For and eigenvalue equation in Sturm-Liouville form

with scalar product

we examine the first part of the operator

We require

We start from

using integration by parts we have

then using integration by parts again we find

The second term clearly is , so we require the first term for self-adjoint form.

So we find

    for an operator to be in self-adjoint form.

The variation theorem states that given a system with a Hamiltonian , then if is any normalised, well-behaved function that satisfies the boundary conditions of the Hamiltonian, then

For un-normalised wavefunctions

Example 1: Schrödinger Equation in quantum mechanics

We will consider the Hamiltonian

The normal scalar product to use is

where is the weight function.

Therefore, to calculate   we have

using integration by parts on the first term gives

And for self-adjoint operators, which this must be, the surface term vanishes. Giving

This makes calculation easier.

So 

Any second order linear ordinary differential equation

     or      

where is a second order linear operator, can be rewritten in Sturm-Liouville form

Method:

Start with a general 2nd order linear ODE

Divide by A(x) to obtain

Now we let ,  and 

Now we use the integrating factor, p(x), to reformulate the 2nd order differential part, such that

The equation now becomes

where we now define and . So the original equation is now cast in Sturm-Loiuville form.

This can also be applied to eigenvalue problems...

Any second order linear eigenvalue equation

     or      

where is a second order linear operator, can be rewritten in Sturm-Liouville form

The same proceedure is followed as the non eigenvalue problem. The only difference is the introduction of the weight function, , which is defined as

Note: we need   on the interval

and we require

For a derivation of this see this post.

This ensures all eigenvalues, , are real and corresponding eigenfunctions, , are orthogonal relative to the weight function w(x).

Mathematically:

   (orthogonal)

and

   (real eigenvalues)

Example 1: Legendre's Equation

Divide by , which gives

Now we compute the integrating factor,

So the Sturm-Liouville form is

So we see the weight function, .

If an Linear operator   has a Green's function, , the following relationship can be written down.

This can be used to solve inhomogeneous differential equations of the form

The Green's function is a sort of right inverse of L such that it knowledge of yields the function u(x) by direct integration. This proof is as follows.

From 

Multiply both sided by and integrate with respect to

Using the properties of the delta function

Since the operator L on the left hand side is in it can be pulled out the front of the integral, which gives

Therefore we have found the equation for which is

A linear operator preserves

    additivity
     homogeneity of degree 1

In the context of ordinary differential equations the linear operator L is of the form

Therefore a linear differntial equation would be

   inhomogenous

   homogeneous

And if the coefficients are constant, i.e. not functions of the ode is said to have constant coefficients.

Some simple examples:

     homogeneous with constant coefficients

      homogeneous with non constant coefficients

      inhomogeneous with non constant coefficients

       inhomogeneous with constant coefficients