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

[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?

 

[Orodruin]

The first eigenfunction is $n=0$, not $n=1$.