Solving $y'' + \lambda y = 0$, $y(-L) = y(L)$ and $y'(-L) = y'(L)$

Question

I am working on solving the ODE: $y'' + \lambda y = 0$, $y(-L) = y(L)$ and $y'(-L) = y'(L)$.

After applying the first condition, I get: $2c_2 \sin(\sqrt{\lambda} L) = 0$. I avoid letting $c_2 = 0$ to avoid a trivial case. Then $\sqrt{\lambda} = \frac{n \pi}{L}$. Now, I still have to apply the second condition. After doing that I get: $-2\sqrt{\lambda}c_1 \cdot \sin(\sqrt{\lambda} L) = 0$. If I let $c_1 \neq 0$ then I get $\sqrt{\lambda} = \frac{n\pi}{L}$ again .

Am I supposed to let $c_1 = 0$ or no?

Here is my full work:

Answer 1

The solution is trivial only if $c_2$ and $c_1$ are zero, so neither condition by itself has you choose between $\sqrt{\lambda} = \frac{ n\pi}{L}$ or the trivial solution; both conditions are need for that.

The first boundary condition tells you that either $c_2=0$ or $\sqrt{\lambda} = \frac{ n\pi}{L}$. The second boundary condition tells you that either $c_1=0$ or $\sqrt{\lambda} = \frac{ n\pi}{L}$. Thus, we have (($\sqrt{\lambda} = \frac{ n\pi}{L}$) or ( $c_2=0$ and $c_1=0$)). But if we have $c_2=0$ and $c_1=0$, then we can take $\sqrt{\lambda}$ to be $\frac{ n\pi}{L}$; after all, if both coefficients are zero, then the value of $\lambda$ is irrelevant. Thus, we can take $\sqrt{\lambda}$ to be $\frac{ n\pi}{L}$ and $c_2$ and $c_1$ to be free.

So if $\sqrt{\lambda} = \frac{ n\pi}{L}$ for some integer n, then the full solution is $y(x)=c_1cos(\omega x)+c_2sin(\omega x)$, where $\omega ^2={\lambda}$. Otherwise, the only solution is the trivial one.

Answer 2

You have solved the problem. Specifically, $$y(x) = c_1 \cos(\sqrt{\lambda}x) + c_2 \sin(\sqrt{\lambda}x).$$ If $\sqrt{\lambda} \neq n\pi/L$ then $c_1=c_2=0$. That is, $y(x)=0$. Otherwise, $c_1$ and $c_2$ can take on arbitrary values.