The following is an exercise from the book Dynamics and Bifurcations by Hale/Koçak.

Let $\phi(t, t_0, x_0)$ denote the solution of the initial value problem $\dot x = f(t, x), x(t_0) = x_0$. Furthermore assume the existence and uniqueness of solutions.

Consider a 1-periodic scalar differential equation $\dot x = f(t, x)$ satisfying $f(t, 0) = 0$. If there is an $r\gt 0$ such that for $|x_0|\lt r$ the solutions $\phi(t, t_0, x_0) \to 0$ as $t \to +\infty$, then show that the zero solution $x = 0$ is stable.

I am a little confused because I believe I gave a proof of this claim without using the 1-periodicity of the differential equation but I can't find a mistake in my proof. I would appreciate it if someone could tell me whether it is correct.

Here is my proof:

We will only show stability for positive $x$ values because the proof is similar for the negative case. Let $y_0$ be an arbitrary point in $(0, r)$ and choose $\varepsilon \gt 0$. Since $\phi(t, 0, y_0)$ converges to $0$, there exists $t_0 \gt 0$ such that if $t \gt t_0$, then $|\phi(t, 0, y_0)| \lt \varepsilon$. Now define $\delta = \phi(t_0, 0, y_0)$. We know that $\delta$ is positive because uniqueness doesn't allow the solution through $y_0$ to cross $0$. We now show that for any $x_0$ in $(0, \delta)$ the solution $\phi(t, t_0, x_0)$ stays in $(0, \varepsilon)$ for all $t \ge t_0$. It follows from uniqueness that if $x_0 \lt \delta$, then $\phi(t, t_0, x_0) \lt \phi(t, t_0, \delta)$. But we also know due to uniqueness that $\phi(t, t_0, \delta) = \phi(t, t_0, \phi(t_0, 0, y_0)) = \phi(t, 0, y_0) \lt \varepsilon$. This concludes the proof.



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