A Global Description of the Positive Solutions of Sublinear Second-Order Discrete Boundary Value Problems

Let be an integer with , , . We study the global structure of positive solutions of the problem

(1.1)

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 sign-changing 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.

Let denote the Banach space defined by

(1.2)

equipped with the norm

(1.3)

Let denote the Banach space defined by

(1.4)

equipped with the norm

(1.5)

Define an operator by

(1.6)

To state our main results, we need the spectrum theory of the linear eigenvalue problem

(1.7)

Lemma 1.1 ([5, 11]).

Let (A1) hold. Then there exists a sequence satisfying that

1. (i)

   is the set of eigenvalues of (1.7);

    
2. (ii)

   for ;

    
3. (iii)

   for , is one-dimensional subspace of ;

    
4. (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

(1.8)

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

(1.9)

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

(1.10)

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

(2.1)

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

1. (i)

   there exist , , and , such that ;

    
2. (ii)

   , where ;

    
3. (iii)
   for every , is a relatively compact set of , where

   

   (2.2)
    

Then there exists an unbounded component in and .

Let

(2.3)

It is easy to see that

(2.4)

Denote the cone in by

(2.5)

Now we define a map by

(2.6)

Define an operator by

(2.7)

Then the operator satisfies .

For , let

(2.8)

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

(2.9)

where

(2.10)

Proof.

Since for , it follows that

(2.11)

Define by

(3.1)

Then with

(3.2)

By (A3), it follows that

(3.3)

To apply the global bifurcation theorem, we extend to be an odd function by

(3.4)

Similarly we may extend to be an odd function for each .

Now let us consider the auxiliary family of the equations

(3.5)

Let be such that

(3.6)

Then

(3.7)

Let us consider

(3.8)

as a bifurcation problem from the trivial solution .

Equation (3.8) can be converted to the equivalent equation

(3.9)

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

(3.10)

Then . This together with the fact

(3.11)

implies that

(3.12)

Since , we have that

(3.13)

Set . Then

(3.14)

Now, choosing a subsequence and relabelling if necessary, it follows that there exists with

(3.15)

such that

(3.16)

Moreover, using (3.13), (3.12), and the assumption , it follows that

(3.17)

and consequently, for . This contradicts (3.15). Therefore

(3.18)

Step 2.

We show that .

Assume on the contrary that . Let be such that

(3.19)

Since , for any , we have from (2.6) that

(3.20)

(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

(3.21)

Proof.

Assume on the contrary that there exists a sequence such that

(3.22)

Then, (3.11), (3.12), and (3.13) hold. Set , then

(3.23)

Now, choosing a subsequence and relabeling if necessary, it follows that there exists with

(3.24)

such that

(3.25)

Moreover, from (3.13), (3.12), and the assumption , it follows that

(3.26)

and consequently, for . This contradicts (3.24). Therefore

(3.27)

Lemma 3.3.

Let (A1)–(A4) hold. Then there exits such that

(3.28)

Proof.

Assume on the contrary that there exists such that . Then

(3.29)

Set , then

(3.30)

and for all ,

(3.31)

where ). Let

(3.32)

Then , and

(3.33)

which contradicts (3.30). Therefore, there exists , such that

(3.34)

Proof of Theorem 1.2.

Take . Let be as in Lemma 3.3, and let be a fixed constant satisfying and

(3.35)

where

(3.36)

It is easy to see that there exists , such that

(3.37)

This implies that

(3.38)

(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 ,

(3.39)

This together with the compactness of implies that there exists , such that

(3.40)

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

(3.41)

For with , we define a connected subset in satisfying

1. (1)

   ;

    
2. (2)

   joins with infinity in .

    

We claim that satisfies all of the conditions of Lemma 2.4.

Since

(3.42)

we have from Lemmas 3.1–3.3 and (3.40) that for and ,

(3.43)

Thus, there exists , such that , and accordingly, condition (i) in Lemma 2.4 is satisfied. Obviously,

(3.44)

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. (1)

   ;

    
2. (2)

   joins with infinity in ,

    

and the superior limit of contains a component joining with infinity in .

It is easy to verify that

(3.45)

Now, for each , let be a connected component containing . Let

(3.46)

Set

(3.47)

then since

(3.48)

From Lemma 2.4, it follows that is closed in , and furthermore, is compact in .

Let

(3.49)

then

(3.50)

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

(3.51)

(3.52)

where and are the boundary and closure of in , respectively.

Evidently, the following family of the open sets of :

(3.53)

is an open covering of . Since is compact set in , there exist such that , and the family of open sets in :

(3.54)

is a finite open covering of . There is

(3.55)

Let

(3.56)

Then is a bounded open set in ,

(3.57)

and by (3.52), we have

(3.58)

where and are the boundary and closure of in , respectively.

Equation (3.58) together with (3.55) and (3.57) implies that

(3.59)

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

(3.60)

Then

(3.61)

which implies

(3.62)

This is impossible by (A3) and the assumption .

Step 2.

We show that .

Suppose on the contrary that there exists with and

(3.63)

for some constant , then

(3.64)

Thus

(3.65)

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

(3.66)

for some constant , then

(3.67)

Thus

(3.68)

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 .