joshmccraney said:
Yes, this is how I began to accept the theorem, but what about that ##\sigma(x)##? That's what threw me off.
The general theory of linear operators on an inner product space then says that if an operator is self-adjoint with respect to the inner product then its eigenvalues (if any) are real, and that eigenvectors corresponding to different eigenvalues are orthogonal with respect to the inner product. By definition, f and g are orthogonal with respect to an inner product \langle \cdot, \cdot \rangle if and only if \langle f, g \rangle = 0, and the operator A is self-adjoint if and only if \langle A(f), g \rangle = \langle f, A(g) \rangle for every f and g in the space.
Now with suitable boundary conditions it can be shown that the second order linear differential operator \mathcal{L}_1(y) = (p(x)y')' - q(x)y is self-adjoint with respect to the inner product <br />
\langle f, g \rangle_1 = \int_a^b f(x)g(x)\,dx.<br />
Consider now the most general linear second-order differential operator \mathcal{L}(y) = y'' + B(x)y' + C(x)y. This can be put into self-adoint form by taking <br />
y'' + B(x)y' + C(x)y = \frac{1}{w(x)}\left( \frac{d}{dx}(p(x)y') - q(x)y\right)<br /> where w, p and q can be found in terms of B and C as <br />
w(x) = \exp\left( \int B(x)\,dx\right), \\<br />
p(x) = \exp\left( \int B(x)\,dx \right), \\<br />
q(x) = - C(x) \exp\left( \int B(x)\,dx \right).<br />
It follows that \mathcal{L} = (1/w(x))\mathcal{L}_1 is self-adjoint with respect to the inner product <br />
\langle f, g \rangle_2 = \int_a^b w(x)f(x)g(x)\,dx.<br /> (I suppose one should at some point prove that if w(x) > 0 on (a,b) then indeed this is an inner product). Thus if you include a weight function then you can apply Sturm-Liouville theory to any second-order linear differential operator, and the most frequent example in practice (aside from Fourier series) are the
Bessel functions which arise in separable solutions of Laplace's equation in cylindrical polar coordinates and are the solutions of <br />
y'' + \frac{y'}{x} - \frac{\alpha^2}{x^2}y = \lambda y,<br /> where \alpha is a parameter and for which we find w(x) = x.