Every solution of $x''+x+x^3=0$ is set on $\mathbb{R}$ (proof)

Question

Show that every solution of $x'' + x + x^3 = 0$ is set on $\mathbb R$.

Proof:

If we transform this higher order ode into a system of first order ode's, we get: $$\vec{x}'=\vec{F}(x)$$ with $$ F_1(x)=x_2, \quad F_2(x)=-x_1-x_1^3$$

Now, since all the partial derivatives of $F$ are continuous within an interval $I=(a,b)$ and thus bounded in $(a,b)$, $\,F$ satisfies a local lipschitz condition in $(a,b)$.

Given that, Picard theorem says that there exists a solution $x(t) \in A$, $t \in(c,d)$ with $A$ a compact set of $\mathbb{R^2}$.

Since $F$ is continuous in $\mathbb{R^2}$, $x(t)$ is defined everywhere: $$(c,d)=(-\infty,\infty)$$

Answer 1

The only problem is the last sentence, nothing in what is before allows you to conclude that.

The usual approach in such problems is to find an integral, such as in this case $$ E(x,y)=\frac12 y^2+\frac12 x^2+\frac14 x^4 $$ (the sum of the kinetic and potential energies). Notice that indeed $$\frac{d}{dt} E(x(t),y(t))=0$$ along solutions. Now you only need to note that each level curve of $E(x,y)$ is a bounded set and so all solutions are global (solutions starting in a level curve remain in the same level curve).