1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stability of system

  1. Dec 14, 2011 #1
    For the system

    [itex]\dot{x}[/itex]=y2
    [itex]\dot{y}[/itex]=x2

    Both the eigenvalues are zero. How do I
    find the eigenvectors so that I can sketch
    the phase portrait and how do I classify
    the stability of the fixed point (0,0)?
     
  2. jcsd
  3. Dec 15, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, obviously, both [itex]x^2[/itex] and [itex]y^2[/itex] are positive for all non-zero x and y so (0, 0) is unstable.
     
  4. Dec 18, 2011 #3
    Yes, that is true. Thank you. How do I find the eigenvectors though?
     
  5. Dec 19, 2011 #4

    mbp

    User Avatar

    It is not necessary to compute eigenvectors. This system is Hamiltonian (conservative). On dividing one equation by the other you get
    \begin{equation}
    \frac{dx}{dy} = \frac{y^2}{x^2}
    \end{equation}
    Separating variables and integrating you find the Hamiltonian
    \begin{equation}
    H(x,y) = \frac{1}{3} (x^3-y^3)
    \end{equation}
    The level sets \begin{equation}H = constant\end{equation} define the phase portrait.
     
  6. Dec 21, 2011 #5
    oh my god, that make life so easy. Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stability of system
Loading...