on the linear and nonlinear discrete second-order neumann boundary value problems
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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 nondecreasinglimu!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
lkrq2< 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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