Is x=0 a stable equilibrium for x´´= -x^5?

[Mr.Brown]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

now I am into trying to start with the basic epsilon-delta criterion and try to work around that but I am not really getting anywhere any hint would be appriciated :)

 

[HallsofIvy]

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
\frac{dV}{dt}= \frac{\partial V}{dx} f(x)
is negative semi-definite (asymptotically stable if that derivative is negative definite).

Here's a more detailed explanation
http://www.personal.rdg.ac.uk/~shs99vmb/notes/anc/lecture2.pdf

 

[Mr.Brown]

hehe sweet i got it done :)
thanks