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Show whether the IVP $y'(t) = y(t)^2 , y(0) = 1$ satisfy the Picard-Lindelöf theorem


So we need to show $$y(t)^2$$ is continuous for all $(t,y)$. If $f(t,y)=y^2,$ then this holds clearly for all $(t,y)$.

We also need to check that it is Lipschitz continuous, i.e $$\lim_{t_2 \to t_1}\frac{y(t_1)-y(t_2)}{t_1-t_2}\leq \alpha$$ for all $\alpha$ real. We can simply take the derivative with respect to $y$. So $$\frac{d}{dy} f(t,y) = 2y.$$ This derivative is bounded on $[1 - \delta, 1 + \delta]$ for all $\delta > 0$, thus it is Lipschitz continuous around the initial value (IV) $y(0)=1$.

Thus the Picard-Lindelöf theorem holds in this case.

Is what I did correct ? Thanks for your feedback !

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  • $\begingroup$ Please state your version of Picard-Lindelöf. There are variants of the theorem for several non-equivalent situations. // What has the difference quotient of $y(t)$ (which does not yet exist at that point in the theorem) to do with the Lipschitz continuity of $f$? $\endgroup$ Commented Mar 12, 2020 at 16:54
  • $\begingroup$ @LutzLehmann Sorry, I'm an engineering student. I don't really know much about the theorem. They taught us that to check the theorem, we need to check if a) the function is continuous b) to check if it is Lipschitz continuous, take the derivative of the function and see if it is defined around the initial value $\endgroup$
    – user446410
    Commented Mar 12, 2020 at 17:53

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Per your comments, you are using the practical version of the theorem.

  1. Check that $f$ is continuous in both variables,
  2. Check that $f$ is continuously differentiable in $y$.

If the $y$ derivative exists and is continuous, then it is bounded over bounded sets and each bound is a Lipschitz constant and thus the more "elementary" form of the theorem can be applied, as you noted.

Your limit in the middle makes no sense, there is no apparent role for $α$, any or all. This might be marked as an error. Everything before and after that is correct.

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  • $\begingroup$ Thanks for the corrections ! The limit was supposed to mean that the derivative never blows toward infinity. $\endgroup$
    – user446410
    Commented Mar 12, 2020 at 21:36

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