text:
\section{Introduction}

This paper deals with the equation
\begin{equation}
\label{1}
\ddot{x}(t) +a(t)\dot{x}(t)+b(t)x(t)=f(t),~ t\in[0,\omega],
\end{equation}
with locally summable coefficients $a,b,f,$ which together with nonlinear
equation
\begin{equation}
\label{2}
\ddot{x}(t)=g(t,x(t),\dot{x}(t)),\;t\in [0,\omega ],
\end{equation}
continue to attract attention of many mathematicians due to their
significance in applications. In this paper we obtain two group of
results: on nonoscillation and on exponential stability. To obtain
stability results we will use nonoscillatory equations as the so-called
model equations for the left or right regularization and then use the
classical Bohl-Perron theorem \cite{DK,H}.
Another approach to stability study for periodic
equations in the present paper is based on lower and upper estimates of
the distance between two adjacent zeros (i.e., of nonoscillation
intervals) for nontrivial solutions of homogeneous equations. The
foundations of this approach can be found in the work by Zhukovskii
\cite{zhukovskii}, Kre\u{\i}n \cite{krein} and Yakubovich
\cite{yakubovich1}.
Zones of Lyapunov's stability can be also studied on this basis.
This explains why we connect the different
areas together as well as the fact that actually our approach develops
applications of the classical nonoscillation area. Note that we obtain new
exponential stability conditions for equations with measurable
coefficients. In most stability conditions it was assumed that $b(t)\equiv
b>0$ \cite{H1,H2,LN,PS,S},
$b(t)\geq 0$ is a differentialble function \cite{BH,H3,I,KG,MV}
or some restrictions like slow varying coefficients \cite{DIS,G1,G2}
were imposed. We
consider here equation (\ref{1}) without usual restrictions on parameters of
the equations, the coefficients are even not required to be continuous.

Let us describe nonoscillation in general in order to understand how the
results of the present paper develop also many other topics. The
nonoscillation area
consists of many topics which seem to have no relevance to each
others but they are deeply connected.

The classical de la Vall\'{e}e-Poussin theorem claims that existence of a
positive function $v$ such that $v^{\prime \prime }(t)+p(t)v(t)\leq 0$
for $t\in [ 0,\omega ]$ impiles that $[0,\omega ]$ is a
nonoscillation interval. The idea of theorems on differential
inequalities can be formulated as follows: under certain
conditions solutions of inequalities are
greater or less than the solution of the equation. The idea to construct
an approximate
integration method for the numerical solution of differential equations
based on the comparison of solutions of equations and inequalities first
appeared in the works of famous Russian mathematician
Chaplygin~\cite{chaplygin}
and later was developed by other famous Russian mathematician Luzin~\cite
{luzin}. Concerning our object we can formulate the differential
inequality theorem in the form: {\it under certain conditions the
inequalities}
\begin{equation}
\label{3}
(\pounds y)(t)\geq (\pounds x)(t),~t\in [ 0,\omega ],\,\;y(0)\geq
x(0),\;y^{\prime }(0)\geq x^{\prime }(0)\;
\end{equation}
{\it imply }$y(t)\geq x(t)${\it \ for }$t\in [0,\omega ].$

Independly Azbelev~\cite{azbelev1}, Beckenbach,
Bellman~\cite{bellman} and Wilkins~\cite{wilkins} established that
(\ref{3}) can be applied in a nonoscillation interval $[0,\omega ]$ only.
The general solution of equation (\ref{1}) has the following
representation
\begin{equation}
\label{4}
x(t)=\int_{0}^{t}C(t,s)f(s)ds+x_{1}(t)x_{0}+x_{2}(t)x_{0}^{\prime },
\end{equation}
where
$\ x_{1}$ and $x_{2}$ are the solutions of the homogeneous equation
\begin{equation}
\label{5}
\pounds x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=0,\;t\in [0,\omega ],
\end{equation}
satisfying the initial conditions $x_{1}(0)=1$, $x_{1}^{\prime }(0)=0$ and
$x_{2}(0)=0$, $x_{2}^{\prime }(0)=1$, respectively,
$x_{0}$ and$\ x_{0}^{\prime }$ are corresponding constants.
The kernel $C(t,s)$ of the integral in solution's representation (\ref{4})
is called {\em the Cauchy function} of equation (\ref{1}). {\em The
fundamental function} $X(t,s)$ of (\ref{1}) is defined as follows:
$X(t,s)$ for each fixed $s\geq 0$ as a function of $t$ satisfies
\begin{equation}
\label{6}
\ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=0,\;t\in [ s,\omega],
\end{equation}
\begin{equation}
\label{7}
x(s)=0,\;x^{\prime }(s)=1.
\end{equation}
For equation (\ref{1}) the Cauchy function and the fundamental function
$X(t,s)$ coincide  \cite{AMR}.
We assume that $X(t,s)=C(t,s)=0$ for $0\leq t<s.$

If the solution of the initial value problem $
(\pounds x)(t)=0,\;x(0)=0,\;x^{\prime }(0)=1$ does not vanish at the point $%
t=\omega $\ , then the boundary value problem
\begin{equation}
\label{8}
\pounds x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=f(t),\;t\in [ 0,\omega ],
x(0)=0,\;x(\omega )=0,%
\end{equation}
is uniquely solvable and its solution has the representation
\begin{equation}
\label{9}
x(t)=\int_{0}^{\omega }G(t,s)f(s)ds,
\end{equation}
where the kernel of the integral representation $G(t,s)$ is called the
Green's function of the problem (\ref{8}). Differential inequality
theorems are actually results on positivity or negativity of corresponding
Green's functions.

The equivalence of nonoscillation and the unique solvability of the
interpolation problems
\begin{equation}
\label{10}
\begin{array}{l}{\displaystyle
\pounds x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=f(t),~~t\in
[0,\omega ],}\\
x(t_{1})=0,~~
\;x(t_{2})=0, ~~~0\leq t_{1}<t_{2}\leq \omega ,
\end{array}
\end{equation}
is obvious \cite{levin}. Let us say that the Green's function of problem
(\ref{10}) behaves regularly if
\begin{equation}
\label{11}
G(t,s)(t-t_{1})(t-t_{2})>0,\;t,s\in (0,\omega ),\;t\neq t_{1},\;t\neq t_{2}.%
\end{equation}
It was first proven in \cite{chichkin} that nonoscillation of the equation $
\pounds x=0$\ on the interval $[0,\omega ]$\ is necessary and sufficient for
the regular behavior of  Green's functions of interpolation problem
(\ref{10})
(see also the well known paper by Levin \cite{levin}).

Let us denote by $C_{[0,\omega ]}$ the space of continuous functions $
x:[0,\omega ]\rightarrow \hbox{I\kern-.2em\hbox{R}}$ with the norm $\left\| x\right\| =\max_{t\in
[0,\omega ]}\left| x(t)\right| ,$ and by $D_{[0,\omega ]}$ the linear
space of functions $x:[0,\omega ]\rightarrow \hbox{I\kern-.2em\hbox{R}}$ with absolute
continuous
derivatives. Let $b(t)=b^{+}(t)-b^{-}(t),$ where $b^{+}(t)\geq
0,\;b^{-}(t)\geq 0,$ and
\begin{equation}
\label{12}
\pounds _{0}x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t)-b^{-}(t)x(t)=0,\;t\in [
0,\omega ].
\end{equation}

It was proven in \cite{AzbelevDom2} that the boundary value problem
\begin{equation}
\label{13}
\pounds _{0}x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t) - b^{-}(t)x(t)=f(t),
~~t\in [ 0,\omega ],
x(0)=0, ~x(\omega)=0,
\end{equation}
is uniquely solvable and
its Green's function $G_{0}(t,s)$ is negative
$G_{0}(t,s)<0$ for $t,s\in (0,\omega )$. In the
space $C_{[0,\omega ]}$ let us define the integral operator
$K:C_{[0,\omega]}\rightarrow C_{[0,\omega ]}$ by the equality
\begin{equation}
\label{14}
(Kx)(t)=-\int_{0}^{\omega }G_{0}(t,s)b^{+}(s)x(s)ds.
\end{equation}
Note that the operator $K$ actually maps $C_{[0,\omega]}$ into $
D_{[0,\omega ]}$ due to the properties of Green's function $G_{0}(t,s).$
Hence the equation $x=Kx+g$, where $g(t)=\int_{0}^{\omega }G_{0}(t,s)f(s)ds$,
is equivalent to the boundary value problem (\ref{8}).

The above argument can be summarized in the form of the statement
on six equivalences.

{\bf Theorem A} \cite{AzbelevDom2}. {\it The following assertions are
equivalent:}

{\it 1) there is }$v\in D_{[0,\omega ]}$
{\it \ such that }$v(t)\geq 0${\it \ and }$(
\pounds v)(t)\leq 0${\it \ for }$t\in [ 0,\omega ]${\it \ and }$%
v(0)+v(\omega )-\int_{0}^{\omega }(\pounds v)(t)dt>0;$

{\it 2) }$C(t,s)>0${\it \ for }$0\leq s\leq t\leq \omega ;$

{\it 3) problem (\ref{8}) has a unique solution for each summable }
$f${\it \ and }$G(t,s)<0${\it \ for }$t,s\in (0,\omega );$

{\it 4) each nontrivial solution of the homogeneous equation \pounds }$x=0$%
{\it \ has at most one zero on }$[0,\omega ];$

{\it 5) the spectral radius of the operator }$K:C_{[0,\omega ]}\rightarrow
C_{[0,\omega ]}${\it \ is less than one;}

{\it 6) problem (\ref{10}) has a unique solution for each summable }$f${\it \ and
its Green's function behaves regularly for }$t,s\in (0,\omega ).$

\bigskip

The Polia-Mammana decomposition \cite{Mammana, Polia} is a possibility for
representation of the operator $\pounds :D_{[0,\omega ]}\rightarrow
L_{[0,\omega ]}$\ in the form of products of the first order differential
operators
\begin{equation}
\label{15}
\pounds =h_{2}\frac{d}{dt}h_{1}\frac{d}{dt}h_{0},
\end{equation}
where the real valued functions $h_{i}$ do not have zeros and are smooth
enough. This decomposition is possible if and only if $[0,\omega ]$ is a
nonoscillation interval. In particular, representation (\ref{15}) allows
us to obtain the generalized Rolle's theorem \cite{levin}: {\em if the
solution $x$ of equation (\ref{1}) has more than 3 zeros on the
nonoscillation interval, then the function $f$ has at least one
zero on this interval}.

The theory of oscillatory kernels plays an important role in oscillation of
mechanical systems \cite{Gantmakher}. The oscillatory kernel $G(t,s)$
is characterized by the inequalities
\begin{equation}
\label{16}
G(t,s)>0,\;\det \left| G(t_{i},s_{j})\right| _{1}^{m}\geq
0,\;0<t_{1}<...<t_{m}<\omega ,\;0<s_{1}<...<s_{m}<\omega ,\;m=1,2,...
\end{equation}
while for $t_{i}=s_{i}$\ $(1=1,2,,...,m)$ the inequality
in (\ref{16}) has to be strict.
In \cite{Gantmakher} it was demonstrated that actually the fact
that the kernel is oscillatory means that the integral operator
$G:L_{[0,\omega]}\rightarrow D_{[0,\omega ]}$ of the form
$(Gf)(t)=\int_{0}^{\omega
}G(t,s)f(s)ds$\ does not increase the number of sign's changes. If we
consider the integral operator, where the kernel $G(t,s)$
is the Green's function of problem (\ref{8}), then in this case the inverse
operator $\pounds :D_{[0,\omega ]}\rightarrow L_{[0,\omega ]}$\ defined on
the functions satisfying the conditions $x(0)=0,\;x(\omega )=0$ should not
decrease the number of sign's changes. Although a direct verification of
an infinite number of inequalities (\ref{16}) is possible only for very
simple kernels and cannot be implemented in most interesting for
applications cases, non-decreasing of the number of
sign's changes for the integral operators with Green's functions as
kernels can be checked
through the Polia-Mammana decomposition and the generalized Rolle's theorem.
This connection of oscillatory kernels and Polia-Mammana decomposition was
discovered by M.G.Kre\u{\i}n.

Thus nonoscillation solves in many important
cases the problem of checking oscillatory kernels.
As a conclusion we note that each new nonoscillation result or test
develops all these directions.

\section{Preliminaries}
\setcounter{equation}{0}

Let us start with the following simple corollaries of Theorem A.
Choosing $v(t)=\exp (\lambda t)$ in the first assertion of Theorem A, we
obtain the following result.

\begin{corollary}
 If there exists such a real constant $\lambda $ such that
\begin{equation}
\label{19}
\lambda ^{2}+a(t)\lambda +b(t)\leq 0,
\end{equation}
 then for each $\omega $  assertion 2)-6) of Theorem A are true.
If in addition $b(t)\geq \beta >0$ and this $\lambda < 0,$  then
equation (\ref{1}) is  exponentially stable (the fundamental function has an
exponential estimate).
\end{corollary}
\bigskip

Solving the equation
\begin{equation}
\label{20}
\lambda ^{2}+a(t)\lambda +b(t)=0,
\end{equation}
for all $t,$ we get $\lambda _{1}(t)=-\frac{a(t)}{2}-\sqrt{\frac{a^{2}(t)}{4}%
-b(t)}$ and $\lambda _{2}(t)=-\frac{a(t)}{2}+\sqrt{\frac{a^{2}(t)}{4}-b(t)}.$

\begin{corollary}$ \cite{levin}$
 If for sufficiently large $t\geq
t_{0}$ there exist $\nu _{0},\nu _{1}$ and $\nu _{2}$ such
that $\lambda _{1}(t)$  and $\lambda _{2}(t)$  are real functions
and $\nu _{0}\leq \lambda _{1}(t)\leq \nu _{1}\leq \lambda
_{2}(t)\leq \nu _{2},$ where $\nu _{0}<\nu _{1}<\nu _{2},$ then
the fundamental system of equation (\ref{1}) satisfies the inequalities
\begin{equation}
\label{21}
c_{i}\exp (v_{i-1}t)\leq x_{i}(t)\leq d_{i}\exp
(v_{i}t),\;i=1,2\;(c_{i},d_{i}>0;\;t\geq t_{0}).
\end{equation}
If in addition $\nu_2 <0,$  then equation (\ref{1}) is  exponentially stable.
\end{corollary}

In order to  study stability properties, we consider the scalar
differential equation of the  second order (\ref{1}) under the following
conditions:

(a1) $a(t), b(t)$ are Lebesgue measurable and
essentially bounded functions on $[0,\infty)$.

(a2) $f:[t_0,\infty)\rightarrow R $ is a Lebesgue measurable locally
essentially bounded function.

\noindent

\noindent
{\bf Definition}. Eq.~(\ref{1}) is {\em (uniformly) exponentially stable},
if the fundamental function  $X(t,s)$ of (\ref{1}) {\em has
an exponential estimate}
if there exist
positive numbers
$K>0, \lambda>0$, such that
\begin{equation}
\label{17}
|X(t,s)|\leq K~e^{-\lambda (t-s)},~~t\geq s\geq 0.
\end{equation}

Consider the equation
\begin{equation}
\label{18}
\ddot{x}(t) +a\dot{x}(t)+bx(t)=0,
\end{equation}
where $a>0, b>0$ are positive numbers.
This equation is exponentially stable.
Denote by $Y(t,s)$ the fundamental function of  (\ref{18}).

\begin{uess}
Let $a>0, b>0$.

1) If  $a^2>4b$ then ${\displaystyle \int_0^t|Y(t,s)|ds\leq\frac{1}{b}, ~~
\int_0^t|Y^{'}_t(t,s)|ds\leq\frac{2a}{\sqrt{a^2-4b}(a-\sqrt{a^2-4b})} }$.
\vspace{2mm}

2) If $a^2<4b$ then ${\displaystyle \int_0^t|Y(t,s)|ds\leq
\frac{4}{a\sqrt{4b-a^2}}, ~~
\int_0^t|Y^{'}_t(t,s)|ds\leq \frac{2(a+\sqrt{4b-a^2})}{a\sqrt{4b-a^2}} }$.
\vspace{2mm}

3)  If  $a^2=4b$ then ${\displaystyle \int_0^t|Y(t,s)|ds\leq \frac{1}{b},
~~
\int_0^t|Y^{'}_t(t,s)|ds\leq \frac{2}{\sqrt{b}}}$.
\end{uess}
{\bf Proof.}
If $a^2>4b$ then the characteristic equation $\lambda^2+a\lambda+b=0$
has two negative roots
$$
\lambda_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2}, ~0\geq \lambda_1>\lambda_2.
$$
By simple calculation we have
$$
0<Y(t,s)=\frac{1}{\sqrt{a^2-4b}}\left(e^{\lambda_1(t-s)}-e^{\lambda_2(t-s)}
\right).
$$
Since ${\displaystyle
\frac{1}{\lambda_1}e^{\lambda_1 t} < \frac{1}{\lambda_2}e^{\lambda_2 t} }$
then
$$
\int_0^t Y(t,s)ds\leq \frac{1}{\sqrt{a^2-4b}}
\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right)=\frac{1}{b}.
$$
We have $Y_t^{'}(t,s)=\frac{1}{\sqrt{a^2-4b}}\left(\lambda_1e^{\lambda_1(t-s)}-\lambda_2e^{\lambda_2(t-s)}\right)$.
Hence
${\displaystyle |Y_t^{'}(t,s)|\leq
\frac{|\lambda_1|+|\lambda_2|}{\sqrt{a^2-4b}}e^{\lambda_1(t-s)} }$
$$=\frac{a}{\sqrt{a^2-4b}}e^{\lambda_1(t-s)} \mbox{~~and~~~}
\int_0^t|Y^{'}_t(t,s)|ds\leq \frac{2a}{\sqrt{a^2-4b}(a-\sqrt{a^2-4b})}.$$

If $a^2-4b<0$, then the characteristic equation has two complex roots
and the fundamental function has the form
$$
Y(t,s)=\frac{2}{\sqrt{4b-a^2}}e^{-\frac{a}{2}(t-s)}
\sin\left( \frac{\sqrt{4b-a^2}}{2}(t-s) \right).
$$
Hence
${\displaystyle
|Y(t,s)|\leq \frac{2}{\sqrt{4b-a^2}}e^{-\frac{a}{2}(t-s)}
}$
and
${\displaystyle
\int_0^t|Y(t,s)|ds\leq \frac{4}{a\sqrt{4b-a^2}}.
}$
The second inequality in 2) is proven in a similar way as the second
inequality in 1).

If $a^2=4b$, then the characteristic equation has the double root
$\lambda=-\frac{a}{2}$  and
$$
Y(t,s)=(t-s)e^{-\frac{a}{2}(t-s)}.
$$
Since
${\displaystyle \int_0^{\infty} s e^{-as/2}= \frac{4}{a^2}=\frac{1}{b} }$
then the first inequality in 3) holds. The second inequality in 3) is
proven similarly to the previous cases.
\qed

Let us introduce some functional spaces on a semi-axis. Denote
by $L_{\infty}[t_0,\infty)$ the space of all essentially bounded on
$[t_0,\infty)$
functions and by $C[t_0,\infty)$ the space of all continuous
bounded on  $[t_0,\infty)$ functions
with the supremum norm.

\begin{uess}$\cite{DK,H}$
Suppose  (a1)-(a2) hold and there exists $t_0\geq 0$
such that for every
$f\in L_{\infty}[t_0,\infty)$
both the solution $x$ of the problem
$$
\ddot{x}(t) +a(t)\dot{x}(t)+b(t)x(t)=f(t),~ t\geq t_0,
$$$$
x(t)=0, \dot{x}(t)=0,~t\leq t_0,
$$
and its derivative $\dot{x}$ belong to $C[t_0,\infty)$. Then
equation (\ref{1}) is exponentially
stable.
\end{uess}

\section{Integro-differential equation}
\setcounter{equation}{0}

To obtain positiveness conditions for the fundamental function of equation
(\ref{1}) we consider first a similar problem for the following
integro-differential equation
\begin{equation}
\label{22}
\dot{y}(t)+\int_0^t e^{-\int _s^ta(\xi)d \xi}b(s)y(s)ds=0,
\end{equation}
for which we assume that condition (a1) holds.

Together with (\ref{8}) we consider
for each $t_0\geq 0$ the initial value problem
\begin{equation}
\label{23}
\dot{y}(t)+\int_{t_0}^t e^{-\int _s^ta(\xi)d \xi}b(s)y(s)ds=f(t),
\end{equation}
\begin{equation}
\label{24}
y(t_0)=y_0.
\end{equation}
We assume that condition (a2) holds for the function $f(t)$.

\noindent
{\bf Definition.} A locally absolutely
continuous on $[t_0,\infty)$ function $y:\hbox{I\kern-.2em\hbox{R}} \rightarrow \hbox{I\kern-.2em\hbox{R}}$ is called
{\em a solution} of  problem (\ref{23}), (\ref{24})
if it satisfies equation (\ref{23}) for almost every
$t\in [t_0, \infty)$, and equality (\ref{24}) for $t= t_0$.

\noindent
{\bf Definition.} For each $s\geq 0$ the solution $Y(t,s)$ of the problem
\begin{equation}
\label{25}
\dot{y}(t)+\int_{s}^t e^{-\int _s^ta(\xi)d \xi}b(\tau)y(\tau)d\tau=0, ~~
t>s, ~~ y(s)=1,
\end{equation}
is called {\em a fundamental function} of equation (\ref{22}).

We assume $Y(t,s)=0, t<s$.

\begin{uess}$\cite{H}$
Let (a1)-(a2) hold. Then there exists one and only one
solution of  problem (\ref{23}), (\ref{24}) that can be presented in the form
\begin{equation}
\label{26}
y(t)=Y(t,t_0)y_0+\int_{t_0}^t Y(t,s)f(s)ds.
\end{equation}
\end{uess}

Let us obtain conditions under which equation (\ref{22}) has a
positive solution.
We remark that
the theorem remains true if the zero initial point is replaced by any
$t_0 \geq 0$.

In future we will apply the following result (\cite{AMR}, Theorem 1.2.1).
Let $L[0,T]$ be the space of all integrable on $[0,T]$ functions with the
norm $\|f\|=\int^{T}_{0}|f(s)|ds$.

\begin{uess}$\cite{AMR}$
\label{lemmaAzb}
Suppose that a function $r(t,s)$ is measurable over the square
$[a,b]\times[a,b]$, for almost every $s$
the function $r(t,s)$ as a function of $t$ has
finite one-sided limits
at each point $t$
and there exists a function
$v\in L[0,T]$ such that $|r(\cdot,s)|\leq v(\cdot)$. Then the integral
operator
$$
(Ky)(t)=\int_a^b r(t,s)y(s)ds
$$
maps  L[0,T] to L[0,T] and is a compact operator in this space.
\end{uess}
\begin{guess}
Suppose $a(t)\geq 0, b(t)\geq 0$. Then
the following conditions are equivalent for equation (\ref{22}).

1) There exists a positive solution of the inequality
\begin{equation}
\label{27}
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)y(s)ds\leq 0, ~t \geq 0.
\end{equation}

2) There exists a locally essentially bounded  function $u(t)\geq 0$ such that
\begin{equation}
\label{28}
u(t)\geq \int_0^t e^{-\int_s^t [a(\xi)-u(\xi)]d \xi}b(s)ds, ~t \geq 0.
\end{equation}

3) $Y(t,s)>0, t\geq s\geq 0$.

4) There exists a positive solution of  equation (\ref{8}) for $t \geq 0$.
\end{guess}
{\bf Proof.}
1) $\Longrightarrow$ 2).
Suppose $y(t)>0$ is a solution of (\ref{27}).
Hence on any bounded interval $y(t)\geq \alpha>0$, then
${\displaystyle u(t)=-\frac{\dot{y}(t)}{y(t)}\geq 0}$  and is an
essentially locally bounded function. We also have
$$
y(t)=y(0)e^{-\int_0^tu(\xi)d \xi}, ~~
\dot{y}(t)=-y(0)u(t)e^{-\int_0^tu(\xi)d \xi}.
$$
Substituting $y, \dot{y}$ in (\ref{27}) we have
\begin{eqnarray*}
& & \dot{y}(t)+  \int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)y(s)ds
\\
& = & -y(0)u(t)e^{-\int_0^t u(\xi)d \xi}+y(0)\int_0^t
e^{-\int_s^t a(\xi)d \xi}b(s)e^{-\int_0^su(\xi)d \xi}ds
\\
& = & -y(0)e^{-\int_0^tu(\xi))d \xi}\left[u(t)-
\int_0^t e^{-\int_s^t [a(\xi)-u(\xi)]d \xi}b(s)ds\right]\leq 0.
\end{eqnarray*}
Hence $u(t)\geq 0$ is a solution of inequality (\ref{28}).

2) $\Longrightarrow$ 3).
Consider the nonhomogeneous equation (\ref{27})
\begin{equation}
\label{29}
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)y(s)ds=f(t),~~ y(0)=0.
\end{equation}
A solution of  (\ref{29})  will
be sought in the form
\begin{equation}
\label{30}
y(t)=\int_{0}^t e^{-\int_s^t u(\xi)d \xi}z(s)ds,~~
\dot{y}(t)=z(t)-u(t)\int_{0}^t e^{-\int_s^t u(\xi)d \xi}z(s)ds,
\end{equation}
so $\dot{y}(t)+u(t)y(t)=z(t), y(0)=0.$

After substituting (\ref{30}) into (\ref{29}) we have
\begin{equation}
\label{31}
z(t)-u(t)\int_{0}^t e^{-\int_s^t u(\xi)d \xi}z(s)ds
+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)
\int_{0}^s e^{-\int_{\tau}^s u(\xi)d \xi}z(\tau)d\tau ds=f(t).
\end{equation}
Since
\begin{eqnarray*}
& & \int_{0}^t \left[ e^{-\int_s^t a(\xi)d \xi}b(s)\int_{0}^s
e^{-\int_{\tau}^s
u(\xi)d \xi}z(\tau)d\tau \right] ds
\\
& = & \int_{0}^t \left[ \int_{\tau}^t e^{-\int_s^t a(\xi)d \xi}b(s)
e^{-\int_{\tau}^su(\xi)d \xi}ds \right] z(\tau)d\tau
\\
& = & \int_{0}^t e^{-\int_{\tau}^t u(\xi)d \xi} \left[ \int_{\tau}^t
e^{-\int_s^t (a(\xi)-u(\xi))d \xi }b(s)ds \right] z(\tau)d\tau,
\end{eqnarray*}
then equation (\ref{31}) can be rewritten as
\begin{equation}
\label{32}
z(t)-\int_{0}^t e^{-\int_{\tau}^t u(\xi)d \xi}
\left[u(t)-\int_{\tau}^t e^{-\int_s^t (a(\xi)-u(\xi))d \xi }b(s)ds\right]
z(\tau)d\tau=f(t).
\end{equation}
On every finite interval $[0,T]$ equation (\ref{32}) has the form
\begin{equation}
\label{33}
z-Hz=f, ~t\in [0,T].
\end{equation}
Operator $H:L[0,T]\rightarrow L[0,T]$ is bounded.
In order to show that this operator is compact we apply
Lemma \ref{lemmaAzb}.
Operator $H$ can be rewritten in the form $H=PH_1-H_2$, where
$$
(Pz)(t)=u(t)z(t),~(H_1z)(t)=\int_{0}^t e^{-\int_{\tau}^t
a(\xi)d \xi}z(\tau)d\tau,
$$$$
(H_2z)(t)=\int_{0}^t
\left[ e^{-\int_{\tau}^t a(\xi)d \xi}\int_{\tau}^t e^{-\int_s^t
(a(\xi)-u(\xi))d \xi }b(s)ds \right] z(\tau)d\tau.
$$
It is easy to see that for operators $H_1, H_2$ all conditions of Lemma 5 hold.
Then these operators are compact.
Operator $P$ is bounded operator,
hence operator $H$ is a compact Volterra integral operator
with spectral radius $r(T)=0$ \cite{R}.
Hence for the solution of equation (\ref{33}) we have
$z=(I-H)^{-1}f$, where $I$ is the identity operator.

Since
$$
u(t)-\int_{\tau}^t e^{-\int_s^t (a(\xi)-u(\xi))d \xi }b(s)ds
\geq u(t)-\int_0^t e^{-\int_s^t (a(\xi)-u(\xi))d \xi }b(s)ds\geq 0,
$$
then $H$ is a positive operator.
Hence $(I-H)^{-1}= 1+H+H^2+H^3+ \dots $
is also a positive operator.

Suppose now that in the equation (\ref{33}) we have $f(t)\geq 0$. Then for
the solution of (\ref{33})
we have $z(t)\geq 0$. Equality (\ref{16}) implies that for every
right-hand side $f(t)\geq 0$ the solution of equation (\ref{29}) $y(t)\geq 0$.
But $y(t)=\int_0^t Y(t,s) f(s)ds$. Hence $Y(t,s)\geq 0,~ 0\leq s\leq t\leq T$.
Since $T>0$ is an arbitrary number then $Y(t,s)\geq 0,~ 0\leq s\leq t<\infty$.
We only have to prove that the strong inequality for $Y(t,s)>0$ holds.

After substituting $y(t)=e^{-\int_0^tu(\xi)~d \xi} $ in the left-hand
side of
equation (\ref{29}) we see that this function is a solution of (\ref{29})
with $f(t)<0$. By the solution representation formula (\ref{26}) we have
$$
y(t)=Y(t,0)+\int_0^t Y(t,s) f(s)ds.
$$
Hence  $Y(t,0)\geq y(t)>0$. The general case $Y(t,s)>0$ is considered similarly.

Implications 3)$\Longrightarrow$4) and 4)$\Longrightarrow$1) are evident.
\qed

\noindent
{\bf Remark.} Nonoscillation conditions for general integro-differential
equations with a bounded memory were obtained in \cite{BB}.
However, these results are not applicable to equation (\ref{22}).
Nonoscillation results for integro-differential
equation can also be found in \cite{DG2001}.

\section{Positive Solutions}
\setcounter{equation}{0}

The following lemma gives a connection between
equations (\ref{1}) and (\ref{22}).

\begin{uess}
Denote by $x_1(t), x_2(t)$ and $X(t,s)$
the fundamental system and the fundamental function of (\ref{1}),
respectively, by $Y(t,s)$
the fundamental function of (\ref{8}). Then
$$
x_1(t)=Y(t,0), ~x_2(t)=\int_0^t Y(t,\tau)e^{-\int_0^{\tau}a(\xi)d\xi}d\tau,
~X(t,s)=\int_s^t Y(t,\tau)e^{-\int_s^{\tau}a(\xi)d\xi}d\tau.
$$
\end{uess}
{\bf Proof.}
For the solution of equation (\ref{1}) we have
$$
\dot{x}(t)=e^{-\int_0^{t}a(\xi)d\xi}\dot{x}(0)
-\int_0^{t}e^{-\int_s^{t}a(\xi)d\xi}b(s)x(s)ds.
$$
Hence
$$
\dot{x}(t)+\int_0^{t}e^{-\int_s^{t}a(\xi)d\xi}b(s)x(s)ds
=e^{-\int_0^{t}a(\xi)d\xi}\dot{x}(0)
$$
and
$$
x(t)=Y(t,0)x(0)+\left[ \int_0^t
Y(t,\tau)e^{-\int_0^{\tau}a(\xi)d\xi}d\tau\right] \dot{x}(0).
$$
But for the solution of (\ref{1}) we have another representation:
$$
x(t)=x_1(t)x(0)+x_2(t)\dot{x}(0).
$$
The equalities for the fundamental system of (\ref{1}) are proven.
Since $X(t,s)=x_2(t,s)$, then the proof of the equality for $X(t,s)$ is
similar.
\qed

\begin{corollary}
If the fundamental function $Y(t,s)$ of (\ref{22}) is positive then
the fundamental system and the fundamental function of (\ref{1}) are positive.
\end{corollary}
\begin{corollary}
Suppose $a(t)\geq 0, b(t)\geq 0,$ and the fundamental function of (\ref{8}) is positive.
Then
\begin{equation}
\label{34}
0\leq \int_{t_0}^t X(t,s)b(s)ds\leq 1,
\end{equation}
where $X(t,s)$ is the fundamental function of (\ref{1}).
\end{corollary}
{\bf Proof.}
The function $x(t)\equiv 1$ is the solution of (\ref{2}) with $f(t)=b(t)$.
By the solution representation formula we have
$$
1=x_1(t)+\int_{t_0}^t X(t,s)b(s)ds.
$$
Corollary 3 implies $x_1(t)>0, X(t,s)>0.$ Hence the inequality (\ref{34})
is valid.
\qed

Together with (\ref{1}) consider the following equation
\begin{equation}
\label{35}
\ddot{x}(t) +a_1(t)\dot{x}(t)+b_1(t)x(t)=0,
\end{equation}
where for $a_1(t), b_1(t)$ condition (a1) holds.
\begin{corollary}
Suppose $a_1(t)\geq a(t)\geq 0, b(t)\geq b_1(t)\geq 0,$
and equation  (\ref{22}) has a positive solution.
Then the fundamental function and the fundamental system of  (\ref{35})
 are positive.
\end{corollary}
{\bf Proof.}
If (\ref{22}) has a positive solution, then inequality (\ref{28}) has
a nonnegative solution $u(t)\geq 0$. This function is a nonnegative solution of
inequality (\ref{28}) where $a(t)$ and $b(t)$ are replaced by $a_1(t)$ and
$b_1(t)$.
Corollary 3 implies this corollary.
\qed

\begin{guess}
Suppose $a_1(t)\geq a(t)\geq 0, b(t)\geq b_1(t), b(t)\geq 0$.
If the fundamental function of (\ref{22}) is positive,
then the fundamental function and the fundamental system of (\ref{35})
are positive.
\end{guess}
{\bf Proof.}
Consider the equation
\begin{equation}
\label{36}
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a_1(\xi)d \xi}b_1(s)y(s)ds=f(t),~~
y(0)=0,
\end{equation}
which can be rewritten in the form
\begin{eqnarray}
& & \dot{y}(t)+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)y(s)ds
\nonumber
\\
& = &\int_{0}^t \left( e^{-\int_s^t a(\xi)d \xi}-e^{-\int_s^t
a_1(\xi)d \xi}\right)b(s)y(s)ds
\nonumber
\\
\label{37}
& + & \int_{0}^t e^{-\int_s^t a_1(\xi)d \xi}(b(s)-b_1(s))y(s)ds+f(t),
 ~~ y(0)=0.
\end{eqnarray}
Hence (\ref{37})  is equivalent to the equation
\begin{equation}
\label{38}
y(t)=\int_{0}^t Y(t,s)\int_{0}^s \left(e^{-\int_{\tau}^s
a(\xi)d \xi}-e^{-\int_{\tau}^s
a_1(\xi)d \xi}\right)b(\tau)y(\tau)d\tau ~ ds
\end{equation}
$$
+\int_{0}^t Y(t,s)\int_{0}^s e^{-\int_{\tau}^s
a_1(\xi)d \xi}(b(\tau)-b_1(\tau))y(\tau)d\tau ~ ds + g(t),
$$
where $g(t)=\int_{0}^t Y(t,s) f(s)ds\geq 0$.

Equation (\ref{38}) has the form $y=Hy+g$, where $H$ is a positive compact
Volterra operator in the space $L{[0,T]}$
for every $T>0$.
Hence the spectral radius $r(H)=0$
in the space $L{[0,T]}$ and then for the solution of (\ref{38})
we have $y(t)=((I-T)^{-1}g)(t)\geq 0$. The solution of
the equation (\ref{36}) has the form
$y(t)=\int_{0}^t Y_1(t,s)f(s)ds$, where $Y_1(t,s)$ is the
fundamental function of the equation
$$
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a_1(\xi)d \xi}b_1(s)y(s)ds=0.
$$
We obtained that for every $f(t)\geq 0$ the solution of (\ref{36})
is also nonnegative.
Hence the inequality $Y_1(t,s)\geq 0$ holds.
Now, let us prove the strict inequality $Y_1(t,s)> 0$.

The function $Y_1(t,0)$ is the solution of the equation
\begin{equation}
\label{39}
\dot{y}(t)+\int_0^t e^{-\int_s^t a_1(\xi)d \xi}b_1(s)y(s)ds=0,~~ y(0)=1.
\end{equation}
After rewriting (\ref{39}) in the form
\begin{eqnarray}
& & \dot{y}(t) +\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)y(s)ds
\nonumber
\\
& = & \int_{0}^t \left(e^{-\int_s^t a(\xi)d \xi}-e^{-\int_s^t
a_1(\xi)d \xi}\right)b(s)y(s)ds
\nonumber
\\
\label{40}
& + & \int_{0}^t e^{-\int_s^t a_1(\xi)d \xi}(b(s)-b_1(s))y(s)ds,~~ y(0)=1,
\end{eqnarray}
equation (\ref{40}) has the form
$$
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)y(s)ds =f(t), y(0)=1,
$$
where $f(t)\geq 0$. Hence
$$
Y_1(t,0)=Y(t,0)+\int_{0}^t Y(t,s)f(s)ds.
$$
Then $Y_1(t,0)\geq Y(t,0)>0$.

The general case $Y_1(t,s)>0$ is considered similarly.
Now, Lemma 6 implies the statement of the theorem.
\qed

\noindent
{\bf Remark.} The theorem remains true if we replace the zero initial
point $0$ by any $t_0>0$.

Denote $b^+=\max\{b,0\}$.
\begin{corollary}
Suppose $a(t)\geq 0$ and the fundamental
function of the equation
$$
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b^+(s)y(s)ds=0
$$
is positive. Then the fundamental function and the fundamental system of equation (\ref{1})
are positive.
\end{corollary}

Now let us proceed to explicit sufficient conditions when
equation (\ref{1}) has a positive solution.

\begin{guess}
Suppose  at least one of the following conditions holds
\vspace{2mm}

1) $ a(t)\geq \int_{t_0}^t b^+(s)ds$,
\vspace{2mm}

2) there exists $\lambda>0$ such that  $a(t)\geq \lambda
b^+(t)+\frac{1}{\lambda},~ t\geq t_0$,
\vspace{2mm}

3) $a(t)\geq 0$ and there exists $\lambda>0$ such that $\int_{t_0}^t e^{-\int_s^t
(a(\xi)-\lambda)d\xi}b^+(s)ds\leq \lambda, t\geq t_0$.

Then the fundamental function and the fundamental system of  (\ref{1}) are positive
for $t\geq t_0$.
\end{guess}
{\bf Proof.} It is sufficient to prove the theorem for the case $b(t)\geq 0$.

1) The function $u(t)=a(t)$  is a solution of inequality (\ref{28}).

2) The function $u(t)=a(t)-\lambda b(t)$  is a solution of inequality (\ref{28}).

3) The function $u(t)=\lambda$  is a solution of inequality (\ref{28}).
\qed

\begin{corollary}
Suppose $a(t)\geq 0$ and at least one of the following conditions holds
\vspace{2mm}

1) $a(t)\geq a>0, ~\int_0^{\infty}b^+(s)ds<\infty$,
\vspace{2mm}

2) $a^2(t)\geq 4B$, where
${\displaystyle B=\limsup_{t\rightarrow\infty}b^+(t)}$,
\vspace{2mm}

3) ${\displaystyle
\inf_{\lambda>0}\limsup_{t\rightarrow\infty}\frac{1}{\lambda}
\int_0^t e^{-\int_s^t(a(\xi)-\lambda)d\xi}b^+(s)ds<1}$.

Then there exists $t_0\geq 0$ such that the fundamental function
and the fundamental system of  (\ref{1}) are positive for $t\geq t_0$.
\end{corollary}
{\bf Proof.} The proof of 1) and 3) is evident.
To prove 2) we assume $\lambda=1/\sqrt{B}$.
\qed

Now let us compare solutions of the same equation with different right
hand sides and initial conditions, as well as  of different
equations.

\begin{guess}
Suppose $a(t)\geq 0, b(t)\geq 0$, the fundamental function of (\ref{22}) is positive.
Denote by $x(t)$ the solution of the problem
$$
\ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=f(t),~t\geq t_0,
$$$$
x(t_0)=x_0,~ x'(t_0)=x^{'}_0,
$$
by $v(t)$  the solution of the above problem , where $f(t), x_0,x_0^{'}$
are replaced with $f_1(t), v_0,v_0^{'}$. If $f(t)\geq f_1(t), x_0\geq
v_0,x_0^{'}\geq v_0^{'}$,
then $x(t)\geq v(t)$.
\end{guess}
{\bf Proof.}
By Corollary 1 the fundamental system and the fundamental function
of (\ref{1}) are positive.
The statement of the theorem follows from solution representation
formula (\ref{4}).
\qed

\begin{guess}
Suppose $a_1(t)\geq a(t)\geq 0, b(t)\geq b_1(t)\geq 0$,
the fundamental function of (\ref{22}) is positive.
Denote by $x_1(t),x_2(t), X(t,s)$ the fundamental system and the fundamental
function of (\ref{1});
by $v_1(t),v_2(t), V(t,s)$ the fundamental system and the fundamental
function of (\ref{35}).

Then $x_1(t)\leq v_1(t), x_2(t)\leq v_2(t), X(t,s)\leq V(t,s)$.
\end{guess}
{\bf Proof.}
Denote by $Y(t,s)$ the fundamental function of (\ref{8}), by $Y_1(t,s)$
the fundamental function
of (\ref{22}) where $a(t), b(t)$ is replaced by $a_1(t), b_1(t)$,
respectively.
By Corollary 1 it is sufficient to prove that $Y_1(t,s)\geq Y(t,s)$.

The function $Y_1(t,0)$ is the solution of the equation
\begin{equation}
\label{41}
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a_1(\xi)d \xi}b_1(s)y(s)ds=0, ~ y(0)=1,
\end{equation}
which can be rewritten equation in the form
\begin{equation}
\label{42}
\dot{y}(t)+\int_{0}^t e^{-\int_s^t a_1(\xi)d \xi}b(s)y(s)ds=
\int_{0}^t e^{-\int_s^t a(\xi)d \xi}(b(s)-b_1(s))y(s)ds
\end{equation}
$$
+\int_{0}^t \left(e^{-\int_s^t a(\xi)d \xi}-e^{-\int_s^t a_1(\xi)d \xi}\right)b_1(s)y(s)ds, ~ y(0)=1.
$$
By formula (\ref{26}) for the solution of (\ref{42}) we have
$$
Y_1(t,0)=Y(t,0)+\int_0^t Y(t,s)\int_{0}^s e^{-\int_{\tau}^s a(\xi)d \xi}(b(\tau)-b_1(\tau))y(\tau)d\tau ds
$$$$
+\int_0^t Y(t,s)\int_{0}^s \left(e^{-\int_{\tau}^s a(\xi)d \xi}-e^{-\int_{\tau}^s a_1(\xi)d \xi}\right)b_1(\tau)y(\tau)d\tau ds.
$$
Since $a(t)-a_1(t)\geq 0,~ b(t)-b_1(t)\geq 0$, then $Y_1(t,0)\geq Y(t,0)$.
The inequality $Y_1(t,s)\geq Y(t,s)$ is obtained similarly.
\qed

\begin{corollary}
Suppose
$$
a(t)\equiv a>0, ~~ b(t)\geq b>0, ~~ a^2-4b\geq 0.
$$
Then the fundamental function $X(t,s)$ of (\ref{1}) is positive and for this function
we have the following estimation
$$
\int_0^t X(t,s)ds\leq\frac{1}{b}.
$$
\end{corollary}
{\bf Proof.}
By Theorem 5 we have $0<X(t,s)\leq V(t,s)$, where $V(t,s)$ is the
fundamental function of equation (\ref{18}).
Application of Lemma 2 (parts 1 and 3) completes the proof.
\qed

\section{Stability}
\setcounter{equation}{0}

\begin{guess}
Suppose $a(t) \geq \alpha>0, b(t)\geq \beta>0$. If the fundamental function of
(\ref{1})
 is positive then equation (\ref{1}) is exponentially stable.
\end{guess}
{\bf Proof.}
Consider (\ref{1}) with initial conditions $x(0)=0, \dot{x}(0)=0$.
Suppose $f\in L_{\infty}[0,\infty)$. For the solution of this problem we
have
$x(t)=\int_0^t X(t,s)f(s)ds$.
By Corollary 4
$$
|x(t)|\leq \int_0^t X(t,s)|f(s)|ds=\int_0^t X(t,s)b(s)\frac{|f(s)|}{b(s)}ds
\leq \frac{\|f\|}{\beta} \int_0^t X(t,s)b(s)ds\leq
\frac{\|f\|}{\beta},
$$
where $\| \cdot \|$ is the sup-norm in $C[0,\infty)$.
Then $x\in C[0,\infty)$.
We have \\
${\displaystyle
\dot{x}(t)=\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)x(s)ds
+\int_{0}^t e^{-\int_s^t a(\xi)d \xi}b(s)f(s)ds}$,
hence $\dot{x}\in C[0,\infty)$.
By Lemma 3  equation (\ref{1}) is
exponentially stable.
\qed

\begin{corollary}
Suppose $a(t)\geq \alpha >0, b(t)\geq \beta>0$ and at least one of
the conditions of Corollary 7 holds.
Then equation (\ref{1}) is exponentially stable.
\end{corollary}

Denote
$$
\alpha=\liminf_{t\rightarrow\infty}a(t), \beta=\liminf_{t\rightarrow\infty}b(t),
B=\limsup_{t\rightarrow\infty}b(t).
$$

\begin{guess}
Suppose ${\displaystyle 0<\beta\leq B<\frac{1}{2}{\alpha}^2}$.
Then equation (\ref{1}) is exponentially stable.
\end{guess}
{\bf Proof.}
Without loss of generality we can assume that $b(t)\geq \beta$ for any
$t\geq 0$ and there exists $\epsilon>0$ such that
\begin{equation}
\label{43}
\frac{\epsilon}{4} {\alpha}^2<b(t)<\left(\frac{1}{2}-
\frac{\epsilon}{4}\right) {\alpha}^2.
\end{equation}
Consider (\ref{1}) with initial conditions $x(0)=0, \dot{x}(0)=0$.
Suppose $f\in L_{\infty}[0,\infty)$ and let us prove that  $x\in
C[0,\infty)$.

Denote by $X_0(t,s)$ the fundamental function of the equation
\begin{equation}
\label{44}
\ddot{x}(t)+a(t)\dot{x}(t)+\frac{1}{4}{\alpha}^2x(t)=0.
\end{equation}
By Corollary 7(2), $X_0(t,s)>0, t\geq s\geq 0$.
Equation (\ref{1}) can be rewritten in the form
in the form
\begin{equation}
\label{45}
\ddot{x}(t)+a(t)\dot{x}(t)+\frac{1}{4}{\alpha}^2x(t)
+ \left( b(t)-\frac{1}{4}{\alpha}^2 \right) x(t)=f(t),
\end{equation}
with the initial conditions $x(0)=0, \dot{x}(0)=0$.
Equation (\ref{45}) is equivalent to
\begin{equation}
\label{46}
x(t)+\int_0^t X_0(t,s) \left( b(s)-\frac{1}{4}{\alpha}^2
\right) x(s)ds=g(t),
\end{equation}
where
$g(t)=\int_0^t X_0(t,s) f(s)ds$.
By Theorem 6  equation (\ref{44})
is exponentially stable hence $X_0(t,s)$ has an exponential
estimation. Then $g\in L_{\infty}[0,\infty)$.

Equation (\ref{46}) has the form $x+Px=g$. Corollary 2 implies
\begin{eqnarray*}
|(Px)(t)| & \leq &
\int_0^t X_0(t,s) \left| b(s)-\frac{1}{4}{\alpha}^2 \right| |x(s)|~ds
\\
& \leq & \int_0^t X_0(t,s) \frac{1}{4}{\alpha}^2 \frac
{|b(s)-\frac{1}{4}{\alpha}^2|}{\frac{1}{4}{\alpha}^2}ds \|x\|
\\
& \leq & \frac{\sup_{t\geq
0}|b(t)-\frac{1}{4}{\alpha}^2|}{\frac{1}{4}{\alpha}^2}
\|x\|.
\end{eqnarray*}
Inequality
${\displaystyle
\frac{\sup_{t\geq 0}|b(t)-\frac{1}{4}{\alpha}^2|}{\frac{1}{4}{\alpha}^2} \leq
1-\epsilon
}$
is equivalent to (\ref{43}). Hence $\|P\|<1$,  where the norm is
in $C[0,\infty)$. Then $x\in C[0,\infty)$ for a solution of
(\ref{46}) and therefore for a solution of (\ref{1}). Similar to the proof of Theorem 6 we have $\dot{x}\in
C[0,\infty)$.
By Lemma 3 equation (\ref{1}) is exponentially stable.
\qed

\noindent
{\bf Example 1.} Consider the following equation
\begin{equation}
\label{47}
\ddot{x}(t)+a\dot{x}(t)+(1+0.99\sin{t})x(t)=0.
 \end{equation}
If  $a> \sqrt{3.98}$
then by Theorem 7 equation (\ref{47})  is exponentially stable.
\vspace{2mm}

In all previous results we obtain stability conditions for
equations with a positive fundamental function and for "small"
perturbations of such equations. Below we will give stability conditions
for equations without any positiveness assumptions.

\begin{guess}
Suppose $a(t)\geq \alpha>0, b(t)\geq 0$ and there exist $A>0, B>0$
such that the following condition holds
\begin{equation}
\label{48}
||a(t)-A||\left\|\frac{b}{a}\right\|+||b(t)-B||<
\left\{\begin{array}{ll}
B ~~ ,&A^2\geq 4B,\\
\frac{A\sqrt{4B-A^2}}{4},&A^2< 4B,
\end{array}\right.
\end{equation}
where $\| \cdot \|$ is the norm in the space $L_{\infty}[t_0,\infty)$
for some $t_0\geq 0$. Then equation  (\ref{1}) is exponentially stable.
\end{guess}
{\bf Proof.}
Without loss of generality we can assume $t_0=0$.
Consider equation (\ref{1}) with the initial conditions $x(0)=0,
\dot{x}(0)=0$.
Suppose $f\in L_{\infty}[0,\infty)$. Let us prove that  $x, \dot{x}\in
C[0,\infty)$.
To this end, rewrite equation (\ref{1}) in the form
\begin{equation}
\label{49}
\ddot{x}(t)+A\dot{x}(t)+Bx(t)+(a(t)-A)\dot{x}(t)
+(b(t)-B)x(t)=f(t), ~x(0)=\dot{x}(0)=0.
\end{equation}
Hence
\begin{equation}
\label{50}
x(t)+\int_0^t Y(t,s) \left[ (a(s)-A)\dot{x}(s)+(b(s)-B)x(s)
\right]~ds=f_1(t),
\end{equation}
where $Y(t,s)$ is the fundamental function of
\begin{equation}
\label{51}
\ddot{x}(t)+A\dot{x}(t)+Bx(t)=0,
\end{equation}
$f_1(t)=\int_0^t Y(t,s) f(s) ds.$
Equation (\ref{51}) is exponentially stable
thus $f_1\in L_{\infty}[0,\infty)$.

Equation (\ref{1}) can be rewritten in a different form
\begin{equation}
\label{52}
\dot{x}(t) +\int_0^t e^{-\int_s^t a(\xi)d \xi}b(s)x(s)ds=r(t), ~ t\geq 0,
\end{equation}
where $r(t)=\int_0^t e^{-\int_s^t a(\xi)d \xi}f(s)ds$.
Since $a(t)\geq \alpha>0$, then $r\in L_{\infty}[0,\infty)$.
Substituting $\dot{x}(t)$ from (\ref{52}) to (\ref{50}) we have
\begin{equation}
\label{53}
x(t)-\int_0^t Y(t,s) \left[(a(s)-A)\int_0^s e^{-\int_{\tau}^s
a(\xi)d \xi}b(\tau)x(\tau)d\tau-(b(s)-B)x(s)\right]ds=f_2(t),
\end{equation}
where
$f_2(t)=f_1(t)+\int_0^t Y(t,s) r(s)ds$. Evidently
$f_2\in L_{\infty}[0,\infty)$.

Equation (\ref{53}) has the form $x-Hx=f_2$, where
\begin{eqnarray*}
\|(Hx)(t)\| & \leq & \int_0^t Y(t,s) \left[ |a(s)-A|\int_0^s
e^{-\int_{\tau}^s
a(\xi)d \xi}a(\tau)\frac{b(\tau)}{a(\tau)}d\tau+| b(s)-B| \right]ds ~\|x\|
\\
& \leq & K \left( \|a(t)-A \|\left\|\frac{b}{a}\right\|+ \|b(t)-B\|
\right)\|x\|,
\end{eqnarray*}
where by Lemma 2 the constant $K$ is
$$
K=\sup_{t\geq 0}\int_0^t
|Y(t,s)|ds=\left\{\begin{array}{ll}
\frac{1}{B},& A^2\geq 4B,\\
\frac{4}{A\sqrt{4B-A^2}}, & A^2<4B.\\
\end{array}\right.
$$
Then $\|H \|<1$, hence $x\in C[0,\infty)$ for the solution $x$ of
(\ref{53}) and therefore of (\ref{1}).
As in the proof of Theorem 7, $\dot{x}\in C[0,\infty)$.
By Lemma 3 equation (\ref{1}) is exponentially stable.
\qed

\begin{corollary}
Suppose there exist $a>0, B>0$ such that
\begin{equation}
\label{54}
B-\frac{a\sqrt{4B-a^2}}{4}<m\leq M<B+\frac{a\sqrt{4B-a^2}}{4},
\end{equation}
where ${\displaystyle m=\liminf_{t\rightarrow\infty}b(t)>0,~
M=\limsup_{t\rightarrow\infty}b(t)}$. Then the equation
\begin{equation}
\label{55}
\ddot{x}(t)+a\dot{x}(t)+b(t)x(t)=0
\end{equation}
is exponentially stable.
\end{corollary}
{\bf Proof}  follows from the second inequality (\ref{38}) if we let $A=a$.
\qed

Application of the first inequality (\ref{38})
to equation (\ref{55}) gives the same stability conditions
which were obtained by application of Theorem 7.

{\bf Example 2.}
Consider the equation
\begin{equation}
\label{56}
\ddot{x}(t)+(10+\alpha(t)\sin t)\dot{x}(t)+(26+\beta(t)\cos t)x(t)=0,
\end{equation}
where $\alpha(t), \beta(t)$ are measurable functions, such that
$|\alpha(t)|\leq 1, |\beta(t)|\leq 1$.

If we take $A=10, B=26$, then all condition of Theorem 8 hold.
Then equation (\ref{56}) is exponentially stable.
\vspace{2mm}

We will obtain new stability conditions using the derivative of the
fundamental function of comparison equations.

\begin{guess}
Suppose $a(t)\geq \alpha>0, b(t)\geq 0$ and there exist $a>0,
b>0$ such that at least one of the following conditions holds:

1)
${\displaystyle
a^2>4b, ~\|a(t)-a\|\frac{2a}{\sqrt{a^2-4b}(a-\sqrt{a^2-4b})}+\|b(t)-b\|\frac{1}{b}<1.
}$

2)
${\displaystyle
a^2<4b, ~\|a(t)-a\|\frac{2(a+\sqrt{4b-a^2})}{a\sqrt{4b-a^2}}+\|b(t)-b\|\frac{4}{a\sqrt{4b-a^2}}<1.
}$

3)
${\displaystyle
a^2=4b, ~\|a(t)-a\|\frac{2}{\sqrt{b}}+\|b(t)-b\|\frac{1}{b}<1,
}$
\\
where $\| \cdot \|$ is the norm in the space $L_{\infty}[t_0,\infty)$
for some $t_0\geq 0$.
Then equation  (\ref{1}) is exponentially stable.
\end{guess}
{\bf Proof.}
Without loss of generality we can assume $t_0=0$.
Suppose $x(t)$ is a solution of (\ref{1})
with initial conditions $x(0)=x'(0)=0$. Denote $z(t)=\ddot{x}(t)+a\dot{x}(t)+bx(t), ~Y(t,s)$
is the fundamental function of the equation $\ddot{x}(t)+a\dot{x}(t)+bx(t)=0$. Then
\begin{equation}
\label{57}
x(t)=\int_0^t Y(t,s)z(s)ds,~ \dot{x}(t)=\int_0^t Y_t^{'}(t,s)z(s)ds.
\end{equation}
Equation (\ref{1}) is equivalent to the equation
\begin{equation}
\label{58}
\ddot{x}(t)+a\dot{x}(t)+b(t)x(t)+(a(t)-a)\dot{x}(t)+(b(t)-b)x(t)=f(t).
\end{equation}
After substituting $z(t)$ and (\ref{57}) into (\ref{58})
we have the following equation
\begin{equation}
\label{59}
z(t)+(a(t)-a)\int_0^t Y_t^{'}(t,s)z(s)ds+(b(t)-b)\int_0^t Y(t,s)z(s)ds=f(t).
\end{equation}
Equation (\ref{59}) has the form $z+Hz=f$. For the norm of the operator
$H$ in the space
$L_{\infty}[0,\infty)$ we have
$$
\|H\|\leq \|a(t)-a\|\left\|\int_0^t
|Y_t^{'}(t,s)|ds\right\|+\|b(t)-b\|\left\|\int_0^t |Y(t,s)|ds\right\|.
$$
By Lemma 2 conditions 1)-3) of the theorem imply $\|H\|<1$. Hence for
the solution of (\ref{59})
we have $z\in L_{\infty}[0,\infty)$. Equation $\ddot{x}(t)+a\dot{x}(t)+bx(t)=0$ is exponentially stable.
Equalities (\ref{57}) imply that the solution of (\ref{1}) and its derivative are bounded functions.
Then by Lemma 3 equation (\ref{1}) is exponentially stable.
\qed

\noindent
{\bf Example 3.} Consider the following equation
\begin{equation}
\label{60}
\ddot{x}(t)+\dot{x}(t)+(b+\sin{t})x(t)=0.
\end{equation}
If $b>4.25$ then by Theorem 9(2) ($a=1$, $b$ is the same as in the
equation)
equation (\ref{60})  is exponentially stable.
Theorem 8 gives the same result, if we take $a=1, B=b$.
Theorem 7 does not give any stability condition for this equation.
\vspace{2mm}

\section{Zones of Lyapunov's Stability}

In the following, let us assume that $a(t)$
is an absolutely continuous function.

\setcounter{equation}{0}

It is known that the substitution
\begin{equation}
\label{61}
x(t)=z(t)\exp \left\{ -\frac{1}{2}\int_{0}^{t}a(s)ds\right\}
\end{equation}
transforms the homogeneous equation
\begin{equation}
\label{62}
\pounds x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=0,\;t\in [ 0,\infty ),
\end{equation}
into the equation
\begin{equation}
\label{63}
\ddot{z}(t)+p(t)z(t)=0,\;t\in [ 0,\infty ),
\end{equation}
where
\begin{equation}
\label{64}
p(t)=b(t)-\frac{a^{2}(t)}{4}-\frac{a^{\prime }(t)}{2}.
\end{equation}
Obviously coefficient $p(t)$ can be presented as $
p(t)=p^{+}(t)-p^{-}(t),$ where $p^{+}(t)\geq 0$ and $p^{-}(t)\geq 0.$

Consider now equation (\ref{62}) with an $\omega$-periodic
coefficients $a(t)$ and $b(t).$

It is known from the works of the well known mathematicians
Zhukovskii \cite{zhukovskii}, Kre\u{\i}n \cite{krein} and Yakubovich
\cite{yakubovich1} that there is a deep connection between the problem of
the Lyapunov's stability and the nonoscillation intervals. We propose the
following statement.

\begin{guess} Assume that $a(t+\omega )=a(t),\;b(t+\omega )=b(t)\;$%
for $t\in [0,\infty )$ and
\begin{equation}
\label{65}
\int_{0}^{\omega }p(t)dt>0,
\end{equation}
where $p(t)$ is defined in (\ref{64}),
and at least one of the following three conditions holds:

1) $a(t)\geq \int_{0}^{t}b^{+}(s)ds$  for $t\in [0,\omega ];$

2) $a(t)\geq 0$ and there exists $\lambda >0$ such that $
\int_{0}^{t}\exp \left\{ -\int_{s}^{t}(a(\xi )-\lambda )d\xi
\right\}
b^{+}(s)ds\leq \lambda $  for $t\in [0,\omega ];$

3) $\int_{0}^{\omega }p^{+}(t)dt\leq \frac{4}{\omega }.$

Then all solutions of homogeneous equation (\ref{62}) tend to
zero when $t\to \infty$, if $\int_{0}^{\omega }a(t)dt>0$,
and all solutions are bounded if $\int_{0}^{\omega }a(t)dt=0$.
\end{guess}
{\bf Proof.} It is known \cite{levin} that if $[0,\omega ]$ is a
nonoscillation interval for (\ref{62}),
where $\omega $ is the period of the coefficient $p(t)$, then
condition (\ref{65}) garantees that all solutions of equation (\ref{62})
are bounded. Each of the conditions 1)-3) yields that $[0,\omega ]$ is a
nonoscillation interval.
The conditions on the integral of the function $a(t)$ and reference to
the substitution (\ref{61}) completes the proof.
\qed

\section{Floquet Theory and Stability}
\setcounter{equation}{0}

Consider now the equation
\begin{equation}
\label{66}
\pounds x(t)\equiv \ddot{x}(t)+a(t)\dot{x}(t)+b(t)x(t)=0,\;t\in [0,\infty ),
\end{equation}
with $\omega$-periodic coefficients $a(t+\omega
)=a(t)$, $b(t+\omega)=b(t)$. For this equation there exist solutions
satisfying the condition
\begin{equation}
\label{67}
x(t+\omega )=\lambda x(t).
\end{equation}

The foundations and applications of the Floquet theory were presented in the
book by Yakubovich and Starzhinski\u{\i} \cite{yakubovich}.
Using the Floquet theory for ordinary differential equations write the
equation for $\lambda $:
\begin{equation}
\label{68}
\lambda ^{2}-(x_{1}(\omega )+x_{2}^{\prime }(\omega ))\lambda +W(\omega )=0,%
\end{equation}
where $x_{1}$ and $x_{2}$ are two solutions of the equation
(\ref{66})  such that $x_{1}(0)=1$, $x_{1}^{\prime }(0)=0$, $x_{2}(0)=0$,
$x_{2}^{\prime}(0)=1$. Denote by
$$
W(t)=\det \left(
\begin{array}{cc}
x_{1}(t) & x_{2}(t) \\
x_{1}^{\prime }(t) & x_{2}^{\prime }(t)
\end{array}
\right)
$$
the Wronskian of the fundamental system of (\ref{66}). Obviously
$W(0)=1$.

If $\lambda _{1}$ is a real root of equation (\ref{67}), then the
corresponding solution
of equation (\ref{66})
has the representation
\begin{equation}
\label{69}
y(t)=g(t)\exp \left( \frac{\ln |\lambda _{1}|}{\omega }t\right) ,
\end{equation}
where $g$ is $\omega$-periodic if $\lambda _{1}>0$ and is $
2\omega $-periodic function if $\lambda _{1}<0$. If equation (\ref{68}) has two
complex roots $\lambda _{1}=|\lambda _{1}|\exp (i\theta )$ and $\lambda
_{2}=|\lambda _{1}|\exp (-i\theta )$, then the corresponding solutions
of equation (\ref{66}) have the form
\begin{equation}
\label{70}
y_{1}(t)=\left( g_{1}(t)\cos \frac{\theta t}{\omega }-g_{2}(t)\sin \frac{%
\theta t}{\omega }\right) \exp \left( \frac{\ln |\lambda _{1}|}{\omega }%
t\right) ,
\end{equation}
\begin{equation}
\label{71}
y_{2}(t)=\left( g_{2}(t)\cos \frac{\theta t}{\omega }+g_{1}(t)\sin \frac{%
\theta t}{\omega }\right) \exp \left( \frac{\ln |\lambda _{1}|}{\omega }%
t\right),
\end{equation}
where $g_{1}$ and $g_{2}$ are $\omega $-periodic functions.

\begin{guess} Assume that equation (\ref{66}) is oscillatory and the
distance between zeros of its solutions is different from $2\omega$.
Then the following statements are valid.

a) Equation (\ref{66}) is exponentially stable if
\begin{equation}
\label{72}
\int_{0}^{\omega }a(t)dt>0.
\end{equation}

b) The fundamental solutions of equation (\ref{66}) are of the form
(\ref{69}), where $\left| \lambda _{1}\right| >1$  if
\begin{equation}
\label{73}
\int_{0}^{\omega }a(t)dt<0.
\end{equation}

c) If
\begin{equation}
\label{74}
\int_{0}^{\omega }a(t)dt=0
\end{equation}
then the fundamental solutions of equation (\ref{66}) are bounded.
\end{guess}
{\bf Proof.}
It follows from the classical formula of Ostrogradskii that condition (\ref{72})
implies the inequality $W(\omega )<1,$ condition (\ref{73}) implies the
inequality $W(\omega )>1,$ and condition (\ref{74}) implies that $W(\omega )=1.$
The condition that the distance between zeros of solutions of (\ref{66}) is
different from $2\omega $ excludes the existence of real roots of equation
(\ref{68}). In this case the inequality $W(\omega )<1$ implies that $\left|
\lambda _{1}\right| <1,$ the equality $W(\omega )=1$ implies that $\left|
\lambda _{1}\right| =1,$ and the inequality $W(\omega )>1$ implies that
$\left| \lambda _{1}\right| >1.$ Now the representation of solutions
(\ref{70}),(\ref{71}) completes the
proof.
\qed

\noindent
{\bf Remark.} The condition that the distance between zeros
are different from $2\omega $\ is essential as the following example
demonstrates.

\noindent{\bf Example 4.}
Consider the equation
\begin{equation}
\label{75}
\pounds x(t)\equiv x^{\prime \prime }(t)+\frac{2\sin ^{2}t+\cos t\sin t}
{1+\cos t\sin t}x^{\prime }(t)+\frac{\sin ^{2}t-\cos t\sin t}{1+\cos t\sin
t}x(t)=0,\;t\in [ 0,\infty ).
\end{equation}
Inequality (\ref{72}) for the coefficient
$a(t)=\frac{2\sin ^{2}t + \cos t\sin t}
{1+\cos t\sin t}$\ is fulfilled with $\omega=\pi$, but this equation is
not exponentially stable: its fundamental system is $x_{1}=e^{-t}\cos t$
and $x_{2}=\sin t$.

\bigskip

Using the substitution (\ref{61}), we again obtain the equation
\begin{equation}
\label{76}
z^{\prime \prime }(t)+p(t)z(t)=0,~~t\in [0,\infty ),
\end{equation}
where
\begin{equation}
\label{77}
p(t)=b(t)-\frac{a^{2}(t)}{4}-\frac{a^{\prime }(t)}{2},
\end{equation}
Evidently zeros of the solution $x$ of equation (\ref{66}) and the
corresponding solution $z$ of the equation (\ref{63}) coinside. Let us denote
\begin{equation}
\label{78}
P=ess\inf_{t\in [0,\omega ]}p(t),\;Q=ess\sup_{t\in [0,\omega
]}p(t).
\end{equation}
Estimating distances  between two adjacent zeros (i.e. nonoscillation
intervals) from below and from above we get the following result.

\begin{guess} Suppose $P>0,$ there exists a positive integer $k$ such that
$\frac{k-1}{k}<\sqrt{\frac{P}{Q}}$ and
\begin{equation}
\label{79}
\omega \in \left( 0,\frac{\pi }{2\sqrt{Q}}\right] \cup \left(
\frac{1}{2}\frac{\pi }{%
\sqrt{P}},\frac{\pi }{\sqrt{Q}}\right) \cup ...\cup \left( \frac{k-1}{2}%
\frac{\pi }{\sqrt{P}},\frac{k\pi }{2\sqrt{Q}}\right) .
\end{equation}
Then equation (\ref{66}) is oscillatory and distance between zeros of its
solutions is different from $2\omega .$
\end{guess}
{\bf Proof. }If equation (\ref{68}) has real roots, then there exist such
zeros $t_{0},t_{1}$ of a solution $x(t)$ that the distance between
$t_{0}$ and $t_{1}$ equals $\omega $ or $2\omega$. We will reject this
possibility, since the distance between zeros of $g(t)$ in (\ref{69})
cannot be $2\omega$.

Assume that $x(t_{0})=0.$ We use the functions $v=\sin \sqrt{Q}%
(t-t_{0})$ in the first assertion of Theorem A to get that the
spectral radius of the operator $K_{t_{0},t_{0}+\frac{\pi }{2\sqrt{Q}}%
}:C_{[t_{0},t_{0}+\frac{\pi }{2\sqrt{Q}}]}\rightarrow C_{[t_{0},t_{0}+\frac{%
\pi }{2\sqrt{Q}}]}$ defined by the equality
\begin{equation}
\label{80}
K_{t_{0},t_{0}+\frac{\pi }{2\sqrt{Q}}}x(t)=-\int_{t_{0}}^{t_{0}+\frac{\pi }{2%
\sqrt{Q}}}G_{t_{0},t_{0}+\frac{\pi }{2\sqrt{Q}}}(t,s)p(s)x(s)ds.
\end{equation}
where $G_{t_{0,}t_{0}+\frac{\pi }{2\sqrt{Q}}}(t,s)$\ is the Green's function
of the problem
\begin{equation}
\label{81}
x^{\prime \prime }(t)+a(t)x^{\prime }(t)+b(t)x(t)=f(t),\;t\in
[0,\omega ],
x(t_{0})=0,\;x\left( t_{0}+\frac{\pi }{2\sqrt{Q}}\right)
=0,
\end{equation}
is less than one.

Applying Theorem 5.4 in \cite{krasnoselskii}, p. 81, we obtain that the
spectral radius of the operator
$
K_{t_{0},t_{0}+\frac{\pi }{2\sqrt{P}}}:C_{[t_{0},t_{0}+\frac{\pi }{2\sqrt{P}}
]}\rightarrow C_{[t_{0},t_{0}+\frac{\pi }{2\sqrt{P}}]},$ defined by the
equality
\begin{equation}
\label{82}
K_{t_{0},t_{0}+\frac{\pi }{2\sqrt{P}}}x(t)=-\int_{t_{0}}^{t_{0}+\frac{\pi }{2
\sqrt{P}}}G_{t_{0},t_{0}+\frac{\pi }{2\sqrt{P}}}(t,s)p(s)x(s)ds.
\end{equation}
where $G_{t_{0},t_{0}+\frac{\pi }{2\sqrt{P}}}(t,s)$ is the Green's function
of the problem
\begin{equation}
\label{83}
x^{\prime \prime }(t)+a(t)x^{\prime }(t)+b(t)x(t)=f(t),\;t\in [0,\omega ],
x(t_{0})=0,\;x\left( t_{0}+\frac{\pi }{2\sqrt{P}}\right)
=0,
\end{equation}
is greater or equal to one. Moving the point $t_{0}$ we obtain that there
are no zeros in the zones defined by (\ref{79}).
\qed

{\bf Acknowlegment}: the research of the first and the third author were
supported by The Israel Science Foundation (grant No. 828/07), the second
author was partially supported by National Sciences and Engineering
Research Council of Canada.

redpajama_set_name: RedPajamaArXiv


If your beach holiday dream involves rickety thatch bungalows and grumpy, surly staff, Moon Resort is the ticket.
Though new brick bungalows are sprouting up all over Long Beach, Moon has eschewed solid walls for a rustic, earthy feel though the bungalows, which are dotted throughout the jungley property, have been upgraded with a solid concrete base. Backpackers can now look forward to comforts such as terracotta tiles underfoot and a fairly new tiled ensuite private bathroom. Rooms are equipped with a fan, mosquito net and bed with a proper mattress. The front porch, which comes with a few chairs and a table, is a pleasant enough spot to hang out.
We're doubtful these thatch huts would do well against the elements during rainy season, and the doors seem pretty flimsy so if you're worried about security, opt for the concrete cottages which are located closer to the beach – they come with the same amenities. They are sturdier, blander and some have air-con. Moon Resort also has a beach bar/restaurant and loungers with umbrellas for guests. For a backpacker on a budget, this is poor value for what you get but this remains one of the cheapest beachfront properties on Long Beach and it also happens to be in the liveliest section. The room rate certainly doesn't include any smiles from the sullen reception.


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The goal of sustainable agriculture is to meet society's food and textile needs in the present without compromising the ability of future generations to meet their own needs. Practitioners of sustainable agriculture seek to integrate three main objectives into their work: a healthy environment, economic profitability, and social and economic equity.
What should a ranch plan include?
Therefore, a successful ranch plan should include enterprise calendars of operations, stock flows, forage flows, labor needs, herd production records and land use plans to give the manager control and a means of monitoring progress toward goals. Animal Selection.
What are the different types of ruminants?
There is a wide range of breeds available in each of the major ruminant species, i.e., cattle, sheep and goats. Hardier breeds that, in general, have lower growth and milk production potential, are better adapted to less favorable environments with sparse or highly seasonal forage growth. Animal nutrition.
What is a systems approach?
A systems approach gives us the tools to explore the interconnections between farming and other aspects of our environment. Everyone plays a role in creating a sustainable food system. A systems approach also implies interdisciplinary efforts in research and education.
How do consumers contribute to the food system?
Through their purchases, they send strong messages to producers, retailers and others in the system about what they think is important. Food cost and nutritional quality have always influenced consumer choices.
What is the principal resource that has helped agriculture and society to prosper?
Water . Water is the principal resource that has helped agriculture and society to prosper, and it has been a major limiting factor when mismanaged. Water supply and use. In California, an extensive water storage and transfer system has been established which has allowed crop production to expand to very arid regions.
Additional Science Or Mathematics
One additional course (minimum 3 credits) from BIOL, CHEM, GSCI, MARN, or PHYS, or one extra MATH or STATcourse beyond those required for general education requirements.
See more on catalog.uconn.edu
36 Credit Group
Agriculture and Natural Resources majors must meet all the requirements listed under the 36 Credit requirements for all CAHNR majors, which must include at least 24 credits (combined total) from departments in CAHNR. These credits must include at least 3 credits of course work from each of four (4) distinct departments in CAHNR.
Writing Competency
Students must pass one 2000-level or above W course in any department of the College of Agriculture, Health and Natural Resources.
Career Statement
Agriculture and Natural Resources majors must submit a statement describing how courses relate to their desired career. This statement and courses for the major must be approved by advisor and College of Agriculture, Health and Natural Resources Associate Dean as early as possible in order to confirm the courses approved for the final Plan of Study. A minor in Agricult…

redpajama_set_name: RedPajamaCommonCrawl


Craftster Community > JEWELRY AND TRINKETS > Trinkets and Jewelry: Completed Projects: General (Moderators: I Sew Cute, Mistress Jennie) > ALice bands and bows- Image heavy!!
ALice bands and bows- Image heavy!!
And here it is on my head, which is where it is currently residing!
Hope you like them, Im thinking of setting up an Etsy shop and these are so quick and easy to make- and Ive now found a place to get cheap wholesale bands. Woo for wholesale!
The cure for boredom is curiosity. There is no cure for curiosity- Dorothy Parker.
Re: ALice bands and bows- Image heavy!!


Reflections of a Grady Doctor: And Justice for all.
Oh, just wanted to follow up. Here's a picture of Isaiah this morning with the replacement Jake Justice. Getting the replacement Jake wasn't so easy either. Oh, did I tell you? Turns out the dude has been discontinued. Discontinued! Well hallelujah for used and collectors items on Amazon. Lawdy Miss Claudie! First the Green Lantern situation and now Jake Justice gets axed from the Rescue Heroes?! Seriously? Seriously.
Don't worry. I didn't tell Isaiah about the diss-continuation. Besides, I'm told that there's been some sort of crazy ethnic remix on Spiderman and trust me--that'll be confusing enough.
Labels: black like me, i don't make this stuff up, my F.P.
Thanks for the follow up. This made my day! Give the boy a hug from me!
I think I mentioned this before when you posted it earlier, but we have piles of those guys up in our attic -- both my boys were nuts for them. I'm sure there's a Jake Justice as well -- so if you ever need a replacement, let me know.
I thought of you the other day when I was at a Target (Prado Sandy Springs) and saw a Jake Justice!
justice has been served. in one sense anyway. your little one's face is priceless!
The hands that prepared it.
Faces of Grady: Hepatitis, see?
Grady Verbatim: What have you done for me lately?


Take a leap of faith with this atmospheric and beautiful platformer.
Shoot & pop balls in the coolest puzzle game ever! It's super fun & addicting!
Cookie Rush is a yummy match 3 game full of new, fun elements & features!


Update [10/2/2016]: It appears that Google has finally put a stop to this trick.
Update [11/7/2016]: Looks like Jonathan in the comments has figured a new way to do it!
Update [2/7/2017]: The website http://gdurl.com works too!
Imagine you want to use a QR code so guests at Parents' Night can scan the code and stream video or audio straight to their mobile device. If you share the files in Google Drive, users will be taken to a Google page where they have to download the file first. But if you can create a link that points directly to the file, users can access the content right in their mobile browser.
Turns out you can do this while still using Google Drive. Any public folder in Drive can host files and provide direct links to the files.
This will provide a folder that will give direct links to files inside the folder.
Note: hosting view will not display files created in Google Docs.
Once in the hosting folder, you can right click on any of the files to get a direct link to the file.
Is this still working for you, Dale? I'm receiving a 404 error when I try to append the folder ID to the URL.
Answered my own question. The folder needs to be shared publicly on the web, otherwise the folder won't work.
Any idea why? Folder and files are set to be public.
Like Casu, this has stopped working for me.
The Drive UI still shows a "Hosting" URL – but it yields a 403 (permission deinied).
i share and .mkv in piblic the streaming not work and the download said 404 error!
You can also just plug the file ID itself into the hosting URL scheme (sharing must be set to either Anyone with link or Public on the web).
@Josh Yes. That's what I wrote above.
It only works for a certain time (calls) with "/host".
After that it redirects to a 404 page!
@alex, I still don't have trouble with this. It works every time.
OMG. Thank you. I finally got it to work!
My experience is that the direct link stopped working sometime after Google's official shutdown of hosting on September 1, 2016.
but now I get a 502 error.
Is the direct link definitely still working for anyone out there? If so, how are you doing it?
For me also stopped working, I am desperate because I really need this. Do you know any alternative?
I write integration scripts for a living so if you need professional help, do reach out as well.
I tried your technique in my Google Apps for Education account. No luck.
Yup. It looks like this trick has been stopped by Google.
but this only displays a (directory) list of the files. Getting them to display is another issue. Google drive knows how to display many types, but uses the Drive builtin viewer. Not quite the same as streaming the file to the mobile browser for display.
Not sure how long this will work, but it seems to do the trick.
Google is now redirecting those https://drive.google.com/uc?id= URIs. How annoying!
Also does anyone know a way to get the filename from the fileid?
Any suggestions on how to get a viewable link for this index.html file?
The URL opens a download dialog, instead of processing the HTML stream.
The https://docs.google.com example does the same: tries to download.
Google has put a stop to this. Dropbox stopped hosting HTML files too.


There are various factors that help build a better website design. However, there are certain factors that ensure that a website design is more than just a perfect design – One that makes people wonder and leave them awestruck. Such is their influence.
Web designers use mind-controlling techniques to create such influential designs that leave an impact on the visitors. Those techniques may sound like common factors but as you dig in deeper, you'll realize the art of exploiting the extra-ordinary out of the commonalities that exist.
Let's take a look at these web designing tricks that control the mind of the individuals, guiding them just the way they need to be guided.
Color combinations are often overlooked when designing a website. The sooner you realize their importance, the better. When planning a website, color combos play an important role in driving the traffic by playing with the psyche of the visitors. Colors are to be used to portray the content in a certain light but it does not, however, compromise the brand identity in any way. For e.g. blue is a color that is admired by all and is so used by banks and even social media to keep that admiration intact. Greens usually represent nature while yellow or orange have haste about them and can be used for limited time offers. If you have a luxury item in place, use black as it stands for sophistication.
If you know the art of designing a clever layout then you must be really good at guiding the users just where you want them to focus on. Where would you put the important text on the website? Considering the reading patterns of the users, they read in a Z: starting from top left, a glance in the middle and stop at bottom right. Don't forget to keep the visuals and content in complexities. If the user gets confused, he is sure to flee away.
For any information that you seek from the visitor, there needs to be clarity behind the purpose. A tip here is when asking for payments use a legitimate third-party source for protection of the transaction. In case something goes wrong, the contact information that you have provided should be working. It shows how much approachable you are thus making you very much real to the customer.
Since humans often connect to situations emotionally, leverage the obvious to connect with them with your content. If you tend to sway them to your viewpoint by showing the relevant images and concepts to justify, they are sure to end up believing you. If you connect with them emotionally with a positive intact, they are more inclined to taking action and share the content. Take the social media for example, say if you are having a great day and a brand posts about how thankful you should be to God for giving you this day, you are tempted to post it on your network. It really works.
Since customer-oriented businesses are the real game today, it has become essential for the experts to focus on what customers want rather than what they have to provide. Another approach is to offer the existing product in a way that it becomes a need for the customer. Same is the case with websites. This calls for exquisitely appealing design as well as everything else in places as customers want them to be.


text:
«Война пуговиц» () — кинофильм. Экранизация произведения, автор которого — Луи Перго.

Сюжет
Фильм рассказывает о мальчишках из маленькой ирландской деревушки, враждующих со сверстниками из соседней деревни. В ходе стычек они отрезают у противников пуговицы, так как именно это наиболее жестоко наказывается родителями обеих деревень. В центре картины мальчик, сын бедных родителей, которого наказывают с чрезмерной суровостью. Он сбегает из дома, чтобы избежать жестокого наказания, но в итоге его всё равно отправляют в приют для трудных подростков.

В ролях

Ссылки

Семейные фильмы Великобритании
Фильмы-драмы Великобритании
Фильмы-драмы Франции
Фильмы-драмы Японии
Приключенческие фильмы Великобритании
Приключенческие фильмы Франции
Семейные фильмы Франции
Семейные фильмы Японии
Экранизации романов Франции

redpajama_set_name: RedPajamaWikipedia


I am staying in the house of a woman whose only child died. How do you get over something like that? My guess is, you never do. Why are we so preoccupied with "getting over" things?
She has stunning artwork and sculptures ... Rocco is in love with the "T-Rex egg" in the backyard. Really hope he doesn't break it.
Love this. It's like two people are countries. At a kissing point, with their own laws and policies and customs.
Even the tree is amazing. It grew out of the earth until something big happened, which made it change course entirely. It went in a completely new direction but the knot in its trunk where it all changed and shifted is probably the strongest part of the whole tree.
I've looked at this picture the most. My heart breaks for a mother who could not save her daughter, as death looks on.
It's a beautiful house, I'm sure the owner is beautiful too. There's grief and strength in every room.
You have such a way with words Eden.
I consider myself lucky. I survived the death of my child and then got to mother two more.
I can't even begin to comprehend what all those mothering emotions and copious amounts of love would've been redirected to or in which other form they could have taken. I shudder to think.
Thanks Eden for saying what's on your mind and in your heart.
Too scary a concept for me to allow my mind to focus on.
I was with a best friend when she received the call telling her That her brother had shot himself dead.
It was a strange experience - her world had stopped but the world continued around us.
When I was on uni break I stayed with her and her parents on the coast. I hadn't met her parents before but I could see how much the grief had changed them. Heartbreaking. I slept in her brothers room, in his bed, everything was the way he left it. At night when everyone was asleep her dad would come into the room And play the piano. Her mother would never say his name - only referring to him as 'someone that used to live here'.
Thoughts are with your friend.


text:
store
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test to upload local repository

redpajama_set_name: RedPajamaGithub


According to Yahoo! Education, the accounting profession is the number one fastest growing field for recent college graduates. The demand for accountants has increased every year for last decade and there are no signs of this trend changing. With new tax laws, public company oversight regulations, and a more complex business environment, accountants will always be needed in increasing numbers. It's no wonder why more college students are changing career paths and graduating with accounting degrees.
Now is a great time to get into the field of accounting. Completing an accounting program and earning a degree in accounting can open so many job and careers paths. The profession isn't what it was 50 years ago. Today there are many different career paths available to accountants from tax, audit, and assurance work to advisory roles and industry management positions. There's really no limit to want an accountant can do in the modern world of business.
All of these career paths start with one thing: education. Having a great education is the foundation to a productive career. A great education can differentiate you from other candidates and set you apart from co-workers during a promotion process. There are many different educational options available today.
You can attend an online accounting program or a formal masters in accounting degree. It's all available to you, but what should you actually pursue and what school is best for you? Here's some information about the top 10 accounting schools and degree programs in the US and online.
Typically during your education process, you will stir towards some type of accounting certification. Each has its own level of educational requirements and work experience qualifications that must be met before even attempting the standardized exams.
Keep in mind that these are just professional certifications, licenses, and designations. This isn't a list of jobs or careers. A CPA can hold many different jobs and use his or her certification in many careers from public accounting to private industry to governmental accounting. It's important to keep in mind that the world of accounting and finance is huge.
Here's some more information about the most popular accounting certifications.
There are tons of job and career opportunities out there for accountants as long as you have the proper education. The accounting profession ranges from simple bookkeeping to complex tax and audit compliance work. There are four main sectors that offer a wide variety different jobs and careers. Here's a list of each main sector and the jobs commonly found in that sector.
As with any career, annual salary and benefits depend not only on job descriptions, but also geographic location and company size. For example, a bookkeeper for a small, private company will typically make much less than one for a large company located in a major city.
With greater salary comes the expectation of greater knowledge and ability. Smaller companies may require at least a two-year associate's degree for a bookkeeping position. Larger companies will often require a bachelor's degree or higher. Careers in public accounting and other high level positions like controllers usually require at least a bachelor's degree in accounting or finance as well as some type professional certification and license.
A career in public accounting is one of the most obvious starting points for any college student. Public accounting firms offer a wide variety of positions and a career paths as well as a great learning opportunity. Landing a job with one of the big 4 accounting firms can propel your career beyond public accounting and into corporate finance and executive management.
Here's some information about the big four accounting firms and other top firms that are great places to work for.


Home » Defence & Security » Equipment » Why do peacetime naval accidents keep happening?
This article, first published by Save The Royal Navy on 14th November, takes a look at a question that many people seem to be asking about naval accidents these days. How do these vessels, with relatively large and well-trained crews, and some of the best naviagtion and detection equipment available today, still manage to run aground or bump into other ships, often in the open sea and in calm weather? The answers are fascinating and once again show that where something seemingly minor can go wrong – it generally will!
2017 was an especially bad year for fatal naval accidents. Then last week the Norwegian frigate KNM Helge Ingstad collided with an oil tanker and is was run onto rocks in a vain attempt to prevent her from sinking. There have always been serious accidents involving warships and submarines during peacetime operations but with the advent of modern navigation technologies, there is some surprise that these incidents keep happening. Here we look briefly at the circumstances of some of the accidents and what might be learned from them.
The first warning that there were really systemic problems in the US Pacific Fleet came in May 2017. Cruiser USS Lake Champlain collided with a fishing vessel Nam Yang 502 in broad daylight in the Sea of Japan. The cruiser was not using AIS and the navigation radar display on the bridge was malfunctioning. Her bridge watch team was found to have been slow to react and made the wrong calls in attempting to avoid a collision.
The morning after – wounded USS Fitzgerald listing heavily limps into Yokosuka, Japan. She has been returned to the US for complex repairs totaling $398 Million.
At around 0426 on 8 November, the Norwegian AEGIS frigate collided with the 62,000-ton oil Tanker MV Sola in Hjeltefjorden. The oil tanker was undamaged but the frigate had a large hole torn down her starboard quarter. She suffered a loss of propulsion and was in imminent danger of sinking. Miraculously only 7 sailors suffered minor injuries, power supplies on board were maintained and the entire crew was safely evacuated. The KNM Helge Ingstat completed several weeks of Flag Officer Sea Training (FOST) with the Royal Navy in February 2018. This training is more focused on warfare than basic navigation skills but the quick response by the crew to a sudden crisis instilled by FOST may have helped saved lives.
Tugs from the nearby oil terminals were quickly on the scene and the Captian made the tough call to request they run her onto the nearest shoreline in an attempt to prevent her sinking. Unfortunately, the Norwegian Fjords mostly consist of steep rocky shores, far from ideal for grounding a ship. Tugs battled to keep the unstable ship upright against the shore but she gradually listed to starboard. Attempts were made to secure the ship to the shore with steel cables to prevent it sliding into the deeper water. this proved unsuccessful and by 12th November she had sunk deeper and was almost entirely submerged.
The cause of the accident is unknown and it will take some time before the results of the investigation are public. In Norway, there is consternation as to how such an accident could happen in this area. There is a 24-hour marine traffic centre with radar coverage of the area and the ship was in waters she knew well, heading for the Haakonsvern Naval Base. Unconfirmed media reports say the frigate was warned she was on a collision course on VHF radio by both the tanker and the traffic control centre which she acknowledged, just minutes before the collision. Other reports say the frigate was doing a very brisk 17 knots at the time of the collision and her navigator has somehow misread the ship's position, mentioning navigation marks some way from their actual location. Either way, something went very badly wrong on the bridge of the frigate with every indication of serious human error.
The RN has also suffered significant incidents in the 21st Century. HMS Nottingham hit rocks off Australia on July 7th, 2002. This was caused by accumulated navigational and situational awareness error by the bridge team. The subsequent damage control response and leadership were excellent and saved the ship. Having been recently refitted and upgraded, Nottingham was considered too valuable to write off. The ship was returned to the UK by barge, repairs took two years and cost £39m.
In November 2002, while conducting Perisher training, HMS Trafalgar, traveling at 50m depth at speed of 14 knots hit rocks off the Isle of Skye. Three sailors were injured and repair cost £5 Million. Charts in the control room had been obscured leading to the accident.
It is clear that even the best trained and equipped 'tier-1' navies have accidents. It is a statistical certainty there have been other accidents or very near misses that are unreported or covered up. This is especially likely to be the case for the navies of nations under control of repressive regimes, which may give the false impression Western navies are more accident prone. Both the US and Royal Navy offer a measure of transparency by publishing detailed Board of Enquiry reports after the events. Vessels belonging to top-tier navies also tend to spend a far greater proportion of their time at sea, increasing their chances of accidents. It would be surprising if the Chinese Navy which is expanding very rapidly has not had experienced accidents that have not been made public. The Soviet/Russian Navy also has a long and spectacular record of disasters both in port and at sea, many of which are now in the public domain.
There has been increasing speculation that cyber attacks on navigational equipment or GPS spoofing could be behind some of the more recent accidents. So far none of the officials enquires has found this to be the case, or even hinted at it. Warships were making navigational mistakes long before the technology for cyber attack existed. Although it is a theoretical possibility, to conduct such an attack on a naval vessel would be challenging as navies are increasingly vigilant about cyber defence and have multiple back up systems. The technology on board merchant ships probably presents more vulnerable targets.
Reactions to the crisis by the crews were usually good. In most cases, there was no panic and naval training paid off. Prompt damage control measures can often prevent the complete loss of the ship or further loss of life.
Naval vessels routinely have to conduct maneuvers that entail some risk. This may involve concealing one's presence by dousing lights, turning off AIS and restricting electronic emissions. There is an inherent tension between safety-first peacetime procedures and realistic training to maintain an operational mindset and posture.
Comments on Why do peacetime naval accidents keep happening?


Christine and I have spent the last two weekends getting the house in a livable condition.
Our first task was to paint the second bedroom. We only changed our minds on the color scheme once (okay, maybe twice)… After two (or is it three now?) weekends, the room is painted, and we've moved the furniture into it.
I finally got around to installing the water filter on the line running to the ice-maker on the refrigerator. But, after installing it, I realized that the brass pipe running to the fridge was a) connected to the hot-water (bad according to the filter paper-work) and b) was crimped, meaning the pressure through that line was too low. So, after two or three trips to Home Depot, I finally got all of the pieces that I needed to tap a new line into the cold-water, and run a new line to the fridge filter… Good news: we have filtered water and ice again.
As spring rolls around, I'm looking to get a ton of top-soil or so delivered, so I can spread it across the yard… The yard is generally in great shape, but there are a few bare spots in the back under the trees, and I figure a good coat of new top-soil will help the grass grow in better this summer.
The house is finally taking shape, and feels like a home (instead of a small old house filled with boxes). We still haven't found our silver-ware, and so, we're eating with plastic forks and knives, but they work, and they're pretty easy to clean up.
One last thing we'll have to do soon is buy some dainty 'grandmother' type furniture for the front living room… the house just seems to want an antique couch or chair in the room.


The Russian National Oil and Gas Forum and the 18th edition of the Neftegaz International Exhibition will run on 17-18 and 16-19 April 2018 at Expocentre Fairgrounds in Moscow.
Technological trends in the development of the oil and gas industry, digitalization of the world economy and global competition for markets and resources forms a new list of the energy-related issues demanding practical solutions from associated events of the Forum and the Exhibition.
In 2018 the associated events of the National Oil and Gas Forum and the Neftegaz Exhibition include technological conferences, foresight sessions, professional round tables, and workshops. They are devoted to a wide range of relevant issues such as digital transformation of the oil and gas sector; upgrade of the oil and gas machine building companies and its export potential; strategies of science and technology development of the industry and state support measures to introduce breakthrough technologies; localization of production in some oil and gas related sectors; technology transfer of high-tech products of the fuel and energy sector and international cooperation; robotization and artificial intelligence; creation of engineering centres, clusters and technology parks; development of oil and gas startups and innovative projects; exploitation of difficult to recover and unconventional hydrocarbons; inner drivers and export potential of the oil and gas industry.
The Neftegaz international exhibition is a high-profile world-class industry event whose participants are representatives of hundreds of leading oil and gas companies from dozens of countries. The exhibitors and visitors are nearly all key companies of the oil and gas sector, their major suppliers and contractors. Participation in the trade show provides small and new companies with a unique opportunity to draw attention to their developments and proposals of potential customers. The Neftegaz 2018 product sectors will be devoted to automation, petrochemistry, and latest developments of service providers and engineering and transport companies. There will be widely presented relevant industry trends such as oil and gas exploration; oil and gas well construction, development and operation; gathering, transportation and storage of oil, gas and oil products; oil and gas processing and petrochemical industry, equipment and technologies; locking and regulating fittings; power and electrotechnical equipment for oil and gas industry, cables, welding equipment; ACS and process telemechanization, instrumentation, means of measurement assurance; equipment and pipeline corrosion prevention; ecology, systems of environmental monitoring; labour protection and safety systems; innovations made in Russia.
Today, the Russian National Oil and Gas Forum and the Neftegaz exhibition are major events of the oil and gas industry annually featuring dozens of relevant events: professional conferences, round tables, technology seminars, focus exhibitions, different public and press events.
You can learn more about the Forum program from the Organizing Committee at +7 (495) 640-34-64, 8-800-333-05-15 or at mail@oilandgasforum.ru. The Forum official website is at www.oilandgasforum.ru.


Ingredients: * Whole Milk Powder, ** Raw Cane Sugar, ** Cocoa Butter, ** Chocolate Liquor, ** Unrefined Whole Cane Sugar, * Ground Hazelnuts, ** Ground Vanilla Beans.
*Fair Trade ingredient. By weight 70% Fair Trade content. **:Organic.
Made with organic hazlenuts. May contain traces of peanuts, almonds, cashews, pistachios and pecans. Soy- and gluten-free.
CO-OP INFO: CONACADO was founded in 1988 to link small-scale cacao farmers facing major market challenges. The co-op, now 9,500 members, focuses on sustainable cacao production.
Through our Fair Trade partnerships with co-ops, small-scale farmers gain crucial market access they wouldn't have otherwise. The co-ops are democratically run by the farmers, so that together they decide how to invest in their communities.
This biodegradable packaging is produced with renewable materials made from sustainably harvested wood.


OLDER TANKS STILL HAVE SERVICE LIFE REMAINING - CONTINUING IN DELIVERY SERVICE, SPARE TRUCK, CHANGED OVER TO OTHER SERVICE SUCH AS BULK OIL. WHATEVER THE PLAN, WE CAN HELP BRING THE TANK BACK TO "NEAR- NEW" CONDITION.
There is intense preparation and planning that goes into swinging a tank to a newer refurbished. GTTC representatives will meet to determine refurbished needs, measure the tank dimensions, evaluate the tank and piping systems, and then provide a detailed price proposal. Typically, the new refurbished is prepped with new PTO's, pumps and axel location finalized ahead of time.
Upon arrival the old tank is removed and work is commenced to give a quick turnaround to minimize downtime. Feel free to give us a call so we can discuss and possibly develop a long-term fuel upgrade plan, schedule units for swings or upgrade accordingly.
Tanks are typically blasted, cleaned, primed, sanded and painted - just like a new unit.
GTTC is an R-STAMP facility authorized to perform pressure vessel repairs.
All layout changes are reviewed with a Design Certified Engineer (DCE) to ensure compliance.
Existing decks can be removed, repainted or replaced.
We will work with your schedule in an attempt to minimize down time.


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Champions Challenge might refer to two different field hockey tournaments organized by the International Hockey Federation (FIH):

Hockey Champions Challenge I: introduced in 2001 as Hockey Champions Challenge, its name changed in 2009 to the current one.
Hockey Champions Challenge II: introduced in 2009.


How to do on-page SEO using Yoast WordPress Plug-in?
If you are using any existing SEO plugin like All In One SEO (https://wordpress.org/plugins/all-in-one-seo-pack/), , All In One SEO (Old), Headspace 2 (https://urbangiraffe.com/plugins/headspace2/), or any other such plugin, or if you are using any plugins by importing your settings into this plugin.
Go to Dashboard > Features > Advances setting pages & enable this feature.
If you are installing Yoast SEO on a WordPress blog, where you were using some other SEO plugin or theme SEO settings, then after activating advanced settings you should go to SEO > Tools> Import & Export > Import from other SEO plugins in the left panel. Select plugin from which you would like to import your settings.
The first step to getting started with this plugin is to submit your site to Google, Alexa, Bing, and Yandex. After installing plugin, click on SEO > Dashboard > Webmaster Tools to access this page.
Once you have done this, let's move to the more advanced parts of this plugin.
This is one of the most important parts of this plugin. Once you click on "Titles & Metas", click on "General" tab first to see what separator is being used and what that separator does. This will assist you to quickly set up title and metas SEO aspect of the plugin.
Title template: The home title that you want to show in Google search.
Meta description template: Your homepage Meta description for search engines.
This is where you will configure SEO for all of your post types.
By default, you will have Pages, Posts & Media types. If you are using more custom types in WordPress, you will see its section here. Here you can define SEO for the post, page, and media sections of your WordPress blog.
If you are running a news blog and want to show the date for your posts in search engines, you can check "Date in Snippet Preview" box.
You should prefer keeping tags and categories as noindex and dofollow. However, you should set this up depending upon your own SEO plan.
If you would like to use my settings, simply check the "Noindex", "Follow", and "Hide" boxes. Once again, you can configure the title template according to your needs. Since we are not indexing tags and categories, you can add more user and brand-centric title templates.
Here you can set indexing for your author and date-based archive. Also, you can disable or enable date and author archive. I suggest that you check "noindex" and "follow" boxes, and leave everything else as it is.
This is the last one under "Titles & Metas" settings. You can ignore "Use meta keywords tag?" Google already has said that they do not bother about keyword Meta tags.
This completes the "Metas & Titles" settings of WordPress SEO by Yoast. The final options will keep your WordPress blog head section clean by hiding information such as posted short links, WLW manifest links, and others.
In SEO by Yoast 3.0, they have added Google Knowledge Graph feature http://searchengineland.com/library/google/google-knowledge-graph). This will show your social media profile in search engines whenever someone searches for your brand name.
The configuration is easy. Simply add all of your social media profiles in the format shown in the image below.
You should add a default picture, which can help Facebook show an image when your posts have no image & are liked or shared on Facebook.
In the Yoast SEO plugin, you can configure advanced permalink settings. Usually 2-3 settings are used from here to avoid parameters being indexed in Google searches. With this plugin in particular, it is easier to get rid of the replytocom parameter (https://www.shoutmeloud.com/how-to-fix-replytocom-links-issue-in-wordpress.html), which is quite a headache.
This setting is under Advanced > Permalinks.
Breadcrumbs (https://yoast.com/wordpress/plugins/breadcrumbs/) are useful for getting search engines in order to understand about your site structure, and help users to navigate your site.
This is another useful feature of WordPress SEO by Yoast plugin. You can edit WordPress .htaccess and robots.txt file from the dashboard.
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