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### Non-Linear Systems

#### Discussion

Laplace transform techniques work well for constant-coefficient linear systems. However, many situations in practice have non-linear systems. We can approximate the answers numerically using the techniques of chapter 2. After all, in that chapter we first converted a second-order equation to a first-order system and then applied the numerical methods to the system. Those techniques work for any first-order system, whether it comes from a higher-order equation or starts life as a system naturally. But if you want to know the range of behaviors a system can manifest, just finding numerical approximations, even for a variety of starting points, may not be enough. In this section we will discuss two techniques for determining long-term behavior of solutions to non-linear systems. We will stick to autonomous systems since those are the simpler (and more common) variety of non-linear systems.

#### Equililbria

For first-order autonomous equations we could quickly deduce long-term behavior from geometric principles. The solutions converge to (stable) equilibria or head off to $\pm\infty$ (or stay precariously at an unstable equilibria when given exactly the right initial condition). We can develop a similar set of rules for first-order systems. The first step of course is to find the equilibria. And we do this just the way we do with first order equations, we ask where the derivatives are 0.*Definition:*Given the autonomous system $$ \begin{align} \frac{dx}{dt}&=f(x,y) \\ \frac{dy}{dt}&=g(x,y) \end{align} $$ the equilibria are points $(x_0,y_0)$ such that $f(x_0,y_0)=0$ and $g(x_0,y_0)=0$.

*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

In the studio on systems you will discover rules for how solutions to first-order linear systems behave about equilibria. Those same rules (almost) apply to non-linear systems. The key is to "linearize" the non-linear system so you can apply the rules. For the system $$ \begin{align} \frac{dx}{dt}&=f(x,y) \\ \frac{dy}{dt}&=g(x,y) \end{align} $$ with an equilibrium point at $(x_0,y_0)$, we do this by replacing $f(x,y)$ by its tangent-plane approximation about $(x_0,y_0)$ and similarly replacing $g(x,y)$ by its tangent-plane approximation about $(x_0,y_0)$. Then we have
*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

Find and classify the equilibria for the autonomous system $$ \begin{align} \frac{dx}{dt}&=4x-2xy \\ \frac{dy}{dt}&=xy-3y \end{align} $$*Step 1:*Find the equilibria by solving the simultaneous equations $$ \begin{align} 4x-2xy&=0 \\ xy-3y&=0 \end{align} $$ This gives us the two equilibria $(0,0)$ and $(3,2)$.

*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.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett