First Eigenfunction of Simple Equation Consider the interval $[-a,a]$ and the following problem: $$\phi'' + \lambda\phi=0$$ $$ \phi(\pm a) = 0. $$ The obvious sequence of orthogonal eigenfunctions seems to be $\sin(\frac{\pi n}{a}x)$ with eigenvalues $(\frac{\pi n}{a})^2$. But we also know for symmetric operators such as the Laplacian, the first eigenfunction is always positive everywhere in the interior of the domain. What's happening?

You're forgetting another set of eigenfunctions: $\cos(\frac{(2k+1)\pi}{2a}x)$ for $k\in\mathbb{N}$. In fact, the full set of eigenfunctions are $$\cos\left(\frac{n\pi}{2a}x\right) ~~ \text{for odd } n = 1,3,5,\ldots$$ $$\sin\left(\frac{n\pi}{2a}x\right) ~~ \text{for even } n = 2,4,6,\ldots$$ As you can see, the first eigenfunction $\cos(\pi x/2a)$ is positive everywhere in $(-a,a)$.