#### Abstract

Let *X* be a Banach space and a family of connected subsets of . We prove the existence of unbounded components in superior limit of , denoted by , which have prescribed shapes. As applications, we investigate the global behavior of the set of positive periodic solutions to nonlinear first-order differential equations with delay, which can be used for modeling physiological processes.

#### 1. Introduction and the Main Results

The connectivity result on the fixed set of a -parameter family of maps, which goes back to Leray and Schauder [1] and was proved in its full generality by Browder [2], is a useful tool in the study of global continua of solutions on nonlinear differential equations. Costa and Gonçalves [3] stated and proved a suitable version for the study of nonlinear boundary value problems at resonance. Massabò and Pejsachowicz [4] generalized the main results of [1, 2] to the -parameter family of compact vector fields. The above results were established when the parameter(s) changes in a bounded set. Sun and Song [5] proved the existence of unbounded connected component of -parameter family of compact vector fields, where the parameter varies on whole real line. All of these results play important roles in the study of nonlinear functional analysis and nonlinear differential equations.

For clearly reading, we firstly recall Kuratowski's definitions and notations in [6].

Let be a metric space. Let be a family of subsets of . Then the* superior limit * of is defined by

A* component* of a set means a maximal connected subset of .

*Definition 1. *Let be a Banach space with the norm . Let be a component of solutions in . meets and infinity means that there existed a sequence such that as .

For , let us denote

Let be a family of connected subsets of . The purpose of this paper is to study the existence of unbounded components in which have prescribed shapes.

More precisely, we will prove the following theorems.

Theorem 2. *Let be a Banach space and let be a family of connected subsets of . Assume that*(A1)*there exist and , such that
*(A2)* and meets and infinity;*(A3)*for every , is a relatively compact set of , where
**Then there exists a component in satisfying*(a)* meets and infinity;*(b)*. *

Theorem 3. *Let be a constant. Let be a Banach space, and let be a family of connected subsets of . Assume that*(H1)*;
*(H2)*there exist and , such that
*(H3)* for all , and meets and infinity in ;*(H4)*for every , is a relatively compact set of .** Then there exists a component in such that*(a)*both and are unbounded;*(b)*.*

#### 2. Proofs of the Main Results

To prove Theorems 2 and 3, we need the following preliminary result, which is proved by Ma and An [7] by using Whyburn lemma, and the method of Sun and Song to prove [6, Lemma 2.2].

Lemma 4 (see [7, Lemma 2.2]). *Let be a Banach space with the norm . 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
**Then there exists an unbounded component in and .*

*Proof of Theorem 2. *(a) It is a direct consequence of Lemma 4.

(b) Assume, on the contrary, that the conclusion is not true. Then there exists with and . Hence, there exists , such that
Thus, there exists , such that, for ,
However, this contradicts (3).

*Proof of Theorem 3. *(a) Since , we may assume that
So, it follows from conditions (ii) and (iii) that meets and infinity in .

For each , let be a component containing . Let

Set
Then since
From Lemma 4, it follows that is closed in , and, furthermore, is compact in .

Let
By Lemma 4, contains a component which meets and infinity in . Obviously

If for some , then Theorem 3 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
and for every ,
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 . This implies that

Let
Then is a bounded open set in ,
and by (16), we have
where and are the boundary and closure of in , respectively.

Now, (22) together with (19) and (21) implied that
However, this contradicts .

Therefore, there exists such that which is unbounded in both and .

(b) By a fully analogous argument as in the proof of Theorem 2(b) (with minor modifications), one can immediately obtain the desired results.

#### 3. Application to Functional Differential Equations

In recent years, there has been considerable interest in the existence of -periodic solutions of the equation where are -periodic functions and is a continuous -periodic function. Equation (24) has been proposed as a model for a variety of physiological processes and conditions including production of blood cells, respiration, and cardiac arrhythmias. See, for example, [8–20] and the references therein.

Recently, Wang [18] used the fixed point index [20, 21] to study the existence, multiplicity, and nonexistence of positive solutions of (24) under the following assumptions.(C1) are -periodic functions, , on ; is -periodic functions.(C2) are continuous. for , are given positive constants. for .

Let and denote His results provide no any information about the global behavior of the set of positive solutions of (24).

In this section, we will use Theorems 2 and 3 to establish several results on the global behavior of the set of positive solutions of (24), and, accordingly, we get some existence and multiplicity results of positive solutions of (24).

We will work essentially in the Banach space - with sup norm .

By a positive solution of (24), we mean a pair , where and is a solution of (24) with on .

Let be the closure of the set of positive solutions of (24).

We extend the function to a continuous function defined on in such a way that for all . For , we then look at arbitrary solutions of the eigenvalue problem It was shown in [18] that (26) is equivalent to where By the positivity of Green's function , , and , such solutions are positive. Therefore, the closure of the set of nontrivial solutions of (24) in is exactly .

Next, we consider the spectrum of the linear eigenvalue problem

Lemma 5. *Let be a positive constant. Then the linear problem (29) has a unique eigenvalue , which is positive and simple, and the corresponding eigenfunction is of one sign.*

*Proof. *Define an operator by
Let be the cone
Then it follows from [18, Lemma 2.2] that is strongly positive and completely continuous. Thus, the desired result is a direct consequence of Krein-Rutman Theorem, cf. [22, Theorem 19.3].

Theorem 6. *Let (C1)-(C2) hold. Assume that*(C3)*. **Then contains a component satisfying*(1)* meets and ;*(2)*for every , there exists , such that
*

Corollary 7. *Let (C1)–(C3) hold. Then there is a constant such that (24) has at least two positive solutions as .*

*Proof. *Let
Then . It is easy to see from Theorem 6 that (24) has at least two positive solutions as .

Denote the cone in by and for , let

Define an operator by

Lemma 8 (see [18]). *Assume that (C1)-(C2) hold. Then is completely continuous.*

Lemma 9. *Let (C1)-(C2) hold. If , then
**
where .*

*Proof. *It is well known from Wang [18] that
This together with the fact that is -periodic and on implies that

To prove above Theorem 6, we define by Then with By (C3), it follows that

Now let us consider the auxiliary family of the equations Let be such that Note that Define a linear operator with From [18], it follows that is compact and continuous.

Now (43) can be rewritten to the form where It is easy to check that It is easy to check that (48) is equivalent to Let us consider (48) as a bifurcation problem from the trivial solution .

Since , the results of Nonlinear Krein-Rutman Theorem (see Dancer [22] and Zeidler [23, Corollary 15.12]) for (48) can be stated as follows: there exists a continuum of positive solutions of (48) joining to infinity in . Moreover, and is the only bifurcation point of (48) lying on trivial solutions line .

*Proof of Theorem 6. *Let us verify that satisfies all of the conditions of Theorem 2.

It follows from (42) that as . Therefore, (A2) holds.

Let be fixed. Then there exists , such that
Thus
From this and Lemma 9, it follows that there exists with , such that (48) has no solution with
Since is arbitrary, we see that (A1) is satisfied.(A3)can be deduced directly from the Arzela-Ascoli Theorem and the definition of .

Therefore, the superior limit of contains an unbounded component with .

Moreover, .

Finally, we show that meets and .

Let with
We claim that .

Assume on the contrary that . Let
Then
So on as . This contradicts (58). Thus (55) implies that

Assume on the contrary that ( after taking a subsequence and relabeling if necessary).

Since on , it follows from (59) that uniformly on . This together with implies on as . This contradicts (58) again.

Theorem 10. *Let (C1)-(C2) hold. Assume that*(C4)*. **Then contains a component satisfying the following.*(1)*For given ,
*(2)*For every , there exists , such that
*

Corollary 11. *Let (C1)-(C2) and (C4) hold. Then there is a constant such that (24) has at least two positive solutions as .*

To prove Theorem 10, we define as in (40). Notice that (C4) implies that Let be the function satisfying (44)-(45).

Now (43) can be rewritten to the form where It is easy to check that Let us consider (63) as a bifurcation problem from the trivial solution .

By similar method to deal with (48) in which , we have from and [23, Corollary 15.12]) that there exists a continuum of positive solutions of (63) joining to infinity in . Moreover, and is the only bifurcation point of (48) lying on trivial solutions line .

Lemma 12. *Assume that (C1)-(C2) hold. If , then
**
where
*

*Proof. *Since and for , it follows that

Lemma 13. *Assume that (C1)-(C2) and (C4) hold, and let be a compact subinterval of . Then
**
for some positive constant , independent of , and .*

*Proof. *Assume on the contrary that there exists a sequence such that
Set . Then
By (72), is bounded in . This together with the fact that implies that there exists with
such that
(after choosing a subsequence and relabeling if necessary). Since is bounded in , is bounded in , and subsequently, for some . By the standard method, we can prove that
Moreover, combining (75) and (76) with the assumption and the corresponding integral equations of (72) and (73) and using Lebesgue dominated convergence theorem, we conclude that
Note that and on . This means that . However, this contradicts (78).

Now, we are in the position to prove Theorem 10.

*Proof of Theorem 10 (sketched). *fixed . From Lemma 12, it follows that (63) has no solution if

Applying Lemma 13, it is easy to verify that satisfies all of the conditions of Theorem 3. So, there exists a component in such that(a)both and are unbounded;(b). Lemma 13 guarantees that satisfies

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. The research was supported by the NSFC (nos. 11301059 and 11361054) and HSSF of Ministry of Education of China (no. 13YJA790078).