# Local convergence implies stability for periodic scalar differential equations

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.