Find the expansion of $f (x) = x, 0 ≤ x ≤ \pi$ in a series of eigenfunctions of the Sturm–Liouville system $y''+\lambda y = 0, y (0) = 0, y (\pi)=0.$

Question

I need to expand the given function in terms of eigen functions series, firstly i find the solutions to the SLP and from there i get my eigen functions $n$ values, and then i find the coefficient Cn and simply multiply it with my eigen function to get the series! But in this particular question, i dont know how to find the series if i have gotten more than one eigen function? P.S : I havent yet studied fourier series in depth, just have studied orthogonal series and the definition to generalised fourier series.

Answer 1

The eigenfunctions are $\{ \sin(nx) \}_{n=1}^{\infty}$, and the eigenfunction expansion is the Fourier sine series for $f$: $$ f \sim \sum_{n=1}^{\infty}\frac{\int_0^\pi f(x')\sin(nx)dx'}{\int_0^\pi \sin^2(nx')dx'}\sin(nx) $$ You can determine the normalization constants $$ \int_0^\pi \sin^2(nx')dx' = \frac{1}{2}\int_{-\pi}^{\pi}\sin^2(nx')dx \\ = \frac{1}{4}\int_{-\pi}^{\pi}\sin^2(nx')+\cos^2(nx')dx'=\frac{\pi}{2} $$ So $$ f\sim\sum_{n=1}^{\infty}\frac{2}{\pi}\int_0^{\pi}x'\sin(nx')dx' \sin(nx). $$ The remaining integrals are brute force.