
#1
Nov2712, 03:28 PM

P: 2

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. 



#2
Nov2712, 03:32 PM

P: 2





#3
Dec512, 06:32 AM

HW Helper
P: 774

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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