Global existence of ODE $f'(x)=-f(x)^3$, $f(0)=a$ I am studying a PDE and as simplification I begin with studying the ODE
\begin{equation*}
f(t)'=-f(t)^3,  f(0)=a.
\end{equation*} I want to develop global existence for this ODE, .i.e. for any $T>0$ $$\sup_{t\in[0,T]}|f(t)|<\infty.$$
Of course we can just write down the solution which is given by $f(t)=\pm \frac{1}{\sqrt{1/a^2+2t}}$ with + for $a>0$ and - for $a<0$. However such an explicit solution can't be hoped for in the PDE case and thus I want to know why global solutions exist. This clearly depends on the minus sign in the equation as $g(t)'=g(t)^3$ blows-up in finite time. However any Lipschitz argument or norm bound will treat both equations in the same manner and not account for the negativity.
Of course the idea would be to use some bounds to repetitively apply the local solution theory and glue solutions together, however one needs to show that these solution intervals don't decrease in size.
I can see how the equations behave differently from a plot but I am lacking a convincing argument. If we view the equation as $f(t)'=F(f(t))$ with $F(x)=-x^3$ and assume that $a>0$ then have have $F(a)<0$ and thus $f(0)'<0$ thus for some (small) $t_1>0$ $f(t_1)<f(0)$. If we assume we get a sequence $(t_n)_{n\in\mathbb{N}}$ for which we always have $0<f(t_n)$ we get a decreasing sequence and our solution won't blow-up in finite time. While this argument is intuitive it isn't rigours at all, but it shows that the local Lipschitz constant in a uniqueness proof wouldn't increase but even decrease.
Could someone develop a rigours argument, show how the solutions can be glued together and why the solution intervals don't decrease or point at a different approach to the global existence.

Consider $V(y)=y^2$, then
$$
\frac{d}{dt}V(f(t))=2f(t)f'(t)=-2f(t)^4<0.
$$
Thus this simple Lyapunov function is decreasing, which implies that it and with it $f$ is bounded.
