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Understanding stability of fixed points in 2D maps.

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1

$begingroup$

I'm trying to understand the stability analysis for a map of the form $$(x_n+1, y_n+1) = A(x_n,y_n)$$

Where A is a 2x2 matrix - assumed to be diagonalisable and with distinct eigenvalues. I understand there are two cases to consider:

• If $lambda_1$ and $lambda_2$ are real then it's clear that the general solution is $v_n = (lambda)^n v_0$ and this is clearly stable iff $|lambda|$ are both less than 1.


• If If $lambda_1$ and $lambda_2$ are a complex conjugate pair then we can write the general solution in polar coordinates (as seen here http://math.colgate.edu/~wweckesser/math312Spring05/handouts/Linear2DMaps.pdf) to a vector proportional to $r^n$ where $r = |lambda|$ and again this is stable iff $r<1$.

Now my confusion comes from looking at the stability of an area preserving map (E.g. the standard map http://mathworld.wolfram.com/StandardMap.html). In such a case the characteristic polynomial is: $$p(lambda) = lambda^2 - Tr(A)lambda + 1$$

It's claimed in the mathworld article that $Tr(A) < 2$ i.e. the real parts of the conjugate pair are less than 1. Whereas with the stability analysis outlined above it would seem that the fixed points of an area preserving map are marginally stable as their eigenvalues lie on the unit disk?

Thanks for any help with understanding this difference!

linear-algebra dynamical-systems nonlinear-system

share|cite|improve this question

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

asked May 24 '15 at 10:57

WoosterWooster

1,405935

$endgroup$

add a comment | 

1

$begingroup$

I'm trying to understand the stability analysis for a map of the form $$(x_n+1, y_n+1) = A(x_n,y_n)$$

Where A is a 2x2 matrix - assumed to be diagonalisable and with distinct eigenvalues. I understand there are two cases to consider:

• If $lambda_1$ and $lambda_2$ are real then it's clear that the general solution is $v_n = (lambda)^n v_0$ and this is clearly stable iff $|lambda|$ are both less than 1.


• If If $lambda_1$ and $lambda_2$ are a complex conjugate pair then we can write the general solution in polar coordinates (as seen here http://math.colgate.edu/~wweckesser/math312Spring05/handouts/Linear2DMaps.pdf) to a vector proportional to $r^n$ where $r = |lambda|$ and again this is stable iff $r<1$.

Now my confusion comes from looking at the stability of an area preserving map (E.g. the standard map http://mathworld.wolfram.com/StandardMap.html). In such a case the characteristic polynomial is: $$p(lambda) = lambda^2 - Tr(A)lambda + 1$$

It's claimed in the mathworld article that $Tr(A) < 2$ i.e. the real parts of the conjugate pair are less than 1. Whereas with the stability analysis outlined above it would seem that the fixed points of an area preserving map are marginally stable as their eigenvalues lie on the unit disk?

Thanks for any help with understanding this difference!

linear-algebra dynamical-systems nonlinear-system

share|cite|improve this question

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

asked May 24 '15 at 10:57

WoosterWooster

1,405935

$endgroup$

add a comment | 

$begingroup$

I'm trying to understand the stability analysis for a map of the form $$(x_n+1, y_n+1) = A(x_n,y_n)$$

Where A is a 2x2 matrix - assumed to be diagonalisable and with distinct eigenvalues. I understand there are two cases to consider:

• If $lambda_1$ and $lambda_2$ are real then it's clear that the general solution is $v_n = (lambda)^n v_0$ and this is clearly stable iff $|lambda|$ are both less than 1.


• If If $lambda_1$ and $lambda_2$ are a complex conjugate pair then we can write the general solution in polar coordinates (as seen here http://math.colgate.edu/~wweckesser/math312Spring05/handouts/Linear2DMaps.pdf) to a vector proportional to $r^n$ where $r = |lambda|$ and again this is stable iff $r<1$.

Now my confusion comes from looking at the stability of an area preserving map (E.g. the standard map http://mathworld.wolfram.com/StandardMap.html). In such a case the characteristic polynomial is: $$p(lambda) = lambda^2 - Tr(A)lambda + 1$$

It's claimed in the mathworld article that $Tr(A) < 2$ i.e. the real parts of the conjugate pair are less than 1. Whereas with the stability analysis outlined above it would seem that the fixed points of an area preserving map are marginally stable as their eigenvalues lie on the unit disk?

Thanks for any help with understanding this difference!

linear-algebra dynamical-systems nonlinear-system

share|cite|improve this question

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

asked May 24 '15 at 10:57

WoosterWooster

1,405935

$endgroup$

share|cite|improve this question

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

asked May 24 '15 at 10:57

WoosterWooster

1,405935

share|cite|improve this question

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

asked May 24 '15 at 10:57

WoosterWooster

1,405935

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

edited May 26 '15 at 22:14

Lukas Geyer

13.8k1556

asked May 24 '15 at 10:57

WoosterWooster

1,405935

asked May 24 '15 at 10:57

WoosterWooster

1,405935

1 Answer
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$begingroup$

In linear area-preserving maps the product of the eigenvalues (i.e. the determinant) lies on the unit circle. This does not imply that all the eigenvalues lie on the unit circle. You can have two real values, one within the unit circle and one outside of it. The sum of the eigenvalues (the trace of the matrix) will tell you whether or not this is the case.

share|cite|improve this answer

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313

$endgroup$

add a comment | 

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0

$begingroup$

In linear area-preserving maps the product of the eigenvalues (i.e. the determinant) lies on the unit circle. This does not imply that all the eigenvalues lie on the unit circle. You can have two real values, one within the unit circle and one outside of it. The sum of the eigenvalues (the trace of the matrix) will tell you whether or not this is the case.

share|cite|improve this answer

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313

$endgroup$

add a comment | 

0

$begingroup$

In linear area-preserving maps the product of the eigenvalues (i.e. the determinant) lies on the unit circle. This does not imply that all the eigenvalues lie on the unit circle. You can have two real values, one within the unit circle and one outside of it. The sum of the eigenvalues (the trace of the matrix) will tell you whether or not this is the case.

share|cite|improve this answer

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313

$endgroup$

add a comment | 

$begingroup$

In linear area-preserving maps the product of the eigenvalues (i.e. the determinant) lies on the unit circle. This does not imply that all the eigenvalues lie on the unit circle. You can have two real values, one within the unit circle and one outside of it. The sum of the eigenvalues (the trace of the matrix) will tell you whether or not this is the case.

share|cite|improve this answer

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313

$endgroup$

share|cite|improve this answer

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313

answered Jun 1 '15 at 10:05

citronrosecitronrose

497313