Differential "Freshman's dream" for Laplacian operator.

Question

Today I encountered quite an interesting phenomenon. There is an exercise in multivariable calculus that asks students to prove the identity $$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = e^{-2\xi} \left( \frac{\partial^2 f}{\partial \xi^2} + \frac{\partial^2 f}{\partial \theta^2} \right), $$ where the coordinates transformation is given by $(x,y) = F(\xi,\theta) = (e^\xi \cos(\theta), e^\xi \sin(\theta))$, assuming $f \in C^2$. I have seen a person misunderstood the question and proved $$ \left(\frac{\partial f}{\partial x} \right)^2 + \left(\frac{\partial f}{\partial y} \right)^2 = e^{-2\xi} \left( \left(\frac{\partial f}{\partial \xi} \right)^2 + \left(\frac{\partial f}{\partial \theta} \right)^2 \right) $$ instead. To my surprise, his proof contains no mistake and the misinterpreted equation is actually true!

This got me into thinking about the generalization of the above "Freshman's dream" for Laplacian operator:

Which coordinates transformation $(x,y) = F(\xi,\theta)$ (or, equivalently, $(\xi,\theta) = G(x,y)$ )has the property that for any $f\in C^2$, we have $$\begin{align} \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} &= h(\xi,\theta) \left( \frac{\partial^2 f}{\partial \xi^2} + \frac{\partial^2 f}{\partial \theta^2} \right) \\ &\text{if and only if} \\ \left(\frac{\partial f}{\partial x} \right)^2 + \left(\frac{\partial f}{\partial y} \right)^2 &= h(\xi,\theta) \left( \left(\frac{\partial f}{\partial \xi} \right)^2 + \left(\frac{\partial f}{\partial \theta} \right)^2 \right) \end{align}$$ on an open domain $D\subset\Bbb R^2$ for some (sufficiently smooth) function $h>0$?

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Here's my thought so far:

Suppose that our coordinates transformation is given by $(\xi,\theta) = G(x,y) = (G_1(x,y),G_2(x,y))$. By some calculation (that I shall skip), we can compute that the Laplacian $\Delta = \partial_x^2 + \partial_y^2$ in the coordinate $(\xi,\theta)$ can be written as $$ \partial_x^2 + \partial_y^2 = (\Delta G_1)\partial_\xi + (\Delta G_2)\partial_\theta + |\nabla G_1|^2 \partial_\xi^2 + |\nabla G_2|^2 \partial_\theta^2 + 2(\nabla G_1 \cdot \nabla G_2)\partial_\xi \partial_\theta, $$ hence $\partial_x^2 + \partial_y^2 = h(\partial_\xi^2 + \partial_\theta^2)$ if and only if $$ |\nabla G_1|^2 = |\nabla G_2|^2 = h, \quad \Delta G_1 = \Delta G_2 = 0, \quad \text{and}\quad \nabla G_1 \cdot \nabla G_2 = 0. $$ On the other hand, we have $$ \begin{pmatrix}\partial_x f &\partial_y f \end{pmatrix} = \begin{pmatrix}\partial_\xi f &\partial_\theta f \end{pmatrix} \begin{pmatrix}\partial_x G_1 &\partial_y G_1 \\ \partial_x G_2 &\partial_y G_2 \end{pmatrix}, $$ hence in order that $(\partial_x f)^2 + (\partial_y f)^2 = h ((\partial_\xi f)^2 + (\partial_\theta f)^2 )$, we need the Jacobian matrix to be of the form $$ \begin{pmatrix}\partial_x G_1 &\partial_y G_1 \\ \partial_x G_2 &\partial_y G_2 \end{pmatrix} = \sqrt{h}\ M, $$ where $M$ is an orthogonal matrix at each point on $D$. In particular, this is equivalent to $$ |\nabla G_1|^2 = |\nabla G_2|^2 = h \quad \text{and}\quad \nabla G_1 \cdot \nabla G_2 = 0. $$ It seems like the "miracle" we have seen earlier is a little bit less surprising than expected!

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From my calculation above (unless I made some mistakes), it seems like $$\begin{align} \partial_x^2 f + \partial_y^2 f &= h(\partial_\xi^2 f + \partial_\theta^2 f) \\ &\text{if and only if} \\ (\partial_x f)^2 + (\partial_y f)^2 &= h ((\partial_\xi f)^2 + (\partial_\theta f)^2 ) \ \ \text{and} \ \ \Delta G_1 = \Delta G_2 = 0, \end{align}$$ i.e. $G$ is a harmonic function coordinate-wise. In particular, this is true for our original coordinate transformation function since we can rewrite $(x,y) = (e^\xi \cos(\theta), e^\xi \sin(\theta))$ as $(\xi,\theta) = G(x,y)$, where $$ G(x,y) = (G_1(x,y) , G_2(x,y)) = \left( \frac12 \ln(x^2+y^2), \arctan\left( \frac{y}{x} \right) \right). $$ Here $G_1$ and $G_2$ are indeed harmonic (on an appropriate domain). I want to know if we can find more interesting examples like this? I have a feeling that this should be related to holomorphic functions and harmonic conjugate but my knowledge of complex analysis is pretty limited.

Is there a general theory that would allow us to construct a coordinate transformation function $G$ satisfying the above properties?

Answer 1

Summary. The orthogonality of the matrix $M$ also implies that $\Delta G_1 = \Delta G_2 = 0$, so the two conditions are equivalent. That is, the conditions $|\nabla G_1|^2 = |\nabla G_2|^2 = h > 0$ and $\nabla G_1 \cdot \nabla G_2 = 0$ are quite restrictive and guarantee that either $G_1, G_2$ or $G_1(x, -y), -G_2(x,-y)$ are harmonic conjugates. In either case, $G$ is a conformal change of coordinates. This is part of the proof below, which answers your main question about the class of such coordinate transformations.

Discussion. The starting condition you've specified, namely that $$ \begin{equation} \tag{*}\label{eq:conformal-lap} \partial_x^2 + \partial_y^2 = h(\xi, \theta) \left( \partial_{\xi}^2 + \partial_{\theta}^2 \right) \end{equation} $$ with $G(x, y) = (\xi , \theta) \in C^2$ and $h > 0$, is equivalent to the condition that $(\xi, \theta) = G(x,y)$ is a conformal change of coordinates. Geometrically, conformality means that the coordinate transformation locally preserves angles, which analytically means that as a complex map, $G : D \to \mathbb{C}$ is (anti-)holomorphic and $\partial_x G, \partial_y G \neq 0$ on $D$.

Proof. ("$\Longleftarrow$") To see that this implies the transformation $\eqref{eq:conformal-lap}$, start by noting that by (anti-)holomorphicity, $G_1$ and $G_2$ satisfy either the Cauchy-Riemann equations $\partial_x G_1 = \partial_y G_2$ and $\partial_y G_1 = - \partial_x G_2$, or $\partial_x G_1 = -\partial_y G_2$ and $\partial_y G_1 = \partial_x G_2$. Then in either case, $\nabla G_1 \cdot \nabla G_2 = 0$ and together with $\partial_x G, \partial_y G \neq 0$ we have, $$ |\nabla G_1|^2 = (\partial_x G_1)^2 + (\partial_y G_1)^2 = (\partial_y G_2)^2 + (\partial_x G_2)^2 = |\nabla G_2|^2 =: h > 0. $$ Also straightforwardly, $\Delta G_1, \Delta G_2 = 0$. Combined with your computation of $\partial_x^2 + \partial_y^2$ in the new coordinates, we recover the form $\eqref{eq:conformal-lap}$.

("$\Longrightarrow$") In the opposite direction, again as you've noted, $|\nabla G_1|^2 = |\nabla G_2|^2 = h > 0$ and $\nabla G_1 \cdot \nabla G_2 = 0$ together imply that the Jacobian matrix of $G$ is $\sqrt{h} M$ with $M$ an orthogonal matrix on $D$. Since all $2 \times 2$ orthogonal matrices are either of the form $\left( \begin{smallmatrix} a & b \\ b & -a \end{smallmatrix} \right)$ or $\left( \begin{smallmatrix} a & -b \\ b & a \end{smallmatrix} \right)$, this implies either the Cauchy-Riemann equations or $\partial_x G_1 = - \partial_y G_2$ and $\partial_y G_1 = \partial_x G_2$ and since $h > 0$ we have $\partial_x G, \partial_y G \neq 0$ on $D$.

$\square$

Remarks.

1. Since the coordinate transformations are from Euclidean coordinates, these are also known as isothermal coordinates, which speaks to the harmonicity condition on $G_1$ and $G_2$. There is some discussion on MathOverflow that might be of interest for generalisations to Riemannian manifold settings.

2. In the Riemannian setting, letting $g$ be the Euclidean metric in $(x,y)$ coordinates, $G$ induces a change of metric by its pullback: $$ \begin{align*} \tilde{g} &:= G_* g \\ &= \begin{pmatrix} (\partial_x G_1)^2 + (\partial_x G_2)^2 & \partial_x G_1 \partial_y G_1 + \partial_x G_2 \partial_y G_2 \\ \partial_x G_1 \partial_y G_1 + \partial_x G_2 \partial_y G_2 & (\partial_y G_1)^2 + (\partial_y G_2)^2 \end{pmatrix} \end{align*} $$ and the harmonic conjugate relations (that is, the Cauchy-Riemann equations) imply that $\tilde{g} = h \left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$. Then, $\eqref{eq:conformal-lap}$ can be written $\Delta_g = h \Delta_{\tilde{g}}$, which is a form that holds for conformal changes of metric on any Riemannian surface (smooth Riemannian manifolds of dimension two).