Fourier Series of ODE

Question

I am having trouble finding the Fourier series of a 2nd order ODE. Should I be using the piecewise function as well to set up the range for t?

Solve $ y'' + \omega^2 𝑦 = r(t) $, where $ r(t) = |t|, -\pi < t < \pi $ by using Fourier series.

So far I have set up the ODE and set equal to r[t]

r[t] = y''[t] + ω^2 y[t]
Plot(r[t], {t, -π, π}]

Any help with the Mathematica code would be greatly appreciated. How can I find An, Bn with the function being an ODE?

Answer 1

Develop the driving force Abs[t] into FourierSeries and solve diff equation for the general coefficients.

FourierSeries[Abs[t], t, 3, FourierParameters -> {1, 1}]

(*   -((2 E^(-I t))/\[Pi]) - (2 E^(I t))/\[Pi] - (2 E^(-3 I t))/(
      9 \[Pi]) - (2 E^(3 I t))/(9 \[Pi]) + \[Pi]/2   *)  

fc[n_] = FourierCoefficient[Abs[t], t, n, 
             FourierParameters -> {1 , 1 }]

Treat n==0 separately and proofing identity yields True

FourierSeries[Abs[t], t, 9, FourierParameters -> {1, 1}] == 
   fc[0] + Sum[fc[n] (E^(I n t) + E^(-I n t)), {n, 1, 9}] // Simplify

for comparison solve numericaly for Abs[t], since DSolve gave the wrong result in version 8.0, even for Sqrt[t^2]. And solve for n==0 separately.

ynr[w_] := 
   y /. First@
NDSolve[{y[0] == 0, y'[0] == 0, y''[t] + w^2 y[t] == Abs[t]}, 
y, {t, -Pi, 3 Pi}]

ysol[0, w_] = 
   y /. First@
   DSolve[{y[0] == 0, y'[0] == 0, y''[t] + w^2 y[t] == fc[0]}, y, t]

ysol[n_ /; n >= 1, w_] = 
   y /. First@
DSolve[{y[0] == 0, y'[0] == 0, 
  y''[t] + w^2 y[t] == fc[n] (E^(I n t) + E^(-I n t))}, y, t] // 
ExpToTrig

Manipulate[
  Plot[Evaluate[{ynr[2][t], 
   ysol[0, 2][t] + Sum[ysol[n, 2][t], {n, 1, nmax, 2}]}], {t, -Pi, 
   3 Pi}, GridLines -> Automatic, PlotStyle -> {Red, Blue}, 
PlotRange -> {0, 1.5}], {nmax, 1, 7, 1, Appearance -> "Labeled"
}]

Developing to n== 5 seems sufficient.