Research Article | Open Access
Wei Song, Sheng Chen, Xunbo Yin, "Strictly Decreasing Solutions of a Class of Iterative Equations on the Unit Circle", Discrete Dynamics in Nature and Society, vol. 2010, Article ID 207832, 14 pages, 2010. https://doi.org/10.1155/2010/207832
Strictly Decreasing Solutions of a Class of Iterative Equations on the Unit Circle
We discuss the strictly decreasing solutions of a class of iterative equations on the unit circle . The conditions for the existence, uniqueness, and stability of such solutions are presented.
Let be the set of all continuous self-mappings on a topological space . Let . For integer , define the th iterate of by and , where denotes the identity mapping on and denotes the composition of mappings. An equation with iteration as its main operation is simply called an iterative equation. Founded on the problem of iterative roots, the problem of invariant curves and some problems originating from dynamical systems, and the iterative equation where and are given functions and is the unknown function, have been investigated extensively [1, 2].
Let denote the unit circle and denote the set of all continuous maps from to itself. Let denote the point in the complex plane so as to distinguish it from . The iteration on the unit circle is also important. Many scholars had got a lot of results on iterative roots and iteration groups, such as [3–7]. In 2007, Zdun and Zhang  discussed the solutions of (1.1) on , that is, the equation in the class of homeomorphisms In , the authors lift from the circle to and from the -dimensional torus to . They apply ingenious techniques of restricting and extending to these lifts so that the problem can be reduced and discussed on the compact interval . However, they only discussed the strictly increasing solutions of (1.2). In this paper, we will discuss the strictly decreasing solutions of a special form of (1.2), that is, the following equation:
2. Basic Definitions and Lemmas
Considering a nonempty subset , a map is called to be increasing (strictly increasing) if (, resp.) for all with . Similarly, a map is called to be decreasing (strictly decreasing) if (, resp.) for all with . A homeomorphism is said to be orientation preserving (orientation reversing) if it is strictly increasing (strictly decreasing, resp.).
If with , there exist with and . Define the oriented arc Obviously, the map is strictly increasing (strictly decreasing) if yields (, resp.).
2.1. The Lifts of Maps on
Let . A continuous map is said to be a lift of if The following properties are well known.
Lemma 2.1 (see ). (i) Each has a lift . (ii) There exists a constant such that each lift of satisfies . The constant is said to be the degree of and denoted by , that is, . (iii) For and , one has and If is a homeomorphism, then . (iv) If is a lift of , then for each , the map is a lift of and every lift of can be expressed in this form, where denotes the set of integers.
Lemma 2.2 (see ).
(i) If is a continuous map such that is an integer , then can be extended to a map with
(ii) Suppose is continuous, is an integer, and Then there exists a unique such that is a lift of .
A map is said to be Lipschitzian if there exists a lift of satisfying
for a constant . By (iv) of Lemma 2.1, the constant is independent of the choice of , and is said to be a Lipschitzian constant of
In the sequel one will discuss the strictly increasing (strictly decreasing) maps by means of their lifts.
Lemma 2.3. A lift of is strictly monotonic (monotonic) if and only if there exists a such that is strictly monotonic (monotonic).
The proof is trivial. So we omit it.
Lemma 2.4 (see ). A map is strictly increasing (strictly decreasing) if and only if () and its lift is strictly increasing (strictly decreasing) in .
Thus we get the following theorem.
Theorem 2.5. A map is strictly increasing (strictly decreasing) if and only if it is an orientation-preserving (orientation-reversing) homeomorphism.
Proof. We only need to show that has an inverse mapping. Let be the lift of . The map is the inverse mapping, where is the inverse function of .
2.2. Induced Maps of Maps on
Firstly, we define where or . Let ; the map is a homeomorphism from to For , define which is a self-map on . We extend to the compact interval . Let For convenience, we call the induced map of , which is a self-map on .
Lemma 2.6. The induced map of an orientation-reversing homeomorphism is continuous and strictly decreasing on . It can be extended to a lift of . This lift satisfies and is unique on this property.
Proof. Obviously, is continuous on the interval . We only need to prove and .
Concerning the continuity at , we note that By the continuity of at we have . Let be any lift of . By , we must have where denotes the set of integers. By (iv) of Lemma 2.1, let be a lift of with . Since , we have We can easily verify in the same way. Note that is strictly decreasing on , so is strictly decreasing on .
Let , and is the integer with . Define . We can easily get that . Suppose that is another lift of which maps into itself and By (iv) of Lemma 2.1, we have for some integer . Obviously, Then
The following lemma is a converse of Lemma 2.6.
Lemma 2.7. Suppose that is strictly decreasing and satisfies Then the map belongs to the class . Moreover, can be extended to a lift of .
Proof. Clearly, , and is an orientation-reversing homeomorphism of . Then is orientation reversing on the whole . We only need to show the continuity of at Let () denote approaches to in clockwise (counter-clockwise) direction. Note that
By the same argument, we have .
Given and integer with , define One can easily show that .
2.3. Induced Maps of Maps on
For simplicity, let
For , define Then (1.4) can be written as
As in , we also have the following two remarks.
Remark 2.8. maps into
Remark 2.9. Let denote the range of . If and (1.4) has a solution on , then and
2.3.1. The Case:
The following important and interesting lemma is proved in .
Lemma 2.10 (see ). If is continuous and , then there exists a unique continuous function such that Moreover, for each , there exists an such that
The map is said to be the lift of , and is defined to be the degree of and is denoted by .
Theorem 2.12. Suppose that is continuous, , is continuous, , and (1.4) has a solution . Let . Then
Definition 2.13. Suppose is continuous. Let , and let be fixed.(1) One says is increasing with respect to the th variable if the map is increasing.(2) One says is constant with respect to the th variable if there is a such that the map
Corollary 2.14. Let , and is continuous with If is increasing with respect to each variable and is nonconstant in at least two variables, then (1.4) has no solutions in .
2.3.2. The Case
The following assumptions (H1) and (A) are cited from .(H1) is continuous, (A) There exists a constant such that for we have , and for we have
Under assumptions (H1) and (A), as in , define The function defined on is said to be the induced map of Let us note that
Lemma 2.15 (see ). Under assumptions (H1) and (A), the induced map of is continuous.
Lemma 2.17 (see ). If is differentiable in with respect to each variable and for every there exist such that , then satisfies (H2).
3. Existence of Strictly Decreasing Solutions
Let denote the set of all continuous maps on . It is well known that equipped with the supremum norm is a Banach space. Let denote the set of all continuous self-mappings on ; is a compact closed subset of .
The following lemma is useful in the proof of Theorem 3.2.
Lemma 3.1 (see ). Let , and suppose that is a self-homeomorphism of such that for all , where is a constant. Then(i) for all ;(ii)
Theorem 3.2. Suppose (H1), (H2) hold and with a Lipschitz constant . Equation (1.4) has a solution with a Lipschitz constant .
Proof. Let and be the induced maps of and defined in Section 2. By Lemma 2.16 (1.4) is equivalent to the following equation:
By Lemma 2.3, the map is strictly decreasing The map can be extended to a lift of , and
Let . Note that , since Define an operator by , where By (H2), we have for any Hence, is increasing, , and Thus
Define by where is defined in (3.5) and is the induced map of For convenience, denote Then, we have . Since is increasing, we have that for all with Hence,
Furthermore, for any , we have where Lemma 3.1 and (H2) are applied. Finally we get that Hence maps continuously into itself. By Schauder's fixed point theorem, has a fixed . By Lemma 2.2, the map can be extended to a map such that , and there exists a unique such that is a lift of . Note that for any with , there is an integer and a nonnegative integer such that and . Thus Obviously and is a solution of (1.4) with a Lipschitz constant .
4. Uniqueness and Stability
Let and be their lifts, respectively. By (iv) of Lemma 2.1, we can define
Definition 4.1. For a given small constant , one says that is —close to if
Proof. By Banach contraction theorem and (3.10), the solution is unique.
Suppose that both satisfy the conditions in Theorem 3.2 and are the unique solutions of (1.4) corresponding to the given , respectively. Assume that be the lifts of , respectively. Let be the restrictions on of Let be the induced map of defined in (2.27). By the proof of Theorem 3.2, we have where is defined as (3.5) and By (3.10), we have Thus
For any , there is an integer with such that This means that By the same argument, we have Thus By the discussion above, we know that (1.4) is stable.
Example 5.1. Consider the equation
The map has a lift and . Moreover, has a Lipschitz constant . The induced map of is
It is easy to verify that both (H1) and (H2) hold. The constants are By Theorem 3.2, (5.1) has a solution with a Lipschitz constant
Note that . We have Thus by Theorem 4.3, the solution is unique and the equation is stable.
The authors would like to thank the referees for their valuable comments and suggestions which led to truly significant improvement of the paper. This work is part of Project 11001064 supported by National Natural Science Foundation of China, Project HITC200706 and HITC200701 supported by Science Research Foundation in Harbin Institute of Technology.
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