\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 180, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/180\hfil Analytic solutions]
{Analytic solutions for iterative functional differential equations}

\author[P. Zhang\hfil EJDE-2012/180\hfilneg]
{Pingping Zhang}  % in alphabetical order

\address{Pingping Zhang \newline
 Department of Mathematics, Sichuan University\\
 Chengdu, Sichuan 610064, China}
\email{zhangpingpingmath@163.com}

\thanks{Submitted June 7, 2012. Published October 16, 2012.}
\thanks{Supported by grants J12L59, ZR2012AQ028, ZR2011AL001 and BZXYL1005.}
\subjclass[2000]{34K05, 39B12, 39B32}
\keywords{Iterative differential equation; existence; analytic
solution; \hfill\break\indent polynomial solution}

\begin{abstract}
 Because of its technical difficulties the existence of analytic
 solutions to the iterative differential equation
 $x'(z)=x(az+bx(z)+c x'(z))$ is a source of open problems.
 In this article we obtain analytic solutions, using Schauder's fixed
 point theorem. Also we present a unique solution which is a nonconstant
 polynomial in the complex field.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

The study of differential equations with variable delays
\begin{equation}
x'(t)=f(t,x(t-\tau(t))), \label{f}
\end{equation}
began in 60's. Then both theory and applications have been considered;
see the monographs \cite{RB,JH,HS}. 
In recent years equations with complicated deviating arguments 
$\tau(x(t))$ have attracted the attention of mathematicians, since these
equations often appear in applied sciences. For example, Cooke
established the following equation, related to a genetic
phenomenon \cite{KC},
$$
x'(t)+ax(t-h(t,x(t)))=F(t),\quad t\in\mathbb{R}.
$$
Periodic solutions in periodic systems for the population model
$$
x'(t)=f(t,x(t-\tau(t,x(t)))),\quad t\in\mathbb{R}
$$
were investigated in \cite{YL}, where $f$ and $\tau$ have the same
period. 
On the other hand, the analytic solutions of the equation $x'(z)=x(z-\tau(z))$, 
such as $x'(z)=x(az+bx(z))$ and $x'(z)=x(az+bx'(z))$, even the
more complicated equation $\alpha z+\beta x'(z)=x(az+bx''(z))$,
have been  investigated in the complex field; 
see for example \cite{EE,ZJ,LL}, \cite{SC1}-\cite{HS}, \cite{WS}-\cite{PZ}.
In these studies, the Schr\"{o}der transformation is an important tool 
which can reduce the equation involving iteration of unknown functions
to an auxiliary differential equation without the iterative operation.
To the best of our knowledge, there are only a few results on analytic
solutions of the iterative differential equation that includes state and
state derivative simultaneously in $\tau(z)$. 
The main reason is that this equation being
rewritten as another one which does not involve $\tau(z)$ is 
difficult to analyze; there the Schr\"{o}der transformation, a very useful
method in the eliminating iteration of unknown function, does not
work.

In this article we study analytic solutions of the equation
\begin{equation}
x'(z)=x(az+bx(z)+c x'(z)),\quad z\in\mathbb{C} \label{main}
\end{equation}
by using Schauder's fixed point theorem, in which $a,b,c$ are
complex constants. Moreover, we find the unique nonconstant
polynomial solution in complex field.

\section{Analytic solution}

In this section we aim to find analytic solutions of equation \eqref{main}
in a neighborhood of the origin by using Schauder's fixed point
theorem. In addition, a non-existence result on analytic solutions
is also given, which is generalized to the equations involving higher
order derivatives.

\begin{theorem} \label{thm1}
Suppose that $|a|<1$ and $(1-|a|)|x_0|+|a|+|b|+|c|<1$, then
 \eqref{main} has an analytic solution of the form
\begin{equation}
x(z)=x_0+\sum_{n=1}^\infty a_n z^n \label{analytic}
\end{equation}
in a neighborhood of the origin. 
\end{theorem}


\begin{proof}
 It suffices to find the solution of an integral equation
\begin{equation}
x(z)=x_0 + \int_{0}^{z} x(as+bx(s)+cx'(s)) ds. \label{integral}
\end{equation}
Since $(1-|a|)|x_0|+|a|+|b|+|c|<1$, there exists positive value
$h$ satisfying
\begin{equation}
\frac{(|b|+|c|)|x_0|}{1-|a|-|b|-|c|}< h< 1-|x_0|. \label{23}
\end{equation}
 For constants $M$ and $L$ satisfy
\begin{equation}
    \frac{|x_0|}{1-h}\leq M\leq 1
\qquad {\rm and} \qquad
    L>\frac{1+M|b|}{1-M|c|},\label{24}
\end{equation}
  we define functional subspace
\begin{align*}
X_h =\Big\{&x \in C^1(\bar{B}_h(0),\mathbb{C}):  x(0)=x_{0},\;
    |x(z_1)-x(z_2)| \le M|z_1 -z_2|,\\
   & \|x'\|_{C^0}\leq |x_0|+Mh, \|{x_1}'-{x_2}'\|_{C^0}\leq
    L\|x_1-x_2\|_{C^0},\\
   & \forall z_1,z_2 \in \bar{B}_h(0),\forall x_1,x_2 \in
  C^1(\bar{B}_h(0),\mathbb{C})\Big\},
\end{align*}
where $\bar{B}_h(0)$ is the closure of open ball $B_h(0)$
centered at $0$ with radius $h$. Clearly, $X_h$ is a convex and
closed subset of $C^1(\bar{B}_h(0),\mathbb{C})$, a Banach space
with the standard norm $\|x\|_C^0=\|x\|_{C^0} +\|x'\|_{C^0}$,
where $\|x\|_{C^0} =\max_{z\in\bar{B}_h(0)} |x(z)|$ and
$\|x'\|_{C^0} =\max_{z\in\bar{B}_h(0)} |x'(z)|$. Moreover,
we have $|x_n(z)|\leq |x_0|+Mh$ for any $x_n \in X_h$. So $\{x_n
(z)\}$ is compact in $\bar{B}_h(0)$ by Ascoli-Arzela Lemma.
Therefore, $X_h$ is a compact and convex subset of the Banach space
$C^1(\bar{B}_h(0),\mathbb{C})$.

Now we define the operator $\mathcal{J}$ on $X_h$ by
\begin{equation}
\mathcal{J} x(z) := x_0 + \int_{0}^{z} x(as+bx(s)+cx'(s)) ds.
\label{map}
\end{equation}
Then a fixed point of map $\mathcal{J}$ is a solution of the
integral equation \eqref{integral}. From \eqref{23}, we have
\begin{align*}
|az+bx(z)+cx'(z)|
&\leq |a|h+|b|\cdot(|x_0|+Mh)+|c|\cdot(|x_0|+Mh)\\
&\leq |a|h+|b|\cdot(|x_0|+h)+|c|\cdot(|x_0|+h) \\
&\leq  h,\quad \forall z\in\bar{B}_h(0);
\end{align*}
i.e., $x(az+bx(z)+cx'(z))$ is well-defined for $x \in X_h$.

To verify $\mathcal{J} (X_h) \subset X_h$, we note that for
arbitrary $x \in X_h$,
\[
\mathcal{J} x(0) =    x_{0}+\int_{0}^{0} x(as+bx(s)+cx'(s))ds=x_0,
\]
and from \eqref{24},
\begin{align*} 
| \mathcal{J} x(z_2)-\mathcal{J} x(z_1)| 
&\leq  | \int_{z_1}^{z_2} x(as+bx(s)+cx'(s))ds |\\
&\leq  (|x_0|+Mh) |z_2-z_1|\\
&\leq  M |z_2-z_1|.
\end{align*}
We also have $ |(\mathcal{J} x)'(z)|\leq |x_0|+Mh$; i.e.,
$\|(\mathcal{J} x)'\|_{C^0}\leq |x_0|+Mh$.
Moreover, for any $x_1, x_2 \in X_h$,
\begin{equation}
\begin{split}
&|(\mathcal{J} x_1)'(z)-(\mathcal{J} x_2)'(z)| 
\\
&=  |x_1 (az+bx_1 (z)+c{x_1}'(z))-x_2 (az+bx_2 (z)+c{x_2}'(z))| 
\\
&\leq  |x_1 (az+bx_1 (z)+c{x_1}'(z))-x_2 (az+bx_1 (z)+c{x_1}'(z))| 
\\
&\quad + |x_2 (az+bx_1 (z)+c{x_1}'(z))-x_2 (az+bx_2 (z)+c{x_2}'(z))|
\\
&\leq \|x_1 -x_2 \|_{C^0}+
        M(|b|\cdot\|x_1 -x_2 \|_{C^0}+|c|\cdot\|{x_1}' -{x_2}' \|_{C^0})
\\
&\leq  \|x_1 -x_2 \|_{C^0}+
        M(|b|+|c|L)\cdot\|x_1 -x_2 \|_{C^0}
\\
&=(1+M(|b|+|c|L))\|x_1-x_2\|_{C^0}
\\
&\leq     L\|x_1-x_2\|_{C^0}.
\end{split} \label{J}
\end{equation}
These relations imply that $\mathcal{J} x \in X_h$; i.e.,
$\mathcal{J}$ is a self-mapping.

Furthermore, from \eqref{J} we also see that
\begin{align*}
    &|\mathcal{J} x_1(z)-\mathcal{J} x_2(z)|
\\
&\leq   | \int_{0}^z  |x_1 (as+bx_1 (s)+c{x_1}'(s))-x_2 (as+bx_2 (s)+c{x_2}'(s))|ds
\\
&\leq     | \int_{0}^z  L\|x_1 -x_2 \|_{C^0}ds
\\
&\leq   hL\|x_1-x_2\|_{C^0},
\end{align*}
implying that $\mathcal{J}$ is continuous.

Therefore we can conclude from Schauder's Fixed Point Theorem that
$\mathcal{J}$ has a fixed point $x \in X_h$; i.e., 
$\mathcal{J} x = x$, which gives a solution for equation \eqref{main} 
associated with $x(0)=x_0$ on the domain $|z|\leq h$, naturally, it is an analytic
solution of equation \eqref{main} with the form
$x(z)=x_0+\sum_{n=1}^\infty a_n z^n$ in complex field $|z|\leq h$.
This completes the proof. 
\end{proof}

\begin{example} \label{examp1}\rm
 Consider equation
\begin{equation}
x'(z)=x(az+bx(z)+c x'(z)), \label{27}
\end{equation}
where $a=\sqrt{2}/{4}+i\sqrt{2}/{4}$,
$b=c=\sqrt{2}/{16}+i\sqrt{2}/{16}$.

Since $|a|=1/2$, $|b|=|c|=1/8$, then for arbitrary $|x_0|<1/2$,
equation \eqref{27} has an analytic solution with the form as
$x(z)=x_0+\sum_{n=1}^\infty a_n z^n$ in a neighborhood of the
origin.
\end{example}

Next, we obtain a result on non-existence of analytic
solution for equation \eqref{main}, which can be generalized to the
equations with high order derivatives.

\begin{theorem} \label{thm2}
Equation \eqref{main} has no nonzero analytic solution of the form
\begin{equation}
x(z)=\sum_{n=2}^\infty a_n z^n. \label{main2}
\end{equation}
\end{theorem}

\begin{proof} 
Substituting \eqref{main2} into \eqref{main}, we have
\begin{equation}
\sum_{n=1}^\infty (n+1)a_{n+1} z^n =  \sum_{n=2}^\infty a_{n} (q (z))^n,
\label{3a}
\end{equation}
where 
$$
q(z)=(a+2a_2 c)z+\sum_{n=2}^\infty (a_n b+(n+1)a_{n+1} c)z^n.
$$ 
Comparing coefficient of $z$ in \eqref{3a}, we have 
$2a_2=0$; that is, $ a_2 =0$. In general we obtain
$a_n =0$, $n=3,4,\dots$,
as coefficient of $z^n$ in \eqref{3a}. Therefore, equation  \eqref{main}
has only the zero analytic solution of the form \eqref{main2}.
\end{proof}

To extend the above result, we have the
non-existence of nonzero analytic solutions to the more general
equations.


\begin{theorem} \label{thm3}
The equation
\begin{equation}
x'(z)=x(az+bx(z)+c x^{(m)}(z)),\quad  m=2,3,\dots, \label{4}
\end{equation}
has no nonzero analytic solution of the form
\[
x(z)=\sum_{n=m+1}^\infty a_n z^n.
\]
\end{theorem}

The proof of the above theorem is similar to Theorem \ref{thm2} and is omitted.

\section{Polynomial solution}

In this section we find the unique nonconstant complex polynomial
solution to \eqref{main}. Furthermore, the iterative differential
equations including high order derivatives have also unique
nonconstant complex polynomial solution which is independent of the
order of derivative.

\begin{theorem} \label{thm4}
Suppose that $a\neq 0,1$ and $b\neq0$, then  \eqref{main} has a
unique nonconstant complex polynomial solution
\begin{equation}
x(z)=\frac{a^2 c+ab}{(a-1)b^2}-\frac{a}{b}z. \label{main1}
\end{equation}
\end{theorem}


\begin{proof} Let
\begin{equation}
x(z)=\sum_{i=0}^n a_i z^i. \label{abc}
\end{equation}
Substituting \eqref{abc} into \eqref{main}, we have
\begin{equation}
\begin{split}
&a_1 +2a_2 z+ 3a_3 z^2 +\dots +na_n z^{n-1} \\
&= a_0 +a_1 p_n (z)+a_2 (p_n (z))^2+\dots +a_n (p_n (z))^n,
\end{split}\label{a}
\end{equation}
where
\begin{equation}
\begin{split}
p_n (z)&= (a_0 b+a_1 c)+(a+a_1 b+2a_2 c)z+(a_2 b+3a_3 c)z^2 \\
&\quad +(a_3 b+4a_4 c)z^3+\dots +(a_{n-1} b+na_n c)z^{n-1}+a_n bz^n.
\end{split}\label{b}
\end{equation}
Comparing coefficient of $z^{n^2}$, we have
$$
a_n (a_n b)^n=0;
$$
that is, $a_n =0$. Then \eqref{a} and \eqref{b} are simplified into
\begin{equation}
\begin{split}
&a_1 +2a_2 z+ 3a_3 z^2 +\dots +(n-1)a_{n-1} z^{n-2} \\
&= a_0 +a_1 p_{n-1} (z)+a_2 (p_{n-1} (z))^2+\dots +a_{n-1} (p_{n-1} (z))^{n-1},
\end{split}\label{aa}
\end{equation}
and
\begin{equation}
\begin{split}
p_{n-1} (z)
&= (a_0 b+a_1 c)+(a+a_1 b+2a_2 c)z+(a_2 b+3a_3 c)z^2
+(a_3 b+4a_4 c)z^3 \\
&\quad +\dots +(a_{n-2} b+(n-1)a_{n-1} c)z^{n-2}+a_{n-1} bz^{n-1},
\end{split}\label{bb}
\end{equation}
respectively. Comparing coefficient of $z^{(n-1)^2}$, we obtain
$$
a_{n-1} (a_{n-1} b)^{n-1}=0;
$$
that is, $a_{n-1}=0$. Thus, repeating the same procedure as before,
we can obtain the simplest forms of \eqref{a} and \eqref{b}
respectively as
\begin{gather*}
a_1 = a_0 +a_1 p_{1} (z),
\\
p_{1} (z)=(a_0 b+a_1 c)+(a+a_1 b)z.
\end{gather*}
Namely,
\begin{equation}
a_1 = a_0 +a_1 ((a_0 b+a_1 c)+(a+a_1 b)z). \label{aaaa}
\end{equation}
From which, we obtain a nonconstant complex polynomial solution
\eqref{main1}, which also is unique since it is determined only by
constants $a,b,c$ in equation \eqref{main}.
\end{proof}

\begin{example} \label{examp2} \rm
Consider the equation
\begin{equation}
x'(z)=x(az+bx(z)+cx'(z)), \label{38}
\end{equation}
where $a=1+i$, $b=1+2i$, $c=3+i$. Then  \eqref{38} has a unique
complex polynomial solution $x(z)=(3-9i)/(4+3i)-(1+i)z/(1+2i)$.
\end{example}

For the iterative differential equations involving high-order
derivatives, we have the following result.

\begin{theorem} \label{thm5}
Suppose that $a\neq 0,1$ and $b\neq0$, then the equation
\begin{equation}
x'(z)=x(az+bx(z)+c x^{(m)}(z)),\ \ m=2,3,\dots , \label{2}
\end{equation}
has a unique nonconstant complex polynomial solution
\[
x(z)=\frac{a}{(a-1)b}-\frac{a}{b}z.
\]
\end{theorem}
 
The proof of the above theorem is similar to Theorem \ref{thm4},
and is omitted. 

\begin{example} \label{examp3}\rm
 Consider the equation
\begin{equation}
x'(z)=x(az+bx(z)+cx^{(m)}(z)) \label{39}
\end{equation}
for arbitrary $m\in\{2,3,\dots \}$, where $a=2+i$,
$b=1+i$, $c=3+3i$.
Then \eqref{39} has a unique complex polynomial solution
$x(z)=(1-2i)/2-(2+i)z/(1+i)$.
\end{example}

\subsection*{Acknowledgments}
The author is grateful to the anonymous referees for their detailed 
comments and remarks which helped to clarify this article.

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\end{document}