Eigenfunction Expansions

Eigenfunction expansions are the computational engine behind separation of variables for the heat, wave, and Laplace equations on bounded domains — the spatial eigenfunctions of the Laplacian (or of a Sturm–Liouville operator obtained after separating variables) supply the basis in which the time-dependent or harmonic problem is solved mode by mode. In quantum mechanics, the time-independent Schrödinger equation is literally a Sturm–Liouville eigenvalue problem, and any wavefunction is expanded in energy eigenstates — the same $c_n = \langle f,\phi_n\rangle$ formula computes transition amplitudes. In acoustics and structural engineering, normal modes of a vibrating beam or membrane are Sturm–Liouville eigenfunctions, and forced vibration problems are solved by exactly the resonance analysis of Case 2 above — a driving frequency equal to a natural frequency produces unbounded resonant response unless the forcing is orthogonal to that mode. In numerical analysis and scientific computing, spectral methods approximate PDE solutions by truncating an eigenfunction expansion, achieving exponential convergence for smooth data. The Fredholm alternative appearing in Case 2 recurs throughout functional analysis, integral equations, and the study of Green's Functions for Boundary Value Problems, where the same solvability dichotomy governs invertibility of compact self-adjoint operators.

:::quiz[Which quantity determines whether the inhomogeneous problem $L[y]-\mu r y = f$ has a unique solution, no solution, or infinitely many?]

• The sign of $f$ on $[a,b]$
• Whether $\mu$ coincides with an eigenvalue $\lambda_k$, and if so whether $\langle f,\phi_k\rangle_r=0$ ✓
• Whether $p(x)$ is bounded
• The parity (even/odd) of the boundary conditions

:::quiz[The classical Fourier sine series on $[0,L]$ with Dirichlet conditions is a special case of eigenfunction expansion with]

• $p\equiv1$, $q\equiv0$, $r\equiv1$, and Dirichlet boundary conditions giving $\phi_n=\sin(n\pi x/L)$ ✓
• $p(x)=x$, $q\equiv0$, $r\equiv1/x$
• $q(x) = -\lambda$ chosen so that $r\equiv 0$
• Any weight $r$ so long as $p$ is negative

Ask the tutor: How does the eigenfunction expansion method change if the boundary conditions are periodic instead of separated, and what happens to the Fredholm alternative in that case?

Think of $L^2_r(a,b)$ — square-integrable functions on $[a,b]$ weighted by $r$ — as an infinite-dimensional vector space with inner product $\langle\, ,\rangle_r$. The eigenfunctions $\phi_n$ of a self-adjoint operator $L$ play exactly the role that an orthogonal eigenbasis $\{v_n\}$ plays for a symmetric matrix $A$: every vector decomposes as $x = \sum c_n v_n$, and $Ax = \sum \lambda_n c_n v_n$. Solving $Ax = b$ becomes solving $\lambda_n c_n = b_n$ coordinate by coordinate — trivial, provided no $\lambda_n$ is zero. The eigenfunction expansion method is this exact linear-algebra fact, lifted to function space, where "coordinates" are Fourier coefficients and "vectors" are functions on $[a,b]$. The classical Fourier sine and cosine series are the special case $p\equiv 1$, $q\equiv 0$, $r\equiv 1$, with Dirichlet or Neumann conditions — the eigenfunctions are $\sin(n\pi x/L)$ or $\cos(n\pi x/L)$. What makes the general Sturm–Liouville framework powerful is that the same three-step recipe — expand the source, divide by the eigenvalue, resum — works for any self-adjoint second-order operator on any bounded interval, with any separated boundary conditions, and any weight. This is precisely why the method of separation of variables, applied to the heat, wave, or Laplace equation on a finite rod, disk, or box, always terminates in a Fourier-type series: the spatial part of the separated PDE is always a Sturm–Liouville problem.

Example 1 (Dirichlet rod with a non-symmetric, non-resonant load). Solve

on $[0,1]$, using eigenfunction expansion, and compare to direct integration. Here $L[y] = -y''$, so the equation is $L[y] - \mu y = f$ with $\mu = 1$ and $f(x)=-x(1-x)^2$ (rewriting $y''+y=x(1-x)^2$ as $-y'' - (-1)y = -x(1-x)^2$, i.e. $-y'' -\mu y = f$ with $\mu=-1$; to avoid sign bookkeeping we instead work directly with the operator as given). Take $p\equiv1,q\equiv0,r\equiv1$ on $[0,1]$ with Dirichlet conditions. The eigenpairs of $-\phi'' = \lambda \phi$, $\phi(0)=\phi(1)=0$, are the classical ones:

The ODE $y''+y = x(1-x)^2$ is $-y'' - y = -x(1-x)^2$, i.e. in the form $L[y]-\mu y = f$ with $\mu = 1$ and $f(x) = -x(1-x)^2$. Since $\mu=1$ is not of the form $n^2\pi^2$ for any integer $n$ (as $\pi^2\approx 9.87$, so $n^2\pi^2 \in \{9.87, 39.5,\dots\}$, never equal to $1$), we are in Case 1: a unique solution exists. Compute the Fourier coefficients of $f(x) = -x(1-x)^2 = -x + 2x^2 - x^3$:

Using the standard moments $\int_0^1 x\sin(n\pi x)\,dx = \dfrac{-\cos(n\pi)}{n\pi} = \dfrac{(-1)^{n+1}}{n\pi}$, and (by integration by parts twice more, or tables) the exact value

(a direct computation via three integrations by parts of $x(1-x)^2 = x - 2x^2+x^3$ against $\sin(n\pi x)$, tracking boundary terms which vanish since $\sin(n\pi x)$ and its relevant antiderivative structure vanish appropriately at $0,1$ for the polynomial parts — the algebra is routine and each step is a standard reduction formula for $\int x^k \sin(n\pi x)\,dx$), we get

By Case 1 of the theorem, with $\lambda_n - \mu = n^2\pi^2 - 1$,

and the solution is

Consistency check. For $n$ even, $1-(-1)^n = 0$, so $y_n = \dfrac{2}{n\pi(n^2\pi^2-1)}$, which decays like $n^{-3}$; for $n$ odd, $y_n = \dfrac{-2}{n\pi(n^2\pi^2-1)} + \dfrac{16}{n^3\pi^3(n^2\pi^2-1)}$, dominant term also $O(n^{-3})$. Since $f$ is a polynomial vanishing at both endpoints (so effectively $C^\infty$ with matching boundary values), the coefficients decay like $n^{-3}$ as expected for a function whose second derivative does not itself vanish at the boundary — consistent with the general Fourier-decay heuristics, and the series converges absolutely and uniformly by comparison with $\sum n^{-3}$, confirming a genuine $C^2$ classical solution as guaranteed by the theorem.

Example 2 (Robin condition at one end, weight $r(x)=x$ — a genuinely non-symmetric Sturm–Liouville setting, resonant case). Consider the Cauchy–Euler-type Sturm–Liouville problem coming from $L[y] = -(xy')'$ on $[1,e]$ with weight $r(x) = 1/x$, i.e. the eigenvalue problem

Substituting $x = e^t$ turns $-(xy')'$ into $-y_{tt}$ (the standard reduction for Euler-type operators: with $t=\ln x$, $x\frac{d}{dx} = \frac{d}{dt}$, and one checks $(xy')' \cdot x = x^2 y'' + xy' = y_{tt}$ under this substitution, so $-(xy')' = -y_{tt}/x$, matching $\lambda y/x$ exactly when $-y_{tt} = \lambda y$), and $t$ ranges over $[0,1]$ as $x$ ranges over $[1,e]$. With $y(1)=0,y(e)=0$ becoming $y(t=0)=0, y(t=1)=0$, we recover

as eigenfunctions of the original problem in $x$, with weight $r(x)=1/x$:

Now solve the resonant inhomogeneous problem at $\mu = \lambda_2 = 4\pi^2$:

Here $f(x) = \cos(\pi\ln x)/x$; in the $t$-variable $f(t) = \cos(\pi t)$ (matching the weight $r=1/x$ correctly, since the pairing $\langle f,\phi_n\rangle_r = \int_1^e f\phi_n\,\frac{dx}{x} = \int_0^1 \cos(\pi t)\sin(n\pi t)\,dt$). Compute $f_2 = \langle f,\phi_2\rangle_r / \|\phi_2\|_r^2$:

So $\langle f,\phi_2\rangle_r = 4/(3\pi) \neq 0$. By Case 2 of the theorem with $k=2$: since $f_2 \ne 0$ (equivalently $\langle f, \phi_2\rangle_r \ne 0$), the solvability condition $\langle f,\phi_k\rangle_r = 0$ fails, so this resonant problem has no solution at all — the eigenfunction expansion method correctly detects and certifies non-existence, which direct integration would only reveal after a much longer computation (or would obscure entirely). Contrast this with $f(x) = \sin(2\pi \ln x)/x$, i.e. $f(t) = \sin(2\pi t) = \phi_2(t)$ exactly: then $\langle f,\phi_2\rangle_r = \int_0^1 \sin^2(2\pi t)\,dt = 1/2 \ne 0$ also fails, so that choice is resonant and unsolvable too — whereas $f(t) = \sin(3\pi t)$ gives $\langle f,\phi_2\rangle_r = \int_0^1 \sin(3\pi t)\sin(2\pi t)\,dt = 0$ (orthogonal distinct modes), so the solvability condition holds, and the problem $-(xy')' - 4\pi^2 y/x = \sin(3\pi\ln x)/x$, $y(1)=y(e)=0$, has the family of solutions

with $f_3 = 1$ contributing the term $\dfrac{1}{9\pi^2-4\pi^2}\sin(3\pi\ln x) = \dfrac{1}{5\pi^2}\sin(3\pi\ln x)$ and all other $f_n=0$, so explicitly