Fundamental Solutions of the Laplacian in Constant Curvature

differential geometryhyperbolic geometry

Explicit fundamental solutions in Euclidean, spherical, and hyperbolic space.

Here’s a quick computation of the fundamental solution to the laplacian Δf=δ\Delta f = \delta in the constant curvature geometries. This is well known but I always have to hunt a bit when I need them or end up re-deriving so figured I’d write it down.

Laplacian in Constant Curvature

If XnX^n is an nn-dimensional constant curvature geometry, O∈XO\in X a point and r=dist(−,O)r=\mathrm{dist}(-,O) is the Riemannian distance from OO, then the fundamental solution for the Laplacian based at OO is F(r)=1ωn−1∫drk(r)n−1F(r)=\frac{1}{\omega_{n-1}}\int\frac{dr}{k(r)^{n-1}} For ωn−1\omega_{n-1} the volume of the unit n−1n-1 sphere, and k(r)k(r) is equal to sin⁡(r)\sin(r), rr or sinh⁡(r)\sinh(r) depending on if XX is spherical, Euclidean or hyperbolic respectively.

And here’s some specific ones in low dimensions

The fundamental solutions to the laplacian in constant curvature are proportional to the following functions, where rr is the geodesic distance from the chosen origin.

Two Dimensions:

  • E2\mathbb{E}^2: F(r)=12πlog⁡(r)F(r)=\frac{1}{2\pi}\log(r)
  • S2\mathbb{S}^2: F(r)=12πlog⁡∣tan⁡r2∣F(r)=\frac{1}{2\pi} \log\left|\tan\frac{r}{2}\right|
  • H2\mathbb{H}^2: F(r)=12πlog⁡∣tanh⁡r2∣F(r)=\frac{1}{2\pi} \log \left|\tanh\frac{r}{2}\right|

Three Dimensions:

  • E3\mathbb{E}^3: F(r)=−14π1rF(r)=\frac{-1}{4\pi}\frac{1}{r}
  • S3\mathbb{S}^3: F(r)=−14π1tan⁡rF(r)=\frac{-1}{4\pi} \frac{1}{\tan r}
  • H3\mathbb{H}^3: F(r)=−14π1tanh⁡rF(r)=\frac{-1}{4\pi} \frac{1}{\tanh r}

The Trick

In Euclidean, hyperbolic and spherical geometry we work in polar coordinates, where the laplacian factors additively into two components, one differentiating along the radius and the other along n−1n-1 spheres, the level sets of constant rr.

ΔEnf=1rn−1∂∂r(rn−1∂f∂r)+1r2ΔSn−1\Delta_{\mathbb{E}^n} f = \frac{1}{r^{n-1}}\frac{\partial}{\partial r}\left(r^{n-1}\frac{\partial f}{\partial r}\right)+\frac{1}{r^2}\Delta_{\mathbb{S}^{n-1}}

ΔHnf=1sinh⁡n−1(r)∂∂r(sinh⁡n−1(r)∂f∂r)+1sinh⁡2(r)ΔSn−1\Delta_{\mathbb{H}^n} f = \frac{1}{\sinh^{n-1}(r)}\frac{\partial}{\partial r}\left(\sinh^{n-1}(r)\frac{\partial f}{\partial r}\right)+\frac{1}{\sinh^2(r)}\Delta_{\mathbb{S}^{n-1}}

ΔSnf=1sin⁡n−1(r)∂∂r(sin⁡n−1(r)∂f∂r)+1sin⁡2(r)ΔSn−1\Delta_{\mathbb{S}^n} f = \frac{1}{\sin^{n-1}(r)}\frac{\partial}{\partial r}\left(\sin^{n-1}(r)\frac{\partial f}{\partial r}\right)+\frac{1}{\sin^2(r)}\Delta_{\mathbb{S}^{n-1}}

Looking at these coordinate expressions we take advantage of a certain similarity - they are all the same up to a choice of sin⁡(r)\sin(r), rr or sinh⁡(r)\sinh(r). Thus, we can give a uniform expression abstracting these differences into a function k(r)k(r):

Δf=1k(r)n−1∂∂r(k(r)n−1∂f∂r)+1k(r)2ΔSn−1\Delta f = \frac{1}{k(r)^{n-1}}\frac{\partial}{\partial r}\left(k(r)^{n-1}\frac{\partial f}{\partial r}\right)+\frac{1}{k(r)^2}\Delta_{\mathbb{S}^{n-1}}

This still looks a bit scary, but because everything is symmetric under rotations we might think to look for solutions sharing the same symmetry group. We’ll call such solutions radial as they depend only on rr. In this case things simplify drastically

Radial Solutions

In any of the constant curvature geometries, if ff is a radial solution to Δf=0\Delta f = 0 then f≡f(r)f\equiv f(r) is proporitional to the following integral: f(r)∝∫1k(r)n−1 drf(r)\propto \int \frac{1}{k(r)^{n-1}}\,dr

If f=f(r)f=f(r) then it is constant along n−1n-1 spheres, and so its restriction to the unit sphere satisfies ΔSn−1f=0\Delta_{\mathbb{S}^{n-1}}f =0. But if this term is zero, then the entire expression for the Laplacian collapses to

Δf=1k(r)n−1∂∂r(k(r)n−1∂f∂r)=0\Delta f = \frac{1}{k(r)^{n-1}}\frac{\partial}{\partial r}\left(k(r)^{n-1}\frac{\partial f}{\partial r}\right)=0

Cancelling the prefactor of 1/kn−11/k^{n-1} we see

∂∂r(k(r)n−1∂f∂r)=0\frac{\partial}{\partial r}\left(k(r)^{n-1}\frac{\partial f}{\partial r}\right)=0

and as this derivative is identically zero, the quantity differentiated must be constant. That is, for some C∈RC\in\mathbb{R} we have

k(r)n−1∂f∂r=Ck(r)^{n-1}\frac{\partial f}{\partial r}=C

And dividing through by kn−1k^{n-1} produces a differential equation for f(r)f(r), which upon integrating yields our formula

f(r)=∫Ck(r)n−1 drf(r)=\int \frac{C}{k(r)^{n-1}}\,dr

Our goal is to promote a radial solution to a fundamental solution, which means finding the correct multiplying factor CC. In practice, the easiest way for me to do this has been to find a function with suitably normalized derivative and then scale by the volume of the unit n−1n-1 sphere.

Radial to Fundamental

If F~(r)\tilde{F}(r) any radial solution to ΔF~=0\Delta \tilde{F}=0, then F(r)=1vol(SX(1))F~(r)F~′(1)F(r)=\frac{1}{\mathrm{vol}(\mathbb{S}_X(1))} \frac{\tilde{F}(r)}{\tilde{F}^\prime(1)} is a fundamental solution to the laplacian, where SX(1)\mathbb{S}_X(1) is the sphere of radius 11 in XX centered at r=0r=0.

Let XX be one of the constant curvature geometries, and dVdV be its Riemannian volume form. If ΔF\Delta F is to act like the delta distribution under integration, then for any region Ω⊂X\Omega\subset X we must have

∫ΩΔFdvol={0O∉Ω1O∈Ω\int_\Omega \Delta F d\mathrm{vol}=\begin{cases}0 & O\not\in \Omega\\ 1 & O\in\Omega \end{cases}

Where OO is the point δ\delta is based at (r=0r=0 in our coordinate system). On any region not containing OO this is trivial as our radial solution F~\tilde{F} has constantly zero Laplacian everywhere it is defined. So, we only need to concern ourselves with regions containing OO. For specificity, we can consider the unit ball Ω=BX(1)⊂X\Omega=B_X(1)\subset X. Here we compute using the divergence theorem and set the result equal to 1 (since O∈ΩO\in \Omega):

∫BX(1)ΔF dV=∫BX(1)div⋅gradF dV=∫∂BX(1)⟨gradF,n^⟩dS=∫SX(1)⟨gradF,∂r⟩dS=∫SX(1)∂F∂r dS\begin{align} \int_{B_X(1)}\Delta F\,dV&=\int_{B_X(1)}\mathrm{div}\cdot\mathrm{grad} F\, dV\\ &=\int_{\partial B_X(1)}\langle \mathrm{grad}F,\hat{n}\rangle dS\\ &=\int_{\mathbb{S}_X(1)}\langle \mathrm{grad}F,\partial_r\rangle dS\\ &=\int_{\mathbb{S}_X(1)} \frac{\partial F}{\partial r}\,dS \end{align}

Where dSdS is the Riemannian n−1n-1 volume form induced on the sphere SX(1)\mathbb{S}_X(1) by the overall volume dvold\mathrm{vol}. Since FF (and hence F′F^\prime) are functions of rr alone they are constant on the sphere SX(1)\mathbb{S}_X(1), and restricting to the sphere is the same as plugging in r=1r=1. This allows the one term above to be pulled out of the integral, giving

=∂F∂r(1)∫SX(1)dS=∂F∂r(1)vol(SX(1))=\frac{\partial F}{\partial r}(1)\int_{\mathbb{S}_X(1)}dS=\frac{\partial F}{\partial r}(1)\mathrm{vol}(\mathbb{S}_X(1))

This quantity is supposed to be 11. If it is not, we can simply divide the current definition of FF by it, and notice this constant multiple carries all the way through the calculation unscathed.

In fact, we can even do a little better than this since we know the functions k(r)k(r) that we are integrating. By the fundamental theorem of calculus

F′(r)=∂∂rF(r)=∂∂r∫1k(r)n−1 dr=1k(r)n−1F^\prime(r)=\frac{\partial}{\partial r}F(r)=\frac{\partial}{\partial r}\int\frac{1}{k(r)^{n-1}}\,dr =\frac{1}{k(r)^{n-1}}

Thus F′(1)F^\prime(1) is none other than 1/k(1)n−11/k(1)^{n-1}, and we may as well incorporate this directly:

If XX is a constant curvature geometry, then a fundamental solution FF to the laplacian is given by F(r)=k(1)n−1vol(SX(1))∫drk(r)n−1F(r)=\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\int\frac{dr}{k(r)^{n-1}}

FF is a scalar multiple of ∫drk(r)n−1\int\frac{dr}{k(r)^{n-1}} which we know to be harmonic on X∖OX\smallsetminus O. And, as we saw before integrating the radial derivative over the unit sphere gives

∫SX(1)∂F∂rdS=F′(1)∫SX(1)dS=F′(1)vol(S1(X))\begin{align} \int_{\mathbb{S}_X(1)}\frac{\partial F}{\partial r}dS &= F^\prime(1)\int_{\mathbb{S}_X(1)}dS\\ &=F^\prime(1) \mathrm{vol}(\mathbb{S}_1(X))\\ \end{align}

But now computing F′(1)F^\prime(1) we see we have correctly arranged things:

∂∂r∣r=1F(r)=∂∂r∣r=1k(1)n−1vol(SX(1))∫drk(r)n−1=k(1)n−1vol(SX(1))∂∂r∣r=1∫drk(r)n−1=k(1)n−1vol(SX(1))1k(1)n−1=1vol(SX(1))\begin{align} \frac{\partial}{\partial r}\Big|_{r=1}F(r)&=\frac{\partial}{\partial r}\Big|_{r=1}\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\int\frac{dr}{k(r)^{n-1}}\\ &=\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\frac{\partial}{\partial r}\Big|_{r=1}\int\frac{dr}{k(r)^{n-1}}\\ &=\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\frac{1}{k(1)^{n-1}}\\ &=\frac{1}{\mathrm{vol}(\mathbb{S}_X(1))} \end{align}

Plugging this back into the original yields

∫SX(1)∂F∂rdS=F′(1)vol(S1(X))=1vol(SX(1))vol(S1(X))=1\begin{align} \int_{\mathbb{S}_X(1)}\frac{\partial F}{\partial r}dS&=F^\prime(1) \mathrm{vol}(\mathbb{S}_1(X))\\ &=\frac{1}{\mathrm{vol}(\mathbb{S}_X(1))}\mathrm{vol}(\mathbb{S}_1(X))\\ &=1 \end{align}

But its even a little better than this! Because we are in the very special situation of constant curvature, we can actually compute the “volume” (ie surface area) of these spheres directly. For each nn let ωn\omega_n be the volume of the unit nn-sphere in En\mathbb{E}^n. Then for all radii RR

En:volSE(R)=ωn−1Rn−1=ωn−1k(r)n−1\mathbb{E}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{E}}(R)}=\omega_{n-1} R^{n-1}=\omega_{n-1} k(r)^{n-1} Sn:volSS(R)=ωn−1sin⁡(R)n−1=ωn−1k(r)n−1\mathbb{S}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{S}}(R)}=\omega_{n-1} \sin(R)^{n-1}=\omega_{n-1} k(r)^{n-1} Hn:volSH(R)=ωn−1sinh⁡(R)n−1=ωn−1k(r)n−1\mathbb{H}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{H}}(R)}=\omega_{n-1} \sinh(R)^{n-1}=\omega_{n-1} k(r)^{n-1}

Thus uniform across all geometries we have vol(SX(1))=ωn−1k(1)n−1\mathrm{vol}(\mathbb{S}_X(1))=\omega_{n-1}k(1)^{n-1} and plugging this into our corollary we see that the factors of k(1)n−1k(1)^{n-1} always cancel out. This gives our main theorem:

Laplacian in Constant Curvature

If XnX^n is an nn-dimensional constant curvature geometry, O∈XO\in X a point and r=dist(−,O)r=\mathrm{dist}(-,O) is the Riemannian distance from OO, then the fundamental solution for the Laplacian based at OO is F(r)=1ωn−1∫drk(r)n−1F(r)=\frac{1}{\omega_{n-1}}\int\frac{dr}{k(r)^{n-1}} For ωn−1\omega_{n-1} the volume of the unit n−1n-1 sphere, and k(r)k(r) is equal to sin⁡(r)\sin(r), rr or sinh⁡(r)\sinh(r) depending on if XX is spherical, Euclidean or hyperbolic respectively.

Euclidean Space

Here are the explicit solutions for Euclidean spaces of low dimension, with k(r)=rk(r)=r.

2D

F(r)=12πlog⁡rF(r)=\frac{1}{2\pi}\log r

First we compute the integral ∫drk(r)=∫1r  dr=log⁡(r)\int\frac{dr}{k(r)}=\int \frac{1}{r}\;dr=\log(r) And then the normalizing factor: the “volume” of 1 dimensional sphere is just the cirumference of the unit circle, or 2π2\pi. Thus F(r)=12πlog⁡(r)F(r)=\frac{1}{2\pi}\log(r)

3D

F(r)=−14π1rF(r)=\frac{-1}{4\pi}\frac{1}{r}

We first compute the integral to find a radial solution: ∫drk(r)2=∫1r2 dr=−1r\int\frac{dr}{k(r)^2}=\int \frac{1}{r^2}\,dr=\frac{-1}{r} And then normalize by 4π4\pi, the surface area of the unit sphere F(r)=−14π1rF(r)=\frac{-1}{4\pi}\frac{1}{r}

The same pattern continues in higher dimensions,

n-Dimensions

Spherical Space

Here are the explicit solutions for spherical spaces of low dimension, with k(r)=sin⁡(r)k(r)=\sin(r).

2D

F(r)=12πlog⁡∣tan⁡r2∣F(r)= \frac{1}{2\pi}\log\left|\tan\frac{r}{2}\right|

To find a radial solution we must compute ∫drsin⁡r=∫1sin⁡(r) dr=∫csc⁡(r) dr\int\frac{dr}{\sin r}=\int\frac{1}{\sin(r)}\,dr =\int\csc(r)\,dr This is one of those annoying integrals you learn in calculus and then forget, because deriving it needs a trick. Anyway, ∫csc⁡(r) dr=ln⁡∣csc⁡(r)−cot⁡(r)∣\int\csc(r)\,dr = \ln |\csc(r)-\cot(r)| There are many trig identites one could use to “simplify” here: for example this is equal to −ln⁡∣csc⁡r+cot⁡r∣-\ln|\csc r+\cot r| (which I didn’t know until wolfram alpha told me so). For us, its nicest to simplify as csc⁡r−cot⁡r=1sin⁡r−cos⁡rsin⁡r=1−cos⁡rsin⁡r=tan⁡r2\csc r-\cot r = \frac{1}{\sin r}-\frac{\cos r}{\sin r}=\frac{1-\cos r}{\sin r}=\tan\frac{r}{2} where the last equality is the tangent half angle formula.

We are in 2 dimensions so the normalizing factor is the unit 1-sphere’s volume or 2π2\pi: F(r)=12πlog⁡∣tan⁡r2∣F(r)=\frac{1}{2\pi}\log\left|\tan\frac{r}{2}\right|

3D

F(r)=14π1tan⁡(r)F(r)=\frac{1}{4\pi}\frac{1}{\tan(r)}

Computing a radial solution to start, we recognize an easier integral from calculus involving the cotangent ∫drk(r)2=∫1sin⁡2(r)  dr=∫csc⁡2(r) dr=−cot⁡(r)=−1tan⁡(r)\int\frac{dr}{k(r)^2}=\int\frac{1}{\sin^2(r)}\;dr =\int\csc^2(r)\, dr = -\cot(r)=\frac{-1}{\tan(r)} For the normalizing constant in dimension 3 we look to the unit 2-sphere, whose area is 4π4\pi. F(r)=−14π1tan⁡rF(r)=\frac{-1}{4\pi}\frac{1}{\tan r}

Unlike in Euclidean space the calculations don’t stabilize in difficulty beyond n=3n=3, and instead get more challenging as powers of kk grow in the integrand.

4D

F(r)=−14π2(cos⁡rsin⁡2r−log⁡∣tan⁡r2∣) F(r)=\frac{-1}{4\pi ^2}\left(\frac{\cos r}{\sin^2 r}-\log\left|\tan\frac{r}{2}\right|\right)

The radial solution is ∫drk(r)3=∫1sin⁡3(r)  dr=∫csc⁡3(r) dr\int\frac{dr}{k(r)^3}=\int\frac{1}{\sin^3(r)}\;dr =\int\csc^3(r)\, dr This is an integral you might see in calculus if you’re looking for really tricky examples at the end of the chapter (perhaps the more commonly seen case is its sibling sec⁡3(x)\sec^3(x)):

∫csc⁡3(r) dr=−csc⁡rcot⁡r+log⁡∣csc⁡r−cot⁡r∣2\int\csc^3(r)\,dr = \frac{-\csc r\cot r+\log|\csc r-\cot r|}{2}

We can simplify this a bit by (i) factoring out the constant −1/2-1/2, (ii) expanding cosecant and cotangent in terms of sines and cosines, and treating the last term with the half-angle identity, as we previously did. =−12(cos⁡rsin⁡2r−log⁡∣tan⁡r2∣)=\frac{-1}{2}\left(\frac{\cos r}{\sin^2 r}-\log\left|\tan\frac{r}{2}\right|\right)

Here the normalizing constant is the surface area of the unit 3-sphere. We can find this if we know the volume of 4−4-balls grows as 12π2R4\frac{1}{2}\pi^2 R^4 by differentiation, 2π2R32\pi^2 R^3, so the unit sphere has volume 2π22\pi^2. Dividing by this gives the answer

−14π2(cos⁡rsin⁡2r−log⁡∣tan⁡r2∣)\frac{-1}{4\pi ^2}\left(\frac{\cos r}{\sin^2 r}-\log\left|\tan\frac{r}{2}\right|\right)

In higher dimensions, we continue having to confront the integral of 1/sin⁡N(r)1/\sin^N(r) for various NN. Such integrals are doable (though not fun), with the tricks basically sorting by even and odd dimensions.

Hyperbolic Space

Here are the explicit solutions for hyperbolic spaces of low dimension, with k(r)=sinh⁡(r)k(r)=\sinh(r).

2D

F(r)=12πlog⁡∣tan⁡r2∣F(r)=\frac{1}{2\pi}\log\left|\tan\frac{r}{2}\right|

Very analogous to the spherical case, we have the integral ∫drk(r)=∫1sinh⁡r dr=∫csch⁡r dr\int\frac{dr}{k(r)}= \int\frac{1}{\sinh r}\,dr = \int \operatorname{csch}r\,dr This integral can be accomplished by substitution and partial fractions unpacking the definition of sinh⁡\sinh in terms of the exponential…or just remembered in analogy with the spherical trigonometric case: =log⁡∣csch⁡r−coth⁡r∣=\log\left|\operatorname{csch}r - \operatorname{coth}r\right| Writing in terms of sinh⁡\sinh and cosh⁡\cosh lets us recognize the hyperbolic tangent half angle identity: coth⁡r−csch⁡=cosh⁡r−1sinh⁡r=tanh⁡r2\operatorname{coth}r - \operatorname{csch}=\frac{\cosh r-1}{\sinh r}=\tanh\frac{r}{2} Where we have switched the order of the terms from the previous lines, as they occur inside an absolute value so it does not affect anything. Finally dividing by the normalizing constant of 2D, or 2π2\pi yields the answer F(r)=12πlog⁡∣tan⁡r2∣F(r)=\frac{1}{2\pi}\log\left|\tan\frac{r}{2}\right|

3D

F(r)=−14π1tanh⁡rF(r)= \frac{-1}{4\pi}\frac{1}{\tanh r}

The radial solution is calculable as ∫drk(r)2=∫1sinh⁡2r dr=∫csch⁡2r dr=−coth⁡r=−1tanh⁡r\int\frac{dr}{k(r)^2}=\int\frac{1}{\sinh^2 r}\,dr =\int\operatorname{csch}^2r \, dr=-\coth r=\frac{-1}{\tanh r} The normalizing constant in three dimensions is 4π4\pi, so F(r)=−14π1tanh⁡rF(r)=\frac{-1}{4\pi}\frac{1}{\tanh r}

4D

Higher Dimensions:

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