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The exercise is:

Approximate the solutions of $x'' + x + ε*x^3 = 0$, the initial value is $x(0) = 1$ and $x'(0) = 0$ up to order one. (Hint: It is not necessary to convert this second order equation to a first order system.)

To solve this, we have to use Regular perturbation theory: we have to get it with the equation: $\varphi(t, \epsilon) = \varphi_{0}(t) + \varphi_{1}(t)\epsilon + o(\epsilon)$ where $\varphi$ is the solution of the ode and $\epsilon$ is the perturbation.

$\varphi_{0}(t) = \varphi(t, 0)$ is given as the solution of the unperturbed equation where $\epsilon = 0$

$$\varphi_{1}(t) = \frac{\partial \varphi(t,\epsilon)}{\partial \epsilon} \bigg|_{\epsilon=0}$$

how can I compute $\varphi_{1}(t)$? or is this the wrong approach?

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  • $\begingroup$ Why are you taking a derivative with respect to $\epsilon$? $\endgroup$
    – Paul
    Apr 26, 2017 at 18:29
  • $\begingroup$ i think it is the second taylor-term of the approximation, it was defined in the script we use in our lecture, I dont really know why in respect to $\epsilon$ $\endgroup$
    – Pferd
    Apr 28, 2017 at 17:54

1 Answer 1

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First you solve the unperturbed equation $x''+x=0$ with the given initial conditions to get $\phi_0(t)$. Can you do that? Then you use that to approximate the cubic term. You need to find the solution to $\phi_1(t)''+\phi_1(t)+\epsilon \phi_0(t)^3=0$. You are imagining that the effect of the cubic term is small so the solution will be approximately $\phi_0(t)$. That gives you a nice second order differential equation to solve.

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  • $\begingroup$ so I found the term $\phi_0(t) = cos(t)$ and I put it in the equation $\phi_1(t)''=-\phi_1(t)-\epsilon\phi_0(t)^3=-\phi_1(t)-\epsilon\cos(t)^3$, $\phi_1(t)$ is also periodic, so in generaly it looks like $a*cos(t) + b*sin(t)$?, how can I find $a$ and $b$ ? thank you for your answer $\endgroup$
    – Pferd
    Apr 27, 2017 at 21:16
  • $\begingroup$ There is an identity relating $\cos^3 t$ to $\cos t$ and $\cos 3t$ You can plug that in. You get a forced oscillator with a forcing frequency three times higher than resonance. The $\cos t$ term is right on resonance, so will generate a solution that involves $t \cos t$, which I believe one usually argues is non-physical, but the $\cos 3t$ term changes the shape of the resonance. $\endgroup$
    – Ross Millikan
    Apr 27, 2017 at 23:53
  • $\begingroup$ great explanation, thank you for your help. $\endgroup$
    – Pferd
    Apr 28, 2017 at 18:19

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