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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 130, 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 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/130\hfil Fourth-order periodic boundary value
problems]
{Periodic boundary-value problems for fourth-order
differential equations with delay}

\author[S. A. Iyase\hfil EJDE-2011/130\hfilneg]
{Samuel A. Iyase}

\address{Samuel A. Iyase \newline
Department of Mathematics,
Computer Science and Information Technology,
Igbinedion University, Okada,
P.M.B. 0006, Benin City, Edo State, Nigeria}
\email{driyase2011@yahoo.com, iyasesam@gmail.com}

\thanks{Submitted June 3, 2011. Published October 11, 2011.}
\subjclass[2000]{34B15}
\keywords{Periodic solution; uniqueness, uniqueness; \hfill\break\indent
Carathoeodory conditions;  fourth order ODE; delay}

\begin{abstract}
 We study the periodic boundary-value problem
 \begin{gather*}
 x^{(iv)}(t)+f(\ddot{x})\dddot{x}(t)+b\ddot{x}(t)
 +g(t,\dot{x}(t-\tau))+dx=p(t)\\
 x(0)=x(2\pi),\quad \dot{x}(0)=\dot{x}(2\pi),\quad
 \ddot{x}(0)=\ddot{x}(2\pi),\quad \dddot{x}(0)=\dddot{x}(2\pi),
 \end{gather*}
 Under some resonant conditions on the asymptotic behaviour of the
 ratio $g(t,y)/(by)$ for $|y|\to\infty$. Uniqueness of periodic
 solutions is also examined.
\end{abstract}

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

\section{Introduction}

In this article we study the periodic boundary-value problem
\begin{equation}\label{1.1}
\begin{gathered}
x^{(iv)}(t)+f(\ddot{x})\dddot{x}(t)+b\ddot{x}
 +g(t,\dot{x}(t-\tau))+dx=p(t)\\
x(0)=x(2\pi),\quad \dot{x}(0)=\dot{x}(2\pi),\quad
\ddot{x}(0)=\ddot{x}(2\pi),\quad \dddot{x}(0)=\dddot{x}(2\pi),
\end{gathered}
\end{equation}
with fixed delay $\tau\in[0,2\pi)$,
$f:\mathbb{R}\to\mathbb{R}$ is a continuous function,
$P:[0,2\pi]\to\mathbb{R}$ and $g:[0,2\pi]\times\mathbb{R}\to\mathbb{R}$
are $2\pi-$periodic in $t$ and $g$ satisfies Caratheodory conditions
with $b$ and $d$ real constants. The unknown function
$x:[0,2\pi]\to\mathbb{R}$ is defined for $0<t\leq\tau$ by
$x(t-\tau)=[2\pi-(t-\tau)]$. We are concerned  with the existence
and  uniqueness of periodic solution of equation \eqref{1.1} under
some resonant conditions on $g$.

It is pertinent to note that fourth-order differential equations
with time delay are used to model problems in engineering and
biological or physiological systems. For instance, the oscillatory
movements of muscles that occur as a result of the interaction of
a muscle with its load (see \cite{a5}). For other papers dealing
with the study of fourth order differential equations with time
delay see \cite{a2,a3} and references therein.

In what follows, we shall use the spaces $C([0,2\pi]),
C^{k}([0,2\pi])$ and $L^{k}([0,2\pi])$ of continuous, $k$ times
continuously differentiable or measurable real functions whose
$kth$ power of the absolute value are lebesgue integrable. We
shall use the following Sobolev spaces:
\begin{align*}
W^{4,2}_{2\pi}
&=\big\{x:[0,2\pi]\to\mathbb{R}: x,\dot{x},\ddot{x},\dddot{x}
\text{are absolutely continuous on $[0,2\pi]$ and}\\
&\quad x(0)=x(2\pi),\;\dot{x}(0)=\dot{x}(2\pi),\;\ddot{x}(0)=\ddot{x}(2\pi),\;
\dddot{x}(0)=\dddot{x}(2\pi)\big\}
\end{align*}
with the norm
$$
|x|^2_{W^{4,2}_{2\pi}}=\sum^{4}_{i=0}\frac{1}{2\pi}
\int^{2\pi}_0|x^{i}(t)|^2dt
$$
and
\[
H^{1}_{2\pi}=\{x:[0,2\pi]\to\mathbb{R}:x
\text{ is absolutely continuous on $[0,2\pi]$ and
$\dot{x}\in L^2_{2\pi}$}\}
\]
with the norm
$$
|x|^2_{W^{4,2}_{2\pi}}=\Big(\frac{1}{2\pi}\int^{2\pi}_0x(t)dt\Big)
^2+\frac{1}{2\pi}\int^{2\pi}_0|\dot{x}|^2dt.
$$

\section{The Linear Problem}

We consider here the linear delay equation
\begin{equation}\label{2.1}
\begin{gathered}
x^{(iv)}(t)+a\dddot{x}(t)+b\ddot{x}(t)+c\dot{x}(t-\tau))+dx=0\\
 x(0)=x(2\pi),\quad \dot{x}(0)=\dot{x}(2\pi),\quad
\ddot{x}(0)=\ddot{x}(2\pi),\quad \dddot{x}(0)=\dddot{x}(2\pi),
\end{gathered}
\end{equation}
where $c$ is a real constant.

\begin{lemma}\label{lem:2.1}
Let $b<0$, $d>0$ and
\begin{equation}\label{2.2}
0<\frac{c}{b}<n
\end{equation}
where $n$ is an integer $n \geq 1$. Then  \eqref{2.1} has no  non-trivial
periodic solution for any fixed $\tau\in[0,2\pi)$.
\end{lemma}

\begin{proof}
We consider a solution of the form $x(t)=e^{\lambda t}$ where
$\lambda=in$ with $i^2=-1$. Then Lemma \ref{lem:2.1} will follow if
$$
\psi(n,\tau)=n^{4}-bn^2+cn\sin n\tau+d\neq 0
$$
for all $n\geq 1$ and $\tau\in[0,2\pi)$.
By \eqref{2.2}, we obtain
\begin{equation*}\begin{split}
b_{-1}\psi(n,\tau)
&=\frac{n^{4}}{b}-n^2+\frac{c}{b}n\sin n\tau+\frac{d}{b}\\
&\leq \frac{n^{4}}{b}-n^2+\frac{c}{b}n+\frac{d}{b}\\
&<\frac{n^{4}}{b}+\frac{d}{b}<0.
\end{split}
\end{equation*}
Therefore, $\psi(n,\tau)\neq 0$ and the result follows.
If $x\in L^{1}_{2\pi}$ we shall write
$$
\bar{x}=\frac{1}{2\pi}\int^{2\pi}_0x(t)dt,\quad
\tilde{x}(t)=x(t)-\bar{x}
$$
such that $\int^{2\pi}_0\tilde{x}(t)dt=0$.
\end{proof}

We consider next the delay equation
\begin{equation}\label{2.3}
\begin{gathered}
x^{(iv)}(t)+a\dddot{x}(t)+b\ddot{x}(t)+c(t)\dot{x}(t-\tau))+dx=0\\
x(0)=x(2\pi),\quad \dot{x}(0)=\dot{x}(2\pi),\quad
\ddot{x}(0)=\ddot{x}(2\pi),\quad \dddot{x}(0)=\dddot{x}(2\pi),
\end{gathered}
\end{equation}
where $a,b$ are constants and $c(t)\in L^2_{2\pi}$.

\begin{theorem}\label{thm:2.1}
Let $b<0$, $d>0$ and $\Gamma(t)=b^{-1}c(t)\in L^2_{2\pi}$.
 Suppose that
\begin{equation}\label{2.4}
0 < \Gamma(t)< 1\,.
\end{equation}
Then  \eqref{2.3} has no non-trivial periodic solution for every
fixed $\tau\in[0,2\pi)$.
\end{theorem}

\begin{proof}
Let $x(t)$ be any solution of \eqref{2.3}. Then
\begin{align*}
0&=\frac{1}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}(t)
 \Big[\frac{b^{-1}}{2\pi}\Big\{x^{(iv)}+a\dddot{x}+dx+\{\ddot{x}
 +\Gamma(t)\dot{x}(t-\tau )\}\Big\}\Big]dt\\
&=-\frac{b^{-1}}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}^2(t)dt
-\frac{db^{-1}}{2\pi}\int^{2\pi}_0\dot{\tilde{x}}^2(t)dt
+\frac{1}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}(t)[\dddot{x}(t)
+\Gamma(t)\dot{x}(t-\tau)]dt\\&\geq \frac{1}{2\pi}
\int^{2\pi}_0\ddot{\tilde{x}}(t)[\ddot{x}(t)
+\Gamma(t)\dot{x}(t-\tau)]dt\\
&=\int^{2\pi}_0[\ddot{\tilde{x}}^2(t)
 +\Gamma(t)\ddot{\tilde{x}}(t)\dot{x}(t-\tau)]dt\\
&=\frac{1}{2\pi}\int^{2\pi}_0\Big[\ddot{\tilde{x}}^2(t)
-\frac{\Gamma(t)}{2}\ddot{\tilde{x}}^2(t)
-\frac{\Gamma(t)}{2}\dot{\tilde{x}}^2(t-\tau)\Big]dt\\
&\quad +\frac{1}{2\pi}\int^{2\pi}_0\frac{\Gamma(t)}{2}
\Big[\ddot{\tilde{x}}(t)+\dot{x}(t-\tau)\Big]^2dt.
\end{align*}
In the above expression we used  the equality
$$
ab=\big(\frac{a+b}{2}\big)^2-\frac{a^2}{2}-\frac{b^2}{2}.
$$
From the periodicity of $\dot{x}(t)$, it follows that
$$
\frac{1}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}^2(t)dt
=\frac{1}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}^2(t-\tau)dt.
$$
Hence,
\begin{align*}
0&\geq\frac{1}{2}\Big[\frac{1}{2\pi}\int^{2\pi}_0[\ddot{\tilde{x}}^2(t)
-\Gamma(t)\ddot{\tilde{x}}^2(t)]dt\Big]\\
&= \frac{1}{2}\Big[\frac{1}{2\pi}\int^{2\pi}_0[\ddot{\tilde{x}}^2
(t-\tau)-\Gamma(t)\dot{\tilde{x}}^2(t-\tau)]dt\\
&\geq\delta|\dot{\tilde{x}}|^2_{H^{1}_{2\pi}}
=\delta|\dot{x}|_{H^{1}_{2\pi}}.
\end{align*}
By \cite[Lemma 1]{a4} where $\delta>0$ is a constant.
This implies that $x$ is  constant a.e. But since $d\neq 0$
we must have $x =0$, a. e.
\end{proof}

\section{The non-linear problem}

We  shall  consider here  a preliminary Lemma  which  will enable
us  obtain  a priori estimates required  for  our results.

\begin{lemma}\label{lem:3.1}
Let all the conditions of Lemma \ref{lem:2.1} hold and let
$\delta$ be related to $\Gamma(t)$ by Theorem \ref{thm:2.1}.
Suppose that $v\in L^2_{2\pi}$ and
$$
0<v(t)<\Gamma(t)+\epsilon\quad\text{a.e. }t\in[0,2\pi]
$$
holds for any $v\in L^2_{2\pi}$, where $\epsilon >0$.
Then
$$
\frac{1}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}(t)\Big[b^{-1}\{x^{(iv)}
+a\dddot{x}+dx\}+\ddot{x}+\Gamma(t)\dot{x}(t-\tau)\Big]dt
\geq(\delta-\epsilon)|\dot{x}|^2_{H^{1}_{2\pi}}.
$$
\end{lemma}

\begin{proof}
From the proof of Theorem \ref{thm:2.1}, we have
\begin{align*}
&\frac{1}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}(t)
\Big[b^{-1}\{x^{(iv)}+a\dddot{x}+dx\}+\ddot{x}+v(t)\dot{x}(t-\tau)
 \Big]dt\\
&\geq\frac{1}{2}\Big[\frac{1}{2\pi}\int^{2\pi}_0[\ddot{\tilde{x}}^2(t)
 -\Gamma(t)\ddot{\tilde{x}}^2(t)]dt\Big]
 +\frac{1}{2}\Big[\frac{1}{2\pi}\int^{2\pi}_0[\ddot{\tilde{x}}^2
 (t-\tau)-\Gamma(t)\dot{\tilde{x}}^2(t-\tau)]dt\Big]\\
&\quad -\epsilon\frac{1}{2\pi}\int^{2\pi}_0(\dot{\tilde{x}}^2
 (t-\tau)+\ddot{\tilde{x}}^2(t))dt\\
& \geq\frac{1}{2}\Big[\frac{1}{2\pi}\int^{2\pi}_0[\ddot{\tilde{x}}^2
 (t-\tau)-\Gamma(t)\dot{\tilde{x}}^2 (t-\tau)]dt\Big]
 -\frac{\epsilon}{2\pi}\int^{2\pi}_0\dot{x}^2 (t-\tau)\\
&\quad -\frac{\epsilon}{2\pi}\int^{2\pi}_0\ddot{\tilde{x}}^2(t-\tau)dt\\
&\geq\delta|\dot{\tilde{x}}|^2_{H^{1}_{2\pi}}
 -\epsilon|\ddot{\tilde{x}}|^2_{H^{1}_{2\pi}}\\
&\geq(\delta-\epsilon)|\dot{\tilde{x}}|^2_{H^{1}_{2\pi}}.
\end{align*}
\end{proof}

We shall consider the non-linear delay equation
\begin{equation}\label{3.1}
x^{(iv)}+f(\ddot{x})\dddot{x}+b\ddot{x}+g(t,\dot{x}(t-\tau))+dx=p(t)
\end{equation}
where $f:\mathbb{R}\to\mathbb{R}$ is a continuous function and
$g: [0 ,2\pi]\times \mathbb{R}\to\mathbb{R}$ are $2\pi$  periodic
 in $t$ and $g$ satisfies  Caratheodory condition; that
is, $g(\cdot, x)$ is measurable  on $[0, 2\pi]$ for each
$x\in\mathbb{R}$ and $g(t,\cdot)$ is continuous  on $\mathbb{R}$
for almost each  $t\in [0, 2\pi]$. We assume moreover that for
$r > 0$ there exists $Y_{r}\in L^2_{2\pi}$ such that
$|g(t,y)|\leq Y_{r}(t)$ for  a.e. $t\in[0, 2\pi]$ and $x\in [-r, r]$.

\begin{theorem}\label{thm:3.1}
Let $b<0$ and $d>0$. Suppose that $g$ is Caratheodory function
satisfying the inequality
\begin{gather}\label{3.2}
 g(t,y)\geq 0,\quad |y|\leq r \\
 \label{3.3} \lim_{|y|\to\infty}\sup\frac{g(t,y)}{by}
 \leq\Gamma (t)
\end{gather}
uniformly a.e., $t\in[0,2\pi]$ where $r>0$ is a constant and
$\Gamma(t)\in L^2_{2\pi}$ is such that
\begin{equation}\label{3.4}
0<\Gamma(t)<1
\end{equation}
Then for arbitrary continuous function $f$, the boundary-value
problem \eqref{3.1} has at least one  $2\pi$-periodic solution.
\end{theorem}

\begin{proof}
Let $\delta>0$ be associated to the function  $\Gamma$ by
Theorem \ref{thm:2.1}. Then by \eqref{3.2}, \eqref{3.3} there
exists a constant $R_1 > 0$ such that
\begin{equation}\label{3.5}
0\leq\frac{g(t,y)}{by}<\Gamma(t)+\frac{\delta}{2}
\end{equation}
if $|y|\geq R_1$ for a. e., $t\in[0,2\pi]$ and all $y\in\mathbb{R}$.
Define $\bar{Y}:[0,2\pi]\times\mathbb{R}\to\mathbb{R}$ by
\begin{equation}\label{3.6}
\overline{Y}=\begin{cases}
y^{-1}g(t,y), &|y|\geq R_1\\
R^{-1}g(t,R), & 0<y<R_1\\
 -R^{-1}_1g(t,-R_1),& -R_1<y<0\\
\Gamma(t),& y=0.
\end{cases}
\end{equation}
Then by \eqref{3.5}, we have
\begin{equation}\label{3.7}
0\leq\overline{Y}(t,y)<\Gamma(t)+\frac{\delta}{2}
\end{equation}
for a. e. $t\in[0,2\pi]$ for all $y\in\mathbb{R}$.
Moreover the function $\overline{Y}(t,y)$ satisfies Caratheodory
conditions and
$$
\tilde{g}(t,\dot{x}(t-\tau))=b^{-1}g(t,\dot{x}(t-\tau))
-\overline{Y}(t,\dot{x}(t-\tau))\dot{x}(t-\tau)
$$
is such that a. e.
$t\in[0,2\pi]$ and all $x\in\mathbb{R}$, we have
\begin{equation}\label{3.8}
 |\tilde{g}(t,\dot{x}(t-\tau))|\leq\alpha (t)
\end{equation}
for some $\alpha(t)\in L^2_{2\pi}$.
To prove  that  \eqref{3.1} has at  least  one  periodic  solution,
it  suffices to show  that the  possible  solution  of the family
of equations
\begin{equation}\label{3.9}
\begin{split}
&b^{-1}[x^{(iv)}+\lambda f(\ddot{x})\dddot{x}]+\ddot{x}
+(1-\lambda)\Gamma(t)\dot{x}(t-\tau)+\lambda Y(t,\dot{x}(t-\tau))\\
&+b^{-1}dx+\lambda\tilde{g}(t,\dot{x}(t-\tau))
+ \overline{Y}(t,\dot{x}(t-\tau))=\lambda b^{-1}p(t)
\end{split}
\end{equation}
are a-priori bounded in $W^{4,2}_{2\pi}$ independently of
$\lambda\in[0,1]$. By inequality \eqref{3.7} one has
\begin{equation}\label{3.10}
0\leq(1-\lambda)\Gamma(t)+\lambda\overline{Y}(t,\dot{x}(t-\tau))
\leq\Gamma(t)+\frac{\delta}{2}
\end{equation}
for a. e. $t\in[0,2\pi]$ and all $x\in \mathbb{R}$.
From Theorem \ref{thm:2.1}, we can derive that for $\lambda=0$
equation \eqref{3.9} has only the trivial solution.
Then using Lemma \ref{lem:3.1} and Cauchy Schwarz inequality
we obtain
\begin{align*}
&0=\frac{1}{2\pi}\int^{2\pi}_0\ddot{x}\Big\{b^{-1}[x^{(iv)}
 +f(\ddot{x})\dddot{x}]+\ddot{x}+(1-\lambda)\Gamma(t)\dot{x}(t-\tau)\\
&\quad +\lambda\overline{Y}(t,\dot{x}(t-\tau))\dot{x}(t-\tau)
 +\lambda\tilde{g}(t,\dot{x}(t-\tau))+b^{-1}dx-\lambda p(t)\Big\}dt\\
&\geq\frac{\delta}{2}|\dot{x}|^2_{H^{1}_{2\pi}}-(|\alpha|_2
 +|b^{-1}||p|_2)|\dot{\tilde{x}}|_2+|b^{-1}|d|\dot{\tilde{x}}|_2\\
&\geq\frac{\delta}{2}|\dot{x}|^2_{H^{1}_{2\pi}}
 -\beta|\dot{x}|_{H^2_{2\pi}}-b^{-1}|\dddot{x}|^2_{2\pi}\\
&\geq \frac{\delta}{2}|\dot{x}|^2_{H^{1}_{2\pi}}
 -\beta|\dot{x}|_{H^{1}_{2\pi}}
\end{align*}
for some $\beta>0$. Hence,
\begin{equation}\label{3.11}
|\dot{x}|_{H^{1}_{2\pi}}\leq\frac{2\beta}{\delta}=c_1,
\end{equation}
with $c_1>0$. This implies
\begin{gather}
\label{3.12} |\ddot{x}|_2\leq c_2 \\
\label{3.13} |\ddot{x}|_{\infty}\leq c_3
\end{gather}
where $c_2>0$ and $c_3>0$.
 Using Wirtinger's inequality in \eqref{3.12}, we obtain
\begin{equation}\label{3.14}
|\dot{x}|_2\leq c_4
\end{equation}
with $c_4>0$. Multiplying \eqref{3.9} by $-\ddot{x}(t)$ and
integrating over $[0,2\pi]$, we obtain
$$
|\dddot{x}|^2_2\leq|\ddot{x}|^2_2|1+\frac{\delta}{2}|\ddot{x}|_2
+|\alpha|_2+d|\dot{x}|_2+|p|_2|\ddot{x}|_2
$$
Applying Wirtingers inequality we obtain
\begin{equation}\label{3.15}
|\dddot{x}|^2_2\leq c_5
\end{equation}
with $c_5>0$ and hence
$$
|\ddot{x}|_{\infty}\leq c_6
$$
with $c_6>0$. We multiply \eqref{3.9} by $x^{(iv)}(t)$ and integrate
over $[0,2\pi]$ to get
\begin{align*}
-b^{-1}|x^{(iv)}|^2_2
&\leq |f(\ddot{x})|_{\infty}|\ddot{x}|_2|x^{(iv)}|_2|b^{-1}|
+|\ddot{x}|_2|x^{(iv)}|_2+|1+\frac{\delta}{2}||\dot{x}|_2|x^{i(iv)}|_2\\
&\quad +|b^{-1}||d||\ddot{x}|_2+|\alpha|_2|x^{(iv)}|_2
 +|p|_2|x^{i(iv)}|_2\\
&\leq|f(\ddot{x})|_{\infty}|\ddot{x}|_2|x^{(iv)}|_2|b^{-1}|
 +|\ddot{x}|_2|x^{(iv)}|_2\\
&\quad +|1+\frac{\delta}{2}||\dot{x}|_2|x^{(iv)}|_2|b^{-1}|d|x^{(iv)}|_2
 +|\alpha|_2|x^{(iv)}|_2+|p|_2|x^{i(iv)}|_2|b^{-1}|,
\end{align*}
where we used the Wirtinger's inequality.
Thus
\begin{equation}\label{3.16}
|x^{(iv)}|_2\leq c_7
\end{equation}
with $c_7>0$. Finally multiplying \eqref{3.9} by $x(t)$ and
integrating over $[0,2\pi]$ we obtain
\begin{equation}\label{3.17}
|x|_2\leq c_8
\end{equation}
with $c_8>0$. Hence,
$$
|x|_{W^{4,2}_{2\pi}}=|x|_2+|\dot{x}|_2+|\ddot{x}|_2
+|\dddot{x}|_2+|x^{(iv)}|_2\leq c_8+c_4+c_2+c_5+c_7=C_{9}
$$
Taking $R>C_{9}>0$, the required a priori bound in
$W^{4,2}_{2\pi}$ is obtained independently of $x$ and $\lambda$.
\end{proof}

\section{Uniqueness Result}

For $f(x) = a$, $a$ constant, in \eqref{1.1},  we have the
following uniqueness result.

\begin{theorem}\label{thm:4.1}
Let $a, b, d$ be constants with $b <0$ and $d >0$. Suppose $g$
is a Caratheodory function satisfying
\begin{equation}\label{4.1}
0<\frac{g(t,\dot{x}_1)-g(t,\dot{x}_2)}{b(\dot{x}_1-\dot{x}_2)}
<\Gamma(t)
\end{equation}
for all $x_1,x_2\in\mathbb{R}$ with $x_1\neq x_2$ where
$\Gamma(t)\in L^2_{2\pi}$ is such that $0<\Gamma(t)<1$.
Then for all arbitrary constant $a$ and every $\tau\in[0,2\pi)$ the
 boundary-value  problem
\begin{equation}\label{4.2}
\begin{gathered}
 x^{(iv)}(t)+a\dddot{x}+b\ddot{x}+g(t,\dot{x}(t-\tau))+dx=p(t)\\
 x(0)=x(2\pi),\dot{x}(0)=\dot{x}(2\pi),\ddot{x}(0)=\ddot{x}(2\pi),
 \dddot{x}(0)=\dddot{x}(2\pi),
\end{gathered}
\end{equation}
has at most one solution.
\end{theorem}

\begin{proof}
Let $x_1,x_2$ be any two solutions of \eqref{4.2}.
Set $x =x_1-x_2$. Then $x$ satisfies the boundary value problem
\begin{gather*}
b^{-1}x^{(iv)}(t)+a\dddot{x}+\Gamma(t)\dot{x}(t-\tau)+b^{-1}dx=0\\
x(0)=x(2\pi),\quad \dot{x}(0)=\dot{x}(2\pi),\quad
\ddot{x}(0)=\ddot{x}(2\pi),\quad \dddot{x}(0)=\dddot{x}(2\pi),
\end{gather*}
where the function $\Gamma(t)\in L^2_{2\pi}$ is defined by
$$
\Gamma(t)=\begin{cases}
\frac{g(t,\dot{x}_1(t-\tau))-g(t,\dot{x}_2(t-\tau))}{\dot{x}(t)}
&\text{if }\ddot{x}(t)\neq 0\\
\frac{1}{2} &\text{if }\ddot{x}(t)=0
\end{cases}
$$
if $\dot{x}(t)$ on every subset of $[0, 2\pi]$ of positive measure,
then $x$ is  constant Since $d\neq 0$ we must have $x =0$ and
hence $x_1 = x_2$ a.e.
Suppose on the other hand that $\dot{x}(t)\neq 0$ on a certain
subset of $[0, 2\pi]$ of  positive measure, then  using
the arguments  of Theorem \ref{thm:2.1} we  obtain that $x =0$
and hence $x_1 = x_2$ a .e.
\end{proof}

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 Wurzburg (1982) Lecture Note Math No.  1017.
 Springer Verlag Berlin,  (1983).

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