Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mu$ be a function on the complex plane with the property $\mu(z) = \overline{\mu(\bar{z})}$, such that $\mu(z) = \epsilon e^{-2\pi i \bar{z}}$ on the upper-half plane, where $\epsilon$ is a complex number such that $|\epsilon|<1$.

I'm interested in the solution to the following Beltrami equation with this Beltrami coefficient:

\begin{equation} \mu \frac{\partial f}{\partial z} = \frac{\partial f}{\partial \bar{z}} \end{equation}

Now, since $|\mu| = |\epsilon|e^{-2\pi y} < 1$ (since $y >0$), there exists a unique quasiconformal solution $f$, satisfying $f(\bar{z})=\overline{f(z)}$ and fixing the points 0, 1, and infinity.

I'm basically trying to understand how $f_z$ grows.

Now, because $\mu$ is such a simple function, I feel like one should be able to explicitly write a solution in this case. I've tried many different approaches, but I have been very unsuccessful. I tried looking in Ahlfors and a few other references, but I haven't been able to find anything too useful.

Any ideas?

share|improve this question
I was not aware of the existence of such a function $f$, if $|\mu|<1$. Where can I find a reference for this fact? –  plusepsilon.de Nov 15 '10 at 10:39
I hope I didn't quote the theorem wrong. There are many references, such as Ahlfors's Lectures on Quasiconformal Mapping, but I personally like Lehto the best. (MR0867407) –  BrainDead Nov 15 '10 at 15:30
I don't understand your statement that $|\mu|=\epsilon$. Your $\mu$ is unbounded in the plane! –  Alexandre Eremenko Oct 10 '12 at 18:08
@Alexandre Eremenko - $|\mu|=\epsilon$ is incorrect, so I fixed it. But $\mu$ is bounded. On the upperhalf, it is equal to $\mu(z)=\epsilon e^{−2\pi i x}e^{−2\pi y}$. The values of $\mu$ on the lower half plane is $\mu(z)=\bar{\epsilon} e^{2\pi i x}e^{2\pi y}$. –  BrainDead Feb 17 '13 at 6:02
add comment

1 Answer

Sorry in advance for responding to an old question, especially since this isn't even a solution. In my defence, this is my first post and I haven't seen anything in the FAQ for etiquette forbidding answering old posts?

I've been thinking about very similar questions lately, and stumbled upon this post during a Google search for ways of solving the Beltrami equation. There is currently a construction (attributed to Gauss) detailed on the "Beltrami equation" Wikipedia entry for locally solving the Beltrami equation at a point (the default is $z=0$) when $\mu$ is (locally) expressible as a power series in $z$ and $\bar{z}$. I've tried the procedure but it hasn't quite produced a correct function, and this has prompted me to try to solve it myself. I believe that I have a working algorithm for finding general solutions to the Beltrami equation for a sufficiently nice $\mu$ - such as the one you've specified.

Caveat: I realize that you're not simply wanting to solve the Beltrami equation, but would like to find the unique normalized solution $f$ such that $f$ is a homeomorphism of the upper half plane, but all that I'm able to furnish is an (explicit) family of solutions which probably contains the desired guy. I have no idea how you might be able to track down the unique normalized quasiconformal solution in this haystack.

A pretty simple non-zero solution to the Beltrami equation for your specified coefficient is:

$$f(z):=z-\frac{\epsilon}{2\pi i}\exp(-2\pi i\bar{z})+\frac{\epsilon}{2\pi i}.$$

Note that although $f:0\mapsto 0,1\mapsto1,\infty\mapsto\infty$ and is smooth, it doesn't satisfy that $f(\bar{z})=\bar{f(z)}$ and isn't the quasiconformal map we want.

First observe that $f(z)$ satisfies the Beltrami equation over all of $\mathbb{C}$ (not just the upper half plane).

Next observe that post-composing $f$ by a holomorphic function $\varphi(z)$ also results in a solution $\varphi\circ f(z)$ that also satisfies the Beltrami equation. Equivalently, any power series of $f$ is also a solution over its domain of convergence.

Since $f(z)$ contains a $z$ term, it should be reasonably easy to show that any solution to the Beltrami equation defined at $z=0$ that's also expressible as a power series in $z$ and $\bar{z}$ will agree with some power series in $f(z)$ in a neighborhood around $z=0$.

This results in a (countably) infinite dimensional family of solutions to the Beltrami equation, but it's not clear to me how one might be able to extract the quasiconformal solution. In fact, I'm not familiar enough with quasiconformal maps to know if the desired solution can be written as a power series in $z$ and $\bar{z}$ around $z=0$, although this is not, I feel, an unreasonable expectation.

Being overly optimistic, I feel that the $f(z)$ above isn't a terrible approximation for the actual quasiconformal solution, especially if one simply wanted some growth-rate intuition. In particular, the approximation behaves best where the imaginary value of the input is large, which is somewhat unsurprising because the Beltrami coefficient is nearly $0$ when the imaginary part of $z$ is high, so the solution should be roughly the identity map around these parts - the identity function, of course, is the dominant term in $f(z)$ for $\mathrm{Im}(z)\gg0$.

Just in case you want a way of constructing a solution to a given Beltrami equation that is "generative", in the sense that any other solution defined at $z=0$ should be a power series of this one, I believe that the following procedure should work:

Step 1: We adopt the notation that the Beltrami coefficent $\mu(z)$ - which by assumption can be written as a power series in $z$ and $\bar{z}$, is written as $\mu(z,\bar{z})$. Then solve for

$$\frac{\partial F}{\partial z}(z,w)=-\mu(w, F(z,w)),\; F(0,w)=w.$$

This is purely motivated by the aforementioned Wikipedia entry, although the order of the inputs for $\mu$ do differ from Wikipedia's.

Step 2: Find $G(z,w)$ such that $G(z,F(z,w))=w$. That is, the functions $$(z,w)\mapsto (z,F(z,w))\text{ and }(z,w)\mapsto (z,G(z,w))$$ are inverse to each other.

I'd be surprised if $G(z,w)$ always exists over the entire domain of $\mu$, but it should for "nice" examples of $\mu$ (It's existed for all the examples that I've checked by hand, perhaps something along the lines of $\mu$ being defined over a simply connected domain and expressible as a power series in $z$ and $\bar{z}$ might suffice)?

Step 3: A "generative" solution is given by: $$f(z)=G(\bar{z},z).$$

Here's hoping that someone'll take this mess and figure out how to build the normalized quasiconformal map. =)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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