Quite a long ago, I tried to work out explicitly the content of the Newlander-Nirenberg theorem. My aim was trying to understand wether a direct proof could work in the simplest possible case, namely that of surfaces. The result is that the most explicit statement I could get is a PDE I was not able to solve.


Assume a quasi-complex structure $J$ is given on the surface $S$; we want to prove that this is induced by a complex one (in this case there are no compatibility conditions). This can be easily transformed in the problem of local existence for a second order PDE, as follows.


We look for local charts on $S$ which are holomorphic (with respect to the quasi-complex structure on $S$). Two such charts are then automatically compatible. So the problem is local.


Fix a small open set $U \subset S$ and identify it with a neighboorhood of $0 \in \mathbb{R}^2$ via a differentiable chart. Locally we can write $J = \left(\begin{matrix}[a | b] \\ [c | d ]\end{matrix}\right)$ for some functions $a, \cdots, d$ (pretend it is a two by two matrix).


A chart is given by a complex valued function $f = u + iv$. The condition that the differential is $\mathbb{C}$-linear can be verified on a basis of the tangent space; moreover if it is true for a vector v, it remains true for Jv, which is linearly independent. Here we have used that $J^2 = -1$.


So we need only to check it for the vector $\partial\_x$. Since $J \partial\_x = a \partial\_x + c \partial\_y$, the condition says


$-v\_x = a u\_x + c u\_y$


$u\_x = a v\_x + c v\_y$


Hence we need to solve this system, with $f = u + i v$ non singular ($f$ will be then locally invertible). Since $a$ and $c$ do not vanish simultaneously, we can assume $c(0) \neq 0$, hence $c \neq 0$ on $U$ provided $U$ is small.


We can then solve for $u\_y$ and get the equivalent system


$u\_x = a v\_x + c v\_y$


$-u\_y = \frac{1 + a^2}{c}v\_x +a v\_y$


Moreover the Jacobian $J\_f = u\_x v\_y + u\_y v\_x = \frac{1}{c}(v\_x^2 + (a v\_x + c v\_y)^2)$, so $f$ is nonsigular if $v$ is. By Poincaré's lemma, the system admits a local solution if and only if


$\frac{\partial}{\partial\_y} \left( a v\_x + c v\_y \right) - \frac{\partial}{\partial\_x} \left( \frac{1 + a^2}{c}v\_x +a v\_y \right) = 0$.


Hence we are looking for a local solution of the last equation with $(v\_x(0), v\_y(0)) \neq (0, 0)$.


So my question is:



> 
> Is there a simple way to prove local existence for a nonsingular solution of the last displayed equation?
> 
> 
> 


I should make clear that I'm not looking for a proof of Newlander-Nirenberg; of this there are plenty. I am more interested in seeing what Newlander-Nirenberg becomes in terms of PDE in the simplest possible case, and then see that the PDE thus obtained is solvable. According to the answer of Andy, the equation which comes out is the Beltrami equation, so I will have a look at it. Still, I'm curious if any standard PDE technique can solve the equation I derived in the most stupid way above.



Newlander-Nirenberg for surfaces
I'm not sure if the following is elementary enough, but it does only use standard PDE machinery (plus some basic Riemannian geometry). It's also nice in that it suggests an approach to proving the uniformization theorem (via metrics of constant curvature).


Say you have an almost-complex structure on the unit disk. Your goal is to find a conformal isomorphism of this disk (or at least some neighborhood of the origin) with an open subset of the complex plane. To do it, first choose a metric $g$ on the disk which is compatible with the given complex structure. Let $K$ be the curvature of this metric. If you can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then you'll be done, since the exponential map with respect to $\tilde{g}$ will be an isometry, hence also a conformal isomorphism. 


So, multiply $g$ by an arbitrary positive function $e^f$, and compute the curvature of the new metric. You'll find that it's given by the formula:


$\tilde{K} = e^{-2f}(K - \Delta f) $ 


where $\Delta$ is the Laplacian with respect to the metric $g$. Setting the left hand side equal to zero, you have reduced to solving the Laplace equation, which can be done locally using standard PDE techniques.


As for a more "direct" approach...


The equation you wrote down should reduce to solving the Laplace equation as well, using the notion of conjugate harmonic functions. However, solving your equation will inevitably be a bit more subtle due to your requirement that the solution have nonvanishing differential at the origin. There is a proof along the lines you're suggesting in Taylor's PDE book, chapter 5, section 11, and I think there's a similar one in Jost's "Postmodern analysis". Basically the idea is to rescale your coordinate system so that the metric is nearly flat, in which case you should have a conformal map that is close to the identity map in a high enough sobolev space, and therefore has a nonvanishing derivative at the origin.


