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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 108, pp. 1--12.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/108\hfil Large time behavior of solutions]
{Large time behavior of solutions to second-order differential equations
with $p$-Laplacian}

\author[M. Medve\v d, E. Pek\'arkov\'a\hfil EJDE-2008/108\hfilneg]
{Milan Medve\v d, Eva Pek\'arkov\'a} % in alphabetical order

\address{Milan Medve\v d \newline
Department of Mathematical Analysis and Numerical
Mathematics, Faculty of Mathematics, Physics and Informatics,
Comenius University, Ml\'ynsk\'a dolina, 842 48 Bratislava, Slovakia}
\email{medved@fmph.uniba.sk}

\address{Eva Pek\'arkov\'a \newline
Department of Mathematics and Statistics, Faculty of Science,
Masaryk University, Jan\'a\v ckovo n\'am. 2a, CZ-602 00 Brno,
Czech Republic}
\email{pekarkov@math.muni.cz}

\thanks{Submitted June 9, 2008. Published August 11, 2008.}
\subjclass[2000]{34C11}
\keywords{Second order differential equation; $p$-Laplacian;
Bihari's inequality; \hfill\break\indent asymptotic properties;
 Dannan's inequality}

\begin{abstract}
 We study asymptotic properties of solutions for certain second-order
 differential equation with $p$-Laplacian. The main purpose is to
 investigate when all global solutions  behave at infinity like
 nontrivial linear functions. Making use of Bihari's inequality and
 its Dannan's version, we obtain results for differential equations
 with $p$-Laplacian analogous which extend those known in the
 literature concerning ordinary second order differential equations.
\end{abstract}

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

\section{Introduction}

In this paper, we study asymptotic properties of the second-order
differential equation with $p$-Laplacian
\begin{equation}\label{e:1}
 (|u'|^{p-1}u')'+f(t,u,u')=0,\quad  p \geq 1.
\end{equation}

In the sequel, it is assumed that all solutions of  \eqref{e:1}
are continuously extendable throughout the entire real axis. We
refer to such solutions as to global solutions. We shall prove
sufficient conditions under which all global solutions are
asymptotic to $at + b$, as $ t \to + \infty$, where $a, b$ are
real numbers. The problem for ordinary second order differential
equations without $p$-Laplacian has been studied by many authors,
e. g. by  Cohen \cite{c1},  Constantin \cite{c2}, Dannan \cite{d1},
Kusano and Trench \cite{k1,k2},  Rogovchenko \cite{r2},
  Rogovchenko \cite{r3},
Tong \cite{t1} and  Trench \cite{t2}. Our results are more close to those
obtained in the papers \cite{r2,r3}. The main tool of the proofs are
the Bihari's and Dannan's integral inequalities. We remark that
sufficient conditions on the existence of global solutions for
second order differential equations and   second order
functional-differential equations with $p$-Laplacian are proved in
the papers \cite{b1,b2,b3,b4,m1}. Many references concerning
differential equations with $p$-Laplacian can be found in the
paper by  Rachunkov\' a,  Stan\v ek and  Tvrd\' y \cite{r1}, where
boundary value problems for such equations are treated.

Let
\begin{equation}\label{e:2}
 u(t_0)=u_0, \quad u'(t_0)=u_1,
\end{equation}
where $u_0, u_1 \in \mathbb{R}$ be initial condition for solutions
of \eqref{e:1}.

We say that a solution $u(t)$ of \eqref{e:1} possesses the property
(L) if $u(t)=at+b+o(t)$ as $t \to \infty$, where $a$, $b$
are real constants.

\section{Main results}\label{s:2}

%Theorem 1
\begin{theorem}\label{theorem1}
Let $p \geq 1$, $r > 0$ and $t_0 > 0$.
Suppose that the following conditions are satisfied:
\begin{itemize}
\item[(1)] $f(t,u,v)$ is a continuous function in
$D=\{(t,u,v): t \in <t_0,\infty), u, v \in \mathbb{R}\}$, where $t_0 > 0$
\item[(2)] There exist continuous functions
$h,g:\mathbb{R}_{+}=<0,\infty) \to \mathbb{R}_{+}$ such that
\begin{equation*} %\label{e:3}
 |f(t,u,v)|\leq h(t)g\Big(\big[\frac{|u|}{t}\big]^{r}\Big)|v|^{r},  \,(t, u, v) \in D,
\end{equation*}
where for $s>0$ the function $g(s)$ is positive and nondecreasing,
\begin{equation*} %\label{e:4}
\int_{t_0}^{\infty} h(s) \mathrm{d}s < \infty,
\end{equation*}
and if we denote
\begin{equation*} %\label{e:5}
G(x)=\int_{t_0}^{x} \frac{\mathrm{d}s}{s^{r/p}g(s^{r/p})},
\end{equation*}
then
$$
G(\infty)= \int_{t_0}^{\infty} \frac{\mathrm{d}s}{s^{r/p}g(s^{r/p})}
= \frac{p}{r}\int_{a}^{\infty} \frac{\tau^{\frac{p}{r}- 1}\mathrm{d}
\tau}{\tau g(\tau)} = \infty,
$$
where $a = (t_0)^{r/p}$.
\end{itemize}
Then any global solution $u(t)$ of the equation (1) possesses the
property (L).
\end{theorem}

\begin{proof}\label{proof theorem1}
Without loss of generality we may assume $t_0 = 1$. Let $u(t)$ be a
solution of \eqref{e:1}, \eqref{e:2}. Then
\begin{equation}\label{e:6}
 (u'(t))^{p} \leq |u'(t)|^{p-1} u'(t) \leq c_2
+ \int_{1}^{t}|f(s,u(s),u'(s))|\mathrm{d}s,
\end{equation}
where $c_2 = |u_1|^{p}$. Let $w(t)$ be the right-hand side of
inequality \eqref{e:6}. Then
\begin{equation*} %\label{e:7}
u'(t) \leq w(t)^{1/p}
\end{equation*}
and
\begin{equation}\label{e:8}
u(t) \leq c_1 + \int_{1}^{t} w(s)^{1/p} \mathrm{d}s \leq c_1 + (t-1)w(t)^{1/p}
\leq t[c_1 + w(t)^{1/p}],
\end{equation}
where $c_1 = |u_0|$, i.e.
\begin{equation*} %\label{e:9}
u(t) \leq t[c_1+w(t)^{1/p}], t \geq 1.
\end{equation*}
Applying the inequality $(A+B)^{p} \leq 2^{p-1}(A^{p}+B^{p}), A, B \geq 0$
and the assumption (2) of Theorem \ref{theorem1} we obtain from \eqref{e:8}:
\begin{equation} \label{e:11}
\begin{aligned}
\Big( \frac{|u(t)|}{t}\Big)^{p} &\leq 2^{p-1}c_{1}^{p}+2^{p-1}w(t) \\
&\leq 2^{p-1}c_{1}^{p}+2^{p-1}\Big(c_2+\int_{0}^{t} h(s)g
\Big(\big[\frac{|u(s)|}{s}\big]^{r} \Big) |u'(s)|^{r}\Big) \mathrm{d}s.
\end{aligned}
\end{equation}
Let
\begin{equation}\label{e:12}
d=2^{p-1}(c_{1}^{p}+c_{2}), \quad H(t)=2^{p-1}h(t).
\end{equation}
Then
\begin{equation}\label{e:13}
\Big(\frac{|u(t)|}{t} \Big)^{p} \leq d
+ \int_{1}^{t} H(s) g\Big(\big[\frac{|u(s)|}{s}\big]^{r}\Big)|u'(s)|^{r} \mathrm{d}s
 := z(t);
\end{equation}
i.e.,
\begin{equation*} %\label{e:14}
\Big(\frac{|u(t)|}{t} \Big)^{r} \leq z(t)^{r/p}.
\end{equation*}
From the assumption (2) of Theorem \ref{theorem1} and the
inequality \eqref{e:6} it follows
\begin{equation*} %\label{e:15}
|u'(t)|^{p} \leq u_{1}^{p}+\int_{1}^{t} h(s) g \Big(\big[\frac{|u(s)|}{s}
\big]^{r}\Big) |u'(s)|^{r} \mathrm{d}s \leq z(t);
\end{equation*}
i.e. we have
\begin{equation*} %\label{e:16}
|u'(t)|^{p} \leq z(t).
\end{equation*}
Since $g(s)$ is nondecreasing, the inequality \eqref{e:11} yields
\begin{equation*} %\label{e:17}
g \Big(\big[\frac{|u(t)|}{t}\big]^{r} \Big) \leq g(z(t)^{r/p})
\end{equation*}
and so we conclude for $t \geq 1$,
\begin{equation*} %\label{e:18}
z(t) \leq d + \int_{1}^{t} H(s) g(z(t)^{r/p}) z(t)^{r/p} \mathrm{d}s.
\end{equation*}
From the assumption (2) of Theorem \ref{theorem1} it follows that the
inverse $G^{-1}$ of $G$ is defined on the interval $(G(+0),\infty)$.
Applying the Bihari theorem (see \cite{b5}) we obtain
\begin{equation*} %\label{e:19}
z(t) \leq G^{-1}\Big( G(d)+2^{p-1} \int_{1}^{\infty} h(s)\mathrm{d}s \Big)
:=K < \infty.
\end{equation*}
Therefore the inequality \eqref{e:12} yields
\begin{equation*} %\label{e:20}
|u'(t)| \leq L:=K^{1/p}
\end{equation*}
and from \eqref{e:11} we have
\begin{equation*} %\label{e:21}
\frac{|u(t)|}{t} \leq L.
\end{equation*}
Since
\begin{equation*} %\label{e:22}
\int_{1}^{t} |f(s,u(s),u'(s))|\mathrm{d}s \leq \int_{1}^{t} h(s)
g (\Big(\frac{|u(s)|}{s} \Big)^{r})|u'(s)|^{r} \mathrm{d}s \leq z(t) \leq K
\end{equation*}
for $t \geq 1$, the integral $\int_{1}^{\infty} |f(s,u(s),u'(s))|\mathrm{d}s$ exists. From \eqref{e:13} it follows that there exists $a \in \mathbb{R}$ such that
\begin{equation*} %\label{e:23}
\lim_{t \to \infty} u'(t) = a.
\end{equation*}
By the l'Hospital rule, we can conclude that
\begin{equation*} %\label{e:24}
\lim_{t \to \infty} \frac{u(t)}{t}=\frac{u_{1}+\int_{1}^{t}
u'(\tau)\mathrm{d}\tau}{t}=\lim_{t \to \infty} u'(t)=a.
\end{equation*}
Therefore there exist $b \in \mathbb{R}$ such that $u(t)=at+b+o(t)$.
\end{proof}



\subsection*{Example 1}
Let $t_0 = 1,\,p \geq r > 0$,
\begin{equation}
f(t, u, u') = \eta(t)t^{1 - \alpha}e^{- t}\Big (\frac{u}{t}\Big )^{p - r}
\ln \Big [2 + \Big (\frac{|u|}{t}\Big )^r\Big ](u')^r,\,t \geq 1,
\end{equation}
where $0 < \alpha < 1, \ \eta(t)$ is a continuous function on interval
$\langle 1, \infty)$ with $K = \sup_{t \geq 1}|\eta(t)|<\infty$.

The function $f(t, u, u')$ can be written in the form
\begin{equation}
f(t, u, u') = h(t)g\Big ([\frac{u}{t}]^r\Big )(u')^r,
\end{equation}
where $h(t) = \eta(t)t^{1 - \alpha}e^{- t},\, g(u) = u^{\frac{p}{r}- 1}\ln (2 + |u|)$.
Obviously $g(u)$ is positive, continuous and nondecreasing function,
$\int_1^\infty |h(s)|ds < K\Gamma (\alpha) = K\int_0^\infty s^{1 - \alpha}
e^{- s}\mathrm{d}s$
and
\begin{equation}
\int_1^\infty \frac{\tau^{\frac{p}{r}- 1}\mathrm{d}\tau}{\tau g(\tau)}
= \int_1^\infty \frac{\mathrm{d}\tau}{\tau \ln (2 + \tau)}
 > \int_1^\infty \frac{\mathrm{d}\tau}{(2 + \tau )\ln (2 + \tau)} = \infty.
\end{equation}
Thus we have proved that all conditions of Theorem 1 are satisfied.
This means that for every solution $u(t)$ of the initial value
problem \eqref{e:1}, \eqref{e:2} there exist numbers $a, b$ such
that $u(t) = at + b + o(t)$ as  $t \to \infty$.

\begin{theorem}\label{theorem2}
Let $p \geq 1, r > 0$ and $t_0 > 0$. Suppose the following conditions
are satisfied:
\begin{itemize}
\item[(1)] The function $f(t,u,v)$ is continuous in
 $\mathrm{D}=\{(t,u,v):t \in <t_0,\infty), u,v \in \mathbb{R}\}$,
\item[(2)] There exist continuous functions
$h_{1}, h_{2}, h_{3}, g_{1}, g_{2}: \mathbb{R}_{+} \to \mathbb{R}_{+}$ such that
\begin{equation*} %\label{e:25}
|f(t,u,v)| \leq h_{1}(t)g_{1}\Big(\big[\frac{|u|}{t}\big]^{r}\Big)+ h_{2}(t)g_{2}(|v|^{r})+h_{3}(t),\,(t, u, v) \in D,
\end{equation*}
where $H_{i}:=\int_{t_0}^{\infty} h_{i}(s) \mathrm{d}s < \infty$, $i = 1,2,3$,
for $s > 0$ the functions $g_{1}(s)$, $g_{2}(s)$ are nondecreasing and if
\begin{equation*} %\label{e:26}
G(x)=\int_{t_0}^{x} \frac{\mathrm{d}s}{g_{1}(s^{r/p})+g_{2}(s^{r/p})}
\end{equation*}
then
\begin{equation*}
G(\infty)= \int_{t_0}^{\infty} \frac{\mathrm{d}s}{g_{1}(s^{r/p})+g_{2}(s^{r/p})} = \frac{p}{r}\int_{a}^{\infty} \frac{\tau^{\frac{p}{r}- 1}\mathrm{d}\tau}{g_{1}(\tau)+g_{2}(\tau)} = \infty,
\end{equation*}
where $a = (t_0)^{r/p}$.
\end{itemize}
Then any global solution $u(t)$ of the equation (1) possesses the
property (L).
\end{theorem}

\begin{proof}\label{proof theorem2}
Without loss of generality we may assume $t_0 = 1$.
By the standard existence results, it follows from the continuity of
the function $f$ that equation \eqref{e:1} has solution $u(t)$
corresponding to the initial data $u(1)=u_{0}$, $u'(1)=u_{1}$.
Two times of integration \eqref{e:1} from $1$ to $t$, yields for $t \geq 1$
\begin{gather}\label{e:27}
(u'(t))^{p} \leq |u'(t)|^{p-1}u'(t)=u_{1}^{p}-\int_{1}^{t}f(s,u(s),u'(s)) \mathrm{d}s,
\\ \label{e:28}
u(t) \leq u_{0}+(t-1)\Big[u_{1}^{p}-\int_{1}^{t}f(s,u(s),u'(s)) \mathrm{d}s \Big]^{1/p}.
\end{gather}
It follows from \eqref{e:27} and \eqref{e:28} that for $t \geq 1$,
\begin{gather*} %\label{e:29}
|u'(t)| \leq w(t)^{1/p}, \\ %\label{e:30}
|u(t)| \leq t\big(c_{1}+w(t)^{1/p}\big),
\end{gather*}
where $c_{1}=|u_{0}|$, $c_{2}=|u_{1}|^{p}$, $w(t)=c_{2}+\int_{1}^{t}|f(s,u(s),u'(s))| \mathrm{d}s$.
Using the assumption (2) we obtain for $t \geq 1$
\begin{align*} %\label{e:31}
|u'(t)|
&\leq \Big[c_{2}+\int_{1}^{t} h_{1}(s)g_{1}
\Big(\big[\frac{|u(s)|}{s} \big]^{r}\Big)\mathrm{d}s \\
&\quad + \int_{1}^{t}h_{2}(s)
g_{2}(|u'(s)|^{r})\mathrm{d}s + \int_{1}^{t}h_{3}(s)\mathrm{d}s \Big]^{1/p},
\end{align*}
\begin{align*} %\label{e:32}
\frac{|u(t)|}{t}
&\leq c_{1}+\Big[c_{2}+\int_{1}^{t} h_{1}(s)g_{1}
\Big(\big[\frac{|u(s)|}{s} \big]^{r}\Big)\mathrm{d}s \\
&\quad + \int_{1}^{t}h_{2}(s)
g_{2}(|u'(s)|^{r})\mathrm{d}s + \int_{1}^{t}h_{3}(s)\mathrm{d}s \Big]^{1/p}.
\end{align*}
Applying the inequality $(A+B)^{p} \leq 2^{p-1}(A^{p}+B^{p})$,
where $A,B \geq 0$, we obtain
\begin{equation} \label{e:33}
\begin{aligned}
\big(\frac{|u(t)|}{t}\big)^{p}
&\leq d +\int_{1}^{t} H_{1}(s)g_{1}\Big(\big[\frac{|u(s)|}{s} \big]^{r}\Big)\mathrm{d}s\\
&\quad  + \int_{1}^{t}H_{2}(s) g_{2}(|u'(s)|^{r})\mathrm{d}s
 + \int_{1}^{t}H_{3}(s)\mathrm{d}s.
\end{aligned}
\end{equation}
where $d = 2^{p-1}(c_{1}^{p}+c_{2})$, $H_{i}(t)=2^{p-1}h_{i}(t)$,
$i=1,2,3$.
Denote by $z(t)$ the right-hand side inequality \eqref{e:33}
\begin{equation} \label{e:34}
|u'(t)|^{r} \leq z(t)^{r/p},  \quad
\Big(\frac{|u(t)|}{t}\Big)^{r} \leq z(t)^{r/p}.
\end{equation}
Since the function $g_{1}(s)$ and $g_{2}(s)$ are nondecreasing for $s > 0$,
we obtain
\[ %\label{e:35}
g_{1}\Big(|u'(t)|^{r}\Big) \leq g_{1}\Big(z(t)^{r/p}\Big),\quad
g_{1}\Big(\Big[\frac{|u(t)|}{t}\Big]^{r}\Big) \leq g_{2}\Big(z(t)^{r/p}\Big).
\]
Thus, for $t \geq 1$,
\begin{equation}\label{e:36}
z(t) \leq d + \int_{1}^{t} H_{1}(s)g_{1}(z(s)^{r/p})\mathrm{d}s
+ \int_{1}^{t}H_{2}(s) g_{2}(z(s)^{r/p})\mathrm{d}s
+ \int_{1}^{t}H_{3}(s)\mathrm{d}s.
\end{equation}
Furthermore, due to evident inequality
\begin{equation}\label{e:37}
H_{1}(s)g_{1}(z(s)^{r/p}) + H_{2}(s) g_{2}(z(s)^{r/p}) \leq
(H_{1}(s)+H_{2}(s)) (g_{1}(z(s)^{r/p}) + g_{2}(z(s)^{r/p}))
\end{equation}
By \eqref{e:37}, we have
\begin{equation*} %\label{e:38}
z(t) \leq d+ \bar{H}_{3}+ \int_{1}^{t}(H_{1}(s)+H_{2}(s))(g_{1}(z(s)^{r/p})
+ g_{2}(z(s)^{r/p}))\mathrm{d}s;
\end{equation*}
i.e.,
\begin{equation}\label{e:39}
z(t) \leq d+ 2^{p-1}\bar{h}_{3}+ 2^{p-1}\int_{1}^{t}(h_{1}(s)+h_{2}(s))
(g_{1}(z(s)^{r/p}) + g_{2}(z(s)^{r/p}))\mathrm{d}s.
\end{equation}
Applying Bihari's inequality (see \cite{b5}) to \eqref{e:39}, we obtain,
for $t \geq 1$,
\begin{equation*} %\label{e:40}
z(t) \leq G^{-1}\Big(G(d+2^{p-1}\bar{h}_{3})+2^{p-1}
\int_{1}^{t}(h_{1}(s)+h_{2}(s))\mathrm{d}s \Big),
\end{equation*}
where
\begin{equation*} %\label{e:41}
G(x)=\int_{1}^{x}\frac{\mathrm{d}s}{g_{1}(s^{r/p})+g_{2}(s^{r/p})},
\end{equation*}
and $G^{-1}(x)$ is the inverse function for $G(x)$ defined for
$x \in (G(+0),\infty)$. Note that $G(+0) < 0$,
and $G^{-1}(x)$ is increasing.

Now, let
$$
K=G(d+2^{p-1}\bar{h}_{3})+2^{p-1}(\bar{h}_{1}+\bar{h}_{2}) < \infty.
$$
Since $G^{-1}(x)$ is increasing, we have
\begin{equation*} %\label{e:42}
z(t) \leq G^{-1}(K) < \infty;
\end{equation*}
so it yields
\begin{equation*} %\label{e:43}
\frac{|u(t)|}{t} \leq G^{-1}(K), \quad  |u'(t)| \leq G^{-1}(K).
\end{equation*}
Using assumption (2) of the Theorem \ref{theorem2}, we have
\begin{align*} %\label{e:44}
\int_{1}^{t}|f(s,u(s),u'(s))|\mathrm{d}s
&\leq h_{1}(t)g_{1}\Big(\big[\frac{|u|}{t}\big]^{r}\Big)
 + h_{2}(t)g_{2}(|u'(s)|^{r})+h_{3}(t) \\
&\leq z(t) \leq G^{-1}(K),
\end{align*}
where $t \geq 1$, the integral $\int_{1}^{t}|f(s,u(s),u'(s))|\mathrm{d}s$ converges, and there exists an $a \in \mathbb{R}$ such that
\begin{equation*} %\label{e:45}
\lim_{t \to \infty}u'(t) = a.
\end{equation*}
\end{proof}

\subsection*{Example 2}
Let $t_0 = 1, p \geq r > 0$,
\begin{align*}
f(t, u, v) &= \eta_1(t)t^{1 - \alpha_1}e^{-t}
\big(\frac{u}{t}\big)^{p-r}\ln \big[2 +\big(\frac{u}{t}\big)^{r}\big]\\
 &\quad +\eta_2(t)t^{1 - \alpha_2}e^{-t}v^{p-r}\ln(3 +  v^{r})
 + \eta_3(t)t^{1 - \alpha_3}e^{-t}
\end{align*}
where $0 < \alpha_i < 1$, $\eta_i(t)$ are continuous functions on
$[ 1, \infty)$,  $K_i = \sup_{t \geq 1}|\eta_i(t)|
 < \infty$, $i = 1, 2, 3$. Then $f(t, u, u')$ can be written as
$$
f(t,u,v) = h_1(t)g_1\big([\frac{u}{t}]^{r}\big) + h_2(t)g_2(v^r) + h_3(t),
$$
where $h_i(t) = \eta_i(t) t^{1-\alpha_i}e^{-t}$,
$i = 1, 2, 3$, $g_1(u) = u^{\frac{p}{r}}\ln(2 + u)$,
$g_2(u) = u^{\frac{p}{r}}\ln(2 + u)$.
Then
$$
|f(t, u, v)| \leq |h_1(t)|g_1\Big([\frac{u}{t}]^{r}\Big)
+ |h_2(t)|g_2(|v|^r) + |h_3(t)|,
$$
where $(t, u, v) \in D = \{(t, u, v): t \in \langle 1, \infty),
u, v \in \mathbb{R}\}$, $|h_i(t)| \leq K_i\Gamma(\alpha_i)$,
$i = 1, 2, 3$ and obviously we have
\begin{align*}
G(\infty) &= \int_1^\infty\frac{\tau^{\frac{p}{r}- 1}\mathrm{d}\tau}{g_1(\tau)
+ g_2(\tau)} \\
&= \int_1^\infty\frac{\tau^{\frac{p}{r}- 1}\mathrm{d}\tau}{\tau^{\frac{p}{r}}
 [\ln(2 + \tau) + \ln(3 + \tau)]} \\
&\geq \frac{1}{2}\int_1^\infty\frac{\mathrm{d}\tau}{(3 + \tau)\ln(3 + \tau)}
= \infty.
\end{align*}
This means that all assumptions of Theorem \ref{theorem2} are satisfied
and thus any global solution $u(t)$ of the equation (1) possesses the
 property (L).


\begin{theorem}\label{theorem3}
Let $t_0 > 0$. Suppose that the following assumptions hold:
\begin{itemize}
\item[(i)] there exist nonnegative continuous function
$h_{1}, h_{2}, g_{1}, g_{2}: \mathbb{R_{+}}\to \mathbb{R_{+}}$ such that
    \begin{equation*}
    |f(t,u,v)| \leq  h_{1}(t) g_{1}\Big(\big[\frac{|u|}{t}\big]^{r}\Big)
+h_{2}(t)g_{2}(|v|^{r});
    \end{equation*}


\item[(ii)] for $s>0$ the function $g_{1}(s)$, $g_{2}(s)$ are nondecreasing,
and
\begin{equation*} %\label{e:46}
g_{1}(\alpha u) \leq \psi_{1}(\alpha)g_{1}(u), \qquad g_{2}(\alpha u)
\leq \psi_{2}(\alpha)g_{2}(u)
\end{equation*}
for $\alpha \geq 1$, $u \geq 0$, where the functions $\psi_{1}(\alpha)$,
$\psi_{2}(\alpha)$ are continuous for $\alpha \geq 1$;


\item[(iii)] $\int_{t_0}^{\infty}h_{i}(s) \mathrm{d}s = H_{i} < \infty$, $i = 1,2$.
\end{itemize}
Assume that there exists a constant $K \geq 1$ such that
\begin{align*} %\label{e:47}
K^{-1}(\psi_{1}(K)+\psi_{2}(K))2^{p-1}(H_{1}+H_{2})
&\leq \int_{t_0}^{+\infty} \frac{\mathrm{d}s}{g_{1}(s^{r/p})+g_{2}(s^{r/p})} \\
&= \frac{p}{r}\int_{a}^{+\infty} \frac{\tau^{\frac{p}{r} - 1}
\mathrm{d}\tau}{g_{1}(\tau)+g_{2}(\tau)},
\end{align*}
where $a = (t_0)^{r/p}$.
Then any global solution $u(t)$ of the equation \eqref{e:1} with initial
data $u(t_0)=u_{0}$, $u'(t_0)=u_{1}$ such that
$(|u_{0}|+ |u_{1}|)^{p} \leq K$ possesses the property (L).
\end{theorem}

\begin{proof}\label{proof theorem3}
Without loss of generality we may assume $t_0=1$. Arguing in the same
way as in Theorem \ref{theorem1}, we obtain by assumption (i) of
Theorem \ref{theorem3}
\begin{equation}\label{e:48}
|u'(t)|\leq \Big[|u_{1}|^{p}+\int_{1}^{t}h_{1}(s) g_{1}\Big([\frac{u(s)}{s}]^{r}\Big) \mathrm{d}s
+ \int_{1}^{t}h_{2}(s) g_{2}(|u'(s)|^{r}) \mathrm{d}s \Big]^{1/p}
\end{equation}
\begin{equation}\label{e:49}
\frac{|u(t)|}{t} \leq |u_{0}|+\Big[|u_{1}|^{p}+\int_{1}^{t}h_{1}(s) g_{1}\Big([\frac{u(s)}{s}]^{r}\Big) \mathrm{d}s + \int_{1}^{t}h_{2}(s) g_{2}(|u'(s)|^{r}) \mathrm{d}s \Big]^{1/p}
\end{equation}
where $t \geq 1$.
\begin{equation}\label{e:50}
\Big(\frac{|u(t)|}{t}\Big)^{p} \leq K+2^{p-1}\Big(\int_{1}^{t}h_{1}(s) g_{1}\Big([\frac{u(s)}{s}]^{r}\Big) \mathrm{d}s + \int_{1}^{t}h_{2}(s) g_{2}(|u'(s)|^{r}) \mathrm{d}s\Big),
\end{equation}
where $K = 2^{p-1}(|u_{0}|^{p}+|u_{1}|^{p}) \geq (|u_{0}|+|u_{1}|)^{p}$.
Denoting by $z(t)$ the right-hand side of inequality \eqref{e:50} we
have by \eqref{e:48} and \eqref{e:50}
\begin{equation} \label{e:51}
|u'(t)|^{r}\leq z(t)^{r/p}, \quad
\Big(\frac{|u(t)|}{t}\Big)^{r}  \leq z(t)^{r/p}.
\end{equation}
Since the function $g_{1}(s)$, $g_{2}(s)$ are nondecreasing for $s>0$,
for $t \geq 1$, \eqref{e:51} yields
\begin{equation}\label{e:52}
z(t)\leq K + 2^{p-1}\Big(\int_{1}^{t}h_{1}(s) g_{1}\big(z(s)^{r/p}\big) \mathrm{d}s
+ \int_{1}^{t}h_{2}(s) g_{2}(z(s)^{r/p})\Big) \mathrm{d}s.
\end{equation}
By assumption (ii) of Theorem \ref{theorem3}, the functions $g_{1}(u)$,
$g_{2}(u)$ belong to the class $\mathbb{H}$. Furthermore, if $g_{1}(u)$
and $g_{2}(u)$ belong to the class $\mathbb{H}$ with corresponding
multiplier function $\psi_{1}(\alpha)$, $\psi_{2}(\alpha)$ respectively,
then the sum $g_{1}(u)+g_{2}(u)$. Applying Bihari's Theorem
(see \cite{b5}) to \eqref{e:52}, we have for $t \geq 1$
\begin{equation}\label{e:53}
z(t)\leq KW^{-1}(K^{-1}(\psi_{1}(K)+\psi_{2}(K)))2^{p-1}\int_{1}^{t}
(h_{1}(s)+h_{2}(s))\mathrm{d}s,
\end{equation}
where
$$
W(u)=\int_{1}^{u}\frac{\mathrm{d}s}{g_{1}\big(s^{r/p}\big)
+g_{2}\big(s^{r/p}\big)},
$$
and $W^{-1}(u)$ is inverse function for $W(u)$.
Inequality \eqref{e:53} holds for all $t \geq 1$ because
\begin{equation*}
(K^{-1}(\psi_{1}(K)+\psi_{2}(K))2^{p-1}(H_{1}+H_{2})=L < \infty.
\end{equation*}
Since $W^{-1}(u)$ is increasing, we get
$$
z(t) \leq KW^{-1}(L)< \infty,
$$
so it follows from \eqref{e:51}, \eqref{e:52} that
$$
\frac{|u(t)|}{t} \leq KW^{-1}(L), \quad
 |u'(t)| \leq KW^{-1}(L).
$$
The rest of the proof is similar to that of Theorem \ref{theorem2}
and thus it is omitted.
\end{proof}


\subsection*{Example 3}

Let $t_0 > 0$. Consider  \eqref{e:1} with $p \geq 1, \frac{p}{q} = 2$,
\begin{equation}
f(t, u, v)= h_1(t)u^2 = h_2(t)v^2,
\end{equation}
where $h_1(t) = \frac{\eta_1(t)}{t^2}t^{1 - \alpha_1}e^{- t}$,
$h_2(t)= \eta_2(t)t^{1 - \alpha_2}e^{- t}$,
$0 < \alpha_i \leq 1$, $\eta_i(t)$, $i = 1, 2$ are continuous functions
on the interval $\langle 0, \infty)$ with
$K_i = \sup_{t \geq t_0}|\eta_i(t)| < \infty$. Then we can write
\begin{equation}
f(t, u, v) = \eta_1(t)t^{1-\alpha_1}e^{- t}\big(\frac{u}{t}\big)^2
+ \eta_2(t)t^{1-\alpha_2}e^{- t}v^2
\end{equation}
and
\begin{equation}
|f(t, u, u')| \leq K_1\Gamma(\alpha_1)g_1(u) + K_2\Gamma(\alpha_2)g_2(u'),
\end{equation}
where $g_1(u) = u^2$, $g_2(u') = (u')^2$ . The functions $g_1, g_2$
satisfy the condition (ii) of Theorem \ref{theorem3} with
$\psi_1(\alpha) = \psi_2(\alpha) = \alpha^2$ and
\begin{equation}
\int_{t_0}^\infty\frac{\tau^{\frac{p}{r}- 1}\mathrm{d}\tau}{g_1(\tau) + g_2(\tau)}
= \int_{t_0}^\infty\frac{\mathrm{d}\tau}{\tau} = \infty.
\end{equation}
Thus all assumptions of Theorem \ref{theorem3} are satisfied and
therefore any global solution $u(t)$ of the equation \eqref{e:1}
(independently on the initial values $u_0, u_1$) possesses the property (L).


\begin{theorem}\label{theorem4}
Let $t_0 > 0$. Suppose that the assumptions (i) and (iii) of
Theorem \ref{theorem3} hold, while (ii) is replaced by
\begin{itemize}
\item[(ii')] for $s > 0$ the functions $g_{1}(s)$, $g_{2}(s)$
are nonnegative, continuous and nondecreasing, $g_{1}(0)=g_{2}(0)=0$
and satisfy a Lipschitz condition
$$
|g_{1}(u+v)-g_{1}(u)| \leq \lambda_{1}v, \quad |g_{2}(u+v)-g_{2}(u)|
\leq \lambda_{2}v,
$$
 where $\lambda_{1}, \lambda_{2}$ are positive constants.
\end{itemize}
Then any global solution $u(t)$ of  \eqref{e:1} with initial data
$u(t_0)=u_{0}$, $u'(t_0)=u_{1}$ such that $|u_{0}|^{p}+|u_{1}|^{p} \leq K$
possesses property (L).
\end{theorem}


\begin{proof}
Applying \cite[Corollary 2]{d1} to \eqref{e:52}, we have for $t \geq 1$
\begin{align*}
z(t) &\leq K + 2^{p-1}\int_{t_0}^{t} (h_{1}(s)+h_{2}(s))(g_{1}(K)+g_{2}(K)) \\
&\quad\times  \exp{\Big(2^{p-1}\int_{t_0}^{t}(\lambda_{1}+\lambda_{2})
 (h_{1}(\tau)+h_{2}(\tau))\mathrm{d}\tau}\Big)\mathrm{d}s \\
&\leq K+2^{p-1}(H_{1}+H_{2})(g_{1}(K)+g_{2}(K))
 \exp{\Big(2^{p-1}(\lambda_{1}+\lambda_{2})(H_{1}+H_{2})\Big)}\\
& < +\infty.
\end{align*}
The proof can be completed with the same argument as in
Theorem \ref{theorem2}.
\end{proof}

\begin{theorem}\label{theorem5}
Let $t_0 > 0$. Suppose that there exist continuous functions
$h, g_{1},g_{2}:\mathbb{R}_{+} \to \mathbb{R}_{+}$ such that
$$
|f(t,u,v)| \leq h(t) g_{1}\Big(\big[\frac{|u|}{t}\big]^{r}
\Big)g_{2}(|v|^{r}),
$$
where for $s > 0$ the functions $g_{1}(s)$, $g_{2}(s)$ are nondecreasing;
$$
\int_{t_0}^{\infty} h(s) \mathrm{d}s < \infty,
$$
 and if we denote
$$
G(x)=\int_{t_0}^{x}\frac{\mathrm{d}s}{g_{1}(s^{r/p})g_{2}(s^{r/p})}
$$
then $G(+\infty)=\frac{p}{r}\int_{a}^{\infty}
\frac{\tau^{\frac{p}{r}- 1}}{g_{1}(\tau)g_{2}(\tau)}\mathrm{d}\tau = +\infty$,
where $a =(t_0)^\frac{r}{p}$.
Then any global solution $u(t)$ of the equation \eqref{e:1} possesses
the property (L).
\end{theorem}


\begin{proof}\label{proof theorem5}
Without loss of generality we may assume $t_0 = 1$. Arguing as in the
proof of Theorem \ref{theorem2}, we obtain for $t \geq 1$
\begin{equation} \label{e:55}
\begin{gathered}
|u'(t)|\leq \Big[|u_{1}|^{p}+\int_{1}^{t}h(s)g_{1}\Big([\frac{u(s)}{s}]^{r}
\Big)g_{2}(|u'(s)|^{r}) \mathrm{d}s \Big]^{1/p},\\
\frac{|u(t)|}{t} \leq  |u_{0}|+\Big[|u_{1}|^{p}+\int_{1}^{t}h(s) g_{1}
\Big([\frac{u(s)}{s}]^{r}\Big)g_{2}(|u'(s)|^{r}) \mathrm{d}s \Big]^{1/p},\\
\Big(\frac{|u(t)|}{t}\Big)^{p} \leq C+2^{p-1}\int_{1}^{t}h(s)
g_{1}\Big([\frac{u(s)}{s}]^{r}\Big)g_{2}(|u'(s)|^{r}) \mathrm{d}s,
\end{gathered}
\end{equation}
where $C = 2^{p-1}(|u_{0}|^{p}+|u_{1}|^{p}) \geq (|u_{0}|+|u_{1}|)^{p}$.
Denoting by $z(t)$ the right-hand side of inequality \eqref{e:55}
and using the assumptions of the Theorem \ref{theorem5}, we have
for $t \geq 1$
\begin{equation}\label{e:56}
z(t) \leq 1+C + 2^{p-1}\int_{1}^{t}h(s)g_{1}(z^{r/p})g_{2}(z^{r/p})\mathrm{d}s.
\end{equation}
Applying Bihari's inequality (see \cite{b5}) to \eqref{e:56}, for $t \geq 1$,
we obtain
\begin{equation*} %\label{e:57}
z(t) \leq G^{-1}\Big(G(1+C)+2^{p-1}\int_{1}^{t} h(s) \mathrm{d}s\Big) \leq G^{-1}(K),
\end{equation*}
where
$$
G(w)=\int_{1}^{w}\frac{\mathrm{d}s}{g_{1}(s^{r/p})g_{2}(s^{r/p})},
$$
and $G^{-1}(w)$ is the inverse function for $G(w)$.
The function $G^{-1}(w)$ is defined for $w \in (G(+0), \infty)$,
where $G(+0)<0$, it is increasing, and
$$
K=G(1+C)+2^{p-1}\int_{1}^{\infty}h(s)\mathrm{d}s < \infty.
$$
The rest of proof is similar that of Theorem \ref{theorem2} and thus
is omitted.
\end{proof}

\subsection*{Example 4}
Let $t_0 = 1$, $p \geq r > 0$,
\begin{equation*}
f(t, u, v) = \eta(t)t^{1-\alpha}e^{-t}\Big[\Big(\frac{u}{t}\Big )^{p-r}
\ln \big[2 + \Big (\frac{u}{t}\Big )^r\big] \Big ]^{\frac{3}{4}}
\cdot \Big [v^{p-r}\ln(2 + v^r)\Big ]^{\frac{1}{4}},
\end{equation*}
where $\eta(t)$ is a continuous function on $\langle 1, \infty)$ with $K = \sup_{t \in \langle 1, \infty)}\eta(t) < \infty$.
Let
\begin{equation*}
g_1(u) = \Big [u^{\frac{p}{r}-1}\ln (2 + u)\Big ]^{3/4},\quad
g_2(v) = \Big [v^{\frac{p}{r}-1}\ln (2 + v)\Big ]^{1/4},\quad
h(t) = \eta(t)t^{1-\alpha}e^{-t}.
\end{equation*}
Then
\begin{equation*}
f(t, u, v) = h(t)g_1\Big([\frac{u}{t}]^r\Big)g_2(v^r)
\end{equation*}
and
\begin{align*}
G(+\infty)
&=\frac{p}{r}\int_1^{\infty} \frac{\tau^{\frac{p}{r}- 1}}{g_{1}(\tau)
 g_{2}(\tau)}\mathrm{d}\tau = \frac{p}{r}\int_1^\infty
 \frac{\mathrm{d}\tau}{\tau\ln(2 + \tau)} \\
&> \frac{p}{r}\int_1^\infty \frac{\mathrm{d}\tau}{(2 + \tau)\ln(2 + \tau)}
= +\infty.
\end{align*}
Obviously $|f(t, u, v)|$ can be estimated as in Theorem \ref{theorem5}.
Thus all assumptions of Theorem \ref{theorem5} are satisfied and this
 means that any global solution of the equation \eqref{e:1} possesses
the property (L).


\begin{theorem}\label{theorem6}
Let $t_0 > 0$. Suppose that the following conditions hold:
\begin{itemize}
\item[(i)] there exist nonnegative continuous functions
 $h, g_{1}, g_{2}: \mathbb{R}_{+} \to \mathbb{R}_{+}$
such that
$$
|f(t,u,v)| \leq h(t) g_{1}\Big(\Big[\frac{|u(t)|}{t}
 \Big]^{r}\Big)g_{2}(|v|^{r})$$

\item[(ii)] for $s > 0$ the functions $g_{1}(s), g_{2}(s)$ are
 nondecreasing; and
$$
g_{1}(\alpha u)\leq \psi_{1}(\alpha)g_{1}(u), \quad g_{2}(\alpha u)
\leq \psi_{2}(\alpha)g_{2}(u)
$$
for $\alpha \geq 1, u \geq 0$, where the functions
$\psi_{1}(\alpha), \psi_{2}(\alpha)$ are continuous for $\alpha \geq 1$;

\item[(iii)]$\int_{t_0}^{\infty}h(s)\mathrm{d}s = H < +\infty$.
\end{itemize}
Assume also that there exists a constant $K \geq 1$ such that
\begin{equation}\label{e:58}
K^{-1}H\psi_{1}(K)\psi_{2}(K) \leq \int_{1}^{\infty} \frac{\mathrm{d}s}{g_{1}(s^{r/p})g_{2}(s^{r/p})} = \\
\frac{p}{r}\int_{a}^{\infty} \frac{\tau^{\frac{p}{r}- 1}\mathrm{d}\tau}{g_{1}(\tau)g_{2}(\tau)},
\end{equation}
 where $a =(t_0)^\frac{r}{p}$.
Then any global solution $u(t)$ of the equation \eqref{e:1} with initial
 data $u(t_0)=u_{0}, u'(t_0)=u_{1}$ such that
$2^{p-1}(|u_{0}|^{p}+|u_{1}|^{p}) \leq K$ possesses the property (L).
\end{theorem}

\begin{proof}
Without loss of generality we assume that $t_0 = 1$. With the same
argument as in Theorem \ref{theorem2},  for $t \geq 1$, we have
\begin{gather*} %\label{e:59}
|u'(t)|\leq \Big[ |u_{1}|^{p}+\int_{1}^{t} h(s) g_{1}
\Big(\big[\frac{|u(s)|}{s}\big]^{r}\Big)g_{2}(|u'(s)|^{r})\mathrm{d}s\Big]^{1/p},\\
\frac{|u(t)|}{t} \leq |u_{0}|+\Big[ |u_{1}|^{p}
+\int_{1}^{t} h(s) g_{1}\Big(\big[\frac{|u(s)|}{s}\big]^{r}\Big)
g_{2}(|u'(s)|^{r})\mathrm{d}s\Big]^{1/p}.
\end{gather*}
Applying the inequality $(A+B)^{p} \leq 2^{p-1}(A^{p}+B^{p})$, $A,B \geq 0$
we obtain
\begin{equation}\label{e:61}
\Big(\frac{|u(t)|}{t}\Big)^{p} \leq 2^{p-1}(|u_{0}|^{p}+|u_{1}|^{p})
+2^{p-1}\Big[\int_{1}^{t} g_{1}\Big(\big[\frac{|u(s)|}{s}\big]^{r}
\Big)g_{2}(|u'(s)|^{r})\mathrm{d}s\Big].
\end{equation}
Denoting by $z(t)$ the right-hand side of inequality \eqref{e:61},
 for $t \geq 1$, we obtain
\begin{equation}\label{e:62}
z(t) \leq K + \int_{1}^{t}H(s)g_{1}(z(s)^{r/p})g_{2}(z(s)^{r/p})\mathrm{d}s,
\end{equation}
where $K=2^{p-1}(|u_{0}|^{p}+|u_{1}|^{p})$ and $H(t)=2^{p-1}h(t)$.
Assumption (ii) implies that the functions $g_{1}(u)$, $g_{2}(u)$ belong
to the class $\mathbb{H}$. Furthermore, it follows from
\cite[Lemma 1]{c1}
that if $g_{1}(u)$ and $g_{2}(u)$ belong to the class $\mathbb{H}$
with the corresponding multiplier functions $\psi_{1}(\alpha)$
and $\psi_{2}(\alpha)$ respectively, then the product $g_{1}(u)g_{2}(u)$
also belongs to $\mathbb{H}$ and the corresponding multiplier function
is $\psi_{1}(\alpha)\psi_{2}(\alpha)$. Thus, applying \cite[Theorem 1]{d1}
to \eqref{e:62},  for $t \geq 1$, we have
\begin{equation}\label{e:63}
z(t) \leq KW^{-1}\Big( K^{-1}\psi_{1}(K)\psi_{2}(K)\int_{1}^{t}H(s)\mathrm{d}s\Big),
\end{equation}
where
\begin{equation}\label{e:64}
W(u)=\int_{1}^{u}\frac{\mathrm{d}s}{g_{1}(s^{r/p})g_{2}(s^{r/p})},
\end{equation}
and $W^{-1}(u)$ is the inverse function for $W(u)$. Evidently,
inequality \eqref{e:63} holds for all $t \geq 1$ since by \eqref{e:58}
\begin{equation}\label{e:65}
K^{-1}\psi_{1}(K)\psi_{2}(K)\int_{1}^{t}H(s)\mathrm{d}s \in Dom(W^{-1})
\end{equation}
for all $t \geq 1$. The rest of the proof is analogous to that of
Theorem \ref{theorem2} and is omitted.
\end{proof}

\begin{theorem}\label{theorem7}
Let $t_0 > 0$. Suppose that  assumptions (i) and (iii) of
Theorem \ref{theorem6} hold, while (ii) is replaced by
\begin{itemize}
\item[(ii')] for $s > 0$ the functions $g_{1}(s)$, $g_{2}(s)$ are
 continuous and nondecreasing, $g_{1}(0)=g_{2}(0)=0$, and satisfy
 a Lipschitz condition
$$
|g_{1}(u+v)-g_{1}(u)|\leq \lambda_{1}v, \quad |g_{2}(u+v)-g_{2}(u)|
\leq \lambda_{2}v,
$$
where $\lambda_{1}, \lambda_{2}$ are positive constants.
\end{itemize}
Then any global solution $u(t)$ of the equation \eqref{e:1} with initial
data $u(t_0)=u_{0}$, $u'(t_0)=u_{1}$ such that
$|u_{0}|^{p}+|u_{1}|^{p} \leq K$ possesses the property (L).
\end{theorem}

\begin{proof}
Without loss of generality we may assume $t_{0}=1$.
Applying \cite[Corollary 2]{d1} to \eqref{e:62}, we have for $t \geq 1$
\begin{align*} %\label{e:66}
z(t)
&\leq K + g_{1}(K)g_{2}(K) \int_{1}^{t}H(s)\exp{\Big(\lambda_{1}\lambda_{2}
 \int_{1}^{t}H(\tau)\mathrm{d}\tau}\Big)\mathrm{d}s \\
&\leq K +  \bar{H}g_{1}(K)g_{2}(K)\exp{(\lambda_{1}\lambda_{2}\bar{H})}
< +\infty.
\end{align*}

The proof of the above theorem can be completed with the same argument
as in Theorem \ref{theorem2}.
\end{proof}

\subsection*{Acknowledgements}
The first author was supported by  Grant No. 1/0098/08 from
 the Slovak Grant Agency VEGA-SAV-MŠ.
The second author was supported by Grant No. 201/08/0469  from
Grant Agency of the Czech Republic.


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\end{document}



\vskip 0.3cm $^*$Department of Mathematical Analysis and Numerical
Mathematics, Faculty of Mathematics, Physics and Informatics,
Comenius University, Mlynsk\' a dolina, 842 48 Bratislava, Slovakia,

 e-mail : medved$\@$fmph.uniba.sk \vskip 0.3cm

 $^{**}$  Deparment of Mathematics, Masaryk University,
Jan\´ a\v ckovo n\´ am 2a, CZ-602 00 Brno, Czech Republic,

 e-mail: pekarkov$\@$mail.muni.cz