## 1. Introduction

Let . Let

(1.1)

and for , let

(1.2)

Let

(1.3)

and for , let

(1.4)

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 second-order discrete problems with minimum and maximum:

(1.5)
(1.6)

where is a continuous function, are fixed numbers satisfying and satisfying  .

Functional boundary value problem has been studied by several authors [17]. But most of the papers studied the differential equations functional boundary value problem [16]. As we know, the study of difference equations represents a very important field in mathematical research [812], 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

(1.7)

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 right-hand 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

(1.8)

where satisfies the condition that there exists a nondecreasing function satisfying , such that

(1.9)

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 second-order difference equation boundary value problem (1.5), (1.6) when satisfies

(H 1) is continuous, and there exist , such that

(1.10)

where .

In our paper, we assume , if .

## 2. Preliminaries

Definition 2.1.

Let be a functional. is increasing if

(2.1)

Set

(2.2)

Remark 2.2.

Obviously, belong to . Now, if we take

(2.3)

then boundary condition (1.6) is equal to

(2.4)

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

(2.5)

Furthermore, one has

(2.6)

Proof.

Without loss of generality, we suppose .

1. (i)

For , we have

(2.7)

Then

(2.8)

Furthermore,

(2.9)

which implies

(2.10)
2. (ii)

For , we get

(2.11)

Then

(2.12)

Furthermore,

(2.13)

which implies

(2.14)
3. (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

(2.20)

In particular, one has

(2.21)

Lemma 2.5.

Suppose . If satisfies

(2.22)

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

(2.23)

Lemma 2.6.

Suppose is a solution of (1.5) and . Then

(2.24)

Proof.

Let

(2.25)

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

(2.26)

At first, we prove the inequality

(2.27)

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

(2.28)

Combining this with

(2.29)

we have

(2.30)

At the same time, for , we have and

(2.31)

For , we get

(2.32)

So, for ,

(2.33)

Thus

(2.34)

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

(2.35)

For , we have

(2.36)

So, we get

(2.37)

At the same time, for ,

(2.38)

Combining this with , we have

(2.39)

for .

Also, for , we have and

(2.40)

Similarly, we get

(2.41)

By (2.39) and (2.41), for ,

(2.42)

Then

(2.43)

Subsubcase 1.2.2 ().

By Lemma 2.3 (we take ), it is easy to obtain that

(2.44)

At the same time, for ,

(2.45)

Together with , we have

(2.46)

Thus

(2.47)

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

(2.48)

For , we have

(2.49)

Together with and , for , we get

(2.50)

for .

Also, for , we have

(2.51)

This being combined with , we get

(2.52)

From (2.50) and (2.52),

(2.53)

At last, from Case 1 and Case 2, we obtain

(2.54)

Then by the definition of and (2.54),

(2.55)

Similarly, we can prove

(2.56)

From (2.26), (2.55), and (2.56), the assertion is proved.

Remark 2.7.

It is easy to see that is continuous, and

(2.57)

Lemma 2.8.

Let be a positive constant as in (2.3), as in (2.3), as in (2.23). Set

(2.58)

Define :

(2.59)

Then

(2.60)

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

(2.61)

For ,

(2.62)

By Borsuk theorem, to prove , we only need to prove that the following hypothesis holds.

1. (a)

is an odd operator on , that is,

(2.63)
2. (b)

is a completely continuous operator;

3. (c)

for .

First, we take , then

(2.64)

Thus is asserted.

Second, we prove .

Let be a sequence. Then for each and the fact . The Bolzano-Weiestrass theorem and is finite dimensional show that, going if necessary to subsequences, we can assume . Then

(2.65)

Since and are continuous, is a continuous operator. Then is a completely continuous operator.

At last, we prove (c).

Assume, on the contrary, that

(2.66)

for some . Then

(2.67)
(2.68)
(2.69)

By (2.67) and Lemma 2.5 (take ), there exists , such that . Also from (2.67), we have , then we get

(2.70)
(2.71)

Case 1.

If , then . Now, we claim . In fact, and (2.68) show that there exists satisfying . This being combined with ,

(2.72)

So, , which contradicts with .

Case 2.

If , then from (2.67), and the definition of , we have

(2.73)

Together with (2.69), we get , and

(2.74)

Furthermore, shows that is strictly increasing. From (2.68) and Lemma 2.5, there exist satisfying . Thus, . It is not difficult to see that

(2.75)

that is,

(2.76)

Similarly, , then we get and , which contradicts with .

Case 3.

If , then from (2.67), we get and

(2.77)

By (2.69), we have

(2.78)

If , then . Furthermore, , which contradicts with .

If , then . Furthermore, , which contradicts with .

Then (c) is proved.

From the above discussion, the conditions of Borsuk theorem are satisfied. Then, we get

(2.79)

Set

(2.80)

Similarly, we can prove

(2.81)

## 3. The Main Results

Theorem 3.1.

Suppose holds. Then (1.5) and (1.6) have at least two different solutions when and

(3.1)

Proof.

Let . Consider the boundary conditions

(3.2)
(3.3)

Suppose is a solution of (1.5). Then from Remark 2.7,

(3.4)

Now, if (1.5) and (3.2) have a solution , then Lemma 2.6 and (3.2) show that and

(3.5)

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

(3.6)

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

(3.7)

Define operator ,

(3.8)

Obviously,

(3.9)

Consider the parameter equation

(3.10)

Now, we prove (3.10) has a solution, when .

By Lemma 2.8, . Now we prove the following hypothesis.

1. (a)

is a completely continuous operator;

2. (b)
(3.11)

Since is finite dimensional, is a completely continuous operator.

Suppose (b) is not true. Then,

(3.12)

for some . Then

(3.13)
(3.14)
(3.15)

From (3.13), is a solution of second-order 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

1. (i)

for all,

(3.16)
2. (ii)

For all,

(3.17)

Case 2.

If , . Set

(3.18)
1. (i)

For , if , then

(3.19)

that is,

(3.20)

For ,

(3.21)

then

(3.22)
2. (ii)

Similarly, we can prove for .

Combining Case 1 with Case 2, we get

(3.23)

Moreover, , so,

(3.24)

Similarly, consider the operator ,

(3.25)

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

(3.26)

Proof.

Obviously, . Set

(3.27)

Then Define continuous function ,

(3.28)

Then

(3.29)

Set . Then satisfies .

By Theorem 3.1,

(3.30)
(3.31)

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

(3.32)

are two different solutions of (1.5) and (2.4), then are the two different solutions of (1.5) and (1.6).