\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 138, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/138\hfil Convergence of solutions]
{Convergence of solutions for a fifth-order nonlinear
differential equation}

\author[O. A. Adesina, A. S. Ukpera\hfil EJDE-2007/138\hfilneg]
{Olufemi Adeyinka Adesina, Awar Simon Ukpera}  % in alphabetical order

\address{Olufemi Adeyinka Adesina \newline
Department of Mathematics, Obafemi Awolowo University \\
Ile-Ife, Nigeria}
\email{oadesina@oauife.edu.ng}

\address{Awar Simon Ukpera \newline
Department of Mathematics, Obafemi Awolowo University \\
Ile-Ife, Nigeria}
\email{aukpera@oauife.edu.ng}

\thanks{Submitted May 7, 2007. Published October 17, 2007.}
\subjclass[2000]{34D20}
\keywords{Convergence of solutions; nonlinear fifth order equations;
\hfill\break\indent
 Routh-Hurwitz interval; Lyapunov functions}

\begin{abstract}
 In this paper, we present sufficient conditions for all solutions of
 a fifth-order nonlinear differential equation to converge. In this context,
 two solutions converge to each other if their difference and those of their
 derivatives up to order four approach zero as time approaches infinity.
 The nonlinear functions involved are not necessarily differentiable,
 but satisfy certain increment ratios that lie in the closed
 sub-interval of the Routh-Hurwitz interval.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Nonlinear differential equations of higher order have been
extensively studied with high degree of generality. In particular,
there have been interesting works on asymptotic behaviour,
boundedness, periodicity, almost periodicity and stability of
solutions for fifth-order nonlinear differential equations.
Authors that have worked in this direction include Abou-El-Ela and
Sadek \cite{a1,a2,a3}, Adesina \cite{a4,a5,a6},
Afuwape and Adesina \cite{a9,a10}, Chukwu \cite{c1,c2},
 Sadek \cite{s1} and Tunc \cite{t2,t3,t4,t5,t6}, to mention a few.
Most of the nonlinear functions involved in these works were assumed to be
differentiable, specially, the restoring terms. Specifically, in
1975 and 1976 respectively, Chukwu \cite{c1,c2} discussed the
boundedness and stability of the solutions of the differential
equations
\begin{equation}
x^{(v)}+f_{1}(x,x',x'',{x'''},x^{(iv)})x^{(iv)}+b{x'''}
+f_3(x'')+f_4(x')+f_{5}(x)=p(t)\label{e1.1}
\end{equation}
and
\begin{equation}
x^{(v)}+ax^{(iv)}+f_{2}{(x''')}+cx''+f_4(x')+f_{5}(x)=
p(t,x,x',x'',x''',x^{(iv)}).\label{e1.2}
\end{equation}
Later, Yu \cite{y1} studied the boundedness and asymptotic stability
of the solutions of the differential equation
\begin{equation}
x^{(v)}+\phi(x,x',x'',{x'''},x^{(iv)})x^{(iv)}+b{x'''}
+h(x'')+g(x')+f(x)=p(t,x,x',x'',x''',x^{4}).\label{e1.3}
\end{equation}

Other interesting results on the boundedness and stability of
solutions for equations of the form \eqref{e1.3} were obtained by
Abou-El-Ela and Sadek \cite{a1}, Tiryaki and Tunc \cite{t1}
 and Tunc \cite{t2}. In
the case where the fifth order differential equations were
non-autonomous, the asymptotic behaviour of solutions were treated
by Abou-El-Ela and Sadek \cite{a3}, Sadek \cite{s1} and
Tunc \cite{t3,t4,t5,t6}. Some of
the results in these works have been generalized to real vector
differential equations, see for instance Abou-El-Ela and Sadek
\cite{a2}. All the above mentioned works were done by using the
Lyapunov's second method except for the works of Adesina \cite{a4,a5,a6} and
Afuwape and Adesina \cite{a9,a10}, where the frequency domain technique
was employed to study some qualitative behaviour of solutions.

However, the problem of convergence of solutions to these
equations in which the nonlinear terms are not necessarily required
to be differentiable, has so far remained intractable.
The purpose of this paper therefore is to tackle this problem.
Motivation for this study comes from the works of Afuwape \cite{a7,a8}
and Ezeilo \cite{e1} where sufficient conditions for the convergence
of solutions of fourth and third order equations were proved respectively.

\noindent\textbf{Definition}
Two solutions $x_1(t)$, $x_2(t)$ of the equation \eqref{e1.4}
are said to converge (to each other) if $x_1-x_2\to 0$, ${x_1}'-{x_2}'\to 0$,
${x_1}''-{x_2}''\to 0$,
${x_1}'''-{x_2}'''\to 0$, ${x_1}^{(iv)}-{x_2}^{(iv)}\to 0$ as
$t\to \infty$.

In this paper, we shall investigate the convergence of solutions
for equation
\begin{equation}
x^{(v)}+ax^{(iv)}+bx'''+f(x'')+g(x')+h(x)=
p(t,x,x',x'',x''',x^{(iv)}),\label{e1.4}
\end{equation}
where $a$, $b$ are positive constants, functions $f$, $g$, $h$ and $p$
are real valued and continuous in their respective arguments such that the
uniqueness theorem is valid, and the solutions are continuously
dependent on the initial conditions. Moreover, $f(0)=g(0)=h(0)=0$.
Our results assert the existence of convergence of solutions with
the functions $f$, $g$, and $h$ not necessarily differentiable.
Here, the functions $h$ and $g$ are only required to
satisfy the increment ratios
\begin{gather*}
\frac{h(\zeta+\eta)-h(\zeta)}{\eta}\in I_0, \\
\frac{g(\zeta+\eta)-g(\zeta)}{\eta}\in I_1,
\end{gather*}
where $I_0$ and $I_1$ are closed sub-intervals of the Routh-Hurwitz interval.
Our results generalize, to fifth-order equations, the results in
\cite{a7,a8}.
Some existing results on fifth-order nonlinear differential equations
are also generalized.

\section{Assumptions and Main Results}

\noindent\textbf{Assumptions:}
\begin{enumerate}
\item  The function $p(t,x,x',x'',x''',x^{(iv)})$ is equal to
$q(t)+r(t,x,x',x'',x''',x^{(iv)})$ with $r(t,0,0,0,0,0)=0$ for all $t$;

\item For some positive constants $a$, $b$, $\alpha$, $\beta$ and $\Delta_0$,
$(ab-\alpha)\alpha-a^2\beta>0$, $(ab-\alpha)\alpha+a\Delta_0 >0$,
$(ab-\alpha)>0$ and $b^2>\beta$;

\item For some positive constants $a$, $b$, $\alpha$, $\beta$, $\Delta_0$,
$\Delta_1$ $K_0$ and $K_1$, the intervals
\begin{gather*}
I_0\equiv \Big[\Delta_0,K_0\big[\frac{[(ab-\alpha)\alpha-a^2\beta]}{a}\big]\Big],
\\
I_1\equiv \Big[\Delta_1,K_1\big[\frac{[(ab-\alpha)\alpha+a\Delta_0]}{a^2}
\big]\Big]
\end{gather*}
are in  the Routh-Hurwitz interval.
\end{enumerate}

The following results are proved.

\begin{theorem} \label{thm2.1}
In addition to the basic assumptions and 1--3 above, we assume that
\begin{itemize}
\item[(i)] there are positive constants $\alpha$, $\alpha_0$, $\beta$ and
$\beta_0$ such that
\begin{gather}
\alpha\leq\frac{f(z_2)-f(z_1)}{z_2-z_1}\leq\alpha_0,\quad z_2\not=z_1,
\label{e2.1} \\
\beta\leq\frac{g(y_2)-g(y_1)}{y_2-y_1}\leq\beta_0,\quad y_2\not=y_1;
\label{e2.2}
\end{gather}

\item[(ii)] for any $\zeta, \eta$, $(\eta\not=0)$, the increment ratios
for $h$ and $g$ satisfy
\begin{gather*}
\frac{h(\zeta+\eta)-h(\zeta)}{\eta}\in I_0, \\
\frac{g(\zeta+\eta)-(\zeta)}{\eta}\in I_1;
\end{gather*}

\item[(iii)] there is a continuous function $\phi(t)$ such that
\begin{equation}
\begin{aligned}
& |r(t, x_1, y_1, z_1, u_1, v_1)-r(t, x_2, y_2, z_2,
u_2, v_2)| \\
& \leq \phi(t)(|x_1-x_2|+|y_1-y_2|+|z_1-z_2|+|u_1-u_2|+|v_1-v_2|)
\end{aligned}\label{e2.3}
\end{equation}
holds for arbitrary $t, x_1, y_1, z_1, u_1, v_1, x_2, y_2, z_2, u_2, v_2$.
\end{itemize}
Then if there exists a constant ${D}_1$ such that
\begin{equation}
\int_0^t{\phi^{\varrho}(\tau)d\tau}\leq {D}_1\label{e2.4}
\end{equation}
for some $\varrho$ with $1\leq\varrho\leq 2$,
then all solutions of  \eqref{e1.4} converge.
\end{theorem}

\begin{theorem} \label{thm2.2}
Assume the conditions in the Theorem \ref{thm2.1} are satisfied.
Let $x_1(t)$, $x_2(t)$ be any two  solutions of
\eqref{e1.4}. Then for each fixed $\varrho$, $1\leq\varrho\leq2$,
there are constants $D_2$, $D_3$ and $D_4$ such that for $t_2\geq t_1$,
\begin{equation}
S(t_2)\leq
D_{2}S(t_1)\exp\big\{-D_3(t_2-t_1)+D_4\int_{t_1}^{t_2}{{\phi}^{\varrho}
(\tau)d\tau}\big\},\label{e2.5}
\end{equation}
where
\begin{equation}
\begin{aligned}
S(t)&=(x_2(t)-x_1(t))^2+({x_2}'(t)-{x_1}'(t))^2+({x_2}''(t)-{x_1}''(t))^2\\
&\quad +({x_2}'''(t)-{x_1}'''(t))^2+({x_2}^{(iv)}(t)-{x_1}^{(iv)}(t))^2.
\end{aligned} \label{e2.6}
\end{equation}
\end{theorem}

\begin{remark} \label{rmk2.3} \rm
If $p=0$ and the hypotheses (i) and (ii) of the Theorem \ref{thm2.1}
hold for arbitrary $\eta\not=0$, then the trivial solution of  \eqref{e1.4}
is exponentially stable.
\end{remark}

\begin{remark} \label{rmk2.4} \rm
If $p=0$ and the hypotheses (i) and (ii) of the Theorem \ref{thm2.1}
hold for arbitrary $\eta\not=0$, and $\zeta=0$, then there exists
a constant $D_5>0$ such that every solution $x(t)$ of  \eqref{e1.4} satisfies
\[
|x(t)|\leq D_5;\quad |x'(t)|\leq D_5;\quad
|x''(t)|\leq D_5;\quad |x'''(t)|\leq D_5;\quad
|x^{(iv)}(t)|\leq D_5.
\]
\end{remark}

For the rest of this article, $D_1, D_2, D_3, \dots$ and the ${D^*}$'s
stand for positive constants. Their identities are preserved throughout
this paper.

\section {Proof of Main Results}

\begin{proof}[Proof of Theorem \ref{thm2.2}]
It is convenient here to consider  \eqref{e1.4} as the equivalent system
\begin{equation}
\begin{aligned}
 x' &=  y, \\ y' &=  z,\\
z' &=  u, \\ u' &=  v+Q(t), \\
v' &= -av-bu-f(z)-g(y)-h(x)+r(t,x,x',x'',x''',x^{(iv)})-aQ(t),
\end{aligned} \label{e3.1}
\end{equation}
where
$Q(t)=\int_0^t{q(\tau) d\tau}$.
Let $x_i(t), y_i(t), z_i(t), u_i(t), v_i(t)$, $(i=1,2)$, be two solutions of
\eqref{e1.4}, such that inequalities \eqref{e2.1}, \eqref{e2.2},
\begin{gather*}
\Delta_0\leq \frac{h(x_2)-h(x_1)}{x_2-x_1} \leq
K_0\big[\frac{[(ab-\alpha)\alpha-a^2\beta]}{a}\big],\\
\Delta_1\leq \frac{g(y_2)-g(y_1)}{y_2-y_1} \leq
K_1\big[\frac{[(ab-\alpha)\alpha+a\Delta_0]}{a^2}\big]
\end{gather*}
are satisfied.
The main tool in the proofs of the convergence theorems will be the
function
\begin{equation}
\begin{aligned}
2V =& \beta^2[1-\epsilon]x^2+[\alpha^2+\frac{b\beta(\alpha
  +\alpha(1-\epsilon))}{1-\epsilon}+\frac{a\beta^2\epsilon}{\alpha(1
  -\epsilon)}]y^2\\
&\quad +[b^2+\frac{(b^2-\beta)\epsilon}{1-\epsilon}+\frac{{\epsilon}^2\beta}{1
 -\epsilon}+\epsilon
 \beta\frac{(ab-\alpha)}{\alpha(1-\epsilon)}]z^2+a^2u^2
 +[1+\frac{\epsilon}{1-\epsilon}]v^2\\
&\quad +2\alpha\beta(1-\epsilon)xy+2b\beta(1-\epsilon)xz
 +2a\beta(1-\epsilon)xu+2\beta(1-\epsilon)xv\\
&\quad  +2(b\alpha+a\beta\epsilon)yz+2[a\alpha+\beta\epsilon
 +\frac{\beta\epsilon}{1-\epsilon}]yu
 +2\alpha yv +2ab(1-\epsilon)zu\\
&\quad +2[b+\frac{(a\alpha+b)\epsilon}{1-\epsilon}]zv+2auv,
\end{aligned} \label{e3.2}
\end{equation}
where $0<\epsilon<1$, $ab-\alpha>0$ and $b^2>\beta$.
Indeed we can rearrange the terms in \eqref{e3.2} to obtain
\begin{equation}
2V=2V_1+2V_2+2V_3+2V_4+2V_5+2V_6,\label{e3.3}
\end{equation}
where
\begin{gather*}
2V_1 =[\beta(1-\epsilon)x+\alpha y+bz+\frac{au}{2}+v]^2
 +\frac{{\epsilon}^2}{1-\epsilon}z^2
 +2\frac{(a\beta+b)}{1-\epsilon}zv+\frac{\epsilon}{2(1-\epsilon)}v^2;\\
2V_2 =\beta^2(1-\epsilon)\epsilon x^2+a\beta(1-\epsilon)xu
 +\frac{1}{8}a^2u^2;\\
2V_3 =b\beta\frac{(\epsilon+\epsilon(1-\epsilon))}{1-\epsilon}y^2
 +2[\frac{a\alpha}{2}+\frac{\beta\epsilon}{1-\epsilon}+\beta\epsilon]yu
 +\frac{1}{8}a^2u^2;\\
2V_4 =\frac{a\beta^2\epsilon}{\alpha(1-\epsilon)}y^2+2a\beta\epsilon yz
 +\epsilon \beta\frac{(ab-\alpha)}{\alpha(1-\epsilon)}z^2;\\
2V_5 =\frac{(b^2-\beta)\epsilon}{2(1-\epsilon)}z^2
 +\frac{(ab-2ab\epsilon)}{2}zu+\frac{1}{8}a^2u^2;\\
2V_6 =\frac{a^2u^2}{4}+auv+\frac{\epsilon}{2(1-\epsilon)}v^2.
\end{gather*}
We note that $V_1$ is obviously positive definite.
This follows from the condition $0<\epsilon<1$. Also $V_i, i=2, 3, \dots , 6$
 regarded as quadratic forms in $x$ and $u$, $y$ and $u$, $y$ and $z$, $z$
 and $u$, $z$ and $v$, $u$ and $v$ respectively is always positive.
Let us recall that a real $2\times2$ matrix
\[
\begin{pmatrix}a_1&a_2\cr a_3&a_4\end{pmatrix}
\]
 is positive definite if and only if it is symmetric, and the elements
$a_1$, $a_4$ and $a_1a_4-a_2a_3$ are non negative. Thus we can rearrange
the terms in $V_2$ as
\[
\begin{pmatrix}x ,&u\end{pmatrix}
\begin{pmatrix}\beta^2(1-\epsilon)&a\beta\frac{(1-\epsilon)}{2}\\
 a\beta\frac{(1-\epsilon)}{2}&\frac{a^2}{8} \end{pmatrix}
\begin{pmatrix}x\\ u \end{pmatrix},
\]
from which we have $\frac{2}{3}<\epsilon<1$ as a condition for the positive
semi-definiteness. Similarly, for $V_3$, we have
\[
a^2b\beta\epsilon\frac{(2-\epsilon)}{1-\epsilon}\geq
\big[\frac{a\alpha}{2}+\frac{\beta\epsilon}{1-\epsilon}+\beta\epsilon\big]^2
\]
as a condition for its positive semi-definiteness.
As for $V_4$ and $V_5$, we have
\[
(ab-\alpha)\beta\geq a\alpha^2(1-\epsilon)^2\quad\mbox{and}\quad
a^2(b^2-\beta)\geq (ab-2ab\epsilon)^2\frac{(1-\epsilon)}{\epsilon}
\]
as conditions for the positive semi-definiteness. The condition for
positive semi-definiteness of $V_6$ is the same as that for $V_1$.
Hence $V$ is positive definite. We can therefore find a constant $D_6>0$,
such that
\begin{equation}
D_6(x^2+y^2+z^2+u^2+v^2)\leq V.\label{e3.4}
\end{equation}
Furthermore, by using the Schwartz inequality
$|x||u|\leq \frac{1}{2}(x^2+u^2)$, then $2|V_2|\leq {D_1}^{*}(x^2+u^2)$
for some ${D_1}^{*}={D_1}^{*}(a,\beta,\epsilon)>0$. Similarly, we obtain
the following estimates:
\begin{gather*}
2|V_3|\leq {D_2}^{*}(y^2+u^2),\quad
{D_2}^{*}={D_2}^{*}(a,b,\alpha,\beta,\epsilon)>0,\\
2|V_4|\leq {D_3}^{*}(y^2+z^2),\quad
{D_3}^{*}={D_3}^{*}(a,b,\alpha,\beta,\epsilon)>0,\\
2|V_5|\leq {D_4}^{*}(z^2+u^2),\quad
{D_4}^{*}={D_4}^{*}(a,b,\alpha,\beta,\epsilon)>0, \\
2|V_6|\leq {D_5}^{*}(u^2+v^2),\quad
{D_5}^{*}={D_5}^{*}(a,\epsilon)>0.
\end{gather*}
 Thus there exists a constant $D_7>0$, such that
\begin{equation}
 V \leq D_7(x^2+y^2+z^2+u^2+v^2),\label{e3.5}
\end{equation}
where
\[
D_7=\max\big\{{D_1}^{*};{D_2}^{*};{D_3}^{*};{D_4}^{*};{D_5}^{*}\big\}.
\]
Using inequalities \eqref{e3.4} and \eqref{e3.5}, we obtain
\begin{equation}
D_6(x^2+y^2+z^2+u^2+v^2)\leq V\leq D_7(x^2+y^2+z^2+u^2+v^2).\label{e3.6}
\end{equation}
\end{proof}

The following result can be easily verified for $W\equiv V$.

\begin{lemma} \label{lem3.1}
Let the function $W(t)=W(x_2-x_1, y_2-y_1, z_2-z_1, u_2-u_1, v_2-v_1)$
be defined by
\begin{align*}
2W &= \beta^2[1-\epsilon](x_2-x_1)^2+[\alpha^2
  +\frac{b\beta(\alpha+\alpha(1-\epsilon))}{1-\epsilon}
  +\frac{a\beta^2\epsilon}{\alpha(1-\epsilon)}](y_2-y_1)^2\\
&\quad +\big[b^2+\frac{(b^2-\beta)\epsilon}{1-\epsilon}+\frac{{\epsilon}^2\beta}{1-\epsilon}+\epsilon
 \beta\frac{(ab-\alpha)}{\alpha(1-\epsilon)}\big](z_2-z_1)^2
 +a^2(u_2-u_1)^2\\
&\quad +[1+\frac{\epsilon}{1-\epsilon}](v_2-v_1)^2
 +2\alpha\beta(1-\epsilon)(x_2-x_1)(y_2-y_1)\\
&\quad +2b\beta(1-\epsilon)(x_2-x_1)(z_2-z_1)
 +2a\beta(1-\epsilon)(x_2-x_1)(u_2-u_1)\\
&\quad +2\beta(1-\epsilon)(x_2-x_1) (v_2-v_1)
 +2(b\alpha+a\beta\epsilon)(y_2-y_1)(z_2-z_1)\\
&\quad +2[a\alpha+\beta\epsilon+\frac{\beta\epsilon}{1-\epsilon}](y_2-y_1)
  (u_2-u_1)
 +2\alpha(y_2-y_1)(v_2-v_1) \\
&\quad +2ab(1-\epsilon)(z_2-z_1)(u_2-u_1)
 +2[b+\frac{(a\alpha+b)\epsilon}{1-\epsilon}](z_2-z_1)(v_2-v_1) \\
&\quad +2a(u_2-u_1)(v_2-v_1),
\end{align*} %\label{e3.7}
where $0<\epsilon<1$ and $W(0,0,0,0,0)=0$, then there exist finite
constants $D_6>0$, $D_{7}>0$ such that
\begin{equation}
\begin{aligned}
 & D_6\big\{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2+(u_2-u_1)^2+(v_2-v_1)^2\big\}\\
&\leq W\\
&\leq D_{7}\big\{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2+(u_2-u_1)^2+(v_2-v_1)^2
\big\}.
\end{aligned} \label{e3.8}
\end{equation}
\end{lemma}

\begin{proof} These inequalities follows from the verification of $W$
as a Lyapunov function and the fact that the solutions
$(x_i, y_i, z_i, u_i, v_i+Q(t))$, $(i=1,2)$,
satisfy the system \eqref{e3.1}. Then $S(t)$ as defined in  \eqref{e2.6}
becomes
\begin{align*}
S(t)&=\big\{(x_2(t)-x_1(t))^2+(y_2(t)-y_1(t))^2+(z_2(t)-z_1(t))^2\\
&\quad +(u_2(t)-u_1(t))^2+(v_2(t)-v_1(t))^2\big\}.
\end{align*}
\end{proof}

Next we prove a result on the derivative of $W(t)$
with respect to $t$.

\begin{lemma} \label{lem3.2}
Let the hypotheses (i) and (ii) of the Theorem \ref{thm2.1} hold, then
there exist positive constants $D_8$ and $D_9$ such that
\begin{equation}
\frac{dW}{dt}\leq -2D_{8}S+D_{9}S^{1/2}|\theta| \label{e3.9}
\end{equation}
where
$\theta=r(t,x_2, y_2, z_2, u_2, v_2+Q)-r(t,x_1, y_1, z_1, u_1, v_1+Q)$.
\end{lemma}

\begin{proof}
Using the system \eqref{e3.1}, a direct computation of $\frac{dW}{dt}$
gives after simplification
\begin{equation}
\frac{dW}{dt}=-W_1+W_2,\label{e3.10}
\end{equation}
where
\begin{align*}
W_1 & = \beta(1-\epsilon)H(x_2,x_1)(x_2-x_1)^2+\alpha\beta\epsilon(y_2-y_1)^2\\
&\quad +\frac{1}{\alpha(1-\epsilon)}(b\alpha+a\beta\epsilon)
 (\alpha(1-\epsilon)-1)(z_2-z_1)^2\\
&\quad +ab\epsilon(u_2-u_1)^2+\frac{1}{1-\epsilon}a\epsilon(v_2-v_1)^2+
 (G(y_2,y_1)-\beta)[\beta(1-\epsilon)(x_2-x_1)\\
&\quad +\alpha(y_2-y_1) -\frac{b\alpha+a\beta\epsilon}{\alpha(1-\epsilon)}
 (z_2-z_1)+a(u_2-u_1)+\frac{(v_2-v_1)}{1-\epsilon}](y_2-y_1)\\
&\quad +\big(F(z_2,z_1)-\alpha\big)\Big[\beta(1-\epsilon)(x_2-x_1)+\alpha(y_2-y_1)
 +a(u_2-u_1)\\
&\quad +\frac{(v_2-v_1)}{1-\epsilon}\Big](z_2-z_1)\\
&\quad +2(\beta\epsilon+ab\alpha)(y_2-y_1)(u_2-u_1)-\frac{\beta\epsilon(1
 +(1-\epsilon))}{1-\epsilon}(z_2-z_1)(u_2-u_1)\\
&\quad +\frac{1}{1-\epsilon}[\alpha-\frac{1}{\alpha}(ab\alpha(1-(1-\epsilon))
 -a^2\beta\epsilon)](z_2-z_1)(v_2-v_1);
\end{align*} %\label{e3.11}
\begin{equation}
\begin{aligned}
W_2&=\theta(t)[\beta(1-\epsilon)(x_2-x_1)+\alpha(y_2-y_1)-\frac{a\beta\epsilon
+b\alpha}{\alpha(1-\epsilon)}(z_2-z_1)\\
&\quad +a(u_2-u_1)+\frac{(v_2-v_1)}{1-\epsilon}];
\end{aligned}\label{e3.12}
\end{equation}
\begin{gather*}
F(z_2,z_1) =\frac{f(z_2)-f(z_1)}{z_2-z_1},\quad z_2\not=z_1;\\
G(y_2,y_1) =\frac{g(y_2)-g(y_1)}{y_2-y_1},\quad y_2\not=y_1;\\
H(x_2,x_1) =\frac{h(x_2)-f(x_1)}{x_2-x_1},\quad x_2\not=x_1.
\end{gather*}
Let $\chi_1=G(y_2,y_1)-\beta$ and $\chi_2=F(z_2,z_1)-\alpha$.
Furthermore let $H(x_2,x_1)$ be denoted simply by $H$, and define
\[
\sum_{i=1}^3\lambda_i=1; \quad
\sum_{i=1}^7\mu_i=1;\quad \sum_{i=1}^6\nu_i=1;\quad
\sum_{i=1}^4\tau_i=1;\quad \sum_{i=1}^3\gamma_i=1,
\]
where $\lambda_i>0$, $\mu_i>0$, $\nu_i>0$, $\tau_i>0$ and
$\gamma_i>0$. Then $W_1$ can be re-arranged as
\begin{equation}
W_1=W_{11}+W_{12}+W_{13}W_{14}+W_{15}+W_{16}+W_{17}+W_{18}
+W_{19}+W_{21}+W_{22},\label{e3.13}
\end{equation}
where
\begin{align*}
W_{11}& =\lambda_1\beta(1-\epsilon)H(x_2-x_1)^2+\alpha(\mu_1\beta\epsilon
+\chi_1)(y_2-y_1)^2\\
&\quad +\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}{\alpha(1
-\epsilon)}(z_2-z_1)^2+\tau_1ab\epsilon(u_2-u_1)^2\\
 &\quad +\gamma_1\frac{a\epsilon}{1-\epsilon}(v_2-v_1)^2;
\\
W_{12} & = \lambda_2\beta(1-\epsilon)H(x_2-x_1)^2+\chi_1\beta(1
-\epsilon)(x_2-x_1)(y_2-y_1)\\
&\quad +\mu_2\alpha\beta\epsilon (y_2-y_1)^2;
\\
W_{13} &= \lambda_3\beta(1-\epsilon)H(x_2-x_1)^2
  +\chi_2\beta(1-\epsilon)(x_2-x_1)(z_2-z_1)\\
&\quad +\nu_2\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}{\alpha(1
 -\epsilon)}(z_2-z_1)^2;\\
W_{14}&=\mu_3\alpha\beta\epsilon(y_2-y_1)^2+\frac{1}{\alpha(1
  -\epsilon)}\chi_1(b\alpha+a\beta\epsilon)(y_2-y_1)(z_2-z_1)\\
&\quad +\nu_3\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}
  {\alpha(1-\epsilon)}(z_2-z_1)^2;\\
W_{15}&=\mu_4\alpha\beta\epsilon(y_2-y_1)^2+a\chi_1(y_2-y_1)(u_2-u_1)
  +\tau_2ab\epsilon(u_2-u_1)^2;\\
W_{16}&=\mu_5\alpha\beta\epsilon(y_2-y_1)^2+\frac{1}{1-\epsilon}\chi_1(y_2
  -y_1)(v_2-v_1)+\frac{1}{1-\epsilon}\gamma_2a\epsilon(v_2-v_1)^2;\\
W_{17}&=\nu_4\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}
   {\alpha(1-\epsilon)}(z_2-z_1)^2+\chi_2\alpha(z_2-z_1)(y_2-y_1)\\
&\quad +\mu_6\alpha\beta\epsilon(y_2-y_1)^2;\\
W_{18} &=\nu_5\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}
  {\alpha(1-\epsilon)}(z_2-z_1)^2+a\chi_1(z_2-z_1)(u_2-u_1)\\
  &\quad +\tau_3ab\epsilon(u_2-u_1)^2;\\
W_{19} &=\nu_6\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}
  {\alpha(1-\epsilon)}(z_2-z_1)^2\\
  &\quad +\frac{(1+\alpha)-(ab\alpha(1-(1-\epsilon)^2)-a^2\beta\epsilon)}
  {\alpha(1-\epsilon)}(z_2-z_1)(v_2-v_1)\\
 &\quad +\frac{1}{1-\epsilon}\gamma_3a\epsilon(v_2-v_1)^2;\\
W_{21} &=\tau_4ab\epsilon(u_2-u_1)^2+2(\beta\epsilon+ab\alpha)(u_2-u_1)
  (y_2-y_1)+\mu_7\alpha\beta\epsilon(y_2-y_1)^2;\\
W_{22} &=\nu_6\frac{(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)}
   {\alpha(1-\epsilon)}(z_2-z_1)^2-2\beta\epsilon
   \frac{(1+(1-\epsilon))}{1-\epsilon}(z_2-z_1)(u_2-u_1) \\
  &\quad +\tau_4ab\epsilon(u_2-u_1)^2.
\end{align*}
Since each $W_{1i}$, $(i=1,2,\dots ,9)$, $W_{21}$ and $W_{22}$ are
quadratic in their respective variables, then by using the fact that
any quadratic of the form $Ap^2+Bpq+Cq^2$ is non negative if
$4AC-B^2\geq 0$, it follows that
\begin{gather*}
 W_{12}\geq 0 \quad\text{if }H\leq
\frac{4}{\Delta_1(1-\epsilon)}\lambda_2\mu_2\alpha\epsilon;\\
W_{13}\geq 0 \quad\text{if }{\chi_2}^2\leq
\frac{4}{\alpha\beta(1-\epsilon)^2}\lambda_3\Delta_0\nu_2(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1);\\
W_{14}\geq 0 \quad \mbox{if } {\chi_1}^2\leq4\mu_3b\epsilon\nu_3\alpha^2(1-\epsilon)(\alpha(1-\epsilon)-1);\\
W_{15}\geq 0 \quad\text{if }{\chi_1}^2\leq 4\mu_4\alpha b^2\epsilon^2\tau_2;\\
W_{16}\geq 0 \quad\text{if }{\chi_1}^2\leq 4\mu_5a\alpha\beta\epsilon^2\gamma_2;\\
W_{17}\geq 0 \quad\text{if }{\chi_2}^2\leq
\frac{4}{\alpha^2}\nu_4(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)\mu_6\beta\epsilon;\\
W_{18}\geq 0 \quad\text{if }{\chi_1}^2\leq \frac{4}{a\alpha(1-\epsilon)}\nu
_5(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)b\tau_3\epsilon.
\end{gather*}
Thus $W_1\geq W_{11}$ provided that the above inequalities are satisfied
in addition to
\begin{gather}
\begin{aligned}
0\leq {\chi_1}^2\leq & 4\min\big\{
\mu_3b\epsilon\nu_3\alpha^2(1-\epsilon)(\alpha(1-\epsilon)-1);\mu_4\alpha
b^2\epsilon^2\tau_2;\\
& \mu_5a\alpha\beta\epsilon^2\gamma_2;\frac{\nu_5}{a\alpha(1-\epsilon)}
 (b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)b\tau_3\epsilon\big\};
\end{aligned} \label{e3.14}
\\
0\leq {\chi_2}^2\leq
\frac{4}{\alpha}(b\alpha+a\beta\epsilon)(\alpha(1-\epsilon)-1)\min
\big\{\frac{\Delta_0\nu_2\lambda_3}{\beta(1-\epsilon)^2};
\frac{\nu_4\mu_6\beta\epsilon}{\alpha}\big\}; \label{e3.15}
\end{gather}
with $H$ lying in
\begin{equation}
I_0\equiv [\Delta_0,K_0[\frac{[(ab-\alpha)\alpha-a^2\beta]}{a}]]\label{e3.16}
\end{equation}
and $G$ lying in
\begin{equation}
I_1\equiv [\Delta_1,K_1[\frac{[(ab-\alpha)\alpha+a\Delta_0]}{a^2}]],
\label{e3.17}
\end{equation}
where $I_0$ and $I_1$ are sub-interval of the
 Routh-Hurwitz intervals $(0,\frac{[(ab-\alpha)\alpha-a^2\beta]}{a})$
and $(0,\frac{[(ab-\alpha)\alpha+a\Delta_0]}{a^2})$ respectively, and
\begin{gather*}
K_0=\frac{a}{(ab-\alpha)\alpha-a^2\beta}\times\frac{4}{\Delta_1(1-\epsilon)}
\lambda_2\mu_2\alpha\epsilon;
\\
\begin{aligned}
K_1 & =4\frac{e}{(ab-\alpha)\alpha-a^2\beta}\times
\min\big\{\mu_3 b\epsilon\nu_3\alpha^2(1-\epsilon)(\alpha(1-\epsilon)-1);\\
& \quad \mu_4\alpha b^2\epsilon^2\tau_2;\mu_5a\alpha\beta\epsilon^2\gamma_2;
 \frac{\nu_5}{a\alpha(1-\epsilon)}(b\alpha+a\beta\epsilon)(\alpha(1
 -\epsilon)-1)b\tau_3\epsilon\big\}.
\end{aligned}
\end{gather*}
On choosing
\[
2D_{8}=\min\big\{\beta(1-\epsilon);\alpha;\frac{-(a\beta\epsilon+b\alpha)}
{\alpha(1-\epsilon)};a;\frac{1}{1-\epsilon}\big\},
\]
we have
\begin{equation}
W_1\geq W_{11}\geq 2D_{8}S, \label{e3.18}
\end{equation}
and if
\[
D_9=\max\big\{\beta(1-\epsilon);\alpha;\frac{-(a\beta\epsilon+b\alpha)}
{\alpha(1-\epsilon)};a;\frac{1}{1-\epsilon}\big\},
\]
then
\begin{equation}
W_2\leq D_{9}S^{1/2}|\theta|.\label{e3.19}
\end{equation}
Combining \eqref{e3.18} and \eqref{e3.19} in \eqref{e3.10},
inequality \eqref{e3.9} is obtained.
At last the conclusion to the proof of the Theorem \ref{thm2.1} will now be given.
For this purpose, let $\varrho$ be any constant in the range
$1\leq\varrho\leq 2$ and set $\sigma=1-\frac{\varrho}{2}$ so that
$0\leq\sigma\leq\frac{1}{2}$. Then, on rearranging
inequality \eqref{e3.9} we have
\begin{equation}
\frac{dW}{dt}+D_{8}S \leq D_{9}S^{1/2}|\theta|-D_{8}S,\label{e3.20}
\end{equation}
from which
\[
\frac{dW}{dt}+D_{8}S\equiv D_{10}S^{\sigma}W^{*},
\]
where
\begin{equation}
W^{*}=S^{(\frac{1}{2}-\sigma)}[|\theta|-D_{11}S^{1/2}],\label{e3.21}
\end{equation}
with $D_{11}=\frac{D_8}{D_{10}}$.
If $|\theta|<D_{11}S^{1/2}$, then $W^{*}<0$. On the other hand, if
$|\theta|\geq D_{11}S^{1/2}$, then the definition of $W^{*}$ in the
equation \eqref{e3.21} gives at least
\[
W^{*}\leq S^{(\frac{1}{2}-\sigma)}|\theta|
\]
and also $S^{1/2}\leq \frac{|\theta|}{D_{11}}$. Thus
\[
S^{\frac{1}{2}(1-2\sigma)}\leq [\frac{|\theta|}{D_{11}}]^{(1-2\sigma)},
\]
and from this together with $W^{*}$ follows
\[
W^{*}\leq D_{12}{|\theta|}^{2(1-\sigma)},
\]
where $D_{12}={D_{11}}^{(\sigma-1)}$. On using the estimate on $W^{*}$ in
inequality \eqref{e3.20}, we obtain
\begin{equation}
\frac{dW}{dt}+D_{8}S\leq D_{10}D_{12}S^{\sigma}{|\theta|}^{2(1-\sigma)}\leq
D_{13}S^{\sigma}{\phi}^{2(1-\sigma)}S^{(1-\sigma)}\label{e3.22}
\end{equation}
which follows from
\begin{align*}
& |r(t, x_1, y_1, z_1, u_1, v_1)-r(t, x_2, y_2, z_2, u_2, v_2)|\\
& \leq \phi(t)(|x_1-x_2|+|y_1-y_2|+|z_1-z_2|+|u_1-u_2|+|v_1-v_2|).
\end{align*}
In view of the fact that $\varrho=2(1-\sigma)$, we obtain
\[
\frac{dW}{dt}\leq -D_{8}S+D_{13}{\phi}^{\varrho}S,
\]
 and on using inequalities \eqref{e3.8}, we have
\begin{equation}
\frac{dW}{dt}+[D_{14}-D_{15}{\phi}^{\varrho}(t)]W\leq 0,\label{e3.23}
\end{equation}
for some constants $D_{14}$ and $D_{15}$.
On integrating the estimate \eqref{e3.23} from $t_1$ to $t_2$ $(t_1\leq t_2)$,
we obtain
\begin{equation}
W(t_2)\leq
W(t_1)\exp\big\{-D_{14}(t_2-t_1)+D_{15}\int_{t_1}^{t_2}{{\phi}^{\varrho}(\tau)d
\tau}\big\}.\label{e3.24}
\end{equation}
On using Lemma \ref{lem3.1}, we obtain inequality \eqref{e2.5}, with
$D_2=\frac{D_7}{D_6}$;
$D_3=D_{14}$ and $D_4=D_{15}$. This completes the proof of the
Theorem \ref{thm2.1}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
 The proof follows from the inequality \eqref{e2.5} and the
condition \eqref{e2.4} on $\phi(t)$. On choosing
${D_2}=\frac{D_3}{D_4}$ in inequality
\eqref{e2.5}, then as $t=t_2-t_1\to \infty$, $S(t)\to 0$ which proves
that $x_2-x_1\to 0$, $y_2-y_1\to 0$, $z_2-z_1\to 0$, $u_2-u_1\to 0$,
$v_2-v_1\to 0$ as $t\to \infty$.
\end{proof}

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\end{document}
