MHB Fourier series damped driven oscillator ODE

Dustinsfl
Messages
2,217
Reaction score
5
$$
-\sum_{n = 0}^{\infty}n^2\omega^2C_ne^{in\omega t} + 2\beta\sum_{n = 0}^{\infty}in\omega C_ne^{in\omega t} + \omega_0^2\sum_{n = 0}^{\infty}C_ne^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}
$$
How can I justify removing the summations and solving for $C_n$?
$$
-n^2\omega^2C_ne^{in\omega t} + 2\beta in\omega C_ne^{in\omega t} + \omega_0^2C_ne^{in\omega t} = f_ne^{in\omega t}
$$
 
Physics news on Phys.org
dwsmith said:
$$
-\sum_{n = 0}^{\infty}n^2\omega^2C_ne^{in\omega t} + 2\beta\sum_{n = 0}^{\infty}in\omega C_ne^{in\omega t} + \omega_0^2\sum_{n = 0}^{\infty}C_ne^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}
$$
How can I justify removing the summations and solving for $C_n$?
$$
-n^2\omega^2C_ne^{in\omega t} + 2\beta in\omega C_ne^{in\omega t} + \omega_0^2C_ne^{in\omega t} = f_ne^{in\omega t}
$$
Combining the summations, you can write this as $$\sum_{n = 0}^{\infty}\bigl(-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n \bigr) e^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}.$$
Now use the fact that a (reasonably well-behaved) function has a unique Fourier expansion to conclude that the coefficients on each side must be the same, to conclude that $-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n = f_n$ for each $n.$
 
Opalg said:
Combining the summations, you can write this as $$\sum_{n = 0}^{\infty}\bigl(-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n \bigr) e^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}.$$
Now use the fact that a (reasonably well-behaved) function has a unique Fourier expansion to conclude that the coefficients on each side must be the same, to conclude that $-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n = f_n$ for each $n.$

Could I also just multiple through by $\frac{1}{2\pi}\overline{e^{in\omega t}}$, and by Sturm-Liouville, the summation integrates to 1?
 
Last edited:
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top