Consider the ODE given by
$$\left\{ \begin{aligned} z'(t) &= z(t) \cdot (z(t) - a) \cdot (1-z(t)) && t>0\\ z(0) &= z_0 \end{aligned} \right.$$
where $z_0 \in (0,a)$ and $a \in (0,1)$.
I would like to show that
$$\lim_{t\to \infty} z(t) = 0$$
in this scenario - preferably whilst avoid solving the ODE explicitly (since I have reason to believe it should be an "obvious" result).
As context, this comes up in solving the nonlinear PDE
$$\left\{ \begin{aligned} &u_t - \Delta u = u(u-a)(1-u) &&(x,t) \in \Omega \times (0,\infty) \\ &\partial_n u \equiv 0 &&(x,t) \in \partial \Omega \times (0,\infty) \\ &u(x,0) = \varphi(x) &&x \in \Omega \end{aligned} \right.$$
wherein:
- $\Omega \subseteq \mathbb{R}^n$
- $\varphi(x) \ge 0$
- $\varphi\not \equiv 0$
- $\| \varphi\|_\infty < a$
which is Example $3.4$ in Mingxin Wang's Nonlinear Second Order Parabolic Equations (ISBN-10: 0367711982).
It is claimed without proof that
$$ z_0 < a \implies z(t;z_0) \xrightarrow[\text{uniformly}]{t \to +\infty} 0$$
and
$$ z_0 > a \implies z(t;z_0) \xrightarrow[\text{uniformly}]{t \to +\infty} 1 $$
but I cannot for the life of me see why.
Clearly, $z_0 < a$ gives $z'(t) < 0$ at $t=0$, and similarly $z_0 > a$ gives $z'(t) > 0$ there. However, I'm not sure what to do with this information; what does it imply, if anything, about the future times? Ideally we'd be able to bound $z$ above/below as needed and use some sort of monotone convergence theorem to establish the desired result.
If we do calculate an explicit solution to the ODE (Wolfram link), then we get a result that is implicit and uses logarithms: without accounting for $z_0$, we'd have
$$C + t = \frac{1}{a(a-1)} \Big( a \ln(1 - z(t)) - (a-1) \ln(z(t)) - \ln(z(t) - a) \Big)$$
and if we desire this to be well-defined over the reals, then we see that
$$0 < a < z(t) < 1$$
in the scenario I wish to focus on (where $z_0 < a$). This at least gives upper and lower bounds on $z$, but it's not clear if those are the tightest (or if, indeed, some amount of information is lost in solving the ODE).
Does anyone have any ideas? Hopefully I'm just overlooking something obvious.
Let $a \in (0, 1).$
Suppose $z(t)$ solves the initial value problem,
$$\left\{ \begin{aligned} z'(t) &= z(t)\,(z(t) - a)\,(1-z(t)) && t>0\\ z(0) &= z_0 \end{aligned} \right.$$
Define the open set $U = (-\infty, a)$ and the classic Lyapunov candidate $$\begin{aligned}V: U &\to \mathbb{R}\\ z &\mapsto z^2.\end{aligned}$$ Observe that,
$$\begin{aligned} \dot{V}(z(t)) &= 2\,z(t)\,\dot{z}(t)\\ &= 2\,z(t)\,[ z(t)\,(z(t) - a)\,(1-z(t)) ], \\ &= 2\,V(z(t))\,(z(t) - a)\,(1-z(t)).\end{aligned}$$
If $z(t) \in U,$ then $z(t) - a < 0$ and $1 - z(t) > 0.$ As a result, their product $(z(t) - a)\,(1 - z(t)) < 0.$ Moreover, $V(z(t)) > 0$ for all $z(t) \in U \setminus \{0\}.$ Combine these data to find, $$\dot{V}(z(t)) < 0,$$ when $z(t) \neq 0.$ This establishes asymptotic stability of solutions initialized in $U$ by Lyapunov's Second Method.
In fact, this is the maximal stable region since $z = a$ is an invariant set that solutions can't cross over.