Kansas State University

Discussion

Equililbria

Theorem: If $(x_0,y_0)$ is an equilibrium point for the autonomous system $$ \begin{align} \frac{dx}{dt}&=f(x,y) \\ \frac{dy}{dt}&=g(x,y) \end{align} $$ then $x(t)=x_0$, $y(t)=y_0$ is a solution of the system.

Proof: This follows immediately from $\displaystyle\frac{dx}{dt}=f(x(t),y(t))=f(x_0,y_0)=0$ and the similar calculation that $\displaystyle\frac{dy}{dt}=0$.

Conversely, any constant solution must be an eqilibria. Of course, now that we've found these equilibria we want to decide if solution curves head toward them or away from them. Actually, while those were the two possibilities for first-order equations, we have additional possibilities for systems. In a first-order equation we are moving in just one dimension. With systems we can move in more than one dimension, and that allows for some new sorts of curves.

Linearization

Theorem: If a non-linear system has an equilibrium point at $(x_0,y_0)$ and the linearized system at that equilibrium point has

• solutions that diverge straight out away from the origin
• solutions that converge directly to the origin (possibly with a twist)
• a saddle point at the origin where solutions converge in one direction and diverge away in a different direction
• solutions that spiral away from the origin
• solutions that spiral in toward the origin

We won't give a careful proof of this theorem, but the basic idea is the following. As long as you stay sufficient close to the equilibrium point $(x_0,y_0)$, the tangent-plane approximations will be close enough to the non-linear functions $f(x,y)$ and $g(x,y)$ that the solutions will behave similarly. One key point to note is that the theorem above covers 5 different sorts of behavior. In studio we had a 6^th possible behavior, that the solutions spin around the origin along ellipses, neither converging toward the origin nor diverging away from it. Unfortunately, if the linearized system has this behavior, we are unable to conclude what the non-linear system does. The small differences between the non-linear system and the linearized system could easily be enough to move the solutions off the perfect ellipses so they don't return perfectly to the same point each cycle. The non-linear system will certainly spin around the origin but it might spiral in, spiral out, or keep its distance. When we encounter this case, we will need a more detailed analysis.

With this theorem we are now ready for a paradigm.

Paradigm

Step 2: For each equilibrium point, find the linearized system near that point.

The tangent-plane approximation to any function $f(x,y)$ near a point $(x_0,y_0)$ is given by $$ f(x,y) \approx f(x_0,y_0) + \frac{\partial f}{\partial x}(x_0,y_0)\times(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)\times(y-y_0) $$ So about the point $(0,0)$ the tangent-plane approximations are $$ \begin{align} 4x-2xy &\approx 0 + \left.(4-2y)\right|_{(x,y)=(0,0)}\times(x-0) + \left.(-2x)\right|_{(x,y)=(0,0)}\times(y-0) \\ &=4x \\ xy-3y &\approx 0 + \left.(y)\right|_{(x,y)=(0,0)}\times(x-0) + \left.(x-3)\right|_{(x,y)=(0,0)}\times(y-0) \\ &=-3y \end{align} $$ so near $(0,0)$ our linearized system becomes $$ \begin{align} \frac{dx}{dt}&=4x \\ \frac{dy}{dt}&=-3y \end{align} $$ Repeating this for $(3,2)$ we get $$ \begin{align} 4x-2xy &\approx 0 + \left.(4-2y)\right|_{(x,y)=(3,2)}\times(x-0) + \left.(-2x)\right|_{(x,y)=(3,2)}\times(y-0) \\ &=-6y \\ xy-3y &\approx 0 + \left.(y)\right|_{(x,y)=(3,2)}\times(x-0) + \left.(x-3)\right|_{(x,y)=(3,2)}\times(y-0) \\ &=2x \end{align} $$ and near $(3,2)$ our linearized system becomes $$ \begin{align} \frac{dx}{dt}&=-6y \\ \frac{dy}{dt}&=2x \end{align} $$ Step 3: Compute the trace and determinant and classify each critical point according to the rules developed in studio.

Near $(0,0)$ our linearized system is $$ \begin{align} \frac{dx}{dt}&=4x \\ \frac{dy}{dt}&=-3y \end{align} $$ which has trace 1 and determinant -12, and so forms a saddle point. By our theorem above, that means $(0,0)$ is a saddle-point for the original non-linear system.

Near $(3,2)$ our linearized system is $$ \begin{align} \frac{dx}{dt}&=-6y \\ \frac{dy}{dt}&=2x \end{align} $$ which has trace 0 and determinant 12, and so the solutions to the linearized system spin around the origin on ellipses, neither converging toward the origin nor diverging away from it. But now we run into a problem. This is the one case where we can't conclude what the behavior of the non-linear system is from the linearized system. So we list $(3,2)$ as "unable to classify."

This has just given a taste of this subject. As we move to higher dimensions even more issues arise. The general study of the behavior of non-linear systems leads to chaos theory, a complex and beautiful subject even in two dimensions. By the way, for the particular paradigm problem given above, the solutions spin around the equilibrium at $(3,2)$ neither spiraling in nor out (though moving on more complicated curves than ellipses). The key to showing this is a trick that is in the written homework.