First Eigenfunction of Simple Equation

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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?

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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 $(-a,a)$.