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Stability of ODE solution

  1. Jun 25, 2007 #1
    1. The problem statement, all variables and given/known data
    ok i got the following one i have x´´= -x^5, show that the point x=0 is a stable equilibrium.

    I´m given the hint to use the function V(x, x´) = x´^2/2 + x^6/6

    2. Relevant equations

    3. The attempt at a solution

    surly i tried linearization but that doesn´t work ( ok everyone knew that :) )

    now im into trying to start with the basic epsilon-delta criterion and try to work around that but im not really getting anywhere any hint would be appriciated :)
  2. jcsd
  3. Jun 25, 2007 #2


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    Staff Emeritus
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    First, of course, "equilibrium" means that the function x= 0 for all t (not point- you may mean the point in the phase plane (0,0)) is a constant solution to the differential equation- that should be obvious. The "stable' part means that if x(0) is close to 0, then as t goes to infinity, x stays close to 0 (or asymptotically stable if x goes to 0).

    Suppose x(1)> 0. What sign does x' have? What does that tell you about what happens to x as t increases? Suppose x(0)< 0. What is x'? What does that tell you about what happens to x as t increases>

    Actually, from the hint, it looks like you are expected to use 'Liapunov's Direct Method'. Do you know what that is? If this is a homework question, it should be in your book! Essentially it says that if x=0 is an equilibriums solution to x'= f(x) then it is a stable equilibrium solution if there exist a continuously differentiable function V(x) which is positive definite and such that
    [tex]\frac{dV}{dt}= \frac{\partial V}{dx} f(x)[/tex]
    is negative semi-definite (asymptotically stable if that derivative is negative definite).

    Here's a more detailed explanation
  4. Jun 25, 2007 #3
    hehe sweet i got it done :)
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