the oscillation theorem
Look at the modes of a guitar string in order: the fundamental is a single hump with no interior node, the first overtone has one node in the middle, the next has two, and so on. The oscillation theorem says this pattern is universal for every Sturm-Liouville problem — counting the wiggles tells you exactly which mode you are looking at.
Precisely: for a regular Sturm-Liouville problem, order the eigenvalues lambda_1 < lambda_2 < lambda_3 < ... starting from the smallest. Then the n-th eigenfunction u_n has exactly n-1 zeros strictly inside the interval (a, b). The ground state u_1 never crosses zero (it has one sign throughout), u_2 crosses once, u_3 crosses twice. The number of interior zeros increases by exactly one at each step up the spectrum — there are no gaps and no surprises.
This is more than a curiosity. It gives you a way to identify and order eigenfunctions by inspection: count the nodes. It guarantees the lowest eigenfunction has constant sign, which is why the ground state of a quantum well or the slowest-decaying heat mode is nodeless and positive. And it is the rigorous content behind 'higher modes are more oscillatory'. The proof rests on the Sturm comparison theorem: each successive lambda is larger, acting as a stronger restoring coefficient, forcing in exactly one more zero.
On [0, L] with X(0) = X(L) = 0, the eigenfunctions are X_n = sin(n*pi*x/L). Count the interior zeros: sin(pi*x/L) has none, sin(2*pi*x/L) has one (at x = L/2), sin(3*pi*x/L) has two (at L/3 and 2L/3). The n-th has n-1 — exactly the theorem.
Count interior zeros to read off a mode's rank in the spectrum.
The count is interior zeros only — the forced zeros at Dirichlet endpoints do not count. The clean 'n-1 zeros' statement is for regular separated boundary conditions; periodic problems (where eigenvalues can be double) follow a slightly modified nodal-counting rule.