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I'm suck with the following question

I'm asked to find all possible real values of $a$ for which the solution to the following ODE is periodic. The ODE is: $$x''=-\nabla U(x),\quad x\in\mathbb R^2.$$ We have then two cases:

a) $U(x)=x_1^2+ax_2^2$ and

b) $U(x)=x_1^2+x_2^2+ax_1x_2.$

For case a) the equation is equivalent to $x_1''-2x_1=0$ which leads to $x_1=C_1\cos(\sqrt2t)+C_2\sin(\sqrt2 t)$ and $x_2''-2ax_2=0,$ which forces $a$ to be grater than $0$ for the component $x_2$ to be periodic. All this leads to $x_2=C_3\cos(\sqrt{2a}t)+C_4\sin(\sqrt{2a}t).$ For $x=(x_1,x_2)$ to be periodic, it should be true, even if i'm not sure, that the ratio of the two periods must be rational. If this were true then i should have $\sqrt a\in\mathbb Q$, but this forces $a\in\mathbb Q^2$. Am i right?

For case b) i can write the equation as $$\left(\begin{array}{c}x_1''\\ x_2''\end{array}\right)=\left(\begin{array}{cc}2 & a\\ a & 2\end{array}\right)\cdot\left(\begin{array}{c}x_1\\ x_2\end{array}\right).$$ Where to go from here, i've never found myself in a similar situation. Do i have to diagonalize the matrix? And then? Really, I am new to these things. Any help is greatly appreciated.

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Small note: constant functions are also considered periodic, though they have no "least period". So you may consider $a=0$ as well. – Shaun Ault Sep 24 '11 at 14:30
up vote 1 down vote accepted

Consider the general periodic solution

$\left(\begin{array}{c}x_1\\ x_2\end{array}\right)=\left(\begin{array}{c}X_1\\ X_2\end{array}\right)\sin(\omega t)+\left(\begin{array}{c}C_1\\ C_2\end{array}\right)$

and plug it into your ode(s) to find what values of $a$ solve this.

$\left(\begin{array}{c}x_1''\\ x_2''\end{array}\right) = -\omega^2 \left(\begin{array}{c}x_1\\ x_2\end{array}\right) $

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is it sufficient? my question is serious because i'm new to this subject.. Coul you post more details please? – uforoboa Sep 24 '11 at 18:16
To be more precise.. Does $\omega$ have to be the same for both the equations or do I have to have a relation of the type $\omega_1=\mathbb Q\omega_2$ to fulfill the requirements of the problem? thanks for your attention – uforoboa Sep 24 '11 at 18:17
Yes, the system shares the same frequency ($\omega$), and multiple displacements ($X_1, X_2$) – ja72 Sep 26 '11 at 20:05

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