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.