Eigenvalues of y''+λy=0

**Combinatus** · March 31st 2012, 08:30 AM

The problem is stated as follows:

"Show that $y''+\lambda y=0$ with the initial conditions $y(0)=y(\pi)+y'(\pi)=0$ has an infinite sequence of eigenfunctions with distinct eigenvalues. Identify the eigenvalues explicitly."

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$\lambda \le 0$ seems to yield the trivial solution, so $\lambda > 0$ . The general solution is then $y=A\sin{(\sqrt{\lambda} x)} + B\cos{(\sqrt{\lambda} x)}$ . The first initial condition gives $y=A\sin{(\sqrt{\lambda} x)}$ , and then the second gives $A(\sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}) = 0.$

I've tried to solve $\sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}= 0$ for $\lambda$ by writing the LHS as a single sine function, and separately by dividing with $\cos{(\sqrt{\lambda} \pi)}$ , but neither approach seems to give a good way of giving an explicit formula for $\lambda$ .

(The best I've been able to do with the single sine function is to write $\sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}$ as $\sqrt{1+\lambda} \sin{(\sqrt{\lambda} x + \delta)}$ . Consequently, $\lambda = \left(\dfrac{\pi k - \delta}{\pi}\right)^2, k \in \mathbb{N}$ . But then $\delta = \arccos \frac{1}{\sqrt{1+\lambda}} = \arcsin \frac{\sqrt{\lambda}}{\sqrt{1+\lambda}}$ , so I wouldn't call this explicit.)

Thoughts?

**Ulysses** · April 1st 2012, 04:03 PM

Try this:
$\sin (\sqrt{\lambda}\pi)+\sqrt{\lambda}\cos (\sqrt{\lambda}\pi)=0$
$\sqrt{\lambda}=-\frac{\sin (\sqrt{\lambda}\pi)}{\cos (\sqrt{\lambda}\pi)}$

Your eigenvalues are given by the intersection of this curves.

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