So if I'm given a Strum Liouville equation in a form similar to the one I've mentioned above, which method gives me the correct eigenvalues and eigenfunctions?
Should I be using $$\lambda= 0, \lambda> 0, \lambda< 0$$ or, $$\lambda + 1 = 0, \lambda + 1 > 0, \lambda + 1 < 0$$ and then setting $\lambda + 1 = k^2$ or something?
I've seen both be mentioned in multiple places on MathSE, but not in any textbooks or the like, so at this point, I'm genuinely just really confused.
Does it matter which one I use? Or does it depend on something?
I'd really appreciate if anyone could answer! Thanks!
The ODE $y''+(\lambda+1)y=0,\ y(0) = 0,\ y(L) = 0$ admits as trivial solution $y=0$. As we known the general solution, assuming $\lambda+1>0$ is given by $y=c_1\cos(\sqrt{\lambda+1}t)+c_2\sin(\sqrt{\lambda+1}t)$ so we have at the boundary
$$ \cases{ y(0) = c_1 = 0\\ y(L) = c_1\cos(\sqrt{\lambda+1}L)+c_2\sin(\sqrt{\lambda+1}L)=0 } $$
Non trivial solutions are obtained when $\sin(\sqrt{\lambda+1}L)=0$ or when
$$ \sqrt{\lambda+1}L=2k\pi\Rightarrow \lambda = \left(\frac{2k\pi}{L}\right)^2-1 $$
Analogous procedure can be applied when $\lambda+1<0$