### Author Topic: Table 9.3.1 on page 523 of the textbook  (Read 3456 times)

#### Sarah Dunning

• Jr. Member
•  • Posts: 6
• Karma: 2 ##### Table 9.3.1 on page 523 of the textbook
« 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

#### Victor Ivrii ##### Re: Table 9.3.1 on page 523 of the textbook
« Reply #1 on: November 28, 2014, 05:34:35 PM »
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 non-linear 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 on-linear 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.

#### Sarah Dunning

• Jr. Member
•  • Posts: 6
• Karma: 2 ##### Re: Table 9.3.1 on page 523 of the textbook
« Reply #2 on: November 30, 2014, 09:09:06 AM »
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?

#### Victor Ivrii ##### Re: Table 9.3.1 on page 523 of the textbook
« Reply #3 on: November 30, 2014, 10:23:56 AM »
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'=-x-yf( 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 $r=e^{c-t}$, $\theta'=1/(c-t)$ and $\theta =-\ln (t-c) +c_2$ so it makes an infinite number of rotations around $0$ as $t\to +\infty$ (but this rotation slows down).

« Last Edit: December 01, 2014, 09:52:14 PM by Victor Ivrii »