Note: Laplacian Preharmonic

Steve Trettel

|

$$\newcommand{\RR}{\mathbb{R}}$$

Finding fundamental solutions for the Laplace operator on Riemannian manifolds can be challenging. In the case of spherical symmetry things work out particularly nicely, as we are able to use the radial function on the manifold to reduce our problem to a 1-dimensional ODE.

Here’s a possible means of generalizing this trick: first find a function that has the same level sets as a fundamental solution, and then rescale that function get an actual harmonic function (by solving a 1-variable ODE). I’ll give this a bit of terminology:

(Preharmonic Function) Let $(M,g)$ be a Riemannian manifold and $\Delta$ be its laplacian. Then a real valued function $F\colon M\to\RR$ is called pre-harmonic if there is a function $f\colon\RR\to\RR$ such that the composition $f\circ F$ is harmonic.

This note proves two things: first, we give a characterization of preharmonic functions which makes it simple to detect them

(Preharmonic Detector) A function $F\colon M\to\RR$ is preharmonic if and only if $\Delta F/|\nabla F|^2$ is constant on the level sets of $F$. Equivalently, if $$|\nabla F|^2, d(\Delta F, dF) = \Delta f, d(|\nabla F|^2, dF)$$

and second, if $F$ is preharmonic, we give an explicit procedure for upgrading it to a harmonic function.

(Upgrading Preharmonics) Let $F\colon M\to\RR$ be a preharmonic function, so $\frac{\Delta F}{|\nabla F|^2}=h\circ F$ for some $h\colon\RR\to\RR$ and define $f\colon\RR\to\RR$ as $$f(x)=\int e^{-\int h(x) dx}dx$$ Then $f\circ F$ is harmonic.

I don’t know if this tool will end up proving useful for anything, but wanted to record it nonetheless.

Detecting Preharmonics

(Detecting Preharmonics) A function $F\colon M\to\RR$ is preharmonic if and only if $\Delta F/|\nabla F|^2$ is constant on the level sets of $F$.

If two functions $M\to\RR$ share the same level sets, then we can factor one through the other by a function $\RR\to\RR$ which just ‘relables’ the level sets with their new values. In our case, $\Delta F/|\nabla F|^2$ having the same level sets as $F$ means that we may write $$\frac{\Delta F}{|\nabla F|^2}=h\circ F$$ for some such function $h\colon\RR\to\RR$. With this in mind, we prove the equivalence in two steps, doing one direction at a time. The proof will require we recall the chain rule for the Laplacian, in order to differentiate a composition $f\circ F$ of functions $f\colon\RR\to\RR$ and $F\colon M\to\RR$.

$$\Delta (f\circ F)= f^{\prime\prime}(F)|\nabla F|^2 + g^\prime(F)\Delta F$$

First, we show this condition is sufficient:

Assume that $F\colon M\to\RR$ is a function where $\Delta F/|\nabla F|^2=h\circ E$ for a function $h$ of a real variable. Let $f\colon\RR\to\RR$ a smooth function, and using the Laplacian chain rule we compute $$\Delta(f\circ F)=f^{\prime\prime}(F)|\nabla F|^2 + f^\prime(F)\Delta F$$ Setting this to zero gives a constraint on $g$ implying the composition is harmonic: $$f^{\prime\prime}(F)|\nabla F|^2 + f^\prime(F)\Delta F=0$$ $$\implies f^{\prime\prime}(F) + f^\prime(F)\frac{\Delta F}{|\nabla F|^2}=0$$ $$\implies f^{\prime\prime}(F) + f^\prime(F)h(F)=0$$ While this is an equation relating functions on $M$, all functions involved are composed with $F\colon M\to\RR$. Thus we may rewrite the left side as $$(f^{\prime\prime}-f^\prime h)\circ F =0$$ The only way for this composition to be constantly equal to zero is if $f^{\prime\prime}-f^\prime h=0$ for every real number that comes out of $F$. But this is just a linear second order differential equation for $f$ on the real line, which has solutions by existence and uniqueness theorems. Any such solution provides a harmonic function $f\circ F$, proving the theorem.

Now we show its in fact necessary:

The proof follows the same logic. Now assume $F$ is preharmonic, so there exists some $f$ with $\Delta(f\circ F)=0$. Using the chain rule $$f^{\prime\prime}(F)|\nabla F|^2 + f^\prime(F)\Delta F=0$$ We can re-group terms moving everything involving $f$ to one side, and the rest to the other: $$\frac{f^{\prime\prime}(F)}{f^\prime(F)}=\frac{\Delta F}{|\nabla F|^2}$$ Now the left hand side of this equation is explicitly a function of $F$ alone: its the function $f^{\prime\prime}{f^\prime}\colon\RR\to\RR$ composed with $F\colon M\to\RR$. So, it is constant on the level sets of $F$. But its equal to the right side, so that must also be constant on the level sets of $F$, completing the proof.

Upgrading Preharmonics

The previous theorem ended with a condition on $g$ determined by an ODE to which we appealed to existence & uniqueness theorems. But in fact this equation can be solved by hand, giving us a means of actually upgrading $F$ to a harmonic function

(Upgrading Preharmonics) Assume $F\colon M\to\RR$ satisfies $$\frac{\Delta F}{|\nabla F|^2}=h\circ F$$ for some $h\colon\RR\to\RR$, and define $f\colon\RR\to\RR$ as $$f(x)=\int e^{-\int h(x) dx}dx$$ Then $f\circ F$ is harmonic.
As seen in the previous proofs using the Laplacian chain rule we know $f$ and $h$ satisfy the equation $$f^{\prime\prime}(x)+h(x)f^\prime(x)=0$$ Re-arranging, $$\frac{f^{\prime\prime}(x)}{f^\prime(x)}=-h(x)$$ we recognize the left hand side as a logarithmic derivative $$\left(\log f^\prime(x)\right)^\prime = -h(x)$$ Integrating and then exponentiating gives $f^\prime$: $$\log f^\prime = -\int h dx,,\implies f^\prime = e^{-\int h ,dx}$$ Integrating one more time yields $g$: $$f(x)=\int e^{-\int h,dx},dx$$ (Note the integrals here are general antiderivatives, so each allows an additive constant)

A Better Detector

We have a robust means of actually upgrading preharmonic functions to actual solutions, but our means to detect preharmonics is a bit difficult to use. Its not an equation we can attempt to solve, but rather a condition on a candidate function $E$: given a ‘guess’ we ‘check’ by computing $\Delta F / |\nabla F|^2$ and see if the result depends only on $F$.

To improve upon this we would like to find an equation that preharmonic functions $E$ satisfy. One first idea is to note that if $F$ and $\Delta F / |\nabla F|^2$ have the same level sets, then their differentials are linearly dependent and so the wedge of their differentials must vanish:

$$d\left(\frac{\Delta F}{|\nabla F|^2}\right)\wedge dF = 0$$

This is a bit of a complicated looking equation. To save me from having to do this algebra time and again, here’s the result of expanding this out:

$$ \begin{align} d\left(\frac{\Delta F}{|\nabla F|^2}\right)&= \frac{ |\nabla F|^2, d(\Delta F) - \Delta f, d(|\nabla F|^2)}{|\nabla F|^4} \end{align} $$

Wedging with $dF$ and setting equal to zero, we can clear the denominator which leaves

$$\left(|\nabla F|^2, d(\Delta F) - \Delta f, d(|\nabla F|^2)\right)\wedge dF=0$$

Finally, distributing and moving some terms to the other side gives a simpler equation where computing both sides for a proposed form of $F$ and equating components gives a system of differential equations the proposal must satisfy to be preharmonic.

$$|\nabla F|^2, d(\Delta F)\wedge dF = \Delta f, d(|\nabla F|^2)\wedge dF$$

This is usually the best form to calculate with, but for greater memorability one can factor out the differentials to

$$|\nabla F|^2, d(\Delta F,dF) = \Delta f, d(|\nabla F|^2,dF)$$

This works because (for example) $$d(\Delta FdF)=(d\Delta F)\wedge dF + \Delta f d^2F = (d\Delta F)\wedge dF +0$$

A Simplifying Special Case

The above was all general: holding for any preharmonic. But as our goal is to guess and checK one way to get a simpler equation is to ask for a stronger condition. Instead of asking for $\Delta f/|\nabla f|^2$ to have the same level sets as $F$, why don’t we ask that both $\Delta f$ and $|\nabla f|^2$ each do?

This instead imposes the two equations

$$( d\Delta f)\wedge dF = 0\hspace{1cm}(d|\nabla F|^2)\wedge dF =0$$

which are both less complicated, and give two rather than one constraints to work with.

◀ Note: Laplacian Discovering Preharmonics
Note: Laplacian Spherical Symmetry ▶