1. Introduction

An important form of iterative equations is the polynomial-like iterative equation

(1.1)

where is a given function, is an unknown function, and is the th iterate of that is, The case of all constant was considered in [1–10]. In 2000, W. N. Zhang and J. A. Baker first discussed the continuous solutions of such an iterative equation with variable coefficients which are all continuous in interval In 2001, J. G. Si and X. P. Wang furthermore gave the continuously differentiable solution of such equation in the same conditions as in [11]. In this paper, we continue the works of [11, 12], and consider the series-like iterative equation with variable coefficients

(1.2)

where are given continuous functions and We improve the methods given by the authors in [11, 12], and the conditions of [11, 12] are weakened by constructing a new structure operator.

2. Preliminaries

Let , clearly is a Banach space, where , for in .

Let , then is a Banach space with the norm , where, for in .

Being a closed subset, defined by

(2.1)

is a complete space.

The following lemmas are useful, and the methods of proof are similar to those of paper [4], but the conditions are weaker than those of [4].

Lemma 2.1.

Suppose that and

(2.2)
(2.3)

where and are positive constants.Then

(2.4)

for any in , where denotes .

Lemma 2.2.

Suppose that satisfy (2.2).Then

(2.5)

Lemma 2.3.

Suppose that satisfy (2.2) and (2.3).Then

(2.6)

for where as and as .

3. Main Results

For given constants and , let

(3.1)

Theorem 3.1 (existence).

Given positive constants and if there exists constants and , such that

,

,

then (1.2) has a solution in .

Proof.

For convenience, let

Define such that , where

(3.2)

Since , it is easy to see that for all , and for all It follows from that is uniformly convergent. Then is continuous for . Also we have

(3.3)

thus .

For any , we have

(3.4)

By condition , we see that is convergent, therefore is uniformly convergent for , this implies that is continuously differentiable for . Moreover

(3.5)

By Lemma 2.1,

(3.6)

Thus .

Define as follows:

(3.7)

where . Because , and are continuously differentiable for all , is continuously differentiable for all . By conditions and , for any in we have

(3.8)

We furthermore have

(3.9)

Thus is a self-diffeomorphism.

Now we prove the continuity of under the norm . For arbitrary ,

(3.10)

Let

(3.11)

Then we have

(3.12)

which gives continuity of .

It is easy to show that is a compact convex subset of . By Schauder's fixed point theorem, we assert that there is a map** such that

(3.13)

Let we have as a solution of (1.2) in . This completes the proof.

Theorem 3.2 (Uniqueness).

Suppose that (P1) and (P2) are satisfied, also one supposes that

then for arbitrary function in , (1.2) has a unique solution .

Proof.

The existence of (1.2) in is given by Theorem 3.1, from the proof of Theorem 3.1, we see that is a closed subset of , by (3.12) and , we see that is a contraction. Therefore has a unique fixed point in , that is, (1.2) has a unique solution in , this proves the theorem.

4. Example

Consider the equation

(4.1)

where It is easy to see that

For any in ,

(4.2)

thus By condition , we can choose and by condition , we can choose . Then by Theorem 3.1, there is a continuously differentiable solution of (4.1) in .

Remark 4.1.

Here is not monotone for , hence it cannot be concluded by [11, 12].