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### Geometric Methods for 2 Dimensional Systems

#### Discussion

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.

Note that you can tell from the arrows that the solutions will spin around $(-1,-1).$ However, it is not always possible to decide whether the solution curves are spiraling in or out from the arrows alone, hence the inclusion of step 4. Also note that as you can see from the sketch, some solutions will spin into $(-1,-1)$ while others will be swept away to $(\infty,\infty),$ with the split depending on how the curves split across the saddle point at $(1,1).$ Obviously it is of interest to find the line that separates initial values that go to $(-1,-1)$ and those that go to $(\infty,\infty).$ Finding this line is a delicate question that we won't go into in this class.

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©2010, 2014 Andrew G. Bennett