# Why do periodic boundary conditions imply no negative eigenvalues?

Im looking at the boundary value problem $$\Theta''+\lambda\Theta=0$$ with periodic boundary conditions $\Theta(\theta)=\Theta(\theta+2\pi)$ and $\Theta'(\theta)=\Theta'(\theta+2\pi)$. For the case when $\lambda < 0$, setting $\alpha=\sqrt{-\lambda}$, we get a general solution $$\Theta(\theta)=A\cosh(\alpha\theta)+B\sinh(\alpha\theta)$$ for constants $A,B$ and so $\Theta'(\theta)=A\alpha\sinh(\alpha\theta)+B\alpha\cosh(\alpha\theta)$. The boundary conditions then imply that $$A\cosh(\alpha\theta)+B\sinh(\alpha\theta)=A\cosh(\alpha(\theta+2\pi))+B\sinh(\alpha(\theta+2\pi))$$ $$A\alpha\sinh(\alpha\theta)+B\alpha\cosh(\alpha\theta)=A\alpha\sinh(\alpha(\theta+2\pi))+B\alpha\cosh(\alpha(\theta+2\pi))$$ How does this imply only the trivial solution, that $A=B=0$?

• Try plugging in $\theta = 0$ (and maybe $\theta = -2\pi$) and solving the system of equations. Commented Apr 23, 2016 at 2:42
• @CameronWilliams : you need exploiting $\Theta(0) = \Theta(2\pi)$ and $\Theta'(0) = \Theta'(2\pi)$, both conditions tell us that the only solution is $\Theta(\theta) = 0$ for $\Theta''(\theta)+\lambda \Theta(\theta) = 0$ with $\lambda < 0$ Commented Apr 23, 2016 at 2:48
• @user1952009 I meant into both equations, not just one. Commented Apr 23, 2016 at 2:49
• @CameronWilliams : since the solutions are of the form $\Theta(\theta) = a e^{\sqrt{-\lambda} \theta} + b e^{-\sqrt{-\lambda} \theta}$, two constants, two equations needed Commented Apr 23, 2016 at 2:50
• @user1952009 Yes that is what I was saying. I was not disagreeing with you, merely clarifying what I meant. Commented Apr 23, 2016 at 2:51

1. Assume that $\Theta$ is twice differentiable, periodic, and not identically zero.
2. Multiply OP's ODE $$\Theta^{\prime\prime} +\lambda\Theta ~=~0 \tag{1}$$ with $-\bar{\Theta}$ on both sides and integrate over a period: $$0~=~-\int_0^{2\pi}\! d\theta \left( \Theta^{\prime\prime} +\lambda\Theta \right)\bar{\Theta}~\stackrel{\text{int. by parts}}{=}~\underbrace{\int_0^{2\pi}\! d\theta |\Theta^{\prime}|^2}_{\geq 0}-\lambda\underbrace{\int_0^{2\pi}\! d\theta|\Theta|^2}_{> 0} . \tag{2}$$ The integration by parts yields no boundary terms since $\Theta$ is assumed to be periodic.
3. Conclude that $\lambda\geq 0$ is non-negative.