Nonlinear system phase portraits

  • Thread starter Thread starter schmiggy
  • Start date Start date
  • Tags Tags
    Nonlinear System
schmiggy
Messages
37
Reaction score
0

Homework Statement


See attached image


Homework Equations


Classification of critical points chart (unless you remember it)


The Attempt at a Solution


See attached.

Now, I'm not entirely sure what exactly I'm doing. With linear systems, the goal is to find the eigenvectors, eigenvalues and thus the general solution. Using the general solution you can then derive the attributes of the phase portrait for that particular system to aid in sketching (such as behaviour as t -> +- infinity). Is that what I'm supposed to be doing here?

So basically, taking the critical points and substituting them into the matrix and then finding the eigenvalues and vectors for the respective matrix? Then do the same again for the next critical point? Is the goal to end up with as many general solutions as there are critical points and then combine them into one sketch?

I'm not really sure what the "let X=x, Y=y" is for either, I was trying to follow an example in my lecture notes on how to linearise the system but that's where it stops - thus why I thought the goal might be to linearise the system and then find the respective general system (for each critical point).

Also, I feel like I've done something wrong making the matrix and made up some pretend maths to get my lambda for the (0,0) critical point. What a mess! Any direction would be appreciated, thanks :)
 

Attachments

  • Image.jpg
    Image.jpg
    36.3 KB · Views: 455
Physics news on Phys.org
You have a first order nonlinear system, which can be written as X' = f(X), where X is a vector.
In such systems, the phase portrait around a hyperbolic equilibrium point (an equilibrium point X* such that the real part of every eigenvalue of Df(X*) is nonzero) is the same as the phase portrait of the linear system X' = Df(X*).X, in a neighbourhood of X*.

This means that in a vicinity of X* the phase portraits are the same, and you can deduce the stability by analysing the linear system.

Remark: Df(X*) is the jacobian matrix evaluated at the equilibrium point X*.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top