Stable and unstable fixed points of $y' = 500y^2(1 - y)$

Question

$y' = 500y^2(1 - y)$ has fixed points at $y^*=0$ and $y^*=1$

$y(0) =0$

There are 2 definitions;

1 says that:

A fixed point $y^*$ is asymptotically stable if $$ \sigma(Df(y^*)) \subset \Bbb C^- ~~\text{ where }~~ \Bbb C^- := {z \in \Bbb C : Re(z) < 0} ; $$ $\sigma(Df(y^*))$ is the set of eigenvalues of the matrix $Df(y^*)$.

2 says that:

A fixed point $y^*$ is asymptotically stable (or attractive) if there exists a ball $B_\delta(y^*)$ (of radius $\delta> 0$ and centered at $y^*$) such that, whenever $y_0 \in B_\delta(y^*)$, the solution to $y' = f (y)$, $y(0) = y_0$ satisfies the limit $y(t) \to y^*$ for $t \rightarrow \infty$.

I am convinced the first definition cannot be applied and so I need to show that for the fixed point $y^*=1$ there exists a $\delta$ such that whenever $|y_0 - 1|< \delta $, the solution to $y' = 500y^2(1-y), y(0)=0$ satisfies $y(t) = 1$ as $t \rightarrow \infty$.

How would i go about doing this

Answer 1

As the equation is scalar, you can more directly explore the behavior near the stationary solutions.

• For $y\approx y^*=0$, the equation is dominated by $y'=500y^2$, $y(0)=y_0$, which has a solution $y(t)=\frac{y_0}{1-500y_0t}$ which moves away from $0$ for $y_0>0$ and $t<(500y_0)^{-1}$, and towards $0$ for $y_0<0$. This shows that in the case $f'(y^*)=0$ the stability can not be determined from the derivative value alone, as both stable and unstable behavior are possible.

• For $y\approx y^*=1$ the ODE is dominated by $y'=-500(y-1)$, $y(0)=y_0$, which has a solution $y(t)=1+(y_0-1)e^{-500t}$. This moves toward $1$ for all initial values $y_0\approx y^*$. This demonstrates the claim of Theorem 1 (not a definition) that $f'(y^*)=-500$ means that $y^*$ is stable.

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For scalar autonomous equations you can do a very quick qualitative analysis of the problem by determining segments where $f$ has constant sign. The signs around the roots of $f$ then determine the stability of that stationary point. Here you get positive sign for $y<1$ and negative sign for $y>1$. This already tells you that $1$ is a stable stationary point. All trajectories starting in $(0,\infty)$ converge towards $1$, so that you could take $δ=1$.