# Difference of period between cartesian and polar eigenvalue representation

1. Aug 15, 2012

### williamrand1

The solution to a linear differential equation is, y=exp(ax). If a is complex ,say a=b+ic, then the period is T=2pi/c. My question is, if a is in polar form, a=r*exp(iθ), how is the period then T=2pi/θ.

Any help would be great,

Thank,

Will

2. Aug 15, 2012

### Ibix

It isn't. Consider a trivial case where $a=-1=exp(i\pi)$. Obviously the period of y is infinite because there is no complex component to a. Therefore the period is not 2, which your latter formula would suggest.

3. Aug 15, 2012

### williamrand1

Thanks Ibix.

I understand ur point.

Im reading a paper and in it a linear stability analysis is done on a model. The characteristic eqt turns out to be, a^2 -(2-d)*a +1=0, a is the eigenvalue and d is a constant from the jacobian matrix. In the paper they say the period of the cycle of the linearised form is T=2pi/arctan(sqrt(4b-b^2)/(2-b)). The denominator of T is equal to θ.

Thanks,

Will

4. Aug 15, 2012

### williamrand1

Sorry, in the equation for T it should be d not b. Any help would be great. Thanks..

5. Aug 16, 2012

### Ibix

Obviously I haven't seen the paper so I might be missing something, but that does not make sense to me.

One, the range of values for T depends on the choice of range of arctan, which is absurd. Two, the right-hand side of the expression for T is dimensionless when it should have dimensions of time.

My money is on a typo in the paper. Might there be an erratum notice somewhere? Is there any numerical data in the paper that you can check, or subsequent derivations?

You could try posting a link to the paper. I'm afraid I shall be out of touch for a few days, but someone else might be able to help.

Last edited: Aug 16, 2012
6. Aug 16, 2012

### williamrand1

Thanks Ibix.

I emailed the author of the paper, hopefully he can explained it to me.

Ill let you know the result....

7. Aug 16, 2012

### williamrand1

Problem Solved!!

The problem used a discrete system, not a continuous one. So the linearized solution is y=a^k. so in polar form a=r*exp(iθ) then y=r^k * exp(iθk) so the period would be T=2pi/θ.