Fundamental Solutions of the Laplacian with Spherical Symmetry

differential geometry

The fundamental solution is the integral of reciprocal surface area.

This note redoes the calculation of fundamental solutions to the laplacian in constant curvature geometries, in a more abstract and general framework. The key to those calculations was spherical symmetry, and here we work with general Riemannian manifolds that have spherical symmetry about some point O∈M\providecommand{\Isom}{\operatorname{Isom}} O\in M, finding the fundamental solution for all of them.

Theorem

Let (Mn,g)\providecommand{\Isom}{\operatorname{Isom}} (M^n,g) be a spherically symmetric manifold with radial function r\providecommand{\Isom}{\operatorname{Isom}} r, and A(r)\providecommand{\Isom}{\operatorname{Isom}} A(r) the surface area of its level sets. Then the following function is a fundamental solution for the Laplacian on M\providecommand{\Isom}{\operatorname{Isom}} M, f(r)=∫drA(r)\providecommand{\Isom}{\operatorname{Isom}} f(r)=\int \frac{dr}{A(r)}

The entire content of the theorem is that the geometry enters through one function, the area of the geodesic sphere, and the potential is recovered from it by a single integration. In constant curvature that one function is ωn−1k(r)n−1\providecommand{\Isom}{\operatorname{Isom}} \omega_{n-1}k(r)^{n-1}, and the three familiar answers are just the three ways k\providecommand{\Isom}{\operatorname{Isom}} k can behave:

Reading the two panels against each other is reading the theorem. Hyperbolic area runs away exponentially, so 1/A\providecommand{\Isom}{\operatorname{Isom}} 1/A is integrable and the potential saturates at a finite ceiling. Euclidean area sits at the borderline: in the plane ∫dr/r\providecommand{\Isom}{\operatorname{Isom}} \int dr/r diverges and the potential climbs forever, while from three dimensions on it converges. And spherical area collapses back to zero at r=π\providecommand{\Isom}{\operatorname{Isom}} r=\pi, so the potential turns around and runs off a second time - which is the antipodal singularity discussed at the end of this note.

Proving this theorem requires two calculations that were done in recent notes, reviewed below.

Spherical Symmetry

First we need a precise definition of spherical symmetry to work with. If (M,g)\providecommand{\Isom}{\operatorname{Isom}} (M,g) is a Riemannian manifold, we write Isom⁡(M)\providecommand{\Isom}{\operatorname{Isom}} \operatorname{Isom}(M) for the group of all isometries M→M\providecommand{\Isom}{\operatorname{Isom}} M\to M. If S⊂M\providecommand{\Isom}{\operatorname{Isom}} S\subset M is a subset, we write Isom⁡(M;S)\providecommand{\Isom}{\operatorname{Isom}} \operatorname{Isom}(M;S) for the subgroup of isometries sending S→S\providecommand{\Isom}{\operatorname{Isom}} S\to S.

DefinitionSpherical Symmetry

A Riemannian manifold (Mn,g)\providecommand{\Isom}{\operatorname{Isom}} (M^n,g) is spherically symmetric if there is a point O∈M\providecommand{\Isom}{\operatorname{Isom}} O\in M where Isom⁡(M;O)\providecommand{\Isom}{\operatorname{Isom}} \operatorname{Isom}(M;O) contains a copy of Isom⁡(Sn−1)\providecommand{\Isom}{\operatorname{Isom}} \operatorname{Isom}(\mathbb{S}^{n-1}).

This definition picks out a particular point O\providecommand{\Isom}{\operatorname{Isom}} O in the manifold as special, so its natural to look at the distance function from O\providecommand{\Isom}{\operatorname{Isom}} O:

DefinitionRadial Function

If (M,g)\providecommand{\Isom}{\operatorname{Isom}} (M,g) is spherically symmetric about O\providecommand{\Isom}{\operatorname{Isom}} O, the radial function for M\providecommand{\Isom}{\operatorname{Isom}} M measures the geodesic distance from O\providecommand{\Isom}{\operatorname{Isom}} O: r(−)=dist(O,−)\providecommand{\Isom}{\operatorname{Isom}} r(-)=\mathrm{dist}(O,-)

We call a function f ⁣:M→R\providecommand{\Isom}{\operatorname{Isom}} f\colon M\to\RR radial if it factors through r ⁣:M→R\providecommand{\Isom}{\operatorname{Isom}} r\colon M\to \RR. This radial function is not smooth at O\providecommand{\Isom}{\operatorname{Isom}} O (the graph nearby looks like a “cone”), and in general it may not be smooth at other points (such as the antipodal point to O\providecommand{\Isom}{\operatorname{Isom}} O if M\providecommand{\Isom}{\operatorname{Isom}} M is the n\providecommand{\Isom}{\operatorname{Isom}} n-sphere). For the rest of this note we will work with spaces where r\providecommand{\Isom}{\operatorname{Isom}} r is assumed smooth: so we delete points like the antipode if needed.

Finally a small technical point worth noting: with what we’ve introduced there are two natural families of hypersurfaces one could define geometrically in a spherically symmetric manifold: the group orbits of Isom⁡(Sn−1)\providecommand{\Isom}{\operatorname{Isom}} \Isom(\mathbb{S}^{n-1}), and the level sets of r\providecommand{\Isom}{\operatorname{Isom}} r. But these are the same

Proposition

The group orbits of Isom⁡(Sn−1)\providecommand{\Isom}{\operatorname{Isom}} \Isom(\mathbb{S}^{n-1}) coincide with the level sets of r\providecommand{\Isom}{\operatorname{Isom}} r.

Proof

If two points p,q\providecommand{\Isom}{\operatorname{Isom}} p,q lie in the same group orbit, q=Φ(p)\providecommand{\Isom}{\operatorname{Isom}} q=\Phi(p) for some isometry Φ∈Isom⁡(Sn−1)\providecommand{\Isom}{\operatorname{Isom}} \Phi\in\Isom(\mathbb{S}^{n-1}). Because this isometry fixes O\providecommand{\Isom}{\operatorname{Isom}} O we have r(q)=dist(O,q)=dist(O,Φ(p))=dist(Φ(O),Φ(p))=dist(O,p)=r(p)\providecommand{\Isom}{\operatorname{Isom}} r(q)=\mathrm{dist}(O,q)=\mathrm{dist}(O,\Phi(p))=\mathrm{dist}(\Phi(O),\Phi(p))=\mathrm{dist}(O,p)=r(p) So p,q\providecommand{\Isom}{\operatorname{Isom}} p,q lie in the same level set of r\providecommand{\Isom}{\operatorname{Isom}} r, and orbits are subsets of level sets.

To get the reverse inclusion, let p,q\providecommand{\Isom}{\operatorname{Isom}} p,q lie at the same distance r\providecommand{\Isom}{\operatorname{Isom}} r from O\providecommand{\Isom}{\operatorname{Isom}} O, and let γp,γq\providecommand{\Isom}{\operatorname{Isom}} \gamma_p,\gamma_q be the geodesics starting from O\providecommand{\Isom}{\operatorname{Isom}} O ending at each after time r\providecommand{\Isom}{\operatorname{Isom}} r. Then the initial tangent vectors γp˙,γq˙\providecommand{\Isom}{\operatorname{Isom}} \dot{\gamma_p},\dot{\gamma_q} are unit vectors in TOM\providecommand{\Isom}{\operatorname{Isom}} T_OM, and the assumed Isom⁡(Sn−1)=O(n)\providecommand{\Isom}{\operatorname{Isom}} \Isom(\mathbb{S}^{n-1})=\mathrm{O}(n) isometries of M\providecommand{\Isom}{\operatorname{Isom}} M descend by differentiation to an action on TOM\providecommand{\Isom}{\operatorname{Isom}} T_OM by isometries. Since O(n)\providecommand{\Isom}{\operatorname{Isom}} \mathrm{O}(n)‘s action on the unit sphere in Rn\providecommand{\Isom}{\operatorname{Isom}} \RR^n is transitive, there is some Φ\providecommand{\Isom}{\operatorname{Isom}} \Phi whose derivative satisfies dΦOγp˙=γq˙\providecommand{\Isom}{\operatorname{Isom}} d\Phi_O \dot{\gamma_p}=\dot{\gamma_q}. But as isometries preserve geodesics and geodesics are determined by their initial conditions, this implies that γq=Φ∘γp\providecommand{\Isom}{\operatorname{Isom}} \gamma_q = \Phi\circ\gamma_p and so q=γq(r)=Φγp(r)=Φp\providecommand{\Isom}{\operatorname{Isom}} q=\gamma_q(r)=\Phi\gamma_p(r)=\Phi p Thus p,q\providecommand{\Isom}{\operatorname{Isom}} p,q lie in the same group orbit.

Finding Fundamental Solutions

Here we carry out the explicit computations to find a fundamental solution f ⁣:M→R\providecommand{\Isom}{\operatorname{Isom}} f\colon M\to \RR to Δf=δO\providecommand{\Isom}{\operatorname{Isom}} \Delta f=\delta_O. The radial function will play an important role so we start with a quick lemma computing its Laplacian.

Lemma

Let (Mn,g)\providecommand{\Isom}{\operatorname{Isom}} (M^n,g) be a spherically symmetric manifold with radial function r\providecommand{\Isom}{\operatorname{Isom}} r, and A(r)\providecommand{\Isom}{\operatorname{Isom}} A(r) the surface area of its level sets. Then Δr=A′(r)A(r)\providecommand{\Isom}{\operatorname{Isom}} \Delta r = \frac{A^\prime(r)}{A(r)}

Proof

We proceed to calculate using Δr=⋆d⋆dr\providecommand{\Isom}{\operatorname{Isom}} \Delta r = \star d\star d r, building up one operation at a time. The first to understand is ⋆dr\providecommand{\Isom}{\operatorname{Isom}} \star dr. By definition this is the n−1\providecommand{\Isom}{\operatorname{Isom}} n-1 form such that dr∧⋆dr=⟨dr,dr⟩vol\providecommand{\Isom}{\operatorname{Isom}} dr\wedge\star dr =\langle dr,dr\rangle \mathrm{vol} For vol\providecommand{\Isom}{\operatorname{Isom}} \mathrm{vol} the Riemannian volume form on M\providecommand{\Isom}{\operatorname{Isom}} M. The metric on 1-forms is induced from vectors via the musical isomorphism so ⟨dr,dr⟩=⟨∇r,∇r⟩\providecommand{\Isom}{\operatorname{Isom}} \langle dr,dr\rangle =\langle \nabla r,\nabla r\rangle and by general theorems of Riemannian geometry, since r\providecommand{\Isom}{\operatorname{Isom}} r is a distance function ∇r\providecommand{\Isom}{\operatorname{Isom}} \nabla r is a unit vector field. Thus dr∧⋆dr=vol\providecommand{\Isom}{\operatorname{Isom}} dr\wedge \star dr = \mathrm{vol}

Again by general theorems of Riemannian geometry, ∇r\providecommand{\Isom}{\operatorname{Isom}} \nabla r is orthogonal to the geodesic spheres centered at O\providecommand{\Isom}{\operatorname{Isom}} O, so we can orthogonally decompose the volume form of M\providecommand{\Isom}{\operatorname{Isom}} M into an r\providecommand{\Isom}{\operatorname{Isom}} r component, and the volume form ωr\providecommand{\Isom}{\operatorname{Isom}} \omega_r of the level set Sr\providecommand{\Isom}{\operatorname{Isom}} S_r. Thus by definition we must have ⋆dr=ωr\providecommand{\Isom}{\operatorname{Isom}} \star dr = \omega_r. Because each of these level sets is a round sphere, we can write their volume forms uniformly in terms of the unit volume form ω\providecommand{\Isom}{\operatorname{Isom}} \omega of the n−1\providecommand{\Isom}{\operatorname{Isom}} n-1-sphere, scaled by the area A∘r\providecommand{\Isom}{\operatorname{Isom}} A\circ r of the geodesic sphere in question. Putting this together, ⋆dr=(A∘r)ω\providecommand{\Isom}{\operatorname{Isom}} \star dr = (A\circ r)\omega Now we differentiate with the product rule d(⋆dr)=d((A∘r)ω)=d(A∘r)∧ω+(A∘r)dω\providecommand{\Isom}{\operatorname{Isom}} d(\star dr)=d\left((A\circ r)\omega\right)=d(A\circ r)\wedge \omega + (A\circ r)d\omega Since ω\providecommand{\Isom}{\operatorname{Isom}} \omega is a closed form on the sphere, dω=0\providecommand{\Isom}{\operatorname{Isom}} d\omega =0 and the first term simplifies as d(A∘r)=(A′∘r)dr\providecommand{\Isom}{\operatorname{Isom}} d(A\circ r)=(A^\prime \circ r)dr so d⋆dr=(A′∘r)dr∧ω\providecommand{\Isom}{\operatorname{Isom}} d\star dr = (A^\prime \circ r) dr\wedge \omega Because this result is a top-dimensional form it must be some multiple of the volume form on M\providecommand{\Isom}{\operatorname{Isom}} M, and it will make our next step easier if we rewrite it in that form. Recall we know an orthogonal decomposition of the volume form vol=dr∧ωr\providecommand{\Isom}{\operatorname{Isom}} \mathrm{vol}=dr\wedge \omega_r, and ωr=(A∘r)ω\providecommand{\Isom}{\operatorname{Isom}} \omega_r=(A\circ r)\omega. Computing (and dropping all composition with r\providecommand{\Isom}{\operatorname{Isom}} r from the notation for brevity) dr∧ω=dr∧AAω=1A(dr∧Aω)=1Adr∧ωr=1Avol\providecommand{\Isom}{\operatorname{Isom}} dr\wedge\omega = dr\wedge \frac{A}{A}\omega=\frac{1}{A}\left(dr\wedge A\omega\right)=\frac{1}{A} dr\wedge \omega_r=\frac{1}{A}\mathrm{vol} Substituting this in, d⋆dr=(A′∘r)dr∧ω=A′∘rA∘rvol\providecommand{\Isom}{\operatorname{Isom}} d\star dr = (A^\prime\circ r) dr\wedge \omega = \frac{A^\prime\circ r}{A\circ r}\mathrm{vol} Now we apply the final hodge star, completing the proof ⋆d⋆dr=⋆(A′∘rA∘rvol)=A′∘rA∘r⋆vol=A′∘rA∘r\providecommand{\Isom}{\operatorname{Isom}} \star d\star d r = \star\left(\frac{A^\prime\circ r}{A\circ r}\mathrm{vol}\right)=\frac{A^\prime\circ r}{A\circ r}\star\mathrm{vol} =\frac{A^\prime\circ r}{A\circ r}

Now we come to our main theorem:

Theorem

Let (Mn,g)\providecommand{\Isom}{\operatorname{Isom}} (M^n,g) be a spherically symmetric manifold with radial function r\providecommand{\Isom}{\operatorname{Isom}} r, and A(r)\providecommand{\Isom}{\operatorname{Isom}} A(r) the surface area of its level sets. Then the following function is a fundamental solution for the Laplacian on M\providecommand{\Isom}{\operatorname{Isom}} M, φ(r)=∫drA(r)\providecommand{\Isom}{\operatorname{Isom}} \varphi(r)=\int \frac{dr}{A(r)}

We prove this in two steps. First we see that f∘r\providecommand{\Isom}{\operatorname{Isom}} f\circ r really is harmonic on M∖O\providecommand{\Isom}{\operatorname{Isom}} M\smallsetminus O, and then we show (as a distribution) it has the correct behavior on neighborhoods of O\providecommand{\Isom}{\operatorname{Isom}} O.

Proposition

Let r\providecommand{\Isom}{\operatorname{Isom}} r be the radial distance function and f ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} f\colon\RR\to\RR defined by f(x)=∫∗xdtA(t)\providecommand{\Isom}{\operatorname{Isom}} f(x)=\int_\ast^x \frac{dt}{A(t)} Then the composition φ=f∘r\providecommand{\Isom}{\operatorname{Isom}} \varphi=f\circ r is harmonic on M∖O\providecommand{\Isom}{\operatorname{Isom}} M\smallsetminus O.

Proof

If f ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} f\colon\RR\to\RR is any function such that f∘r\providecommand{\Isom}{\operatorname{Isom}} f\circ r is harmonic, the chain rule for the Laplacian implies 0=Δ(f∘r)=f′′∥∇r∥2+f′Δr\providecommand{\Isom}{\operatorname{Isom}} 0=\Delta(f\circ r)=f^{\prime\prime}\|\nabla r\|^2+f^\prime \Delta r Since r\providecommand{\Isom}{\operatorname{Isom}} r is a Riemannian distance ∥∇r∥=1\providecommand{\Isom}{\operatorname{Isom}} \|\nabla r\|=1, and Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r we calculated above to be the logarithmic derivative of the area, Δr=A′(r)A(r)\providecommand{\Isom}{\operatorname{Isom}} \Delta r = \frac{A^\prime(r)}{A(r)}, for A\providecommand{\Isom}{\operatorname{Isom}} A the surface area of geodesic spheres about O\providecommand{\Isom}{\operatorname{Isom}} O. Thus, for every p∈M\providecommand{\Isom}{\operatorname{Isom}} p\in M we have f′′(r(p))+f′(r(p))A′(r(p))A(r(p))=0\providecommand{\Isom}{\operatorname{Isom}} f^{\prime\prime}(r(p))+f^\prime(r(p))\frac{A^\prime(r(p))}{A(r(p))}=0

This is satisfied for all p∈M\providecommand{\Isom}{\operatorname{Isom}} p\in M if and only if f ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} f\colon\RR\to\RR and A ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} A\colon\RR\to\RR obey the prescribed 1-dimensional ODE f′′+f′A′A=0\providecommand{\Isom}{\operatorname{Isom}} f^{\prime\prime}+f^\prime\frac{A^\prime}{A}=0 for all r(p)=x∈R\providecommand{\Isom}{\operatorname{Isom}} r(p)=x\in\RR. Clearing denominators reveals the left hand side to be a product rule:

0=Af′′+A′f′=(Af′)′\providecommand{\Isom}{\operatorname{Isom}} 0=A f^{\prime\prime}+A^\prime f^\prime= \left(A f^\prime\right)^\prime

Thus Af′\providecommand{\Isom}{\operatorname{Isom}} Af^\prime is constant. This provides a differential equation for f\providecommand{\Isom}{\operatorname{Isom}} f: A(x)f′(x)=C   ⟹  f′(x)=CA(x)\providecommand{\Isom}{\operatorname{Isom}} A(x)f^\prime(x) = C\,\implies f^\prime(x) = \frac{C}{A(x)} The case C=1\providecommand{\Isom}{\operatorname{Isom}} C=1 yields the function we are after by quadrature f(x)=∫∗xdtA(t)\providecommand{\Isom}{\operatorname{Isom}} f(x)=\int_\ast^x\frac{dt}{A(t)} Where ∗∈R\providecommand{\Isom}{\operatorname{Isom}} \ast\in\RR is arbitrary, setting a constant of integration.

In situations like this we will often allow ourselves the following abuse of notation: if the constant of integration is arbitrary then we might like to write the integral as indefinite with x\providecommand{\Isom}{\operatorname{Isom}} x as the dummy variable of integration. And, when the end goal is to compose with another function like r\providecommand{\Isom}{\operatorname{Isom}} r, we may switch the dummy variable to this and write

f∘r ⁣:p↦∫∗r(p)dtA(t)=(∫∗r(−)dtA(t))(p)=(∫drA(r))(p)\providecommand{\Isom}{\operatorname{Isom}} f\circ r\colon p\mapsto \int_\ast^{r(p)}\frac{dt}{A(t)}=\left(\int_\ast^{r(-)}\frac{dt}{A(t)}\right)(p)=\left(\int\frac{dr}{A(r)}\right)(p)   ⟹  φ:=∫drA(r)\providecommand{\Isom}{\operatorname{Isom}} \implies\varphi:=\int\frac{dr}{A(r)}

Next we see that integration of Δφ\providecommand{\Isom}{\operatorname{Isom}} \Delta \varphi on domains Ω⊂M\providecommand{\Isom}{\operatorname{Isom}} \Omega\subset M behaves like the δ\providecommand{\Isom}{\operatorname{Isom}} \delta distribution. Note - all integrals over regions containing O\providecommand{\Isom}{\operatorname{Isom}} O should be considered distributionally.

Proposition

Let Ω⊂M\providecommand{\Isom}{\operatorname{Isom}} \Omega\subset M be a domain for integration. Then

∫ΩΔφ vol={0O∉Ω1O∈Ω\providecommand{\Isom}{\operatorname{Isom}} \int_\Omega \Delta \varphi \,\mathrm{vol} =\begin{cases} 0 & O\not \in \Omega\\ 1 & O\in \Omega \end{cases}
Proof

If O∉Ω\providecommand{\Isom}{\operatorname{Isom}} O\not\in\Omega, then φ\providecommand{\Isom}{\operatorname{Isom}} \varphi is well defined and smooth on all of Ω\providecommand{\Isom}{\operatorname{Isom}} \Omega, and Δφ=0\providecommand{\Isom}{\operatorname{Isom}} \Delta \varphi = 0 on Ω\providecommand{\Isom}{\operatorname{Isom}} \Omega. Thus ∫ΩΔφ vol=∫Ω0 vol=0\providecommand{\Isom}{\operatorname{Isom}} \int_\Omega \Delta \varphi \,\mathrm{vol}=\int_\Omega 0\, \mathrm{vol}=0 So we need only be concerned with domains where O∈Ω\providecommand{\Isom}{\operatorname{Isom}} O\in\Omega. First we see the value of the integral is independent of the particular domain. If Ω1,Ω2\providecommand{\Isom}{\operatorname{Isom}} \Omega_1,\Omega_2 are two domains containing O\providecommand{\Isom}{\operatorname{Isom}} O in their interior, its possible to find a small geodesic ball B\providecommand{\Isom}{\operatorname{Isom}} B about O\providecommand{\Isom}{\operatorname{Isom}} O contained in both of them. Then we may write Ωi=(Ωi∖B)∪B\providecommand{\Isom}{\operatorname{Isom}} \Omega_i = (\Omega_i\smallsetminus B)\cup B which gives a decomposition of the corresponding integrals ∫ΩiΔφ vol=∫Ωi∖BΔφ vol+∫BΔφ vol\providecommand{\Isom}{\operatorname{Isom}} \int_{\Omega_i}\Delta \varphi\,\mathrm{vol}=\int_{\Omega_i\smallsetminus B}\Delta \varphi\,\mathrm{vol}+\int_{B}\Delta \varphi\,\mathrm{vol} But the first integral in this sum is over a domain not containing O\providecommand{\Isom}{\operatorname{Isom}} O, so is zero by the previous argument. Thus both the integral over Ω1\providecommand{\Isom}{\operatorname{Isom}} \Omega_1 and Ω2\providecommand{\Isom}{\operatorname{Isom}} \Omega_2 equal the integral over B\providecommand{\Isom}{\operatorname{Isom}} B, and so are equal to one another.

Now let’s evaluate such an integral. Since the value is independent of domain, we can without loss of generality fix some geodesic ball B\providecommand{\Isom}{\operatorname{Isom}} B about O\providecommand{\Isom}{\operatorname{Isom}} O, and note that by the definition of the Hodge dual ∫BΔφ vol=∫B⋆Δφ\providecommand{\Isom}{\operatorname{Isom}} \int_B \Delta \varphi\,\mathrm{vol}=\int_B \star \Delta \varphi and unpacking Δφ\providecommand{\Isom}{\operatorname{Isom}} \Delta \varphi allows some cancellation as ⋆2=id\providecommand{\Isom}{\operatorname{Isom}} \star^2=\mathrm{id} on top-dimensional forms ⋆Δφ=⋆⋆d⋆dφ=d⋆dφ\providecommand{\Isom}{\operatorname{Isom}} \star \Delta \varphi = \star\star d\star d \varphi = d\star d \varphi Now that our form begins with exterior differentiation we can apply (a distributional version of) Stokes’ theorem: ∫Bd⋆dφ=∫∂B⋆dφ\providecommand{\Isom}{\operatorname{Isom}} \int_B d\star d \varphi=\int_{\partial B} \star d \varphi Since B\providecommand{\Isom}{\operatorname{Isom}} B is a geodesic ball centered at O\providecommand{\Isom}{\operatorname{Isom}} O, its boundary is a sphere: ∂B=S\providecommand{\Isom}{\operatorname{Isom}} \partial B=S To continue, we must bring in our definition of φ=∫drA(r)\providecommand{\Isom}{\operatorname{Isom}} \varphi =\int\frac{dr}{A(r)}.

φ=f∘rforf(x)=∫∗xdtA(t)\providecommand{\Isom}{\operatorname{Isom}} \varphi = f\circ r\hspace{0.5cm}\textrm{for}\hspace{0.5cm}f(x)=\int_\ast^x \frac{dt}{A(t)}

Differentiating, dφ=d(f∘r)=(f′∘r)dr=1A∘rdr\providecommand{\Isom}{\operatorname{Isom}} d\varphi = d(f\circ r)=(f^\prime\circ r) dr=\frac{1}{A\circ r}dr ⋆dφ=⋆(1A∘rdr)=1A∘r⋆dr\providecommand{\Isom}{\operatorname{Isom}} \star d\varphi = \star\left(\frac{1}{A\circ r}dr\right)=\frac{1}{A\circ r}\star dr

We know from the calculation of Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r how to work with ⋆dr\providecommand{\Isom}{\operatorname{Isom}} \star dr: at any fixed r\providecommand{\Isom}{\operatorname{Isom}} r, this is just the volume form ωr\providecommand{\Isom}{\operatorname{Isom}} \omega_r of the sphere of that radius about O\providecommand{\Isom}{\operatorname{Isom}} O, and ωr=(A∘r)ω\providecommand{\Isom}{\operatorname{Isom}} \omega_r = (A\circ r)\omega for ω\providecommand{\Isom}{\operatorname{Isom}} \omega the unit volume (integrating to 1\providecommand{\Isom}{\operatorname{Isom}} 1 on any level set Sr\providecommand{\Isom}{\operatorname{Isom}} S_r) ⋆dφ=1A∘r⋆dr=1A∘rωr=1A∘r(A∘r)ω=ω\providecommand{\Isom}{\operatorname{Isom}} \star d\varphi =\frac{1}{A\circ r}\star dr = \frac{1}{A\circ r}\omega_r=\frac{1}{A\circ r}(A\circ r)\omega=\omega

Putting this all together, we have successfully computed the integral:

∫BΔφ vol=∫Bd⋆dφ=∫∂B=S⋆dφ=∫Sω=1\providecommand{\Isom}{\operatorname{Isom}} \int_B \Delta \varphi\,\mathrm{vol}=\int_B d\star d\varphi=\int_{\partial B=S} \star d \varphi =\int_S \omega = 1

Constant Curvature Geometries

With the general theorem in hand we now specialize the result to the familiar cases of constant curvature. We do not need to choose any coordinates; the only geometric quantity needed from each geometry is the “area” of the embedded (n−1)\providecommand{\Isom}{\operatorname{Isom}} (n-1) spheres of radius r\providecommand{\Isom}{\operatorname{Isom}} r:

Fact

The surface area of the geodesic sphere of radius r\providecommand{\Isom}{\operatorname{Isom}} r in the n\providecommand{\Isom}{\operatorname{Isom}} n-dimensional spaces of constant curvature are En ⁣: ωn−1rn−1Sn ⁣: ωn−1sin⁡(r)n−1Hn ⁣: ωn−1sinh⁡(r)n−1\providecommand{\Isom}{\operatorname{Isom}} \mathbb{E}^{n}\colon\, \omega_{n-1} r^{n-1}\hspace{1cm}\mathbb{S}^{n}\colon\,\omega_{n-1}\sin(r)^{n-1}\hspace{1cm}\mathbb{H}^{n}\colon\, \omega_{n-1}\sinh(r)^{n-1} Where ωk\providecommand{\Isom}{\operatorname{Isom}} \omega_k is the size of the unit k\providecommand{\Isom}{\operatorname{Isom}} k sphere of curvature 1: ω1=2πω2=4πω3=2π2…\providecommand{\Isom}{\operatorname{Isom}} \omega_1=2\pi\hspace{0.5cm}\omega_2= 4\pi\hspace{0.5cm}\omega_3= 2\pi^2\ldots

This immediately gives integral forms of the fundamental solutions in each constant curvature geometry:

Corollary

The fundamental solutions to Δf=δO\providecommand{\Isom}{\operatorname{Isom}} \Delta f =\delta_O in constant curvature are En ⁣: f(r)=1ωn−1∫drrn−1\providecommand{\Isom}{\operatorname{Isom}} \mathbb{E}^{n}\colon\, f(r)=\frac{1}{\omega_{n-1}}\int\frac{dr}{r^{n-1}} Sn ⁣: 1ωn−1∫drsin⁡(r)n−1\providecommand{\Isom}{\operatorname{Isom}} \mathbb{S}^{n}\colon\,\frac{1}{\omega_{n-1}}\int\frac{dr}{\sin(r)^{n-1}} Hn ⁣: 1ωn−1∫drsinh⁡(r)n−1\providecommand{\Isom}{\operatorname{Isom}} \mathbb{H}^{n}\colon\, \frac{1}{\omega_{n-1}}\int\frac{dr}{\sinh(r)^{n-1}}

These are exactly the formulas the constant curvature note arrived at by separating the Laplacian in polar coordinates; here they come out of a single lemma about Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r, and the function k(r)∈{r,sin⁡r,sinh⁡r}\providecommand{\Isom}{\operatorname{Isom}} k(r)\in\{r,\sin r,\sinh r\} that had to be introduced by hand there is now visibly just A(r)/ωn−1\providecommand{\Isom}{\operatorname{Isom}} A(r)/\omega_{n-1}.

For Euclidean geometry, all of these integrals are easily computable by hand: n=2\providecommand{\Isom}{\operatorname{Isom}} n=2 results in a logarithm and the rest directly follow from the ‘power rule’ fE2(r)=12πlog⁡(r)fEn=−1(n−2)ωn−11rn−2\providecommand{\Isom}{\operatorname{Isom}} f_{\mathbb{E}^2}(r)=\frac{1}{2\pi}\log(r)\hspace{1.5cm}f_{\mathbb{E}^n}=\frac{-1}{(n-2)\omega_{n-1}}\frac{1}{r^{n-2}}

For hyperbolic and spherical geometry, we can compute dimension 2\providecommand{\Isom}{\operatorname{Isom}} 2 by hand

fS2(r)=12πlog⁡∣tan⁡r2∣fH2(r)=12πlog⁡∣tanh⁡r2∣\providecommand{\Isom}{\operatorname{Isom}} f_{\mathbb{S}^2}(r)=\frac{1}{2\pi}\log \left|\tan\frac{r}{2}\right|\hspace{1cm}f_{\mathbb{H}^2}(r)=\frac{1}{2\pi}\log \left|\tanh\frac{r}{2}\right|

An Aside on the Sphere

The spherical answers deserve a word of caution, and it is the same caution as the remark above about deleting the antipode. On a closed manifold there is no function with Δf=δO\providecommand{\Isom}{\operatorname{Isom}} \Delta f=\delta_O at all: integrating both sides over M\providecommand{\Isom}{\operatorname{Isom}} M, Stokes’ theorem sends the left side to 0\providecommand{\Isom}{\operatorname{Isom}} 0 while the right side is 1\providecommand{\Isom}{\operatorname{Isom}} 1. So something has to give, and what gives is the assumption that the singularity is alone.

Look at what the formula actually produces. As r→0\providecommand{\Isom}{\operatorname{Isom}} r\to0 we have tan⁡r2≈r2\providecommand{\Isom}{\operatorname{Isom}} \tan\frac{r}{2}\approx\frac{r}{2}, so fS2≈12πlog⁡r\providecommand{\Isom}{\operatorname{Isom}} f_{\mathbb{S}^2}\approx\frac{1}{2\pi}\log r and we do get a source of strength 1\providecommand{\Isom}{\operatorname{Isom}} 1 at O\providecommand{\Isom}{\operatorname{Isom}} O. But as r→π\providecommand{\Isom}{\operatorname{Isom}} r\to\pi, writing s=π−r\providecommand{\Isom}{\operatorname{Isom}} s=\pi-r gives tan⁡r2=cot⁡s2≈2s\providecommand{\Isom}{\operatorname{Isom}} \tan\frac{r}{2}=\cot\frac{s}{2}\approx\frac{2}{s}, so fS2≈−12πlog⁡s\providecommand{\Isom}{\operatorname{Isom}} f_{\mathbb{S}^2}\approx-\frac{1}{2\pi}\log s: a sink of strength 1\providecommand{\Isom}{\operatorname{Isom}} 1 at the antipode. The same happens in S3\providecommand{\Isom}{\operatorname{Isom}} \mathbb{S}^3, where −14πtan⁡r\providecommand{\Isom}{\operatorname{Isom}} -\frac{1}{4\pi\tan r} blows up at r=π\providecommand{\Isom}{\operatorname{Isom}} r=\pi with the opposite sign. What the construction returns on a sphere is a dipole, ΔfSn=δO−δ−O\providecommand{\Isom}{\operatorname{Isom}} \Delta f_{\mathbb{S}^n}=\delta_O-\delta_{-O} which is the closest thing to a fundamental solution the sphere admits, and its total integral is correctly zero. (The other standard fix is to solve Δf=δO−1vol(M)\providecommand{\Isom}{\operatorname{Isom}} \Delta f=\delta_O-\frac{1}{\mathrm{vol}(M)}, spreading the compensating charge uniformly instead of concentrating it at one point.) Euclidean and hyperbolic space are non-compact and escape the issue: there A(r)→∞\providecommand{\Isom}{\operatorname{Isom}} A(r)\to\infty fast enough that the second singularity runs off to infinity.

There is a sharper way to see that nothing gradual is going on. The flux of φ\providecommand{\Isom}{\operatorname{Isom}} \varphi through the circle of radius r\providecommand{\Isom}{\operatorname{Isom}} r is A(r)φ′(r)=2πsin⁡r⋅12πsin⁡r=1\providecommand{\Isom}{\operatorname{Isom}} A(r)\varphi^\prime(r)=2\pi\sin r\cdot\frac{1}{2\pi \sin r}=1 independent of r\providecommand{\Isom}{\operatorname{Isom}} r. So the charge enclosed by the geodesic ball is exactly 1\providecommand{\Isom}{\operatorname{Isom}} 1 for every r∈(0,π)\providecommand{\Isom}{\operatorname{Isom}} r\in(0,\pi) - it never leaks away as the ball grows. It is still exactly 1\providecommand{\Isom}{\operatorname{Isom}} 1 when the bounding circle has shrunk back to a point, at which moment the ball is all of S2\providecommand{\Isom}{\operatorname{Isom}} \mathbb{S}^2 and the enclosed charge is obliged to be 0\providecommand{\Isom}{\operatorname{Isom}} 0. The entire discrepancy is the −1\providecommand{\Isom}{\operatorname{Isom}} -1 sitting at the antipode.

Whereas getting explicit formulae for higher dimensions relies on the reduction formulae for integrating csc⁡n(r)\providecommand{\Isom}{\operatorname{Isom}} \csc^n(r) and csch⁡n(r)\providecommand{\Isom}{\operatorname{Isom}} \operatorname{csch}^n(r):

Fact

The integrals of cosecant and its hyperbolic analog satisfy the following recurrences: ∫csc⁡nx dx=−cot⁡xcsc⁡n−2xn−1+n−2n−1∫csc⁡n−2x dx\providecommand{\Isom}{\operatorname{Isom}} \int\csc^n x\,dx = \frac{-\cot x\csc^{n-2}x}{n-1}+\frac{n-2}{n-1}\int\csc^{n-2}x\,dx ∫csch⁡nx dx=−coth⁡xcsch⁡n−2xn−1−n−2n−1∫csch⁡n−2x dx\providecommand{\Isom}{\operatorname{Isom}} \int\operatorname{csch}^nx\,dx = \frac{-\coth x\operatorname{csch}^{n-2}x}{n-1}-\frac{n-2}{n-1}\int\operatorname{csch}^{n-2}x\,dx

Note the sign of the second term differs between the two, which is why the hyperbolic answers are not simply the spherical ones with r↦ir\providecommand{\Isom}{\operatorname{Isom}} r\mapsto ir term by term.

The General Case:

What if we are given a spherically symmetric metric in cartesian coordinates, where geometric quantities like the geodesic distance r\providecommand{\Isom}{\operatorname{Isom}} r are not immediately apparent? Here we record the necessary computations. For brevity we will write ℓ=x12+⋯+xn2\providecommand{\Isom}{\operatorname{Isom}} \ell=\sqrt{x_1^2+\cdots+x_n^2} to be the ‘coordinate length’. Consider the metric g=σ(ℓ)2gEn(⋅,⋅)\providecommand{\Isom}{\operatorname{Isom}} g = \sigma(\ell)^2g_{\mathbb{E}^n}(\cdot,\cdot) Where σ ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} \sigma\colon\RR\to\RR is some smooth function. Some computation yields the following characterization:

Theorem

If g=σ(ℓ)2dsEn2\providecommand{\Isom}{\operatorname{Isom}} g = \sigma(\ell)^2ds^2_{\mathbb{E}^n} for ℓ=∣(x,y,…)∣\providecommand{\Isom}{\operatorname{Isom}} \ell=|(x,y,\ldots)|, then the fundamental solution to the Laplacian on (Rn,g)\providecommand{\Isom}{\operatorname{Isom}} (\RR^n,g) is f(ℓ)=1ωn−1∫dℓℓ[ℓσ(ℓ)]n−2\providecommand{\Isom}{\operatorname{Isom}} f(\ell) = \frac{1}{\omega_{n-1}}\int\frac{d\ell}{\ell\left[\ell\sigma(\ell)\right]^{n-2}}

This metric has spherical symmetry given by the usual action of O(n)\providecommand{\Isom}{\operatorname{Isom}} \mathrm{O}(n) on Rn\providecommand{\Isom}{\operatorname{Isom}} \RR^n and so symmetry arguments1 imply that radial lines are geodesics, and so we can compute the radial function r\providecommand{\Isom}{\operatorname{Isom}} r on Rn\providecommand{\Isom}{\operatorname{Isom}} \RR^n as a function of ℓ\providecommand{\Isom}{\operatorname{Isom}} \ell by integrating along the geodesic γ(t)=(t,0,…)\providecommand{\Isom}{\operatorname{Isom}} \gamma(t)=(t,0,\ldots) to get a function ρ ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} \rho\colon\RR\to\RR:

ρ(l)=∫0lσ(t) dt\providecommand{\Isom}{\operatorname{Isom}} \rho(l)=\int_0^l \sigma(t)\,dt

Then the radial function r ⁣:Rn→R\providecommand{\Isom}{\operatorname{Isom}} r\colon\RR^n\to\RR is given by ρ∘ℓ\providecommand{\Isom}{\operatorname{Isom}} \rho\circ \ell. We are going to need to be careful with bounds, so write f(x)=∫∗xdtA(t)\providecommand{\Isom}{\operatorname{Isom}} f(x)=\int_\ast^x \frac{dt}{A(t)} where ∗∈R\providecommand{\Isom}{\operatorname{Isom}} \ast\in\RR is an arbitrary real number (fixing the constant of integration). Then the fundamental solution to the Laplacian is φ=f∘r=f∘ρ∘ℓ\providecommand{\Isom}{\operatorname{Isom}} \varphi = f\circ r = f\circ\rho\circ\ell: φ ⁣:q↦∫∗r(q)dtA(t)=∫∗ρ(ℓ(q))dtA(t)\providecommand{\Isom}{\operatorname{Isom}} \varphi \colon q\mapsto \int_\ast^{r(q)}\frac{dt}{A(t)}=\int_\ast^{\rho(\ell(q))}\frac{dt}{A(t)}

We can rewrite this integral using the substitution t=ρ(l)\providecommand{\Isom}{\operatorname{Isom}} t=\rho(l) and simplify using ρ′(l)=σ(l)\providecommand{\Isom}{\operatorname{Isom}} \rho^\prime(l)=\sigma(l) from differentiating the previous integral:

∫∗ρ(ℓ(q))dtA(t)=∫∗ℓ(q)d[ρ(l)]A(ρ(l))=∫∗ℓ(q)σ(l)dlA(ρ(l))\providecommand{\Isom}{\operatorname{Isom}} \int_\ast^{\rho(\ell(q))}\frac{dt}{A(t)}=\int_\ast^{\ell(q)}\frac{d[\rho(l)]}{A(\rho(l))}=\int_\ast^{\ell(q)}\frac{\sigma(l) dl}{A(\rho(l))}

where A(ρ(l))\providecommand{\Isom}{\operatorname{Isom}} A(\rho(l)) is the surface area of the sphere with coordinate length l\providecommand{\Isom}{\operatorname{Isom}} l. We can directly calculate this from the metric: at l\providecommand{\Isom}{\operatorname{Isom}} l every vector is stretched by the uniform factor σ(l)\providecommand{\Isom}{\operatorname{Isom}} \sigma(l), so the (n−1)\providecommand{\Isom}{\operatorname{Isom}} (n-1) directions spanning the unit (n−1)\providecommand{\Isom}{\operatorname{Isom}} (n-1) sphere are stretched by this same factor. Thus the overall area is scaled by σ(l)n−1\providecommand{\Isom}{\operatorname{Isom}} \sigma(l)^{n-1} relative to the area that same sphere would have in the standard Euclidean metric. Since this would be AE=ωn−1ln−1\providecommand{\Isom}{\operatorname{Isom}} A_\mathbb{E}=\omega_{n-1}l^{n-1}, we see

A(ρ(l))=σ(l)n−1AE=ωn−1ln−1σ(l)n−1\providecommand{\Isom}{\operatorname{Isom}} A(\rho(l))=\sigma(l)^{n-1}A_{\mathbb{E}}=\omega_{n-1}l^{n-1}\sigma(l)^{n-1}

Substituting this in,

φ ⁣:q↦∫∗ℓ(q)σ(l)dlωn−1ln−1σ(l)n−1=1ωn−1∫∗ℓ(q)dlln−1σ(l)n−2\providecommand{\Isom}{\operatorname{Isom}} \varphi\colon q\mapsto \int_\ast^{\ell(q)}\frac{\sigma(l)dl}{\omega_{n-1}l^{n-1}\sigma(l)^{n-1}}=\frac{1}{\omega_{n-1}}\int_\ast^{\ell(q)}\frac{dl}{l^{n-1}\sigma(l)^{n-2}}

Now we can return to our standard abuse of notation: with the upper bound directly just ℓ(p)\providecommand{\Isom}{\operatorname{Isom}} \ell(p) with arbitrary lower bound, we use ℓ\providecommand{\Isom}{\operatorname{Isom}} \ell as the dummy variable in our indefinite integral and interpret the resulting function of ℓ\providecommand{\Isom}{\operatorname{Isom}} \ell as a function of a real variable composed with the coordinate length function ℓ\providecommand{\Isom}{\operatorname{Isom}} \ell:

f(ℓ)=1ωn−1∫dℓℓ[ℓσ(ℓ)]n−2\providecommand{\Isom}{\operatorname{Isom}} f(\ell) = \frac{1}{\omega_{n-1}}\int\frac{d\ell}{\ell\left[\ell\sigma(\ell)\right]^{n-2}}

Appendix: Proving Existence without Computing

Here’s an argument in the same spirit as the explicit one above, that proves the existence of a radial harmonic function without actually constructing it (essentially, it just avoids computing Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r). Totally unnecessary for our purposes here but I want to record it in case its of future use.

Proposition

There exists a radial harmonic function on M∖O\providecommand{\Isom}{\operatorname{Isom}} M\smallsetminus O.

Proof

A radial harmonic function is some φ ⁣:M→R\providecommand{\Isom}{\operatorname{Isom}} \varphi\colon M\to\RR which satisfies Δφ=0\providecommand{\Isom}{\operatorname{Isom}} \Delta \varphi=0 and factors φ=f∘r\providecommand{\Isom}{\operatorname{Isom}} \varphi=f\circ r for some f ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} f\colon\RR\to\RR. The chain rule for the Laplacian gives an equation for f\providecommand{\Isom}{\operatorname{Isom}} f 0=Δ(f∘r)=f′′∥∇r∥2+f′Δr\providecommand{\Isom}{\operatorname{Isom}} 0=\Delta(f\circ r)=f^{\prime\prime}\|\nabla r\|^2+f^\prime \Delta r Since r\providecommand{\Isom}{\operatorname{Isom}} r is a Riemannian distance ∥∇r∥=1\providecommand{\Isom}{\operatorname{Isom}} \|\nabla r\|=1 and this simplifies giving an equation that must hold for all p∈M\providecommand{\Isom}{\operatorname{Isom}} p\in M: f′′(r(p))+f′(r(p))Δr(p)=0\providecommand{\Isom}{\operatorname{Isom}} f^{\prime\prime}(r(p))+f^\prime(r(p)) \Delta r(p) = 0 If Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r is a radial function (so Δr=h∘r\providecommand{\Isom}{\operatorname{Isom}} \Delta r = h\circ r for some h ⁣:R→R\providecommand{\Isom}{\operatorname{Isom}} h\colon\RR\to\RR) then this becomes f′′(r(p))+f′(r(p))h(r(p))=0\providecommand{\Isom}{\operatorname{Isom}} f^{\prime\prime}(r(p))+f^\prime(r(p)) h(r(p)) = 0, which only holds on M\providecommand{\Isom}{\operatorname{Isom}} M if for all x\providecommand{\Isom}{\operatorname{Isom}} x in the range of r\providecommand{\Isom}{\operatorname{Isom}} r we have f′′(x)+f′(x)h(x)=0\providecommand{\Isom}{\operatorname{Isom}} f^{\prime\prime}(x)+f^\prime(x)h(x)=0 This is an ordinary differential equation on the real line which is easily solved via integrating factors, and any such solution provides a harmonic radial function. Thus it suffices to prove that Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r is radial: equivalently that Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r is constant on the level sets of r\providecommand{\Isom}{\operatorname{Isom}} r.

Let p,q∈M\providecommand{\Isom}{\operatorname{Isom}} p,q\in M lie on the same level set r(p)=r(q)\providecommand{\Isom}{\operatorname{Isom}} r(p)=r(q). Because r\providecommand{\Isom}{\operatorname{Isom}} r-level sets = Isom⁡(Sn−1)\providecommand{\Isom}{\operatorname{Isom}} \Isom(\mathbb{S}^{n-1}) orbits, there is an isometry Φ\providecommand{\Isom}{\operatorname{Isom}} \Phi with Φ(p)=q\providecommand{\Isom}{\operatorname{Isom}} \Phi(p)=q. Thus Δr(q)=Δr(Φ(p))=[(Δr)∘Φ](p)\providecommand{\Isom}{\operatorname{Isom}} \Delta r(q)=\Delta r(\Phi(p))=[(\Delta r)\circ \Phi](p)

Because Φ\providecommand{\Isom}{\operatorname{Isom}} \Phi is an isometry it pulls out of the Laplacian: (Δr)∘Φ=Δ(r∘Φ)\providecommand{\Isom}{\operatorname{Isom}} (\Delta r)\circ \Phi=\Delta(r\circ \Phi). And as r\providecommand{\Isom}{\operatorname{Isom}} r is the Riemannian distance from O\providecommand{\Isom}{\operatorname{Isom}} O, r∘Φ(x)=dist(O,Φ(x))=dist(Φ(O),Φ(x))=dist(O,x)=r(x)\providecommand{\Isom}{\operatorname{Isom}} r\circ\Phi(x)=\mathrm{dist}(O,\Phi(x))=\mathrm{dist}(\Phi(O),\Phi(x))=\mathrm{dist}(O,x)=r(x)

Putting these together, Δr=Δ(r∘Φ)=(Δr)∘Φ\providecommand{\Isom}{\operatorname{Isom}} \Delta r = \Delta(r\circ \Phi)=(\Delta r)\circ \Phi, and so

Δr(q)=[(Δr)∘Φ](p)=Δr(p)\providecommand{\Isom}{\operatorname{Isom}} \Delta r(q)=[(\Delta r)\circ \Phi](p)=\Delta r(p)

So Δr\providecommand{\Isom}{\operatorname{Isom}} \Delta r is constant on the level sets of r\providecommand{\Isom}{\operatorname{Isom}} r: its radial, as required.

Footnotes

  1. If a geodesic starts with initial tangent in the fixed plane of a reflection isometry, then it must remain in that fixed plane for all time. Radial lines are the intersection of n−1\providecommand{\Isom}{\operatorname{Isom}} n-1 independent reflections, so geodesics starting out radially are confined to them. ↩

← All notes