- #1
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?
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?