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?


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.

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  • $\begingroup$ Thank you very much, but if $f$ is not an harmonic function, (i.e. $\Delta f= a$ with $a$ constant not null), what's happen? $\endgroup$ – exxxit8 Aug 7 '19 at 12:14
  • $\begingroup$ what do you think will happen? I'll take guesses. $\endgroup$ – uniquesolution Aug 7 '19 at 12:16
  • $\begingroup$ I don't think change anything $\endgroup$ – exxxit8 Aug 7 '19 at 12:17

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