Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
2  
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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.