Fundamental Solutions of the Laplacian with Spherical Symmetry

The fundamental solution is the integral of reciprocal surface area.

$\newcommand{\Isom}{\operatorname{Isom}}$

This not 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 $O\in M$ , finding the fundamental solution for all of them.

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

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

The Laplacian Chain Rule: if $F\colon M\to\RR$ and $f\colon\RR\to\RR$ then $\Delta(f\circ F)=f^{\prime\prime} \|\nabla F\|^2+f^\prime\Delta F$ .
Laplacians and Isometries: if $\Phi\in\mathrm{Isom}(M)$ and $F\colon M\to\RR$ then $\Delta(F\circ \Phi)=(\Delta F)\circ\Phi$ .

Spherical Symmetry

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

Spherical Symmetry

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

This definition picks out a particular point $O$ in the manifold as special, so its natural to look at the distance function from $O$ :

Radial Function

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

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

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

The group orbits of $\Isom(\mathbb{S}^{n-1})$ coincide with the level sets of $r$ .

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

To get the reverse inclusion, let $p,q$ lie at the same distance $r$ from $O$ , and let $\gamma_p,\gamma_q$ be the geodesics starting from $O$ ending at each after time $r$ . Then the initial tangent vectors $\dot{\gamma_p},\dot{\gamma_q}$ are unit vectors in $T_OM$ , and the assumed $\mathrm{SO}(n)$ isometries of $M$ descend by differentiation to an action on $T_OM$ by isometries. Since $\mathrm{SO}(n)$ ‘s action on the unit sphere in $\RR^n$ is transitive, there is some $\Phi_\star$ with $\Phi_\star \dot{\gamma_p}=\dot{\gamma_q}$ . But as isometires preserve geodesics and geodesics are determined by their initial conditions, this implies that $\gamma_q = \Phi\circ\gamma_p$ and so $q=\gamma_q(r)=\Phi\gamma_p(r)=\Phi p$ Thus $p,q$ lie in the same group orbit.

Finding Fundamental Solutions

Here we carry out the explicit computations to find a fundamental solution $f\colon M\to \RR$ to $\Delta f=\delta_O$ . The radial function will play an important role so we start with a quick lemma computing its Laplacian.

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

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

Again by general theorems of Riemannian geometry, $\Delta r$ is orthogonal to the geodesic spheres centered at $O$ , so we can orthogonally decompose the volume form of $M$ into an $r$ component, and the volume form $\omega_r$ of the level set $S_r$ . Thus by definition we must have $\star dr = \omega_r$ . Because each of these level sets is a round sphere, we can write their volume forms uniformly in terms of the unit volume form $\omega$ of the $n-1$ -sphere, scaled by the area $A\circ r$ of the geodesic sphere in question. Putting this together, $\star dr = (A\circ r)\omega$ Now we differentiate with the product rule $d(\star dr)=d\left((A\circ r)\omega\right)=d(A\circ r)\wedge \omega + (A\circ r)d\omega$ Since $\omega$ is a closed form on the sphere, $d\omega =0$ and the first term simplifies as $d(A\circ r)=(A^\prime \circ r)dr$ so $d\star dr = (A^\prime \circ r) dr\wedge \omega$ Because this result is a top-dimensional form it must be some multiple of the volume form on $M$ , 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 $\mathrm{vol}=dr\wedge \omega_r$ , and $\omega_r=(A\circ r)\omega$ . Computing (and dropping all composition with $r$ from the notation for brevity) $dr\wedge\omega = dr\wedge \frac{A}{A}\omega=\frac{1}{A}\left(dr\wedge A\omega\right)=\frac{1}{A} dr\wedge \omega_r=\frac{1}{A}\mathrm{vol}$ Substituting this in, $d\star dr = (A^\prime\circ r) dr\wedge \omega = \frac{A^\prime\circ r}{A\circ r}\mathrm{vol}$ Now we apply the final hodge star, completing the proof $\star d\star d r = \star\left(\frac{A^\prime\circ r}{A\circ r}\mathrm{vol}\right)=\frac{A^\prime\circ r}{A\circ r}\star\mathrm{vol} =\frac{A^\prime\circ r}{A\circ r}$

Now we come to our main theorem:

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

Let $r$ be the radial distance function and $f\colon\RR\to\RR$ defined by $f(x)=\int_\ast^x \frac{dt}{A(t)}$ Then the composition $\varphi=f\circ r$ is harmonic on $M\smallsetminus O$ .

If $f\colon\RR\to\RR$ is any function such that $f\circ r$ is harmonic, the chain rule for the Laplacian implies $0=\Delta(f\circ r)=f^{\prime\prime}\|\nabla r\|^2+f^\prime \Delta r$ Since $r$ is a Riemannian distance $\|\nabla r\|=1$ and $\Delta r$ we calculated above to be the radial function $\Delta r = \frac{A^\prime(r)}{A(r)}$ for $A$ the surface area of geodesic spheres about $O$ . Thus, for every $p\in M$ we have $f^{\prime\prime}(r(p))+f^\prime(r(p))\frac{A^\prime(r(p))}{A(r(p))}=0$

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

0=A f^{\prime\prime}+A^\prime f^\prime= \left(A f^\prime\right)^\prime

Thus $Af^\prime$ is constant. This provides a differential equation for $f$ : $A(x)f^\prime(x) = C\,\implies f^\prime(x) = \frac{C}{A(x)}$ The case $C=1$ yields the function we are after by quadrature $f(x)=\int_\ast^x\frac{dt}{A(t)}$ Where $\ast\in\RR$ 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 $x$ as the dummy variable of integration. And, when the end goal is to compose with another function like $r$ , we may switch the dummy variable to this and write

$f\circ r\colon p\mapsto \int_\ast^{r(p)}\frac{dt}{A(t)}=\left(\int_\ast^{r(-)}\frac{dt}{A(t)}\right)(p)=\left(\int\frac{dr}{A(r)}\right)(p)$ $\implies\varphi:=\int\frac{dr}{A(r)}$

Next we see that integration of $\Delta \varphi$ on domains $\Omega\subset M$ behaves like the $\delta$ distribution. Note - all integrals over regions containing $O$ should be considered distributionally.

Let $\Omega\subset M$ be a domain for integration. Then

\int_\Omega \Delta \varphi \,\mathrm{vol} =\begin{cases} 0 & O\not \in \Omega\\ 1 & O\in \Omega \end{cases}

If $O\not\in\Omega$ , then $\varphi$ is well defined and smooth on all of $\Omega$ , and $\Delta \varphi = 0$ on $\Omega$ . Thus $\int_\Omega \Delta \varphi \,\mathrm{vol}=\int_\Omega 0\, \mathrm{vol}=0$ So we need only be concerned with domains where $O\in\Omega$ . First we see the value of the integral is independent of the particular domain. If $\Omega_1,\Omega_2$ are two domains containing $O$ in their interior, its possible to find a small geodesic ball $B$ about $O$ contained in both of them. Then we may write $\Omega_i = (\Omega_i\smallsetminus B)\cup B$ which gives a decomposition of the corresponding integrals $\int_{\Omega_i}\Delta \varphi\,\mathrm{vol}=\int_{\Omega_i\smallsetminus B}\Delta \varphi\,\mathrm{vol}+\int_{B}\Delta \varphi\,\mathrm{vol}$ But the first integral in this sum is over a domain not containing $O$ , so is zero by the previous argument. Thus both the integral over $\Omega_1$ and $\Omega_2$ equal the integral over $B$ , 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 $B$ about $O$ , and note that by the definition of the Hodge dual $\int_B \Delta \varphi\,\mathrm{vol}=\int_B \star \Delta \varphi$ and unpacking $\Delta \varphi$ allows some cancellation as $\star^2=\mathrm{id}$ on top-dimensional forms $\star \Delta \varphi = \star\star d\star d \varphi = d\star d \varphi$ Now that our form begins with exterior differentiation we can apply (a distributional version of) Stokes’ theorem: $\int_B d\star d \varphi=\int_{\partial B} \star d \varphi$ Since $B$ is a geodesic ball centered at $O$ , its boundary is a sphere: $\partial B=S$ To continue, we must bring in our definition of $\varphi =\int\frac{dr}{A(r)}$ .

$\varphi = f\circ r\hspace{0.5cm}\textrm{for}\hspace{0.5cm}f(x)=\int_\ast^x \frac{dt}{A(t)}$

Differentiating, $d\varphi = d(f\circ r)=(f^\prime\circ r) dr=\frac{1}{A\circ r}dr$ $\star d\varphi = \star\left(\frac{1}{A\circ r}dr\right)=\frac{1}{A\circ r}\star dr$

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

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

$\int_B \Delta \varphi\,\mathrm{vol}=\int_B d\star d\varphi=\int_{\partial B=S} \star d f =\int_S \omega = 1$

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 $(n-1)$ spheres of radius $r$ :

The surface area of $n$ spheres of radius $r$ in the $n+1$ -dimensional spaces of constant curvature are $\mathbb{E}^{n+1}\colon\, \omega_n r^n\hspace{1cm}\mathbb{S}^{n+1}\colon\,\omega_n\sin(r)^n\hspace{1cm}\mathbb{H}^{n+1}\colon\, \omega_n\sinh(r)^n$ Where $\omega_n$ is the size of the unit $n$ sphere of curvature 1: $\omega_1=2\pi\hspace{0.5cm}\omega_2= 4\pi\hspace{0.5cm}\omega_3= 2\pi^2\ldots$

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

The fundamental solutions to $\Delta f =\delta_O$ in constant curvature are $\mathbb{E}^{n}\colon\, f(r)=\frac{1}{\omega_{n-1}}\int\frac{dr}{r^{n-1}}$ $\mathbb{S}^{n+1}\colon\,\frac{1}{\omega_{n-1}}\int\frac{dr}{\sin(r)^{n-1}}$ $\mathbb{H}^{n+1}\colon\, \frac{1}{\omega_{n-1}}\int\frac{dr}{\sinh(r)^{n-1}}$

For Euclidean geometry, all of these integrals are easily computable by hand: $n=2$ results in a logarithm and the rest directly follow from the ‘power rule’ $f_{\mathbb{E}^2}(r)=\frac{1}{2\pi}\log(r)\hspace{1.5cm}f_{\mathbb{E}^n}=\frac{-1}{(n-2)\omega_{n-1}}\frac{1}{r^{n-2}}$

For hyperbolic and spherical geometry, we can compute dimension $2$ by hand

$f_{\mathbb{S}^2}(r)=\frac{1}{2\pi}\log \left|\tan\frac{r}{2}\right|\hspace{1cm}f_{\mathbb{H}^2}(r)=\frac{1}{2\pi}\log \left|\tanh\frac{r}{2}\right|$

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

The integrals of secant and its hyperbolic analog satisfy the following recurrences: $\int\csc^n x\,dx = \frac{-\cot x\csc^{n-2}x}{n-1}+\frac{n-2}{n-1}\int\csc^{n-2}x\,dx$ $\int\operatorname{csch}^nx\,dx =$

The General Case:

What if we are given a spherically symmetric metric in cartesian coordinates, where geometric quantities like the geodesic distance $r$ are not immediately apparent? Here’s we record the necessary computations. For brevity we will write $\ell=\sqrt{x^2+y^2+z^2}$ to be the ‘coordinate length’. Consider the metric $g = \sigma(\ell)^2g_{\mathbb{E}^n}(\cdot,\cdot)$ Where $\sigma\colon\RR\to\RR$ is some smooth function. Some computation yields the following characterization:

If $g = \sigma(\ell)^2ds^2_{\mathbb{E}^n}$ for $\ell=|(x,y,\ldots)|$ , then the fundamental solution to the Laplacian on $(\RR^n,g)$ is $f(\ell) = \frac{1}{\omega_{n-1}}\int\frac{d\ell}{\ell\left[\ell\sigma(\ell)\right]^{n-2}}$

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

$\rho(l)=\int_0^l \sigma(t)\,dt$

Then the radial function $r\colon\RR^n\to\RR$ is given by $\rho\circ \ell$ . We are going to need to be careful with bounds, so write $f(x)=\int_\ast^x \frac{dt}{A(t)}$ where $\ast\in\RR$ is an arbitrary real number (fixing the constant of integration). Then the fundamental solution to the Laplacian is $\varphi = f\circ r = f\circ\rho\circ\ell$ : $\varphi \colon q\mapsto \int_\ast^{r(q)}\frac{dt}{A(t)}=\int_\ast^{\rho(\ell(q))}\frac{dt}{A(t)}$

We can rewrite this integral using the substitution $t=\rho(l)$ and simplify using $\rho^\prime(l)=\sigma(l)$ from differentiating the previous integral:

$\int_\ast^{\rho(\ell(q))}\frac{dt}{A(t)}=\int_\ast^{\ell(q)}\frac{d[\rho(l)]}{A(\rho(l))}=\int_\ast^{\ell(q)}\frac{\sigma(l) dl}{A(\rho(l))}$

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

$A(\rho(l))=\sigma(l)^{n-1}A_{\mathbb{E}}=\omega_{n-1}l^{n-1}\sigma(l)^{n-1}$

Substituting this in,

$\varphi\colon q\mapsto \int_\ast^{\ell(q)}\frac{\sigma(l)dl}{\omega_{n-1}l^{n-1}\sigma(l)^{n-1}}=\frac{1}{\omega_{n-1}}\int_\ast^{\ell(q)}\frac{dl}{l^{n-1}\sigma(l)^{n-2}}$

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

$f(\ell) = \frac{1}{\omega_{n-1}}\int\frac{d\ell}{\ell\left[\ell\sigma(\ell)\right]^{n-2}}$

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 $\Delta r$ ). Totally unnecessary for our purposes here but I want to record it in case its of future use.

There exists a radial harmonic function on $M\smallsetminus O$ .

A radial harmonic function is some $\varphi\colon M\to\RR$ which satisfies $\Delta \varphi=0$ and factors $\varphi=f\circ r$ for some $f\colon\RR\to\RR$ . The chain rule for the Laplacian gives an equation for $f$ $0=\Delta(f\circ r)=f^{\prime\prime}\|\nabla r\|^2+f^\prime \Delta r$ Since $r$ is a Riemannian distance $\|\nabla r\|=1$ and this simplifies giving an equation that must hold for all $p\in M$ : $f^{\prime\prime}(r(p))+f^\prime(r(p)) \Delta r(p) = 0$ If $\Delta r$ is a radial function (so $\Delta r = h\circ r$ for some $h\colon\RR\to\RR$ ) then this becomes $f^{\prime\prime}(r(p))+f^\prime(r(p)) h(r(p)) = 0$ , which only holds on $M$ if for all $x$ in the range of $r$ we have $f^{\prime\prime}(x)+f^\prime(x)h(x)=0$ 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 $\Delta r$ is radial: equivalently that $\Delta r$ is constant on the level sets of $r$ .

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

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

Putting these together, $\Delta r = \Delta(r\circ \Phi)=(\Delta r)\circ \Phi$ , and so

$\Delta r(q)=[(\Delta r)\circ \Phi](p)=\Delta r(p)$

So $\Delta r$ is constant on the level sets of $r$ : its radial, as required.

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 $n-1$ independent reflections, so geodesics starting out radially are confined to them. ↩

Spherical Symmetry

Spherical Symmetry

Radial Function

Finding Fundamental Solutions

Constant Curvature Geometries

The General Case:

Appendix: Proving Existence without Computing

Footnotes