Consider the Regular Sturm Liouville(RSLP) problem \(\displaystyle{\left({p}{\left({x}\right)}{y}’\right)}’+{\left({q}{\left({x}\right)}+\lambda{r}{\left({x}\right)}\right)}{y}={0}\)

Reuben Brennan

Reuben Brennan

Answered question

2022-03-28

Consider the Regular Sturm Liouville(RSLP) problem
(p(x)y)+(q(x)+λr(x))y=0

Answer & Explanation

Nathanial Carey

Nathanial Carey

Beginner2022-03-29Added 12 answers

Step 1
If your boundary conditions are
y(a)=y(b) and y(a)=y(b) with p(a)=p(b) then
1) there are infinitely many eigenvalues
<λ0<λ1λ2
2) For λ0 there is unique eigenfuntion ϕ0 If λ2i+1<λ2i+2 then there are unique eigenfunctions ϕ2i+1,ϕ2i+2
3) If λ2i+1=λ2i+2 then there are two independent eigenfunctions ϕ2i+1,ϕ2i+2.
4)If ϕi and ϕj correspond to distinct λi and λj then they are orthogonal.
5) The set of all eigenfunctions forms a basis of corresponding space.
6)Furthermore, ϕ0 has no roots in [a, b], ϕ2i+1, ϕ2i+2 have exactly 2i+1 roots in [a,b).
Proof is given (as well as some more general cases) in Coddington and Levinson, Theory of ODE, in Chapter 8 and exercises.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?