Phase Plane, real unequal eigenvalues of the same sign

• Aug 27th 2012, 08:58 AM
csstudent5000
Phase Plane, real unequal eigenvalues of the same sign
I have the following eigenvalues: r_1 = 4, r_2 = 2, which gives me the eigenvectors v_1 = [1, 1] and v_2 = [1,3] for r_1 and r_2 respectively. How should I draw the phase plane for this? This gives me two vectors that both approach infinity when t -> infinity? I've tried to find an example problem that draws the phase plane for this with no success. If anyone could either explain and/or point me to an example with this type of eigenvalues/vectors then that would be great. Thanks. :-)
• Aug 27th 2012, 09:19 AM
HallsofIvy
Re: Phase Plane, real unequal eigenvalues of the same sign
First, draw the lines corresponding to the eigenvectors. If both eigenvectors are positive, this is a "source", if negative, a "sink". Every line through the equilibrium point is a solution. In order to show the difference in eigenvalues, one thing you can do is draw the lines denser (closer together) close to the eigenvector with larger eigenvalue, farther apart close to the eigenvector with smaller eigenvalue.
• Aug 27th 2012, 09:32 AM
csstudent5000
Re: Phase Plane, real unequal eigenvalues of the same sign
Quote:

Originally Posted by HallsofIvy
First, draw the lines corresponding to the eigenvectors. If both eigenvectors are positive, this is a "source", if negative, a "sink". Every line through the equilibrium point is a solution. In order to show the difference in eigenvalues, one thing you can do is draw the lines denser (closer together) close to the eigenvector with larger eigenvalue, farther apart close to the eigenvector with smaller eigenvalue.

Thank you for the input. However what I don't know how to do is how to draw the actual solutions, i.e. the lines that are drawn between the eigenvectors. What does it look like? Do you know what I could search for to find an example of how to do this?