Is it possible to prove unboundedness of 3rd order ODE? Consider the 3rd order ODE
$$\dddot{x}+A\ddot{x}-\dot{x}^{2}+x=0$$ where $\dot{x}\equiv \frac{dx}{dt},\ddot{x}\equiv \frac{d^{2}x}{dt^{2}}, etc$. $A$ is a constant.
If we multiply this equation by $\ddot{x}$ and integrate we can convert it into
$$\frac{1}{2}\ddot{x}^{2}-\frac{1}{3}\dot{x}^{3}+x\dot{x}=C+\int_{0}^{t}(\dot{x}^{2}-A\ddot{x}^{2})ds$$
where $C$ is a constant and $t>0$. If $A\leq 0$ the integral diverges as $t\rightarrow \infty$ and at least one of $\dot{x},\ddot{x},x$ must also diverge. 
By numerically integrating it seems that for $0<A<1.98$ the solution also diverges very quickly. 
I have been wondering whether there is a way of proving this, i.e. that for all $A<A*, A*=1.97...$ (or some other constant $A*>0$) the motion is unbounded (with the exception of $x(t)=0$).  
 A: I suspect there may be periodic solutions.  For $A = 1$, numerically plotting the solution with initial conditions $$x(0)=0, \dot{x}(0) = 0.442091320614410, \ddot{x}(0) = 0.774949154843236$$ I get this:

This looks to me like an approximation of an unstable periodic orbit.
A: Actually, no matter what $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded.  Here is why:  First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x_0$, $\dot x = x_1$, and $\ddot x = x_2$.  Then the equation becomes the first order system
$$
\begin{aligned}
\dot x_0 &= x_1\,,\\
\dot x_1 &= x_2\,,\\
\dot x_2 &= -x_0 + {x_1}^2 - A\,x_2\,.
\end{aligned}
$$
This vector field (i.e., ODE) has one singular point, $(x_0,x_1,x_2) = (0,0,0)$,
and its linearization at this point has the matrix
$$
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
-1 & 0 & -A
\end{pmatrix}
$$
The eigenvalues of this matrix are the roots of  $\lambda^3+A\lambda^2 + 1 = 0$, and there is always at least one negative real root.  Hence the stable manifold of $(0,0,0)$ has dimension at least $1$, so there will always be a nonzero solution that decays to zero (in infinite time).
