I showed the following theorem:
If $\varphi$ is a maximal solution of $\begin{cases}x' = f(t,x) \\ x(t_0) = x_0\end{cases}$, $f:]a,b[ \times \mathbb{R}^d \to \mathbb{R}^d$ continuous $-\infty \le a < b \le +\infty$ with $V \in C^1(\mathbb{R}^d,\mathbb{R}^+_0)$ coercive, $m,n \in C(\mathbb{R},\mathbb{R}^+_0)$ such that $\forall (t,x) \in D. \dot V(t,x) \le m(t) V(x) + n(t)$ then $\omega = b$.
This has a direct application to show that the solutions of $x''+cx'+x^3 = 0$ are defined on $\mathbb{R}$. In particular if the domain of the maximal solution is $]\alpha,\omega[$. Then $\omega = +\infty$. The problem is whether $\alpha = -\infty$ and whether this can be stablished with the theorem above?
I tried to show it using the dual IVP but it didn't work. So perhaps there is a counterexample?
Best Answer
Your equation, for $c>0$, describes the mechanical system of a particle on a line at position $x$ moving in the potential field $V(x)=\frac14x^4$ with friction or energy dissipation with coefficient $c$. Physical intuition dictates that the total energy of the particle continually lowers, moving towards the stable rest position at $x=0$.
That is, set $$E=\frac12x'^2+\frac14x^4$$ the total energy, then $\frac{dE}{dt}=-c x'^2$, so that $E$ is falling in time as long as the particle is moving. However, as $$\frac{dE}{dt}\ge -2cE\implies E(t)\ge E(0)e^{-2ct},$$ the rest position is not reached in finite time. For $t<0$ one similarly gets $E(t)\le E(0)e^{-2ct},$ so that the energy and thus the state is always finite at finite times, so that the solution has maximal domain $\Bbb R$.