explain why I'm wrong

[Kuciwalker]

Problem: A solution X(t) of a system is called recurrent if {X(tn)} -> X(0) for some sequence {tn} -> infinity. Prove that a gradient dynamical system* has no nonconstant recurrent solutions.

My answer: I'm pretty sure this isn't true. Let X(t) be a periodic solution of such a system with period T. Define t0 = 0, tn = n*T - 1/n. Clearly {tn} -> infinity. {X(tn)} = {X(n*T - 1/n)} = {X(-1/n)}, which is clearly nonconstant and converges to X(0).

My only thought is that maybe the definition of recurrent should have stated "for all sequences" rather than "for some sequence"...

*i.e. a system X' = F(X) where F is the gradient of some function Rn -> R

[Wezil]

You forgot to carry the 4.

[Kuciwalker]

It occurs to me I might not even need the -1/n, since even if the sequence is constant the solution would not be (except in the trivial case).

[Japher]

Because you're a democrat.

[KrazyHorse]

Problem: A solution X(t) of a system is called recurrent if {X(tn)} -> X(0) for some sequence {tn} -> infinity. Prove that a gradient dynamical system* has no nonconstant recurrent solutions.

My answer: I'm pretty sure this isn't true. Let X(t) be a periodic solution of such a system with period T. Define t0 = 0, tn = n*T - 1/n. Clearly {tn} -> infinity. {X(tn)} = {X(n*T - 1/n)} = {X(-1/n)}, which is clearly nonconstant and converges to X(0).

My only thought is that maybe the definition of recurrent should have stated "for all sequences" rather than "for some sequence"...

*i.e. a system X' = F(X) where F is the gradient of some function Rn -> R

How the hell is a periodic function a gradient dynamical system? I've never heard the term before, but no nonconstant periodic differentiable function can satisfy the definition of a gradient dynamical system as far as I can tell. Intuitively speaking, set g(0) = 0. Now g'(x) > 0 and g(x) > 0 for some x (to force g nonconstant). Then there has to be a y > x s.t. g(y) = g(x) AND g'(y) < 0 (to force g back to the origin, by periodicity of g). However, this makes the posited F double valued because:

1) g'(x) = F(g(x))
2) g'(y) = F(g(y))

but g(x) = g(y)

????

EDIT: fixed typo: should have read "set g(0) = 0", not "set g(x) = 0"

[loinburger]

KH took the words right out of my mouth

[Japher]

2) g'(y) = F(g(y))



This is what I see:

gay = fagy

ooh, math is so funny

[Kuciwalker]

How the hell is a periodic function a gradient dynamical system?

It would be a solution to a gradient dynamical system.

[loinburger]

2) g'(y) = F(g(y))



This is what I see:

gay = fagy

ooh, math is so funny

You probably also laugh at Ϲuntz Algebra

[KrazyHorse]

Kuci

I have no idea what you mean by making that distinction, but you need to read what I wrote more carefully, for content rather than syntax.

[KrazyHorse]

You have some function g:R->R. If it's periodic and differentiable then it needs to go through the same value traveling upward and traveling downward. This isn't allowed by the definition of a gradient dynamical system AFAICT

[Kuciwalker]

Oh, I'm stupid. My thought process was "gradient dynamical system" -> "conservative force" -> "ooh, a pendulum".

[Kuciwalker]

****, now I have to actually work out a semirigorous proof.

[KrazyHorse]

The difference between a conservative force and what you've defined here is that a conservative force would give you X'' = F(X)

Second deriv, not first.

[Kuciwalker]

Yes, I figured that out when trying to actually construct the system a moment ago.