Sturm–Liouville problems and Green's functions

In one sentence

A Sturm–Liouville problem has real, simple eigenvalues increasing to infinity with eigenfunctions orthogonal in a weighted inner product, and the Green's function inverts the operator exactly when zero is not among those eigenvalues.

Why the exam asks it

Because it is the ODE topic with the most structure — everything is forced, so questions can ask precise things — and because the weight function and the Fredholm alternative are two places where a plausible-sounding option is wrong.

The idea

The Sturm–Liouville problem is

$$\big(p(x)y'\big)' + q(x)y + \lambda w(x)y = 0 \quad\text{on } [a,b],$$

with separated boundary conditions, $p, w > 0$ and everything continuous. That is the regular case; if $p$ or $w$ vanishes at an endpoint, or the interval is infinite, it is singular and the conclusions weaken.

For a regular problem:

• Eigenvalues are real, simple, and form an increasing sequence $\lambda_1 < \lambda_2 < \cdots \to \infty$.
• Eigenfunctions for distinct eigenvalues are orthogonal with respect to the weight $w$: $\int_a^b y_m y_n w\,dx = 0$.
• The eigenfunctions form a complete orthogonal basis of $L^2_w[a,b]$.
• The $n$-th eigenfunction has exactly $n-1$ zeros inside $(a,b)$.

The reality of the eigenvalues comes from self-adjointness, exactly as for Hermitian matrices in the linear algebra unit — this is that theorem in infinite dimensions.

The example everything is checked against

$$y'' + \lambda y = 0, \qquad y(0) = y(\pi) = 0.$$

Here $p = w = 1$ and $q = 0$. Solutions $\sin(\sqrt\lambda\, x)$ satisfy the left condition; the right one forces $\sqrt\lambda\,\pi = n\pi$, so

$$\lambda_n = n^2, \qquad y_n = \sin(nx), \qquad n = 1,2,3,\dots$$

Real, simple, increasing to infinity. Orthogonal: $\int_0^\pi \sin(mx)\sin(nx)\,dx = 0$ for $m\ne n$, and $\pi/2$ when $m = n$. And $\sin(nx)$ has exactly $n-1$ interior zeros. That is the Fourier sine series, and it is a Sturm–Liouville expansion.

Changing the boundary conditions to $y'(0) = y'(\pi) = 0$ gives $\lambda_n = n^2$ with $y_n = \cos(nx)$ and $n = 0$ now allowed — so $\lambda = 0$ becomes an eigenvalue, which matters below.

Green's functions

To solve $Ly = f$ with homogeneous boundary conditions, write

$$y(x) = \int_a^b G(x,t)f(t)\,dt .$$

$G$ is built from two homogeneous solutions, $u$ satisfying the left condition and $v$ the right:

$$G(x,t) = \frac{u(x_<)\,v(x_>)}{p(t)\,W(t)},$$

with $x_< = \min(x,t)$ and $x_> = \max(x,t)$. Properties worth carrying:

• $G$ is symmetric, $G(x,t) = G(t,x)$, because the operator is self-adjoint.
• $G$ is continuous in $x$, and $\partial G/\partial x$ jumps by $1/p(t)$ across $x = t$.

Existence is the Fredholm alternative. $G$ exists exactly when the homogeneous problem has only the zero solution — that is, when $\lambda = 0$ is not an eigenvalue. If it is, then $Ly = f$ is solvable only for $f$ orthogonal to the null space, and the solution is not unique.

So for $y'' = f$ with $y(0)=y(\pi)=0$ a Green's function exists, since the eigenvalues are $n^2 \ge 1$. With Neumann conditions it does not, since $\lambda = 0$ is an eigenvalue with constant eigenfunction.

Where intuition breaks

"Eigenfunctions are orthogonal in the usual $L^2$ inner product." With respect to the weight $w$. When $w = 1$ the two coincide, which is why the qualifier is easy to lose — and questions set $w \ne 1$ precisely to test it.

"Eigenvalues could be complex." Real, for a regular problem, by self-adjointness. Options offer complex eigenvalues as a plausible-looking possibility.

"Eigenvalues could repeat." Simple, in the regular separated case — one eigenfunction each up to scalars. Periodic boundary conditions are the exception, where $\sin$ and $\cos$ share an eigenvalue.

"A Green's function always exists." Only when zero is not an eigenvalue. Neumann conditions on $y'' = f$ are the standard failure.

"$G$ is smooth." Continuous, with a jump in its $x$-derivative across the diagonal. That jump is what produces the delta function, and it is often the thing being asked for.

"The eigenfunctions span everything." They span $L^2_w$, which is the right statement — completeness in the weighted space, not pointwise convergence for every function.

"Singular problems behave the same." Bessel and Legendre are singular, and the conclusions need care: the spectrum may be continuous, and boundedness at the singular endpoint replaces a boundary condition.

The exam's angle

1. Put the equation into Sturm–Liouville form by finding the integrating factor; then $p$, $q$ and $w$ are readable and the weight is settled.
2. Solve the constant-coefficient case and impose the boundary conditions. Almost every examinable problem is $y'' + \lambda y = 0$ with some pair of conditions.
3. Count zeros to identify which eigenfunction is which — the $n$-th has $n-1$ interior zeros.
4. For Green's function existence, ask whether $\lambda = 0$ is an eigenvalue. That is the whole test.
5. Use orthogonality to compute coefficients — the expansion coefficients are inner products with the weight.

The night before

• Regular Sturm–Liouville: $(py')' + qy + \lambda wy = 0$, $p, w > 0$, separated conditions.
• Eigenvalues real, simple, increasing to $\infty$. Eigenfunctions orthogonal with weight $w$, complete in $L^2_w$, the $n$-th with $n-1$ interior zeros.
• $y'' + \lambda y = 0$, $y(0)=y(\pi)=0$: $\lambda_n = n^2$, $y_n = \sin nx$, $\int_0^\pi \sin^2(nx) = \pi/2$.
• Self-adjointness is why the eigenvalues are real — the Hermitian theorem in infinite dimensions.
• $G(x,t) = u(x_<)v(x_>)/(pW)$; symmetric; continuous with a derivative jump $1/p(t)$.
• Green's function exists $\iff$ $\lambda = 0$ is not an eigenvalue. Fails for Neumann on $y'' = f$.
• Singular problems (Bessel, Legendre) need separate care.