3805 Structural Stability for Non linear Systems

In the preceding note we discussed the structural stability of a linear system. How does it apply to non-linear systems?
Suppose our non-linear system has a critical point at $P$, and we want to study its trajectories near $P$ by linearizing the system at $P$.
This linearization is only an approximation to the original system, so if it turns out to be a borderline case, i.e., one sensitive to the exact value of the coefficients, the trajectories near $P$ of the original system can look like any of the types obtainable by slightly changing the coefficients of the linearization.
It could also look like a combination of types. For instance, if the linearized system had a critical line (i.e., one eigenvalue zero), the original system could have a sink node on one half of the critical line, and an unstable saddle on the other half. (This actually occurs.)
In other words, the method of linearization to analyze a non-linear system near a critical point doesn't fail entirely, but we don't end up with a definite picture of the non-linear system near P; we only get a list of possibilities. In general one has to rely on computation or more powerful analytic tools to get a clearer answer. The first thing to try is a computer picture of the non-linear system, which often will give the answer.
Example.
$$\begin{aligned} x'=y-x^2\\y'=-x+y^2 \end{aligned}$$ Jacobian: $$J=\begin{pmatrix} -2x&1\\-1&2y \end{pmatrix}$$ Crititcal points: $$y-x^2=0\rArr y=x^2$$ $$-x+y^2=-x+x^4=0\rArr x=0, 1$$ so $(0,0)$ and $(1,1)$ are the critical points.
$J(1,1)=<script type="math/tex; mode=display">\begin{pmatrix}-2&1\\-1&2\end{pmatrix}$ :
characteristic equation: $\lambda^2-3=0 \rArr \lambda=\plusmn\sqrt{3}$
linearized system has a saddle.
This is structurally stable $\rArr$ the nonlinear system has a saddle at $(1,1)$.
$J(0,0)=<script type="math/tex; mode=display">\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ :
eigenvalues: $\lambda=\plusmn i$
a linearized center.
This is not structurally stable. The nonlinear system could be any one of a center, spiral out or spiral in. Using a computer program it appears that $(0,0)$ is in fact a center. (This can be proven using more advanced methods.)
We can show the trajectories near $(0,0)$ are not spirals by exploiting the symmetry of the picture. First note, if $(x(t), y(t))$ is a solution then so is $(y(-t), x(-t))$. That is, the trajectory is symmetric in the line $x = y$. This implies it can't be a spiral. Since the only other choice choice is that the critical point $(0,0)$ is a center, the trajectories must be closed.

The following two examples show that a linearized center might also be a spiral in or a spiral out in the nonlinear system.
Example a. $x'=y,y'=-x-y^3$
Example b. $x'=y,y'=-x+y^3$
In both examples the only critical point is $(0,0)$.
$$J(0,0)=\begin{pmatrix} 0&1\\-1&-3y^2 \end{pmatrix}=\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}\rArr \lambda=\plusmn i\rArr \text{linearized center}$$ This is not structurally stable.
In example a the critical point turns out to be a spiral sink. In example b it is a spiral source.
Below are computer-generated pictures. Because the $y^3$ term causes the spiral to have a lot of turns we 'improved' the pictures by using the power 1.1 instead.