# Green's function on sphere

Consider radial (normal) coordinates on a sphere $S^n, n \geq 2$. Let the "origin" be the north pole $(0, 0,..., 1)$ and the coordinates be denoted by $(r, \theta)$. We know that the Laplacian $\Delta_{S^n}$ can be written as $$\Delta_{S^n} = \partial^2_r + (n - 1)\text{cot }r\partial_r + \frac{1}{\text{sin}^2\text{ }r}\Delta_\theta.$$ I was wondering, is there a radial Green's function for $\Delta_{S^n}$, and how to find it out. Or how about the Laplacian in the upper hemisphere $\{x = (x_1,...,x_{n + 1}) \in \mathbb{R}^{n + 1} : x_{n + 1} \geq 0\}$, and imposing the Dirichlet boundary conditions on the boundary $\{ x \in S^n : x_{n + 1} = 0\}$? Any help in appreciated.

• 1/sin(r) is a solution Commented Jan 18, 2017 at 17:07
• Slightly related: mathoverflow.net/q/303128 Commented Mar 15, 2019 at 18:05
• I think you usually put Fubini-Study metric on it for the complex case... Commented Sep 27, 2019 at 20:45

A good reference for Green's function for the Laplacian on compact manifolds (with or without boundary) is the book "Some nonlinear problems in Riemannian geometry" by Aubin, page 108ff. The Green's function for the Laplacian on a compact manifold $$M$$ without boundary is unique up to an additive constant. If we fix $$y\in M$$, then all Green's functions $$G_y$$ at $$y$$ satisfy $$\Delta G_y=\delta_y-\frac{1}{\mathrm{vol}(M)}$$ in the sense of distributions. The term $$-\frac{1}{\mathrm{vol}(M)}$$ appears since one has to project to the orthogonal complement of the kernel of the Laplacian. In particular, $$G_y$$ are smooth on $$M\setminus\{y\}$$. In the example of the standard sphere with geodesic polar coordinates at a point $$y$$ with radial variable $$r$$ all Green's functions at $$y$$ are functions of $$r$$ only (they must be invariant under rotations fixing $$y$$ since the Laplacian is invariant under such rotations). In order to get an explicit formula for the Green's function in your coordinate system on the sphere you therefore have to solve $$f''(r)+(n-1)\cot(r)f'(r)=-\frac{1}{\mathrm{vol}(S^n)}$$ for $$r>0$$ and look for a solution which is singular at $$r=0$$ and has a limit as $$r\to\pi$$. Similar things hold for the Green's function on the upper hemisphere with Dirichlet boundary condition (see e.g. Aubin's book).
• I didn't follow all the steps leading up to it, but the final differential equation looks quite tractable. Multiply both sides by $\sin^{n-1} r$, so the left hand side is $\tfrac{d}{dr} \left( f'(r) \sin^{n-1}(r) \right)$. Since we don't have a singularity at $r=\pi$, we have $\lim_{r \to \pi^-} f'(r) \sin^{n-1}(r) = 0$, so $f'(r) \sin^{n-1}(r) = \tfrac{1}{\mathrm{vol}} \int_{s=r}^{\pi} \sin^{n-1} s ds$ and thus $f(r) =\tfrac{1}{\mathrm{vol}} \int \frac{\int_{s=r}^{\pi} \sin^{n-1} s ds}{\sin^{n-1} r} dr$. A computer algebra system can do these integrals. Commented Sep 7, 2020 at 4:46
• Actually, I am not sure why the integrals can be done in elementary form when $n$ is odd, since then $\int \sin^{n-1}(x)$ is of the form $cx + \mathrm{polynomial}(\sin x)$, and it isn't clear to me why $\int \tfrac{x}{\sin^{n-1}(x)}$ should be elementary, but experimentation suggests that it is. If $n$ is even, then $\int \sin^{n-1}(x)$ is polynomial in $\sin x$, so the outer integral is the integral of a rational function of $\sin x$, and hence elementary. Commented Sep 9, 2020 at 14:46
• Okay, I see why this is happening. Integrating by parts, $\int x \csc^{2k}(x) = \int \int \csc^{2k}(x) = \int \int f(\cot x) \csc^2 x$ where $f$ is some even polynomial. Substituting $u = \cot x$, we get $\int g(\cot x)$ for some odd polynomial $g$. Then we can write that integral as $a \int \cot x dx + \int h(\cot x) \csc^2 x dx$ for another odd polynomial $g$ and a scalar $a$. We have $\int \cot x dx = \log \sin x$, and the second integral can be done by $u = \cot x$ again to give a polynomial in $\cot(x)$. Commented Sep 9, 2020 at 15:04
In particular, for the two-sphere, $$G(x,y)=\frac1{2\pi}\log\|x-y\|$$ (up to an additive constant), where $$\|\cdot\|$$ is the Euclidean norm of $$\mathbb R^3$$. Higher dimensional spheres have more involved Green functions unfortunately...
You could watch this video Here, whose title is 'Green’s function for the Laplace-Beltrami operator on a toroidal surface', at the beginning, he introduced the Green function on sphere, as Prof.Hardy said, the Green function of Laplace on sphere is $$\frac{1}{2 \pi} \ln \frac{1}{|x-y|}+C$$ where the distance is the Euclead diastance. And the strict proof can be found in Here.