MAT244-2014F > MAT244 Math--Lectures

Table 9.3.1 on page 523 of the textbook

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**Sarah Dunning**:

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**:

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**:

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**:

--- Quote from: Sarah Dunning 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?

--- End quote ---

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).

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