Expanding the function $f(x) = 1, 0 < x < 1$ in a series of the eigenfunctions?

Question

I have a problem that asks me to expand the function $ f(x) = 1, 0 < x < 1$ in a series of the eigenfunctions of the given problem. For example, one given problem is $y'' + 2y' + ( \lambda + 1)y = 0, y(0) = 0, y(1) = 0$.

What is meant by this ("expand the function in a series of the eigenfunctions of the given problem")? I don't understand what it is asking to do?

Thanks for any help.

Answer 1

This is an eigenvalue/eigenfunction problem. Disregarding the endpoint conditions for the moment, the solutions of your problem have the form $$ y(x)=Ae^{-x}\cos(\sqrt{\lambda}x)+Be^{-x}\sin(\sqrt{\lambda}x). $$ The solutions that satisfy $y(0)=y(1)=0$ are constant multiples of $$ y_n(x) = e^{-x}\sin(n\pi x),\;\; \lambda_n=n^2\pi^2,\;\;n=1,2,3,\cdots. $$ In order to expand the constant function $f(x)=1$ in the eigenfunctions $\{ y_n \}$, you must find coefficients $C_n$ such that $$ 1 = \sum_{n=1}^{\infty}C_n e^{-x}\sin(n\pi x) \\ e^{x} = \sum_{n=1}^{\infty}C_n \sin(n\pi x). $$ Using the orthogonality of the $\sin(n\pi x)$ functions on $[0,1]$ with respect to the integral gives $C_n$: $$ \int_{0}^{1}e^{x}\sin(n\pi x)dx = C_n\int_{0}^{1}\sin^2(n\pi x)dx \\ C_n = \frac{\int_{0}^{1}e^{x}\sin(n\pi x)dx}{\int_{0}^{1}\sin^2(n\pi x)dx}. $$ The series will converge to $1$ except at $x=0,1$, where it obviously converges to $0$.

Answer 2

Generically, this sort of problem has a sequence of eigenfunctions $\{ f_n \}_{n=1}^\infty$; expanding a given function $f$ in terms of the eigenfunctions means determining scalars $c_n$ such that $f=\sum_{n=1}^\infty c_n f_n$. Strictly speaking this is an infinite linear system; to solve it, one should hope to find an orthogonal system of eigenfunctions with respect to some inner product $\langle \cdot,\cdot \rangle$, so that the $c_n$ are given by $\frac{\langle f,f_n \rangle}{\| f_n \|^2}$.