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\begin{document}
\centerline{Math 322 lecture for Friday, Week 11}
\bigskip

Recall the notation from last time.  We are studying a planar system
\begin{align}\label{eqn:11frisys}
  x'&= P(x,y)\\
  y'&= Q(x,y)
\end{align}
where~$P$ and~$Q$ are polynomials.  We embedded our system in the plane~$z=1$
in~$\R^3$ and projected the flow along lines centered at the origin onto the
unit sphere~$S$ centered at the origin. This flow induced a flow along the
equator of the sphere. We are interested in critical points of this flow along
the equator, where~$z=0$.  To find those we saw that we should \ldots
\bigskip

\textsc{Step 1.} Clear denominators in the equation
\[
  yP\left( \frac{x}{z},\frac{y}{z}\right)-xQ\left(  \frac{x}{z},\frac{y}{z}
  \right)=0,
\]
and then set~$z=0$. 
\bigskip

The result is several points of the form~$(a,b,0)$ on
the equator of the sphere.  

\textsc{Step 2.} To analyze these, we will project the flow on
the sphere to either the plane~$x=1$ or the plane~$y=1$.  If~$a,b$ are both
nonzero, then either plane will do.  If~$a=0$, then the point in question
is~$(0,1,0)$, and we'd need to project to the plane~$y=1$, and if~$b=0$, the
point is~$(1,0,0)$, and we'd need to project to~$x=1$.  We'll consider these
cases separately:
\medskip

{\bf Projection to the plane~$x=1$.}  We project the point~$(x,y,1)$ along a
line through the origin, i.e., we scale this point, to get a point on the
plane~$x=1$:
\[
  (x,y,1)\rightsquigarrow \left( 1,\frac{y}{x},\frac{1}{x} \right).
\]
Let
\[
  u :=\frac{y}{x}\quad\text{and}\quad v:=\frac{1}{x}.
\]
We use the coordinates to identify the plane~$x=1$ with the ordinary
plane~$\R^2$.  It follows that
\[
  x=\frac{1}{v}\quad\text{and}\quad y=\frac{u}{v}.
\]
Plug these into system~\ref{eqn:11frisys}:
\begin{align*}
  x'=P(x,y)&\quad\Rightarrow\quad \left( \frac{1}{v} \right)'=P\left( \frac{1}{v},\frac{u}{v} \right)\\[10pt]
	   &\quad\Rightarrow\quad v'=-v^2P\left( \frac{1}{v},\frac{u}{v} \right)
\end{align*}
Similarly,
\allowdisplaybreaks
\begin{align*}
  y'=Q(x,y)&\quad\Rightarrow\quad \left( \frac{u}{v} \right)'=Q\left( \frac{1}{v},\frac{u}{v} \right)\\[10pt]
	   &\quad\Rightarrow\quad u'v-uv' =v^2Q\left( \frac{1}{v},\frac{u}{v}
	   \right)\\[10pt]
	   &\quad\Rightarrow\quad u'v =v^2Q\left( \frac{1}{v},\frac{u}{v} \right)+uv'\\[10pt]
	   &\quad\Rightarrow\quad u'v =v^2Q\left( \frac{1}{v},\frac{u}{v} \right)
	   -uv^2P\left( \frac{1}{v},\frac{u}{v}\right)\\[10pt]
	   &\quad\Rightarrow\quad u' =v\left(Q\left( \frac{1}{v},\frac{u}{v} \right)
	   -uP\left( \frac{1}{v},\frac{u}{v}\right)\right).
\end{align*}
So the system in the~$u,v$-plane is
\begin{align}\label{11fri_original}
  u' &= v\left(Q\left( \frac{1}{v},\frac{u}{v} \right) -uP\left( \frac{1}{v},\frac{u}{v}\right)\right)\\[10pt]
  v'&= -v^2P\left( \frac{1}{v},\frac{u}{v} \right).\nonumber
\end{align}
The problem is that it is likely this system is not defined where~$v=0$
(at the equator).  To get the induced flow on the equator, we
need to clear denominators (thus, scaling the vector field but not changing
its direction at any point).  Define
\[
  d:=\max\left\{ \deg P,\deg Q \right\}
\]
To clear denominators we scale the vector field in~\eqref{11fri_original} by~$v^{d-1}$ to get the system
\begin{empheq}[box=\fbox]{align*}
  u' &= v^d\left(Q\left( \frac{1}{v},\frac{u}{v} \right) -uP\left( \frac{1}{v},\frac{u}{v}\right)\right)\\[10pt]
  v'&= -v^{d+1}P\left( \frac{1}{v},\frac{u}{v} \right).
\end{empheq}
We analyze the point~$\left( \frac{b}{a},0 \right)$, since this is the point
corresponding to~$(a,b,0)$ in the~$u,v$-plane.
\medskip

\textbf{Important points:}  The right side of the system~\eqref{11fri_original} 
defines the vector field whose trajectories we would like to determine.  What
effect does scaling that vector field by~$v^{d-1}$ have on the solution
trajectories?  The vector field gives the tangent vector for a solution
trajectory.  So one effect is to scale the speed of the trajectory by the
magnitude $|v|^{d-1}$.  Note that~$v=1/x$ where~$x$ comes from the
point~$(x,y,1)$ in the~$z=1$ plane.  As we go ``out to infinity'' in the $z=1$
plane, by taking~$x$ larger, the scaling factor $|v^{d-1}|$ decreases in
magnitude.  We have chosen~$d$ just write so that the resulting vector field
does not blow up on the equator and is also not identically the zero on the equator.

What about the direction of the trajectory?  Since both
components of the vector field are scaled the same amount, there are two
choices: (i) if $v^{d-1}>0$, the direction is the same, and (ii) if $v^{d-1}<0$,
the direction is reversed.  Next, what significance does this have for analyzing
critical points at the equator?  Suppose we are interesting in a trajectory
corresponding containing a point $(x,y,1)$ in the original~$z=1$ plane.
If~$x>0$, then since~$v=1/x>0$, it follows that~$v^{d-1}>0$, and the direction
does not change.  On the other hand, if~$x<0$, then~$v=1/x<0$.  If~$d$ is odd,
then $v^{d-1}>0$, and if~$d$ is even, then~$v^{d-1}<0$. So in the latter case,
in which~$d$ is even, the direction of the vector field and hence the direction of it
solution trajectories is reversed.
\medskip

{\bf Projection to the plane~$y=1$.} By a similar analysis (which will be
assigned for homework), if~$b\neq0$, we can project to the~$y=1$ plane and
derive an analogous system of equations:
\begin{empheq}[box=\fbox]{align*} u' &= v^d\left(P\left( \frac{u}{v},\frac{1}{v} \right) -uQ\left( \frac{u}{v},\frac{1}{v}\right)\right)\\[10pt] v'&= -v^{d+1}Q\left(\frac{u}{v}, \frac{1}{v} \right). \end{empheq}
We are interested in the point~$\left( \frac{a}{b},0 \right)$ in this plane.
\bigskip

\textbf{Global phase portrait}
A couple of lectures ago, we introduced the global phase portrait of a planar
system.  It is the central projection of the flow of the vector field onto the
top half of the sphere.  In order to compute it, find and analyze all critical points of the
planar system.  Next, find all critical points of the system at infinity.  These
come in antipodal pairs: $(a,b,0)$ and~$(-a,-b,0)$.  Without loss of generality,
suppose~$a>0$.  Then we analyze the scaled system, scaling by~$v^{d-1}$
with~$v=1/x$ in the plane~$x=1$ at the
point~$\left(\frac{b}{a},0\right)$.  At the antipodal point, we analyze the same
system but scaled by~$(-1/x)^{d-1}$ and at the
point~$\left(\frac{-b}{-a},0\right)$, i.e., at the same point.  This means the systems at antipodal points
are either the same when projected to the $x=1$ plane up to a possible reversal
of directions (which happens exactly in the case~$d$ is even.
\newpage

\centerline{\textbf{Exercises}}
\bigskip

In the following exercises, we will analyze the critical points, both
finite and at~$\infty$ for the system
\vskip 0.3in

\begin{align}\label{eqn: 12mon1}
  x'&= x^2+y^2-1\\
  y'&= 5xy-5.\nonumber
\end{align}

{\bf Problem 1.} Find all critical points of the system, including critical
points at~$\infty$.
\bigskip

{\bf Problem 2.} Analyze each point at~$\infty$ by projecting to the plane~$x=1$. Draw the flow
in the plane~$x=1$.
\bigskip

{\bf Problem 3.} Try to reconcile your results from Problem 2 with the flow of
the original system displayed below:
\begin{center}
  \includegraphics[height=2.5in]{11fri.png}
\end{center}
Try to draw a global phase portrait.
\end{document}

