Stability of the solution y=0 of the differential equation $y'=y^2$

Question

I have to discuss the stability of the solution $y(t)=0$ of the nonlinear differential equation $y'=y^2$. y is increasing irrespective of the sign of the initial condition $y(0)$ because the RHS of the differential equation is $y^2$. When $y(0)$ is negative as t approaches infinity, it seems the solutions are approaching $y-0$. But on the other hand, $y(0)>0$, the solutions are going away from $y=0$. I think the solution is unstable. But not sure. Can someone share their ideas?

Answer 1

This is clearly not stable. Given $\varepsilon > 0$, the unique solution (Picard-Lindelöf Theorem) to $$ y'=y^2, \quad y(0)=\varepsilon $$ is given by $$ y(t) = \frac{1}{\frac{1}{\varepsilon}-t}. $$ this does not even exist on $[0, \infty)$, so we can not talk about stabilty. It even explodes in finite time.

On the other hand, take $\delta <0$ and let $y$ be the solution to $$ y' = y^2, \quad y(0)=\delta. $$ There is a qualitative argument as to why $\lim_{t \rightarrow \infty} y(t) = 0$. Because of uniqueness, $y(t)$ can not cross $0$, so it stays negative. Also, it is always increasing because of $y' = y^2\geq 0$. Hence, a solution for every $t \geq 0$ exists and $\ell := \lim_{t \rightarrow \infty} y(t)$ exists, too. Assume that $\ell < 0$. In that case, $y$ stays in $[\delta, \ell]$. Moreover: $$ y(t) = \delta + \int^t_0 y(s)^2~\mathrm{d}s \geq \delta + \int^t_0 \ell^2~\mathrm{d}s = \delta + t \ell^2 $$ So: $$ \lim_{t \rightarrow \infty} y(t) \geq \lim_{t \rightarrow \infty} \delta + t\ell^2 = \infty $$ Contradiction, and therefore $\ell = 0$.