Are There Infinite Eigenfunctions with Distinct Eigenvalues for y''+λy=0?

[Combinatus]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

\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?

 

[LCKurtz]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

\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?

That looks well done. You don't need the $\sqrt{1+\lambda}$ in the answer though. You can just write your final eigenfunction$$
y_n = \sin(\frac{n\pi - \delta}{\pi}x)$$When you write something like this up, it smooths things out a bit if you do this case as $\lambda = \mu^2 >0$ and carry the $\mu$ through to the end. It reads nicer without all the square root signs.

 

[Combinatus]

That looks well done. You don't need the $\sqrt{1+\lambda}$ in the answer though. You can just write your final eigenfunction$$
y_n = \sin(\frac{n\pi - \delta}{\pi}x)$$When you write something like this up, it smooths things out a bit if you do this case as $\lambda = \mu^2 >0$ and carry the $\mu$ through to the end. It reads nicer without all the square root signs.

Thanks for your input! I'm still a bit uncomfortable with u_n = \sin\left(\dfrac{n\pi - \delta}{\pi}x\right) as \delta depends on \lambda, though. I tried a few additional approaches, but I haven't been able to get anything "better" than that.