The analytic techniques discussed previously work well for determining
behavior near an equilibrium. But in many cases you want to know how a
system behaves more generally. In such circumstances geometric and
numerical techniques may add useful information. In this section we will
deal
with geometric techniques applied to autonomous systems in two dimensions.
Techniques such as these get much more complicated in higher dimensions.
Paradigm
We will sketch some solutions of the system
$$ \begin{align}
\frac{dx}{dt}&=y-x^3 \\
\frac{dy}{dt}&=x^2+y^2-2
\end{align} $$
Step 1: Sketch the curves $x'=0$ and $y'=0.$
In this case, these
are the curves $y-x^3=0$ for $x'=0$ (sketched in blue) and $x^2+y^2-2=0$
for $y'=0$ (sketched in red).
Step 2: Draw horizontal arrows showing where $x' > 0$ and where
$x' < 0.$ These will be on opposite sides of the curve $x'=0.$
In this
case, we have $x'=y-x^3$ which is 0 along the curve $y=x^3.$ Above this
curve, $y$ has increased and so $x'=y-x^3> 0.$ Therefore we draw the
arrows pointing toward increasing $x$ in the region above the curve. Below
the curve $y$ is decreased so $x' < 0$ and the arrows go to the left.
If you have trouble recognizing whether the values should be positive or
negative, you can always check values at a representative point. For
example, the point $(0,1)$ lies above the curve $y=x^3.$ And at $(0,1)$
$x'=1-0^3=1 > 0.$ Since $x'$ can't change sign without crossing the
value 0, that means $x' > 0$ at all points above $y=x^3.$ Similarly you
can check the value of $x'$ at $(0,-1)$ to see $x' < 0$ below the
curve.
Step 3: Draw vertical arrows showing where $y' > 0$ and where
$y' < 0.$ As above, these will be on opposite sides of the curve
$y'=0.$
In this case, we have $y'=x^2+y^2-2$ which is 0 along the circle
$x^2+y^2=2.$ Outside this
circle, $x^2+y^2$ has increased and so $y'=x^2+y^2-2> 0.$ Therefore we
draw the
arrows pointing up toward increasing $y$ in the region outside the circle.
Inside the circle
$x^2+y^2$ is decreased so $y' < 0$ and the arrows go down. As before,
you can check values at a representative point if you prefer. For
example, the point $(0,0)$ lies inside the circle. And at $(0,0)$
$y'=0^2+0^2-2=-2 < 0.$ Since $xy'$ can't change sign without crossing
the
value 0, that means $y' < 0$ at all points inside the circle. Similarly
you
can check the value of $x'$ at $(2,2)$ to see $y' > 0$ outside the
circle.
Step 4: Identify the behavior around each equilibrium point.
Do this using the analytical techniques we have used previously.
The equilibrium points occur where the red and blue curves cross, at
$(1,1)$ and $(-1,-1)$ in this case. Using the techniques of the previous
section we classify $(1,1)$ as a saddle point while curves will spiral in
to $(-1,-1).$
Step 5: Sketch solution curves by following the arrows.
So if we start at a point outside the circle and below the curve $y=x^3$
then the solution will move northwesterly (with $x$ decreasing since below
the curve and $y$ increasing since outside the circle). The solution will
continue to move northwesterly until it crosses either $x'=0$ or $y'=0.$
So looking at the curve that starts outside the circle and just above the
x-axis we see it goes up until it crosses into the circle, At that point
it starts moving southwest as now $y' < 0.$ The curve then moves down
toward $(-1,-1)$ and then spirals into that equilibrium. This curve and
others are sketched below.
