Invariant curves for a delay differential equation with a piecewise constant argument

Abstract

In order to understand the dynamics of a second order delay differential equation with a piecewise constant argument, we investigate invariant curves of the derived planar mapping from the equation. All invariant curves are given in this paper.

1 Introduction

The study of differential equations with piecewise constant argument (EPCA) initiated in [1, 2]. These equations represent a hybrid of continuous and discrete dynamical systems and combine the properties of both differential and difference equations, hence, they are of importance in control theory and in certain biomedical models [3]. In this paper the second order delay differential equation with a piecewise constant argument

$$ x''(t)+g\bigl(x\bigl([t]\bigr)\bigr)=0, \quad t\in \mathbb{R}, x\in\mathbb{R}, $$

(1)

where $x''(t)$ denotes the second order derivative of $x(t)$, $[t]$ denotes the greatest integer less than or equal to t, and $g:\mathbb {R}\rightarrow \mathbb{R}$ is a continuous or at least piecewise continuous function, is considered. In 1987, Aftabizadeh et al. discussed the oscillatory and periodic properties of the solutions of (1) in [4]. In 1989, Gyori and Ladas investigated linearized oscillations of the solutions of (1) in [5]. Later, Wiener and Cooke considered oscillations of the solutions of systems of two differential equations with piecewise constant arguments in [6].

The invariant curve [7–11] is another interesting problem in the study of dynamics because it can be used to reduce a system to a 1-dimensional one. The problem of invariant curves is actually a part of the research on invariant manifolds. In 1997, Ng and Zhang studied the nonlinear $C^{1}$ invariant curve of planar mapping $G:\mathbb {R}^{2}\rightarrow \mathbb{R}^{2}$,

$$ G(x, y)=\biggl(y, 2y-x-\frac{1}{2}\bigl(g(y)+g(x)\bigr)\biggr), $$

(2)

derived from (1) in [12] when g is nonlinear and gave the conditions that G has linear invariant curves when g is linear. In 2003, Yang et al. investigated nonlinear $C^{0}$ invariant curves of (2) when g is nonlinear in [13]. So far, nonlinear invariant curves of (2) when g is linear have not been studied. So it is very interesting to look for nonlinear invariant curves of (2) when g is linear. In this paper all the invariant curves of the planar mapping G are given including the linear and nonlinear ones when g is linear.

2 Main results

We discuss invariant curves of the form $y=f(x)$ for the planar mapping (2). Its invariant curves of the form $y=f(x)$ satisfy $f(y)=2y-x-\frac {1}{2}(g(y)+g(x))$, which leads to the iterative functional equation

$$ f\bigl(f(x)\bigr)=2f(x)-x-\frac{1}{2}\bigl(g\bigl(f(x)\bigr)+g(x)\bigr), \quad \forall x\in\mathbb{R}. $$

(3)

Considering linear g and $g(x)=ax+b$, we compute that

$$ f\bigl(f(x)\bigr)-\biggl(2-\frac{a}{2}\biggr)f(x)+\biggl(1+ \frac{a}{2}\biggr)x=-b,\quad \forall x\in\mathbb{R}. $$

(4)

Thus, the invariant curves of planar mapping G with $g(x)=ax+b$ can be obtained by solving functional (4). We mainly discuss the generic cases $a \notin\{-2, 4\}$, but leave the special cases $a=-2$ and $a=4$ to the last part of this section. For generic $a\notin\{-2, 4\}$, (4) with $b=0$ is of the form discussed in [14, 15]. In order to apply the results of [14], we let

$$ r_{1}:=\frac{(4-a)-(a^{2}-16a)^{\frac{1}{2}}}{4},\qquad r_{2}:=\frac{(4-a)+(a^{2}-16a)^{\frac{1}{2}}}{4}, $$

(5)

which are the roots of the characteristic polynomial $P(r):=r^{2} -(2-\frac{a}{2})r +1+\frac{a}{2}$.

From (5) we see that the characteristic roots $r_{1}$, $r_{2}$ of (4) have the following possibilities:

• (C1) $0< r_{1}<1<r_{2}$, if and only if $-2< a<0$.

• (C2) $r_{1}=r_{2}=1$, if and only if $a=0$.

• (C3) $r_{1}<0<r_{2}\neq1$ and $r_{1}\ne-r_{2}$, if and only if $a<-2$.

• (C4) $r_{1}=r_{2}<0$, if and only if $a=16$.

• (C5) $r_{1}< r_{2}<-1$, if and only if $a>16$.

Note that the case $r_{2}>r_{1}>1$ is not listed because the case $r_{2}>r_{1}>1$ implies $\frac{(4-a)-(a^{2}-16a)^{\frac{1}{2}}}{4}>1$, i.e., $-a>(a^{2}-16a)^{\frac{1}{2}}$, which does not hold, and that the case $0< r_{1}<r_{2}<1$ is not listed because $0< r_{1}<r_{2}<1$ implies $0<\frac{(4-a)+(a^{2}-16a)^{\frac{1}{2}}}{4}<1$, i.e., $0< a<4$, which contradicts the requirement that $\Delta=a^{2}-16a\geq0$, and that the case $0< a<16$ is not listed because in this case (4) with $b=0$ has no continuous solutions, neither $r_{1}$ nor $r_{2}$ is real, by [14]. Since we consider $a\notin\{-2, 4\}$, none of the case $r_{1}=0$, the case $r_{2}=0$, and the case $r_{1}=-r_{2}\ne0$ is listed. Corresponding to the above list, we have the following results.

Theorem 2.1

(i) If $-2< a<0$, then a continuous solutions ϕ of (4) with $b=0$ is either of the piecewise linear form that $f(x):= r_{i}x$ for $x>0$, or $:=0$ for $x=0$, or $:=r_{j}x$ for $x<0$, where $i,j=1, 2$, or given by

$$ f(x):=\left \{ \begin{array}{l@{\quad}l} f_{n}(x), &x \in[x_{n}, x_{n+1}), n=0, 1, 2, \ldots, \\ f_{-n}^{-1}(x), &x \in[x_{-n}, x_{-n+1}), n=1, 2, \ldots, \end{array} \right . $$

where $x_{n}=\frac{r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1}+r_{2}x_{0})$, $n \in\mathbb{Z}$, with an arbitrarily chosen $x_{0}\in(-\infty, +\infty)$ and $x_{1} \in[r_{1}x_{0}, r_{2}x_{0}]$, and $f_{n}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{n-1}^{-1}(x)$ for all $x\in[x_{n}, x_{n +1})$, $n=1, 2, \ldots$ , $f_{-n-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}}f^{-1}_{-n}(x)$ for all $x\in[x_{-n}, x_{-n+1})$, $n=1, 2, \ldots$ , and $f_{-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}}f_{0}(x)$, $x\in[x_{0}, x_{1})$, with the arbitrarily chosen functions $f_{0}$ such that $f_{0}(x_{0})=x_{1}$, $f_{0}(x_{1})=x_{2}$, and $r_{1}\leq\frac{f_{0}(x)-f_{0}(y)}{x-y}\leq r_{2}$ for all $x, y \in [x_{0}, x_{1})$. (ii) If $a=0$, then (4) with $b=0$ has a unique continuous solution f and $f(x)=x+\beta$, where $\beta\in\mathbb{R}$ is an arbitrary constant.

Proof

The proof is a simple application of well-known results in [14]. The result (i) is given by Theorem 2 of [14], where the characteristic roots $r_{1}$, $r_{2}$ satisfy $r_{2}>1>r_{1}>0$ as shown in (C1). We can deduce the result (ii) from Theorem 8 of [14], where $r_{1}=r_{2}=1$ as shown in (C2). The proof is completed. □

Theorem 2.2

(i) If $a<-2$, then (4) with $b=0$ only has two continuous solutions f and $f(x)=r_{1}x$ or $r_{2}x$. (ii) If $a=16$, (4) with $b=0$ just has a continuous solution $f(x)=-3x$. (iii) If $a>16$, all continuous solutions f of (4) with $b=0$ are given by

$$ f(x): = \left \{ \begin{array}{l@{\quad}l} f_{2n}(x), &x\in[x_{-2n},x_{-2n+2}),n=0, 1, 2, \ldots, \\ f_{2n+1}(x), &x\in[x_{-2n+3},x_{-2n+1}), n=0, 1, 2, \ldots, \\ 0, &x=0, \\ f_{-2n}^{-1}(x), & x\in[x_{2n}, x_{2n+2}), n=1, 2, \ldots, \\ f_{-2n+1}^{-1}(x), & x\in[x_{2n+3}, x_{2n+1}), n=1, 2, \ldots, \end{array} \right . $$

where the sequence $\{x_{n}\}$ is defined by $x_{n}=\frac {r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1}+r_{2}x_{0})$, $n\in\mathbb{Z}$, with an arbitrarily chosen $x_{0}\in(0, +\infty)$ and $x_{1} \in[r_{1}x_{0}, r_{2}x_{0}]$, and $f_{2n-1}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n-2}^{-1}(x)$, $x\in [x_{-2n+5},x_{-2n+3})$, $n=1, 2, \ldots$ , $f_{2n}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n-1}^{-1}(x)$, $x\in [x_{-2n},x_{-2n+2})$, $n=1, 2, \ldots$ , $f_{-2n}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac{1}{r_{1}r_{2}} f^{-1}_{-2n+1}(x)$, $x \in[x_{2n+3}, x_{2n+1})$, $n=1, 2,\ldots$ , $f_{-2n-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}} f^{-1}_{-2n}(x)$, $x \in[x_{2n}, x_{2n+2})$, $n=1, 2,\ldots$ , and $f_{-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac{1}{r_{1}r_{2}} f_{0}(x)$, $x\in[x_{0}, x_{2})$, with an arbitrarily chosen continuous function $f_{0}$ on $[x_{0},x_{2})$ such that $f_{0}(x_{0})=x_{1}$, $f_{0}(x_{2})=x_{3}$, and $r_{1}\leq\frac {f_{0}(x)-f_{0}(y)}{x-y}\leq r_{2}$, $\forall x, y \in[x_{0}, x_{2})$.

Proof

Firstly, we consider (i). By Theorem 5 in [14], (4) with $b=0$ only has two continuous solutions f and $f(x)=r_{1}x$ or $r_{2}x$, where the characteristic roots $r_{1}$, $r_{2}$ satisfy $r_{1}<0<r_{2}\neq1$ and $r_{1}\ne-r_{2}$ as shown in (C3). Next, we consider (ii). By Theorem 6 in [14], (4) with $b=0$ just has a continuous solution $f(x)=-3x$, where the characteristic roots $r_{1}$, $r_{2}$ satisfy $r_{1}=r_{2}=-3$ as shown in (C4). Finally, we consider (iii). In order to piecewise construct all solutions of (4) with $b=0$ we need a partition for the interval $(-\infty, \infty)$. For this purpose we consider a homogeneous linear difference equation

$$ x_{n+2}-\biggl(2-\frac{a}{2}\biggr)x_{n + 1}+\biggl(1+ \frac{a}{2}\biggr)x_{n}=0, $$

(6)

which has the same coefficients as (4) with $b=0$ correspondingly. Its characteristic equation is

$$ r^{2}-\biggl(2-\frac{a}{2}\biggr)r+1+\frac{a}{2}=0, $$

(7)

which has two characteristic roots $r_{1}$ and $r_{2}$ satisfying $r_{1}< r_{2}<-1$ as shown in (C5). Thus, (4) and (6) can be, respectively, rewritten as

$$\begin{aligned}& f\bigl(f(x)\bigr)-(r_{1}+r_{2})f(x)+r_{1}r_{2}x=0, \end{aligned}$$

(8)

$$\begin{aligned}& x_{n+2}-(r_{1}+r_{2})x_{n+1}+r_{1}r_{2}x_{n}=0, \quad n = 0, 1, 2,\ldots. \end{aligned}$$

(9)

If f is a solution of (4) with $b=0$, we easily see that f is invertible. In fact, if $f(x_{1})=f(x_{2})$, then $f(f(x_{1}))=f(f(x_{2}))$. Thus, $x_{1} = x_{2}$ by (4) because $a\neq-2$, which implies that f is one to one. Next we only need to show that $f(x)\rightarrow-\infty$ as $x\rightarrow+\infty$ and $f(x)\rightarrow+\infty$ as $x \rightarrow-\infty$ because $f(x)\rightarrow\pm\infty$ as $x \rightarrow\pm\infty$, then the left-hand side of (4) with $b=0$ tends to ±∞ by $a>16$, but the right-hand side is equal to 0. Otherwise, $f(x)$ has a finite limit as $x \rightarrow\infty$, then $f(f(x))-(2-\frac{a}{2})f(x)$ converges to a finite limit by the continuity of f on the whole of ℝ, but $(1+\frac{a}{2})x$ does not, which contradicts the requirement that $f(f(x))-(2-\frac{a}{2})f(x)=-(1+\frac{a}{2})x$. Thus, we rewrite (4) in the following equivalent form:

$$ f^{-1}\bigl(f^{-1}(x)\bigr)-\biggl(2-\frac{a}{2} \biggr)f^{-1}(x)+\frac{a}{2}x=0, $$

(10)

which is called the dual equation to (4) with $b=0$. Solving the homogeneous linear difference (9) with arbitrarily chosen real initial values $x_{0}$ and $x_{1}$, we obtain

$$ x_{n}=\frac{r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1} +r_{2}x_{0}),\quad n\in \mathbb{Z}. $$

(11)

Let $x_{0}=x$ and $x_{n+1}=f(x_{n})$ in (11), we have

$$ f^{n}(x) = \frac{r_{2}^{n}}{r_{2}-r_{1}}\bigl(f(x)-r_{1}x\bigr)+ \frac{r_{1}^{n}}{r_{2}-r_{1}}\bigl(-f(x)+r_{2}x\bigr),\quad n \in\mathbb{Z}. $$

Furthermore, we can obtain

$$\begin{aligned}& \Delta f^{n}(x, y) = \frac {r_{2}^{n}}{r_{2}-r_{1}}\bigl(\Delta f(x, y)-r_{1}\bigr)+ \frac{r_{1}^{n}}{r_{2}-r_{1}}\bigl(-\Delta f(x, y)+r_{2} \bigr), \end{aligned}$$

(12)

$$\begin{aligned}& f^{n+1}(x)-f^{n}(x)= r_{2}^{n} \frac {r_{2}-1}{r_{2}-r_{1}}\bigl(f(x)-r_{1}x\bigr)+ r_{1}^{n} \frac{r_{1}-1}{r_{2}-r_{1}}\bigl(-f(x)+r_{2}x\bigr), \end{aligned}$$

(13)

where $\Delta f^{n}(x, y)= \frac{f^{n}(x)-f^{n}(y)}{x - y}$ for any $x\neq y$ and $n\in\mathbb{Z}$. From (12) we can see that

$$\begin{aligned}& \lim_{n\rightarrow+\infty}\frac{\Delta f^{n}(x, y)}{r_{2}^{n}}= \frac{(\Delta f(x, y) - r_{1})}{r_{2} - r_{1}}, \\& \lim_{n\rightarrow-\infty}\frac{\Delta f^{n}(x, y)}{r_{1}^{n}}= \frac{(-\Delta f(x, y) + r_{2})}{r_{2} - r_{1}}. \end{aligned}$$

Since f is strictly monotonic, $\Delta f^{n}(x, y)>0$ for even n, which implies $\Delta f(x, y)-r_{1}\geq0$ and $-\Delta f(x, y)+r_{2}\geq0$, that is,

$$ r_{1}\leq\Delta f(x, y)\leq r_{2}. $$

(14)

Moreover, we can see that $f(0)=0$ from (13). In what follows, we arbitrarily choose $x_{0}\in(0, +\infty)$ and $x_{1}\in[r_{1}x_{0}, r_{2}x_{0}]$ and define a sequence $\{x_{n}\}$, $n\in\mathbb{Z}$, by (11). The sequences $\{x_{2n}\}$, $\{x_{2n+1}\}$, $\{x_{-2n}\}$ and $\{x_{-2n+1}\}$, where $n=0, 1, 2,\ldots$ , are strictly monotone such that $x_{2n}\rightarrow+\infty$, $x_{2n+1}\rightarrow-\infty$, $x_{-2n} \rightarrow0$, and $x_{-2n+1} \rightarrow0$ as $n \rightarrow\infty$. Thus, the sequence $\{x_{n}\}$, $n\in\mathbb{Z}$, is a partition of the interval $(-\infty, \infty)$. Next we arbitrarily choose a continuous function defined in the interval $[x_{0}, x_{2})$, satisfying $f_{0}(x_{0})=x_{1}$, $f_{0}(x_{2})=x_{3}$, and condition (14). We can recursively define the homeomorphisms $f_{2n-1}: [x_{-2n+5}, x_{-2n+3})\rightarrow [x_{-2n}, x_{-2n+2})$, $n=1, 2,\ldots$ , and $f_{2n}: [x_{-2n}, x_{-2n+2})\rightarrow [x_{-2n+3}, x_{-2n+1})$, $n=1, 2,\ldots$ , such that

$$\begin{aligned}& f_{2n-1}(x_{-2n+5}) = x_{-2n+2},\qquad f_{2n-1}(x_{-2n+3})=x_{-2n}, \end{aligned}$$

(15)

$$\begin{aligned}& f_{2n}(x_{-2n}) = x_{-2n+1}, \qquad f_{2n}(x_{-2n+2})=x_{-2n+3}, \end{aligned}$$

(16)

$$\begin{aligned}& r_{1} \leq \Delta f_{2n-1}(x, y) \leq r_{2},\quad \forall x, y \in[x_{-2n+5}, x_{-2n+3}), \end{aligned}$$

(17)

$$\begin{aligned}& r_{1} \leq \Delta f_{2n}(x, y) \leq r_{2}, \quad \forall x, y \in[x_{-2n}, x_{-2n+2}). \end{aligned}$$

(18)

In fact, for $f_{2n}$ defined satisfying (16) and (18), we let

$$ f_{2n+1}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n}^{-1}(x), \quad \forall x \in [x_{-2n+3}, x_{-2n+1}). $$

Obviously, $f_{2n+1}(x_{-2n+3})=x_{-2n}$ and $f_{2n+1}(x_{-2n+1})=x_{-2n-2}$. Making use of (18), we have $\frac{1}{r_{2}} \leq\frac{f^{-1}_{2n}(x) - f^{-1}_{2n}(y)}{x-y} \leq\frac{1}{r_{1}}$ for $x, y \in[x_{-2n}, x_{-2n+2})$. It is easy to deduce that

$$ r_{1} \leq\Delta f_{2n+1}(x, y) \leq r_{2},\quad \forall x, y \in [x_{-2n+3}, x_{-2n+1}). $$

Furthermore, we again let

$$ f_{2n+2}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n+1}^{-1}(x), \quad \forall x \in [x_{-2n-2}, x_{-2n}). $$

By the same argument we can see that

$$\begin{aligned}& f_{2n+2}(x_{-2n-2})=x_{-2n-1}, \qquad f_{2n+2}(x_{-2n})=x_{-2n+1}, \end{aligned}$$

(19)

$$\begin{aligned}& r_{1}\leq\Delta f_{2n+2}(x, y)\leq r_{2},\quad \forall x, y \in [x_{-2n-2}, x_{-2n}). \end{aligned}$$

(20)

By induction both $f_{2n-1}$ and $f_{2n}$ are well defined. Similarly, we can also recursively define the homeomorphisms $f_{-2n+1}: [x_{2n-2}, x_{2n}) \rightarrow [x_{2n+3},x_{2n+1})$, $n = 1, 2, \ldots$ , and $f_{-2n}: [x_{2n+3}, x_{2n+1}) \rightarrow [x_{2n},x_{2n+2})$, $n = 1, 2, \ldots$ . By the properties of the dual (10) we can obtain

$$\begin{aligned}& f_{-2n+1}(x_{2n-2}) = x_{2n+1}, \qquad f_{-2n+1}(x_{2n})=x_{2n+3}, \\& f_{-2n}(x_{2n+3}) = x_{2n+2}, \qquad f_{-2n}(x_{2n+1})=x_{2n}, \\& \frac{1}{r_{2}} \leq \Delta f_{-2n+1}(x, y) \leq \frac{1}{r_{1}}, \quad \forall x, y \in[x_{2n-2}, x_{2n}), \\& \frac{1}{r_{2}} \leq \Delta f_{-2n}(x, y) \leq\frac {1}{r_{1}}, \quad \forall x, y \in[x_{2n+3}, x_{2n+1}). \end{aligned}$$

Therefore,

$$\begin{aligned}& f^{-1}_{-2n+1}(x_{2n+1}) = x_{2n-2},\qquad f^{-1}_{-2n+1}(x_{2n+3})=x_{2n}, \\& f^{-1}_{-2n}( x_{2n+2}) = x_{2n+3}, \qquad f^{-1}_{-2n}(x_{2n})=x_{2n+1}, \\& r_{1} \leq \Delta f^{-1}_{-2n+1}(x, y) \leq r_{2}, \quad \forall x, y \in[x_{2n+3}, x_{2n+1}), \\& r_{1} \leq \Delta f^{-1}_{-2n}(x, y) \leq r_{2}, \quad \forall x, y \in[x_{2n}, x_{2n+2}). \end{aligned}$$

Thus, we can define

$$ f(x): = \left \{ \begin{array}{l@{\quad}l} f_{2n}(x), &x\in[x_{-2n},x_{-2n+2}), n=0, 1, 2, \ldots, \\ f_{2n+1}(x), &x\in[x_{-2n+3},x_{-2n+1}), n=0, 1, 2, \ldots, \\ 0, &x=0, \\ f_{-2n}^{-1}(x), & x\in[x_{2n}, x_{2n+2}), n=1, 2, \ldots, \\ f_{-2n+1}^{-1}(x), & x\in[x_{2n+3}, x_{2n+1}), n=1, 2, \ldots. \end{array} \right . $$

f is continuous on ℝ because $f_{2n}(x_{-2n})=x_{-2n+1}=f_{2n+2}(x_{-2n})$, $f_{2n+1}(x_{-2n+1})=x_{-2n-2}=f_{2n+3}(x_{-2n+1})$, where $n=0, 1, 2, \ldots$ , $f_{1}^{-1}(x_{0}) = f_{-1}(x_{0})$, $f_{-2n}^{-1}(x_{2n+2})=x_{2n+3}=f_{-2n-2}^{-1}(x_{2n+2})$, and $f_{-2n+1}^{-1}(x_{2n+3})=x_{2n+2}=f_{-2n-1}^{-1}(x_{2n+3})$, where $n=1, 2, 3, \ldots$ . we can easily check that f defined in Theorem 2.2 satisfies (4) with $b=0$ in ℝ. In fact, if $x\in[x_{-2n},x_{-2n+2})$, $n=0, 1, 2, \ldots$ , $f^{2}(x)=f_{2n+1}(f_{2n}(x))=(r_{1}+r_{2})f_{2n}(x)-r_{1}r_{2}x=(r_{1}+r_{2})f(x)-r_{1}r_{2}x$, i.e., $f^{2}(x)-(r_{1}+r_{2})f(x)-r_{1}r_{2}x=0$. Similarly, we can also check that f satisfies (4) with $b=0$ for $x\in[x_{-2n+3},x_{-2n+1})$, $x\in[x_{2n+3}, x_{2n+1})$, $x\in[x_{2n}, x_{2n+2})$ and $x=0$, where $n=1, 2, 3, \ldots$ . The proof is completed. □

Remark

In the case that $b\ne0$, as indicated in [16] for (2) therein, (4) can be reduced equivalently to the equation

$$ \tilde{f}\bigl(\tilde{f}(x)\bigr)-\biggl(2-\frac{a}{2}\biggr)\tilde {f}(x)+\biggl(1+\frac{a}{2}\biggr)x=0, $$

(21)

the same type of equation as the one considered in Theorems 2.1 and 2.2 with vanishing b, by the replacement $\tilde{f}(x)=f(x+\xi)-\xi$, where $\xi=\frac{-b}{(1-r_{1})(1-r_{2})}$, if its characteristic roots $r_{1}$, $r_{2}$ are both real but neither of them is equal to 1. In this case solutions can be found from Theorems 2.1 and 2.2. So (4) with $b\neq0$ can be reduced to (21) except for the case $a=0$. For the case of $a=0$ and $b\neq0$, (4) has no real continuous solutions. In fact, by induction and (4) we can obtain $f^{n}(x)=nf(x)-(n-1)x-\frac{n(n+1)}{2}b$, $n\in\mathbb{Z}$. Furthermore, we have $f^{n+1}(x)-f^{n}(x)=f(x)-x-(n+1)b$, $n\in\mathbb{Z}$. For an arbitrary $x\in\mathbb{R}$, $f^{n+1}(x)-f^{n}(x)$ has the same sign when n takes the values N and −N, where N is a large positive integer, because f is strictly monotonic. But $f(x)-x-(n+1)b$ has not, which contradicts the requirement $f^{n+1}(x)-f^{n}(x)=f(x)-x-(n+1)b$, $n\in\mathbb{Z}$.

In what follows, we consider the case that either $a=-2$ or $a=4$, which is not generic.

For $a=-2$, (4) is of the form $f^{2}(x)-3f(x)=-b$, from which we get with the replacement $y=f(x)$: $f(x)=3x-b$.

For $a=4$, (4) is of the form

$$ f^{2}(x)=-3x-b, $$

(22)

which is the problem of iterative roots of the linear function $F(x):=-3x-b$. By the theory of iterative roots, as shown in [8], we know (22) has no real continuous solutions.