- #1

Runei

- 193

- 17

I'm trying to analyze a system of elastically coupled oscillators, whose masses are all the same, using Fourier expansion. So the differential equation I am looking at right now is of the form

[itex]m\frac{d^2\hat{y}_k}{dt^2} + \gamma\frac{d\hat{y}_k}{dt} - \kappa\Delta^2\hat{y}_k = \hat{f}_k(t)[/itex]

Here the [itex]\hat{y}_k[/itex] is a particular Fourier coefficient of a position vector with all of its components being the position function y(t) of each of the particles. The [itex]\hat{f}_k[/itex] is the corresponding Fourier coefficient for the driving force. So basically

[itex]\vec{y} = <y_0(t), y_1(t), y_2(t), ..., y_{M-1}(t)>[/itex]

and

[itex]\vec{f} = <f_0(t), f_1(t), f_2(t), ..., f_{M-1}(t)> [/itex]

And these two vectors have then been decomposed into Fourier representation. After taking the inner product it was possible to create M differential equations of the form above, which I am now working with.

Now -- I am trying to show that a particular solution to the differential equation is given by

[itex]\hat{y}_k(t) = A(\nu(k), \omega(k, \gamma))\sin(\nu(k)t + \phi)[/itex]

When the Fourier coefficients for the force are oscillating like

[itex]\hat{f}_k(t) = \hat{f}_k(0)sin(\nu(k)t) = \frac{e^{i\nu(k)t} - e^{-i\nu(k)t}}{2i}\hat{f}_k(0)[/itex]

It might be me whose blind (or tired after hours of work), but when I insert the particular solution (or what should be), I get to the point where I have the following:

[itex]A(\nu(k), \omega(\gamma, k))\left[-m\nu(k)^2sin(\nu(k)t+\phi) + \gamma\nu(k)\cos(\nu(k)t+\phi) - \kappa\lambda(k)\sin(\nu(k)t+\phi)\right] = \frac{e^{i\nu(k)t} - e^{-i\nu(k)t}}{2i}[/itex]

I've tried several things such as taking out the sine, and converting it using the addition trigonometric identity. I also tried converting the sines and cosines to complex exponentials, but I just can't seem to see where I should go.

So basically I'm out of creativity and a almost out of coffee. So any help, hints, tips or guidance would be very much appreciated.

Thank you.