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Home International News India blasted for meddling in Chinese firms' energy projects in Sri Lanka
India blasted for meddling in Chinese firms' energy projects in Sri Lanka
Mon, 6 Dec 2021, 2:50 PM
India blasted for meddling in Chinese firms' energy projects in Sri Lanka
BEIJING, Dec 6 (APP): Chinese experts have criticized India for meddling in Chinese companies' energy projects in Sri Lanka, saying that New Delhi's interference poses a serious threat to development and prosperity in South Asian countries such as Sri Lanka.
The comments came after Indian media reported that a project involving the installation of hybrid energy projects on some Sri Lankan islands had been shelved, owing to "security concerns from a third party."
Sino Soar Hybrid Technology, a Beijing-based high-tech renewable energy company, is responsible for the project.
"It is ridiculous that India uses security reasons to meddle in the project. The Sri Lankan project is small, supplying electricity to villages on three small islands. Since the islands and the main territory of Sri Lanka are separated by the sea, it is very likely that the power grid will not be connected to other places in Sri Lanka," a source close to the company told the Global Times.
All major renewable energy projects in the world are installed with data collection systems to collect weather information, such as wind speed and light intensity, according to the source.
However, since 2010 this information has been required to be saved at local data centers as countries around the world have strengthened data security, the source said.
"India has always taken neighbouring countries such as Sri Lanka as its backyard, and our help in Sri Lanka will be regarded as China moving India's cheese, for China has been regarded as the potential enemy for India," Wang Dehua, a senior South Asian affairs expert in Shanghai, told the Global Times.
India took a heavy blow after the deadly Galwan Valley border clash last year, and it has gotten more sensitive on South Asian countries regarding Chinese issues. India is pursuing head-to-head competition with China in the Indian Ocean, by fair means or foul, said Liu Zongyi, secretary-general of the Research Center for China-South Asia Cooperation at the Shanghai Institutes for International Studies.
Experts warned that India's meddling in projects will hurt the interests of Sri Lanka, as the project being smashed by India has started again in the Maldives.
The Maldives' Minister of Environment Shauna Aminath last week said in a tweet that the country is installing a total of 2.5 MW of solar photovoltaic power in all the inhabited islands in the Thaa Atoll.
Chinese Embassy in Sri Lanka said in an earlier tweet that Sino Soar Hybrid Technology, after being suspended from building a Hybrid Energy system in three northern islands due to "security concerns" from a third party, signed a contract with the Maldivian government to establish solar power plants on 12 islands in the Maldives.
"If the project is operated successfully in Sri Lanka, it could avoid over 2,000 tons of carbon dioxide emissions per year," the source said.
Sinosoar's Solar Hybrid System for Sri Lanka was to have a capacity of 1,700 kilowatts (KW) at peak solar power and 530 KW of wind power. The planned capacity for the battery energy storage system was 2,350 KW and that of the generators was 2,050 KW, according to an article from the company on Sina Weibo in August.
It is not the first time for India to put pressure on Sri Lanka over a Chinese investment project.
Sri Lanka Ports Authority said it signed an agreement with India's Adani Group to build a new terminal next to a $500-million Chinese-run jetty in Colombo, a move seen as countering China's rising influence in the region.
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IESCO workers busy in fixing road lights at Khayaban e Shorwardi...

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MGM Resorts to have minority role in Japan scheme: CEO
Jul 31, 2020 Newsdesk Japan, Latest News, Top of the deck
U.S.-based casino operator MGM Resorts International is to have a minority stake – of up to 45 percent – in a mooted resort scheme in Japan, said on Thursday the firm's new chief executive and president, Bill Hornbuckle (pictured in a file photo).
"We will have between a 40- and 45-percent stake", Mr Hornbuckle said, in reference to MGM Resorts' partnership with Japanese financial services group Orix Corp for the development of a casino resort in the Japanese city of Osaka. He added that the stake was "relative to the overall investment and the risk-reward [premium]" related to the development of casino resort in Japan. Mr Hornbuckle did not further elaborate on the topic.
His comments were made during a conference call with investment analysts following MGM Resorts second quarter results' announcement.
Mr Hornbuckle stressed that MGM Resorts would only pursue a casino resort in Japan if the firm considered it to be a "prudent" investment, that is "going to pay the kind of returns it needs to pay."
He added: "There is a long way to go… We like that we are not fully 'all-in' on this investment. We like the fact there is probably going to be a delay [in the casino licensing process] and a reopening of some of the conversations, that will hopefully make this a better investment for anybody interested in it, most notably us."
In February, the Osaka authorities said the only qualified applicant for its request-for-proposal (RFP) phase for a casino scheme was the MGM Resorts-Orix consortium.
Despite being the sole qualified applicant, the consortium still needs to undergo all the remaining stages of the Osaka RFC process until an official decision on the selected private-sector partner is announced by the local authorities.
In June, the Osaka prefecture and city authorities confirmed they had extended the deadline for the RFP process in their tilt at having a casino resort. No specific cut-off date in relation to the refreshed deadline was mentioned at the time by the authorities.
Japan has in recent months been tackling its own Covid-19 outbreak stemming from the pandemic. An MGM Resorts Japan LLC representative told GGRAsia in June that the Covid-19 pandemic had "affected business travel and the ability of members of [MGM's] Las Vegas team to travel to and from Japan to meet and continue discussions with Osaka prefecture/city."
In his Thursday's comments, Mr Hornbuckle confirmed that Osaka's RFP process had "been stopped".
He added: "We do believe it will be delayed, presumably to the first part of next year, but officially we don't know that yet."
Local governments in Japan wishing to host a casino resort will need to apply to the central authorities for the right to have one. A maximum of three resorts will be permitted in a first phase of liberalisation. The final basic policy document on integrated resorts – due to set out the criteria by which casino applications will be assessed by the national government – is yet to be disclosed publicly.
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Two crystallizing dishes are filled with Na2SO4; a salt bridge and an attachment to a power supply connect them.
Both dishes contain a universal indicator: red, acidic; blue, basic; green, neutral.
The current is allowed to run through.
The two dishes are combined in a beaker.
The water turns red at the anode and blue at the cathode. When the dishes are mixed together, the water turns green.
Electrolysis of water produces equivalent amounts of acid and base. Notice that the cathode equation should be multiplied by two in order to conserve the charge in the chemical reaction.
When the dishes are mixed together, the solution is neutral again. Balancing the above equations with respect to the electrons also balances it with respect to the water's ions, so the overall sum of the two dishes is an electrically neutral and pH neutral solution.


text:
\section{\bf Introduction}
\label{sec1}

The present paper continues the
work on the classification of Lie point symmetries described in the previous
paper\cite{pandey:2007} of the second order ordinary differential equation
of the Li\'enard type,
\begin{eqnarray}
A(x,\dot{x},\ddot{x})\equiv \ddot {x} + f(x)\dot {x} + g(x)= 0,
\label{1.1}
\end{eqnarray}
where $f$ and $g$ are smooth functions of $x$ and over-dot
denotes differentiation with respect to $t$.  In the first part\cite{pandey:2007}
while solving the determining equations we have assumed the case in
which one of the symmetry functions $b(x) = 0$ and identified the corresponding
equations.  The question now naturally
arises as to what are the invariant equations when $b(x) \neq 0$. In
this part we present the answer to this question.

On solving the determining equations with $b(x) \neq 0$ we find that
a class of equations possess eight parameter Lie-point
symmetries.  The forms of $f$ and $g$ which lead to the maximum
number of symmetry generators are as follows (in increasing order of
generality):
\begin{align}
\mbox{(i)~~} & f  =  0, &&  g  =  \lambda_1x+\lambda_2,
\nonumber\\
\mbox{(ii)~~} &  f  =  k, &&  g  = \lambda_1x+\lambda_2,\nonumber\\
\mbox{(iii)~~} &  f  =  kx, &&   g  = \frac{1}{9}k^2 x^3+\lambda_1x +
\lambda_2,  \nonumber\\
\mbox{(iv)~~} & f  =  k_1x+k_2, &&  g  =  \frac{1}{9}k_1^2x^3
+\frac{1}{3}k_1k_2x^2+\lambda_1x +\lambda_2,
\label{s1}
\end{align}
where $k,k_1,k_2,\lambda_1$ and $\lambda_2$ are constants.  The
first case includes the equation of motion of the
free particle and simple
harmonic oscillator. It is well known that both the equations
admit eight symmetry generators \cite{agu:1984,Hydon:2000,wulf:1976} satisfying an
$sl(3,R)$ algebra.
Similarly, the case (ii) which corresponds to the damped harmonic
oscillator which has also been shown to admit $sl(3,R)$ symmetry algebra
\cite{cervil:1984a}. From case (iii), by restricting the
parameters $\lambda_1=\lambda_2=0$, one can get the modified Emden
type equation which is the first nonlinear ODE shown by Mahomed and Leach
to admit a rich set of Lie point symmetry generators\cite{mahomed:1985}.
Here, we show that the general modified Emden type equation
with linear term and constant external forcing also admits eight
point symmetry generators. We also report the class of equations
which falls under case (iv) admitting eight symmetry generators
and this may be considered as the most general form of
(\ref{1.1}).  It is important to note that  all these equations
are linearizable under point transformations.  The explicit forms of the infinitesimal
symmetries and their associated generators for the general cases
are also reported for the first time, as far as the authors'
knowledge goes.

In all the above
cases,  it turns out that $f_{xx} =0$.  By considering the next higher
degree polynomial in the variable $x$ for $f$, say quadratic, $
f  = k_1x^2+k_2x+k_3$, so that $f_{xx} \neq 0$, we get
\begin{eqnarray}
g  = \frac{1}{9}k_1^2 x^5+\frac{5}{18}k_1
k_2x^4+\lambda_1x^3 +\lambda_2x^2+\lambda_3x+\lambda_4,\nonumber
\end{eqnarray}
which leads to one symmetry generator only.  In general, we
prove that for $f_{xx} \neq 0$
one obtains lesser parameter symmetry group only.

The plan of the paper is as follows. In Sec.~II we recall the
derivation of the determining equations for the infinitesimal
symmetries and describe a procedure to solve them.  In Sec.~III by
demanding all the four symmetry functions to be non-zero, we construct
systematically the class of functions $f$ and $g$ which admit
eight symmetry generators.  We start our analysis with
$f=0, g=0$ in Eq. (\ref{1.1}), which lead to the equation of motion of the free
particle, and derive all earlier known equations from (\ref{1.1})
which admit eight point symmetries. Proceeding further by
assuming the function $f$ to be  a constant we deduce damped
harmonic oscillator equation and its variant equations.  We also
present the infinitesimal symmetries and their associated generators
for these equations.  Next, in Sec. IV, by considering the function $f$ to be
linear in $x$, we deduce two new
family of nonlinear ODEs which admit eight symmetry generators.  In
Sec. V, we prove from the symmetry determining equations that when $f_{xx} \neq 0$
only lesser parameter symmetries
can exist.  We also show the equivalence of our results
with the Lie's linearization criterion.
Finally in Sec.~VI we provide a summary of our results. The solutions
of the determining equations encountered in the symmetry analysis of
new equations are presented in Appendices A and B.
\section{\bf Symmetry determining equation of (\ref{1.1})
with $ {b \neq 0}$} \label{ec3}

To make this paper self contained we recall the essential determining
equations very briefly here.  The invariance of Eq. (\ref{1.1}) under
one parameter Lie group of infinitesimal transformations,
$\tilde{t}=t+\epsilon \xi(t,x)+O(\epsilon^2), \;\;
\tilde{x}=x+\epsilon \eta(t,x)+O(\epsilon^2), \;\epsilon \ll 1,$
leads us to four determining equations for the functions $\xi$ and $\eta$
(vide Eqs. (8)-(11) in Paper I).  The
first two determining equations can be integrated straightforwardly to yield
$\xi = a(t)+b(t)x, \;\;
\eta = \dot{b}{x}^2-2b\Im(x) +c(t)x+d(t)$,
where
$\Im_x = F(x) = \int_0^{x}f(x')dx'\;\mbox{and}\;
\Im_{xx}=f(x)$.  The
remaining two determining equations, after substituting the forms of
$\xi$ and $\eta$,  read
\begin{eqnarray}
(\dot{b} x^2-2b\Im+cx+d)f_x+(\dot{a}+\dot{b}x)f+3bg-4\dot{b}F+
3\ddot{b}x+2\dot{c}-\ddot{a}=0, \label{2.14}
\end{eqnarray}
and
\begin{eqnarray}
&&(\dot{b}x^2-2b\Im+cx+d)g_x-(c-2\dot{a}-2bF)g - 2\ddot{b}\Im
\nonumber\\
&&\qquad\qquad\qquad\qquad
+(\ddot{b} x^2-2\dot{b}\Im + \dot{c}x
+\dot{d})f +\dddot{b}x^2+\ddot{c}x +\ddot{d}  = 0. \label{2.15}
\end{eqnarray}
Since $g$ is linear in (\ref{2.14}) we can rewrite (\ref{2.14})
to obtain
\begin{eqnarray}
g=\frac{1}{3b} [-(\dot{b}x^2-2b\Im+cx+d)f_x-(\dot{a}+\dot{b}x)f
+4\dot{b}F-3\ddot{b}x-2\dot{c}+\ddot{a} ], \; b \neq 0.
\label{2.16}
\end{eqnarray}
While deriving Eq. (\ref{2.16}) we assumed that $b\neq 0$.  On the
other hand one may also consider the case $b=0$. In fact this was
the case of the Paper I.  Now we solve the determining equations
(\ref{2.14}) and (\ref{2.15}) with $b\neq 0$ and identify the functions
that are invariant under eight parameter Lie point symmetries.
\section{\bf Maximal Lie point symmetries of Li\'enard  type systems: Linear
ODES }
To begin with we consider polynomial forms for $f$ in the variable $x$ with
increasing degrees and derive the
corresponding explicit forms of $g$ through Eq. (\ref{2.16}). Substituting
the expressions
of $f$ and $g$ into Eq. (\ref{2.15}) and
solving the resultant equation we obtain the infinitesimal
symmetries $\xi$ and $\eta$
which in turn leads to the classification of the class of second order ODEs
which admit maximum number of symmetry generators.  We find that the
maximal number (eight) of symmetries exists only when  $f_{xx}=0$.
When $f_{xx} \neq 0$, lesser parameter Lie point symmetries only exist.
\subsection{\bf Linear Undamped Systems: $ f=0$}
To start with we consider the simple form $f=0$. Correspondingly from Eq.
(\ref{2.16}) we get
\begin{eqnarray}
g=\lambda_1 x +\lambda_2,
\label{3.1}
\end{eqnarray}
where $\lambda_1$ and $\lambda_2$ are arbitrary constants.
Substituting Eq.~(\ref{3.1}) into (\ref{2.15}) and solving it consistently
we can find the infinitesimal
symmetries. However, to classify the results systematically we consider the
following subcases, \emph{viz.}, $(i)~\lambda_1$ = $\lambda_2$ = $0$,
$(ii)~\lambda_2 \neq 0$,~$\lambda_1$ = $0$,
 $(iii)~\lambda_1 \neq 0,~ \lambda_2$ = $0$ and
$(iv)~\lambda_1 \neq \lambda_2 \neq 0$.
\subsubsection{\bf Free particle motion ($\lambda_1 = \lambda_2 = 0)$}
When both $\lambda_1$ and $\lambda_2$ are equal to zero, the
function $g$  also becomes zero and so Eq. (\ref{1.1})
takes the form
\begin{eqnarray}
\ddot{x}=0
\label{3.2}
\end{eqnarray}
which is obviously the equation of motion of a free particle. Substituting $f
= g = 0$ into Eq. (\ref{2.15}) we obtain the equation $\dddot{b}x^2+
\ddot{c}x+\ddot{d}=0$.  Equating the coefficients of $x^i,\;i=0,1,2$, to zero
separately one
obtains $\dddot{b}=0$, $\ddot{c}=0$ and $\ddot{d}=0$.  Solving the later
two equations
one obtains $c=c_1+c_2t$ and $d=d_1+d_2t$, where $c_i's$ and
$d_i's,\; i=1,2$, are arbitrary constants.  Similarly from Eq. (\ref{2.14}),
we obtain $\ddot b=0$ and $\ddot a=2\dot c$. Solving these two equations
one obtains $b=b_1+b_2t$ and $a=a_1+a_2t+c_2t^2$, where $b_i's$ and
$a_i's,\; i=1,2$, are arbitrary constants. Inserting the forms
$a(t),b(t),c(t)$ and $d(t)$ in $\xi = a(t)+b(t)x$
and $\eta=\dot{b}{x}^2+c(t)x+d(t)$ one gets
\begin{eqnarray}
\xi =  a_1+a_2t+c_2t^2+(b_1+b_2t)x,\;\;
\eta  =  d_1+d_2t+(c_1+c_2t)x+b_2x^2.
\label{3.3}
\end{eqnarray}
The associated eight infinitesimal generators take the following form
\begin{eqnarray}
X_1 = \frac{\partial}{\partial t}, \;\;
X_2 = t\frac{\partial}{\partial t},\;\;
X_3=x\frac{\partial}{\partial t},\;\;
X_4 = x\left(t\frac{\partial}{\partial t}
+x\frac{\partial}{\partial x}\right),
\nonumber\\
X_5 = x\frac{\partial}{\partial x},\;\;
X_6 = t\left(t\frac{\partial}{\partial t}+x\frac{\partial}{\partial
x}\right),\;\;
X_7 = \frac{\partial}{\partial x},\;\;
X_8 = t\frac{\partial}{\partial x},
\label{3.4}
\end{eqnarray}
which satisfy an $sl(3, R)$ algebra.  The symmetries and their
generators (\ref{3.4}) coincide exactly with the known ones
given in Refs.\onlinecite{agu:1984,Hydon:2000}.
\subsubsection{\bf Free falling particle ($\lambda_2,\;\neq 0,\; \lambda_1 = 0)$}
If we choose $\lambda_1=0$ and $\lambda_2$ as arbitrary in Eq. (\ref{3.1}), then from
Eq. (\ref{1.1}) we get
\begin{eqnarray}
\ddot{x}+\lambda_2=0,
\label{3.6}
\end{eqnarray}
which is the equation of motion of a free falling particle. Now solving the
determining Eq.
(\ref{2.15}) with $f = 0$ and $g = \lambda_2$ we obtain \cite{agu:1984}
\begin{eqnarray}
\xi  & = & a_1+a_2t+(c_2+\frac{3}{2}\lambda_2b_1)t^2+
\frac{\lambda_2}{2}b_2t^3+(b_1+b_2t)x, \nonumber\\
\eta & = & d_1+d_2t+\frac{\lambda_2}{2}(c_1-2a_2)t^2-
\lambda_2(\lambda_2b_1+\frac{1}{2}c_2)t^3
-\frac{\lambda_2^2}{4}b_2t^4
+(c_1+c_2t)x+b_2x^2,
\label{3.7}
\end{eqnarray}
where $a_i,b_i,c_i$ and $d_i,\; i=1,2$, are arbitrary constants.
Here too, we have the following eight generators \cite{agu:1984}
\begin{eqnarray}
\quad
X_1 &  =&   \frac{\partial}{\partial t},\;\;
X_2 = t\frac{\partial}{\partial t}-\lambda_2t^2\frac{\partial}
{\partial x}, \;\;
X_3 = (x+\frac{3}{2}\lambda_2t^2)\frac{\partial}{\partial t}
-\lambda_2^2t^3\frac{\partial}{\partial x},
\nonumber\\\quad
X_4  & = &  t(x+\frac{\lambda_2}{2}t^2)\frac{\partial}{\partial t}
+(x^2-\frac{\lambda_2^2}{4}t^4)\frac{\partial}{\partial x},\;\;
X_5 = (x+\frac{\lambda_2}{2}t^2)\frac{\partial}{\partial x},
\nonumber\\
 \quad
X_6  & = &  t^2\frac{\partial}{\partial t}+
t(x-\frac{\lambda_2}{2}t^2)\frac{\partial}{\partial x},\;\;
X_7=\frac{\partial}{\partial x},\;\;
X_8=t\frac{\partial}{\partial x},
\label{3.8}
\end{eqnarray}
which also form an $sl(3, R)$ algebra.
\subsubsection{\bf Free linear harmonic oscillator ($\lambda_1
\neq 0,~\lambda_2 = 0)$}
We consider $\lambda_1 $ as arbitrary and $\lambda_2 = 0$
so that Eq. (\ref{1.1}) becomes
\begin{equation}
\ddot{x}+\lambda_1x = 0,
\label{3.10}
\end{equation}
which is nothing but the equation of a linear harmonic oscillator (for
$\lambda_1 >0$).  Solving Eq. (\ref{2.15}) with $f=0$ and $g=\lambda_1x$ we
get the following well known infinitesimal symmetry
transformations \cite{agu:1984}
\begin{eqnarray}
\xi&=&a_1+a_2\sin 2\alpha t+a_3\cos 2\alpha t
+(b_1\sin \alpha t+b_2\cos \alpha t)x,  \label{3.11a}\\
\eta&=&d_1\sin \alpha t+d_2\cos \alpha t+(c_1+a_2\alpha\cos 2\alpha t
-a_3\alpha\sin 2\alpha t)x
+\alpha(b_1\cos\alpha t-b_2\sin \alpha t)x^2, \nonumber
\end{eqnarray}
for Eq. (\ref{3.10})
where $\alpha^2 = \lambda_1$ and
$a_i,b_j,d_j$,$\;i = 1,2,3$, $j = 1,2$ and $c_1$ are arbitrary constants.
The associated generators forming the $sl(3,R)$ algebra are \cite{agu:1984}
\begin{eqnarray}
 X_1 &  = &  \frac{\partial}{\partial t}, \;\;
X_2 = \sin 2\alpha t\frac{\partial}{\partial t}+\alpha x\cos 2\alpha t
\frac{\partial}{\partial x},
 X_3  =   \cos 2\alpha t\frac{\partial}{\partial t}-\alpha x\sin 2\alpha t
\frac{\partial}{\partial x},\nonumber \\
X_4 & = & x(\sin \alpha t\frac{\partial}{\partial t}
+\alpha x\cos \alpha t)
\frac{\partial}{\partial x},
 X_5  =  x(\cos \alpha t\frac{\partial}{\partial t}
 -\alpha x\sin \alpha t)
\frac{\partial}{\partial x},\nonumber \\
X_6 & = & x\frac{\partial}{\partial x},\;\;
 X_7  =  \sin \alpha t\frac{\partial}{\partial x},\;\;
X_8 = \cos \alpha t\frac{\partial}{\partial x}.
\label{3.12}
\end{eqnarray}
On the other hand in the case $\lambda_1<0$, Eq. (\ref{3.10}) becomes the
repulsive harmonic oscillator which is invariant under the following
infinitesimal symmetries
\begin{eqnarray}
 \quad
\xi & =  & a_1+a_2e^{2\alpha t}+a_3e^{-2\alpha t}
+(b_1e^{\alpha t}+b_2e^{-\alpha t})x, \nonumber \\
 \quad
\eta  & = &  d_1e^{\alpha t}+d_2e^{-\alpha t}+(c_1+\alpha(a_2e^{2\alpha t}
-a_3e^{-2\alpha t}))x
+\alpha(b_1e^{\alpha t}-b_2e^{-\alpha t})x^2,
\label{3.11}
\end{eqnarray}
The associated $sl(3,R)$ symmetry generators are
\begin{eqnarray}
 X_1 &  = &  \frac{\partial}{\partial t}, \;\;
X_2 = e^{2\alpha t}(\frac{\partial}{\partial t}+\alpha x
\frac{\partial}{\partial x}),\;\;
 X_3  =   e^{-2\alpha t}(\frac{\partial}{\partial t}-\alpha x
\frac{\partial}{\partial x}),\;\;
X_4  =  xe^{\alpha t}(\frac{\partial}{\partial t}+\alpha x
\frac{\partial}{\partial x}),\nonumber \\
 X_5 & = & xe^{-\alpha t}(\frac{\partial}{\partial t}-\alpha x
\frac{\partial}{\partial x}),\;\;
X_6   =  e^{\alpha t}\frac{\partial}{\partial x},\;\;
 X_7  =  e^{-\alpha t}\frac{\partial}{\partial x},\;\;
X_8 = x\frac{\partial}{\partial x}.
\label{3.12}
\end{eqnarray}

\subsubsection{\bf Displaced linear harmonic oscillator
($\lambda_1\neq \lambda_2 \neq 0)$}
Finally, we consider  both $\lambda_1 $ and $\lambda_2$ to be
arbitrary (not equal to zero) and  we have an equation of the
form
\begin{equation}
\ddot{x}+\lambda_1 x +\lambda_2 = 0,
\label{3.14}
\end{equation}
which corresponds to a displaced simple harmonic motion.  Eq.~(\ref{3.14})
can be transformed to (\ref{3.10}) by a simple transformation,
$x \rightarrow x' $$= x+(\frac{\lambda_2}{\lambda_1})$ and consequently the
infinitesimal symmetries of the former can be derived from that of Eq.~(\ref{3.11a}).

\subsection{\bf Linear Damped Systems: $f = constant = k$}
In the previous subsection, III A, we considered the case $f=0$ and derived the form of $g$.
Now we fix $f=\mbox{constant}=k$ and deduce the associated form of $g$.
Substituting  $f = k$ in Eq. (\ref{2.16}), we again obtain $g$ as a linear function
in $x$, that is
\begin{eqnarray}
g=\lambda_1x+\lambda_2,
\label{3.15}
\end{eqnarray}
where $\lambda_1$ and $\lambda_2$ are arbitrary constants. Consequently one
can solve Eq. (\ref{2.15}) with (\ref{3.15}) to obtain the infinitesimal symmetries. However,
to relate the present results with those of
the literature, we consider the following four cases separately:
$(i)~\lambda_1 = \lambda_2 = 0$,
$(ii)~\lambda_1 \neq 0,\;\lambda_2 = 0$, $(iii)~\lambda_2 \neq
0,\;\lambda_1 = 0 , $ and $(iv)~\lambda_1$$\neq \lambda_2 \neq 0$.

\subsubsection{\bf Free particle in a viscous medium ($\lambda_1 =
\lambda_2 = 0)$}
Let $\lambda_1=\lambda_2=0$. Then from Eq. (\ref{3.15})  we obtain  $g = 0$.
Substituting $f = k$ and $g = 0$ in Eq. (\ref{1.1}), we get
\begin{equation}
\ddot{x}+k\dot{x}=0,
\label{3.16}
\end{equation}
which corresponds to the equation of a free particle in a viscous medium. Solving
Eq. (\ref{2.15}) with $f=k$ and $g=0$ we get
\begin{eqnarray}
\xi & = & a_1+\frac{a_2}{k}e^{kt}+\frac{c_2}{k^2}e^{-kt}
+(b_1+\frac{b_2}{k}e^{kt})x, \nonumber\\
\eta & = & d_1+\frac{d_2}{k}e^{-kt}+(c_1-\frac{c_2}{k}e^{-kt})x
-kb_1x^2,
\label{3.17}
\end{eqnarray}
where $a_i$, $b_i$, $c_i$ and $d_i$, $i=1,2$, are arbitrary constants.
The corresponding generators satisfying the  $sl(3,R)$ algebra are\cite{prince:1979}
\begin{eqnarray}
 X_1 & = & \frac{\partial}{\partial t},\;\;
 X_2 = \frac{1}{k}e^{kt}\frac{\partial}{\partial t}, \;\;
 X_3  =   x\left(\frac{\partial}{\partial t}
-kx\frac{\partial}{\partial x}\right),\;\;
 X_4 =  \frac{x}{k}e^{kt}\frac{\partial}{\partial t},\nonumber\\
 X_5 & = & x\frac{\partial}{\partial x},\;\;
 X_6 =  \frac{1}{k}e^{-kt}\left(\frac{1}{k}\frac{\partial}{\partial t}
-x\frac{\partial}{\partial x}\right),\;\;
 X_7 =  \frac{\partial}{\partial x}, \;\;
 X_8  =  \frac{1}{k}e^{-kt}\frac{\partial}{\partial x}.
\label{3.18}
\end{eqnarray}

\subsubsection{\bf Damped linear harmonic oscillator
($\lambda_1\neq 0,~\lambda_2 = 0)$}
Considering  $\lambda_1$ as  arbitrary and $\lambda_2=0$ in Eq. (\ref{3.15}), we obtain from (\ref{1.1}),  the equation of a damped harmonic oscillator,
\begin{equation}
\ddot{x}+k\dot{x}+\lambda_1 x =0.
\label{3.20}
\end{equation}
The invariance property of this equation has been discussed in detail in
Ref.\onlinecite{cervil:1984a}. Following our procedure we find Eq. (\ref{3.20}) to be
invariant under the infinitesimal transformation,
\begin{eqnarray}
\xi  &=& a_1+a_2\sin2\beta t-a_3\cos2\beta t
+e^{\alpha t}(b_1\cos\beta t+b_2\sin\beta t)x, \nonumber\\
\eta &=& \left[b_3+a_2(\beta\cos2\beta t-\alpha\sin2\beta t)
 +a_3(\beta\sin 2\beta t - \alpha\cos 2\beta t)\right]x
 \nonumber\\
 & & \left.- \left[b_1e^{\alpha t}(\alpha\cos\beta t \right.
 +\beta\sin\beta
t)+b_2e^{\alpha t} (\alpha\sin\beta t -\beta\cos\beta t) \right]x^2
\nonumber\\
 & & + e^{-\alpha t}(d_1\cos\beta t+d_2\sin\beta t),
\label{3.21}
\end{eqnarray}
where $\alpha=\frac{k}{2}$, $\beta=
\frac{1}{2}(4\lambda_1-k^2)^\frac{1}{2}$ and
$a_i,b_i,d_j,\; i=1,2,3,\;j=1,2$, are arbitrary constants.

The $sl(3,R)$ symmetry generators \cite{cervil:1984a} are
\begin{eqnarray}
 X_1 & = & \frac{\partial}{\partial t},\;\;
X_2 = \sin2\beta t\frac{\partial}{\partial t}+x(\beta\cos2\beta t-\alpha
\sin2\beta t)
\frac{\partial}{\partial x},\nonumber\\
 X_3 & = & -\cos2\beta t\frac{\partial}{\partial t}+x(\beta\sin2\beta t-
\alpha\cos2\beta t)
\frac{\partial}{\partial x}, \nonumber\\
 X_4 & = & x e^{\alpha t}\left(\cos\beta t\frac{\partial}{\partial
t}-x(\alpha\cos\beta t +\beta\sin\beta t)
\frac{\partial}{\partial x}\right),\nonumber\\
 X_5 & = & x e^{\alpha t}\left(\sin\beta t\frac{\partial}{\partial
t}+x(\beta\cos\beta t -\alpha\sin\beta t)
\frac{\partial}{\partial x}\right),\nonumber\\
 X_6 & = & x\frac{\partial}{\partial x},\;\;
 X_7 = e^{-\alpha t}\cos\beta t\frac{\partial}{\partial x},\;\;
 X_8 = e^{-\alpha t}\sin\beta t\frac{\partial}{\partial x}.
\label{3.22}
\end{eqnarray}
In the above derivation, $4\lambda_1 \le k^2 $. Then for the choice
$4\lambda_1 = k^2$, we have the following infinitesimal symmetries:
\begin{eqnarray}
\xi = &&a_1+a_2 t+a_3 t^2
+(b_1+(b_2+b_3t)e^{-\frac{k}{2}t})x,\label{3.21a} \\
\eta = && -\left(\frac{3}{2}(b_2+b_3t)-b_3\right)
x^2e^{-\frac{k}{2}t}
+\left(c_1+(a_3-\frac{k}{2}a_2)t-\frac{k}{2}a_3t^2\right)x
+(d_1+d_2t)e^{\frac{-k}{2}t}.
\nonumber
\end{eqnarray}
The respective infinitesimal vector fields forming an $sl(3,R)$ algebra are
\begin{eqnarray}
 X_1 & = & \frac{\partial}{\partial t},\;\;
X_2 =  t\frac{\partial}{\partial t}-\frac{k}{2}tx
\frac{\partial}{\partial x},\;\;
 X_3  =   t^2\frac{\partial}{\partial t}+xt(1-\frac{k}{2} t)
\frac{\partial}{\partial x}, \nonumber\\
 X_4 & = & x e^{kt}\frac{\partial}{\partial t},\;\;
 X_5  =  x e^{-\frac{k}{2}t}\left(\frac{\partial}{\partial t}
 -\frac{3}{2}kx\frac{\partial}{\partial x}\right),\;\;
 X_6  =  xe^{-\frac{k}{2}t}\left(t\frac{\partial}{\partial t}
 -\frac{3}{2}kx\frac{\partial}{\partial x}\right),\nonumber\\
 X_7 & = & x\frac{\partial}{\partial x},\;\;
 X_8 = e^{-\frac{k}{2}t}\frac{\partial}{\partial x}.
\label{3.21b}
\end{eqnarray}
Finally the choice $4\lambda_1> k^2$ leads to the following infinitesimal
symmetries:
\begin{eqnarray}
\xi  & = & a_1+\frac{a_2}{\beta}e^{\beta t}- \frac{a_3}{\beta}e^{-\beta t}
+(b_1e^{(\alpha+\beta)t}+b_2e^{(\alpha-\beta)t})x, \nonumber\\
\eta & = & \left((\beta-\alpha)b_1e^{(\alpha+\beta)t}
-(\beta+\alpha)b_2e^{(\alpha-\beta)t}\right)x^2
+d_1e^{(-\alpha+\beta)t}
\nonumber\\
& & +\left(c_1+\frac{1}{2}(1-\frac{2\alpha }{\beta})a_2 e^{\beta t}
+\frac{1}{2}(1+\frac{2\alpha}{\beta})a_3 e^{-\beta t}\right)x
+d_2e^{-(\alpha+\beta)t},
\label{3.21c}
\end{eqnarray}
where $\alpha = \frac{k}{2},\;
\beta = \frac{1}{2}(k^2-4\lambda_1)^{\frac{1}{2}}$. The associated vector
fields  forming an $sl(3,R)$ algebra are
\begin{eqnarray}
 X_1 & = & \frac{\partial}{\partial t},\;\;
X_2 =  e^{\beta t}\left(\frac{1}{\beta}\frac{\partial}{\partial t}
+\frac{1}{2}(1-\frac{2\alpha}{\beta})x\frac{\partial}{\partial x}\right),\nonumber\\
 X_3  & = &  e^{-\beta t}\left(-\frac{1}{\beta}\frac{\partial}{\partial t}
+\frac{1}{2}(1+\frac{2\alpha}{\beta})x\frac{\partial}{\partial x}\right), \;\;
 X_4  =  e^{(\alpha+\beta) t}\left(x \frac{\partial}{\partial t}
 +(\beta-\alpha)x^2\frac{\partial}{\partial x}\right),\label{3.21d}\\
 X_5 & = & e^{(\alpha-\beta) t}\left(x \frac{\partial}{\partial t}
 -(\alpha+\beta)x^2\frac{\partial}{\partial x}\right),\;\;
 X_6  =  x\frac{\partial}{\partial x},\;\;
 X_7 = e^{(-\alpha+\beta)t}\frac{\partial}{\partial x},\;\;
 X_8 = e^{-(\alpha+\beta)t}\frac{\partial}{\partial x}.
\nonumber
\end{eqnarray}

It is well known that equation (\ref{3.20}) admits a time dependent Hamiltonian
of the form $H=\frac{p^2}{2m}e^{-kt}+\frac{\lambda_1}{2}x^2e^{kt}$.  However,
very recently we have proved that the system (\ref{3.20}) also admits a conservative
Hamiltonian description \cite{Sekar:2007} for all the values of $k$ and
$\lambda_1$. The explicit form of the Hamiltonian can be deduced from
equation (71) of paper I by fixing $\alpha=k$,
$\beta=\lambda_1$ and $q=1$ in it.

\subsubsection{\bf Falling particle in a viscous medium
$(\lambda_2 \ne 0,~ \lambda_1 = 0)$}
Choosing  $\lambda_1=0$ and $\lambda_2$ to be  arbitrary in Eq. (\ref{3.15}),
we obtain from Eq. (\ref{1.1})
\begin{equation}
\ddot{x}+k\dot{x}+\lambda_2=0,
\label{3.24}
\end{equation}
which is the equation of a falling particle in a viscous medium.
Eq.~(\ref{3.24}) can be transformed into (\ref{3.16}) through the
transformation $x \rightarrow x' = x + (\frac{\lambda_2}{k})t $, and so its infinitesimal
symmetries and generators can be derived from (\ref{3.17}) and
(\ref{3.18}) by appropriately replacing the variable $x$.

\subsubsection{\bf Displaced damped harmonic oscillator
$(\lambda_1\ne \lambda_2 \ne 0)$}
Let  $\lambda_1\neq \lambda_2\neq 0$ in Eq. (\ref{3.15}).
Then Eq. (\ref{1.1}) becomes
\begin{eqnarray}
\ddot{x}+k\dot{x}+\lambda_1x +\lambda_2=0.
\label{3.28}
\end{eqnarray}
Here too one can transform Eq. (\ref{3.28}) to the
damped harmonic oscillator (\ref{3.20}) through the transformation
$x \rightarrow x' = x +\frac{\lambda_2}{\lambda_1}$ and so the invariance
properties can be analyzed from the infinitesimal symmetries of the
damped harmonic oscillator.
\section{\bf Maximal Lie point symmetries of Li\'enard  type systems: Nonlinear ODES }
\subsection{\bf Modified Emden Equations: $f = kx$}
From the above study,  we find that the choices $f = 0$ and $f = \mbox{constant} $
lead to {\it linear} ODEs only. Thus to generate nonlinear
ODEs  which admit maximum number of symmetry generators one needs to consider
linear/higher degree polynomial/nonpolynomial forms for the function $f(x)$.
To begin with we consider $f$ to be a linear function of $x$, say $f=kx$,
where $k$ is a constant. As a consequence we get (from Eq. (\ref{2.16}))
\begin{eqnarray}
g=\frac{1}{9} k^2 x^3 +\lambda_1x+\lambda_2,
\label{3.29}
\end{eqnarray}
where $\lambda_1$ and $\lambda_2$ are arbitrary constants. Substituting the
explicit forms of $f$ and $g$ into Eq. (\ref{2.15}) and solving it consistently
we obtain the symmetry generators of Eq.~(\ref{1.1}). For this purpose,
we consider the following four cases, $(i)~\lambda_1 = \lambda_2
= 0$, $(ii)~\lambda_1 \ne 0,~\lambda_2 = 0$, $(iii)~\lambda_2 \ne 0,~\lambda_1 = 0
$ and $(iv)~\lambda_1 \ne \lambda_2 \ne 0$, separately.

\subsubsection{\bf Modified Emden equation: $\lambda_1 = \lambda_2 = 0$}
For $\lambda_1=\lambda_2=0$, we obtain from Eq. (\ref{1.1}) the
following nonlinear ODE
\begin{equation}
\ddot{x}+kx\dot{x}+\frac{k^2}{9}x^3=0,
\label{3.30}
\end{equation}
whose linearization, invariance and integrability properties
have been widely discussed in the recent literature
\cite{mahomed:1985,sarlet:1987,Duarte:1987a,leach:88,nls1,Sekar:2004a}. From
our above analysis, Eq. (\ref{3.30}) is found to be invariant under the
Lie point symmetries
\begin{eqnarray}
\xi & = & a_1+a_2t+(c_2+\frac{k}{2}d_1)t^2-\frac{k}{6}d_2t^3+
\Biggr [b_1+b_2t-\frac{k}{6}(c_1+a_2)t^2
 \nonumber\\
& &
-\frac{k}{6}(c_2+\frac{k}{3}d_1)t^3+\frac{k^2}{36}d_2t^4\Biggr]x,\nonumber\\
\eta  & = & d_1+d_2t+(c_1+c_2 t-\frac{k}{2}d_2 t^2)x+
\left[b_2-\frac{k}{3}(c_1+a_2)t-\frac{k}{2}(c_2+\frac{k}{3}d_1)t^2
 +\frac{k^2}{9}d_2t^3
\right]x^2\nonumber\\
& &
-\frac{k}{3}\left[b_1+b_2t-\frac{k}{6}(c_1+a_2)t^2
- \frac{k}{6}(c_2+\frac{k}{3}d_1)t^3
 +\frac{k^2}{36}d_2t^4\right]x^3,
\label{3.31}
\end{eqnarray}
where $a_i$, $b_i$, $c_i$ and $d_i$, $i=1,2$, are arbitrary constants.
These infinitesimal symmetries lead to the symmetry generators
\begin{eqnarray}
X_1 & = & \frac{\partial}{\partial t},\;
X_2 = t(1-\frac{k}{6}xt)\frac{\partial}{\partial t}-
\frac{k}{3}x^2t(1-\frac{k}{6}xt)\frac{\partial}{\partial x},\;
X_3 = x\frac{\partial}{\partial t}-\frac{k}{3}x^3
\frac{\partial}{\partial x},
\nonumber\\
 X_4 & = & xt\frac{\partial}{\partial t}
+x^2(1-\frac{k}{3}xt)\frac{\partial}{\partial x},\;
X_5 = -\frac{k}{6}xt^2\frac{\partial}{\partial t}
+x(1-\frac{k}{3}xt+\frac{k^2}{18}x^2t^2)
\frac{\partial}{\partial x},
\nonumber\\
X_6 & = & t^2(1-\frac{k}{6}xt)\frac{\partial}{\partial t}+xt(1-
\frac{k}{2}xt+\frac{k^2}{18}x^2t^2)\frac{\partial}{\partial x},
\nonumber
\end{eqnarray}
\begin{eqnarray}
 X_7 & = & \frac{k}{2}t^2(1-\frac{k}{9}xt)
\frac{\partial}{\partial t}+(1-\frac{k^2}{6}t^2x^2+\frac{k^3}{54}
t^3x^3)\frac{\partial}{\partial x},
\nonumber\\
X_8 & = & -\frac{k}{6}t^3(1-\frac{k}{6}xt)
\frac{\partial}{\partial t}+t(1-\frac{k}{2}xt+\frac{k^2}{9}x^2t^2
-\frac{k^3}{108}x^3t^3)\frac{\partial}{\partial x},
\label{3.32}
\end{eqnarray}
which also satisfy an $sl(3,R)$ algebra \cite{mahomed:1985}.
Eq. (\ref{3.30}) has the general solution
\begin{align}
x(t) = \frac{t+I_1}{\frac{k}{6}t^2+\frac{I_1k}{3}t+I_2},
\label{3.33}
\end{align}
where $I_1$ and $I_2$ are two integrals of motion with the explicit forms
\begin{align}
I_1  =  -t+\frac{x}{\frac{k}{3}x^2+\dot{x}},\quad
I_2  =  \frac{k}{6}t^2+\frac{1-\frac{k}{3}tx}{\frac{k}{3}x^2+\dot{x}}.
\label{3.34}
\end{align}
\subsubsection{\bf Modified Emden
equation with linear term $(\lambda_1 \neq 0,~ \lambda_2 = 0)$}
Choosing  $\lambda_1 $ as arbitrary and $\lambda_2= 0$ in Eq. (\ref{3.29}), we obtain
\begin{eqnarray}
g=\frac{k^2}{9}x^3+\lambda_1 x.
\label{3.35}
\end{eqnarray}
As a result,  we have a nonlinear ODE of the form
\begin{eqnarray}
\ddot{x}+kx\dot{x}+\frac{k^2}{9}x^3+\lambda_1 x=0, \quad \lambda_1 \neq 0.
\label{3.36}
\end{eqnarray}
Using the forms of  $f$ and $g$ in Eq. (\ref{2.15}),
we obtain the following set of coupled ODEs
for the four arbitrary functions $a$, $b$, $c$ and $d$:
\begin{eqnarray}
& & 3 \ddot{b}+3 \lambda_1 b+ k\dot{a}+kc = 0,\quad
2\dot{c}-\ddot{a}+kd = 0,\quad
\ddot{c}+2 \lambda_1 \dot{a}+ k\dot{d} = 0,
\nonumber\\
& & 3\dddot{b}+3\lambda_1\dot b+3k\dot{c}+k^2d = 0,
\quad \ddot{d}+\lambda_1 d = 0.
\label{3.37}
\end{eqnarray}
Note that the fourth equation in (\ref{3.37}) can be obtained
from first and second equations and one has effectively only
four equations for the four
unknowns. Solving equations (\ref{3.37}) we find that eight parameter Lie point
symmetry groups exist for both the choices $(i)\;\lambda_1>0$ and $(ii)\;\lambda_1<0$.
In the first case the infinitesimal symmetries read
\begin {eqnarray}
\xi  & = & a_1-\frac{c_2}{2\alpha^2}\sin 2\alpha t-\frac{c_3}{2\alpha^2}\cos 2\alpha t-\frac{k}{3\alpha^2}d_1
\sin\alpha t -\frac{k}{3\alpha^2}d_2\cos\alpha t+(b_1\sin \alpha t
\nonumber\\
 & & +b_2\cos \alpha t+\frac{k}{6\alpha^2}a_3
\cos 2 \alpha t +\frac{k}{6\alpha^2}a_2\sin 2\alpha
t-\frac{kc_1}{3\alpha^2})x,\nonumber\\
\eta  & = & (c_1+\frac{a_3}{2}\cos2\alpha t+\frac{a_2}{2}\sin2\alpha t
+\frac{k}{3\alpha}d_1\cos\alpha t-
\frac{k}{3\alpha}d_2\sin\alpha t)x
\nonumber\\
& & +(b_1 \alpha \cos\alpha t- b_2 \alpha  \sin\alpha t-\frac{k}{3\alpha}a_3
 \sin2\alpha t+\frac{k}{3\alpha}a_2\cos2\alpha t)x^2
  \nonumber\\
 & &   -\frac{1}{3}(b_1\sin\alpha t+ b_2 \cos\alpha t
 +\frac{k}{6\alpha^2}a_3\cos2\alpha t
 +\frac{k}{6\alpha^2}a_2\sin2\alpha t-\frac{kc_1}{3\alpha^2})x^3
 \nonumber\\
 & & +d_1\sin\alpha t +d_2\cos\alpha t,
\label{3.338}
\end{eqnarray}
where $a_1,b_i,c_j,d_i $, $i=1,2,\;j=1,2,3$, are arbitrary constants and
$\alpha = \sqrt{\lambda_1}$.

The corresponding symmetry generators are
\begin{eqnarray}
X_1 & = & \frac{\partial}{\partial t}, \;
X_2  = (\sin 2\alpha t+\frac{k}{3\alpha}x\cos 2\alpha t)
\frac{\partial}{\partial t}+x(\alpha \cos 2\alpha t-\frac{2k}
{3}x\sin 2\alpha t\nonumber\\
&&-\frac{k^2}{9\alpha}x^2 \cos2\alpha t)
\frac{\partial}{\partial x},\nonumber\\
 X_3  & = & (\cos2\alpha t-\frac{k}{3\alpha}x\sin2\alpha t)
\frac{\partial}{\partial t}-x(\alpha\sin2\alpha t+\frac{2k}{3}x
\cos2\alpha t\nonumber\\
&&-\frac{k^2}{9\alpha}x^2\sin2\alpha t)
\frac{\partial}{\partial x},\nonumber\\
 X_4 & = & x\sin\alpha t\frac{\partial}{\partial t}+
x^2(\alpha\cos\alpha t-\frac{k}{3}x\sin\alpha t)
\frac{\partial}{\partial x},\nonumber \\
 X_5 & = & x\cos\alpha t \frac{\partial}{\partial t}-x^2
(\alpha\sin\alpha t+\frac{k}{3}x\cos\alpha t)
\frac{\partial}{\partial x},\nonumber\\
X_6 & = & x\frac{\partial}{\partial
t}-(\frac{3\alpha^2 x}{k}+\frac{1}{3}x^3)
\frac{\partial}{\partial x}, \;\;
 X_7 =  -\frac{k}{3\alpha^2}\sin\alpha t
\frac{\partial}{\partial t}+(\sin\alpha t+\frac{k}
{3\alpha}x\cos\alpha t)\frac{\partial}{\partial x},\nonumber\\
 X_8 & = & -\frac{k}{3\alpha^2}\cos\alpha t
\frac{\partial}{\partial t}+(\cos\alpha t-\frac{k}
{3\alpha}x\sin\alpha t)\frac{\partial}{\partial x}.
\label{3.39}
\end{eqnarray}
The vector fields (\ref{3.39}) can be shown to form an $sl(3,R)$ algebra.  For the
case (ii) the infinitesimal symmetries involve exponential functions of
$t$, that is
\begin{eqnarray}
\xi & = & a_1+a_2e^{2\alpha t}+a_3e^{-2\alpha t}+{\displaystyle \frac{k}{3\alpha^2}d_1}e^{\alpha t}
+{\displaystyle \frac{k}{3\alpha^2}d_2}e^{-\alpha t}
+[{\displaystyle \frac{k}{3\alpha^2}c_1}+{\displaystyle b_1}e^{\alpha t}
+{\displaystyle b_2}e^{-\alpha t}\nonumber\\
& & -{\displaystyle \frac{k}{3\alpha}}
(a_2e^{2 \alpha t}-a_3e^{-2\alpha t})]x,\label{3.338a}\\
 \eta  & = & -{\displaystyle \frac{1}{3}\Big[\frac{kc_1}{3\alpha ^2}}+{\displaystyle b_1}e^{\alpha t}+{\displaystyle b_2}e^{-\alpha t}-{\displaystyle \frac{k}{3\alpha}}(a_2e^{2\alpha t}
-a_3e^{-2\alpha t})\Big]x^3
+[{\displaystyle \alpha b_1} e^{\alpha t}
-{\displaystyle \alpha b_2} e^{-\alpha t}
-{\displaystyle \frac{2k}{3}}(a_2e^{2\alpha t}
\nonumber\\
& & + a_3e^{-2\alpha t})]x^2
+[ c_1+\alpha(a_2e^{2\alpha t}-{\displaystyle a_3e^{-2\alpha t})}-{\displaystyle \frac{k}{3\alpha}}(d_1e^{\alpha t}
-d_2e^{-\alpha t})]x
+{\displaystyle d_1}e^{\alpha t}+{\displaystyle d_2}e^{-\alpha t},
\nonumber
\end{eqnarray}
where $a_i$, $c_1$,$b_j, d_j$, $i = 1, 2, 3$, $j = 1, 2$, are arbitrary constants.

The corresponding symmetry generators, which also satisfy an $sl(3,R)$ algebra, are
\begin{eqnarray}
& &X_1 =  \frac{\partial}{\partial t},\;\;\;
X_2 =  e^{2\alpha t}\Big[\Big(1-\frac{k}{3\alpha}x\Big)\frac{\partial}{\partial t}+
\Big[\Big(\frac{k}{9\alpha}x^3-\frac{2k}{3}x^2
+\alpha x\Big)\frac{\partial}{\partial x}\Big],\nonumber\\
& &X_3   =  e^{-2\alpha t}\Big[\Big(1+\frac{k}{3\alpha}x\Big)\frac{\partial}{\partial t}-
\Big(\frac{k}{9\alpha}x^3+\frac{2k}{3}x^2
+\alpha x\Big)\frac{\partial}{\partial x}\Big],\nonumber\\
& &X_4  = x\frac{\partial}{\partial t}-
\Big(\frac{k}{3}x^3-\frac{3\alpha^2}{k}x\Big)\frac{\partial}{\partial x},\qquad \qquad \quad
 X_5  =  e^{\alpha t}\Big[x\frac{\partial}{\partial t}-\Big(\frac{k}{3}x^3-\alpha x^2\Big)\frac{\partial}{\partial x}\Big],\nonumber\\
& &X_6  =  e^{-\alpha t}\Big[x\frac{\partial}{\partial t}-\Big(\frac{k}{3}x^3+\alpha x^2\Big)\frac{\partial}{\partial x}\Big],\qquad \quad
X_7  = e^{\alpha t}\Big[\frac{\partial}{\partial t}-
\Big(\alpha x-\frac{3\alpha^2}{k}\Big)\frac{\partial}{\partial x}\Big],\nonumber\\
& &X_8  =  e^{-\alpha t}\Big[\frac{\partial}{\partial t}+\Big(\alpha x+ \frac{3\alpha^2}{k}\Big)\frac{\partial}{\partial x}\Big].
\label{3.39a}
\end{eqnarray}

The nonlinear oscillator equation (\ref{3.36}) admits several unusual dynamical
properties.  In fact the dissipative system (\ref{3.36}) admits a
time independent first integral and a conservative
Hamiltonian description\cite{vkc1:2005} for all values of $k$ and $\lambda$, that is
\begin{align}
H=\frac{9\lambda ^2}{2k^2}
\bigg(2-2(1-\frac{2kp}{3\lambda })^{\frac{1}{2}}
+\frac{k^2x^2}{9\lambda }-\frac{2kp}{3\lambda }-\frac{2k^3x^2p}{27\lambda ^2}
\bigg),
\label{lam123}
\end{align}
where
\begin{eqnarray}
p=\frac{ \partial L}{\partial \dot{x}}=-\frac{27\lambda ^3}{2k}
\bigg(\frac{1}{(k\dot{x}+\frac{k^2}{3}x^2+3\lambda )^2}\bigg)
+\frac{3\lambda }{2k}. \label{lam122}
\end{eqnarray}

For $\lambda > 0$, the system (\ref{3.36}) admits explicit amplitude independent sinusoidal periodic
solution of the form
\begin{align}
x(t) = \frac{A\sin(\omega t+\delta)}{1-\frac{k}{3\omega}A\cos(\omega t+\delta)},
\quad 0 \le A \le \frac{\omega}{k}, \;\; \omega = \sqrt{\lambda}
\label{lam119a}
\end{align}
and when $\lambda \le 0$, the solution turn out to be a front like
one\cite{vkc1:2005}, that is
\begin{align}
x(t)=\frac{3\beta(I_1e^{2\beta t}-1)}
{kI_1I_2e^{\beta t}+k(1+I_1e^{\beta t})},
\label{lam119b}
\end{align}
where $\beta=\sqrt{|\lambda |}\;$ and $A, \delta, I_1$, $I_2$ are constants.

\subsubsection{\bf Modified
Emden equation with constant external forcing $(\lambda_2 \neq 0,~\lambda_1 = 0)$}

In Eq. (\ref{3.29}) let us put  $\lambda_1 = 0$ and $\lambda_2 = \mbox{arbitrary}$
so that Eq. (\ref{1.1}) becomes
\begin{eqnarray}
\ddot{x}+kx\dot{x}+\frac{k^2}{9} x^3+\lambda_2=0.
\label{3.41}
\end{eqnarray}

Substituting $f=kx$ and $g=\frac{k^2}{9}x^3+\lambda_2$ in (\ref{2.15}), we obtain the following determining equations for the arbitrary functions $a$, $b$, $c$ and $d$:
\begin{eqnarray}
& & 3\ddot{b}+k\dot{a}+kc  = 0,\quad
2\dot{c}-\ddot{a}+3\lambda_2 b +kd  =  0,\quad
\ddot{c}+k\dot{d}  =  0, \nonumber\\
& &\ddot{d}-\lambda_2c+2\lambda_2\dot{a}  =  0.
\label{addcon1}
\end{eqnarray}
Solving them consistently one obtains
\begin{eqnarray}
a(t) & = & a_1-k(m_1 b_2e^{p_1t}+m_2 b_3e^{p_2t}+m_2 b_4e^{p_3t}
+m_1 b_5e^{p_4t}+m_1 b_6e^{p_5t}+m_2 b_7e^{p_6t}),
\nonumber \\
b(t) & = & b_1+\frac{b_2}{p_1}e^{p_1t}+\frac{b_3}{p_2}e^{p_2t}
+\frac{b_4}{p_3}e^{p_3t}+\frac{b_5}{p_4}e^{p_4t}+\frac{b_6}{p_5}e^{p_5t}
+\frac{b_7}{p_6}e^{p_6t},
\nonumber \\
c(t) & = & -k(m_2p_1b_2e^{p_1t}+m_1p_2b_3e^{p_2t}
+m_1p_3b_4e^{p_3t}+m_2p_4b_5e^{p_4t}
+m_2p_5b_6e^{p_5t}+m_1p_6b_7e^{p_6t}),
\nonumber \\
d(t) & = & -\frac{3\lambda_2}{k}b_1+r_1 b_2e^{p_1t}
+r_2 b_3e^{p_2t}+r_3 b_4e^{p_3t}
+r_4 b_5e^{p_4t}+r_5 b_6e^{p_5t}
+r_6 b_7e^{p_6t}.
\label{addcon2}
\end{eqnarray}
The above derivation is presented in Appendix A.
Here $a_1$ and $b_i$, $i=1,\ldots,7$, are integration constants
which are also the eight symmetry parameters.
The constants $p_i$ and $\gamma_i,$ where $i=1,\ldots,6$, are introduced for simplicity which depend on the
system parameters and can be fixed from the relations (\ref{app116}),
 and (\ref{app121}).  Finally, the constants $m_1$ and $m_2$ are
defined in the expression (\ref{app120}).

Eq. (\ref{addcon2}) provides us the infinitesimal symmetries of the form
\begin{eqnarray}
\xi & = & a_1 - k(m_1b_2e^{p_1t} + m_2b_3e^{p_2t}
 + m_2b_4e^{p_3t} + m_1b_5e^{p_4t} + m_1b_6e^{p_5t}
+ m_2b_7e^{p_6t})
\nonumber\\
& & +\left(b_1+\frac{b_2}{p_1}e^{p_1t}+\frac{b_3}{p_2}e^{p_2t}
+\frac{b_4}{p_3}e^{p_3t}
+\frac{b_5}{p_4}e^{p_4t}+\frac{b_6}{p_5}e^{p_5t}
+\frac{b_7}{p_6}e^{p_6t}\right)x,
\nonumber \\
\eta & = & (b_2e^{p_1t}+b_3e^{p_2t}+b_4e^{p_3t}+b_5e^{p_4t}+b_6e^{p_5t}
+b_7e^{p_6t})x^2 - \frac{k}{3}\left(b_1+\frac{b_2}{p_1}e^{p_1t}
\right. \nonumber \\
& & \left.  +\frac{b_3}{p_2}e^{p_2t}+\frac{b_4}{p_3}e^{p_3t}
+\frac{b_5}{p_4}e^{p_4t}+\frac{b_6}{p_5}e^{p_5t}
+\frac{b_7}{p_6}e^{p_6t}\right)x^3
-\left(m_2p_1b_2e^{p_1t}
\right. \nonumber \\
& & \left.
 +m_1p_2b_3e^{p_2t}+m_1p_3b_4e^{p_3t}+m_2p_4b_5e^{p_4t}+m_2p_5b_6e^{p_5t}
 +m_1p_6b_7e^{p_6t}\right)x
\nonumber\\
& & -\frac{3\lambda_2}{k}b_1+r_1 b_2e^{p_1t}+r_2 b_3e^{p_2t}+r_3 b_4e^{p_3t}+r_4 b_5e^{p_4t}
+r_5 b_6e^{p_5t}+r_6 b_7e^{p_6t}.
\label{addcon3}
\end{eqnarray}

The associated infinitesimal generators are
\begin{eqnarray}
X_1 & = & \frac{\partial}{\partial t},\;\;\;
X_2 = x\frac{\partial}{\partial t}-\left(\frac{3}{k}\lambda_2
 +\frac{k}{3}x^3\right)\frac{\partial}{\partial x},
\nonumber\\
X_3 & = & e^{p_1t}\Biggr[\left(-km_1+\frac{x}{p_1}\right)
\frac{\partial}{\partial t}
+\left(r_1-m_2p_1x+x^2-\frac{k}{3p_1}x^3\right)
\frac{\partial}{\partial x}\Biggr],
\nonumber\\
X_4 & = &  e^{p_2t}\Biggr[\left(-km_2+\frac{x}{p_2}\right)
\frac{\partial}{\partial t}
+\left(r_2-m_1p_2x+x^2-\frac{k}{3p_2}x^3\right)
\frac{\partial}{\partial x}\Biggr],
\nonumber\\
 X_5 & = & e^{p_3t}\Biggr[\left(-km_2+\frac{x}{p_3}\right)
\frac{\partial}{\partial t}
+\left(r_3-m_1p_3x+x^2-\frac{k}{3p_3}x^3\right)
\frac{\partial}{\partial x}\Biggr],\nonumber\\
 X_6 & = & e^{p_4t}\Biggr[\left(-km_1+\frac{x}{p_4}\right)
\frac{\partial}{\partial t}
+\left(r_4-m_2p_4x+x^2-\frac{k}{3p_4}x^3\right)
\frac{\partial}{\partial x}\Biggr],
\nonumber\\
X_7 & = & e^{p_5t}\Biggr[\left(-km_1+\frac{x}{p_5}\right)
\frac{\partial}{\partial t}
+\left(r_5-m_2p_5x+x^2-\frac{k}{3p_5}x^3\right)
\frac{\partial}{\partial x}\Biggr],
\nonumber\\
  X_8 & = & e^{p_6t}\Biggr[\left(-km_2+\frac{x}{p_6}\right)
\frac{\partial}{\partial t}
+\left(r_6-m_1p_6x+x^2-\frac{k}{3p_6}x^3\right)
\frac{\partial}{\partial x}\Biggr].
\label{55}
\end{eqnarray}
The general solution of Eq. (\ref{3.41}) can be deduced explicitly
following Ref. \onlinecite{Sekar:2005} as
\begin{align}
x(t) = \frac{-3\alpha}{k}+\frac{1}{3k}\left(
\frac{18\alpha^2(1-I_1e^{\pm i\alpha\sqrt{3}t})}
{3\alpha(1-I_1e^{\pm i\alpha\sqrt{3}t})\mp 2i\alpha \sqrt{3}I_1I_2
e^{\frac{3\alpha+i\alpha \sqrt{3}}{2}t}\pm i \alpha \sqrt{3}(1+I_1e^{\pm i\alpha\sqrt{3}t})}
\right).
\end{align}
where $\alpha$ is solution of the equation $\alpha^3+\lambda_1 \alpha
-\frac{k}{3}\lambda_2 = 0$ and $I_1$ and $I_2$ are two integrals of motion
whose explicit forms have also been reported in Ref.\onlinecite{Sekar:2005}.
\subsubsection{\bf Modified
Emden equation with linear term and  constant external forcing
$(\lambda_1 \neq 0,~ \lambda_2 \neq 0)$}
Finally, we choose $\lambda_1 \neq$ $\lambda_2 \neq 0$ in
Eq. (\ref{3.29}), so that
\begin{equation}
g=\frac{k^2}{9} x^3 +\lambda_1x+\lambda_2.\nonumber
\end{equation}
For this case Eq. (\ref{1.1}) becomes
\begin{equation}
\ddot{x}+kx\dot{x}+\frac{1}{9}k^2 x^3+\lambda_1 x+\lambda_2=0.
\label{3.42}
\end{equation}
The determining equations for the infinitesimal symmetries become
\begin{eqnarray}
& & 3\ddot{b}+k\dot{a}+3\lambda_1 b+kc  =  0,\;\;\;\;
2\dot{c}-\ddot{a}+3\lambda_2 b +kd  =  0,\nonumber\\
& & \ddot{c}+2\lambda_1\dot{a}+k\dot{d}  =  0,,\;\;\;\;
\ddot{d}-\lambda_2c+2\lambda_2\dot{a}+\lambda_1d  =  0.
\label{addcon10}
\end{eqnarray}
Even though Eq.~(\ref{addcon10}) looks similar to (\ref{addcon1}), the
appearance of additional terms  makes it  tedious to solve.
We solve Eq.~(\ref{addcon10}) with the assumption that
$k,\lambda_1,\lambda_2 \neq 0$ to obtain
\begin{eqnarray}
a(t) & = & a_1-\frac{2\lambda_1^2}{k^2\lambda_2}b_1+\beta_1b_2e^{\alpha_1 t}
+\beta_2b_3e^{-\alpha_1t}
+ b_4e^{\alpha_2t}(\beta_3\cos\alpha_3t
+\beta_4\sin\alpha_3 t)\nonumber\\
& &  + b_5e^{\alpha_2t}(\beta_3\sin\alpha_3 t
-\beta_4\cos\alpha_3t)+ b_6e^{-\alpha_2t}(\beta_5\cos\alpha_3t
+\beta_4\sin\alpha_3 t)\nonumber\\
 & &  +b_7e^{-\alpha_2t}(\beta_4\cos\alpha_3t
-\beta_{5}\sin\alpha_3 t), \nonumber\\
b(t) & = & b_1+\frac{b_2}{\alpha_1}e^{\alpha_1 t}
-\frac{b_3}{\alpha_1}e^{-\alpha_1t}
+ \frac{b_4e^{\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_2\cos\alpha_3t
+\alpha_3\sin\alpha_3 t)\nonumber\\
& &
 + \frac{b_5e^{\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}
(\alpha_2\sin\alpha_3t-\alpha_3\cos\alpha_3 t)
+ \frac{b_6e^{-\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_3\sin\alpha_3 t
-\alpha_2\cos\alpha_3t)
 \nonumber\\
& &
+ \frac{b_7e^{-\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}
(\alpha_2\sin\alpha_3t+\alpha_3\cos\alpha_3 t),
\nonumber\\
c(t) & = & -\frac{3\lambda_1}{k}b_1+\gamma_1b_2e^{\alpha_1 t}
+\gamma_2b_3e^{-\alpha_1t}
+ b_4e^{\alpha_2t}(\gamma_3\cos\alpha_3t
+\gamma_4\sin\alpha_3 t)\nonumber\\
& &
 + b_5e^{\alpha_2t}(\gamma_3\sin\alpha_3t-\gamma_4\cos\alpha_3t)
+ b_6e^{-\alpha_2t}(\gamma_5\cos\alpha_3t
+\gamma_6\sin\alpha_3 t)\nonumber\\
& &
 +b_7e^{-\alpha_2t}(\gamma_6\cos\alpha_3t
-\gamma_{5}\sin\alpha_3 t),
\nonumber\\
d(t) & = & -\frac{3\lambda_2}{k}b_1+\delta_1b_2e^{\alpha_1 t}
+\delta_2b_3e^{-\alpha_1t}
+ b_4e^{\alpha_2t}(\delta_3\cos\alpha_3t
+\delta_4\sin\alpha_3 t)\nonumber\\
& &
 + b_5e^{\alpha_2t}(\delta_3\sin\alpha_3t-\delta_4\cos\alpha_3t)
+ b_6e^{-\alpha_2t}(\delta_5\cos\alpha_3t
+\delta_6\sin\alpha_3 t)\nonumber\\
& &
 +b_7e^{-\alpha_2t}(\delta_6\cos\alpha_3t
-\delta_{5}\sin\alpha_3 t).
\label{addcon11}
\end{eqnarray}
Here $a_1$ and $b_i$, $i=1,\ldots,7$, are
the arbitrary
constants arising from the integration of Eq.~(\ref{addcon10}) which fix
the eight symmetry generators.  The other constants are included for
simplicity sake.  A detailed analysis on the above derivation and the relation
between these constants and system parameters are given in Appendix B.

The infinitesimal symmetries turn out to be
\begin{eqnarray}
\xi & = & a_1-\frac{2\lambda_1^2}{k^2\lambda_2}b_1+\beta_1b_2e^{\alpha_1 t}
+\beta_2b_3e^{-\alpha_1t}
+ b_4e^{\alpha_2t}(\beta_3\cos\alpha_3t
+\beta_4\sin\alpha_3 t)+ b_5e^{\alpha_2t}(\beta_3\sin\alpha_3 t
 \nonumber\\
& & -\beta_4\cos\alpha_3t)
+ b_6e^{-\alpha_2t}(\beta_5\cos\alpha_3t
+\beta_4\sin\alpha_3 t)+b_7e^{-\alpha_2t}(\beta_4\cos\alpha_3t
-\beta_{5}\sin\alpha_3 t)
\nonumber\\
& & +b_7e^{-\alpha_2t}(\beta_4\cos\alpha_3t
-\beta_{5}\sin\alpha_3 t)
+\left[b_1+\frac{b_2}{\alpha_1}e^{\alpha_1 t}
-\frac{b_3}{\alpha_1}e^{-\alpha_1t}
+ \frac{b_4e^{\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_2cos\alpha_3t
\right. \nonumber\\
& &
\left. +\alpha_3sin\alpha_3 t)
+ \frac{b_5e^{\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_2\sin\alpha_3t
-\alpha_3\cos\alpha_3 t)+\frac{b_6e^{-\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}
(\alpha_3\sin\alpha_3 t-\alpha_2\cos\alpha_3t)
\right. \nonumber\\
& &
\left.+ \frac{b_7e^{-\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_2\sin\alpha_3t
+\alpha_3\cos\alpha_3 t)\right]x,
\nonumber
\end{eqnarray}
\begin{eqnarray}
\eta & = & \frac{k}{3}\left[b_1+\frac{b_2}{\alpha_1}e^{\alpha_1 t}
-\frac{b_3}{\alpha_1}e^{-\alpha_1t}
+ \frac{b_4e^{\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_2\cos\alpha_3t
+\alpha_3\sin\alpha_3 t)
+\frac{b_5e^{\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_2\sin\alpha_3t
\right. \nonumber \\
& &
\left. -\alpha_3\cos\alpha_3 t)
+\frac{b_6e^{-\alpha_2t}}{(\alpha_2^2+\alpha_3^2)}(\alpha_3\sin\alpha_3 t
  -\alpha_2\cos\alpha_3t)+\frac{b_7e^{-\alpha_2t}}
{(\alpha_2^2+\alpha_3^2)}(\alpha_2\sin\alpha_3t
\right.
\nonumber \\
  & & \left.
  +\alpha_3\cos\alpha_3 t)\right]x^3
 +\Biggr[b_2e^{\alpha_1t}+b_3e^{-\alpha_1t}
+e^{\alpha_2t}(b_4\cos\alpha_3t+b_5\sin\alpha_3t)
+e^{-\alpha_2t}(b_6\cos\alpha_3t
  \nonumber \\
& &
-b_7\sin\alpha_3t)\Biggr]x^2
+\Biggr[-\frac{3\lambda_1}{k}b_1+\gamma_1b_2e^{\alpha_1 t}
+\gamma_2b_3e^{-\alpha_1t}
+ b_4e^{\alpha_2t}(\gamma_3\cos\alpha_3t
+\gamma_4\sin\alpha_3 t)
\nonumber\\
& &
 + b_5e^{\alpha_2t}(\gamma_3\sin\alpha_3t-\gamma_4\cos\alpha_3t)
+ b_6e^{-\alpha_2t}(\gamma_5\cos\alpha_3t
+\gamma_6\sin\alpha_3 t)
+b_7e^{-\alpha_2t}(\gamma_6\cos\alpha_3t
\nonumber\\
& & -\gamma_{5}\sin\alpha_3 t)\Biggr]x
-\frac{3\lambda_2}{k}b_1+\delta_1b_2e^{\alpha_1 t}
+\delta_2b_3e^{-\alpha_1t}
+ b_4e^{\alpha_2t}(\delta_3\cos\alpha_3t
+\delta_4\sin\alpha_3 t)
\nonumber\\
& & + b_5e^{\alpha_2t}(\delta_3\sin\alpha_3t-\delta_4\cos\alpha_3t)
+ b_6e^{-\alpha_2t}(\delta_5\cos\alpha_3t
+\delta_6\sin\alpha_3 t)
\nonumber\\
& & +b_7e^{-\alpha_2t}(\delta_6\cos\alpha_3t
-\delta_{5}\sin\alpha_3 t).\label{addcon11a}
\end{eqnarray}

The associated generators constituting a $sl(3,R)$ algebra are
\begin{eqnarray}
X_1 & = & \frac{\partial}{\partial t},\;
X_2 = \left(x-\frac{2\lambda_1^2}{k^2\lambda_2^2}\right)
\frac{\partial}{\partial t}+\left(-\frac{3\lambda_2}{k}-
\frac{3\lambda_1}{k}x+\frac{k}{3}x^3\right)
\frac{\partial}{\partial x},
\nonumber\\
X_3 & = & \left[\left(\beta_1+\frac{x}{\alpha_1}\right)
\frac{\partial}{\partial t}+\left(\delta_1+\gamma_1x+x^2+\frac{k}{3\alpha_1}x^3
\right)\frac{\partial}{\partial x}\right]e^{\alpha_1t},
\nonumber\\
X_4 & = & \left[\left(\beta_2-\frac{x}{\alpha_1}\right)
\frac{\partial}{\partial t}+\left(\delta_2+\gamma_2x+x^2-\frac{k}{3\alpha_1}x^3
\right)\frac{\partial}{\partial x}\right]e^{-\alpha_1t},
\nonumber
\end{eqnarray}
\begin{eqnarray}
X_5 & = & \Biggr[\left(\beta_3\cos\alpha_3t+\beta_4\sin\alpha_3t
+\frac{1}{(\alpha_2^2+\alpha_3^3)}(\alpha_2\cos\alpha_3t
+\alpha_3\sin\alpha_3t)x\right)\frac{\partial}{\partial t}
\nonumber\\
& &
+\left(\delta_3\cos\alpha_3t+\delta_4\sin\alpha_3t
+(\gamma_3\cos\alpha_3t+\gamma_4\sin\alpha_3t)x+\cos\alpha_3t x^2
\right.
\nonumber\\
 & &
\left. +\frac{1}{(\alpha_2^2+\alpha_3^2)}
(\alpha_2\cos\alpha_3t+\alpha_2\sin\alpha_3t)x^3\right)
\frac{\partial}{\partial x}\Biggr]e^{\alpha_2t},
\nonumber\\
X_6 & = & \Biggr[\left(\beta_3\sin\alpha_3t-\beta_4\cos\alpha_3t
+\frac{1}{(\alpha_2^2+\alpha_3^3)}(\alpha_2\sin\alpha_3t
-\alpha_3\cos\alpha_3t)x\right)\frac{\partial}{\partial t}
\nonumber\\
& &
+\Biggr(-\delta_4\cos\alpha_3t+\delta_3\sin\alpha_3t
+(\gamma_3\sin\alpha_3t-\gamma_4\cos\alpha_3t)x+\sin\alpha_3t x^2
\nonumber\\
& &   +\frac{1}{(\alpha_2^2+\alpha_3^2)}
(\alpha_2\sin\alpha_3t-\alpha_3\cos\alpha_3t)x^3\Biggr)
\frac{\partial}{\partial x}\Biggr]e^{\alpha_2t},
\nonumber\\
X_7 & = & \Biggr[\left(\beta_5\cos\alpha_3t+\beta_4\sin\alpha_3t
+\frac{1}{(\alpha_2^2+\alpha_3^3)}(\alpha_3\sin\alpha_3t
-\alpha_2\cos\alpha_3t)x\right)\frac{\partial}{\partial t}
\nonumber\\
& &
+\Biggr(\delta_5\cos\alpha_3t+\delta_6\sin\alpha_3t
+(\gamma_5\cos\alpha_3t+\gamma_6\sin\alpha_3t)x+\cos\alpha_3t x^2
\nonumber\\
& &
 +\frac{1}{(\alpha_2^2+\alpha_3^2)}
(\alpha_3\sin\alpha_3t-\alpha_2\cos\alpha_3t)x^3\Biggr)
\frac{\partial}{\partial x}\Biggr]e^{-\alpha_2t},
\nonumber\\
X_8 & = & \Biggr[\left(\beta_4\cos\alpha_3t
-\beta_{5}\sin\alpha_3t
+\frac{1}{(\alpha_2^2+\alpha_3^3)}(\alpha_2\sin\alpha_3t
+\alpha_3\cos\alpha_3t)x\right)\frac{\partial}{\partial t}
\nonumber\\
& &
+\Biggr(\delta_6\cos\alpha_3t-\delta_{5}\sin\alpha_3t
+(\gamma_6\cos\alpha_3t-\gamma_{5}\sin\alpha_3t)x-\sin\alpha_3t x^2
\nonumber\\
& &
 +\frac{1}{(\alpha_2^2+\alpha_3^2)}
(\alpha_2\sin\alpha_3t+\alpha_3\cos\alpha_3t)x^3\Biggr)
\frac{\partial}{\partial x}\Biggr]e^{-\alpha_2t}.
\label{addcon11aa}
\end{eqnarray}
As far the authors knowledge goes {\it the above symmetry generators are
being reported for the first time}.  Eq. (\ref{3.42})
is also linearizable one as it admits eight parameter Lie
point symmetries.  The general solution\cite{Sekar:2004b} of Eq. (\ref{3.42}) reads as
\begin{align}
x(t) = \frac{-3\alpha}{k}+\frac{6}{k}\left(
\frac{(3\alpha^2+\lambda_1)(1-I_1e^{\pm i\hat{\alpha}t})}
{3(1-I_1e^{\pm \hat{\alpha}t})-2(3\alpha^2+\lambda_1)I_2e^{(-3\alpha\pm
\frac{\hat{\alpha}}{2})t}
\pm \hat{\alpha} (1-I_1e^{\pm \hat{\alpha}t})}
\right)
\label{mgen1}
\end{align}
where
\begin{align}
\alpha^3+\alpha \lambda_1-k\lambda_2 = 0,\;\;and \;\;
\hat{\alpha} = \sqrt{-3\alpha^2-4\lambda_1}.
\end{align}
For more details about the derivation of the solution (\ref{mgen1}) one
may refer to Ref.\onlinecite{Sekar:2004b}.

\subsection{\bf Generalized Modified Emden
Equations: $f = k_1x + k_2$}
Next we consider the case where $f$
as the most general linear function of $x$, that is $f=k_1 x +k_2$,
where $k_1$ and $k_2$ are arbitrary constants.  As a consequence we arrive at
\begin{eqnarray}
g=\frac{k_1^2}{9}x^3 +\frac{k_1k_2}{3}x^2+\lambda_1x+\lambda_2,
\label{3.43}
\end{eqnarray}
where $\lambda_1$ and $\lambda_2$ are  arbitrary constants.

Proceeding in a similar fashion as earlier, we consider the following
four cases:
$(i)~\lambda_1 = \lambda_2 = 0$, $(ii)~\lambda_1
\neq 0,\; \lambda_2 = 0$ $(iii)~\lambda_2
\neq 0, ~\lambda_1 = 0,$ and
$(iv)~\lambda_1 \neq \lambda_2 \neq 0$, separately.  The corresponding
equations of motion, respectively, are
\begin{eqnarray}
&& \ddot{x}+(k_1x+k_2)\dot{x}+\frac{k_1k_2}{3}x^2+\frac{k_1^2}{9}x^3 = 0.
 \label{3.45a} \\
&& \ddot{x}+(k_1x+k_2)\dot{x}+\frac{k_1k_2}{3}x^2+\frac{k_1^2}{9}x^3+
 \lambda_1 x = 0.
 \label{3.45b} \\
 &&\ddot{x}+(k_1x+k_2)\dot{x}+\frac{k_1k_2}{3}x^2+\frac{k_1^2}{9}x^3+
 \lambda_2 = 0.
 \label{3.45c} \\
&& \ddot{x}+(k_1x+k_2)\dot{x}+\frac{k_1k_2}{3}x^2+\frac{k_1^2}{9}x^3+
 \lambda_1 x + \lambda_2 = 0.
 \label{3.45d}
\end{eqnarray}
However, all the above equations can be transformed straightforwardly to the
equations discussed in the previous subsections.  For example with the subtitution,
$X=x+\frac{k_2}{k_1}$, Eq. ~(\ref{3.45a}) can be transformed to the form
\begin{eqnarray}
\ddot{X}+k_1X\dot{X}+\frac{k_1^2}{9}X^3-\frac{k_2^2}{3}X+\frac{2}{9} \frac{k_2^3}{k_1} = 0.
\label{3.45e}
\end{eqnarray}
Equation (\ref{3.45e}) coincides exactly with (\ref{3.42}) by restricting the constants $\lambda_1=-\frac{k_2^2}{3}$ and $\lambda_2=\frac{2}{9} \frac{k_2^3}{k_1}$ in the latter. As a result the infinitesimal symmetries and their
associated
symmetry generators can also be derived from the Eqs.~(\ref{addcon11a}) and
(\ref{addcon11aa}).

Eq.~(\ref{3.45b}) can be cast into the form,
\begin{eqnarray}
\ddot{X}+k_1X\dot{X}+\frac{k_1^2}{9}X^3+\frac{(3\lambda_1-k_2^2)}{3}X+
\frac{(2k_2^3-9k_2\lambda_1)}{9k_1} = 0,
\label{3.45f}
\end{eqnarray}
through the transformation $X=x+\frac{k_2}{k_1}$.  Eq. (\ref{3.45f}) is of the form
(\ref{3.42}).  Therefore, the infinitesimal symmetries and the associated symmetry generators of equation (\ref{3.45f}) can also be derived from the Eqs.~(\ref{addcon11a}) and
(\ref{addcon11aa}) by appropriately fixing the constants.

In a similar way, one can transform Eqs.~(\ref{3.45c}) and (\ref{3.45d}) to the
following form:
\begin{eqnarray}
&\ddot{X}+k_1X\dot{X}+\frac{\displaystyle{k_1^3}}{\displaystyle{9}}X^3
-\displaystyle{\frac{k_2^2}{3}X+\frac{2k_2^3}{9k_1}
+\frac{k_1\lambda_2}{\lambda_1}}
 = 0.
\label{3.45g}\\
&\ddot{X}+k_1X\dot{X}+\frac{\displaystyle{k_1^3}}{\displaystyle{9}}X^3
+\displaystyle{(\lambda_1-\frac{k_2^2}{3})X
+\frac{(2k_2^3+9k_1\lambda_2-9k_2\lambda_1)}{9k_1^2}}
 = 0.
\label{3.45h}
\end{eqnarray}
respectively.  Eqs. (\ref{3.45g}) and (\ref{3.45h}) are also of the form
(\ref{3.42}) and hence we can derive the infinitesimal symmetries directly
from the Eq. (\ref{addcon11a}).

\section{Non-maximal Lie point symmetries for $f_{xx} \neq 0 $}
\subsubsection{\bf General forms of $\mathbf{f(x)}$}
One can consider higher degree polynomials or other forms for $f$ such that $(f_{xx}\neq0)$ and classify the equations and their underlying Lie point symmetries. However, our analysis shows that one can get only lesser number of Lie point symmetries than eight whenever
$f_{xx} \neq0$.
This can be proved in a rather general way as given below.
\par Now we assume $f=k_1x+k_2+l(x)$, where $l(x)$ is an arbitrary function and investigate the outcome when $l(x) \neq 0$. With this choice, $g$ takes the form (vide Eq. (\ref{2.16}))
\begin{equation}
g=\frac{1}{3b}\Big[-(\dot b x^2-2bf_2+cx+d)f_x-(\dot a+\dot b x)+4\dot bf_1-3\ddot b x-2\dot c+\ddot a\Big].\label{3.45i}
\end{equation}
In the above we introduced the notation $f_1=\int{f(x)dx}$ and $f_2$=$\int{f_1(x)dx}$=$\int({\int{f(x)dx})dx}$.
Substituting $g$ and its derivative into Eq. (\ref{2.15}) and expanding the latter we arrive at an expression exclusively in terms of $l(x)$, that is
\begin{eqnarray}
& & \frac{4}{3}{bl\Big(l_4}+\frac{k_2}{2}x^2+\frac{k_1}{6}x^3\Big)^2-\frac{ c^2lx^2}{3b}-\frac{2cdlx}{3b}-\frac{ld^2}{3b}+\frac{4}{3} l\Big(cx+d\Big) \Big(\frac{k_2}{2}x^2+\frac{k_1}{6} x^3+l_4\Big)  \nonumber\\
& & \quad +\frac{c\dot a}{3b} \Big(k_2+k_1 x + l_2\Big)-\frac{2\dot a}{3}\Big(k_2x+\frac{k_1}{2}x^2+l_3\Big)\Big(k_2+k_1x+l_2\Big)-\frac{\dot a}{b} \Big(k_1+l_1\Big)\Big(cx+d\Big)\nonumber\\
& & \quad +2\Big(\frac{k_2}{2} x^2+\frac{k_1}{6} x^3+l_4\Big)\Big(k_1+l_1\Big) \dot a
-\frac{2 \dot a}{3b}\Big(k_2+k_1x+l_2\Big)-\frac{4 \dot b c}{3b}\Big(k_2x+\frac{k_1}{2} x^2+l_3\Big)-\frac{2\dot{a}\ddot{b}x}{b} \nonumber\\
& & \quad +\frac{8}{3} \dot{b}\Big(k_2x+\frac{k_1}{2} x^2+l_3\Big)^2+\frac{\dot {b}}{3b}\Big(k_2+k_1x+l_2\Big)\Big(4 c x+3 d\Big)-\frac{\dot{b}cx^2}{b} \Big(k_1+l_1\Big)-\frac{\dot{b}dx}{b} \Big(k_1+l_1\Big)\nonumber\\
& & \quad -4\Big(\frac{k_2}{2} x^2+\frac{k_1}{6} x^3+l_4\Big)  \Big(k_2+k_1x+l_2\Big)\dot{b}  -\frac{2}{3} x \Big(k_2 x+\frac{k_1}{2} x^2+l_3\Big)\Big(k_2+k_1x+l_2\Big) \dot{b}
-\frac{4\dot{a}\dot{c}}{3b}\nonumber\\
& & \quad + 2x\Big(\frac{k_2}{2}x^2+\frac{k_1}{6} x^3+l_4\Big)\Big(k_1+l_1\Big)\dot b-\frac{2}{3}\dot{b}x^2 l\Big(cx+d\Big)+\frac{4}{3} x^2l\dot{b} \Big(\frac{k_2}{2} x^2+\frac{k_1}{6} x^3+l_4\Big)-\frac{x^2 \dot{b}\ddot{b}}{b}\nonumber\\
& & \quad +\frac{8 \dot{a}\dot{b}}{3b}\Big(k_2x+\frac{k_1}{2}x^2+l_3\Big)-\frac{2\dot{a}\dot {b}}{3b}\Big(k_2+k_1x+l_2\Big)x-\frac{\dot{a}\dot{b}}{b}\Big(k_1+l_1\Big)x^2+\frac{\dot {b}^2}{b}\Big(k_2+k_1 x+l_2\Big)x^2\nonumber\\
& & \quad -\frac{\dot{b}^2x^3}{b}\Big(k_1+l_1\Big)-\frac{l\dot{b}^2x^4}{3b}+\frac{2c\dot c}{3b}-\frac{4}{3}\dot{c}\Big(k_2 x+\frac{k_1}{2}x^2+l_3\Big)+\dot{c}x\Big(k_2+k_1 x+l_2\Big)\nonumber\\
& & \quad +\Big(k_2+k_1 x+l_2\Big)\dot{d}-\frac{c\ddot{a}}{3b}+\frac{2}{3} \Big(k_2x+\frac{k_1}{2}x^2+l_3\Big)\ddot{a}+\frac{2\dot{a}\ddot{a}}{3b}-\frac{d\ddot {b}}{b}-2\ddot{b}\Big(k_2x+k_1x^2+l_3\Big)x\nonumber\\
& & \quad  + x^2 \Big(k_2+k_1 x+l_2\Big) \ddot{b}+\ddot{c}x+\ddot{d}
+\dddot{b}x^2=0.
\label{gen1}
\end{eqnarray}

Again in (\ref{gen1}), for the sake of simplicity, we have used the notation $l_1(x)=\int{l(x)dx}$, $l_2(x)=\int{[\int{l(x)dx}]dx}$ and so on.  It is clear from the
above expression that the first term on the left hand side of Eq. (\ref{gen1}), namely
$\frac{4}{3}{bl\Big(l_4}+\frac{k_2}{2}x^2+\frac{k_1}{6}x^3\Big)^2$ cannot be balanced by
no other term.  This in turn leads us to
\begin{equation}
\frac{4}{3}{bl\Big(l_4}+\frac{k_2}{2}x^2+\frac{k_1}{6}x^3\Big)^2 \Longrightarrow
b =0.
\end{equation}
When $l(x)\neq0$ the above expression clearly indicates that the
symmetry function $b(t)$ should be zero.  Since one of the symmetry functions
becomes zero one can get ultimately lesser Lie point symmetries only. This has in fact been
analyzed in detail in Sec. III of paper I.

On the other
hand restricting $l(x)=0$, that is $f = k_1x+k_2$, then one obtains maximal Lie point symmetries as shown in the earlier sections.

\subsection*{\bf 2. Relationship between the symmetries and the forms
of $\mathbf{f(x)}$ and $\mathbf{g(x)}$}

In our discussions so far, we find that all the examples dealt with in Secs. III and
IV admit eight symmetry generators, and each one of the sets form $sl(3,R)$ algebra. It is a
well known fact that second order ODEs with eight symmetry generators are linearizable
by point transformation\cite{ibra:1999,ma}, which was pointed out by Lie himself \cite{Lie}.  In
fact the theorem that provides criteria for testing the linearization of a scalar
second order ODE $\ddot{x}=f(t,x,\dot{x})$ via a point transformation has the cubic in derivative form \cite{ibra:1999,ma}
\begin{eqnarray}
\ddot{x}+A(t,x)\dot{x}^3+B(t,x)\dot{x}^2+C(t,x)\dot{x}+D(t,x) = 0,
\label{lin1}
\end{eqnarray}
with the coefficients $A,B,C$ and $D$ satisfying  the
following two invariant conditions,
\begin{eqnarray}
&&3A_{tt}+3A_tC-3A_xD+3AC_t+C_{xx}-6AD_x+BC_x-2BB_t-2B_{tx}=0,
\label{lin2}\\
&&6A_tD-3B_xD+3AD_t+B_{tt}-2C_{tx}-3BD_x+3D_{xx}+2CC_x-CB_t=0,
\label{lin3}
\end{eqnarray}
where suffixes refer to partial derivatives. On comparing Eq. (\ref{1.1})
with Eq. (\ref{lin1}), we find that the functions $A(t,x) = B(t,x) = 0$
and $C(t,x)$ and $D(t,x)$ are $f(x)$ and $g(x)$ respectively which are
independent of time. Substituting these forms of $C(t,x)$ and $D(t,x)$
along with $A(t,x) = B(t,x) = 0$ in Eqs. (\ref{lin2}) and (\ref{lin3}),
we obtain
\begin{eqnarray}
f_{xx} = 0, \quad 3g_{xx}-2ff_x = 0.
\label{lin4}
\end{eqnarray}
Integrating Eq. (\ref{lin4}), we obtain
\begin{eqnarray}
f = k_1x+k_2, \quad g = \frac{1}{9}k_1k_2x^3+\frac{1}{3}k_2^2x^2+
\lambda_1 x+\lambda_2,
\label{lin5}
\end{eqnarray}
where $\lambda_1,\lambda_2$ are integration constants. Thus for the function
$f_{xx}\ne0$, one obtains lesser parameter symmetries only . However, our analysis leads
to a more general form from which one can determine systems exhibiting lesser
parameter symmetry groups. Thus this analysis broadens our knowledge on both
integrable
and linearizable equations belonging to Eq. (\ref{1.1}).

Note that the above relation (\ref{lin5}) is the same as the one obtained from the analysis of
the determining equations, Eq. (21) for the Lie symmetries in Sec. II A of paper I for the case
$f_{xx} =0$.

\section{Discussion and Conclusions}

In this paper, we have
studied in some detail the symmetry properties of the Li\'enard type equation
(1), where none of the symmetry functions is equal to zero.  We
have also isolated the class of equations which possess eight Lie point
symmetries.  For the sake of completeness we have also given the solutions
for all the nonlinear ODEs which we have identified as linearizable
in this procedure.   We have also indicated the explicit Hamiltonian forms
for these equations.

A question now naturally arises is about the Noether symmetries of these
equations.   As it is well known, Noether's theorem is based upon the invariance
of action integral and essentially one needs a Lagrangian to start the
analysis.  Since our recent studies show that even dissipative systems
can have time independent Lagrangian \cite{pcsl} and so it is worth exploring
Noether symmetries of these systems as well.  Finally, it is also of importance
to study the detailed symmetry structure when the functions $f$ and $g$
in (1) are also time dependent.  These are being pursued currently.
\section*{Acknowledgments}
One of us (SNP) is grateful to the Centre for Nonlinear Dynamics,
Bharathidasan University, Tiruchirappalli, for warm
hospitality.
 The work of SNP forms part of a
Department of Science and Technology, Government of India
sponsored research project. The work of MS forms part of a
research project sponsored by the National Board for Higher
Mathematics, Government of India.  The work of ML forms part of a
Department of Science and Technology (DST), Ramanna Fellowship and is
also supported by a DST-IRHPA research project.

redpajama_set_name: RedPajamaArXiv


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Tau Kē Tēnei Wiki: Cambridge University Engineering Photo Competition
by Heather Hastie | Nov 15, 2015 | Auē a Tau Kē, Science & Nature | 0 comments
These are the winning photos from Cambridge University's 2015 Engineering photo competition. The annual competition is sponsored by Zeiss and always has many beautiful images.
This year's winner, by Rachel Garsed, is called Shot in the Dark. It's of a bullet hole pattern in liquid crystal. Second prize went to Andrew Payne for his image of a titanium "comet," which he called Titanium Comet Diaspora. Dilek Ozgit and Andrea De Luca jointly submitted the third-prize winning image, which is of multi-walled carbon nanotubes.
There were also other prize categories. The Zeiss SEM prize for a micrograph captured using an electron microscope was won by Kenichi Nakanishi. His image, called Under the Covers, was of cave-like formations in graphene. The Head of Department's prize went to Alex Kendall. Kendall developed a system that was able to take a video from a smartphone or similar and translate it so a robot would be able to navigate using the images.
All images described above are in the video below.
There were nine other beautiful photographs that won prizes in the competition, and many other impressive entries, which can be seen on the University's website. Some of my favourites are:
Anthony Rubinstein-Baylis – 3D spirograph
Dilek Ozgit – Nanomaterial Enhanced Rechargeable Zinc-Silver Oxide Batteries
Antonios Kanellopoulos – Cement paste cavity
Who would have though cement or batteries could be beautiful?
From the University of Cambridge website:
The panel of judges included Kenneth Png from ZEISS, and the Department of Engineering's Professor Roberto Cipolla, Dr Allan McRobie, Head of Department Professor David Cardwell, and Director of Research Philip Guildford. Guildford said that the judges were once again impressed by the quality of the images they received. "I love the way in which the essence of engineering can be captured in a single beautiful image – these intriguing works of art convey wonderful stories of determined engineers battling to crack real world problems and finding the most elegant answers," he said.
I didn't know engineers could wax lyrical so well either!
one × one =


text:
China Orders 13 Tech Companies to Overhaul Fintech Operations
By Hu Yue and Han Wei
The government has grown increasingly concerned over the companies' expanding influence over every aspect of Chinese life
China's financial regulators summoned 13 tech companies Thursday including Tencent Holdings Ltd. and ByteDance Ltd., imposing a raft of requirements on their financial businesses in an expanded effort to rein in the fast-growing fintech sector.
Earlier this month, financial regulators met with Jack Ma's Ant Group Co., outlining a rectification plan consisting of similar requirements for the company to follow, including correcting unfair competition practices in its payment business, breaking down the information monopoly and applying to become a financial holding company.
The latest meeting with more internet companies signals that regulators are expanding the campaign to rein in tech titans' aggressive foray into the financial sector. The government has grown increasingly concerned over the companies' expanding influence over every aspect of Chinese life as well as the vast amounts of data they've amassed through providing services like online shopping, chatting and ride-hailing.
"China's fintech industry participants will have to adjust to tighter regulations that will have a broader and stronger oversight, leading to a period of moderating fintech growth and investment in the sector," Yan Li, an analyst with Moody's Investors Service, said in a report. The new regulations and rules will bring risks for fintech companies "as the tightened regulatory environment would require significant changes to the business models of their financial services businesses."
China has rolled out a stream of regulations and rules targeting the tech sector, including measures to curb market concentration in online payments and prevent big tech companies from abusing their market dominance.
Companies called to the meeting Thursday also included the financial units of Baidu Inc., JD.com, Meituan, Didi Chuxing, Sina Corp., Suning.com and Lufax Holding Ltd., according to a statement (link in Chinese) published by the People's Bank of China.
Pan Gongsheng, a deputy governor of the central bank, chaired the meeting, with representatives from the China Banking and Insurance Regulatory Commission, the China Securities Regulatory Commission and the State Administration of Foreign Exchange.
Regulators issued a list of requirements regarding big internet companies' financial service operations, including stricter scrutiny of shareholders and transactions with connected parties, stricter compliance when listing abroad and curbs on information monopolies and the gathering of personal data.
Companies must obtain licenses to launch financial services, sever "improper links" between their payment services and other financial products, comply with rules when making investment in financial institutions and restructure their financial assets into holding companies to bring the businesses under proper supervision, the regulators said.
The meeting followed a series of regulatory measures targeting Ant, which scrapped a planned record initial public offering amid a tightening regulatory environment.
The rectification order will dramatically revamp Ant's business and put it under similar supervision as a bank. The overhaul means Ant will have to cut off any improper linking of payments with other financial products including its Jiebei and Huabei lending services, blowing up a model in which the company used the digital payment function as a port of entry to attract users into more lucrative areas such as consumer lending.
Luo Meihan and Bloomberg contributed to this story.
Contact reporter Han Wei (weihan@caixin.com) and editor Bob Simison (bobsimison@caixin.com)
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In Depth: The Rectification and Remaking of Ant Group
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Call of Cthulhu: The Official Videogame promises to bring creepy, Lovecraftian horror to your Xbox One and PS4 and this is the first gameplay footage. In it you play a private investigator sent to Darkwater Island to investigate the murder of an artist, but as is custom for Lovecraft there are also weird cults, sea monsters and your own fragile sanity to contend with.
The gameplay in the video above shows how Call of Cthulhu plans to mess with your head, both in the more obviously creepy stuff, but also in more subtle ways that will make you question what's real and what isn't. Hint: we think the terrifying ancient tentacle monsters might actually be real, so avoid those.
Call of Cthulhu is scheduled to come out in 2018 on Xbox One, PS4 and PC, with a release date yet to be confirmed.


The Silvermoon has two, beautiful, wood-floored lobbies that can be used with our main meeting spaces or can be their own nicely appointed destination. These lovely spaces have hosted receptions, dinners, weddings, and much more.
The Entrance Gallery at 215 W. Broad, with its special lighting, can host collections of artwork or photography, as well as specially featured sculpture or seasonal displays from time to time.
The adjacent lobby of the Silvermoon Theatre at 217 West Broad is a bit smaller but has its own charm, special lighting, and art display capabilities.
Unless there is a conflict in the theatre space, the two lobby spaces can be used for extending reception capabilities of the Great Hall by adding as many as 100 more guests that can be adequately hosted for such a use. These are the only spaces with the rich wood flooring salvaged from the structural beams that held these buildings together for a century. Take a look at these Lobby images.


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text:
Q: What gives best precision, exponential of the difference or the quotient of the exponentials? Given a typical programming language.
I have two numbers floating point numbers a and b which are close to one another (i.e. their difference is much smaller, in absolute value, than their average).

|a-b| << |a+b|/2

Mathematically, we have that

exp(a-b) = exp(a)/exp(b)
  .

But when you programme, you can choose to first calculate (a-b) and then exponentiate it, or exponentiate a, then b, then divide them.
If a and b are really close to one another, then the precision on (a-b) might be bad.
Example

(1+pi*10^-20) -(1+1.1*pi*10^-20) = -pi*10^-21

but if you are using floating points with only 19 decimal points of precision. You will get zero as an answer, which is a bad precision. You could have gotten a better precision by reordering the operation as follows

(1-1) + (pi*10^-20 -1.1*pi*10^-20) = -pi*10^-21

which will give you -pi*10^-21 with 19 decimal points of precision.
Therefore my question is, given a finite floating point precision,
which way of calculating exp(a-b) gives the better precision?
The exponential of the difference:

exp(a-b)

or the quotient of the exponentials

exp(a)/exp(b)

?

A:
If a and b are really close to one another, then the precision on (a-b) might be bad.

On the contrary, if a and b are close, then a - b is exact (Sterbenz lemma). Therefore exp(a-b) involves only one rounding step (assume a correctly rounded exp function to simplify reasoning). As a consequence, exp(a-b) gives the correctly rounded result for the expression you are trying to compute. Any other way will never be better than this.
Circumstances in which exp(a-b) is dramatically better than the alternative are when exp(a) and exp(b) individually overflow, resulting in NaN for the division. By contrast, exp(a-b) only produces +inf when that's the best floating-point approximation of the mathematical result, and never produces NaN for finite a and b.

Note: the problem that you may be facing is that a and b have been computed so approximately that the relative accuracy of a - b is terrible. This is an inherent problem of floating-point (cancellation), but at the time of computing exp(a-b) in any way, it is too late to try to solve this problem: the information has already been lost. You can only compute exp(a-b) for the a and the b you have. And the correct way to do this is with exp(a-b).

For simple expressions for which one expects extra precision bits to provide extra accuracy for the end result, it is also possible to test these things empirically by computing a reference result with extra precision. If I had approached the question that way, I might have written the C program (for my platform where FLT_EVAL_METHOD is defined by the compiler as 0 and where long double is IEEE 754 80-bit double-extended):
#include <math.h>
#include <stdlib.h>
#include <stdio.h>

double best(double a, double b) {
  return exp(a-b);
}

double other(double a, double b) {
  return exp(a) / exp(b);
}

long double reference(long double a, long double b) {
  // assume the method doesn't matter so much with
  // long double computations, use any method:
  return expl(a-b);
}

int main(void) {
  for (int i = 0; i < 10; i++) {
    double a = rand() ^ ((long long)rand())<<16 ^ ((long long)rand()) << 32;
    a /= 0x1.0p64;
    double b = a * (1 + (double)rand() / (5.0 * RAND_MAX));
    printf("a=%a\nb=%a\n", a, b);

    double be = best(a, b);
    double o = other(a, b);

    long double r = reference(a, b);

    printf("error sub then exp:%La\n", fabsl(r - be));
    printf("error exp then div:%La\n", fabsl(r - o));
  }
}

A run of the above program produces the results:

~ $ gcc ex.c && ./a.out
a=0x1.82def03cebc5p-2
b=0x1.a65bc869d479cp-2
error sub then exp:0xa.dp-60
error exp then div:0xa.dp-60
a=0x1.8164b7a7be6dep-6
b=0x1.b5b82c63fa0bcp-6
error sub then exp:0x8.1p-58
error exp then div:0x8.1p-58
a=0x1.88b779e295f18p-3
b=0x1.b1836e39a5186p-3
error sub then exp:0x8.5p-59
error exp then div:0xd.ecp-57
a=0x1.b5f437ec6fc4ap-6
b=0x1.e459d0a1d64b9p-6
error sub then exp:0xa.1p-59
error exp then div:0xd.7cp-57
a=0x1.889e1b20a7c9dp-3
b=0x1.8dddc5217d12fp-3
error sub then exp:0x9.4p-61
error exp then div:0xf.6cp-57
a=0x1.2d8f07147984cp-2
b=0x1.65acc944015e6p-2
error sub then exp:0x8.ep-58
error exp then div:0x8.ep-58
a=0x1.78b8432157465p-5
b=0x1.a9fd15e131562p-5
error sub then exp:0x9.4p-61
error exp then div:0xf.6cp-57
a=0x1.d214f566a0e1ep-2
b=0x1.0c90cb63e50c4p-1
error sub then exp:0xd.54p-58
error exp then div:0xd.54p-58
a=0x1.78dfa1c5c2ac4p-2
b=0x1.919d3723ae61p-2
error sub then exp:0x9.e8p-59
error exp then div:0x9.e8p-59
a=0x1.fb68c3c57fdd3p-2
b=0x1.103e0374df0bbp-1
error sub then exp:0xd.54p-58
error exp then div:0x9.56p-57

redpajama_set_name: RedPajamaStackExchange


Our assignment was to create a typeface on a 12-square grid for a client we created by choosing a profession and 3 adjectives. My client was a passionate, fun, and optimistic racecar driver. I created 2 spreads in a made-up Auto magazine to showcase the typeface. I also used my own automotive photography for the imagery.
Our assignment was to create a typeface on a 12-square grid for a client we created by choosing a profession and 3 adjectives. My client was a passionate, fun, and optimistic racecar driver. I created 2 spreads in a made-up Auto magazine to showcase the typeface and used my own automotive photography for the imagery.


Barack Obama and his administration can claim that the economy is recovering all they want, but I won't be convinced until I see the unemployment rate begin to decline. The fact is that hundreds of thousands of people continue to lose their jobs every month, which means that finding a job is becoming harder every day. For most people, survival mode comes into play, and just finding something to keep them afloat is necessary.
Submit your resume to Career Builder and Monster. The "employers" that contact you will either be staffinig agencies trying to place you in a job that no one else wants or an employer who is only willing to offer you a "commission only" position.
Be attracted be a job posting that is very vague and/or does not require any experience and/or guarantees a certain monthly salary.
Think that a good job is going to come to you. If you want a job that you enjoy, you need to pursue it, network heavily, and take a step towards getting another job every single day. Not one day can pass without thinking about getting a job that you like.
Identify three different job fields you would like to pursue based on your interests, skill set, and who you know.
Network with the right people. Networking with anyone will get you a job, networking with people in the job fields you have identified will get you a job that you enjoy. Find clubs and organizations, then attend the meetings.
Take full advantage of social media. Don't just build a LinkedIn, Twitter, and Facebook profile with a lot of connections and followers. Produce content on these social networking sites that focuses on the fields you are looking to get into. Quality followers and connections should be more important to you than quantity.
Volunteer at organizations that you like. Volunteering will keep you fresh, build experience, and it could turn into a part-time or full-time gig.
If you don't have an emergency fund cushion to pay your bills, didn't get severance pay, and/or can't apply for unemployment benefits, you should not spend more than a couple of weeks trying to find a job that you enjoy. Your family and your finances come before your wants and desires. If you can't wait long in between jobs before the bills start piling up, quickly get a couple of part-time jobs that mean nothing to you. Don't spend three or four months looking for "the right job" when you can't afford to. But if you can afford it, try the tips I mentioned above, and work hard at it. You'll thank yourself if you land something you enjoy, rather than something that just pays the bills.


To have your beloved angel posted on this page, please fill out the New Angel form below.
PETIE – Because of Petie's gentleness and manners, we called him the Southern Gentleman. He loved his walks and taught us to take time to smell the roses! In just two short years, he tore out a piece of our hearts. Be pain free and run happily, Petie!
ARTHUR – Arthur became my foster dog on 1/26/17 at 12 years old. Found as a stray, deaf and blind. But for 10 months he had a home that loved him. He left for an easier time on 10/19/17.
IVORY – She became my first foster failing experience. Arrive in my home 9/5/11 at 7 years old. Crossed over the Rainbow Bridge on her birthday 9/5/16.
BUDDY HOLMAN – Buddy was my son and best friend. This is my favorite Buddy picture. I call it his class picture.
BECK – Rescued February 6, 2010; said goodbye February 9, 2018.
GUNNER – Noble, sweet and much loved, left this world at 14 years. He gave so much joy to all who knew him. He will be loved and cherished always.
JAKE – You taught us patience, you taught us love, heaven-sent from above. Kindness, courage, that was you. Never was a friend so true. Now, we ask who rescued who? We are full of gratitude. What was started came full 'round; you were lost, we were found.
SKYE and OPAL – You are missed immensely and will forever be in our hearts. we were fortunate to have you in our lives.
PRINCE – Our little old man. Even through our time was brief, you will be forever in our hearts.
DEVON, 2002-2017 , He rescued us. Beautiful loving soul.
PHEOBE (7/6/1997- 11/12/2010)-You will be greatly missed by all of us.
DEACON – Only with us a short time but we loved his gentle spirit and wagging tail. He'll be missed!
GAGE LOUIE – You were our favorite 'hello' and our hardest 'goodbye.' Cherished by the Louie family, 10/5/2008-6/2/2016. Always remembered, forever loved.
MAGGIE SNELL-Maggie loved playing a game of chase around the kitchen island, which was one of the humans walking around the island while Maggie gleefully trotted behind. Her feline brother loved her dearly and always made sure she was ok. If Maggie sneezed, Mac would run to her side to check on her. Maggie never missed a meal but Mac always stood guard to make sure no one would steal a kibble from his girl. She loved standing in the sun and letting the wind blow through her feathers and of course, her favorite thing to do, nap! She was loved by all and dearly missed.
WINSLOW STANLEY- 12/21/2018 Many times I have referred to Winslow as the best dog in the world and to me that is so true. He had a huge heart and his goal in life was to be close (like, in your lap) to the people who loved him. He made me a better person for having known him for three years. I love you more than ever, Wins.
CODY STURGIS – 3/2/19 Sweet dreams dear boy.


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Review: Star Trek: Harlan Ellison's The City on the Edge of Forever: The Original Teleplay
Star Trek: Harlan Ellison's The City on the Edge of Forever: The Original Teleplay by Harlan Ellison
"The City on the Edge of Forever" is widely regarded as one of the best — if not the best — episode of the original Star Trek series. But as every Trekkie worth his replicator credits knows, the version that got filmed was substantially different from the teleplay that science-fiction writer Harlan Ellison turned in. The notoriously prickly Ellison didn't take too kindly to being rewritten, and he's griped for years about how Gene Roddenberry screwed him and his story over, and how much better his version was than the one that viewers saw. Now his original teleplay has been brought to life in a form that gives us an idea of how it might have looked on the small screen if it'd been made the way Ellison wrote it. This graphic novel adaptation, with scripting by brothers Scott and David Tipton and artwork by J.K. Woodward, is an impressive piece of work. Woodward's art is particularly noteworthy, a highly realistic painted style that captures the likenesses of actors William Shatner, Leonard Nimoy, Joan Collins, and Grace Lee Whitney with eerie accuracy. And it's fascinating to see both the parallels and departures from the more familiar television version of the story. There's only one problem: I personally don't think Ellison's version of the story is better (or even as good as) the revised one.
Oh, his ideas were grander than the revision's, to be certain. His Guardians of Forever would've been much cooler than the "stone donut" the TV producers came up with as a cost-saving substitute, and his story features some poignant moments and themes that arguably shouldn't have been left out. But Ellison's teleplay also includes some really hackneyed space pirates, a lot of unnecessary characters (which of course would've cost money in the form of additional actors who need to be paid), and some cringe-worthy "far-out" sci-fi jargon that sounds like it came straight out of the rocket-ship movies of the 1950s instead of the more naturalistic style Star Trek was going for in the 1960s.
Also, I was deeply troubled by Ellison's misunderstanding of the familiar characters. While it was great to see Yeoman Rand do some butt-kicking instead of playing the helpless female she so often was in the TV series, Spock comes across as a condescending, peevish, frankly kind of bitchy antagonist to Kirk. To be fair, Ellison probably wrote this before the series had really nailed down Spock's characterization, but with the benefit of hindsight, this version of Spock is just flat-out wrong… except in the final scene when he tries to console his heartbroken captain. That scene works beautifully. But in general, Ellison's teleplay, while entertaining and emotionally effective, feels more like an episode of The Outer Limits (which Ellison also wrote for) than Star Trek. It's not that it's bad, because it's not… it's very good as a science-fiction story. It's just not very good Star Trek, if that makes sense.
Still, this graphic novel is well worth checking out for the artwork and the glimpse of what might have been. It includes all the variant covers by Juan Ortiz and Paul Shipper from the single-issue comic run, as well as an afterword that reveals all the "Easter eggs" the writers and artist slipped in. (Watch for an appearance by Ellison himself as "Trooper," a character I'm deeply ambivalent about, because I don't think the story needed him — as Ellison has always claimed — but I do like him and his interactions with Kirk.)
Posted in: Reviews, Star Trek, The Bookshelf
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We all come in to this world, magically, some would say, with EXPECTATIONS! Where did they come from anyways. I have taken some time this summer to observe the world from an outsiders perspective and noticed a common theme. Everyone has expectations, and many people have disappointment over these expectations not being met.
Disappointment, anger, depression, stress, anxiety, all stem from somewhere right??? Yes, brain chemistry has something to do with it. Of course, there is the theory we learn maladaptive thinking patterns from our family as well. Although, it seems all of this is exacerbated by EXPECTATIONS!
I often hear from my beloved clients these words "I didn't think it was going to be like this", " I thought it was going to be easier", "It shouldn't have turned out this way". Followed by depression, anxiety, and of course STRESS...with more expectations after all of that, and you have a vicious cycle my dear friend.
Why do we believe it was supposed to be more than it is. Why do we make these assumptions? It can sometimes be a set up for disappointment. Or maybe a driving force for improvement? Let's think it over and find ways to balance this concept out.
CBT Re-frame- I have new problems now, although I am more in control this time and have more life experience to get through them easier that I did before.
This is destination happiness thinking GRRR haha we all do that right?