Solving sturm-liouville equations

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1. Oct 2, 2015

ognik

Forgive me if I'm just having a mental block about this, but I'm teaching myself the sturm-liouville method - I'm happy with getting the general ODE to this (self-adjoint) form after adjusting for any weighting function):
$$\mathcal{L}y=\lambda y$$ the operator is of the form $$\mathcal{L}= (p(x) \frac{d}{dx})'+q(x)$$ p, q real polynomials.

None of the material I have read, or examples, show what the next step toward a solution is; examples just claim some solution and go on to discuss things like hermitian operator properties. I don't know where they get either the eigenvalues or eigenvectors from, could someone please give me a hint?

I could go back to the general linear homogeneous 2nd order ODE form and use something like Frobenius to solve that, but then what would have been the point of getting it into the S-L form?

2. Oct 2, 2015

Geofleur

Once you know that you have a Sturm-Liouville problem, there are various theorems available regarding the solutions, for example, that they form a complete set in terms of which other functions can be expanded. Studying the Sturm-Liouville problem is a way of studying several ODEs all at once.

3. Oct 2, 2015

ognik

Thanks, was kind of aware of that, but which theorem for example would help me find just 1 eigenvalue OR eigenvector?

4. Oct 9, 2015

ognik

So S-L theory doesn't apply to solving ODE's, we still use the normal methods (separation, frobenius etc) to sove an ODE; we use S-L theory to study sets of related ODEs? Is this restricted to solutions of the same equation, EG showing them orthogonal? Or does it extend further?

5. Oct 10, 2015

pasmith

Theory tells us that the general solution of $\mathcal{L}y = \lambda y$ is $y(x) = a_1y_1(x) + a_2y_2(x)$ for linearly independent solutions $y_1$ and $y_2$ and arbitrary constants $a_1$ and $a_2$. Then we need to apply the self-adjoint boundary conditions, which state that a given linear combination of $y$ and $y'$ must vanish at each end point. If that forces $a_1 = a_2 = 0$ then $\lambda$ is not an eigenvalue. Otherwise $\lambda$ is an eigenvalue.

If for some reason there is no easy way to determine the values of $y_1$ and $y_2$ and their derivatives at the end points then it may necessary to use a numerical shooting method to find an eigenvalue and its corresponding eigenfunction simultaneously.

Occasionally we can make more analytical progress and deduce an algebraic equation which the eigenvalues must satisfy, but finding the solutions of that equation may require numerical root-finding techniques.