`Abstract and Applied AnalysisVolume 2014 (2014), Article ID 731863, 8 pageshttp://dx.doi.org/10.1155/2014/731863`
Research Article

## Non-Single-Valley Solutions for -Order Feigenbaum’s Type Functional Equation

College of Science, China University of Petroleum, Qingdao, Shandong 266555, China

Received 19 March 2014; Accepted 16 June 2014; Published 10 July 2014

Copyright © 2014 Min Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

This work deals with Feigenbaum’s functional equation , , , , where is an integer, is the -fold iteration of , and is a strictly increasing continuous function on that satisfies , , . Using a constructive method, we discuss the existence of non-single-valley continuous solutions of the above equation.

#### 1. Introduction

In 1978, Feigenbaum [1, 2] and independently Couliet and Tresser [3] introduced the notion of renormalization for real dynamical systems. In 1992, Sullivan [4] proved the uniqueness of the fixed point for the period-doubling renormalization operator. This fixed point of renormalization satisfies a functional equation known as the Cvitanović-Feigenbaum equation:

As mentioned above, this equation and its solution play an important role in the theory initiated by Feigenbaum [1, 2]. However, it is difficult to find an exact solution of the above equation in general. This problem can be studied in classes of smooth functions or of continuous functions. In classes of smooth functions, the existence of smooth solutions for (1) has been established in [58] and references therein. As far as we know, continuous solutions of (1) in classes of continuous functions have been relatively little researched. In this direction, we refer to [9, 10]. In particular, Yang and Zhang [9] demonstrated the existence of a single-valley continuous solution for the following equation: which is called the second type of Feigenbaum’s functional equations. In the last years, a number of authors considered the more general equation where is an integer and is the -fold iteration of . It is easy to see that (2) is a special case of (3). For large enough, Eckmann et al. [11] showed that there exists a solution of (3) similar to the function . For any , Zhang et al. [12] and Liao et al. [13] proved that (3) has single-valley-extended continuous solutions.

In the present paper, we will consider Feigenbaum’s functional equations where is a strictly increasing continuous function on that satisfies , , . We will prove the existence of single-valley-extended non-single-valley continuous solutions of (4) by the constructive method. Obviously, let ; then (4) is (3).

#### 2. Basic Definitions and Lemmas

In this section, we will give some characterizations of single-valley-extended non-single-valley continuous solutions of (4); they will be proved in the Appendix.

Definition 1. One calls a single-valley-extended continuous solution of (4) if (i)   is a continuous solution of (4); (ii) there exists such that is strictly decreasing on and strictly increasing on .

Definition 2. One calls a single-valley-extended non-single-valley continuous solution of (4) if (i)   is a single-valley-extended continuous solution of (4); (ii)   has at least an extreme point on .

In the following, we always let and define the sets Obviously, and, from the fact that and are, respectively, strictly decreasing and , , then

Lemma 3. Suppose that is a single-valley-extended non-single-valley continuous solution of (4) and is the extreme point of in . Then the following conclusions hold:(i) has a unique minimum point with ;(ii) is a recurrent but not periodic point of ;(iii) is an extreme point of and ;(iv) has a unique fixed point on , and (v)for and , then if and only if and .

Lemma 4. Suppose that is a single-valley-extended non-single-valley continuous solution of (4). Let , , and ; then the following conclusions hold:(i) are pairwise disjoint;(ii)for all , then is a homeomorphism.

Lemma 5. Suppose that is a single-valley-extended non-single-valley continuous solution of (4). Then for all , has infinite many extreme points and ; is the maximum point of on ; is the minimum point of on .

Lemma 6. Suppose that is a single-valley-extended non-single-valley continuous solution of (4). Then the equation has only one solution on .

Lemma 7. Let , be two single-valley-extended non-single-valley continuous solutions of (4). If then on .

#### 3. Constructive Method of Solutions

In this section, we will prove constructively the existence of single-valley-extended non-single-valley continuous solutions of (4).

Theorem 8. Fix a strictly increasing continuous function on with , . Denote . If is a continuous function on and satisfies the following conditions,(i)there exists an such that and is strictly decreasing on and strictly increasing on ;(ii), ;(iii)there exists an with , ; denote , ; then(a) are pairwise disjoint,(b)for all , then is a homeomorphism,(c) is an endpoint of and ;(iv)the equation has only one solution on ,then there exists a uniquely single-valley-extended non-single-valley continuous function satisfying the equation And has infinitely many extreme points. Conversely, if is the restriction on of a single-valley-extended non-single-valley continuous solution to (4), then conditions above must hold.

Proof. Suppose that satisfies conditions (i)–(iv). Define . By condition (iii), we have that is a homeomorphism. And by we know that is strictly increasing.
Firstly, we define on by induction. Obviously, is well defined on . Suppose that is well defined as on and strictly increasing and decreasing, respectively, on and for all , where is a certain integer. Let then is well defined as on and strictly increasing and decreasing, respectively, on and . Thereby is well defined as a continuous function on and strictly increasing and decreasing, respectively, on and for all by induction. And
Secondly, we prove that and have the same value on the common endpoint of and for all . From condition (ii) we have And letting , in (11) we get That is, and have the same value on the common endpoint of and . Suppose that where is a certain integer. Let in (11); then we have That is, and have the same value on the common endpoint of and for all by induction. Therefore, we can let Since is continuous on and increasing and decreasing, respectively, on and for and (14), (15), and (16), we have that is a non-single-valley continuous function and has infinitely many extreme points on .
Thirdly, we prove that is continuous at as follows. It is trivial that is strictly decreasing and . We prove is strictly increasing on by induction as follows. Since and from (11), we get Suppose that , where is a certain integer. Therefore, by (11) and the fact that is strictly increasing, we have that Thereby, is strictly increasing in by induction. Let ; then . From (12), we have that Let ; we get . By condition (iv), we know . This proves that is continuous at . Thereby, is a continuous function on . We have that satisfies (10) by (12) and is unique from Lemma 7.
Obviously, if is the restriction on of a single-valley-extended non-single-valley continuous solution to (4), then conditions (i)–(iv) must hold by the lemmas in Section 2.

Example 9. Let be defined by Obviously, satisfies the conditions of Theorem 8 with , , and , . Hence it is the restriction to of a single-valley-extended continuous solution to (4). Since has the minimum point and , is a single-valley-extended non-single-valley continuous solution. Its graph is depicted in Figure 1.

Figure 1: The graph of non-single-valley solution.

Example 10. Let be defined by Obviously, satisfies the conditions of Theorem 8 with , , and . Hence it is the restriction to of a single-valley-extended continuous solution to (4). Since has the minimum point and , , is a single-valley-extended non-single-valley continuous solution. Its graph is similar to Figure 1.

#### Appendix

Proof of Lemma 3. (i) Suppose that is a minimum point of . By (4) we have And from we know . If , then for all we have . Thus there exists such that . And by (4) we have This contradicts , . Thus . And from the definition of we know and .
(ii) We now prove that, for all and each , we have Obviously, (A.3) holds for by (4). Suppose that (A.3) holds for , where is a certain integer. Therefore, by induction and (4), we have that that is, (A.3) holds for . Thereby, (A.3) is proved by induction. Let in (A.3); we have that And it is trivial that is strictly decreasing and . Thereby, we have that that is, we proved that is a recurrent but not periodic point of .
(iii) Firstly, we prove that is an extreme point of . By the fact that is strictly increasing on and (4), we have that is strictly increasing in . Thereby is strictly monotone in . Suppose that is not an extreme point; then is strictly decreasing on . We prove is strictly decreasing on , respectively, as follows.
Obviously, is strictly decreasing on . Suppose that is strictly decreasing on for all , where is a certain integer. By (4), we have that is strictly decreasing on . Thereby is strictly monotone on . Suppose that is strictly increasing on ; then we claim that We first prove that Suppose that there exists such that . And from (A.3) we get
Similarly, we have . If , then This contradicts that is not a periodic point. That is, we proved (A.8). Thereby, we have , . And from the fact that is strictly increasing on , we get . Suppose that , where is a certain integer. If , and by the fact that is strictly monotone on and (A.3), we get or This contradicts that . That is, . Thereby, we proved (A.7) by induction. If , then by (A.3) we have This contradicts conclusion (ii). Thereby, we get , . This contradicts that . Thereby, is strictly decreasing on . Furthermore, we proved that is strictly decreasing on , respectively, by induction; that is, is single-valley solution of (4). This contradicts the condition that is a non-single-valley solution of (4). That is, we proved that is an extreme point of .
Secondly, we prove that . We claim that Suppose that ; by (A.5) we get . And from , we have that . Let obviously, is a limited set and , . By (A.5), we have that , . This contradicts that is limited. Thereby, we proved that . Suppose that ; from (A.3), we have This contradicts that is not a periodic point of . That is, (A.14) holds. Suppose that . Since is an extreme point of , we have that is strictly increasing on . Furthermore, we have . We claim that It is trivial that (A.17) holds for . Suppose that (A.17) holds for , where is a certain integer. By (A.3), we know is strictly monotone on . If , then there exists such that . Thus, is an extreme point of on . That is, is not monotone on . Furthermore, is not monotone on . This contradicts that is monotone on . Thus . Similarly, we have . That is, (A.17) holds for . Thereby, (A.17) holds by induction. This contradicts that . Thus .
(iv) Firstly, suppose that is a fixed point of ; then by (A.5), we have . And by , we have . If , then . And by induction, for all , we have . Specially, This contradicts (A.5). Thereby, .
Secondly, there exists at least one fixed point by , . We prove is the unique fixed point as follows. We claim that Suppose that there exists such that ; then by (A.3), we have that is, . And from , this contradicts that does not have a fixed point on . Thus, we have , . Suppose that there exists such that ; then by (A.3), we have Thereby, . That is, is a fixed point of . Since has no fixed point on and , we have that must be strictly increasing on . And has the fixed point ; thus is strictly increasing on . But by (4) and the fact that is decreasing on , we have that is strictly decreasing on . This contradicts. Thus, we have . Thereby, we prove (A.19). Thus, we have . And since is decreasing on , we have that is the unique fixed point.
(v) By (4), we have . And since is the unique minimum point of , it follows that . Thus the sufficiency is proved.
We prove the necessity as follows. Suppose that for some and . Firstly, we claim that Suppose that ; then is not monotone on . Thereby, is not monotone on . This contradicts that is monotone on by (A.3). Thus . We have similarly that . Thereby we prove (A.22); that is, or , .
Secondly, we prove that Suppose that for some ; then . And from (A.5), we have . This contradicts that is not a periodic point of . Thus we prove . Suppose that for some ; then . And from (A.3), we have This contradicts that is a recurrent point of . Thus we prove . Thereby (A.23) holds. That is, we get and .
Lastly, we prove . Suppose that there exists , such that . Then we can suppose that and This contradicts that is not a periodic point of . Thus we prove . Thereby we have and .

Proof of Lemma 4. (i) Firstly, we prove that, for all , we have ; that is, . We claim that Suppose that there exists such that . And from Lemma 3(v) we have . This contradicts Lemma 3(v). Thus (A.26) holds. We claim that It is trivial that by is a recurrent point of . Suppose that there exists such that . And from we have that there exists such that . And from the fact that is an extreme point of , we get that is not monotone on . This contradicts that is strictly monotone on . Thus, (A.27) holds. We claim that Suppose that there exists and such that . And from (4) we get . This contradicts (A.27). Thus . Suppose that there exists and such that . And from (A.27) we have that there exists such that . And from (4) we get . This contradicts (A.27). Thus (A.28) holds. That is, .
Secondly, we prove that , , are pairwise disjoint. Suppose that there exists , such that . Let ; then there exist , such that . Thereby we have . This contradicts . Thus we proved that are pairwise disjoint.
(ii) For all , then is a homeomorphism by Lemmas 3(i) and 3(v). Thereby is also a homeomorphism.

Proof of Lemma 5. is a maximum point of on by Lemma 3(iii). Suppose that is the maximum point of on , where is some certain integer. Firstly, we prove is an extreme point of . If otherwise, is strictly increasing on . Thus and is strictly increasing on . We claim that is strictly increasing on for all . It holds obviously when . Suppose that is strictly increasing on for all , where is some certain integer. is strictly increasing on by (4). Thereby is strictly monotone on . If is strictly decreasing on , then we claim that Its proof is similar to (A.7) and we omit it here. Thereby , . This contradicts . Thus is strictly increasing on . Thereby is strictly increasing on by induction. Furthermore . This contradicts . Thus is an extreme point of .
Secondly, we prove is the minimum point of on . Suppose that there exists such that or ; then we claim that or The proof is similar to (A.7) and we omit it here. Thereby . This contradicts . Thus is the minimum point of on .
Thirdly, we prove is an extreme point of . is strictly increasing on by (4) and is strictly decreasing on . And from Lemma 4(ii) we have that is strictly increasing on . Suppose that is not an extreme point of ; then is strictly decreasing on . And from (4) and the fact that is strictly increasing on , we get that is strictly increasing on . Thereby is strictly decreasing on . This contradicts that is strictly increasing on . Thus is the extreme point of .
Lastly, we have that is the maximum point of on . The proof is similar to and we omit it here. Thereby is the maximum point of on and is the minimum point of on for all by induction.

Proof of Lemma 6. It is trivial that is a solution of the equation by (4). Suppose that is an arbitrary solution of this equation; that is, . Since , we have that there exists , such that . We claim that Obviously, (A.32) holds for . Suppose that (A.32) holds for , where is a certain integer. Therefore, by (4) we have that Since , we have . And since is strictly increasing on by , we have . That is, (A.32) holds for . Thereby, (A.32) is proved by induction. By the fact that is strictly decreasing and , we have .

Proof of Lemma 7. There exist such that by (8). Denote and , . We prove that as follows. It is trivial that , and by Lemmas 3(v) and 4. Since is strictly monotone on , we have that . Let ; then and . Define . By Lemmas 3(v) and 4, we have that is a homeomorphism. And by , we know that is strictly increasing.
We prove on for all by induction as follows.
Obviously, holds on . Suppose that holds on for all , where is a certain integer. Let Since for and from (4) we have Thereby Thus we have on . Thus we get on for all by induction.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This paper is partially supported by the National Natural Science Foundation of China, Tian Yuan Foundation (no. 11326129), and the Fundamental Research Funds for the Central Universities (no. 14CX02152A).

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