text:
\section{Introduction}
In this part of our study on the integrability and linearization of
nonlinear ordinary differential equations (ODEs) we focus our
attention on the theoretical formulation and applications of the
modified Prelle-Singer (PS) procedure (Prelle \& Singer 1983; Duarte
\textit{et al.} 2001;
Chandrasekar \textit{et al.} 2005$a$; 2006) to a set of two
coupled second
order ODEs. The need for this demonstration is due to the fact that
classifying and studying two degrees of freedom dynamical systems are
highly nontrivial problems in the theory of nonlinear dynamical
systems. Historically, several techniques have been proposed to
identify and obtain general solutions of two coupled second
order ODEs.  To cite a few we mention Painlev\'e analysis, Lie
symmetry analysis, generalized Noether symmetries technique, direct
method and so on (Ramani \textit{et al.} 1989; Lakshmanan \&
Sahadevan 1993; Bluman \& Anco 2002; Lakshmanan \& Rajasekar 2003). Each of these methods have their
own advantages and limitations.
For example, among the above, certain methods fulfil necessary
conditions alone whereas the others guarantee only sufficient
conditions for the complete integrability of the system concerned.
This factor alone is a motivating factor to search for
more and more powerful methods to isolate and classify integrable
and non-integrable dynamical systems. In this direction, by
extending the PS procedure and its applications to coupled second
order ODEs we argue that the PS method can be used as a stand-alone
technique to solve a wide class of ODEs of any order irrespective of
whether it is a single or coupled equation.

We mention here that the present analysis is not a straightforward
extension of the scalar case.  In fact by prolonging the theoretical
formulation to the coupled second order ODEs, we deduce the
determining equations for the integrating factors and null forms
appropriately such that one can obtain the aforementioned functions
in a more efficient and straightforward way.  Thus the method of
obtaining the integrating factors for the given equation is also
augmented in this procedure in an efficient manner. Further, while
studying the coupled dynamical systems one may face both isotropic
and non-isotropic cases.  Our method covers both of them in a
natural way. In addition to the above, in this paper, we also
introduce a new method to transform two coupled second
order ODEs to two uncoupled second order ODEs.
Thus, the PS procedure inherits several remarkable features both at the theoretical foundations as well as in the range of applications which we have listed out already in Chandrasekar \textit{et al.} (2005$a$). Finally, we note
that we have carefully fixed the examples so that the basic features
associated with this method and the results which it leads to could be
explained in an efficient way.

The plan of the paper is as follows. In \S2 we describe the PS
method applicable for coupled second-order ODEs and indicate the new
features in finding the integrating factors and integrals of motion.
In \S3 we establish a connection between integrating factors and the
form of equations. In \S4, the uncoupled equations are briefly considered.
In \S5,  we discuss elaborately the method of constructing integrals and
general solutions for the coupled nonlinear ODEs. We support the theory with two nontrivial examples which are being discussed in the contemporary literature.
We also briefly discuss the
application of our procedure for the case of Liouville integrable
systems in \S5d.  We devote \S6 to demonstrate yet another method to identify transformation variables from the first integrals which can be effectively used to rewrite the system of coupled ODEs into uncoupled ones so that one can integrate the resultant equation easily and obtain the general solution. We present our
conclusions in \S7.

\section{Prelle-Singer method for coupled second order ODEs}
\label{sec2}
\subsection{General Theory}
Let us consider a system of two coupled second order ODEs of the form
\begin{eqnarray}
\ddot{x}=\frac{d^2x}{dt^2}=\frac{P_1}{Q_1}, \quad
\ddot{y}={\frac{d^2y}{dt^2}}={\frac{P_2}{Q_2}}, \quad { P_i,Q_i}\in
C{[t,x,y,\dot{x},\dot{y}]},\; i=1,2,\label {cso01}
\end{eqnarray}
where $P_i$ and $Q_i$ are analytic functions of the variables $t,x,y,\dot{x}$
and $\dot{y}$. Let us suppose that the system (\ref{cso01}) admits a
first integral of the form $I(t,x,y,\dot{x},\dot{y})=C$
 with $C$ constant on the solutions so that the total differential gives
\begin{eqnarray}
dI={I_t}{dt}+{I_{x}}{dx}+{I_{y}}{dy}+{I_{\dot{x}}}{d\dot{x}}
+{I_{\dot{y}}}{d\dot{y}}=0, \label {cso02}
\end{eqnarray}
where subscript denotes partial differentiation with respect to that variable.
Rewriting ({\ref{cso01}) in the form
\begin{eqnarray}
\frac{P_1}{Q_1}dt-d\dot{x}=0,\qquad
\frac{P_2}{Q_2}dt-d\dot{y}=0 \label {cso03}
\end{eqnarray}
and adding null terms
$s_1(t,x,y,\dot{x},\dot{y})\dot{x}dt
- s_1(t,x,y,\dot{x},\dot{y})dx$ and $s_2(t,x,y,\dot{x},\dot{y})\dot{y}dt
- s_2(t,x,y,\dot{x},\dot{y})dy$
 with the first equation in (\ref{cso03}), and
$u_1(t,x,y,\dot{x},\dot{y})\dot{x}dt  -
u_1(t,x,y,\dot{x},$ $\dot{y})$ $dx $ and $u_2(t,x,y,\dot{x},\dot{y})\dot{y}dt -
u_2(t,x,y,\dot{x},\dot{y})dy $ with the second equation in (\ref{cso03}),
respectively, we obtain that, on the solutions, the 1-forms
\begin{subequations}
\begin{eqnarray}
&&(\frac{P_1}{Q_1}+s_1\dot{x}+s_2\dot{y})dt-s_1dx-s_2dy-d\dot{x}=0,\label {cso04}\\
&&(\frac{P_2}{Q_2}+u_1\dot{x}+u_2\dot{y})dt-u_1dx-u_2dy-d\dot{y}=0.\label {cso05}
\end{eqnarray}
\label {cso06}
\end{subequations}

At this stage, we wish to point out that one can also analyse the coupled second order ODEs (\ref{cso01}) by rewriting them as a set of four coupled first order ODEs of the form $\dot{x}=x_1,\;\dot{x}_1=P_1/Q_1,\;\dot{y}=y_1$, $\dot{y}_1=P_2/Q_2$ and its equivalent one forms. By introducing four integrating factors, one can deduce the relevant determining equations by following the procedure given by us in the earlier paper, Part-III (Chandrasekar \textit{et al.} 2008), to the above system of first order ODEs. However, after examining several examples we find that it is more advantageous to solve the system (\ref{cso01}) in the second order form itself rather then introducing more number of variables. The procedure is as follows.

Now, on the solutions, the 1-forms (\ref{cso02}) and
(\ref{cso06}) must be proportional. Multiplying (\ref{cso04}) by the factor
$ R(t,x,y,\dot{x},\dot{y})$ and (\ref{cso05}) by
the factor $ K(t,x,y,\dot{x},\dot{y})$, which act as the integrating
factors for (\ref{cso04}) and (\ref{cso05}), respectively, we have on the
solutions that
\begin{eqnarray}
dI=R(\phi_1+S\dot{x})dt+K(\phi_2+U\dot{y})dt-RSdx
-KUdy-Rd\dot{x}-Kd\dot{y}=0,\;\;\label {cso07}
\end{eqnarray}
where $ \phi_i\equiv {P_i}/{Q_i},\; i=1,2,$ and
$S=\frac{Rs_1+Ku_1}{R}$ and $U=\frac{Rs_2+Ku_2}{K}$. Comparing
equations (\ref{cso07}) and (\ref{cso02}) we have, on the solutions,
the relations
\begin{eqnarray}
 I_t  =R(\phi_1+S\dot{x})+K(\phi_2+U\dot{y}),\; I_{x}  = -RS,
  I_{y} =-KU,\;
 I_{\dot{x}}  =-R,\;
  I_{\dot{y}} =-K.
 \label {cso08}
\end{eqnarray}
The compatibility conditions between the equations (\ref{cso08}), namely $I_{tx}=I_{xt},\;I_{t\dot{x}}=I_{\dot{x}t},\;I_{ty}=I_{yt},\;I_{t\dot{y}}=I_{\dot{y}t},\;I_{xy}=I_{yx},\;I_{x\dot{x}}=I_{\dot{x}x},\;I_{y\dot{y}}=I_{\dot{y}y},\;I_{x\dot{y}}=I_{\dot{y}x},\;I_{y\dot{x}}=I_{\dot{x}y}$ and $I_{\dot{x}\dot{y}}=I_{\dot{y}\dot{x}}$, provide us
the conditions,
\begin{eqnarray}
D[S] &=&-\phi_{1x}-\frac{K}{R} \phi_{2x}
    +\frac{K}{R}S\phi_{2\dot{x}}
           +S\phi_{1\dot{x}}+S^2,  \label {eq23}\\
D[U] &=&-\phi_{2y}-\frac{R}{K}\phi_{1y}
    +\frac{R}{K}U\phi_{1\dot{y}}
           +U\phi_{2\dot{y}}+U^2,  \label {eq24}\\
D[R]  &=&-(R\phi_{1\dot{x}}+K\phi_{2\dot{x}}+R S),  \label {eq25}\\
D[K]  &=&-(K\phi_{2\dot{y}}+R\phi_{1\dot{y}}+K U), \label {eq26}\\
SR_{y} &=&-RS_{y}+UK_{x}+KU_{x},\;\;R_{x} =SR_{\dot{x}}+RS_{\dot{x}},
  \label {eq31}\\
R_{y} &=&UK_{\dot{x}}+KU_{\dot{x}},\qquad\qquad \;
K_{x} =SR_{\dot{y}}+RS_{\dot{y}},  \label {eq29}\\
K_{y}&=&UK_{\dot{y}}+KU_{\dot{y}},\qquad\qquad\;
R_{\dot{y}} =K_{\dot{x}}.\label {eq32}
\end{eqnarray}
Here the total differential operator, $D$, is defined by
$
D =\frac{\partial}{\partial{t}}+\dot{x}\frac{\partial}{\partial{x}}
+\dot{y}\frac{\partial}{\partial{y}}
+\phi_1\frac{\partial}{\partial{\dot{x}}}+\phi_2\frac{\partial}
{\partial{\dot{y}}}$. Integrating equations~(\ref{cso08}), we obtain the
integral of motion,
\begin{eqnarray}
I=r_1+r_2+r_3+r_4
-\int\bigg[K+\frac{d}{d\dot{y}}\bigg(r_1+r_2+r_3+r_4\bigg)\bigg]
d\dot{y},
\label {cso09}
\end{eqnarray}
where
\begin{eqnarray}
r_1&=\int\bigg(R(\phi_1+S\dot{x})+K(\phi_2+U\dot{y})\bigg)dt,
\qquad
r_2=-\int\bigg(RS+\frac{d}{dx}(r_1)\bigg)dx,&\qquad\nonumber\\
r_3&=-\int\bigg(KU+\frac{d}{dy}(r_1+r_2)\bigg)dy,\qquad
r_4=-\int\bigg[R+\frac{d}{d\dot{x}}\bigg(r_1+r_2+r_3\bigg)\bigg]d\dot{x}.&
\nonumber
\end{eqnarray}
Solving the determining equations,
(\ref{eq23})-(\ref{eq32}), consistently we can obtain expressions for the
functions $(S,U,R,K)$.
Substituting them into (\ref{cso09}) and evaluating the integrals we can constract
the associated integrals of motion.
It is also clear that equation~(\ref{cso01}) can be
considered as a completely integrable system once we obtain
four independent integrals of motion through this procedure.

\section{Connection between integrating factors and nature of equations}
We note that equations (\ref{eq23})-(\ref{eq32}) constitute an
overdetermined system for the four unknowns, namely $S,U,R$ and $K$.
Among these equations, the first four equations,
(\ref{eq23})-(\ref{eq26}), constitute the evolution equations for
the variables $S,U,R$ and $K$. We mention here that combining
equations (\ref{eq23})-(\ref{eq26}) one can get the following two
identities, namely
\begin{eqnarray}
D{[R S]} =-(R\phi_{1x}+K\phi_{2x}),\quad D{[KU]} =-(R\phi_{1y}+K\phi_{2y}).
\label {eq33}
\end{eqnarray}
Any solution $(S,U,R,K)$ which satisfies the equations
(\ref{eq23})-(\ref{eq26}) also satisfies the equation (\ref{eq33}).
Once the functions $S,\;U,\;R$ and $K$ are fixed then the rest of
the problem is to verify whether these functions satisfy the `extra
determining equations', that is (\ref{eq31})-(\ref{eq32}), or not.
If these functions satisfy the extra determining equations then they
form a compatible set of solution and one can proceed to construct the
associated integral of motion from (\ref{cso09}). On the other hand
if the functions do
not satisfy the extra determining equations  then one has to look
for alternate ways to obtain compatible solutions. In fact, in
practice, one often meets the case in which  certain solution(s)
which satisfies(satisfy) the evolutionary determining equations
(\ref{eq23})-(\ref{eq26}) does(do) not satisfy the extra determining
equations.   More specifically, for a class of problems one often
gets one or two or even three  sets of $(S,U,R,K)$ by solving
(\ref{eq23})-(\ref{eq26}) which does(do) satisfy the rest and
another(other) set(s) does(do) not satisfy the later equations. In
this situation we find an interesting fact that one can use the
integral(s) derived from the  set(s) which satisfies(satisfy) all
the equations (\ref{eq23})-(\ref{eq32})  and deduce the other
compatible solution(s) $(S,U,\hat{R},\hat{K})$  (definition of
$\hat{R}$ and $\hat{K}$ follows). For example, let the set
$(S_4,U_4,R_4,K_4)$ be a  solution of the evolution
equations~(\ref{eq23})-(\ref{eq26}) and not of the  extra
determining equations~(\ref{eq31})-(\ref{eq32}). After analyzing
several examples we find that one can make the set
$(S_4,U_4,R_4,K_4)$  compatible by modifying the form of $R_4$ and
$K_4$ as
\begin{eqnarray}
\hat{R}  = F(t,x,y,\dot{x},\dot{y})R,\qquad \hat{K}  =
G(t,x,y,\dot{x},\dot{y})K, \label{tc03}
\end{eqnarray}
where $F$ and $G$ are functions to be determined. However, after
making these modifications, $S_4$ and $U_4$ remain the same. A
motivation to do this type of modification came from our earlier
work on the applicability of the PS procedure to scalar second order
ODEs (Chandrasekar \textit{et al.} 2005$a$) . However, unlike the
scalar case, now we have to choose two functions, namely $F$ and
$G$, appropriately, such that the compatible forms of $\hat{R}$ and
$\hat{K}$ can be fixed. Since we have to choose two functions one
might ask the question whether these two  functions are necessarily
different or the same. In the following we answer this question.

Since $\hat{R}$ and $\hat{K}$ should
satisfy equations~(\ref{eq25}) and (\ref{eq26}), we have
\begin{align}
D[F]R+FD[R]=-(FR\phi_{1\dot{x}}+GK\phi_{2\dot{x}}+FRS),
\label {tc04}\\
D[G]K+GD[K]=-(FR\phi_{1\dot{y}}+GK\phi_{2\dot{y}}+GKU).
\label {tc05}
\end{align}
Substituting the expressions for $D[R]$ and $D[K]$ (vide (\ref{eq25}) and
(\ref{eq26})) in (\ref{tc04}) and (\ref{tc05}) and
simplifying the resultant equations we find that
\begin{align}
D[F]R=(F-G)K\phi_{2\dot{x}}, \quad D[G]K=(G-F)R\phi_{1\dot{y}}.
\label {tc06}
\end{align}
On the other hand substituting the modified forms (\ref{tc03}) in the
integrability conditions (\ref{eq33}), we get
\begin{align}
D[F]RS=(F-G)K\phi_{2x}, \quad D[G]KU=(G-F)R\phi_{1y}.
\label {tc07}
\end{align}
Combining equations (\ref{tc06}) and (\ref{tc07}) we obtain
\begin{align}
(F-G)K(S\phi_{2\dot{x}}-\phi_{2x})=0, \quad
(G-F)R(U\phi_{1\dot{y}}-\phi_{1y})=0.
\label {tc07a}
\end{align}
From equation (\ref{tc07a}) we can conclude that either
$F\neq G$ or $F=G$. In the first case, we have uncoupled equations, that is,
\begin{align}
F\neq G \Rightarrow
\phi_{2\dot{x}}=\phi_{1\dot{y}}=\phi_{2x}=\phi_{1y}=0\quad
(\hat{R}=FR,\;\hat{K}=GK)
\label {tc07aa}
\end{align}
 and in the second case we have coupled equations, that is,
\begin{align}
F= G \Rightarrow
\phi_{2\dot{x}},\phi_{1\dot{y}},\phi_{2x},
\phi_{1y}\neq0\quad
(\hat{R}=FR,\;\hat{K}=FK).
\label {tc07ab}
\end{align}
Note that in the case of $F\neq G$, the other possible solution
$S=\phi_{2x}/\phi_{2\dot{x}}$, $U=\phi_{1y}/\phi_{1\dot{y}}$, when
used in (\ref{eq23}) and (\ref{eq24}) lead to inconsistencies and so
this choice is not considered. Further, the case $R=0$ and $K=0$ leads to
the trivial solution. The above analysis clearly shows that
for the uncoupled equations one should choose $F$ and $G$ as
different and for the coupled equations one should choose them as
the same.

Finally, we mention the important point that the functions $F$ and $G$
are nothing but functions of integrals of motion. To show this,
substituting back the relations (\ref{tc07aa}) and (\ref{tc07ab}) in
(\ref{tc06}) and (\ref{tc07}), respectively, we get
\begin{align}
D[F]R=0,\; D[G]K=0 \; \mbox{and}\; D[F]RS=0,\; D[G]KU=0,\;
\Rightarrow D[F]=0= D[G].\label {tc06b}
\end{align}
Thus, irrespective of whether the given equations are coupled ones
or uncoupled ones the functions $F$ and $G$ can always be taken as
integrals of motion or functions of them and the problem now is how
to determine them. In the following, we discuss this in detail.

\section{Case 1: $F\neq G$ (Uncoupled Equations) }
\subsection{Theory}
In this case, we have the following form of equation of motion,
namely
\begin{eqnarray}
\ddot{x}=\phi_1(t,x,\dot{x}), \quad
\ddot{y}=\phi_2(t,y,\dot{y}).\label {tc08}
\end{eqnarray}
As a consequence the evolution equations for the functions $S,U,R$ and $K$ (vide equations (\ref{eq23})-(\ref{eq26})) are simplified to the forms
\begin{eqnarray}
D[S] &=&-\phi_{1x}+S\phi_{1\dot{x}}+S^2,  \label {eq23a}\\
D[U] &=&-\phi_{2y}+U\phi_{2\dot{y}}+U^2,  \label {eq24a}\\
D[R]  &=&-R S-R\phi_{1\dot{x}},  \label {eq25a}\\
D[K]  &=&-K U-K\phi_{2\dot{y}}. \label {eq26a}
\end{eqnarray}
Solving equations (\ref{eq23a}) and (\ref{eq24a}) one can get explicit
forms of $S$ and $U$. Substituting the known forms of $S$ and $U$ into the
equations (\ref{eq25a}) and (\ref{eq26a}) and integrating the resultant
equations one can fix the forms of $R$ and $K$. From the known forms of
$(S,U,R,K)$ one can fix the integrals of motion from (\ref{cso09}).
One can also consider equation (\ref{tc08}) as two independent
second order ODEs and solve the equations independently
by adopting the procedure given in our previous paper
(Chandrasekar \textit{et al.} 2005$a$). To avoid repetition we do
not discuss the uncoupled case further here.

\section{Case 2: $F=G$ (Coupled equations)}
\subsection{Theory}
Now we focus on the case $F=G$. To obtain the forms $(S,U,R,K)$ one
needs to solve the compatibility conditions (\ref{eq23}) -
(\ref{eq26}). For this purpose, we rewrite
equations~(\ref{eq23})-(\ref{eq26}) in terms of two variables alone, namely $R$
and $K$, by eliminating $S$ and $U$, and analyse the resulting
coupled second order partial differential equations (PDEs) and obtain expressions for $R$ and $K$.
From the latter we deduce the forms of $S$ and $U$ through the
relations (\ref{eq25}) and (\ref{eq26}).

To rewrite equations~(\ref{eq23})-(\ref{eq26}) interms of $R$ and $K$ let us
take a total derivative of equations (\ref{eq25}) and (\ref{eq26}). Doing so we get
\begin{align}
D^2[R]=-D[R\phi_{1\dot{x}}+K\phi_{2\dot{x}}+RS],\quad
D^2[K]=-D[K\phi_{2\dot{y}}+R\phi_{1\dot{y}}+KU].
\label {3.1}
\end{align}
Using the identities (\ref{eq33}), equation (\ref{3.1}) can be
rewritten in a coupled form for $R$ and $K$ as
\begin{align}
D^2[R]+D[R\phi_{1\dot{x}}+K\phi_{2\dot{x}}]=R\phi_{1x}+K\phi_{2x},
\label {tc01}\\
D^2[K]+D[K\phi_{2\dot{y}}+R\phi_{1\dot{y}}]=R\phi_{1y}+K\phi_{2y}.
\label {tc02}
\end{align}
One can note that the above determining equations (\ref{tc01}) and
(\ref{tc02}) form a system of linear PDEs. To solve
equations~(\ref{tc01}) and (\ref{tc02}) one may assume a specific ansatz either polynomial or rational in $\dot{x}$ and $\dot{y}$ for $R$ and $K$ and substituting
the known expressions of $\phi_1$ and $\phi_2$ and their derivatives
into (\ref{tc01}) and
(\ref{tc02}) and solving them one can obtain expressions for the
integrating factors $R$ and $K$. Once $R$ and $K$ are known then the
functions $(S,U)$ can be fixed through the relation
(\ref{eq25})-(\ref{eq26}). Knowing $S,U,R$ and $K$, one has to make
sure that this set $(S,U,R,K)$ does also satisfy the remaining
compatibility conditions (\ref{eq31}) - (\ref{eq32}). The set
$(S,U,R,K)$ which satisfies all the equations (\ref{eq23}) -
(\ref{eq32}) is then the acceptable solution and one can then
determine the associated integral $I$ using the relation
(\ref{cso09}). For complete integrability we require four
independent compatible sets $(S_i,U_i,R_i,K_i),\;i=1,2,3,4$.

As discussed in \S3, suppose the sets
$(S_i,U_i,R_i,K_i),\;i=1,2,3,$ are found to satisfy the
equations~(\ref{eq23}) - (\ref{eq32}) and the fourth set $(S_4,U_4,$
$R_4,K_4)$ does not satisfy equations (\ref{eq31}) - (\ref{eq32}).
In this case to identify the correct form of $\hat{R_4}$ and
$\hat{K_4}$ one may assume that $\hat{R_4} = F(I_1,I_2,I_3)R_4$ and
$\hat{K_4} = F(I_1,I_2,I_3)K_4$, where $F(I_1,I_2,I_3)$ is a
function of the integrals $I_1,\;I_2$ and $I_3$. To determine the
explicit form of $F(I_1,I_2,I_3)$ we proceed as follows.
Substituting
\begin{eqnarray}
\hat{R_4}  = F(I_1,I_2,I_3)R_4 ,\quad \hat{K_4}  = F(I_1,I_2,I_3)K_4
\label{ext06}
\end{eqnarray}
into equations~(\ref{eq31})-(\ref{eq32}), we get the following
set of six relations:
\begin{align}
\frac{(a_1F_1'+b_1F_2'+c_1F_3')}{d_1}=F,\quad
\frac{(a_2F_1'+b_2F_2'+c_2F_3')}{d_2}=F,\nonumber\\
\frac{(a_3F_1'+b_{3}F_2'+c_{3}F_3')}{d_{3}}=F,\quad
\frac{(a_4F_1'+b_{4}F_2'+c_{4}F_3')}{d_{4}}=F,\nonumber\\
\frac{(a_5F_1'+b_{5}F_2'+c_{5}F_3')}{d_{5}}=F,\quad
\frac{(a_6F_1'+b_{6}F_2'+c_{6}F_3')}{d_{6}}=F,
\label{meth02}
\end{align}
where
\begin{eqnarray}
&&a_1=R(I_{1x}-SI_{1\dot{x}}),\quad
b_1=R(I_{2x}-SI_{2\dot{x}}),\quad
c_1=R(I_{3x}-SI_{3\dot{x}}),
\nonumber\\
&&d_1=(SR_{\dot{x}}+RS_{\dot{x}}-R_{x}),\quad
a_2=K(I_{1y}-UI_{1\dot{y}}),\quad
b_2=K(I_{2y}-UI_{2\dot{y}}),
\nonumber\\
&&c_2=K(I_{3y}-UI_{3\dot{y}}),\quad
d_2=(UK_{\dot{y}}+KU_{\dot{y}}-K_{y}),\quad
a_3=(RI_{1y}-UKI_{1\dot{x}}),
\nonumber\\
&&b_{3}=(RI_{2y}-UKI_{2\dot{x}}),\quad
c_{3}=(RI_{3y}-UKI_{3\dot{x}}),\quad
d_{3}=(UK_{\dot{x}}+KU_{\dot{x}}-R_{y}),
\nonumber\\
&&a_{4}=(KI_{1x}-SRI_{1\dot{y}}),\quad
b_{4}=(KI_{2x}-SRI_{2\dot{y}}),\quad
c_{4}=(KI_{3x}-SRI_{3\dot{y}}),
\nonumber\\
&&d_{4}=(SR_{\dot{y}}+RS_{\dot{y}}-K_{x}),\quad
a_{5}=(RI_{1\dot{y}}-KI_{1\dot{x}}),\quad
b_{5}=(RI_{2\dot{y}}-KI_{2\dot{x}}),
\nonumber\\
&&c_{5}=(RI_{3\dot{y}}-KI_{3\dot{x}}),\quad
d_{5}=(K_{\dot{x}}-R_{\dot{y}}),\quad
a_{6}=(SRI_{1y}-UKI_{1x}),
\nonumber\\
&&b_{6}=(SRI_{2y}-UKI_{2x}),\quad
c_{6}=(SRI_{3y}-UKI_{3x}),
\nonumber\\
&&d_{6}=(UK_{x}+KU_{x}-SR_{y}-RS_{y})\nonumber
\label{meth03}
\end{eqnarray}
are all known functions of $t,x,y,\dot{x}$ and $\dot{y}$ and
$F_{i}'=\partial F/\partial I_i$.

Equation~(\ref{meth02}) represents an over determined  system of
equations for the unknown $F$. A simple way to solve this equation
is to uncouple it for $F_i'$, ($=\frac{\partial F}{\partial I_i}$)
$i=1,2,3,$ and solve the resultant equations. For example,
eliminating $F_2'$ and $F_3'$ from the first three relations in
equation~(\ref{meth02}) we obtain an equation for $F_1'$ in the form
\begin{align}
\frac{F_1'}{F}=\frac{(d_1c_2-c_1d_2)(b_1c_{3}-b_{3}c_1)
-(d_1c_3-c_1d_3)(b_1c_{2}-b_{2}c_1)}{(a_1c_2-c_1a_2)(b_1c_{3}-b_{3}c_1)
-(a_1c_3-c_1a_3)(b_1c_{2}-b_{2}c_1)}.
\label{meth04}
\end{align}
On the other hand eliminating $F_1'$ and $F_3'$ from equation~(\ref{meth02})
(again from the first three relations)
we arrive at the following equations for $F_2'$ in the form
\begin{align}
\frac{F_2'}{F}=\frac{(d_1c_2-c_1d_2)(a_1c_{3}-a_{3}c_1)
-(d_1c_3-c_1d_3)(a_1c_{2}-a_{2}c_1)}{(b_1c_2-c_1b_2)(a_1c_{3}-a_{3}c_1)
-(b_1c_3-c_1b_3)(a_1c_{2}-a_{2}c_1)}.
\label{meth05}
\end{align}
In the similar way one obtains the following equation for $F_3'$
\begin{align}
\frac{F_3'}{F}=\frac{(d_1b_2-b_1d_2)(a_1b_{3}-a_{3}b_1)
-(d_1b_3-b_1d_3)(a_1b_{2}-a_{2}b_1)}{(c_1b_2-b_1c_2)(a_1b_{3}-a_{3}b_1)
-(c_1b_3-b_1c_3)(a_1b_{2}-a_{2}b_1)}.
\label{meth06}
\end{align}

One can easily check that the combination of other relations in equation
(\ref{meth02}) along with the forms (\ref{meth04})-(\ref{meth06}) gives rise to
relations which are effectively nothing but the constraint
equations~(\ref{eq31})-(\ref{eq32}) and so no new constraint actually arises.
Consequently, equations (\ref{meth04})-(\ref{meth06}) can be written as
\begin{align}
\frac{\partial F}{\partial I_1}=g_1(I_1,I_2,I_3)F,\quad
\frac{\partial F}{\partial I_2}=g_2(I_1,I_2,I_3)F \quad  \mbox{and}\quad
\frac{\partial F}{\partial I_3}=g_3(I_1,I_2,I_3)F,
\label{meth07}
\end{align}
where $g_i$'s, $i=1,2,3,$ are functions of $I_1$, $I_2$ and $I_3$.
Now solving equations (\ref{meth07}) one can obtain the
explicit form of $F(I_1,I_2,I_3)$.  Once $F$ is known we can obtain the
complete solution $\hat{R_4}$ and $\hat{K_4}$ from which, along with $S_4$ and
$U_4$, the fourth integral $I_4$ can be constructed using the expression
(\ref{cso09}).

Finally, we note that in some cases one can meet the situation that
the sets $(S_i,U_i,R_i,K_i),\;i=1,2,$ alone are found to satisfy the
equations~(\ref{eq23}) - (\ref{eq32}) and the third set
$(S_3,U_3,R_3,K_3)$ (as well as the fourth set) does not satisfy
equations (\ref{eq31}) - (\ref{eq32}). In this case $F$ may be a
function of the integrals $I_1$ and $I_2$ which can be derived from
the sets $(S_i,U_i,R_i,K_i),\;i=1,2$. We need to find the explicit
form of $F(I_1,I_2)$ in order to obtain the compatible solution
$(S_3,U_3,\hat{R_3},\hat{K_3})$. To recover the complete form of
$\hat{R_3}$ and $\hat{K_3}$ one may assume that $\hat{R_3} =
F(I_1,I_2)R_3$ and $\hat{K_3} = F(I_1,I_2)K_3$ and proceed as before
and obtain the determining equations for $F$. Since $F$ is a function
of $I_1$ and $I_2$ alone one essentially gets the same form (\ref{meth02})
but without the factor $F_3'$ ($=\frac{\partial F}{\partial I_3}$).
Since $c_i$'s $i=1,...6$, are also exclusive functions of $I_3$ and
their derivatives they do not appear in the determining equations.
Solving the
resultant determining equations one can fix the form of $F$ which in turn
provides us $\hat{R}_3$ and $\hat{K}_3$ from which one can
constract the third integral $I_3$ for the given problem.
Now from the known
forms $I_1,I_2$ and $I_3$, one can proceed as before to obtain the
fourth compatible set $(S_4,U_4,\hat{R}_4,\hat{K}_4)$ and thereby
obtain the fourth integral $I_4$ also.

Similarly, if the set $(S_1,U_1,R_1,K_1)$ alone is found to satisfy
the equations~(\ref{eq23}) - (\ref{eq32}) and the second set
$(S_2,U_2,R_2,K_2)$ does not satisfy equations (\ref{eq31}) -
(\ref{eq32}) then, in this case the determining equations for $F$
takes the form (\ref{meth02}) with $F_2=F_3=b_i=c_i=0,\;i=1,...6$.
Following the procedure mentioned above one can first derive
the second integral. From a knowledge of $(I_1,I_2)$, one can
again follow the above procedures and construct $I_3$ and $I_4$.
In the following we illustrate the theory with specific examples.

\subsection{Example 2: Coupled modified Emden equation}
The modified Emden equation (MEE) and its variants arise in different branches of
physics (Erwin \textit{et al.} 1984; Dixon \& Tuszynski 1990) and
considerable attention has  been paid recently to explore the mathematical
and geometrical properties of MEE and its variant equations (see for example,
Mahomed \& Leach 1985; Steeb 1993; Chandrasekar
\textit{et al.} 2005$b$; Euler \textit{et al.} 2007). The equation reads
\begin{eqnarray}
\ddot{x}+3kx\dot{x}+kx^3+\lambda x=0,
\label {MEE01}
\end{eqnarray}
where $k$ and $\lambda$ are arbitrary parameters,
which can be completely integrated in terms of elementary functions
(Chandrasekar \textit{et al.} 2005$b$).
It exhibits some unusual features such as amplitude independent frequency,
nonstandard Lagrangian and time independent Hamiltonian forms and
transition from periodic to front-like solutions as the sign of the
parameter changes (Chandrasekar \textit{et al.} 2005$b$).

We now consider a coupled two dimensional version of (\ref{MEE01})
and demonstrate how it can be integrated out by following the theory
given in the previous section. The coupled system of equations reads
\begin{eqnarray}
&&\ddot{x}+2(k_1x+k_2y)\dot{x}+(k_1\dot{x}+k_2\dot{y})x+(k_1x+k_2y)^2x
+\lambda_1 x=0,\nonumber\\
&&\ddot{y}+2(k_1x+k_2y)\dot{y}+(k_1\dot{x}+k_2\dot{y})y+(k_1x+k_2y)^2y
+\lambda_2 y=0,
\label {case32}
\end{eqnarray}
where $k_i$ and $\lambda_i$, $i=1,2$, are arbitrary parameters.

To determine the integrating factors we seek a rational form of ansatz for
$R$ and $K$ in the form (suggested by the equation (\ref{MEE01}))
\begin{align}
R=\frac{a_1+a_2\dot{x}+a_3\dot{y}}
{(a_4+a_5\dot{x}+a_6\dot{y})^q},\qquad K=\frac{b_1+b_2\dot{x}+b_3\dot{y}}
{(a_4+a_5\dot{x}+a_6\dot{y})^r}
 \label{case32a},
\end{align}
where $q$ and $r$ are arbitrary numbers (to be fixed) and $a_i$'s, $i=1,2...6$
and $b_i$'s, $i=1,2,3,$ are arbitrary functions of $t,\;x$ and $y$. Substituting
(\ref{case32a}) into (\ref{tc01}) and (\ref{tc02}) and equating the
coefficients of different powers of $\dot{x}$ and $\dot{y}$ to zero we get a
set of linear PDEs for the variables $a_i$'s,
$i=1,2...6,$ and $b_j$'s, $j=1,2,3$. Solving the resultant equations we find the
following sets of particular solutions for $R$ and $K$:
\begin{eqnarray}
&&R_1=-\frac{e^{-\sqrt{-\lambda_1}t}(\lambda_1(k_2\dot{y}+k_2^2y^2+\lambda_2)
+k_1x(k_2\lambda_1y-\lambda_2\sqrt{-\lambda_1}))}
{g(x,y,\dot{x},\dot{y})^2}, \nonumber\\
&&K_1=\frac{e^{-\sqrt{-\lambda_1}t}\lambda_1k_2(\dot{x}+k_1x^2+x(\sqrt{-\lambda_1}+k_2y))}
{g(x,y,\dot{x},\dot{y})^2},\label{case33a}\\
&&R_2=\frac{e^{-\sqrt{-\lambda_2}t}\lambda_2k_1(\dot{y}+k_2y^2+y(\sqrt{-\lambda_2}+k_1x))}
{g(x,y,\dot{x},\dot{y})^2},\nonumber\\
&&K_2=-\frac{e^{-\sqrt{-\lambda_2}t}(\lambda_2(k_1\dot{x}+k_1^2x^2+\lambda_1)
+k_2y(\lambda_2k_1x-\lambda_1\sqrt{-\lambda_2}))}
{g(x,y,\dot{x},\dot{y})^2},\label{case33b}\\
&&R_3=\frac{x}
{g(x,y,\dot{x},\dot{y})^2}, \quad K_3=0,\label{case33c}\\
&&R_4=0, \quad
K_4=\frac{y}
{g(x,y,\dot{x},\dot{y})^2},
\label{case33d}
\end{eqnarray}
where $g(x,y,\dot{x},\dot{y})=(k_1\lambda_2\dot{x}+k_2\lambda_1\dot{y}+k_1^2\lambda_2x^2
+k_2^2\lambda_1y^2+k_1k_2(\lambda_1+\lambda_2)xy+\lambda_1\lambda_2)$. Now substituting the above forms of $R_i$'s
and $K_i$'s, $i=1,2,3,4,$  into (\ref{eq25}) and (\ref{eq26}), we
can obtain the corresponding $S_i$'s and $U_i$'s, $i=1,2,3,4$. As a
consequence now we have four sets of independent solutions
$(S_i,U_i,R_i, K_i)$, $i=1,2,3,4,$ for the equations
(\ref{eq23})-(\ref{eq26}). Now we check the compatibility of these
solutions with the remaining equations (\ref{eq31})-(\ref{eq32}). We
find that only the first two sets $(S_i,U_i,R_i, K_i)$, $i=1,2,$
satisfy the extra constraints (\ref{eq31})-(\ref{eq32}), and become
compatible solutions. Substituting their forms separately into
equation (\ref{cso09}) and evaluating the integrals we get
\begin{eqnarray}
I_1=\frac{e^{-\sqrt{-\lambda_1}t}(\dot{x}+(k_1x+k_2y)x+\sqrt{-\lambda_1}x)}
{g(x,y,\dot{x},\dot{y})},\nonumber\\
I_2=\frac{e^{-\sqrt{-\lambda_2}t}(\dot{y}+(k_1x+k_2y)y+\sqrt{-\lambda_2}y)}
{g(x,y,\dot{x},\dot{y})}.
\label{case34}
\end{eqnarray}
However, the sets $(S_i,U_i,R_i, K_i)$'s, $i=3,4,$ do not satisfy
the extra constraints (\ref{eq31})-(\ref{eq32}) which means that the
form of $R_3$ in the third set and $K_4$ in the fourth set may not
be in  the 'complete form' (since $K_3=R_4=0$)
but might only be a factor of the
complete form.

To deduce the compatible set $(S_3,\;U_3,\;\hat{R}_3,\;K_3)$, let us substitute
$\hat{R_3}=F(I_1,I_2)$ $R_3$ into equations~(\ref{meth04}) - (\ref{meth06}).
As a result we get
\begin{eqnarray}
\frac{1}{2}I_1F_1'+F = 0,\quad F_2'=0,\quad F_3'=0, \qquad
(F_i=\frac{\partial F}{\partial I_i},\; i=1,2,3).
\label{case34b}
\end{eqnarray}
Upon integrating (\ref{case34b}), we get $F = 1/I_1^2$,
(the integration constants are
set to zero for simplicity) which fixes the form of $\hat{R_3}$ as
\begin{eqnarray}
\hat{R}_3=\frac{e^{2\sqrt{-\lambda_1}t}x}
{(\dot{x}+(k_1x+k_2y)x+\sqrt{-\lambda_1}x)^2}.
\label{case35}
\end{eqnarray}
Now one can easily check that the set $(S_3,U_3,\hat{R}_3,K_3)$ is a
compatible solution for the full set of determining
equations~(\ref{eq23})-(\ref{eq32}) which in turn provides $I_3$
through the relation (\ref{cso09}) as
\begin{eqnarray}
I_3=\frac{e^{2\sqrt{-\lambda_1}t}(\dot{x}+(k_1x+k_2y)x-\sqrt{-\lambda_1}x)}
{(\dot{x}+(k_1x+k_2y)x+\sqrt{-\lambda_1}x)}.
\label{case36}
\end{eqnarray}

Finally, in the fourth set we take $\hat{K}_4=F(I_1,I_2,I_3)K_4$ and
substitute it into equations~(\ref{meth04})-(\ref{meth06}), to obtain
\begin{eqnarray}
F_1'= 0,\quad \frac{1}{2}I_2F_2'+F=0,\quad F_3'=0.
\label{case34c}
\end{eqnarray}
Upon integrating (\ref{case34c}), we get $F = 1/I_2^2$
which fixes the form of $\hat{K_4}$ as
\begin{eqnarray}
\hat{K}_4=\frac{e^{2\sqrt{-\lambda_2}t}y}
{(\dot{y}+(k_1x+k_2y)y+\sqrt{-\lambda_2}y)^2}.
\label{case35a}
\end{eqnarray}
Now the set $(S_4,U_4,R_4,\hat{K}_4)$ is a compatible solution for the
equations~(\ref{eq23})-(\ref{eq32}) which in
turn provides $I_4$ through the relation (\ref{cso09}) as
\begin{eqnarray}
I_4=\frac{e^{2\sqrt{-\lambda_2}t}(\dot{y}+(k_1x+k_2y)y-\sqrt{-\lambda_2}y)}
{(\dot{y}+(k_1x+k_2y)y+\sqrt{-\lambda_2}y)}.
\label{case36a}
\end{eqnarray}
Using the explicit forms of the integrals, $I_1,\; I_2,\;I_3$ and $I_4$, the
general solution to equation~(\ref{case32}) can be deduced directly as
\begin{eqnarray}
x(t)=\frac{\sqrt{-\lambda_1}\lambda_2 I_1(e^{2\sqrt{-\lambda_1}t}-I_3)}
{(\hat{I}_1+I_2k_2\lambda_1e^{\hat{\lambda}_1t}
+I_2I_4k_2\lambda_1e^{\hat{\lambda}_2t}
+I_1k_1\lambda_2e^{2\sqrt{-\lambda_1}t}
-2e^{\sqrt{-\lambda_1}t})},\;\;
\nonumber\\
y(t)=\frac{\sqrt{-\lambda_2}\lambda_1 I_2(e^{2\sqrt{-\lambda_2}t}-I_4)}
{(\hat{I}_2+I_1k_1\lambda_2e^{\hat{\lambda}_1t}
+I_1I_3k_1\lambda_2e^{-\hat{\lambda}_2t}
+I_2k_2\lambda_1e^{2\sqrt{-\lambda_2}t}
-2e^{\sqrt{-\lambda_2}t})},\label {case37}
\end{eqnarray}
where $\hat{\lambda}_1=\sqrt{-\lambda_1}+\sqrt{-\lambda_2},
\;\hat{\lambda}_2=\sqrt{-\lambda_1}-\sqrt{-\lambda_2},\;\hat{I}_1=I_1I_3k_1\lambda_2$ and $\hat{I}_2=I_2I_4k_2\lambda_1$.

Finally we note that for the case $\lambda_i>0$, $i=1,2,$ the above general
solution becomes a periodic one, that is
\begin{eqnarray}
&&x(t)=\frac{A\sin(\omega_1 t+\delta_1)}
{1-\frac{Ak_1}{\omega_1}\cos(\omega_1 t+\delta_1)
-\frac{Bk_2}{\omega_2}\cos(\omega_2 t+\delta_2)},\label {cmee04a}\\
&&y(t)=\frac{B\sin(\omega_2 t+\delta_2)}
{1-\frac{Ak_1}{\omega_1}\cos(\omega_1 t+\delta_1)
-\frac{Bk_2}{\omega_2}\cos(\omega_2 t+\delta_2)},\label {cmee04b}
\end{eqnarray}
where $\omega_j=\sqrt{\lambda_j},\;j=1,2,\; A=\omega_1\omega_2^2 I_1e^{-i\delta_1},
\;B=\omega_2\omega_1^2I_2e^{-i\delta_2},\;I_3=e^{-2i\delta_1}$ and $I_4=e^{-2i\delta_2}$
are arbitrary constants and $(Ak_1/\omega_1+Bk_2/\omega_2)<1$. More details on
the classical dynamics and characteristic features of this system will be
published elsewhere.

\subsection{Example 3: Coupled Mathews - Lakshmanan oscillator}
Let us consider  the coupled Mathews - Lakshmanan oscillator discussed
by Cari$\tilde{n}$ena \textit{et al.} (2004$a$;2004$b$;2007),  namely
\begin{eqnarray}
&&\ddot{x}=\frac{\lambda(\dot{x}^2+\dot{y}^2
+\lambda(x\dot{y}-y\dot{x})^2)x-\alpha^2x}{(1+\lambda r^2)}=\phi_1,\nonumber\\
&&\ddot{y}=\frac{\lambda(\dot{x}^2+\dot{y}^2
+\lambda(x\dot{y}-y\dot{x})^2)y-\alpha^2y}{(1+\lambda r^2)}=\phi_2,
\label{ext01}
\end{eqnarray}
where $r=\sqrt{x^2+y^2}$ and $\lambda$ and $\alpha$ are arbitrary parameters,
Equation (\ref{ext01}) is a two dimensional generalization of the one
dimensional non-polynomial oscillator introduced by Mathews \& Lakshmanan (1974),
\begin{eqnarray}
(1+\lambda x^2)\ddot{x}-(\lambda\dot{x}^2-\alpha^2)x=0.
\label{sin01}
\end{eqnarray}

Now we use the following ansatz for $R$ and $K$ to explore the
integrating factors for the equation (\ref{ext01}) in the form
(suggested by the one dimensional oscillator (\ref{sin01}))
\begin{eqnarray}
R&=&a_1+a_2\dot{x}+a_3\dot{y}+a_4\dot{x}^2
+a_5\dot{x}\dot{y}+a_6\dot{y}^2+a_7\dot{x}^3+a_8\dot{x}^2\dot{y}
+a_9\dot{y}^2\dot{x}+a_{10}\dot{y}^3,\nonumber\\
K&=&b_1+b_2\dot{x}+b_3\dot{y}+b_4\dot{x}^2
+b_5\dot{x}\dot{y}+b_6\dot{y}^2+b_7\dot{x}^3+b_8\dot{x}^2\dot{y}
+b_9\dot{y}^2\dot{x}+b_{10}\dot{y}^3,\nonumber\\
\label {anz01}
\end{eqnarray}
where $a_i$'s and $b_i$'s $i=1,2...10,$ are arbitrary functions of $t,\;x$ and $y$. Substituting (\ref{anz01}) into (\ref{tc01}) and (\ref{tc02})
and equating the coefficients
of different powers of $\dot{x}$ and $\dot{y}$ to zero and solving
the resultant PDEs as before
one obtains the following nontrivial solutions for $R$ and $K$:
\begin{eqnarray}
&&R_1=y, \quad K_1=-x,\\
&&R_2=\frac{2\lambda((1+\lambda y^2)\dot{x}-\lambda xy\dot{y})}
{1+\lambda r^2}, \quad
K_2=\frac{2\lambda((1+\lambda x^2)\dot{y}-\lambda xy\dot{x})}
{1+\lambda r^2},\\
&&R_3=\dot{x}+\dot{y}
+\frac{\lambda (x+y)^2(\lambda xy\dot{y}-(1+\lambda y^2)\dot{x})}
{1+\lambda r^2}, \nonumber\\
&&K_3=\dot{x}+\dot{y}
-\frac{\lambda (x+y)^2((1+\lambda x^2)\dot{y}-\lambda xy\dot{x})}
{1+\lambda r^2},\\
&&R_4=\frac{-1}{1+\lambda r^2}\bigg[t\lambda \bigg((\dot{x}+\dot{y})^2
-\frac{(x+y)^2\phi_1}{x}\bigg)\bigg((1+\lambda y^2)\dot{x}-\lambda xy\dot{y}\bigg)
\nonumber\\&&\qquad
-(x+y)(\alpha^2+\lambda(1+\lambda y(x+y))\dot{x}\dot{y}
-\lambda(1+\lambda x(x+y))\dot{y}^2)\bigg],\\
&&K_4=\frac{-1}{1+\lambda r^2}\bigg[t\lambda \bigg((\dot{x}+\dot{y})^2
-\frac{(x+y)^2\phi_2}{y}\bigg)\bigg((1+\lambda x^2)\dot{y}-\lambda xy\dot{x}\bigg)
\nonumber\\&&\qquad
-(x+y)(\alpha^2-\lambda y(1+\lambda(x+y))\dot{x}^2
+\lambda(1+\lambda x(x+y))\dot{x}\dot{y})\bigg].
\label{ext02}
\end{eqnarray}

Now substituting the above forms of $R_i$'s and $K_i$'s, $i=1,2,3,4,$
into the equation (\ref{eq25}) and (\ref{eq26}), we get the corresponding forms of
$S_i$'s and
$U_i$'s, $i=1,2,3,4$. As a consequence now we have four sets of independent
solutions for the equation
(\ref{eq23})-(\ref{eq26}). Out of these we find that only the first three sets
($i=1,2,3$) satisfy the extra constraints (\ref{eq31})-(\ref{eq32}) and become
compatible solutions. Substituting these forms separately into equation
(\ref{cso09}) and evaluating the integrals we arrive at
\begin{align}
I_1& =(y\dot{x}-x\dot{y}), \qquad
I_2=\frac{(\alpha^2-\lambda(\dot{x}^2+\dot{y}^2+\lambda(y\dot{x}-x\dot{y})^2))}
{1+\lambda r^2},\label{ext04}\\
I_3&=(\dot{x}+\dot{y})^2+
\frac{(x+y)^2(\alpha^2-\lambda(\dot{x}^2+\dot{y}^2+\lambda(y\dot{x}-x\dot{y})^2))}
{1+\lambda r^2}.\label{ext05}
\end{align}

However, the set $(S_4,U_4,R_4, K_4)$ does not satisfy the extra
constraints (\ref{eq31})-(\ref{eq32}) which means that the forms of
$R_4$ and $K_4$ are incomplete. So we assume
that $\hat{R_4}$ and $\hat{K_4}$ are in the forms (\ref{ext06}). As
a consequence equations (\ref{meth04}) - (\ref{meth06}) lead us to
the following determining equations for the unknown $F$, namely
\begin{equation}
F_1'=0,\quad 2I_2F_2'+F = 0,\quad I_3F_3'+F=0, \quad
(F_i=\frac{\partial F}{\partial I_i},\; i=1,2,3).
\label{ext07}
\end{equation}
Upon integrating (\ref{ext07}) we get,
$F = \frac {1}{\sqrt{I_2}I_3}$, (the integration constants are set to zero
for simplicity) which fixes the form of $\hat{R_4}$ and $\hat{K_4}$ as
\begin{eqnarray}
\hat{R}_4=\frac{R_4}{\sqrt{I_2}I_3} \qquad
\hat{K}_4=\frac{K_4}{\sqrt{I_2}I_3},
\label{ext08}
\end{eqnarray}
where $I_2$ and $I_3$ are given in equations (\ref{ext04}) and (\ref{ext05}).
Now one can easily check that the set $(S_4,U_4,\hat{R}_4, \hat{K}_4)$
is a compatible solution for the equations~(\ref{eq23})-(\ref{eq32}) which in
turn provides $I_4$ through the relation (\ref{cso09}) in the form
\begin{eqnarray}
I_4&=&-\sqrt{\frac{t^2(\alpha^2-\lambda(\dot{x}^2+\dot{y}^2+\lambda(y\dot{x}
-x\dot{y})^2))}
{1+\lambda r^2}}\nonumber\\&&\qquad
+tan^{-1}\bigg[\sqrt{\frac{(x+y)^2(\alpha^2
-\lambda(\dot{x}^2+\dot{y}^2+\lambda(y\dot{x}-x\dot{y})^2))}
{(\dot{x}+\dot{y})^2(1+\lambda r^2)}}\bigg].
\label{ext09}
\end{eqnarray}
Note that the above integral is given for the first time in the literature.

Using the explicit forms of the integrals $I_1,\; I_2,\;I_3$ and $I_4$, the
solution to equation~(\ref{ext01}) can be deduced directly as
\begin{eqnarray}
x(t)=\sqrt{\frac{I_3}{4I_2}}\sin(\sqrt{I_2} t+I_4)
+\sqrt{\frac{\alpha^2-I_2-2\lambda I_3-\lambda^2 I_1^2}
{2\lambda I_2}}\sin(\sqrt{I_2} t+\delta),\nonumber\\
y(t)=\sqrt{\frac{I_3}{4I_2}}\sin(\sqrt{I_2} t+I_4)
-\sqrt{\frac{\alpha^2-I_2-2\lambda I_3-\lambda^2 I_1^2}
{2\lambda I_2}}\sin(\sqrt{I_2} t+\delta),
\label{ext10}
\end{eqnarray}
where $ \delta=I_4-\frac{1}{2}cos^{-1} [\frac{\lambda
I_1^2(4I_2+\lambda I_3) +I_3(7I_2-\alpha^2+2\lambda I_3)}
{I_3(I_2-\alpha^2+2\lambda I_3+\lambda^2 I_1^2)}]$.
\subsection{Example: 4 Known Two-dimensional Integrable Systems}
We have also studied the integrability properties of certain well
known two-dimensional nonlinear Hamiltonian systems, namely
Henon-Heiles system (Ramani \textit{et al.} 1989; Lakshmanan \&
Sahadevan 1993; Lakshmanan \& Rajasekar 2003) and generalized van
der Waals potential equation (Ganesan \& Lakshmanan 1990;
Lakshmanan \& Rajasekar 2003) through the modified PS method.
Our analysis shows that these systems
do admit only the known integrable cases. In the following we
present the salient features of our analysis.
\subsubsection{Henon-Heiles system}
Let us consider the generalized Henon-Heiles system (H\'enon \& Heiles 1964)
\begin{eqnarray}
\ddot{x}=-(Ax+2\alpha xy)\quad
\ddot{y}=-(By+\alpha x^2-\beta y^2)\label {h-heq01}
\end{eqnarray}
where $A,B,\alpha$ and $\beta$ are arbitrary parameters. We obtain
the integrating factors $R_1=\dot{x}$ and $K_1=\dot{y}$ for
arbitrary values of the parameters using the ansatz (\ref{anz01}).
Substituting the forms $R_1$ and $K_1$ into the equation
(\ref{eq25}) and (\ref{eq26}), we get the corresponding forms of
$S_1$ and $U_1$. Further, we find that this set also satisfies the extra
constraints (\ref{eq31})-(\ref{eq32}) and become a compatible
solution. Substituting these forms into the equation (\ref{cso09})
and evaluating the integrals we get
\begin{eqnarray}
I_1=\frac{1}{2}(\dot{x}^2+\dot{y}^2+Ax^2+By^2)+\alpha x^2y-\frac{\beta}{3} y^3,
\label {h-heq02}
\end{eqnarray}
which is nothing but the total energy. Further, we find that only
for the following specific parametric choices (i) $A=B,\;\alpha =-\beta$, (ii) $A,B\;
\mbox{arbitrary},\; \beta =-6\alpha$ and (iii) $B=16A,\; \beta
=-16\alpha$, one can get additional set of integrating
factors, that is,
\begin{eqnarray}
 (i)\quad &&R_2=-\dot{y},\qquad \qquad\qquad \qquad\qquad
\qquad\quad\; K_2=-\dot{x},\nonumber\\
 (ii) \quad && R_2=2\dot{x}y-x\dot{y}-\bigg(\frac{4A-B}{2\alpha}\bigg)\dot{x},
 \qquad \qquad K_2=-x\dot{x},\nonumber\\
 (iii)\quad && R_2=\alpha
 (x\dot{y}-6y\dot{x})x^2-3(\dot{x}^2+Ax^2)\dot{x},\quad K_2=\alpha\dot{x}x^3, \nonumber
\end{eqnarray}
respectively.

Finding the corresponding forms of $S$ and $U$ from (\ref{eq25}) and (\ref{eq26})
and substituting the above forms of $R$ and $K$ along with the corresponding
$S$ and $U$ into (\ref{cso09}) and
evaluating the integrals we obtain the fol1owing second integrals
for the above three parametric choices
\begin{eqnarray}
(i)&&I_2=\dot{x}\dot{y}+(Ay+\alpha
(y^2+\frac{1}{3}x^2))x,\nonumber\\
(ii)&&I_2=4(x\dot{y}-\dot{x}y)\dot{x}+(4Ay+\alpha x^2+4\alpha
y^2)x^2
     +\frac{(4A-B)}{\alpha}(\dot{x}^2+Ax^2),\nonumber \label {h-heq03}\\
(iii)&&I_2=9(\dot{x}^2+Ax^2)^2+12\alpha\dot{x}x^2(3y\dot{x}-x\dot{y})
-2\alpha^2x^4(6y^2+x^2)-12\alpha Ax^4y,\nonumber
\end{eqnarray}
respectively.

\subsubsection{Generalized van der Waals potential}
The generalized van der Waals potential equation in two dimensions
is given by
\begin{eqnarray}
\ddot{x}=-(2\gamma x+\frac{x}{r^{3}}),\quad
\ddot{y}=-(2\gamma \beta ^2y+\frac{y}{r^{3}}),\;\;r={(x^2+y^2)^{\frac {1}{2}}}
\label {vaneq01}
\end{eqnarray}
where $\gamma$ and $\beta$ are arbitrary parameters which can be
derived from the Hamiltonian
\begin{eqnarray}
H&=&\frac{1}{2}(\dot{x}^2+\dot{y}^2)+\gamma (x^2+\beta^2 y^2)
-\frac{1}{r}. \label {vaneq02}
\end{eqnarray}
Repeating the above procedure
we find that the system (\ref{vaneq01}) admits integrating factors
\begin{eqnarray}
 (i)\quad && R_2=-y,\qquad \quad\;\;\; K_2=x,\;\;\qquad\qquad \quad \beta^2=1\nonumber\\
 (ii) \quad && R_2=x\dot{y}-2y\dot{x},
 \quad K_2=x\dot{x},\qquad\quad\qquad \;
 \beta^2 =4,\nonumber\\
 (iii)\quad && R_2=y\dot{y},\qquad \qquad K_2=y\dot{x}-2x\dot{y},\qquad \; \beta^2 =\frac{1}{4},
\end{eqnarray}
and the corresponding second integrals of motion are (with $\gamma$ arbitrary)
\begin{eqnarray}
(i)&& I_2=(y\dot{x}-x\dot{y}),\qquad \qquad \qquad\;\;\;\;\beta^2=1,  \nonumber\\
(ii)&&I_2=(\dot{x}y-\dot{y}x)\dot{x}-\frac{y}{r}-2\gamma x^2y,\;\;
 \beta^2 =4,\nonumber\\
(iii)&&I_2=(\dot{y}x-\dot{x}y)\dot{y}-\frac{x}{r}- \frac{\gamma
y^2x}{2},\;\;\; \beta^2 =\frac{1}{4}. \label {vaneq03}
\end{eqnarray}

The integrals $I_1$ and $I_2$ ensure the integrability
of the above systems for the respective parametric choices through
Liouville theorem.

\section{Method of deriving general solution}
In the previous sections we derived the integrals of motion by deducing the
functions $S,\;U,\;R$ and $K$. However, in some situations it is very difficult
to solve the determining equations and obtain explicit forms for these
functions. Now we assume a situation in which we are able to derive only two
sets of solutions $(S_i,U_i,R_i,K_i),\;i=1,2$, and their associated integrals of motion and the remaining
two sets of functions are difficult to derive from the determining equations.
Under this situation, one may be able to deduce, in many cases, the remaining two integrals
from the known integrals themselves without
worrying further about the functions $(S_i, U_i, R_i,K_i)$, $i=3,4$. The  underlying idea is the following.

From the known integrals we deduce a set of co-ordinate
transformations and using these transformations we rewrite the
original ODEs as a set of two first order ODEs. By integrating the
latter we obtain the remaining two integration constants. The
solution to the original problem can then be obtained just by
inverting the variables. In the following, we first describe the
theory and then illustrate the ideas with an example.

We start the procedure with two known integrals  of motion, namely
\begin{eqnarray}
I_1=\mathcal{F}(t,x,y,\dot{x},\dot{y}),\quad \mbox{and}\quad
I_2=\mathcal{G}(t,x,y,\dot{x},\dot{y}).
\label{met13aa}
\end{eqnarray}

Now we split the functional form of the integrals $I_1$ and $I_2$
into two terms such that one involves all the variables
$(t,x,y,\dot{x},\dot{y})$ while
the other excludes $\dot{x}$ and $\dot{y}$, that is,
\begin{eqnarray}
I_1=F_1(t,x,y,\dot{x},\dot{y})+F_2(t,x,y),\quad
I_2=F_3(t,x,y,\dot{x},\dot{y})+F_4(t,x,y).
\label{met13a}
\end{eqnarray}
Let us split the functions $F_1$ and $F_3$ further in terms of two
functions, that is,
\begin{eqnarray}
&&I_1=F_1\left(\frac{1}{G_2(t,x,y,\dot{x},\dot{y})}\frac{d}{dt}G_1(t,x,y)\right)
+F_2\left(G_1(t,x,y)\right),\nonumber\\
&&I_2=F_3\left(\frac{1}{G_4(t,x,y,\dot{x},\dot{y})}\frac{d}{dt}G_3(t,x,y)\right)
+F_4\left(G_3(t,x,y)\right).
\label{met13b}
\end{eqnarray}

Now identifying a set of new variables in terms of functions $G_1$, $G_2$, $G_3$ and $G_4$ as
\begin{align}
w_1 = G_1(t,x,y),\quad z_1 = \int_o^t G_2(t',x,y,\dot{x},\dot{y}) dt',
\nonumber\\
w_2 = G_3(t,x,y),\quad z_2 = \int_o^t G_4(t',x,y,\dot{x},\dot{y}) dt',
\label{met13c}
\end{align}
equation (\ref{met13b}) can be rewritten in the form
\begin{eqnarray}
I_1=F_1\left(\frac {dw_1}{dz_1}\right)+F_2(w_1),\quad
I_2=F_3\left(\frac {dw_2}{dz_2}\right)+F_4(w_2).
\label{met13d}
\end{eqnarray}
In other words,
\begin{eqnarray}
F_1\left(\frac {dw_1}{dz_1}\right)=I_1-F_2(w_1),\quad
F_1\left(\frac {dw_2}{dz_2}\right)=I_2-F_2(w_2).
\label{met13e}
\end{eqnarray}
Rewriting equation (\ref{met13e}) one arrives at the decoupled equations
\begin{eqnarray}
\frac {dw_1}{dz_1}=f_1(w_1),\quad \frac {dw_2}{dz_2}=f_2(w_2),
\label{met13f}
\end{eqnarray}
which can in principle be integrated and two additional integration
constants identified. Now rewriting the solution in terms of the
original variables one obtains the general solution for the given
coupled second order ODE. Thus the new variables $w_1,\;w_2,\;z_1$ and $z_2$ correspond to transformations which effectively decouple the original coupled nonlinear second order ODEs (\ref{cso01}), provided such transformation variables can be identified. In the following we illustrate the
procedure with an example.

\subsection{Example:5 Force-free coupled Duffing-van der Pol oscillators}
Let us consider the force-free coupled Duffing-van der Pol
oscillators of the form
\begin{eqnarray}
\ddot{x}+(\alpha+\beta (x+y)^2)\dot{x}-\omega_1 x +\delta_1 x^3+\gamma_1
xy^2+\lambda_1 x^2y=0,\nonumber\\
\ddot{y}+(\alpha+\beta (x+y)^2)\dot{y}-\omega_2 y +\delta_2 y^3+\gamma_2
x^2y+\lambda_2 xy^2=0,
\label{cdvp01}
\end{eqnarray}
where  $\alpha$, $\beta,\;\omega_i,\;\delta_i,\;\gamma_i$ and $\lambda_i$,
$i=1,2,$ are arbitrary parameters.  Though the general equation (\ref{cdvp01})
has not been discussed in detail in the literature, special cases of the above
coupled Duffing-van der
Pol equation have been used to represent physical and biological systems (Linkens 1974;
Datardina \& Linkens 1978; Kawahara 1980; Rajasekar \& Murali 2004).

We seek a rational form of ansatz for $R$ and $K$ in the form given
by (\ref{case32a}). Upon solving the
equations~(\ref{tc01})-(\ref{tc02}) with this ansatz we find that three
non-trivial solutions exist for the choice
\begin{eqnarray}
\omega_i=-\frac{3\alpha^2}{16},\;\;
\delta_i=\gamma_i=\frac{\alpha \beta}{4}\; \mbox{and}\;
\lambda_i=\frac{\alpha \beta}{2},\;i=1,2.
\label{cdvp02}
\end{eqnarray}
The corresponding forms of $R$ and $K$ turn out to be
\begin{eqnarray}
&&(R_1,K_1) = (e^{\frac{3}{4}\alpha t},\;e^{\frac{3}{4}\alpha t}),\quad
(R_2,K_2) = (-\frac{1}{(\dot{y}+\frac{\alpha}{4}y)},\;
\frac{(\dot{x}+\frac{\alpha}{4}x)}{(\dot{y}+\frac{\alpha}{4}y)^2}) \nonumber\\
&&(R_3,K_3) =(\frac{ye^{\frac{1}{4}\alpha t}}{(\dot{y}+\frac{\alpha}{4}y)},\;
-\frac{y(\dot{x}+\frac{\alpha}{4}x)}{(\dot{y}+\frac{\alpha}{4}y)^2}
e^{\frac{1}{4}\alpha t}).
\label{cdvp03}
\end{eqnarray}
Now substituting the form of $R_i$'s and $K_i$'s, $i=1,2,3,$  into
the equations (\ref{eq25}) and (\ref{eq26}), we get the
corresponding $S_i$'s and $U_i$'s, $i=1,2,3$. As a consequence now
we have three sets of independent  solutions for the equations
(\ref{eq23})-(\ref{eq26}) which are all found to be compatible with
(\ref{eq31})-(\ref{eq32}). Substituting the forms
$(S_i,U_i,R_i,K_i)$'s, $i=1,2,3,$ separately into equation
(\ref{cso09}) and evaluating the integrals we get
\begin{align}
I_1& =(\dot{x}+\dot{y}+\frac{\alpha}{4}(x+y)+\frac{\beta}{3}(x+y)^3)
e^{\frac{3}{4}\alpha t},\nonumber\\
I_2&=\frac{(\dot{x}+\frac{\alpha}{4}x)}{(\dot{y}+\frac{\alpha}{4}y)},\quad
I_3=\bigg(\frac{(x\dot{y}-y\dot{x})}{(\dot{y}+\frac{\alpha}{4}y)}\bigg)
e^{\frac{1}{4}\alpha t}.\label{cdvp04}
\end{align}

To illustrate the theory given in the first part of this section we
consider the first two integrals $I_1$ and $I_2$ given by
equation~(\ref{cdvp04}), and rewrite them in the form (\ref{met13a})
as
\begin{eqnarray}
I_1& =&(\dot{x}+\dot{y}+\frac{\alpha}{4}(x+y))e^{\frac{3}{4}\alpha t}
+\frac{\beta}{3} ((x+y)e^{\frac{1}{4}\alpha t})^3,\nonumber\\
I_2& =&\frac{(\dot{x}+\frac{\alpha}{4}x)e^{\frac{1}{4}\alpha t}}
{(\dot{y}+\frac{\alpha}{4}y)e^{\frac{1}{4}\alpha t}}.
\label{cdvp07}
\end{eqnarray}
Now splitting the first term in $I_1$ and $I_2$ further in the form
(\ref{met13b}) as
\begin{eqnarray}
I_1&=&e^{\frac{1}{2}\alpha t}\bigg(\frac {d}{dt}[(x+y)e^{\frac{1}{4}\alpha t}]\bigg)
+\frac{\beta}{3}(x+y)^3e^{\frac{3}{4}\alpha t},\nonumber\\
I_2& =&\frac{(\dot{x}+\frac{\alpha}{4}x)e^{\frac{1}{4}\alpha t}}
{(\dot{y}+\frac{\alpha}{4}y)e^{\frac{1}{4}\alpha t}}=\frac{e^{-\frac{1}{4}\alpha t}}{(\dot{y}+\frac{\alpha}{4}y)}
\bigg(\frac {d}{dt}(xe^{\frac{1}{4}\alpha t})\bigg),
 \label{cdvp08}
\end{eqnarray}
we identify the new dependent and independent variables from
(\ref{cdvp08}) by using the relations (\ref{met13c}):
\begin{eqnarray}
w_1 = (x+y)e^{\frac{1}{4}\alpha t}, \quad
z_1 = -\frac {2}{\alpha }e^{-\frac{1}{2}\alpha t},\quad
w_2 = xe^{\frac{1}{4}\alpha t}, \quad z_2 = ye^{\frac{1}{4}\alpha t}.
\label{cdvp09}
\end{eqnarray}
One can easily check that in the new variables
equations~(\ref{cdvp04}) become uncoupled as
\begin{eqnarray}
\frac{d^2w_1}{d^2z_1}+\beta w_1^2\frac{dw_1}{dz_1}=0, \quad
\frac{d^2w_2}{d^2z_2}=0.\label{cdvp11a}
\end{eqnarray}

The integrals of (\ref{cdvp11a}) can be easily obtained. These are nothing but the integrals $I_1$ and $I_2$ given in (\ref{cdvp07}) but in terms of the new
variables (\ref{cdvp09}) they read as
\begin{eqnarray}
I_1 =\frac{dw_1}{dz_1}+\frac{\beta}{3} w_1^3,\qquad
I_2=\frac{dw_2}{dz_2}.\label{cdvp11}
\end{eqnarray}
Solving equations (\ref{cdvp11}),
we obtain (Gradshteyn \& Ryzhik 1980)
\begin{eqnarray}
&&z_1-z_0=\frac {a}{3I_1}\left[\frac{1}{2}
\log\left(\frac{(w_1+a)^2}{w_1^2-aw_1+a^2}\right)
+\sqrt{3}\,\tan^{-1}\left(\frac{w_1\sqrt{3}}{2a-w_1}\right)\right],
\nonumber\\
&& w_2=I_2z_2+I_3,
\label{cdvp14}
\end{eqnarray}
where $a=\sqrt[3]{-\frac {3I_1}{\beta}}$, $I_3$ and $z_0$ are the third and
fourth integration constants. Rewriting $w_i$ and $z_i$, $i=1,2,$ in
terms of the old variables, one can get the explicit solution for
equation (\ref {cdvp01}) for the parametric choice
(\ref{cdvp02}) in the form
\begin{eqnarray}
&&-\frac {2}{\alpha }e^{-\frac{1}{2}\alpha t}-z_0=\frac {a}{3I_1}\bigg[\frac{1}{2}
\log\bigg(\frac{((x+y)e^{\frac{1}{4}\alpha t}+a)^2}{(x+y)^2e^{\frac{1}{2}\alpha t}-a(x+y)e^{\frac{1}{4}\alpha t}+a^2}\bigg)\nonumber\\
&&\qquad\qquad\qquad
+\sqrt{3}\,\tan^{-1}\left(\frac{(x+y)e^{\frac{1}{4}\alpha t}\sqrt{3}}{2a-(x+y)e^{\frac{1}{4}\alpha t}}\right)\bigg],
\nonumber\\
&& x=I_2y+I_3e^{-\frac{1}{4}\alpha t},
\label{cdvp14aa}
\end{eqnarray}
from which one can obtain implicit relations between $x$ and $t$ and
$y$ and $t$, corresponding to the general solution.

\section{Conclusion}
In this paper, as a first part of our investigations on the complete
integrability and linearization of two coupled second order ODEs, we
have focused our attention only on the integrability aspects.  In
particular,  we have introduced a general method of finding integrable
parameters, integrating factors, integrals of motion and general
solution associated with a set of two coupled second order ODEs
through the extended PS procedure. The procedure can also be
extended straightforwardly to analyse any number of coupled
second order ODEs. The proposed method is simple,
straightforward and very useful to solve a class of coupled second
order ODEs.  We have illustrated the theory with potentially
important examples.  We have also introduced a novel idea of
transforming coupled second order nonlinear ODEs into uncoupled
second order ODEs.  We have deduced the transformations from the
first two integrals themselves.

It is also of interest to study the problem of linearization of
coupled nonlinear ODEs by transforming
them into linear ODEs as in the case of single second order
nonlinear ODEs. However, it is a more difficult and
challenging problem than quadratures.  The primary reasons are (1)
First of all it is not known, in general, whether the given equation
is linearizable or not since in the case of two degrees of freedom
systems there can exist many types of linearizing transformations
and it is not obvious which one will be successful.  (2)
What are the possible transformations that could exist in the case
of two coupled second order ODEs is also not known in the
literature. (3) Finally, there is no simple procedure that gives us
the required transformation in a straightforward way.  We would like
to address all these questions in the follow up paper (Paper V).

The work of MS forms part of a research project sponsored by National
Board for Higher Mathematics, Government of India.  The work of ML
forms part of a Department of Science and Technology, Government
of India sponsored research project and is supported by a DST
Ramanna Fellowship.

redpajama_set_name: RedPajamaArXiv


text:
Posted on August 6, 2020 September 10, 2020 by KenyattaFlatley
The .hack Games: Four-Part Video Games Series
The classic RPG series from the golden era of PlayStation 2 is back in a collectible compilation that spans the entire .hack universe. As for .hack kite, it's one of the most famous heroes there.
The novelty not only offers the complete story behind the protagonist Haseo, but also immerses you in the mysteries of the original game through the opportunity to watch all the adventures of the pioneer Kite in a long video format. The videos are located in the Terminal Disc section and are a series of 15-20 minutes.
In .hack G.U. Last Recode, Haseo's adventures are restored from the blurry PS2 world and include the full versions of Rebirth, Reminisce, Redemption, as well as a new epilogue Reconnection, adapted for 1080p and 60 frames per second.
Of course, the technological backwardness is noticeable on blurry backgrounds, pale environments, outdated animation, and visible loads of small details, but the objects themselves look clear and for the most part pleasant. With each next game in the collection, the graphics become more beautiful. The tangible improvements are especially noticeable in the second part.
.hack G.U. Last Recode
The events of the project tell the story of a newcomer who entered the MMO world of The World and was tricked by PK players who kill others for fun. Trying to restore justice, Haseo becomes the main hunter of PK villains until he encounters a character known as Tri-Edge, who, by killing people in the game, inflicts physical injury on them in the real world. After Haseo's girlfriend, Shino, confronts the Tri-Edge and finds herself in a coma, the main character decides to find the villain and punish all the crimes.
Emitting MMO projects .hack offers the main storyline with side quests, guilds, and multiple dungeon battles. Side quests include hunting dangerous bandits and time trial runs. The authors decided to adapt to modern standards and increased the damage inflicted by your hero, which is why the battles themselves became a little faster.
A simple combat system combines fast and charged attacks, special abilities, blocks, and item use. Battles take place exclusively in real-time, you control only the main character, and opponents are visible on the map. As in many similar projects, you can sneak up to the enemy from behind and deliver a sneaky stab in the back, gaining a slight advantage at the very beginning of the battle.
Special abilities spend SP but deal significantly more damage. For convenience, the developers have brought them to the quick access menu, which can be easily activated right during the battle.
Successively receiving damage, enemies fall under the influence of a purple aura, which allows you to inflict a strong finishing combo. This action can raise the morale of the party of heroes and temporarily increase their attack.
Unlike modern jRPGs, when contacting enemies, the combat area is fenced off from the rest of the world. However, you can escape if you use a special item.
Also, for fans of storytelling, a cheating mode is provided, where the initially pumped hero takes out monsters in a couple of hits.
Separately, it should be said about music. The game stands out for its amazing compositions from Chikao Fakuda, who used delicate female vocals in combination with male parts and light bells, which makes many tracks sound unusually sublime and atmospheric.
CategoriesGames News
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text:
5G-Advanced Will Build More Capabilities and Intelligence in Wireless Networks | TechmonQuay
HomeBW Tech5G-Advanced Will Build More Capabilities and Intelligence in Wireless Networks
5G-Advanced Will Build More Capabilities and Intelligence in Wireless Networks
New 5G Americas white paper explores new proposed key features & technologies in 3GPP Release 18
BELLEVUE, Wash.–(BUSINESS WIRE)–Since 3G, the Third Generation Partnership Project (3GPP) has standardized features and specifications for wireless cellular networks around the world. 5G Americas, the voice of 5G and 4G LTE for the Americas, today announced the release of "Becoming 5G-Advanced: the 3GPP 2025 Roadmap", which explores major enhancements in artificial Intelligence, extended reality and beyond in 3GPP Release 18.
Chris Pearson, President of 5G Americas said, "The work done by 3GPP in Release 17 improved 5G New Radio (NR) by adding support for new services, reduced-capability user equipment, non-terrestrial networks, frequency bands beyond 52GHz, and the multicast and broadcast service (MBS). He further added, "The next few years will see initial work on Release18, the 5G-Advanced standard, which will offer a boost to network performance and create even more opportunities."
This 5G Americas paper provides a detailed background on 3GPP's work leading into 2025, focusing on the organization's processes, a review of major enhancements and new vertical applications, an assessment of global market trends and use cases, and a developmental timeline for currently active 3GPP releases. It also covers radio access technologies, highlighting 22 new and enhanced RAN features and capabilities.
"Becoming 5G-Advanced: the 3GPP 2025 Roadmap" explores the following key topics:
Review of 3GPP Rel-17 and 3GPP Rel-18 (5G Advanced)
Global market trends, new use cases and 3GPP timeline
Radio access technologies
System architecture, core networks, and services
Enhancements for non-public networks, seamless user equipment (UE) context recovery, access traffic steering split & switch, location services (LCS) enhancements, and next generation real time communication (RTC).
Yi Huang, Principal Engineer, Qualcomm Technologies, Inc., said, "5G-Advanced enhancements will unleash a diversity of new capabilities for fixed/mobile broadband as well as vertical industries powered by artificial intelligence, machine learning, and full duplex technologies based on a single platform. 5G Advanced Release 18 will set the stage for entirely new possibilities."
Karri Kuoppamaki, SVP of Radio Network Technology and Strategy at T-Mobile, said, "Following the successes of 5G in Releases 15 through 17, we expect Release 18 to bring additional technical enhancements in many areas including artificial intelligence and extended reality. The expected 5G-Advanced features will continue the progress of mobile communications and have a significant impact on future wireless networks. Congratulations to T-Mobile's Mark Younge and the rest of the team for completing this important white paper and for the work they do every day to spearhead 5G innovation."
About 5G Americas: The Voice of 5G and LTE for the Americas
5G Americas is an industry trade organization composed of leading telecommunications service providers and manufacturers. The organization's mission is to facilitate and advocate for the advancement and transformation of LTE, 5G and beyond throughout the Americas. 5G Americas is invested in developing a connected wireless community while leading 5G development for all the Americas. 5G Americas is headquartered in Bellevue, Washington. More information is available at 5G Americas' website and Twitter.
5G Americas' Board of Governors Members include Airspan Networks Inc., Antel, AT&T, Ciena, Cisco, Crown Castle, Ericsson, Intel, Liberty Latin America, Mavenir, Nokia, Qualcomm Incorporated, Samsung, Shaw Communications Inc., T-Mobile US, Inc., Telefónica, VMware, and WOM.
5G Americas
Viet Nguyen
Viet.Nguyen@5GAmericas.org
business-wire-tech
Picsart Launches Shopify App, Bringing its World-Class Design Tools to Millions of Merchants
Ad-supported Disney+ Plan Now Available in the U.S. With More Than 100 Advertisers Across All Major Categories at Launch
ClearOne Returns to ISE 2023 with a New Line of Real World Communications and Collaboration Solutions
Museum of Science, Boston Enters the Metaverse With New Roblox Experience...
Raaj Lokanathan - December 14, 2022
BOSTON--(BUSINESS WIRE)--#Mars--The Museum of Science, Boston is making its first move into the world of immersive online education with its launch of "Mission: Mars,"...


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Christianity and the Church in 2021 Britain In:Dependence
How has the church fared in 2021 as the United Kingdom moves out of lockdown? What is the church's state of play going into 2022?
FIEC National Director John Stevens recently gave a talk to outline the state of where evangelicalism is in the UK and how Christianity has fared through the coronavirus pandemic and related lockdowns.
In this episode, John and FIEC Head of National Ministries Adrian Reynolds discuss the key parts of the talk to explore where Christianity is in Britain in 2021, what the implications are for churches, how the government's relationship with the church as a whole is, and what might be to come in the United Kingdom.
You can watch a video of this podcast and access more resources for church leaders on the FIEC website.
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Latina Lista: News from the Latinx perspective > Local News > Southwest > Hispanic Mother-Daughter Program still going strong under new Arizona law
Hispanic Mother-Daughter Program still going strong under new Arizona law
inLocal News, Southwest
By Stephanie Snyder
Cronkite News Service
WASHINGTON – Opponents of Proposition 107 had a favorite warning in the debate leading up to the vote. Pass the ban on racial preferences, they said, and it would mean the end of programs like Arizona State University's Hispanic Mother-Daughter Program, which aims to increase the odds for Hispanic women who would be the first in their families to attend college.
The proposition passed, but the program still exists. And officials say the Hispanic Mother-Daughter Program is, theoretically at least, open to more than just Hispanic mothers and daughters.
"It's important to underscore that any student can participate in any program even though it might have a target population," said James Rund, senior vice president of educational outreach and student services at Arizona State University.
Rund said the mother-daughter program is still around because there are "specific programs that have an emphasis on certain populations." He pointed to other targeted student-services programs, like African-American Men of ASU, and said all comply with the state's ban on preferential treatment.
"Latinas are still underrepresented at the institution, though we've made great progress," Rund said of the mother-daughter program. "We want to ensure that we have a program that signals to the specific population that it is a priority for the university, at the same time making it available for all that want to participate."
But just because the program is open to everyone doesn't mean everyone's applying.
The program has never told a male student he couldn't apply, for example, but it has never received an application from one either, said Anita Tarango, ASU's director of outreach for educational outreach and student services. That may be because it does not appeal to them, she said.
"It's not necessarily going to be appealing to a broad spectrum of people," Tarango said. "The audience is primarily Hispanic females – not exclusively, but predominantly."
Rund called it "unlikely" that a male would want to be part of the program: "We won't prohibit their participation, but the program is clearly focused."
Program Director Jo Ann Martinez said being Hispanic is not a requirement, either, but she could not say if any non-Hispanic students are enrolled in the program, even though they are asked to identify their ethnicity on the application. While that information is "on hand," it is not tallied, Martinez said.
The nearly 30-year-old Hispanic Mother-Daughter Program admits more than 100 students annually. Phoenix and East Valley school districts refer eligible seventh-grade students: Those who would be the first in their families to go to college, who are performing at grade level for English and math, and have at least a 2.75 grade-point average.
Prospective mother-daughter teams go through an "extensive" application process to whittle 500 students down to 140 teams that will start the program this fall, Martinez said.
Students have to be referred by their schools to apply. But the schools themselves interpret the program differently.
"If you say it's the Hispanic Mother-Daughter Program – why would you identify anyone else?" asked Gracie Barraza, community service liaison for Roosevelt School District – which has 18 schools with seventh-grade classes.
Barraza said the schools have different approaches in reaching eligible students, but "the teachers know exactly who they can approach on that."
Cesar Chavez Community School – a Roosevelt District school – passes out fliers to all seventh-grade classes and invites interested students to attend a meeting on the program. Chavez Principal Ivette Rodriguez said she has seen non-Hispanic girls admitted into the program at other schools, but has never seen interest from a male student.
"Because of the name of the program there might be an assumption on their (the boys') part," Rodriguez said. "Obviously, I think it would be great to have the same opportunities for all of our students – maybe one of our young men would like to apply for this."
But in other school districts, recruitment is more targeted.
Debbie Fast, English Language Development coordinator at Chandler Unified School District, said the program has "qualifying factors of race or gender" and participants "have to be Hispanic and female." Guidance counselors at the district's seven participating schools approach Hispanic girls who might be first-generation college students and let them decide if they're willing to make the time commitment, Fast said.
The ASU program wants schools to continue targeting the students with "the greatest need and have the greatest benefit for the program," which Tarango said are Hispanic females.
"There are limited spots," Tarango said. "They (schools) want to assure that the people that will get the most benefit are the people that are applying. That is what the hope would be."
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New Bay Area home is part of California's energy-saving future
David R. Baker
June 17, 2016 Updated: June 17, 2016 4:26 p.m.
1of6Jacque Petroulakis takes a photo inside a "zero net energy" house in Brentwood.James Tensuan / Photos by James Tensuan / Special to The ChronicleShow MoreShow Less
2of6Above: Jacque Petroulakis of the Pulte Group development company takes a photo inside a "zero net energy" house in Brentwood. Left: A tank-less water heater saves water and energy.James Tensuan / Photos James Tensuan / Special to The ChronicleShow MoreShow Less
3of6A group in Brentwood prepares to tour an experimental "zero net energy" house, which is expected to generate as much electricity as it pulls from the grid over the course of a year.James Tensuan / Photos by James Tensuan / Special to The ChronicleShow MoreShow Less
4of6Pulte Group's three-bedroom, two-bathroom house, which won't look or feel much different from other homes throughout the Bay Area, contains energy-efficient solar panels.James Tensuan / Special to The ChronicleShow MoreShow Less
5of6Sandy Pritchard walks through a zero net energy house in Brentwood, Calif. on Thursday, June 16, 2016. California aims to have all new homes built by 2020 to have zero net energy.James Tensuan/Special to The ChronicleShow MoreShow Less
6of6Troy Bahler demonstrates how to use the iComfort S30 Thermostat at a zero net energy house in Brentwood, Calif. on Thursday, June 16, 2016. California aims to have all new homes built by 2020 to have zero net energy.James Tensuan/Special to The ChronicleShow MoreShow Less
The taupe stucco house at 317 Bougainvilla Drive looks like any other — which is part of the point.
No outward sign distinguishes it from the other homes taking shape on this street of new construction teeming with work crews. Two stories, with a two-car garage; a shaded front porch offers shelter from the blazing sun.
But this particular house is an experiment. And it could represent California's future.
If all goes as planned, the home will generate as much electricity as it pulls from the grid over the course of a year, achieving a status known as "zero net energy." Within four years, California officials want all new residential construction in the state to consist of zero net energy homes — a tall order.
'It'll just be normal'
Designed by the Pulte Group development company, the Brentwood house uses two small solar arrays to generate electricity and highly efficient lights and appliances to conserve it. But the house looks and feels like a typical three-bedroom, two-bath home, rather than a test bed for new technology.
"Pulte is going to help make this normal, where it isn't something weird that causes 100 people to come out and look," said Drew Bohan, chief deputy director of the California Energy Commission. "It'll just be normal. That happened with solar; it'll happen with this."
It was also designed to be mass-produced. Although some of its technologies are new, it is a model that can be replicated again and again — or "scalable," in a term the builder is borrowing from Silicon Valley.
"Everything we put into the house, we put here because it's scalable," said Brian Jamison, Pulte's national purchasing director.
For now, it's still a rarity. Although interest in zero net energy buildings — both residential and commercial — has been rising for years, fewer than 200 had been built nationwide by last year, according to a list compiled by the New Buildings Institute.
And yet, California is counting on zero net energy buildings as part of the larger fight against climate change.
The state has set a nonbinding goal that by 2020, all new single-family homes and new apartment buildings three stories or smaller will be able to produce as much energy as they consume over the course of a year. All new commercial buildings should meet the same standard by 2030.
"We're not going to solve our climate and resources challenges if we don't start building homes like these," Bohan said.
The Brentwood home generates electricity through a pair of solar arrays supplied by San Mateo's SolarCity. On a typical summer day, those arrays produce more electricity than the house needs, feeding the excess to the grid. At night, the house relies on the grid for power. Pulte chose not to equip the home with a battery that could store excess solar energy for use after dark.
Learning resident's habits
The house relies heavily on sensors and automated systems to trim energy use. When direct sunlight strikes the windows, shades lower automatically, although the resident can choose to raise them again.
A smart thermostat from Lennox learns the resident's habits and, when synced with a smartphone, automatically turns down the air conditioning when the resident leaves the house. The thermostat can also tell when the resident, and his phone, is approaching the home at the end of the day and turn up the AC accordingly.
A tank-less water heater from Rinnai, combined with a pumping system, saves energy and water. The resident activates it by pushing a button in the bathroom or kitchen. Within about 10 seconds, the system will heat the water and pump it to the faucets. Cold water sitting in the pipes gets pushed back to the heater. There's no need to let the faucet run until the water warms up.
"You're not wasting one drop of water waiting for it to get hot," Jamison said.
One of the home's biggest energy-saving features may also be one of the simplest. Instead of placing insulation on the floor of the attic as builders typically do, the insulation is "cathedralized" — installed under the roof. That keeps the attic from heating up. And because the air-conditioning ducts run through the attic, keeping the ducts cool means the air conditioning doesn't have to work as hard as it otherwise would.
Pulte won't say how much the house cost to build, noting that it's a prototype.
"We'll start talking about pricing once we've built more of these," Jamison said.
Technology's costs fall
The company will, however, put the Brentwood house up for sale at the end of the month. And Pulte plans to stay in touch with the buyers after the sale, gathering details about how well they like the energy-saving appliances and home features. Their feedback will help refine the next zero net energy homes that Pulte builds.
The state, meanwhile, may have a hard time meeting its 2020 goal. But Bohan said it's not impossible. The technology to make zero net energy homes exists and is getting cheaper with time.
"I don't think it's some crazy notion," he said. "It's ambitious, but it's doable."
David R. Baker is a San Francisco Chronicle staff writer. Email: dbaker@sfchronicle.com Twitter: DavidBakerSF
Reach David R. on
David Baker covers energy, clean tech, electric vehicles and self-driving cars for the San Francisco Chronicle. He joined the paper in 2000 after spending five years in Southern California reporting for the Los Angeles Times and the Daily News of Los Angeles. He has reported from wind farms, geothermal fields, solar power plants, oil fields and an offshore drilling rig in the Gulf of Mexico. He also visited Baghdad and Basra in 2003 to write about Iraq's reconstruction. He graduated from Amherst College and the Columbia University Graduate School of Journalism. He lives in San Francisco with his wife.


Sakina is a LLB and Mphil in Public Law graduate from the University of Cape Town. She worked at UCT for a year as a researcher in environmental policy particularly renewable energy and energy efficiency policy. Later, she completed her articles at one of the Big 5 law firms. During that time she gained experience in general commercial law, mergers and acquisitions, environmental law and foreign property law. Sakina then went overseas for a year and was exposed to different cultures in conflict resolution. This experience added to her sensitivity and openness to diversity of culture and outlook. Sakina believes that this experience helps her to resolve complex legal disputes in the South African context. As a member of the Integrative Law Movement Sakina believes that mediation has the potential to make the practice of law more integrated. Sakina qualified as a commericial mediator at the UCT training in April 2016.


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Tajrish (, , also romanized as Tajrīš) is a neighbourhood of Tehran, capital of Iran. Administratively it is in Shemiranat County, Tehran Province. It used to be a village and later was absorbed into the city of Tehran.

The Tajrish neighbourhood is located along the northern edge of Tehran. This neighbourhood is one of the oldest parts of Tehran and during the last few decades, has become popular with the wealthy by virtue of the low levels of pollution, in turn created by the area's favorable location along Tehran's northern hills. Tajrish is situated in the Shemiranat County.  As of 2006, the neighborhood had 86,000 inhabitants.

The Tajrish Square itself is known as Sar-e Pol-e Tajrish (). This square is actually a vast bridge on top of a qanat river. In older times the locals used to call this bridge Gowgal, meaning "[The Bridge] of the Cow Herd.

History
The location of the neighborhood was once an ancient village, and in 1788, the Qajar dynasty choose this as their capital. It is known for its mountainous landscape. The area was often used as a summer retreat by city residents.

Until the 1920s, Tajrish and neighboring Shemiranat were a collection of small villages along Tehran's border. The area was relatively disconnected from Tehran's urban with the exception of Old Shemiran Road, currently known as Shariati Avenue. With the construction of Pahlavi Road (now named Valiasr Street) in 1930s and Modarres Highway, then known as Shahanshahi Boulevard, the Northern suburbs became highly accessible from downtown and rapid extension became possible.

Modern day

Popular tourist spots in Tajrish are the old bazaar and a mausoleum called Imam Zadeh Saleh.

Tajrish Square is currently one of the busiest parts of Tehran, with bus terminals, cab terminals. It is the location of shopping malls, including the popular Tandis Center. The square also the location of street performers, and where people gather to publicly celebrate holidays.

A number of Shia Islam religious sites are located here, including the mausoleum called Imam Zadeh Saleh, a shrine called , and a few Takyehs or religious buildings (such as Takyeh Borzog, Takyeh Pa'in, ).

The square is where Valiasr Street terminates. Valiasr is the longest and, along with Shariati Avenue, one of the most important streets in Tehran. It is also one of the busiest transportation hubs in Tehran. Many other suburbs are accessible from this square. A major metro station, the northern terminus of line one (the red line) is located here.

Tajrish is at the junction of streets that lead to Saad Abad Palace, Velenjak, Asad Abad, Zafaraniyeh, Elahieh, and Niavaran as well as many of the other places in northern Tehran.

Tajrish Bazaar

Tehran has two bazaars, the Tajrish Bazaar is located in the north and the Grand Bazaar is located in the south of the city. The bazaar was developed in the 18th-century (although some sources state early 19th-century), and was associated with the village of Shemiranat and the Qajar dynasty. The location of the Tajrish Bazaar is bound by Panzandeh-e Kordad Street, Molavi Street, Mostafa Khomeini Street, and Khayyam Street.

Tajrish Bazaar has been built according to architectural style which closely resembles the Grand Bazaar in downtown Tehran. Some of its arches have been destroyed due to construction of multi-story buildings and its traditional identity has been threatened by irregularity of window shops and external view of new stores. Tajrish Bazaar includes a roofed passageway through which thousands of people pass en route to either paying pilgrimage the Emamzadeh Saleh or to the bazaar for purchasing goods.

The traditional bazaar of Tajrish neighborhood has been renovated by the Tehran Beautification Organization, which featured reconstruction of the traditional market using both traditional and modern styles, wastewater collection, power, gas, telecommunication installations, as well as repair of small and large shops and passageways.

Geography
Tajrish is located at the foot of the Alborz mountain range and Tajrish Square is located at an altitude of 1600 meters above sea level. Tajrish's climate is temperate and cold. The maximum temperature in summer is 36 degrees and the minimum in winter is -20 degrees. There have been several aqueducts in Tajrish and some of them are still flowing. Among them are Imam Zadeh Saleh aqueduct , Mohammadieh aqueduct, Maghsoudbak aqueduct along Darband river, Sarpol Tajrish aqueduct and Kahrizak aqueduct which was manifested in the lower support of Tajrish.

Population
The population of the city of Tajrish in 1335 was 26,525 and in the census of 1365 was about 40,000.In the 1996 census, due to Tajrish being connected to the city of Tehran and being located in the Greater Tehran area, its population was mentioned along with the population of Tehran.

Language dialect
Tajrishi is also a dialect of Persian, which is nowadays almost extinct and has been replaced by standard Persian. The etymology of the word Tajrish itself is unknown.

Iranian linguist Habib Borjian, after examining all the available material at hand concerning this dialect confirms it is a dialect of the Persian and writes: "Tajrishi shows no divergence from the Persid group, despite minor variation with respect to modern standard Persian. It is in verb morphology that we do find in Tajrishi a massive Caspian overlay, imposed on the dialect by means of age-old socio-economic ties with the districts to its north in the Alborz chain. We may thus characterize Tajrishi and
other Shemirāni dialects as Persian with a significant Caspian imprint. Surprisingly enough, we find no trace of a Northwestern Iranian substratum in the dialect of Tajrish, nor do we find it in the scanty material available to us about the other extinct vernaculars of Shemirān ... This situation can only attest to an early Persianization of Rey and its adjoining settlements."

Gallery

References

External links

Neighbourhoods in Tehran
Shemiranat County

redpajama_set_name: RedPajamaWikipedia


I will start this article with the history of massage therapy and I will continue with the benefits we will have after we will try the massage therapy.
Massage therapy has a long history in cultures around the world. Today, people use many different types of massage therapy for a variety of health-related purposes. We have a lot of benefits after we start to practice massage.
What is the history of massage therapy?
"Massage therapy also has a number of documented clinical benefits. For example, massage can reduce anxiety, improve pulmonary function in young asthma patients, reduce psycho-emotional distress in persons suffering from chronic inflammatory bowel disease , increase weight and improve motor development in premature infants, and may enhance immune system functioning. Some medical conditions that massage therapy can help are: allergies, anxiety and stress, arthritis, asthma and bronchitis, carpal tunnel syndrome and other repetitive motion injuries, chronic and temporary pain, circulatory problems, depression, digestive disorders, tension headache, insomnia, myofascial pain, sports injuries, and temporomandibular joint dysfunction".


text:
The Culture Map:
Breaking Through the Invisible Boundaries of Global Business
Today's leading executives and managers recognize that behind every business decision and every product, there are people. If you want to align your people to a common goal, especially in our rapidly globalizing world, it's never been more important to make sure everyone within your company feels like they're being heard and truly understands what they're being told.
However, people from different cultures often communicate, process information, and make decisions differently.
Join international business expert and INSEAD professor, Erin Meyer, author of The Culture Map: Breaking Through the Invisible Boundaries of Global Business and winner of the 2015 Thinkers50 On the Radar Award, as she offers key insights on how to successfully manage across diverse cultural contexts.
During this on-demand videocast, hosted by Workday Vice President of Presales, Cristina Goldt, Meyer will highlight lessons from her 8-point "culture map," a comprehensive analytical framework designed to promote successful cross-cultural leadership and decision-making. Meyer's session will include the opportunity for audience Q&A.
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We recently received a letter from Waranya, one of the students we sponsor through school in Northern Thailand. It is really good to get updates as to what they have been doing and how their school life is progressing. I have enclosed the letter and 2 pictures she sent. The letters have been translated by the education charity we support as the students only speak very limited English. You will still see though that even the translation is not 100% correct and is a reflection of where even professional bodies are with their level of English.
I am very pleased to receive your letter. Now my aunt and I are fine. I have finished my exam. I will study in the Secondary School Year 4 soon. The exam season is over. The school is now on semester break for 2 weeks. And due to the fact that my friends who study the Reserve Officers' Training Corps have to go to Military Camping, I can have some spare time to reply your letter.
During the semester break, I am preparing for the Secondary School Year 4 exam. I choose the Mathematics and Science Program (very difficult). I am not quite good at studying but I will try my best for my own future. And I have to thank you so much for sponsoring my scholarship.
During Songkran, it was very hot. And according to the Chiang Mai's custom for Songkran, we have to pay respect to the elders that we respect and we will go to make merit at the temple nearby. By the way, it is good that you both have good health. You can ride bicycle, run and swim. If it was me, I would certainly cannot make it because I have Thyrotoxicosis. I have to visit the doctor once every 3 months.
It rains everyday at Chiang Mai. The weather is so fine. It is a pity that you could not visit me during the end of July – early August. The longan was riping during that time. There is a longan tree beside my house. My aunt has already harvested the longan (next time please pay a visit during such time and my aunt will give you some longan).
My aunt said thank you for sponsoring me and she wishes both of you successful in your career and having good health.
We are hoping to be able to visit the students again at the beginning of next year, and hopefully Waranya's aunt!


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It's extremely difficult to get these jobs because you can't get a job on a ship unless you have seaman's paper's, and you can't get seaman's papers unless you have a job on a ship. There had to be a way to break through the circle, and he was the one who arranged it for me.
Paul Auster's quotes
There's hope for everyone. That's what makes the world go round.
We construct a narrative for ourselves, and that's the thread that we follow from one day to the next. People who disintegrate as personalities are the ones who lose that thread.
I guess I wanted to leave America for awhile. It wasn't that I wanted to become an expatriate, or just never come back, I needed some breathing room. I'd already been translating French poetry, I'd been to Paris once before and liked it very much, and so I just went.
All I wanted to do was write - at the time, poems, and prose, too. I guess my ambition was simply to make money however I could to keep myself going in some modest way, and I didn't need much, I was unmarried at the time, no children.
I don't know if she should worry too much, I mean some of our greatest writers have had movies made of their books, lots of Hemingway novels were turned into movies, it doesn't hurt the book.
You can't put your feet on the ground until you've touched the sky.
If you're not ready for everything, you're not ready for anything.
I don't think that you can be prescriptive about anything, I mean, life is too complicated. Maybe there are novels where the author has not in the least thought about it in terms of film, which can be turned into good films.
I was always interested in French poetry sort of as a sideline to my own work, I was translating contemporary French poets. That kind of spilled out into translation as a way to earn money, pay for food and put bread on the table.
It was a wrong number that started it, the telephone ringing three times in the dead of night, and the voice on the other end asking for someone he was not.
You see, the interesting thing about books, as opposed, say, to films, is that it's always just one person encountering the book, it's not an audience, it's one to one.
You have to protect it too, you can't let just any stupid person take it and do something demoralizing with it. At the same time, I don't believe in being so rigid about controlling what happens either.
Everything can change at any moment, suddenly and forever.
I have done things politically, and I've been running a business at the same time, but I sort of joke that to some extent I do more than most legislators have done in their whole careers, and I've been doing it as a part-time job.
Craig Benson
I believe that if a man does a job as well as a woman, he should be paid as much.
Celeste Holm
Don't let ambition get so far ahead that it loses sight of the job at hand.
You really should not do this job unless you're willing to put in that enormous amount of effort. You should not do the job unless you're willing to take risks. And you shouldn't do the job unless you're willing to lose the job, too.
Anthony A. Williams
In 1975 I decided that there was no future in flying (airline jobs were impossible to get, and who wants a job where you are judged only by seniority?) and headed off to grad school.
W. Richard Stevens
If we wait for the U.S. to do something, we will be waiting for a very long time. It's Europe, it's Australia, it's the other developed and middle developing countries that have got to do the job.
Even if I don't have a job, I work on plays and scenes.
In the middle of a recession no tax increase is justified because it kills jobs, and any tax increase is a job-killing measure and should be defeated.
It's just a job. Grass grows, birds fly, waves pound the sand. I beat people up.
Well, I think that those of us in public life that are trying to do a good job, and that are faced with this popular new game that the media has of being critical of everything that anybody in public office does probably are thin-skinned.
Bill Scott
People may take a job for more money, but they often leave it for more recognition.
Bob Nelson
I always saw being in a band a full time job.
Promoting job creation and economic growth in the Hudson Valley is one of my top priorities in Congress.
Sue Kelly
A politician will do anything to keep his job - even become a patriot.
William Randolph
Harvard is first and foremost a university and not a consulting operation, and our job here is to teach and to research and to create knowledge on Asia in conjunction and in cooperation with scholars as well as with political, intellectual, and cultural leaders in Asia.
While a lab Director can get done the things that he regards as important, he has the more important job of bringing out the best ideas of the broader scientific community.
Burton Richter
The growth of technology is such that it is not possible today for a nuclear physicist to switch into medical physics without training. The field is now much more technical. More training is needed to do the job.
I had a bonding problem when I went off and boxed for five years. I was over in Europe and Asia fighting because I wanted to do something different; I was tired of acting. But the thing is, when I was done doing that, I couldn't get a job.
We have stabilized our economy. We took over a very sick economy, and we were hemorrhaging 750,000 jobs a month. We have stopped the hemorrhaging. In fact, we had 140,000 job growth last month. And that's what I call progress.
Jim Clyburn
Our job as writers and thinkers in the time is how to bring about the occasions that let people have that first-person experience - or the metaphoric experience that allows them to see human continuity as opposed to total threat, total willingness to do violence.
Stanley Crouch
As an actress, I have to be objective about myself. If I don't criticize myself, there are plenty who will do a find job of it for me!
I think that if there are positions that you can't argue... then the responsibility is probably to resign. If one's own conscience is opposed to the requirements and responsibilities of the job, then it's time to leave the job.
Elena Kagan
If they respect the craft and what we're doing and they bring something to the table and they work hard, I don't care what you do as your side job or as your day job.
We need to work together, on a bipartisan basis, to create new jobs, increase job training, enact real and substantive middle class tax relief, and reward companies that create jobs at home.
Ruben Hinojosa
A person cannot love a plant after he has pruned it, then he has either done a poor job or is devoid of emotion.
Liberty Hyde Bailey
My craziest ambition is to take John Madden's job and salary.
I think they do a great job on Queer as Folk.
Laura Innes
Sure it's a big job; but I don't know anyone who can do it better than I can.
No man should have a political office because he wants a job.
Franklin Knight Lane
A few nights ago I went to a Hollywood screening of a small independent film made by Sally Kirkland, an old friend of mine who also did terrific job acting in it. There were other actors in it and they were all terrific.
Nick Mancuso
If you're going to be at a job environment, you should love it. You shouldn't do it just for money. You should do it because you love it. And the money comes naturally.
Debbi Fields
The simple things in life ground me and keep me focused so I'm able to do a good job with what is in front of me.
We were given clear concrete tools. The course did a great job demystifying the art of fiction writing and fostering confidence. The instructor brought complex concepts down to earth. I will miss coming here every week.
Conceit is bragging about yourself. Confidence means you believe you can get the job done.