How can I show that $\tilde x=0$ is asymptotically stable for
$x'=-t \cos x, t \in \mathbb R$
I guess I need to find a Lyapunov function but I'm not sure
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa Jun 05 '19 at 18:12
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These posts may help: Lyapunov stability, Finding Lyapunov Function, How to find out a suitable Lyapunov function?, Finding a Lyapunov function for a given system. – dantopa Jun 05 '19 at 18:16
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Or this: Using a Lyapunov function to classify stability and sketching a phase portrait – dantopa Jun 05 '19 at 18:16
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1One problem with this question is that $0$ is not an equilibrium for $\dot x = -t\cos x$ for all $t$; on the other hand, if we consider $\dot x = -t \sin x$ . . . or look at the equilibria $x = (2n + 1)\pi / 2$ – Robert Lewis Jun 05 '19 at 18:25
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1@RobertLewis How would it work for $x'=-t \sin x$? – user679927 Jun 05 '19 at 18:41
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@user679927: give me a little while and I'll see what I can come up with. Cheers! – Robert Lewis Jun 05 '19 at 18:43
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@user679927: still at it; got a bit distracted!;) – Robert Lewis Jun 07 '19 at 19:05