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


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 $x(t)=0$).

  • $\begingroup$ @YCor I welcome a better suggestion. Maybe "How to prove unboundness for a range of A?" $\endgroup$ – user2175783 Jan 23 '19 at 16:05
  • 2
    $\begingroup$ 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. $\endgroup$ – Christian Remling Jan 23 '19 at 17:52
  • $\begingroup$ @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. $\endgroup$ – user2175783 Jan 23 '19 at 19:14
  • 4
    $\begingroup$ There is always at least a one-parameter family of unbounded solutions, namely $x(t) = t^2/4 + c t + c^2 - A/2$. $\endgroup$ – 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).

  • 1
    $\begingroup$ To find the stable manifold/stable paths I assume would be possible only numerically? $\endgroup$ – user35202 Jan 25 '19 at 16:54
  • 2
    $\begingroup$ @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. $\endgroup$ – Robert Bryant Jan 25 '19 at 21:16
  • 2
    $\begingroup$ @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 $0$. $\endgroup$ – Robert Bryant Jan 26 '19 at 14:47
  • $\begingroup$ Thank you for explaining it so clearly. Very interesting. $\endgroup$ – user35202 Jan 26 '19 at 18:41
  • 2
    $\begingroup$ @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 $n\ge 2$. $\endgroup$ – 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:

enter image description here

This looks to me like an approximation of an unstable periodic orbit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.