# Strictly Decreasing Solutions of a Class of Iterative Equations on

Authors: Wei Song; Sheng Chen; Xunbo Yin
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We discuss the strictly decreasing solutions
of a class of iterative equations on the unit circle𝕋1. The conditions for the
existence, uniqueness, and stability of such solutions are presented.


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## Body

## 1. Introduction

LetC0(X,X) be the set of all continuous self-mappings on a topological space X. Let f∈C0(X,X). For integer n≥0, define the nth iterate of f by fn=f∘fn-1 and f0=id, where id denotes the identity mapping on X 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(1.1)Φ(f(x),f2(x),…,fn(x))=F(x),x∈X,
where F and Φ are given functions and f is the unknown function, have been investigated extensively [1, 2].Let𝕋1={z∈ℂ:z=e2πit,t∈ℝ} denote the unit circle and C0(𝕋1,𝕋1) denote the set of all continuous maps from 𝕋1 to itself. Let 1 denote the point (1,0) in the complex plane ℂ so as to distinguish it from 1∈ℝ. The iteration on the unit circle 𝕋1 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 [8] discussed the solutions of (1.1) on 𝕋1, that is, the equation(1.2)Φ(f(z),f2(z),…,fn(z))=F(z),z∈𝕋1,
in the class of homeomorphisms (1.3)H10(𝕋1,𝕋1)={f∈C0(𝕋1,𝕋1):f(𝕋1)=𝕋1homeomorphicallyandf(1)=1}.
In [8], the authors lift F from the circle 𝕋1 to ℝ and Φ from the n-dimensional torus 𝕋n to ℝn. They apply ingenious techniques of restricting and extending to these lifts so that the problem can be reduced and discussed on the compact interval [0,1]. 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:(1.4)Φ(f(z),f3(z),…,f2n-1(z))=F(z),z∈𝕋1.


## 2. Basic Definitions and Lemmas

As in [8], let h:t∈ℝ↦e2πit∈𝕋1. If v,w,z∈𝕋1, there exist unique t1,t2∈[0,1) such that wh(t1)=z and wh(t2)=v. The cyclic order on 𝕋1 is defined in [9], that is, (2.1)v≺w≺z⇔0<t1<t2,(2.2)v⪯w⪯z⇔t1≤t2ort2=0.
Obviously, the relations v≺w≺z, w≺z≺v and z≺v≺w are equivalent.Considering a nonempty subsetA⊆𝕋1, a map f:A→𝕋1 is called to be increasing (strictly increasing) if f(v)⪯f(w)⪯f(z) (f(v)≺f(w)≺f(z), resp.) for all v,w,z∈A with v≺w≺z. Similarly, a map f:A→𝕋1 is called to be decreasing (strictly decreasing) if f(z)⪯f(w)⪯f(v) (f(z)≺f(w)≺f(v), resp.) for all v,w,z∈A with v≺w≺z. A homeomorphism h:𝕋1→𝕋1 is said to be orientation preserving (orientation reversing) if it is strictly increasing (strictly decreasing, resp.).Ifv,z∈𝕋1 with v≠z, there exist tv,tz∈ℝ with tv<tz<tv+1 and h(tv)=v,h(tz)=z. Define the oriented arc (2.3)(v,z)⃗={w=h(t):t∈(tv,tz)}.
Obviously, the map f is strictly increasing (strictly decreasing) if w∈(v,z)⃗ yields f(w)∈(f(v),f(z))⃗ (f(w)∈(f(z),f(v))⃗, resp.).


### 2.1. The Lifts of Maps on𝕋1

Letf∈C0(𝕋1,𝕋1). A continuous map f̃:ℝ→ℝ is said to be a lift of f if (2.4)h∘f̃=f∘h.
The following properties are well known.Lemma 2.1 (see [10]).
(i) Eachf∈C0(𝕋1,𝕋1) has a lift f̃. (ii) There exists a constant k∈ℤ such that each lift f̃ of f satisfies f̃(t+1)-f̃(t)=k,forallt∈ℝ. The constant k is said to be the degree of f and denoted by deg(f), that is, deg(f)=f̃(t+1)-f̃(t),forallt∈ℝ. (iii) For f,g∈C0(𝕋1,𝕋1) and id:𝕋1→𝕋1, one has deg(g∘f)=deg(g)·deg(f) and deg(id)=1. If f∈C0(𝕋1,𝕋1) is a homeomorphism, then |deg(f)|=1. (iv) If f̃ is a lift of f, then for each j∈ℤ, the map f̃+j is a lift of f and every lift of f can be expressed in this form, where ℤ denotes the set of integers.Lemma 2.2 (see [10]).
(i) Ifg:[0,1]→ℝ is a continuous map such that g(1)-g(0) is an integer k, then g can be extended to a map g̃:ℝ→ℝ with
(2.5)g̃(t+1)-g̃(t)=k,∀t∈ℝ.
(ii) Supposef̃:ℝ→ℝ is continuous, k is an integer, and f̃(t+1)-f̃(t)=k,forallt∈ℝ. Then there exists a unique f∈C0(𝕋1,𝕋1) such that f̃ is a lift of f.A mapf∈C0(𝕋1,𝕋1) is said to be Lipschitzian if there exists a lift f̃ of f satisfying (2.6)|f̃(t1)-f̃(t2)|≤M|t1-t2|,∀t1,t2∈ℝfor a constantM>0. By (iv) of Lemma 2.1, the constant  M is independent of the choice of f̃, and M is said to be a Lipschitzian constant of f.In the sequel one will discuss the strictly increasing (strictly decreasing) mapsf∈C0(𝕋1,𝕋1) by means of their lifts.Lemma 2.3.
A liftf̃ of f∈C0(𝕋1,𝕋1) is strictly monotonic (monotonic) if and only if there exists a t0∈ℝ such that f̃|[t0,t0+1) is strictly monotonic (monotonic).The proof is trivial. So we omit it.Lemma 2.4 (see [11]).
A mapf∈C0(𝕋1,𝕋1) is strictly increasing (strictly decreasing) if and only if deg(f)=1 (deg(f)=-1) and its lift f̃ is strictly increasing (strictly decreasing) in ℝ.Thus we get the following theorem.Theorem 2.5.
A mapf∈C0(𝕋1,𝕋1) is strictly increasing (strictly decreasing) if and only if it is an orientation-preserving (orientation-reversing) homeomorphism.Proof.
We only need to show thatf has an inverse mapping. Let f̃ be the lift of f. The map f-1=h∘f̃-1 is the inverse mapping, where f̃-1 is the inverse function of f̃.


### 2.2. Induced Maps of Maps on𝕋1

Firstly, we define(2.7)H10(𝕋1,k)={f∣f∈H10(𝕋1,𝕋1),deg(f)=k},
where k=1 or -1. Let h*:=h|(0,1); the map h* is a homeomorphism from (0,1) to 𝕋1∖{1}. For f∈H10(𝕋1,-1), define f̃*:=h*-1∘f∘h* which is a self-map on (0,1). We extend f̃* to the compact interval I=[0,1]. Let(2.8)G(t)={1,t=0,f̃*(t),t∈(0,1),0,t=1.
For convenience, we call G the induced map of f, which is a self-map on [0,1].Lemma 2.6.
The induced mapG of an orientation-reversing homeomorphism f∈H10(𝕋1,-1) is continuous and strictly decreasing on I. It can be extended to a lift f̃ of f. This lift satisfies f̃(0)=1,f̃(1)=0 and is unique on this property.Proof.
Obviously,G is continuous on the interval (0,1). We only need to prove limt→0+G(t)=1 and limt→1-G(t)=0.
Concerning the continuity at0, we note that
(2.9)limt→0+G(t)=limt→0+h*-1∘f∘h*(t)=limt→0+h*-1∘f(e2πit).
By the continuity of f at 1, we have limt→0+f(e2πit)=1. Let F̃ be any lift of f. By h∘F̃=f∘h, we must have F̃(0)=k0∈ℤ, where ℤ denotes the set of integers. By (iv) of Lemma 2.1, let F̃ be a lift of f with F̃(0)=1. Since f∘h*(t)=h*∘F̃(t),t∈(0,1), we have
(2.10)limt→0+h*-1∘f∘h*(t)=limt→0+F̃(t)=1.
We can easily verify limt→1-G(t)=0 in the same way. Note that G is strictly decreasing on (0,1), so G is strictly decreasing on [0,1].
Lett∈ℝ, and k is the integer with t∈[k,k+1). Define f̃:=G(t-k)+k. We can easily get that f∘h=h∘f̃. Suppose that g̃ is another lift of f which maps [0,1] into itself and g̃(0)=1,g̃(1)=0. By (iv) of Lemma 2.1, we have f̃=g̃+j for some integer j. Obviously, j=f̃(0)-g̃(0)=0. Then f̃=g̃.The following lemma is a converse of Lemma2.6.Lemma 2.7.
Suppose thatG∈C0(I,I) is strictly decreasing and satisfies G(0)=1,G(1)=0. Then the map
(2.11)f:={h*∘G∘h*-1(z),z∈𝕋1∖{1},1,z=1.
belongs to the class H10(𝕋1,-1). Moreover, G can be extended to a lift of f.Proof.
Clearly,f(1)=1, and f|𝕋1∖{1} is an orientation-reversing homeomorphism of 𝕋1∖{1}. Then f is orientation reversing on the whole 𝕋1. We only need to show the continuity of f at 1. Let z→1+ (z→1-) denote z approaches to 1 in clockwise (counter-clockwise) direction. Note that
(2.12)limz→1+f(z)=limt→0+h*∘G∘h*-1(e2πit)=limt→0+h*∘G(t)=1.
By the same argument, we have limz→1-f(z)=1.
Givent∈ℝ and integer k with t∈[k,k+1), define
(2.13)f̃(t)=G(t-k)+k.
One can easily show that f∘h=h∘f̃.


### 2.3. Induced Maps of Maps on𝕋n

For simplicity, let(2.14)𝕋n=𝕋1×⋯×𝕋1︷n,1n=(1,…,1︷n).Forf∈H10(𝕋1,-1), define (2.15)Hf(z):=(f(z),f3(z),…,f2n-1(z)).
Then (1.4) can be written as (2.16)Φ∘Hf(z)=F(z),z∈𝕋1.As in [8], we also have the following two remarks.Remark 2.8.
Hf maps 𝕋1 into (𝕋1∖1)n∪{1n}.Remark 2.9.
LetRang(Φ) denote the range of Φ. If F∈H10(𝕋1,-1) and (1.4) has a solution f∈H10(𝕋1,-1) on F, then RangΦ=𝕋1 and Φ(1n)=1.LetDom(Φ) denote the domain of Φ. As in [8], Dom(Φ) is very important for (1.4). Now, we discuss the relationship of Dom(Φ) and (1.4). The first case is that the domain of Φ is 𝕋n.


#### 2.3.1. The Case:Dom(Φ)=𝕋n

The following important and interesting lemma is proved in [8].Lemma 2.10 (see [8]).
IfΦ:𝕋n→𝕋1 is continuous and Φ(1n)=1, then there exists a unique continuous function Φ̃:ℝn→ℝ such that
(2.17)Φ(h(t1),…,h(tn))=e2πiΦ̃(t1,…,tn),Φ̃(0,…,0)=0.
Moreover, for each k∈{1,…,n}, there exists an mk∈ℤ such that
(2.18)Φ̃(t1,…,tk+1,…,tn)=Φ̃(t1,…,tk,…,tn)+mk,∀t1,…,tk∈ℝ.The mapΦ̃ is said to be the lift of Φ, and (m1,…,mn) is defined to be the degree of Φ and is denoted by deg(Φ).Lemma 2.11.
LetF̃ be the lift of F with F̃(1)=0, and let Φ̃ be the lift of Φ with Φ̃(0,…,0)=0. Let f∈H10(𝕋1,-1) be a solution of (1.4), and let  f̃ be a lift of f with f̃(1)=0. Then (1.4) is equivalent to
(2.19)Φ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t),t∈I.Proof.
Note thatfj(h(t))=h(f̃j(t)) for all t∈ℝ. For z=e2πit,t∈I and (1.4) is equivalent to
(2.20)Φ(e2πif̃(t),e2πif̃3(t),…,e2πif̃2n-1(t))=e2πiF̃(t).
By Lemma 2.10, e2πiΦ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=e2πiF̃(t). This means that for each t∈I, we have
(2.21)Φ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t)+k(t),
where k(t)∈ℤ. Since f̃(1)=0,F̃(1)=0, and Φ̃(0,…,0)=0, we get k(t)=0. By the continuity of f̃,F̃, and Φ̃, we get that k(t)≡0.Theorem 2.12.
Suppose thatΦ is continuous, Φ(1n)=1, F is continuous, F(1)=1, and (1.4) has a solution f∈H10(𝕋1,-1). Let deg(Φ)=(m1,…,mn). Then deg(F)=-(m1+⋯+mn).Proof.
LetF̃ be the lift of F with F̃(1)=0, and let Φ̃ be the lift of Φ with Φ̃(0,…,0)=0. Let f∈H10(𝕋1,-1) be the solution of (1.4), and let f̃ be its lift with f̃(1)=0. Note that
(2.22)f̃(t+1)=f̃(t)-1,t∈ℝ.
Let H̃(t)=(f̃(t),f̃3(t),…,f̃2n-1(t)). Since deg(Φ)=(m1,…,mn), then by (2.18), we have
(2.23)Φ̃(-1,…,-1)=Φ̃(0,…,0)-m1-⋯-mn.
Thus, we get that
(2.24)F̃(t+1)=ϕ̃(H̃(t+1))=Φ̃(H̃(t)+(-1,…,-1))=Φ̃(H̃(t))-m1-⋯-mn=F̃(t)-m1-⋯-mn.Definition 2.13.
SupposeΦ:𝕋n→𝕋1 is continuous. Let 1≤i≤n, and let w1,…,wi-1,wi+1,…,wn∈ℂ be fixed.(1)
One saysΦ is increasing with respect to the ith variable if the map
(2.25)Φ(w1,…,wi-1,zi,wi+1,…,wn)
is increasing.(2)
One saysΦ is constant with respect to the ith variable if there is a z0∈𝕋1 such that the map(2.26)Φ(w1,…,wi-1,zi,wi+1,…,wn)≡z0.Corollary 2.14.
LetF∈H10(𝕋1,-1), and Φ is continuous with Φ(1n)=1. If Φ is increasing with respect to each variable and Φ is nonconstant in at least two variables, then (1.4) has no solutions in H10(𝕋1,-1).Proof.
Letdeg(Φ)=(m1,…,mn). Hence, by Lemma 2.3, we have m1≥0,…,mn≥0 and mk=0 if and only if Φ is constant with respect to the kth variable. If (1.4) has a solution f in H10(𝕋1,-1), then by Theorem 2.12, we get that deg(F)=-(m1+⋯+mn)≤-2. This a contradiction.From the discussion above, we know thatdeg(F) and deg(Φ) are interrelated. This is too severe for (1.4). So as in [8], by Remark 2.8, it is also natural to suppose that Dom(Φ)=(𝕋1∖1)n∪{1n}.


#### 2.3.2. The CaseDom(Φ)=(𝕋1∖1)n∪{1n}

The following assumptions (H1) and (A) are cited from [8].(H1)
Φ:(𝕋1∖{1})n∪{1n}→𝕋1 is continuous, Φ((𝕋1∖{1})n)=𝕋1∖{1}.(A)
There exists a constantδ>0 such that for 0<tk<δ,k=1,…,n, we have Φ(h*(t1),…,h*(tn))∈(1,i)⃗, and for 1-δ<tk<1,k=1,…,n, we have Φ(h*(t1),…,h*(tn))∈(-i,1)⃗.Under assumptions (H1) and (A), as in [8], define(2.27)Ψ(t1,…,tn):=h*-1(Φ(h*(t1),…,h*(tn))),tj∈(0,1),j=1,…,n,Ψ(0,…,0)=0,Ψ(1,…,1)=1.
The function Ψ defined on (0,1)n∪{(0,…,0),(1,…,1)} is said to be the induced map of Φ. Let us note that (2.28)limtj→0,j=1,…,nΨ(t1,…,tn)=0,limtj→1,j=1,…,nΨ(t1,…,tn)=1.Lemma 2.15 (see [8]).
Under assumptions (H1) and (A), the induced mapΨ of Φ is continuous.Lemma 2.16.
LetF̃ be the lift of F with F̃(1)=0, and let Ψ be the induced map of Φ. Let f∈H10(𝕋1,-1) be a solution of (1.4), and let f̃ be a lift of f with f̃(1)=0. Then (1.4) is equivalent to
(2.29)Ψ(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t),t∈I.Lemma2.16 can be proved by means of the method which is used to prove Lemma 2.11.As in [8], we also need the following assumption.(H2)
[8] There are nonnegative real constants αj,βj,j=1,…,n with β1≥α1>0,βj≥αj≥0,j=2,…,n, such that(2.30)∑j=1nαj(tj-sj)≤Ψ(t1,…,tn)-Ψ(s1,…,sn)≤∑j=1nβj(tj-sj)
for all tj≥sj in (0,1),j=1,…,n.Lemma 2.17 (see [8]).
IfΨ is differentiable in (0,1)n with respect to each variable and for every k there exist αk,βk such that α1>0,0≤αk≤∂Ψ/∂tk≤βk, then Ψ satisfies (H2).


## 2.1. The Lifts of Maps on𝕋1

Letf∈C0(𝕋1,𝕋1). A continuous map f̃:ℝ→ℝ is said to be a lift of f if (2.4)h∘f̃=f∘h.
The following properties are well known.Lemma 2.1 (see [10]).
(i) Eachf∈C0(𝕋1,𝕋1) has a lift f̃. (ii) There exists a constant k∈ℤ such that each lift f̃ of f satisfies f̃(t+1)-f̃(t)=k,forallt∈ℝ. The constant k is said to be the degree of f and denoted by deg(f), that is, deg(f)=f̃(t+1)-f̃(t),forallt∈ℝ. (iii) For f,g∈C0(𝕋1,𝕋1) and id:𝕋1→𝕋1, one has deg(g∘f)=deg(g)·deg(f) and deg(id)=1. If f∈C0(𝕋1,𝕋1) is a homeomorphism, then |deg(f)|=1. (iv) If f̃ is a lift of f, then for each j∈ℤ, the map f̃+j is a lift of f and every lift of f can be expressed in this form, where ℤ denotes the set of integers.Lemma 2.2 (see [10]).
(i) Ifg:[0,1]→ℝ is a continuous map such that g(1)-g(0) is an integer k, then g can be extended to a map g̃:ℝ→ℝ with
(2.5)g̃(t+1)-g̃(t)=k,∀t∈ℝ.
(ii) Supposef̃:ℝ→ℝ is continuous, k is an integer, and f̃(t+1)-f̃(t)=k,forallt∈ℝ. Then there exists a unique f∈C0(𝕋1,𝕋1) such that f̃ is a lift of f.A mapf∈C0(𝕋1,𝕋1) is said to be Lipschitzian if there exists a lift f̃ of f satisfying (2.6)|f̃(t1)-f̃(t2)|≤M|t1-t2|,∀t1,t2∈ℝfor a constantM>0. By (iv) of Lemma 2.1, the constant  M is independent of the choice of f̃, and M is said to be a Lipschitzian constant of f.In the sequel one will discuss the strictly increasing (strictly decreasing) mapsf∈C0(𝕋1,𝕋1) by means of their lifts.Lemma 2.3.
A liftf̃ of f∈C0(𝕋1,𝕋1) is strictly monotonic (monotonic) if and only if there exists a t0∈ℝ such that f̃|[t0,t0+1) is strictly monotonic (monotonic).The proof is trivial. So we omit it.Lemma 2.4 (see [11]).
A mapf∈C0(𝕋1,𝕋1) is strictly increasing (strictly decreasing) if and only if deg(f)=1 (deg(f)=-1) and its lift f̃ is strictly increasing (strictly decreasing) in ℝ.Thus we get the following theorem.Theorem 2.5.
A mapf∈C0(𝕋1,𝕋1) is strictly increasing (strictly decreasing) if and only if it is an orientation-preserving (orientation-reversing) homeomorphism.Proof.
We only need to show thatf has an inverse mapping. Let f̃ be the lift of f. The map f-1=h∘f̃-1 is the inverse mapping, where f̃-1 is the inverse function of f̃.


## 2.2. Induced Maps of Maps on𝕋1

Firstly, we define(2.7)H10(𝕋1,k)={f∣f∈H10(𝕋1,𝕋1),deg(f)=k},
where k=1 or -1. Let h*:=h|(0,1); the map h* is a homeomorphism from (0,1) to 𝕋1∖{1}. For f∈H10(𝕋1,-1), define f̃*:=h*-1∘f∘h* which is a self-map on (0,1). We extend f̃* to the compact interval I=[0,1]. Let(2.8)G(t)={1,t=0,f̃*(t),t∈(0,1),0,t=1.
For convenience, we call G the induced map of f, which is a self-map on [0,1].Lemma 2.6.
The induced mapG of an orientation-reversing homeomorphism f∈H10(𝕋1,-1) is continuous and strictly decreasing on I. It can be extended to a lift f̃ of f. This lift satisfies f̃(0)=1,f̃(1)=0 and is unique on this property.Proof.
Obviously,G is continuous on the interval (0,1). We only need to prove limt→0+G(t)=1 and limt→1-G(t)=0.
Concerning the continuity at0, we note that
(2.9)limt→0+G(t)=limt→0+h*-1∘f∘h*(t)=limt→0+h*-1∘f(e2πit).
By the continuity of f at 1, we have limt→0+f(e2πit)=1. Let F̃ be any lift of f. By h∘F̃=f∘h, we must have F̃(0)=k0∈ℤ, where ℤ denotes the set of integers. By (iv) of Lemma 2.1, let F̃ be a lift of f with F̃(0)=1. Since f∘h*(t)=h*∘F̃(t),t∈(0,1), we have
(2.10)limt→0+h*-1∘f∘h*(t)=limt→0+F̃(t)=1.
We can easily verify limt→1-G(t)=0 in the same way. Note that G is strictly decreasing on (0,1), so G is strictly decreasing on [0,1].
Lett∈ℝ, and k is the integer with t∈[k,k+1). Define f̃:=G(t-k)+k. We can easily get that f∘h=h∘f̃. Suppose that g̃ is another lift of f which maps [0,1] into itself and g̃(0)=1,g̃(1)=0. By (iv) of Lemma 2.1, we have f̃=g̃+j for some integer j. Obviously, j=f̃(0)-g̃(0)=0. Then f̃=g̃.The following lemma is a converse of Lemma2.6.Lemma 2.7.
Suppose thatG∈C0(I,I) is strictly decreasing and satisfies G(0)=1,G(1)=0. Then the map
(2.11)f:={h*∘G∘h*-1(z),z∈𝕋1∖{1},1,z=1.
belongs to the class H10(𝕋1,-1). Moreover, G can be extended to a lift of f.Proof.
Clearly,f(1)=1, and f|𝕋1∖{1} is an orientation-reversing homeomorphism of 𝕋1∖{1}. Then f is orientation reversing on the whole 𝕋1. We only need to show the continuity of f at 1. Let z→1+ (z→1-) denote z approaches to 1 in clockwise (counter-clockwise) direction. Note that
(2.12)limz→1+f(z)=limt→0+h*∘G∘h*-1(e2πit)=limt→0+h*∘G(t)=1.
By the same argument, we have limz→1-f(z)=1.
Givent∈ℝ and integer k with t∈[k,k+1), define
(2.13)f̃(t)=G(t-k)+k.
One can easily show that f∘h=h∘f̃.


## 2.3. Induced Maps of Maps on𝕋n

For simplicity, let(2.14)𝕋n=𝕋1×⋯×𝕋1︷n,1n=(1,…,1︷n).Forf∈H10(𝕋1,-1), define (2.15)Hf(z):=(f(z),f3(z),…,f2n-1(z)).
Then (1.4) can be written as (2.16)Φ∘Hf(z)=F(z),z∈𝕋1.As in [8], we also have the following two remarks.Remark 2.8.
Hf maps 𝕋1 into (𝕋1∖1)n∪{1n}.Remark 2.9.
LetRang(Φ) denote the range of Φ. If F∈H10(𝕋1,-1) and (1.4) has a solution f∈H10(𝕋1,-1) on F, then RangΦ=𝕋1 and Φ(1n)=1.LetDom(Φ) denote the domain of Φ. As in [8], Dom(Φ) is very important for (1.4). Now, we discuss the relationship of Dom(Φ) and (1.4). The first case is that the domain of Φ is 𝕋n.


### 2.3.1. The Case:Dom(Φ)=𝕋n

The following important and interesting lemma is proved in [8].Lemma 2.10 (see [8]).
IfΦ:𝕋n→𝕋1 is continuous and Φ(1n)=1, then there exists a unique continuous function Φ̃:ℝn→ℝ such that
(2.17)Φ(h(t1),…,h(tn))=e2πiΦ̃(t1,…,tn),Φ̃(0,…,0)=0.
Moreover, for each k∈{1,…,n}, there exists an mk∈ℤ such that
(2.18)Φ̃(t1,…,tk+1,…,tn)=Φ̃(t1,…,tk,…,tn)+mk,∀t1,…,tk∈ℝ.The mapΦ̃ is said to be the lift of Φ, and (m1,…,mn) is defined to be the degree of Φ and is denoted by deg(Φ).Lemma 2.11.
LetF̃ be the lift of F with F̃(1)=0, and let Φ̃ be the lift of Φ with Φ̃(0,…,0)=0. Let f∈H10(𝕋1,-1) be a solution of (1.4), and let  f̃ be a lift of f with f̃(1)=0. Then (1.4) is equivalent to
(2.19)Φ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t),t∈I.Proof.
Note thatfj(h(t))=h(f̃j(t)) for all t∈ℝ. For z=e2πit,t∈I and (1.4) is equivalent to
(2.20)Φ(e2πif̃(t),e2πif̃3(t),…,e2πif̃2n-1(t))=e2πiF̃(t).
By Lemma 2.10, e2πiΦ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=e2πiF̃(t). This means that for each t∈I, we have
(2.21)Φ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t)+k(t),
where k(t)∈ℤ. Since f̃(1)=0,F̃(1)=0, and Φ̃(0,…,0)=0, we get k(t)=0. By the continuity of f̃,F̃, and Φ̃, we get that k(t)≡0.Theorem 2.12.
Suppose thatΦ is continuous, Φ(1n)=1, F is continuous, F(1)=1, and (1.4) has a solution f∈H10(𝕋1,-1). Let deg(Φ)=(m1,…,mn). Then deg(F)=-(m1+⋯+mn).Proof.
LetF̃ be the lift of F with F̃(1)=0, and let Φ̃ be the lift of Φ with Φ̃(0,…,0)=0. Let f∈H10(𝕋1,-1) be the solution of (1.4), and let f̃ be its lift with f̃(1)=0. Note that
(2.22)f̃(t+1)=f̃(t)-1,t∈ℝ.
Let H̃(t)=(f̃(t),f̃3(t),…,f̃2n-1(t)). Since deg(Φ)=(m1,…,mn), then by (2.18), we have
(2.23)Φ̃(-1,…,-1)=Φ̃(0,…,0)-m1-⋯-mn.
Thus, we get that
(2.24)F̃(t+1)=ϕ̃(H̃(t+1))=Φ̃(H̃(t)+(-1,…,-1))=Φ̃(H̃(t))-m1-⋯-mn=F̃(t)-m1-⋯-mn.Definition 2.13.
SupposeΦ:𝕋n→𝕋1 is continuous. Let 1≤i≤n, and let w1,…,wi-1,wi+1,…,wn∈ℂ be fixed.(1)
One saysΦ is increasing with respect to the ith variable if the map
(2.25)Φ(w1,…,wi-1,zi,wi+1,…,wn)
is increasing.(2)
One saysΦ is constant with respect to the ith variable if there is a z0∈𝕋1 such that the map(2.26)Φ(w1,…,wi-1,zi,wi+1,…,wn)≡z0.Corollary 2.14.
LetF∈H10(𝕋1,-1), and Φ is continuous with Φ(1n)=1. If Φ is increasing with respect to each variable and Φ is nonconstant in at least two variables, then (1.4) has no solutions in H10(𝕋1,-1).Proof.
Letdeg(Φ)=(m1,…,mn). Hence, by Lemma 2.3, we have m1≥0,…,mn≥0 and mk=0 if and only if Φ is constant with respect to the kth variable. If (1.4) has a solution f in H10(𝕋1,-1), then by Theorem 2.12, we get that deg(F)=-(m1+⋯+mn)≤-2. This a contradiction.From the discussion above, we know thatdeg(F) and deg(Φ) are interrelated. This is too severe for (1.4). So as in [8], by Remark 2.8, it is also natural to suppose that Dom(Φ)=(𝕋1∖1)n∪{1n}.


### 2.3.2. The CaseDom(Φ)=(𝕋1∖1)n∪{1n}

The following assumptions (H1) and (A) are cited from [8].(H1)
Φ:(𝕋1∖{1})n∪{1n}→𝕋1 is continuous, Φ((𝕋1∖{1})n)=𝕋1∖{1}.(A)
There exists a constantδ>0 such that for 0<tk<δ,k=1,…,n, we have Φ(h*(t1),…,h*(tn))∈(1,i)⃗, and for 1-δ<tk<1,k=1,…,n, we have Φ(h*(t1),…,h*(tn))∈(-i,1)⃗.Under assumptions (H1) and (A), as in [8], define(2.27)Ψ(t1,…,tn):=h*-1(Φ(h*(t1),…,h*(tn))),tj∈(0,1),j=1,…,n,Ψ(0,…,0)=0,Ψ(1,…,1)=1.
The function Ψ defined on (0,1)n∪{(0,…,0),(1,…,1)} is said to be the induced map of Φ. Let us note that (2.28)limtj→0,j=1,…,nΨ(t1,…,tn)=0,limtj→1,j=1,…,nΨ(t1,…,tn)=1.Lemma 2.15 (see [8]).
Under assumptions (H1) and (A), the induced mapΨ of Φ is continuous.Lemma 2.16.
LetF̃ be the lift of F with F̃(1)=0, and let Ψ be the induced map of Φ. Let f∈H10(𝕋1,-1) be a solution of (1.4), and let f̃ be a lift of f with f̃(1)=0. Then (1.4) is equivalent to
(2.29)Ψ(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t),t∈I.Lemma2.16 can be proved by means of the method which is used to prove Lemma 2.11.As in [8], we also need the following assumption.(H2)
[8] There are nonnegative real constants αj,βj,j=1,…,n with β1≥α1>0,βj≥αj≥0,j=2,…,n, such that(2.30)∑j=1nαj(tj-sj)≤Ψ(t1,…,tn)-Ψ(s1,…,sn)≤∑j=1nβj(tj-sj)
for all tj≥sj in (0,1),j=1,…,n.Lemma 2.17 (see [8]).
IfΨ is differentiable in (0,1)n with respect to each variable and for every k there exist αk,βk such that α1>0,0≤αk≤∂Ψ/∂tk≤βk, then Ψ satisfies (H2).


## 2.3.1. The Case:Dom(Φ)=𝕋n

The following important and interesting lemma is proved in [8].Lemma 2.10 (see [8]).
IfΦ:𝕋n→𝕋1 is continuous and Φ(1n)=1, then there exists a unique continuous function Φ̃:ℝn→ℝ such that
(2.17)Φ(h(t1),…,h(tn))=e2πiΦ̃(t1,…,tn),Φ̃(0,…,0)=0.
Moreover, for each k∈{1,…,n}, there exists an mk∈ℤ such that
(2.18)Φ̃(t1,…,tk+1,…,tn)=Φ̃(t1,…,tk,…,tn)+mk,∀t1,…,tk∈ℝ.The mapΦ̃ is said to be the lift of Φ, and (m1,…,mn) is defined to be the degree of Φ and is denoted by deg(Φ).Lemma 2.11.
LetF̃ be the lift of F with F̃(1)=0, and let Φ̃ be the lift of Φ with Φ̃(0,…,0)=0. Let f∈H10(𝕋1,-1) be a solution of (1.4), and let  f̃ be a lift of f with f̃(1)=0. Then (1.4) is equivalent to
(2.19)Φ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t),t∈I.Proof.
Note thatfj(h(t))=h(f̃j(t)) for all t∈ℝ. For z=e2πit,t∈I and (1.4) is equivalent to
(2.20)Φ(e2πif̃(t),e2πif̃3(t),…,e2πif̃2n-1(t))=e2πiF̃(t).
By Lemma 2.10, e2πiΦ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=e2πiF̃(t). This means that for each t∈I, we have
(2.21)Φ̃(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t)+k(t),
where k(t)∈ℤ. Since f̃(1)=0,F̃(1)=0, and Φ̃(0,…,0)=0, we get k(t)=0. By the continuity of f̃,F̃, and Φ̃, we get that k(t)≡0.Theorem 2.12.
Suppose thatΦ is continuous, Φ(1n)=1, F is continuous, F(1)=1, and (1.4) has a solution f∈H10(𝕋1,-1). Let deg(Φ)=(m1,…,mn). Then deg(F)=-(m1+⋯+mn).Proof.
LetF̃ be the lift of F with F̃(1)=0, and let Φ̃ be the lift of Φ with Φ̃(0,…,0)=0. Let f∈H10(𝕋1,-1) be the solution of (1.4), and let f̃ be its lift with f̃(1)=0. Note that
(2.22)f̃(t+1)=f̃(t)-1,t∈ℝ.
Let H̃(t)=(f̃(t),f̃3(t),…,f̃2n-1(t)). Since deg(Φ)=(m1,…,mn), then by (2.18), we have
(2.23)Φ̃(-1,…,-1)=Φ̃(0,…,0)-m1-⋯-mn.
Thus, we get that
(2.24)F̃(t+1)=ϕ̃(H̃(t+1))=Φ̃(H̃(t)+(-1,…,-1))=Φ̃(H̃(t))-m1-⋯-mn=F̃(t)-m1-⋯-mn.Definition 2.13.
SupposeΦ:𝕋n→𝕋1 is continuous. Let 1≤i≤n, and let w1,…,wi-1,wi+1,…,wn∈ℂ be fixed.(1)
One saysΦ is increasing with respect to the ith variable if the map
(2.25)Φ(w1,…,wi-1,zi,wi+1,…,wn)
is increasing.(2)
One saysΦ is constant with respect to the ith variable if there is a z0∈𝕋1 such that the map(2.26)Φ(w1,…,wi-1,zi,wi+1,…,wn)≡z0.Corollary 2.14.
LetF∈H10(𝕋1,-1), and Φ is continuous with Φ(1n)=1. If Φ is increasing with respect to each variable and Φ is nonconstant in at least two variables, then (1.4) has no solutions in H10(𝕋1,-1).Proof.
Letdeg(Φ)=(m1,…,mn). Hence, by Lemma 2.3, we have m1≥0,…,mn≥0 and mk=0 if and only if Φ is constant with respect to the kth variable. If (1.4) has a solution f in H10(𝕋1,-1), then by Theorem 2.12, we get that deg(F)=-(m1+⋯+mn)≤-2. This a contradiction.From the discussion above, we know thatdeg(F) and deg(Φ) are interrelated. This is too severe for (1.4). So as in [8], by Remark 2.8, it is also natural to suppose that Dom(Φ)=(𝕋1∖1)n∪{1n}.


## 2.3.2. The CaseDom(Φ)=(𝕋1∖1)n∪{1n}

The following assumptions (H1) and (A) are cited from [8].(H1)
Φ:(𝕋1∖{1})n∪{1n}→𝕋1 is continuous, Φ((𝕋1∖{1})n)=𝕋1∖{1}.(A)
There exists a constantδ>0 such that for 0<tk<δ,k=1,…,n, we have Φ(h*(t1),…,h*(tn))∈(1,i)⃗, and for 1-δ<tk<1,k=1,…,n, we have Φ(h*(t1),…,h*(tn))∈(-i,1)⃗.Under assumptions (H1) and (A), as in [8], define(2.27)Ψ(t1,…,tn):=h*-1(Φ(h*(t1),…,h*(tn))),tj∈(0,1),j=1,…,n,Ψ(0,…,0)=0,Ψ(1,…,1)=1.
The function Ψ defined on (0,1)n∪{(0,…,0),(1,…,1)} is said to be the induced map of Φ. Let us note that (2.28)limtj→0,j=1,…,nΨ(t1,…,tn)=0,limtj→1,j=1,…,nΨ(t1,…,tn)=1.Lemma 2.15 (see [8]).
Under assumptions (H1) and (A), the induced mapΨ of Φ is continuous.Lemma 2.16.
LetF̃ be the lift of F with F̃(1)=0, and let Ψ be the induced map of Φ. Let f∈H10(𝕋1,-1) be a solution of (1.4), and let f̃ be a lift of f with f̃(1)=0. Then (1.4) is equivalent to
(2.29)Ψ(f̃(t),f̃3(t),…,f̃2n-1(t))=F̃(t),t∈I.Lemma2.16 can be proved by means of the method which is used to prove Lemma 2.11.As in [8], we also need the following assumption.(H2)
[8] There are nonnegative real constants αj,βj,j=1,…,n with β1≥α1>0,βj≥αj≥0,j=2,…,n, such that(2.30)∑j=1nαj(tj-sj)≤Ψ(t1,…,tn)-Ψ(s1,…,sn)≤∑j=1nβj(tj-sj)
for all tj≥sj in (0,1),j=1,…,n.Lemma 2.17 (see [8]).
IfΨ is differentiable in (0,1)n with respect to each variable and for every k there exist αk,βk such that α1>0,0≤αk≤∂Ψ/∂tk≤βk, then Ψ satisfies (H2).


## 3. Existence of Strictly Decreasing Solutions

LetC0(I) denote the set of all continuous maps on I=[0,1]. It is well known that C0(I) equipped with the supremum norm ∥g∥=sup{|g(t)|:t∈I} is a Banach space. Let C0(I,I) denote the set of all continuous self-mappings on I; C0(I,I) is a compact closed subset of C0(I).The following lemma is useful in the proof of Theorem3.2.Lemma 3.1 (see [12]).
Leti=1,2, and suppose that gi is a self-homeomorphism of I such that |gi(x)-gi(y)|≤N|x-y| for all x,y∈I, where N>0 is a constant. Then(i)
∥g1n-g2n∥≤(∑i=0n-1Ni)∥g1-g2∥ for all n∈ℤ;(ii)
∥g1-g2∥≤N∥g1-1-g2-1∥.For0≤l≤L, define (3.1)𝔉(I,l,L)={g∈C0(I,I)∣g(0)=0,g(1)=1,l(t-s)≤g(t)-g(s)≤L(t-s),∀s≤t∈I},𝔉′(I,l,L)={g∈C0(I,I)∣g(0)=1,g(1)=0,l(t-s)≤g(s)-g(t)≤L(t-s),∀s≤t∈I}.
Both 𝔉(I,l,L) and 𝔉′(I,l,L) are compact convex subsets of C0(I) [1, 13].Theorem 3.2.
Suppose (H1), (H2) hold andF∈H10(𝕋1,-1) with a Lipschitz constant M>α1. Equation (1.4) has a solution f∈H10(𝕋1,-1) with a Lipschitz constant M/α1.Proof.
LetG and Ψ be the induced maps of F and Φ defined in Section 2. By Lemma 2.16 (1.4) is equivalent to the following equation:
(3.2)Ψ(g(t),g3(t),…,g2n-1(t))=G(t),t∈I.
By Lemma 2.3, the map G∈C0(I) is strictly decreasing G(0)=1,andG(1)=0. The map G can be extended to a lift of F, and
(3.3)|G(t1)-G(t2)|≤M|t1-t2|,∀t1,t2∈I.
LetM0=∑j=1nβj(α1-1M)2j-2. Note that 𝔉′(I,0,α1-1M)≠ϕ, since M>α1. Define an operator
(3.4)𝔏:𝔉′(I,0,α1-1M)→C0(I)
by g↦𝔏g, where
(3.5)𝔏g(t)=Ψ(t,g2(t),…,g2n-2(t)),t∈I.
By (H2), we have for any t≥s∈I,(3.6)𝔏g(t)-𝔏g(s)=Ψ(t,g2(t),…,g2n-2(t))-Ψ(s,g2(s),…,g2n-2(s))≥α1(t-s)+∑j=2nαj(g2j-2(t)-g2j-2(s))≥α1(t-s),𝔏g(t)-𝔏g(s)=Ψ(t,g2(t),…,g2n-2(t))-Ψ(s,g2(s),…,g2n-2(s))≤β1(t-s)+‍∑j=2nβj(g2j-2(t)-g2j-2(s))≤M0(t-s).
Hence, 𝔏g is increasing, 𝔏g(0)=0,𝔏g(1)=1, and 𝔏g∈𝔉(I,α1,M0). Thus 𝔏g-1∈𝔉(I,M0-1,α1-1).
Define𝔗:𝔉′(I,0,α1-1M)→C0(I) by
(3.7)𝔗(g)=𝔏g-1∘G,
where 𝔏g is defined in (3.5) and G is the induced map of F. For convenience, denote 𝔗(g)=𝔗g. Then, we have 𝔗g(0)=1,𝔗g(1)=0. Since 𝔏g-1 is increasing, we have that for all s,t∈I with t≥s,(3.8)0≤𝔗g(s)-𝔗g(t)=𝔏g-1∘G(s)-𝔏g-1∘G(t)≤α1-1(G(s)-G(t))≤α1-1M(t-s).
Hence, 𝔗(g)∈𝔉′(I,0,α1-1M).
Furthermore, for anyg1,g2∈𝔉′(I,0,α1-1M), we have
(3.9)∥𝔗g1-𝔗g2∥=∥𝔏g1-1∘G-𝔏g2-1∘G∥=∥𝔏g1-1-𝔏g2-1∥≤α1-1∥𝔏g1-𝔏g2∥,∥𝔏g1-𝔏g2∥≤supt∈I|Ψ(t,g12(t),…,g12n-2(t))-Ψ(t,g22(t),…,g22n-2(t))|≤∑j=2nβj∥g12j-2-g22j-2∥≤∑j=2nβj∑i=12j-2(α1-1M)i-1∥g1-g2∥,
where Lemma 3.1 and (H2) are applied. Finally we get that
(3.10)∥𝔗g1-𝔗g2∥≤∑j=2nβj∑i=12j-2α1-iMi-1∥g1-g2∥.
Hence 𝔗 maps 𝔉′(I,0,α1-1M) continuously into itself. By Schauder's fixed point theorem, 𝔗 has a fixed g∈𝔉′(I,0,α1-1M). By Lemma 2.2, the map g can be extended to a map g̃ such that g̃(t+1)-g̃(t)=-1,forallt∈ℝ, and there exists a unique f∈C0(𝕋1,𝕋1) such that g̃ is a lift of f. Note that for any t,s∈ℝ with t<s, there is an integer k and a nonnegative integer m such that t∈[k,k+1) and s∈[k+m,k+m+1). Thus
(3.11)|g̃(t)-g̃(s)|≤|g̃(t)-g̃(k+1)|+∑j=1m|g̃(k+j)-g̃(k+j+1)|+|g̃(s)-g̃(k+m)|≤|g̃(t-k)-g̃(1)|+(m-1)|g̃(0)-g̃(1)|+|g̃(s-k-m)-g̃(0)|≤Mα1[1-(t-k)+m-1+s-k-m]=Mα1(s-t).
Obviously f∈H10(𝕋1,-1) and f is a solution of (1.4) with a Lipschitz constant M/α1.


## 4. Uniqueness and Stability

LetF1,F2∈C0(𝕋1,𝕋1) and F̃1,F̃2 be their lifts, respectively. By (iv) of Lemma 2.1, we can define (4.1)∥F1-F2∥ℝ:=supt∈ℝ|F̃1(t1)-F̃2(t2)|.Definition 4.1.
For a given small constantϵ>0, one says that F1 is ϵ-C0—close to F2 if
(4.2)∥F1-F2∥ℝ<ϵ.Now as in [8], we give the definition that (1.4) is stable.Definition 4.2.
LetF1,F2∈C0(𝕋1,𝕋1) and f1,f2 be corresponding solutions of (1.4) on F1,F2. One says that (1.4) is stable if for all ɛ>0 there is a δ>0 such that if F1 is δ-C0—close to F2, then f1 is ɛ-C0— close to f2.Theorem 4.3.
Suppose that the conditions in Theorem3.2 still hold, and
(4.3)r=∑j=2nβj∑i=12j-2α1-iMi-1<1.
Then (1.4) has a unique solution f∈H10(𝕋1,-1) with a Lipschitz constant M/α1. Moreover, (1.4) is stable.Proof.
By Banach contraction theorem and (3.10), the solution f is unique.
Suppose thatF1,F2∈H10(𝕋1,-1) both satisfy the conditions in Theorem 3.2 and f1,f2 are the unique solutions of (1.4) corresponding to the given Fj,j=1,2, respectively. Assume that F̃i,f̃i be the lifts of Fi,fi,i=1,2, respectively. Let Gi,gi be the restrictions on [0,1] of F̃i,f̃i,i=1,2. Let Ψ be the induced map of Φ defined in (2.27). By the proof of Theorem 3.2, we have
(4.4)gi=𝔏gi-1∘Gi,i=1,2,
where 𝔏gi is defined as (3.5) and 𝔏gi-1∈𝔉(I,M0-1,α1-1),i=1,2. By (3.10), we have
(4.5)∥g1-g2∥=∥𝔏g1-1∘G1-𝔏g2-1∘G2∥≤∥𝔏g1-1∘G1-𝔏g1-1∘G2∥+∥𝔏g1-1∘G2-𝔏g2-1∘G2∥≤α1-1∥G1-G2∥+r∥g1-g2∥.
Thus
(4.6)∥g1-g2∥≤[(1-r)α1]-1∥G1-G2∥.
For anyt∈ℝ, there is an integer k with t∈[k,k+1) such that
(4.7)|f̃1(t)-f̃2(t)|=|f̃1(t-k)+k-f̃2(t-k)-k|=|g1(t-k)-g2(t-k)|.
This means that
(4.8)∥f̃1-f̃2∥ℝ=∥g1-g2∥.
By the same argument, we have
(4.9)∥F̃1-F̃2∥ℝ=∥G1-G2∥.
Thus
(4.10)∥f̃1-f̃2∥ℝ≤[(1-r)α1]-1∥F̃1-F̃2∥ℝ.
By the discussion above, we know that (1.4) is stable.


## 5. Example

Example 5.1.
Consider the equation(5.1)(f(z))359/360(f3(z))(1/720πi)ln(f3(z))=exp(2πi(e-z1/2πi)e-1).
Obviously
(5.2)F(z)=exp(2πi(e-z1/2πi)e-1),Φ(z1,z2)=z1359/360z2(1/720πi)ln(z2).
The map F has a lift F̃ and F̃|[0,1]=(e-et)/(e-1). Moreover, F has a Lipschitz constant 2. The induced map of Φ is
(5.3)Ψ(t1,t2)=359360t1+1360t22,0<tj<1,j=1,2,Ψ(0,0)=0,Ψ(1,1)=1.
It is easy to verify that both (H1) and (H2) hold. The constants are α1=β1=359/360,α2=0,β2=1/180. By Theorem 3.2, (5.1) has a solution f∈H10(𝕋1,-1) with a Lipschitz constant 720/359.
Note thatα1-1=359/360<1. We have
(5.4)r=β2∑i=12α1-iMi-1<1180∑i=122i-1=160.
Thus by Theorem 4.3, the solution is unique and the equation is stable.



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*Source: 207832-2010-10-04.xml*