Discovering Preharmonic Functions via Separable Ansatze

Separation of variables as a systematic tool for finding near-harmonic functions.

The goal of defining preharmonic functions and producing a means of upgrading them to actual harmonics is to simplify the problem of finding solutions to $\Delta \varphi=0$ on a manifold $(M,g)$. Inside the space of preharmonic functions, the actual harmonics have infinite codimension, so we’ve in some sense taken care of an infinite number of degrees of freedom. Unfortunately this still leaves infinitely many more, as the preharmonics are also infinite codimension in the smooth functions on $M$. The goal of this note is to illustrate one technique of finding preharmonic functions, by guessing a simple ansatz, and plug this into the preharmonic condition to get some equations to solve.

We illustrate this technique on the Euclidean plane, where we begin in cartesian coordinates to obscure the rotational symmetry. Our nice ansatze are that there is a preharmonic function that is additively separable (so $g(x,y)=a(x)+b(y)$) or multiplicatively separable (so $g(x,y)=a(x)b(y)$). Using the preharmonic condition, we find explicit solutions of both of these forms, and use them to derive the fundamental solution of the Laplacian. This is of note because the fundamental solution itself is neither additively nor multiplicatively separable. Thus, this technique gives strictly more power to the ansatz-guesser than trying to work with the Laplacian directly.

The Setup

Recall the general theory. A function $F\colon M\to\RR$ is preharmonic if and only if it has the same level sets as a true harmonic function. With some work this reduces to the following equation: $F$ is preharmonic if and only if $\|\nabla F\|^2\, d(\Delta F\, dF) = \Delta F\, d(\|\nabla F\|^2\, dF)$

Then, if $F$ is preharmonic function, write $\frac{\Delta F}{\|\nabla F\|^2}=h\circ F$ for some $h\colon\RR\to\RR$. Then $f\circ F$ is harmonic, for $f\colon\RR\to\RR$ below: $f(x)=\int e^{-\int h(x) dx}dx$

Euclidean Geometry

For this example we work in the Euclidean plane $\mathbb{E}^2$. We define coordinates by choosing a point $O\in\EE^2$ and two orthogonal geodesics $\mathrm{vert}$ and $\mathrm{horiz}$ through $O$. We then define the following two functions $\EE^2\to\RR$:

$X(-)=\mathrm{dist}(-,\mathrm{vert})\hspace{1cm}Y(-)=\mathrm{dist}(-,\mathrm{horiz})$

The differentials $dX,dY$ are unit length as $X,Y$ are distance functions. The level sets of these functions are equidistant curves to the vertical and horizontal geodesics, which themselves are orthogonal geodesics (since $\EE^2$ is flat). Thus, $\mathrm{vol}=dX\wedge dY$ and putting these facts together yields $\langle dX, dY\rangle = 0\hspace{1cm}|dX|=|dY|=1$ $\star dX = dY\hspace{1cm}\star dY=-dX$

From this its straightforward to compute the Laplacians of $X,Y$. We do $X$, the $Y$ case is analogous. $\Delta X = \star d\star d X = \star d(\star dX)=\star d(dY)=\star (d^2Y)=\star 0 = 0$

These facts together with the two-variable chain rule for the Laplacian let us compute $\Delta g(X,Y)$ for a function $g\colon\RR^2\to\RR$ (the cross term $2g_{12}\langle\nabla X,\nabla Y\rangle$ drops out because $dX$ and $dY$ are orthogonal):

$$\begin{align} \Delta g(X,Y)&=g_{11}(X,Y)\|dX\|^2+g_{22}(X,Y)|dY|^2+g_1(X,Y)\Delta X + g_2(X,Y)\Delta Y\\ &=g_{11}(X,Y)(1)+g_{22}(X,Y)(1)+g_1(X,Y)(0) + g_2(X,Y)(0)\\ &=g_{11}(X,Y)+g_{22}(X,Y) \end{align}$$

(Of course, we already know this that in Cartesian coordinates $\Delta = \partial_x^2+\partial_y^2$ but…)

Additively Separable

Here we propose to find a preharmonic function $g(X,Y)$ which is additively separable meaning there are two functions $a,b\colon\RR\to\RR$ such that $g(X,Y)=a(X)+b(Y)$ The preharmonic condition on $g(X,Y)$ requires we compute $\Delta g$ and $\|\nabla g\|^2=\|dg\|^2$ so we begin with those:

$$\begin{align} dg(X,Y)&=d\left(a(X)+b(Y)\right)\\ &=d(a(X))+d(b(Y))\\ &=a^\prime(X)dX+b^\prime(Y)dY \end{align}$$

Then as $dX,dY$ form an orthonormal basis at each point,

$$\begin{align} \|dg\|^2=\langle dg,dg\rangle &= (a^\prime)^2 \|dX\|^2+2 a^\prime b^\prime\langle dX,dY\rangle + (b^\prime)^2\|dY\|^2\\ &= (a^\prime)^2+(b^\prime)^2 \end{align}$$

And, using our formula for the laplacian chain rule,

$\Delta g(X,Y)=a^{\prime\prime}+b^{\prime\prime}$

The preharmonic condition on $g$ requires that

$\|d g\|^2\,\left[ d\Delta g\wedge dg\right] = \Delta g\,\left[ d\|d g\|^2\wedge dg\right]$

The easiest way to satisfy this equation would be for both sides to constantly equal zero. And, since we are looking for simple solutions, let’s impose that!

$$\begin{cases} d\Delta g\wedge dg =0\\ d\|d g\|^2\wedge dg=0 \end{cases}$$

Computing the first equation: $d(\Delta g )=d(a^{\prime\prime}+b^{\prime\prime})=a^{\prime\prime\prime}dX + b^{\prime\prime\prime}dY$ So

$$\begin{align} d(\Delta g)\wedge dg &= \left(a^{\prime\prime\prime}dX + b^{\prime\prime\prime}dY\right)\wedge\left(a^\prime dX+b^\prime dY\right)\\ &= a^{\prime\prime\prime }b^\prime dX \wedge dY +b^{\prime\prime\prime}a^\prime dY\wedge dX\\ &= \left( a^{\prime\prime\prime }b^\prime-b^{\prime\prime\prime}a^\prime\right)\mathrm{vol} \end{align}$$

For the second equation,

$d\|dg\|^2= d\left((a^\prime)^2+(b^\prime)^2\right)=2a^\prime a^{\prime\prime}dX + 2b^\prime b^{\prime\prime}dY$

$$\begin{align} d\|dg\|^2\wedge dg &= \left(2a^\prime a^{\prime\prime}dX + 2b^\prime b^{\prime\prime}dY\right)\wedge\left(a^\prime dX+b^\prime dY\right)\\ &= 2a^\prime a^{\prime\prime}b^\prime dX \wedge dY + 2 b^\prime b^{\prime\prime }a^\prime dY\wedge dX\\ &=\left(2a^\prime a^{\prime\prime}b^\prime-2 b^\prime b^{\prime\prime }a^\prime\right)dX\wedge dY\\ &=2a^\prime b^\prime (a^{\prime\prime}-b^{\prime\prime})\mathrm{vol} \end{align}$$

If these are both simultaneously equal to zero, we are looking for real valued functions $a,b$ which satisfy the following system of ODEs:

$$\begin{cases} a^{\prime\prime\prime }b^\prime-b^{\prime\prime\prime}a^\prime=0\\ 2a^\prime b^\prime (a^{\prime\prime}-b^{\prime\prime})=0 \end{cases}$$

Its easiest to begin with the second equation. So long as $a,b$ are not constants (which leads to the trivially harmonic $g=\mathrm{const}$) we know $a^\prime$ and $b^\prime$ are not constantly zero and so we must have $a^{\prime\prime}-b^{\prime\prime}=0$, or $a^{\prime\prime}(X)=b^{\prime\prime }(Y)$ But since these are functions of different variables which are always equal, we recognize a favorite trick - each side must individually be constant! Thus, there is some $K$ such that $a^{\prime\prime}=b^{\prime\prime }=K$ Integrating each we get

$a(X)=\frac{K}{2}X^2+C_1 X+C_2$ $b(Y)=\frac{K}{2}Y^2+C_3 Y+ C_4$

Since both of these are quadratic, their third derivatives are identically zero and the first equation in the system is automatically satisfied. Thus for any constants $K,A,B,C$ the following is preharmonic

$g(X,Y)=K(X^2+Y^2)+AX+ BY+C$

This is the sum of a function $K(X^2+Y^2)$ with a function $AX+BY+C$ which is harmonic on all of $\mathbb{E}^2$. Since we are in search of fundamental solutions, (which are undefined at the origin) we can ignore this ‘trivial’ harmonic piece, and normalizing to $K=1$, we’ve found a nontrivial preharmonic function $g(X,Y)=X^2+Y^2$

Upgrading

We now use the general theory to upgrade this to a harmonic function on $\EE^2\smallsetminus O$. Computing, $\Delta g = 2+2=4$ $\|dg\|^2= (2X)^2+(2Y)^2=4(X^2+Y^2)$ So, $\frac{\Delta g}{\|d g\|^2}=\frac{1}{X^2+Y^2}$ and so defining $h(s)=1/s$ we have $h\circ g = \frac{\Delta g}{\|dg\|^2}$. Computing the rescaling via integrating factor,

$$\begin{align} f(s)&=\int e^{-\int h\,ds}\,ds\\ &=\int e^{-\int\frac{ds}{s}}\,ds\\ &=\int e^{-\log s}\,ds\\ &=\int e^{\log\frac{1}{s}}\,ds\\ &=\int \frac{1}{s}\,ds\\ &=\log s \end{align}$$

Thus, $f\circ g(X,Y)$ is harmonic on $\EE^2\smallsetminus O$: $f\circ g(X,Y)=\log g(X,Y)=\log(X^2+Y^2)$

From here its just a simple rescaling to the fundamental solution!

$\varphi(X,Y)=\frac{1}{2\pi}\log(r)=\frac{1}{4\pi}\log(X^2+Y^2)$

Note that nowhere did we have to know that the answer was radial. The ansatz was additive separability in cartesian coordinates, and the rotational symmetry fell out of the ODEs on its own.

Multiplicatively Separable

Now we propose to find a preharmonic function $g(X,Y)$ which is multiplicatively separable meaning there are two functions $a,b\colon\RR\to\RR$ such that $g(X,Y)=a(X)b(Y)$ The preharmonic condition on $g(X,Y)$ requires we compute $\Delta g$ and $\|\nabla g\|^2=\|dg\|^2$ so we begin with those:

$$\begin{align} dg(X,Y)&=d\left(a(X)b(Y)\right)\\ &=d(a(X))b(Y)+ a(X)d(b(Y))\\ &=a^\prime(X)b(Y)dX+a(X)b^\prime(Y)dY \end{align}$$

Then as $dX,dY$ form an orthonormal basis at each point,

$$\begin{align} \|dg\|^2=\langle dg,dg\rangle &= (a^\prime b)^2 \|dX\|^2+2 (a^\prime b) (a b^\prime)\langle dX,dY\rangle + (ab^\prime)^2\|dY\|^2\\ &= (a^\prime b)^2+(a b^\prime)^2 \end{align}$$

And, using our formula for the laplacian chain rule,

$\Delta g(X,Y)=a^{\prime\prime} b + a b^{\prime\prime}$

Again, the preharmonic condition on $g$ requires that

$\|d g\|^2\,\left[ d\Delta g\wedge dg\right] = \Delta g\,\left[ d\|d g\|^2\wedge dg\right]$

Same as above; the easiest way to satisfy this equation would be for both sides to constantly equal zero. And, since we are looking for simple solutions, let’s impose that!

$$\begin{cases} d\Delta g\wedge dg =0\\ d\|d g\|^2\wedge dg=0 \end{cases}$$

Expanding the first

$$\begin{align} d\left(\Delta g\right) &= d\left(a^{\prime\prime} b + a b^{\prime\prime}\right)\\ &= (a^{\prime\prime\prime}b+a^\prime b^{\prime\prime})dX + (a^{\prime\prime}b^\prime + ab^{\prime\prime\prime})dY \end{align}$$

and wedging with $dg=a^\prime b dX + ab^\prime dY$,

$$\begin{align} d\left(\Delta g\right)\wedge dg &= (a^{\prime\prime\prime}b+a^\prime b^{\prime\prime})(a b^\prime) dX\wedge dY + (a^{\prime\prime}b^\prime + ab^{\prime\prime\prime})(a^\prime b) dY\wedge dX\\ &= \left[(a^{\prime\prime\prime}b+a^\prime b^{\prime\prime})ab^\prime-(a^{\prime\prime}b^\prime + ab^{\prime\prime\prime})a^\prime b\right]\mathrm{vol} \end{align}$$

Repeating with the second equation,

$$\begin{align} d\|dg\|^2&= d\left((a^\prime b)^2+(a b^\prime)^2\right)\\ &= 2a^\prime b d(a^\prime b)+ 2ab^\prime d(ab^\prime)\\ &= 2a^\prime b \left(a^{\prime\prime}b \, dX + a^\prime b^\prime dY\right)+2ab^\prime\left(a^\prime b^\prime dX + a b^{\prime\prime}dY\right)\\ &=2\left(a^\prime a^{\prime\prime}b^2 + a a^\prime (b^\prime)^2\right)dX +2\left((a^\prime)^2b b^\prime + a^2 b^\prime b^{\prime\prime}\right)dY \end{align}$$ $$\begin{align} (d\|dg\|^2)\wedge dg &= 2\left(a^\prime a^{\prime\prime}b^2 + a a^\prime (b^\prime)^2\right)\left(ab^\prime\right) dX\wedge dY + 2\left((a^\prime)^2b b^\prime + a^2 b^\prime b^{\prime\prime}\right)\left(a^\prime b\right)dY\wedge dX\\ &= 2\left [\left(a^\prime a^{\prime\prime}b^2 + a a^\prime (b^\prime)^2\right)ab^\prime-\left((a^\prime)^2b b^\prime + a^2 b^\prime b^{\prime\prime}\right) a^\prime b\right]\mathrm{vol} \end{align}$$

Together these result in the following system of equations:

$$\begin{cases} (a^{\prime\prime\prime}b+a^\prime b^{\prime\prime})ab^\prime-(a^{\prime\prime}b^\prime + ab^{\prime\prime\prime})a^\prime b=0\\ \left(a^\prime a^{\prime\prime}b^2 + a a^\prime (b^\prime)^2\right)ab^\prime-\left((a^\prime)^2b b^\prime + a^2 b^\prime b^{\prime\prime}\right) a^\prime b=0 \end{cases}$$

Every solution to this system is a multiplicatively separable preharmonic (though this may not cover all such functions - we are looking at a special case to simplify things after all). As in the additive case, the second equation provides us with the easiest line of attack: factoring out a copy of $a^\prime b^\prime$ and noting these are zero only in the trivial case where $g$ is constant,

$\left( a^{\prime\prime}b^2 + a (b^\prime)^2\right)a-\left((a^\prime)^2b + a^2 b^{\prime\prime}\right)b =0$

Multiplying through we collect terms with $a^2$ on one side and $b^2$ on the other:

$\left[a^{\prime\prime}a -(a^\prime)^2\right] b^2= \left[b^{\prime\prime}b-(b^\prime)^2\right]a^2$

Dividing through gives a separation of variables: thus each side must equal the same constant

$\frac{a^{\prime\prime}a -(a^\prime)^2}{a^2}=\frac{b^{\prime\prime}b-(b^\prime)^2}{b^2}=K$

Every term has a product of two copies of $a$ or $b$ (and their derivatives). This suggests proposing an ansatz where $a^\prime$ is a multiple of $a$, say $a^\prime(X)=u(X)a(X)$ for some function $u$ (we use $u$ rather than $f$, which is already spoken for as the upgrading function). This implies $a^{\prime\prime}=(ua)^\prime = u^\prime a + ua^\prime = u^\prime a + u^2a = (u^\prime + u^2)a$ Plugging this in,

$K=\frac{a^{\prime\prime}a -(a^\prime)^2}{a^2}=\frac{(u^\prime + u^2)a^2-u^2a^2}{a^2}=u^\prime$

Thus $u^\prime=K$ and so $u(X)=KX + C$ is affine. Looking at the linear case $u(s)=Ks$ for simplicity, we have $a^\prime = ua \implies \frac{a^\prime}{a}=u\implies \left[\log a\right]^\prime = u$

$$\begin{align} a(s)&=\exp\left(\int u\,ds\right)\\ &=\exp\left(\int K s\,ds \right)\\ &=\exp\left(\frac{K}{2} s^2\right) \end{align}$$

The same computation with $b$ gives the same function of $Y$, so our candidate is

$g(X,Y)=a(X)b(Y)=\exp\left(\frac{K}{2}X^2\right)\exp\left(\frac{K}{2}Y^2\right)=\exp\left(\frac{K}{2}(X^2+Y^2)\right)$

Finally, we need to see that this solution also satisfies the first equation. The quickest route is not to expand the third derivatives but to notice that both quantities appearing in the system are functions of $s=X^2+Y^2$ alone. Writing $g=e^{Ks/2}$ and using $a^{\prime\prime}=K(1+KX^2)a$,

$\Delta g = a^{\prime\prime}b+ab^{\prime\prime} = K\left[2+K(X^2+Y^2)\right]g=K(2+Ks)e^{Ks/2}$ $\|dg\|^2 = (a^\prime b)^2+(ab^\prime)^2 = K^2(X^2+Y^2)g^2 = K^2s\,e^{Ks}$

Both are functions of $s$, as is $g$ itself, so all three are constant on the circles $s=\mathrm{const}$. Their differentials are therefore all proportional to $ds$, and any wedge of two of them vanishes — in particular $d(\Delta g)\wedge dg = 0$, which is the first equation.

So it does, and we have our separable preharmonic. We can see directly from here that $X^2+Y^2$ has the same level sets as what we found, and so the multiplicative ansatz has landed on the same function as the additive one, just relabeled. Upgrading proceeds exactly as in the additive case and returns the same $\log(X^2+Y^2)$ — which is the point of the preharmonic formalism: it only ever sees the level sets, so two quite different-looking ansatze that pick out the same foliation are doing the same work.