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Homework Help: Linear systems modeling dynamics

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    R' = aJ
    J' = bR

    What happens to the graphs of R(t) and J(t)?

    3. The attempt at a solution

    I made the matrix {{0, a}{b, 0}} and then got the equation (L=lambda) L^2 - 0L + (a+b) after computing the trace and determinant of that matrix. I then solved for the eigenvalues using the quadratic and got L1 = i sqrt(a+b) and L2 = -i sqrt(a+b)

    But I didn't think I was supposed to get imaginary numbers when dealing with eigenvalues because then I can't graph the eigenvectors on the real plane. Did I do something wrong?

  2. jcsd
  3. Sep 27, 2008 #2


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    Yes, the determinant of [[-L,a],[b,-L]] is L^2-ab. How did you get this (a+b) stuff?
  4. Sep 27, 2008 #3
    Good call. I have no idea how I got (a+b) when it should be -ab.
    Last edited: Sep 27, 2008
  5. Sep 27, 2008 #4
    Another question related to this problem. For the eigenvectors I got


    e1 = {{b},{L-a}} = {{0},{b}}
    e2 = {{L-d},{c}} = {{-b},{a}}

    Now how do I use those eigenvectors to sketch a phase portrait of the system? My book doesn't explain it well.

  6. Sep 27, 2008 #5


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    I don't think those are the eigenvectors either. Let's go back to the first question. What are the eigenvalues?
  7. Sep 28, 2008 #6
    My bad. My last post was referring to a different problem.

    Here's the other problem I'm working on:

    R' = 0
    J' = aR + bJ

    {{0, 0},{a,b}}

    L^2 - bL

    L1 = (b+sqrt(b^2))/2 = b
    L2 = (b-b)2/ = 0

    e1 = {{b},{L-a}} = {{0},{b}}
    e2 = {{L-d},{c}} = {{-b},{a}}

    Now how do I go from those eigenvectors to drawing asymptotes and graphing it? I thought the eigenvectors were used to find the slope of the asymptotes, but they contain variables in this case. Then how do I know where to draw the curves and how are points moving along them in time?

  8. Sep 28, 2008 #7


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    The eigenvectors are linear solutions to the system. If one eigenvector is <0, b>, which points in the direction of the line from (0,0) to (0,b), x= 0, then the L- axis, R= 0, is itself a solution. (That should be obvious from the fact that one equation is R'= 0. In fact, it not at all difficult to just solve the two equations.) The fact that the corresponding eigenvalue is b tells you that the value of y, as t increases, increases with "speed" b which, here, is a constant. The fact that <-b, a>, which points in the direction of the line, which points in the direction of the line from (0,0) to (-b,a), y= -(b/a)x, is an eigenvector tells you that the line y= -(b/a)x is also a solution. The fact that the corresponding eigenvalue is 0 tells you that there is no "motion": each point on that line is a constant solution. In cases where neither of the lines has 0 eigenvalue, other phase lines would be curves that do not cross them or each other and have "motion" the same direction as the curves. In this special case, where one eigenvalue is 0, you will have lines parallel to the line with non-zero eigenvalue and the same direction simply crossing the line with eigenvalue 0.
  9. Sep 29, 2008 #8
    Thanks, that helped me out a lot.
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