Let $ S $ be a non-empty set and $ f:S\to S $ be a self-map. For each positive integer $ n $, $ f^n $, the $ n $-fold composition of $ f $ with itself, also known as the $ n $-th iterate of $ f $, is defined recursively by $ f^1 = f $, and $ f^n = f\circ f^{n-1} $.

Let $ S $ be a non-empty subset of the real line $ \mathbb{R} $ and $ F:S\to \mathbb{R} $ be given. The polynomial-like iterative equation

$ \begin{equation} \lambda_1f(x)+\lambda_2f^2(x)+\ldots+\lambda_n f^n(x) = F(x), \quad \forall x\in S, \end{equation} $

(1.1)

where $ \lambda_i, i = 1, 2, \ldots, n $ are constants, has been discussed in many papers under various settings. Si [1] obtained results on $ C^2 $ solutions with $ S = [a, b] $, a finite closed interval, and $ F $, a $ \mathcal{C}^2 $ self-map on [a, b], $ F(a) = a $, $ F(b) = b $. His work is based on Zhang's paper [2] in which the results cover the existence, the uniqueness, and the stability of the differentiable solutions. In [3] invertible solutions are obtained in some local neighbourhoods of the fixed points of the functions. Monotonic solutions and convex solutions are discussed by Nikodem, Xu, and Zhang in [4, 5]. Recently, Ng and Zhao [6] studied the periodic and continuous solutions of Eq. (1.1). For some properties of solutions for polynomial-like iterative functional equations, we refer the interested readers to [7]–[10].

In 2000, Zhang and Baker [11] studied the continuous solutions of

$ \begin{equation} \lambda_1(x)f(x)+\lambda_2(x)f^2(x)+\ldots+\lambda_n(x) f^n(x) = F(x). \end{equation} $

(1.2)

Later, Xu [12] considered the analytic solutions of Eq. (1.2). In this paper, we continue to consider the continuous and periodic solutions of Eq. (1.2). In fact, if $ \lambda_i(x), i = 1, 2, \ldots, n $ are constants, then the conclusions in [6] can be derived by our results.

Notations, preliminaries, and the current setting.

$ \begin{align*} &C(\mathbb{R}, \mathbb{R}) \rm{ - the\;linear\;space\;of\;all\;continuous\;self-maps\;on\; \mathbb{R} }.\\ &{\mathcal{P}}_T = \{f\in C(\mathbb{R}, \mathbb{R})\, :\, f(x+T) = f(x), \ \forall x\in \mathbb{R}\}\, (T>0) \\&\quad \rm{ - a\;Banach\;space\;with\;the\;norm}\;\, \, \|f\| = \max\limits_{x\in [0, T]}|f(x)| = \max\limits_{x\in \mathbb{R}}|f(x)|.\\ &{\mathcal{P}}_T(L, M) = \{f\in {\mathcal{P}}_T\, :\, \|f\|\leq L, \ |f(y)-f(x)|\leq M|y-x|, \ \forall x, y\in \mathbb{R} \}. \\&\quad {\rm{- a\;closed\;convex\;nonempty\;subset\;of}}\;{\mathcal{P}}_T ({\rm{constants}}\;L\ge 0, M\ge 0 ). \end{align*} $

Assume that $ \lambda_i\in {\mathcal{P}}_T(L_i, M_i), |\lambda_1(x)|\geq k_1>0, \forall x\in [0, T] $ and $ F\in {\mathcal{P}}_T(L', M') $, and we seek solutions $ f\in {\mathcal{P}}_T(L, M) $.

We shall find sufficient conditions on the constants $ L' $, $ M' $, $ L $ and $ M $ under which the existence of a solution is assured. The Schauder fixed point theorem being a main tool, its statement is included, and will be applied with $ \Omega: = {\mathcal{P}}_T(L, M) $ in the Banach space $ {\mathcal{P}}_T $.

Theorem 2.1 Schauder([13]) Let $ \Omega $ be a closed convex and nonempty subset of a Banach space $ (\mathbb{B}, \|\cdot\|) $. Suppose that $ A $ maps $ \Omega $ into $ \Omega $ and is compact and continuous. Then there exists $ z\in \Omega $ with $ z = Az $.

Let $ A: {\mathcal{P}}_T(L, M)\rightarrow {\mathcal{P}}_T $ be defined by

$ \begin{eqnarray} Af(x) = \frac{1}{\lambda_1(x)}F(x)-\frac{1}{\lambda_1(x)}\sum\limits_{i = 2}^n\lambda_i(x) f^i(x), \ \forall f\in {\mathcal{P}}_T(L, M). \end{eqnarray} $

(2.1)

The space $ {\mathcal{P}}_T $ is closed under composition. The range of $ A $ is clearly contained in $ {\mathcal{P}}_T $. Fixed points of $ A $ correspond to the solutions of (1.2). Hence we seek conditions under which the assumptions in Schauder's theorem are met.

Lemma 2.2 (see [2]) For any $ f, g \in {\mathcal{P}}_T(L, M), x, y\in \mathbb{R} $, the following inequalities hold for every positive integer $ n $.

$ \begin{align} &|f^{n}(y)-f^{n}(x)|\leq M^n|y-x|, \end{align} $

(2.2)

$ \begin{align} & ||f^{n}-g^{n}||\leq \sum\limits_{j = 0}^{n-1}M^j\|f-g\|. \end{align} $

(2.3)

Lemma 2.3 Suppose $ \lambda_i\in {\mathcal{P}}_T(L_i, M_i), |\lambda_1(x)|\geq k_1>0, \forall x\in [0, T] $, then operator $ A $ is continuous and compact on $ {\mathcal{P}}_T(L, M) $.

Proof For any $ f, g\in {\mathcal{P}}_T(L, M), x\in \mathbb{R} $, by (2.3) we get

$ \begin{eqnarray*} |(Af)(x)-(Ag)(x)|&\leq& \frac{1}{|\lambda_1(x)|}\sum\limits_{i = 2}^n|\lambda_i(x)| |f^i(x)-g^i(x)| \nonumber\\&\leq& \frac{1}{k_1}\sum\limits_{i = 2}^n\sum\limits_{j = 0}^{i-1}L_iM^j\|f-g\|. \end{eqnarray*} $

Thus

$ \begin{eqnarray} \|Af-Ag\|&\leq& \frac{1}{k_1}\sum\limits_{i = 2}^n\sum\limits_{j = 0}^{i-1}L_iM^j\|f-g\|. \end{eqnarray} $

(2.4)

This proves that $ A $ is continuous. $ {\mathcal{P}}_T(L, M) $ is closed, uniformly bounded and equicontinuous on $ \mathbb{R} $. By the Arzelà-Ascoli theorem ([14], page 28), applied to a sufficiently large closed interval of $ \mathbb{R} $ and taking the periodicity of the functions into account, it is compact. Since continuous functions map compact sets to compact sets, $ A $ is a compact map.

Theorem 2.4 Suppose that $ F\in {\mathcal{P}}_T(L', M'), \lambda_i\in{\mathcal{P}}_T(L_i, M_i), i = 1, 2, \ldots, n $ and $ |\lambda_1(x)|\geq k_1>0 $, $ \forall x\in [0, T] $. If the constants $ L', M', L, M $ satisfy the conditions

$ \begin{eqnarray} L'\leq \Big(k_1-\sum\limits_{i = 2}^nL_i\Big)L, \ M'\leq k_1M-\frac{L'}{k_1}M_1-\sum\limits_{i = 2}^n\Big(L_iM^i+L M_i+\frac{LL_i}{k_1}M_1\Big), \end{eqnarray} $

(2.5)

then (1.2) has a solution in $ {\mathcal{P}}_T(L, M) $.

Proof By (2.1), for all $ x\in \mathbb{R} $, we have

$ \begin{eqnarray*} |(Af)(x)|&\leq&\frac{1}{|\lambda_1(x)|}|F(x)|+\frac{1}{|\lambda_1(x)|}\sum\limits_{i = 2}^n|\lambda_i(x)| |f^i(x)| \nonumber\\&\leq& \frac{1}{|\lambda_1(x)|} L'+\frac{1}{|\lambda_1(x)|}\sum\limits_{i = 2}^n|\lambda_i(x)| L \nonumber\\&\leq& \frac{1}{k_1} L'+\frac{L}{k_1}\sum\limits_{i = 2}^nL_i. \end{eqnarray*} $

Thus the first condition of (2.5) assures

$ \begin{eqnarray*} \|Af\|\leq L. \end{eqnarray*} $

For all $ x, y\in \mathbb{R} $, by (2.1), (2.2), we have

$ \begin{eqnarray*} &&|(Af)(y)-(Af)(x)|\\& = &\Big|\frac{1}{\lambda_1(y)}F(y)-\frac{1}{\lambda_1(y)}\sum\limits_{i = 2}^n\lambda_i(y) f^i(y)-\frac{1}{\lambda_1(x)}F(x)+\frac{1}{\lambda_1(x)}\sum\limits_{i = 2}^n\lambda_i(x) f^i(x)\Big|\\ &\leq&\Big|\frac{1}{\lambda_1(y)}F(y)-\frac{1}{\lambda_1(x)}F(x)\Big|+\sum\limits_{i = 2}^n \Big| \frac{\lambda_i(y)}{\lambda_1(y)}f^i(y)-\frac{\lambda_i(x)}{\lambda_1(x)}f^i(x)\Big|\\ &\leq&\frac{1}{|\lambda_1(y)|}|F(y)-F(x)|+\Big|\frac{1}{\lambda_1(y)}-\frac{1}{\lambda_1(x)}\Big||F(x)|\\ &&+\sum\limits_{i = 2}^n \frac{|\lambda_i(y)|}{|\lambda_1(y)|}|f^i(y)-f^i(x)|+\sum\limits_{i = 2}^n \Big|\frac{\lambda_i(y)}{\lambda_1(y)}-\frac{\lambda_i(x)}{\lambda_1(x)}\Big||f^i(x)| \\ &\leq&\frac{1}{k_1}M'|y-x|+\Big|\frac{\lambda_1(x)-\lambda_1(y)}{\lambda_1(y)\lambda_1(x)}\Big||F(x)| \\&&+\sum\limits_{i = 2}^n\frac{L_i}{k_1}M^i|y-x|+\sum\limits_{i = 2}^n\Big|\frac{\lambda_i(y)\lambda_1(x)-\lambda_i(x)\lambda_1(y)} {\lambda_1(y)\lambda_1(x)}\Big||f^i(x)|\\ &\leq&\frac{1}{k_1}M'|y-x|+\frac{M_1 L'}{k_1^2}|y-x|+\frac{1}{k_1}\sum\limits_{i = 2}^nL_i M^i|y-x| \\&&+\sum\limits_{i = 2}^n\frac{|\lambda_1(x)||\lambda_i(y)-\lambda_i(x)|+|\lambda_i(x)||\lambda_1(x)-\lambda_1(y)|} {|\lambda_1(y)\lambda_1(x)|}L \\ &\leq&\frac{1}{k_1}M'|y-x|+\frac{M_1 L'}{k_1^2}|y-x|+\frac{1}{k_1}\sum\limits_{i = 2}^nL_i M^i|y-x| \\&&+\frac{L }{k_1}\sum\limits_{i = 2}^nM_i|y-x|+\frac{LM_1}{k_1^2}\sum\limits_{i = 2}^nL_i|y-x| \\& = &\Big(\frac{1}{k_1}M'+\frac{L'M_1}{k_1^2}+\frac{1}{k_1}\sum\limits_{i = 2}^nL_i M^i+\frac{L }{k_1}\sum\limits_{i = 2}^nM_i+\frac{LM_1}{k_1^2}\sum\limits_{i = 2}^nL_i\Big)|y-x| \end{eqnarray*} $

The second condition of (2.5) assures

$ \begin{eqnarray*} |(Af)(y)-(Af)(x)|\leq M|y-x|. \end{eqnarray*} $

Therefore $ Af\in {\mathcal{P}}_T(L, M) $. This proves that $ A $ maps $ {\mathcal{P}}_T(L, M) $ into itself. All conditions of Schauder's fixed point theorem are satisfied. Thus there exists an $ f $ in $ {\mathcal{P}}_T(L, M) $ such that $ f = Af $. This is equivalent to that $ f $ is a solution of (1.2) in $ {\mathcal{P}}_T(L, M) $.

In this section, uniqueness and stability of (1.2) will be proved.

Theorem 3.1 (i) Suppose that $ A $ is defined by (2.1) and

$ \begin{eqnarray} \alpha: = \frac{1}{k_1}\sum\limits_{i = 2}^n\sum\limits_{j = 0}^{i-1}L_iM^j<1. \end{eqnarray} $

(3.1)

Then $ A $ is contractive with contraction constant $ \alpha<1 $. It has at most one fixed point and (1.2) has at most one solution in $ {\mathcal{P}}_T(L, M) $.

(ii) Suppose that (2.5) and (3.1) are satisfied. Then (1.2) has a unique solution in $ {\mathcal{P}}_T(L, M) $.

Proof (i) According to (2.4), $ \|Af-Ag\|\leq \alpha\|f-g\| $ and so $ \alpha $ is a contraction constant for $ A $. Let $ f, g \in {\mathcal{P}}_T(L, M) $ be fixed points of $ A $. Then $ \|f-g\| = \|Af-Ag\|\leq \alpha\|f-g\| $. Hence $ \alpha<1 $ yields $ \|f-g\| = 0 $, resulting in $ f = g $. This proves that there is at most one fixed point. (ii) The existence is cared for by Theorem 2.4.

Theorem 3.2 Let $ {\mathcal{P}}_T(L, M) $ and $ {\mathcal{P}}_T(L', M') $ be fixed and allow $ F(x) $ and $ \lambda_i(x) $ in (1.2) to vary. The unique solution obtained in Theorem 3.1, part (ii), depends continuously on $ F(x) $ and $ \lambda_i(x)\ (i = 1, 2, \ldots, n) $.

Proof Under the assumptions of Theorem 3.1, part (ii), we consider any two functions $ F, G $ in $ {\mathcal{P}}_T(L', M') $ and a parallel pair of unique $ f, g $ in $ {\mathcal{P}}_T(L, M) $ satisfying

$ \begin{equation*} \begin{split} &\lambda_1(x)f(x)+\lambda_2(x)f^2(x)+\ldots+\lambda_n(x) f^n(x) = F(x), \\&\mu_1(x)g(x)+\mu_2(x)g^2(x)+\ldots+\mu_n(x) g^n(x) = G(x). \end{split} \end{equation*} $

Then

$ \begin{eqnarray*} &&\|f-g\|\\& = &\max\limits_{x\in [0, T]}\Big|\frac{1}{\lambda_1(x)}F(x)-\frac{1}{\lambda_1(x)}\sum\limits_{i = 2}^n\lambda_i(x) f^i(x)-\frac{1}{\mu_1(x)}G(x)+\frac{1}{\mu_1(x)}\sum\limits_{i = 2}^n\mu_i(x) g^i(x)\Big|\\ &\leq&\max\limits_{x\in [0, T]}\Big|\frac{1}{\lambda_1(x)}F(x)-\frac{1}{\mu_1(x)}G(x)\Big|+\sum\limits_{i = 2}^n\max\limits_{x\in [0, T]}\Big|\frac{\lambda_i(x)}{\lambda_1(x)} f^i(x)-\frac{\mu_i(x)}{\mu_1(x)} g^i(x)\Big|\\ &\leq&\max\limits_{x\in [0, T]}\frac{1}{|\lambda_1(x)|}|F(x)-G(x)|+\max\limits_{x\in [0, T]}\Big|\frac{1}{\lambda_1(x)}-\frac{1}{\mu_1(x)}\Big|| G(x) | \\&&+\sum\limits_{i = 2}^n\max\limits_{x\in [0, T]}\Big|\frac{\lambda_i(x)}{\lambda_1(x)}\Big||f^i(x)-g^i(x)| +\sum\limits_{i = 2}^n\max\limits_{x\in [0, T]}\Big|\frac{\lambda_i(x)}{\lambda_1(x)}-\frac{\mu_i(x)}{\lambda_1(x)}\Big| | g^i (x)| \\&&+\sum\limits_{i = 2}^n\max\limits_{x\in [0, T]}\Big|\frac{\mu_i(x)}{\lambda_1(x)}-\frac{\mu_i(x)}{\mu_1(x)}\Big| | g^i (x)| \\&\leq&\frac{1}{k_1}\|F-G\|+\frac{L'}{k_1^2}\|\lambda_1-\mu_1\|+\frac{1}{k_1}\sum\limits_{i = 2}^n\sum\limits_{j = 0}^{i-1}L_i M^j\|f-g\| \\&&+\frac{L}{k_1}\sum\limits_{i = 2}^n\|\lambda_i-\mu_i\|+\frac{L}{k_1^2}\sum\limits_{i = 2}^nL_i\|\lambda_1-\mu_1\|. \end{eqnarray*} $

Thus

$ \begin{eqnarray} &&\Big(1-\frac{1}{k_1}\sum\limits_{i = 2}^n\sum\limits_{j = 0}^{i-1}L_i M^j\Big)\|f-g\|\\&\leq& \frac{1}{k_1}\|F-G\|+\frac{1}{k_1^2}\Big(L'+L\sum\limits_{i = 2}^nL_i\Big)\|\lambda_1-\mu_1\|+\frac{L}{k_1}\sum\limits_{i = 2}^n\|\lambda_i-\mu_i\|, \ \ \end{eqnarray} $

(3.2)

where, by (3.1), $ 1-\frac{1}{k_1}\sum\limits_{i = 2}^n\sum\limits_{j = 0}^{i-1}L_i M^j>0. $ Hence $ \|f-g\| $ tends to $ 0 $ when $ G(x) $ tends to $ F(x) $ in norm and $ \mu_i(x) $ tends to $ \lambda_i(x) $.

In this section, some examples are provided to illustrate that the assumptions of Theorem 2.4 and 3.1 do not self-contradict.

Example 4.1 Consider the equation

$ \begin{eqnarray} (4+\cos^2(x))f(x)+f(f(x)) = \sin (x), \quad \forall x\in \mathbb{R}. \end{eqnarray} $

(4.1)

It corresponds to the case of $ \lambda_1(x) = 4+\cos^2(x), \lambda_2(x) = 1, F(x) = \sin (x) $. Taking $ k_1 = 4, \ L_1 = 5, \ L = M = L' = M' = M_1 = L_2 = M_2 = 1 $. A simple calculation yields

$ \begin{eqnarray*} L' = 1\leq 3 = (k_1-L_2)L, \end{eqnarray*} $

and

$ \begin{eqnarray*} M' = 1\leq \frac{3}{2} = k_1M-\frac{L'}{k_1}M_1-\Big(L_2M^2+L M_2+\frac{LL_2}{k_1}M_1\Big) \end{eqnarray*} $

then (2.5) is satisfied. Theorem 2.4 gives a continuous periodic solution for Eq. (4.1) in $ {\mathcal{P}}_{2\pi}(1, 1) $. Noting

$ \begin{eqnarray*} \alpha = \frac{1}{k_1}L_2(1+M) = \frac{1}{2}<1, \end{eqnarray*} $

(3.1) is satisfied, hence by Theorem 3.1, we know the continuous periodic solution is the unique one in $ {\mathcal{P}}_{2\pi}(1, 1) $.

Example 4.2 Consider the equation

$ \begin{eqnarray} (4+\cos^2(x))f(x)+\delta f(f(x)) = \sin (x), \quad \forall x\in \mathbb{R}. \end{eqnarray} $

(4.2)

As Example 4.1, $ \lambda_1(x) = 4+\cos^2(x), \lambda_2(x) = \delta, F(x) = \sin (x) $, where $ \delta $ is a parameter. So taking $ k_1 = 4, \ L_1 = 5, \ L_2 = M_2 = \delta, \ L = M = L' = M' = M_1 = 1 $. In order to apply (2.5) in Theorem 2.4, we need

$ \begin{eqnarray} L' = 1\leq 4-\delta = (k_1-L_2)L, \end{eqnarray} $

(4.3)

and

$ \begin{eqnarray} M' = 1\leq \frac{15}{4}-\frac{9}{4}\delta = k_1M-\frac{L'}{k_1}M_1-\Big(L_2M^2+L M_2+\frac{LL_2}{k_1}M_1\Big). \end{eqnarray} $

(4.4)

Then we see $ \delta\leq \frac{11}{9} $ by (4.3) and (4.4).

From (3.1), we have

$ \begin{eqnarray} \alpha = \frac{1}{k_1}L_2(1+M) = \frac{\delta}{2}<1. \end{eqnarray} $

(4.5)

From (4.3)–(4.5), by Theorem 3.1, we know (4.2) has an unique continuous periodic solution in $ {\mathcal{P}}_{2\pi}(1, 1) $ with $ \delta\leq \frac{11}{9} $. This improves estimates of Example 4.1.