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{\bf Question}

Consider the Henon map with $b=0.4$.  Verify the following facts:

\begin{description}
\item[(i)]
For $-0.09<a<0.27$ there is one sink fixed point and one saddle
fixed point.

\item[(ii)]
As $a$ increases through 0.27, the largest magnitude eigenvalue of
the first fixed point passes through -1, and an attracting 2-cycle
is created.

\item[(iii)]
This 2-cycle ceases to be attracting as $a$ increases through
0.85.
\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
\underline {Fixed points} $(x,y$ given by $x^2+(1-b)x-a=0,\ y=x$.
Eigenvalues of $Df(y)$ given by $\lambda^2+2x\lambda-6=0,\
\lambda=-x\pm\sqrt{x^2+b}$: real as $b>$%gap?%

As $a$ increases from -0.09 to 0.27 the x-coordinate of the fixed
points spread out from -0.3 (repeated) to $x=-0.9,\ +0.3$.

For $x<-0.3$ we have $x^2+b>0.49$ so $+\sqrt{x^2+b}>0.7$ and
therefore on eigenvalue $\lambda>1$: \underline{saddle} (the other
eigenvalue stays inside the unit circle).

For $-0.3<x<0.3$ we have $x^2+b<0.49$ so both $|\lambda|<1$:
\underline{sink}.

\item[(ii)]
As $x$ increases through 0.3 (that is $a$ increases through 0.27)
we see $\lambda=-x-\sqrt{x^2+b}$ decreases through-  \ \ \ \ We %gap?%
expect this to lead to creation of a 2-cycle, but check:

2-cycle is created \ \ \ \ \ as a increases through

$\ds\frac{3}{4}(1-b)^2=\ds\frac{3}{4}(0.6)^2=0.27. \surd$

Moreover this occurs at point (x,x) where
$x=\ds\frac{1}{2}(1-b)=0.3.\surd$

The eigenvalues of $Df^2$ at a per-2 point are
$\lambda=t\pm\sqrt{t^2-b^2}$ where (with $b=0.4$) we have
$t=1.12-2a$.  For a just $>0.27$ we have $t<0.58$ and
$t^2-b^2<0.1764$ so $+\sqrt{t^2-b^2}<0.42$: thus
$\lambda=t+\sqrt{t^2-b^2}$ just<1, so the 2-cycle is attracting.

\item[(iii)]
As $a$ increases thourgh 0.85 we see that:

$t$ decreases through -0.58

$\sqrt{t^2-b^2}$ increases through 0.42.

So the eigenvalue
$\lambda=t-\sqrt{t^2-b^2}$ decreases through -1; 2-cycle no longer
attracting.

\end{description}
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