# Differentiable Solutions of Equations Involving Iterated Functional Series

Authors: Wei Song
Journal: Abstract and Applied Analysis
 (2008)

---

## Abstract

The nonmonotonic differentiable solutions of equations
involving iterated functional series are investigated. Conditions for the existence,
uniqueness, and stability of such solutions are given. These extend earlier results due to Murugan and Subrahmanyam.


---

## Body

## 1. Introduction

Letf be a self-mapping on a topological space X and fm denote the mth iterate of f,
that is, fm=f∘fm−1,f0=id,m=1,2,…. Let C(X,X) be the set of all continuous self-mappings on X.
Equations having iteration as their main operation, that is, including iterates
of the unknown mapping, are called iterative equations. It is one of the most
interesting classes of functional equations [1–4], because it concludes the problem of iterative roots
[1, 5, 6], that is, finding f∈C(X,X) such that fn is identical to a given F∈C(X,X).
As a natural generalization of the problem of iterative roots, a class of
iterative equations named as polynomial-like iterative equation(1.1)λ1f(x)+λ2f2(x)+⋯+λnfn(x)=F(x),x∈I=[a,b],had fascinated many scholars,
such as Dhombres [7], Zhao [8], Mukherjea and Ratti [9]. Despite their nice
constructive proofs, the classical methods prevented them from obtaining more
fruitful results. In 1986, Zhang [10] constructed an interesting operator called “structural operator” for (1.1) and used
the fixed point theory in Banach space to get the solutions of (1.1).
Hence he overcame the difficulties encountered by the formers. By means of this
method, Zhang and Si made a series of work concerning these
qualitative problems, such as [11–15]. In 2002, Kulczycki and Tabor [16] improved Zhang's method and
investigated the existence of Lipschitzian solutions of the iterative
functional equation(1.2)∑n=1∞λnfn(x)=F(x),x∈B,where B is a compact convex subset of ℝn and F:B→B are a given Lipschitz function. It is easy to
see that (1.1) is the special case of (1.2) with λi=0,  i=n+1,… and B=[a,b].Recently Zhang et al. [17] and Xu et al. [18] developed this method and they have got the
nonmonotonic, convex, and decreasing continuous solutions of (1.1). In
fact they have answered the open problem 2 which was proposed by J. Zhang et
al. [19].The problem of differentiable solutions of iterative
equation had also fascinated many scholars' attentions. In Zhang
[12] and Si
[15], the C1 and C2 solutions of (1.1) are considered. In
Wang and Si [20] the differentiable solutions of the below
equation(1.3)H(x,ϕn1(x),…,ϕni(x))=F(x),x∈I=[a,b] are considered.
Murugan and Subrahmanyam [21, 22] discussed the existence and uniqueness of C1 solutions of the more general
equations(1.4)∑i=1∞λiHi(fi(x))=F(x),x∈I=[a,b],(1.5)∑i=1∞λiHi(x,ϕai1(x),…,ϕaini(x))=F(x),x∈I=[a,b],which involve iterated
functional series. All the above references only got the increasing differentiable
solutions for the above equations because they only considered the case that F is increasing. Li and Deng
[23] considered the C1 solutions of the (1.2). In [24] C1 solutions of the equation(1.6)∑n=1∞λn(x)fn(x)=F(x),x∈B,where B is a compact convex subset of ℝn and λn(x):B→R
are discussed. Li and Deng [23] and Li [24] work in higher dimensional case, they do not require
monotonicity. It should be pointed out that Mai and Liu [25] made an important
contribution to Cm solutions of iterative equations. Mai
and Liu proved the existence, uniqueness of Cm solutions of a relatively general kind of
iterative equations(1.7)G(x,f(x),…,fn(x))=0,x∈J,where J is a connected closed subset of ℝ and G∈Cm(Jn+1,ℝ),n≥2. Here Cm(Jn+1,ℝ) denotes the set of all Cm mappings from Jn+1 to ℝ.Inspired by the above work, we will investigate (1.4) and extend earlier results due to Murugan and
Subrahmanyam in two directions. In [21] the authors only get increasing solutions of
(1.4), so the present paper will investigate the nonmonotonic differentiable
solutions of (1.4) and give conditions for the existence, uniqueness, and
stability of such solutions. In [21] the authors require not only all coefficients are
nonnegative but also Hi,i=2,3,… are all increasing, but we will find that
those conditions are not necessary.


## 2. Preliminaries

LetI=[a,b] and J=[c,d] be two compact intervals. Let C1(I,J) be the set of all continuously differentiable
functions from I to J.
Then C1(I,J) is a closed subset of the Banach space C1(I,ℝ) consisting of all continuously differentiable
functions from I to ℝ with norm ∥⋅∥c1.
Here the norm ∥⋅∥c1 is defined by ∥φ∥c1=∥φ∥c0+∥φ′∥c0,φ∈C1(I,ℝ), where ∥φ∥c0=maxx∈I|φ(x)| and φ′ is the derivative of φ.
Following Zhang [12], we define the families of functions(2.1)A(I,J,m,M,N)={φ∈C1(I,J):φ(a)=c,φ(b)=d,m≤|φ′(x)|≤M,|φ′(x1)−φ′(x2)|≤N|x1−x2|,∀x,x1,x2∈I},A′(I,J,m,M,N)={φ∈C1(I,J):φ(a)=d,φ(b)=c,m≤|φ′(x)|≤M,|φ′(x1)−φ′(x2)|≤N|x1−x2|,∀x,x1,x2∈I},where 0≤m<M, N>0 are all constants.Lemma 2.1.
Both𝔄(I,J,m,M,N) and 𝔄′(I,J,m,M,N) are compact convex subsets of C1(I,J).The Lemma above can be proved by a method which is
contained in the proof of  Theorem 3.1 in  [12].Lemma 2.2 (see [12]).
Suppose thatφ,ϕ∈𝔄(I,J,m,M,N) (or φ,ϕ∈𝔄′(I,J,m,M,N)). Then for n=1,2,…,(2.2)|(φn)′(x)|≤Mn,∀x∈I,|(φn)′(x1)−(φn)′(x2)|≤N(∑i=n−12n−2Mi)|x1−x2|,∀x1,x2∈I,∥φn−ϕn∥c0≤(∑i=1nMi−1)∥φ−ϕ∥c0,∥(φn)′−(ϕn)′∥c0≤nMn−1∥φ′−ϕ′∥c0+Q(n)N(∑i=1n−1(n−i)Mn+i−2)∥φ−ϕ∥c0,where Q(1)=0,Q(m)=1 as m=2,3,… and (φn)′ denotes dφn/dx.We can get the following Lemma from [12]. In [12] the author proved that the
lemma is valid for C1(I,I),
but we find it is also valid for C1(I,J).Lemma 2.3 (see [12]).
Suppose thatf∈C1(I,J) satisfies that(2.3)0<δ≤f′(x),∀x∈I,|f′(x1)−f′(x2)|≤M*|x1−x2|,∀x1,x2∈I,where δ, M* are positive constants. Then(2.4)|(f−1)′(y1)−(f−1)′(y2)|≤M*δ3|y1−y2|,∀y1,y2∈J.Lemma 2.4 (see [18]).
If bothfi:I→J,i=1,2 are homeomorphisms from I to J such that(2.5)|fi(x1)−fi(x2)|≤K|x1−x2|,∀x1,x2∈I,where K is a positive constant. Then(2.6)∥f1−f2∥c0≤K∥f1−1−f2−1∥c0.


## 3. Differentiable Solutions of (1.4)

### 3.1. Existence of Solutions

Let{λi}i=1∞ be coefficients of (1.4) and I=[a,b].
For any f∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) define A=∑i=1∞λiHi(fi−1(a)) and B=∑i=1∞λiHi(fi−1(b)).
It is easy to see that both the convergence and the value of A,B have nothing to do with the choice of f.Theorem 3.1.
SupposeM>1, L are positive constants and H1∈𝔄(I,I,m1,M1,N1),Hi∈𝔄(I,I,0,Mi,Ni)orHi∈𝔄'(I,I,0,Mi,Ni) for i=2,3,…, where m1>0 and Mi≥1,Ni are positive constants for i=1,2,…. Assume further the following
conditions:(3.1)K0=λ1m1−∑i=2∞|λi|MiMi−1>0,K1=1(M−1)∑i=1∞|λi+1|Mi+1Mi−1(Mi−1)<∞,K0−K1M2>0,K2=∑i=1∞|λi|NiMi−1(Mi−1)<∞,−∞<A<B<∞,  hold. Then for any given F∈𝔄(I,J,0,K0M,L)(or𝔄′(I,J,0,K0M,L)), (1.4) has a solution f∈𝔄(I,I,0,M,M*)(or𝔄′(I,I,0,M,M*)), where M*≥(L+K4M2)/(K0−K1M2), K4=∑i=1∞|λi|NiM2(i−1) and I=[a,b], J=[A,B].As in  [22], firstly we give the following three Lemmas which
lead directly to the proof of Theorem 3.1. In the sequel we denote(3.2)N=(L+K4M2)(K0−K1M2).Lemma 3.2.
Under the assumptions of Theorem3.1
the following series: (3.3)K3=λ1M1+∑i=2∞|λi|MiMi−1,K4=∑i=1∞|λi|NiM2(i−1),K5=∑k=1∞|λk+1|Mk+1(∑i=1k(Mi−1)),K6=∑k=1∞|λk+1|Mk+1kMk−1,K7=∑k=1∞|λk+1|Nk+1Mk(∑i=1kMi−1),K8=∑K=2∞|λk+1|Mk+1∑j=1k−1(k−j)Mk+j−2,  are all convergent.Proof.
The convergence ofK3, K4, K5, K6, and K7 is easy to be verified. As mentioned in
[21], the equality(3.4)∑i=1n−1(n−i)xn+i−2=xn−1(xn−1(x−1)2−nx−1),x≠1holds. We get that(3.5)K8=∑K=2∞|λk+1|Mk+1(Mk−1(Mk−1(M−1)2−kM−1)).By the convergence of K1 and K6, K8 is also convergent.Lemma 3.3.
Under the assumptions of Theorem3.1,
for each f∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) the mapping Lf:I→ℝ defined by(3.6)Lf(x)=∑i=1∞λiHi(fi−1(x)),x∈Ihas the following
properties:
(i)
Lf∈𝔄(I,J,K0,K3,K1N+K4);(ii)
Lf−1∈𝔄(J,I,1/K3,1/K0,(K1N+K4)/(K0)3),
where I=[a,b] and J=[A,B].Proof.
For anyf∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)),
we have(3.7)Lf(a)=∑i=1∞λiHi(fi−1(a))<Lf(b)=∑i=1∞λiHi(fi−1(b)).It is easy to see that for any x∈I(3.8)0<λ1m1−∑i=2∞|λi|MiMi−1≤∑i=1∞λiHi′(fi−1(x))(fi−1)′(x)≤λ1M1+∑i=2∞|λi|MiMi−1.We have for any x∈I(3.9)0<K0≤L′f(x)≤K3,and for any y∈J(3.10)0<1K3≤(Lf−1)′(y)≤1K0.Thus Lf:I→J is an orientation-preserving diffeomorphism.By Lemma2.2 we can see that for any x1,x2∈I,  (3.11)|L′f(x1)−L′f(x2)|=|∑i=1∞λiHi′(fi−1(x1))(fi−1)′(x1)−∑i=1∞λiHi′(fi−1(x2))(fi−1)′(x2)|≤∑i=1∞|λi|{|Hi′(fi−1(x1))|⋅|(fi−1)′(x1)−(fi−1)′(x2)|+|Hi′(fi−1(x1))−Hi′(fi−1(x2))|⋅|(fi−1)′(x2)|}≤{∑i=2∞|λi|MiN(∑j=i−22(i−2)Mj)+∑i=1∞|λi|NiM2(i−1)}|x1−x2|=(K1N+K4)|x1−x2|.By (3.9), (3.11), and Lemma 2.3 we get
for any y1,y2∈J:(3.12)|(Lf−1)′(y1)−(Lf−1)′(y2)|≤K1N+K4(K0)3|y1−y2|.Lemma 3.4.
Under the assumptions of Theorem3.1,
for each f1,f2∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) the mappings Lf1,Lf2:I→J satisfy the following inequalities:
(i)
∥Lf1−1−Lf2−1∥c0≤K5/K0∥f1−f2∥c0;(ii)
∥(Lf1−1)′−(Lf2−1)′∥c0≤(((K1N+K4)K5+(K7+NK8)K0)/(K0)3)∥f1−f2∥c0+K6/(K0)2∥f1′−f2′∥c0.Proof.
Firstly we have(3.13)∥Lf1−Lf2∥c0≤∑k=1∞|λk+1|⋅∥Hk+1∘f1k−Hk+1∘f2k∥c0≤(2.2)∑k=1∞|λk+1|Mk+1(∑i=1kMi−1)⋅∥f1−f2∥c0,and thus(3.14)∥Lf1−1−Lf2−1∥c0≤(Lemma⁢(2.4))1K0∥Lf1−Lf2∥c0≤K5K0∥f1−f2∥c0.Secondly we get(3.15)∥(Lf1−1)′−(Lf2−1)′∥c0=maxy∈J{|1L′f1(Lf1−1(y))−1L′f2(Lf2−1(y))|}≤(3.9)maxy∈J{|L′f2(Lf2−1(y))−L′f1(Lf1−1(y))|(K0)2}≤1(K0)2⋅maxy∈J{|L′f2(Lf2−1(y))−L′f2(Lf1−1(y))|}+1(K0)2⋅maxy∈J{|L′f2(Lf1−1(y))−L′f1(Lf1−1(y))|}≤(3.11)K1N+K4(K0)2⋅maxy∈J{|Lf2−1(y)−Lf1−1(y)|}+1(K0)2⋅maxx∈I{|L′f2(x)−L′f1(x)|}=K1N+K4(K0)2⋅∥Lf2−1−Lf1−1∥c0+1(K0)2⋅∥L′f2−L′f1∥c0.Notice that(3.16)∥L′f2−L′f1∥c0≤∑k=1∞|λk+1|⋅∥(Hk+1∘f2k)′−(Hk+1∘f1k)′∥c0and for k=1,2,…,(3.17)∥(Hk+1∘f2k)′−(Hk+1∘f1k)′∥c0=∥(Hk+1'∘f2k)⋅(f2k)′−(Hk+1'∘f1k)⋅(f1k)′∥c0≤∥Hk+1'∘f2k−Hk+1'∘f1k∥c0⋅∥(f2k)′∥c0+∥Hk+1'∘f1k∥c0∥(f2k)′−(f1k)′∥c0≤MkNk+1∑i=1kMi−1∥f2−f1∥c0+Mk+1kMk−1∥f2′−f1′∥c0+Mk+1Q(k)N∑i=1k−1(k−i)Mk+i−2∥f2−f1∥c0,then(3.18)∥L′f2−L′f1∥c0≤(K7+NK8)∥f2−f1∥c0+K6∥f2′−f1′∥c0.Finally we get(3.19)∥(Lf1−1)′−(Lf2−1)′∥c0≤((K1N+K4)K5+(K7+NK8)K0(K0)3)∥f1−f2∥c0+K6(K0)2∥f1′−f2′∥c0.Proof of Theorem3.1..
For anyf∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) we define Θ(f) as follows:(3.20)Θ(f)=Lf−1∘F,and denote Θ(f)=g for convenience. Clearly g∈C1(I,I), g(a)=a, g(b)=b,  (or g(a)=b,g(b)=a),
and (3.10) yields that for any x∈I,  (3.21)|g′(x)|=|(Lf−1)′(F(x))⋅F′(x)|≤K0MK0=M.Furthermore by (3.12) we get that
for any x1,x2∈I,  (3.22)|g′(x1)−g′(x2)|≤|(Lf−1)′(F(x1))|⋅|F′(x1)−F′(x2)|+|(Lf−1)′(F(x1))−(Lf−1)′(F(x2))|⋅|F′(x2)|≤1K0L|x1−x2|+K0M⁢⋅K1N+K4(K0)3⁢⋅K0M|x1−x2|=N|x1−x2|.So g∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)),
which means that Θ(𝔄(I,I,0,M,N))⊂𝔄(I,I,0,M,N)(orΘ(𝔄′(I,I,0,M,N))⊂𝔄′(I,I,0,M,N)).
Secondly we prove that(3.23)Θ:A(I,I,0,M,N)⟶A(I,I,0,M,N)(or⁢Θ:A′(I,I,0,M,N)⟶A′(I,I,0,M,N)) is continuous. For any f1,f2∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) we denote gi=Θ(fi),i=1,2. It is easy to see that(3.24)∥g1−g2∥c0=∥Lf1−1∘F−Lf2−1∘F∥c0≤∥Lf1−1−Lf2−1∥c0≤K5K0∥f1−f2∥c0.By Lemma 3.4 we get(3.25)∥g′1−g′2∥c0=∥((Lf1−1)′∘F)⋅F′−((Lf2−1)′∘F)⋅F′∥c0≤K0M⁢⋅∥((Lf1−1)′−(Lf2−1)′)∥c0≤((K1N+K4)K5M+(K7+NK8)K0M(K0)2)∥f1−f2∥c0+MK6K0∥f′1−f′2∥c0.By the discussion above we
get(3.26)∥g1−g2∥c1=∥g1−g2∥c0+∥g′1−g′2∥c0≤{K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2}∥f2−f1∥c0+MK6K0∥f′2−f′1∥c0≤E∥f1−f2∥c1,where(3.27)E=max{K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2,⁢MK6K0}.HenceΘ:𝔄(I,I,0,M,N)→𝔄(I,I,0,M,N)(orΘ:𝔄′(I,I,0,M,N)→𝔄′(I,I,0,M,N)) is continuous. By Schauder fixed point theorem,
there exists a function f∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) such that(3.28)f=Θ(f)=Lf−1∘F.That means f is a solution of (1.4) in 𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)).


## 3.1. Existence of Solutions

Let{λi}i=1∞ be coefficients of (1.4) and I=[a,b].
For any f∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) define A=∑i=1∞λiHi(fi−1(a)) and B=∑i=1∞λiHi(fi−1(b)).
It is easy to see that both the convergence and the value of A,B have nothing to do with the choice of f.Theorem 3.1.
SupposeM>1, L are positive constants and H1∈𝔄(I,I,m1,M1,N1),Hi∈𝔄(I,I,0,Mi,Ni)orHi∈𝔄'(I,I,0,Mi,Ni) for i=2,3,…, where m1>0 and Mi≥1,Ni are positive constants for i=1,2,…. Assume further the following
conditions:(3.1)K0=λ1m1−∑i=2∞|λi|MiMi−1>0,K1=1(M−1)∑i=1∞|λi+1|Mi+1Mi−1(Mi−1)<∞,K0−K1M2>0,K2=∑i=1∞|λi|NiMi−1(Mi−1)<∞,−∞<A<B<∞,  hold. Then for any given F∈𝔄(I,J,0,K0M,L)(or𝔄′(I,J,0,K0M,L)), (1.4) has a solution f∈𝔄(I,I,0,M,M*)(or𝔄′(I,I,0,M,M*)), where M*≥(L+K4M2)/(K0−K1M2), K4=∑i=1∞|λi|NiM2(i−1) and I=[a,b], J=[A,B].As in  [22], firstly we give the following three Lemmas which
lead directly to the proof of Theorem 3.1. In the sequel we denote(3.2)N=(L+K4M2)(K0−K1M2).Lemma 3.2.
Under the assumptions of Theorem3.1
the following series: (3.3)K3=λ1M1+∑i=2∞|λi|MiMi−1,K4=∑i=1∞|λi|NiM2(i−1),K5=∑k=1∞|λk+1|Mk+1(∑i=1k(Mi−1)),K6=∑k=1∞|λk+1|Mk+1kMk−1,K7=∑k=1∞|λk+1|Nk+1Mk(∑i=1kMi−1),K8=∑K=2∞|λk+1|Mk+1∑j=1k−1(k−j)Mk+j−2,  are all convergent.Proof.
The convergence ofK3, K4, K5, K6, and K7 is easy to be verified. As mentioned in
[21], the equality(3.4)∑i=1n−1(n−i)xn+i−2=xn−1(xn−1(x−1)2−nx−1),x≠1holds. We get that(3.5)K8=∑K=2∞|λk+1|Mk+1(Mk−1(Mk−1(M−1)2−kM−1)).By the convergence of K1 and K6, K8 is also convergent.Lemma 3.3.
Under the assumptions of Theorem3.1,
for each f∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) the mapping Lf:I→ℝ defined by(3.6)Lf(x)=∑i=1∞λiHi(fi−1(x)),x∈Ihas the following
properties:
(i)
Lf∈𝔄(I,J,K0,K3,K1N+K4);(ii)
Lf−1∈𝔄(J,I,1/K3,1/K0,(K1N+K4)/(K0)3),
where I=[a,b] and J=[A,B].Proof.
For anyf∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)),
we have(3.7)Lf(a)=∑i=1∞λiHi(fi−1(a))<Lf(b)=∑i=1∞λiHi(fi−1(b)).It is easy to see that for any x∈I(3.8)0<λ1m1−∑i=2∞|λi|MiMi−1≤∑i=1∞λiHi′(fi−1(x))(fi−1)′(x)≤λ1M1+∑i=2∞|λi|MiMi−1.We have for any x∈I(3.9)0<K0≤L′f(x)≤K3,and for any y∈J(3.10)0<1K3≤(Lf−1)′(y)≤1K0.Thus Lf:I→J is an orientation-preserving diffeomorphism.By Lemma2.2 we can see that for any x1,x2∈I,  (3.11)|L′f(x1)−L′f(x2)|=|∑i=1∞λiHi′(fi−1(x1))(fi−1)′(x1)−∑i=1∞λiHi′(fi−1(x2))(fi−1)′(x2)|≤∑i=1∞|λi|{|Hi′(fi−1(x1))|⋅|(fi−1)′(x1)−(fi−1)′(x2)|+|Hi′(fi−1(x1))−Hi′(fi−1(x2))|⋅|(fi−1)′(x2)|}≤{∑i=2∞|λi|MiN(∑j=i−22(i−2)Mj)+∑i=1∞|λi|NiM2(i−1)}|x1−x2|=(K1N+K4)|x1−x2|.By (3.9), (3.11), and Lemma 2.3 we get
for any y1,y2∈J:(3.12)|(Lf−1)′(y1)−(Lf−1)′(y2)|≤K1N+K4(K0)3|y1−y2|.Lemma 3.4.
Under the assumptions of Theorem3.1,
for each f1,f2∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) the mappings Lf1,Lf2:I→J satisfy the following inequalities:
(i)
∥Lf1−1−Lf2−1∥c0≤K5/K0∥f1−f2∥c0;(ii)
∥(Lf1−1)′−(Lf2−1)′∥c0≤(((K1N+K4)K5+(K7+NK8)K0)/(K0)3)∥f1−f2∥c0+K6/(K0)2∥f1′−f2′∥c0.Proof.
Firstly we have(3.13)∥Lf1−Lf2∥c0≤∑k=1∞|λk+1|⋅∥Hk+1∘f1k−Hk+1∘f2k∥c0≤(2.2)∑k=1∞|λk+1|Mk+1(∑i=1kMi−1)⋅∥f1−f2∥c0,and thus(3.14)∥Lf1−1−Lf2−1∥c0≤(Lemma⁢(2.4))1K0∥Lf1−Lf2∥c0≤K5K0∥f1−f2∥c0.Secondly we get(3.15)∥(Lf1−1)′−(Lf2−1)′∥c0=maxy∈J{|1L′f1(Lf1−1(y))−1L′f2(Lf2−1(y))|}≤(3.9)maxy∈J{|L′f2(Lf2−1(y))−L′f1(Lf1−1(y))|(K0)2}≤1(K0)2⋅maxy∈J{|L′f2(Lf2−1(y))−L′f2(Lf1−1(y))|}+1(K0)2⋅maxy∈J{|L′f2(Lf1−1(y))−L′f1(Lf1−1(y))|}≤(3.11)K1N+K4(K0)2⋅maxy∈J{|Lf2−1(y)−Lf1−1(y)|}+1(K0)2⋅maxx∈I{|L′f2(x)−L′f1(x)|}=K1N+K4(K0)2⋅∥Lf2−1−Lf1−1∥c0+1(K0)2⋅∥L′f2−L′f1∥c0.Notice that(3.16)∥L′f2−L′f1∥c0≤∑k=1∞|λk+1|⋅∥(Hk+1∘f2k)′−(Hk+1∘f1k)′∥c0and for k=1,2,…,(3.17)∥(Hk+1∘f2k)′−(Hk+1∘f1k)′∥c0=∥(Hk+1'∘f2k)⋅(f2k)′−(Hk+1'∘f1k)⋅(f1k)′∥c0≤∥Hk+1'∘f2k−Hk+1'∘f1k∥c0⋅∥(f2k)′∥c0+∥Hk+1'∘f1k∥c0∥(f2k)′−(f1k)′∥c0≤MkNk+1∑i=1kMi−1∥f2−f1∥c0+Mk+1kMk−1∥f2′−f1′∥c0+Mk+1Q(k)N∑i=1k−1(k−i)Mk+i−2∥f2−f1∥c0,then(3.18)∥L′f2−L′f1∥c0≤(K7+NK8)∥f2−f1∥c0+K6∥f2′−f1′∥c0.Finally we get(3.19)∥(Lf1−1)′−(Lf2−1)′∥c0≤((K1N+K4)K5+(K7+NK8)K0(K0)3)∥f1−f2∥c0+K6(K0)2∥f1′−f2′∥c0.Proof of Theorem3.1..
For anyf∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) we define Θ(f) as follows:(3.20)Θ(f)=Lf−1∘F,and denote Θ(f)=g for convenience. Clearly g∈C1(I,I), g(a)=a, g(b)=b,  (or g(a)=b,g(b)=a),
and (3.10) yields that for any x∈I,  (3.21)|g′(x)|=|(Lf−1)′(F(x))⋅F′(x)|≤K0MK0=M.Furthermore by (3.12) we get that
for any x1,x2∈I,  (3.22)|g′(x1)−g′(x2)|≤|(Lf−1)′(F(x1))|⋅|F′(x1)−F′(x2)|+|(Lf−1)′(F(x1))−(Lf−1)′(F(x2))|⋅|F′(x2)|≤1K0L|x1−x2|+K0M⁢⋅K1N+K4(K0)3⁢⋅K0M|x1−x2|=N|x1−x2|.So g∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)),
which means that Θ(𝔄(I,I,0,M,N))⊂𝔄(I,I,0,M,N)(orΘ(𝔄′(I,I,0,M,N))⊂𝔄′(I,I,0,M,N)).
Secondly we prove that(3.23)Θ:A(I,I,0,M,N)⟶A(I,I,0,M,N)(or⁢Θ:A′(I,I,0,M,N)⟶A′(I,I,0,M,N)) is continuous. For any f1,f2∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) we denote gi=Θ(fi),i=1,2. It is easy to see that(3.24)∥g1−g2∥c0=∥Lf1−1∘F−Lf2−1∘F∥c0≤∥Lf1−1−Lf2−1∥c0≤K5K0∥f1−f2∥c0.By Lemma 3.4 we get(3.25)∥g′1−g′2∥c0=∥((Lf1−1)′∘F)⋅F′−((Lf2−1)′∘F)⋅F′∥c0≤K0M⁢⋅∥((Lf1−1)′−(Lf2−1)′)∥c0≤((K1N+K4)K5M+(K7+NK8)K0M(K0)2)∥f1−f2∥c0+MK6K0∥f′1−f′2∥c0.By the discussion above we
get(3.26)∥g1−g2∥c1=∥g1−g2∥c0+∥g′1−g′2∥c0≤{K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2}∥f2−f1∥c0+MK6K0∥f′2−f′1∥c0≤E∥f1−f2∥c1,where(3.27)E=max{K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2,⁢MK6K0}.HenceΘ:𝔄(I,I,0,M,N)→𝔄(I,I,0,M,N)(orΘ:𝔄′(I,I,0,M,N)→𝔄′(I,I,0,M,N)) is continuous. By Schauder fixed point theorem,
there exists a function f∈𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)) such that(3.28)f=Θ(f)=Lf−1∘F.That means f is a solution of (1.4) in 𝔄(I,I,0,M,N)(or𝔄′(I,I,0,M,N)).


## 4. Uniqueness and Stability of Solutions

Theorem 4.1.
Let{λi}i=1∞ be coefficient of (1.4) and M>1, N>0  positive constants. Suppose that the conditions in (3.1)
are valid. Further one assumes that(4.1)E=max{K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2,⁢MK6K0}<1.Then for any F∈𝔄(I,J,0,K0M,L)(or𝔄′(I,J,0,K0M,L)),
there exists a unique function f∈𝔄(I,I,0,M,M*)(or𝔄′(I,I,0,M,M*)) satisfying (1.4), where M*≥(L+K4M2)/(K0−K1M2). Furthermore the solution f depends continuously on the given function F.Proof.
IfE<1,
then by (3.26) the map Θ defined in Theorem 3.1 becomes a strict
contraction. The fix point of Θ,
which is a solution of (1.4), is unique by Banach's contraction
principle.
Letf1,f2 be the solutions of (1.4) for the
corresponding functions F1,F2. First, since(4.2)fi=Lfi−1∘Fi,i=1,2,we get(4.3)∥f1−f2∥c0=∥Lf1−1∘F1−Lf2−1∘F2∥c0≤∥Lf1−1∘F1−Lf1−1∘F2∥c0+∥Lf1−1∘F2−Lf2−1∘F2∥c0≤(Lemma(3.4))1K0∥F1−F2∥c0+K5K0∥f1−f2∥c0.Second, we have(4.4)∥f′1−f′2∥c0=∥((Lf1−1)′∘F1)⋅F′1−((Lf2−1)′∘F2)⋅F′2∥c0≤∥((Lf1−1)′∘F1)⋅F′1−((Lf1−1)′∘F1)⋅F′2∥c0+∥((Lf1−1)′∘F1)⋅F′2−((Lf1−1)′∘F2)⋅F′2∥c0+∥((Lf1−1)′∘F2)⋅F′2−((Lf2−1)′∘F2)⋅F′2∥c0≤(3.12)1K0∥F′1−F′2∥c0+M(K1N+K4)(K0)2∥F1−F2∥c0+K0M∥(Lf1−1)′−(Lf2−1)′∥c0≤(Lemma(3.4))1K0∥F′1−F′2∥c0+M(K1N+K4)(K0)2∥F1−F2∥c0+{(K1N+K4)K5M+(K7+NK8)K0M(K0)2}∥f1−f2∥c0+MK6K0∥f′2−f′1∥c0.
By the above discussion we get(4.5)∥f1−f2∥c1=∥f1−f2∥c0+∥f′1−f′2∥c0≤E∥f1−f2∥c1+{M(K1N+K4)(K0)2+1K0}∥F1−F2∥c1,which means that(4.6)∥f1−f2∥c1≤11−E{M(K1N+K4)(K0)2+1K0}∥F1−F2∥c1.So the solution f depends continuously on the given function F.


### 4.1. Examples

Example 4.2.
LetM=10,L=20 and I=[0,1]. The equation(4.7)1001(ef(x)−1)1000(e−1)−(f2(x))21000=x+12sin(2πx),where x∈[0,1],
has a unique solution f∈𝔄([0,1],[0,1],0,10,900).Proof.
It is easy to
see thatH1(x)=(ex−1)/(e−1)∈𝔄(I,I,1/2,2,2),H2(x)=x2∈𝔄(I,I,0,2,2) and F(x)=x+1/2sin(2πx)∈𝔄(I,I,0,4.5,20).
By simple calculation we get that(4.8)K0=9612000,K1=1500,K2=9999500,K3=20221000,K4=1101500,K5=K6=1500,K7=150,K8=0,K0M=961200,K0−K1M2=5612000,A=0<B=1,N=(L+K4M2)(K0−K1M2)=480400561<900,  by Theorem 3.1 the equation has a
solution f∈𝔄([0,1],[0,1],0,10,900). Further we get that(4.9)K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2<804961,MK6K0=40961,this means E<1.
By Theorem 4.1 the solution is unique.By similar discussion we have the following example.Example 4.3.
LetM=10,L=20, and I=[0,1]. The equation(4.10)1001(ef(x)−1)1000(e−1)−1−(f2(x))21000=1−x−12sin(2πx),where x∈[0,1]
has a unique solution f∈𝔄'([0,1],[0,1],0,10,900).


## 4.1. Examples

Example 4.2.
LetM=10,L=20 and I=[0,1]. The equation(4.7)1001(ef(x)−1)1000(e−1)−(f2(x))21000=x+12sin(2πx),where x∈[0,1],
has a unique solution f∈𝔄([0,1],[0,1],0,10,900).Proof.
It is easy to
see thatH1(x)=(ex−1)/(e−1)∈𝔄(I,I,1/2,2,2),H2(x)=x2∈𝔄(I,I,0,2,2) and F(x)=x+1/2sin(2πx)∈𝔄(I,I,0,4.5,20).
By simple calculation we get that(4.8)K0=9612000,K1=1500,K2=9999500,K3=20221000,K4=1101500,K5=K6=1500,K7=150,K8=0,K0M=961200,K0−K1M2=5612000,A=0<B=1,N=(L+K4M2)(K0−K1M2)=480400561<900,  by Theorem 3.1 the equation has a
solution f∈𝔄([0,1],[0,1],0,10,900). Further we get that(4.9)K5K0+(K1N+K4)K5M+(K7+NK8)K0M(K0)2<804961,MK6K0=40961,this means E<1.
By Theorem 4.1 the solution is unique.By similar discussion we have the following example.Example 4.3.
LetM=10,L=20, and I=[0,1]. The equation(4.10)1001(ef(x)−1)1000(e−1)−1−(f2(x))21000=1−x−12sin(2πx),where x∈[0,1]
has a unique solution f∈𝔄'([0,1],[0,1],0,10,900).



---

*Source: 636843-2009-03-08.xml*