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A Global Description of the Positive Solutions of Sublinear SecondOrder Discrete Boundary Value Problems
Advances in Difference Equations volume 2009, Article number: 671625 (2009)
Abstract
Let be an integer with , , . We consider boundary value problems of nonlinear secondorder difference equations of the form , , , where , and, for , and , , . We investigate the global structure of positive solutions by using the Rabinowitz's global bifurcation theorem.
1. Introduction
Let be an integer with , , . We study the global structure of positive solutions of the problem
Here is a positive parameter, and are continuous. Denote and .
There are many literature dealing with similar difference equations subject to various boundary value conditions. We refer to Agarwal and Henderson [1], Agarwal and O'Regan [2], Agarwal and Wong [3], Rachunkova and Tisdell [4], Rodriguez [5], Cheng and Yen [6], Zhang and Feng [7], R. Ma and H. Ma [8], Ma [9], and the references therein. These results were usually obtained by analytic techniques, various fixed point theorems, and global bifurcation techniques. For example, in [8], the authors investigated the global structure of signchanging solutions of some discrete boundary value problems in the case that . However, relatively little result is known about the global structure of solutions in the case that , and no global results were found in the available literature when . The likely reason is that the Rabinowitz's global bifurcation theorem [10] cannot be used directly in this case.
In the present work, we obtain a direct and complete description of the global structure of positive solutions of (1.1) under the assumptions:

(A1)
;

(A2)
is continuous and for ;

(A3)
, where ;

(A4)
, where .
Let denote the Banach space defined by
equipped with the norm
Let denote the Banach space defined by
equipped with the norm
Define an operator by
To state our main results, we need the spectrum theory of the linear eigenvalue problem
Let (A1) hold. Then there exists a sequence satisfying that

(i)
is the set of eigenvalues of (1.7);

(ii)
for ;

(iii)
for , is onedimensional subspace of ;

(iv)
for each , if , then has exactly simple generalized zeros in .
Let denote the closure of set of positive solutions of (1.1) in .
Let be a subset of . A component of is meant a maximal connected subset of , that is, a connected subset of which is not contained in any other connected subset of .
The main results of this paper are the following theorem.
Theorem 1.2.
Let (A1)–(A4) hold. Then there exists a component in which joins with , and
for some . Moreover, there exists such that (1.1) has at least two positive solutions for .
We will develop a bifurcation approach to treat the case directly. Crucial to this approach is to construct a sequence of functions which is asymptotic linear at and satisfies
By means of the corresponding auxiliary equations, we obtain a sequence of unbounded components via Rabinnowitz's global bifurcation theorem [10], and this enables us to find an unbounded component satisfying
2. Some Preliminaries
In this section, we give some notations and preliminary results which will be used in the proof of our main results.
Definition 2.1 (see [12]).
Let be a Banach space, and let be a family of subsets of . Then the superior limit of is defined by
Definition 2.2 (see [12]).
A component of a set is meant a maximal connected subset of .
Lemma 2.3 ([12, Whyburn]).
Suppose that is a compact metric space, and are nonintersecting closed subsets of , and no component of interests both and . Then there exist two disjoint compact subsets and , such that , , .
Using the above Whyburn's lemma, Ma and An [13] proved the following lemma.
Lemma 2.4 ([13, Lemma ]).
Let be a Banach space, and let be a family of connected subsets of . Assume that

(i)
there exist , , and , such that ;

(ii)
, where ;

(iii)
for every , is a relatively compact set of , where
(2.2)
Then there exists an unbounded component in and .
Let
It is easy to see that
Denote the cone in by
Now we define a map by
Define an operator by
Then the operator satisfies .
For , let
Using the standard arguments, we may prove the following lemma.
Lemma 2.5.
Assume that (A1)–(A2) hold. Then and is completely continuous.
Lemma 2.6.
Assume that (A1)–(A2) hold. If , then
where
Proof.
Since for , it follows that
3. Proof of the Main Results
Define by
Then with
By (A3), it follows that
To apply the global bifurcation theorem, we extend to be an odd function by
Similarly we may extend to be an odd function for each .
Now let us consider the auxiliary family of the equations
Let be such that
Then
Let us consider
as a bifurcation problem from the trivial solution .
Equation (3.8) can be converted to the equivalent equation
Further we note that for near in .
The results of Rabinowitz [10] for (3.8) can be stated as follows. For each integer , , there exists a continuum of solutions of (3.8) joining to infinity in . Moreover,
Lemma 3.1.
Let (A1)–(A4) hold. Then, for each fixed , joins to in .
Proof.
We divide the proof into two steps.
Step 1.
We show that .
Assume on the contrary that . Let be such that
Then . This together with the fact
implies that
Since , we have that
Set . Then
Now, choosing a subsequence and relabelling if necessary, it follows that there exists with
such that
Moreover, using (3.13), (3.12), and the assumption , it follows that
and consequently, for . This contradicts (3.15). Therefore
Step 2.
We show that .
Assume on the contrary that . Let be such that
Since , for any , we have from (2.6) that
(where ), which yields that is bounded. However, this contradicts (3.19).
Therefore, joins to in .
Lemma 3.2.
Let (A1)–(A4) hold and let be a closed and bounded interval. Then there exists a positive constant , such that
Proof.
Assume on the contrary that there exists a sequence such that
Then, (3.11), (3.12), and (3.13) hold. Set , then
Now, choosing a subsequence and relabeling if necessary, it follows that there exists with
such that
Moreover, from (3.13), (3.12), and the assumption , it follows that
and consequently, for . This contradicts (3.24). Therefore
Lemma 3.3.
Let (A1)–(A4) hold. Then there exits such that
Proof.
Assume on the contrary that there exists such that . Then
Set , then
and for all ,
where ). Let
Then , and
which contradicts (3.30). Therefore, there exists , such that
Proof of Theorem 1.2.
Take . Let be as in Lemma 3.3, and let be a fixed constant satisfying and
where
It is easy to see that there exists , such that
This implies that
(see (2.10) for the definition of ), and accordingly, we may choose which is independent of . From Lemma 2.6 and (3.35), it follows that for ,
This together with the compactness of implies that there exists , such that
Notice that satisfies all conditions in Lemma 2.4, and consequently, contains a component which is unbounded. However, we do not know whether joins with or not. To answer this question, we have to use the following truncation method.
Set
For with , we define a connected subset in satisfying

(1)
;

(2)
joins with infinity in .
We claim that satisfies all of the conditions of Lemma 2.4.
Since
we have from Lemmas 3.1–3.3 and (3.40) that for and ,
Thus, there exists , such that , and accordingly, condition (i) in Lemma 2.4 is satisfied. Obviously,
that is, condition (ii) in Lemma 2.4 holds. Condition (iii) in Lemma 2.4 can be deduced directly from the Arzelà Ascoli theorem and the definition of . Therefore, the superior limit of contains a component joining with infinity in .
Similarly, for each , we may define a connected subset, , in satisfying

(1)
;

(2)
joins with infinity in ,
and the superior limit of contains a component joining with infinity in .
It is easy to verify that
Now, for each , let be a connected component containing . Let
Set
then since
From Lemma 2.4, it follows that is closed in , and furthermore, is compact in .
Let
then
If for some , , then Theorem 1.2 holds.
Assume on the contrary that for all .
For every , let be the component in which contains . Using the standard method, we can find a bounded open set in , such that
where and are the boundary and closure of in , respectively.
Evidently, the following family of the open sets of :
is an open covering of . Since is compact set in , there exist such that , and the family of open sets in :
is a finite open covering of . There is
Let
Then is a bounded open set in ,
and by (3.52), we have
where and are the boundary and closure of in , respectively.
Equation (3.58) together with (3.55) and (3.57) implies that
However, this contradicts (3.50).
Therefore, there exists such that which is unbounded in both and .
Finally, we show that joins with . This will be done by the following three steps.
Step 1.
We show that .
Suppose on the contrary that there exists with
Then
which implies
This is impossible by (A3) and the assumption .
Step 2.
We show that .
Suppose on the contrary that there exists with and
for some constant , then
Thus
where . By (A2), it follows that . Obviously, (3.65) implies that is bounded. This is a contradiction.
Step 3.
We show that .
Suppose on the contrary that there exists with
for some constant , then
Thus
where . By (A2), it follows that . Obviously, (3.68) implies that is bounded. This is a contradiction.
To sum up, there exits a component which joins and .
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Acknowledgments
This work was supported by the NSFC (no. 10671158), the NSF of Gansu Province (no. 3ZS051A25016), NWNUKJCXGC0317, the SpringSun program (no. Z2004162033), SRFDP (no. 20060736001), and the SRF for ROCS, SEM (2006 ).
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Keywords
 Banach Space
 Open Covering
 Global Structure
 Connected Subset
 Global Bifurcation