Another proof that solutions to ẋ=f(x) can't oscillate

Question

Let $ẋ=f(x)$ be a vector field on the line. Use the existence of a potential function $V(x)$ to show that solutions $x(t)$ cannot oscillate.

I know from the textbook (Nonlinear Dynamics and Chaos, Strogatz) that there are no periodic solutions to $ẋ=f(x)$. I really am not sure how to think or go about this problem. If someone would kindly nudge me in the right direction I would greatly appreciate it–thanks in advance!

Try looking at the integral $\int_0^T (\dot x)^2 dt$ over a solution to $\dot x=V'(x)$. — Khanickus, Feb 18 '17 at 08:29

score 0 · Answer 1 · answered Feb 18 '17 at 08:58

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Hint: Let us denote $\gamma$ - a solution of your equation at some starting point $x_0$. What if we are interested in length of $\gamma$ in different times $t$? If we could estimate this lenght such that for any finite time $t < \infty$ this value was finite then this would be a proof to a non-oscillation.

Hint to hint: When you are going to estimate use your potentiality equations for velocity vector $\dot x$.

answered Feb 18 '17 at 08:58

Byobe

523

@Khanickus argument can simplify proof, but it is more geometrical argument of mine. Hope you enjoy with this estimation! – Byobe Feb 18 '17 at 08:59

Another proof that solutions to ẋ=f(x) can't oscillate

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