Let $U$ be a bounded domain in $R^2$ and let $n : U \to S^2$. Which (necessary/sufficient) conditions must $n$ satisfy in order that there exist an immersion $f : U \to R^3$ such that $n(x)$ is the normal to $f$ at the point $f(x)$, for all $x\in U$ ? The ideal answer would be of the form: "such $f$ exists if and only if $n$ satisfies the following PDE: ..."



Conditions on a unit vector field to be the Gauss map of some surface immersed in R^3?
As Alexandre Eremenko points out, there's no PDE that $n$ would have to satisfy, at least for local solvability, which is believable when you think of it in heuristic terms: There are $3$ unknowns involved in specifying an immersion $f:U\to\mathbb{R}^3$ and only $2$ arbitrary functions needed to specify $n:U\to S^2$, so it's an underdetermined problems and there 'ought' to be many local solutions, in particular, there shouldn't be any equations that $n$ would have to satisfy in order to be the Gauss mapping of some immersion.


More explicitly, you can see how to write down all the possible $f$s for a given $n$ in the 'generic' case in which $n:U\to S^2$ is itself an immersion. [As Eremenko points out, in this case, a particular solution is just to take $f = n$.] 


Here's how to get the 'general' solution: Let $x$ and $y$ be the standard coordinates on $\mathbb{R}^2$ and note that the assumption that $n$ be an immersion is equivalent to the condition that $n\_x$ and $n\_y$ be linearly independent on $U$. Now, consider a possible solution $f$ and set $c = f\cdot n$, so that $f = c\ n + g$, where $g:U\to\mathbb{R}^3$ satisfies $n\cdot g = 0$. Then one has
$$
0 = n\cdot df = n\cdot (c\ dn + n\ dc + dg) = dc + n\cdot dg = dc - g\cdot dn + d(g\cdot n)
= dc - g\cdot dn
$$
(note that I have used that $n\cdot dn = \tfrac12d(n\cdot n) = 0$). This latter equation uncouples as the pair of equations 
$$
g\cdot n\_x = c\_x\qquad\text{and}\qquad g\cdot n\_y = c\_y\ ,\tag{1}
$$
which, together with $g\cdot n = 0$, determines $g$ completely in terms of $c$. 


Conversely, if we start with any function $c:U\to \mathbb{R}$ and define $g:U\to\mathbb{R}^3$ to be the unique vector field that satisfies $g\cdot n = 0$ and the equations (1), then setting $f = c\ n + g$ will yield a mapping $f:U\to\mathbb{R}^3$ with the property that $n\cdot df = 0$. Thus, wherever $f$ is an immersion, its Gauss map will be $n$. It is easy to see that for any $p\in U$, the generic choice of $c$ on a neighborhood of $p$ will give an $f$ that is an immersion, so there are many local solutions to this problem, and they essentially depend on one arbitrary function (i.e., $c$) of $2$ variables. 


How to determine the conditions that $c$ defined on $U$ should satisfy such that the resulting $f$ will be an immersion everywhere on $U$ is a global question that doesn't have anything to do with solving PDE.


In the case that $n$ is not an immersion, there are singularity issues where $dn$ changes rank, but the cases when the rank is constant are easy to deal with.


