What functions can be represented as a series of eigenfunctions

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?

3 Answers

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)

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.

• I liked these endpoint considerations. That is a very nice introduction to the fact that pointwise Fourier convergence must be trickier than the L^2 one. – Giuseppe Negro Apr 16 '19 at 16:15

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.