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Sturm-Liouville Problems

A regular Sturm-Liouville problem on an interval is of the form

where f(x) is the function to be found and p(x)>0, q(x), and

are given functions.

Except for the various sign requirements, the distinguishing feature is that the coefficient of the f' term (-p') is the derivative of the coefficient for the f'' term (-p). Starting with an arbitrary second order linear O.D.E., you can achieve such a form by multiplying the entire O.D.E. with a suitable factor.

The boundary conditions may either be periodic ones,

or they can be homogeneous of the form

where A, B, C, and D are given constants. Note the important fact that a Sturm-Liouville problem must be completely homogeneous: f=0 must be a solution.

If you have a Sturm-Liouville problem, simply (well, simply ...) solve it. The solutions only exists for certain values of . Make sure you find all solutions, or you are in trouble. They will form an infinite sequence of `eigenfunctions', say f₁(x), f₂(x), f₃(x), ... with corresponding `eigenvalues' , ,, ... that go off to positive infinity.

Note: If you have to know why they are called eigenfunctions and eigenvectors, if you consider functions like f to play the role of (infinite-dimensional) vectors, and the operator

to be the matrix, (transforming one ``vector'' into another function or ``vector''), then the O.D.E. takes the form of an eigenvalue problem..

Any arbitrary function, say g(x), on the interval [a,b] can always be represented as a sum of the found eigenfunctions:

If you know g(x), you can find its Fourier coefficients C_n from the following formula:

where

is the function appearing in the Sturm-Liouville differential equation.

Note: The above formula is referred to as orthogonality. If you have to know why, assume that our function g(x) happens to be an eigenfunction, say f₅. In that case, C₅ has to be one (which is automatic - a good way to check that you have your formula correct), and all other C_n need to be zero. From the formula, you see that then f₅ must integrate to zero against any other function f_n. Identifying such an integral as an inner product between ``vectors'', f₅ would be orthogonal to all other eigenfunctions.

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