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Consider the sphere $S^2 = \lbrace (x,y,z) :\ x^2 + y^2 + z^2 = 1 \rbrace$. This is a smooth manifold in $\mathbb{R}^3$, and for a given point $s \in S^2$, one can consider its coordinate neighborhood.

There are many ways to put a smooth structure on $S^2$, but all require at least two coordinate neighborhoods. One simple way is to make use of the local coordinates $\theta, \phi$ (azimuthal angle and inclination angle). For $s \in U = S^2 \setminus {(0,0,1), (0,0,-1)}$, define a coordinate map by

$ g : (0,2\pi) \times (0,\pi) \rightarrow U $

with formula given by

$ g(\theta, \phi) = (\sin(\phi) \cos(\theta), \sin(\phi) \sin(\theta), \cos(\phi)) $

This map is one-to-one, onto, and bicontinuous, so it is a homeomorphism. One can construct a similar map onto $V = S^2 \setminus {(1,0,0),(-1,0,0)}$, so the whole sphere is covered. The fact that this defines a differentiable structure is easy to work out.

The question I want to ask is, are there any non-constant solutions to the equation

$ \frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0 $

where $f : S^2 \rightarrow R$ is a smooth function.

Furthermore, what sorts of boundary conditions are necessary in order to ensure uniqueness.

share|cite|improve this question
But your expression $\partial^2 f/\partial \theta^2 + \partial^2 f/\partial \phi^2$ is not the Laplacian on a sphere! So your equation is not consistent with your title. (Also, you probably want to ask about whether there are non-constant solutions to the equation in question.) – Rahul Nov 1 '10 at 10:15
Where do you want the solution to live? Since $\theta,\phi$ is only defined in a coordinate chart, do you mean for the function $f$ to be a map from $U \to \mathbb{C}$? Do you require the function to extend to the north/south poles in any particular way? Continuously? With continuous derivatives? – Willie Wong Nov 1 '10 at 10:20
These are both good comments and I will edit my question. – Eric Haengel Nov 1 '10 at 10:46
up vote 6 down vote accepted

Since $f$ is a smooth function from $S^2\to \mathbb{R}$, it is a smooth function on your coordinate chart. It is easy to see that by periodicity, you also have that $f$ is a smooth function from $S^1\times(0,\pi)\to\mathbb{R}$.

Since $f$ is smooth on $S^2$, the limits

$$ \lim_{\phi \to 0, \pi} \frac{1}{\sin \phi} \frac{\partial}{\partial \theta} f $$

need to exist and be finite: this is the requirement that $f$ is smooth in "the other chart". Therefore you have that $\partial_\theta f \to 0$ as $\phi$ approach either boundary.

Your equation, however, requires that $\partial_\theta f$ be a harmonic function on $S^1\times (0,\pi)$, a compact domain with boundary. Using the maximum principle, you immediately have that $\partial_\theta f = 0$ everywhere. Plugging back into the equation, this implies $\partial^2f/\partial \phi^2 = 0$.

And so $f$ must be a linear function in $\phi$. And we now apply the smoothness constraint yet again:

At the north pole, smoothness will require that

$$ \lim_{\phi \to 0} \partial_\phi f(\theta,\phi) = \lim_{\phi \to 0} \frac{1}{\sin \phi} \partial_\theta f(\theta + \pi/2, \phi) = 0$$

which implies that the constant slope of the linear function $f(\phi)$ is 0. And hence $f$ must be the constant function.

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If I understand your question correctly, you are aware that $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2}$ is not actually the Laplacian on a sphere, but you want to know whether there are any nontrivial smooth functions on $S^2$ which are harmonic with respect to this modified Laplacian. The answer is no.

A function $f(\theta,\phi)$ which satisfies $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$ is really a harmonic function on the rectangle $[0,2\pi] \times [0,\pi]$, which is mapped onto the sphere via $g^{-1}$. Continuity on the sphere requires that $f(\theta,0) = \text{const}$ (continuity at the north pole), $f(\theta,\pi) = \text{const}$ (continuity at the south pole), and $f(0,\phi) = f(2\pi,\phi) = h(\phi)$ for some $h:[0,\pi]\rightarrow \mathbb{R}$ (continuity across the prime meridian). For any $h$, the boundary conditions are independent of $\theta$, so $f$ is a function of $\phi$ alone. But then $\frac{\partial^2 f}{\partial \phi^2} = 0$ implies that $f$ is linear in $\phi$. If the slope is nonzero, $f$ on the sphere fails to be smooth at the poles. Therefore, for $f$ to be smooth on the sphere and satisfy $\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial \phi^2} = 0$, it has to be constant.

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Ah, you beat me to an answer :) – Willie Wong Nov 1 '10 at 12:23
@Willie: But yours is more precisely expressed :) – Rahul Nov 1 '10 at 18:09

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