Note: Laplacian Constant Curvature
Finding Fundamental Solutions in Euclidean, Spherical and Hyperbolic Geometry
Steve Trettel
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Here’s a quick computation of the fundamental solution to the laplacian $\Delta f = \delta$ 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.
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 $r$ is the geodesic distance from the chosen origin.
Two Dimensions:
- $\mathbb{E}^2$: $F(r)=\frac{1}{2\pi}\log(r)$
- $\mathbb{S}^2$: $F(r)=\frac{1}{2\pi} \log\left|\tan\frac{r}{2}\right|$
- $\mathbb{H}^2$: $F(r)=\frac{1}{2\pi} \log \left|\tanh\frac{r}{2}\right|$
Three Dimensions:
- $\mathbb{E}^3$: $F(r)=\frac{-1}{4\pi}\frac{1}{r}$
- $\mathbb{S}^3$: $F(r)=\frac{-1}{4\pi} \frac{1}{\tan r}$
- $\mathbb{H}^3$: $F(r)=\frac{-1}{4\pi} \frac{1}{\tanh r}$
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 $n-1$ spheres, the level sets of constant $r$.
$$\Delta_{\mathbb{E}^n} f = \frac{1}{r^{n-1}}\frac{\partial}{\partial r}\left(r^{n-1}\frac{\partial f}{\partial r}\right)+\frac{1}{r^2}\Delta_{\mathbb{S}^{n-1}}$$
$$\Delta_{\mathbb{H}^n} f = \frac{1}{\sinh^{n-1}(r)}\frac{\partial}{\partial r}\left(\sinh^{n-1}(r)\frac{\partial f}{\partial r}\right)+\frac{1}{\sinh^2(r)}\Delta_{\mathbb{S}^{n-1}}$$
$$\Delta_{\mathbb{S}^n} f = \frac{1}{\sin^{n-1}(r)}\frac{\partial}{\partial r}\left(\sin^{n-1}(r)\frac{\partial f}{\partial r}\right)+\frac{1}{\sin^2(r)}\Delta_{\mathbb{S}^{n-1}}$$
Looking at these coordinate expressions we take advantage of a certain similarity - they are all the same up to a choice of $\sin(r)$, $r$ or $\sinh(r)$. Thus, we can give a uniform expression abstracting these differences into a function $k(r)$:
$$\Delta f = \frac{1}{k(r)^{n-1}}\frac{\partial}{\partial r}\left(k(r)^{n-1}\frac{\partial f}{\partial r}\right)+\frac{1}{k(r)^2}\Delta_{\mathbb{S}^{n-1}}$$
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 $r$. In this case things simplify drastically
If $f=f(r)$ then it is constant along $n-1$ spheres, and so its restriction to the unit sphere satisfies $\Delta_{\mathbb{S}^{n-1}}f =0$. But if this term is zero, then the entire expression for the Laplacian collapses to
$$\Delta f = \frac{1}{k(r)^{n-1}}\frac{\partial}{\partial r}\left(k(r)^{n-1}\frac{\partial f}{\partial r}\right)=0$$
Cancelling the prefactor of $1/k^{n-1}$ we see
$$\frac{\partial}{\partial r}\left(k(r)^{n-1}\frac{\partial f}{\partial r}\right)=0$$
and as this derivative is identically zero, the quantity differentiated must be constant. That is, for some $C\in\mathbb{R}$ we have
$$k(r)^{n-1}\frac{\partial f}{\partial r}=C$$
And dividing through by $k^{n-1}$ produces a differential equation for $f(r)$, which upon integrating yields our formula
$$f(r)=\int \frac{C}{k(r)^{n-1}},dr$$
Our goal is to promote a radial solution to a fundamental solution, which means finding the correct multiplying factor $C$. 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 $n-1$ sphere.
Let $X$ be one of the constant curvature geometries, and $dV$ be its Riemannian volume form. If $\Delta F$ is to act like the delta distribution under integration, then for any region $\Omega\subset X$ we must have
$$\int_\Omega \Delta F d\mathrm{vol}=\begin{cases}0 & O\not\in \Omega\
1 & O\in\Omega
\end{cases}$$
Where $O$ is the point $\delta$ is based at ($r=0$ in our coordinate system). On any region not containing $O$ this is trivial as our radial solution $\tilde{F}$ has constantly zero Laplacian everywhere it is defined. So, we only need to concern ourselves with regions containing $O$. For specificity, we can consider the unit ball $\Omega=B_X(1)\subset X$. Here we compute using the divergence theorem and set the result equal to 1 (since $O\in \Omega$):
$$
\begin{align}
\int_{B_X(1)}\Delta F,dV&=\int_{B_X(1)}\mathrm{div}\cdot\mathrm{grad} F, dV\
&=\int_{\partial B_X(1)}\langle \mathrm{grad}F,\hat{n}\rangle dS\
&=\int_{\mathbb{S}_X(1)}\langle \mathrm{grad}F,\partial_r\rangle dS\
&=\int_{\mathbb{S}_X(1)} \frac{\partial F}{\partial r},dS
\end{align}
$$
Where $dS$ is the Riemannian $n-1$ volume form induced on the sphere $\mathbb{S}_X(1)$ by the overall volume $d\mathrm{vol}$. Since $F$ (and hence $F^\prime$) are functions of $r$ alone they are constant on the sphere $\mathbb{S}_X(1)$, and restricting to the sphere is the same as plugging in $r=1$. This allows the one term above to be pulled out of the integral, giving
$$=\frac{\partial F}{\partial r}(1)\int_{\mathbb{S}_X(1)}dS=\frac{\partial F}{\partial r}(1)\mathrm{vol}(\mathbb{S}_X(1))$$
This quantity is supposed to be $1$. If it is not, we can simply divide the current definition of $F$ 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 $k(r)$ that we are integrating. By the fundamental theorem of calculus
$$F^\prime(r)=\frac{\partial}{\partial r}F(r)=\frac{\partial}{\partial r}\int\frac{1}{k(r)^{n-1}},dr =\frac{1}{k(r)^{n-1}}$$
Thus $F^\prime(1)$ is none other than $1/k(1)^{n-1}$, and we may as well incorporate this directly:
$F$ is a scalar multiple of $\int\frac{dr}{k(r)^{n-1}}$ which we know to be harmonic on $X\smallsetminus O$. And, as we saw before integrating the radial derivative over the unit sphere gives
$$\begin{align}
\int_{\mathbb{S}_X(1)}\frac{\partial F}{\partial r}dS &= F^\prime(1)\int_{\mathbb{S}_X(1)}dS\
&=F^\prime(1) \mathrm{vol}(\mathbb{S}_1(X))\
\end{align}$$
But now computing $F^\prime(1)$ we see we have correctly arranged things:
$$
\begin{align}
\frac{\partial}{\partial r}\Big|_{r=1}F(r)&=\frac{\partial}{\partial r}\Big|_{r=1}\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\int\frac{dr}{k(r)^{n-1}},dr\
&=\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\frac{\partial}{\partial r}\Big|_{r=1}\int\frac{dr}{k(r)^{n-1}},dr\
&=\frac{k(1)^{n-1}}{\mathrm{vol}(\mathbb{S}_X(1))}\frac{1}{k(1)^{n-1}}\
&=\frac{1}{\mathrm{vol{\mathbb{S}_X(1)}}}
\end{align}
$$
Plugging this back into the original yields
$$\begin{align}
\int_{\mathbb{S}_X(1)}\frac{\partial F}{\partial r}dS&=F^\prime(1) \mathrm{vol}(\mathbb{S}_1(X))\
&=\frac{1}{\mathrm{vol{\mathbb{S}_X(1)}}}\mathrm{vol}(\mathbb{S}_1(X))\
&=1
\end{align}$$
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 $n$ let $\omega_n$ be the volume of the unit $n$-sphere in $\mathbb{E}^n$. Then for all radii $R$
$$\mathbb{E}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}{\mathbb{E}}(R)}=\omega{n-1} R^{n-1}=\omega_{n-1} k(r)^{n-1}$$ $$\mathbb{S}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{S}}(R)}=\omega_{n-1} \sin(R)^{n-1}=\omega_{n-1} k(r)^{n-1}$$ $$\mathbb{H}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{H}}(R)}=\omega_{n-1} \sinh(R)^{n-1}=\omega_{n-1} k(r)^{n-1}$$
Thus uniform across all geometries we have $\mathrm{vol}(\mathbb{S}X(1))=\omega{n-1}k(1)^{n-1}$ and plugging this into our corollary we see that the factors of $k(1)^{n-1}$ always cancel out. This gives our main theorem:
Euclidean Space
Here are the explicit solutions for Euclidean spaces of low dimension, with $k(r)=r$.
The same pattern continues in higher dimensions,
Spherical Space
Here are the explicit solutions for spherical spaces of low dimension, with $k(r)=\sin(r)$.
To find a radial solution we must compute $$\int\frac{dr}{\sin r}=\int\frac{1}{\sin(r)},dr =\int\csc(r),dr$$ This is one of those annoying integrals you learn in calculus and then forget, because deriving it needs a trick. Anyway, $$\int\csc(r),dr = \ln |\csc(r)-\cot(r)|$$ There are many trig identites one could use to “simplify” here: for example this is equal to $-\ln|\csc r+\cot r|$ (which I didn’t know until wolfram alpha told me so). For us, its nicest to simplify as $$\csc r-\cot r = \frac{1}{\sin r}-\frac{\cos r}{\sin r}=\frac{1-\cos r}{\sin r}=\tan\frac{r}{2}$$ 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 $2\pi$: $$F(r)=\frac{1}{2\pi}\log\left|\tan\frac{r}{2}\right|$$
Unlike in Euclidean space the calculations don’t stabilize in difficulty beyond $n=3$, and instead get more challenging as powers of $k$ grow in the integrand.
The radial solution is $$\int\frac{dr}{k(r)^3}=\int\frac{1}{\sin^3(r)};dr =\int\csc^3(r), dr$$ 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 $\sec^3(x)$):
$$\int\csc^3(r),dr = \frac{-\csc r\cot r+\log|\csc r-\cot r|}{2}$$
We can simplify this a bit by (i) factoring out the constant $-1/2$, (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. $$=\frac{-1}{2}\left(\frac{\cos r}{\sin^2 r}-\log\left|\tan\frac{r}{2}\right|\right)$$
Here the normalizing constant is the surface area of the unit 3-sphere. We can find this if we know the volume of $4-$balls grows as $\frac{1}{2}\pi^2 R^4$ by differentiation, $2\pi^2 R^3$, so the unit sphere has volume $2\pi^2$. Dividing by this gives the answer
$$\frac{-1}{4\pi ^2}\left(\frac{\cos r}{\sin^2 r}-\log\left|\tan\frac{r}{2}\right|\right)$$
In higher dimensions, we continue having to confront the integral of $1/\sin^N(r)$ for various $N$. 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 $k(r)=\sinh(r)$.
Higher Dimensions: