v ∈ P ∩ 𝜕Ω1 ) and the second part of the proof of Theorem 1.1.4 (with the set Ω3 , for which ‖Q(v)‖ ≥ ‖v‖ for all v ∈ P ∩ 𝜕Ω3 ).

1.1.3 An example Let N = 20, L = 2, p = 2, q = 1, ξ1 = 5, ξ2 = 15, a1 = 2, a2 = 31 , η1 = 10, and b1 = 21 . We consider the difference equation

20 � 1 Second-order difference equations Δ2 un−1 − 2un + f (n, un ) = 0,

n = 1, 19,

(1.19)

with the multi-point boundary conditions 1 u0 = 2u5 + u15 , 3

1 u20 = u10 . 2 p

(1.20) p

We obtain A = 2 + √3, d0 ≈ 2.73999 × 1011 > 0, ∑i=1 ai Aξi ≈ 1.26502 × 108 > 1, ∑i=1 ai A1ξi ≈ q 3.63956 × 10−12 , ∑i=1 bi Aηi

q 1, ∑i=1 bi A1ηi 11

≈ 9.53882 × 10−7 > A120 ≈ ≈ 262, 087 < 0.00276244 < 20 A ≈ 2.74758 × 10 . Therefore, assumption (H1) is satisfied. In addition, we have g(n, j) =

(A2 ×{

A − 1)(A20 − A−20 )

(Aj − A−j )(A20−n − An−20 ), (An − A−n )(A20−j − Aj−20 ),

1 ≤ j < n ≤ 20, 0 ≤ n ≤ j ≤ 19,

G(n, j) = g(n, j) + +

1 1 1 1 1 1 1 [An ( 10 − 20 ) + n (A20 − A10 )](2g(5, j) + g(15, j)) d0 2A A 2 3 A 1 1 1 1 1 1 1 [An (1 − 2 5 − ) + n (2A5 + A15 − 1)] g(10, j), d0 3 A15 A 3 2 A

n = 0, 20, j = 1, 19,

A (Aj − A−j )(A20−j − Aj−20 ), j = 1, 19, (A2 − 1)(A20 − A−20 ) 1 min{An − A−n , A20−n − An−20 }, n = 0, 20. k(n) = 19 A − A−19 h(j) =

We deduce Λ ≈ 3.84003043 and k0 ≈ 4.7053 × 10−11 . We consider cn = ln n+3 > 0 for all n n = 1, 19, and then we have a := define the function

Λ2 k0

11 ∑N−1 j=1 h(j)cj ≈ 6.535016 × 10 . We denote b = a + 1 and

(u + 1)1/2 n { { + ln , u ∈ [0, b], n = 1, 19, { { n(n + 4) n +3 { { { { { (b + 1)1/2 n { { {u − b + + ln , u ∈ (b, 2b], n = 1, 19, n(n + 4) n +3 f (n, u) = { { { { (u + 3)2 (2b + 3)2 (b + 1)1/2 n { { { − + + b + ln , { { n(n + 2) n(n + 2) n(n + 4) n +3 { { { { u ∈ (2b, ∞), n = 1, 19. The function f is a continuous one, and it satisfies the inequality f (n, u) ≥ −cn for all u ∈ ℝ+ and n = 1, 19. Then assumption (H2) is also satisfied. After some computations, we obtain S := maxn=0,20 (∑19 j=1 G(n, j)) ≈ 0.499998. Therefore, for r = b we deduce Φ(r) ≈ r 12 ̃ is satisfied. Because limu→∞ f (n,u) = ∞ 161, 679 < ≈ 1.30701 × 10 , so assumption (H3) S

u

1.2 A difference equation with a positive parameter � 21

uniformly for n = 1, 19, by Theorem 1.1.6, we conclude that problem (1.19), (1.20) has at least one positive solution u = (un )n=0,20 . Remark 1.1.2. The results presented in this section were published in the paper [53].

1.2 A second-order difference equation with a linear term, a sign-changing nonlinearity, and a positive parameter We consider the nonlinear second-order difference equation Δ2 un−1 − Lun + λf (n, un ) = 0,

n = 1, N − 1,

(1.21)

with the multi-point boundary conditions p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi ,

(1.22)

i=1

where N ∈ ℕ, N > 2, p, q ∈ ℕ, ai ∈ ℝ, ξi ∈ ℕ for all i = 1, p, bi ∈ ℝ, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, L is a positive constant, λ is a positive parameter, and f is a sign-changing nonlinearity. Under some assumptions on the nonlinearity f , we present intervals for the parameter λ such that problem (1.21), (1.22) has positive solutions (un )n=0,N with un > 0 for all n = 0, N. The problem (1.21), (1.22) with L = 0 and ai = 0 for all i = 1, p was studied in the paper [52]. As in Section 1.1 we consider the second-order difference equation Δ2 un−1 − Lun + yn = 0,

n = 1, N − 1,

(1.23)

with the multi-point boundary conditions (1.22), where yn ∈ ℝ for all n = 1, N − 1. We also consider the number A the biggest solution of the characteristic equation r 2 − (L + 2

2)r+1 = 0 associated with equation (1.23), namely, A = L+2+ 2L +4L , d0 the number defined by (1.4), g and G the functions defined by (1.8) and (1.10), respectively, h(j), j = 1, N − 1, and k(n), n = 0, N, the finite sequences defined in Lemma 1.1.4, and the constant Λ defined by (1.11). In this section we will use the auxiliary results Lemmas 1.1.3, 1.1.5, and 1.1.6 from Section 1.1. In the proof of the main existence results we will apply the Guo–Krasnosel’skii fixed point theorem (Theorem 1.1.1) and the nonlinear alternative of Leray–Schauder type presented below. √

Theorem 1.2.1. Let X be a Banach space with Ω ⊂ X closed and convex. Assume U is a relatively open subset of Ω with 0 ∈ U, and let S : U → Ω be a completely continuous operator. Then either (1) S has a fixed point in U, or (2) there exist u ∈ 𝜕U and ν ∈ (0, 1) such that u = νSu.

22 � 1 Second-order difference equations 1.2.1 Existence of positive solutions We present firstly the basic assumptions that we will use in the existence results. p p (I1) We have ai ≥ 0 for all i = 1, p, bj ≥ 0 for all j = 1, q, ∑i=1 ai Aξi ≥ 1, ∑i=1 ai A1ξi ≤ 1, q

q

∑i=1 bi A1ηi ≥ A1N , ∑i=1 bi Aηi ≤ AN , d0 > 0, and L > 0. (I2) The function f : {1, . . . , N − 1} × ℝ+ → ℝ is continuous, and there exist cn ≥ 0, n = 1, N − 1, such that f (n, u) ≥ −cn for all n = 1, N − 1, u ∈ ℝ+ .

We denote by (sn )n=0,N the solution of problem (1.23), (1.22) with yn = λcn for all n = 1, N − 1, namely, the solution of the boundary value problem Δ2 un−1 − Lun + λcn = 0, n = 1, N − 1, { { { p q { { {u0 = ∑ ai uξi , uN = ∑ bi uηi , i=1 i=1 { where cn , n = 0, N, are given in (I2). So, by using the Green function G and Lemma 1.1.3, we have sn = λ ∑N−1 j=1 G(n, j)cj for all n = 0, N. We consider the difference equation Δ2 wn−1 − Lwn + λ(f (n, (wn − sn )∗ ) + cn ) = 0,

n = 1, N − 1,

(1.24)

with the multi-point boundary conditions p

w0 = ∑ ai wξi , i=1

q

wN = ∑ bi wηi ,

(1.25)

i=1

where y∗ = y if y ≥ 0 and y∗ = 0 if y < 0. We deduce easily that the sequence (un )n=0,N is a positive solution of problem (1.21), (1.22) (un > 0 for all n = 0, N) if and only if (wn )n=0,N , wn = un + sn , n = 0, N, is a solution of the boundary value problem (1.24), (1.25) with wn > sn for all n = 0, N. By using Lemma 1.1.3, we also obtain that the sequence (wn )n=0,N is a solution of problem (1.24), (1.25) if and only if (wn )n=0,N is a solution of the problem N−1

wn = λ ∑ G(n, j)(f (j, (wj − sj )∗ ) + cj ), j=1

n = 0, N.

(1.26)

We consider the Banach space X = ℝN+1 = {w = (wn )n=0,N , wn ∈ ℝ, n = 0, N}, endowed with the maximum norm ‖w‖ = maxn=0,N |wn |, and we define the operator T : X → X, T(w) = (Tn (w))n=0,N , where N−1

Tn (w) = λ ∑ G(n, j)(f (j, (wj − sj )∗ ) + cj ), j=1

n = 0, N, w = (wn )n=0,N .

1.2 A difference equation with a positive parameter

� 23

By (I2), the operator T is completely continuous. We also define the cone P = {w ∈ X, w = (wn )n=0,N , wn ≥

k(n) ‖w‖, ∀ n = 0, N}. Λ

By using Lemma 1.1.6, we deduce that T(P) ⊂ P. In addition, we remark that the sequence (wn )n=0,N is a solution of problem (1.26) if and only if (wn )n=0,N is a fixed point of operator T. So, the existence of positive solutions of problem (1.21), (1.22) is reduced to the fixed point problem of operator T in the cone P. We denote k0 = min{k(n), n = 1, N − 1}. Because N > 2, we have k0 ∈ (0, 1). Theorem 1.2.2. Assume that (I1) and (I2) hold. Additionally, we make the following assumption. (I3) We have limu→∞ minn=1,N−1 f (n,u) = ∞. u Then there exists a constant λ∗ > 0 such that for any λ ∈ (0, λ∗ ] the problem (1.21), (1.22) has at least one positive solution. Proof. We choose a positive number R1 >

Λ2 k0

∑N−1 j=1 h(j)cj , and we define the set Ω1 =

−1 {w ∈ X, ‖w‖ < R1 }. We introduce the constant λ∗ = min{1, R1 (ΛM0 ∑N−1 j=1 h(j)) } with M0 = max{maxn=1,N−1,u∈[0,R ] {f (n, u) + cn }, 1}. 1

Let λ ∈ (0, λ∗ ]. Because sn ≤ λΛ ∑N−1 j=1 h(j)cj for all n = 0, N, we deduce for any w ∈ P ∩ 𝜕Ω1 that [wn − sn ]∗ ≤ wn ≤ ‖w‖ = R1 ,

∀ n = 0, N,

and wn − sn ≥

N−1 k(n) k(n) ‖w‖ − sn ≥ ‖w‖ − λΛ ∑ h(j)cj Λ Λ j=1

≥

N−1 N−1 k0 R1 k R − λΛ ∑ h(j)cj ≥ 0 1 − λ∗ Λ ∑ h(j)cj Λ Λ j=1 j=1

≥

N−1 k0 R1 − Λ ∑ h(j)cj > 0, Λ j=1

∀ n = 1, N − 1.

Then for any w ∈ P ∩ 𝜕Ω1 , we conclude that N−1

Tn (w) ≤ λΛ ∑ h(j)[f (j, (wj − sj )∗ ) + cj ] j=1

N−1

≤ λ∗ ΛM0 ∑ h(j) ≤ R1 = ‖w‖, j=1

∀ n = 0, N.

24 � 1 Second-order difference equations Hence, we obtain 󵄨 󵄩 󵄨 󵄩󵄩 󵄩󵄩T(w)󵄩󵄩󵄩 = max 󵄨󵄨󵄨Tn (w)󵄨󵄨󵄨 ≤ ‖w‖, n=0,N

∀ w ∈ P ∩ 𝜕Ω1 .

(1.27)

−1 Now we consider the constant L1 = 2Λ(λk02 ∑N−1 j=1 h(j)) . By (H3) we deduce that there exists a constant M1 > 0 such that

f (n, u) ≥ L1 u,

∀ n = 1, N − 1, u ≥ M1 .

(1.28)

2

1 2Λ , k ∑N−1 We define R2 > R1 with R2 ≥ max{ 2ΛM j=1 h(j)cj } and let Ω2 = {w ∈ X, ‖w‖ < R2 }. k0 0 Then for any w ∈ P ∩ 𝜕Ω2 , we have

N−1

N−1

j=1

j=1

wn − sn = wn − λ ∑ G(n, j)cj ≥ wn − ∑ G(n, j)cj N−1

≥ wn − Λ ∑ h(j)cj ≥ wn − j=1

= wn [1 −

wn Λ2 N−1 ∑ h(j)cj ‖w‖k(n) j=1

w Λ2 N−1 ∑ h(j)cj ] ≥ n ≥ 0, R2 k0 j=1 2

∀ n = 1, N − 1.

So, we obtain 1 [wn − sn ]∗ = wn − sn ≥ wn 2 k k(n) ≥ ‖w‖ ≥ 0 R2 ≥ M1 , 2Λ 2Λ

∀ n = 1, N − 1.

(1.29)

Then for any w ∈ P ∩ 𝜕Ω2 , by (1.28) and (1.29) we deduce that f (n, [wn − sn ]∗ ) ≥ L1 ([wn − sn ]∗ ) ≥

L1 w , 2 n

∀ n = 1, N − 1.

Therefore, for any w ∈ P ∩ 𝜕Ω2 and n = 1, N − 1 we conclude that N−1

N−1

Tn (w) ≥ λ ∑ G(n, j)L1 ([wj − sj ]∗ ) ≥ λ ∑ G(n, j) j=1

j=1

N−1

≥

λL1 k0 R2 ∑ k(n)h(j) ≥ 2Λ j=1

L1 k0 R2 2Λ

λL1 k02 R2 N−1 2Λ

∑ h(j) = R2 . j=1

Hence, we obtain 󵄩󵄩 󵄩 󵄩󵄩T(w)󵄩󵄩󵄩 ≥ ‖w‖,

∀ w ∈ P ∩ 𝜕Ω2 .

(1.30)

1.2 A difference equation with a positive parameter

� 25

By (1.27), (1.30), and Theorem 1.1.1 (i), we deduce that operator T has a fixed point w1 = (wn1 )n=0,N ∈ P with R1 ≤ ‖w1 ‖ ≤ R2 . In addition, we have N−1

un1 = wn1 − sn = wn1 − λ ∑ G(n, j)cj j=1

N−1

≥ wn1 − λΛ ∑ h(j)cj ≥ j=1

≥

N−1 k(n) 󵄩󵄩 1 󵄩󵄩 󵄩󵄩w 󵄩󵄩 − λΛ ∑ h(j)cj Λ j=1

N−1 k0 R1 − Λ ∑ h(j)cj > 0, Λ j=1 p

∀ n = 1, N − 1, q

u01 = w01 − s0 = ∑ ai uξ1i > 0,

uN1 = wN1 − sN = ∑ bi uη1 i > 0.

i=1

i=1

Then u1 = (un1 )n=0,N is a positive solution of problem (1.21), (1.22). Theorem 1.2.3. Assume that (I1) and (I2) hold. Additionally, we make the following assumption. (I4) We have ∑N−1 i=1 ci > 0, lim infu→∞ minn=1,N−1 f (n, u) > L0 , with L0 =

2Λ2 ∑N−1 j=1 h(j)cj k02 ∑N−1 j=1 h(j)

and

lim max

u→∞ n=1,N−1

|f (n, u)| = 0. u

Then there exists a constant λ∗ > 0 such that for any λ ≥ λ∗ , the problem (1.21), (1.22) has at least one positive solution. Proof. By (H4) we obtain that there exists M2 > 0 such that f (n, u) ≥ L0 for all n = −1 1, N − 1 and u ≥ M2 . We define λ∗ = M2 (Λ ∑N−1 j=1 h(j)cj ) . We assume that λ ≥ λ∗ . Let 2

R3 = 2λΛ ∑N−1 j=1 h(j)cj > 0 and Ω3 = {w ∈ X, ‖w‖ < R3 }. Then for any w ∈ P ∩ 𝜕Ω3 , we k0 deduce that wn − sn ≥ ≥

N−1 k(n) ‖w‖ − λ ∑ G(n, j)cj Λ j=1 N−1 N−1 k R k(n) ‖w‖ − λΛ ∑ h(j)cj ≥ 0 3 − λΛ ∑ h(j)cj Λ Λ j=1 j=1 N−1

N−1

j=1

j=1

= λΛ ∑ h(j)cj ≥ λ∗ Λ ∑ h(j)cj = M2 > 0, Therefore, for any w ∈ P ∩ 𝜕Ω3 we conclude that

∀ n = 1, N − 1.

26 � 1 Second-order difference equations N−1

Tn (w) ≥ λk(n) ∑ h(j)(f (j, [wj − sj ]∗ ) + cj ) j=1

N−1

≥ λk0 ∑ h(j)L0 = ‖w‖, j=1

∀ n = 1, N − 1.

So, we obtain 󵄩 󵄩󵄩 󵄩󵄩T(w)󵄩󵄩󵄩 ≥ ‖w‖,

∀ w ∈ P ∩ 𝜕Ω3 .

(1.31)

−1 Next we consider the positive number ε = (2λΛ ∑N−1 j=1 h(j)) . Then by (I4), we deduce that there exists M3 > 0 such that |f (n, u)| ≤ εu for all n = 1, N − 1 and u ≥ M3 . So, we obtain |f (n, u)| ≤ M4 + εu for all n = 1, N − 1 and u ≥ 0, where M4 = maxn=1,N−1,u∈[0,M ] |f (n, u)|. We define now R4 > R3 with R4 ≥ 2λΛ ∑N−1 j=1 h(j)(M4 + cj ) 3 and Ω4 = {w ∈ X, ‖w‖ < R4 }. Then for any w ∈ P ∩ 𝜕Ω4 , we have wn − sn > M2 for all n = 1, N − 1 and N−1

Tn (w) ≤ λΛ ∑ h(j)[f (j, [wj − sj ]∗ ) + cj ] j=1

N−1

≤ λΛ ∑ h(j)[M4 + ε[wj − sj ]∗ + cj ] j=1

N−1

N−1

j=1

j=1

≤ λΛ ∑ h(j)(M4 + cj ) + λεΛR4 ∑ h(j) ≤

R4 R4 + = R4 = ‖w‖, 2 2

∀ n = 0, N.

Therefore, 󵄩󵄩 󵄩 󵄩󵄩T(w)󵄩󵄩󵄩 ≤ ‖w‖,

∀ w ∈ P ∩ 𝜕Ω4 .

(1.32)

By (1.31), (1.32), and Theorem 1.1.1 (ii), we deduce that operator T has a fixed point w1 ∈ P, w1 = (wn1 )n=0,N , with R3 ≤ ‖w1 ‖ ≤ R4 . Besides, we conclude that N−1

un1 = wn1 − sn ≥ wn1 − λΛ ∑ h(j)cj ≥ j=1

N−1 k(n) 󵄩󵄩 1 󵄩󵄩 󵄩󵄩w 󵄩󵄩 − λΛ ∑ h(j)cj Λ j=1

N−1

≥

N−1 k0 R3 − λΛ ∑ h(j)cj = λΛ ∑ h(j)cj Λ j=1 j=1 N−1

≥ λ∗ Λ ∑ h(j)cj = M2 > 0, j=1

p

u01 = w01 − s0 = ∑ ai uξ1i > 0, i=1

∀ n = 1, N − 1, q

uN1 = wN1 − sn = ∑ bi uη1 i > 0. i=1

1.2 A difference equation with a positive parameter � 27

Then u1 = (un1 )n=0,N is a positive solution of problem (1.21), (1.22). In a similar manner as in the proof of Theorem 1.2.3, we obtain the following result. Theorem 1.2.4. Assume that (I1) and (I2) hold. Additionally, we make the following assumption. ̃ We have ∑N−1 ci > 0, (I4) i=1 lim min f (n, u) = ∞,

u→∞ n=1,N−1

and

lim max

u→∞ n=1,N−1

|f (n, u)| = 0. u

Then there exists a constant λ̃∗ > 0 such that for any λ ≥ λ̃∗ , the problem (1.21), (1.22) has at least one positive solution. Theorem 1.2.5. Assume that (I1) and (I2) hold. Additionally, we make the following assumption. (I5) We have f (n, 0) > 0 for all n = 1, N − 1. Then there exists a constant λ0 > 0 such that for any λ ∈ (0, λ0 ] the problem (1.21), (1.22) has at least one positive solution. Proof. Let δ ∈ (0, 1) be fixed. By using (I2) and (I5), there exists R0 ∈ (0, 1] such that f (n, u) ≥ δf (n, 0) > 0 for all n = 1, N − 1 and u ∈ [0, R0 ]. We define f (R0 ) =

max

{f (n, u) + cn }

n=1,N−1, u∈[0,R0 ]

≥ max {δf (n, 0) + cn } > 0, n=1,N−1

−1

N−1

λ0 = R0 (2Λf (R0 ) ∑ h(j)) . j=1

We will show that for any λ ∈ (0, λ0 ], problem (1.21), (1.22) has at least one positive solution. Let λ ∈ (0, λ0 ] be arbitrary, but fixed for the moment. We define the set U = {w ∈ P, w = (wn )n=0,N , ‖w‖ < R0 }. We suppose that there exist w ∈ 𝜕U (‖w‖ = R0 ) and ν ∈ (0, 1) such that w = νT(w). We deduce that [wn − sn ]∗ = wn − sn ≤ wn ≤ R0 , [wn − sn ] = 0, ∗

if

if

wn − sn < 0,

wn − sn ≥ 0, n = 0, N.

Then by Lemma 1.1.5, we have N−1

wn = νTn (w) ≤ Tn (w) ≤ λΛ ∑ h(j)f (R0 ) j=1

28 � 1 Second-order difference equations N−1

≤ λ0 Λf (R0 ) ∑ h(j) = j=1

R0 . 2

R

Therefore, we obtain R0 = ‖w‖ ≤ 20 , which is a contradiction. Hence, by Theorem 1.2.1 (with Ω = P) we conclude that operator T has a fixed point 0 w = (wn0 )n=0,N ∈ U. That is, w0 = T(w0 ) or wn0 = Tn (w0 ), n = 0, N, and ‖w0 ‖ ≤ R0 with wn0 ≥

k(n) ‖w0 ‖ λ

for all n = 0, N. Moreover, by Lemma 1.1.5 we deduce that N−1

wn0 = Tn (w0 ) ≥ λ ∑ G(n, j)(δf (j, 0) + cj ) j=1

N−1

N−1

≥ λk0 ∑ h(j)f (j, 0) + λ ∑ G(n, j)cj j=1

j=1

N−1

≥ λk0 min f (j, 0) ∑ h(j) + sn > sn , j=1,N−1

j=1

∀ n = 1, N − 1, p

so un0 = wn0 − sn > 0 for all n = 1, N − 1, and u00 = w00 − s0 = ∑i=1 ai uξ0i > 0, uN0 = wN0 − sN = q

∑i=1 bi uη0i > 0. Then u0 = (un0 )n=0,N is a positive solution of problem (1.21), (1.22).

Theorem 1.2.6. Assume that (I1), (I2), (I3), and (I5) hold. Then the boundary value problem (1.21), (1.22) has at least two positive solutions for λ > 0 sufficiently small. Proof. By Theorem 1.2.2 (in which we choose R1 > 1) and Theorem 1.2.5, we deduce that for 0 < λ ≤ min{λ∗ , λ0 }, problem (1.21), (1.22) has at least two positive solutions u1 = (un1 )n=0,N and u0 = (un0 )n=0,N with ‖u1 + ̃s‖ > 1 and ‖u0 + ̃s‖ ≤ 1, where ̃s = (sn )n=0,N . 1.2.2 Examples Let N = 20, L = 2, p = 2, q = 1, ξ1 = 5, ξ2 = 15, a1 = 2, a2 = 31 , η1 = 10, and b1 = 21 . We consider the difference equation Δ2 un−1 − 2un + λf (n, un ) = 0,

n = 1, 19,

(1.33)

with the multi-point boundary conditions 1 u0 = 2u5 + u15 , 3

1 u20 = u10 . 2 p

(1.34) p

We obtain A = 2 + √3, d0 ≈ 2.73999 × 1011 > 0, ∑i=1 ai Aξi ≈ 1.26502 × 108 > 1, ∑i=1 ai A1ξi ≈ q 1, ∑i=1 bi A1ηi 11

q 3.63956 × 10−12 , ∑i=1 bi Aηi

0.00276244 < ≈ 9.53882 × 10−7 > A120 ≈ ≈ 262, 087 < 20 A ≈ 2.74758 × 10 . Therefore, assumption (I1) is satisfied. As in Section 1.1.3, we find

1.2 A difference equation with a positive parameter

g(n, j) =

(A2 ×{

A − 1)(A20 − A−20 )

(Aj − A−j )(A20−n − An−20 ), (An − A−n )(A20−j − Aj−20 ),

� 29

1 ≤ j < n ≤ 20, 0 ≤ n ≤ j ≤ 19,

G(n, j) = g(n, j) + +

1 1 1 1 1 1 1 [An ( 10 − 20 ) + n (A20 − A10 )](2g(5, j) + g(15, j)) d0 2A A 2 3 A 1 1 1 1 1 1 1 [An (1 − 2 5 − ) + n (2A5 + A15 − 1)] g(10, j), d0 3 A15 A 3 2 A

n = 0, 20, j = 1, 19,

A (Aj − A−j )(A20−j − Aj−20 ), j = 1, 19, (A2 − 1)(A20 − A−20 ) 1 min{An − A−n , A20−n − An−20 }, n = 0, 20. k(n) = 19 A − A−19 h(j) =

We also obtain Λ ≈ 3.84003043 and k0 ≈ 4.7053 × 10−11 . Example 1.2.1. We consider the function f (n, u) =

n (u + 1)4/3 + ln , n(n + 2) n+3

∀ n = 1, 19, u ∈ [0, ∞).

Here we have cn = ln n+3 ≥ 0 for all n = 1, 19, and we obtain f (n, u) ≥ −cn for all n n = 1, 19 and u ∈ [0, ∞). Because limu→∞ f (n, u)/u = ∞, assumptions (I2) and (I3) are satisfied. We choose R1 = 6.53502 × 1011 , which satisfies the condition from the beginning of the proof of Theorem 1.2.2. Then M0 ≈ 1.89034 × 1015 and λ∗ ≈ 3.09645 × 106 . By Theorem 1.2.2 we deduce that the problem (1.33), (1.34) has at least one positive solution for any λ ∈ (0, λ∗ ]. Example 1.2.2. We consider the function f (n, u) = (u − 1)(u − 2),

n = 1, 19, u ∈ [0, ∞).

̃0 > 0 (M ̃0 = 1 ) such that f (n, u) + M ̃0 ≥ 0 for all n = 1, 19 and u ≥ 0 Then there exists M 4 1 ̃0 = for all n = 1, 19) and f (n, 0) = 2 > 0 for all n = 1, 19. So, assumptions (I2) and (cn = M 4 (I5) are satisfied. Let δ = 83 < 1 and R0 = 21 . Then f (n, u) ≥ δf (n, 0) for all n = 1, 19 and u ∈ [0, 21 ]. Besides, f (R0 ) = 49 , and therefore we obtain λ0 ≈ 9.95213 × 108 . By Theorem 1.2.5, for any λ ∈ (0, λ0 ], we deduce that problem (1.33), (1.34) has at least one positive solution. Remark 1.2.1. The results presented in this section were published in the paper [54].

2 Systems of second-order finite difference equations with uncoupled multi-point boundary conditions In this chapter we investigate the existence and multiplicity of positive solutions for two systems of nonlinear second-order difference equations supplemented with uncoupled multi-point boundary conditions. The nonlinearities from the systems are nonnegative functions and satisfy some assumptions containing concave functions, or they are signchanging functions.

2.1 Nonnegative nonlinearities We consider in this section the system of nonlinear second-order difference equations Δ2 un−1 + f (n, un , vn ) = 0, { 2 Δ vn−1 + g(n, un , vn ) = 0,

n = 1, N − 1,

(2.1)

n = 1, N − 1,

subject to the uncoupled multi-point boundary conditions p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi , i=1

r

v0 = ∑ ci vζi , i=1

l

vN = ∑ di vρi , i=1

(2.2)

where N ∈ ℕ, N ≥ 2, p, q, r, l ∈ ℕ, ξi ∈ ℕ for all i = 1, p, ηi ∈ ℕ for all i = 1, q, ζi ∈ ℕ for all i = 1, r, ρi ∈ ℕ for all i = 1, l, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, 1 ≤ ζ1 < ⋅ ⋅ ⋅ < ζr ≤ N − 1, and 1 ≤ ρ1 < ⋅ ⋅ ⋅ < ρl ≤ N − 1. Under sufficient conditions on the nonnegative nonlinearities f and g which contain some concave functions, we investigate the existence and multiplicity of positive solutions of problem (2.1), (2.2) by using the fixed point index theory. By a positive solution of (2.1), (2.2) we mean a pair of sequences (u, v) = ((un )n=0,N , (vn )n=0,N ) satisfying (2.1) and (2.2) with un ≥ 0 and vn ≥ 0 for all n = 0, N, and un > 0 for all n = 1, N − 1 or vn > 0 for all n = 1, N − 1. The existence and nonexistence of nonnegative and nontrivial solutions (u, v) (un ≥ 0, vn ≥ 0 for all n = 0, N, and (u, v) ≠ (0, 0)) of the system Δ2 un−1 + λsn f (n, un , vn ) = 0, { 2 Δ vn−1 + μtn g(n, un , vn ) = 0,

n = 1, N − 1, n = 1, N − 1,

where λ and μ are positive parameters, sn , tn ≥ 0 for all n = 1, N − 1, and there exist i0 , j0 ∈ {1, . . . , N − 1} such that si0 > 0, tj0 > 0, supplemented with the boundary conditions (2.2), were studied in the papers [35] and [36] under various assumptions on the nonnegative nonlinearities f and g. In the proof of the existence theorems we used the https://doi.org/10.1515/9783111040370-002

2.1 Nonnegative nonlinearities

� 31

Guo–Krasnosel’skii fixed point theorem. In the paper [43], by applying some theorems from the fixed point index theory, the authors investigated the existence and multiplicity of positive solutions (u, v) (un ≥ 0, vn ≥ 0 for all n = 0, N, maxn=0,N un > 0, and maxn=0,N vn > 0) of the system Δ2 un−1 + f (n, vn ) = 0, { 2 Δ vn−1 + g(n, un ) = 0,

n = 1, N − 1, n = 1, N − 1,

subject to the boundary conditions (2.2). The assumptions used in [43] for the nonnegative continuous functions f and g are different than those we will use in this section for problem (2.1), (2.2). In [37] we studied the existence and nonexistence of positive solutions (u, v) (un ≥ 0, vn ≥ 0 for all n = 0, N, un > 0 for all n = 0, N − 1, and vn > 0 for all n = 1, N) of the system Δ2 un−1 + sn f (vn ) = 0, { 2 Δ vn−1 + tn g(un ) = 0,

n = 1, N − 1,

(2.3)

n = 1, N − 1,

subject to the boundary conditions p

u0 = ∑ ai uξi + a0 , i=1

q

uN = ∑ bi uηi , i=1

r

v0 = ∑ ci vζi , i=1

l

vN = ∑ di vρi + b0 , i=1

where a0 and b0 are positive constants. For the existence result we applied the Schauder fixed point theorem. Similar results were obtained for the system (2.3) supplemented with the following boundary conditions: p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi + a0 , i=1

r

v0 = ∑ ci vζi + b0 , i=1

l

vN = ∑ di vρi i=1

or p

u0 = ∑ ai uξi + a0 , i=1

q

uN = ∑ bi uηi , i=1

r

v0 = ∑ ci vζi + b0 , i=1

l

vN = ∑ di vρi i=1

or p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi + a0 , i=1

r

v0 = ∑ ci vζi , i=1

l

vN = ∑ di vρi + b0 , i=1

where a0 and b0 are positive constants. The results obtained in the above papers ([35– 37, 43]) were also presented in Chapter 3 of the monograph [38].

32 � 2 Systems of difference equations with uncoupled BCs 2.1.1 Auxiliary results We begin this section with a result from [35] related to the following system of secondorder difference equations, Δ2 un−1 + zn = 0,

n = 1, N − 1,

(2.4)

subject to the multi-point boundary conditions p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi ,

(2.5)

i=1

where p, q ∈ ℕ, ξi ∈ ℕ for all i = 1, p, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and zn ∈ ℝ for all n = 1, N − 1. q p p q We denote d1 = (1 − ∑i=1 bi ) ∑i=1 ai ξi + (1 − ∑i=1 ai )(N − ∑i=1 bi ηi ). Lemma 2.1.1 ([35]). If d1 ≠ 0, then the solution (un )n=0,N of problem (2.4), (2.5) is given by un = ∑N−1 j=1 𝒢1 (n, j)yj for all n = 0, N, where the Green function 𝒢1 is defined by 𝒢1 (n, j) = g0 (n, j) q

q

p

+

1 [(N − n)(1 − ∑ bk ) + ∑ bi (N − ηi )] ∑ ai g0 (ξi , j) d1 i=1 i=1 k=1

+

1 [n(1 − ∑ ak ) + ∑ ai ξi ] ∑ bi g0 (ηi , j), d1 i=1 i=1 k=1

p

∀ n = 0, N,

p

q

j = 1, N − 1,

(2.6)

and g0 (n, j) =

j(N − n), 1 { n(N − j), N

1 ≤ j ≤ n ≤ N, 0 ≤ n ≤ j ≤ N − 1.

(2.7)

Next we will present some properties of the function g0 and the Green function 𝒢1 . Lemma 2.1.2. The function g0 given by (2.7) has the following properties: (a) g0 (n, j) ≥ 0, for all n = 0, N, j = 1, N − 1; (b) g0 (n, j) ≤ h(j), for all n = 0, N, j = 1, N − 1, where h(j) = g0 (j, j) =

1 j(N − j), N

∀ j = 1, N − 1;

(c) g0 (n, j) ≥ k(n)h(j), for all n = 0, N, j = 1, N − 1, where k(n) =

1 n(N − n), N(N − 1)

∀ n = 0, N.

2.1 Nonnegative nonlinearities

� 33

Proof. For the proofs of (a) and (b), see [9]. For (c), if 1 ≤ j ≤ n ≤ N, then we have 1 1 1 j(N − n) ≥ n(N − n) j(N − j) N N(N − 1) N N(N − 1) ≥ n(N − j),

⇔

the last inequality being satisfied for all j = 1, N − 1 and n = 0, N. If 0 ≤ n ≤ j ≤ N − 1, then we obtain 1 1 1 n(N − j) ≥ n(N − n) j(N − j) N N(N − 1) N N(N − 1) ≥ j(N − n),

⇔

the last inequality being satisfied for all j = 1, N − 1 and n = 0, N. p

q

Lemma 2.1.3. If ai ≥ 0 for all i = 1, p, ∑i=1 ai < 1, bi ≥ 0 for all i = 1, q, ∑i=1 bi < 1, then the Green function 𝒢1 of problem (2.4), (2.5) given by (2.6) satisfies the following inequalities: (a) 𝒢1 (n, j) ≤ Ah(j), for all n = 0, N, j = 1, N − 1, where A=1+

q

p

1 (N − ∑ bi ηi )(∑ ai ) d1 i=1 i=1 p

+

q

1 (N − ∑ ai (N − ξi ))(∑ bi ) > 0; d1 i=1 i=1

(b) 𝒢1 (n, j) ≥ k(n)h(j), for all n = 0, N, j = 1, N − 1. Proof. By the assumptions on the coefficients ai , i = 1, p, and bj , j = 1, q, we can easily see that d1 > 0 and A > 0. By using Lemma 2.1.2, for all n = 0, N and j = 1, N − 1, we deduce that 𝒢1 (n, j) ≤ h(j){1 +

q

p

+

q

p

1 [N(1 − ∑ bk ) + ∑ bi (N − ηi )](∑ ai ) d1 i=1 i=1 k=1 p

q

1 [N(1 − ∑ ak ) + ∑ ai ξi ](∑ bi )} = Ah(j), d1 i=1 i=1 k=1

and 𝒢1 (n, j) ≥ g0 (n, j) ≥ k(n)h(j),

that is, we obtain inequalities (a) and (b).

34 � 2 Systems of difference equations with uncoupled BCs p

Lemma 2.1.4. Assume that ai ≥ 0 for all i = 1, p, ∑i=1 ai < 1, bi ≥ 0 for all i = 1, q, q ∑i=1 bi < 1, and zn ≥ 0 for all n = 1, N − 1. Then the solution (un )n=0,N of problem (2.4), (2.5) satisfies the inequality un ≥ A1 k(n)um for all n, m = 0, N.

Proof. By using Lemmas 2.1.1–2.1.3, we deduce that N−1

N−1

N−1

j=1

j=1

un = ∑ 𝒢1 (n, j)zj ≥ ∑ k(n)h(j)zj ≥ ∑ j=1

N−1

=

1 1 k(n) ∑ 𝒢1 (m, j)zj = k(n)um , A A j=1

1 𝒢 (m, j)k(n)zj A 1 ∀ n, m = 0, N.

We can also formulate similar results as Lemmas 2.1.1–2.1.4 for the discrete boundary value problem Δ2 vn−1 + ̃zn = 0, r

v0 = ∑ ci vζi , i=1

n = 1, N − 1,

(2.8)

l

vN = ∑ di vρi ,

(2.9)

i=1

where r, l ∈ ℕ, ci ≥ 0 for all i = 1, r, ∑ri=1 ci < 1, ζi ∈ ℕ for all i = 1, r, di ≥ 0 for all i = 1, l, ∑li=1 di < 1, ρi ∈ ℕ for all i = 1, l, 1 ≤ ζ1 < ⋅ ⋅ ⋅ < ζr ≤ N − 1, 1 ≤ ρ1 < ⋅ ⋅ ⋅ < ρl ≤ N − 1, and ̃zn ≥ 0 for all n = 1, N − 1. We denote l

r

r

l

i=1

i=1

i=1

i=1

d2 = (1 − ∑ di ) ∑ ci ζi + (1 − ∑ ci )(N − ∑ di ρi ) > 0, 𝒢2 (n, j) = g0 (n, j)

+

l l r 1 [(N − n)(1 − ∑ dk ) + ∑ di (N − ρi )] ∑ ci g0 (ζi , j) d2 i=1 i=1 k=1

+

r r l 1 [n(1 − ∑ ck ) + ∑ ci ζi ] ∑ di g0 (ρi , j), d2 i=1 i=1 k=1

∀ n = 0, N, B=1+ +

j = 1, N − 1,

l r 1 (N − ∑ di ρi )(∑ ci ) d2 i=1 i=1

r l 1 (N − ∑ ci (N − ζi ))(∑ di ) > 0. d2 i=1 i=1

Then we deduce the inequalities 𝒢2 (n, j) ≤ Bh(j) and 𝒢2 (n, j) ≥ k(n)h(j) for all n = 0, N, j = 1, N − 1. In addition, the solution (vn )n=0,N of problem (2.8), (2.9) satisfies the

inequality vn ≥ B1 k(n)vm for all n, m = 0, N.

2.1 Nonnegative nonlinearities

� 35

We recall now some theorems concerning the fixed point index theory. Let E be a real Banach space with the norm ‖ ⋅ ‖ and let P ⊂ E be a cone, with “≤” indicating the partial ordering defined by P and θ being the zero element in E. For ϱ > 0, let Bϱ = {u ∈ E, ‖u‖ < ϱ} be the open ball of radius ϱ centered at θ, and let its boundary be 𝜕Bϱ = {u ∈ E, ‖u‖ = ϱ}. The proofs of our results are based on the following fixed point index theorems (see [4, 5, 29, 49]). Theorem 2.1.1. Let A : Bϱ ∩ P → P be a completely continuous operator which has no fixed points on 𝜕Bϱ ∩ P. If ‖Au‖ ≤ ‖u‖ for all u ∈ 𝜕Bϱ ∩ P, then i(A, Bϱ ∩ P, P) = 1. Theorem 2.1.2. Let A : Bϱ ∩ P → P be a completely continuous operator. If there exists u0 ∈ P \ {θ} such that u − Au ≠ λu0 for all λ ≥ 0 and u ∈ 𝜕Bϱ ∩ P, then i(A, Bϱ ∩ P, P) = 0. Theorem 2.1.3. Let Ω ⊂ E be a bounded open set with θ ∈ Ω. Assume that A : Ω ∩ P → P is a completely continuous operator. (a) If u ≰ Au for all u ∈ 𝜕Ω ∩ P, then the fixed point index i(A, Ω ∩ P, P) = 1. (b) If Au ≰ u for all u ∈ 𝜕Ω ∩ P, then the fixed point index i(A, Ω ∩ P, P) = 0.

2.1.2 Existence and multiplicity of positive solutions In this section we present sufficient conditions on the functions f and g such that problem (2.1), (2.2) has positive solutions with respect to a cone. We present the assumptions that we shall use in the sequel. (K1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, ci ≥ 0, ζi ∈ ℕ for all i = 1, r, 1 ≤ ζ1 < ⋅ ⋅ ⋅ < ζr ≤ N − 1, di ≥ 0, ρi ∈ ℕ for all i = 1, l, 1 ≤ ρ1 < ⋅ ⋅ ⋅ < ρl ≤ N − 1, and p q ∑i=1 ai < 1, ∑i=1 bi < 1, ∑ri=1 ci < 1, ∑li=1 di < 1. (K2) The functions f , g : {1, . . . , N − 1} × ℝ+ × ℝ+ → ℝ+ are continuous. (K3) There exist functions a, b ∈ C(ℝ+ , ℝ+ ) such that (a) a(⋅) is concave and strictly increasing on ℝ+ with a(0) = 0; (b) f (n, u, v) { f0i = lim inf ∈ (0, ∞], { { v→0+ a(v) { { { { { uniformly with respect to (n, u) ∈ {1, . . . , N − 1} × ℝ+ , and { { { g(n, u, v) { { g0i = lim inf ∈ (0, ∞], { { u→0+ b(u) { { { uniformly with respect to (n, v) ∈ {1, . . . , N − 1} × ℝ+ ; { (c) limu→0+

a(Cb(u)) u

= ∞ exists for any constant C > 0.

36 � 2 Systems of difference equations with uncoupled BCs (K4) There exist α1 , α2 > 0 with α1 α2 ≤ 1 such that f (n, u, v) s { f∞ = lim sup ∈ [0, ∞), { { vα1 v→∞ { { { { { { uniformly with respect to (n, u) ∈ {1, . . . , N − 1} × ℝ+ , and { { {g s = lim g(n, u, v) = 0 { { ∞ { u→∞ { uα2 { { exists uniformly with respect to (n, v) ∈ {1, . . . , N − 1} × ℝ+ . { (K5) There exist functions c, d ∈ C(ℝ+ , ℝ+ ) such that (a) c(⋅) is concave and strictly increasing on ℝ+ ; (b) f (n, u, v) i { ∈ (0, ∞], f∞ = lim inf { v→∞ { c(v) { { { { { uniformly with respect to (n, u) ∈ {1, . . . , N − 1} × ℝ+ , and { { { g(n, u, v) i { { g∞ = lim inf ∈ (0, ∞], { { u→∞ d(u) { { { uniformly with respect to (n, v) ∈ {1, . . . , N − 1} × ℝ+ ; { (c) limu→∞ c(Cd(u)) = ∞ exists for any constant C > 0. u (K6) There exist β1 , β2 > 0 with β1 β2 ≥ 1 such that f (n, u, v) f0s = lim sup ∈ [0, ∞), { { { vβ1 { v→0+ { { { { { uniformly with respect to (n, u) ∈ {1, . . . , N − 1} × ℝ+ , and { { g(n, u, v) { { g s = lim =0 { { { 0 u→0+ uβ2 { { exists uniformly with respect to (n, v) ∈ {1, . . . , N − 1} × ℝ+ . { (K7) The functions f (n, u, v) and g(n, u, v) are nondecreasing with respect to u and v, and there exists N0 > 0 such that 3N0 and (N 2 − 1) max{A, B} 3N0 g(n, N0 , N0 ) < , (N 2 − 1) max{A, B} f (n, N0 , N0 )
0, C2 > 0, and a sufficiently small r1 > 0 such that f (n, u, v) ≥ C1 a(v),

g(n, u, v) ≥ C2 b(u),

∀ (n, u) ∈ {1, . . . , N − 1} × ℝ+ , v ∈ [0, r1 ],

∀ (n, v) ∈ {1, . . . , N − 1} × ℝ+ , u ∈ [0, r1 ],

(2.11)

and a(C3 b(u)) ≥

72C3 max{A, B}N 3 u , C1 C2 (N 2 − 1)2

2 where C3 = max{ (N−1)C h(j), j = 1, N − 1}. N

∀ u ∈ [0, r1 ],

(2.12)

38 � 2 Systems of difference equations with uncoupled BCs We will show that (𝒬1 (u, v), 𝒬2 (u, v)) ≰ (u, v) for all (u, v) ∈ 𝜕Br1 ∩ 𝒫 . We suppose that there exists (u, v) ∈ 𝜕Br1 ∩ 𝒫 , that is, ‖(u, v)‖Y = r1 , such that (𝒬1 (u, v), 𝒬2 (u, v)) ≤ (u, v). Then u ≥ 𝒬1 (u, v) and v ≥ 𝒬2 (u, v). By using the monotonicity and concavity of a(⋅), the Jensen inequality, Lemma 2.1.3, and the relations (2.11) and (2.12), we obtain N−1

un ≥ 𝒬1n (u, v) = ∑ 𝒢1 (n, i)f (i, ui , vi ) N−1

i=1

N−1

≥ C1 ∑ h(i)k(n)a(vi ) ≥ C1 k(1) ∑ h(i)a(vi ) i=1

N−1

i=1

N−1

=

C1 ∑ h(i)a( ∑ 𝒢2 (i, j)g(j, uj , vj )) N i=1 j=1

≥

N−1 C1 N−1 ∑ h(i)a(C2 ∑ 𝒢2 (i, j)b(uj )) N i=1 j=1

≥

N−1 C1 N−1 ∑ h(i)a(C2 ∑ h(j)k(i)b(uj )) N i=1 j=1

≥

N−1 C1 N−1 ∑ h(i)a(C2 k(1) ∑ h(j)b(uj )) N i=1 j=1

=

C1 N−1 C N−1 ( ∑ h(i))a( 2 ∑ h(j)b(uj )) N i=1 N j=1

≥

C1 (N 2 − 1) N−1 (N − 1)C2 h(j)b(uj )) ∑ a( 6N(N − 1) j=1 N

=

C1 (N + 1) N−1 (N − 1)C2 h(j) ⋅ C3 b(uj )) ∑ a( 6N NC3 j=1

≥

C1 (N + 1) N−1 (N − 1)C2 h(j)a(C3 b(uj )) ∑ 6N NC3 j=1

≥

72C3 max{A, B}N 3 C1 C2 (N 2 − 1) N−1 uj ∑ h(j) 2 6N C3 C1 C2 (N 2 − 1)2 j=1

≥

12N max{A, B} N−1 1 ∑ h(j) k(j)‖u‖ A N2 − 1 j=1

≥

12N max{A, B} N−1 1 ‖u‖ ≥ 2‖u‖, ∑ h(j) AN N2 − 1 j=1

∀ n = 1, N − 1.

So, ‖u‖ ≥ maxn=1,N−1 un ≥ 2‖u‖, and then ‖u‖ = 0.

(2.13)

2.1 Nonnegative nonlinearities

� 39

In a similar manner, we deduce that N−1

a(vi ) ≥ a(𝒬2i (u, v)) = a( ∑ 𝒢2 (i, j)g(j, uj , vj )) j=1

N−1

≥

1 ∑ a((N − 1)𝒢2 (i, j)g(j, uj , vj )) N − 1 j=1

≥

1 N−1 ∑ a((N − 1)h(j)k(i)C2 b(uj )) N − 1 j=1

≥

1 N−1 C2 (N − 1) h(j)b(uj )) ∑ a( N − 1 j=1 N

=

1 N−1 C2 (N − 1) h(j) ⋅ C3 b(uj )) ∑ a( N − 1 j=1 NC3

≥

1 N−1 C2 (N − 1) h(j)a(C3 b(uj )) ∑ N − 1 j=1 NC3

≥

72C3 max{A, B}N 3 C2 N−1 uj ∑ h(j) NC3 j=1 C1 C2 (N 2 − 1)2

=

N−1 72N 2 max{A, B} N−1 h(j)( ∑ ∑ 𝒢1 (j, θ)f (θ, uθ , vθ )) (N 2 − 1)2 C1 j=1 θ=1

≥

N−1 72N 2 max{A, B} N−1 h(j)( ∑ ∑ h(θ)k(j)C1 a(vθ )) (N 2 − 1)2 C1 j=1 θ=1

≥

N−1 72N max{A, B} N−1 ( ∑ h(j))( ∑ h(θ)a(vθ )) 2 2 (N − 1) j=1 θ=1

=

12N max{A, B} N−1 ∑ h(θ)a(vθ ) N2 − 1 θ=1

≥

12N max{A, B} N−1 1 ∑ h(θ)a( k(θ)‖v‖) 2 B N −1 θ=1

≥

1 12N max{A, B} N−1 ( ∑ h(θ))a( ‖v‖) BN N2 − 1 θ=1

≥ 2N max{A, B}

1 a(‖v‖) ≥ 2a(‖v‖), BN

∀ i = 1, N − 1.

Then we conclude that a(‖v‖) = a(supi=0,N vi ) ≥ a(v1 ) ≥ 2a(‖v‖), and hence a(‖v‖) = 0. By (K3) (a), we obtain ‖v‖ = 0.

(2.14)

40 � 2 Systems of difference equations with uncoupled BCs Therefore, by (2.13) and (2.14), we deduce that ‖(u, v)‖Y = 0, which is a contradiction. Hence, (𝒬1 (u, v), 𝒬2 (u, v)) ≰ (u, v) for all (u, v) ∈ 𝜕Br1 ∩ 𝒫 . By Theorem 2.1.3 (b), we conclude that the fixed point index i(𝒬, Br1 ∩ 𝒫 , 𝒫 ) = 0.

(2.15)

On the other hand, by (K4) we deduce that there exist C4 > 0, C5 > 0, and C6 > 0 such that f (n, u, v) ≤ C4 vα1 + C5 , α2

g(n, u, v) ≤ ε1 u + C6 ,

∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ ,

∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ ,

(2.16)

with ε1 = min{

α

2 3 6α2 +1 ) , }. B(N 2 − 1) 4AC4 (N 2 − 1) 8B(AC4 )α2 (N 2 − 1)α2 +1

6

(

Then, by (2.16) we have N−1

N−1

i=1

i=1

α

𝒬1n (u, v) = ∑ 𝒢1 (n, i)f (i, ui , vi ) ≤ ∑ Ah(i)(C4 vi 1 + C5 ) N−1

α

= AC4 ∑ h(i)vi 1 + Ξ1 , N−1

i=1

∀ n = 0, N, N−1

𝒬2n (u, v) = ∑ 𝒢2 (n, i)g(i, ui , vi ) ≤ ∑ i=1

N−1

i=1

α

= Bε1 ∑ h(i)ui 2 + Ξ2 , i=1

α Bh(i)(ε1 ui 2

(2.17) + C6 )

∀ n = 0, N,

where Ξ1 = AC5 (N 2 − 1)/6 and Ξ2 = BC6 (N 2 − 1)/6. We consider now the functions p̃, q̃ : ℝ+ → ℝ+ defined by p̃(w) = q̃(w) = Because limw→∞

̃(w) p w

α

α

2 1 AC4 (N 2 − 1) 3w [( ) + Ξ2 ] + Ξ1 , 2 6 4AC4 (N − 1)

α

2 AC4 (N 2 − 1) α1 6α2 w + Ξ ) ( + Ξ2 . 1 6 8(AC4 (N 2 − 1))α2

= limw→∞

q̃(w) w

ists R1 > r1 such that p̃(w) ≤

1 w, 4

={

0, if α1 α2 < 1, we conclude that there ex1/8, if α1 α2 = 1,

q̃(w) ≤

1 w, 4

∀ w ≥ R1 .

(2.18)

2.1 Nonnegative nonlinearities

� 41

We will show that (u, v) ≰ (𝒬1 (u, v), 𝒬2 (u, v)) for all (u, v) ∈ 𝜕BR1 ∩𝒫 . We suppose that there exists (u, v) ∈ 𝜕BR1 ∩ 𝒫 , that is, ‖(u, v)‖Y = R1 , such that (u, v) ≤ (𝒬1 (u, v), 𝒬2 (u, v)). So, by (2.17) we obtain N−1

α

∀ n = 0, N,

α

∀ n = 0, N.

un ≤ 𝒬1n (u, v) ≤ AC4 ∑ h(i)vi 1 + Ξ1 , i=1

N−1

vn ≤ 𝒬2n (u, v) ≤ Bε1 ∑ h(i)ui 2 + Ξ2 , i=1

Then for all n = 0, N we deduce that N−1

N−1

i=1

j=1

un ≤ AC4 ∑ h(i)(Bε1 ∑

α h(j)uj 2

α1

+ Ξ2 ) + Ξ1 α1

N−1 N2 − 1 α (Bε1 ∑ h(j)uj 2 + Ξ2 ) + Ξ1 = AC4 6 j=1 α1

N−1 N2 − 1 (Bε1 ∑ h(j)‖u‖α2 + Ξ2 ) + Ξ1 ≤ AC4 6 j=1

= AC4 ≤ AC4 ≤ AC4

α

1 N2 − 1 N 2 − 1 α2 (Bε1 ‖u‖ + Ξ2 ) + Ξ1 6 6

α

α

α

α

2 1 N2 − 1 3‖u‖ [( ) + Ξ ] + Ξ1 2 6 4AC4 (N 2 − 1) 2 1 3‖(u, v)‖Y N2 − 1 [( ) + Ξ ] + Ξ1 2 6 4AC4 (N 2 − 1)

(2.19)

and N−1

α

vn ≤ Bε1 ∑ h(i)ui 2 + Ξ2 i=1

N−1

N−1

i=1

j=1

α

α2

≤ Bε1 ∑ h(i)(AC4 ∑ h(j)vj 1 + Ξ1 ) + Ξ2 ≤ Bε1 ≤

α

2 N 2 − 1 α1 N −1 (AC4 ‖v‖ + Ξ1 ) + Ξ2 6 6

2

α

2 6α2 N 2 − 1 α1 (AC ‖v‖ + Ξ ) + Ξ2 4 1 6 8(AC4 (N 2 − 1))α2

α

≤

2 N 2 − 1 󵄩󵄩 6α2 󵄩󵄩α1 (AC (u, v) + Ξ ) + Ξ2 . 󵄩 󵄩 4 1 󵄩Y 6 󵄩 8(AC4 (N 2 − 1))α2

(2.20)

42 � 2 Systems of difference equations with uncoupled BCs By using (2.19), (2.20), and (2.18), we conclude that un ≤ 41 ‖(u, v)‖Y and vn ≤ 41 ‖(u, v)‖Y for all n = 0, N. Therefore, we obtain ‖(u, v)‖Y ≤ 21 ‖(u, v)‖Y , so ‖(u, v)‖Y = 0, which is a contradiction, because ‖(u, v)‖Y = R1 > 0. So, (u, v) ≰ (𝒬1 (u, v), 𝒬2 (u, v)) for all (u, v) ∈ 𝜕BR1 ∩ 𝒫 . By Theorem 2.1.3 (a), we deduce that the fixed point index i(𝒬, BR1 ∩ 𝒫 , 𝒫 ) = 1.

(2.21)

Because 𝒬 has no fixed points on 𝜕Br1 ∪ 𝜕BR1 , by (2.15) and (2.21), we conclude that i(𝒬, (BR1 \ Br1 ) ∩ 𝒫 , 𝒫 ) = i(𝒬, BR1 ∩ 𝒫 , 𝒫 ) − i(𝒬, Br1 ∩ 𝒫 , 𝒫 ) = 1. So, the operator 𝒬 has at least one fixed point (u1 , v1 ) ∈ (BR1 \ Br1 ) ∩ 𝒫 , with r1 < ‖(u1 , v1 )‖Y < R1 , that is, ‖u1 ‖ > 0 or ‖v1 ‖ > 0. Because u1 ∈ 𝒫1 and v1 ∈ 𝒫2 , we obtain un1 > 0 for all n = 1, N − 1 or v1n > 0 for all n = 1, N − 1. Theorem 2.1.5. Assume that (K1), (K2), (K5), and (K6) hold. Then the problem (2.1), (2.2) has at least one positive solution. Proof. By (K5) there exist Ci > 0, i = 7, . . . , 11, such that f (n, u, v) ≥ C7 c(v) − C8 ,

g(n, u, v) ≥ C9 d(u) − C10 ,

∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ ,

(2.22)

and c(C12 d(u)) ≥ where C12 = max{

C9 (N−1) h(i), N

72C12 N 3 max{A, B}u − C11 , C7 C9 (N 2 − 1)2

∀ u ∈ ℝ+ ,

(2.23)

i = 1, N − 1} > 0. Then we obtain

N−1

N−1

i=1

i=1

𝒬1n (u, v) = ∑ 𝒢1 (n, i)f (i, ui , vi ) ≥ ∑ 𝒢1 (n, i)(C7 c(vi ) − C8 ) N−1

N−1

≥ ∑ h(i)k(n)(C7 c(vi ) − C8 ) ≥ ∑ h(i)k(1)(C7 c(vi ) − C8 ) i=1

=

i=1

N−1

1 ∑ h(i)(C7 c(vi ) − C8 ), N i=1

N−1

N−1

i=1

i=1

∀ n = 1, N − 1, (2.24)

𝒬2n (u, v) = ∑ 𝒢2 (n, i)g(i, ui , vi ) ≥ ∑ 𝒢2 (n, i)(C9 d(ui ) − C10 ) N−1

N−1

≥ ∑ h(i)k(n)(C9 d(ui ) − C10 ) ≥ ∑ h(i)k(1)(C9 d(ui ) − C10 ) i=1

=

N−1

1 ∑ h(i)(C9 d(ui ) − C10 ), N i=1

i=1

∀ n = 1, N − 1.

2.1 Nonnegative nonlinearities

� 43

We will prove that the set U = {(u, v) ∈ 𝒫 , (u, v) = 𝒬(u, v) + λ(φ1 , φ2 ), λ ≥ 0} is bounded, where (φ1 , φ2 ) ∈ 𝒫 \ {(0, 0)}. Indeed, (u, v) ∈ U implies that u ≥ 𝒬1 (u, v), v ≥ 𝒬2 (u, v) for some φ1 , φ2 ≥ 0. By (2.24), we obtain un ≥ 𝒬1n (u, v) ≥ vn ≥ 𝒬2n (u, v) ≥

C7 N−1 ∑ h(i)c(vi ) − C13 , N i=1

C9 N−1 ∑ h(i)d(ui ) − C14 , N i=1

∀ n = 1, N − 1,

(2.25)

∀ n = 1, N − 1,

(2.26)

where C13 = C8 (N 2 − 1)/(6N), C14 = C10 (N 2 − 1)/(6N). By the monotonicity and concavity of c(⋅) and the Jensen inequality, inequality (2.26) implies that c(vn + C14 ) ≥ c( ≥ = ≥ =

C9 N−1 ∑ h(i)d(ui )) N i=1

1 N−1 C9 (N − 1) h(i)d(ui )) ∑ c( N − 1 i=1 N

1 N−1 C9 (N − 1) h(i) ⋅ C12 d(ui )) ∑ c( N − 1 i=1 NC12 1 N−1 C9 (N − 1) h(i)c(C12 d(ui )) ∑ N − 1 i=1 NC12 C9 N−1 ∑ h(i)c(C12 d(ui )), NC12 i=1

∀ n = 1, N − 1.

(2.27)

Since c(vn ) ≥ c(vn + C14 ) − c(C14 ), by the relations (2.25), (2.26), and (2.27), we deduce that un ≥ ≥

C7 N−1 ∑ h(i)c(vi ) − C13 N i=1

C7 N−1 ∑ h(i)[c(vi + C14 ) − c(C14 )] − C13 N i=1

=

C7 N−1 ∑ h(i)c(vi + C14 ) − C15 N i=1

≥

C N−1 C7 N−1 ∑ h(i)[ 9 ∑ h(j)c(C12 d(uj ))] − C15 N i=1 NC12 j=1

=

C7 C9 (N 2 − 1) N−1 ∑ h(j)c(C12 d(uj )) − C15 6N 2 C12 j=1

44 � 2 Systems of difference equations with uncoupled BCs

≥

C7 C9 (N 2 − 1) N−1 72C12 N 3 max{A, B} h(j)( uj − C11 ) − C15 ∑ 6N 2 C12 C7 C9 (N 2 − 1)2 j=1

=

12N max{A, B} N−1 ∑ h(j)uj − C16 ≥ 2‖u‖ − C16 , N2 − 1 j=1

∀ n = 1, N − 1,

where C15 =

C7 c(C14 )(N 2 − 1) + C13 , 6N

C7 C9 C11 (N 2 − 1)2 + C15 . 36N 2 C12

C16 =

Therefore, ‖u‖ ≥ u1 ≥ 2‖u‖ − C16 , and hence ‖u‖ ≤ C16 .

(2.28)

1 1 Since c(vn ) ≥ c( B1 k(n)‖v‖) ≥ c( BN ‖v‖) ≥ BN c(‖v‖) for all n = 1, N − 1, by the relations (2.22), (2.25), (2.26), and (2.27), we obtain

c(vn ) ≥ c(vn + C14 ) − c(C14 ) ≥ ≥

C9 N−1 ∑ h(i)c(C12 d(ui )) − c(C14 ) NC12 i=1

C9 N−1 72C12 N 3 max{A, B} ui − C11 ) − c(C14 ) ∑ h(i)( NC12 i=1 C7 C9 (N 2 − 1)2

=

72N 2 max{A, B} N−1 ∑ h(i)ui − C17 C7 (N 2 − 1)2 i=1

≥

C N−1 72N 2 max{A, B} N−1 ∑ h(i)( 7 ∑ h(j)c(vj ) − C13 ) − C17 2 2 N j=1 C7 (N − 1) i=1

=

12N max{A, B} N−1 ∑ h(j)c(vj ) − C18 N2 − 1 j=1

≥

12N max{A, B} N−1 1 c(‖v‖) − C18 ∑ h(j) BN N2 − 1 j=1

≥ 2c(‖v‖) − C18 ,

∀ n = 1, N − 1,

where C17 =

C9 C11 (N 2 − 1) + c(C14 ), 6NC12

C18 =

12C13 N 2 max{A, B} + C17 . C7 (N 2 − 1)

Then c(‖v‖) ≥ c(v1 ) ≥ 2c(‖v‖) − C18 , so c(‖v‖) ≤ C18 . By (K5) (a) and (c), we deduce that limv→∞ c(v) = ∞. Thus, there exists C19 > 0 such that

2.1 Nonnegative nonlinearities

‖v‖ ≤ C19 .

� 45

(2.29)

By (2.28) and (2.29), we conclude that ‖(u, v)‖Y ≤ C16 + C19 for all (u, v) ∈ U. That is, the set U is bounded. Then there exists a sufficiently large R2 > 0 such that (u, v) ≠ 𝒬(u, v) + λ(φ1 , φ2 ) for all (u, v) ∈ 𝜕BR2 ∩ 𝒫 and λ ≥ 0. By Theorem 2.1.2 we deduce that i(𝒬, BR2 ∩ 𝒫 , 𝒫 ) = 0.

(2.30)

On the other hand, by (K6) there exist C20 > 0 and a sufficiently small r2 > 0 (r2 < R2 , r2 ≤ 1) such that f (n, u, v) ≤ C20 vβ1 , β2

g(n, u, v) ≤ ε2 u ,

∀ (n, u) ∈ {1, . . . , N − 1} × ℝ+ , v ∈ [0, r2 ], ∀ (n, v) ∈ {1, . . . , N − 1} × ℝ+ , u ∈ [0, r2 ],

(2.31)

2

where ε2 = (2ABβ1 C20 ( N 6−1 )β1 +1 )−1/β1 > 0. We will show that (u, v) ≰ Q(u, v) for all (u, v) ∈ 𝜕Br2 ∩ 𝒫 . We suppose that there exists (u, v) ∈ 𝜕Br2 ∩ 𝒫 , that is, ‖(u, v)‖Y = r2 ≤ 1, such that (u, v) ≤ (𝒬1 (u, v), 𝒬2 (u, v)), or u ≤ 𝒬1 (u, v) and v ≤ 𝒬2 (u, v). Then by (2.31) we obtain N−1

N−1

β

un ≤ 𝒬1n (u, v) = ∑ 𝒢1 (n, i)f (i, ui , vi ) ≤ AC20 ∑ h(i)vi 1 i=1

N−1

N−1

i=1

j=1

β1

i=1

≤ AC20 ∑ h(i)( ∑ 𝒢2 (i, j)g(j, uj , vj )) N−1

N−1

i=1

j=1

≤ AC20 ∑ h(i)(B ∑

β h(j)ε2 uj 2 )

β1

β1

β

ABβ1 C20 (N 2 − 1)ε2 1 N−1 ≤ ( ∑ h(j)) ‖u‖β1 β2 6 j=1 = ABβ1 C20 (

β1 +1

N2 − 1 ) 6

β

≤ ABβ1 C20 ε2 1 (

β

ε2 1 ‖u‖β1 β2

β1 +1

N2 − 1 ) 6

1 ‖u‖ = ‖u‖, 2

∀ n = 0, N.

Therefore, ‖u‖ ≤ 21 ‖u‖, so ‖u‖ = 0. In addition, N−1

vn ≤ 𝒬2n (u, v) = ∑ 𝒢2 (n, i)g(i, ui , vi ) i=1

(2.32)

46 � 2 Systems of difference equations with uncoupled BCs N−1

β

≤ B ∑ h(i)ε2 ui 2 ≤ i=1

Bε2 (N 2 − 1) β2 ‖u‖ , 6

∀ n = 0, N.

(2.33)

By (2.32) and (2.33) we deduce that ‖v‖ = 0, and then ‖(u, v)‖Y = 0, which is a contradiction, because ‖(u, v)‖Y = r2 > 0. Then (u, v) ≰ 𝒬(u, v) for all (u, v) ∈ 𝜕Br2 ∩ 𝒫 . By Theorem 2.1.3 (a) we conclude that i(𝒬, Br2 ∩ 𝒫 , 𝒫 ) = 1.

(2.34)

Because 𝒬 has no fixed points on 𝜕Br2 ∪ 𝜕BR2 , by (2.30) and (2.34) we deduce that i(𝒬, (BR2 \ Br2 ) ∩ 𝒫 , 𝒫 ) = i(𝒬, BR2 ∩ 𝒫 , 𝒫 ) − i(𝒬, Br2 ∩ 𝒫 , 𝒫 ) = −1. So, the operator 𝒬 has at least one fixed point (u2 , v2 ) ∈ (BR2 \ Br2 ) ∩ 𝒫 , with r2 < ‖(u2 , v2 )‖Y < R2 , which is a positive solution for problem (2.1), (2.2). Theorem 2.1.6. Assume that assumptions (K1), (K2), (K3), (K5), and (K7) hold. Then the problem (2.1), (2.2) has at least two positive solutions. Proof. By using (K7), for any (u, v) ∈ 𝜕BN0 ∩ 𝒫 , we obtain N−1

𝒬1n (u, v) ≤ A ∑ h(i)f (i, N0 , N0 ) i=1


N0 such that

2.2 Sign-changing nonlinearities and positive parameters

i(𝒬, Br1 ∩ 𝒫 , 𝒫 ) = 0,

i(𝒬, BR2 ∩ 𝒫 , 𝒫 ) = 0.

� 47

(2.36)

Because 𝒬 has no fixed points on 𝜕Br1 ∪ 𝜕BR2 ∪ 𝜕BN0 , by the relations (2.35) and (2.36), we obtain i(𝒬, (BR2 \ BN0 ) ∩ 𝒫 , 𝒫 ) = i(𝒬, BR2 ∩ 𝒫 , 𝒫 ) − i(𝒬, BN0 ∩ 𝒫 , 𝒫 ) = −1,

i(𝒬, (BN0 \ Br1 ) ∩ 𝒫 , 𝒫 ) = i(𝒬, BN0 ∩ 𝒫 , 𝒫 ) − i(𝒬, Br1 ∩ 𝒫 , 𝒫 ) = 1.

Then 𝒬 has at least one fixed point (u1 , v1 ) ∈ (BR2 \ BN0 ) ∩ 𝒫 and at least one fixed point (u2 , v2 ) ∈ (BN0 \ Br1 ) ∩ 𝒫 . Therefore, problem (2.1), (2.2) has two distinct positive solutions (u1 , v1 ), (u2 , v2 ). Remark 2.1.1. In (K3), if a(v) = vp with p ≤ 1 and b(u) = uq with q > 0, the condition from (K3) (c) is satisfied if pq < 1. In (K5), if c(v) = vp with p ≤ 1 and d(u) = uq with q > 0, the condition from (K5) (c) is satisfied if pq > 1. n Examples. (1) We consider f (n, u, v) = n+1 (1 + e−(u+v) ) and g(n, u, v) = (1 + e−n )uθ for (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ . For a(v) = vp with p ≤ 1 and b(u) = uq for q > 0 and pq < 1, assumptions (K3) and (K4) are satisfied if q > θ and α2 > θ. For example, if θ = 45 , p = 31 , q = 43 , α1 = 31 , and α2 = 3, we can apply Theorem 2.1.4, and we deduce that the problem (2.1), (2.2) has at least one positive solution. (2) We consider f (n, u, v) = (1 + e−u )vθ1 and g(n, u, v) = (1 + e−v )uθ2 for (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ . For c(v) = vp with p ≤ 1, d(u) = uq for q > 0, and pq > 1, assumptions (K5) and (K6) are satisfied if p < θ1 , q < θ2 , β1 < θ1 , and β2 < θ2 . For example, if θ1 = 4, θ2 = 2, p = 35 , q = 95 , β1 = 3, and β2 = 31 , we can apply Theorem 2.1.5, and we conclude that the problem (2.1), (2.2) has at least one positive solution.

Remark 2.1.2. The results presented in this section were published in the paper [3].

2.2 Sign-changing nonlinearities and positive parameters We consider in this section the system of nonlinear second-order difference equations Δ2 un−1 + λf (n, un , vn ) = 0, { 2 Δ vn−1 + μg(n, un , vn ) = 0,

n = 1, N − 1,

(2.37)

n = 1, N − 1,

subject to the uncoupled multi-point boundary conditions p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi , i=1

r

v0 = ∑ ci vζi , i=1

l

vN = ∑ di vρi , i=1

(2.38)

where N ∈ ℕ, N ≥ 2, p, q, r, l ∈ ℕ, ξi ∈ ℕ for all i = 1, p, ηi ∈ ℕ for all i = 1, q, ζi ∈ ℕ for all i = 1, r, ρi ∈ ℕ for all i = 1, l, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1,

48 � 2 Systems of difference equations with uncoupled BCs 1 ≤ ζ1 < ⋅ ⋅ ⋅ < ζr ≤ N − 1, 1 ≤ ρ1 < ⋅ ⋅ ⋅ < ρl ≤ N − 1, λ and μ are positive parameters, and f and g are sign-changing functions. We will present some assumptions on the functions f and g and intervals for the parameters λ and μ such that problem (2.37), (2.38) has positive solutions. So, problem (2.37), (2.38) is a semipositone discrete boundary value problem. By a positive solution of (2.37), (2.38) we mean a pair of sequences (u, v) = ((un )n=0,N , (vn )n=0,N ) satisfying (2.37) and (2.38) with un ≥ 0 and vn ≥ 0 for all n = 0, N, and un > 0 and vn > 0 for all n = 1, N − 1. In this section we will use the auxiliary results from Section 2.1.1. In the proof of the main result we will apply the nonlinear alternative of Leray–Schauder type (Theorem 1.2.1). 2.2.1 Main result We present the assumptions that we shall use in this section. (H1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, ci ≥ 0, ζi ∈ ℕ for all i = 1, r, 1 ≤ ζ1 < ⋅ ⋅ ⋅ < ζr ≤ N − 1, di ≥ 0, ρi ∈ ℕ for all i = 1, l, 1 ≤ ρ1 < ⋅ ⋅ ⋅ < ρl ≤ N − 1, and p q ∑i=1 ai < 1, ∑i=1 bi < 1, ∑ri=1 ci < 1, ∑li=1 di < 1, and λ, μ > 0. (H2) The functions f , g : {1, . . . , N − 1} × ℝ+ × ℝ+ → ℝ are continuous, and there exist sn ≥ 0, tn ≥ 0, n = 1, N − 1, such that f (n, u, v) ≥ −sn

and

g(n, u, v) ≥ −tn ,

∀ n = 1, N − 1, u, v ∈ ℝ+ .

(H3) We have f (n, 0, 0) > 0 and g(n, 0, 0) > 0 for all n = 1, N − 1. We consider now the system of nonlinear second-order difference equations Δ2 xn−1 + λ(f (n, [xn − pn ]∗ , [yn − qn ]∗ ) + sn ) = 0,

{

n = 1, N − 1,

Δ2 yn−1 + μ(g(n, [xn − pn ]∗ , [yn − qn ]∗ ) + tn ) = 0,

n = 1, N − 1,

(2.39)

with the multi-point boundary conditions p

q

x0 = ∑ ai xξi ,

xN = ∑ bi xηi ,

i=1

i=1

r

y0 = ∑ ci yζi , i=1

l

yN = ∑ di yρi , i=1

(2.40)

where z∗n = zn if zn ≥ 0, z∗n = 0 if zn < 0 for n = 1, N − 1, and sn , tn , n = 1, N − 1, are given in (H2). Here ((pn )n=0,N , (qn )n=0,N ) is the solution of the system of second-order difference equations Δ2 un−1 + λsn = 0,

n = 1, N − 1,

Δ vn−1 + μtn = 0,

n = 1, N − 1,

{

2

(2.41)

2.2 Sign-changing nonlinearities and positive parameters

� 49

with the boundary conditions p

u0 = ∑ ai uξi , i=1

q

uN = ∑ bi uηi , i=1

r

v0 = ∑ ci vζi , i=1

l

vN = ∑ di vρi . i=1

(2.42)

By using Lemma 2.1.1 and the Green functions 𝒢1 , 𝒢2 from Section 2.1.1, the solution of problem (2.41), (2.42) is given by N−1

{ { pn = λ ∑ 𝒢1 (n, i)si , { { { { i=1 { N−1 { { { { {qn = μ ∑ 𝒢2 (n, i)ti , i=1 {

n = 0, N, n = 0, N.

Under assumptions (H1) and (H2), we have pn ≥ 0 and qn ≥ 0 for all n = 0, N. We shall prove that there exists a solution ((xn )n=0,N , (yn )n=0,N ) for the boundary value problem (2.39), (2.40) with xn ≥ pn and yn ≥ qn for all n = 0, N, and xn > pn and yn > qn for all n = 1, N − 1. In this case, ((un )n=0,N , (vn )n=0,N ), where un = xn − pn and vn = yn − qn for all n = 0, N, represents a positive solution of the boundary value problem (2.37), (2.38). Indeed, by (2.39)–(2.42), we find Δ2 un−1 = Δ2 (xn−1 − pn−1 ) = Δ2 xn−1 − Δ2 pn−1 = −λf (n, [xn − pn ]∗ , [yn − qn ]∗ ) − λsn + λsn 2

= −λf (n, un , vn ),

∀ n = 1, N − 1,

2

Δ vn−1 = Δ (yn−1 − qn−1 ) = Δ2 yn−1 − Δ2 qn−1 = −μg(n, [xn − pn ]∗ , [yn − qn ]∗ ) − μtn + μtn = −μg(n, un , vn ),

∀ n = 1, N − 1,

p

p

p

p

i=1 q

i=1 q

i=1 q

i=1 q

i=1 r

r

u0 = x0 − p0 = ∑ ai xξi − ∑ ai pξi = ∑ ai (xξi − pξi ) = ∑ ai uξi , uN = xN − pN = ∑ bi xηi − ∑ bi pηi = ∑ bi (xηi − pηi ) = ∑ bi uηi , i=1 r

i=1

r

i=1

v0 = y0 − q0 = ∑ ci yζi − ∑ ci qζi = ∑ ci (yζi − qζi ) = ∑ ci vζi , i=1

i=1

l

l

i=1

l

l

i=1

i=1

i=1

i=1

i=1

vN = yN − qN = ∑ di yρi − ∑ di qρi = ∑ di (yρi − qρi ) = ∑ di vρi . Hence, in what follows, we shall study the boundary value problem (2.39), (2.40). By using Lemma 2.1.1, the pair ((xn )n=0,N , (yn )n=0,N ) is a solution of problem (2.39), (2.40) if and only if ((xn )n=0,N , (yn )n=0,N ) is a solution of the system

50 � 2 Systems of difference equations with uncoupled BCs N−1

{ { xn = λ ∑ 𝒢1 (n, i)(f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + si , { { { { i=1 { N−1 { { { ∗ ∗ { {yn = μ ∑ 𝒢2 (n, i)(g(i, [xi − pi ] , [yi − qi ] ) + ti , i=1 {

n = 0, N, n = 0, N.

We consider the Banach space X = ℝN+1 = {x = (x0 , x1 , . . . , xN ), xi ∈ ℝ, i = 0, N}, with the maximum norm ‖ ⋅ ‖, ‖x‖ = maxn=0,N |xn |, and the Banach space Y = X × X with the norm ‖(x, y)‖Y = ‖x‖ + ‖y‖. We define the cones 1 k(n)‖x‖, ∀ n = 0, N} ⊂ X, A 1 𝒫2 = {y ∈ X, y = (yn )n=0,N , yn ≥ k(n)‖y‖, ∀ n = 0, N} ⊂ X, B 𝒫1 = {x ∈ X, x = (xn )n=0,N , xn ≥

and 𝒫 = 𝒫1 × 𝒫2 ⊂ Y , where A and B are defined in Section 2.1.1 (Lemma 2.1.3 and the corresponding one for 𝒢2 ). For λ, μ > 0 we introduce the operators ℱ1 , ℱ2 : Y → X and ℱ : Y → Y defined by ℱ1 (x, y) = (ℱ1n (x, y))n=0,N , ℱ2 (x, y) = (ℱ2n (x, y))n=0,N , N−1

ℱ1n (x, y) = λ ∑ 𝒢1 (n, i)(f (i, [xi − pi ] , [yi − qi ] ) + si ), ∗

∗

i=1

N−1

ℱ2n (x, y) = μ ∑ 𝒢2 (n, i)(g(i, [xi − pi ] , [yi − qi ] ) + ti ), ∗

∗

i=1

n = 0, N, n = 0, N,

and ℱ (x, y) = (ℱ1 (x, y), ℱ2 (x, y)) for (x, y) = ((xn )n=0,N , (yn )n=0,N ). We deduce that ((xn )n=0,N , (yn )n=0,N ) is a solution of problem (2.39), (2.40) if and only if ((xn )n=0,N , (yn )n=0,N ) is a fixed point of operator ℱ in the space Y . Lemma 2.2.1. If (H1) and (H2) hold, then the operator ℱ : 𝒫 → 𝒫 is a completely continuous operator. Proof. Let (x, y) ∈ 𝒫 be an arbitrary element. Because ℱ1 (x, y) and ℱ2 (x, y) satisfy the problem (2.4), (2.5) for zn = λ(f (n, [xn − pn ]∗ , [yn − qn ]∗ ) + sn ), n = 1, N − 1, and the problem (2.8), (2.9) for ̃zn = μ(g(n, [xn − pn ]∗ , [yn − qn ]∗ ) + tn ), n = 1, N − 1, by Lemma 2.1.4 and the corresponding one for problem (2.8), (2.9), we obtain 1 k(n) max ℱ1m (x, y) = A m=0,N 1 ℱ2n (x, y) ≥ k(n) max ℱ2m (x, y) = B m=0,N ℱ1n (x, y) ≥

1 󵄩 󵄩 k(n)󵄩󵄩󵄩ℱ1 (x, y)󵄩󵄩󵄩, A 1 󵄩 󵄩 k(n)󵄩󵄩󵄩ℱ2 (x, y)󵄩󵄩󵄩, B

∀ n = 0, N, ∀ n = 0, N.

Then ℱ1 (x, y) ∈ 𝒫1 , ℱ2 (x, y) ∈ 𝒫2 , so (ℱ1 (x, y), ℱ2 (x, y)) ∈ 𝒫 . Hence, ℱ (𝒫 ) ⊂ 𝒫 . By using standard arguments, we conclude that the operators ℱ1 and ℱ2 are completely continuous, and therefore ℱ is a completely continuous operator.

2.2 Sign-changing nonlinearities and positive parameters

� 51

Theorem 2.2.1. We suppose that (H1), (H2), and (H3) hold. Then there exist constants λ1 > 0 and μ1 > 0 such that for any λ ∈ (0, λ1 ] and μ ∈ (0, μ1 ], the boundary value problem (2.37), (2.38) has at least one positive solution. Proof. Let τ0 ∈ (0, 1) be fixed. By (H2) and (H3) there exists r0 ∈ (0, 1) such that f (n, u, v) ≥ τ0 f (n, 0, 0),

g(n, u, v) ≥ τ0 g(n, 0, 0),

∀ n = 1, N − 1, u, v ∈ [0, r0 ].

(2.43)

We define f0 = g0 = Λ1 =

max

{f (n, u, v) + sn },

max

{g(n, u, v) + tn },

n=1,N−1, u,v∈[0,r0 ] n=1,N−1, u,v∈[0,r0 ]

A(N 2 − 1) , 6

Λ2 =

B(N 2 − 1) , 6

λ1 =

r0 , 4f0 Λ1

μ1 =

r0 . 4g0 Λ2

We see that f0 ≥ max {τ0 f (n, 0, 0) + sn } > 0, n=1,N−1

g0 ≥ max {τ0 g(n, 0, 0) + tn } > 0. n=1,N−1

We will show that for any λ ∈ (0, λ1 ] and μ ∈ (0, μ1 ], the problem (2.37), (2.38) has at least one positive solution. For this, let λ ∈ (0, λ1 ] and μ ∈ (0, μ1 ] be arbitrary, but fixed for the moment. We define the set V = {(x, y) ∈ 𝒫 , x = (xn )n=0,N , y = (yn )n=0,N , ‖(x, y)‖Y < r0 }. We assume that there exist (x, y) ∈ 𝜕V (‖(x, y)‖Y = r0 or ‖x|| + ‖y‖ = r0 ) and ν ∈ (0, 1) such that (x, y) = νℱ (x, y), so x = νℱ1 (x, y) and y = νℱ2 (x, y). We obtain [xi − pi ]∗ = xi − pi ≤ xi ≤ r0 , [xi − pi ] = 0, ∗

if xi − pi < 0, ∀ i = 0, N,

[yi − qi ] = yi − qi ≤ yi ≤ r0 , ∗

[yi − qi ] = 0, ∗

if xi − pi ≥ 0,

if yi − qi ≥ 0,

if yi − qi < 0, ∀ i = 0, N.

Then by Lemma 2.1.3 and the corresponding one for 𝒢2 , for all n = 0, N we find xn = νℱ1n (x, y) ≤ ℱ1n (x, y) N−1

= λ ∑ 𝒢1 (n, i)(f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + si ) i=1

N−1

N−1

i=1

i=1

≤ λf0 ∑ Ah(i) ≤ λ1 f0 A ∑ h(i) = λ1 f0 A

N 2 − 1 r0 = , 6 4

52 � 2 Systems of difference equations with uncoupled BCs yn = νℱ2n (x, y) ≤ ℱ2n (x, y) N−1

= μ ∑ 𝒢2 (n, i)(g(i, [xi − pi ]∗ , [yi − qi ]∗ ) + ti ) i=1

N−1

N−1

i=1

i=1

≤ μg0 ∑ Bh(i) ≤ μ1 g0 B ∑ h(i) = μ1 g0 B r

r

N 2 − 1 r0 = . 6 4 r

Then ‖x‖ ≤ 40 and ‖y‖ ≤ 40 . So, r0 = ‖(x, y)‖Y = ‖x‖ + ‖y‖ ≤ 20 , which is a contradiction. Hence, by Lemma 2.2.1 and Theorem 1.2.1, we deduce that ℱ has a fixed point (x 1 , y1 ) ∈ V . That is, (x 1 , y1 ) = ℱ (x 1 , y1 ) or x 1 = ℱ1 (x 1 , y1 ), y1 = ℱ2 (x 1 , y1 ) (xn1 = ℱ1n (x 1 , y1 ), y1n = ℱ2n (x 1 , y1 ), n = 0, N, x 1 = (xn1 )n=0,N , y1 = (y1n )n=0,N ), and ‖x 1 ‖ + ‖y1 ‖ ≤ r0 with xn1 ≥ 1 k(n)‖x 1 ‖ A

and y1n ≥ B1 k(n)‖y1 ‖ for all n = 0, N. Besides, by (2.43), we conclude that N−1

xn1 = ℱ1n (x 1 , y1 ) ≥ λ ∑ 𝒢1 (n, i)(τ0 f (i, 0, 0) + si ) N−1

i=1

≥ λ ∑ 𝒢1 (n, i)si = pn , i=1

N−1

xn1 > λ ∑ 𝒢1 (n, i)si = pn , i=1

∀ n = 0, N, ∀ n = 1, N − 1,

N−1

y1n = ℱ2n (x 1 , y1 ) ≥ μ ∑ 𝒢2 (n, i)(τ0 g(i, 0, 0) + ti ) N−1

i=1

≥ μ ∑ 𝒢2 (n, i)ti = qn , i=1

N−1

y1n > μ ∑ 𝒢2 (n, i)ti = qn , i=1

∀ n = 0, N, ∀ n = 1, N − 1.

Let u1 = (un1 )n=0,N , v1 = (v1n )n=0,N , un1 = xn1 − pn , v1n = y1n − qn for all n = 0, N. Then un1 ≥ 0, v1n ≥ 0 for all n = 0, N, and un1 > 0, v1n > 0 for all n = 1, N − 1. Hence, (u1 , v1 ) is a positive solution of (2.37), (2.38).

2.2.2 An example Let N = 20, p = 2, q = 3, r = 1, l = 2, a1 = 1/2, a2 = 1/3, ξ1 = 4, ξ2 = 16, b1 = 1/3, b2 = 1/4, b3 = 1/5, η1 = 5, η2 = 10, η3 = 15, c1 = 3/4, ζ1 = 10, d1 = 1/3, d2 = 1/5, ρ1 = 3, and ρ2 = 18. We consider the system of second-order difference equations Δ2 un−1 + λf (n, un , vn ) = 0, { 2 Δ vn−1 + μg(n, un , vn ) = 0,

n = 1, 19, n = 1, 19,

(2.44)

2.2 Sign-changing nonlinearities and positive parameters

� 53

with the multi-point boundary conditions { {u0 = { { {v0 =

1 u + 2 4 3 v , 4 10

1 u , 3 16 v20

1 1 1 u20 = u5 + u10 + u15 , 3 4 5 1 1 = v3 + v18 . 3 5

(2.45)

47 671 < 1, ∑1i=1 ci = 43 < 1, and ∑2i=1 di = 158 < 1, d1 = 180 , We have ∑2i=1 ai = 65 < 1, ∑3i=1 bi = 60 d2 = 147 . Hence, assumption (H1) is satisfied. In addition, we obtain A ≈ 6.11028316 and 20 B ≈ 3.47845805. We consider the functions

f (n, u, v) = (u + v)3 + cos u,

g(n, u, v) = (u + v)1/5 + cos v,

∀ n = 1, 19, u, v ∈ ℝ+ .

(2.46)

We have here sn = tn = 1 for all n = 1, 19, and f (n, u, v) ≥ cos u ≥ −1, g(n, u, v) ≥ cos v ≥ −1 for all n = 1, 19 and u, v ≥ 0. Hence, assumption (H2) is satisfied. Assumption (H3) is also satisfied because f (n, 0, 0) = 1 > 0 and g(n, 0, 0) = 1 > 0 for all n = 1, 19. We consider τ0 = 1/2 < 1 and r0 = 1. Then f (n, u, v) ≥ τ0 f (n, 0, 0) = 21 and g(n, u, v) ≥ τ0 g(n, 0, 0) = 21 for all n = 1, 19 and u, v ∈ [0, 1]. Besides, we obtain f0 = g0 = In addition, we find Λ1 = 1 266g0 B

max

{f (n, u, v) + sn } ≈ 9.54030231,

max

{g(n, u, v) + tn } ≈ 3.01750724.

n=1,19, u,v∈[0,1] n=1,19, u,v∈[0,1] 133A , 2

Λ2 =

133B , 2

and hence λ1 =

1 266f0 A

≈ 6.45 ⋅ 10−5 and μ1 =

≈ 35.82 ⋅ 10−5 . By Theorem 2.2.1, for any λ ∈ (0, λ1 ] and μ ∈ (0, μ1 ], we deduce that problem (2.44), (2.45) with the nonlinearities (2.46) has at least one positive solution ((un )n=0,20 , (vn )n=0,20 ).

3 Systems of second-order finite difference equations with nonnegative nonlinearities, coupled multi-point boundary conditions, and positive parameters In this chapter we investigate the existence and nonexistence of positive solutions for two systems of nonlinear second-order difference equations subject to coupled multipoint boundary conditions, with positive parameters in the systems or in the boundary conditions. The nonlinearities of the systems are nonnegative functions and satisfy various assumptions.

3.1 Positive parameters in the system of difference equations In this section we consider the system of nonlinear second-order difference equations Δ2 un−1 + λf (n, un , vn ) = 0,

{ 2 Δ vn−1 + μg(n, un , vn ) = 0,

n = 1, N − 1, n = 1, N − 1,

(3.1)

subject to the coupled multi-point boundary conditions u0 = 0,

p

uN = ∑ ai vξi , i=1

v0 = 0,

q

vN = ∑ bi uηi , i=1

(3.2)

where N ∈ ℕ, N ≥ 2, p, q ∈ ℕ, ai ∈ ℝ, ξi ∈ ℕ for all i = 1, p, bi ∈ ℝ, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, and 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1. We will present sufficient conditions on the parameters λ, μ and on the functions f , g such that positive solutions of (3.1), (3.2) exist. In the proof of the main existence results we apply the Guo–Krasnosel’skii fixed point theorem (Theorem 1.1.1). In addition, some nonexistence results for the positive solutions of the above problem are also given. By a positive solution of problem (3.1), (3.2) we mean a pair of sequences (u, v) = ((un )n=0,N , (vn )n=0,N ) satisfying (3.1) and (3.2) with un ≥ 0, vn ≥ 0 for all n = 0, N and (u, v) ≠ (0, 0).

3.1.1 Preliminary results In this section, we present some auxiliary results related to the following system of second-order difference equations: https://doi.org/10.1515/9783111040370-003

3.1 Positive parameters in the system of difference equations

Δ2 un−1 + xn = 0, { 2 Δ vn−1 + yn = 0,

n = 1, N − 1,

� 55

(3.3)

n = 1, N − 1,

with the coupled multi-point boundary conditions (3.2), where xi , yi ∈ ℝ for all i = p q 1, N − 1. We denote d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ). Lemma 3.1.1. If d ≠ 0, then the unique solution of (3.3), (3.2) is given by N−1 n {N ∑ (N − j)xj d j=1

n−1

un = − ∑ (n − j)xj + j=1

p

ξi −1

i=1

j=1

p

N−1

i=1

j=1

− N ∑ ai ( ∑ (ξi − j)yj ) + (∑ ai ξi )( ∑ (N − j)yj ) p

q

ηi −1

i=1

i=1

j=1

− (∑ ai ξi )[∑ bi ( ∑ (ηi − j)xj )]}, (3.4)

n−1

N−1 n vn = − ∑ (n − j)yj + {N ∑ (N − j)yj d j=1 j=1 q

ηi −1

i=1

j=1

q

N−1

i=1

j=1

− N ∑ bi ( ∑ (ηi − j)xj ) + (∑ bi ηi )( ∑ (N − j)xj ) q

p

ξi −1

i=1

i=1

j=1

− (∑ bi ηi )[∑ ai ( ∑ (ξi − j)yj )]}, for all n = 0, N. Proof. For system (3.3), for u1 and v1 fixed, we obtain n−1

{ { un = nu1 − (n − 1)u0 − ∑ (n − j)xj , { { { { j=1 { n−1 { { { { {vn = nv1 − (n − 1)v0 − ∑ (n − j)yj , j=1 {

n = 2, N, n = 2, N,

and for u0 = v0 = 0 (by (3.2)), we deduce that n−1

un = nu1 − ∑ (n − j)xj , j=1

p

n−1

vn = nv1 − ∑ (n − j)yj , j=1

q

n = 2, N.

By (3.2), we have uN = ∑i=1 ai vξi , vN = ∑i=1 bi uηi . From the above relations, we conclude that

56 � 3 Systems of difference equations with coupled BCs ξ −1

p

N−1 i { { { Nu − ∑ (N − j)xj = ∑ ai [ξi v1 − ∑ (ξi − j)yj ], { 1 { { { j=1 i=1 j=1 { η q { N−1 i −1 { { { {Nv1 − ∑ (N − j)yj = ∑ bi [ηi u1 − ∑ (ηi − j)xj ], { j=1 i=1 j=1 {

or equivalently p

ξ −1

p

N−1 i { { { Nu − ( a ξ )v = (N − j)x − a ( (ξi − j)yj ), ∑ ∑ ∑ ∑ { 1 i i 1 j i { { { i=1 j=1 i=1 j=1 { ηi −1 q q { N−1 { { { { − (∑ bi ηi )u1 + Nv1 = ∑ (N − j)yj − ∑ bi ( ∑ (ηi − j)xj ). { i=1 j=1 i=1 j=1 {

p

(3.5)

The above system in the unknowns u1 and v1 has the determinant d = N 2 − q

(∑i=1 ai ξi )(∑i=1 bi ηi ). By the assumptions of this lemma, we have d ≠ 0. System (3.5) has a unique solution, namely,

u1 =

p

ξ −1

N−1 i 1 {N ∑ (N − j)xj − N ∑ ai ( ∑ (ξi − j)yj ) d j=1 i=1 j=1 p

N−1

i=1

j=1

+ (∑ ai ξi )( ∑ (N − j)yj ) p

q

ηi −1

i=1

j=1

− (∑ ai ξi )[∑ bi ( ∑ (ηi − j)xj )]}, i=1

v1 =

q

N−1

η −1

i 1 {N ∑ (N − j)yj − N ∑ bi ( ∑ (ηi − j)xj ) d j=1 i=1 j=1

q

N−1

i=1

j=1

+ (∑ bi ηi )( ∑ (N − j)xj ) q

p

ξi −1

i=1

i=1

j=1

− (∑ bi ηi )[∑ ai ( ∑ (ξi − j)yj )]}.

Therefore, we deduce expression (3.4) for the solution of problem (3.3), (3.2). Here

we consider that ∑0i=1 zi = 0 and ∑−1 i=1 zi = 0.

Lemma 3.1.2. If d ≠ 0, the solution ((un )n=0,N , (vn )n=0,N ) of problem (3.3), (3.2) can be

written as

3.1 Positive parameters in the system of difference equations N−1

N−1

j=1

j=1

N−1

N−1

j=1

j=1

un = ∑ G1 (n, j)xj + ∑ G2 (n, j)yj ,

� 57

n = 0, N, (3.6)

vn = ∑ G3 (n, j)yj + ∑ G4 (n, j)xj ,

n = 0, N,

where G1 (n, j) = g0 (n, j) + G2 (n, j) =

p

p

q

n (∑ a ξ )(∑ bi g0 (ηi , j)), d i=1 i i i=1

nN ∑ a g (ξ , j), d i=1 i 0 i q

p

n G3 (n, j) = g0 (n, j) + (∑ bi ηi )(∑ ai g0 (ξi , j)), d i=1 i=1 G4 (n, j) =

(3.7)

q

nN ∑ b g (η , j), d i=1 i 0 i

and g0 (n, j) =

1 j(N − n), { N n(N − j),

1 ≤ j ≤ n ≤ N, 0 ≤ n ≤ j ≤ N − 1,

for all n = 0, N and j = 1, N − 1. Proof. By Lemma 3.1.1 and relation (3.4), we have un =

N−1 N−1 1 n−1 [ ∑ j(N − n)xj + ∑ n(N − j)xj − ∑ n(N − j)xj ] N j=1 j=1 j=n p

+

ξ −1

N−1 i n {N ∑ (N − j)xj − N ∑ ai ( ∑ (ξi − j)yj ) d j=1 i=1 j=1 p

N−1

i=1

j=1

+ (∑ ai ξi )( ∑ (N − j)yj ) p

q

ηi −1

i=1

i=1

j=1

− (∑ ai ξi )[∑ bi ( ∑ (ηi − j)xj )]} =

N−1 1 n−1 [ ∑ j(N − n)xj + ∑ n(N − j)xj ] N j=1 j=n

+

N−1 1 {−N 2 ∑ n(N − j)xj Nd j=1

(3.8)

58 � 3 Systems of difference equations with coupled BCs p

q

N−1

i=1

j=1

+ (∑ ai ξi )(∑ bi ηi )( ∑ n(N − j)xj ) i=1

N−1

p

ξi −1

j=1

i=1

j=1

+ nN 2 ∑ (N − j)xj − nN 2 ∑ ai ( ∑ (ξi − j)yj ) p

N−1

i=1

j=1

+ nN(∑ ai ξi )( ∑ (N − j)yj ) p

q

i=1

i=1

ηi −1

− nN(∑ ai ξi )[∑ bi ( ∑ (ηi − j)xj )]} n−1

=

j=1

N−1

1 [ ∑ j(N − n)xj + ∑ n(N − j)xj ] N j=1 j=n p

+

q

N−1 n {(∑ ai ξi )[∑ bi ηi ( ∑ (N − j)xj ) Nd i=1 i=1 j=1 ηi −1

q

− ∑ bi ( ∑ (ηi − j)xj )] i=1

j=1

p

ξi −1

N−1

i=1

j=1

j=1

+ N ∑ ai [− ∑ N(ξi − j)yj + ∑ ξi (N − j)yj ]}. Because q

η −1

q

N−1 i 1 {∑ bi ηi [ ∑ (N − j)xj ] − ∑ bi [ ∑ N(ηi − j)xj ]} N i=1 j=1 i=1 j=1 η −1

q

=

N−1 i 1 ∑ bi [ ∑ ηi (N − j)xj − ∑ N(ηi − j)xj ] N i=1 j=1 j=1

=

N−1 i 1 ∑ bi [ ∑ j(N − ηi )xj + ∑ ηi (N − j)xj ] N i=1 j=1 j=η

η −1

q

i

q

N−1

i=1

j=1

= ∑ bi ( ∑ g0 (ηi , j)xj ) and ξ −1

p

N−1 i 1 {∑ ai [− ∑ N(ξi − j)yj + ∑ ξi (N − j)yj ]} N i=1 j=1 j=1 p

=

ξ −1

ξ −1

i i 1 {∑ ai [− ∑ N(ξi − j)yj + ∑ ξi (N − j)yj N i=1 j=1 j=1

3.1 Positive parameters in the system of difference equations

N−1

+ ∑ ξi (N − j)yj ]} j=ξi

ξ −1

p

=

N−1 i 1 {∑ ai [ ∑ j(N − ξi )yj + ∑ ξi (N − j)yj ]} N i=1 j=1 j=ξ i

p

N−1

i=1

j=1

= ∑ ai ( ∑ g0 (ξi , j)yj ), we obtain N−1

un = ∑ g0 (n, j)xj + j=1

p

q

N−1 n (∑ ai ξi )[∑ bi ( ∑ g0 (ηi , j)xj )] d i=1 i=1 j=1

p

+

N−1 nN [∑ ai ( ∑ g0 (ξi , j)yj )] d i=1 j=1

N−1

= ∑ g0 (n, j)xj + j=1

p

q

N−1 n (∑ ai ξi )[ ∑ (∑ bi g0 (ηi , j))xj ] d i=1 j=1 i=1

p

+

nN N−1 [ ∑ (∑ a g (ξ , j))yj ] d j=1 i=1 i 0 i

N−1

N−1

j=1

j=1

= ∑ G1 (n, j)xj + ∑ G2 (n, j)yj ,

n = 0, N,

where g0 and G1 , G2 are given in (3.8) and (3.7), respectively. For vn we deduce vn =

N−1 N−1 1 n−1 [ ∑ j(N − n)yj + ∑ n(N − j)yj − ∑ n(N − j)yj ] N j=1 j=1 j=n q

+

η −1

N−1 i n {N ∑ (N − j)yj − N ∑ bi ( ∑ (ηi − j)xj ) d j=1 i=1 j=1 q

N−1

i=1

j=1

+ (∑ bi ηi )( ∑ (N − j)xj ) q

p

ξi −1

i=1

i=1

j=1

− (∑ bi ηi )[∑ ai ( ∑ (ξi − j)yj )]} =

N−1 1 n−1 [ ∑ j(N − n)yj + ∑ n(N − j)yj ] N j=1 j=n

+

N−1 1 {−N 2 ∑ n(N − j)yj Nd j=1

� 59

60 � 3 Systems of difference equations with coupled BCs p

q

N−1

i=1

j=1

+ (∑ ai ξi )(∑ bi ηi ) ∑ n(N − j)yj i=1

N−1

q

ηi −1

j=1

i=1

j=1

+ nN 2 ∑ (N − j)yj − N 2 n ∑ bi ( ∑ (ηi − j)xj ) q

N−1

i=1

j=1

+ nN(∑ bi ηi )( ∑ (N − j)xj ) q

p

i=1

i=1

ξi −1

− nN(∑ bi ηi )[∑ ai ( ∑ (ξi − j)yj )]} n−1

=

j=1

N−1

1 [ ∑ j(N − n)yj + ∑ n(N − j)yj ] N j=1 j=n q

p

N−1 n {(∑ bi ηi )[∑ ai ξi ( ∑ (N − j)yj ) Nd i=1 i=1 j=1

+

ξi −1

p

− ∑ ai ( ∑ N(ξi − j)yj )] i=1

j=1

q

ηi −1

N−1

i=1

j=1

j=1

+ N ∑ bi [− ∑ N(ηi − j)xj + ∑ ηi (N − j)xj ]}. Because p

ξ −1

p

N−1 i 1 [∑ ai ξi ( ∑ (N − j)yj ) − ∑ ai ( ∑ N(ξi − j)yj )] N i=1 j=1 i=1 j=1 ξ −1

p

=

N−1 i 1 {∑ ai [ ∑ ξi (N − j)yj − ∑ N(ξi − j)yj ]} N i=1 j=1 j=1

=

N−1 i 1 ∑ ai [ ∑ j(N − ξi )yj + ∑ ξi (N − j)yj ] N i=1 j=1 j=ξ

ξ −1

p

i

p

N−1

i=1

j=1

= ∑ ai ( ∑ g0 (ξi , j)yj ) and η −1

q

N−1 i 1 {∑ bi [− ∑ N(ηi − j)xj + ∑ ηi (N − j)xj ]} N i=1 j=1 j=1 q

=

η −1

i 1 {∑ bi [− ∑ N(ηi − j)xj N i=1 j=1

3.1 Positive parameters in the system of difference equations ηi −1

� 61

N−1

+ ∑ ηi (N − j)xj + ∑ ηi (N − j)xj ]} j=1

j=ηi

ηi −1

q

=

N−1 1 [∑ bi ( ∑ j(N − ηi )xj + ∑ ηi (N − j)xj )] N i=1 j=1 j=η i

q

N−1

i=1

j=1

= ∑ bi ( ∑ g0 (ηi , j)xj ), we obtain N−1

vn = ∑ g0 (n, j)yj + j=1

q

p

N−1 n (∑ bi ηi )[∑ ai ( ∑ g0 (ξi , j)yj )] d i=1 i=1 j=1

q

+

N−1 nN [∑ bi ( ∑ g0 (ηi , j)xj )] d i=1 j=1

N−1

= ∑ g0 (n, j)yj + j=1

q

p

N−1 n (∑ bi ηi )[ ∑ (∑ ai g0 (ξi , j))yj ] d i=1 j=1 i=1

q

+

nN N−1 [ ∑ (∑ b g (η , j))xj ] d j=1 i=1 i 0 i

N−1

N−1

j=1

j=1

= ∑ G3 (n, j)yj + ∑ G4 (n, j)xj ,

n = 0, N,

where G3 , G4 are given in (3.7). Lemma 3.1.3 ([9], see also [35]). The function g0 given in (3.8) has the following properties: (a) g0 (n, j) ≥ 0, ∀ n = 0, N, j = 1, N − 1; (b) g0 (n, j) ≤ g0 (j, j) = j(N − j)/N, ∀ n = 0, N, j = 1, N − 1; c (c) For every c ∈ {1, . . . , [[N/2]]}, we have minn=c,N−c g0 (n, j) ≥ N−1 g0 (j, j) for all j = 1, N − 1, where [[N/2]] is the largest integer not greater than N/2. Lemma 3.1.4. If ai ≥ 0, ξi ∈ ℕ for all i = 1, p, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < p q ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) > 0, then the Green functions Gi , i = 1, 4, given by (3.7) satisfy Gi (n, j) ≥ 0 for all n = 0, N, j = 1, N − 1, i = 1, 4. Moreover, if xn ≥ 0, yn ≥ 0 for all n = 1, N − 1, then the solution ((un )n=0,N , (vn )n=0,N ) of problem (3.3), (3.2) satisfies un ≥ 0, vn ≥ 0 for all n = 0, N. Proof. By using the assumptions of this lemma, we have Gi (n, j) ≥ 0 for all n = 0, N, j = 1, N − 1, i = 1, 4, so un ≥ 0, vn ≥ 0 for all n = 0, N.

62 � 3 Systems of difference equations with coupled BCs Lemma 3.1.5. Assume that ai ≥ 0, ξi ∈ ℕ for all i = 1, p, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d > 0. Then the functions Gi , i = 1, 4, satisfy the following inequalities: (a1 ) G1 (n, j) ≤ I1 (j), for all n = 0, N, j = 1, N − 1, where I1 (j) = g0 (j, j) +

p

q

N (∑ a ξ )(∑ bi g0 (ηi , j)); d i=1 i i i=1

(a2 ) for every c ∈ {1, . . . , [[N/2]]}, we have minn=c,N−c G1 (n, j) ≥ (b1 ) G2 (n, j) ≤ I2 (j), ∀ n = 0, N, j = 1, N − 1, where I2 (j) =

for all j = 1, N − 1;

c I (j), N 2

for all j = 1, N − 1;

p

N2 ∑ a g (ξ , j); d i=1 i 0 i

(b2 ) for every c ∈ {1, . . . , [[N/2]]}, we have minn=c,N−c G2 (n, j) ≥ (c1 ) G3 (n, j) ≤ I3 (j), for all n = 0, N, j = 1, N − 1, where I3 (j) = g0 (j, j) +

c I (j), N 1

q

p

N (∑ b η )(∑ ai g0 (ξi , j)); d i=1 i i i=1

(c2 ) for every c ∈ {1, . . . , [[N/2]]}, we have minn=c,N−c G3 (n, j) ≥ (d1 ) G4 (n, j) ≤ I4 (j), ∀ n = 0, N, j = 1, N − 1, where I4 (j) =

c I (j), N 3

for all j = 1, N − 1;

c I (j), N 4

for all j = 1, N − 1.

q

N2 ∑ b g (η , j); d i=1 i 0 i

(d2 ) for every c ∈ {1, . . . , [[N/2]]}, we have minn=c,N−c G4 (n, j) ≥

Proof. Inequalities (a1 ), (b1 ), (c1 ), (d1 ) are evident. For the other inequalities, for c ∈ {1, . . . , [[N/2]]} and n = c, N − c, j = 1, N − 1, we deduce G1 (n, j) ≥

p

p

≥

q

c c g0 (j, j) + (∑ ai ξi )(∑ bi g0 (ηi , j)) N −1 d i=1 i=1 q

c N c [g (j, j) + (∑ ai ξi )(∑ bi g0 (ηi , j))] = I1 (j), N 0 d i=1 N i=1 p

G2 (n, j) ≥

cN c ∑ a g (ξ , j) = I2 (j), d i=1 i 0 i N

G3 (n, j) ≥

c c g (j, j) + (∑ bi ηi )(∑ ai g0 (ξi , j)) N −1 0 d i=1 i=1

q

p

3.1 Positive parameters in the system of difference equations q

≥ G4 (n, j) ≥

� 63

p

N c c [g0 (j, j) + (∑ bi ηi )(∑ ai g0 (ξi , j))] = I3 (j), N d i=1 N i=1 q

cN c ∑ bi g0 (ηi , j) = I4 (j). d i=1 N

Therefore, we obtain inequalities (a2 ), (b2 ), (c2 ), (d2 ) of this lemma. Lemma 3.1.6. Assume that ai ≥ 0, ξi ∈ ℕ for all i = 1, p, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, d > 0, c ∈ {1, . . . , [[N/2]]}, and xn , yn ≥ 0 for all n = 1, N − 1. Then the solution of problem (3.3), (3.2) satisfies the inequalities min un ≥

n=c,N−c

c max u , N m=0,N m

min vn ≥

n=c,N−c

c max v . N m=0,N m

Proof. For c ∈ {1, . . . , [[N/2]]}, n = c, N − c, m = 0, N, we have N−1

N−1

j=1

j=1

un = ∑ G1 (n, j)xj + ∑ G2 (n, j)yj N−1

≥

c c N−1 ∑ I1 (j)xj + ∑ I (j)y N j=1 N j=1 2 j

≥

c N−1 c N−1 c ∑ G1 (m, j)xj + ∑ G2 (m, j)yj = um , N j=1 N j=1 N N−1

N−1

j=1

j=1

vn = ∑ G3 (n, j)yj + ∑ G4 (n, j)xj N−1

≥

c c N−1 ∑ I3 (j)yj + ∑ I (j)xj N j=1 N j=1 4

≥

c N−1 c c N−1 ∑ G3 (m, j)yj + ∑ G (m, j)xj = vm . N j=1 N j=1 4 N

Then we deduce the conclusion of this lemma.

3.1.2 Existence of positive solutions In this section, we shall give sufficient conditions on λ, μ, f , and g such that positive solutions with respect to a cone for our problem (3.1), (3.2) exist. We present the assumptions that we shall use in the sequel. (A1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < p q ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) > 0. (A2) The functions f , g : {1, . . . , N − 1} × ℝ+ × ℝ+ → ℝ+ are continuous.

64 � 3 Systems of difference equations with coupled BCs For c ∈ {1, . . . , [[N/2]]}, we introduce the following extreme limits: f (n, u, v) , n=1,N−1 u + v

g(n, u, v) , n=1,N−1 u + v

f0s = lim sup max

g0s = lim sup max

f (n, u, v) f0i = lim inf+ min , u+v→0 n=c,N−c u + v

g0i = lim inf+ min

u+v→0+

s f∞ i f∞

f (n, u, v) , = lim sup max u+v→∞ n=1,N−1 u + v

f (n, u, v) = lim inf min , u+v→∞ n=c,N−c u + v

u+v→0+

u+v→0 n=c,N−c

g(n, u, v) , u+v

g(n, u, v) , n=1,N−1 u + v

s g∞ = lim sup max u+v→∞

i g∞

= lim inf min

u+v→∞ n=c,N−c

g(n, u, v) . u+v

In the definitions of the extreme limits above, the variables u and v are nonnegative. By using the functions Gi , i = 1, 4, from Section 3.1.1 (Lemma 3.1.2), our problem (3.1), (3.2) can be written equivalently as the following system: N−1

N−1

{ { un = λ ∑ G1 (n, i)f (i, ui , vi ) + μ ∑ G2 (n, i)g(i, ui , vi ), { { { { i=1 i=1 { N−1 N−1 { { { { {vn = μ ∑ G3 (n, i)g(i, ui , vi ) + λ ∑ G4 (n, i)f (i, ui , vi ), i=1 i=1 {

n = 0, N, n = 0, N.

We consider the Banach space X = ℝN+1 = {u = (u0 , u1 , . . . , uN ), ui ∈ ℝ, i = 0, N} with the maximum norm ‖ ⋅ ‖, ‖u‖ = maxn=0,N |un |, and the Banach space Y = X × X with the norm ‖(u, v)‖Y = ‖u‖ + ‖v‖. For c ∈ {1, . . . , [[N/2]]}, we define the cone P ⊂ Y by P = {(u, v) ∈ Y ; u = (un )n=0,N , v = (vn )n=0,N , un ≥ 0, vn ≥ 0, ∀ n = 0, N and

min (un + vn ) ≥

n=c,N−c

c 󵄩󵄩 󵄩 󵄩(u, v)󵄩󵄩󵄩Y }. N󵄩

For λ, μ > 0, we introduce the operators 𝒯1 , 𝒯2 : Y → X and 𝒯 : Y → Y defined by 𝒯1 (u, v) = (𝒯1n (u, v))n=0,N , 𝒯2 (u, v) = (𝒯2n (u, v))n=0,N , N−1

N−1

i=1

i=1

𝒯1n (u, v) = λ ∑ G1 (n, i)f (i, ui , vi ) + μ ∑ G2 (n, i)g(i, ui , vi ), N−1

N−1

i=1

i=1

T2n (u, v) = μ ∑ G3 (n, i)g(i, ui , vi ) + λ ∑ G4 (n, i)f (i, ui , vi ),

n = 0, N, n = 0, N,

and 𝒯 (u, v) = (𝒯1 (u, v), 𝒯2 (u, v)), (u, v) = ((un )n=0,N , (vn )n=0,N ) ∈ Y , where Gi , i = 1, 4, are the functions defined in Section 3.1.2 (relation (3.7)). The solutions of problem (3.1), (3.2) coincide with the fixed points of the operator 𝒯 in the space Y .

3.1 Positive parameters in the system of difference equations

� 65

Lemma 3.1.7. If (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, then 𝒯 : P → P is a completely continuous operator. Proof. Let (u, v) ∈ P be an arbitrary element. Because 𝒯1 (u, v) and 𝒯2 (u, v) satisfy problem (3.3), (3.2) for xn = λf (n, un , vn ), n = 1, N − 1, and yn = μg(n, un , vn ), n = 1, N − 1, by Lemma 3.1.6 we obtain c max 𝒯1m (u, v) = N n=c,N−c m=0,N c min 𝒯2n (u, v) ≥ max 𝒯 (u, v) = N m=0,N 2m n=c,N−c min 𝒯1n (u, v) ≥

c 󵄩󵄩 󵄩 󵄩𝒯 (u, v)󵄩󵄩󵄩, N󵄩 1 c 󵄩󵄩 󵄩 󵄩𝒯 (u, v)󵄩󵄩󵄩. N󵄩 2

Hence, we conclude min [𝒯1n (u, v) + 𝒯2n (u, v)]

n=c,N−c

≥ min 𝒯1n (u, v) + min 𝒯2n (u, v) n=c,N−c

n=c,N−c

c󵄩 󵄩 c󵄩 󵄩 c󵄩 󵄩 ≥ 󵄩󵄩󵄩𝒯1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒯2 (u, v)󵄩󵄩󵄩 = 󵄩󵄩󵄩𝒯 (u, v)󵄩󵄩󵄩Y . N N N By Lemma 3.1.4, (A1), and (A2), we obtain 𝒯1n (u, v) ≥ 0, 𝒯2n (u, v) ≥ 0 for all n = 0, N, so we deduce that 𝒯 (u, v) ∈ P. Hence, we get 𝒯 (P) ⊂ P. By using standard arguments, we can easily show that 𝒯1 and 𝒯2 are completely continuous, and then 𝒯 is a completely continuous operator. N−1 N−1 For c ∈ {1, . . . , [[N/2]]}, we denote A = ∑N−1 j=1 I1 (j), B = ∑j=1 I2 (j), C = ∑j=1 I3 (j), N−c N−c N−c N−c ̃ ̃ ̃ ̃ D = ∑N−1 j=1 I4 (j), A = ∑j=c I1 (j), B = ∑j=c I2 (j), C = ∑j=c I3 (j), D = ∑j=c I4 (j), where Ii , i = 1, 4, are defined in Section 3.1.1 (Lemma 3.1.5). i i For f0s , g0s , f∞ , g∞ ∈ (0, ∞) and numbers α1 , α2 ∈ [0, 1], α3 , α4 ∈ (0, 1), a ∈ [0, 1], and b ∈ (0, 1), we define the numbers

L1 = max{

bα (1 − b)α4 aα1 N 2 (1 − a)α2 N 2 , }, L2 = min{ s 3 , }, 2 i 2 i ̃ ̃ f f0s D c f A c f D 0A ∞

L3 = max{

2

∞

a(1 − α1 )N (1 − a)(1 − α2 )N 2 , }, ̃ c2 g i B c2 g i C̃ ∞

∞

b(1 − α3 ) (1 − b)(1 − α4 ) L4 = min{ , }. g0s B g0s C

i i Theorem 3.1.1. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0s , g0s , f∞ , g∞ ∈ (0, ∞), α1 , α2 ∈ [0, 1], α3 , α4 ∈ (0, 1), a ∈ [0, 1], b ∈ (0, 1), L1 < L2 , and L3 < L4 . Then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2).

66 � 3 Systems of difference equations with coupled BCs Proof. For c given in Theorem 3.1.1, we consider the above cone P ⊂ Y and the operators 𝒯1 , 𝒯2 , and 𝒯 . Let λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ), and let ε > 0 be a positive number such i i that ε < f∞ , ε < g∞ , and aα1 N 2

i − ε)A ̃ c2 (f∞

a(1 − α1 )N 2 ≤ μ, ̃ c2 (g i − ε)B

≤ λ,

∞

(1 − a)(1 − α2 )N 2 ≤ μ, c2 (g i − ε)C̃

(1 − a)α2 N 2 ≤ λ, ̃ c2 (f i − ε)D ∞

∞

b(1 − α3 ) ≥ μ, (g0s + ε)B

bα3 ≥ λ, (f0s + ε)A

(1 − b)α4 ≥ λ, (f0s + ε)D

(1 − b)(1 − α4 ) ≥ μ. (g0s + ε)C

By using (A2) and the definitions of f0s and g0s , we deduce that there exists K1 > 0 such that f (n, u, v) ≤ (f0s + ε)(u + v) and g(n, u, v) ≤ (g0s + ε)(u + v) for all n = 1, N − 1 and u, v ∈ ℝ+ with 0 ≤ u + v ≤ K1 . We define the set Ω1 = {(u, v) ∈ Y , ‖(u, v)‖Y < K1 }. Now let (u, v) ∈ P ∩ 𝜕Ω1 , that is, (u, v) ∈ P with ‖(u, v)‖Y = K1 , or equivalently ‖u‖ + ‖v‖ = K1 . Then un + vn ≤ K1 for all n = 0, N, and by Lemma 3.1.5 we obtain N−1

N−1

𝒯1n (u, v) = λ ∑ G1 (n, i)f (i, ui , vi ) + μ ∑ G2 (n, i)g(i, ui , vi ) i=1

N−1

N−1

i=1

i=1

i=1

≤ λ ∑ I1 (i)f (i, ui , vi ) + μ ∑ I2 (i)g(i, ui , vi ) N−1

N−1

≤ λ ∑ I1 (i)(f0s + ε)(ui + vi ) + μ ∑ I2 (i)(g0s + ε)(ui + vi ) i=1

i=1

N−1

N−1

≤ λ(f0s + ε) ∑ I1 (i)(‖u‖ + ‖v‖) + μ(g0s + ε) ∑ I2 (i)(‖u‖ + ‖v‖) i=1

󵄩 󵄩 = [λ(f0s + ε)A + μ(g0s + ε)B]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y 󵄩 󵄩 󵄩 󵄩 ≤ [bα3 + b(1 − α3 )]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = b󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ,

i=1

∀ n = 0, N.

Therefore, ‖𝒯1 (u, v)‖ ≤ b‖(u, v)‖Y . In a similar manner, we conclude N−1

N−1

𝒯2n (u, v) = μ ∑ G3 (n, i)g(i, ui , vi ) + λ ∑ G4 (n, i)f (i, ui , vi ) i=1

N−1

N−1

i=1

i=1

i=1

≤ μ ∑ I3 (i)g(i, ui , vi ) + λ ∑ I4 (i)f (i, ui , vi ) N−1

N−1

≤ μ ∑ I3 (i)(g0s + ε)(ui + vi ) + λ ∑ I4 (i)(f0s + ε)(ui + vi ) i=1

i=1

3.1 Positive parameters in the system of difference equations N−1

� 67

N−1

≤ μ(g0s + ε) ∑ I3 (i)(‖u‖ + ‖v‖) + λ(f0s + ε) ∑ I4 (i)(‖u‖ + ‖v‖) [μ(g0s

i=1

λ(f0s

i=1

󵄩 󵄩 = + ε)C + + ε)D]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y 󵄩 󵄩 󵄩 󵄩󵄩 ≤ [(1 − b)(1 − α4 ) + (1 − b)α4 ]󵄩󵄩(u, v)󵄩󵄩󵄩Y = (1 − b)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ,

∀ n = 0, N.

Therefore, ‖𝒯2 (u, v)‖ ≤ (1 − b)‖(u, v)‖Y . Then, for (u, v) ∈ P ∩ 𝜕Ω1 , we deduce 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝒯 (u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩𝒯1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒯2 (u, v)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ b󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + (1 − b)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

(3.9)

i i i −ε)(u+v) By the definitions of f∞ and g∞ , there exists K 2 > 0 such that f (n, u, v) ≥ (f∞ i and g(n, u, v) ≥ (g∞ − ε)(u + v) for all u, v ≥ 0 with u + v ≥ K 2 and n = c, N − c. We consider K2 = max{2K1 , K 2 N/c}, and we define Ω2 = {(u, v) ∈ Y , ‖(u, v)‖Y < K2 }. Then for (u, v) ∈ P with ‖(u, v)‖Y = K2 , we obtain

un + vn ≥ min (un + vn ) n=c,N−c

≥

c 󵄩󵄩 c 󵄩 󵄩(u, v)󵄩󵄩󵄩Y = K2 ≥ K 2 , N󵄩 N

∀ n = c, N − c.

Then, by Lemma 3.1.5, we conclude 𝒯1c (u, v) ≥

≥ ≥ ≥

μc N−1 λc N−1 ∑ I1 (i)f (i, ui , vi ) + ∑ I (i)g(i, ui , vi ) N i=1 N i=1 2

μc N−c λc N−c ∑ I1 (i)f (i, ui , vi ) + ∑ I (i)g(i, ui , vi ) N i=c N i=c 2

μc N−c λc N−c i i − ε)(ui + vi ) + − ε)(ui + vi ) ∑ I1 (i)(f∞ ∑ I (i)(g∞ N i=c N i=c 2

μc2 N−c λc2 N−c 󵄩 󵄩 󵄩 󵄩 i i − ε)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + 2 ∑ I2 (i)(g∞ − ε)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ∑ I1 (i)(f∞ 2 N i=c N i=c

2 λc2 i ̃ + μc (g i − ε)B] ̃ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 (f − ε) A 󵄩 󵄩Y N2 ∞ N2 ∞ 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 ≥ [aα1 + a(1 − α1 )]󵄩󵄩(u, v)󵄩󵄩Y = a󵄩󵄩(u, v)󵄩󵄩Y .

=[

So, ‖𝒯1 (u, v)‖ ≥ 𝒯1c (u, v) ≥ a‖(u, v)‖Y . In a similar manner, we deduce 𝒯2c (u, v) ≥

≥

μc N−1 λc N−1 ∑ I3 (i)g(i, ui , vi ) + ∑ I (i)f (i, ui , vi ) N i=1 N i=1 4

μc N−c λc N−c ∑ I3 (i)g(i, ui , vi ) + ∑ I (i)f (i, ui , vi ) N i=c N i=c 4

68 � 3 Systems of difference equations with coupled BCs

≥ ≥

μc N−c λc N−c i i − ε)(ui + vi ) + − ε)(ui + vi ) ∑ I3 (i)(g∞ ∑ I (i)(f∞ N i=c N i=c 4

μc2 N−c λc2 N−c 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 i i + I (i)(g (u, v) − ε) ∑ 󵄩 󵄩 3 ∞ 󵄩Y N 2 ∑ I4 (i)(f∞ − ε)󵄩󵄩(u, v)󵄩󵄩Y 󵄩 N 2 i=c i=c

2 μc2 i ̃ + λc (f i − ε)D] ̃ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 (g − ε) C ∞ 󵄩Y 󵄩 N2 N2 ∞ 󵄩 󵄩 󵄩 󵄩 ≥ [(1 − a)(1 − α2 ) + (1 − a)α2 ]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = (1 − a)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

=[

So, ‖𝒯2 (u, v)‖ ≥ 𝒯2c (u, v) ≥ (1 − a)‖(u, v)‖Y . Hence, for (u, v) ∈ P ∩ 𝜕Ω2 , we obtain 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝒯 (u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩𝒯1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒯2 (u, v)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≥ a󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + (1 − a)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

(3.10)

By using (3.9), (3.10), Lemma 3.1.1, and Theorem 1.1.1, we conclude that 𝒯 has a fixed point (u, v) ∈ P ∩ (Ω2 \ Ω1 ) such that K1 ≤ ‖u‖ + ‖v‖ ≤ K2 . i i We investigate now the extreme cases, where f0s , g0s could be 0, or f∞ , g∞ could be ∞. b 1−b b 1−b ′ ′ If we denote L2 = min{ f s A , f s D } and L4 = min{ g s B , g s C }, then by using similar arguments 0 0 0 0 as those used in the proof of Theorem 3.1.1, we obtain the following results. i i Theorem 3.1.2. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0s = 0, g0s , f∞ , g∞ ∈ ′ (0, ∞), α1 , α2 ∈ [0, 1], a ∈ [0, 1], b ∈ (0, 1), and L3 < L4 . Then for each λ ∈ (L1 , ∞) and μ ∈ (L3 , L′4 ), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). i i Theorem 3.1.3. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, g0s = 0, f0s , f∞ , g∞ ∈ ′ ′ (0, ∞), α1 , α2 ∈ [0, 1], a ∈ [0, 1], b ∈ (0, 1), and L1 < L2 . Then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , ∞), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2).

Theorem 3.1.4. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0s = g0s = 0, i i f∞ , g∞ ∈ (0, ∞), α1 , α2 ∈ [0, 1], a ∈ [0, 1]. Then for each λ ∈ (L1 , ∞) and μ ∈ (L3 , ∞), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). i Theorem 3.1.5. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0s , g0s , f∞ ∈ i s s i i s s i i (0, ∞), g∞ = ∞} or {f0 , g0 , g∞ ∈ (0, ∞), f∞ = ∞} or {f0 , g0 ∈ (0, ∞), f∞ = g∞ = ∞}, α3 , α4 ∈ (0, 1), b ∈ (0, 1). Then for each λ ∈ (0, L2 ) and μ ∈ (0, L4 ), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). i Theorem 3.1.6. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0s = 0, g0s , f∞ ∈ i s i s i s s i (0, ∞), g∞ = ∞} or {f0 = 0, f∞ = ∞, g0 , g∞ ∈ (0, ∞)} or {f0 = 0, g0 ∈ (0, ∞), f∞ = i g∞ = ∞}, b ∈ (0, 1). Then for each λ ∈ (0, ∞) and μ ∈ (0, L′4 ), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2).

3.1 Positive parameters in the system of difference equations

� 69

i Theorem 3.1.7. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0s , f∞ ∈ s i s i s i s (0, ∞), g0 = 0, g∞ = ∞} or {f0 , g∞ ∈ (0, ∞), g0 = 0, f∞ = ∞} or {f0 ∈ (0, ∞), g0s = i i 0, f∞ = g∞ = ∞}, b ∈ (0, 1). Then for each λ ∈ (0, L′2 ) and μ ∈ (0, ∞), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). i Theorem 3.1.8. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0s = g0s = 0, f∞ ∈ i s s i i s s i (0, ∞), g∞ = ∞} or {f0 = g0 = 0, f∞ = ∞, g∞ ∈ (0, ∞)} or {f0 = g0 = 0, f∞ = i g∞ = ∞}. Then for each λ ∈ (0, ∞) and μ ∈ (0, ∞), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2).

Remark 3.1.1. Each of Theorems 3.1.1–3.1.4 above contains nine subcases, because α1 , α2 can be 0, 1, or between 0 and 1. s s In what follows, for f0i , g0i , f∞ , g∞ ∈ (0, ∞) and numbers α1 , α2 ∈ [0, 1], α3 , α4 ∈ (0, 1), a ∈ [0, 1], b ∈ (0, 1), we define the numbers 2 2 ̃ 1 = max{ aα1 N , (1 − a)α2 N }, L ̃ 2 = min{ bα3 , (1 − b)α4 }, L s A s D i i 2 2 ̃ ̃ f∞ f∞ c f0 A c f0 D

2 2 ̃ 3 = max{ a(1 − α1 )N , (1 − a)(1 − α2 )N }, L i i 2 2 ̃ c g B c g C̃ 0

0

̃ 4 = min{ b(1 − α3 ) , (1 − b)(1 − α4 ) }. L s B s C g∞ g∞ s s Theorem 3.1.9. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0i , g0i , f∞ , g∞ ∈ ̃1 < L ̃ 2 , and L ̃3 < L ̃ 4 . Then (0, ∞), α1 , α2 ∈ [0, 1], α3 , α4 ∈ (0, 1), a ∈ [0, 1], b ∈ (0, 1), L ̃1, L ̃ 2 ) and μ ∈ (L ̃3, L ̃ 4 ), there exists a positive solution ((un ) for each λ ∈ (L n=0,N , (vn )n=0,N ) for (3.1), (3.2).

Proof. For c given in the theorem, we consider again the above cone P ⊂ Y and the ̃1, L ̃ 2 ) and μ ∈ (L ̃3, L ̃ 4 ), and let ε > 0 be a positive operators 𝒯1 , 𝒯2 , and 𝒯 . Let λ ∈ (L i i number such that ε < f0 , ε < g0 and aα1 N 2

̃ c2 (f0i − ε)A

≤ λ,

2

(1 − a)α2 N ≤ λ, ̃ c2 (f i − ε)D 0

bα3 ≥ λ, s + ε)A (f∞ (1 − b)α4 ≥ λ, s + ε)D (f∞

a(1 − α1 )N 2 ≤ μ, ̃ c2 (g i − ε)B 0

(1 − a)(1 − α2 )N 2 ≤ μ, c2 (g i − ε)C̃ 0

b(1 − α3 ) ≥ μ, s + ε)B (g∞ (1 − b)(1 − α4 ) ≥ μ. s + ε)C (g∞

By using (A2) and the definitions of f0i and g0i , we deduce that there exists K3 > 0 such that f (n, u, v) ≥ (f0i − ε)(u + v), g(n, u, v) ≥ (g0i − ε)(u + v) for all u, v ≥ 0 with 0 ≤ u + v ≤ K3

70 � 3 Systems of difference equations with coupled BCs and n = c, N − c. We denote Ω3 = {(u, v) ∈ Y , ‖(u, v)‖Y < K3 }. Let (u, v) ∈ P with ‖(u, v)‖Y = K3 , that is, ‖u‖ + ‖v‖ = K3 . Because un + vn ≤ ‖u‖ + ‖v‖ = K3 for all n = 0, N, by using Lemma 3.1.5 we obtain 𝒯1c (u, v) ≥

≥ ≥ ≥

μc N−1 λc N−1 ∑ I1 (i)f (i, ui , vi ) + ∑ I (i)g(i, ui , vi ) N i=1 N i=1 2

μc N−c λc N−c ∑ I1 (i)f (i, ui , vi ) + ∑ I (i)g(i, ui , vi ) N i=c N i=c 2

μc N−c λc N−c ∑ I1 (i)(f0i − ε)(ui + vi ) + ∑ I (i)(g0i − ε)(ui + vi ) N i=c N i=c 2

N−c N−c μc2 i λc2 i 󵄩󵄩 󵄩 󵄩󵄩 󵄩 (f − ε) I (i) (g − ε) (u, v) + ∑ ∑ I2 (i)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y 󵄩 󵄩 1 󵄩 󵄩Y N 2 0 N2 0 i=c i=c

2 λc2 i ̃ + μc (g i − ε)B] ̃ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 (f − ε) A 󵄩 󵄩Y N2 0 N2 0 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 ≥ [aα1 + a(1 − α1 )]󵄩󵄩(u, v)󵄩󵄩Y = a󵄩󵄩(u, v)󵄩󵄩Y .

=[

Therefore, ‖𝒯1 (u, v)‖ ≥ 𝒯1c (u, v) ≥ a‖(u, v)‖Y . In a similar manner, we conclude 𝒯2c (u, v) ≥

≥ ≥ ≥

μc N−1 λc N−1 ∑ I3 (i)g(i, ui , vi ) + ∑ I (i)f (i, ui , vi ) N i=1 N i=1 4

μc N−c λc N−c ∑ I3 (i)g(i, ui , vi ) + ∑ I (i)f (i, ui , vi ) N i=c N i=c 4

μc N−c λc N−c ∑ I3 (i)(g0i − ε)(ui + vi ) + ∑ I (i)(f0i − ε)(ui + vi ) N i=c N i=c 4

N−c N−c μc2 i λc2 i 󵄩󵄩 󵄩 󵄩󵄩 󵄩 (g − ε) I (i) (f − ε) (u, v) + ∑ ∑ I4 (i)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y 󵄩 󵄩 3 󵄩 󵄩Y N 2 0 N2 0 i=c i=c

2 μc2 i ̃ + λc (f i − ε)D] ̃ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 (g − ε) C 0 󵄩 󵄩Y N2 N2 0 󵄩 󵄩 󵄩 󵄩 ≥ [(1 − a)(1 − α2 ) + (1 − a)α2 ]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = (1 − a)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

=[

So, ‖𝒯2 (u, v)‖ ≥ 𝒯2c (u, v) ≥ (1 − a)‖(u, v)‖Y . Thus, for an arbitrary element (u, v) ∈ P ∩ 𝜕Ω3 , we deduce 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝒯 (u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩𝒯1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒯2 (u, v)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≥ a󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + (1 − a)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

(3.11)

Now we define the functions f ∗ , g ∗ : {1, . . . , N − 1} × ℝ+ → ℝ+ , f ∗ (n, x) = max0≤u+v≤x f (n, u, v), g ∗ (n, x) = max0≤u+v≤x g(n, u, v), x ∈ ℝ+ , n = 1, N − 1. Then

3.1 Positive parameters in the system of difference equations

� 71

f (n, u, v) ≤ f ∗ (n, x), g(n, u, v) ≤ g ∗ (n, x) for all n = 1, N − 1, u ≥ 0, v ≥ 0, and u + v ≤ x. The functions f ∗ (n, ⋅), g ∗ (n, ⋅) are nondecreasing for every n ∈ {1, . . . , N − 1}, and they satisfy the conditions f ∗ (n, x) s = f∞ , x n=1,N−1

lim sup max x→∞

g ∗ (n, x) s = g∞ . x n=1,N−1

lim sup max x→∞

Therefore, for ε > 0 there exists K 4 > 0 such that for all x ≥ K 4 and n = 1, N − 1, we have f ∗ (n, x) f ∗ (n, x) s ≤ lim sup max + ε = f∞ + ε, x x x→∞ n=1,N−1

g ∗ (n, x) g ∗ (n, x) s ≤ lim sup max + ε = g∞ + ε, x x x→∞ n=1,N−1 s s so f ∗ (n, x) ≤ (f∞ + ε)x and g ∗ (n, x) ≤ (g∞ + ε)x. We consider K4 = max{2K3 , K 4 } and we denote Ω4 = {(u, v) ∈ Y , ‖(u, v)‖Y < K4 }. Let (u, v) ∈ P ∩ 𝜕Ω4 . By the definitions of f ∗ and g ∗ , for any i = 1, N − 1 and n = 0, N, we obtain

󵄩 󵄩 f (i, un , vn ) ≤ f ∗ (i, un + vn ) ≤ f ∗ (i, ‖u‖ + ‖v‖) = f ∗ (i, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ), 󵄩 󵄩 g(i, un , vn ) ≤ g ∗ (i, un + vn ) ≤ g ∗ (i, ‖u‖ + ‖v‖) = g ∗ (i, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ). Then, for all n = 0, N, we deduce N−1

N−1

i=1

i=1

𝒯1n (u, v) ≤ λ ∑ I1 (i)f (i, ui , vi ) + μ ∑ I2 (i)g(i, ui , vi ) N−1

N−1

󵄩 󵄩 󵄩 󵄩 ≤ λ ∑ I1 (i)f ∗ (i, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ) + μ ∑ I2 (i)g ∗ (i, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ) i=1

i=1

N−1

N−1

󵄩 󵄩 󵄩 󵄩 s s ≤ λ(f∞ + ε) ∑ I1 (i)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + μ(g∞ + ε) ∑ I2 (i)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y s [λ(f∞

i=1

s μ(g∞

󵄩 󵄩 = + ε)A + + ε)B]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y 󵄩 󵄩 󵄩 󵄩 ≤ [bα3 + b(1 − α3 )]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = b󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

i=1

Therefore, ‖𝒯1 (u, v)‖ ≤ b‖(u, v)‖Y . In a similar manner, we conclude for all n = 0, N N−1

N−1

i=1

i=1

𝒯2n (u, v) ≤ μ ∑ I3 (i)g(i, ui , vi ) + λ ∑ I4 (i)f (i, ui , vi ) N−1

N−1

󵄩 󵄩 󵄩 󵄩 ≤ μ ∑ I3 (i)g ∗ (i, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ) + λ ∑ I4 (i)f ∗ (i, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ) i=1

i=1

72 � 3 Systems of difference equations with coupled BCs N−1

N−1

󵄩 󵄩 󵄩 󵄩 s s ≤ μ(g∞ + ε) ∑ I4 (i)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + ε) ∑ I3 (i)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + λ(f∞ s [μ(g∞

i=1

s λ(f∞

i=1

󵄩 󵄩 = + ε)C + + ε)D]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y 󵄩 󵄩 󵄩 󵄩 ≤ [(1 − b)(1 − α4 ) + (1 − b)α4 ]󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = (1 − b)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y . So, ‖𝒯2 (u, v)‖ ≤ (1 − b)‖(u, v)‖Y . Then for any (u, v) ∈ P ∩ 𝜕Ω4 , it follows that 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝒯 (u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩𝒯1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒯2 (u, v)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ b󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + (1 − b)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y .

(3.12)

By using (3.11), (3.12), Lemma 3.1.7, and Theorem 1.1.1, we deduce that 𝒯 has a fixed point (u, v) ∈ P ∩ (Ω4 \ Ω3 ) such that K3 ≤ ‖(u, v)‖Y ≤ K4 . s s We investigate now the extreme cases, where f∞ , g∞ could be 0, or f0i , g0i could be 1−b b b ′ ′ ̃ = min{ s , 1−b ̃ = min{ s , s } and L ∞. If we denote L s C }, then by using similar 4 2 f∞ A f∞ D g∞ B g∞ arguments as those used in the proof of Theorem 3.1.9, we obtain the following results. s Theorem 3.1.10. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0i , g0i , f∞ ∈ s ′ ′ ̃ ̃ ̃ ̃ (0, ∞), g∞ = 0, α1 , α2 ∈ [0, 1], a ∈ [0, 1], b ∈ (0, 1), L1 < L2 . Then for each λ ∈ (L1 , L2 ) ̃ 3 , ∞), there exists a positive solution ((un ) and μ ∈ (L n=0,N , (vn )n=0,N ) for (3.1), (3.2). s Theorem 3.1.11. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0i , g0i , g∞ ∈ s ′ ̃ ̃ ̃ (0, ∞), f∞ = 0, α1 , α2 ∈ [0, 1], a ∈ [0, 1], b ∈ (0, 1), and L3 < L4 . Then for each λ ∈ (L1 , ∞) ̃3, L ̃ ′ ), there exists a positive solution ((un ) and μ ∈ (L 4 n=0,N , (vn )n=0,N ) for (3.1), (3.2).

Theorem 3.1.12. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, f0i , g0i ∈ (0, ∞), s s ̃ 1 , ∞) and μ ∈ (L ̃ 3 , ∞), there f∞ = g∞ = 0, α1 , α2 ∈ [0, 1], a ∈ [0, 1]. Then for each λ ∈ (L exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). Theorem 3.1.13. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0i = ∞, g0i , s s s s s s f∞ , g∞ ∈ (0, ∞)} or {f0i , f∞ , g∞ ∈ (0, ∞), g0i = ∞} or {f0i = g0i = ∞, f∞ , g∞ ∈ (0, ∞)}, ̃ ̃ α3 , α4 ∈ (0, 1), b ∈ (0, 1). Then for each λ ∈ (0, L2 ) and μ ∈ (0, L4 ), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). s Theorem 3.1.14. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0i = ∞, g0i , f∞ ∈ s i s i s i i s s (0, ∞), g∞ = 0} or {f0 , f∞ ∈ (0, ∞), g0 = ∞, g∞ = 0} or {f0 = g0 = ∞, f∞ ∈ (0, ∞), g∞ = ̃ ′ ) and μ ∈ (0, ∞), there exists a positive solution 0}, b ∈ (0, 1). Then for each λ ∈ (0, L 2 ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). s Theorem 3.1.15. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0i = ∞, g0i , g∞ ∈ s i s i s i i s s (0, ∞), f∞ = 0} or {f0 , g∞ ∈ (0, ∞), g0 = ∞, f∞ = 0} or {f0 = g0 = ∞, f∞ = 0, g∞ ∈ ̃ ′ ), there exists a positive solution (0, ∞)}, b ∈ (0, 1). Then for each λ ∈ (0, ∞) and μ ∈ (0, L 4 ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2).

3.1 Positive parameters in the system of difference equations

� 73

Theorem 3.1.16. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}, {f0i = ∞, g0i ∈ s s s s s (0, ∞), f∞ = g∞ = 0} or {f0i ∈ (0, ∞), g0i = ∞, f∞ = g∞ = 0} or {f0i = g0i = ∞, f∞ = s g∞ = 0}. Then for each λ ∈ (0, ∞) and μ ∈ (0, ∞), there exists a positive solution ((un )n=0,N , (vn )n=0,N ) for (3.1), (3.2). 3.1.3 Nonexistence of positive solutions We present in this section intervals for λ and μ for which there exists no positive solution of problem (3.1), (3.2). s Theorem 3.1.17. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}. If f0s , f∞ , g0s , s g∞ < ∞, then there exist positive constants λ0 , μ0 such that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ), the boundary value problem (3.1), (3.2) has no positive solution. s s Proof. From the definitions of f0s , f∞ , g0s , g∞ , which are finite, we deduce that there exist M1 , M2 > 0 such that

f (i, u, v) ≤ M1 (u + v),

g(i, u, v) ≤ M2 (u + v),

∀ i = 1, N − 1, u, v ≥ 0.

We define λ0 = min{

1 1 , }, 4M1 A 4M1 D

μ0 = min{

1 1 , }, 4M2 B 4M2 C

N−1 N−1 N−1 where A = ∑N−1 j=1 I1 (j), B = ∑j=1 I2 (j), C = ∑j=1 I3 (j), D = ∑j=1 I4 (j). We shall show that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ), problem (3.1), (3.2) has no positive solution. Let λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ). We suppose that (3.1), (3.2) has a positive solution ((un )n=0,N , (vn )n=0,N ). Then, by using Lemma 3.1.5, we obtain N−1

N−1

un = 𝒯1n (u, v) = λ ∑ G1 (n, i)f (i, ui , vi ) + μ ∑ G2 (n, i)g(i, ui , vi ) i=1

N−1

i=1

N−1

≤ λ ∑ I1 (i)f (i, ui , vi ) + μ ∑ I2 (i)g(i, ui , vi ) i=1

N−1

i=1

N−1

≤ λM1 ∑ I1 (i)(ui + vi ) + μM2 ∑ I2 (i)(ui + vi ) i=1

N−1

i=1

N−1

≤ λM1 ∑ I1 (i)(‖u‖ + ‖v‖) + μM2 ∑ I2 (i)(‖u‖ + ‖v‖) i=1

󵄩 󵄩 = (λM1 A + μM2 B)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ,

i=1

∀ n = 0, N.

Therefore, we conclude 󵄩 󵄩 ‖u‖ ≤ (λM1 A + μM2 B)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y

74 � 3 Systems of difference equations with coupled BCs 1󵄩 󵄩 󵄩 󵄩 < (λ0 M1 A + μ0 M2 B)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ≤ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y . 2

(3.13)

In a similar manner, we obtain N−1

N−1

vn = 𝒯2n (u, v) = μ ∑ G3 (n, i)g(i, ui , vi ) + λ ∑ G4 (n, i)f (i, ui , vi ) i=1

N−1

i=1

N−1

≤ μ ∑ I3 (i)g(i, ui , vi ) + λ ∑ I4 (i)f (i, ui , vi ) i=1

i=1

N−1

N−1

≤ μM2 ∑ I3 (i)(ui + vi ) + λM1 ∑ I4 (i)(ui + vi ) i=1

i=1

N−1

N−1

≤ μM2 ∑ I3 (i)(‖u‖ + ‖v‖) + λM1 ∑ I4 (i)(‖u‖ + ‖v‖) i=1

󵄩 󵄩 = (μM2 C + λM1 D)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ,

i=1

∀ n = 0, N.

Therefore, we deduce 󵄩 󵄩 ‖v‖ ≤ (μM2 C + λM1 D)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y

1󵄩 󵄩 󵄩 󵄩 < (μ0 M2 C + λ0 M1 D)󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ≤ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y . 2

(3.14)

Hence, by (3.13) and (3.14), we conclude 󵄩󵄩 󵄩 󵄩󵄩(u, v)󵄩󵄩󵄩Y = ‖u‖ + ‖v‖
0 and f (i, u, v) > 0 for all i = c, N − c, u ≥ 0, v ≥ 0, u + v > 0, then there exists a positive constant λ̃0 such that for every λ > λ̃0 and μ > 0, the boundary value problem (3.1), (3.2) has no positive solution.

Proof. From the assumptions of the theorem, we deduce that there exists m1 > 0 such that f (i, u, v) ≥ m1 (u + v) for all i = c, N − c and u, v ≥ 0. We define N2 N2 , }, λ̃0 = min{ ̃ c2 m1 D ̃ c2 m1 A ̃ = ∑N−c I1 (i) and D ̃ = ∑N−c I4 (i). We shall show that for every λ > λ̃0 and μ > 0, where A i=c i=c problem (3.1), (3.2) has no positive solution. Let λ > λ̃0 and μ > 0. We suppose that (3.1), (3.2) has a positive solution ((un )n=0,N , (vn )n=0,N ).

3.1 Positive parameters in the system of difference equations

N2 ̃, c2 m1 A

̃ ≥ D, ̃ then λ̃0 = If A

� 75

and therefore we obtain N−1

N−1

uc = 𝒯1c (u, v) = λ ∑ G1 (c, i)f (i, ui , vi ) + μ ∑ G2 (c, i)g(i, ui , vi ) i=1

N−1

N−c

i=1

≥ λ ∑ G1 (c, i)f (i, ui , vi ) ≥ λ ∑ G1 (c, i)f (i, ui , vi ) i=1

i=c

N−c

≥ λm1 ∑ G1 (c, i)(ui + vi ) ≥ i=c

2

=

λcm1 N−c c ∑ I (i) (‖u‖ + ‖v‖) N i=c 1 N

̃󵄩 λc m1 A 󵄩󵄩(u, v)󵄩󵄩󵄩 . 󵄩 󵄩Y 2 N

Then, we conclude ̃󵄩 ̃󵄩 λ̃ c2 m A λc2 m1 A 󵄩 󵄩 󵄩 󵄩 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y > 0 2 1 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 N N

‖u‖ ≥ uc ≥

so ‖(u, v)‖Y = ‖u‖ + ‖v‖ ≥ ‖u‖ > ‖(u, v)‖Y , which is a contradiction. 2 ̃ < D, ̃ then λ̃0 = 2N , and therefore we deduce If A ̃ c m1 D

N−1

N−1

vc = 𝒯2c (u, v) = μ ∑ G3 (c, i)g(i, ui , vi ) + λ ∑ G4 (c, i)f (i, ui , vi ) i=1

N−1

N−c

i=1

≥ λ ∑ G4 (c, i)f (i, ui , vi ) ≥ λ ∑ G4 (c, i)f (i, ui , vi ) i=1

N−c

≥ λm1 ∑ G4 (c, i)(ui + vi ) ≥ 2

=

i=c

i=c

λcm1 N−c c ∑ I (i) (‖u‖ + ‖v‖) N i=c 4 N

̃󵄩 λc m1 D 󵄩󵄩(u, v)󵄩󵄩󵄩 . 󵄩 󵄩Y 2 N

Then, we conclude ‖v‖ ≥ vc ≥

̃󵄩 ̃󵄩 λ̃ c2 m D λc2 m1 D 󵄩 󵄩 󵄩 󵄩 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y > 0 2 1 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 N N

so ‖(u, v)‖Y = ‖u‖ + ‖v‖ ≥ ‖v‖ > ‖(u, v)‖Y , which is a contradiction. Therefore, the boundary value problem (3.1), (3.2) has no positive solution. i Theorem 3.1.19. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}. If g0i , g∞ > 0 and g(i, u, v) > 0 for all i = c, N − c, u ≥ 0, v ≥ 0, u + v > 0, then there exists a positive ̃ 0 such that for every μ > μ ̃ 0 and λ > 0, the boundary value problem (3.1), (3.2) constant μ has no positive solution.

76 � 3 Systems of difference equations with coupled BCs Proof. From the assumptions of the theorem, we deduce that there exists m2 > 0 such that g(i, u, v) ≥ m2 (u + v) for all i = c, N − c and u, v ≥ 0. We define ̃ 0 = min{ μ

N2

,

N2

̃ c2 m2 C̃ c2 m2 B

},

̃ = ∑N−c I2 (i) and C̃ = ∑N−c I3 (i). We shall show that for every μ > μ ̃ 0 and λ > 0, where B i=c i=c problem (3.1), (3.2) has no positive solution. ̃ 0 and λ > 0. We suppose that (3.1), (3.2) has a positive solution ((un )n=0,N , Let μ > μ (vn )n=0,N ). 2 ̃ then μ ̃ ≥ C, ̃ 0 = 2N , and therefore we obtain If B ̃ c m2 B

N−1

N−1

uc = 𝒯1c (u, v) = λ ∑ G1 (c, i)f (i, ui , vi ) + μ ∑ G2 (c, i)g(i, ui , vi ) i=1

N−1

N−c

i=1

≥ μ ∑ G2 (c, i)g(i, ui , vi ) ≥ μ ∑ G2 (c, i)g(i, ui , vi ) i=1

i=c

N−c

≥ μm2 ∑ G2 (c, i)(ui + vi ) ≥ i=c

2

=

μcm2 N−c c ∑ I (i) (‖u‖ + ‖v‖) N i=c 2 N

̃󵄩 μc m2 B 󵄩󵄩(u, v)󵄩󵄩󵄩 . 󵄩 󵄩Y 2 N

Then we conclude ‖u‖ ≥ uc ≥

̃󵄩 ̃󵄩 ̃ c2 m B μ μc2 m2 B 󵄩 󵄩 󵄩 󵄩 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y > 0 2 2 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 N N

so ‖(u, v)‖Y = ‖u‖ + ‖v‖ ≥ ‖u‖ > ‖(u, v)‖Y , which is a contradiction. 2 ̃ then μ ̃ < C, ̃ 0 = N , and therefore we deduce If B c2 m2 C̃

N−1

N−1

vc = T2c (u, v) = μ ∑ G3 (c, i)g(i, ui , vi ) + λ ∑ G4 (c, i)f (i, ui , vi ) i=1

N−1

N−c

i=1

≥ μ ∑ G3 (c, i)g(i, ui , vi ) ≥ μ ∑ G3 (c, i)g(i, ui , vi ) i=1

N−c

≥ μm2 ∑ G3 (c, i)(ui + vi ) ≥ i=c

μc2 m2 C̃ 󵄩󵄩 󵄩 = 󵄩(u, v)󵄩󵄩󵄩Y . N2 󵄩

i=c

μcm2 N−c c ∑ I3 (i) (‖u‖ + ‖v‖) N i=c N

Then we conclude ‖v‖ ≥ vc ≥

̃ 0 c2 m2 C̃ 󵄩 μ μc2 m2 C̃ 󵄩󵄩 󵄩󵄩 󵄩󵄩(u, v)󵄩󵄩󵄩 = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 , (u, v) > 󵄩 󵄩 󵄩 󵄩 󵄩Y 󵄩 󵄩Y Y N2 N2 󵄩

3.1 Positive parameters in the system of difference equations

� 77

so ‖(u, v)‖Y = ‖u‖ + ‖v‖ ≥ ‖v‖ > ‖(u, v)‖Y , which is a contradiction. Therefore, the boundary value problem (3.1), (3.2) has no positive solution. i Theorem 3.1.20. Assume that (A1) and (A2) hold and c ∈ {1, . . . , [[N/2]]}. If f0i , f∞ , g0i , i g∞ > 0 and f (i, u, v) > 0, g(i, u, v) > 0 for all i = c, N − c, u ≥ 0, v ≥ 0, u + v > 0, then there exist positive constants λ̂0 and μ̂ 0 such that for every λ > λ̂0 and μ > μ̂ 0 , the boundary value problem (3.1), (3.2) has no positive solution.

Proof. From the assumptions of the theorem, we deduce that there exist m1 , m2 > 0 such that f (i, u, v) ≥ m1 (u + v), g(i, u, v) ≥ m2 (u + v), for all i = c, N − c and u, v ≥ 0. We define λ̂0 =

N2

̃ 2c2 m1 A

,

μ̂ 0 =

N2

2c2 m2 C̃

,

̃ = ∑N−c I1 (i) and C̃ = ∑N−c I3 (i). Then for every λ > λ̂0 and μ > μ̂ 0 , problem (3.1), where A i=c i=c (3.2) has no positive solution. Indeed, let λ > λ̂0 and μ > μ̂ 0 . We suppose that (3.1), (3.2) has a positive solution ((un )n=0,N , (vn )n=0,N ). In a similar manner as that used in the proofs of Theorems 3.1.18 and 3.1.19, we obtain ‖u‖ ≥ uc ≥

̃󵄩 λc2 m1 A 󵄩󵄩(u, v)󵄩󵄩󵄩 , 󵄩 󵄩Y 2 N

‖v‖ ≥ vc ≥

μc2 m2 C̃ 󵄩󵄩 󵄩 󵄩(u, v)󵄩󵄩󵄩Y , N2 󵄩

so 2 ̃ ̃󵄩 λc2 m1 A 󵄩󵄩 󵄩 󵄩󵄩(u, v)󵄩󵄩󵄩 + μc m2 C 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 󵄩󵄩(u, v)󵄩󵄩󵄩Y = ‖u‖ + ‖v‖ ≥ 󵄩 󵄩 󵄩 󵄩Y Y 2 2 N N 2 ̂λ c2 m A ̃ ̃󵄩 μ̂ c m C 󵄩 󵄩 󵄩 > 0 2 1 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + 0 2 2 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y N N

1󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 2

which is a contradiction. Therefore, the boundary value problem (3.1), (3.2) has no positive solution. We can also define λ̂′0 =

N2

̃ 2c2 m1 D

,

μ̂ ′0 =

N2

̃ 2c2 m2 B

,

̃ = ∑N−c I2 (i) and D ̃ = ∑N−c I4 (i). Then for every λ > λ̂′ and μ > μ̂ ′ , problem (3.1), where B i=c i=c 0 0 (3.2) has no positive solution. Indeed, let λ > λ̂′0 and μ > μ̂ ′0 . We suppose that (3.1), (3.2) has a positive solution ((un )n=0,N , (vn )n=0,N ). In a similar manner as that used in the proofs of Theorems 3.1.18 and 3.1.19, we obtain ‖v‖ ≥ vc ≥

̃󵄩 λc2 m1 D 󵄩 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 N

‖u‖ ≥ uc ≥

̃󵄩 μc2 m2 B 󵄩 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 N

78 � 3 Systems of difference equations with coupled BCs so 2 ̃󵄩 ̃ μc2 m2 B 󵄩 󵄩󵄩 󵄩󵄩(u, v)󵄩󵄩󵄩 + λc m1 D 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩 󵄩󵄩(u, v)󵄩󵄩󵄩Y = ‖u‖ + ‖v‖ ≥ 󵄩 󵄩Y 󵄩 Y N2 N2 󵄩 ̃󵄩 ̃󵄩 μ̂ ′ c2 m B λ̂′ c2 m D 󵄩 󵄩 > 0 2 2 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + 0 2 1 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y N N 1󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 2 2

which is a contradiction. Therefore, the boundary value problem (3.1), (3.2) has no positive solution. Remark 3.1.2. Under the assumptions of Theorem 3.1.20, we have the following observations: ̃ Theorem 3.1.20 gives some supplementary information ̃≥D ̃ and B ̃ ≤ C, (a) In the case A for the domains of λ and μ for which there is no positive solution of (3.1), (3.2) in ̃λ ̃ μ comparison to Theorems 3.1.18 and 3.1.19, because λ̂0 = 20 and μ̂ 0 = 20 . ̃ Theorem 3.1.20 gives some supplementary information ̃≤D ̃ and B ̃ ≥ C, (b) In the case A for the domains of λ and μ for which there is no positive solution of (3.1), (3.2) in ̃λ ̃ μ comparison to Theorems 3.1.18 and 3.1.19, because λ̂′0 = 20 and μ̂ ′0 = 20 . 3.1.4 Examples Let N = 30, p = 3, q = 2, a1 = 3, a2 = 1, a3 = 1/2, ξ1 = 5, ξ2 = 15, ξ3 = 25, b1 = 1, b2 = 1/2, η1 = 10, η2 = 20. We consider the system of second-order difference equations Δ2 un−1 + λf (n, un , vn ) = 0,

n = 1, 29,

Δ vn−1 + μg(n, un , vn ) = 0,

n = 1, 29,

{

2

(3.15)

with the coupled multi-point boundary conditions u0 = 0,

u30 = 3v5 + v15 + v25 /2, p

v0 = 0,

v30 = u10 + u20 /2.

q

We have d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) = 50 > 0. The functions g0 and Ii , i = 1, 4, are given by g0 (n, j) =

1 j(30 − n), 1 ≤ j ≤ n ≤ 30, { 30 n(30 − j), 0 ≤ n ≤ j ≤ 29,

1 (1, 335j − 2j2 ), { { { 60 1 I1 (j) = { 60 (15, 300 − 195j − 2j2 ), { { 1 2 { 30 (15, 300 − 480j − j ),

1 ≤ j ≤ 10, 11 ≤ j ≤ 20, 21 ≤ j ≤ 29,

(3.16)

3.1 Positive parameters in the system of difference equations

� 79

111 j, 1 ≤ j ≤ 5, { 2 { { { 3 { { (180 + j), 6 ≤ j ≤ 15, I2 (j) = { 32 { (360 − 11j), 16 ≤ j ≤ 25, { { 2 { {3 { 2 (510 − 17j), 26 ≤ j ≤ 29, 1 (1, 140j − j2 ), { 30 { { { { { 1 (5, 400 + 60j − j2 ), I3 (j) = { 301 { (10, 800 − 300j − j2 ), { { 30 { {1 2 { 30 (15, 300 − 480j − j ),

15j, { { { I4 (j) = {180 − 3j, { { {360 − 12j,

1 ≤ j ≤ 5, 6 ≤ j ≤ 15, 16 ≤ j ≤ 25, 26 ≤ j ≤ 29,

1 ≤ j ≤ 10, 11 ≤ j ≤ 20, 21 ≤ j ≤ 29.

We take c = 4, and then A ≈ 3, 974.83333333, B = 5, 962.5, C ≈ 4, 124.83333333, D = ̃ = 2, 538. ̃ ≈ 3, 734.26666666, B ̃ = 5, 476.5, C̃ ≈ 3, 789.76666666, D 2, 700, A Example 1. We consider the functions j(u + v)2 (2 + sin v) , u+v+1 j2 (u + v)2 (3 + cos u) g(j, u, v) = , u+v+1 f (j, u, v) =

i i for all j = 1, 29 and u, v ≥ 0. We have f0s = g0s = 0, f∞ = 4, g∞ = 32. For a = α1 = α2 = 1/2, −3 −4 we obtain L1 ≈ 1.3852 ⋅ 10 , L3 ≈ 1.1596 ⋅ 10 . Then, by Theorem 3.1.4, we deduce that for each λ ∈ (L1 , ∞) and μ ∈ (L3 , ∞), there exists a positive solution ((un )n=0,30 , (vn )n=0,30 ) for problem (3.15), (3.16). s s Because f0s = g0s = 0 < ∞, f∞ = 87 < ∞, g∞ = 3, 364 < ∞, we can also apply Theorem 3.1.17. So, we conclude that there exist λ0 , μ0 > 0 such that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ), problem (3.15), (3.16) has no positive solution. By Theorem 3.1.17, the positive constants λ0 and μ0 are given by λ0 = min{ 4M1 A , 4M1 D } = 4M1 A and

μ0 = min{ 4M1 B , 4M1 C } = 2

2

and μ0 ≈ 1.24639 ⋅ 10−8 .

1 . 4M2 B

1

1

1

Then, we obtain M1 = 87, M2 = 3, 364, λ0 ≈ 7.22939 ⋅ 10−7 ,

Example 2. We consider the functions f (j, u, v) =

j p1 (u2 + v2 ), 1 + j2

g(j, u, v) =

j2 p (eu+v − 1), 1+j 2

i for all j = 1, 29 and u, v ≥ 0, where p1 , p2 > 0. We have f0s = 0, g0s = 841 p , f i = g∞ = ∞. 30 2 ∞ 1 1 1 15 ′ For b = 1/2, we obtain L4 = min{ 2g s B , 2g s C } = 2g s B = 841p B . Then by Theorem 3.1.6, 0

0

0

2

we conclude that for each λ ∈ (0, ∞) and μ ∈ (0, L′4 ), there exists a positive solution

80 � 3 Systems of difference equations with coupled BCs ((un )n=0,30 , (vn )n=0,30 ) for problem (3.15), (3.16). For example, if p2 = 1, we obtain L′4 ≈ 2.99135 ⋅ 10−6 . Remark 3.1.3. The results presented in this section were published in the paper [39].

3.2 Positive parameters in the boundary conditions In this section we consider the system of nonlinear second-order difference equations Δ2 un−1 + sn f (vn ) = 0, { 2 Δ vn−1 + tn g(un ) = 0,

n = 1, N − 1, n = 1, N − 1,

(3.17)

with the coupled multi-point boundary conditions u0 = 0,

p

uN = ∑ ai vξi + a0 , i=1

v0 = 0,

q

vN = ∑ bi uηi + b0 , i=1

(3.18)

where N ∈ ℕ, N ≥ 2, p, q ∈ ℕ, ai ∈ ℝ, ξi ∈ ℕ for all i = 1, p, bi ∈ ℝ, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and a0 and b0 are positive constants. We present sufficient conditions for the functions f and g and intervals for the parameters a0 and b0 such that problem (3.17), (3.18) has at least one positive solution or it has no positive solutions. A positive solution of (3.17), (3.18) is a pair of sequences ((un )n=0,N , (vn )n=0,N ) which satisfy (3.17) and (3.18) with un > 0, vn > 0 for all n = 1, N. We will use in this section the auxiliary results from Section 3.1. Our main existence result is based on the Schauder fixed point theorem, which we present now. Theorem 3.2.1. Let X be a Banach space and let Y ⊂ X be a nonempty, bounded, convex, and closed subset. If the operator A : Y → Y is completely continuous, then A has at least one fixed point.

3.2.1 Main results We present first the assumptions that we shall use in the sequel. (H1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < p q ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) > 0. (H2) The constants sn , tn ≥ 0 for all n = 1, N − 1, and there exist i0 , j0 ∈ {1, . . . , N − 1} such that si0 > 0, tj0 > 0. (H3) The functions f , g : ℝ+ → ℝ+ are continuous and there exists c0 > 0 such that c c f (u) < L0 , g(u) < L0 for all u ∈ [0, c0 ], where

3.2 Positive parameters in the boundary conditions

N−1

N−1

N−1

N−1

i=1

i=1

i=1

i=1

� 81

L = max{ ∑ si I1 (i) + ∑ ti I2 (i), ∑ ti I3 (i) + ∑ si I4 (i)} where Ii , i = 1, 4, are defined in Lemma 3.1.5. (H4) The functions f , g : ℝ+ → ℝ+ are continuous and satisfy the conditions lim

u→∞

f (u) = ∞, u

lim

u→∞

g(u) = ∞. u

By assumptions (H1) and (H2) and Lemma 3.1.5, we deduce that the constant L from assumption (H3) is positive. Now we consider the following system of second-order difference equations: Δ2 hn−1 = 0,

{

n = 1, N − 1,

2

Δ kn−1 = 0,

(3.19)

n = 1, N − 1,

with the coupled boundary conditions h0 = 0,

p

hN = ∑ ai kξi + a0 ,

k0 = 0,

i=1

q

kN = ∑ bi hηi + b0 , i=1

(3.20)

with a0 > 0 and b0 > 0. The problem (3.19), (3.20) has the solution (h, k) = ((hn )n=0,N , (kn )n=0,N ) with p

n (a N + b0 ∑ ai ξi ), d 0 i=1

n = 0, N,

n kn = (a0 ∑ bi ηi + b0 N), d i=1

n = 0, N.

hn =

q

(3.21)

By assumption (H1), we obtain hn > 0 and kn > 0 for all n = 1, N. We use the sequences (hn )n=0,N and (kn )n=0,N and we make a change of unknown sequences for problem (3.17), (3.18) such that the new boundary conditions have no positive parameters. For a solution (u, v) = ((un )n=0,N , (vn )n=0,N ) of problem (3.17), (3.18), we define the sequences (xn )n=0,N and (yn )n=0,N by xn = un − hn ,

yn = vn − kn ,

n = 0, N.

Then problem (3.17), (3.18) can be equivalently written as the system of difference equations Δ2 xn−1 + sn f (yn + kn ) = 0, { 2 Δ yn−1 + tn g(xn + hn ) = 0,

n = 1, N − 1, n = 1, N − 1,

(3.22)

82 � 3 Systems of difference equations with coupled BCs with the boundary conditions without parameters x0 = 0,

p

xN = ∑ ai yξi , i=1

y0 = 0,

q

yN = ∑ bi xηi . i=1

(3.23)

Using the Green functions Gi , i = 1, 4, from Section 3.1, a pair (x, y) = ((xn )n=0,N , (yn )n=0,N ) is a solution of problem (3.22), (3.23) if and only if it is a solution for the problem N−1

{ { xn = ∑ G1 (n, i)si f (yi + ki ) { { { { i=1 { { { { N−1 { { { + ∑ G2 (n, i)ti g(xi + hi ), { { { { i=1 { N−1 { { { { yn = ∑ G3 (n, i)ti g(xi + hi ) { { { { i=1 { { { { N−1 { { { { + ∑ G4 (n, i)si f (yi + ki ), { i=1

n = 0, N, (3.24)

n = 0, N.

We consider the Banach space X = ℝN+1 with the maximum norm ‖u‖ = maxn=0,N |un |, u = (un )n=0,N and the space Y = X × X with the norm ‖(x, y)‖Y = ‖x‖ + ‖y‖. We define the set E = {(xn )n=0,N , 0 ≤ xn ≤ c0 , ∀ n = 0, N} ⊂ X. We also define the operators 𝒮1 , 𝒮2 : E × E → X and 𝒮 : E × E → Y by 𝒮1 (x, y) = (𝒮1n (x, y))n=0,N , 𝒮2 (x, y) = (𝒮2n (x, y))n=0,N , with N−1

𝒮1n (x, y) = ∑ G1 (n, i)si f (yi + ki ) i=1

N−1

+ ∑ G2 (n, i)ti g(xi + hi ), i=1

n = 0, N,

N−1

𝒮2n (x, y) = ∑ G3 (n, i)ti g(xi + hi ) i=1

N−1

+ ∑ G4 (n, i)si f (yi + ki ), i=1

n = 0, N,

and 𝒮 (x, y) = (𝒮1 (x, y), 𝒮2 (x, y)), for (x, y) = ((xn )n=0,N , (yn )n=0,N ) ∈ E × E. We easily see that (x, y) is a solution of system (3.24) if and only if (x, y) is a fixed point of operator 𝒮 . Therefore, we will determine the fixed points of operator 𝒮 . The first result is the following existence theorem for problem (3.17), (3.18).

3.2 Positive parameters in the boundary conditions

�

83

Theorem 3.2.2. We assume that assumptions (H1), (H2), and (H3) are satisfied. Then there exist a1 > 0 and b1 > 0 such that for any a0 ∈ (0, a1 ] and b0 ∈ (0, b1 ], the problem (3.17), (3.18) has at least one positive solution. Proof. By assumption (H3) we deduce that there exist s0 > 0 and t0 > 0 such that f (u) ≤ c c0 for all u ∈ [0, c0 + s0 ] and g(u) ≤ L0 for all u ∈ [0, c0 + t0 ]. We define now a1 and b1 as L follows: p q – If ∑i=1 ai ξi ≠ 0 and ∑i=1 bi ηi ≠ 0, then a1 = min{ –

p

ds0 dt , 02 }, q 2N ∑i=1 bi ηi 2N q

dt ds0 , 20 }, q 2N ∑i=1 bi ηi N

p

b1 =

ds0 . 2N 2

q

If ∑i=1 ai ξi ≠ 0 and ∑i=1 bi ηi = 0, then a1 =

–

ds0 dt0 , }. 2 2N 2N ∑pi=1 ai ξi

If ∑i=1 ai ξi = 0 and ∑i=1 bi ηi ≠ 0, then a1 = min{

–

b1 = min{

p

dt0 , 2N 2

b1 = min{

ds0 dt0 , }. 2 N 2N ∑pi=1 ai ξi

q

If ∑i=1 ai ξi = 0 and ∑i=1 bi ηi = 0, then a1 =

dt0 , N2

b1 =

ds0 . N2

Let a0 ∈ (0, a1 ] and b0 ∈ (0, b1 ]. Then for ((xn )n=0,N , (yn )n=0,N ) ∈ E we have yn + kn ≤ c0 +

q

N (a ∑ b η + b0 N) d 0 i=1 i i q

≤ c0 + xn + hn ≤ c0 +

N (a ∑ b η + b1 N) ≤ c0 + s0 , d 1 i=1 i i p

N (a N + b0 ∑ ai ξi ) d 0 i=1 p

≤ c0 +

N (a N + b1 ∑ ai ξi ) ≤ c0 + t0 , d 1 i=1

for all n = 0, N, so f (yn + kn ) ≤

c0 , L

g(xn + hn ) ≤

c0 , L

∀ n = 0, N.

(3.25)

84 � 3 Systems of difference equations with coupled BCs Then, by using Lemma 3.1.4 we obtain 𝒮1n (x, y) ≥ 0, S2n (x, y) ≥ 0 for all n = 0, N and x = (xn )n=0,N , y = (yn )n=0,N ∈ E. By Lemma 3.1.5 and the definition of L from (H3), for all (x, y) ∈ E × E, we find N−1

N−1

S1n (x, y) ≤ ∑ I1 (i)si f (yi + ki ) + ∑ I2 (i)ti g(xi + hi ) i=1

i=1

N−1

≤

N−1

c0 ( ∑ s I (i) + ∑ ti I2 (i)) ≤ c0 , L i=1 i 1 i=1

N−1

∀ n = 0, N,

N−1

S2n (x, y) ≤ ∑ I3 (i)ti g(xi + hi ) + ∑ I4 (i)si f (yi + ki ) i=1

i=1

N−1

≤

N−1

c0 ( ∑ t I (i) + ∑ si I4 (i)) ≤ c0 , L i=1 i 3 i=1

∀ n = 0, N.

Therefore, 𝒮 (E × E) ⊂ E × E. By using a standard method, we conclude that 𝒮 is a completely continuous operator. Then, by Theorem 3.2.1, we deduce that 𝒮 has a fixed point (x, y) = ((xn )n=0,N , (yn )n=0,N ) ∈ E × E, which is a nonnegative solution for problem (3.24) or equivalently for problem (3.22), (3.23). Hence, (u, v) = ((un )n=0,N , (vn )n=0,N ), where un = xn + hn , vn = yn + kn , n = 0, N, is a positive solution of problem (3.17), (3.18). This p p solution satisfies the conditions dn (a0 N + b0 ∑i=1 ai ξi ) ≤ un ≤ c0 + dn (a0 N + b0 ∑i=1 ai ξi ) q q n n and d (a0 ∑i=1 bi ηi + b0 N) ≤ vn ≤ c0 + d (a0 ∑i=1 bi ηi + b0 N) for all n = 0, N. The second result is the following nonexistence theorem for the boundary value problem (3.17), (3.18). Theorem 3.2.3. We assume that assumptions (H1), (H2), and (H4) are satisfied. Then there exist a2 > 0 and b2 > 0 such that for any a0 ≥ a2 and b0 ≥ b2 , the problem (3.17), (3.18) has no positive solution. Proof. By assumption (H2) there exists c ∈ {1, 2, . . . , [[N/2]]} such that i0 , j0 ∈ {c, . . . , N − N−c c}. Then ∑i=c si I1 (i) > 0 and ∑N−c i=c ti I3 (i) > 0. We define the number R0 = max{(

−1

−1

c2 N−c c2 N−c ∑ si I1 (i)) , ( 2 ∑ ti I3 (i)) } > 0. 2 N i=c N i=c

By using (H4), for R0 defined as above, we conclude that there exists M0 > 0 such that f (u) > 2R0 u, g(u) > 2R0 u for all u ≥ M0 . We define now a2 > 0 and b2 > 0 as follows: p q – If ∑i=1 ai ξi ≠ 0 and ∑i=1 bi ηi ≠ 0, then a2 = max{

dM0 dM0 , }, 2cN 2c ∑qi=1 bi ηi

b2 = max{

dM0 dM0 , }. p 2c ∑i=1 ai ξi 2cN

3.2 Positive parameters in the boundary conditions

–

p

q

p

dM0 dM0 , }, cN 2c ∑qi=1 bi ηi

b2 =

dM0 . 2cN

q

If ∑i=1 ai ξi ≠ 0 and ∑i=1 bi ηi = 0, then a2 =

–

85

If ∑i=1 ai ξi = 0 and ∑i=1 bi ηi ≠ 0, then a2 = max{

–

�

p

dM0 , 2cN

b2 = max{

dM0 dM0 , }. p 2c ∑i=1 ai ξi cN

q

If ∑i=1 ai ξi = 0 and ∑i=1 bi ηi = 0, then a2 =

dM0 , cN

b2 =

dM0 . cN

Let a0 ≥ a2 and b0 ≥ b2 . We assume that ((un )n=0,N , (vn )n=0,N ) is a positive solution of (3.17), (3.18). Then (x, y) = ((xn )n=0,N , (yn )n=0,N ) with xn = un − hn , yn = vn − kn , n = 0, N, is a solution for problem (3.22), (3.23), where (h, k) = ((hn )n=0,N , (kn )n=0,N ) is the solution of problem (3.19), (3.20) (given by (3.21)). By using Lemma 3.1.4, we have xn ≥ 0, yn ≥ 0 for all n = 0, N, and by Lemma 3.1.6, we obtain minn=c,N−c xn ≥ Nc ‖x‖ and minn=c,N−c yn ≥ c ‖y‖. N Using now (3.21), we deduce that hc ‖h‖ = hN n=c,N−c k min kn = kc = c ‖k‖ = k n=c,N−c N min hn = hc =

c ‖h‖, N

c ‖k‖. N

Therefore, we obtain c c ‖x‖ + ‖h‖ = N N c c min (yn + kn ) ≥ ‖y‖ + ‖k‖ = N N n=c,N−c min (xn + hn ) ≥

n=c,N−c

c (‖x‖ + ‖h‖) ≥ N c (‖y‖ + ‖k‖) ≥ N

c ‖x + h‖, N c ‖y + k‖. N

Besides, we have min (xn + hn ) ≥

n=c,N−c

p

c c ‖h‖ = (a0 N + b0 ∑ ai ξi ) N d i=1 p

≥ min (yn + kn ) ≥

n=c,N−c

c (a N + b2 ∑ ai ξi ) ≥ M0 , d 2 i=1 q

c c ‖k‖ = (a0 ∑ bi ηi + b0 N) N d i=1

86 � 3 Systems of difference equations with coupled BCs q

≥

c (a ∑ b η + b2 N) ≥ M0 . d 2 i=1 i i

By Lemma 3.1.6 and the above inequalities we find N−1

N−1

i=1

i=1

xc = ∑ G1 (c, i)si f (yi + ki ) + ∑ G2 (c, i)ti g(xi + hi ) N−c

≥ ∑ G1 (c, i)si f (yi + ki ) ≥ i=c

≥ ≥

c N−c ∑ I (i)s f (y + ki ) N i=c 1 i i

N−c 2cR0 N−c 2cR0 min (yi + ki ) ∑ si I1 (i) ∑ I1 (i)si (yi + ki ) ≥ N i=c N i=c,N−c i=c

2cR0 M0 N−c ∑ si I1 (i) > 0. N i=c

We deduce that ‖x‖ ≥ xc > 0. In a similar manner we obtain N−1

N−1

i=1

i=1

yc = ∑ G3 (c, i)ti g(xi + hi ) + ∑ G4 (c, i)si f (yi + ki ) N−c

≥ ∑ G3 (c, i)ti g(xi + hi ) ≥ i=c

≥ ≥

c N−c ∑ I (i)t g(x + hi ) N i=c 3 i i

N−c 2cR0 N−c 2cR0 min (xi + hi ) ∑ ti I3 (i) ∑ I3 (i)ti (xi + hi ) ≥ N i=c N i=c,N−c i=c

2cR0 M0 N−c ∑ ti I3 (i) > 0. N i=c

We conclude that ‖y‖ ≥ yc > 0. In addition, from the above inequalities we have xc ≥ ≥

2cR0 N−c ∑ s I (i)(yi + ki ) N i=c i 1

N−c 2c2 R0 ‖y + k‖ ∑ si I1 (i) ≥ 2‖y + k‖ ≥ 2‖y‖. N2 i=c

Hence, 1 1 ‖y‖ ≤ xc ≤ ‖x‖. 2 2 In a similar manner we deduce

(3.26)

3.2 Positive parameters in the boundary conditions

yc ≥ ≥

� 87

2cR0 N−c ∑ t I (i)(xi + hi ) N i=c i 3

N−c 2c2 R0 ‖x + h‖ ∑ ti I3 (i) ≥ 2‖x + h‖ ≥ 2‖x‖. N2 i=c

Therefore, 1 1 ‖x‖ ≤ yc ≤ ‖y‖. 2 2

(3.27)

Hence, by (3.26) and (3.27) we conclude that ‖x‖ ≤ 21 ‖y‖ ≤ 41 ‖x‖, which is a contradiction (we saw before that ‖x‖ > 0). Therefore, problem (3.17), (3.18) has no positive solution. Similar results as Theorems 3.2.2 and 3.2.3 can be obtained if instead of boundary conditions (3.18) we have p

u0 = a0 ,

uN = ∑ ai vξi , i=1

v0 = 0,

q

vN = ∑ bi uηi + b0 , i=1

(3.28)

or u0 = 0,

p

uN = ∑ ai vξi + a0 , i=1

q

v0 = b0 ,

vN = ∑ bi uηi , i=1

(3.29)

or u0 = a0 ,

p

uN = ∑ ai vξi , i=1

v0 = b0 ,

q

vN = ∑ bi uηi , i=1

(3.30)

where a0 and b0 are positive constants. For problem (3.17), (3.28), instead of the pair of sequences (hn )n=0,N and (kn )n=0,N from (3.21), we consider the solution of system Δ2 h̃n−1 = 0, Δ2 k̃n−1 = 0,

{

n = 1, N − 1, n = 1, N − 1,

with the coupled boundary conditions h̃0 = a0 ,

p

h̃N = ∑ ai k̃ξi , i=1

k̃0 = 0,

q

k̃N = ∑ bi h̃ηi + b0 , i=1

namely, p

p

q

1 h̃n = {n[b0 (∑ ai ξi ) + a0 [(∑ ai ξi )(∑ bi ) − N]] d i=1 i=1 i=1

88 � 3 Systems of difference equations with coupled BCs p

q

i=1

i=1

+ a0 [N 2 − (∑ ai ξi )(∑ bi ηi )]}, q

n k̃n = [a0 ∑ bi (N − ηi ) + b0 N], d i=1

n = 0, N,

n = 0, N.

By assumption (H1) we obtain h̃n > 0 for all n = 0, N − 1 and k̃n > 0 for all n = 1, N. For problem (3.17), (3.29), instead of the pair of sequences (hn )n=0,N and (kn )n=0,N from (3.21), we consider the solution of system Δ2 ȟ n−1 = 0, { 2 Δ ǩn−1 = 0,

n = 1, N − 1, n = 1, N − 1,

with the coupled boundary conditions ȟ 0 = 0,

p

ȟ N = ∑ ai ǩξi + a0 ,

q

ǩN = ∑ bi ȟ ηi ,

ǩ0 = b0 ,

i=1

i=1

namely, p

n ȟ n = [a0 N + b0 ∑ ai (N − ξi )], d i=1 q

n = 0, N, q

p

1 ǩn = {n[a0 (∑ bi ηi ) + b0 [(∑ bi ηi )(∑ ai ) − N]] d i=1 i=1 i=1 p

q

i=1

i=1

+ b0 [N 2 − (∑ ai ξi )(∑ bi ηi )]},

n = 0, N.

By assumption (H1) we obtain ȟ n > 0 for all n = 1, N and ǩn > 0 for all n = 0, N − 1. For problem (3.17), (3.30), instead of the pair of sequences (hn )n=0,N and (kn )n=0,N from (3.21), we consider the solution of system Δ2 ĥ n−1 = 0, { 2 Δ k̂n−1 = 0,

n = 1, N − 1, n = 1, N − 1,

with the coupled boundary conditions ĥ 0 = a0 ,

p

ĥ N = ∑ ai k̂ξi , i=1

k̂0 = b0 ,

q

k̂N = ∑ bi ĥ ηi , i=1

namely, q

p

p

1 ĥ n = {n[−a0 N + a0 (∑ bi )(∑ ai ξi ) + b0 ∑ ai (N − ξi )] d i=1 i=1 i=1

3.2 Positive parameters in the boundary conditions p

q

i=1

i=1

+ a0 [N 2 − (∑ ai ξi )(∑ bi ηi )]}, p

q

�

89

n = 0, N, q

1 k̂n = {n[−b0 N + b0 (∑ ai )(∑ bi ηi ) + a0 ∑ bi (N − ηi )] d i=1 i=1 i=1 p

q

i=1

i=1

+ b0 [N 2 − (∑ ai ξi )(∑ bi ηi )]},

n = 0, N.

By assumption (H1) we obtain ĥ n > 0 and k̂n > 0 for all n = 0, N − 1. Therefore, we also obtain the following results. Theorem 3.2.4. Assume that assumptions (H1), (H2), and (H3) hold. Then there exist a1 > 0 and b1 > 0 such that for any a0 ∈ (0, a1 ] and b0 ∈ (0, b1 ], the problem (3.17), (3.28) has at least one positive solution (un > 0 for all n = 0, N − 1 and vn > 0 for all n = 1, N). Theorem 3.2.5. Assume that assumptions (H1), (H2), and (H4) hold. Then there exist a2 > 0 and b2 > 0 such that for any a0 ≥ a2 and b0 ≥ b2 , the problem (3.17), (3.28) has no positive solution (un > 0 for all n = 0, N − 1 and vn > 0 for all n = 1, N). Theorem 3.2.6. Assume that assumptions (H1), (H2), and (H3) hold. Then there exist a1 > 0 and b1 > 0 such that for any a0 ∈ (0, a1 ] and b0 ∈ (0, b1 ], the problem (3.17), (3.29) has at least one positive solution (un > 0 for all n = 1, N and vn > 0 for all n = 0, N − 1). Theorem 3.2.7. Assume that assumptions (H1), (H2), and (H4) hold. Then there exist a2 > 0 and b2 > 0 such that for any a0 ≥ a2 and b0 ≥ b2 , the problem (3.17), (3.29) has no positive solution (un > 0 for all n = 1, N and vn > 0 for all n = 0, N − 1). Theorem 3.2.8. Assume that assumptions (H1), (H2), and (H3) hold. Then there exist a1 > 0 and b1 > 0 such that for any a0 ∈ (0, a1 ] and b0 ∈ (0, b1 ], the problem (3.17), (3.30) has at least one positive solution (un > 0 and vn > 0 for all n = 0, N − 1). Theorem 3.2.9. Assume that assumptions (H1), (H2), and (H4) hold. Then there exist a2 > 0 and b2 > 0 such that for any a0 ≥ a2 and b0 ≥ b2 , the problem (3.17), (3.30) has no positive solution (un > 0 and vn > 0 for all n = 0, N − 1).

3.2.2 An example We consider N = 30, p = 3, q = 2, a1 = 3, a2 = 1, a3 = 1/2, ξ1 = 5, ξ2 = 15, ξ3 = 25, b1 = 1, b2 = 1/2, η1 = 10, η2 = 20, sn = tn = 1 for all n = 1, 29. We also consider the functions f , g : ℝ+ → ℝ+ , f (x) =

ãx α1 , 2x + 3

g(x) =

̃ α2 bx , 3x + 1

∀ x ∈ [0, ∞),

90 � 3 Systems of difference equations with coupled BCs with ã, b̃ > 0 and α1 , α2 > 2. We have limx→∞ f (x)/x = limx→∞ g(x)/x = ∞. Therefore, we consider the system of second-order difference equations α

ãvn1 { Δ2 un−1 + = 0, { { { (2vn + 3) { ̃ α2 { { {Δ2 v + bun = 0, n−1 (3un + 1) {

n = 1, 29,

(3.31)

n = 1, 29,

with the coupled multi-point boundary conditions 1 u30 = 3v5 + v15 + v25 + a0 , 2 1 v30 = u10 + u20 + b0 , 2

{ {u0 = 0, { { {v0 = 0,

(3.32)

where a0 and b0 are positive constants. p q We obtain d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) = 50 > 0. The functions Ii , i = 1, 4, from Lemma 3.1.5 are given in Section 3.1.4. Hence, we deduce that assumptions (H1), (H2), and (H4) are satisfied. In addition, by using the functions Ii , i = 1, 4, we find A := ∑29 i=1 I1 (i) ≈ 29 29 29 3, 974.8333333, B := ∑i=1 I2 (i) = 5, 962.5, C := ∑i=1 I3 (i) ≈ 4, 124.83333333, D := ∑i=1 I4 (i) = 2, 700, and then L = max{A + B, C + D} = A + B ≈ 9, 937.33333333. We choose c0 = 1 and if we select ã, b̃ satisfying the conditions ã < L5 , b̃ < L4 , then we conclude that f (x) < 1/L, g(x) < 1/L for all x ∈ [0, 1]. For example, if ã ≤ 5 ⋅ 10−4 and b̃ ≤ 4 ⋅ 10−4 , then the above conditions for f and g are satisfied. So, assumption (H3) is also satisfied. By Theorem 3.2.2 we deduce that there exist a1 > 0 and b1 > 0 such that for any a0 ∈ (0, a1 ] and b0 ∈ (0, b1 ], the problem (3.31), (3.32) has at least one positive solution. By Theorem 3.2.3 we conclude that there exist a2 > 0 and b2 > 0 such that for any a0 ≥ a2 and b0 ≥ b2 , the problem (3.31), (3.32) has no positive solution. We will next determine the constants a1 , b1 , a2 , and b2 for some values of the parameters. We consider ã = 5 ⋅ 10−4 , b̃ = 4 ⋅ 10−4 , α1 = 3, and α2 = 4. The sequences n (12a0 + 17b0 ) and kn = n5 (3b0 + 2a0 ) for all (hn )n=0,30 and (kn )n=0,30 from (3.21) are hn = 20 n = 0, 30. We obtain here s0 = f −1 ( L1 ) − 1 ≈ 0.00242 and t0 = g −1 ( L1 ) − 1 ≈ 0.00194, and

then we deduce that a1 = min{ 240 , 360 } ≈ 5.37 ⋅ 10−5 and b1 = min{ 360 , 510 } ≈ 3.79 ⋅ 10−5 . Now we choose c = 4 (the constant from the beginning of the proof of Theorem 3.2.3), and 26 we find Ξ1 := ∑26 i=4 si I1 (i) ≈ 3, 734.26666667, Ξ2 := ∑i=4 ti I3 (i) ≈ 3, 789.76666667, and then 16 Ξ1 −1 16 Ξ2 −1 ≥ R and R0 = max{( 900 ) , ( 900 ) } ≈ 0.01506319. For R := 2R0 +0.1, the inequalities f (u) u s

g(v) v

t

s

t

≥ R are satisfied for u ≥ M0′ ≈ 522.0012899 and v ≥ M0′′ ≈ 31.40551765, respectively.

We take M0 = max{M0′ , M0′′ } = M0′ , and then we conclude that a2 = max{

163.1254 and b2 =

5M 5M max{ 340 , 240 }

≈ 108.7503.

5M0 5M0 , 16 } 24

Remark 3.2.1. The results presented in this section were published in the paper [44].

≈

4 Systems of second-order finite difference equations with nonnegative nonlinearities, coupled multi-point boundary conditions, without parameters In this chapter we investigate the existence and multiplicity of positive solutions for two systems of nonlinear second-order difference equations supplemented with coupled multi-point boundary conditions, without parameters. The nonlinearities of the systems are nonnegative functions and satisfy various assumptions.

4.1 Some particular nonlinearities In this section we consider the system of nonlinear second-order difference equations Δ2 un−1 + f (n, vn ) = 0, { 2 Δ vn−1 + g(n, un ) = 0,

n = 1, N − 1, n = 1, N − 1,

(4.1)

with the coupled multi-point boundary conditions u0 = 0,

p

uN = ∑ ai vξi , i=1

v0 = 0,

q

vN = ∑ bi uηi , i=1

(4.2)

where N ∈ ℕ, N ≥ 2, p, q ∈ ℕ, ai ∈ ℝ, ξi ∈ ℕ for all i = 1, p, bi ∈ ℝ, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, and 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1. Under sufficient conditions on the functions f and g, we study the existence and multiplicity of positive solutions of problem (4.1), (4.2) by using the fixed point index theory. By a positive solution of problem (4.1), (4.2) we mean a pair of sequences (u, v) = ((un )n=0,N , (vn )n=0,N ) satisfying (4.1) and (4.2), with un ≥ 0, vn ≥ 0 for all n = 0, N and (u, v) ≠ (0, 0). In this section we will use the auxiliary results from Section 3.1. We will also use Theorem 2.1.1, Theorem 2.1.2, and the following theorem (Lemma 2.3 from [65]). Theorem 4.1.1 ([65]). Let A : Bϱ ∩ P → P be a completely continuous operator which has no fixed point on 𝜕Bϱ ∩ P (P ⊂ E is a cone and E is a real Banach space). If there exists a linear operator L : P → P and u0 ∈ P \ {θ} such that (i) u0 ≤ Lu0 ,

(ii) Lu ≤ Au,

∀ u ∈ 𝜕Bϱ ∩ P,

then i(A, Bϱ ∩ P, P) = 0. In the above theorem, “≤” is the partial ordering defined by P and θ is the zero element in E. https://doi.org/10.1515/9783111040370-004

92 � 4 Systems of difference equations without parameters 4.1.1 Main results We present firstly the basic assumptions that we shall use in the sequel. (A1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < p q ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) > 0. (A2) The functions f , g : {1, . . . , N − 1} × ℝ+ → ℝ+ are continuous. By using the functions Gi , i = 1, 4, from Section 3.1 (Lemma 3.1.2), the problem (4.1), (4.2) can be written equivalently as the following system: N−1

N−1

{ { un = ∑ G1 (n, i)f (i, vi ) + ∑ G2 (n, i)g(i, ui ), { { { { i=1 i=1 { N−1 N−1 { { { { { vn = ∑ G3 (n, i)g(i, ui ) + ∑ G4 (n, i)f (i, vi ), i=1 i=1 {

n = 0, N, n = 0, N.

We consider the Banach space X = ℝN+1 = {u = (u0 , u1 , . . . , uN ), ui ∈ ℝ, i = 0, N} with the maximum norm ‖ ⋅ ‖, ‖u‖ = maxn=0,N |un |, and the Banach space Y = X × X with the norm ‖(u, v)‖Y = ‖u‖ + ‖v‖. We define the cone Q ⊂ Y by Q = {(u, v) ∈ Y ; u = (un )n=0,N , v = (vn )n=0,N , un ≥ 0, vn ≥ 0, ∀ n = 0, N}.

We introduce the operators 𝒜1 , 𝒜2 : Y → X and 𝒜 : Y → Y defined by 𝒜1 (u, v) = (𝒜1n (u, v))n=0,N , 𝒜2 (u, v) = (𝒜2n (u, v))n=0,N , N−1

N−1

i=1

i=1

𝒜1n (u, v) = ∑ G1 (n, i)f (i, vi ) + ∑ G2 (n, i)g(i, ui ), N−1

N−1

𝒜2n (u, v) = ∑ G3 (n, i)g(i, ui ) + ∑ G4 (n, i)f (i, vi ), i=1

i=1

n = 0, N, n = 0, N,

and 𝒜(u, v) = (𝒜1 (u, v), 𝒜2 (u, v)), (u, v) = ((un )n=0,N , (vn )n=0,N ) ∈ Y . Under assumptions (A1) and (A2), it is easy to see that the operator 𝒜 : Q → Q is completely continuous. Thus, the existence and multiplicity of positive solutions of problem (4.1), (4.2) are equivalent to the existence and multiplicity of fixed points of operator 𝒜. Theorem 4.1.2. Assume that (A1) and (A2) hold. Additionally, we make the following assumptions. (A3) There exists c ∈ {1, . . . , [[N/2]]} such that i (i) f∞ = lim min

u→∞ n=c,N−c

f (n, u) = ∞; u

i (ii) g∞ = lim min

u→∞ n=c,N−c

g(n, u) = ∞. u

4.1 Some particular nonlinearities

� 93

(A4) There exist p1 ≥ 1 and q1 ≥ 1 such that (i) f0s = lim+ max

u→0 n=1,N−1

f (n, u) = 0; up1

(ii) g0s = lim+ max

u→0 n=1,N−1

g(n, u) = 0. uq1

Then problem (4.1), (4.2) has at least one positive solution ((un )n=0,N , (vn )n=0,N ). Proof. For c given in (A3), we define the cone Q0 = {(u, v) ∈ Q, u = (un )n=0,N , v = (vn )n=0,N , min un ≥

n=c,N−c

c ‖u‖, N

min vn ≥

n=c,N−c

c ‖v‖}. N

From our assumptions and Lemma 3.1.6, for any (u, v) ∈ Q, we deduce that 𝒜(u, v) = (𝒜1 (u, v), 𝒜2 (u, v)) ∈ Q0 , that is, 𝒜(Q) ⊂ Q0 . We consider the sequences u0 = (un0 )n=0,N , v0 = (v0n )n=0,N , defined by N−1

N−1

{ { un0 = ∑ G1 (n, i) + ∑ G2 (n, i), { { { { i=1 i=1 { N−1 N−1 { { { 0 { {vn = ∑ G3 (n, i) + ∑ G4 (n, i), i=1 i=1 {

n = 0, N, n = 0, N,

that is, (u0 , v0 ) is the solution of problem (3.3), (3.2) with x 0 = (xn0 )n=1,N−1 , y0 = (y0n )n=1,N−1 , xn0 = 1, y0n = 1 for all n = 1, N − 1. Hence, (u0 , v0 ) = 𝒜(x 0 , y0 ) ∈ Q0 . We define the set M = {(u, v) ∈ Q, there exists λ ≥ 0 such that (u, v) = 𝒜(u, v) + λ(u0 , v0 )}. We will show that M ⊂ Q0 and M is a bounded set of Y . If (u, v) ∈ M, then there exists λ ≥ 0 such that (u, v) = 𝒜(u, v) + λ(u0 , v0 ), or equivalently N−1

un = ∑ G1 (n, i)(f (i, vi ) + λ) i=1

N−1

+ ∑ G2 (n, i)(g(i, ui ) + λ), i=1

n = 0, N,

N−1

vn = ∑ G3 (n, i)(g(i, ui ) + λ) i=1

N−1

+ ∑ G4 (n, i)(f (i, vi ) + λ), i=1

n = 0, N.

94 � 4 Systems of difference equations without parameters

and

By Lemma 3.1.6, we obtain (u, v) = ((un )n=0,N , (vn )n=0,N ) ∈ Q0 , and hence M ⊂ Q0 ,

‖u‖ ≤

N min u , c n=c,N−c n

From (A3), we conclude that for ε1 =

‖v‖ ≤ 2N cm4

N min v , c n=c,N−c n

and ε2 =

f (n, u) ≥ ε1 u − C1 ,

2N , cm2

∀ (u, v) ∈ M.

(4.3)

there exist C1 , C2 > 0 such that

g(n, u) ≥ ε2 u − C2 ,

∀ n = c, N − c, u ∈ [0, ∞),

(4.4)

where mi = ∑N−c j=c Ii (j), i = 2, 4, and Ii , i = 2, 4, are defined in Lemma 3.1.5. For (u, v) = ((un )n=0,N , (vn )n=0,N ) ∈ M and n = c, N − c, by using Lemma 3.1.5 and relations (4.4), it follows that un = 𝒜1n (u, v) + λun0

N−1

N−1

≥ 𝒜1n (u, v) = ∑ G1 (n, i)f (i, vi ) + ∑ G2 (n, i)g(i, ui ) i=1

N−c

≥ ∑ G2 (n, i)g(i, ui ) ≥ i=c

N−c

i=1

c ∑ I (i)(ε2 ui − C2 ) N i=c 2

cm2 C2 cε m ≥ 2 2 min ui − = 2 min ui − C3 , N i=c,N−c N i=c,N−c

vn = 𝒜2n (u, v) + λv0n

N−1

N−1

≥ 𝒜2n (u, v) = ∑ G3 (n, i)g(i, ui ) + ∑ G4 (n, i)f (i, vi ) i=1

N−c

≥ ∑ G4 (n, i)f (i, vi ) ≥ i=c

i=1

N−c

c ∑ I (i)(ε1 vi − C1 ) N i=c 4

cε m cm4 C1 ≥ 1 4 min vi − = 2 min vi − C4 , N i=c,N−c N i=c,N−c where C3 = cm2 C2 /N and C4 = cm4 C1 /N. Therefore, we deduce min ui ≤ C3 ,

i=c,N−c

min vi ≤ C4 ,

i=c,N−c

∀ (u, v) = ((un )n=0,N , (vn )n=0,N ) ∈ M. From relations (4.3) and (4.5), we obtain ‖u‖ ≤ NC3 c

NC4 c

NC3 , c

‖v‖ ≤

NC4 , c

(4.5) and then ‖(u, v)‖Y =

‖u‖ + ‖v‖ ≤ + =: C5 , for all (u, v) ∈ M, that is, M is a bounded set of Y . Besides, there exists a sufficiently large R1 > 1 such that (u, v) ≠ 𝒜(u, v) + λ(u0 , v0 ),

∀ (u, v) ∈ 𝜕BR1 ∩ Q, ∀ λ ≥ 0.

4.1 Some particular nonlinearities

� 95

By Theorem 2.1.2, we deduce that the fixed point index of operator 𝒜 over BR1 ∩ Q with respect to Q is i(𝒜, BR1 ∩ Q, Q) = 0.

(4.6)

1 Next, from assumption (A4), we conclude that for ε3 = min{ 4M , 4M1 } and ε4 = 1

min{ 4M1 , 4M1 }, there exists r1 ∈ (0, 1] such that 2

4

3

f (n, u) ≤ ε3 up1 ,

g(n, u) ≤ ε4 uq1 ,

∀ n = 1, N − 1, u ∈ [0, r1 ],

(4.7)

where Mi = ∑N−1 j=1 I1 (j), i = 1, 4.

By using (4.7), we deduce that for all (u, v) ∈ Br1 ∩ Q and n = 0, N N−1

p

N−1

q

𝒜1n (u, v) ≤ ∑ I1 (i)ε3 vi 1 + ∑ I2 (i)ε4 ui 1 i=1

i=1

p1

≤ ε3 M1 ‖v‖ + ε4 M2 ‖u‖q1 ≤ N−1

q

N−1

1 1 1󵄩 󵄩 ‖v‖ + ‖u‖ = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 4 4 4 p

𝒜2n (u, v) ≤ ∑ I3 (i)ε4 ui 1 + ∑ I4 (i)ε3 vi 1 i=1

i=1

q1

≤ ε4 M3 ‖u‖ + ε3 M4 ‖v‖p1 ≤

1 1󵄩 1 󵄩 ‖u‖ + ‖v‖ = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y . 4 4 4

These imply that ‖𝒜1 (u, v)‖ ≤ 41 ‖(u, v)‖Y , ‖𝒜2 (u, v)‖ ≤ 41 ‖(u, v)‖Y , so 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝒜(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩𝒜1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒜2 (u, v)󵄩󵄩󵄩 ≤

1 󵄩󵄩 󵄩 󵄩(u, v)󵄩󵄩󵄩Y , 2󵄩

∀ (u, v) ∈ 𝜕Br1 ∩ Q.

By Theorem 2.1.1 we conclude that the fixed point index of operator 𝒜 over Br1 ∩ Q with respect to Q is i(𝒜, Br1 ∩ Q, Q) = 1.

(4.8)

Combining (4.6) and (4.8), we obtain i(𝒜, (BR1 \ Br1 ) ∩ Q, Q) = i(𝒜, BR1 ∩ Q, Q) − i(𝒜, Br1 ∩ Q, Q) = −1. We deduce that 𝒜 has at least one fixed point (u, v) ∈ (BR1 \Br1 )∩Q, that is, r1 < ‖(u, v)‖Y < R1 , which is a positive solution of problem (4.1), (4.2). Theorem 4.1.3. Assume that (A1) and (A2) hold. Additionally, we make the following assumptions. (A5) s (i) f∞ = lim max

u→∞ n=1,N−1

f (n, u) = 0; u

s (ii) g∞ = lim max

u→∞ n=1,N−1

g(n, u) = 0. u

96 � 4 Systems of difference equations without parameters (A6) There exist c ∈ {1, . . . , [[N/2]]}, p2 ∈ (0, 1], and q2 ∈ (0, 1] such that (i) f0i = lim+ min

u→0 n=c,N−c

f (n, u) = ∞; up2

(ii) g0i = lim+ min

u→0 n=c,N−c

g(n, u) = ∞. uq2

Then problem (4.1), (4.2) has at least one positive solution ((un )n=0,N , (vn )n=0,N ). 1 Proof. From assumption (A5), we deduce that for ε5 = min{ 4M , 4M1 } and ε6 = 1

min{ 4M1 , 4M1 } there exist C6 , C7 > 0 such that 2

4

3

f (n, u) ≤ ε5 u + C6 ,

g(n, u) ≤ ε6 u + C7 ,

∀ n = 1, N − 1, u ∈ [0, ∞).

Hence, for (u, v) ∈ Q, by using (4.9), we obtain N−1

N−1

𝒜1n (u, v) ≤ ∑ I1 (i)(ε5 vi + C6 ) + ∑ I2 (i)(ε6 ui + C7 ) i=1

i=1

N−1

N−1

i=1

i=1

≤ ε5 ‖v‖ ∑ I1 (i) + C6 ∑ I1 (i) N−1

N−1

+ ε6 ‖u‖ ∑ I2 (i) + C7 ∑ I2 (i) i=1

i=1

= ε5 ‖v‖M1 + C6 M1 + ε6 ‖u‖M2 + C7 M2 ≤

1 1 1󵄩 󵄩 ‖v‖ + ‖u‖ + C8 = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + C8 , 4 4 4

N−1

∀ n = 0, N,

N−1

𝒜2n (u, v) ≤ ∑ I3 (i)(ε6 ui + C7 ) + ∑ I4 (i)(ε5 vi + C6 ) i=1

i=1

N−1

N−1

i=1

i=1

≤ ε6 ‖u‖ ∑ I3 (i) + C7 ∑ I3 (i) N−1

N−1

+ ε5 ‖v‖ ∑ I4 (i) + C6 ∑ I4 (i) i=1

i=1

= ε6 ‖u‖M3 + C7 M3 + ε5 ‖v‖M4 + C6 M4 ≤

1 1󵄩 1 󵄩 ‖u‖ + ‖v‖ + C9 = 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y + C9 , 4 4 4

∀ n = 0, N,

where C8 := C6 M1 + C7 M2 and C9 := C7 M3 + C6 M4 . So, we deduce that 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝒜(u, v)󵄩󵄩󵄩Y = 󵄩󵄩󵄩𝒜1 (u, v)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒜2 (u, v)󵄩󵄩󵄩 ≤ where C10 := C8 + C9 .

1 󵄩󵄩 󵄩 󵄩(u, v)󵄩󵄩󵄩Y + C10 , 2󵄩

(4.9)

4.1 Some particular nonlinearities

� 97

Then there exists a sufficiently large R2 ≥ max{4C10 , 1} such that 3󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝒜(u, v)󵄩󵄩󵄩Y ≤ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , 4

󵄩 󵄩 ∀ (u, v) ∈ Q, 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y ≥ R2 .

Hence, ‖𝒜(u, v)‖Y < ‖(u, v)‖Y for all (u, v) ∈ 𝜕BR2 ∩ Q and by Theorem 2.1.1, we have i(𝒜, BR2 ∩ Q, Q) = 1.

(4.10)

On the other hand, from (A6) we conclude that for ε7 = there exists r2 ∈ (0, 1) such that f (n, u) ≥ ε7 up2 ,

g(n, u) ≥ ε8 uq2 ,

N c(m3 +m4 )

and ε8 =

∀ n = c, N − c, u ∈ [0, r2 ],

N c(m1 +m2 )

(4.11)

where mi = ∑N−c j=c Ii (j), i = 1, 4. From (4.11) and Lemma 3.1.5, we deduce for any (u, v) ∈ Br2 ∩ Q N−c

N−c

i=c

i=c

𝒜1n (u, v) ≥ ∑ G1 (n, i)f (i, vi ) + ∑ G2 (n, i)g(i, ui ) N−c

N−c

p

q

≥ ε7 ∑ G1 (n, i)vi 2 + ε8 ∑ G2 (n, i)ui 2 i=c

i=c

N−c

N−c

i=c

i=c

≥ ε7 ∑ G1 (n, i)vi + ε8 ∑ G2 (n, i)ui =: L1n (u, v), N−c

N−c

i=c

i=c

∀ n = 0, N,

𝒜2n (u, v) ≥ ∑ G3 (n, i)g(i, ui ) + ∑ G4 (n, i)f (i, vi ) N−c

N−c

q

p

≥ ε8 ∑ G3 (n, i)ui 2 + ε7 ∑ G4 (n, i)vi 2 i=c

i=c

N−c

N−c

i=c

i=c

≥ ε8 ∑ G3 (n, i)ui + ε7 ∑ G4 (n, i)vi =: L2n (u, v),

∀ n = 0, N.

Hence, 𝒜(u, v) ≥ L(u, v),

∀ (u, v) ∈ 𝜕Br2 ∩ Q,

(4.12)

where the linear operator L : Q → Q is defined by L(u, v) = (L1 (u, v), L2 (u, v)), L1 (u, v) = (L1n (u, v))n=0,N , L2 (u, v) = (L2n (u, v))n=0,N . We consider now (ũ 0 , ṽ0 ) ∈ Q \ {(0, 0)} with ũ 0 = (ũn0 )n=0,N and ṽ0 = (ṽ0n )n=0,N defined by N−c

N−c

i=c

i=c

ũn0 = ∑ G1 (n, i) + ∑ G2 (n, i),

n = 0, N,

98 � 4 Systems of difference equations without parameters N−c

N−c

i=c

i=c

ṽ0n = ∑ G3 (n, i) + ∑ G4 (n, i),

n = 0, N.

Then L(ũ 0 , ṽ0 ) = (L1 (ũ 0 , ṽ0 ), L2 (ũ 0 , ṽ0 )) and N−c

N−c

N−c

i=c

j=c

j=c

L1n (ũ 0 , ṽ0 ) = ε7 ∑ G1 (n, i)( ∑ G3 (i, j) + ∑ G4 (i, j)) N−c

N−c

N−c

i=c

j=c

j=c

+ ε8 ∑ G2 (n, i)( ∑ G1 (i, j) + ∑ G2 (i, j)) N−c

N−c

i=c

j=c

≥ ε7 ∑ G1 (n, i)( ∑ N−c

N−c c c I3 (j) + ∑ I4 (j)) N N j=c

N−c

+ ε8 ∑ G2 (n, i)( ∑ i=c

j=c

N−c c c I1 (j) + ∑ I2 (j)) N N j=c

N−c

=

N−c ε c ε7 c (m3 + m4 ) ∑ G1 (n, i) + 8 (m1 + m2 ) ∑ G2 (n, i) N N i=c i=c

N−c

N−c

i=c

i=c

= ∑ G1 (n, i) + ∑ G2 (n, i) = ũn0 ,

∀ n = 0, N,

N−c

N−c

N−c

i=c

j=c

j=c

L2n (ũ 0 , ṽ0 ) = ε8 ∑ G3 (n, i)( ∑ G1 (i, j) + ∑ G2 (i, j)) N−c

N−c

N−c

+ ε7 ∑ G4 (n, i)( ∑ G3 (i, j) + ∑ G4 (i, j)) i=c

j=c

N−c

N−c

i=c

j=c

≥ ε8 ∑ G3 (n, i)( ∑ N−c

c c I (j) + ∑ I2 (j)) N 1 N j=c

N−c

+ ε7 ∑ G4 (n, i)( ∑ i=c

j=c

N−c

j=c

N−c c c I3 (j) + ∑ I4 (j)) N N j=c

N−c

=

N−c ε8 c εc (m1 + m2 ) ∑ G3 (n, i) + 7 (m3 + m4 ) ∑ G4 (n, i) N N i=c i=c

N−c

N−c

i=c

i=c

= ∑ G3 (n, i) + ∑ G4 (n, i) = ṽ0n ,

∀ n = 0, N.

So, L(ũ 0 , ṽ0 ) ≥ (ũ 0 , ṽ0 ).

(4.13)

We may suppose that 𝒜 has no fixed point on 𝜕Br2 ∩ Q (otherwise the proof is finished). From (4.12), (4.13), and Theorem 4.1.1, we conclude that

4.1 Some particular nonlinearities

i(𝒜, Br2 ∩ Q, Q) = 0.

� 99

(4.14)

Therefore, from (4.10) and (4.14), we have i(𝒜, (BR2 \ Br2 ) ∩ Q, Q) = i(𝒜, BR2 ∩ Q, Q) − i(𝒜, Br2 ∩ Q, Q) = 1. Then 𝒜 has at least one fixed point in (BR2 \ Br2 ) ∩ Q, that is, r2 < ‖(u, v)‖Y < R2 . Thus, problem (4.1), (4.2) has at least one positive solution (u, v) ∈ Q. Theorem 4.1.4. Assume that (A1), (A2), (A3), and (A6) hold. Additionally, assume the functions f and g satisfy the following condition. (A7) For each n = 1, N − 1, f (n, u) and g(n, u) are nondecreasing with respect to u and there exists a constant R0 > 0 such that f (n, R0 )
1, 0 < β0 < 1, γ0 > 1, 0 < δ0 < 1, a0 , b0 > 0. We have d = N 2 − p q (∑i=1 ai ξi )(∑i=1 bi ηi ) = 50 > 0. The functions Ii , i = 1, 4, from Lemma 3.1.5 are given in

4.2 General nonlinearities

� 101

29 Section 3.1.4. We also deduce M1 = ∑29 j=1 I1 (j) ≈ 3, 974.83333333, M2 = ∑j=1 I2 (j) = 5, 962.5,

29 M3 = ∑29 j=1 I3 (j) ≈ 4, 124.83333333, and M4 = ∑j=1 I4 (j) = 2, 700. Then m0 = maxi=1,4 Mi = M2 . The functions f (n, u) and g(n, u) are nondecreasing with respect to u, for any n = 1, 29, and for p2 = q2 = 1 and c ∈ {1, . . . , 15}, assumptions (A3) and (A6) are sati i isfied; indeed, we obtain f∞ = ∞, g∞ = ∞, f0i = ∞, and g0i = ∞. We take R0 = 1 and then f (n, R0 ) = 2a0 , g(n, R0 ) = 2b0 for all n = 1, 29. If a0 < 8m1 and b0 < 8m1 , then 0

0

assumption (A7) is satisfied. For example, if a0 ≤ 2.096 ⋅ 10−5 and b0 ≤ 2.096 ⋅ 10−5 , then by Theorem 4.1.4, we deduce that problem (4.17), (4.18) has at least two positive solutions. Remark 4.1.1. The results presented in this section were published in the paper [40].

4.2 General nonlinearities In this section we consider the system of second-order nonlinear difference equations Δ2 un−1 + f (n, un , vn ) = 0, { 2 Δ vn−1 + g(n, un , vn ) = 0,

n = 1, N − 1, n = 1, N − 1,

(4.19)

with the coupled multi-point boundary conditions u0 = 0,

p

uN = ∑ ai vξi , i=1

v0 = 0,

q

vN = ∑ bi uηi , i=1

(4.20)

where N ∈ ℕ, N ≥ 2, p, q ∈ ℕ, ai ∈ ℝ, ξi ∈ ℕ for all i = 1, p, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, bi ∈ ℝ, ηi ∈ ℕ for all i = 1, q, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N −1, and f , g are nonnegative functions. Under sufficient conditions on the functions f and g, we study the existence of at least one or two positive solutions for problem (4.19), (4.20), by using the fixed point index theory. A positive solution of (4.19), (4.20) is a pair of sequences (u, v) = ((un )n=0,N , (vn )n=0,N ) which satisfy (4.19) and (4.20) with un ≥ 0, vn ≥ 0 for all n = 0, N and un > 0 for all n = 1, N or vn > 0 for all n = 1, N.

4.2.1 Auxiliary results In this section we will use the auxiliary results from Section 3.1. In addition, we will prove another two lemmas. We consider the system of second-order linear difference equations Δ2 un−1 + x̃n = 0, { 2 Δ vn−1 + ̃yn = 0,

n = 1, N − 1, n = 1, N − 1,

(4.21)

102 � 4 Systems of difference equations without parameters with the coupled multi-point boundary conditions (4.20), where x̃i , ỹi ∈ ℝ for all i = p q 1, N − 1. Let us denote d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ). We recall firstly Lemma 3.1.2. Lemma 4.2.1. If d ≠ 0, then the unique solution of problem (4.21), (4.20) is given by N−1

N−1

j=1

j=1

N−1

N−1

j=1

j=1

un = ∑ G1 (n, j)x̃j + ∑ G2 (n, j)̃yj ,

n = 0, N, (4.22)

vn = ∑ G3 (n, j)̃yj + ∑ G4 (n, j)x̃j ,

n = 0, N,

where G1 (n, j) = g0 (n, j) + G2 (n, j) =

p

p

q

n (∑ a ξ )(∑ bi g0 (ηi , j)), d i=1 i i i=1

nN ∑ a g (ξ , j), d i=1 i 0 i q

p

n G3 (n, j) = g0 (n, j) + (∑ bi ηi )(∑ ai g0 (ξi , j)), d i=1 i=1 G4 (n, j) =

(4.23)

q

nN ∑ b g (η , j), d i=1 i 0 i

and g0 (n, j) =

1 j(N − n), { N n(N − j),

1 ≤ j ≤ n ≤ N, 0 ≤ n ≤ j ≤ N − 1,

for all n = 0, N and j = 1, N − 1. p

Lemma 4.2.2. We suppose that ai ≥ 0, ξi ∈ ℕ for all i = 1, p, ∑i=1 ai > 0, bi ≥ 0, ηi ∈ ℕ q for all i = 1, q, ∑i=1 bi > 0, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d > 0. Then the Green functions Gi , i = 1, 4, given by (4.23) satisfy the following inequalities for all n = 0, N and j = 1, N − 1: (a1 ) G1 (n, j) ≤ σ1 h(j), where σ1 = 1 +

p

q

N (∑ a ξ )(∑ bi ) > 0, d i=1 i i i=1

(a2 ) G1 (n, j) ≤ δ1 n, where p

q

1 δ1 = 1 + (∑ ai ξi )(∑ bi ηi ) > 0, d i=1 i=1

4.2 General nonlinearities

(a3 ) G1 (n, j) ≥ ϱ1 nh(j), where ϱ1 =

p

q

σ2 =

N2 ∑ a > 0, d i=1 i

δ2 =

N ∑ a ξ > 0, d i=1 i i

1 (∑ a ξ )(∑ bi k(ηi )) > 0, d i=1 i i i=1

(b1 ) G2 (n, j) ≤ σ2 h(j), where p

(b2 ) G2 (n, j) ≤ δ2 n, where p

(b3 ) G2 (n, j) ≥ ϱ2 nh(j), where ϱ2 =

p

N ∑ a k(ξ ) > 0, d i=1 i i

(c1 ) G3 (n, j) ≤ σ3 h(j), where σ3 = 1 +

q

p

N (∑ b η )(∑ ai ) > 0, d i=1 i i i=1

(c2 ) G3 (n, j) ≤ δ3 n, where q

p

1 δ3 = 1 + (∑ bi ηi )(∑ ai ξi ) > 0 (δ3 = δ1 ), d i=1 i=1 (c3 ) G3 (n, j) ≥ ϱ3 nh(j), where ϱ3 =

q

p

σ4 =

N2 ∑ b > 0, d i=1 i

δ4 =

N ∑ b η > 0, d i=1 i i

1 (∑ b η )(∑ ai k(ξi )) > 0, d i=1 i i i=1

(d1 ) G4 (n, j) ≤ σ4 h(j), where q

(d2 ) G4 (n, j) ≤ δ4 n, where q

� 103

104 � 4 Systems of difference equations without parameters (d3 ) G3 (n, j) ≥ ϱ4 nh(j), where q

N ∑ b k(η ) > 0, d i=1 i i

ϱ4 = where h(j) = g0 (j, j) = k(n) =

1 j(N − j), N

∀ j = 1, N − 1,

1 n(N − n), N(N − 1)

∀ n = 0, N.

Proof. By using Lemma 2.1.2, for all n = 0, N and j = 1, N − 1 we obtain (a1 )

G1 (n, j) ≤ h(j) +

(a2 )

G1 (n, j) ≤ n +

(a3 )

G1 (n, j) ≥

(b1 )

G2 (n, j) ≤

(b2 )

G2 (n, j) ≤

(b3 )

G3 (n, j) ≥

(c1 )

G3 (n, j) ≤ h(j) +

(c2 )

G3 (n, j) ≤ n +

(c3 )

G3 (n, j) ≥

(d1 )

G4 (n, j) ≤

(d2 )

G4 (n, j) ≤

(d3 )

G4 (n, j) ≥

p

q

N (∑ a ξ )(∑ bi h(j)) = σ1 h(j), d i=1 i i i=1 p

q

n (∑ a ξ )(∑ bi ηi ) = δ1 n, d i=1 i i i=1 p

q

n (∑ a ξ )(∑ bi k(ηi )h(j)) = ϱ1 nh(j), d i=1 i i i=1 p

N2 ∑ a h(j) = σ2 h(j), d i=1 i p

nN ∑ a ξ = δ2 n, d i=1 i i p

nN ∑ a k(ξ )h(j) = ϱ2 nh(j), d i=1 i i q

p

N (∑ b η )(∑ ai h(j)) = σ3 h(j), d i=1 i i i=1 q

p

n (∑ b η )(∑ ai ξi ) = δ3 n, d i=1 i i i=1 q

p

n (∑ b η )(∑ ai k(ξi )h(j)) = ϱ3 nh(j), d i=1 i i i=1 q

N2 ∑ b h(j) = σ4 h(j), d i=1 i q

nN ∑ b η = δ4 n, d i=1 i i q

nN ∑ b k(η )h(j) = ϱ4 nh(j). d i=1 i i

4.2 General nonlinearities

� 105

p

Lemma 4.2.3. We suppose that ai ≥ 0, ξi ∈ ℕ for all i = 1, p, ∑i=1 ai > 0, bi ≥ 0, ηi ∈ ℕ q ∑i=1 bi

for all i = 1, q, > 0, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, d > 0, and x̃n , ̃yn ≥ 0 for all n = 1, N − 1. Then the solution of problem (4.21), (4.20) given by (4.22) satisfies the inequalities un ≥ γ1 num and vn ≥ γ2 nvm for all n, m = 0, N, where γ1 = min{

ϱ1 ϱ2 , }>0 σ1 σ2

and

γ2 = min{

ϱ3 ϱ4 , } > 0. σ3 σ4

Proof. By using the inequalities for the Green functions Gi , i = 1, 4, from Lemma 4.2.2, we obtain

N−1

N−1

j=1

j=1

un = ∑ G1 (n, j)x̃j + ∑ G2 (n, j)̃yj N−1

N−1

j=1

j=1

≥ ϱ1 ∑ nh(j)x̃j + ϱ2 ∑ nh(j)̃yj N−1

≥

ϱ1 n ϱ n N−1 ∑ G1 (m, j)x̃j + 2 ∑ G2 (m, j)̃yj σ1 j=1 σ2 j=1

≥ min{

N−1 N−1 ϱ1 ϱ2 , }n( ∑ G1 (m, j)x̃j + ∑ G2 (m, j)̃yj ) σ1 σ2 j=1 j=1

= γ1 num ,

∀ n, m = 0, N,

and N−1

N−1

j=1

j=1

vn = ∑ G3 (n, j)̃yj + ∑ G4 (n, j)x̃j N−1

N−1

j=1

j=1

≥ ϱ3 ∑ nh(j)̃yj + ϱ4 ∑ nh(j)x̃j N−1

≥

ϱ3 n ϱ n N−1 ∑ G3 (m, j)̃yj + 4 ∑ G4 (m, j)x̃j σ3 j=1 σ4 j=1

≥ min{

N−1 N−1 ϱ3 ϱ4 , }n( ∑ G3 (m, j)̃yj + ∑ G4 (m, j)x̃j ) σ3 σ4 j=1 j=1

= γ2 nvm ,

∀ n, m = 0, N.

106 � 4 Systems of difference equations without parameters 4.2.2 Existence of positive solutions We present in this section sufficient conditions on the functions f and g such that problem (4.19), (4.20) has positive solutions with respect to a cone. We introduce first the basic assumptions that we shall use in the sequel. p (H1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, ∑i=1 ai > 0, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, q ∑i=1 bi > 0, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, and d = p q N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) > 0. (H2) The functions f , g : {1, . . . , N − 1} × ℝ+ × ℝ+ → ℝ+ are continuous. By using Lemma 4.2.1 and the Green functions Gi , i = 1, . . . , 4, the problem (4.19), (4.20) can be equivalently written as the following system: N−1

N−1

{ { un = ∑ G1 (n, i)f (i, ui , vi ) + ∑ G2 (n, i)g(i, ui , vi ), { { { { i=1 i=1 { N−1 N−1 { { { { {vn = ∑ G3 (n, i)g(i, ui , vi ) + ∑ G4 (n, i)f (i, ui , vi ), i=1 i=1 {

n = 0, N, n = 0, N.

We consider the Banach space X = ℝN+1 = {u = (u0 , u1 , . . . , uN ), ui ∈ ℝ, i = 0, N}, with the maximum norm ‖ ⋅ ‖, ‖u‖ = max |un |, and the Banach space Y = X × X with the norm ‖(u, v)‖Y = ‖u‖ + ‖v‖. Let us define the cones L1 = {u ∈ X, u = (un )n=0,N , un ≥ γ1 n‖u‖, ∀ n = 0, N} ⊂ X,

L2 = {v ∈ X, v = (vn )n=0,N , vn ≥ γ2 n‖v‖, ∀ n = 0, N} ⊂ X,

where γ1 , γ2 are defined in Lemma 4.2.3 and L = L1 × L2 ⊂ Y . We also introduce the operators 𝒟1 , 𝒟2 : Y → X and 𝒟 : Y → Y defined by 𝒟1 (u, v) = (𝒟1n (u, v))n=0,N , 𝒟2 (u, v) = (𝒟2n (u, v))n=0,N , N−1

N−1

i=1

i=1

𝒟1n (u, v) = ∑ G1 (n, i)f (i, ui , vi ) + ∑ G2 (n, i)g(i, ui , vi ), N−1

N−1

i=1

i=1

𝒟2n (u, v) = ∑ G3 (n, i)g(i, ui , vi ) + ∑ G4 (n, i)f (i, ui , vi ),

n = 0, N, n = 0, N,

and 𝒟(u, v) = (𝒟1 (u, v), 𝒟2 (u, v)) for (u, v) = ((un )n=0,N , (vn )n=0,N ) ∈ Y . The solutions of our problem (4.19), (4.20) coincide with the fixed points of the operator 𝒟 in the space Y . So, we will investigate the existence of the fixed points of operator 𝒟. Under assumptions (H1) and (H2), by using Lemma 4.2.3, we can show that the operator 𝒟 : L → L is completely continuous. Theorem 4.2.1. We suppose that (H1) and (H2) hold. Additionally, we make the following assumptions.

4.2 General nonlinearities

� 107

(H3) There exist functions a, b ∈ C(ℝ+ , ℝ+ ) and constants p1 , p2 ∈ (0, 1] such that: (a)

f0i = lim inf v→0+

f (n, u, v) ∈ (0, ∞], a(v)

uniformly with respect to (n, u) ∈ {1, . . . , N − 1}×ℝ+ , and

g0i = lim inf u→0+

g(n, u, v) ∈ (0, ∞], b(u)

uniformly with respect to (n, v) ∈ {1, . . . , N − 1}×ℝ+ ; (b)

lim

u→0+

a(u) =∞ up1

and

lim

u→0+

b(u) = ∞. up2

(H4) There exist constants α1 , α2 ∈ (0, 1] such that f (n, u, v) = 0, vα1 uniformly with respect to (n, u) ∈ {1, . . . , N − 1} × ℝ+ ,

s f∞ = lim

v→∞

g(n, u, v) = 0, uα2 uniformly with respect to (n, v) ∈ {1, . . . , N − 1} × ℝ+ .

and

s g∞ = lim

u→∞

Then problem (4.19), (4.20) has at least one positive solution ((un )n=0,N , (vn )n=0,N ). Proof. By using assumption (H3), we deduce that there exist C1 , C2 > 0 and a sufficiently small r1 ∈ (0, 1] such that f (n, u, v) ≥ C1 a(v),

∀ (n, u) ∈ {1, . . . , N − 1} × ℝ+ ,

g(n, u, v) ≥ C2 b(u), ∀ (n, v) ∈ {1, . . . , N − 1} × ℝ+ , a(u) ≥ C̃0 up1 , b(u) ≥ C0 up2 , ∀ u ∈ [0, r1 ],

v ∈ [0, r1 ],

u ∈ [0, r1 ],

(4.24)

p p where C0 = 12((N 2 − 1)γ1 2 C2 ϱ2 )−1 , C̃0 = 12((N 2 − 1)γ2 1 C1 ϱ4 )−1 . We will prove that (𝒟1 (u, v), 𝒟2 (u, v)) ≰ (u, v) for all (u, v) ∈ 𝜕Br1 ∩L. We assume that there exists (u, v) ∈ 𝜕Br1 ∩ L, that is, ‖(u, v)‖Y = r1 , such that (𝒟1 (u, v), 𝒟2 (u, v)) ≤ (u, v). Then 𝒟1 (u, v) ≤ u and 𝒟2 (u, v) ≤ v. By using Lemma 4.2.2 and relations (4.24), we obtain N−1

N−1

un ≥ 𝒟1n (u, v) = ∑ G1 (n, i)f (i, ui , vi ) + ∑ G2 (n, i)g(i, ui , vi ) i=1

N−1

N−1

i=1

≥ ∑ G2 (n, i)g(i, ui , vi ) ≥ C2 ϱ2 ∑ nh(i)b(ui ) i=1

i=1

N−1

N−1

i=1

i=1

p

≥ C2 ϱ2 ∑ h(i)b(ui ) ≥ C0 C2 ϱ2 ∑ h(i)ui 2

108 � 4 Systems of difference equations without parameters N−1

p

p

N−1

≥ C0 C2 ϱ2 ∑ h(i)ip2 γ1 2 ‖u‖p2 ≥ C0 C2 ϱ2 γ1 2 ∑ h(i)‖u‖p2 i=1 p

= C0 C2 ϱ2 γ1 2

i=1

2

N − 1 p2 ‖u‖ ≥ 2‖u‖, 6

∀ n = 1, N.

2 In the above inequality we used the relation ∑N−1 i=1 h(i) = (N − 1)/6. Then ‖u‖ ≥ maxn=1,N |un | ≥ 2‖u‖, and hence

‖u‖ = 0.

(4.25)

Arguing as before, we deduce N−1

N−1

vn ≥ 𝒟2n (u, v) = ∑ G3 (n, i)g(i, ui , vi ) + ∑ G4 (n, i)f (i, ui , vi ) i=1

N−1

i=1

N−1

≥ ∑ G4 (n, i)f (i, ui , vi ) ≥ C1 ϱ4 ∑ nh(i)a(vi ) i=1

i=1

N−1

N−1

i=1

i=1

p ≥ C1 ϱ4 ∑ h(i)a(vi ) ≥ C̃0 C1 ϱ4 ∑ h(i)vi 1 N−1

N−1

p p ≥ C̃0 C1 ϱ4 ∑ h(i)ip1 γ2 1 ‖v‖p1 ≥ C̃0 C1 ϱ4 γ2 1 ∑ h(i)‖v‖p1 i=1

2 p N −1 = C̃0 C1 ϱ4 γ2 1 ‖v‖p1 ≥ 2‖v‖, 6

i=1

∀ n = 1, N.

So, ‖v‖ ≥ maxn=1,N |vn | ≥ 2‖v‖, and then ‖v‖ = 0.

(4.26)

By (4.25) and (4.26), we conclude that ‖(u, v)‖Y = 0, which is a contradiction. Hence, (𝒟1 (u, v), 𝒟2 (u, v)) ≰ (u, v) for all (u, v) ∈ 𝜕Br1 ∩ L. By Theorem 2.1.3 (b), we obtain that the fixed point index i(𝒟, Br1 ∩ L, L) = 0.

(4.27)

On the other hand, by (H4) we deduce that there exist C3 > 0 and C4 > 0 such that f (n, u, v) ≤ ε1 vα1 + C3 ,

g(n, u, v) ≤ ε2 uα2 + C4 ,

∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ , with ε1 = 3(8(N 2 − 1) max{σ1 , σ4 })−1 > 0 and ε2 = 3(8(N 2 − 1) max{σ2 , σ3 })−1 > 0.

(4.28)

4.2 General nonlinearities

� 109

Then by (4.28), for any (u, v) ∈ L, we obtain N−1

N−1

α

α

𝒟1n (u, v) ≤ σ1 ∑ h(i)(ε1 vi 1 + C3 ) + σ2 ∑ h(i)(ε2 ui 2 + C4 ) i=1

N−1

i=1

N−1

α

α

= σ1 ε1 ∑ h(i)vi 1 + σ2 ε2 ∑ h(i)ui 2 i=1

i=1

+ (σ1 C3 + σ2 C4 ) N−1

𝒟2n (u, v) ≤ σ3 ∑ i=1

α h(i)(ε2 ui 2

N−1

N2 − 1 , 6

∀ n = 0, N, N−1

+ C4 ) + σ4 ∑ i=1

N−1

α

α h(i)(ε1 vi 1

(4.29) + C3 )

α

= σ3 ε2 ∑ h(i)ui 2 + σ4 ε1 ∑ h(i)vi 1 i=1

+ (σ3 C4 + σ4 C3 )

i=1

2

N −1 , 6

∀ n = 0, N.

Now we define the functions p̃, q̃ : ℝ+ → ℝ+ given by N2 − 1 1 max{α1 ,α2 } w + (σ1 C3 + σ2 C4 ) , 8 6 1 N2 − 1 q̃(w) = wmax{α1 ,α2 } + (σ3 C4 + σ4 C3 ) , 8 6

p̃(w) =

w ∈ ℝ+ , w ∈ ℝ+ .

(4.30)

Because lim

w→∞

0, p̃(w) q̃(w) = lim ={ w→∞ w w 1/8,

if max{α1 , α2 } < 1,

if max{α1 , α2 } = 1,

we deduce that there exists R1 > r1 , R1 ≥ 1 such that p̃(w) ≤

1 w, 4

q̃(w) ≤

1 w, 4

∀ w ≥ R1 .

(4.31)

We will prove that (u, v) ≰ (𝒟1 (u, v), 𝒟2 (u, v)) for all (u, v) ∈ 𝜕BR1 ∩L. We assume that there exists (u, v) ∈ 𝜕BR1 ∩ L, that is, ‖(u, v)‖Y = R1 , such that (u, v) ≤ (𝒟1 (u, v), 𝒟2 (u, v)). So, by (4.29) we deduce that N−1

N−1

α

α

un ≤ 𝒟1n (u, v) ≤ σ1 ε1 ∑ h(i)vi 1 + σ2 ε2 ∑ h(i)ui 2 i=1

N2 − 1 , + (σ1 C3 + σ2 C4 ) 6 N−1

i=1

∀ n = 0, N, α

N−1

α

vn ≤ 𝒟2n (u, v) ≤ σ3 ε2 ∑ h(i)ui 2 + σ4 ε1 ∑ h(i)vi 1 i=1

i=1

110 � 4 Systems of difference equations without parameters

+ (σ3 C4 + σ4 C3 )

N2 − 1 , 6

∀ n = 0, N.

Then we obtain N−1

N−1

un ≤ σ1 ε1 ∑ h(i)‖v‖α1 + σ2 ε2 ∑ h(i)‖u‖α2 i=1

2

i=1

N −1 6 N 2 − 1 󵄩󵄩 󵄩max{α1 ,α2 } ≤ (σ1 ε1 + σ2 ε2 ) 󵄩(u, v)󵄩󵄩󵄩Y 6 󵄩 N2 − 1 + (σ1 C3 + σ2 C4 ) 6 1 󵄩󵄩 N2 − 1 󵄩max{α1 ,α2 } ≤ 󵄩󵄩(u, v)󵄩󵄩󵄩Y + (σ1 C3 + σ2 C4 ) , 8 6 + (σ1 C3 + σ2 C4 )

N−1

α2

N−1

∀ n = 0, N,

α1

(4.32)

vn ≤ σ3 ε2 ∑ h(i)‖u‖ + σ4 ε1 ∑ h(i)‖v‖ i=1

i=1

N2 − 1 + (σ3 C4 + σ4 C3 ) 6 2 N − 1 󵄩󵄩 󵄩max{α1 ,α2 } ≤ (σ3 ε2 + σ4 ε1 ) 󵄩(u, v)󵄩󵄩󵄩Y 6 󵄩 N2 − 1 + (σ3 C4 + σ4 C3 ) 6 1󵄩 N2 − 1 󵄩max{α1 ,α2 } ≤ 󵄩󵄩󵄩(u, v)󵄩󵄩󵄩Y , + (σ3 C4 + σ4 C3 ) 8 6

∀ n = 0, N.

By using (4.30)–(4.32), we deduce that un ≤ 41 ‖(u, v)‖Y and vn ≤ 41 ‖(u, v)‖Y for all n = 0, N. So, we conclude that ‖(u, v)‖Y ≤ 21 ‖(u, v)‖Y , and then ‖(u, v)‖Y = 0, which is a contradiction, because ‖(u, v)‖Y = R1 > 0. Hence, (u, v) ≰ (𝒟1 (u, v), 𝒟2 (u, v)) for all (u, v) ∈ 𝜕BR1 ∩ L. By Theorem 2.1.3 (a), we obtain that the fixed point index i(𝒟, BR1 ∩ L, L) = 1.

(4.33)

Because 𝒟 has no fixed points on 𝜕Br1 ∪ 𝜕BR1 , by (4.27) and (4.33), we deduce that i(𝒟, (BR1 \ Br1 ) ∩ L, L) = i(𝒟, BR1 ∩ L, L) − i(𝒟, Br1 ∩ L, L) = 1. Then the operator 𝒟 has at least one fixed point (u, v) ∈ (BR1 \Br1 )∩L, with r1 < ‖(u, v)‖Y < R1 , that is, ‖u‖ > 0 or ‖v‖ > 0. Because u ∈ L1 and v ∈ L2 , we obtain un > 0 for all n = 1, N or vn > 0 for all n = 1, N, so (u, v) is a positive solution of problem (4.19), (4.20). Remark 4.2.1. Theorem 4.2.1 remains valid if assumptions (H3) and (H4) are replaced by the general conditions (4.24) and (4.28).

4.2 General nonlinearities

� 111

Using similar arguments as those used in the proof of Theorem 4.2.1, we obtain the following result. Theorem 4.2.2. We suppose that (H1) and (H2) hold. Additionally, we make the following assumptions. (H3′ ) (a) There exist a function a ∈ C(ℝ+ , ℝ+ ) and a constant p1 ∈ (0, 1] such that ̃f i = lim inf f (n, u, v) ∈ (0, ∞], uniformly with respect to n ∈ {1, . . . , N − 1}, 0 u+v→0+ a(u + v) u,v≥0

with lim

u+v→0+

u,v≥0

a(u + v) = ∞, (u + v)p1

or (b) there exist a function b ∈ C(ℝ+ , ℝ+ ) and a constant p2 ∈ (0, 1] such that g̃0i = lim inf u+v→0+

u,v≥0

g(n, u, v) ∈ (0, ∞], uniformly with respect to n ∈ {1, . . . , N − 1}, b(u + v)

with lim

u+v→0+

u,v≥0

b(u + v) = ∞. (u + v)p2

(H4′ ) There exist constants α1 , α2 ∈ (0, 1] such that ̃f s = lim f (n, u, v) = 0, ∞ u+v→∞ (u + v)α1 u,v≥0 s and g̃∞ = lim

u+v→∞

u,v≥0

g(n, u, v) = 0, (u + v)α2

uniformly with respect to n ∈ {1, . . . , N − 1}, uniformly with respect to n ∈ {1, . . . , N − 1}.

Then problem (4.19), (4.20) has at least one positive solution ((un )n=0,N , (vn )n=0,N ). Theorem 4.2.3. We suppose that (H1) and (H2) hold. Additionally, we make the following assumptions. (H5) There exist functions c, d ∈ C(ℝ+ , ℝ+ ) and constants q1 , q2 ≥ 1 such that (a)

i f∞ = lim inf v→∞

f (n, u, v) ∈ (0, ∞], c(v)

uniformly with respect to (n, u) ∈ {1, . . . , N − 1}×ℝ+ , and

i g∞ = lim inf u→∞

g(n, u, v) ∈ (0, ∞], d(u)

uniformly with respect to (n, v) ∈ {1, . . . , N − 1}×ℝ+ , and (b)

lim

u→∞

c(u) =∞ uq1

and

lim

u→∞

d(u) = ∞. uq2

112 � 4 Systems of difference equations without parameters (H6) There exist constants β1 , β2 ≥ 1 such that f (n, u, v) = 0, vβ1 uniformly with respect to (n, u) ∈ {1, . . . , N − 1}×ℝ+ ,

f0s = lim

v→0+

g(n, u, v) = 0, uβ2 uniformly with respect to (n, v) ∈ {1, . . . , N − 1}×ℝ+ .

and

g0s = lim

u→0+

Then problem (4.19), (4.20) has at least one positive solution ((un )n=0,N , (vn )n=0,N ). Proof. By (H5) there exist constants Ci > 0, i = 5, . . . , 8, C10 > 0, and C12 > 0 such that f (n, u, v) ≥ C5 c(v) − C6 ,

g(n, u, v) ≥ C7 d(u) − C8 ,

∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ , q1

c(u) ≥ C9 u − C10 ,

q2

d(u) ≥ C11 u − C12 ,

q

(4.34)

∀ u ∈ ℝ+ ,

q

where C9 = 12(C5 ϱ4 γ2 1 (N 2 − 1))−1 , C11 = 12(C7 ϱ2 γ1 2 (N 2 − 1))−1 . Then by relations (4.34), we obtain for all (u, v) ∈ L N−1

𝒟1n (u, v) ≥ ∑ G2 (n, i)g(i, ui , vi ) i=1

N−1

N−1

i=1

i=1

≥ ϱ2 ∑ nh(i)g(i, ui , vi ) ≥ ϱ2 ∑ h(i)(C7 d(ui ) − C8 ) N−1

q

≥ C7 ϱ2 ∑ h(i)(C11 ui 2 − C12 ) − ϱ2 C8 i=1

N−1

N2 − 1 6

q2

≥ C7 ϱ2 C11 ∑ h(i)(iγ1 ‖u‖) i=1

N2 − 1 N2 − 1 − ϱ2 C8 6 6 2 N − 1 q ≥ C7 C11 ϱ2 γ1 2 ‖u‖q2 6 N2 − 1 N2 − 1 − C7 C12 ϱ2 − ϱ2 C8 , 6 6 − C7 C12 ϱ2

∀ n = 1, N.

Arguing as before, we deduce for all (u, v) ∈ L N−1

𝒟2n (u, v) ≥ ∑ G4 (n, i)f (i, ui , vi ) i=1

N−1

N−1

i=1

i=1

≥ ϱ4 ∑ nh(i)f (i, ui , vi ) ≥ ϱ4 ∑ h(i)(C5 c(vi ) − C6 )

(4.35)

4.2 General nonlinearities N−1

q

≥ C5 ϱ4 ∑ h(i)(C9 vi 1 − C10 ) − ϱ4 C6 i=1

N−1

� 113

N2 − 1 6

q1

≥ C5 ϱ4 C9 ∑ h(i)(iγ2 ‖v‖) i=1

N2 − 1 N2 − 1 − ϱ4 C6 6 6 2 q N −1 ≥ C5 C9 ϱ4 γ2 1 ‖v‖q1 6 N2 − 1 N2 − 1 − C5 C10 ϱ4 − ϱ4 C6 , 6 6 − C5 C10 ϱ4

∀ n = 1, N.

(4.36)

We will show that the set B = {(u, v) ∈ L, (u, v) = 𝒟(u, v) + λ(φ1 , φ2 ), λ ≥ 0} is bounded, where (φ1 , φ2 ) ∈ L \ {(0, 0)}. Indeed, let (u, v) ∈ B, which implies that 𝒟1 (u, v) ≤ u and 𝒟2 (u, v) ≤ v for all λ ≥ 0. By the relations (4.35) and (4.36), we deduce that un ≥ 𝒟1n (u, v)

N 2 − 1 q2 ‖u‖ 6 N2 − 1 N2 − 1 − ϱ2 C8 − C7 C12 ϱ2 6 6 q2 = 2‖u‖ − C13 , ∀ n = 1, N, q

≥ C7 C11 ϱ2 γ1 2

(4.37)

and vn ≥ 𝒟2n (u, v)

N 2 − 1 q1 ‖v‖ 6 N2 − 1 N2 − 1 − ϱ4 C6 − C5 C10 ϱ4 6 6 = 2‖v‖q1 − C14 , ∀ n = 1, N, q

≥ C5 C9 ϱ4 γ2 1

(4.38)

where N2 − 1 N2 − 1 + ϱ2 C8 , 6 6 N2 − 1 N2 − 1 C14 = C5 C10 ϱ4 + ϱ4 C6 . 6 6 C13 = C7 C12 ϱ2

If q2 = 1, then by (4.37) we obtain ‖u‖ ≤ C13 . If q2 > 1, then by (4.37) we deduce that there exists M1 > 0 such that ‖u‖ ≤ M1 . If q1 = 1, then by (4.38) we have ‖v‖ ≤ C14 . If q1 > 1, then by (4.38) we obtain that there exists M2 > 0 such that ‖v‖ ≤ M2 . Therefore, we conclude that ‖(u, v)‖Y ≤ M0 (M0 > 0), for all (u, v) ∈ B, that is, the set B is bounded.

114 � 4 Systems of difference equations without parameters Hence, there exists a sufficiently large R2 > 0 such that (u, v) ≠ 𝒟(u, v) + λ(φ1 , φ2 ) for all (u, v) ∈ 𝜕BR2 ∩ L and λ ≥ 0. By Theorem 2.1.2 we deduce that i(𝒟, BR2 ∩ L, L) = 0.

(4.39)

On the other hand, by (H6) there exists a sufficiently small r2 > 0 (r2 < R2 , r2 ≤ 1) such that f (n, u, v) ≤ ε3 vβ1 ,

∀ (n, u) ∈ {1, . . . , N − 1} × ℝ+ , v ∈ [0, r2 ],

g(n, u, v) ≤ ε4 u ,

∀ (n, v) ∈ {1, . . . , N − 1} × ℝ+ , u ∈ [0, r2 ],

β2

(4.40)

where ε3 = 3(4(N 2 − 1) max{σ1 , σ4 })−1 > 0 and ε4 = 3(4(N 2 − 1) max{σ2 , σ3 })−1 > 0. We will prove that (u, v) ≰ 𝒟(u, v) for all (u, v) ∈ 𝜕Br2 ∩ L. We assume that there exists (u, v) ∈ 𝜕Br2 ∩ L, that is, ‖(u, v)‖Y = r2 ≤ 1, such that (u, v) ≤ (𝒟1 (u, v), 𝒟2 (u, v)), or u ≤ 𝒟1 (u, v) and v ≤ 𝒟2 (u, v). Then by (4.40) we obtain N−1

N−1

β

β

un ≤ 𝒟1n (u, v) ≤ ε3 σ1 ∑ h(i)vi 1 + ε4 σ2 ∑ h(i)ui 2 2

i=1

i=1

2

N − 1 β1 N − 1 β2 ‖v‖ + ε4 σ2 ‖u‖ 6 6 N 2 − 1 󵄩󵄩 󵄩min{β1 ,β2 } ≤ (ε3 σ1 + ε4 σ2 ) 󵄩󵄩(u, v)󵄩󵄩󵄩Y 6 1 󵄩󵄩 󵄩min{β1 ,β2 } 1 󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩(u, v)󵄩󵄩Y ≤ 󵄩󵄩(u, v)󵄩󵄩󵄩Y , ∀ n = 0, N, 4 4 ≤ ε3 σ1

N−1

vn ≤ 𝒟2n (u, v) ≤ ε4 σ3 ∑ 2

i=1

β h(i)ui 2

N−1

+ ε3 σ4 ∑

2

i=1

β h(i)vi 1

(4.41)

N − 1 β1 N − 1 β2 ‖u‖ + ε3 σ4 ‖v‖ 6 6 N 2 − 1 󵄩󵄩 󵄩min{β1 ,β2 } ≤ (ε4 σ3 + ε3 σ4 ) 󵄩(u, v)󵄩󵄩󵄩Y 6 󵄩 ≤ ε4 σ3

≤

1 󵄩󵄩 󵄩min{β1 ,β2 } 1 󵄩󵄩 󵄩 ≤ 󵄩󵄩(u, v)󵄩󵄩󵄩Y , 󵄩(u, v)󵄩󵄩󵄩Y 4󵄩 4

∀ n = 0, N.

By the above relations (4.41), we deduce that ‖u‖ ≤ 41 ‖(u, v)‖Y and ‖v‖ ≤ 41 ‖(u, v)‖Y . Therefore, we obtain ‖(u, v)‖Y ≤ 21 ‖(u, v)‖Y , and then ‖(u, v)‖Y = 0, which is a contradiction, because ‖(u, v)‖Y = r2 > 0. Then (u, v) ≰ 𝒟(u, v) for all (u, v) ∈ 𝜕Br2 ∩ L. By Theorem 2.1.3 (a) we conclude that i(𝒟, Br2 ∩ L, L) = 1. Because 𝒟 has no fixed points on 𝜕Br2 ∪ 𝜕BR2 , by (4.39) and (4.42), we deduce that i(𝒟, (BR2 \ Br2 ) ∩ L, L) = i(𝒟, BR2 ∩ L, L) − i(𝒟, Br2 ∩ L, L) = −1.

(4.42)

4.2 General nonlinearities

� 115

So, the operator 𝒟 has at least one fixed point (u, v) ∈ (BR2 \ Br2 ) ∩ L, with r2 < ‖(u, v)‖Y < R2 , which is a positive solution for our problem (4.19), (4.20). Remark 4.2.2. Theorem 4.2.3 remains valid if assumptions (H5) and (H6) are replaced by the general conditions (4.34) and (4.40). Using similar arguments as those used in the proof of Theorem 4.2.3 we obtain the following result. Theorem 4.2.4. We suppose that (H1) and (H2) hold. Additionally, we make the following assumptions. (H5′ ) (a) There exist a function c ∈ C(ℝ+ , ℝ+ ) and a constant q1 ≥ 1 such that ̃f i = lim inf f (n, u, v) ∈ (0, ∞], ∞ u+v→∞ c(u + v) u,v≥0 with

lim

u+v→∞

u,v≥0

uniformly with respect to n ∈ {1, . . . , N − 1},

c(u + v) = ∞, (u + v)q1

or (b) there exist a function d ∈ C(ℝ+ , ℝ+ ) and a constant q2 ≥ 1 such that i g̃∞ = lim inf u+v→∞

u,v≥0

with

g(n, u, v) ∈ (0, ∞], d(u + v) lim

u+v→∞

u,v≥0

uniformly with respect to n ∈ {1, . . . , N − 1},

d(u + v) = ∞. (u + v)q2

(H6′ ) There exist constants β1 , β2 ≥ 1 such that ̃f s = lim f (n, u, v) = 0, 0 u+v→0+ (u + v)β1 u,v≥0 and g̃0s = lim

u+v→0+

u,v≥0

g(n, u, v) = 0, (u + v)β2

uniformly with respect to n ∈ {1, . . . , N − 1}, uniformly with respect to n ∈ {1, . . . , N − 1}.

Then problem (4.19), (4.20) has at least one positive solution ((un )n=0,N , (vn )n=0,N ). Theorem 4.2.5. We suppose that assumptions (H1), (H2), (H3), and (H5) hold. Additionally, we make the following assumption. (H7) The functions f (n, u, v) and g(n, u, v) are nondecreasing with respect to u and v, for any n ∈ {1, . . . , N − 1}, and there exists R0 > 0 such that 3R0 , 2(N 2 − 1) max{σ1 , σ4 } 3R0 g(n, R0 , R0 ) < , 2 2(N − 1) max{σ2 , σ3 } f (n, R0 , R0 )
0 (r1 < R0 ) and a sufficiently large R2 > R0 such that i(𝒟, Br1 ∩ L, L) = 0,

i(𝒟, BR2 ∩ L, L) = 0.

(4.44)

Because 𝒟 has no fixed points on 𝜕Br1 ∪ 𝜕BR2 ∪ 𝜕BR0 , by the relations (4.43) and (4.44), we obtain i(𝒟, (BR2 \ BR0 ) ∩ L, L) = i(𝒟, BR2 ∩ L, L) − i(𝒟, BR0 ∩ L, L) = −1, i(𝒟, (BR0 \ Br1 ) ∩ L, L) = i(𝒟, BR0 ∩ L, L) − i(𝒟, Br1 ∩ L, L) = 1.

Then 𝒟 has at least one fixed point (u1 , v1 ) ∈ (BR2 \ BR0 ) ∩ L and at least one fixed point (u2 , v2 ) ∈ (BR0 \ Br1 ) ∩ L. Therefore, problem (4.19), (4.20) has two distinct positive solutions (u1 , v1 ), (u2 , v2 ).

4.2 General nonlinearities

� 117

In a similar manner as we proved Theorem 4.2.5, we obtain the following theorem. Theorem 4.2.6. We suppose that assumptions (H1), (H2), (H3′ ), (H5′ ), and (H7) hold. Then problem (4.19), (4.20) has at least two positive solutions. Examples. (1) We consider the functions n (1 + e−u ), g(n, u, v) = (1 + e−n )uθ , n+1 ∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ .

f (n, u, v) =

For a(v) = vζ with ζ < 1 and b(u) = uρ with ρ < 1, assumptions (H3) and (H4) are satisfied if ρ > θ, α2 > θ, p1 > ζ , p2 > ρ with p1 ≤ 1 and p2 ≤ 1. For example, if θ = 21 , ζ = 32 , ρ = 43 , 11 19 α1 = 51 , α2 = 43 , p1 = 12 , and p2 = 20 , we can apply Theorem 4.2.1, and we deduce that problem (4.19), (4.20) has at least one positive solution. (2) We consider the functions f (n, u, v) = (1 + e−u )vθ1 ,

g(n, u, v) = (1 + e−v )uθ2 ,

∀ (n, u, v) ∈ {1, . . . , N − 1} × ℝ+ × ℝ+ . For c(v) = vζ with ζ > 1 and d(u) = uρ with ρ > 1, assumptions (H5) and (H6) are satisfied if ζ < θ1 , ρ < θ2 , β1 < θ1 , β2 < θ2 , ζ > q1 , ρ > q2 with q1 ≥ 1 and q2 ≥ 1. For example, if θ1 = 4, θ2 = 5, ζ = 2, ρ = 165 , β1 = 73 , β2 = 3, q1 = 32 , and q2 = 2, we can apply Theorem 4.2.3, and we conclude that problem (4.19), (4.20) has at least one positive solution. Remark 4.2.3. The results presented in this section were published in the paper [41].

5 Systems of second-order finite difference equations with sign-changing nonlinearities and coupled multi-point boundary conditions In this chapter we investigate the existence of positive solutions for a system of nonlinear second-order difference equations with parameters and sign-changing nonlinearities, satisfying multi-point coupled boundary conditions.

5.1 Positive parameters in the system of difference equations In this section we consider the system of nonlinear second-order difference equations Δ2 un−1 + λf (n, un , vn ) = 0, { 2 Δ vn−1 + μg(n, un , vn ) = 0,

n = 1, N − 1, n = 1, N − 1,

(5.1)

with the multi-point coupled boundary conditions u0 = 0,

p

uN = ∑ ai vξi , i=1

v0 = 0,

q

vN = ∑ bi uηi , i=1

(5.2)

where N ∈ ℕ, N ≥ 2, p, q ∈ ℕ, ai ∈ ℝ, ξi ∈ ℕ for all i = 1, p, bi ∈ ℝ, ηi ∈ ℕ for all i = 1, q, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, λ, μ > 0, and f , g are sign-changing functions. We present sufficient conditions for the functions f and g and intervals for the parameters λ and μ such that problem (5.1), (5.2) has positive solutions. Hence, problem (5.1), (5.2) is a semipositone discrete boundary value problem. By a positive solution of (5.1), (5.2) we mean a pair of sequences (u, v) = ((un )n=0,N , (vn )n=0,N ) satisfying (5.1) and (5.2) with un > 0 and vn > 0 for all n = 1, N. In this section we will use the auxiliary results from Section 4.2. In the proofs of the main results we will apply the nonlinear alternative of Leray–Schauder type (Theorem 1.2.1) and the Guo–Krasnosel’skii fixed point theorem (Theorem 1.1.1).

5.1.1 Existence of positive solutions We present firstly the assumptions that we shall use in the sequel. p (H1) We have ai ≥ 0, ξi ∈ ℕ for all i = 1, p, ∑i=1 ai > 0, bi ≥ 0, ηi ∈ ℕ for all i = 1, q, q ∑i=1 bi > 0, 1 ≤ ξ1 < ⋅ ⋅ ⋅ < ξp ≤ N − 1, 1 ≤ η1 < ⋅ ⋅ ⋅ < ηq ≤ N − 1, d = N 2 − p q (∑i=1 ai ξi )(∑i=1 bi ηi ) > 0, and λ, μ > 0. https://doi.org/10.1515/9783111040370-005

5.1 Positive parameters in the system of difference equations

� 119

(H2) The functions f , g : {1, . . . , N − 1} × ℝ+ × ℝ+ → ℝ are continuous, and there exist cn ≥ 0, dn ≥ 0, n = 1, N − 1, such that f (n, u, v) ≥ −cn

and

g(n, u, v) ≥ −dn ,

∀ n = 1, N − 1, u, v ∈ [0, ∞).

(H3) We have f (n, 0, 0) > 0 and g(n, 0, 0) > 0 for all n = 1, N − 1. (H4) We have f∞ = lim min u+v→∞

u,v≥0 n=1,N−1

f (n, u, v) =∞ u+v

or g∞ = lim min u+v→∞

u,v≥0 n=1,N−1

g(n, u, v) = ∞. u+v

(H5) We have ̃f = lim max |f (n, u, v)| = 0 ∞ u+v→∞ u+v u,v≥0 n=1,N−1

|g(n, u, v)| = 0. u+v u,v→≥0 n=1,N−1

and g̃∞ = lim

max

u+v∞

We consider the system of nonlinear second-order difference equations Δ2 xn−1 + λ(f (n, [xn − pn ]∗ , [yn − qn ]∗ ) + cn ) = 0, { 2 Δ yn−1 + μ(g(n, [xn − pn ]∗ , [yn − qn ]∗ ) + dn ) = 0,

n = 1, N − 1, n = 1, N − 1,

(5.3)

with the multi-point coupled boundary conditions p

x0 = 0,

xN = ∑ ai yξi , i=1

y0 = 0,

q

yN = ∑ bi xηi , i=1

(5.4)

where z∗n = zn if zn ≥ 0 and z∗n = 0 if zn < 0, for n = 1, N − 1. Here ((pn )n=0,N , (qn )n=0,N ) is the solution of the system of second-order difference equations Δ2 un−1 + λcn = 0,

n = 1, N − 1,

Δ vn−1 + μdn = 0,

n = 1, N − 1,

{

2

(5.5)

with the boundary conditions u0 = 0,

p

uN = ∑ ai vξi , i=1

v0 = 0,

q

vN = ∑ bi uηi , i=1

(5.6)

where cn , dn , n = 1, N − 1, are given by assumption (H2). By using Lemma 4.2.1 and the Green functions Gi , i = 1, 4, given by (4.23), the solution of problem (5.5), (5.6) is given by

120 � 5 Systems with sign-changing nonlinearities N−1

N−1

{ { pn = λ ∑ G1 (n, i)ci + μ ∑ G2 (n, i)di , { { { { i=1 i=1 { N−1 N−1 { { { { {qn = μ ∑ G3 (n, i)di + λ ∑ G4 (n, i)ci , i=1 i=1 {

n = 0, N, n = 0, N.

Under assumptions (H1) and (H2), we have pn ≥ 0 and qn ≥ 0 for all n = 0, N. We shall prove that there exists a solution ((xn )n=0,N , (yn )n=0,N ) for the boundary value problem (5.3), (5.4) with xn ≥ pn and yn ≥ qn for all n = 0, N and xn > pn and yn > qn for all n = 1, N. In this case, ((un )n=0,N , (vn )n=0,N ), where un = xn − pn and vn = yn − qn for all n = 0, N represents a positive solution of the boundary value problem (5.1), (5.2). Indeed, by (5.3)–(5.6), we have Δ2 un−1 = Δ2 (xn−1 − pn−1 ) = Δ2 xn−1 − Δ2 pn−1 = −λf (n, [xn − pn ]∗ , [yn − qn ]∗ ) − λcn + λcn 2

= −λf (n, un , vn ),

∀ n = 1, N − 1,

2

Δ vn−1 = Δ (yn−1 − qn−1 ) = Δ2 yn−1 − Δ2 qn−1 = −μg(n, [xn − pn ]∗ , [yn − qn ]∗ ) − μdn + μdn = −μg(n, un , vn ), u0 = x0 − p0 = 0,

p

∀ n = 1, N − 1, v0 = y0 − q0 = 0, p

p

p

i=1 q

i=1 q

i=1 q

i=1 q

i=1

i=1

i=1

uN = xN − pN = ∑ ai yξi − ∑ ai qξi = ∑ ai (yξi − qξi ) = ∑ ai vξi , vN = yN − qN = ∑ bi xηi − ∑ bi pηi = ∑ bi (xηi − pηi ) = ∑ bi uηi . i=1

Therefore, in what follows we shall investigate the boundary value problem (5.3), (5.4). By using Lemma 4.2.1, the pair ((xn )n=0,N , (yn )n=0,N ) is a solution of problem (5.3), (5.4) if and only if ((xn )n=0,N , (yn )n=0,N ) is a solution of the system N−1

{ { xn = λ ∑ G1 (n, i)(f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + ci ) { { { { i=1 { { { { N−1 { { { + μ ∑ G2 (n, i)(g(i, [xi − pi ]∗ , [yi − qi ]∗ ) + di ), n = 0, N, { { { { i=1 { N−1 { { { { yn = μ ∑ G3 (n, i)(g(i, [xi − pi ]∗ , [yi − qi ]∗ ) + di ) { { { { i=1 { { { { N−1 { { { { + λ ∑ G4 (n, i)(f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + ci ), n = 0, N. { i=1

(5.7)

5.1 Positive parameters in the system of difference equations

� 121

We consider the Banach space X = ℝN+1 = {x = (x0 , x1 , . . . , xN ), xi ∈ ℝ, i = 0, N} with the maximum norm ‖x‖ = maxn=0,N |xn | and the Banach space Y = X × X with the norm ‖(x, y)‖Y = ‖x‖ + ‖y‖. We define the cones L1 = {x ∈ X, x = (xn )n=0,N , xn ≥ γ1 n‖x‖, ∀ n = 0, N} ⊂ X,

L2 = {y ∈ X, y = (yn )n=0,N , yn ≥ γ2 n‖y‖, ∀ n = 0, N} ⊂ X,

where γ1 , γ2 are defined in Lemma 4.2.3 and L = L1 × L2 ⊂ Y . For λ, μ > 0 we introduce the operators ℰ1 , ℰ2 : Y → X and ℰ : Y → Y defined by ℰ1 (x, y) = (ℰ1n (x, y))n=0,N , ℰ2 (x, y) = (ℰ2n (x, y))n=0,N , N−1

ℰ1n (x, y) = λ ∑ G1 (n, i)(f (i, [xi − pi ] , [yi − qi ] ) + ci ) ∗

∗

i=1

N−1

+ μ ∑ G2 (n, i)(g(i, [xi − pi ]∗ , [yi − qi ]∗ ) + di ), i=1

N−1

n = 0, N,

ℰ2n (x, y) = μ ∑ G3 (n, i)(g(i, [xi − pi ] , [yi − qi ] ) + di ) ∗

∗

i=1

N−1

+ λ ∑ G4 (n, i)(f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + ci ), i=1

n = 0, N,

and ℰ (x, y) = (ℰ1 (x, y), ℰ2 (x, y)), for (x, y) = ((xn )n=0,N , (yn )n=0,N ). By using system (5.7) and the above operator ℰ , we obtain the following lemma. Lemma 5.1.1. Under assumptions (H1) and (H2), the solutions of problem (5.7) (and hence of problem (5.3), (5.4)) coincide with the fixed points of operator ℰ in the space Y . Lemma 5.1.2. If (H1) and (H2) hold, then the operator ℰ : L → L is a completely continuous operator. Proof. Let (x, y) ∈ L be an arbitrary element. Because ℰ1 (x, y) and ℰ2 (x, y) satisfy the problem (4.21), (4.20) for x̃n = λ(f (n, [xn − pn ]∗ , [yn − qn ]∗ ) + cn ) and ̃yn = μ(g(n, [xn − pn ]∗ , [yn − qn ]∗ ) + dn ), n = 1, N − 1, by Lemma 4.2.3 we obtain 󵄩

󵄩

∀ n = 0, N,

󵄩

󵄩

∀ n = 0, N,

ℰ1n (x, y) ≥ γ1 n max ℰ1m (x, y) = γ1 n󵄩󵄩󵄩ℰ1 (x, y)󵄩󵄩󵄩, m=0,N

ℰ2n (x, y) ≥ γ2 n max ℰ2m (x, y) = γ2 n󵄩󵄩󵄩ℰ2 (x, y)󵄩󵄩󵄩, m=0,N

so ℰ1 (x, y) ∈ L1 , ℰ2 (x, y) ∈ L2 , and (ℰ1 (x, y), ℰ2 (x, y)) ∈ L. Hence, ℰ (L) ⊂ L. By using standard arguments, we deduce that the operators ℰ1 and ℰ2 are completely continuous, and then ℰ is a completely continuous operator.

122 � 5 Systems with sign-changing nonlinearities Remark 5.1.1. Our problem (5.1), (5.2) was reduced to the fixed point problem for operator ℰ into three steps: (i) reduction to problem (5.3), (5.4); (ii) reduction to the system (5.7), and (iii) reduction to the fixed point problem for operator ℰ in the cone L. Theorem 5.1.1. Assume that (H1), (H2), and (H3) hold. Then there exist constants λ0 > 0 and μ0 > 0 such that for any λ ∈ (0, λ0 ] and μ ∈ (0, μ0 ], the problem (5.1), (5.2) has at least one positive solution. Proof. Let δ ∈ (0, 1) be fixed. By using (H2) and (H3), there exists R0 ∈ (0, 1] such that f (n, u, v) ≥ δf (n, 0, 0), ∀ n = 1, N − 1,

g(n, u, v) ≥ δg(n, 0, 0),

u, v ∈ [0, R0 ].

(5.8)

We define f ̄(R0 ) = ̄ 0) = g(R

max

{f (n, u, v) + cn },

max

{g(n, u, v) + dn },

n=1,N−1, u,v∈[0,R0 ] n=1,N−1, u,v∈[0,R0 ]

σ1 (N 2 − 1) , 6 σ (N 2 − 1) p̃3 = 3 , 6 R0 λ0 = max{ 8p̃ f ̄(R p̃1 =

1

μ0 = max{

σ2 (N 2 − 1) , 6 σ (N 2 − 1) p̃4 = 4 , 6 R0 , }, ̄ ̃ ) 8 p 0 4 f (R0 ) p̃2 =

R0 R0 , }. ̄ 0 ) 8p̃3 g(R ̄ 0) 8p̃2 g(R

We see that f ̄(R0 ) ≥ max {δf (n, 0, 0) + cn } > 0, n=1,N−1

̄ 0 ) ≥ max {δg(n, 0, 0) + dn } > 0. g(R n=1,N−1

We will show that for any λ ∈ (0, λ0 ] and μ ∈ (0, μ0 ], problem (5.3), (5.4) has at least one positive solution. So, let λ ∈ (0, λ0 ] and μ ∈ (0, μ0 ] be arbitrary, but fixed for the moment. We define the set U = {(x, y) ∈ L, x = (xn )n=0,N , y = (yn )n=0,N , ‖(x, y)‖Y < R0 }. We suppose that there exist (x, y) ∈ 𝜕U (‖x‖ + ‖y‖ = R0 ) and ν ∈ (0, 1) such that (x, y) = νℰ (x, y), or x = νℰ1 (x, y) and y = νℰ2 (x, y). We deduce that [xi − pi ]∗ = xi − pi ≤ xi ≤ R0 ,

[xi − pi ] = 0, ∗

if xi − pi ≥ 0,

if xi − pi < 0, ∀ i = 0, N,

5.1 Positive parameters in the system of difference equations

[yi − qi ]∗ = yi − qi ≤ yi ≤ R0 , [yi − qi ] = 0,

� 123

if yi − qi ≥ 0,

if yi − qi < 0, ∀ i = 0, N.

∗

Then by Lemma 4.2.2, for all n = 0, N we obtain xn = νℰ1n (x, y) ≤ ℰ1n (x, y) N−1

N−1

i=1

i=1

̄ 0) ≤ λσ1 ∑ h(i)f ̄(R0 ) + μσ2 ∑ h(i)g(R ̄ 0) ≤ ≤ λ0 p̃1 f ̄(R0 ) + μ0 p̃2 g(R yn = νℰ2n (x, y) ≤ ℰ2n (x, y)

R0 R0 R0 + = , 8 8 4

N−1

N−1

i=1

i=1

̄ 0 ) + λσ4 ∑ h(i)f ̄(R0 ) ≤ μσ3 ∑ h(i)g(R R R R ̄ 0 ) + λ0 p̃4 f ̄(R0 ) ≤ 0 + 0 = 0 . ≤ μ0 p̃3 g(R 8 8 4 2 Here we used the relation ∑N−1 i=1 h(i) = (N − 1)/6. R R R R Then we obtain ‖x‖ ≤ 40 and ‖y‖ ≤ 40 . Hence, R0 = ‖(x, y)‖Y = ‖x‖ + ‖y‖ ≤ 40 + 40 = R0 , which is a contradiction. 2 Therefore, by Lemma 5.1.2 and Theorem 1.2.1 (with Ω = L), we deduce that ℰ has a fixed point (x 0 , y0 ) ∈ U. That is, (x 0 , y0 ) = ℰ (x 0 , y0 ) or x 0 = ℰ1 (x 0 , y0 ), y0 = ℰ2 (x 0 , y0 ) (⇔ xn0 = ℰ1n (x 0 , y0 ), y0n = ℰ2n (x 0 , y0 ), n = 0, N, x 0 = (xn0 )n=0,N , y0 = (y0n )n=0,N ), and ‖x 0 ‖ + ‖y0 ‖ ≤ R0 with xn0 ≥ γ1 n‖x 0 ‖ and y0n ≥ γ2 n‖y0 ‖ for all n = 0, N. Moreover, by (5.8), we conclude that

xn0 = ℰ1n (x 0 , y0 ) N−1

N−1

≥ λ ∑ G1 (n, i)(δf (i, 0, 0) + ci ) + μ ∑ G2 (n, i)(δg(i, 0, 0) + di ) i=1

N−1

N−1

i=1

i=1

i=1

≥ λ ∑ G1 (n, i)ci + μ ∑ G2 (n, i)di = pn , N−1

N−1

i=1

i=1

xn0 > λ ∑ G1 (n, i)ci + μ ∑ G2 (n, i)di = pn , y0n = ℰ2n (x 0 , y0 ) N−1

n = 0, N, n = 1, N,

N−1

≥ μ ∑ G3 (n, i)(δg(i, 0, 0) + di ) + λ ∑ G4 (n, i)(δf (i, 0, 0) + ci ) i=1

N−1

N−1

i=1

i=1

i=1

≥ μ ∑ G3 (n, i)di + λ ∑ G4 (n, i)ci = qn ,

n = 0, N,

124 � 5 Systems with sign-changing nonlinearities N−1

N−1

i=1

i=1

y0n > μ ∑ G3 (n, i)di + λ ∑ G4 (n, i)ci = qn ,

n = 1, N.

Then xn0 ≥ pn , y0n ≥ qn for all n = 0, N and xn0 > pn , y0n > qn for all n = 1, N. Let u = (un0 )n=0,N , v0 = (v0n )n=0,N , un0 = xn0 − pn , v0n = y0n − qn for all n = 0, N. Then un0 ≥ 0, v0n ≥ 0 for all n = 0, N and un0 > 0, v0n > 0 for all n = 1, N. Therefore, (u0 , v0 ) is a positive solution of (5.1), (5.2). 0

Theorem 5.1.2. Assume that (H1), (H2), and (H4) hold. Then there exist λ∗ > 0 and μ∗ > 0 such that for any λ ∈ (0, λ∗ ] and μ ∈ (0, μ∗ ], the problem (5.1), (5.2) has at least one positive solution. Proof. We choose a positive number R1 such that R1 > max{1,

2 N−1 2 N−1 ∑ (δ1 cn + δ2 dn ), ∑ (δ d + δ4 cn ), γ1 n=1 γ2 n=1 3 n

2 (∑ a ξ ) γ1 γ2 i=1 i i

−1

q

−1

p

2 (∑ b η ) γ1 γ2 i=1 i i

N−1

∑ (δ1 cn + δ2 dn ),

n=1

N−1

∑ (δ3 dn + δ4 cn )},

n=1

and we define the set 󵄩 󵄩 Ω1 = {(x, y) ∈ Y , x = (xn )n=0,N , y = (yn )n=0,N , 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y < R1 }. We introduce the numbers λ∗ = min{1, μ∗ = min{1,

3R1 3R1 , }, 2 2σ1 M1 (N − 1) 2σ4 M1 (N 2 − 1)

3R1 3R1 , }, 2 2σ2 M2 (N − 1) 2σ3 M2 (N 2 − 1)

with M1 = max{

max

{f (n, u, v) + cn }, 1},

M2 = max{

max

{g(n, u, v) + dn }, 1}.

n=1,N−1, u,v≥0, u+v≤R1

n=1,N−1, u,v≥0, u+v≤R1

Let λ ∈ (0, λ∗ ] and μ ∈ (0, μ∗ ]. Then for any (x, y) ∈ L ∩ 𝜕Ω1 , x = (xn )n=0,N , y = (yn )n=0,N , we have [xn − pn ]∗ ≤ xn ≤ ‖x‖,

[yn − qn ]∗ ≤ yn ≤ ‖y‖,

∀ n = 0, N.

5.1 Positive parameters in the system of difference equations

� 125

Then for any (x, y) ∈ L ∩ 𝜕Ω1 , we obtain N−1

󵄩 󵄩󵄩 ∗ ∗ 󵄩󵄩ℰ1 (x, y)󵄩󵄩󵄩 ≤ λσ1 ∑ h(i)[f (i, [xi − pi ] , [yi − qi ] ) + ci ] i=1

N−1

+ μσ2 ∑ h(i)[g(i, [xi − pi ]∗ , [yi − qi ]∗ ) + di ] i=1

N−1

N−1

i=1

i=1

≤ λ∗ σ1 M1 ∑ h(i) + μ∗ σ2 M2 ∑ h(i) R R R ‖(x, y)‖Y ≤ 1 + 1 = 1 = , 4 4 2 2 N−1

󵄩󵄩 󵄩 ∗ ∗ 󵄩󵄩ℰ2 (x, y)󵄩󵄩󵄩 ≤ μσ3 ∑ h(i)[g(i, [xi − pi ] , [yi − qi ] ) + di ] i=1

N−1

+ λσ4 ∑ h(i)[f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + ci ] i=1

N−1

N−1

i=1

i=1

≤ μ∗ σ3 M2 ∑ h(i) + λ∗ σ4 M1 ∑ h(i) R R ‖(x, y)‖Y R . ≤ 1 + 1 = 1 = 4 4 2 2 Therefore, 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩ℰ (x, y)󵄩󵄩󵄩Y = 󵄩󵄩󵄩ℰ1 (x, y)󵄩󵄩󵄩 + 󵄩󵄩󵄩ℰ2 (x, y)󵄩󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y , ∀ (x, y) ∈ L ∩ 𝜕Ω1 . On the other hand, we choose a constant L > 0 such that L ≥ max{

24 24 , , λϱ1 γ1 (N 2 − 1) λϱ4 γ2 (N 2 − 1)

24 24 , }. μϱ2 γ1 (N 2 − 1) μϱ3 γ2 (N 2 − 1)

From (H4), we deduce that there exists a constant M0 > 0 such that f (n, u, v) ≥ L(u + v) or

g(n, u, v) ≥ L(u + v),

∀ n = 1, N − 1, u, v ≥ 0, u + v ≥ M0 . Now we define R2 = max{2R1 ,

4M0 4M0 4 N−1 , , ∑ (δ c + δ2 di ), γ1 γ2 γ1 i=1 1 i

(5.9)

126 � 5 Systems with sign-changing nonlinearities 4 N−1 ∑ (δ d + δ4 ci )} > 0, γ2 i=1 3 i and let 󵄩 󵄩 Ω2 = {(x, y) ∈ Y , x = (xn )n=0,N , y = (yn )n=0,N , 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y < R2 }. First we suppose that f∞ = ∞, that is, f (n, u, v) ≥ L(u + v),

∀n = 1, N − 1, u, v ≥ 0, u + v ≥ M0 .

(5.10)

Then for any (x, y) ∈ L ∩ 𝜕Ω2 , we have ‖(x, y)‖Y = R2 or ‖x‖ + ‖y‖ = R2 . We deduce that ‖x‖ ≥ R2 /2 or ‖y‖ ≥ R2 /2. We suppose firstly that ‖x‖ ≥ R2 /2. Then for any (x, y) ∈ L ∩ 𝜕Ω2 , we obtain N−1

N−1

xn − pn = xn − λ ∑ G1 (n, i)ci − μ ∑ G2 (n, i)di i=1

i=1

N−1

N−1

i=1

i=1

≥ xn − n(δ1 ∑ ci + δ2 ∑ di ) ≥ xn −

N−1

xn ( ∑ (δ c + δ2 di )) γ1 ‖x‖ i=1 1 i

= xn [1 −

N−1 1 ( ∑ (δ1 ci + δ2 di ))] γ1 ‖x‖ i=1

≥ xn [1 −

N−1 2 ( ∑ (δ1 ci + δ2 di ))] γ1 R2 i=1

1 ≥ xn ≥ 0, 2

∀ n = 0, N.

Therefore, we conclude 1 1 [xn − pn ]∗ = xn − pn ≥ xn ≥ γ1 n‖x‖ 2 2 1 1 ≥ γ1 nR2 ≥ γ1 R2 ≥ M0 , ∀ n = 1, N − 1. 4 4 Hence, [xn − pn ]∗ + [yn − qn ]∗ ≥ [xn − pn ]∗ = xn − pn ≥ M0 ,

∀ n = 1, N − 1.

(5.11)

5.1 Positive parameters in the system of difference equations

� 127

Then for any (x, y) ∈ L ∩ 𝜕Ω2 and n = 1, N − 1, by (5.10) and (5.11) we deduce that f (n, [xn − pn ]∗ , [yn − qn ]∗ ) ≥ L([xn − pn ]∗ + [yn − qn ]∗ ) ≥ L([xn − pn ]∗ ) ≥

L x , 2 n

∀ n = 1, N − 1.

It follows that for any (x, y) ∈ L ∩ 𝜕Ω2 and n = 1, N − 1 N−1

ℰ1n (x, y) ≥ λ ∑ G1 (n, i)(f (i, [xi − pi ] , [yi − qi ] ) + ci ) ∗

∗

i=1

N−1 N−1 1 ≥ λ ∑ G1 (n, i)L([xi − pi ]∗ ) ≥ λ ∑ G1 (n, i) Lγ1 R2 4 i=1 i=1 N−1 N−1 1 1 ≥ λ ∑ ϱ1 nh(i) Lγ1 R2 ≥ λ ϱ1 Lγ1 R2 ∑ h(i) ≥ R2 . 4 4 i=1 i=1

Then ‖ℰ1 (x, y)‖ ≥ ‖(x, y)‖Y and 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩ℰ (x, y)󵄩󵄩󵄩Y ≥ 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y ,

∀ (x, y) ∈ L ∩ 𝜕Ω2 .

If ‖y‖ ≥ R2 /2, then for any (x, y) ∈ L ∩ 𝜕Ω2 , we conclude that N−1

N−1

yn − qn = yn − μ ∑ G3 (n, i)di − λ ∑ G4 (n, i)ci i=1

i=1

N−1

N−1

i=1

i=1

≥ yn − n(δ3 ∑ di + δ4 ∑ ci ) ≥ yn −

N−1

yn ( ∑ (δ d + δ4 ci )) γ2 ‖y‖ i=1 3 i

= yn [1 −

N−1 1 ( ∑ (δ3 di + δ4 ci ))] γ2 ‖y‖ i=1

≥ yn [1 −

N−1 2 ( ∑ (δ3 di + δ4 ci ))] γ2 R2 i=1

1 ≥ yn ≥ 0, 2

∀ n = 0, N.

Therefore, we deduce that 1 1 [yn − qn ]∗ = yn − qn ≥ yn ≥ γ2 n‖y‖ 2 2 1 1 ≥ γ2 nR2 ≥ γ2 R2 ≥ M0 , ∀ n = 1, N − 1. 4 4

(5.12)

128 � 5 Systems with sign-changing nonlinearities Hence, [xn − pn ]∗ + [yn − qn ]∗ ≥ [yn − qn ]∗

= yn − qn ≥ M0 ,

∀ n = 1, N − 1.

(5.13)

Then for any (x, y) ∈ L ∩ 𝜕Ω2 and n = 1, N − 1, by (5.10) and (5.13) we obtain f (n, [xn − pn ]∗ , [yn − qn ]∗ ) ≥ L([xn − pn ]∗ + [yn − qn ]∗ ) ≥ L([yn − qn ]∗ ) ≥

L y , 2 n

∀ n = 1, N − 1.

It follows that for any (x, y) ∈ L ∩ 𝜕Ω2 and n = 1, N − 1 N−1

ℰ2n (x, y) ≥ λ ∑ G4 (n, i)(f (i, [xi − pi ] , [yi − qi ] ) + ci ) ∗

∗

i=1

N−1

N−1 1 ≥ λ ∑ G4 (n, i)L([yi − qi ]∗ ) ≥ λ ∑ G4 (n, i) Lγ2 R2 4 i=1 i=1 N−1 N−1 1 1 ≥ λ ∑ ϱ4 nh(i) Lγ2 R2 ≥ λ ϱ4 Lγ2 R2 ∑ h(i) ≥ R2 . 4 4 i=1 i=1

Then ‖ℰ2 (x, y)‖ ≥ ‖(x, y)‖Y and we obtain relation (5.12). We suppose now that g∞ = ∞, that is, g(n, u, v) ≥ L(u + v) for all n = 1, N − 1 and u, v ≥ 0, u + v ≥ M0 . Then for any (x, y) ∈ L ∩ 𝜕Ω2 , we have ‖(x, y)‖Y = R2 . Hence, ‖x‖ ≥ R2 /2 or ‖y‖ ≥ R2 /2. If ‖x‖ ≥ R2 /2, then for any (x, y) ∈ L ∩ 𝜕Ω2 , we deduce in a similar manner as above that xn − pn ≥ 21 xn for all n = 0, N, and N−1

ℰ1n (x, y) ≥ μ ∑ G2 (n, i)(g(i, [xi − pi ] , [yi − qi ] ) + di ) ∗

∗

i=1

N−1 N−1 1 ≥ μ ∑ G2 (n, i)L([xi − pi ]∗ ) ≥ μ ∑ G2 (n, i) Lγ1 R2 4 i=1 i=1 N−1 N−1 1 1 ≥ μ ∑ ϱ2 nh(i) Lγ1 R2 ≥ μ ϱ2 Lγ1 R2 ∑ h(i) 4 4 i=1 i=1

≥ R2 ,

∀ n = 1, N − 1.

Hence, we obtain the relation (5.12). If ‖y‖ ≥ R2 /2, then for any (x, y) ∈ L ∩ 𝜕Ω2 , we deduce in a similar manner as above that yn − qn ≥ 21 yn for all n = 0, N, and N−1

ℰ2n (x, y) ≥ μ ∑ G3 (n, i)(g(i, [xi − pi ] , [yi − qi ] ) + di ) i=1

∗

∗

5.1 Positive parameters in the system of difference equations

� 129

N−1 N−1 1 ≥ μ ∑ G3 (n, i)L([yi − qi ]∗ ) ≥ μ ∑ G3 (n, i) Lγ2 R2 4 i=1 i=1 N−1 N−1 1 1 ≥ μ ∑ ϱ3 nh(i) Lγ2 R2 ≥ μ ϱ3 Lγ2 R2 ∑ h(i) 4 4 i=1 i=1

≥ R2 ,

∀ n = 1, N − 1.

Hence, we obtain again the relation (5.12). Therefore, by Lemma 5.1.2, Theorem 1.1.1, and relations (5.9) and (5.12), we conclude that operator ℰ has a fixed point (x 1 , y1 ) ∈ L ∩ (Ω2 \ Ω1 ), that is, R1 ≤ ‖(x 1 , y1 )‖Y ≤ R2 . Since ‖(x 1 , y1 )‖Y ≥ R1 , we have ‖x 1 ‖ ≥ R1 /2 or ‖y1 ‖ ≥ R1 /2. This time, we suppose firstly that ‖x 1 ‖ ≥ R1 /2. Then we deduce N−1

N−1

xn1 − pn = xn1 − λ ∑ G1 (n, i)ci − μ ∑ G2 (n, i)di i=1

i=1

N−1

N−1

i=1

i=1

≥ xn1 − n(δ1 ∑ ci + δ2 ∑ di ) ≥ xn1 −

xn1

N−1

γ1 ‖x 1 ‖

∑ (δ1 ci + δ2 di ) i=1

N−1

≥ [1 −

2 ∑ (δ c + δ2 di )]xn1 γ1 R1 i=1 1 i

≥ [1 −

2 N−1 󵄩 󵄩 ∑ (δ c + δ2 di )]γ1 n󵄩󵄩󵄩x 1 󵄩󵄩󵄩 γ1 R1 i=1 1 i

≥

R1 2 N−1 [1 − ∑ (δ c + δ2 di )]γ1 n = Λ1 n, 2 γ1 R1 i=1 1 i

∀ n = 0, N,

so xn1 ≥ pn + Λ1 n for all n = 0, N, where Λ1 = q

γ1 R1 N−1 − ∑ (δ1 ci + δ2 di ) > 0. 2 i=1

q

Then y1N = ∑i=1 bi xη1 i ≥ Λ1 ∑i=1 bi ηi > 0 and q

q

q

󵄩󵄩 1 󵄩󵄩 󵄩 1󵄩 γ R 1 1 󵄩󵄩y 󵄩󵄩 ≥ yN = ∑ bi xηi ≥ ∑ bi γ1 ηi 󵄩󵄩󵄩x 󵄩󵄩󵄩 ≥ 1 1 (∑ bi ηi ) > 0. 2 i=1 i=1 i=1 Therefore, we obtain N−1

N−1

i=1

i=1

y1n − qn = y1n − μ ∑ G3 (n, i)di − λ ∑ G4 (n, i)ci

130 � 5 Systems with sign-changing nonlinearities N−1

N−1

i=1

i=1

≥ y1n − n(δ3 ∑ di + δ4 ∑ ci ) y1n N−1 ≥ y1n − ∑ (δ3 di + δ4 ci ) γ2 ‖y1 ‖ i=1 ≥

y1n [1

−1

q

N−1 2 (∑ bi ηi ) ( ∑ (δ3 di + δ4 ci ))] − γ1 γ2 R1 i=1 i=1 q

−1

N−1 2 󵄩 󵄩 ≥ γ2 n󵄩󵄩󵄩y1 󵄩󵄩󵄩[1 − (∑ bi ηi ) ( ∑ (δ3 di + δ4 ci ))] γ1 γ2 R1 i=1 i=1 q

q

−1

γγR 2 ≥ 1 2 1 n(∑ bi ηi )[1 − (∑ bi ηi ) 2 γ γ 1 2 R1 i=1 i=1 N−1

× ( ∑ (δ3 di + δ4 ci ))] = Λ2 n, i=1

∀ n = 0, N,

where Λ2 =

q

N−1 γ1 γ2 R1 (∑ bi ηi ) − ∑ (δ3 di + δ4 ci ) > 0. 2 i=1 i=1

Hence, y1n ≥ qn + Λ2 n for all n = 0, N. If ‖y1 ‖ ≥ R1 /2, then N−1

N−1

y1n − qn = y1n − μ ∑ G3 (n, i)di − λ ∑ G4 (n, i)ci i=1

i=1

N−1

N−1

i=1

i=1

≥ y1n − n(δ3 ∑ di + δ4 ∑ ci ) y1n N−1 ≥ y1n − ∑ (δ3 di + δ4 ci ) γ2 ‖y1 ‖ i=1 ≥ [1 −

2 N−1 ∑ (δ d + δ4 ci )]y1n γ2 R1 i=1 3 i

≥ [1 −

2 N−1 󵄩 󵄩 ∑ (δ d + δ4 ci )]γ2 n󵄩󵄩󵄩y1 󵄩󵄩󵄩 γ2 R1 i=1 3 i

≥

R1 2 N−1 [1 − ∑ (δ d + δ4 ci )]γ2 n = Λ3 n, 2 γ2 R1 i=1 3 i

so y1n ≥ qn + Λ3 n for all n = 0, N, where

∀ n = 0, N,

5.1 Positive parameters in the system of difference equations

Λ3 = p

� 131

γ2 R1 N−1 − ∑ (δ3 di + δ4 ci ) > 0. 2 i=1

p

Then xN1 = ∑i=1 ai y1ξi ≥ Λ3 ∑i=1 ai ξi > 0 and p

p

p

󵄩󵄩 1 󵄩󵄩 󵄩 1󵄩 γ R 1 1 󵄩󵄩x 󵄩󵄩 ≥ xN = ∑ ai yξi ≥ ∑ ai γ2 ξi 󵄩󵄩󵄩y 󵄩󵄩󵄩 ≥ 2 1 (∑ ai ξi ) > 0. 2 i=1 i=1 i=1 Therefore, we obtain N−1

N−1

xn1 − pn = xn1 − λ ∑ G1 (n, i)ci − μ ∑ G2 (n, i)di i=1

i=1

N−1

N−1

i=1

i=1

≥ xn1 − n(δ1 ∑ ci + δ2 ∑ di ) xn1 N−1 ≥ xn1 − ∑ (δ1 ci + δ2 di ) γ1 ‖x 1 ‖ i=1 ≥

xn1 [1

−1

p

N−1 2 (∑ ai ξi ) ( ∑ (δ1 ci + δ2 di ))] − γ1 γ2 R1 i=1 i=1

󵄩 󵄩 ≥ γ1 n󵄩󵄩󵄩x 1 󵄩󵄩󵄩[1 −

p

−1

N−1 2 (∑ ai ξi ) ( ∑ (δ1 ci + δ2 di ))] γ1 γ2 R1 i=1 i=1

p

p

−1

γγR 2 ≥ 1 2 1 n(∑ ai ξi )[1 − (∑ ai ξi ) 2 γ γ 1 2 R1 i=1 i=1 N−1

× ( ∑ (δ1 ci + δ2 di ))] = Λ4 n, i=1

∀ n = 0, N,

where Λ4 =

p

N−1 γ1 γ2 R1 (∑ ai ξi ) − ∑ (δ1 ci + δ2 di ) > 0. 2 i=1 i=1

Hence, xn1 ≥ pn + Λ4 n for all n = 0, N. Let un1 = xn1 − pn and v1n = y1n − qn for all n = 0, N. Then un1 ≥ Λ5 n and v1n ≥ Λ6 n for all n = 0, N, where Λ5 = min{Λ1 , Λ4 } and Λ6 = min{Λ2 , Λ3 }. This completes the proof. Theorem 5.1.3. Assume that (H1), (H2), (H3), and (H4) hold. Then the boundary value problem (5.1), (5.2) has at least two positive solutions for any λ > 0 and μ > 0 sufficiently small.

132 � 5 Systems with sign-changing nonlinearities Proof. By Theorems 5.1.1 and 5.1.2 we deduce that for 0 < λ ≤ min{λ0 , λ∗ } and 0 < μ ≤ min{μ0 , μ∗ }, problem (5.1), (5.2) has at least two positive solutions (u0 , v0 ) and (u1 , v1 ) with ‖(u0 + p̃, v0 + q̃)‖Y ≤ 1 and ‖(u1 + p̃, v1 + q̃)‖Y > 1, where p̃ = (pn )n=0,N , q̃ = (qn )n=0,N . Theorem 5.1.4. Assume that λ = μ, assumptions (H1), (H2), and (H5) hold, and ∑N−1 i=1 ci > 0 or ∑N−1 d > 0. Additionally, we make the following assumption. i=1 i (H6) We have i f∞ = lim inf min f (n, u, v) > L0

or

u+v→∞

u,v≥0 n=1,N−1

i g∞ = lim inf min g(n, u, v) > L0 , u+v→∞

u,v≥0 n=1,N−1

where L0 = max{

4 N−1 4 N−1 ∑ (δ1 ci + δ2 di ), ∑ (δ c + δ4 di ), γ1 i=1 γ2 i=1 3 i q

−1

p

−1

4 (∑ b η ) γ1 γ2 i=1 i i 4 (∑ a ξ ) γ1 γ2 i=1 i i

N−1

∑ (δ3 di + δ4 ci ), i=1

N−1

∑ (δ1 ci + δ2 di )}(min{ϱ1 , ϱ2 } i=1

N2 − 1 ) . 6 −1

Then there exists λ∗ > 0 such that for any λ ≥ λ∗ , problem (5.1), (5.2) (with λ = μ) has at least one positive solution. Proof. By (H6) we conclude that there exists M3 > 0 such that f (n, u, v) ≥ L0

or

∀ n = 1, N − 1, u, v ≥ 0, u + v ≥ M3 .

g(n, u, v) ≥ L0 ,

We define −1

N−1

N−1

−1

λ∗ = max{M3 ( ∑ (δ1 ci + δ2 di )) , M3 ( ∑ (δ3 di + δ4 ci )) } > 0. i=1

i=1

We assume now that λ ≥ λ∗ . Let R3 = max{

4λ N−1 4λ N−1 ∑ (δ1 ci + δ2 di ), ∑ (δ d + δ4 ci ), γ1 i=1 γ2 i=1 3 i q

−1

p

−1

4λ (∑ b η ) γ1 γ2 i=1 i i 4λ (∑ a ξ ) γ1 γ2 i=1 i i

N−1

∑ (δ3 di + δ4 ci ), i=1

N−1

∑ (δ1 ci + δ2 di )} > 0, i=1

5.1 Positive parameters in the system of difference equations

� 133

and 󵄩 󵄩 Ω3 = {(x, y) ∈ Y , 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y < R3 }. i We suppose first that f∞ > L0 , that is, f (n, u, v) ≥ L0 for all n = 1, N − 1 and u, v ≥ 0, u + v ≥ M3 . Let (x, y) ∈ L ∩ 𝜕Ω3 . Then ‖(x, y)‖Y = R3 , so ‖x‖ ≥ R3 /2 or ‖y‖ ≥ R3 /2. We assume that ‖x‖ ≥ R3 /2. Then for all n = 0, N, we deduce that N−1

N−1

i=1

i=1

xn − pn ≥ γ1 n‖x‖ − λδ1 n ∑ ci − λδ2 n ∑ di ≥ n[

N−1

γ1 R3 − λ ∑ (δ1 ci + δ2 di )] 2 i=1 N−1

N−1

≥ n[2λ ∑ (δ1 ci + δ2 di ) − λ ∑ (δ1 ci + δ2 di )] i=1

i=1

N−1

N−1

i=1

i=1

= nλ ∑ (δ1 ci + δ2 di ) ≥ nλ∗ ∑ (δ1 ci + δ2 di ) ≥ M3 n ≥ 0. Therefore, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1, we have [xn − pn ]∗ + [yn − qn ]∗ ≥ [xn − pn ]∗ = xn − pn ≥ M3 n ≥ M3 .

(5.14)

Hence, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1, we conclude that N−1

ℰ1n (x, y) ≥ λ ∑ G1 (n, i)(f (i, [xi − pi ] , [yi − qi ] ) + ci ) i=1

∗

∗

N−1

≥ λϱ1 n ∑ h(i)f (i, [xi − pi ]∗ , [yi − qi ]∗ ) i=1

N−1

N−1

i=1

i=1

󵄩 󵄩 ≥ λL0 ϱ1 n ∑ h(i) ≥ λL0 ϱ1 ∑ h(i) ≥ R3 = 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y . Therefore, we obtain ‖ℰ1 (x, y)‖ ≥ R3 for all (x, y) ∈ L ∩ 𝜕Ω3 , so 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩ℰ (x, y)󵄩󵄩󵄩Y ≥ R3 = 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y ,

∀ (x, y) ∈ P ∩ 𝜕Ω3 .

If ‖y‖ ≥ R3 /2, then for all n = 0, N we deduce N−1

N−1

i=1

i=1

yn − qn ≥ γ2 n‖y‖ − λδ3 n ∑ di − λδ4 n ∑ ci ≥ n[

N−1

γ2 R3 − λ ∑ (δ3 di + δ4 ci )] 2 i=1

(5.15)

134 � 5 Systems with sign-changing nonlinearities N−1

N−1

≥ n[2λ ∑ (δ3 di + δ4 ci ) − λ ∑ (δ3 di + δ4 ci )] i=1

i=1

N−1

N−1

i=1

i=1

= nλ ∑ (δ3 di + δ4 ci ) ≥ nλ∗ ∑ (δ3 di + δ4 ci ) ≥ M3 n ≥ 0. Therefore, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1, we have [xn − pn ]∗ + [yn − qn ]∗ ≥ [yn − qn ]∗ = yn − qn ≥ M3 n ≥ M3 .

(5.16)

Hence, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1, we obtain in a similar manner as above that ℰ1n (x, y) ≥ R3 = ‖(x, y)‖Y . Hence, ‖ℰ1 (x, y)‖ ≥ R3 for all (x, y) ∈ L ∩ 𝜕Ω3 and we deduce again relation (5.15). i We suppose now that g∞ > L0 , that is, g(n, u, v) ≥ L0 , for all n = 1, N − 1 and u, v ≥ 0, u + v ≥ M3 . Let (x, y) ∈ L ∩ 𝜕Ω3 . Then ‖(x, y)‖Y = R3 , so ‖x‖ ≥ R3 /2 or ‖y‖ ≥ R3 /2. i If ‖x‖ ≥ R3 /2, then we obtain in a similar manner as in the first case (f∞ > L0 ) that xn − pn ≥ M3 n ≥ 0 for all n = 0, N. Therefore, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1 we deduce inequalities (5.14). Hence, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1, we conclude that N−1

ℰ1n (x, y) ≥ λ ∑ G2 (n, i)(g(i, [xi − pi ] , [yi − qi ] ) + di ) i=1

∗

∗

N−1

≥ λϱ2 n ∑ h(i)g(i, [xi − pi ]∗ , [yi − qi ]∗ ) i=1

N−1

N−1

i=1

i=1

󵄩 󵄩 ≥ λL0 ϱ2 n ∑ h(i) ≥ λL0 ϱ2 ∑ h(i) ≥ R3 = 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y . Therefore, we obtain ‖ℰ1 (x, y)‖ ≥ R3 , so ‖ℰ (x, y)‖Y ≥ R3 = ‖(x, y)‖Y for all (x, y) ∈ L ∩ 𝜕Ω3 , that is, we have relation (5.15). i If ‖y‖ ≥ R3 /2, then we conclude in a similar manner as in the first case (f∞ > L0 ) that yn − qn ≥ M3 n ≥ 0 for all n = 0, N. Therefore, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1 we deduce inequalities (5.16). Hence, for any (x, y) ∈ L ∩ 𝜕Ω3 and n = 1, N − 1, we obtain in a similar manner as above that ℰ1n (x, y) ≥ R3 = ‖(x, y)‖Y . Hence, ‖ℰ1 (x, y)‖ ≥ R3 and ‖ℰ (x, y)‖Y ≥ ‖(x, y)‖Y for all (x, y) ∈ L ∩ 𝜕Ω3 , that is, we have relation (5.15). On the other hand, we consider the positive number ε=

3 −1 (max{σi , i = 1, 4}) . 2 4λ(N − 1)

5.1 Positive parameters in the system of difference equations

Then by (H5) we deduce that there exists M4 > 0 such that 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨f (n, u, v)󵄨󵄨󵄨 ≤ ε(u + v), 󵄨󵄨󵄨g(n, u, v)󵄨󵄨󵄨 ≤ ε(u + v), ∀ n = 1, N − 1, u, v ≥ 0, u + v ≥ M4 . Therefore, we obtain 󵄨 󵄨󵄨 󵄨󵄨f (n, u, v)󵄨󵄨󵄨 ≤ M5 + ε(u + v), ∀ n = 1, N − 1, u, v ≥ 0,

󵄨 󵄨󵄨 󵄨󵄨g(n, u, v)󵄨󵄨󵄨 ≤ M5 + ε(u + v),

M5 = max{

󵄨󵄨 󵄨 󵄨󵄨f (n, u, v)󵄨󵄨󵄨,

where max

n=1,N−1, u,v≥0, u+v≤M4

max

n=1,N−1, u,v≥0, u+v≤M4

󵄨󵄨 󵄨 󵄨󵄨g(n, u, v)󵄨󵄨󵄨}.

We define now N−1

N−1

R4 = max{2R3 , 8λσ1 ∑ h(i)(M5 + ci ), 8λσ2 ∑ h(i)(M5 + di ), i=1

i=1

N−1

N−1

i=1

i=1

8λσ3 ∑ h(i)(M5 + di ), 8λσ4 ∑ h(i)(M5 + ci )}, and we let 󵄩 󵄩 Ω4 = {(x, y) ∈ Y , 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y < R4 }. For any (x, y) ∈ L ∩ 𝜕Ω4 , we have N−1

ℰ1n (x, y) ≤ λ ∑ σ1 h(i)[f (i, [xi − pi ] , [yi − qi ] ) + ci ] ∗

∗

i=1

N−1

+ λ ∑ σ2 h(i)[g(i, [xi − pi ]∗ , [yi − qi ]∗ ) + di ] i=1

N−1

≤ λσ1 ∑ h(i)[M5 + ε([xi − pi ]∗ + [yi − qi ]∗ ) + ci ] i=1

N−1

+ λσ2 ∑ h(i)[M5 + ε([xi − pi ]∗ + [yi − qi ]∗ ) + di ] i=1

N−1

N−1

i=1

i=1

≤ λσ1 ∑ h(i)(M5 + ci ) + λσ1 εR4 ∑ h(i)

� 135

136 � 5 Systems with sign-changing nonlinearities N−1

N−1

i=1

i=1

+ λσ2 ∑ h(i)(M5 + di ) + λσ2 εR4 ∑ h(i) R R R R R 1󵄩 󵄩 ≤ 4 + 4 + 4 + 4 = 4 = 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y , 8 8 8 8 2 2

∀ n = 0, N,

Y so ‖ℰ1 (x, y)‖ ≤ ‖(x,y)‖ for all (x, y) ∈ L ∩ 𝜕Ω4 . 2 In a similar manner we obtain

N−1

ℰ2n (x, y) ≤ λ ∑ σ3 h(i)[g(i, [xi − pi ] , [yi − qi ] ) + di ] ∗

∗

i=1

N−1

+ λ ∑ σ4 h(i)[f (i, [xi − pi ]∗ , [yi − qi ]∗ ) + ci ] i=1

N−1

≤ λσ3 ∑ h(i)[M5 + ε([xi − pi ]∗ + [yi − qi ]∗ ) + di ] i=1

N−1

+ λσ4 ∑ h(i)[M5 + ε([xi − pi ]∗ + [yi − qi ]∗ ) + ci ] i=1

N−1

N−1

i=1

i=1

≤ λσ3 ∑ h(i)(M5 + di ) + λσ3 εR4 ∑ h(i) N−1

N−1

+ λσ4 ∑ h(i)(M5 + ci ) + λσ4 εR4 ∑ h(i) i=1

i=1

R R R R R 1󵄩 󵄩 ≤ 4 + 4 + 4 + 4 = 4 = 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y , 8 8 8 8 2 2

∀ n = 0, N,

Y so ‖ℰ2 (x, y)‖ ≤ ‖(x,y)‖ for all (x, y) ∈ L ∩ 𝜕Ω4 . 2 Therefore, we deduce that

󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩ℰ (x, y)󵄩󵄩󵄩Y ≤ 󵄩󵄩󵄩(x, y)󵄩󵄩󵄩Y ,

∀ (x, y) ∈ L ∩ 𝜕Ω4 .

(5.17)

By (5.15), (5.17), and Theorem 1.1.1, we conclude that operator ℰ has a fixed point (x 1 , y1 ) ∈ L ∩ (Ω4 \ Ω3 ). Since ‖(x 1 , y1 )‖Y ≥ R3 , we have ‖x 1 ‖ ≥ R3 /2 or ‖y1 ‖ ≥ R3 /2. We suppose that ‖x 1 ‖ ≥ R3 /2. Then xn1 − pn ≥ M3 n for all n = 0, N. Besides, q

q

q

γR 󵄩 󵄩 y1N = ∑ bi xη1 i ≥ γ1 󵄩󵄩󵄩x 1 󵄩󵄩󵄩 ∑ bi ηi ≥ 1 3 ∑ bi ηi > 0, 2 i=1 i=1 i=1 and then q

γR 󵄩󵄩 1 󵄩󵄩 1 󵄩󵄩y 󵄩󵄩 ≥ yN ≥ 1 3 ∑ bi ηi > 0. 2 i=1

5.1 Positive parameters in the system of difference equations

� 137

Therefore, we deduce that for all n = 0, N N−1

N−1

i=1

i=1

y1n − qn ≥ y1n − λδ3 n ∑ di − λδ4 n ∑ ci N−1

󵄩 󵄩 ≥ γ2 n󵄩󵄩󵄩y1 󵄩󵄩󵄩 − λn ∑ (δ3 di + δ4 ci ) i=1

q

≥

N−1 γ1 γ2 R3 n ∑ bi ηi − λn ∑ (δ3 di + δ4 ci ) 2 i=1 i=1 N−1

N−1

i=1

i=1

≥ 2nλ ∑ (δ3 di + δ4 ci ) − nλ ∑ (δ3 di + δ4 ci ) N−1

N−1

= nλ ∑ (δ3 di + δ4 ci ) ≥ λ∗ n ∑ (δ3 di + δ4 ci ) ≥ M3 n. i=1

i=1

If ‖y1 ‖ ≥ R3 /2, then y1n − qn ≥ M3 n for all n = 0, N. Besides, p

p

p

γR 󵄩 󵄩 xN1 = ∑ ai y1ξi ≥ γ2 󵄩󵄩󵄩y1 󵄩󵄩󵄩 ∑ ai ξi ≥ 2 3 ∑ ai ξi > 0, 2 i=1 i=1 i=1 and then p

γR 󵄩󵄩 1 󵄩󵄩 1 󵄩󵄩x 󵄩󵄩 ≥ xN ≥ 2 3 ∑ ai ξi > 0. 2 i=1 Therefore, we conclude that for all n = 0, N N−1

N−1

i=1

i=1

xn1 − pn ≥ xn1 − λδ1 n ∑ ci − λδ2 n ∑ di N−1

󵄩 󵄩 ≥ γ1 n󵄩󵄩󵄩x 1 󵄩󵄩󵄩 − λn ∑ (δ1 ci + δ2 di ) p

≥

i=1

N−1 γ1 γ2 R3 n ∑ ai ξi − λn ∑ (δ1 ci + δ2 di ) 2 i=1 i=1 N−1

N−1

i=1

i=1

≥ 2nλ ∑ (δ1 ci + δ2 di ) − nλ ∑ (δ1 ci + δ2 di ) N−1

N−1

= nλ ∑ (δ1 ci + δ2 di ) ≥ λ∗ n ∑ (δ1 ci + δ2 di ) ≥ M3 n. i=1

i=1

Let un1 = xn1 − pn and v1n = y1n − qn for all n = 0, N. Then un1 ≥ M3 n and v1n ≥ M3 n for all n = 0, N. Hence, we deduce that (u1 , v1 ) is a positive solution of (5.1), (5.2), which completes the proof.

138 � 5 Systems with sign-changing nonlinearities In a similar manner as we proved Theorem 5.1.4, we obtain the following theorems. Theorem 5.1.5. Assume that λ = μ, assumptions (H1), (H2), and (H5) hold, and ∑N−1 i=1 ci > 0 or ∑N−1 d > 0. Additionally, we make the following assumption. i=1 i (H6′ ) We have i ̃0 f∞ = lim inf min f (n, u, v) > L u+v→∞

u,v≥0 n=1,N−1

or

i ̃0, g∞ = lim inf min g(n, u, v) > L u+v→∞

u,v≥0 n=1,N−1

where N−1 N−1 ̃ 0 = max{ 4 ∑ (δ1 ci + δ2 di ), 4 ∑ (δ3 ci + δ4 di ), L γ1 i=1 γ2 i=1 q

−1

p

−1

4 (∑ b η ) γ1 γ2 i=1 i i 4 (∑ a ξ ) γ1 γ2 i=1 i i

N−1

∑ (δ3 di + δ4 ci ), i=1

N−1

∑ (δ1 ci + δ2 di )}(min{ϱ3 , ϱ4 } i=1

N2 − 1 ) . 6 −1

Then there exists λ′∗ > 0 such that for any λ ≥ λ′∗ , problem (5.1), (5.2) (with λ = μ) has at least one positive solution. Theorem 5.1.6. Assume that λ = μ, assumptions (H1), (H2), and (H5) hold, and ∑N−1 i=1 ci > 0 or ∑N−1 d > 0. Additionally, we make the following assumption. i=1 i (H7) We have ̂ = lim min f (n, u, v) = ∞ f∞ u+v→∞

u,v≥0 n=1,N−1

or ĝ∞ = lim min g(n, u, v) = ∞. u+v→∞

u,v≥0 n=1,N−1

Then there exists λ̂∗ > 0 such that for any λ ≥ λ̂∗ , problem (5.1), (5.2) (with λ = μ) has at least one positive solution.

5.1.2 Examples Let N = 25, p = 3, q = 2, a1 = 2, a2 = 1, a3 = 31 , ξ1 = 4, ξ2 = 15, ξ3 = 21, b1 = 1, b2 = 21 , η1 = 10, η2 = 18. We consider the system of second-order difference equations Δ2 un−1 + λf (n, un , vn ) = 0, { 2 Δ vn−1 + μg(n, un , vn ) = 0, with the coupled multi-point boundary conditions

n = 1, 24, n = 1, 24,

(5.18)

5.1 Positive parameters in the system of difference equations

{ {u0 = 0, { { {v0 = 0, p

� 139

1 u25 = 2v4 + v15 + v21 , 3 1 v25 = u10 + u18 . 2

(5.19)

q

We have d = N 2 − (∑i=1 ai ξi )(∑i=1 bi ηi ) = 55 > 0. So, assumption (H1) is satisfied. Besides, we deduce that g0 (n, j) =

1 j(25 − n), { 25 n(25 − j),

1 ≤ j ≤ n ≤ 25, 0 ≤ n ≤ j ≤ 24,

1 j(25 − j), ∀ j = 1, 24, 25 1 k(n) = n(25 − n), ∀ n = 0, 25. 600 h(j) =

In addition, we have σ1 ≈ 21.45454545, δ1 ≈ 11.36363636, ϱ1 ≈ 0.19363636, σ2 ≈ 37.87878788, δ2 ≈ 13.63636364, ϱ2 ≈ 0.26212121, σ3 ≈ 29.78787879, δ3 ≈ 11.36363636, ϱ3 ≈ 0.19921212, σ4 ≈ 17.04545455, δ4 ≈ 8.63636364, ϱ4 ≈ 0.16136364, γ1 ≈ 0.00692, γ2 ≈ 0.00668769. Example 1. We consider the functions f (n, u, v) = (u − 1)(u − 2) + cos(3v),

g(n, u, v) = (v − 2)(v − 3) + sin(2u),

for all n = 1, 24 and u, v ∈ [0, ∞). Then there exists M0 > 0 (M0 = 5/4) such that f (n, u, v)+ M0 ≥ 0, g(n, u, v) + M0 ≥ 0 for all n = 1, 24 and u, v ∈ [0, ∞) (cn = dn = M0 = 45 for all n = 1, 24). So, assumptions (H2) and (H3) are satisfied (f (n, 0, 0) = 3 > 0, g(n, 0, 0) = 6 > 0). Let δ = 41 < 1 and R0 = 21 . Then 3 3 , g(n, u, v) ≥ δg(n, 0, 0) = , 4 2 1 ∀ n = 1, 24, u, v ∈ [0, ]. 2

f (n, u, v) ≥ δf (n, 0, 0) =

Besides, f ̄(R0 ) = ̄ 0) = g(R

max

{f (n, u, v) + cn } = 4.25,

max

{g(n, u, v) + dn } ≈ 8.09147098.

n=1,24, u,v∈[0,R0 ] n=1,24, u,v∈[0,24]

We also obtain p̃1 ≈ 2, 231.27272727, p̃2 ≈ 3, 939.39393939, p̃3 ≈ 3, 097.93939394, p̃4 ≈ 1, 772.72727273, and then λ0 ≈ 8.29562 × 10−6 and μ0 ≈ 2.49332 × 10−6 . By Theorem 5.1.1, for any λ ∈ (0, λ0 ] and μ ∈ (0, μ0 ], we deduce that problem (5.18), (5.19) has at least one positive solution.

140 � 5 Systems with sign-changing nonlinearities Example 2. We consider the functions f (n, u, v) =

(u + v)2 n + ln , n(n + 2) n+1

g(n, u, v) =

2 + sin(u + v) n+1 + ln , n+3 n2

≥ 0, dn = ln n+3 ≥ 0 for for all n = 1, 24 and u, v ∈ [0, ∞). Here we have cn = ln n+1 n n+1 all n = 1, 24, and then evidently we obtain f (n, u, v) ≥ −cn and g(n, u, v) ≥ −dn for all n = 1, 24 and u, v ∈ [0, ∞). Because f∞ = ∞, assumptions (H2) and (H4) are satisfied. We choose R1 = 186, 320, which satisfies the condition from the beginning of the proof of Theorem 5.1.2. Then M1 ≈ 1.15717 × 1010 , M2 = 3, λ∗ ≈ 1.80405 × 10−9 , and μ∗ = 1. By Theorem 5.1.2, we conclude that (5.18), (5.19) has at least one positive solution for any λ ∈ (0, λ∗ ] and μ ∈ (0, μ∗ ]. Remark 5.1.2. The results presented in this section were published in the paper [42].

6 Nonlinear fractional difference equations with nonlocal boundary conditions In this chapter we study the existence of nontrivial solutions, nonnegative solutions, and positive solutions for a class of nonlinear νth-order Atıcı–Eloe fractional difference equations, with left focal boundary conditions or Dirichlet boundary conditions.

6.1 Nonlinear νth-order Atıcı–Eloe fractional difference equations with left focal boundary conditions In this section, for 1 < ν ≤ 2 a real number and T ≥ 3 a natural number, we are concerned with the existence of nontrivial solutions of the nonlinear νth-order Atıcı–Eloe fractional difference equation, Δν u(t) + f (u(t + ν − 1)) = 0,

t ∈ {0, 1, . . . , T},

(6.1)

satisfying the left focal boundary conditions Δu(ν − 2) = 0,

u(ν + T) = 0,

(6.2)

where Δ is the forward difference operator with step size 1, Δν is the Atıcı–Eloe fractional difference, and f : ℝ → ℝ is continuous. Interest in fractional difference equations devoted to both their theoretical development and their applications is currently quite high, with much of this interest spawned by the definitions, in the context of discrete domains, of fractional sums and fractional differences in the pioneering papers by Atıcı and Eloe [7, 8]. Further development and extensions were achieved in the seminal papers by Goodrich [19–21]. Atıcı and Eloe dealt first with a theory for initial value problems for fractional difference equations, followed in their second paper by applications of their definitions in obtaining positive solutions for Dirichlet boundary value problems for fractional difference equations. Their second work centered on a Guo–Krasnosel’skii fixed point argument which required the construction of a Green function for their fractional problem. Other subsequent research by Goodrich that focused on questions expanding the work of Atıcı and Eloe can be found in [22–26]. A couple of recent papers by Henderson [31] and Henderson and Neugebauer [46] were devoted to the existence of local solutions for boundary value problems for Atıcı–Eloe fractional difference equations, and other recent works devoted to Atıcı– Eloe difference equations can be found in, to cite a few, [17, 45, 64].

https://doi.org/10.1515/9783111040370-006

142 � 6 Nonlinear fractional difference equations Discrete fractional calculus and fractional difference equations frequently appear in studies modeling natural processes such as found in the paper by Atıcı and Şengül [10] and in the papers by Magin [55] and Metzler et al. [56]. Especially prominent is the current use of boundary value problems for discrete fractional difference equations in their applications to discrete control processes; see, for example, the monographs devoted to discrete fractional control [14, 15, 57–59]. In this section, we apply the Krasnosel’skii–Zabreiko fixed point theorem [50] in establishing the existence of nontrivial solutions of (6.1), (6.2). Effective use has been made of the Krasnosel’skii–Zabreiko fixed point theorem in showing existence of solutions of boundary value problems (for ordinary differential equations, for difference equations, and for dynamic equations) in the context of when the nonlinearity is almost linear at infinity; see, for example [18, 30, 34, 47, 61].

6.1.1 Some preliminaries and the Green function We begin this subsection with the Atıcı–Eloe definitions of fractional sum and fractional difference in the context of a discrete domain. Definition 6.1.1. Let n ∈ ℕ, let n − 1 < κ ≤ n be a real number, and let a ∈ ℝ. For t ∈ {a + κ, a + κ + 1, . . .}, the κth-order Atıcı–Eloe fractional sum, Δ−κ u, of the function u is defined by Δ−κ u(t) =

1 t−κ ∑ (t − s − 1)(κ−1) u(s), Γ(κ) s=a

Γ(t+1) where t (κ) = Γ(t+1−κ) is the falling function. The κth-order Atıcı–Eloe fractional difference, Δκ u, of the function u is defined by

Δκ u(t) = Δn−(n−κ) u(t) = Δn (Δ−(n−κ) u(t)), where Δ is the forward difference defined by Δu(t) = u(t + 1) − u(t), and Δi u(t) = Δ(Δi−1 u(t)), i = 2, 3, . . . . Remark 6.1.1. We note that for u defined on {a, a + 1, . . .}, Δ−κ u is defined on {a + κ, a + κ + 1, . . .}. In [21], for 1 < ν ≤ 2, Goodrich constructed the Green function, G(t, s), for −Δν y(t) = 0,

t ∈ {0, 1, . . . , T},

(6.3)

satisfying the left focal boundary conditions (6.2). In particular, by direct computation, Goodrich obtained

6.1 Left focal boundary conditions

{ Γ(ν)(ν + T − s − 1)(ν−1) , { { { { { t = ν − 2, ν − 1, { { { { (ν−1) { { + (ν − 1)(ν−2) ](ν + T − s − 1)(ν−1) , {[(2 − ν)t { c(ν, T) { G(t, s) = 0 ≤ t − ν ≤ s ≤ T, { Γ(ν) { { { { [(2 − ν)t (ν−1) + (ν − 1)(ν−2) ](ν + T − s − 1)(ν−1) { { { { 1 { { − c(ν,T) (t − s − 1)(ν−2) , { { { { 0 ≤ s < t − ν ≤ T, {

� 143

(6.4)

with c(ν, T) =

(ν +

T)(ν−2) [(ν

1 . − 1) + (2 − ν)(T + 2)]

Goodrich also obtained the following properties of G(t, s) which will be of importance to us: (a) For each s ∈ {0, . . . , T}, Δt G(ν − 2, s) = 0

and G(ν + T, s) = 0.

(b) We have G(t, s) > 0, for (t, s) ∈ {ν − 2, . . . , ν + T − 1} × {0, . . . , T}. (c) We have maxt∈{ν−2,...,ν+T} G(t, s) = G(s + ν − 1, s), for s ∈ {0, . . . , T}. As stated earlier in this section, we will apply the Krasnosel’skii–Zabreiko fixed point theorem [50] in establishing the existence of nontrivial solutions of (6.1), (6.2). We now state that fixed point theorem. Theorem 6.1.1. Let X be a Banach space and let F : X → X be a completely continuous operator. If there exists a bounded linear operator A : X → X such that 1 is not an eigenvalue and lim

‖u‖→∞

‖F(u) − A(u)‖ = 0, ‖u‖

then F has a fixed point in X. We will apply Theorem 6.1.1 to a nonlinear summation operator whose kernel is the Green function, G(t, s). For our setting, let a Banach space (X, ‖ ⋅ ‖) be defined by X = {h : {ν − 2, . . . , T + ν} → ℝ, Δh(ν − 2) = h(T + ν) = 0},

(6.5)

with norm ‖h‖ =

max

󵄨󵄨 󵄨 󵄨󵄨h(x)󵄨󵄨󵄨.

x∈{ν−2,...,T+ν}

(6.6)

144 � 6 Nonlinear fractional difference equations It is standard that u ∈ X is a fixed point of the completely continuous operator F : X → X defined by T

(Fu)(t) = ∑ G(t, s)f (u(s + ν − 1)), s=0

t ∈ {ν − 2, . . . , T + ν},

(6.7)

if and only if u is a solution of (6.1), (6.2) (for detailed proofs, see the papers [21] and [46]).

6.1.2 Existence results In this subsection, we apply Theorem 6.1.1 to the operator F defined in (6.7) and to an associated linear operator to obtain solutions of (6.1), (6.2). Theorem 6.1.2. Assume f : ℝ → ℝ and lim|r|→∞ |m| < b :=

1

f (r) r

= m. If

∑Ts=0 G(s + ν − 1, s)

,

then the boundary value problem (6.1), (6.2) has a solution u, and moreover, u ≠ 0 when f (0) ≠ 0. Proof. Let the Banach space (X, ‖ ⋅ ‖) and the completely continuous operator F : X → X be as defined in (6.5), (6.6), and (6.7) in Section 6.1.1. Associated with (6.1), (6.2), we consider a linear νth-order equation, Δν u(t) + mu(t + ν − 1) = 0,

t ∈ {0, . . . , T},

(6.8)

satisfying the boundary conditions (6.2), and we define a completely continuous linear operator A : X → X by T

(Au)(t) = m ∑ G(t, s)u(s + ν − 1), s=0

t ∈ {ν − 2, . . . , T + ν}.

(6.9)

Of course, solutions of (6.8), (6.2) are fixed points of A, and conversely. Our first claim is that 1 is not an eigenvalue of A. There are two cases to consider for this claim: (a) m = 0 and (b) m ≠ 0. For (a), if m = 0, since the boundary value problem (6.3), (6.2) has only a trivial solution, it is immediate that 1 is not an eigenvalue of A. For (b), if m ≠ 0 and (6.8), (6.2) has a nontrivial solution, u, then ‖u‖ > 0. So, we have ‖u‖ = ‖Au‖ 󵄨󵄨 T 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 = max 󵄨󵄨m ∑ G(t, s)u(s + ν − 1)󵄨󵄨󵄨 󵄨󵄨 t∈{ν−2,...,T+ν}󵄨󵄨 󵄨 s=0 󵄨

6.1 Left focal boundary conditions

= |m|

� 145

󵄨󵄨 󵄨󵄨 T 󵄨 󵄨󵄨 󵄨󵄨 ∑ G(t, s)u(s + ν − 1)󵄨󵄨󵄨 󵄨 󵄨󵄨󵄨 󵄨 t∈{ν−2,...,T+ν}󵄨 󵄨 󵄨s=0 max T

≤ |m|||u|| ∑ G(s + ν − 1, s) s=0

1 < b‖u‖ b = ‖u‖,

which is a contradiction. So, again, 1 is not an eigenvalue of A. Next, we show that ‖F(u) − A(u)‖ = 0. ‖u‖→∞ ‖u‖ lim

To this end, let ϵ > 0 be given. Since lim|r|→∞ for |r| > M1 ,

f (r) r

= m, there exists an M1 > 0 such that

󵄨󵄨 󵄨 󵄨󵄨f (r) − mr 󵄨󵄨󵄨 < ϵ|r|.

(6.10)

Set 󵄨 󵄨 M = sup 󵄨󵄨󵄨f (r)󵄨󵄨󵄨, |r|≤M1

and let L > M1 be such that M + |m|M1 < ϵ. L If we choose u ∈ X with ‖u‖ > L, then for s ∈ {0, . . . , T}: (i) if |u(s + ν − 1)| ≤ M1 , we have 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨f (u(s + ν − 1)) − mu(s + ν − 1)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨f (u(s + ν − 1))󵄨󵄨󵄨 + |m|󵄨󵄨󵄨u(s + ν − 1)󵄨󵄨󵄨 ≤ M + |m|M1 < ϵL

< ϵ‖u‖, and (ii) if |u(s + ν − 1)| > M1 , from (6.10) we have 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨f (u(s + ν − 1)) − mu(s + ν − 1)󵄨󵄨󵄨 < ϵ󵄨󵄨󵄨u(s + ν − 1)󵄨󵄨󵄨 ≤ ϵ‖u‖. Thus, from (i) and (ii), for all s ∈ {0, . . . , T}, 󵄨󵄨 󵄨 󵄨󵄨f (u(s + ν − 1)) − mu(s + ν − 1)󵄨󵄨󵄨 ≤ ϵ‖u‖.

(6.11)

146 � 6 Nonlinear fractional difference equations It follows from (6.11) that for u ∈ X with ‖u‖ > L, 󵄨󵄨 󵄨󵄨 T 󵄨󵄨 󵄨󵄨 󵄩󵄩 󵄩󵄩 󵄨󵄨 󵄨 G(t, s)[f (u(s + ν − 1)) − mu(s + ν − 1)] = max F(u) − A(u) ∑ 󵄨󵄨 󵄩󵄩 󵄩󵄩 󵄨󵄨 t∈{ν−2,...,T+ν}󵄨󵄨 󵄨󵄨 󵄨s=0 ≤

max

t∈{ν−2,...,T+ν} T

T

󵄨 󵄨 ∑ G(t, s)󵄨󵄨󵄨f (u(s + ν − 1)) − mu(s + ν − 1)󵄨󵄨󵄨

s=0

≤ ϵ‖u‖ ∑ G(s + ν − 1, s) s=0

ϵ = ‖u‖. b As a consequence,

lim

‖u‖→∞

‖F(u) − A(u)‖ = 0. ‖u‖

By Theorem 6.1.1, F has a fixed point u ∈ X, and in turn, u is a desired solution of (6.1), (6.2). Moreover, if in addition f (0) ≠ 0, it is immediate that u ≠ 0. The proof is complete. As a corollary, we will show that when f ≥ 0, (6.1), (6.2) has positive solutions. Corollary 6.1.1. Assume f : [0, ∞) → [0, ∞) is continuous and limr→∞ f (r) = 0. Then r the boundary value problem (6.1), (6.2) has a nonnegative solution u, and moreover, u is a positive solution when f (0) ≠ 0. Proof. Let f : ℝ → ℝ be defined by f (r) = { Then f is continuous on ℝ and lim|r|→∞ fractional equation

f (r), f (−r), f (r) r

r ≥ 0, r < 0.

= 0. It follows from Theorem 6.1.2 that the

Δν u(t) + f (u(t + ν − 1)) = 0,

t ∈ {0, . . . , T},

(6.12)

satisfying the boundary conditions (6.2) has a solution u. In particular, u satisfies T

u(t) = ∑ G(t, s)f (u(s + ν − 1)), s=0

t ∈ {ν − 2, . . . , T + ν},

and hence u(t) ≥ 0, t ∈ {ν − 2, . . . , T + ν}. In view of this, f (u(s + ν − 1)) = f (u(s + ν − 1)), s ∈ {0, . . . , T}, so u satisfies (6.1), (6.2). That is, u is a nonnegative solution of (6.1), (6.2). As before, if f (0) ≠ 0, then u ≠ 0. Remark 6.1.2. The results presented in this section were published in the paper [32].

6.2 Dirichlet boundary conditions

� 147

6.2 Nonlinear νth-order Atıcı–Eloe fractional difference equations with Dirichlet boundary conditions In this section, for 1 < ν ≤ 2 a real number and T ≥ 2 a natural number, we investigate the existence of nontrivial solutions of the nonlinear νth-order Atıcı–Eloe fractional difference equation, Δν u(t) + f (u(t + ν − 1)) = 0,

t ∈ {1, 2, . . . , T + 1},

(6.13)

satisfying the Dirichlet boundary conditions u(ν − 2) = u(ν + T + 1) = 0,

(6.14)

where Δν is the Atıcı–Eloe fractional difference and f : ℝ → ℝ is continuous. In the proof of the main existence result we apply the Krasnosel’skii–Zabreiko fixed point theorem (Theorem 6.1.1). 6.2.1 Preliminary results In [8], for 1 < ν ≤ 2, Atıcı and Eloe constructed the Green function 𝒢 (t, s) for −Δν y(t) = 0,

t ∈ {1, 2, . . . , T + 1}

satisfying the Dirichlet boundary conditions (6.14) by direct computation. It is given by t (ν−1) (ν + T − s)(ν−1) { { , t − ν + 1 ≤ s ≤ T + 1, { { 1 (ν + T + 1)(ν−1) 𝒢 (t, s) = (ν−1) { Γ(ν) { (ν + T − s)(ν−1) {t { − (t − s − 1)(ν−1) , s < t − ν + 1 ≤ T + 1. { (ν + T + 1)(ν−1) They also obtained the following properties of 𝒢 (t, s) which will be of importance to us: (a) For each s ∈ {1, . . . , T + 1}, 𝒢 (ν − 2, s) = 0

and

𝒢 (ν + T + 1, s) = 0.

(b) We have 𝒢 (t, s) > 0, for (t, s) ∈ {ν − 1, . . . , ν + T} × {1, . . . , T + 1}. (c) We have maxt∈{ν−2,...,ν+T+1} 𝒢 (t, s) = 𝒢 (s + ν − 1, s), for s ∈ {1, . . . , T + 1}. We will apply Theorem 6.1.1 to a nonlinear summation operator whose kernel is the Green function, 𝒢 (t, s). For our setting, let a Banach space (Y , ‖ ⋅ ‖) be defined by Y = {h : {ν − 2, . . . , T + ν + 1} → ℝ, h(ν − 2) = h(T + ν + 1) = 0},

148 � 6 Nonlinear fractional difference equations with norm ‖h‖ =

󵄨 󵄨󵄨 󵄨󵄨h(x)󵄨󵄨󵄨.

max

x∈{ν−2,...,T+ν+1}

It is standard that u ∈ Y is a fixed point of the completely continuous operator ℱ : Y → Y defined by T+1

(ℱ u)(t) = ∑ 𝒢 (t, s)f (u(s + ν − 1)),

t ∈ {ν − 2, . . . , T + ν + 1},

s=1

(6.15)

if and only if u is a solution of (6.13), (6.14) (for detailed proofs, see the papers [8] and [31]).

6.2.2 Existence results In this subsection, we apply Theorem 6.1.1 to the operator ℱ defined in (6.15) and to an associated linear operator to obtain solutions of (6.13), (6.14). Using similar arguments as those used in the proof of Theorem 6.1.2, we obtain the following theorem for the boundary value problem (6.13), (6.14). f (r) r

Theorem 6.2.1. Assume f : ℝ → ℝ and lim|r|→∞ |m| < b :=

∑T+1 s=1 𝒢 (s

1

= m. If

+ ν − 1, s)

,

then the boundary value problem (6.13), (6.14) has a solution u, and moreover, u ≠ 0 when f (0) ≠ 0. In the proof of Theorem 6.2.1 we consider the linear νth-order equation Δν u(t) + mu(t + ν − 1) = 0,

t ∈ {1, . . . , T + 1},

(6.16)

satisfying the boundary conditions (6.14), and we consider the completely continuous linear operator 𝒜 : Y → Y by T+1

(𝒜u)(t) = m ∑ 𝒢 (t, s)u(s + ν − 1), s=1

t ∈ {ν − 2, . . . , T + ν + 1}.

The solutions of (6.16), (6.14) are fixed points of 𝒜, and conversely. As a corollary, we can prove (similar to Corollary 6.1.1) that when f ≥ 0, then (6.13), (6.14) has positive solutions.

6.2 Dirichlet boundary conditions

� 149

= 0. Then the Corollary 6.2.1. Assume f : [0, ∞) → [0, ∞) is continuous and limr→∞ f (r) r boundary value problem (6.13), (6.14) has a nonnegative solution u, and moreover, u is a positive solution when f (0) ≠ 0. Remark 6.2.1. The results presented in this section were published in the paper [33].

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Index Atıcı–Eloe fractional difference 141, 142, 147 Atıcı–Eloe fractional difference equation 141, 142, 144, 146–148 Atıcı–Eloe fractional sum 142 Banach space 13, 14, 21, 22, 35, 37, 50, 64, 80, 82, 91, 92, 106, 121, 143, 144, 147 boundary of a set 35 boundary value problem 14, 22, 34, 49, 51, 73–75, 77, 78, 84, 120, 131, 141, 142, 144, 146, 148, 149 bounded operator 143 bounded set 13, 35, 43, 45, 80, 93, 94, 113 characteristic equation 1, 21 closed set 21, 80 compact operator 13 completely continuous operator 13, 14, 21, 23, 35, 37, 50, 65, 80, 84, 91, 92, 106, 121, 143, 144, 148 concave function 30, 35, 36, 38, 43 cone 13, 14, 23, 35, 37, 50, 63, 64, 66, 69, 91–93, 106, 121, 122 continuous function 13, 20, 22, 31, 35, 48, 63, 80, 81, 92, 106, 107, 111, 115, 119, 141, 146, 147, 149 continuous operator 13 convex set 21, 80 coupled multi-point boundary conditions 54, 55, 78, 80–82, 87, 88, 90, 91, 100–102, 118, 119, 138 Dirichlet boundary conditions 5, 141, 147, 148 eigenvalue 143–145 existence of local solutions 141 existence of nonnegative solutions 141, 146, 149 existence of nontrivial solutions 141–144, 147, 148 existence of positive solutions 1, 13–15, 18, 19, 21–23, 25, 27, 29–31, 35, 37, 42, 47, 51, 53, 54, 63, 65, 68, 69, 72, 73, 79, 80, 83, 89–93, 96, 101, 106, 107, 111, 112, 115, 117, 118, 122, 124, 132, 138–141, 146, 149 extreme limit 64 falling function 142 fixed point 13, 14, 16–19, 21, 23, 25, 26, 28, 35, 37, 42, 46, 47, 50, 52, 64, 68, 72, 80, 82, 84, 91, 92, 95, 98–100, 106, 110, 114–116, 121–123, 129, 136, 143, 144, 146, 148 https://doi.org/10.1515/9783111040370-008

fixed point index 30, 31, 35, 40, 42, 45–47, 91, 95, 97–101, 108, 110, 114, 116 forward difference operator 1, 141 Green function 4–6, 11–13, 22, 32, 33, 36, 49, 61, 82, 102, 105, 106, 119, 141–143, 147 Guo–Krasnosel’skii fixed point theorem 1, 13, 21, 31, 54, 118, 141 homogeneous equation 2 initial value problem 141 Jensen inequality 38, 43 Krasnosel’skii-Zabreiko fixed point theorem 142, 143, 147 left focal boundary conditions 141, 142, 144, 146 linear operator 91, 97, 143, 144, 148 linear term 1 maximum norm 14, 22, 37, 50, 64, 82, 92, 106, 121 multi-point boundary conditions 1, 3, 14, 20–22, 28, 32 multiplicity of positive solutions 1, 15, 18, 19, 28, 30, 31, 35, 46, 91, 92, 99, 101, 116, 117, 131 nondecreasing function 36, 71, 99, 101, 115 nonexistence of positive solutions 31, 54, 73–75, 77, 79, 80, 84, 89, 90 nonhomogeneous equation 2 nonlinear alternative of Leray-Schauder type 21, 48, 118 nonlinearities 1, 13, 21, 30, 47, 53, 54, 91, 101, 118, 142 nonnegative function 30, 31, 54, 91, 101 nonnegative solution 146 nonresonant problem 1 nontrivial solution 144 norm 35, 37, 50, 64, 82, 92, 106, 121, 143, 148 open set 13, 35 operator 14, 16–19, 22, 23, 25, 26, 28, 37, 42, 46, 50, 64, 66, 69, 82, 92, 95, 106, 110, 121, 122, 129, 136, 144, 148 parameter 21, 30, 47, 48, 54, 80–82, 90, 91, 118 positive solution 13, 14, 16–19, 21, 22, 25, 27, 28, 30, 35, 46–49, 51, 52, 54, 73–78, 80, 84, 85, 87, 91,

156 � Index

95, 99–101, 110, 115, 116, 118, 120, 122, 124, 132, 137, 141, 146, 148 relatively compact set 13 relatively open set 21 resonant problem 1 Schauder fixed point theorem 31, 80 second-order finite difference equation 1, 14, 19, 21, 22, 28 semipositone problem 48, 118 sequence 3, 14, 21–23, 30, 48, 54, 80, 81, 87, 88, 90, 91, 93, 101, 118 sign-changing function 1, 21, 30, 48, 118

solution 1–6, 12–14, 22, 23, 32, 34, 37, 48, 49, 55, 56, 61, 63, 64, 81, 82, 84, 85, 87, 88, 93, 102, 105, 106, 119–121, 142, 144, 146, 148 strictly increasing function 11, 35, 36 system of second-order finite difference equations 30, 32, 47, 48, 52, 54, 78, 80, 81, 87, 88, 90, 91, 100, 101, 118, 119, 138 unbounded function 1, 13 uncoupled multi-point boundary conditions 30, 31, 47–49, 53 variation of constants method 2

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