# Phase portrait for an ODE (non linear)

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?

• Yes, I already have done that too, forgot to mention Jan 14 '16 at 13:39
• 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. Jan 14 '16 at 14:26

• 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 $U(u)$. Jan 16 '16 at 14:05