TITLE: How to solve $u'' + k u + \epsilon u^3 = 0$?
QUESTION [5 upvotes]: I am looking at the project of my ODE class, there is one problem saying we have to solve $u'' + k u + \epsilon u^3 = 0$. The problem gives us some values of $k$, $\epsilon$ and says you should experiment with different initial values with Euler's method. I have solved the equation using Euler's method. But now I am confused, how can I know I get the right solution, in previous exercises the book gave the answer for the function and we can compare it with the Euler answer, the project page says nothing about the answer. I tried letting $u = e^{rt}$ like in the class to get the characteristics equation if both constants are 1:
$$
r^2 + r + e^{2rt}=0
$$
but I can't solve this. Thanks alot.
The initial values are $u(0)=0$, $u'(0)=1$, and $k=\epsilon =1$.

REPLY [4 votes]: It looks like you are supposed to use perturbation theory to solve this-the $\epsilon$ is a tipoff.  Initially you set $\epsilon$ to zero, which gives an equation you know how to solve.  We designate its solution $u_0$ as the zero order solution and hope that as the other term is small it will not perturb the solution too much.  So $u_0=a \cos (\sqrt k t + \phi)$.  Now you plug this into the perturbation term and solve the new equation, giving $u_1''+ku_1+\epsilon u_0=0$ or $u_1''+ku_1+\epsilon a \cos (\sqrt k t + \phi)=0$.  Now solve this to get $u_1(t)$.  Plug that in to the original equation to get $u_2''+ku_2+\epsilon u_1=0$.  You should get new terms with multiples of the original frequency.  Keep going until you get tired.  As long as $\epsilon$ is small enough, you hope that the corrections are getting smaller and smaller.  
With your edit, I would be worried that $\epsilon =1$ might perturb the solution enough that things don't converge.  Even in the first order, you might get a secular (frequency zero) term representing unbounded motion.

REPLY [2 votes]: With both $k,\epsilon$ positive, it is guaranteed that you get slow oscillation, somewhat like a sine wave or a Bessel function.  Anyway, with $u$ positive $u''$ is strictly negative. The only unpredictable part is that you may get slightly different first derivatives at each new root of $u,$ so the whole thing is not going to be exactly periodic, only nearly so.
I would see about getting numerical estimates for $u'$ at every occurrence of $u=0.$ i think it very likely that, after several oscillations, each such $|u'|$ approaches a specific value, and the curve becomes very close to periodic. But maybe that's just me. 
https://en.wikipedia.org/wiki/Almost_periodic_function