Applied Mathematics and Computation 233 (2014) 62–71

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

On the linear and nonlinear discrete second-order Neumannboundary value problems q

http://dx.doi.org/10.1016/j.amc.2014.01.1120096-3003/� 2014 Elsevier Inc. All rights reserved.

q Supported by NNSF of China (11326127, 11101335).E-mail address: gaokuguo@163.com

Chenghua GaoDepartment of Mathematics, Northwest Normal University, Lanzhou 730070, PR China

a r t i c l e i n f o

Keywords:Discrete Neumann boundary valueproblemsSpectrumGlobal bifurcation techniqueSign-changing solutions

a b s t r a c t

By construct the spectrum of the linear eigenvalue problem, we are concerned with theglobal structure of the set of sign-changing solutions to the discrete second-orderNeumann boundary value problem

�D½pðt � 1ÞDuðt � 1Þ� þ qðtÞuðtÞ ¼ raðtÞf ðuðtÞÞ; t 2 ½1; T�Z;Duð0Þ ¼ DuðTÞ ¼ 0:

Furthermore, we obtain the existence of the sign-changing solutions of the above nonlinearproblem.

� 2014 Elsevier Inc. All rights reserved.

1. Introduction

Let a; b be two integer with a < b, we employ ½a; b�Z to denote the discrete interval given by fa; aþ 1; . . . ; bg. In this paper,we focus on the global structure of the set of sign-changing solutions to the discrete second-order Neumann boundary valueproblem

�D½pðt � 1ÞDuðt � 1Þ� þ qðtÞuðtÞ ¼ raðtÞf ðuðtÞÞ; t 2 ½1; T�Z; ð1:1Þ

Duð0Þ ¼ DuðTÞ ¼ 0; ð1:2Þ

where D denotes the forward difference operator defined by DuðtÞ ¼ uðt þ 1Þ � uðtÞ; T > 1 is an integer, p : ½0; T�Z ! ð0;1Þ;q : ½1; T�Z ! ð0;1Þ; a : ½1; T�Z ! ð0;1Þ and f : ½1; T�Z � R! R is continuous.

Recently, variety methods were adopted to study the existence of solutions of Neumann boundary value problems, see,for example, [1–6] and the references therein. In [1–3], the fixed point theorems or the fixed point index theory on a conewas used to obtain the existence of positive solutions of Neumann boundary value problems of second-order ODEs. In [4],the variation methods were used to discuss the sign-changing solutions to the Sturm–Liouville boundary value problems. Forthe discrete, Anderson et al. [5] used the Schaefer’s theorem to study the existence of solutions of the second-order discreteNeumann boundary value problems, Candito and D’Aguı̀ [6] used the variation methods to discuss the similar problems.Meanwhile, Ma [7], Gao et al. [8] and Ma and Ma [9] used the Rabinowitz bifurcation theorem to discuss the existence ofsign-changing or positive solutions to the similar discrete problems.

C. Gao / Applied Mathematics and Computation 233 (2014) 62–71 63

However, there are few results on the sign-changing solutions to the discrete second-order Neumann problems. Even inthe continuous case, although Zhang et al. obtained the existence of sign-changing solutions, the sign-changing times of thesolutions and the global structure of set of sign-changing solutions are not clear. Thus, in this paper, we try to solve theseproblems, by using the Rabinowitz’s bifurcation theorem. As we know, to use the Rabinowitz’s bifurcation theorem, we needto know the spectrum of the linear eigenvalue problem

�D½pðt � 1ÞDuðt � 1Þ� þ qðtÞuðtÞ ¼ kaðtÞuðtÞ; t 2 ½1; T�Z; ð1:3Þ

Duð0Þ ¼ DuðTÞ ¼ 0: ð1:4Þ

It is worth noticing that there are also several excellent results on the spectrum of discrete linear eigenvalue prob-lems, see, for example, [10–15] and the references therein. However, only [12] discussed some information on thesign-changing of the eigenfunction of the discrete linear eigenvalue problems under the Dirichlet boundaryconditions.

In the Section 2 of this paper, we attempt to obtain some properties of spectrum of (1.3) and (1.4), which contain the sign-changing times of the eigenfunction. To achieve this goal, we first convert (1.3) and (1.4) to a matrix eigenvalue problem, andthen by proving some properties of the generalized Sturm sequence, we obtain (1.3) and (1.4) has T simple, real and positiveeigenvalues, which we denote them by lk, k 2 ½1; T�Z. Moreover, some monotonicity of the eigenvalues, lk, are also obtained.

The first main result we obtain is the following theorem.

Theorem 1.1. (1.3) and (1.4) has T simple eigenvalues lk > 0; k ¼ 1;2; . . . ; T and the corresponding eigenfunction wk to lk hasexactly k� 1 simple generalized zeros in ½1; T�Z.

For the nonlinear problem (1.1) and (1.2), we suppose that the nonlinearity f satisfies:

(H1) f 2 CðR;RÞ with sf ðsÞ > 0 for s – 0;(H2) f0 ¼ limjuj!0f ðuÞ=u 2 ð0;1Þ;(H3) f1 ¼ limjuj!1f ðuÞ=u 2 ð0;1Þ;(H3)0 f1 ¼ limjuj!1f ðuÞ=u ¼ 0;(H3)00 f1 ¼ limjuj!1f ðuÞ=u ¼ 1.

Based on Theorem 1.1 and these assumptions of f, we get the second main result.

Theorem 1.2. Suppose that (H1) and (H2) hold. Then

(i) if (H3) and one of the following two conditions hold, i.e.,

lk

f1< r <

lk

f0ð1:5Þ

or

lk

f0< r <

lk

f1ð1:6Þ

for some k 2 ½1; T�Z, then (1.1) and (1.2) has two solutions uþk and u�k such that uþk and u�k has exactly k� 1 generalized simplezeros in ð1; TÞ, respectively. Moreover, uþk ð0Þ > 0 and u�k ð0Þ < 0.

(ii) if (H3)0 holds, then for k 2 ½1; T�Z, there exists k� : 0 < k� 6 lkf0

, such that (1.1) and (1.2) has two solutions uþk and u�k such

that uþk and u�k for r 2 ðk�;1Þ. Moreover, uþk and u�k has exactly k� 1 generalized simple zeros in ð1; TÞ, respectively. Mean-while, uþk ð0Þ > 0 and u�k ð0Þ < 0.(iii) if (H3)00 holds, then for k 2 ½1; T�Z, there exists q� : q� P lk

f0, such that (1.1) and (1.2) has two solutions uþk and u�k such that

uþk and u�k for r 2 ð0;q�Þ. Moreover, uþk and u�k has exactly k� 1 generalized simple zeros in ð1; TÞ, respectively. Meanwhile,uþk ð0Þ > 0 and u�k ð0Þ < 0.

The rest of this paper is arranged as follows. In Section 2, we discuss the spectrum of (1.3) and (1.4) and prove Theo-rem 1.1. Section 3 is devoted to discuss the global structure of the set of sign-changing solutions to (1.1) and (1.2) and proveTheorem 1.2.

2. Spectrum of second-order linear eigenvalue problems

In this section, we consider the spectrum of (1.3) and (1.4). To get it, we first give a definition of simple generalized zeroswhich will play a very important role in our discussions.

64 C. Gao / Applied Mathematics and Computation 233 (2014) 62–71

Definition 2.1. Suppose that a function y : ½0; T þ 1�Z ! R. If yðtÞ ¼ 0 for some t 2 ½0; T þ 1�Z, then t is a zero of y. If yðtÞ ¼ 0and yðt � 1Þyðt þ 1Þ < 0 for some t 2 ½1; T�Z, then t is a simple zero of y. If yðtÞyðt þ 1Þ < 0 for some t 2 ½0; T�Z, then we say that

y has a node at the point s ¼ tyðtþ1Þ�ðtþ1ÞyðtÞyðtþ1Þ�yðtÞ 2 ðt; t þ 1Þ. The nodes and simple zeros of y are called the simple generalized zeros

of y.

Now, we consider the spectrum properties of (1.3) and (1.4). let us convert (1.3) and (1.4) to a matrix equation

Au ¼ kBu; ð2:1Þ

where u ¼ ðuð1Þ;uð2Þ; . . . ;uðTÞÞ,

B ¼

að1Þ 0 0 � � � 0 0 00 að2Þ 0 � � � 0 0 00 0 að3Þ � � � 0 0 0

..

. ... ..

.� � � ..

. ... ..

.

0 0 0 � � � aðT � 2Þ 0 00 0 0 � � � 0 aðT � 1Þ 00 0 0 � � � 0 0 aðTÞ

0BBBBBBBBBBBB@

1CCCCCCCCCCCCA;

A ¼

pð1Þ þ qð1Þ �pð1Þ 0 0 � � � 0 0 0�pð1Þ cð2Þ �pð2Þ 0 � � � 0 0 0

0 �pð2Þ cð3Þ �pð3Þ � � � 0 0 0

..

. ... . .

. . .. . .

. ... ..

. ...

0 0 0 0 � � � cðT � 2Þ �pðT � 2Þ 00 0 0 0 � � � �pðT � 2Þ cðT � 1Þ �pðT � 1Þ0 0 0 0 � � � 0 �pðT � 1Þ pðT � 1Þ þ qðTÞ

0BBBBBBBBBBBB@

1CCCCCCCCCCCCA;

here cðtÞ ¼ pðtÞ þ pðt � 1Þ þ qðtÞ; t 2 ½2; T � 1�Z. Now, to obtain more properties of eigenvalues of (1.3) and (1.4), let us dosome preliminaries. Let Ai denote the ith principal submatrix of A and Bi the ith principal submatrix of B. Meanwhile, weuse Q iðkÞ denote the ith principal subdeterminant of A� kB and suppose that Q 0ðkÞ ¼ 1, then QTðkÞ ¼ detðA� kBÞ, and it isnot difficult to verify the following generalized Sturm sequence

Q 0ðkÞ ¼ 1;

Q 1ðkÞ ¼ pð1Þ þ qð1Þ � kað1Þ;Q iðkÞ ¼ ðcðiÞ � kaðiÞÞQ i�1ðkÞ � p2ði� 1ÞQ i�2ðkÞ; i ¼ 2;3; . . . ; T � 1;

Q TðkÞ ¼ ðpðT � 1Þ þ qðTÞ � kaðTÞÞQ T�1ðkÞ � p2ðT � 1ÞQT�2ðkÞ:

ð2:2Þ

As we know, to find the eigenvalues of (1.3) and (1.4) is equivalent to find the roots of Q TðkÞ. Thus, it is necessary to discusssome properties of the Sturm sequence (2.2).

Lemma 2.1. The roots of QiðkÞ are real and positive.

Proof. It is easy to see that Ai a symmetric and positive definite matrix. Thus, there exists a nonsingular i� i matrix Ji suchthat J�i AiJi is a i� i identical matrix. Let u ¼ Jiv , then (2.1) equals to

1k

v ¼ J�i BiJiv:

Since J�i BiJi is a real and symmetric matrix, we have all of the eigenvalues of J�i BiJi are real, which implies all of the eigenvaluesof Ai are real. Furthermore, we have the roots of Q iðkÞ are real. Moreover, all eigenvalues of Bi are positive, it follows from theSylvester inertia theorem that all eigenvalues of J�i BiJi are positive. This completes the assertion. h

Lemma 2.2. Two consecutive polynomials Qi�1ðkÞ;Q iðkÞ have no common zeros.

Proof. Suppose on the contrary there exists k ¼ k0 such that Q i�1ðk0Þ ¼ Q iðk0Þ ¼ 0. Then by the recurrence relation (2.2), weget Qi�2ðk0Þ ¼ 0. Furthermore, we can get Q i�3ðk0Þ ¼ � � � ¼ Q 1ðk0Þ ¼ Q0ðk0Þ ¼ 0. However, this contradicts Q 0ðk0Þ ¼ 1. h

Lemma 2.3. Suppose that k ¼ k0 is a root of Q iðkÞ. Then Q i�1ðk0ÞQ iþ1ðk0Þ < 0.

C. Gao / Applied Mathematics and Computation 233 (2014) 62–71 65

Proof. Since Q iðk0Þ ¼ 0, by the Lemma 2.2, we have Qi�1ðk0Þ – 0. Meanwhile, by the recurrence relation (2.2),Q iþ1ðk0Þ ¼ �p2ðiÞQi�1ðk0Þ, which implies that Q iþ1ðk0ÞQi�1ðk0Þ ¼ �p2ðiÞQ2

i�1ðk0Þ < 0. This completes the assertion. h

Lemma 2.4. For i ¼ 1; . . . ; T, the roots of QiðkÞ ¼ 0 are simple, and the roots of Q iðkÞ ¼ 0 and Q iþ1ðkÞ ¼ 0 separate one another.

Proof. From (2.2), the highest term of QiðkÞ is ð�1Þiað1Þ � � � aðiÞ. It follows that QiðkÞ ! þ1 as k! �1 and sgnQ iðkÞ ¼ ð�1Þi

as k large enough.Now, we use the method of mathematical induction to prove the result.First, we deal with the case i ¼ 1. Now, Q1ðkÞ ¼ pð1Þ þ qð1Þ � kað1Þ, which implies k1;1 ¼ ðpð1Þ þ qð1ÞÞ=að1Þ is the only root

of Q1ðkÞ. On the other hand, Q2ðk1;1Þ ¼ ðcð2Þ � k1;1að2ÞÞQ1ðk1;1Þ � p2ð1ÞQ0ðk1;1Þ ¼ �p2ð1Þ < 0;Q2ð�1Þ > 0 and Q2ðþ1Þ > 0.Thus, Q2ðkÞ has two zeros k2;1 2 ð�1; k1;1Þ and k2;2 2 ðk1;1;1Þ. Thus, when i ¼ 1, the result holds.

Second, assume that when i ¼ k� 1, the roots of Qk�1ðkÞ ¼ 0 and QkðkÞ ¼ 0 are simple and the roots of Qk�1ðkÞ ¼ 0 andQkðkÞ ¼ 0 separate one another. For details, we assume that the roots of Qk�1ðkÞ ¼ 0 can be ordered as

0 < kk�1;1 < kk�1;2 < � � � < kk�1;k�1

and the roots of QkðkÞ ¼ 0 can be ordered as

0 < kk;1 < kk;2 < � � � < kk;k�1 < kk;k;

then

0 < kk;1 < kk�1;1 < kk;2 < kk�1;2 < � � � < kk;k�1 < kk�1;k�1 < kk;k:

Now, we consider the case i ¼ k. Since Q k�1ð�1Þ > 0 and Q k�1ðkk�1;jÞ ¼ 0 for j ¼ 1;2; . . . ; k� 1, we get that

ð�1Þ0Qk�1ðkk;1Þ ¼ Q k�1ðkk;1Þ > 0; ð�1Þ1Qk�1ðkk;2Þ > 0; . . . ; ð�1Þk�1Q k�1ðkk;kÞ > 0:

This combines with Lemma 2.3, we get

Q kþ1ð�1Þ > 0; ð�1Þ1Q kþ1ðkk;1Þ > 0; ð�1Þ2Q kþ1ðkk;2Þ > 0; . . . ; ð�1ÞkQkþ1ðkk;kÞ > 0:

Thus, in these kþ 1 interval ð�1; kk;1Þ, ðkk;1; kk;2Þ; . . ., ðkk;k�1; kk;kÞ; ðkk;k;þ1Þ, there is a root of Q kþ1ðkÞ ¼ 0 in each interval. Onthe other hand, Qkþ1ðkÞ ¼ 0 has at most kþ 1 roots, which implies that there is only one root of Qkþ1ðkÞ ¼ 0 in each inter-val. h

From Lemmas 2.3 and 2.4, it is easy to obtain the following comparison results of roots of QiðkÞ ¼ 0.

Corollary 2.1. The roots of QiðkÞ ¼ 0 and Qiþ1ðkÞ ¼ 0 satisfy:

ki;k > kiþ1;k; for i ¼ 1;2; . . . ; T � 1; k ¼ 1; . . . ; i:

ki;k < kiþ1;kþ1; for i ¼ 1;2; . . . ; T � 1; k ¼ 1; . . . ; i:ð2:3Þ

Proof of Theorem 1.1. By Lemma 2.1, we get (1.3) and (1.4) has T positive eigenvalues kT;k. Now, we discuss the sign-changing time of the eigenfunction, wkðtÞ; t 2 ½1; T�Z, to kT;k. Let us consider the Eq. (1.3) with initial conditionuð0Þ ¼ uð1Þ ¼ 1. Then the solution, u, to the initial value problem is the function of t and k, which we denote it by uðt; kÞfor convenience. Now, it is not difficult to see that the eigenvalues, kT;k, of (1.3) and (1.4) are also the roots of the equationuðT; kÞ � uðT þ 1; kÞ ¼ 0. Thus, we only need to show the sign-changing times of the following eigenvector,

uð1; kT;kÞ;uð2; kT;kÞ; . . . ;uðT; kT;kÞ� �

: ð2:4Þ

On the other hand, by the direct computation, we get

Q 0ðkÞ ¼ uð1; kÞ ¼ 1; Q iðkÞ ¼ pð1Þpð2Þ � � �pðiÞuðiþ 1Þ; i ¼ 1;2; . . . ; T � 1:

Thus, we only need to show the sign-changing times of the following sequence

Q 0ðkT;kÞ; Q 1ðkT;kÞ; . . . ;QT�1ðkT;kÞ� �

: ð2:5Þ

First, let us consider the sign-changing times of (2.5) at k ¼ kT;1. Since kT;1 6 kj;k for all j ¼ 1; . . . ; T; k ¼ 1;2; . . . ; T and

sgnQ iðkÞ ¼ sgnQ ið�1Þ ¼ 1; as k < ki;1; ð2:6Þ

we get

sgnQ iðkT;1Þ ¼ 1; for i ¼ 0;1; . . . ; T � 1; ð2:7Þ

which implies the sequence (2.5) changes its sign exactly 0 times.

66 C. Gao / Applied Mathematics and Computation 233 (2014) 62–71

Second, let us consider the sign-changing times of (2.5) at k ¼ kT;T . It is not difficult to see that

sgnQ iðkÞ ¼ sgnQ iðþ1Þ ¼ ð�1Þi; as k > ki;i: ð2:8Þ

This combines with (2.3), we get

Q 0ðkT;TÞ;Q 1ðkT;TÞ; . . . ;Q T�1ðkT;TÞ� �

ð2:9Þ

changes its sign exactly T � 1 times.Finally, we will prove that the sequence (2.5) changes its sign exactly j� 1 times for k ¼ kT;j, j ¼ 2;3; . . . ; T � 1. By Lemmas

2.3 and 2.4, we have

kT;j�1 < kT�1;j�1 < kT�2;j�1 < � � � < kj;j�1 < kj�1;j�1;

kT;j < kT�1;j < kT�2;j < � � � < kj;j;

k1;1 < k2;2 < � � � < kj�2;j�2 < kj�1;j�1:

ð2:10Þ

Case 1. If kj�1;j�1 6 kT;j, then sgnQj�1ðkT;jÞ ¼ ð�1Þj�1 (or sgnQ j�1ðkT;jÞ ¼ 0) and

sgnQ iðkT;jÞ ¼ ð�1Þi; for i ¼ 0;1;2; . . . ; j� 2: ð2:11Þ

Moreover,

sgnQ iðkT;jÞ ¼ ð�1Þj�1; for i ¼ j; . . . ; T � 1: ð2:12Þ

(2.10) and (2.11) imply that the sequence (2.5) changes its sign exactly j� 1 times for k ¼ kT;j.Case 2. If kl1 ;j�1 6 kT;j and kl1�1;j�1 > kT;j for some l1 2 ½j; T�Z, then sgnQl1 ðkT;jÞ ¼ ð�1Þj�1 (or sgnQl1 ðkT;jÞ ¼ 0) and

sgnQ iðkT;jÞ ¼ ð�1Þj�1; for i ¼ l1 þ 1; . . . ; T � 1: ð2:13Þ

On the other hand, by Lemmas 2.3 and 2.4, we have

kT;j�2 < kT�1;j�2 < � � � < kj;j�2 < kj�1;j�2 < kj�2;j�2;

kl1�1;j�1 < kl1�2;j�1 < � � � < kj;j�1 < kj�1;j�1:ð2:14Þ

Since kl1�1;j�1 > kT;j, we divide Case 2 into two sub-cases.Case 2.1. If kT;j P kj�2;j�2, then sgnQj�2ðkT;jÞ ¼ ð�1Þj�2 (or sgnQ j�2ðkT;jÞ ¼ 0) and

sgnQ iðkT;jÞ ¼ ð�1Þj�2; for i ¼ j� 2; . . . ; l1 � 1; sgnQ iðkT;jÞ ¼ ð�1Þi; for i ¼ 1;2; . . . ; j� 3: ð2:15Þ

This combines with (2.13), we get (2.5) changes its sign exactly j� 1 times.Case 2.2. If kl2 ;j�2 6 kT;j and kl2�1;j�2 > kT;j for some l2 2 ½j� 1; l1 � 1�Z, then sgnQl2 ðkT;jÞ ¼ ð�1Þj�2 (or sgnQl2 ðkT;jÞ ¼ 0) and

sgnQ iðkT;jÞ ¼ ð�1Þj�2; for i ¼ l2 þ 1; . . . ; l1 � 1: ð2:16Þ

On the other hand, let us consider the following two inequalities.

kT;j�3 < kT�1;j�3 < � � � < kj�1;j�3 < kj�2;j�3 < kj�3;j�3;

kl2�1;j�2 < kl2�2;j�2 < � � � < kj�1;j�2 < kj�2;j�2:ð2:17Þ

Since kl2�1;j�2 > kT;j, we also can divide Case 2.2 into two sub-cases. The first case is easy to obtain, we do not repeat it again,we only discuss the second case.

Now, repeat the second process at most j� 4 times, we get

kT;1 < kT�1;1 < kT�2;1 < � � � < k2;1 < k1;1;

klj�2 ;2 < klj�2�1;2 < klj�2�2;2 < � � � < k2;2:ð2:18Þ

Moreover,

sgnQ iðkT;jÞ ¼ ð�1Þj�1; for i ¼ l1; . . . ; T � 1;

sgnQ iðkT;jÞ ¼ ð�1Þj�2; for i ¼ l2; . . . ; l1 � 1;

� � �sgnQ iðkT;jÞ ¼ ð�1Þj�ðj�2Þ

; for i ¼ lj�2; . . . ; lj�3:

ð2:19Þ

It is easy to see the case that kT;j P k1;1 can be obtained easily. Now, we only discuss the second case, i.e., If klj�1 ;1 6 kT;j andklj�1�1;1 > kT;j for some lj�1 2 ½2; lj�2 � 1�

Z, then sgnQ lj�1

ðkT;jÞ ¼ ð�1Þ1 (or sgnQ lj�1ðkT;jÞ ¼ 0) and

sgnQ iðkT;jÞ ¼ ð�1Þ1; for i ¼ lj�1 þ 1; . . . ; lj�2 � 1: ð2:20Þ

On the other hand, by klj�1�1;1 > kT;j, we get

C. Gao / Applied Mathematics and Computation 233 (2014) 62–71 67

sgnQ iðkT;jÞ ¼ ð�1Þ0; for i ¼ 0; . . . ; lj�1 � 1: ð2:21Þ

From (2.19)–(2.21), we get the sequence (2.5) changes its sign exactly j� 1 times. h

At the end of this section, let us give a comparison result on the eigenvalues of (1.3) and (1.4) as the weight function aðtÞchanges.

Consider the linear eigenvalue problems

�D½pðt � 1ÞDuðt � 1Þ� þ qðtÞuðtÞ ¼ k~aðtÞuðtÞ; t 2 ½1; T�Z; ð2:22Þ

Duð0Þ ¼ DuðTÞ ¼ 0; ð2:23Þ

where ~aðtÞP aðtÞ for t 2 ½1; T�Z.

Lemma 2.5. Suppose that ~lk and lk are the kth eigenvalues of (1.3), (1.4) and (2.22), (2.23), respectively. Then ~lk 6 lk.

Proof. Similar to the proof of Lemma 2.1, we can convert (2.22) and (2.23) to the Matrix equation Au ¼ keBu. Further, letv ¼ JT u, then

1k

v ¼ J�TeBJTv ð2:24Þ

and

1k

v ¼ J�T BJTv: ð2:25Þ

It is easy to see that the eigenvalue of (2.24) is 1~lk

and the eigenvalue of (2.24) is 1lk

. Since aðtÞ 6 ~aðtÞ, we get B 6 eB, whichimplies J�T BJT 6 J�T eBJT . It is seen from the monotonic behavior of eigenvalues of symmetric matrices ([17], Theorem 3,p. 117), we get

1lk6

1~lk;

which implies lk P ~lk. h

3. Sign-changing solutions of nonlinear eigenvalue problems

Let

E ¼ fu : ½1; T�Z ! RjDuð0Þ ¼ DuðTÞ ¼ 0g:

Then E is a Banach space under the norm

kukE ¼ maxt2½1;T�Z

juðtÞj:

Define L : E! E by

LuðtÞ ¼ �D½pðt � 1ÞDuðt � 1Þ� þ qðtÞuðtÞ; u 2 E:

Let n 2 CðR;RÞ be such that f ðuÞ ¼ f0uþ nðuÞ. Clearly,

limjuj!0

nðuÞu¼ 0:

Let us consider

Lu� kraðtÞf0u� kraðtÞnðuÞ ¼ 0; ð3:1Þ

as a bifurcation problem from the trivial solution u � 0.Eq. (3.1) is equivalent to the following equation

uðtÞ ¼ kL�1½rað�Þf0uð�Þ�ðtÞ þ kL�1½rað�Þnðuð�ÞÞ�ðtÞ ¼ kTuþ Hðk;uÞ:

Obviously, T : E! E is a linear and compact operator. Further, we note that kL�1ðnðuð�ÞÞÞk ¼ oðkukÞ for u near 0 in E, sinceL�1 : E! E is a bounded operator.

Let E ¼ R� E under the product topology. Let Sþk denote the set of functions u 2 E changing its sign k� 1 times in ½1; T�Zand satisfying uð0Þ > 0. Set S�k ¼ �Sþk ; Sk ¼ Sþk [ S�k . Then Sþk and S�k are disjoint and open in X. Finally, let W�k ¼ R� S�k andWk ¼ R� Sk. Let R E be the closure of the set of solutions of (1.1) and (1.2).

68 C. Gao / Applied Mathematics and Computation 233 (2014) 62–71

By Theorem 1.1, lkrf0

is a simple eigenvalue. Now, the result of [16] for (3.1) can be stated as follows: for each m 2 fþ;�g,there exists a continuum Cm

k R joining lkrf0;0

� �to infinity in Wm. Moreover, Cm

k nlkrf0;0

� �n o Wm.

Now, we consider the shape of Cmk, which will be helpful for us to prove Theorem 1.2.

Lemma 3.1. Suppose that (H1)–(H3) hold. Then

(i) if lkf1< r < lk

f0, then

lk

rf1;lk

rf0

� # ProjRC

mk; ð3:2Þ

(ii) if lkf0< r < lk

f1, then

lk

rf0;lk

rf1

� # ProjRC

mk; ð3:3Þ

where ProjRCmk denotes the projection of Cm

k on R.

1juj!1 u 06jsj6u

and

Proof. Let f 2 CðR;RÞ be such that f ðuÞ ¼ f uþ fðuÞ. Clearly, lim fðuÞ ¼ 0. Let ~fðuÞ ¼max jfðsÞj. Then ~f is nondecreasing

limu!1

~fðuÞu¼ 0: ð3:4Þ

Let ðkn; ynÞ 2 Cmk satisfy

kn þ kynkE !1:

We note that kn > 0 for all n 2 N, since ð0;0Þ is the only solution of (3.1) for k ¼ 0 and Cmk \ ðf0g � EÞ ¼ ;.

Case 1.lkrf1

< 1 < lkrf0

.

We divide the proof into two steps.

Step 1. We show that if there exists a constant number M > 0 such that

kn 2 ð0;M�; ð3:5Þ

then (3.2) holds.In this case, it follows that

kynkE !1: ð3:6Þ

We divide the equation

Lyn � knraðtÞf1yn � knraðtÞfðynðtÞÞ ¼ 0 ð3:7Þ

by kynkE and set �yn ¼ ynkynkE

. Since �yn is bounded in E and kn is bounded in R, after taking the subsequence if necessary, we havethat �yn ! �y for some �y 2 E with k�ykE ¼ 1 and kn ! �k for some �k 2 R. Moreover, from (3.4) and the fact that ~f is nondecreasing,we have that

limn!1

jfðynðtÞÞjkynkE

¼ 0; ð3:8Þ

since jfðynðtÞÞjkynkE

6~fðjynðtÞjÞkynkE

6fðkynkEÞkynkE

. Thus,

�yðtÞ ¼XT

s¼1

Gðt; sÞ�kraðtÞf1�yðsÞ;

where Gðt; sÞ is the Green’s function of Lu ¼ 0; Duð0Þ ¼ DuðTÞ ¼ 0. Thus,

L�y� �kraðtÞf1�y ¼ 0: ð3:9Þ

We claim that

�y 2 Cmk:

It is clear that �y 2 Cmk # Cm

k , since Cmk is a closed set in R� E. Moreover, �krf1 ¼ lk, so that

�k ¼ lk

rf1:

C. Gao / Applied Mathematics and Computation 233 (2014) 62–71 69

Thus, (3.2) holds.Step 2. We show that there exists a constant M > 0 such that kn 2 ð0;M� for all n.Since fðkn; ynÞg are the solutions to (3.1), it follows that

Lyn ¼ knraðtÞCnðtÞyn; ð3:10Þ

where CnðtÞ :¼ f ðynÞyn

. From (H1) and (H2), there exist two positive constants q1 and q2, such that

q1 <f ðynÞ

yn< q2: ð3:11Þ

By Lemma 2.5, it concludes that

lk

rq2< kn <

lk

rq1:

Case 2.lkrf0< 1 < lk

rf1.

From Step 2 of Case 1, there exists M > 0 such that for all n 2 N,

kn 2 ð0;M�:

Applying a similar argument to that used in Step 1 of Case 1, after taking a subsequence and relabeling, if necessary, it fol-lows that

kn !lk

rf1; yn !1 as n!1;

which implies that (3.3) holds. h

Proof of Theorem 1.2. (i) Since any solution of (3.1) of the form ð1;uÞ yields a solutions of (1.1) and (1.2), it will be enoughto show that Cm

k crosses the hyperplane f1g � E in R� E. This have been done in Lemma 3.1. h

To obtain Theorem 1.2 (ii), we need to prove the following Lemmas.

Lemma 3.2. Suppose that (H1), (H2) and (H3)0 hold. Let J ¼ ½a; b� be a given compact interval in ð0;1Þ. Then for all k 2 J, thereexists MJ > 0 such that all possible solutions u 2 Cm

k of (1.1) and (1.2) satisfy kukE 6 MJ.

Proof. Suppose on the contrary that there exists a sequence fyng of solutions of (1.1) and (1.2) with fkng J and kynkE !1.Let a 2 ð0;1=rbQÞ, where Q ¼

PTs¼1aðsÞGðs; sÞ. Then, by (H3)0, there exists ua > 0 such that juj > ua implies f ðuÞ=u < a.

Let Ka ¼maxjuj2½0;ua �jf ðuÞj and An ¼ ft 2 ½1; T�Zj0 < jynðtÞj 6 uag, Bn ¼ ft 2 ½1; T�ZjjynðtÞj > uag and Cn ¼ ft 2 ½1; T�ZjjynðtÞj ¼ 0g. Then we have

jynðtÞj ¼ rkn

XT

t¼1

aðtÞGðt; sÞf ðynðsÞÞ

¼ rkn

XAn

aðsÞGðt; sÞf ðynðsÞÞ

þ rkn

XBn

aðsÞGðt; sÞf ðynðsÞÞ

þ rkn

XCn

aðsÞGðt; sÞf ðynðsÞÞ

6 knrKaQ þ knr

XBn

Gðt; sÞaðsÞ f ðynðsÞÞj j

for t 2 ½1; T�Z. Thus,

1kn6

rKaQkynkE

þ rX

Bn

aðsÞGðt; sÞ jf ðynðsÞÞjkynkE

:

On Bn; jynðsÞj > ua implies jf ðynðsÞÞjkynkE

6f ðynðsÞÞ

ynðsÞ< a. Thus,

1kn6

rKaQkynkE

þ aX

Bn

raðsÞGðs; sÞ 6 rKaQkynkE

þ arQ :

Since 0 < a 6 kn 6 b for all n, we have 1kn

P 1b for all n and thus,

1b6

rKaQkynkE

þ arQ :

By the fact kynkE !1 as n!1, we get

1b6 arQ <

1b:

This contradiction completes the proof. h

70 C. Gao / Applied Mathematics and Computation 233 (2014) 62–71

Lemma 3.3. Suppose that (H1), (H2) and (H3)0 hold. Then ProjR Cmk ½

lkrf0;þ1Þ.

Proof. Assume on the contrary that supfkjðk; yÞ 2 Cmkg <1, then there exists a sequence fðkn; ynÞg 2 Cm

k such that

limn!1kynkE ¼ 1; kn < C0 ð3:12Þ

for some positive constant C0 independent of n, since Cmk in unbounded. On the other hand, kn > 0 for all n 2 N, since ð0;0Þ is

the only solution of (3.1) for k ¼ 0 and Cmk \ ðf0g � XÞ ¼ ;. Meanwhile, fðkn; ynÞg satisfy

�D½pðt � 1ÞDynðt � 1Þ� þ qðtÞynðtÞ ¼ rknaðtÞ f ðynðtÞÞynðtÞ

ynðtÞ; t 2 ½1; T�Z;

Dynð0Þ ¼ DynðTÞ ¼ 0:

By (H3)0, there exists a positive constant Lf > 0 such that f ðuÞ 6 Lf u. Now, by Lemma 2.5, we get

lk

rLf6 kn:

This combines with Lemma 3.2, limn!1kn ¼ 1, which contradicts (3.12). Thus,

ProjR Cmk

lk

rf0;1

� : �

Now, By Lemma 3.3, Cmk crosses the hyperplane f1g � E in R� E, and then Theorem 1.1 (ii) holds. To obtain Theorem (iii),

we need prove the following Lemma.

Lemma 3.4. Suppose that (H1), (H2) and (H3)00 hold. Then ProjR Cmk 0; lk

rf0

� �.

Proof. Let fðkn; ynÞg Cmk be such that jknj þ kynkE !1 as n!1. Then

�D½pðt � 1ÞDynðt � 1Þ� þ qðtÞynðtÞ ¼ rknaðtÞf ðynðtÞÞ; t 2 ½1; T�Z;

Dynð0Þ ¼ DynðTÞ ¼ 0:

If fkynkEg is bounded, say, kynkE 6 M1, for some M1 independent of n, then we may assume that

limn!1

kn ¼ 1: ð3:13Þ

Note that

f ðynðtÞÞynðtÞ

P inff ðsÞ

sj0 < jsj 6 M1

� > 0:

Then, there exist two constants M3 > 0; M2 > 0 such that

0 < M2 <f ðynðtÞÞ

ynðtÞ< M3: ð3:14Þ

Combining (3.14) with the relation

�D½pðt � 1ÞDynðt � 1Þ� þ qðtÞynðtÞ ¼ rknaðtÞ f ðynðtÞÞynðtÞ

ynðtÞ ¼ 0; t 2 T; ð3:15Þ

Dynð0Þ ¼ DynðTÞ ¼ 0; ð3:16Þ

we get

lk

rM36 kn 6

lk

rM2:

This contradicts (3.13). So, fkynkEg is bounded uniformly for all n 2 N.Now, taking fðkn; ynÞg 2 Cm

k be such that

kynkE ! þ1; as n! þ1: ð3:17Þ

We show that limn!1

kn ¼ 0.

C. Gao / Applied Mathematics and Computation 233 (2014) 62–71 71

Suppose on the contrary that, choosing a subsequence and relabeling if necessary, kn P b0 for some constant b0 > 0. By(3.17), there exists t0 2 T such that ynðt0Þ ¼ kynkE and ynðt0Þ ! þ1 as n! þ1. Thus,

knf ðynðt0ÞÞ

ynðt0Þ! þ1; as n! þ1: ð3:18Þ

Now, if k < T , let us consider the following linear eigenvalue problems

�D½pðt � 1ÞDvðt � 1Þ� þ qðtÞvðtÞ ¼ aaðtÞvðtÞ; t 2 f1;2; . . . ; kg; ð3:19Þ

Dvð0Þ ¼ vðkþ 1Þ ¼ 0: ð3:20Þ

By Theorem 1.1, we get (3.19), (3.20) has a eigenvalue ak;k, and by Corollary 2.1, we have

knf ðynðtÞÞ

ynðtÞ6

ak;k

r; for t 2 ½1; k�Z uniformly;

which contradicts (3.18).If k ¼ T , we only need substitute (3.20) to the boundary condition Dvð0Þ ¼ DvðTÞ ¼ 0. h

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