Question about $\lambda = 0$ case in Sturm-Liouville problem

Question

Given

$u''(x) + \lambda u(x) = 0, u'(0) = u'(\frac{\pi}{2}) = 0$,

I wish to compute the eigenvalues and corresponding eigenfunctions.

If $\lambda = 0$,

$u''(x) = 0 \implies u'(x) = A \implies u(x) = Ax + B$.

If $\lambda < 0$,

using the characteristic equation $r^2 + \lambda = 0$,

$r^2 + \lambda = 0 \implies r = \pm\sqrt{-\lambda}$.

Then

$u(x) = Ae^{\sqrt{-\lambda}x} + Be^{-\sqrt{-\lambda}x}$

$u'(x) = \sqrt{-\lambda}Ae^{\sqrt{-\lambda}x} + (-\sqrt{-\lambda})Be^{-\sqrt{-\lambda}x}$

$u'(0) = \sqrt{-\lambda}A + (-\sqrt{-\lambda})B = 0 \implies A - B = 0$

$u'(\frac{\pi}{2}) = \sqrt{-\lambda}Ae^{\sqrt{-\lambda}\frac{\pi}{2}} + (-\sqrt{-\lambda})Be^{-\sqrt{-\lambda}\frac{\pi}{2}} = 0 \implies A = B = 0$.

Therefore $u(x) = 0$ and thus there are no negative eigenvalues.

If $\lambda > 0$,

$r^2 + \lambda = 0 \implies r = \pm\sqrt{-\lambda} \implies \pm\sqrt{\lambda}i$.

Then

$u(x) = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$

$u'(x) = -A\sqrt{\lambda}\sin(\sqrt{\lambda}x) + B\sqrt{\lambda}\cos(\sqrt{\lambda}x)$

$u'(0) = B\sqrt{\lambda} = 0 \implies B = 0$

$u'(\frac{\pi}{2}) = -A\sqrt{\lambda}\sin(\sqrt{\lambda}\frac{\pi}{2}) + B\sqrt{\lambda}\cos(\sqrt{\lambda}\frac{\pi}{2}) = 0 \implies -A\sqrt{\lambda}\sin(\sqrt{\lambda}\frac{\pi}{2}) = 0$.

Then

$\frac{\sqrt{\lambda}\pi}{2} = n\pi \implies \sqrt{\lambda} = 2n \implies \lambda = 4n^2$

where $n = 1,2,3,\cdots$.

Therefore $u_n(x) = A_n\cos(2nx)$.

I think I am missing something with the $\lambda = 0$ case. The solution I was provided

states that $u_0(x) = 1$ is the corresponding eigenfunction for $\lambda = 0$ and the

first 2 smallest eigenvalues are $0$ and $4$. When $n = 1$, we have $\lambda = 4$

so I can see where $4$ comes from but am I allowed to let $n = 0$?

I thought $u_n(x) = A_n\cos(2nx)$ applied strictly for $\lambda > 0$.

Thank you very much for reading.

Any help would be greatly appreciated.

Answer 1

You want to solve for solutions $u_{\lambda}(x)$ satisfying $$ u_{\lambda}''(x)+\lambda u_{\lambda}(x)=0 \\ u_{\lambda}'(0)= 0,\; u_{\lambda}'(\pi/2)=0. $$ It is always possible to normalize a non-trivial solution of the above by requiring $u_{\lambda}(0)=1$. If you start with $u_{\lambda}''+\lambda u_{\lambda}=0$ with $u_{\lambda}'(0)=0,\;u_{\lambda}(0)=1$, the solutions are $$ u_{\lambda}(x)=\cos(\sqrt{\lambda}x). $$ So the eigenvalue equation becomes $$ 0= u_{\lambda}'(\pi/2)=-\sqrt{\lambda}\sin(\sqrt{\lambda}\pi/2), $$ which is solved by $\sqrt{\lambda}=0,2,4,6,\cdots$ or $$ \lambda=4n^2,\;\;\; n=0,1,2,3,\cdots. $$ The corresponding eigenfunctions are $\cos(2nx)$. (The case where $n=0$ gives $\cos(0)$, which is the constant function $1$. So this is correct for $n=0$ as well as for $n\ne 0$.)