$y'' + (y')^3 + y = 0$ attractor at (0,0)? This question comes from a true or false. The claim is the following:
All solutions $\varphi(t)$ of $$y'' + (y')^3 + y = 0$$ go to zero as $t\to \infty$ that is $\varphi(t) \to 0$ as $t\to\infty$.
I tried changing to a plane system of the form $$\begin{cases}u' = v\\v' = -v^3-u
\end{cases}$$
where $u = y$, $v = y'$
After that I saw if I could find the implicit equation that gives me the trajectories of the solution curves but the differential equation is to hard to solve.
I tried to linearize the system at $(0,0)$ but the eigenvalues I'm getting are $\pm i$ and because the real part is zero I believe I cannot deduced the type of critical point for the non linear system.
Any help is appreciated
 A: Consider $V:\mathbb{R}^2 \to\mathbb{R}$, $V(x,y)=x^2+y^2$. Then $V$ is a Lyapunov function for the plane system: Let $(u,v):[0,\omega_+) \to \mathbb{R}^2$ be a solution of this system (nonextendable to the right), then
$$
\frac{d}{dt}V(u(t),v(t)) = 2u(t)u'(t)+ 2v(t)v'(t) = 2u(t)v(t)-2v(t)^4-2v(t)u(t)=-2v(t)^4 \le 0.
$$
Hence $t \mapsto u(t)^2+v(t)^2$ is decreasing; in particular $\omega_+= \infty$ (since $(u,v)$ is bounded) and $(0,0)$ is stable. Unfortunately $V$ is not negative definite. At this point I think the invariance principle of La Salle could be applied, however to avoid this, we can do the following:
We first prove that $v(t) \to 0$ $(t \to \infty)$. Assume by contradiction that there is a sequence $(t_n)$ in $[0,\infty)$ with $t_n \to \infty$ and such that $|v(t_n)| \ge c >0$ $(n \in \mathbb{N})$ for some $c > 0$. Since $(u,v)$ is bounded also $v'$ is bounded, hence $v$ is Lipschitz continuous on $[0,\infty)$ with Lipschitz constant $L >0$, say. Now
$$
|v(t)| \ge |v(t_n)|- L|t_n-t|,
$$
so $|v(t)| \ge c/2$ for $t \in [t_n-c/(2L),t_n+c/(2L)]$. Hence
$$
V(u(t),v(t)) = V(u(0),v(0)) -\int_0^t 2v(s)^4 ds \to -\infty \quad (t \to \infty),
$$
a contradiction.
Next, since $t \mapsto u(t)^2+v(t)^2$ is decreasing, it is
convergent to some $r \ge 0$ as $t \to \infty$ and since $v(t) \to 0$ $(t \to \infty)$ we get $u(t)^2 \to r$ $(t \to \infty)$. Since $u$ is continuous we get $u(t) \to q$ with $q \in \{-\sqrt{r},\sqrt{r}\}$ $(t \to \infty)$.
Summing up $(u(t),v(t)) \to (q,0)$ $(t\to \infty)$. But then $(q,0)$ is a critical point of the system, hence $(0,-0^3-q)=(0,0)$, that is $q=0$.
