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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 $x(t)=0$).

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  • $\begingroup$ @YCor I welcome a better suggestion. Maybe "How to prove unboundness for a range of A?" $\endgroup$
    – user2175783
    Commented Jan 23, 2019 at 16:05
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    $\begingroup$ 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
    Commented Jan 23, 2019 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
    Commented Jan 23, 2019 at 19:14
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    $\begingroup$ 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
    Commented Jan 23, 2019 at 19:14

2 Answers 2

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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).

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    $\begingroup$ To find the stable manifold/stable paths I assume would be possible only numerically? $\endgroup$
    – user35202
    Commented Jan 25, 2019 at 16:54
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    $\begingroup$ @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
    Commented Jan 25, 2019 at 21:16
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    $\begingroup$ @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
    Commented Jan 26, 2019 at 14:47
  • $\begingroup$ Thank you for explaining it so clearly. Very interesting. $\endgroup$
    – user35202
    Commented Jan 26, 2019 at 18:41
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    $\begingroup$ @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
    Commented Jan 27, 2019 at 11:13
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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.

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