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Suppose we have to solve the ODE: $$\frac{dx}{dt} = f(x) ,\forall t\geq 0 \text{ and } x(0) = x^{*}$$ where $f(x^{*}) = 0$

If we don't know anything about f(x), one obvious solution is $x(t) = x^{*}, \forall t \geq 0$. It satisfies both the ODE and the initial condition.

Intuitively this should be the only solution: we start from a point at which the rate of change of the state vector is $0$, so we should remain there for ever.

Can this be proven?
Under which circumstances is this true?
Are there any edge cases, maybe depending on f(x) e.g. if it is not continuous etc?

I am aware of the uniqueness-existence theorem, however since this seems to be a special problem, if you could help me with intuition.

Any help is appreciated!

Anonymous
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    Try to violate the hypotheses of the uniqueness theorem, e.g. $f(x)=\sqrt{x}$. The fact that there exists one constant solution does not affect uniqueness. – Miguel Apr 09 '22 at 20:16
  • Maybe relevant – Miguel Apr 09 '22 at 20:21
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    It depends how you interpret your differential equation. If you view it as a dynamical system which you integrate forward in time, then $x^$ is here an equilibrium point. If you start at an equilibrium point, you stay there no matter what. If you do not view it as a dynamical system, and simply as a differential equation, then you can find solutions which start at $x^$ and then move away from that value for other values for some $t$. See e.g. https://math.stackexchange.com/questions/1426755/the-nortons-dome-suggesting-non-determinism-in-classical-physics – KBS Apr 09 '22 at 20:47
  • So, both $f(x) = \sqrt{x}$, with x* = $0$, and Norton's dome prove that autonomous systems with an equilibrium point as initial condition, at least mathematically, can have more than 1 sol'n. – Anonymous Apr 10 '22 at 11:14
  • @KBS could you explain this "If you view it as a dynamical system ... you stay there no matter what." a bit more? Where does it stem from?

    Does a dynamical system imply that derivatives of higher order not contained in the ODE are zero?

    E.g. $f(x) = \sqrt{x}$ has infinite slope at $0+$ and in the example with Norton's dome, someone noted that the acceleration of acceleration was non-zero.

     – Anonymous Apr 10 '22 at 11:23
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    It just stems from what a dynamical system is. You can look at the book by Strogatz, "Nonlinear dynamics and chaos" for more details. Does a dynamical system imply that derivatives of higher order not contained in the ODE are zero? No. It just means that the higher-order derivatives are not contained in the description. – KBS Apr 10 '22 at 11:29
  • OK, thank you ! – Anonymous Apr 10 '22 at 11:32

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