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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?

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  • $\begingroup$ Yes, I already have done that too, forgot to mention $\endgroup$
    – Smurf
    Jan 14 '16 at 13:39
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    $\begingroup$ 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
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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

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  • $\begingroup$ 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

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