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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 39, 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 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/39\hfil Positive periodic solutions]
{Positive periodic solutions for
Li\'enard type $p$-Laplacian equations}

\author[J. Meng\hfil EJDE-2009/39\hfilneg]
{Junxia Meng}

\address{Junxia Meng \newline
College of Mathematics and Information Engineering,
Jiaxing University,  Jiaxing\\
 Zhejiang 314001,  China}
\email{mengjunxia1968@yahoo.com.cn}

\thanks{Submitted February 17, 2009. Published March 19, 2009.}
\thanks{Supported by grant 20070605 from Scientific Research
 Fund of Zhejiang Provincial \hfill\break\indent Education Department}
\subjclass[2000]{34K15, 34C25}
\keywords{$p$-Laplacian; positive periodic solutions;
  Li\'enard equation; \hfill\break\indent topological degree}

\begin{abstract}
 Using topological degree theory, we obtain sufficient conditions
 for the existence and uniqueness  of positive periodic solutions
 for Li\'enard type $p$-Laplacian differential equations.
\end{abstract}

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

\section{Introduction}

In recent years, the existence of periodic solutions for the
Duffing equation, Rayleigh equation and Li\'enard
type equation has  received  a lot of attention.
We refer the reader to \cite{WGQ,lsp,lsp2,zmg,lbw,zfx}
 and the references cited therein.
However, as far as we know,  fewer papers  discuss
the existence and uniqueness of positive periodic solutions for
Li\'enard type $p$-Laplacian differential equation.

 In this paper we study the  existence and uniqueness of positive
 $T$-periodic solutions of the Li\'enard
type $p$-Laplacian differential equation of the form:
\begin{equation}\label{e1}
(\varphi_p(x'(t)))'+f(x(t))x'(t )+g(x(t ))=e(t),
\end{equation}
where $p>1$ and $\varphi_p:\mathbb{R} \to \mathbb{R}$ is given by
 $\varphi_p(s)=|s|^{p-2}s$ for $s\neq0$ and
 $\varphi_p(0)=0$,   $f$ and $g$ are  continuous functions  defined
on $\mathbb{R}$. $e $ is a continuous
periodic function  defined on $\mathbb{R}$ with period $T$, and $T>0$.
By using topological degree theory and some analysis skill, we
establish some sufficient conditions for the existence and uniqueness
of $T$-periodic solutions of \eqref{e1}. The results
of this paper are new and they complement previously  known
results.

 \section{Preliminaries}

For convenience, let us denote
$$
C_{T}^1:=\{x\in C^1(\mathbb{R}, \mathbb{R}): x
\text{ is T-periodic} \},
$$
which is a Banach space endowed with the norm
 $\|x\|=\max\{|x|_{\infty}, |x'|_{\infty}\}$,
and
$$
|x|_\infty=\max_{t\in [0,T]}|x(t)|, \quad
|x'|_\infty=\max_{t\in [0,T]}|x'(t)|, \quad
|x|_k=\Big(\int^T_0|x(t)|^kdt\Big)^{1/k}.
$$
 For the periodic boundary-value problem
 \begin{equation} \label{e2}
 (\varphi_p(x'(t)))'=\widetilde{f}(t,x,x'),\quad
 x(0)=x(T),\quad x'(0)=x'(T)
 \end{equation}
where $\widetilde{f}$ is a continuous function and T$-$periodic in
the first variable,  we have the following result.

\begin{lemma}[\cite{MJM}] \label{lem1}
 Let $\Omega$ be an open bounded set in
 $C_T^1,$ if the following conditions hold
\begin{itemize}
\item[(i)]  For each $\lambda \in(0,1)$ the problem
\[
(\varphi_p(x'(t)))'=\lambda \widetilde{f}(t,x,x') ,\quad
x(0)=x(T),\quad x'(0)=x'(T)
 \]
has no solution on $\partial\Omega$;

\item[(ii)] The equation
\[
F(a):=\frac{1}{T}\int_0^T \widetilde{f}(t,a,0)\,dt=0
\]
has no solution on $\partial \Omega \cap \mathbb{R}$;

\item[(iii)]  The Brouwer degree of $F$ satisfies
\[
\deg (F,\Omega \cap \mathbb{R}, 0)\neq 0,
\]
\end{itemize}
Then the periodic boundary value problem \eqref{e2} has at least
one T$-$periodic solution on $\overline \Omega$.
\end{lemma}

 Set
\begin{equation} \label{e3}
\Psi (x)=\int_{0}^{x}f(u)du,  \quad
 y (t)= \varphi_p(x '(t))+\Psi (x (t)).
\end{equation}
 We can rewrite \eqref{e1} in the  form
\begin{equation}\label{e4}
 \begin{gathered}
  { x'  (t)}   =  |y(t)-\Psi (x(t))|^{q-1}\mathop{\rm sign}(y(t)-\Psi (x(t))), \\
  { y' (t)}=-   g( x (t ))+e(t),
\end{gathered}
\end{equation}
where $q>1$ and $
\frac{1}{p}+\frac{1}{q}=1$.

\begin{lemma} \label{lem2}
Suppose that the following condition  holds.
\begin{itemize}
\item[(A1)]  $g$ is a  continuously differentiable function
defined on $\mathbb{R}$,  and $ g_{x} '(x)<0 $.
\end{itemize}
 Then \eqref{e1} has at most one $T$-periodic    solution.
\end{lemma}

\begin{proof}
 Suppose that $x_{1}(t)$ and $x_{2}(t)$ are two
$T$-periodic solutions of \eqref{e1}.  Then,  from \eqref{e4},  we obtain
\begin{equation} \label{e5}
 \begin{gathered}
  { x_{i}'  (t)}   =  |y_{i}(t)-\Psi (x_{i}(t))|^{q-1}
\mathop{\rm sign}(y_{i}(t)-\Psi (x_{i}(t))), \\
  { y_{i}' (t)}=-   g( x_{i} (t ))+e(t),
\quad i=1,2.
\end{gathered}
 \end{equation}
Set
\begin{equation} \label{e6}
v(t)=x_{1}(t)-x_{2}(t), \quad
u(t)=y_{1}(t)-y_{2}(t),
\end{equation}
it follows from \eqref{e5}
that
\begin{equation} \label{e7}
 \begin{aligned}
  { v'  (t)}   &=  |y_{1}(t)-\Psi (x_{1}(t))|^{q-1}\mathop{\rm sign}
(y_{1}(t)-\Psi (x_{1}(t)))\\
  &\quad -|y_{2}(t)-\Psi (x_{2}(t))|^{q-1}\mathop{\rm sign}(y_{2}(t)-\Psi (x_{2}(t))), \\
  { u' (t)}&=-   [g( x_{1} (t ))-g( x_{2} (t ))],
\end{aligned}
\end{equation}
Now, we prove that
$u(t)\leq 0$  for  all $t\in \mathbb{R}$.
 Contrarily,  in view of $u\in C^{2}[0, T]$ and
$u(t+T)=u(t)$ for  all $t\in \mathbb{R}$, we obtain
$$
\max_{t\in \mathbb{R}} u(t)>0.
$$
Then,   there must exist $t^{*}\in \mathbb{R}$ (for convenience,
we can choose $t^{*}\in (0, T)$) such  that
$$
u(t^{*})=\max_{t\in [0, \ T]} u(t)=\max_{t\in \mathbb{R}} u(t)>0,
$$
which, together with  $g '(x)<0 $,  implies that
\begin{equation} \label{e8}
\begin{gathered}
 u'(t^{*})   =-   [g( x_{1} (t^{*} ))-g( x_{2}
(t^{*} ))]=0, \quad x_{1} (t^{*} )= x_{2} (t^{*} ), \\
\begin{aligned} u''(t^{*}) &=  (-   (g( x_{1} (t ))-g( x_{2} (t
))))' |_{t=t^{*}}\\
&=-[g_{x}'( x_{1} (t^{*} ))x'_{1} (t^{*} )-g_{x}'(
x_{2} (t^{*} ))x'_{2} (t^{*} )]\leq 0.
 \end{aligned}
\end{gathered}
\end{equation}
Then
\begin{equation} \label{e9}
\begin{aligned}
 u''(t^{*})  & = -g_{x}'( x_{1} (t^{*} ))[x'_{1}
(t^{*} )-x'_{2} (t^{*} )] \\
& =  -g_{x}'( x_{1} (t^{*} ))[|y_{1}(t^{*})-\Psi (x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{1}(t^{*})-\Psi (x_{1}(t^{*}))) \\
& \quad -|y_{2}(t^{*})-\Psi
(x_{2}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{2}(t^{*})-\Psi
(x_{2}(t^{*})))] \\
& =  -g_{x}'( x_{1} (t^{*} ))[|y_{1}(t^{*})-\Psi (x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{1}(t^{*})-\Psi (x_{1}(t^{*}))) \\
& \quad -|y_{2}(t^{*})-\Psi
(x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{2}(t^{*})-\Psi
(x_{1}(t^{*})))].
 \end{aligned}
\end{equation}
In view of
\begin{equation} \label{e10}
-g_{x}'( x_{1} (t^{*} ))>0,  \quad
 u(t^{*})=y_{1}(t^{*})-y_{2}(t^{*})>0,
\end{equation}
and
\begin{align*}
& |y_{1}(t^{*})-\Psi (x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{1}(t^{*})
 -\Psi (x_{1}(t^{*}))) \\
&-|y_{2}(t^{*})-\Psi (x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{2}(t^{*})
-\Psi (x_{1}(t^{*})))>0 .
\end{align*}
It follows from \eqref{e9} that
\begin{equation} \label{e11}
\begin{aligned}
 u''(t^{*})  & = -g_{x}'( x_{1} (t^{*} ))[|y_{1}(t^{*})
-\Psi (x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{1}(t^{*})
-\Psi (x_{1}(t^{*}))) \\
& \quad -|y_{2}(t^{*})-\Psi
(x_{1}(t^{*}))|^{q-1}\mathop{\rm sign}(y_{2}(t^{*})-\Psi
(x_{1}(t^{*})))]>0,
  \end{aligned}
\end{equation}
 which contradicts the second equation of \eqref{e8}.
This contradiction implies  that
$$
u(t)=y_{1}(t)-y_{2}(t)\leq 0 \quad \text{ for  all  }
t\in \mathbb{R}.
$$
By using a similar  argument, we can also show that
$$
y_{2}(t)-y_{1}(t)\leq 0 \quad \text{for  all } t\in \mathbb{R}.
$$
Therefore, we obtain
$y_{2}(t)\equiv y_{1}(t)$ for  all  $t\in \mathbb{R}$.
Then, from \eqref{e7}, we get
$$
g( x_{1} (t ))-g( x_{2} (t )) \equiv0  \quad \text{for all  } t\in
\mathbb{R},$$
again from  $ g_{x} '(x)<0 $, which implies
that
$x_{2}(t)\equiv x_{1}(t)$  for  all $t\in \mathbb{R}$.
 Hence, \eqref{e1}  has at most one $T$-periodic
    solution. The proof is  complete.
\end{proof}

 \section{Main Results}

Using   Lemmas \ref{lem1} and  \ref{lem2}, we obtain our main results:

\begin{theorem} \label{thm1}
  Let {\rm (A1)} hold. Suppose that there exists a positive constant
 $d $ such that
\begin{itemize}
\item[(A2)] $ g( x)-e(t) <0$   for $ x >d$ and $t \in \mathbb{R}$,
$ g( x)-e(t) >0$   for $ x \leq 0$ and $t \in \mathbb{R}$.
\end{itemize}
 Then \eqref{e1} has a unique positive $T$-periodic solution.
\end{theorem}

\begin{proof} Consider the homotopic equation of
\eqref{e1} as follows:
\begin{equation}\label{e12}
(\varphi_p(x'(t)))'+\lambda f(x(t ))x'(t ) +\lambda g(
x(t ))=\lambda e(t),\quad \lambda \in(0,1)
\end{equation}
By Lemma \ref{lem2}, and (A1), it is easy to see that
\eqref{e1} has at most one positive $T$-periodic  solution.
Thus, to prove Theorem  \ref{thm1}, it suffices to show that \eqref{e1} has at
least one  $T$-periodic solution. To do this, we shall apply
Lemma  \ref{lem1}.
Firstly, we will claim that the set of all possible
$T$-periodic solutions of    \eqref{e12}  is bounded.

Let $x(t)\in C_{T}^1$ be an arbitrary solution of \eqref{e12}
with period $T$.   By integrating two sides of \eqref{e12} over
$[0,T]$, and noticing that $x'(0)=x'(T)$, we have
\begin{equation}\label{e13}
\int_0^{T}  (g( x(t))-e(t)) \,dt=0.
\end{equation}
As $x(0)=x(T)$,  there exists $t_0\in[0,T]$ such that $x'(t_0)=0$, while
$\varphi_p(0)=0$  we  see
\begin{equation}  \label{e14}
\begin{aligned}
|\varphi_p(x'(t)) |
&=   |\int_{t_0}^t(\varphi_p(x'(s)))'\,ds|\\
&\leq   \lambda\int_0^{T}|f(x(t ))\|x'(t ) |\,dt
  + \lambda\int_0^{T}|g( x(t )) |\,dt +  \lambda\int_0^{T}|e(t )|\,dt,
 \end{aligned}
\end{equation}
where $t \in [t_{0}, t_{0}+T]$.

 From \eqref{e13}, there exists a
$\bar{\xi}\in [0,T]$ such that
$g( x(\bar{\xi} ))-e(\bar{\xi} ) =0$.
In view of (A2), we obtain
$| x(\bar{\xi} )| \leq  d$.
Then, we have
    $$
    |x(t)|=|x(\bar{\xi})+\int^t_{\bar{\xi}}x'(s)ds|\le
    d+ \int^t_{\bar{\xi}}|x'(s)|ds, \ t\in [\bar{\xi}, \quad \bar{\xi}+T],
$$
and
$$
 |x(t)|=|x(t-T)|=|x(\bar{\xi})-\int^{\bar{\xi}}_{ t-T }x'(s)ds|\le
 d+\int^{\bar{\xi}}_{ t-T }|x'(s)|ds,  t\in [\bar{\xi}, \quad
 \bar{\xi}+T].
$$
Combining the above two inequalities, we obtain
\begin{equation} \label{e15}
\begin{aligned}
    |x|_\infty
&=\max_{t\in [0,T]}|x(t)|
=\max_{t\in[\bar{\xi}, \ \bar{\xi}+T]}|x(t)| \\
& \le  \max_{t\in[\bar{\xi}, \ \bar{\xi}+T]}
\{ d+\frac{1}{2}(\int^t_{\bar{\xi}}|x'(s)|ds
    +\int^{\bar{\xi}}_{ t-T }|x'(s)|ds) \} \\
& \le  d+ \frac{1}{2}\int^{T}_{0 }|x'(s)|ds.
\end{aligned}
\end{equation}
Denote
\[
E_1  =   \{t: t\in[0,T],\,|x(t )|>d\}, \quad
E_2  =   \{t:t\in[0,T], |x(t)| \leq d\} .
\]
Since $x (t)$ is   $T$-periodic, multiplying $x (t) $ and \eqref{e12}
 and then integrating it from  $0$ to $T$,  in view of   (A2),  we get
\begin{equation}  \label{e16}
\begin{aligned}
 \int^T_0 |x '(t)|^{p} dt
& =   -\int^T_0 (\varphi_p(x '(t)) )' x (t) dt \\
& =        \lambda\int_{E_1} [g ( x (t ))-e (t )]
 x (t) dt+\lambda\int_{E_2} [g ( x (t ))-e (t )]
 x (t) dt  \\
& \leq   \int^T_0 \max\{|g ( x (t ))-e (t )|:t\in \mathbb{R},
 |x(t)| \leq d \} |x (t)| dt \\
& \leq  D T |x  |_{\infty},
\end{aligned}
\end{equation}
where $D=\max\{|g ( x  )-e (t )|: |x | \leq d , \; t\in \mathbb{R} \}$.

 For $x(t)\in C(\mathbb{R}, \mathbb{R})$   with $x(t+T)=x(t)$,
 and $0< r\le s$, by using H\"older inequality,
    we  obtain
\begin{align*}
\Big(\frac{1}{T}\int^T_0|x(t)|^rdt\Big)^{1/r}
&\le  \Big(\frac{1}{T}(\int^T_0(|x(t)|^{r})^{s/r}dt)^{r/s}
(\int^T_01dt)^{\frac{s-r}{s}} \Big)^{1/r}\\
&=\Big(\frac{1}{T}\int^T_0|x(t)|^s dt\Big)^{1/s},
\end{align*}
this implies that
\begin{equation}\label{e17}
|x|_{r}\leq T^\frac{s-r}{rs}|x|_{s}, \quad \text{for } 0< r\le s.
 \end{equation}
Then,  in view of \eqref{e15}, \eqref{e16} and \eqref{e17},
we can get
\begin{equation} \label{e18}
\begin{aligned}
(\int^{T}_{0 }|x'(t)|dt) ^{p}
&\leq  T^{ p-1 }|x'(t)|^{p}_{p}
  =  T^{p-1} \int^T_0 |x '(t)|^{p} dt   \\
& \leq  T^{p-1}D T |x |_{\infty} \\
& \leq  T^{p }D  (d+ \frac{1}{2}\int^{T}_{0 }|x'(s)|ds).
\end{aligned}
\end{equation}
Since $p>1$,  the above inequality allows as to choose a positive
constant $M_1$ such that
\[
\int^{T}_{0 }|x'(s)|ds\leq M_1, \quad
  |x|_{\infty} \leq d+ \frac{1}{2}\int^{T}_{0}|x'(s)|ds \leq M_1.
 \]
In view of \eqref{e14},  we have
\begin{equation}  \label{e19}
\begin{aligned}
| x' |_{\infty}^{p-1}
&=\max_{t\in[0,T]}\{|\varphi_p(x'(t)) |\}  \\
& = \max_{t\in[t_{0},t_{0}+T]}\{|\int_{t_0}^t(\varphi_p(x'(s)))'\,ds|\} \\
& \leq  \int_0^{T}|f(x(t ))\|x'(t ) |\,dt +  \int_0^{T}|g( x(t )) |\,dt
  + \int_0^{T}|e(t )  |\,dt   \\
&\leq  [\max\{|f ( x  )|:
 |x | \leq  M_1 \} ]M_1+  T[\max\{|g ( x  )|:
 |x | \leq  M_1 \}+|e|_{\infty}]  .
\end{aligned}
\end{equation}
 Thus,   we can get some positive constant $M_2>  M_1+1$ such that
for all $t\in \mathbb{R}$,
$|x'(t)|\leq M_2$.
Set  $\Omega=\{x\in C_{T}^1:\|x\|\leq M_2+1\}$,
then we know that \eqref{e12}
 has no solution on $\partial
\Omega$ as $\lambda \in (0,1)$ and when
$x(t)\in\partial\Omega\cap \mathbb{R}$, $x(t)=M_2+1$ or $x(t)=-M_2-1$, from
$(A_{2})$ , we can see that
\begin{gather*}
\frac{1}{T}\int_0^{T}\{-g( M_2+1)+e(t)\}\,dt
= -\frac{1}{T}\int_0^{T}\{ g( M_2+1)-e(t)\}\,dt>0 , \\
\frac{1}{T}\int_0^{T}\{-g( -M_2-1)+e(t)\}\,dt
= -\frac{1}{T}\int_0^{T}\{g( -M_2-1)-e(t)\}\,dt<0,
\end{gather*}
so condition (ii) is also satisfied. Set
\[
H(x,\mu)=\mu x-(1-\mu)\frac{1}{T}\int_0^{T}\{g( x)-e(t)\}\,dt,
\]
and when $x\in\partial\Omega\cap \mathbb{R}$, $\mu\in[0,1]$ we have
\[
xH(x,\mu)=\mu x^2-(1-\mu)x\frac{1}{T}\int_0^{T}\{g( x)-e(t)\}\,dt>0 .
\]
Thus $H(x,\mu)$ is a homotopic  transformation and
$$
\deg \{F,\Omega\cap \mathbb{R},0\} =
\deg \{-\frac{1}{T}\int_0^{T}\{g(x)-e(t)\}\,dt,\Omega\cap \mathbb{R},0\}
=\deg \{x,\Omega\cap \mathbb{R},0\}\neq 0.
 $$
so condition (iii) is satisfied. In view of the previous
Lemma \ref{lem1}, there exists at least one solution with period $T$.

   Suppose that  $x(t)$ is the $T$-periodic solution of
\eqref{e1}.
Let $\bar{t}$   be  the global  minimum point of $x (t)$
on $[0, T ]$. Then $x '(\bar{t} ) = 0 $ and we claim that
\begin{equation}
(\varphi_p(x'(\bar{t})))' =(|x'(\bar{t})|^{p-2} x'(\bar{t}))'
\geq 0.\label{e20}
\end{equation}
 Assume, by way of contradiction, that \eqref{e20} does not hold. Then
$$
(\varphi_p(x'(\bar{t})))' =(|x'(\bar{t})|^{p-2} x'(\bar{t}))' <0,
$$
and there exists  $ \varepsilon> 0$ such that
$(\varphi_p(x'(t)))' =(|x'(t)|^{p-2} x'(t))' <0$  for
$t\in (\bar{t}-\varepsilon, \bar{t} + \varepsilon)$. Therefore,
$\varphi_p(x'(t)) = |x'(t)|^{p-2} x'(t)  $ is strictly decreasing
for $t\in (\bar{t}-\varepsilon, \bar{t} + \varepsilon)$, which
implies that $x'(t)$ is strictly decreasing for $t\in
(\bar{t}-\varepsilon, \bar{t} + \varepsilon).$ This  contradicts
the definition of  $\bar{t}$. Thus, \eqref{e20} is true.
From  \eqref{e1} and \eqref{e20}, we have
\begin{equation} \label{e21}
g( x(\bar{t} ))-e(\bar{t})\leq 0.
\end{equation}
 In view of  (A2), \eqref{e21}  implies
$x(\bar{t} )>0$.
Thus,
$$
x(t) \geq \min_{t\in [0, \ T]}
     x(t)=x(\bar{t} )>0, \quad\text{for all }t \in \mathbb{R},
$$ % \eqno (23)
which implies that \eqref{e1} has at least one  positive  solution
with period $T$.
 This completes the proof.
\end{proof}

\section{An Example}

As an application, let  us consider the following  equation
\begin{equation} \label{e24}
(\varphi_p x'(t))'+ e^{x(t)} x'(t ) -
  (x^{9}(t )+x(t )-12)    =\cos^{2}{t},
\end{equation}
where   $p=\sqrt{5}$.  We can easily check the conditions (A1)
and  (A2) hold. By  Theorem \ref{thm1}, equation \eqref{e24}
has a unique positive $2\pi$-periodic solution.

Since the periodic solution of p-Laplacian equation \eqref{e24} is positive,
one can easily see that  the results of this paper are essentially new.


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\end{document}