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Sometimes solutions to differential equations can converge to an equilibrium path as time approaches infinity. Also, if you were given a second order differential equation and two different sets of y’(0) and y(0) starter information, it would be possible for those solutions to share the same basic path. As in, if you were to trace the trajectory backwards from one of the curves, you could find the other one’s starter point.

Is it possible to have a differential equation, with two different sets of starter information, and the solutions converging, however you can’t trace one trajectory backward to find the other starter information?

My thought is that if there was such a differential equation, with solutions that converge, but trajectories don’t share a path until equilibrium, it would have to be non-linear and higher than first order.

The van der pol oscillator would almost work as an example, because a lot of solutions from all different starting values converge on the limiting path. However, those solutions converge at infinity. I am wondering if there is an example like that, but the solutions converge in finite time.

math tutor
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    Try $y'(x)=-\sqrt{|y(x)|}$ where all solutions end up on the x axis in finite time. – Lutz Lehmann Feb 01 '17 at 19:10
  • Thanks! This one definitely works. If you plot a flow plot in mathematica, the solutions actually run off the x-axis, because the gradient of y'[t] is undefined when y = 0. Are there any examples you can think of that would have a gradient with no undefined points? Thank you for your help! – math tutor Feb 03 '17 at 18:27
  • As long as you have a locally bounded derivative $f_y$, the ODE $y'=f(x,y)$ satisfies a local Lipschitz condition and has unique solutions forward and backward by Picard-Lindelöf-Cauchy-Lipschitz. – Lutz Lehmann Feb 03 '17 at 22:32
  • We encountered exactly this phenomenon here. – Giuseppe Negro Oct 18 '22 at 14:08

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If you have autonomous diff eq, then any solutions that intersect the same $(y,y')$ coordinate, must completely overlap.

e.g.

$y'' - y = 0\\ y = A\sin \theta + B \cos\theta$

$y_1(0) = 0, y_1'(0) = 1\\ y = \sin\theta$

$y_2(0) = 1, y_2'(0) = 0\\ y_2 = \cos\theta$

$y_1(t) = y_2(t-\frac {\pi}2)$

But it would never be possible for $y_3(t) = 2 \sin t$ to have any point on the path have the same $(y,y')$ coordinate any $(y,y')$ coordinate of $y_1(t)$

One way to think about is that these equations can be "run backward" so if you ever end up at the same place, you must have come along the same path.

If you have a non-autonomous equation, however, different rules apply and you can arrive at the same $(y,y')$ at different times and then head off in different directions.

Doug M
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