Sturm–Liouville Problems
The sines and cosines of Fourier series aren't a lucky accident. Every eigenvalue problem that separation of variables hands you fits one equation, , and its eigenfunctions are always orthogonal (with weight ), so you can expand functions in them exactly the way you did with Fourier series.
What the symbols mean
- Coefficient function
Positive on the interval; for X'' + λX = 0 it's 1.
- Potential
Shifts the eigenvalues; often 0.
- Weight
Positive; it's the weight in the orthogonality integral.
- Eigenvalue
Only special values allow nonzero solutions.
Explanation
A Sturm–Liouville problem is simply a recipe for the "building block" shapes a vibrating or cooling object naturally falls into, like the sines in Fourier series, along with the special numbers (eigenvalues) that go with them. The name just honors Charles Sturm and Joseph Liouville, two French mathematicians who worked out the theory in the 1830s, and "eigenvalue" comes from the German eigen, "own": they're the system's own natural values. It's good for one big reason: it guarantees those building blocks always behave like sines do, so you can break any reasonable function into them. The equation, , looks intimidating, but it's just a few parts: (how stiff or conductive the material is at each point), (an extra push that shifts things around, often zero), (how much each point counts, its weight) and (the special numbers, like frequencies or decay rates, that make a solution possible). It's a lot of letters for one simple idea; bear with it.
Why we need this. Think back to how separation of variables went. Every time, we ended up with plus boundary conditions, got sines or cosines, and then leaned hard on one fact: different sines are orthogonal, so we could pick off each coefficient with an integral. But what if the rod loses heat at one end, or the problem is on a drum instead of a string? The eigenfunctions stop being plain sines. Do we lose orthogonality and have to start over?
No, and Sturm–Liouville theory is the reason. It says the good behavior of sines comes from the shape of the equation, not from sines themselves.
Regular Sturm–Liouville problems (with and boundary conditions of the form at each end) always have:
- Real eigenvalues .
- One eigenfunction per eigenvalue, and has exactly zeros inside the interval.
- Orthogonality with weight .
- Completeness: every piecewise smooth function has a convergent eigenfunction expansion with .
Notice that the coefficient formula is the Fourier coefficient formula with slipped into both integrals. Multiply by and integrate: orthogonality kills every term except , which is exactly the trick you already know.
Examples. is the case , , giving Fourier series. Bessel's equation (drums) and Legendre's equation (spheres) are others, with their own orthogonal families. We'll meet Bessel functions in the drum lesson later in this unit.
Robin conditions like describe an end that leaks heat into the air at a rate proportional to its temperature, somewhere between insulated and held at zero. They give eigenvalues that solve transcendental equations such as , so you find them numerically. The theory still promises they're real, increasing, and come with orthogonal eigenfunctions.
A common mistake. It's tempting to think orthogonality needs whole-number multiples, like and , because that's how the Fourier integrals worked out. It doesn't. The Robin eigenfunctions and have no nice relationship at all, and they're still orthogonal on . What matters is that both solve the same Sturm–Liouville problem with the same boundary conditions.
See it
Worked examples
Find the first eigenvalue of , , .
- 1
The left condition rules out cosine, so with . Plugging into the right condition gives , so .
Practice
Type your answer and check it. Solve them all to complete the lesson.
Which choice of , , gives ?
Bessel's equation in Sturm–Liouville form is . What is the weight?
Find the first eigenvalue of , , .
How many zeros inside the interval does the fourth eigenfunction have?
Evaluate .
A rod on loses heat by convection at , so its slowest mode decays like with (the first eigenvalue in the picture). How long does it take that mode to fall to 1% of its starting size?
Check orthogonality numerically: evaluate for the first two eigenfunctions above, and .
The hardest one here. Get it, and you've got this lesson.
The Rayleigh quotient estimates the first eigenvalue of , , from any trial function that meets the boundary conditions: . Evaluate it for the trial function .
Where it's used
Quantum mechanics is Sturm–Liouville theory: energy levels are eigenvalues. So are the vibration frequencies of beams, the modes of optical fibers, and the stability analysis of structures.