Why Does the First Eigenfunction Have a Zero in Sturm-Liouville Problems?

  • Context: Undergrad 
  • Thread starter Thread starter member 428835
  • Start date Start date
  • Tags Tags
    Eigenfunctions
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
member 428835
Hi PF!

Given ##y''+\lambda^2y=0## and BCs ##y'(0)=y'(1) = 0## we know eigenfunctions are ##y=\cos (n\pi x)##, and for ##n=1## this implies there is one zero on the interval ##x\in(0,1)##. However, I read that for SL problems, the ##jth## eigenfunction has exactly ##j-1## zeros on ##x\in(0,1)##, implying there should be no zeros for ##n=1##, but there is. Can someone reconcile this?
 
Physics news on Phys.org
The first eigenfunction is ##n=0##, not ##n=1##.
 
  • Like
Likes   Reactions: member 428835