### Author Topic: 9.3 problem 18  (Read 3692 times)

#### Brian Bi

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##### 9.3 problem 18
« on: March 25, 2013, 12:32:43 AM »
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
« Last Edit: March 25, 2013, 12:46:22 AM by Brian Bi »

#### Victor Ivrii

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##### Re: 9.3 problem 18
« Reply #1 on: March 25, 2013, 06:57:38 AM »
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

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##### Re: 9.3 problem 18
« Reply #2 on: March 25, 2013, 01:28:57 PM »
So it is a spiral point but I didn't zoom in closely enough?

#### Victor Ivrii

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.