I have a question regarding the following specific differential equation.

$$y'=\left(\begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix}\right)y+\left(\begin{matrix} \sin{(\omega x)} \\ 0 \\ \end{matrix}\right) \;\; \textrm{with}\;\; y(x_{0})=y_{0}$$

I need to show that for every value of $\omega \neq \pm1$ there is at least one periodic solution for $y$. I have found that the solution to the equation is:

$$y(x)=\left(\begin{matrix} \cos{x} & \sin{x} \\ -\sin{x} & \cos{x} \\ \end{matrix}\right)y_{0} + \int^{x}_{x_{0}}\left(\begin{matrix} \cos{(x-\xi)}\sin{(\omega \xi}) \\ -\sin{(x-\xi)}\sin{(\omega \xi)} \\ \end{matrix}\right)d\xi$$

I have a hard time evaluating this integral, but I believe it should be possible to show that for every value $\omega \neq \pm1$ there is at least one periodic solution without evaluating the integral. I fail to see how any irrational number for $\omega$ should give a periodic solution.

Also how would one show that all solutions are periodic if and only if $\omega \in \mathbb{Q}$?

  • $\begingroup$ It seems odd to me that you have $x$ inside your integral whose limits are $x$. $\endgroup$ – Jared Apr 9 '14 at 12:35
  • $\begingroup$ I found the solution to the equation using the following equation, which is similar to the variation of constants. $y(x)=\mathbf{\psi}(x)[\mathbf{\psi}(x_{0})]^{-1}y_{0}+\int^{x}_{x_{0}}\mathbf{\psi}(x)[\mathbf{\psi}(\xi)]^{-1}b(\xi)d\xi$ with $\psi$ the fundamental matrix. $\endgroup$ – Gehaktmolen Apr 9 '14 at 12:40
  • $\begingroup$ @Jared For a more elementary approach to finding the solutions, see my answer here. To get from the last formula in my answer to the one provided by the OP, just factor in the matrix exponential. $\endgroup$ – Git Gud Apr 9 '14 at 12:48
  • $\begingroup$ Isn't it the same question than math.stackexchange.com/questions/744922/… ? If yes, I already gave my answer, thanks to Laplace transforms. Note that the integral exactly corresponds to the last term shown at end of my calculus. $\endgroup$ – JJacquelin Apr 9 '14 at 13:32
  • $\begingroup$ @JJacquelin It is the same question yes. He is a class mate of mine. The answers in that post have helped me a lot, thanks! $\endgroup$ – Gehaktmolen Apr 9 '14 at 14:34

Let $\zeta =y_{1}+iy_{2}$ and $p=-i\partial _{x}$. Then

$$p\zeta (x)=-\zeta (x)-i\sin \omega x.$$

Next we note that $$ (\exp [iap]f)(x)=f(x+a). $$ Thus $$ (\exp [iap]p\zeta )(x)=p\zeta (x+a)=-\zeta (x+a)-i\sin \omega (x+a) $$ Assume $a=\frac{2\pi n}{\omega }$. Then $$ \sin \omega (x+a)=\sin \omega (x+\frac{2\pi n}{\omega })=\sin \omega x, $$ so $\zeta (x)$ and $\zeta (x+\frac{2\pi n}{\omega })$ satify the same equation. In case there is a single solution then $\zeta (x)$ is periodic with period $\frac{2\pi }{\omega }$. Assume that there are two solutions $% \zeta _{1}(x)$ and $\zeta _{2}(x)$. Then their difference $\Delta \zeta (x)$ satisfies the homogeneous equation $$ p\Delta \zeta (x)=-\Delta \zeta (x),\;\Delta \zeta (x_{0})=0, $$ which only has the zero solution.


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