Lyapunov Function: Show 0 is Stable

  • #1
rjcarril
2
0
Assume that f(0) = 0 and Df(0) has eigenvalues with negative real parts. Con-
struct a Lyapunov function to show that 0 is asymptotically stable.
 
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  • #2
rjcarril said:
Assume that f(0) = 0 and Df(0) has eigenvalues with negative real parts. Con-
struct a Lyapunov function to show that 0 is asymptotically stable.

I know it is a strict lyapunov function, but i cannot figure out how to solve it for the general case.
 
  • #3
Consider [itex]V(x) = \|x\|^2 = x \cdot x[/itex]. Then [itex]\nabla V = 2x[/itex] and
[tex]
\dot V = \nabla V \cdot \dot x = \nabla V \cdot f(x) =
2x \cdot (Df(0) \cdot x) + O(\|x\|^3).
[/tex]
What does the condition on the eigenvalues of Df(0) imply about the sign of [itex]x \cdot (Df(0) \cdot x)[/itex]? What does that imply about [itex]\dot V[/itex] for [itex]\|x\|[/itex] sufficiently small?

Can you prove that if [itex]\dot V < 0[/itex] on a neighbourhood of 0 then 0 is asymptotically stable?
 
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