Let $\mu:\mathbb C\to \mathbb C$ be a $C^\infty$ function satisfying $|\mu|\le 1$.
Let us furthermore assume that the function $\mu$ never takes the value $-1$.
Does there exist a $C^\infty$ function

$f:\mathbb C\to \mathbb C$ such that $\frac{\,\partial f/\partial \bar z\,}{\partial f/\partial z}=\mu$ ?
(and $\scriptstyle\partial f/\partial z$ everywhere non-zero)

Note that $f$ is by no means expected to be a homeomorphism. For example, if $\mu\equiv 1$, then a general solution is of the form projection $\mathbb C\to\mathbb R:z\mapsto Re(z)$ followed by any embedding $\mathbb R\to\mathbb C$.

The special case $|\mu|<1$ is the usual Beltrami equation.
More generally, on which domains $U\subset\mathbb C$ can one guarantee that a solution exists?

[The exact problem I care about is defined on the strip $U=\{z\in\mathbb C:0\le Im(z)\le 1\}$, and the function $f$ is furthermore required to be periodic: $f(z)=f(z+1)$.]

Equivalent formulation:
Let $\nu:\mathbb C\to \mathbb C$ be a $C^\infty$ function satisfying $Im(\nu)\ge0$.

$\exists$? $f:\mathbb C\to \mathbb C$ such that $\frac{\,\partial f/\partial y\,}{\partial f/\partial x}=\nu$ ?
(and $\scriptstyle\partial f/\partial x$ everywhere non-zero)

(The two formulations are equivalent via the transformations $\mu=\frac{1+i\nu}{1-i\nu}$ and $\nu=i\frac{1-\mu}{1+\mu}$, which are a version of the Cayley transform. These transformations map $\nu$ in the closed upper half plane to $\mu$ in the closed unit disc minus $-1$.)

Partial result (positive solution in the completely degenerate case):
When $\nu$ is real, equivalently when $|\mu|=1$, then the problem admits a positive solution.

In that case, the vector field $\partial_y-\nu\partial_x$ defines a foliation of $\mathbb C$. That vector field is never horizontal, so all the leaves of the foliation are closed (diffeomorphic to $\mathbb R$). Any function $f:\mathbb C\to\mathbb C$ which is constant on the leaves (i.e., which factors through the 1-dimensional leaf space) is then a solution of the degenerate Beltrami equation $\frac{\,\partial f/\partial y\,}{\partial f/\partial x}=\nu$.

Note that, unlike the leaf space of a general foliation of $\mathbb C$, the leaf space of this specific foliation is a manifold (by which I mean a second-countable possibly non-Hausdorff manifold), and can thus be immersed into $\mathbb C$. That's where the crucial condition $\mu\not=-1$ gets used.


1 Answer 1


A reference for these types of equations is in Astala, Iwaniec and Martin chapter 20. One of the main difficulties is showing that $f$ is a homeomorphism. In general it will not be the case. For example $f(z) = x$ has a Beltrami coefficient $\mu \equiv 1$ but it sends the entire plane into the real line.

  • $\begingroup$ Thank you for the reference; I will have a look. In my cases of interest, $f$ will typically NOT be a homeomorphism. These are exactly the cases I care about. $\endgroup$ Sep 13, 2019 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.