Difference of period between cartesian and polar eigenvalue representation

In summary, the period of a linear differential equation with a complex coefficient can be calculated using T=2pi/θ when the coefficient is in polar form. However, this method only applies to a discrete system and not a continuous one as the linearized solution is different.
  • #1
williamrand1
21
0
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
 
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  • #2
williamrand1 said:
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.
It isn't. Consider a trivial case where [itex]a=-1=exp(i\pi)[/itex]. 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
Ibix said:
It isn't. Consider a trivial case where [itex]a=-1=exp(i\pi)[/itex]. 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.

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
williamrand1 said:
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

Sorry, in the equation for T it should be d not b. Any help would be great. Thanks..
 
  • #5
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:
  • #6
Thanks Ibix.

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

Ill let you know the result...
 
  • #7
williamrand1 said:
Thanks Ibix.

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

Ill let you know the result...

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.
 

Related to Difference of period between cartesian and polar eigenvalue representation

1. What is the difference between cartesian and polar eigenvalue representation?

The difference between cartesian and polar eigenvalue representation lies in the way the eigenvalues and eigenvectors are expressed. In cartesian representation, the eigenvalues are represented as real numbers and the eigenvectors are expressed as column vectors. In polar representation, the eigenvalues are represented as complex numbers and the eigenvectors are expressed as polar coordinates.

2. Which representation is more commonly used in scientific research?

Cartesian eigenvalue representation is more commonly used in scientific research due to its simplicity and ease of calculation. It is also easier to interpret the results in cartesian representation compared to polar representation.

3. What are the advantages of using polar eigenvalue representation?

Polar eigenvalue representation can be advantageous in certain situations, such as when dealing with rotational or circular data. It also allows for a more intuitive understanding of the relationship between eigenvalues and eigenvectors.

4. Can cartesian and polar eigenvalue representations be converted into each other?

Yes, cartesian and polar eigenvalue representations can be converted into each other using mathematical transformations. For example, cartesian eigenvalues can be converted into polar eigenvalues using the complex number representation.

5. Are there any limitations to using cartesian and polar eigenvalue representations?

One limitation of using cartesian and polar eigenvalue representations is that they are only applicable to certain types of data, such as matrices. Other types of data may require different representations. Additionally, complex numbers used in polar eigenvalue representation may be difficult to visualize and interpret for some researchers.

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