% SOLUTIONS.TEX **********************************************
% Chaos and time-series analysis
% J. C. Sprott 2000
% ************************************************************

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\begin{document}

\title{Chaos and Time-Series Analysis}
\author{Julien Clinton Sprott}
\maketitle
\pagenumbering{roman}   % Roman numbering for frontmatter
\setcounter{page}{5}    % Start on page v

\chapter*{Preface}\addcontentsline{toc}{chapter}{Preface}

To be written

\chapter{}

% CHAPTER 2***********
\chapter{}
% ********************
%2.1
\section{}
Change the variable  $X_{n}$ to $X_{n}=\left(
\dfrac{Y_{n}}{A}+\dfrac{1}{2}\right) $. Substitution in (2.4)
yields:\\
$\dfrac{Y_{n+1}}{A}+\dfrac{1}{2}=A\left(
\dfrac{Y_{n}}{A}+\dfrac{1}{2}\right) \left(
1-\dfrac{Y_{n}}{A}-\dfrac{1}{2}\right) =A\left(
\dfrac{1}{4}-\dfrac{Y_{n}^{2}}{A^{2}}\right) $. Multiply both
sides by $A$ to get\ :  $Y_{n+1}=B-Y_{n}^{2}$ with
$B=\dfrac{A^{2}}{4}-\dfrac{A}{2}$

%2.2
\section{}
Fixed point: $X^{\ast }=1-1/2.8\simeq 0.643$ \\
The first ten iterations are: \\
$x_{1}=0.631$ , $x_{2}=0.652$ , $x_{3}=0.635$ , $x_{4}=0.649$ ,
$x_{5}=0.638$ , $x_{6}=0.647$ \\
$x_{7}=0.640$ , $x_{8}=0.645$ , $x_{9}=0.641$ , $x_{10}=0.644$
which indicate an oscillation about $X^{\ast }$

%2.3
\section{}
Let $f(X)=AX(1-X)$ and consider $X_{0}=X^{\ast }+\varepsilon _{0}$
, with $\varepsilon_{0}$ very small and positive. Then
$X_{1}=f(X_{0}) =\allowbreak AX^{\ast }-AX^{\ast 2}-2AX^{\ast
}\allowbreak \varepsilon _{0}+A\varepsilon _{0}-A\varepsilon
_{0}^{2}$.\\
$X^{\ast }$satisfies $X^{\ast }=AX^{\ast }-AX^{\ast 2}$ which
gives $X_{1}=X^{\ast }-2AX^{\ast }\allowbreak
\varepsilon _{0}+A\varepsilon _{0}-A\varepsilon _{0}^{2}$. \\
The $-A\varepsilon _{0}^{2}$ term is smaller than the $\varepsilon
_{0}$ terms and may be neglected. Hence \\
$X_{1}=X^{\ast}+\varepsilon _{0}A\left( 1-2X^{\ast }\right)
\allowbreak =X^{\ast }+\varepsilon _{0}f^{\prime }(X^{\ast })$.
$X_{1}$ may be written as $X_{1}=X^{\ast}+\varepsilon _{1}$ \\
thus giving $\varepsilon _{1}=\varepsilon _{0}f^{\prime }(X^{\ast
})$. After $n$ iterations: $\varepsilon _{n}=\varepsilon
_{0}\left(f^{\prime }(X^{\ast })\right) ^{n}$\\
Hence the fixed point is unstable when $\left| f^{\prime }(X^{\ast
})\right| >1$. For $f^{\prime }(X^{\ast })>1$ all $\varepsilon $'s
are positive which means $X\rightarrow +\infty $ as $n\rightarrow
+\infty $. This is not the case for negative slopes since the
$\varepsilon $'s can both be positive and negative depending
whether $n$ is even or odd.

%2.4
\section{}
See text on page 17.

%2.5
\section{}

The second iterate map is given by $g(X)=f(f(X))$ with
$f(X)=AX(1-X).$Let $p$ and $q$ be the fixed points of the map
(these two points are given by eqn 2.7). \\
$\left. g^{\prime }(X)\right| _{X=p}=\left. f^{\prime
}(f(X))f^{\prime }(X)\right| _{X=p}=f^{\prime }(f(p))f^{\prime
}(p)=f^{\prime }(q)f^{\prime
}(p)$ \\
$=A^{2}(1-2p)(1-2q)=A^{2}\lbrack 1-2(p+q)+4pq\rbrack $ \\
From eqn 2.6 observe that $p+q=\lbrack A(A+1)\rbrack /A^{2}$ and
$pq=(A+1)/A^{2}$ \\
Hence $\left. g^{\prime }(X)\right| _{X=p}=A^{2}\lbrack
1-2A(A+1)/A^{2}+4\left( A+1\right) /A^{2}\rbrack =4+2A-A^{2}$
\\
Eqn 2.8 is found by setting $g^{\prime }(X)=-1$.In general the
2-cycle is stable when $\left| g^{\prime }(X)\right| <1$ ie when
$\ 3<A<1+\sqrt{6}$

%2.6
\section{}
?

%2.7
\section{}
The map  $f(X)=4X(1-X)$ has pre-images $X_{A,B}=\dfrac{1}{2}(1\pm
\sqrt{1-X})$. Then $f^{\prime }(X_{A,B})=\mp 4\sqrt{1-X}$ and
$P(X_{A,B})=\dfrac{2}{\pi \sqrt{X}}$ . Use these two equations in
eqn 2.12 to obtain $P(X)$ as in eqn 2.11

%2.8
\section{}
?

%2.9
\section{}
The pre-images of $X_{n+1}=f(X_{n})=A\min (X_{n},1-X_{n})$ are
$X_{A}=X/A$ and $X_{B}=1-X/A$ . Use these together with the fact
that $\left| f^{\prime }(X_{A,B})\right| =1$ in eqn 2.12 to obtain
$P(X)=1/A.$

%2.10
\section{}
Let $f(X)=A\left( 1-\left| 2X-1\right| ^{\alpha }\right) .$ Then
$f^{\prime }(X)=-2\alpha A$sign$(2X-1)\left| 2X-1\right| ^{\alpha
-1}$. Obviously $f^{\prime }\left( 1/2\right) $ is discontinuous
for $\alpha \leq 1$ (not defined for $\alpha <1$ and having a jump
discontinuity for $\alpha =1$). The second derivative is given by:\\
$f^{\prime \prime }(X)=-2\alpha \left( \alpha -1\right)
A$(sign$(2X-1))^2\left| 2X-1\right| ^{\alpha -2}=-2\alpha \left(
\alpha -1\right) A\left| 2X-1\right| ^{\alpha -2}$.\\
$f^{\prime\prime }(1/2)$ is not defined for $\alpha <2$ , but for
$\alpha \geq 2$, $f^{\prime\prime }$ is continuous everywhere and
therefore $f$ is smooth (Note: sign$\left( x\right) =1$ for $x\geq
0$ and $-1$ for $x<0$).

%2.11
\section{}
From Exercise 2.10 the derivative at 0 is given by $f^{\prime
}(0)=2\alpha A\left| -1\right| ^{\alpha -1}$. Since $\alpha <1/2$
and $0<A<1$ we may conclude that $\left| f^{\prime }(0)\right| <1$
and hence that the point X=0 is stable. The existence of a stable
point eliminates the possibility of presence of chaos in the map.

% CHAPTER 3***********
\chapter{}
% ********************
%3.1
\section{}
The solution to eqn 3.1 is of the form $x_{1}=x_{0}e^{at}$, where
$x_{0}$ is the initial condition. A nearby trajectory starting
from $x_{0}+\varepsilon $ with $\varepsilon $ small is given by
$x_{2}=\left( x_{0}+\varepsilon \right) e^{at}$. Their separation
$\Delta x=x_{2}-x_{1}=\varepsilon e^{at}$ increases exponentially
at a rate $a.$

%3.2
\section{}
The solution $x=\dfrac{1}{1+\left( \dfrac{1}{x_{0}}-1\right)
e^{-at}}$ has derivative $\dfrac{dx}{dt}=\dfrac{a\left(
\dfrac{1}{x_{0}}-1\right) e^{-at}}{\left[ 1+\left(
\dfrac{1}{x_{0}}-1\right) e^{-at}\right] ^{2}}$. Use $x$ and its
derivative in the expression  $\dfrac{dx}{dt}-ax\left( 1-x\right)
$. Verifying that this expression equals 0 proves that eqn 3.3
satisfies eqn 3.2.

%3.3
\section{}
This equation is a separable 1st order ODE. It can be written in
the form $\csc \left( x\right) dx=dt$, but the solution can be
given only in implicit form. \vspace{1pt}Integrating both sides
and applying the initial condition $x\left( 0\right) =x_{0}$ we
obtain
:\\
$\ln \left[ \left( \csc x-\cot x\right) /\left( \csc x_{0}-\cot
x_{0}\right) \right] =t$ \ $\Leftrightarrow \csc x-\cot x=\left(
\csc x_{0}-\cot x_{0}\right) e^{t}$

%3.4
\section{}
$dx/dt=ay$ and $dy/dt=-bx$ . Differentiate the first equation with
respect to time and substitute $dy/dt$ to get
$d^{2}x/dt^{2}+abx=0$. This can have a solution of the form
$x=x_{0}\sin \left( \sqrt{ab}t\right) $. Substitution in the
second equation yields $dy/dt=-bx_{0}\sin \left( \sqrt{ab}t\right)
.$ Integrating once we get $y=x_{0}\sqrt{\tfrac{b}{a}}\cos \left(
\sqrt{ab}t\right) $. The equations for $x$ and $y$ generate points
that lie on an ellipse with $\tfrac{x_{\max }}{y_{\max
}}=\sqrt{\tfrac{a}{b}}$ .

%3.5
\section{}
$v\left( t\right) =\dfrac{dx}{dt}=v_{0}\cos \omega
t+\dfrac{A\Omega }{\omega ^{2}-\Omega ^{2}}\cos \Omega t$ and
$\dfrac{d^{2}x}{dt^{2}}=-\omega v_{0}\sin \omega t-\dfrac{A\Omega
^{2}}{\omega ^{2}-\Omega ^{2}}\sin \Omega t$. Using the
expressions for $x$ and $d^{2}x/dt^{2}$in the left hand side of
ean 3.10 we finally obtain $A\sin \Omega t$ which equals the right
hand side.

%3.6
\section{}
$\tfrac{d^{2}x}{dt^{2}}=-v_{0}\frac{-\omega ^{4}e^{-\omega
^{2}\frac{t}{b}}+b^{4}e^{-bt}}{b^{3}}$ and
$\tfrac{dx}{dt}=v_{0}\frac{-\omega ^{2}e^{-\omega
^{2}\frac{t}{b}}+b^{2}e^{-bt}}{b^{2}}$. Let $\Delta
=\tfrac{d^{2}x}{dt^{2}}+b\tfrac{dx}{dt}+\omega ^{2}x=-v_{0}\omega
^{2}\frac{-\omega ^{2}e^{-\omega
^{2}\frac{t}{b}}+b^{2}e^{-bt}}{b^{3}}$.Then comparing $\Delta $
with each term from eqn 3.18 we get $\left| \Delta
/\tfrac{d^{2}x}{dt^{2}}\right| =\allowbreak \frac{\omega
^{2}e^{-\omega ^{2}\frac{t}{b}}-b^{2}e^{-bt}}{b^{2}e^{-\omega
^{2}\frac{t}{b}}-b^{2}e^{-bt}}\ll \frac{\omega ^{2}e^{-\omega
^{2}\frac{t}{b}}-b^{2}e^{-bt}}{\omega ^{2}e^{-\omega
^{2}\frac{t}{b}}-b^{2}e^{-bt}}=1$ , $\left| \Delta /\left(
b\tfrac{dx}{dt}\right) \right| =\frac{\omega ^{2}}{b^{2}}\ll
\frac{1}{4}$ and $\left| \Delta /\left( \omega ^{2}x\right)
\right| =\frac{-\omega ^{4}e^{-\omega
^{2}\frac{t}{b}}+b^{2}\allowbreak \omega ^{2}e^{-bt}}{-\omega
^{4}e^{-\omega ^{2}\frac{t}{b}}+b^{4}e^{-bt}}\ll \frac{-\omega
^{4}e^{-\omega ^{2}\frac{t}{b}}+b^{4}\allowbreak e^{-bt}}{-\omega
^{4}e^{-\omega ^{2}\frac{t}{b}}+b^{4}e^{-bt}}=1$. Since each term
is much less than 1 we may neglect $\Delta ,$ thus verifying that
the eqn 3.21 satisfies eqn 3.18 .

%3.7
\section{}
$x=v_{0}te^{-bt/2}$,
$v(t)=dx/dt=v_{0}e^{-\frac{1}{2}bt}-\frac{1}{2}v_{0}tbe^{-\frac{1}{2}bt}$
and
$d^{2}x/dt^{2}=-v_{0}be^{-\frac{1}{2}bt}+\frac{1}{4}v_{0}tb^{2}e^{-\frac{1}{2}bt}$
. Substitute these in the right hand side of eqn 3.18 to verify
that it equals zero.

%3.8
\section{}
$x=\left( v_{0}/\omega \right) e^{-\frac{1}{2}bt}\sin \omega t$ ,
$v(t)=dx/dt=-\frac{1}{2}v_{0}e^{-\frac{1}{2}bt}\frac{b\sin \omega
t-2\left( \cos \omega t\right) \omega }{\omega }$ and \\
$d^{2}x/dt^{2}=\frac{1}{4}v_{0}e^{-\frac{1}{2}bt}\frac{b^{2}\sin
\omega t-4b\left( \cos \omega t\right) \omega -4\left( \sin \omega
t\right) \omega ^{2}}{\omega }$. Substitution in eqn 3.18 yields\\
$\Delta =d^{2}x/dt^{2}+bdx/dt+\omega
^{2}x=-\frac{1}{4}v_{0}e^{-\frac{1}{2}bt}b^{2}\frac{\sin \omega
t}{\omega }$. Compare $\Delta $ with each term from eqn 3.18 to
get : $\left| \Delta /\left( \omega ^{2}x\right) \right|
=\dfrac{\frac{1}{4}v_{0}e^{-\frac{1}{2}bt}b^{2}\frac{\sin \omega
t}{\omega }}{v_{0}\omega e^{-\frac{1}{2}bt}\sin \omega
t}=\allowbreak \frac{1}{4}\frac{b^{2}}{\omega ^{2}}\ll 1$, \\
$\left| \dfrac{\Delta }{b\tfrac{dx}{dt}}\right|
=\dfrac{\frac{1}{4}v_{0}e^{-\frac{1}{2}bt}b^{2}\left| \frac{\sin
\omega t}{\omega }\right|
}{\frac{1}{2}v_{0}be^{-\frac{1}{2}bt}\left| \frac{b\sin \omega
t-2\left( \cos \omega t\right) \omega }{\omega }\right|
}<\dfrac{\frac{1}{4}b^{2}\left| \frac{\sin \omega t}{\omega
}\right| }{b}=\allowbreak \frac{1}{4}b\left| \frac{\sin \omega
t}{\omega }\right| \leq \frac{1}{4}\dfrac{b}{\omega }\ll 1$ and
$\left| \Delta /\tfrac{d^{2}x}{dt^{2}}\right|
=\dfrac{\frac{1}{4}v_{0}e^{-\frac{1}{2}bt}b^{2}\left| \frac{\sin
\omega t}{\omega }\right|
}{\frac{1}{4}v_{0}e^{-\frac{1}{2}bt}\left| \frac{b^{2}\sin \omega
t-4b\left( \cos \omega t\right) \omega -4\left( \sin \omega
t\right) \omega ^{2}}{\omega }\right| }<\dfrac{b^{2}\left|
\frac{\sin \omega t}{\omega }\right| }{\left| 4\omega \sin \omega
t\right| }=\allowbreak \frac{1}{4}\frac{b^{2}}{\omega ^{2}}\ll 1$.
Since $\Delta $ is much smaller than each of the terms in eqn 3.18
we may consider the eqn 3.23 roughly satisfying eqn 3.18.

%3.9
\section{}
$x=\frac{A\sin \left( \Omega t-\phi \right) }{\sqrt{\left( \omega
^{2}-\Omega ^{2}\right) ^{2}+b^{2}\Omega ^{2}}}$, $v(t)=dx/dt=$
$\frac{A\Omega \cos \left( \Omega t-\phi \right) }{\sqrt{\left(
\omega ^{2}-\Omega ^{2}\right) ^{2}+b^{2}\Omega ^{2}}}$and
$d^{2}x/dt^{2}=-\frac{A\Omega ^{2}\sin \left( \Omega t-\phi
\right) }{\sqrt{\left( \omega ^{2}-\Omega ^{2}\right)
^{2}+b^{2}\Omega ^{2}}}$. Substitution in the right hand side of
eqn 3.17 yields:\\
 $\dfrac{d^{2}x}{dt^{2}}+b\dfrac{dx}{dt}+\omega
^{2}x=A\frac{\left( \omega ^{2}-\Omega ^{2}\right) \sin \left(
\Omega t-\phi \right) +b\Omega \cos \left( \Omega t-\phi \right)
}{\sqrt{\left( \omega ^{2}-\Omega ^{2}\right) ^{2}+b^{2}\Omega
^{2}}}$ $\ \ \ \ \ \ \ \ \left( \ast \right) $. \\
Use eqn 3.25 and the trigonometric identities $\cos \phi
=1/\sqrt{1+\tan ^{2}\phi }=$ $\tfrac{\omega ^{2}-\Omega
^{2}}{\sqrt{\left( \omega ^{2}-\Omega ^{2}\right) ^{2}+b^{2}\Omega
^{2}}}$and $\sin \phi =\sqrt{1-\cos ^{2}\phi }=\tfrac{b\Omega
}{\sqrt{\left( \omega ^{2}-\Omega ^{2}\right) ^{2}+b^{2}\Omega
^{2}}}$. Eqn $\left( \ast \right) $ can then be written as:\\
$d^{2}x/dt^{2}+bdx/dt+\omega ^{2}x=A\left( \cos \phi \sin \left(
\Omega t-\phi \right) +\sin \phi \cos \left( \Omega t-\phi \right)
\right) =A\sin \Omega t$.

%3.10
\section{}
$x_{m}=\frac{A}{\sqrt{\left( \omega ^{2}-\Omega ^{2}\right)
^{2}+b^{2}\Omega ^{2}}}$ and the derivative
$\dfrac{dx_{m}}{d\omega }=\frac{-A\Omega \left( -2\omega
^{2}+2\Omega ^{2}+b^{2}\right) }{\left( \sqrt{\left( \omega
^{4}-2\omega ^{2}\Omega ^{2}+\Omega ^{4}+b^{2}\Omega ^{2}\right)
}\right) ^{3}}$. It becomes zero for $\Omega
=\frac{1}{2}\sqrt{\left( 4\omega ^{2}-2b^{2}\right) }$ and $\Omega
=-\frac{1}{2}\sqrt{\left( 4\omega ^{2}-2b^{2}\right) }$. $x_{m}$
becomes maximum for $\Omega =\frac{1}{2}\sqrt{\left( 4\omega
^{2}-2b^{2}\right) }$ and the maximum is  $\frac{2A}{b\sqrt{\left(
4\omega ^{2}-b^{2}\right) }}$.

%3.11
\section{}
Find the $\Omega $'s for which
$\dfrac{x_{m}}{A}=\frac{1}{\sqrt{\left( \omega ^{2}-\Omega
^{2}\right) ^{2}+b^{2}\Omega ^{2}}}=\dfrac{1}{\sqrt{2}b\omega
}\Leftrightarrow$ \\
$2b^{2}\omega ^{2}=\left( \omega ^{2}-\Omega ^{2}\right)
^{2}+b^{2}\Omega ^{2}$. Then $\Omega ^{4}+\left( b^{2}-2\omega
^{2}\right) \Omega ^{2}+\omega ^{4}-2b^{2}\omega
^{2}=0\Leftrightarrow \left( \Omega /\omega \right) ^{4}+\left(
\left( b/\omega \right) ^{2}-2\right) \left( \Omega /\omega
\right) ^{2}+1-2\left( b/\omega \right) ^{2}=0$. The $\Omega $'s
satisfy $\left( \tfrac{\Omega _{1}}{\omega }\right) ^{2}+\left(
\tfrac{\Omega _{2}}{\omega }\right) ^{2}=2-\tfrac{1}{Q^{2}}$ and
$\left( \tfrac{\Omega _{1}}{\omega }\right) ^{2}\left(
\tfrac{\Omega _{2}}{\omega }\right)
^{2}=1-\tfrac{2}{Q^{2}}\Rightarrow \tfrac{\Omega _{1}}{\omega
}\tfrac{\Omega _{2}}{\omega }\simeq 1-\tfrac{1}{Q^{2}}$ for large
$Q$. Hence $\left( \tfrac{\Delta \Omega }{\omega }\right)
^{2}=\left( \tfrac{\Omega _{1}}{\omega }\right) ^{2}+\left(
\tfrac{\Omega _{2}}{\omega }\right) ^{2}-2\tfrac{\Omega
_{1}}{\omega }\tfrac{\Omega _{2}}{\omega }\simeq
2-\tfrac{1}{Q^{2}}-2\left( 1-\tfrac{1}{Q^{2}}\right)
=\tfrac{1}{Q^{2}}\Rightarrow \tfrac{\Delta \Omega }{\omega
}=\tfrac{1}{Q}$

%3.12
\section{}
Start with $x^{2}+v^{2}=$ $x^{2}+\left( \tfrac{dx}{dt}\right)
^{2}\simeq 4$ and take the derivative at each side to obtain
$2x\tfrac{dx}{dt}+2\tfrac{dx}{dt}\tfrac{d^{2}x}{dt^{2}}\simeq
0\Leftrightarrow x+\tfrac{d^{2}x}{dt^{2}}\simeq 0$ which
corresponds to eqn 3.26 in the limit as $b\rightarrow 0.$

%3.13
\section{}
Use $v$=$\tfrac{dx}{dt}$ to write eqn3.26 as
$x+\tfrac{dv}{dt}+bv\left( v^{2}+x^{2}-1\right) =0\Rightarrow$ \\
$vx+v\tfrac{dv}{dt}+bv^{2}\left( v^{2}+x^{2}-1\right)
=0\Rightarrow $  $\tfrac{1}{2}\tfrac{d}{dt}\left(
x^{2}+v^{2}\right) +bv^{2}\left( v^{2}+x^{2}-1\right) =0$ \ which
is apparently satisfied by $v^{2}+x^{2}=1,$ the limit cycle.

%3.14
\section{}
?

%3.15
\section{}
SAME AS Prob 3.13

%3.16
\section{}
$f=x-y-x^{3}$ and $g=x-x^{2}y$ . At $\left( 0,0\right) $ we have
$dx/dt=dy/dt=0$ and the Jacobian evaluated there is: \\
$$ J(0,0) = \left(
       \begin{array}{cccc}
        1 & 0 \\
        1 & 0 \\
       \end{array}
     \right) $$
The eigenvalues can then be found to be $\lambda =0$ and $1$ from
which we may conclude that the origin is at an unstable
equilibrium. For solutions approximately on the unit circle ie
satisfying $x^{2}+y^{2}\simeq 1$ we have $dx/dt\simeq x\left(
1-x^{2}\right) -y=xy^{2}-y.$ Multiply $dx/dt$ by $x$ and $dy/dt$
by $y$ to get: \\
$x\left( dx/dt\right) \simeq x^{2}y^{2}-xy$ and $y\left(
dy/dt\right) =yx-x^{2}y^{2}.$ Adding those together we get
$x\left( dx/dt\right) +y\left( dy/dt\right)
=\tfrac{1}{2}\tfrac{d}{dt}\left( x^{2}+y^{2}\right) \simeq 0$ .
Therefore the points near the unit circle tend to remain close to
it since the rate of change of their distance from the origin is
close to zero.

%3.17
\section{}
Write eqn 3.18 as a system of 1st order ODEs ie
$\tfrac{dy}{dt}=-by-\omega ^{2}x=-y-x=g(x,y)$ and
$\tfrac{dx}{dt}=y=f(x,y).$ The equations corresponding to 3.33 and
3.34 with $h=1$ are $x_{n+1}=x_{n}+y_{n}$ and $y_{n+1}=-x_{n}$.
These two can be written in a single equation
$x_{n+1}=x_{n}-x_{n-1}$ \ $\left( \ast \right) $. Starting with
$x_{0}=y_{1}=\beta $ and $x_{1}=\alpha $ iterate $\left( \ast
\right) $ to get $x_{2}=\alpha -\beta $ ,  $x_{3}=-\beta $ ,
$x_{4}=-\alpha $ , $x_{5}=-\alpha +\beta $ , $x_{6}=\beta =$
$x_{0}.$

%3.18
\section{}
Total error: $e(h,t,\epsilon )=th+\epsilon
\sqrt{\tfrac{t}{h}}\Rightarrow \tfrac{\partial e}{\partial
h}=t-\tfrac{1}{2}\epsilon \sqrt{\tfrac{t}{h^{3}}}$ . For minimum
set $\tfrac{\partial e}{\partial h}=0$ to get
$h=\root{3}\of{\tfrac{\epsilon ^{2}}{4t}}$. For this value of $h$
, $\tfrac{\partial ^{2}e}{\partial h^{2}}>0$ which guarantees that
$h=\root{3}\of{\tfrac{\epsilon ^{2}}{4t}}$ gives the minimum.

%3.19
\section{}
The equations become $x_{n+1}=x_{n}+h\left(
\tfrac{y_{n+1}+y_{n}}{2}\right) $ and $y_{n+1}=y_{n}+h\left(
-\tfrac{x_{n+1}+x_{n}}{2}\right) $ which give $x_{1}=x_{0}+h\left(
\tfrac{y_{1}+y_{0}}{2}\right) $ and $y_{1}=y_{0}-h\left(
\tfrac{x_{1}+x_{0}}{2}\right) $ . Solving this system we obtain
$y_{1}=\tfrac{-4h}{4+h^{2}}$ and $x_{1}=\tfrac{4-h^{2}}{4+h^{2}}$.
These give $y_{1}^{2}+x_{1}^{2}=\tfrac{\left( 4-h^{2}\right)
^{2}+16h^{2}}{4+h^{2}}=1$ , so the error is zero.

%3.20
\section{}
$\tfrac{dy}{dt}=-by-\omega ^{2}x=g(x,y)$ \ and
$\tfrac{dx}{dt}=y=f(x,y).$ Then :\\
$x_{n+1}=x_{n}+hf\left(\tfrac{x_{n+1}+x_{n}}{2},\tfrac{y_{n+1}+y_{n}}{2}\right)
$ and $y_{n+1}=y_{n}+hg\left(
\tfrac{x_{n+1}+x_{n}}{2},\tfrac{y_{n+1}+y_{n}}{2}\right) $ which
give $x_{n+1}=x_{n}+h\tfrac{y_{n+1}+y_{n}}{2}$ and
$y_{n+1}=y_{n}+h\left( -b\tfrac{y_{n+1}+y_{n}}{2}-\omega
^{2}\tfrac{x_{n+1}+x_{n}}{2}\right) .$

%3.21
\section{}
If we use the leapfrog method to the $x$ variable once the same
results are obtained as in the text on page 46, namely $x_{1}=1$
and $y_{1}=-h$ which as before gives an error of the order of
$h^{2}$ since $\Delta R=h^{2}/2.$ However when we use the leapfrog
method to the $y$ variable we get $x_{1}=1-h^{2}$ and  $y_{1}=-h$
which give \\
$x_{1}^{2}+y_{1}^{2}=1-h^{2}+h^{4}\simeq 1-h^{2}.$ We
also get an error of the order of $h^{2},$ but in this case
$\Delta R=-h^{2}/2.$

%3.22
\section{}
Let $P_{n}=\left( X_{n},Y_{n}\right) $ and $P_{0}=\left(
1,0\right) .$ This problem is solved in two ways:\\
1. Using the leapfrog method in the $x$ variable:\\
The equations become: $X_{n+1}=X_{n}+Y_{n}$ and $Y_{n+1}=Y_{n}
-X_{n+1}.$ Then \\
$P_{1}=\left( 1,-1\right) $ , $P_{2}=\left( 0,
-1\right) $ , $P_{3}=\left( -1,0\right) $ ,
$P_{4}=\left( -1,1\right) $ , $P_{5}=\left( 0,1\right) $ ,
$P_{6}=P_{0}=\left( 1,0\right) $.\\
2. Using the leapfrog method in the $y$ variable:\\
The equations are: $Y_{n+1}=Y_{n}-X_{n}$ and $X_{n+1}=
X_{n}+Y_{n+1}$ and we get \\
$P_{1}=\left( -1,0\right) $ , $P_{2}=\left( -1,-1\right) $ ,
$P_{3}=\left( 0,-1\right) $ , $P_{4}=\left( 1,0\right) $ ,
$P_{5}=\left( 1,1\right) $ , $P_{6}=P_{0}=\left( 1,0\right) .$\\
In either case we still get a solution of period 6 which is
comparable to the period 2$\pi \simeq 6.28$ of the exact solution.

%3.23
\section{}
?

%3.24
\section{}
?

% CHAPTER 4***********
\chapter{}
% ********************

%4.1
\section{}
\includegraphics[width=2.5in]{fig4-1.eps} \\

%4.2
\section{}
Expand eqn 4.5 to get $\lambda ^{2}-\left( a+d\right) \lambda
+ad-bc=0$ and then use the quadratic formula to obtain the two
eigenvalues.

%4.3
\section{}
DONE IN TEXTBOOK ON PAGE 58

%4.4
\section{}
The Jacobian is given by: $ J = \left(
       \begin{array}{cccc}
        0 & 1 \\
        -1 & 0 \\
       \end{array}
     \right) $
whose characteristic polynomial is $\lambda ^{2}+1=0$ . Hence
$\lambda =\pm i$

%4.5
\section{}
The eigenvalues given by eqn 4.12 can be written as $\lambda
_{1,2}=-\tfrac{b}{2}\left( 1\pm \sqrt{1-\left( \tfrac{2\omega
}{b}\right) ^{2}}\right) $.\\
The term under the square root can never exceed 1, so the two
eigenvalues can either be complex conjugates or real with the same
sign. Thus a saddle point cannot occur since this requires
$\lambda _{1}$ and $\lambda _{2}$ to be real with opposite signs.
For $b>2\omega $ we get stable nodes $\left( \lambda _{1}<\lambda
_{2}<0\right) $ if $b$ is positive and unstable nodes if $b$ is
negative $\left( \lambda _{1}>\lambda _{2}>0\right) \ $. When
$b=2\omega $ we either get a degenerate stable node (for $b>0$ ) \
or a degenerate unstable node (for $b<0$ ). When $b<2\omega $ the
term in the square root becomes negative thus giving complex
solutions. Under this condition we get an unstable focus for $b<0$
and a stable focus for $b>0.$

%4.6
\section{}
If the arrows are in the direction towards the spiral and radial
points, each of these points are stable, otherwise they are
unstable. Doing the same on the saddle point has no effect on its
overall stability.

%4.7
\section{}
In order to find the eigenvectors $v$ one must solve the equation
$\left( J-\lambda I\right) v=0$ ie $ \left(
       \begin{array}{cccc}
        -\lambda & 1 \\
        -\omega^2 & b-\lambda \\
       \end{array}
     \right) $ $v=\left(
       \begin{array}{cccc}
        0 \\
        0 \\
       \end{array}
     \right)$
The solution is $v_{i}=\left(\begin{array}{cccc}
1 \\
\lambda_{i} \\
\end{array}
\right)$ , $i=1,2$. For $b=2.5\omega>0$, $\lambda_{1}=-4b/5$ and
$\lambda_{2}=-b/5$ we obtain the figure shown below which is similar
to the middle plot of Fig 4.1.\\
\includegraphics[width=2.0in]{fig4-7.eps} \\

%4.8
\section{}
The Jacobian $ J = \left(
       \begin{array}{cccc}
        1 & 1 \\
        -2 & 4 \\
       \end{array}
     \right) $
gives the characteristic equation $\ \left( 1-\lambda \right)
\left( 4-\lambda \right) +2=0$ with eigenvalues $\lambda _{1}=3$
and $\lambda _{2}=2$ and eigenvectors $v_{1}=\left(
       \begin{array}{cccc}
        1 \\
        2 \\
       \end{array}
     \right) $ and $v_{2}=\left(
       \begin{array}{cccc}
        1 \\
        1 \\
       \end{array}
     \right) $.

%4.9
\section{}
Add the two solutions given by 4.6 to obtain
$\lambda_{1}+\lambda_{2}=a+d=$trace$(J)$ and multiply them to get
$\lambda_{1}\lambda_{2}=ad-bc=$det$(J)$.

%4.10
\section{}
The Jacobian of the damped harmonic oscillator has trace$\left(
J\right) =-b$ and det$\left( J\right) =\omega ^{2}.$ For the
critically damped case $b=2\omega .$ Hence trace$\left( J\right)
=\pm 2\sqrt{\text{det}\left( J\right) }.$

%4.11
\section{}
Denote by $\tau$ the trace of $J$ and by $\delta$ its determinant.
The eigenvalues given by eqn 4.6 may be written as $\lambda
_{1,2}=\tfrac{\tau \pm \sqrt{\tau ^{2}-4\delta }}{2}.$ The
ultimate fate for each romantic style can be determined by the
values of $\tau $ and $\delta $. Therefore we can have the
following cases:\\
CASE 1: $\ \tau >0$ and $\tau ^{2}>4\delta
>0.$ In this case both $\lambda _{1}$ and $\lambda _{2}$ are
positive which means that the origin is an unstable node and
therefore the couple may end up extremely loving or hating each
other or each of them having opposite feelings for each other.
What decides their final state is a matter of the initial
conditions.\\
CASE 2: $\ \tau >0$ and $\tau ^{2}<4\delta .$ In such a case
$\lambda _{1}$ and $\lambda _{2}$ are complex conjugates with
positive real part which means that the origin is an unstable
spiral. No matter what the initial conditions are the couple will
never be indifferent for each other. They will go through cycles
of growing love and growing hate forever (ie oscillatory feelings
of growing intensity) , never reaching some final feeling
for his/her mate.\\
CASE 3. $\tau =0$ and $\delta>0$. The origin becomes a center. In
this case the couple will pass through never- ending periods of
love and hate, but these feelings are oscillatory with fixed
maximum intensity and do not lead to such an ''explosion'' of
feelings as in the cases 1 and 2. Moreover the couple will never
be indifferent for each other like
in cases 1 and 2.\\
CASE 4: $\delta <0$ The origin is a saddle.Depending on the values
of
the parameters $a,b,c$ and $d$ we can have two cases:\\
4a. The couple will either end up extremely loving each other or
extremely hating each other.\\
4b. The couple will have growing opposite feelings for each other
(eg Romeo extremely loving her and Juliet extremely hating him or
vice versa).\\
Since the eigenvectors may be written as $v_{i}=\left(
       \begin{array}{cccc}
        \tfrac{b}{a-\lambda_{i} } \\
        -1 \\
       \end{array}
     \right)$ with $i=1,2$, we end up in case 4a if for the positive
eigenvalue we have $\tfrac{b}{a-\lambda }<0$ and in case 4b if
this expression is positive for the positive eigenvalue.\\
The outcomes of 4a and 4b are dependent upon the initial feelings
for each other. In both cases 4a and 4b there is a possibility of
that Romeo and Juliet become completely indifferent for each
other. This may happen if and only if initially the couple has
feelings that lie on the eigendirection whose eigenvalue is
negative.\\
CASE 5: $\tau <0$. In this case the couple will end up being
indifferent for one another. The way this happens depends on
whether we have $\tau ^{2}>4\delta >0$ , where we get a stable
node\ or \ $\tau ^{2}<4\delta $, where we get  a stable spiral.

%4.12
\section{}
IMPOSSIBLE TO MAKE SUCH PLOTS. The trace and determinant are not
sufficient to give the nature of the eigenvalues. For the $2\times
2$ case this was possible since knowing the sum and product of the
eigenvalues suffice to fully determine the eigenvalues.

%4.13
\section{}
The equation $\lambda ^{3}+2.14\lambda ^{2}+1=0$ is solved using a
computer and we obtain $\lambda _{1}=-2.33$ \ and \ $\lambda
_{2,3}=0$.$0925\pm 0.649i$ which verifies that the origin is a
spiral saddle of index 2.


%4.14
\section{}
The equilibria can be found by solving $dx/dt=dy/dt=dz/dt=0$ using
eqns 4.33-4.35. It can be shown that the solutions are
$p_{1}^{\ast }=(0,0,0)$ and
$p_{2,3}^{\ast}=(\pm\sqrt{b(r-1)},\pm\sqrt{b(r-1)},r-1)$.
Obviously the first bifurcation occurs when $r=1$ at which the
equilibria  $p_{2}^{\ast}$ and $p_{3}^{\ast}$ are created. The
Jacobian is \\
$ J (x,y,z)= \left(\begin{array}{cccc}
-\sigma & \sigma & 0 \\
r-z & -1 & -x \\
y  & x &  -b \\
\end{array}
\right) $ which gives the characteristic equations \\
$(b+\lambda)(\lambda^2+(\sigma+1)\lambda+\sigma(r-1)=0$  for
$p_{1}^{\ast }$ and
$\lambda^3+(\sigma+b+1)\lambda^2+b(\sigma+r)\lambda+2b\sigma(r-1)=0$
for $p_{2,3}^{\ast }$. For $1<r<r^{\ast}$ the equilibria are
spiral nodes and for $r>r^{\ast}$ they become unstable spiral
saddle points with index 2 (see text on page 69). Since this
happens, at the point where $r=r^{\ast}$ the two of the three
eigenvalues should be complex conjugates with no real part.
Assuming solutions of the form $\lambda=i\omega$ substitute in the
characteristic equation for $p_{2,3}^{\ast }$. The real and
imaginary parts of the left hand side should both equal to zero
which give $\omega ^{2}=b\left( \sigma +r^{\ast }\right) $ and
$\omega^{2}\left( \sigma +b+1\right) = 2b\sigma \left( r^{\ast
}-1\right)$ respectively. Substituting $\omega ^{2}$ in the second
equation and solving for $r^{\ast }$ we obtain $r^{\ast
}=\dfrac{\sigma( b+\sigma +3)}{\sigma -b-1}$

%4.15
\section{}
Solving the system of equations $dx/dt=dy/dt=dz/dt=0$ we obtain
two equilibria, namely $p_{1,2}^{\ast }=\left( \mp 1,\pm
1,0\right) .$ The Jacobian is given by \\
$ J (x,y,z)= \left(
       \begin{array}{cccc}
        -1 & -1 & 0 \\
        -z & 0  & -x \\
        y  & x &  0 \\
       \end{array}
     \right) $.
Hence $ J (p_{1,2}^{\ast })= \left(
       \begin{array}{cccc}
        -1 & -1 & 0 \\
        0 & 0  & \pm 1 \\
        \pm 1  & \mp 1 &  0 \\
       \end{array}
     \right) $. For both $p_{1}^{\ast }$ and $p_{2}^{\ast }$ the
Jacobian has the same characteristic polynomial $\lambda
^{3}+\lambda +\lambda^{2}+2=0$. Solving this equation using a
computer we get $ \lambda_{1} =-1.35$ and $ \lambda_{2,3}
=0.177\pm 1.20i$.

%4.16
\section{}
Use eqns 4.36-4.38 to solve $dx/dt=dy/dt=dz/dt=0$. These give
$x^{\ast }=az^{\ast }$ , $y^{\ast }=-z^{\ast }$ and $z^{\ast
}=\tfrac{c\pm \sqrt{c^{2}-4ab}}{2a}$ . The Jacobian is given by $
J (x,y,z)= \left( \begin{array}{cccc}
0 & -1 & -1 \\
1 & a  & 0 \\
z  & 0 &  x-c \\
\end{array}
\right) $, whose characteristic equation is: \\
$\lambda^{3}-\lambda ^{2}\left( \tfrac{b}{z^{\ast }}+a\right)
+\lambda \left( z^{\ast }+\tfrac{ba}{z^{\ast }}+1\right)
-az-\tfrac{b}{z^{\ast }}=0$.





% CHAPTER 5***********
\chapter{}
% ********************

%5.1
\section{}
Working as in section 5.1.1 on page 81 the result is $\lambda=$ln
10

%5.2
\section{}
Since we have $\left| f^{\prime }(x_{n})\right| =A>0$ \ we get
$\lambda =\ln \left| A\right| $ which is negative for $0<A<1$ so
there exists a stable fixed point for $0<A<1$. For $A>1$ we always
have  $\left| f^{\prime }(x^{\ast })\right| >0$ at $x=x^{\ast }$
so it is impossible to have stable orbits.

%5.3
\section{}
NOT SURE ABOUT REASONING. After many iterations the map settles to
either a fixed point or a cycle. At bifurcation points, all points
which are fixed or belong to a cycle have their derivatives equal
to 1 and hence the logarithm of their derivatives equal to zero.
Since the Lyapunov exponent equals to the average of those
logarithms after many iterations (see eqn 5.3), we may conclude
that the Lypunov exponent equals to zero.

%5.4
\section{}
Since there is only one fixed point $X^{\ast }=1-1/A$ the
probability distribution may take the form $P(X)=\delta (X-X^{\ast
})$ , where $\delta$ is the Dirac delta function. So using
$\lambda =\int_{0}^{1}P(X)\ln \left| A(1-2X)\right| dX$ we obtain
$\lambda =\ln \left| A(1-2X^{\ast })\right| =\ln \left| 2-A\right|
$ .

%5.5
\section{}
Start from $g(X)=f(f(X))=A^{2}X-A^{2}\left( A^{2}+1\right)
X^{2}+2A^{3}X^{3}-A^{3}X^{4}$ which gives $g^{\prime
}(X)=A^{2}-2A^{2}\left( A^{2}+1\right) X+6A^{3}X^{2}-4A^{3}X^{3}$
. Then the probability distribution is given by
$P(X)=\tfrac{1}{2}\left[ \delta (X-X_{+}^{\ast })+\delta
(X-X_{-}^{\ast })\right] $ with $X_{\pm }^{\ast }$ being given in
eqn 2.7. Use $\lambda =\int_{0}^{1}P(X)\ln \left| g^{\prime
}(X^{\ast })\right| dX$ to obtain $\lambda =\tfrac{1}{2}\ln \left|
g^{\prime }(X_{+}^{\ast })g^{\prime }(X_{-}^{\ast })\right| .$ A
lot of algebra would be required to evaluate this expression, but
there is an easier way for doing so if we consider $g^{\prime
}(X_{\pm }^{\ast })=f^{\prime }(f(X_{\pm }^{\ast }))f^{\prime
}(X_{\pm }^{\ast })=f^{\prime }(X_{\mp }^{\ast })f^{\prime
}(X_{\pm }^{\ast })$ $=A^{2}(1-2X_{\mp }^{\ast })(1-2X_{\pm
}^{\ast })=A^{2}\lbrack 1-2(X_{\mp }^{\ast }+X_{\pm }^{\ast
})+4X_{\mp }^{\ast }X_{\pm }^{\ast }\rbrack $ . From equation
(2.6) $X_{+}^{\ast }+X_{-}^{\ast }=\lbrack A(A+1)\rbrack /A^{2}$
and $X_{-}^{\ast }X_{+}^{\ast }=(A+1)/A^{2}$. Hence $g^{\prime
}(X_{\pm }^{\ast })=A^{2}\lbrack 1-2A(A+1)/A^{2}+4\left(
A+1\right) /A^{2}\rbrack =4+2A-A^{2}$ and therefore $\lambda =\ln
\left| 4+2A-A^{2}\right|$.

%5.6
\section{}
The fixed point for $A=2$ is $X^{\ast }=1-1/2=1/2$ . Moreover
since $f\left( X^{\ast }\right) =0$ we may conclude that there
exists a supercycle since these results would force the Lyapunov
exponent to tend to $-\infty $ . Iterating the map starting from
$X_{0}=0.1$ we get  $X_{1}=0.2952$ , $X_{2}=0.41611392$ ,
$X_{3}=0.4859262512$ , $X_{4}=0.4996038592$ , $X_{5}=0.4999996862$
and $X_{6}=0.5000000000$ .

%5.7
\section{}
Substitution of $A=1+\sqrt{5}$ in eqn 2.7 we get $X_{+}^{\ast }=$
$\tfrac{3+\sqrt{5}}{2\left( 1+\sqrt{5}\right) }$and $X_{-}^{\ast
}=\tfrac{1}{2}.$ In order to verify that we have \ supercycle we
have to show that $g^{\prime }\left( X_{-}^{\ast }\right) =0$ with
$g^{\prime }$ being given in the solution of Problem 5.5. With a
bit of algebra one can verify that $g^{\prime }\left( X_{\pm
}^{\ast }\right) =0$.

%5.8
\section{}
???????

%5.9
\section{}
Since the map for those range of values of\ $\alpha $ has a single
fixed point the Lyapunov exponent equals to $\lambda =\ln \left|
f^{\prime }\left( X^{\ast }\right) \right| .$ Since $X^{\ast }=0$
(the fixed point) we may write $\lambda =\ln \left| f^{\prime
}\left( 0\right) \right| =\ln \left( 2\alpha \right) .$

%5.10
\section{}
$\dfrac{\Delta R_{n+1}^{2}}{\Delta R_{n}^{2}}=\dfrac{\left(
a\Delta X_{n}+b\Delta Y_{n}\right) ^{2}+\left( c\Delta
X_{n}+d\Delta Y_{n}\right) ^{2}}{\Delta X_{n}^{2}+\Delta
Y_{n}^{2}}=\dfrac{\left( a+bY_{n}^{^{\prime }}\right) ^{2}+\left(
c+dY_{n}^{^{\prime }}\right) ^{2}}{1+Y_{n}^{^{\prime }2}}$ with
$Y_{n}^{^{\prime }}=\dfrac{\Delta Y_{n}}{\Delta
X_{n}}=\dfrac{c\Delta X_{n-1}+d\Delta Y_{n-1}}{a\Delta
X_{n-1}+b\Delta Y_{n-1}}=\dfrac{c+dY_{n-1}^{^{\prime
}}}{a+bY_{n-1}^{^{\prime }}}$ \ which is the same as eqn 5.12.
Substitution of $\tfrac{\Delta R_{n+1}^{2}}{\Delta R_{n}^{2}}$ in
eqn 5.10 yields eqn 5.11.

%5.11
\section{}
Using the notation of the text we get $\left| a\right| =A$ ,
$b=c=0$ and $\left| d\right|=B .$ Substitution in eqn 5.11 yields
$\lambda _{1}=\lim\limits_{N\rightarrow \infty
}\dfrac{1}{2N}\sum\limits_{n=0}^{N-1}\ln \left[
\tfrac{A^{2}+B^{2}Y_{n}^{^{\prime }2}}{1+Y_{n}^{^{\prime
}2}}\right] =\dfrac{1}{2}\lim\limits_{n\rightarrow \infty }\ln
\left[ B^{2}+\tfrac{A^{2}-B^{2}}{1+Y_{n}^{^{\prime }2}}\right] $
with $\left| Y_{n}^{^{\prime }}\right| =\left|
\tfrac{BY_{n-1}^{^{\prime }}}{A}\right| =\left| Y_{0}\right|
\left| \tfrac{B}{A}\right| ^{n}.$ Then we have :
$\lim\limits_{n\rightarrow \infty }\left|Y_{n}^{^{\prime
}}\right|= \left\{\begin{array}{cccc}
        +\infty & \text for & A<B \\
        0 & \text for  & A>B \\
       \end{array}
     \right.$
Substitution of the limit in the equation for $\lambda _{1\text{
}}$gives $\lambda _{1}=\ln [\max \left( A,B\right)] .$ \ Also use
$\lambda _{1}+\lambda _{2}=\det \left( J\right) =\left| AB\right|
$ \ to determine $\lambda _{2}.$ \\

%5.12
\section{}
A.1.1 Logistic Map     : $J(X)=A(1-2X)$ \\
A.1.2 Sine Map : $J(X)=A\pi \cos \pi X$ \\
A.1.3 Tent Map : $J(X)=A$ sign$(1/2-X)$\\
A.1.4 Linear Congruential Generator : $J(X)=A (\textsf{mod } C)$ \\
A.1.5 Gauss Map : $J(X)=-1/X^2 (\textsf{mod } 1)$ \\
A.2.1 H\`{e}non Map : $ J (X,Y)= \left(
\begin{array}{cccc}
-2aX & b \\
1 & 0  \\
\end{array}
\right) $ \\
A.2.2 Lozi Map : $ J (X,Y)= \left(
\begin{array}{cccc}
-a$ sign$(X) & b \\
1 & 0  \\
\end{array}
\right) $ \\
A.2.3 Tinkerbell Map :  $J (X,Y)= \left(
\begin{array}{cccc}
2X+a & -2Y+b \\
2Y+c & 2X+d  \\
\end{array}
\right) $ \\
A.2.4 Burgers Map :   $J (X,Y)= \left(
\begin{array}{cccc}
a & -2Y \\
Y & X+b  \\
\end{array}
\right) $ \\
A.2.5 Kaplan-Yorke Map : $ J (X,Y)= \left(
\begin{array}{cccc}
a & 0 \\
-4\pi\sin (4\pi X) & b  \\
\end{array}
\right) (\textsf{mod } 1)$ \\
A.2.6 Dissipative Standard Map: $J (X,Y)= \left(
\begin{array}{cccc}
1 & 1 \\
k\cos X (\textsf{mod } 2\pi) & b  \\
\end{array}
\right) $ \\
A.3.1 Chirikov Map : $ J (X,Y)= \left(
\begin{array}{cccc}
1 & 1 \\
k\cos (X) & 1  \\
\end{array}
\right) (\textsf{mod } 2\pi)$ \\
A.3.2 H\`{e}non Area-preserving Quadratic Map : $ J (X,Y)= \left(
\begin{array}{cccc}
\cos \alpha+2X\sin \alpha & - \sin \alpha \\
\sin \alpha -2X\cos \alpha& \cos \alpha  \\
\end{array}
\right) $ \\
A.3.3 Arnold Cat Map: $ J (X,Y)= \left(
\begin{array}{cccc}
1 & 1 \\
1 & k  \\
\end{array}
\right) (\textsf{mod } 1)$ \\
A.3.4 Gingerbreadman Map : $J (X,Y)= \left(
\begin{array}{cccc}
\textsf{sign}(X) & -1 \\
1 & 0  \\
\end{array}
\right) $ \\
A.3.5 Stochastic Web Map : $ J (X,Y)= \left(
\begin{array}{cccc}
\cos \alpha-k\sin \alpha \cos X  & - \sin \alpha \\
\sin \alpha +k\cos \alpha \cos X & \cos \alpha  \\
\end{array}
\right) $ \\
A.3.6 Lorenz Simplest 3-D Map : $ J (x,y,z)= \left(
\begin{array}{cccc}
Y & X & -1 \\
1 & 0  & 0 \\
0  & 1 & 0 \\
\end{array}
\right) $ \\


%5.13
\section{}
Using eqn 5.20 we get $\lambda =a\left( 1-2x\right) $ for the
local Lyapunov exponent. \ Since at $x^{\ast }=1$ we have $f\left(
x^{\ast }\right) =0$ the global Lyapunov exponent is $\lambda
=-a$.

%5.14
\section{}

%5.15
\section{}
Since $ J (x,y)= \left(
\begin{array}{cccc}
0 & 1  \\
-1-2bxy & b(1-x^2) \\
\end{array}
\right) $, we have $\det (J)=2bxy+1$. For area expansion we
require $\det (J)>0$, ie $xy>-1/2b$ and for area contraction we
must have $xy<-1/2b$.

%5.16
\section{}
INCOMPLETE \\
We have $a=d=0$ , $b=c=1$ . From eqn 5.23 we get
$\lambda _{1}+\lambda _{2}=0$ and from eqn 5.21 $\lambda
_{1}=\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}\dfrac{2y^{\prime }}{1+y^{\prime 2}}dt$
with $\dfrac{dy^{\prime }}{dt}=1-y^{\prime 2}\ $\ for \ $\Delta
t\rightarrow 0$

$\int \dfrac{1}{1-y^{2}}dy=\allowbreak \text arctanh\ y$

$\int \dfrac{2\tanh t}{1+\tanh ^{2}t}dt=\allowbreak
-\frac{1}{2}\ln \left( -1+\tanh t\right) -\frac{1}{2}\ln \left(
1+\tanh t\right) +\frac{1}{2}\ln
\left( 1+\tanh ^{2}t\right) $ \\

$\allowbreak \lim\limits_{t\rightarrow \infty }\left(
-\frac{1}{2}\ln \left( -1+\tanh t\right) -\frac{1}{2}\ln \left(
1+\tanh t\right) +\frac{1}{2}\ln \left( 1+\tanh ^{2}t\right)
\right) =\allowbreak \infty $

%5.17
\section{}

%5.18
\section{}
Since $\dfrac{\partial }{\partial x}\left[ \sigma \left(
y-x\right) \right] +\dfrac{\partial }{\partial y}\left[
rx-y-xz\right] +\dfrac{\partial }{\partial z}\left[ xy-bz\right]
=\allowbreak -b-1-\sigma <0$ for all $\sigma ,b>0$ we have volume
contraction with $V(t)\thicksim e^{-\left( b+1+\sigma \right) t}$
.

%5.19
\section{}
A.4.1 Damped Driven Pendulum : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1  & 0\\
-\cos x & -b  & A\cos \Omega z \\
0  & 0 &  0 \\
\end{array}
\right) $$ \\
A.4.2 Driven Van Der Pol Oscillator : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1  & 0\\
-1-2bxy & b(1-x^2)  & A\cos \Omega z \\
0  & 0 &  0 \\
\end{array}
\right) $$ \\
A.4.3 Ueda Attractor : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1 & 0 \\
-3x^{2} & -b  & A\cos \Omega z \\
0  & 0 &  0 \\
\end{array}
\right) $$ \\
A.5.1 Lorenz Attractor : $$ J (x,y,z)= \left(
\begin{array}{cccc}
-\sigma & \sigma & 0 \\
r-z & -1  & -x \\
y  & x &  -b \\
\end{array}
\right) $$ \\
A.5.2 R\"{o}ssler Attractor :$$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & -1 & -1 \\
1 & a  & 0 \\
z  & 0 & x-c \\
\end{array}
\right) $$ \\
A.5.3 Diffusionless Lorenz Attractor :$$ J (x,y,z)= \left(
\begin{array}{cccc}
-1 & -1 & 0 \\
-z & 0  & -x \\
y  & x & 0 \\
\end{array}
\right) $$ \\
A.5.4 Chua's circuit :$$ J (x,y,z)= \left(
\begin{array}{cccc}
\alpha (b-1+1/2(a-b)(\textsf{sign}(x+1)-\textsf{sign}(x-1)) & \alpha & 0 \\
1 & -1  & 1 \\
0  & -\beta & 0 \\
\end{array}
\right) $$ \\
A.5.5 Moore-Spiegel Oscillator : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1 & 0 \\
0 & 0  & 1 \\
-T-2Rxy  & -(T-R+Rx^2) & -1 \\
\end{array}
\right) $$ \\
A.5.6 Thomas' Cyclically Symmetric Attractor : $$ J (x,y,z)=
\left(
\begin{array}{cccc}
-b & \cos y & 0 \\
0 & -b  & \cos z \\
\cos x  & 0 & -\beta \\
\end{array}
\right) $$ \\
A.5.7 Simplest Quadratic Chaotic Flow :$$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1 & 0 \\
0 & 0  & 1 \\
-1  & 2y & -a \\
\end{array}
\right) $$ \\
A.5.8 Simplest Cubic Chaotic Flow : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1 & 0 \\
0 & 0  & 1 \\
y^2-1  & 2yx & -a \\
\end{array}
\right) $$ \\
A.5.9 Simplest Piecewise Linear Chaotic Flow :$$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1 & 0 \\
0 & 0  & 1 \\
$sign$ (x)  & -1 & -a \\
\end{array}
\right) $$ \\
A.6.1 Simplest Driven Chaotic Flow : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1  & 0\\
3x^2 & 0  & \cos \Omega z \\
0  & 0 &  0 \\
\end{array}
\right) $$ \\
A.6.2 Nos\'{e}-Hoover Oscillator : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1  & 0\\
-1 & z  & y \\
0  & -2y &  0 \\
\end{array}
\right) $$ \\
A.6.3 Labyrinth Chaos : $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & \cos y  & 0\\
0 & 0  & \cos z \\
\cos x  & 0 &  0 \\
\end{array}
\right) $$ \\
A.6.4 H\`{e}non-Heilles System : $$ J (x,y,v,w)= \left(
\begin{array}{cccc}
0 & 0  & 1 & 0\\
0 & 0  & 0 & 1 \\
-1-2y  & -2x &  0 & 0\\
-2x  & 2y-1 &  0 & 0\\
\end{array}
\right) $$

%5.20
\section{}
A. $\sum \lambda =\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}zdt=\left\langle z\right\rangle $
Numerical calculation is needed. \\
B,C,E,I,L,M,R,S. $\sum \lambda=\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}-1dt=-1$ \\
D,O. $\sum \lambda=\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}xdt=\left\langle x\right\rangle $
Numerical calculation is needed. \\
F,H,Q. $\sum\lambda =\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}\left( 0.5-1\right) dt=-0.5$ \\
G. $\sum\lambda =\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}\left( 0.4-1\right) dt=-0.6$ \\
J,N. $\sum\lambda =\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}-2dt=-2$ \\
K. $\sum \lambda=\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}\left( y-1+0.3\right) dt=\left\langle
y\right\rangle -0.7$ Numerical calculation is needed. \\
P. $\sum\lambda =\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{0}^{T}\left( 2y\right) dt=2\left\langle
y\right\rangle $Numerical calculation is needed.

%5.21
\section{}
Since one of the eigenvalues is positive we have a strange
attractor of dimension greater than 2.

%5.22
\section{}
$D_{KY}=3+\dfrac{1}{27}\left( 0.11+0.02\right) =\allowbreak
3.\,\allowbreak 004\,8$

%5.23
\section{}
$D_{KY}=4+\dfrac{1}{0.6}\left( 0.5+0.1+0-0.3\right) =\allowbreak
4.\,\allowbreak 5$

%5.24
\section{}
We first have to find the equation of a parabola $y=ax^{2}+bx+c$
which passes through the points $\left( 0,p_{1}\right) ,\left(
1,p_{2}\right) $ and $\left( 2,p_{3}\right) .$ Solving the
resulting system of three equations in three unknowns we get
$$y=\left(\frac{1}{2}p_{1}-p_{2}+\frac{1}{2}p_{3}\right)x^{2}+
\left(4p_{2}-\frac{3}{2}p_{3}-\frac{5}{2}p_{1}\right)x+p_{3}-3p_{2}
+3p_{1}.$$ Hence we get : \\
A.5.1 $p_{1}=p_{2}=0.9056,p_{3}=-13.6667$ and dimension
$\simeq 2.111$ \\
A.5.2 $p_{1}=p_{2}=0.0714,$ $p_{3}=-5.3229$ and dimension
$\simeq 2.025$ \\
A.5.3 $p_{1}=p_{2}=0.2101,$ $p_{3}=-1$ and dimension
$\simeq 2.273$ \\
A.5.4 $p_{1}=p_{2}=0.3271,$ $p_{3}=-2.1926$ and dimension
$\simeq 2.213$ \\
A.5.5 $p_{1}=p_{2}=0.1109,$ $p_{3}=-1$ and dimension
$\simeq 2.171$ \\
A.5.6 $p_{1}=p_{2}=0.0349,$ $p_{3}=-0.54$ and dimension
$\simeq 2.109$ \\
A.5.7 $p_{1}=p_{2}=0.0551,$ $p_{3}=-2.017$ and dimension
$\simeq 2.051$ \\
A.5.8 $p_{1}=p_{2}=0.0837,$ $p_{3}=-2.028$ and dimension
$\simeq 2.074$ \\
A.5.9 $p_{1}=p_{2}=0.0362,$ $p_{3}=-0.6$ and dimension
$\simeq 2.103$ \\
A.6.2 $p_{1}=p_{2}=0.0138,$ $p_{3}=0$ and dimension $=3$ \\
A.6.3 $p_{1}=p_{2}=0.1402,$ $p_{3}=0$ and dimension $=3$ .
%5.25
\section{}

%5.26
\section{}
$\lambda = \ln (a (\textsf{mod} c))$

% CHAPTER 6***********
\chapter{}
% ********************
%6.1
\section{}
In order to determine the inverse map interchange $X_{n}$ and
$X_{n+1}$, $Y_{n}$ and $Y_{n+1}$ to get $Y_{n}=bX_{n+1}$ and
$X_{n}=1+Y_{n+1}-aX_{n+1}^{2}.$ Solving for $Y_{n+1}$ and
$X_{n+1}$ gives the inverse map. Answer: $X_{n+1}=Y_{n}/b$ ,
$Y_{n+1}=X_{n}+aY_{n}^{2}/b-1.$

%6.2
\section{}
The fixed points satisfy $X_{n+1}=X_{n}$ and $Y_{n+1}=Y_{n}$ . By
eqn 6.2, this means that $X_{n+1}=X_{n}=Y_{n+1}=Y_{n}=X^{\ast
}=Y^{\ast }.$ Eqn 6.1 becomes $\left( a_{4}+a_{5}+a_{6}\right)
X^{\ast 2}+\left( a_{2}+a_{3}-1\right) X^{\ast }+a_{1}=0$ .  This
equation has real solutions when \\
$\Delta =\left(a_{2}+a_{3}-1\right)^{2}-4a_{1}\left(
a_{4}+a_{5}+a_{6}\right) \geq 0.$ Under this condition the fixed
points are given by $X^{\ast }=Y^{\ast }=\left( 1-a_{2}-a_{3}\pm
\sqrt{\Delta }\right) /\left( 2a_{4}+2a_{5}+2a_{6}\right) .$

%6.3
\section{}
The fixed points of the H\'{e}non Map are given by the equation
$1.4X^{2}+0.7X-1=0$, whose solution is $X=-1.\,\allowbreak 131\,4$
and $X=0.\,\allowbreak 631\,35\allowbreak .$ Their position
relative to the attractor is
shown in the figure below:\\
\centerline{ \includegraphics[width=2.0in]{fig6-3.eps} } \\

%6.4
\section{}
Interchange $X_{n}$ and $X_{n+1}$, $Y_{n}$ and $Y_{n+1}$ to get :
$X_{n}=a_{1}+a_{2}X_{n+1}+a_{3}Y_{n+1}+a_{4}X_{n+1}^{2}+a_{5}
X_{n+1}Y_{n+1}$ and $X_{n+1}=Y_{n}$ . Solving for $Y_{n+1}$ and
$X_{n+1}$ gives the inverse map. Answer: $X_{n+1}=Y_{n}$ ,
$Y_{n+1}=\tfrac{X_{n}-a_{1}-a_{2}Y_{n}-a_{4}Y_{n}^{2}}{a_{3}-a_{5}
Y_{n}}.$

%6.5
\section{}
As in the previous problem use $X_{n+1}=Y_{n}$ and
$X_{n}=a_{1}+a_{2}X_{n+1}+a_{3}Y_{n+1}+a_{4}X_{n+1}^{2}+a_{5}
X_{n+1}Y_{n+1}+a_{6}Y_{n+1}^{2}.$ The second equation becomes \\
$a_{6}Y_{n+1}^{2}+\left( a_{5}Y_{n}+a_{3}\right)Y_{n+1}+X_{n}-
a_{1}-a_{2}Y_{n}-a_{4}Y_{n}^{2}=0,$ whose solution is: \\
$Y_{n+1}=\dfrac{1}{2a_{6}}\left[ -a_{5}Y_{n}-a_{3}\pm
\sqrt{\left( a_{5}Y_{n}+a_{3}\right) ^{2}-4a_{6}\left(
X_{n}-a_{1}-a_{2}Y_{n}-a_{4}Y_{n}^{2}\right) }\right] $

%6.6
\section{}
Antisymmetry about $r=0.5$ can be verified if we notice that
$f_{i}\left( 0.5-r\right) =-f_{i}\left( 0.5+r\right) $ for
$i=1,2,3.$ It is obvious that  $f_{1}$ is infinite for $r=0$ and
$r=1$ since $\tan \left( \pm \pi /2\right) =\pm \infty $. Morever
, one can verify that eqns 6.4 and 6.5 are also infinite using the
Del' H\^{o}pital's rule. All three functions have half their
values in the range $-1<f<1$ since we have $f(0.5\pm 0.25)=\pm 1$.

%6.7
\section{}

%6.8
\section{}
Eqn 6.6 may be also written as a system of differential equations
of the form:  $\dot{x}=y$ , $\dot{y}=z$ and
$\dot{z}=-\tfrac{3}{5}z-y+\left| x\right| -1.$ Equilibrium points
satisfy $\dot{x}=\dot{y}=0$ . This happens only when $y=z=0$ and
$x=\pm 1.$ The Jacobian takes the form: $$ J (x,y,z)= \left(
\begin{array}{cccc}
0 & 1 & 0 \\
0 & 0  & 1 \\
$sign$ (x)  & -1 & -a \\
\end{array}
\right) $$ \\
When $x=-1$ the eigenvalues are $-0.58846,0.59423 \pm 1.1603i$ and
when $x=1$ the eigenvalues are $0.83555,-0.11778 \pm 1.0876i$.

%6.9
\section{}
The following calculations are valid as long as the $a_{4}$ and
$a_{5}$ are positive numbers. Eqn 6.17 becomes $\dfrac{\alpha
}{\beta ^{2}}\dfrac{dx^{\prime 2}}{dt^{\prime 2}}+a_{1}\left(
\alpha ^{2}x^{2}-a_{4}\right) \dfrac{\alpha }{\beta
}\dfrac{dx^{\prime }}{dt^{\prime }}+a_{5}\alpha x^{\prime
}=a_{2}\sin \left( a_{3}t^{\prime }\right) \Rightarrow $ \\
$\dfrac{dx^{\prime 2}}{dt^{\prime 2}}+a_{1}\beta \alpha ^{2}\left(
x^{2}-\dfrac{a_{4}}{\alpha ^{2}}\right) \dfrac{dx^{\prime
}}{dt^{\prime }}+a_{5}\beta ^{2}x^{\prime }=\dfrac{\beta
^{2}}{\alpha }a_{2}\sin \left( a_{3}\beta t^{\prime }\right) .$
Choose $\beta $ so that $a_{5}\beta ^{2}=1$ ie $\beta =\pm
\sqrt{a_{5}^{-1}}$ and $\alpha =\pm \sqrt{a_{4}}.$ In such a case
the equation becomes $\dfrac{dx^{\prime 2}}{dt^{\prime
2}}+a_{1}^{\prime }\left( x^{2}-1\right) \dfrac{dx^{\prime
}}{dt^{\prime }}+x^{\prime }=a_{2}^{\prime }\sin \left(
a_{3}^{\prime }t^{\prime }\right) $ with $a_{1}^{\prime }=\pm
a_{1}a_{4}\sqrt{a_{5}^{-1}}$ , $a_{2}^{\prime }=\pm
\dfrac{a_{2}}{a_{5}\sqrt{a_{4}}}$ and $a_{3}^{\prime }=\pm
a_{3}\sqrt{a_{5}^{-1}}.$

%6.10
\section{}
In this problem we have to find the $a$ 's and $b$'s for which
$X_{n}=b_{1}\tanh (a_{10}+a_{11}X_{n-1})+b_{2}\tanh
(a_{20}+a_{21}X_{n-1})$. The information we have in order to solve
the problem is that when $X_{n-1}$ is $0$ or $1$ then $X_{n}=0$
and that the maximum should be at $X_{n}=1$ when $X_{n-1}=1/2.$
Since we have six unknowns and 4 equations we may have two free
variables. The equations are: $b_{1}\tanh (a_{10})+b_{2}\tanh
(a_{20})=b_{1}\tanh (a_{10}+a_{11})+b_{2}\tanh (a_{20}+a_{21})=0$
, $b_{1}a_{11}-\vspace{1pt}b_{1}a_{11}\tanh
^{2}(a_{10}+a_{11}/2)+b_{2}a_{21}-b_{2}a_{21}\tanh
^{2}(a_{20}+a_{21}/2)=0$ and $b_{1}\tanh
(a_{10}+a_{11}/2)+b_{2}\tanh (a_{20}+a_{21}/2)=1$

%6.11
\section{}
Cannot determine the LE accurately enough from the diagram in
order to compare them with those in the appendix.

%6.12
\section{}
Since we are interested only for the maximum, fit the parabola
$x=at^{2}+bt+c$ through the points $\left( -1,x_{1}\right) $ ,
$\left( 0,x_{2}\right) $ and $\left( 1,x_{3}\right) .$ In this
case we would get $a=\tfrac{1}{2}$ $\left( x_{1}+x_{3}\right) $ ,
$b=\tfrac{1}{2}$ $\left( x_{3}-x_{1}\right) $ and $c=x_{2}.$ The
maximum of the parabola occurs at $t=-b/\left( 2a\right) $ and
equals to $x_{\max }=c-b^{2}/4a.$

%6.13
\section{}
The intersection with the Poincar\'{e} section occurs only when
$z_{1}z_{2}<0$ . Each point $\left( x,y,z\right) $ on the line
joining  $\left( x_{1},y_{1},z_{1}\right) $ and $\left(
x_{2},y_{2},z_{2}\right) $ is given by $x=x_{1}+\lambda \left(
x_{2}-x_{1}\right) $ , $y=y_{1}+\lambda \left( y_{2}-y_{1}\right)
$ and $z=z_{1}+\lambda \left( z_{2}-z_{1}\right) $ with $0\leq
\lambda \leq 1.$ For $\lambda =\dfrac{z_{1}}{z_{1}-z_{2}}$ we have
$z=0,$ ie an intersection of the line with the plane $z=0.$
Therefore the coordinates of the point $\left( x,y\right) $ on the
Poincar\'{e} section are given by  $x=x_{1}+\dfrac{z_{1}\left(
x_{2}-x_{1}\right) }{z_{1}-z_{2}}$ and $y=y_{1}+\dfrac{z_{1}\left(
y_{2}-y_{1}\right) }{z_{1}-z_{2}}.$

%6.14
\section{}

%6.15
\section{}

%6.16
\section{}

%6.17
\section{}

%6.18
\section{}
The fixed points of the Lozi map satisfy $X^{\ast }=Y^{\ast }$ and
$\ 0.5X^{\ast }=1-1.7\left| X^{\ast }\right| $. This gives either
$0.5X^{\ast }=1-1.7X^{\ast }\Rightarrow X^{\ast }=Y^{\ast
}=0.\,\allowbreak 454\,55$ or $0.5X^{\ast }=1+1.7X^{\ast
}\Rightarrow X^{\ast }=Y^{\ast }=-0.\,\allowbreak 833\,33.$

%6.19
\section{}
The fixed points satisfy $X^{\ast }=10X^{\ast 2}-10Y^{\ast
2}-6Y^{\ast }$ and $X^{\ast }=\dfrac{Y^{\ast }}{4\left( Y^{\ast
}+1\right) }.$ Solving these equations we finally obtain the pairs
$\left( 0,0\right) $ and $\left( 1.\,\allowbreak
027\,6,-1.\,\allowbreak 321\,5\right) .$

%6.20
\section{}
The fixed points satisfy $X=X^{2}-Y^{2}-1$ and $Y(1-2X)=0$
$\Rightarrow $ $\left( i\right) $ $Y=0$ and $\ \left( ii\right) $
$X=1/2$ . Then for $Y=0$ $\left( \text{case }\left( i\right)
\right) $we get the equation $X^{2}-X-1=0$ whose solutions are
$X_{1}^{\ast }=\frac{1}{2}\sqrt{5}+\frac{1}{2}$ and $X_{2}^{\ast
}=\frac{1}{2}-\frac{1}{2}\sqrt{5}.$ The Jacobian of the map has
characteristic polynomial $\left( 2X-\lambda \right)
^{2}+4Y^{2}=0$ and eigenvalues $\lambda =2X\pm 2Yi.$ Therefore
$X_{1}^{\ast }$ is unstable since $\lambda >0$, whereas
$X_{2}^{\ast }$ is stable. For $X=1/2$ we get non-real $Y$ $\left(
\text{case }\left( ii\right) \right) .$ In order to determine the
periodic cycle consider the fixed points of the second iterate
map. For the $Y$ component we get $4XY\left( X^{2}-Y^{2}-1\right)
=Y \Rightarrow Y\left[ 4X\left( X^{2}-Y^{2}-1\right) -1\right]
=0.$ The equation $4X\left( X^{2}-Y^{2}-1\right) -1=0$ together
with the $X$ component of the second iterate map do not have real
solutions. However when $Y=0$, using  $\left( X^{2}-Y^{2}-1\right)
^{2}-4X^{2}Y^{2}-1=X$ we get $X(X^{3}-2X-1)=0$. This gives $X=0$
and $X=-1.$ ,ie in this case we have a 2-period cycle with
oscillations between the points $\left( 0,0\right) $ and $\left(
-1,0\right) .$

%6.21
\section{}
By definition, the derivative of $f$ with respect to $\bar{Z}$ is
$\dfrac{\partial f}{\partial \bar{Z}}=\dfrac{1}{2}\left(
\dfrac{\partial }{\partial X}+i\dfrac{\partial }{\partial
Y}\right) f=\dfrac{1}{2}\left( \dfrac{\partial F}{\partial
X}-\dfrac{\partial G}{\partial Y}\right) +\dfrac{i}{2}\left(
\dfrac{\partial G}{\partial X}+\dfrac{\partial f}{\partial
Y}\right) .$ Since $f$ is a function of $Z$ only we have
$\dfrac{\partial f}{\partial \bar{Z}}=0.$ This condition requires
$\dfrac{\partial F}{\partial X}=\dfrac{\partial G}{\partial Y}$
and $\dfrac{\partial G}{\partial X}=-\dfrac{\partial F}{\partial
Y}.$

%6.22
\section{}
H\'{e}non Map: $\dfrac{\partial F}{\partial X}=-2aX\neq
0=\dfrac{\partial G}{\partial Y}.$ Lozi map: $\dfrac{\partial
F}{\partial Y}=0.5\neq 1=\dfrac{\partial G}{\partial X}.$ Julia
set: $\dfrac{\partial F}{\partial X}=\dfrac{\partial G}{\partial
Y}=2X$ and $\dfrac{\partial G}{\partial X}=-\dfrac{\partial
F}{\partial Y}=2Y.$

%6.23
\section{}
The Cauchy-Riemann equations require$\dfrac{\partial F}{\partial
X}=a_{2}+2a_{4}X+a_{5}Y=0=\dfrac{\partial G}{\partial Y}.$ Hence
we must have $a_{2}=a_{4}=a_{5}=0.$ Also $\dfrac{\partial
F}{\partial Y}=a_{3}+a_{5}X+2a_{6}Y=-1=-\dfrac{\partial
G}{\partial X}$ which gives $a_{3}=-1,a_{5}=a_{6}=0.$ Therefore in
this case we must have $X_{n+1}=a_{1}-Y_{n}$ and $Y_{n+1}=X_{n}$
which is obviously a non-chaotic system.

%6.24
\section{}
Let $F=b_{1}+b_{2}X+b_{3}Y+b_{4}X^{2}+b_{5}XY+b_{6}Y^{2}$ and
$G=c_{1}+c_{2}X+c_{3}Y+c_{4}X^{2}+c_{5}XY+c_{6}Y^{2}.$ The
Cauchy-Riemann equations require$\dfrac{\partial F}{\partial
X}=b_{2}+2b_{4}X+b_{5}Y\equiv c_{3}+c_{5}X+2c_{6}Y=\dfrac{\partial
G}{\partial Y}$ and $\dfrac{\partial F}{\partial
Y}=b_{3}+b_{5}X+2b_{6}Y\equiv
-c_{2}-2c_{4}X-c_{5}Y=-\dfrac{\partial G}{\partial X}$ . Equating
the coefficients with the same powers of $X$ and $Y$ we get :
$b_{2}=c_{3}=a_{2}$ , $b_{4}=c_{5}/2=a_{4}$ ,
$b_{5}/2=c_{6}=a_{5}$ , $b_{3}=-c_{2}=a_{3}$ ,
$b_{5}/2=-c_{4}=a_{5}$ and$\ \ b_{6}=-c_{5}/2=-a_{4}.$ If we also
let $b_{1}=a_{1}$ and $c_{1}=a_{6}$ we obtain eqns 6.18 and 6.19.

% CHAPTER 7***********
\chapter{}
% ********************
%7.1
\section{}
This is done in the textbook on page 132. Eigenvalues?
%7.2
\section{}
Done in textbook (p.133). Eigenvalues?
%7.3
\section{}
When $y=1,$ $\dot{y}=0$ for all values of the parameter $a$. Now,
expand about  $y=1$ by making a change of variable from $y$ to
$v=y-1.$ In this case $v$ is small and we have
$\dot{y}=\dot{v}=a\ln (v+1)+v\thickapprox a\lbrack
v-\tfrac{1}{2}v^{2}+...\rbrack
+v=(a+1)v-\tfrac{1}{2}av^{2}+O(v^{3}).$ Hence a transcritical
bifurcation occurs when $a=-1.$ In order to write the last
equation in normal form, let $v=c\xi $ which transforms it to
$\dot{\xi}=(a+1)\xi -\tfrac{1}{2}ca\xi ^{2}+O(v^{3}).$ Letting
$c=2/a$ and $\mu =(a+1)$ we finally obtain the normal form with \\
$\xi =\tfrac{1}{2}a\left( y-1\right) .$

%7.4
\section{}
The fixed points satisfy $f=\mu x^{\ast }-x^{\ast 3}=0.$ Solving
for $x^{\ast }$ we obtain $x^{\ast }=0$ and $x^{\ast }=\pm
\sqrt{\mu }.$ Then $\tfrac{df}{dx^{\ast }}=\mu -3x^{\ast 2}$ and
for $x^{\ast }=0$ we get $\tfrac{df}{dx^{\ast }}=\mu .$ Thus for
$\mu >0,$ $x^{\ast }$ is unstable and for $\mu <0$ it is stable.
However for the points $x^{\ast }=\pm \sqrt{\mu },$ which exist
only for $\mu >0$ we get $\tfrac{df}{dx^{\ast }}=-2\mu >0$ which
are stable.

%7.5
\section{}
The fixed points satisfy $f=\mu x^{\ast }+x^{\ast 3}=0.$ Solving
for $x^{\ast }$ we obtain $x^{\ast }=0$ and $x^{\ast }=\pm
\sqrt{-\mu }.$ Then $\tfrac{df}{dx^{\ast }}=\mu +3x^{\ast 2}$ and
for $x^{\ast }=0$ we get $\tfrac{df}{dx^{\ast }}=\mu .$ Thus for
$\mu >0,$ $x^{\ast }$ is unstable and for $\mu <0$ it is stable.
However for the points $x^{\ast }=\pm \sqrt{-\mu },$ which exist
only for $\mu <0$ we get $\tfrac{df}{dx^{\ast }}=-2\mu >0$ which
are unstable.

%7.6
\section{}
For $a>1$ the only equilibrium point we have is $x^{\ast }=0$
which is unstable since $\tfrac{df}{dx^{\ast }}=a-\cos x^{\ast
}>0.$ For $0<a<1,$ the equilibrium points satisfy $\dfrac{\sin
x^{\ast }}{x^{\ast }}=a.$ INCOMPLETE-NEEDS FIGURE.

%7.7
\section{}
As shown in the text eqns 7.8 \ and 7.9 may be written in polar
coordinates as $\tfrac{dr}{dt}=r\left( \mu -r^{2}\right) =f(r)$
and $\tfrac{d\theta }{dt}=1.$ The equation $f(r)=0$ is satisfied
for the points $r^{\ast }=\pm \sqrt{\mu }$ which exist for $\mu
>0$ and for $r^{\ast }=0.$ For  $r^{\ast }=0$ we have
$\tfrac{df}{dr}=\mu $ which means that it is stable for $\mu <0$
and unstable for $\mu >0.$ The points satisfying $r^{\ast }=\pm
\sqrt{\mu }$ give $\tfrac{df}{dr}=-\mu $ which is always stable
(given that $\mu >0$ ).

%7.8
\section{}
Multiply eqn 7.8 by $x$ and eqn 7.9 by $y$ and add them together.
Then $x\tfrac{dx}{dt}+y\tfrac{dy}{dt}=\tfrac{1}{2}
\tfrac{dr^{2}}{dt} =r\tfrac{dr}{dt}=r^{2}\left( \mu -r^{2}\right)
$ .

%7.9
\section{}
The equation $\tfrac{d^{2}x}{dt^{2}}+b\left( x^{2}-1\right)
\tfrac{dx}{dt}+x=a$ can be written as $\dot{x}=y$ and
$\dot{y}=a-b\left( x^{2}-1\right) y-x.$ The Jacobian of this
system is thus:\\
\centerline{$J (x,y)= \left(
\begin{array}{cccc}
0 & 1 \\
-2bxy-1 & -b(x^2-1)  \\
\end{array}
\right) $} \\
The fixed points satisfy $\dot{x}=y=0$ and $\dot{y}=0$ which gives
$x=a.$ In such a case the characteristic polynomial is $\lambda
\left[ \lambda +b\left( a^{2}-1\right) \right] +1=0$ which has
solutions $\lambda =\tfrac{-b\left( a^{2}-1\right) \pm
i\sqrt{4-b^{2}\left( a^{2}-1\right) ^{2}}}{2}=A(a,b)\pm iB(a,b).$
We have Hopf bifurcations when $b=0$ since $A(a,0)=0$, $B(a,0)\neq
0$ and $\tfrac{dA}{db}=-\tfrac{\left( a^{2}-1\right) }{2}$ which
is greater than zero provided that $a<1.$ We may also have Hopf
bifurcations when $a=\pm 1$ since $A(\pm 1,0)=0,$ $B(\pm 1,b)\neq
0$ and $\tfrac{dA}{da}=\mp b$ which is greater than zero when $b$
is positive for $a=-1$ and when $b$ is negative for $a=1.$

%7.10
\section{}
The equation $\tfrac{dr}{dt}=r(\mu +r^{2}-r^{4})=f(r)$ gives the
fixed points $r_{1}^{\ast }=0$ and $r_{\pm }^{\ast 2}=\tfrac{1\pm
\sqrt{1+4\mu }}{2}r_{2,3}^{\ast 2}=\tfrac{1\pm \sqrt{1+4\mu
}}{2}.$ To classify their stability compute $\tfrac{df}{dr}=\mu
+3r^{2}-5r^{4}=-4\mu -2r^{2}+5(\mu +r^{2}-r^{4}).$ Obviously for
$r_{1}^{\ast }$ $\tfrac{df}{dr}=\mu $ which is unstable for $\mu
>0$ and stable for $\mu <0.$ We have three cases : (i) $\mu <-1/4$
in which the only fixed point is  $r_{1}^{\ast }$ which is stable.
(ii) $-1/4<\mu <0$ in which \ we find $r_{+}^{\ast }$ to be stable
and  $r_{-}^{\ast }$ to be unstable.(iii)$\mu >0.$ The cycle
$r_{-}^{\ast }$ vanishes, $r_{+}^{\ast }$ is still stable whereas
$r_{1}^{\ast }$ becomes unstable.HYSTERESIS

%7.11
\section{}
HARD

%7.12
\section{}
Hilborn pages 257 and 258. Too hard to be included. At least a
number of hints should be provided.

%7.13
\section{}
Write the second iterate map keeping the terms up to 3rd order :\\
$F(F(X))=-\left( 1+\mu \right) \left[ -\left( 1+\mu \right)
X+X^{3}\right] +\left[ -\left( 1+\mu \right) X+X^{3}\right]
^{3}\simeq $ \\
$X+2X\mu +X\mu ^{2}-2X^{3}-4X^{3}\mu -3X^{3}\mu ^{2}-X^{3}\mu
^{3}=\left( 1+\mu \right) ^{2}X-X^{3}2+4\mu +3\mu ^{2}+\mu
\allowbreak \Rightarrow F(F(X))\simeq \left( 1+\mu \right)
^{2}X-\left( 1+\mu \right) \left( 2+2\mu +\mu ^{2}\right) X^{3},$
which is identical to eqn 7.19. Then taking $F(F(\pm \sqrt{\mu
}))=\pm \left( 1+\mu \right) ^{2}\sqrt{\mu }\mp \mu \left( 1+\mu
\right) \left( 2+2\mu +\mu ^{2}\right) \sqrt{\mu }=\allowbreak \pm
\left[ \sqrt{\mu }-3\left( \sqrt{\mu }\right) ^{5}-3\left(
\sqrt{\mu }\right) ^{7}-\left( \sqrt{\mu }\right) ^{9}\right] .$
Since $\mu $ is small then we may neglect the 5th, 7th and 9th
powers of $\sqrt{\mu }$ to get $F(F(\pm \sqrt{\mu }))\simeq \pm
\sqrt{\mu }$ which shows that $\pm \sqrt{\mu }$ is approximately a
fixed point.


%7.14
\section{}

Substitute $X^{\ast }=\ln A$ in $F(X)=AXe^{-X}$ to get $F(X^{\ast
})=X^{\ast },$ thus showing that $X^{\ast }$ is a fixed point. In
order to show the existence of the flip bifurcation make the
change of variable $Y=X-\ln A$ and define $G(Y)=F(X)-X^{\ast }$to
get $G(Y)=A\left( Y+\ln A\right) e^{-\left( Y+\ln A\right) }-\ln
A=\left( Y+\ln A\right) e^{-Y}-\ln A.$ The map $Y_{n+1}=G(Y_{n})$
has a fixed point at $Y=0.$ Expand $G(Y)$ about $0$ up to order 3
to get : \\
$\vspace{1pt}G(Y)\simeq \left( Y+\ln A\right) \left(
1-Y+\tfrac{1}{2}Y^{2}-\tfrac{1}{6}Y^{3}\right) -\ln A=\left( 1-\ln
A\right) Y+\left( \frac{1}{2}\ln A-1\right) Y^{2}+\left(
\frac{1}{2}-\frac{1}{6}\ln A\right) Y^{3}$ \\
INCOMPLETE \\
$F(X)\simeq \allowbreak \ln A+\left( 1-\ln A\right)
\left( X-\ln A\right) +\frac{1}{2}\left( -2+\ln A\right) \left(
X-\ln A\right) ^{2}+\frac{1}{6}\left( 3-\ln A\right) \left( X-\ln
A\right) ^{3}=\allowbreak \left( 1+\ln A-\frac{1}{2}\ln
^{3}A+\frac{1}{2}\ln ^{2}A\right) X-\left( 1+\ln
A-\frac{1}{2}\left( \ln ^{2}A\right) \right) X^{2}+\left(
\frac{1}{2}-\frac{1}{6}\ln A\right) X^{3}+\frac{1}{6}\ln ^{4}A$


%7.15
\section{}

%7.16
\section{}

%7.17
\section{}

%7.18
\section{}

%7.19
\section{}

%7.20
\section{}

%7.21
\section{}

%7.22
\section{}

%7.23
\section{}

%7.24
\section{}

%7.25
\section{}

%7.26
\section{}

%7.27
\section{}

%7.28
\section{}




% CHAPTER 8***********
\chapter{}
% ********************

%8.1
\section{}
$\tfrac{dH}{dt}=\sum_{i}\left( \tfrac{\partial H}{\partial
p_{i}}\tfrac{dp_{i}}{dt}+\tfrac{\partial H}{\partial
q_{i}}\tfrac{dq_{i}}{dt}\right) =\sum_{i}\left( -\tfrac{\partial
H}{\partial p_{i}}\tfrac{\partial H}{\partial
q_{i}}+\tfrac{\partial H}{\partial q_{i}}\tfrac{\partial
H}{\partial p_{i}}\right) =0$

%8.2
\section{}
$H=\tfrac{p^{2}}{2mL^{2}}+mgL\left( 1-\cos x\right) .$ In this
case we have $q\equiv x$. Eqn 8.5 gives :
$\tfrac{dp}{dt}=-\tfrac{\partial H}{\partial \theta }=-mgL\sin
x=mL^{2}\dfrac{d^{2}x}{dt^{2}}\Rightarrow \dfrac{dv}{dt}=-\omega
^{2}\sin x$ with $v=\dfrac{dx}{dt}$ and $\omega
=\sqrt{\tfrac{g}{L}}.$

%8.3
\section{}
The Jacobian Matrix is given by: $ J (x,v)= \left(
\begin{array}{cccc}
0 & 1 \\
-\omega^2 \cos x & 0 \\
\end{array}
\right) $ which, evaluated at $x=\pi$ , $v=0$ gives $ J (\pi,0)=
\left(\begin{array}{cccc}
0 & 1 \\
\omega^2 & 0 \\
\end{array}
\right) $. The characteristic polynomial is $\lambda^2-\omega^2=0$
which gives $\lambda=\pm \omega \Longrightarrow $ at $x=\pi$ ,
$v=0$ we have a saddle point.

%8.4
\section{}
Start from $E-mgL=\tfrac{mL^{2}}{2}\left( \tfrac{dx}{dt}\right)
^{2}-mgL\cos x$. Let $x_{0}$ be the maximum value of $x$, which
satisfies $\cos x_{0}=-\tfrac{E-mgL}{mgL}$ . Hence
$\tfrac{L}{2g}\left( \tfrac{dx}{dt}\right) ^{2}=\cos x-\cos
x_{0}.$ Solving for $t$ :
$dt=\sqrt{\tfrac{L}{2g}}\tfrac{dx}{\sqrt{\cos x-\cos x_{0}}}$ .
Integrate from $0$ to $x_{0}$ to get 1/4 of the period $T$ which
can be written as
$T=4\sqrt{\tfrac{L}{2g}}\int_{0}^{x_{0}}\tfrac{dx}{\sqrt{\cos
x-\cos
x_{0}}}=2\sqrt{\tfrac{L}{g}}\int_{0}^{x_{0}}\tfrac{dx}{\sqrt{\sin
^{2}\frac{1}{2}x-\sin ^{2}\frac{1}{2}x_{0}}}.$ Make the
substitution $\sin y=\sin ^{2}\frac{1}{2}x/\sin
^{2}\frac{1}{2}x_{0}$ to convert it to
$T=4\sqrt{\tfrac{L}{g}}K(\sin \frac{1}{2}x_{0})$ where
$K(s)=\int_{0}^{\pi /2}\tfrac{ds}{\sqrt{1-k^{2}\sin ^{2}s}}$ is
the complete elliptic integral of the first kind. Since $\sin
\frac{1}{2}x_{0}=\sqrt{\tfrac{1-\cos
x_{0}}{2}}=\sqrt{\tfrac{E}{2mgL}}$, use these formulas to get
$T_{E/mgL}$ : \\
$T_{0.1}=6.364\sqrt{\tfrac{L}{g}}$ ,
$T_{1.0}=7.416\sqrt{\tfrac{L}{g}}$ \ , $T_{2.0}=\infty $.

%8.5
\section{}
In this case we have $q_{1}\equiv x$ , $q_{2}\equiv y$ ,
$p_{1}\equiv v$ , $p_{2}\equiv w$ . Using eqns 8.4 and 8.5 we get
: $\dfrac{dp_{1}}{dt}=\dfrac{dv}{dt}=-\dfrac{\partial H}{\partial
q_{1}}=-\dfrac{\partial H}{\partial x}=-\omega ^{2}\sin x+Ay$ \ ,
$\dfrac{dp_{2}}{dt}=\dfrac{dw}{dt}=-\dfrac{\partial H}{\partial
q_{2}}=-\dfrac{\partial H}{\partial y}=-\Omega ^{2}y+Ax$ ,
$\dfrac{dq_{1}}{dt}=\dfrac{dx}{dt}=\dfrac{\partial H}{\partial
p_{1}}=\dfrac{\partial H}{\partial v}=v$ and
$\dfrac{dq_{2}}{dt}=\dfrac{dy}{dt}=\dfrac{\partial H}{\partial
p_{2}}=\dfrac{\partial H}{\partial w}=w$ .

%8.6
\section{}
We have $q_{1}\equiv x$ , $q_{2}\equiv y$ , $p_{1}\equiv v$ ,
$p_{2}\equiv w$ . Use eqns 8.4 and 8.5 to get :
$\dfrac{dq_{1}}{dt}=\dfrac{dx}{dt}=\dfrac{\partial H}{\partial
p_{1}}=\dfrac{\partial H}{\partial v}=v$ \ ,
$\dfrac{dq_{2}}{dt}=\dfrac{dy}{dt}=\dfrac{\partial H}{\partial
p_{2}}=\dfrac{\partial H}{\partial w}=w$ ,
$\dfrac{dp_{1}}{dt}=\dfrac{dv}{dt}=-\dfrac{\partial H}{\partial
q_{1}}=-\dfrac{\partial H}{\partial x}=-2xy-x$ and
$\dfrac{dp_{2}}{dt}=\dfrac{dw}{dt}=-\dfrac{\partial H}{\partial
q_{2}}=-\dfrac{\partial H}{\partial y}=-y-x^{2}+y^{2}$ .

%8.7
\section{}
LOOK UP FORMULA IN FORD'S REFERENCE

%8.8
\section{}
First note that $x=\dfrac{\chi +\psi }{2}$ , $y=\dfrac{\chi -\psi
}{2}.$ Eqn 8.18 may be written as
$2\dfrac{dv}{dt}=2\dfrac{d^{2}x}{dt^{2}}=\dfrac{d^{2}\chi
}{dt^{2}}+\dfrac{d^{2}\psi }{dt^{2}}=-\chi -\psi -\chi ^{2}+\psi
^{2}$ and eqn 8.19 (with $y^{2}$ replaced by $-y^{2}$) as
$2\dfrac{dw}{dt}=2\dfrac{d^{2}y}{dt^{2}}=\dfrac{d^{2}\chi
}{dt^{2}}-\dfrac{d^{2}\psi }{dt^{2}}=-\chi +\psi -\chi ^{2}-\psi
^{2}.$ Solve for $\dfrac{d^{2}\chi }{dt^{2}}$ and
$\dfrac{d^{2}\psi }{dt^{2}}$ to obtain eqns 8.21 and 8.22. \\
SOLUTION??

%8.9
\section{}
Use eqn 5.24 with $f=y$ , $g=-x+yz$ and $h=1-y^{2},$ from which it
follows that $\dfrac{1}{V}\dfrac{dV}{dt}=\lim\limits_{T \to
\infty}\dfrac{1}{T}\int_{0}^{T}zdt=\left\langle z\right\rangle $

%8.10
\section{}
If we make the change of variables $t$ $\rightarrow $ $-t$ ,
$z\rightarrow -z$ and $y\rightarrow -y$ we still get the same
equations. Therefore the system is time reversible.

%8.11
\section{}
Since performing the change of variables $t$ $\rightarrow $ $-t$ ,
$z\rightarrow -z$ and $x\rightarrow -x$ we still get the same
equations, the system is time reversible. Also , using eqn 5.24
together with $f=-y,$ $g=x+z$ and $h=xz+3y^{2}$ we get for the
time-reversed system :
$\dfrac{1}{V}\dfrac{dV}{dt}=\lim\limits_{T\rightarrow \infty
}\dfrac{1}{T}\int_{T}^{0}xdt=-\left\langle x\right\rangle $ .

%8.12
\section{}
Use $\dfrac{dx}{dt}=y$ , $\dfrac{dy}{dt}=-x+yz$ and
$\dfrac{dz}{dt}=1-y^{2}$ .Since
$\dfrac{dy}{dt}=\dfrac{d^{2}x}{dt^{2}}=-x+\dfrac{dx}{dt}z\Rightarrow
z=\left( \dfrac{d^{2}x}{dt^{2}}+x\right) /\dfrac{dx}{dt}$ . Taking
the derivative of $z$ with respect to $t$ we get
$\dfrac{dz}{dt}=\left(\dfrac{d^{3}x}{dt^{3}}+\dfrac{dx}{dt}\right)
/\dfrac{dx}{dt}-\dfrac{d^{2}x}{dt^{2}}\left(
\dfrac{d^{2}x}{dt^{2}}+x\right) /\left( \dfrac{dx}{dt}\right)
^{2}=1-\left( \dfrac{dx}{dt}\right) ^{2}.$ Simplifying the last
result yields eqn 8.31.

%8.13
\section{}
The differential equation for the simple pendulum can be written
as $\dfrac{dx}{dt}=y=f\left( x,y\right) $ and
$\dfrac{dy}{dt}=k\sin x=g\left( x,y\right) .$ For the case of the
Leapfrog method see solutions to problems 8.20 and 8.21. For the
Runge-Kutta method use $k_{1xn}=Y_{n}$ , $k_{1yn}=k\sin X_{n}$ ,
$k_{2xn}=Y_{n}+\tfrac{1}{2}k\sin X_{n}$ and $k_{2yn}=k\sin
(X_{n}+\tfrac{1}{2}Y_{n}).$ Then
$X_{n+1}=X_{n}+Y_{n}+\tfrac{1}{2}k\sin X_{n}$ and
$Y_{n+1}=Y_{n}+k\sin (X_{n}+\tfrac{1}{2}Y_{n}).$ The Jacobian of
this map has determinant $\det J=\left( 1+\tfrac{1}{2}k\cos
X_{n}\right) \left( 1+\tfrac{1}{2}k\cos
(X_{n}+\tfrac{1}{2}Y_{n})\right) -k\sin
(X_{n}+\tfrac{1}{2}Y_{n})\neq 1.$ Hence the map in this case is
not simplectic (area-preserving).

%8.14
\section{}
Use the Jacobian of the map is given at problem 5.12. whose
determinant is $\det J=\cos ^{2}\alpha +\sin ^{2}\alpha =1.$

%8.15
\section{}
The H\'{e}non area-preserving quadratic map is not invertible.

%8.16
\section{}
The Jacobian is given by $ J(X,Y)= \left(
\begin{array}{cccc}
a_2+2a_4X+a_5Y & a_3+a_5X+2a_6Y  \\
1 & 0   \\
\end{array}
\right) $ \\
The map is area preserving if $|\det J|=|-a_3-a_5X-2a_6Y|=1$ for
all $X$ and $Y$. This is the case only if $|a_3|=1$ and
$a_5=a_6=0$.


%8.17
\section{}
The Jacobian is $ J(X,Y)= \left(
\begin{array}{cccc}
1 & 1  \\
1 & 2   \\
\end{array}
\right) $ whose eigenvalues are $\lambda _{\pm}=\tfrac{3\pm
\sqrt{5}}{2}$ with eigenvectors $v_{\pm}=\left(
\begin{array}{cccc}
1  \\
\lambda _{\pm}-1\\
\end{array}
\right)=\left(
\begin{array}{cccc}
1  \\
\tfrac{1\pm \sqrt{5}}{2}   \\
\end{array}
\right)$ Since $v_+\cdot v_-=0$ we may conclude that the two
manifolds are perpendicular to each other. The unstable manifold
corresponds to $\lambda _{+}$. The tangent of the angle of $v_{+}$
with the $x$-axis is given by the ratio of the y over the x
component of the vector which is obviously equal to the golden
mean.

%8.18
\section{}
As indicated in the text on page 169 if we start from $\left(
-1,3\right) $ $\rightarrow \left( -1,-1\right) \rightarrow \left(
3,-1\right) \rightarrow \left( 5,3\right) \rightarrow \left(
3,5\right) \rightarrow \left( -1,3\right) $ which is a period 5
cycle.

%8.19
\section{}
Use the Jacobian of the map is given at problem 5.12. whose
determinant is $\det J=\cos ^{2}\alpha +\sin ^{2}\alpha =1.$

%8.20
\section{}
he differential equation for the simple pendulum
$\tfrac{d^{2}x}{dt^{2}}+k\sin x=0$ can be transformed to a system
of differential equations :$\tfrac{dx}{dt}=y=f\left( x,y\right) $
and $\tfrac{dy}{dt}=k\sin x=g\left( x,y\right) $. The Leapfrog
method in the y-variable can be written as
$Y_{n+1}=Y_{n}+hg(X_{n},Y_{n})$ and
$X_{n+1}=X_{n}+hf(X_{n},Y_{n+1})$. Since we are dealing with
angles we may take the equations modulo $2\pi $ and if we use
$h=1$ we finally obtain eqns 8.40 and 8.41.

%8.21
\section{}
Use the Jacobian from the solution to problem 5.12 and calculate
its determinant which equals $\det J=1+k\cos X-k\cos X=1$.
Therefore the map is area-preserving.

%8.22
\section{}
The fixed points satisfy $Y_{n}(\textsf{mod}2\pi )=0$ and $k\sin
X_{n}(\textsf{mod}2\pi )=0.$ The fixed points are $Y^{\ast }=0$
and $X^{\ast }=0$ or $\pi .$ The Jacobian of the map is given in
the solution of problem 5.12 with characteristic polynomial
$\lambda ^{2}\allowbreak -\lambda \left( 2+k\cos X\right) +1=0$.
When $X^{\ast }=0$ we get $\lambda ^{2}-\left( 2+k\right) \lambda
+1=0$ whose solution is $\lambda =1+\frac{1}{2}k\pm
\frac{1}{2}k\sqrt{\left( \tfrac{4}{k}+1\right) }.$ So for all $k$
we have a saddle point.When $X^{\ast }=\pi $ we have the equation
$\lambda ^{2}-\left( 2-k\right) \lambda +1=0$ whose solution is
$\lambda =1-\frac{1}{2}k+\frac{1}{2}k\sqrt{\left(
1-\tfrac{4}{k}\right) }$. In this case we have an unstable spiral
point for $0<k<2,$ a stable spiral for $2<k<4$ and a stable fixed
point for $k>4.$

%8.23
\section{}
Start with $X_{0}=0$ and $Y_{0}=\pi .$ The first iteration will
give $Y_{1}=\pi (\textsf{mod}2\pi )=\pi $ and $X_{1}=\pi
(\textsf{mod}2\pi )=\pi $ . The second iteration will give
$Y_{2}=\pi (\textsf{mod}2\pi )=\pi $ and $X_{2}=2\pi
(\textsf{mod}2\pi )=0$ which correspond to the initial point
$\left( X_{0},Y_{0}\right) $ .

%8.24
\section{}
One solution is $(0,\frac{2 \pi}{3}) $ $\rightarrow (\frac{2
\pi}{3},\frac{2 \pi}{3}) \rightarrow (\frac{4 \pi}{3},\frac{2
\pi}{3}) \rightarrow (0,\frac{2 \pi}{3}) $

%8.25
\section{}
The sum of the eigenvalues equals to the logarithm of the
determinant which in this case equals to $\log|\det
J|=\log|b(1+k\cos X)-bk\cos X|=\log|b|$.

%8.26
\section{}
In order to show that $Q$ is conserved we have to show that
$A=(X_{n}^{2}+X_{n-1}^{2}+X_{n-2}^{2}-X_{n}X_{n-1}X_{n-2})-
(X_{n-1}^{2}+X_{n-2}^{2}+X_{n-3}^{2}-X_{n-1}X_{n-2}X_{n-3})=0.$
$A$ simplifies to $A=X_{n}^{2}-X_{n-3}^{2}-X_{n-1}X_{n-2}\left(
X_{n}-X_{n-3}\right) =X_{n}^{2}-\left( X_{n-1}X_{n-2}-X_{n}\right)
^{2}-X_{n-1}X_{n-2}\left( 2X_{n}-X_{n-1}X_{n-2}\right) =0.$ Hence
$Q$ is conserved.

% CHAPTER 09**********
\chapter{}
% ********************



% CHAPTER 10**********
\chapter{}
% ********************
%10.1
\section{}
?
%10.2
\section{}
Assume the polynomial to be of the form $y=az^{2}+bz+c$ and
passing through the points $\left( -1,X_{n-2}\right) $ , $\left(
0,X_{n-1}\right) $ and $\left( 1,X_{n}\right) $. Solving the
resulting system for $a,b$ and $c$ we get
$a=\frac{1}{2}X_{n}+\frac{1}{2}X_{n-2}-X_{n-1}$ ,
$b=\frac{1}{2}X_{n}-\frac{1}{2}X_{n-2}$ and $c=X_{n-1}.$ Using
these coefficients set $z$ equal to 2 to find $X_{n+1}.$
%10.3
\section{}
Assume the polynomial to be of the form $ y=az^{3}+bz^{2}+cz+d $
and passing trough the points $(-2,X_{n-3})$, $(-1,X_{n-2})$ ,
$(0,X_{n-1})$ and $\left(1,X_{n}\right)$.Solving the resulting
system for $a,b,c$ and $d$ we get
$a=-\frac{1}{2}X_{n-1}+\frac{1}{6}X_{n}+\frac{1}{2}X_{n-2}-
\frac{1}{6}X_{n-3}$ ,
$b=-X_{n-1}+\frac{1}{2}X_{n}+\frac{1}{2}X_{n-2}$ ,
$c=\frac{1}{2}X_{n-1}+\frac{1}{3}X_{n}-X_{n-2}+\frac{1}{6}X_{n-3}$
and $d=X_{n-1}$. Using these coefficients set $z$ equal to 2 to
find $X_{n+1}$.

%10.4
\section{}
Assume the polynomial to be of the form
$y=az^{4}+bz^{3}+cz^{2}+dz+e$ and passing through the points
$\left( -2,X_{n-4}\right) $ , $\left( -1,X_{n-3}\right) $ ,
$\left( 0,X_{n-2}\right) $ , $\left( 1,X_{n-1}\right) $ and
$\left( 2,X_{n}\right) $. Solving the resulting system for
$a,b,c,d$ and $e$ we get
$a=\frac{1}{4}X_{n-2}+\frac{1}{24}X_{n}+\frac{1}{24}X_{n-4}-\frac{1}{6}X_{n-1}-\frac{1}{6}X_{n-3}$
,
$b=\frac{1}{12}X_{n}-\frac{1}{12}X_{n-4}-\frac{1}{6}X_{n-1}+\frac{1}{6}X_{n-3}$
,
$c=-\frac{5}{4}X_{n-2}-\frac{1}{24}X_{n}-\frac{1}{24}X_{n-4}+\frac{2}{3}X_{n-1}+\frac{2}{3}X_{n-3}$
,
$d=-\frac{1}{12}X_{n}+\frac{1}{12}X_{n-4}+\frac{2}{3}X_{n-1}-\frac{2}{3}X_{n-3}$
and $e=X_{n-2}.$ Using these coefficients set $z$ equal to 3 to
find $X_{n+1}$.

%10.5
\section{}
Assume the polynomial to be of the form
$y=az^{5}+bz^{4}+cz^{3}+dz^{2}+ez+f$ and passing through the
points $\left( -3,X_{n-5}\right) $, $\left( -2,X_{n-4}\right) $,
$\left( -1,X_{n-3}\right) $, $\left( 0,X_{n-2}\right) $, $\left(
1,X_{n-1}\right) $ and $\left( 2,X_{n}\right) $. Solving the
resulting system for $a,b,c,d,e$ and $f$ we get:
$a=\frac{1}{12}X_{n-2}-\frac{1}{12}X_{n-3}-\frac{1}{24}X_{n-1}+\frac{1}{120}X_{n}-\frac{1}{120}\allowbreak
X_{n-5}+\frac{1}{24}X_{n-4}$ ,\\
$b=\frac{1}{4}X_{n-2}+\frac{1}{24}X_{n}+\frac{1}{24}X_{n-4}-\frac{1}{6}X_{n-1}-\frac{1}{6}X_{n-3}$
,\\
$c=-\frac{5}{12}X_{n-2}+\frac{7}{12}X_{n-3}+\frac{1}{24}X_{n-1}+\frac{1}{24}X_{n}+\frac{1}{24}X_{n-5}-\allowbreak
\frac{7}{24}X_{n-4}$ ,\\
$d=-\frac{5}{4}X_{n-2}+\frac{2}{3}X_{n-3}+\frac{2}{3}X_{n-1}-\frac{1}{24}X_{n}-\frac{1}{24}X_{n-4}\
$,\\ $e=$
$-\frac{1}{20}X_{n}-\frac{1}{30}X_{n-5}+\frac{1}{4}X_{n-4}-X_{n-3}+\frac{1}{3}X_{n-2}+\frac{1}{2}X_{n-1}$
and $f=X_{n-2}.$ \\
Using these coefficients set $z$ equal to 3 to find $X_{n+1}$.

%10.6
\section{}
The values to be predicted are:\\
$X_{11}=\sin 11=\allowbreak -0.\,\allowbreak 99999$, $X_{12}=\sin
12=\allowbreak -0.\,\allowbreak 53657$ and $X_{13}=\sin
13=\allowbreak 0.\,\allowbreak 42017$ Using eqn 10.1 we obtain:\\
$X_{11}=-1.\,\allowbreak 8791$, $X_{12}=-3.\,\allowbreak 593\,1$
and $X_{13}=\allowbreak -5.\,\allowbreak 686$ \\
Using eqn 10.2:\\
$X_{11}=-1.5002$, $X_{12}=-2.4564$ and $X_{13}=-3.4126$\\
Using eqn 10.3:\\
$X_{11}=-1.3483$, $X_{12}=-1.47$ and $X_{13}=-0.3784$ \\
Using eqn 10.4:\\
$X_{11}=-0.\,\allowbreak 51206$, $X_{12}=\allowbreak
2.\,\allowbreak 7113$ and $X_{13}=\allowbreak 12.166$ \\
Using eqn 10.5:\\
$X_{11}= -0.23111$, $X_{12}=4.397$ and $X_{13}=18.035$\\

%10.7
\section{}
Using the original map, the first 6 points are: $X_{1}=0.1,$
$X_{2}=\allowbreak 0.36,$ $X_{3}=0\allowbreak .9216,$
$X_{4}=0\allowbreak .28901$, $X_{5}=0\allowbreak .821\,93$ and
$X_{6}=0.58544$. Knowing the first 3 points we may use eqn 10.1 to
get $X_{4}=1.4832$, $X_{5}=2.0448$ and $X_{6}=2.6064$. Applying
eqn 10.2 to the same set of 3 points we get $X_{4}=1.7848$,
$X_{5}=2.9496$ and $X_{6}=4.416$.

%10.8
\section{}
Set $j=m=1$ in eqns 10.7 and 10.8 and the result follows
immediately.

%10.9
\section{}
$a_{1}=\dfrac{X_{4}X_{3}+X_{3}X_{2}+X_{2}X_{1}}{X_{3}^{2}+X_{2}^{2}+X_{1}^{2}}=\dfrac{32+8+2}{16+4+1}=2.$
Hence $X_{n+1}=2X_{n}.$

%10.10
\section{}
$a_{1}=\dfrac{X_{2}X_{1}}{X_{1}^{2}}=5.$ Hence $X_{n+1}=5X_{n}$ so
$X_{3}=25,$ $X_{4}=125$ and $X_{5}=625$

%10.11
\section{}


% CHAPTER 11**********
\chapter{}
% ********************



% CHAPTER 12**********
\chapter{}
%********************
%12.1
\section{}
$D=\log 3/\log 5=0.6826$

%12.2
\section{}
$D=\log 3/\log 7=0.6826$

%12.3
\section{}
$D=\log 4/\log 3=1.262$

%12.4
\section{}
Answers given in section 11.4

%12.5
\section{}
Answer given in section 11.5

%12.6
\section{}
At the $n$-th stage of the construction we need a minimum of
$2^{m}$ boxes with $L=(1/3)^{m}$, so $D_{0}=-\frac{-\log
2^{m}}{\log (1/3)^{m}}=\frac {\log 2}{\log 3}=0.6309$

%12.7
\section{}

%12.8
\section{}

%12.9
\section{}
By the definition of $C(r)$ we have $C(0)=0$. As $r \rightarrow
\infty$ all Heaviside functions evaluate to 1 since a circle with
infinite radius contains all $X_{n}$'s. In that case the double
sum evaluates to $\sum\limits_{j=1}\limits^{N}
(N-j)=1+2+...+N-1=\tfrac{N(N-1)}{2}$. This gives $C(\infty)=1$.

%12.10
\section{}
$C(r)=\int\limits_{0}\limits^{r}f(r)dr$, where $f(r)$ is the
distribution of points on the real line.

%12.11
\section{}
Referring to problem 12.10,let $f(r)=1$. In this case the integral
is performed on a surface so $C(r)=\int\limits_{0}\limits^{2 \pi
}\int\limits_{0}\limits^{r}r drd\theta=\pi r^2$. Hence
$D_{2}=\frac{d\log (\pi r^2)}{d\log r}=\frac{d [2\log r+\log
\pi]}{d \log r}=2$

%12.12
\section{}

%12.13
\section{}

%12.14
\section{}

%12.15
\section{}
The answers for $D=1,2,...,10$ are $252, 631, 1585, 3982, 10000,
25119, 63096, 1.58\times 10^{5},3.98\times 10^{5},$ $10^{6}$
respectively.

%12.16
\section{}

%12.17
\section{}

%12.18
\section{}
$D=\tfrac {\log \tfrac{494}{298}}{\log \tfrac{5}{3}}=0.989$

%12.19
\section{}
$D=\tfrac {\log \tfrac{244036}{88804}}{\log \tfrac{5}{3}}=1.979$


%12.20
\section{}
$D=\tfrac {\log \tfrac{5(a+b)-2}{3(a+b)-2}}{\log \tfrac{5}{3}}$




\end{document}
