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I want to show that there are countably infinite number of eigenvalues (that can put in ascending order-- with a minimum value) to the 2nd order ODE x''+\lambda x = 0 subjected to boundary conditions $x(0)=0$ and x'(1)+x(1)=0.

My immediate approach would be to solve it.

1) For $\lambda <0$,

general solution: $a \exp(\mu t)+b\exp(-\mu t)$, where $\mu ^2 = -\lambda$;

boundary conditions equations: $a+b=0$ and $(1+\mu)a \exp(\mu )+(1-\mu) b\exp(-\mu)=0$

So no non-trivial solution.

2) For $\lambda =0$, again, easy to show that there are no non-trivial solutions.

3) For $\lambda >0$,

general solution: $a\sin(\mu t)+b\cos(\mu t)$ where $\mu = \sqrt{\lambda}$

Boundary condition: $b=0$ and (so) $a\mu \cos(\mu) +a\sin(\mu) =0$ Giving $\tan(\mu)=-\mu$ but then considering the intersections of the graphs of $y=\tan(z)$ and $y=-z$, there is no minimum $\mu$??!

Also, are there any more elegant way to do this? (may invoke Sturm-Liouville)

Thank you.

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    Since $\lambda$ is real and $\mu^2+\lambda=0$ we have that $\mu$ is either real or strictly imaginary. Take your evaluation of the boundary condition and rearrange it (after setting $a=1,b=-1$ WLOG) so that it reads $\mu(e^\mu+e^{-\mu})+(e^\mu-e^{-\mu})$. Divide through and you get $\mu+\tanh \mu=0$. So you need to find what strictly real/imaginary zeroes this function has.2011-09-24

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