Abstract
Consider the multiplicity of solutions to the nonlinear secondorder discrete problems with minimum and maximum: , , , , where are fixed numbers satisfying are satisfying , ,.
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1. Introduction
Let . Let
and for , let
Let
and for , let
It is clear that the above are norms on and , respectively, and that the finite dimensionality of these spaces makes them Banach spaces.
In this paper, we discuss the nonlinear secondorder discrete problems with minimum and maximum:
where is a continuous function, are fixed numbers satisfying and satisfying .
Functional boundary value problem has been studied by several authors [1–7]. But most of the papers studied the differential equations functional boundary value problem [1–6]. As we know, the study of difference equations represents a very important field in mathematical research [8–12], so it is necessary to investigate the corresponding difference equations with nonlinear boundary conditions.
Our ideas arise from [1, 3]. In 1993, Brykalov [1] discussed the existence of two different solutions to the nonlinear differential equation with nonlinear boundary conditions
where is a bounded function, that is, there exists a constant , such that . The proofs in [1] are based on the technique of monotone boundary conditions developed in [2]. From [1, 2], it is clear that the results of [1] are valid for functional differential equations in general form and for some cases of unbounded righthand side of the equation (see [1, Remark 3 and (5)], [2, Remark 2 and (8)]).
In 1998, Staně k [3] worked on the existence of two different solutions to the nonlinear differential equation with nonlinear boundary conditions
where satisfies the condition that there exists a nondecreasing function satisfying , such that
It is not difficult to see that when we take , (1.8) is to be (1.7), and may not be bounded.
But as far as we know, there have been no discussions about the discrete problems with minimum and maximum in literature. So, we use the Borsuk theorem [13] to discuss the existence of two different solutions to the secondorder difference equation boundary value problem (1.5), (1.6) when satisfies
(H 1) is continuous, and there exist , such that
where .
In our paper, we assume , if .
2. Preliminaries
Definition 2.1.
Let be a functional. is increasing if
Set
Remark 2.2.
Obviously, belong to . Now, if we take
then boundary condition (1.6) is equal to
So, in the rest part of this paper, we only deal with BVP (1.5), (2.4).
Lemma 2.3.
Suppose . If there exist , such that , then
Furthermore, one has
Proof.
Without loss of generality, we suppose .

(i)
For , we have
(2.7)Then
(2.8)Furthermore,
(2.9)which implies
(2.10) 
(ii)
For , we get
(2.11)Then
(2.12)Furthermore,
(2.13)which implies
(2.14) 
(iii)
For , we have
(2.15)Then
(2.16)Furthermore,
(2.17)which implies
(2.18)In particular, it is not hard to obtain
(2.19)
Similarly, we can obtain the following lemma.
Lemma 2.4.
Suppose . If there exists such that , then
In particular, one has
Lemma 2.5.
Suppose . If satisfies
then there exist , such that .
Proof.
We only prove that there exists , such that , and the other can be proved similarly.
Suppose for . Then . Furthermore, , which contradicts with .
Define functional by
Lemma 2.6.
Suppose is a solution of (1.5) and . Then
Proof.
Let
and be the number of elements in the number of elements in .
If , then ; if , then . Equation (2.24) is obvious.
Now, suppose and . It is easy to see that
At first, we prove the inequality
Since , by Lemma 2.5, there exist , such that . Without loss of generality, we suppose .
For any , there exits satisfying one of the following cases:
Case 1.
,
Case 2.
.
We only prove that (2.27) holds when Case 1 occurs, (if Case 2 occurs, it can be similarly proved).
If Case 1 holds, we divide the proof into two cases.
Subcase 1.1.
If , without loss of generality, we suppose , then by Lemma 2.3, we have
Combining this with
we have
At the same time, for , we have and
For , we get
So, for ,
Thus
Subcase 1.2 ().
Without loss of generality, we suppose . Then will be discussed in different situations.
Subsubcase 1.2.1 ().
By Lemma 2.3 (we take ), it is not difficult to see that
For , we have
So, we get
At the same time, for ,
Combining this with , we have
for .
Also, for , we have and
Similarly, we get
By (2.39) and (2.41), for ,
Then
Subsubcase 1.2.2 ().
By Lemma 2.3 (we take ), it is easy to obtain that
At the same time, for ,
Together with , we have
Thus
Case 1.2.3 ().
Without loss of generality, we suppose (when , by Lemma 2.4, it can be proved similarly). Then from Lemma 2.3 (we take ), it is not difficult to see that
For , we have
Together with and , for , we get
for .
Also, for , we have
This being combined with , we get
From (2.50) and (2.52),
At last, from Case 1 and Case 2, we obtain
Then by the definition of and (2.54),
Similarly, we can prove
From (2.26), (2.55), and (2.56), the assertion is proved.
Remark 2.7.
It is easy to see that is continuous, and
Lemma 2.8.
Let be a positive constant as in (2.3), as in (2.3), as in (2.23). Set
Define :
Then
where denotes Brouwer degree, and the identity operator on .
Proof.
Obviously, is a bounded open and symmetric with respect to subset of Banach space .
Define
For ,
By Borsuk theorem, to prove , we only need to prove that the following hypothesis holds.

(a)
is an odd operator on , that is,
(2.63) 
(b)
is a completely continuous operator;

(c)
for .
First, we take , then
Thus is asserted.
Second, we prove .
Let be a sequence. Then for each and the fact . The BolzanoWeiestrass theorem and is finite dimensional show that, going if necessary to subsequences, we can assume . Then
Since and are continuous, is a continuous operator. Then is a completely continuous operator.
At last, we prove (c).
Assume, on the contrary, that
for some . Then
By (2.67) and Lemma 2.5 (take ), there exists , such that . Also from (2.67), we have , then we get
Case 1.
If , then . Now, we claim . In fact, and (2.68) show that there exists satisfying . This being combined with ,
So, , which contradicts with .
Case 2.
If , then from (2.67), and the definition of , we have
Together with (2.69), we get , and
Furthermore, shows that is strictly increasing. From (2.68) and Lemma 2.5, there exist satisfying . Thus, . It is not difficult to see that
that is,
Similarly, , then we get and , which contradicts with .
Case 3.
If , then from (2.67), we get and
By (2.69), we have
If , then . Furthermore, , which contradicts with .
If , then . Furthermore, , which contradicts with .
If , then a contradiction.
Then (c) is proved.
From the above discussion, the conditions of Borsuk theorem are satisfied. Then, we get
Set
Similarly, we can prove
3. The Main Results
Theorem 3.1.
Suppose holds. Then (1.5) and (1.6) have at least two different solutions when and
Proof.
Let . Consider the boundary conditions
Suppose is a solution of (1.5). Then from Remark 2.7,
Now, if (1.5) and (3.2) have a solution , then Lemma 2.6 and (3.2) show that and
So, is a solution of (1.5) and (2.4), that is, is a solution of (1.5) and (1.6).
Similarly, if (1.5), (3.3) have a solution , then and
So, is a solution of (1.5) and (2.4).
Furthermore, since and .
Next, we need to prove BVPs (1.5), (3.2), and (1.5) and (3.3) have solutions, respectively.
Set
Define operator ,
Obviously,
Consider the parameter equation
Now, we prove (3.10) has a solution, when .
By Lemma 2.8, . Now we prove the following hypothesis.

(a)
is a completely continuous operator;

(b)
(3.11)
Since is finite dimensional, is a completely continuous operator.
Suppose (b) is not true. Then,
for some . Then
From (3.13), is a solution of secondorder difference equation . By Remark 2.7, . And from (3.14), there exist , such that . Now, we can prove it in two cases.
Case 1.
If there exists , such that , then

(i)
for all,
(3.16) 
(ii)
For all,
(3.17)
Case 2.
If , . Set

(i)
For , if , then
(3.19)that is,
(3.20)For ,
(3.21)then
(3.22) 
(ii)
Similarly, we can prove for .
Combining Case 1 with Case 2, we get
(3.23)
Moreover, , so,
which contradicts with .
Similarly, consider the operator ,
we can obtain a solution of BVP (1.5) and (3.3).
Theorem 3.2.
Suppose holds. Then (1.5) and (1.6) have at least two different solutions when and
Proof.
Obviously, . Set
Then Define continuous function ,
Then
Set . Then satisfies .
By Theorem 3.1,
have at least two difference solutions . Since is a solution of (3.30), if and only if is a solution of (1.5), we see that
are two different solutions of (1.5) and (2.4), then are the two different solutions of (1.5) and (1.6).
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Acknowledgments
This work was supported by the NSFC (Grant no. 10671158), the NSF of Gansu Province (Grant no. 3ZS051A25016), NWNUKJCXGC, the Springsun Program (no. Z2004162033), SRFDP (Grant no. 20060736001), and the SRF for ROCS, SEM (2006[311]).
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Ma, R., Gao, C. Existence and Multiple Solutions for Nonlinear SecondOrder Discrete Problems with Minimum and Maximum. Adv Differ Equ 2008, 586020 (2008). https://doi.org/10.1155/2008/586020
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DOI: https://doi.org/10.1155/2008/586020