9.3 problem 18

[Brian Bi]

I'm having some trouble getting this problem to work out. There are four critical points: (0,0), (2, 1), (-2, 1), and (-2, -4). At the critical point (-2, -4), the Jacobian is \begin{pmatrix} 10 & -5 \\ 6 & 0 \end{pmatrix} with eigenvalues $5 \pm i\sqrt{5}$. Therefore it looks like it should be an unstable spiral point. However, when I plotted it, it looked like a node. Has anyone else done this problem?

http://www.math.psu.edu/melvin/phase/newphase.html

[Victor Ivrii]

I'm having some trouble getting this problem to work out. There are four critical points: (0,0), (2, 1), (-2, 1), and (-2, -4). At the critical point (-2, -4), the Jacobian is \begin{pmatrix} 10 & -5 \\ 6 & 0 \end{pmatrix} with eigenvalues $5 \pm i\sqrt{5}$. Therefore it looks like it should be an unstable spiral point. However, when I plotted it, it looked like a node. Has anyone else done this problem?

http://www.math.psu.edu/melvin/phase/newphase.html

Explanation:

http://weyl.math.toronto.edu/MAT244-2011S-forum/index.php?topic=48.msg159#msg159

[Brian Bi]

So it is a spiral point but I didn't zoom in closely enough?

[Victor Ivrii]

So it is a spiral point but I didn't zoom in closely enough?

No, the standard spiral remains the same under any zoom. However  your spiral rotates rather slowly in comparison with moving away and as it makes one rotation ($\theta$ increases by $2\pi$) the exponent increases by $5 \times 2\pi/\sqrt{5}\approx 14$ and the radius increases $e^{14}\approx 1.2 \cdot 10^6$ times. If the initial distance was 1 mm, then after one rotation it becomes 1.2 km.

Try plotting $x'=a x- y$, $y'=x+ ay$ for $a=.001,  .1,  .5, 1,  1.5,   2$  to observe that at for some $a$ you just cannot observe rotation.