In particular for the system $$-u''=u^3-\lambda u\text{ , }\lambda>0$$ I start writting it as a first order system $$\begin{matrix}u'&=&v\\-v'&=&u^3-\lambda u\end{matrix}$$ and then calculating the constant solutions $\{0,\sqrt \lambda,-\sqrt \lambda\}$, but from this point, how do I calculate the integral curves going through these points?

Also, is there any software to plot the phase portrait so I can compare with my sketch?

  • $\begingroup$ Yes, I already have done that too, forgot to mention $\endgroup$
    – Smurf
    Jan 14 '16 at 13:39
  • 3
    $\begingroup$ Good news everyone. This is conservative system because it has first integral. Just multiply original second order equation by $u'$ and observe that LHS and RHS are time derivatives of some functions of $u$ and $u'$. This also gives a first integral for first order system (just plut $v$ instead of $u'$). All integral curves lie on level sets of first integral, so studying integral curves of this system is almost the same as studying level sets of first integral. $\endgroup$
    – Evgeny
    Jan 14 '16 at 14:26

Analytical solving of the ODE is possible. But numerical calculus is probably more convenient to answer to the question.

Nevertheless, a closed form for the solutions is derived below, involving the Jacobi amplitude function:

enter image description here

  • $\begingroup$ Neither analytical nor numerical approach is correct here. As @Evgeny mentioned in the comments it is a conservative system of the form $\ddot u=-U'(u)$, whose phase portrait is obtained immediately from the graph of the potential $U(u)$. $\endgroup$
    – Artem
    Jan 16 '16 at 14:05
  • $\begingroup$ Of course. My answer was to people interested in analytical solving of this kind of ODE. But it was not possible to put it in the section "comments". I don't intened to repeat an answer already given. $\endgroup$
    – JJacquelin
    Jan 16 '16 at 14:21

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.