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I have the following problem: In the equation $\ddot{x}+\Omega^2x+\epsilon f(x) = \Gamma \cos t$, $\Omega$ is not close to an odd integer, and $f(x)$ is an odd function of x, with expansion, $$f(a\cos t) = -a_1(a)\cos t - a_3(a)\cos3t-\cdots.$$ Derive a perturbation solution of period $2\pi$, to order $\epsilon$. I have a solution so my question is on the solution of the problem.

Solution:

In the equation $\ddot{x}+\Omega^2x+\epsilon f(x) = \Gamma \cos t$, let $$x = x_0+\epsilon x_1+\cdots$$. The two leading terms satisfy $$\ddot{x_0}+\Omega^2 x_0 = \Gamma \cos t$$ $$\ddot{x_1}+\Omega x_1 = -f(x_0+\epsilon x_1+\cdots)\approx -f(x_0).$$ The $2\pi$ periodic solution of the first equation is $$x_0 = \dfrac{\Gamma}{\Omega^2-1}\cos t$$ provided $\Omega$ is not close to $1$ or any odd integer for higher order terms. The next term $x_1$ in the perturbation satisfies

\begin{align}\ddot{x_1}+\Omega x_1 &= -f\bigg(\dfrac{\Gamma}{\Omega^2-1}\cos t\bigg) \\ &= a_1(k)\cos t+a_3(k)\cos 3t. (1) \end{align} where $k = \dfrac{\Gamma}{\Omega^2-1}$. The $2\pi$ periodic solution is $$x_1 = \dfrac{a_1(k)}{\Omega^2-1}\cos t + \dfrac{a_1(k)}{\Omega^2-9}\cos 3t.$$

My question is at step $(1)$. I understand that they are putting the solution for $x_0$ in the function and expanding it out but why only up to $a_3(k)\cos 3t$?

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  • $\begingroup$ Do we know anything about the $a_i$s? Perhaps they just forgot to write the $\ldots$? I see no reason to truncate that expansion there. $\endgroup$ – Bennett Gardiner Jan 5 '14 at 23:52
  • $\begingroup$ No it doesn't say anything. It even gives an explicit solution for $x_1$. Let me add it. $\endgroup$ – RDizzl3 Jan 5 '14 at 23:55
  • $\begingroup$ What reference are you following, anyway? Is that supposed to be $a_1$ twice in the solution for $x_1$ or did you make a typo? $\endgroup$ – Bennett Gardiner Jan 6 '14 at 1:34
  • $\begingroup$ The is in the solutions manual for Nonlinear Ordinary Differential Equations by D.W Jordan and P. Smith. Problem 5.7. $\endgroup$ – RDizzl3 Jan 6 '14 at 1:45
  • $\begingroup$ Does it say anything about the expansion for $f$? Is it in the limit that $a\rightarrow 0$ $\endgroup$ – Bennett Gardiner Jan 6 '14 at 4:43
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Bennett Gardiner correctly called out a typo in $x_1$: it should read $$x_1 = \dfrac{a_1(k)}{\Omega^2-1}\cos t + \dfrac{a_3(k)}{\Omega^2-9}\cos 3t$$ since this is what we get by solving (1).

The reason for dropping higher frequency terms is that their coefficients are implicitly assumed decaying, probably as $a_{2j+1} = O(\epsilon^j)$. This is something that the authors of the problem failed to specify after writing $$f(a\cos t) = -a_1(a)\cos t - a_3(a)\cos3t-\cdots$$ For example, if $a_{3}$ is zero but $a_5$ is not, then we should have $$x_1 = \dfrac{a_1(k)}{\Omega^2-1}\cos t + \dfrac{a_5(k)}{\Omega^2-25}\cos 5t +\dots$$ instead. And if each $a_j$ is $10^6$ times the preceding one, then all these asymptotics are junk.

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  • $\begingroup$ Thank you for the clarification! $\endgroup$ – RDizzl3 Jan 6 '14 at 6:18

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