Volume 2009, Article ID 173028,7pages doi:10.1155/2009/173028
Research Article
The
C
1
Solutions of the Series-Like Iterative
Equation with Variable Coefficients
Yuzhen Mi, Xiaopei Li, and Ling Ma
Mathematics and Computational School, Zhanjiang Normal University, Zhanjiang, Guangdong 524048, China
Correspondence should be addressed to Xiaopei Li,[email protected]
Received 23 March 2009; Revised 11 June 2009; Accepted 6 July 2009
Recommended by Tomas Dom´ınguez Benavides
By constructing a structure operator quite different from that ofZhang and Baker2000and using the Schauder fixed point theory, the existence and uniqueness of theC1solutions of the series-like
iterative equations with variable coefficients are discussed.
Copyrightq2009 Yuzhen Mi et al. 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.
1. Introduction
An important form of iterative equations is the polynomial-like iterative equation
λ1fx λ2f2x · · ·λnfnx Fx, x∈I: a, b , 1.1
whereFis a given function,fis an unknown function,λi ∈ R1i1,2, . . . , n,andfk k 1,2, . . . , n is the kth iterate of f, that is, f0x x, fkx f ◦ fk−1x. The case of all constantλiswas considered in1–10 . In 2000, W. N. Zhang and J. A. Baker first discussed the continuous solutions of such an iterative equation with variable coefficients λi λix which are all continuous in intervala, b .In 2001, J. G. Si and X. P. Wang furthermore gave the continuously differentiable solution of such equation in the same conditions as in11 . In this paper, we continue the works of11,12 , and consider the series-like iterative equation with variable coefficients
∞
i1
whereλix : I → 0,1 are given continuous functions and∞i1λix 1, λ1x ≥ c >
0∀x∈I, maxx∈Iλix ci.We improve the methods given by the authors in11,12, and the conditions of11,12 are weakened by constructing a new structure operator.
2. Preliminaries
LetC0I,R {f:I → R, f is continuous}, clearlyC0I,R, ·
c0is a Banach space, where
f c0maxx∈I|fx|, forfinC0I,R.
Let C1I,R {f : I → R, f is continuous and continuously differentiable}, then
C1I,Ris a Banach space with the norm ·
c1, where f c1 f c0 f c0, forfinC1I,R. Being a closed subset,C1I, Idefined by
C1I, I f∈C1I, R, fI⊆I, ∀x∈I 2.1
is a complete space.
The following lemmas are useful, and the methods of proof are similar to those of paper4 , but the conditions are weaker than those of4 .
Lemma 2.1. Suppose thatϕ∈C1I, Iand
ϕx≤M, ∀x∈I, 2.2
ϕx1−ϕx2≤M|x
1−x2|, ∀x1, x2∈I, 2.3
whereMandMare positive constants.Then
ϕnx1−ϕnx2≤M 2n−2 in−1
Mi
|x1−x2|, 2.4
for anyx1, x2inI, whereϕndenotesdϕn/dx.
Lemma 2.2. Suppose thatϕ1, ϕ2∈C1I, Isatisfy2.2.Then
ϕn1−ϕn2 c0≤
n
i1
Mi−1
ϕ1−ϕ2 c0. 2.5
Lemma 2.3. Suppose thatϕ1, ϕ2∈C1I, Isatisfy2.2and2.3.Then
ϕk11−ϕk21
c0≤k
1Mkϕ1−ϕ2c0
Qk1M
k
i1
k−i1Mki−1
ϕ1−ϕ2 c0,
2.6
3. Main Results
For given constantsM1>0 andM2>0, let
AM1, M2
ϕ∈C1I, I:ϕx≤M1, ∀x∈I ,
ϕx1−ϕx2≤M
2|x1−x2|, ∀x1, x2∈I
.
3.1
Theorem 3.1existence. Given positive constantsM1, M2 andF ∈ AM1, M2,if there exists constantsN1≥1andN2>0, such that
P1c−∞i2ciN1i−1≥M1/N1,
P2c−∞i2ci2ij−i−21N1j≥M2/N2,
then1.2has a solutionfinAN1, N2.
Proof. For convenience, letdmax{|a|,|b|}.
DefineK:AN1, N2 → C1I, Isuch thatK:f → Kf, where
Kft ∞
i1
λixfit, ∀x, t∈I. 3.2
Sincef ∈ AN1, N2, it is easy to see that|fit| ≤dfor allt∈I, and|λixfit| ≤d|λix|for allx, t ∈ I.It follows from∞i1λix 1 that∞i1λixfitis uniformly convergent. Then
Kftis continuous fort∈I. Also we have
a
∞
i1
λixa≤
∞
i1
λixfit≤
∞
i1
λixbb, 3.3
thusKf ∈C0I, I.
For anyf ∈ AN1, N2, we have
dtdλixfitλixffi−1tfi−1t≤ciN1i. 3.4
By condition P1, we see that ∞i1ciN1i is convergent, therefore∞i1cifit
is uniformly convergent fort∈I, this implies thatKftis continuously differentiable fort∈I. Moreover
dtdKft
≤∞
i1
λixfit≤
∞
i1
ByLemma 2.1,
dtdKft1
− d
dt
Kft2≤ ∞
i1
λixfit1−fit2
≤∞
i1
ci
⎛ ⎝N22i−2
ji−1
N1j ⎞
⎠|t1−t2|:μ2|t1−t2|.
3.6
ThusKf ∈ Aμ1, μ2.
DefineT :AN1, N2 → C1I, Ias follows:
Tft 1
λ1xFt−
1
λ1xKft ft, ∀t, x∈I, 3.7
wheref ∈ AN1, N2. BecauseKf,F,andf are continuously differentiable for allt∈ I,Tf is continuously differentiable for allt∈I. By conditionsP1andP2, for anyt1, t2inI,we have
dtdTft≤ 1 λ1xF
t 1
λ1x
∞
i2
λixfit≤ 1
cM1
1
c
∞
i2
ciN1i
≤ 1
cM1
1
ccN1−M1 N1.
3.8
We furthermore have
dtdTft1− d dt
Tft2≤ 1
λ1xF
t1−Ft2 1
λ1x
∞
i2
ci
fit1−fit2
≤ 1
cM2|t1−t2|
1
c
∞
i2
ciN2
⎛ ⎝2i−2
ji−1
N1j ⎞ ⎠|t1−t2|
≤N2|x1−x2|.
3.9
ThusT :AN1, N2 → AN1, N2is a self-diffeomorphism.
Now we prove the continuity of T under the norm · c1. For arbitrary f1, f2 ∈
Tf1−Tf2 c0 max
t∈I
− 1
λ1xKf1t f1t 1
λ1xKf2t−f2t
≤ 1
cmaxt∈I
∞
i2
λixf1it−
∞
i2
λixf2it
≤ 1 c ∞
i2
cif1i−f2ic0
≤ 1
c
∞
i2
ci i
k1
N1k−1
f1−f2
c0,
dtdTf1− d dtTf2
c0
max t∈I
− 1
λ1x
Kf1t
f1t 1
λ1x
Kf2t
−
f2t
≤ 1
cmaxt∈I
∞
i2
λixf1it−
∞
i2
λixf2it
≤ 1
c
∞
i2
ci
fi
1
−fi 2 c0 ≤ 1 c ∞
i2
ci
iN1i−1f1−f2c0QiN2 i−1
k1
i−kN1ik−2
f1−f2
c0
.
3.10
Let
E1 1
c
∞
i2
ci i
k1
N1k−1QiN2 i−1
k1
i−kNi1k−2 , E2 1 c ∞
i2
ciiN1i−1, Emax{E1, E2}.
3.11
Then we have
Tf1−Tf2
c1Tf1−Tf2c0
Tf1
−
Tf2
c0≤E1f1−f2c0E2f 1−f2c0
≤Ef1−f2c0Ef1−f2c0 Ef1−f2c1,
3.12
which gives continuity ofT.
It is easy to show thatAN1, N2is a compact convex subset ofC1I, I. By Schauder’s fixed point theorem, we assert that there is a mappingf ∈ AN1, N2such that
ft Tft 1
λ1xFt−
1
λ1xKft ft, ∀t∈I. 3.13
Theorem 3.2Uniqueness. Suppose that (P1) and (P2) are satisfied, also one supposes that
P3E <1,
then for arbitrary functionFinAM1, M2,1.2has a unique solutionf∈ AN1, N2.
Proof. The existence of 1.2 in AN1, N2 is given by Theorem 3.1, from the proof of
Theorem 3.1, we see thatAN1, N2is a closed subset ofC1I, I, by3.12andP3, we see
thatT :AN1, N2 → AN1, N2is a contraction. ThereforeT has a unique fixed pointfx inAN1, N2, that is,1.2has a unique solution inAN1, N2, this proves the theorem.
4. Example
Consider the equation
∞
i1
λixfix 1 4x
2, x∈I: −1,1 , 4.1
whereλ1x 33/36 1/36cos2πx/2, λ2x 1/36 1/36sin2πx/2, λ3x 1/36,
λ4x λ5x · · · 0.It is easy to see that 0 ≤λix≤ 1, ∞i1λix 1, c 33/36, c2 2/36, c31/36, c4c5· · ·0.
For anyx, yin−1,1 ,
Fx|0.5x| ≤0.5, Fx−Fy≤ |0.5x|0.5y≤1, 4.2
thusF∈ A0.5,1.By conditionP1, we can chooseN11.1,and by conditionP1, we can chooseN21.5. Then byTheorem 3.1, there is a continuously differentiable solution of4.1 inA1.1,1.5.
Remark 4.1. HereFxis not monotone forx∈−1,1 , hence it cannot be concluded by11,
12 .
Acknowledgments
This work was supported by Guangdong Provincial Natural Science Foundation07301595 and Zhan-jiang Normal University Science Research ProjectL0804.
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