Toronto Math Forum
MAT2442014F => MAT244 MathLectures => Topic started by: Sarah Dunning on November 28, 2014, 12:41:03 PM

Can someone explain how you get the following?
Linear system Type > Locally Linear System Type
Proper Node or Improper Node > Node or Spiral Point
Center > Center or Spiral Point

We don't give you a proof but there is a reasoning: linear system behaviour depend on eigenvalues;
1) we need to distinguish the case of two real disjoint eigenvalues and two complex conjugate ones. This corresponds to positive and negative discriminant and discriminant equal $0$ is a threshold. For linear system it means a proper or improper node but nonlinear terms can "shift" it to a node or a spiral point. Still if eigenvalues are $>0$ system will be unstable, and if they $<0$ stable.
2) for complex conjugate eigenvalues we need to distinguish those with real part $<0$ (stable spiral point), $0$ (center) and $>0$ unstable spiral point and onlinear terms can "shift" the center to stable or unstable spiral.
Example
\begin{equation*}
\left\{
\begin{aligned}
&x'= y + \varepsilon x (x^2+y^2),\\
&y'= x+ \varepsilon y(x^2+y^2)
\end{aligned}
\right.
\end{equation*}
As $varepsilon=0$ we have a linear system with a center, for $\varepsilon \lessgtr 0$ we have unstable/stable spiral point but near origin it is collapsing/expanding very slowly and looks rather like a center (but still is not a center).
This works for general systems, for integrable center remains a center.

For proper node or improper node > node or spiral point how can you shift to get a spiral point? I thought spiral points were for complex eigenvalues. And you can't have only one eigenvalue in the complex case, so how do complex eigenvalues fall under this case?

For proper node or improper node > node or spiral point how can you shift to get a spiral point? I thought spiral points were for complex eigenvalues. And you can't have only one eigenvalue in the complex case, so how do complex eigenvalues fall under this case?
Consider system $x'=f(x,y)$, $y'=g(x,y)$. If $f,g\in C^2$ (twice continuously differentiable, this requirement could be weakened significantly, then rather simple arguments show that node remains a node and its type (PN or IN) is preserved. But if $f,g\in C^1$ (once continuously differentiable) this is no longer true.
Example
\begin{equation}\begin{aligned}
&x'=xyf( r),\\
&y'=y+xf( r)
\end{aligned}\label{K}\end{equation}
with $f(r ) = 1/\ln(r )$ where $\theta, r$ are polar coordinates. Linearized system at $0$ has PN but (\ref{K}) has a spiral point.
Indeed, in polar coordinates (\ref{K}) becomes
\begin{equation}\begin{aligned}
&r'=r,\\
&\theta'=f(r)
\end{aligned}\label{L}\end{equation}
and then $r=e^{ct}$, $\theta'=1/(ct)$ and $\theta =\ln (tc) +c_2$ so it makes an infinite number of rotations around $0$ as $t\to +\infty$ (but this rotation slows down).