What functions can be represented as a series of eigenfunctions

Question

Consider the differential equation:

$y'' = \lambda y$

with the boundary conditions $y(0) = y(2\pi) = 0$.

This equation has eigenfunctions $\mu_n(x) = \sin(\frac{nx}{2})$ with the corresponding eigenvalues $\lambda_n = -\frac{n^2}{4}$ for $n > 0$

• Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series $f(x) = \sum_1^\infty c_n \mu_n(x)$ with coefficients $c_n = \frac{\langle f,\mu_n \rangle}{\langle \mu_n, \mu_n \rangle}$? What conditions those certain functions need to satisfy?

• Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = \lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?

Answer 1

Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2\pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $\sum_{1}^{\infty} \frac{\langle \mu_n,f\rangle}{\langle \mu_n,\mu_n\rangle} \mu_n(x)$ on a subset of $(0,2\pi)$ of measure zero)

Answer 2

Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $\alpha,\beta$: $$ y''=\lambda y,\;\;\; a \le x \le b, \\ \cos\alpha y(a)+\sin\alpha y'(a) = 0 \\ \cos\beta y(b)+\sin\beta y'(b) = 0 $$ This gives rise to a discrete set of eigenvalues $$ \lambda_1 < \lambda_2 < \lambda_3 < \cdots, $$

and associated eigenfunctions $\phi_n(x)$. For any function $f\in L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $x\in(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.

Answer 3

You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining $$ Sf_n:=\sum_{k=1}^n c_n \mu_n, $$ it holds that $$ \int_0^1 |Sf_n(x)-f(x)|^2\, dx\to 0\qquad \text{as }n\to \infty.$$

If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.