Problem about solution of an PDE system For the following system:
$\begin{cases} |\nabla f|^2=1 \\\\ \Delta f=0
\end{cases}$
where $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ and $\nabla f$, $\Delta f$ are Gradient and Laplacian of $f$, respectively.
Are there solutions for this system where the $f$ function is not an affine function?
 A: No, there are no solutions $f$ where the function $f$ is not an affine function. 
To see this, note that differentiation of the identity $|\nabla f|^2=1$, once by $x$ and another time by $y$,
$$|\nabla f|^2=f_x^2+f_y^2\equiv1$$
implies
$$\langle\nabla f,(f_{xy},f_{yy})\rangle = 0$$
and also
$$\langle\nabla f, (f_{xx},f_{xy})\rangle = 0$$
where $\langle\cdot\rangle$ is the Euclidean dot product. We have two vectors perpendicular to the (non-zero) gradient, so they must be proportional, i.e., there exists a number $c(x,y)$ such that
$$(f_{xx},f_{xy})=c(x,y)(f_{xy},f_{yy})\quad\quad (1)$$
Note that if $f_{yy}\neq 0, f_{xx}\neq 0$ at a certain point, then $c(x,y)\neq 0$.
So far we have only used the fact that the gradient has unit length. Now we use the second condition, that the Laplacian vanishes. This means that $f$ is harmonic, and so it is the real part of some analytic function
$$F(x,y)=f(x,y)+ig(x,y)$$
Since $F$ is analytic, the Cauchy-Riemann equations are valid, so that
$$f_x=g_y,\quad f_y=-g_x$$
Therefore
$$f_{xy}=g_{yy}=-g_{xx}$$
Consequently,
$$(f_{xy},f_{yy})=(g_{yy},f_{yy}),\quad (f_{xx},f_{xy})=(f_{xx},-g_{xx})$$
so that (1) implies
$$(f_{xx},-g_{xx})=(f_{xx},f_{xy})=c(x,y)(f_{xy},f_{yy})=c(x,y)(g_{yy},f_{yy})=-c(x,y)(g_{xx},f_{xx})$$
but the vector $(g_{xx},f_{xx})$ is clearly perpendicular to the vector $(-f_{xx},g_{xx})$, so it cannot be proportional to it, unless $c(x,y)=0$, which implies that $f_{xx}=f_{yy}=0$, and subsequently also $f_{xy}=0$,   which implies that $f$ is an affine function.
