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

Question

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

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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 !

Answer 1

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.