# Why are harmonic functions called harmonic functions?

Are they related to harmonic series in any way? Or something else? Wikipedia didn't help.

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Because one can hear the shape of a drum. – Did Mar 23 '12 at 14:48
Actually, one cannot hear the shape of a drum. – Bruno Joyal Jul 27 '12 at 2:39

Consider a sheet of skin stretched into a flat drum head and drummed upon. When the drum head is in vibration, let $f(x,y,t)$ be the height of the drum head at position $(x,y)$ and time $t$. Then $f$ obeys the wave equation: $$\frac{\partial^2}{\partial t^2} f = c^2 \left( \frac{\partial^2}{\partial x^2} f + \frac{\partial^2}{\partial y^2} f \right) \quad (\ast)$$ where $c$ is a physical constant related to things like how tight the skin is stretched and what it is made out of. Such a solution must also obey the physical constraint that there is no motion at the boundary of the drum, where the skin is nailed down.

Every sound can be composed into its overtones. A pure overtone with frequency $\omega$ corresponds to a solution to the wave equation which looks like $f(x,y,t) = g(x,y) \cos(\omega t+b)$ where $$- \frac{\omega^2}{c^2} g =\frac{\partial^2}{\partial x^2} g + \frac{\partial^2}{\partial y^2} g \quad (\ast \ast).$$ Therefore, to understand the sound of a drum, one should figure out for which $\omega$ the PDE $(\ast \ast)$ has solutions which are zero on the boundary of the drum. This is called computing the spectrum of the drum, and a property of the drum which depends only on these $\omega$'s is called a property which one "can hear".

The lowest frequency, which will give the fundamental tone of the drum, will correspond to the lowest nonzero $\omega$ for which $(\ast \ast)$ has solutions. Of course, $(\ast \ast)$ always has the solution that $g$ is a constant and $\omega =0$.

OK, so far that made sense. Now the terminology does something illogical. The name "harmonic" is attached not to the lowest nontrivial frequency, but to the zero frequency. That is to say, $g$ is called "harmonic" if it obeys $$0=\frac{\partial^2}{\partial x^2} g + \frac{\partial^2}{\partial y^2} g \quad (\ast \ast \ast).$$ I don't know the actual history here, but I think of this as a form of mathematical obtuseness. "You musicians want to study the lowest frequency of vibration? Well you can't get lower than zero!"

The actual physical question addressed by $(\ast \ast \ast)$ is "what are the possible stable shapes for a drumhead, if the boundary is not planar? So, if the rim of my drum varies in height, but I tack the drumhead to it anyway, what shape will the drumhead sit at when we're not pounding on it? This is the Dirichlet problem for the Laplace equation; if I give you the values of a harmonic function on the boundary, what does the interior look like?

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Sorry, edited too fast. The second sign was correct, and is now restored; it comes from $\partial^2/(\partial t)^2 \cos(\omega t) = -\omega^2 \cos (\omega t)$. – David Speyer Jul 27 '12 at 17:08
For an idealized string instrument, the wavelength of the fundamental modes of vibrations are $\frac{2L}{n}$, $n\in \mathbb N^+$. These are also the terms of a harmonic series, hence the name of the latter.