# 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 the solution also diverges very quickly.

I have been wondering whether there is a way of proving this, i.e. that for all $$A (or some other constant $$A*>0$$) the motion is unbounded (with the exception of $$x(t)=0$$).

• @YCor I welcome a better suggestion. Maybe "How to prove unboundness for a range of A?" – user2175783 Jan 23 '19 at 16:05
• Of course, $x=0$ is a bounded solution. Are you claiming that all other solutions are unbounded, or that there is at least one unbounded solution? I also don't understand how you conclude that the integral diverges when we don't have any information on the integrand. – Christian Remling Jan 23 '19 at 17:52
• @ChristianRemling Yes I mean all other solutions are unbounded for a certain range of $A$. It does seem like I am too optimistic making the claim about the integral diverging for $A\leq 0$ (the integrand will be positive for $A\leq 0$). I am thinking about it. – user2175783 Jan 23 '19 at 19:14
• There is always at least a one-parameter family of unbounded solutions, namely $x(t) = t^2/4 + c t + c^2 - A/2$. – Robert Israel Jan 23 '19 at 19:14

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).
• @user35202: As far as I know, there's not an explicit parametrization of the (1-dimensional) stable manifold in this case, though there are certainly numerical techniques that will describe it approximately. In cases such as this, there is typically a countable number of values of the parameter $A$ for which the stable manifold is only $C^k$ at the origin (for some finite $k$). Of course, it will be real-analytic everywhere else. That may possibly have some effect on the stability of numerical schemes for describing the stable manifold. – Robert Bryant Jan 25 '19 at 21:16
• @user35202: Here's a little more about this, in case you are interested: If we let $a<0$ be the (unique) negative root of $a^3 + Aa^2+1=0$, i.e., $A = -(a+1/a^2)$, then as long as $a^3\not=(n+1)/n^2$ for any integer $n\ge 2$, there is an analytic function $g_a(\tau)$ on an open neighborhood of $\tau=0$ such that $x(t) = g(e^{at})$ solves the equation when $t>0$ is sufficiently large, where $$g_a(\tau) = \tau + \frac{a^2\,\tau^2}{(3{-}4a^3)}+\frac{2a^4\,\tau^3}{(3{-}4a^3)(4{-}9a^3)}+ \frac{4a^6(13{-}21a^3)\,\tau^4}{3(3{-}4a^3)^2(4{-}9a^3)(5{-}16a^3)} +\cdots .$$ This $x(t)$ converges to $0$. – Robert Bryant Jan 26 '19 at 14:47
• @user35202: You're welcome. By the way, I realized, after I wrote my previous comment (which I can't edit now), that, since $a<0$, we can never have $a^3 = (n{+}1)/n^2$ anyway, so $g_a(\tau)$ is always real-analytic. Meanwhile, the expression above would work to describe unstable solutions associated to the positive roots (if any) of $a^3{+}Aa{+}1=0$ in the regime $t<<0$, but only as long as $a^3 \not= (n{+}1)/n^2$ for some $n\ge 2$. – Robert Bryant Jan 27 '19 at 11:13
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: