Note: Laplacian Spherical Symmetry
Finding Fundamental Solutions in metrics with spherical symmetry
Steve Trettel
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$$\newcommand{\RR}{\mathbb{R}}$$ $$\newcommand{\Isom}{\operatorname{Isom}}$$
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.
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 $S\to S$.
This definition picks out a particular point $O$ in the manifold as special, so its natural to look at the distance function from $O$:
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
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.
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 $O$.
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.
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 $r$:
This immediately gives integral forms of the fundamental solutions in each constant curvature geometry:
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 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:
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.
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 $$\Delta r(q)=\Delta r(\Phi(p))=(\Delta r)\circ \Phi$$
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
$$\Delta r(q)=(\Delta r)\circ \Phi=\Delta r(p)$$
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. ↩︎