coordinate transformations that preserve the laplacian

Question

Consider the Laplacian operator $\Delta=\frac{\partial^2}{\partial x_1^2}+\dots +\frac{\partial^2}{\partial x_n^2} $. We say that a $C^2$ coordinate transformation $\mathbf{f}:\mathbb{R}^n\to \mathbb{R}^n$ is Laplacian-preserving iff for every scalar function $u\in C^{\infty}(\mathbb{R}^n)$, we have

$$ \Delta (u\circ \mathbf{f})= (\Delta u)\circ \mathbf{f}.$$

Example: If $\mathbf{f}(\mathbf{x})=A\mathbf{x}+\mathbf{b}$ where $A$ is an orthogonal matrix and $\mathbf{b}\in\mathbb{R}^n$, then $f$ is Laplacian preserving. See this question.

Question: Are there any other transformations $\mathbf{f}$ that are Laplacian preserving?

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I managed to show that in the case $n=2$ there are no other Laplacian preserving transformations. However, my approach may not work (or could be tedious) in higher dimensions. Here's how I have done it:

Let $\mathbf{f}=(f_1,f_2)$, then (with some abuse of notation)

$$\Delta (u\circ \mathbf{f})= \left( \sum_{i=1}^2 \lVert \nabla f_i\rVert^2 \frac{\partial^2u}{\partial x_i^2} +\Delta f_i \frac{\partial u }{\partial x_i}\right) +2(\nabla f_1\cdot \nabla f_2 )\frac{\partial^2 u }{\partial x_1\partial x_2} $$

From here, I have $\Delta f_i=0$, $\lVert \nabla f_i\rVert^2=1$ for $i\in\{1,2\}$, we take the gradient of the last equation to get $(H f_i)\nabla f_i=\mathbf{0}$, where $Hf_i$ is the Hessian matrix of $f_i$.

However, since $\Delta f_i=0$, the determinant of the Hessian of $f_i$ is $$-\left(\frac{\partial^2 u }{\partial x_1\partial x_2}\right)^2-\left(\frac{\partial^2 u }{\partial x_1^2}\right)^2$$

Therefore, the equation $(H f_i)\nabla f_i=\mathbf{0}$ implies that either $\nabla f_i=\mathbf{0}$, that is, $f_i$ is constant; or $f_i$ is a linear map. It is easy from here to see that $\mathbf{f}(\mathbf{x})=A\mathbf{x}+\mathbf{b}$ where $A$ is orthogonal.

Answer 1

The Laplacian on a core of smooth functions in an inner product space over a Riemann manifold is the dual of

$$ (\nabla f)\star \cdot G \cdot \nabla g \to f^\star (-\nabla^\star \cdot \nabla g) $$

with G the unit matrix in $R^n$ and

$$-\Delta = \nabla^\star \cdot \nabla$$

in cartesian coordinates

A coordinate map introduces the metric matrix $G(x) =J(x)^t \ \cdot J(x)$ of the Jacobian matric $J=\frac{\partial x}{\partial x'}$ and the volume density $$dVol(x) = dx^n \det J = dx^n \sqrt(\det(G(x)))$$ into the inner product assumed to be the Lebesgue integral.

$$ \int \ dVol(x)\ \left((\nabla f\right)^\star \ \cdot \ G(x) \cdot\ \nabla g = \int dVol(x)\ f^\star \ \det{ J^{-1}} \ \ \nabla \cdot \ \left( \det J \ \ G(x) \cdot \ \nabla g \right) $$

It follows, that $\Delta$ is invariant if $G$ is a multiple of the unit matrix and \$det J $ is constant. This defines the euclidean group of global constant translations and rotations.

If the resulting $\Delta$ is of the form $$\Delta = h(x) \sum \partial_i^2 $$ with a position dependent factor, for homogenous problems the Laplace equation looks like the equation in Cartesian coordinates.

These families of systems in Riemann spaces are called conformal equivalent, same coordianate solutions, different densities.

This is generally the case in 2 dimensions, because the Laplacian of all sums of holomorphic functions ($z\to f(z)$ ) and antiholormorphic functions (z\to f(z^\star)) are vanishing.

Answer 2

$$\nabla^2 u(f(x)) = \sum_i ||\nabla f_i||^2 \frac{\partial^2 u}{\partial x_i^2} + 2 \sum_{i,j} (\nabla f_i . \nabla f_j) \frac{\partial^2 u}{\partial x_i x_j} + \sum_i \nabla^2f_i \frac{\partial u}{\partial x_i} .$$

Now fix $f$ and vary $u$,

Now use $u(x) = x_i$ to conclude $\nabla^2 f_i =0$. Now use $u(x)=x_i^2$ to conclude $||\nabla f_i||^2 =1$. Now use $u(x) =x_i x_j$ to conclude that $\nabla f_i . \nabla f_j =0$.

So if you want a function $f$ to universally work for all $u$ then $$[\nabla f_i : 1 \leq i \leq n]$$ must be orthonormal and $\nabla^2 f_i =0$.