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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 51, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/51\hfil Analytic solutions]
{Analytic solutions of a first order functional differential
equation with a state derivative dependent delay}

\author[P. Zhang \hfil EJDE-2009/51\hfilneg]
{Pingping Zhang}

\address{Pingping Zhang \newline
Department of Mathematics, Binzhou University,
  Shandong 256603, China}
\email{zhangpingpingmath@163.com}


\thanks{Submitted September 25, 2008. Published April 13, 2009.}
\subjclass[2000]{34K05, 34A25, 39B32} 
\keywords{Functional differential equation; analytic solution}

\begin{abstract}
  This article  concerns the first-order functional  differential equation
  $$
  x'(z)=x(p(z)+bx'(z))
  $$
  with the distinctive feature that the argument
  of the unknown function depends on the state derivative.
  An  existence theorem is established for analytic solutions and
  systematic methods for deriving explicit solutions are also given.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

 In the past few years there has been a growing
interest in studying  functional differential equations with
state dependent delay. We refer the readers to
Eder \cite{e1}, Elbert \cite{e2}, Feckan \cite{f1}, Stanek \cite{s2},
Wang \cite{w1}.
Qiu \cite{q1} and Stanek \cite{s2}  considered the equation
$$
x'(z)=x\big(p(z)+bx(z)\big),
$$
 and establish sufficient conditions for the existence of analytic solutions.
In this paper, we are concerned with analytic solutions of the
first-order functional differential equation
\begin{equation}
x'(z)=x\big(p(z)+bx'(z)\big) ,\label{e1}
\end{equation}
where $b$ is a nonzero complex number, and $p(z)$ is the given
complex function of a complex variable. A distinctive feature of
\eqref{e1} is that the argument of the unknown function depends on the state
derivative. To construct analytic solution of \eqref{e1} in a
systematic manner, we first let
\begin{equation}
y(z)=p(z)+bx'(z).\label{e2}
\end{equation}
Then for any number $z_{0}$, we have
\begin{equation}
x(z)=x(z_{0})+\frac{1}{b}\int_{z_{0}}^{z}(y(s)-p(s))ds  \label{e3}
\end{equation}
    and so
$$
x(y(z))=x(z_{0})+\frac{1}{b} \int_{z_{0}}^{y(z)}(y(s)-p(s))ds.
$$
Therefore, in view of \eqref{e1} and
      $x'(z)=\frac{1}{b}(y(z)-p(z))$, we have
\begin{equation}
\frac{1}{b}(y(z)-p(z))=x(z_{0})+\frac{1}{b}
      \int_{z_{0}}^{y(z)}(y(s)-p(s))ds.   \label{e4}
\end{equation}
 Furthermore, differentiating both sides of \eqref{e4} with respect
to $z$, we obtain
\begin{equation}
y'(z)-p'(z)=(y(y(z))-p(y(z)))y'(z). \label{e5}
\end{equation}

 To find analytic solutions of \eqref{e5}, we first seek an analytic
 solution $g(z)$ of the auxiliary equation
\begin{equation}
\alpha g'(\alpha z)-p'(g(z))g'(z)=
 \alpha[g(\alpha^{2}z)-p(g(\alpha z))]g'(\alpha z),   \label{e6}
\end{equation}
 satisfying  the  initial value conditions
\begin{equation}
g(0)=0, \quad g'(0)=\eta\neq 0 ,  \label{e7}
\end{equation}
where $\eta$ is a complex number, and $\alpha$ satisfies the
following conditions:
\begin{itemize}
\item[(H1)] $p(z)$ is analytic in a neighborhood of the origin,
 furthermore,\quad$p(0)=\beta\neq-1$ and
$p'(0)=\alpha+\alpha\beta$, where $\alpha,\beta$ are complex
numbers;

\item[(H2)] $0<|\alpha|<1$;

\item[(H3)] $|\alpha|=1, \alpha$ is not a root of unity,
and $\ln|\alpha^{n}-1|^{-1}\leq k \ln n$, $n=2,3,\dots $
for some positive constant $k$.
\end{itemize}

Then we show that \eqref{e5} has an analytic solution of the form
\begin{equation}
y(z)=g(\alpha g^{-1}(z))            \label{e8}
\end{equation}
in a neighborhood of the origin.

\section{Preparatory Lemmas}

We begin with the following preparatory lemma, the proof of which can
be found in cite{s1}.

\begin{lemma} \label{lem1}
 Assume that {\rm (H3)} holds. Then there is  a
 positive number $ \delta$ such that $|\alpha^{n}-1|^{-1}<(2n)^{\delta}$
for $n=1,2,\dots$. Furthermore, the sequence
$\{d_{n}\}_{n=1}^{\infty}$ defined by $d_{1}=1$ and
$$
 d_{n}=|\alpha^{n-1}-1|^{-1}
\max_{{k_{1}+\dots +k_{m}=n}\atop{0\leq k_{1}\leq\dots\leq
k_{m},m\geq 2}}\{d_{k_{1}}\dots d_{k_{m}}\},\quad n=2,3,\dots
$$
 will satisfy
\begin{equation}
d_{n}\leq N^{n-1}n^{-2\delta},\quad n=1,2,\dots, \label{e9}
\end{equation}
where $N=2^{5\delta+1}$.
\end{lemma}

To proceed,  we state and prove two preparatory lemmas which will be
used in the proof of our main result.

\begin{lemma} \label{lem2}
  Suppose  {\rm (H1)--(H2)} hold. Then for any
nonzero complex number  $\eta$,   equation \eqref{e6} has an analytic
solution $g(z)$ in a neighborhood of the origin such that $g(0)=0$
and $g'(0)=\eta$.
\end{lemma}

\begin{proof}
 Because $p(z)$ satisfies (H1),we assume
\begin{equation}
p(z)=\sum_{n=0}^{\infty}p_{n}z^{n},\quad p_{0}=\beta,\quad
p_{1} =\alpha+\alpha\beta . \label{e10}
\end{equation}
Then there exists a positive
constant $\rho$ such that
$$
|p_{n}|\leq \rho^{n-1},\quad n=2,3,\dots.
$$
Introducing new functions
$$
G(z)=\rho g(\rho^{-1}z),\quad
P(z)=\rho p(\rho^{-1}z),
$$
 we obtain from $g(0)=0$ and  $g'(0)=\eta$ that $G(0)=0$ and
$ G'(0) =\eta$ respectively, and by \eqref{e6} we have
$$
\alpha[G(\alpha^{2}z)-P(G(\alpha z))]G'(\alpha z)
=\alpha G'(\alpha z)-P'(G(z))G'(z),$$
which is again  an
equation of the form \eqref{e6}. Here $P$ is of the form
$$
P(z)=\sum_{n=0}^{\infty}P_{n}z^{n},
$$
but $|P_{n}|=|p_{n}\rho^{1-n}|\leq 1$ for $n\geq 2$. Then, without
loss of generality,we assume
\begin{equation}
|p_{n}|\leq 1,\quad n=2,3,\dots. \label{e11}
\end{equation}
Next, we assume that  \eqref{e6} has a formal power series solution
\begin{equation}
 g(z)=\sum_{n=1}^{\infty}c_{n}z^{n}  \label{e12}
\end{equation}
 and substitute \eqref{e10} and \eqref{e12} into \eqref{e6}, we see that the
 sequence $\{c_{n}\}_{n=1}^{\infty}$ is successively determined by
 the condition
\begin{align*}
&\alpha c_{1}+\alpha p_{0}c_{1}-p_{1}c_{1}\\
&+\sum_{n=1}^{\infty}[(n+1)c_{n+1}\alpha^{n+1}+
p_{0}(n+1)c_{n+1}\alpha^{n+1}-p_{1}(n+1)c_{n+1}]z^{n} \\
&=\sum_{n=1}^{\infty}
        \Big(\sum_{k=1}^{n}c_{k}c_{n-k+1}(n-k+1)\alpha^{n+k+1}\Big)z^{n}\\
&\quad+\sum_{n=1}^{\infty}\Big[\sum_{k=1}^{n}{\sum_
      { {l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}}
      ((m+1)p_{m+1}-p_{m}\alpha^{n+1})c_{l_{1}}\dots
       c_{l_{m}}(n-k+1)c_{n-k+1}\Big]z^{n},
\end{align*}
where $ n=1,2,\dots$, in a unique manner.

By comparing  coefficients in both sides, it is easy to see that
the coefficient sequence $\{c_{n}\}_{n=1}^{\infty}$ satisfies
\begin{equation}
(\alpha+\alpha p_{0}-p_{1})c_{1}=0,  \label{e13}
\end{equation}
and
\begin{equation}
\begin{aligned}
&(n+1)(\alpha^{n+1}+p_{0}\alpha^{n+1}-p_{1})c_{n+1}\\
&= \sum_{k=1}^{n}c_{k}c_{n-k+1}(n-k+1)\alpha^{n-k+1}\\
&\quad + \sum_{k=1}^{n}\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots ,
       k}}[(m+1)p_{m+1}-p_{m}\alpha^{n+1}]c_{l_{1}}\dots
       c_{l_{m}}(n-k+1)c_{n-k+1},
 \end{aligned}\label{e14}
\end{equation}
for \quad n=1,2,\dots.

By  (H1), we can ensure $\alpha+\alpha
p_{0}-p_{1}=\alpha+\alpha\beta-\alpha-\alpha\beta=0$. Hence we know
it is suitable for any complex number $c_{1}=\eta\neq 0$. We may now
see by \eqref{e14} that the resulting relation defines
$c_{n}$ ($n=2,3,\dots$) in a unique manner. Next, we want to prove
that the power series \eqref{e12} is convergent in a sufficient small
neighborhood of the origin. With lemma conditions, we can prove
 $\alpha^{n+1}+p_{0}\alpha^{n+1}-p_{1}=\alpha(\alpha^{n}-1)(1+\beta)\neq
0 $ for $n=1,2,\dots$. If not, assuming
$\alpha(\alpha^{n}-1)(1+\beta)=0$,  we can get $\alpha^{n}=1$, so
$|\alpha|=1$ which contradicts condition (H2). To see this, note
that
$$
\lim_{n\to
\infty}\frac{1}{\alpha(\alpha^{n}-1)(1+\beta)}=
-\frac{1}{\alpha(1+\beta)},\quad 0<|\alpha|<1,
$$
 thus there exist some positive number $M$ such that
$|\frac{1}{\alpha(\alpha^{n}-1)(1+\beta)}|\leq M$ for
$n\geq 1$.
By \eqref{e11} and \eqref{e14}, we have
\begin{equation}
|c_{n+1}|\leq M(\sum_{k=1}^{n}|c_{k}||c_{n-k+1}|+\sum_{k=1}^{n}
\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,
k}}(n+2)|c_{l_{1}}|\dots|c_{l_{m}}||c_{n-k+1}|),
\label{e15}
\end{equation}
where $n=1,2,\dots$. If we now  define a positive sequence
$\{q_{n}\}_{n=1}^{\infty}$ by $q_{1}=|\eta|$ and
$$
q_{n+1}= M[\sum_{k=1}^{n} q_{k}q_{n-k+1}+\sum_{k=1}^{n}
\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,
k}}(n+2)q_{l_{1}}\dots q_{l_{m}}q_{n-k+1}],\quad
n=1,2,\dots,
$$
 then it is  easily seen that
$$
|c_{n+1}|\leq q_{n+1},\quad n=0,1,2,\dots.
$$
In other words, the series $\sum _{n=1}^{\infty}q_{n}z^{n}$ is a
majorant series of $\sum _{n=1}^{\infty}c_{n}z^{n}$.  So next we
want to show that the power series $\sum _{n=1}^{\infty}q_{n}z^{n}$
is convergent in a sufficient small neighborhood of the origin. Now
if we define $Q(z)=\sum_{n=1}^{\infty}q_{n}z^{n}$, then
\begin{align*}
Q(z)&=\sum_{n=1}^{\infty}q_{n}z^{n}
=\sum_{n=0}^{\infty}q_{n+1}z^{n+1}\\
&=|\eta|z+\sum_{n=1}^{\infty}q_{n+1}z^{n+1}\\
&= |\eta|z+M\sum_{n=1}^{\infty}(\sum_{k=1}^{n}q_{k}q_{n-k+1})z^{n+1}\\
&\quad +M[\sum_{n=1}^{\infty}(\sum_{k=1}^{n}
\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}
       }(n+2)q_{l_{1}}\dots q_{l_{m}}q_{n-k+1})z^{n+1}]\\
 &= |\eta|z+M(\sum_{n=1}^{\infty}q_{n}z^{n})^{2}+M[\sum_{n=1}^{\infty}(n+2)(Q(z))^{n}]
 (\sum_{n=1}^{\infty}q_{n}z^{n})\\
&= |\eta|z+M[Q(z)]^{2}+M[\sum_{n=1}^{\infty}(n+1)(Q(z))^{n}]Q(z)
+M\sum_{n=1}^{\infty}  (Q(z))^{n+1}\\
&= |\eta|z+M[Q(z)]^{2}+M{(\sum_{n=1}^{\infty}(Q(z))^{n+1})}'Q(z)+
M\frac{Q^{2}(z)}{1-Q(z)}\\
&= |\eta|z+M\frac{4Q^{2}(z)-4Q^{3}(z)+Q^{4}(z)}{[1-Q(z)]^{2}}.
\end{align*}
So the function $Q=Q(z)$ is the implicit function which defined by
the function
$$
Q=|\eta|z+M\frac{4Q^{2}-4Q^{3}+Q^{4}}{(1-Q)^{2}},
$$
or
$$
F(z,Q)\equiv Q-|\eta|z-M\frac{4Q^{2}-4Q^{3}+Q^{4}}{(1-Q)^{2}}=0.
$$
Because the function $F(z,Q)$ is continuous in a neighborhood of the
origin, furthermore,\quad$F(0,0)=0$ and $F_{Q}'(0,0)=1\neq 0
$.\quad By the implicit function theorem, we see that the $Q=Q(z)$
is analytic on a disk with the origin as the center and with a
positive radius. The proof is completed.
\end{proof}

\begin{lemma} \label{lem3}
 Suppose  {\rm (H1), (H3)} hold. Then equation
\eqref{e6} has an  analytic solution $g(z)$ in a neighborhood of the origin
such that $g(0)=0$ and $g'(0)=\eta\neq 0$.
\end{lemma}

\begin{proof}
  Similar to the proof of Lemma \ref{lem2}, we seek a
formal power series solution  \eqref{e12} of equation \eqref{e6} with
$c_{1}=\eta$ and
\begin{equation}
\begin{aligned}
&(n+1)\alpha(\alpha^{n}-1)(1+\beta)c_{n+1}\\
&= \sum_{k=1}^{n}c_{k}c_{n-k+1}(n+1-k)\alpha^{n+k+1}\\
&\quad +\sum_{k=1}^{n}\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots ,
       k}}[(m+1)p_{m+1}-p_{m}\alpha^{n+1}]c_{l_{1}}\dots
       c_{l_{m}}(n-k+1)c_{n-k+1}
\end{aligned}\label{e16}
\end{equation}
 for $n\geq 1$. So next we want to prove that the power series
$\sum_{n=1}^{\infty}c_{n}z^{n}$ is convergent in a sufficient small
neighborhood of the origin.

 The formulation \eqref{e16} can be written as
\begin{equation}
\begin{aligned}
|c_{n+1}|&\leq \frac{1}{|1+\beta|}
|\alpha^{n}-1|^{-1}[\sum_{k=1}^{n}|c_{k}||c_{n+1-k}|\\
 &\quad +\sum_{k=1}^{n}{\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}}
(n+2)|c_{{l}_{1}}|\dots|c_{{l}_{m}}||c_{n-k+1}|],\quad n=1,2,\dots.
\end{aligned} \label{e17}
\end{equation}
If  we now  define a positive sequence $\{v_{n}\}_{n=1}^{\infty}$ by
$v_{1}=|\eta|$ and
$$
v_{n+1}=M|\alpha^{n}-1|^{-1}\Big[\sum_{k=1}^{n}v_{k}v_{n-k+1}+
\sum_{k=1}^{n}{\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}}
(n+2)v_{{l}_{1}}\cdots v_{{l}_{m}}v_{n-k+1}\Big],
$$
 where
$M=1/(|1+\beta|)>0$, $n\geq 1$,  then it is not difficult to show
by induction that
\begin{equation}
|c_{n+1}|\leq v_{n+1}, \quad n=0,1,2,\dots. \label{e18}
\end{equation}
In other words, $V(z)=\sum_{n=0}^{\infty}v_{n}z^{n}$ is a
majorant series of $g(z)=\sum_{n=0}^{\infty}c_{n}z^{n}$. we now only
need to show that $V(z)$ has a positive radius of convergence. To
see this, we define the positive recursive sequence
  $\{q_{n}\}$ by $q_{1}=|\eta|$ and
$$
q_{n+1}=M\Big[\sum_{k=1}^{n}q_{k}q_{n-k+1}+
\sum_{k=1}^{n}{\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}}
(n+2)q_{{l}_{1}}\dots q_{{l}_{m}}q_{n-k+1}\Big],\quad
 n=1,2,\dots.
$$
 Then
\begin{equation}
\begin{aligned}
Q(z)&=\sum_{n=1}^{\infty}q_{n}z^{n}
=\sum_{n=0}^{\infty}q_{n+1}z^{n+1}\\
&= |\eta|z+\sum_{n=1}^{\infty}q_{n+1}z^{n+1}\\
&= |\eta|z+\sum_{n=1}^{\infty}M(\sum_{k=1}^{n}q_{k}q_{n+1-k}
+\sum_{k=1}^{n} \sum_{{l_{1}+\dots+l_{m}= k}\atop{m=1,2,\dots,k}
       }(n+2)q_{l_{1}}\dots q_{l_{m}}q_{n-k+1})z^{n+1}\\
&= |\eta|z+\frac{4Q^{2}(z)-4Q^{3}(z)+Q^{4}(z)}{[1-Q(z)]^{2}}.
\end{aligned}\label{e19}
\end{equation}
Hence by induction, we easily see  by Lemma \ref{lem1} that
\begin{equation}
v_{n+1}\leq q_{n+1}d_{n+1},\quad n=0,1,2,\dots, \label{e20}
\end{equation}
where the sequence $\{d_{n}\}_{n=1}^{\infty}$ is defined by
 Lemma \ref{lem1}.

In fact, if $k=0$, it holds. Assume by induction that $v_{k}\leq
q_{k}d_{k}$ for $k=1,2,\dots,n-1$. Then
\begin{align*}
  v_{n+1}&= M|\alpha^{n}-1|^{-1}\Big[\sum_{k=1}^{n}v_{k}v_{n+1-k}+
\sum_{k=1}^{n}\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}
 (n+2)v_{l_{1}}\dots v_{l_{m}}v_{n-k+1}\Big]\\
&\leq  M|\alpha^{n}-1|^{-1}\Big[\sum_{k=1}^{n}q_{k}d_{k}q_{n+1-k}d_{n+1-k}\\
&\quad +\sum_{k=1}^{n}\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}
 (n+2)q_{l_{1}}\dots q_{l_{m}}d_{l_{1}}\dots d_{l_{m}}q_{n-k+1}d_{n-k+1}\Big]\\
&\leq M\Big[ \sum_{k=1}^{n}q_{k}q_{n-k+1}+
\sum_{k=1}^{n}{\sum_{{l_{1}+\dots+l_{m}=k}\atop{m=1,2,\dots,k}}}
(n+2)q_{{l}_{1}}\dots q_{{l}_{m}}q_{n-k+1}\Big]\\
&\quad\times |\alpha^{n}-1|^{-1}\max_{{l_{1}+\dots+l_{m}=n+1}\atop{m=1,2,\dots,n+1}}
\{d_{l_{1}}\dots d_{l_{m}}\}\\
&=q_{n+1}d_{n+1},\quad n=0,1,2,\dots.
\end{align*}
By equation \eqref{e19}, the  implicit function of $Q=Q(z)$ is defined by
$$
F(z,Q)=Q-|\eta|z-\frac{4Q^{2}-4Q^{3}+Q^{4}}{(1-Q^{2})}.
$$

In view of $F(0,0)=0$ and $F'_{Q}(0,0)=1\neq 0$,  by virtue of
the implicit function theorem there exists a positive constant
$\delta$ such that the function $Q(z)=\sum_{n=1}^{\infty}q_{n}z^{n}$
converges for $|z|<\delta$. So there exists $k>0$ such that
$q_{n}\leq R^{n}$ for $n=1,2,\dots$.

By Lemma \ref{lem1} and  \eqref{e20}, we finally see that
$$
v_{n}\leq R^{n}N^{n-1}n^{-2\delta},\quad n=1,2,\dots,
$$
which implies  that the series  $\sum_{n=1}^{\infty}v_{n}z^{n}$
converges for $|z|<(RN)^{-1}$, therefore, the series \eqref{e12} also
converges for $|z|<(RN)^{-1}$. This completes the proof.
\end{proof}

\section{Existence of Analytic Solutions to \eqref{e1}}

In this section, we state and prove our main result in this article.

\begin{theorem} \label{thm1}
 Suppose the  conditions of Lemma \ref{lem2} or lemma \ref{lem3}
 are satisfied.
Then  \eqref{e5} has an analytic solution
$g(z)$ of the form \eqref{e8} in a neighborhood of the origin, where $g(z)$
is an analytic solution of \eqref{e6}.
\end{theorem}

\begin{proof}
   In view of Lemma \ref{lem2} or lemma \ref{lem3}, we may find a sequence
$\{c_{n}\}_{n=1}^{\infty}$ such that the function $g(z)$ of the form
\eqref{e12} is analytic solution of \eqref{e6} in a neighborhood of
the origin.
Since $g'(0)=\eta\neq 0$, the function $g^{-1}(z)$ is analytic in
a neighborhood of the point $g(0)=0$. If we now define $y(z)$ by
\eqref{e8}, then
$$
y'(z)-p'(z)=\frac{\alpha g'(\alpha g^{-1}(z)
)}{g'(g^{-1}(z))}-p'(z) =\frac{\alpha
g'(\alpha g^{-1}(z))-p'(z)g'(g^{-1}(z))}{g'(g^{-1}(z))},$$
 and
 \begin{align*}
 [y(y(z))-p(y(z))]y'(z)
&= [g(\alpha g^{-1}(g \alpha g^{-1}(z)))-p(g(\alpha g^{-1}(z)
))]\frac{\alpha g'(\alpha
g^{-1}(z))}{g'(g^{-1}(z))}\\
&= \frac{\alpha g'(\alpha g^{-1}(z)
)-p'(z)g'(g^{-1}(z))}{g'(g^{-1}(z))}
\end{align*}
as required. The proof is completed.
\end{proof}

In the last section, we have shown that under the conditions of
lemma \ref{lem2} or  lemma \ref{lem3}, equation \eqref{e5} has an analytic solution
$y(z)=g(\alpha g^{-1}(z))$   in a neighborhood of the point, where $g$
is an analytic solution of \eqref{e6}.
 We now show how to explicitly construct an analytic solution of
\eqref{e1} by \eqref{e5}. In view of
$$
x'(z)=\frac {1}{b}[y(z)-p(z)],
$$
we have
$x'(0)=\frac {1}{b}[y(0)-p(0)]=-\beta/b$.
Furthermore,
$$
x'(0)=x(p(0)+bx'(0))=x(\beta-\beta)=x(0)=
-\frac{\beta}{b}.
$$
 By calculating the derivatives of
both sides of \eqref{e1}, we obtain successively
\begin{gather*}
x''(z)=x'(p(z)+bx'(z))(p'(z)+bx''(z)), \\
x'''(z)=x''(p(z)+bx'(z))(p'(z)+bx''(z))^{2}
+x'(p(z)+bx'(z))(p''(z)+bx'''(z)),
\end{gather*}
 so that
\begin{gather*}
x''(0)=x'(p(0)+bx'(0))(p'(0)+bx''(0)),\\
x'''(0)=x''(0) (p'(0)+bx''(0))^{2}
+x'(0)(p''(0)+bx'''(0));
\end{gather*}
 that is,
$$
x''(0)=-\frac{\alpha\beta}{b},  \quad
x'''(0)= -\frac{\beta(\alpha^{3}+p_{2})}{b(1+\beta)}.
$$
In general, we can show by induction that
\begin{align*}
(x(p(z)+bx'(z)))^{(m)}
&=\sum_{i=1}^{m}p_{im}(p'(z)+bx''(z),
p''(z)+bx'''(z),\dots,
p^{(m)}(z)\\
&\quad +bx^{(m+1)}(z))x^{(i)}(p(z)+bx'(z)),
\end{align*}
where $m=1,2,\dots$, and $p_{im}$ is a polynomial with nonnegative
coefficients. Hence
\begin{align*}
x^{(m+1)}(0)&=\sum_{i=1}^{m}p_{im}(p'(0)+bx''(0),
p''(0)+bx'''(0),\dots, p^{(m)}(0)\\
&\quad +bx^{(m+1)}(0))x^{(i)}(0)=:\Gamma_{m}
\end{align*}
for $m=1,2,\dots$. It is then easy to write out the explicit form of
our solution
$$
x(z)=-\frac{\beta}{b}-\frac{\beta}{b}z
-\frac{\alpha\beta}{2!b}z^{2}-\frac
{\beta(\alpha^{3}+p_{2})}{3!b(1+\beta)}z^{3}+\sum_{m=3}^{\infty}
\frac{\Gamma_{m}}{(m+1)!}z^{m+1}.
$$

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\end{thebibliography}

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