Jtdylktuy

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 : (0,2pi) 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.

Since $f$ is a smooth function from $S^2to mathbb{R}$, 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 $S^1times(0,pi)tomathbb{R}$.

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 $S^1times (0,pi)$, 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.

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 $[0,2pi] times [0,pi]$, 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 $f(0,phi) = f(2pi,phi) = h(phi)$ 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.

Partial differentiation in Laplacian does't works at all in all curved surfaces. SO the answer is No. However I can't say anything about $S^n$ in an continuous series