The Beltrami equation and Neumann series

Question

Let $\mu: \mathbb{C} \to D(0,1).$ A quasiconformal map is a $W^{1}_{2,loc}-$solution to the Beltrami equation $\bar{\partial}f = \mu \partial f$. In this paper, the authors remark that one can formally solve for $f$ in this equation by the Neumann series $$\bar{\partial}f = \mu + \mu T \mu + \mu T \mu T \mu + \cdots$$ where $T$ is the Beurling transform $$Tf(z) = \partial\left(\frac{\bar{\partial}f}{\partial z}\right)^{-1}(z) = -\frac{1}{\pi} p.v \int_{\mathbb{C}}\frac{f(w) \ \mathrm dw}{(z-w)^2}. $$ I cannot see how, even formally, this is true, and no source I can find shows how to do this. What is the heuristic behind this?

Answer 1

The story of the solution of the Betrami equation using the Beurling transformation in $L^2$ has an elegant and elementary explanation in:

Adrien Douady and X. Buff, Le théorème d’intégrabilité des structures presque complexes, The Mandelbrot set, theme and variations, London Math. Soc. Lecture Note Ser., vol. 274, Cambridge Univ. Press, Cambridge, 2000, pp. 307–324. MR 1765096

I can't see how it becomes precisely what you have quoted, but is similar. If we let $f=(I-\mu L)^{-1}\mu$, where $L$ the Beurling transform, then $(I-\mu L)f=\mu$, so $f=\mu+\mu L f$. If $\varphi$ satisfies $\bar\partial \varphi=f$, then $\bar\partial (z+\varphi)=f=\mu+\mu L f=\mu+\mu\bar\varphi=\mu(\bar z+\bar\varphi)$. So $z+\varphi$ is the solution you are looking for, modulo various details explained clearly in that paper.