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I need to find solutions to the Beltrami equation

$$ \frac{\partial w}{\partial\overline{{z}}}=e^{i\varphi(z)}\frac{\partial w}{\partial z} $$

for $w=w(z,\overline{z})$ and $\varphi(z)$ some given, real, harmonic function. So the Beltrami coefficient is just a phase.

Among the almost infinite literature about the Beltrami equation the only thing I've been able to find is that the solution exists and is unique, provided that $w$ is specified along a compact contour on the plane. Solving it numerically is possible, but I'd like to have a deeper understanding of the solutions, and in particular, what are the implications of $\varphi(z)$ being harmonic, and even more, of the Beltrami coefficient being a phase, $|\mu(z)|=1$, a fact that certainty drives the problem.

Any ideas ?

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  • $\begingroup$ Your equation might have a solution but it will not be quasiconformal (I think, its image will be a curve). For quasiconformality you need a Beltrami differential $\mu$ satisfying $||\mu||_\infty<1$. $\endgroup$ Commented Dec 29, 2020 at 21:45
  • $\begingroup$ $\|\mu\|_\infty=1$ implies that $w$ is everywhere degenerate, $J_w\equiv 0$. $\endgroup$ Commented Dec 29, 2020 at 23:57

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Note that, if you take $\phi=0$, then the equation reduces to $w_y =0$, i.e., if $D\subset C$ is the domain of $w$ and $x:D\to\mathbb{R}$ is the projection on the $x$-axis and has connected fibers, then $w= h(x)$ for some $C^1$ function $h:x(D)\to\mathbb{C}$, and this is the general solution on such $D$.

Something similar happens in general: Write $$ \mathrm{d}w = w_z\,\mathrm{d}z + w_{\bar z}\,\mathrm{d}\bar z = w_z\,(\mathrm{d}z + \mathrm{e}^{i\phi(z)}\,\mathrm{d}\bar z) = \mathrm{e}^{i\phi(z)/2}w_z\left(\mathrm{e}^{-i\phi(z)/2}\mathrm{d}z + \mathrm{e}^{i\phi(z)/2}\,\mathrm{d}\bar z\right). $$ Then, setting $\alpha = \mathrm{e}^{-i\phi(z)/2}\mathrm{d}z + \mathrm{e}^{i\phi(z)/2}\,\mathrm{d}\bar z$, we see that $\alpha$ is a real-valued $1$-form, and hence always has a local integrating factor, i.e., it can be written locally in the form $\alpha = f\,\mathrm{d}u$ for some functions real-valued functions $u$ and $f>0$. Thus, if $D\subset\mathbb{C}$ is a domain such that $\alpha$ can be written as $\alpha = f\,\mathrm{d}u$ for some real-valued functions $u$ and $f>0$ on $D$ and the fibers of $u:D\to u(D)\subset \mathbb{R}$ are connected, then any solution of your equation on $D$ can be written in the form $w = h(u)$ for some $C^1$ function $h:u(D)\to\mathbb{C}$, and every such $h$ that is $C^1$ yields a solution. This is because your equation for $w:D\to\mathbb{C}$ reduces to $\mathrm{d}w = p\,\mathrm{d}u$ for some function $p:D\to\mathbb{C}$.

The signficance of $\phi$ being harmonic is not really clear (other than ensuring that $\alpha$ is real-analytic, so that $u$ can be taken to be real-analytic also). Certainly, the behavior of $\phi$ will determine which domains $D\subset\mathbb{C}$ have the right shape to support an integrating factor for $\alpha$, but it is not clear to me that just requiring that $\phi$ be harmonic gives you much easily accessible information along those lines.

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