Sturm-Liouville confusion

[fayled]

So the basic idea is that the differential operator acting on the vector space V of functions f:[a,b]->ℂ, with some weight function w(x)
L≡d/dx[p(x)d/dx]+q(x)
is Hermitian (self-adjoint) if we have that for any two functions u(x), v(x) in this vector space,
[u*(x)p(x)dv/dx]_a^b=0,
thus L has real eigenvalues and it's eigenfunctions form an orthonormal basis for V.

Now my notes seem to be suggesting to me that if we write the SL problem, i.e the eigenvalue equation
Le_n=λ_ne_n
for eigenfunctions e_n and eigenvalues λ_n, then so long as the eigenfunctions satisfy the above boundary condition, i.e [e_n*(x)p(x)de_m/dx]_a^b=0, then L is Hermitian. This is confusing me because as I stated above, I'm sure this boundary condition must apply for all u,v in the space of functions V, not just the eigenfunctions.

There is an example of this form given, which is
d²e_ndx²=λne_n.
It finds the eigenfunctions as exp[(√λ_n)x] and exp[-(√λ_n)x] where the eigenvalues are yet to be determined. Then it says let our interval be [a,b]=[-π,π] and so because p(x)=1, it asks for [e*_mde_n/dx]_-π^π=0 (for any combination of the eigenfunctions), which quantizes the eigenvalues as λ_n=-n² for integer n. However again, as I said above, we are only asking for the eigenfunctions to satisfy this boundary condition, whilst I was under the impression that any two functions, u,v in the space V (which does include, but is not limited to, pairs of the eigenfunctions) must meet this condition, not just the eigenfunctions.

Could anyone clear this up for me? Thankyou in advance :D

 

[fayled]

Note the example should be d²e_n/dx²=λ_ne_n. Anybody please?

 

[fayled]

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[Haborix]

I believe it's the case that the vector space on which L acts is determined by the boundary conditions you place on the problem. Given this, you solve the SL problem and obtain a set of eigenfunctions which form a complete basis for that vector space. So it follows that any function can be written as a sum of eigenfunctions. I hope that makes sense.