PDE with Laplacian and squared of the gradient Let $u$ be a real function in $\mathbb{R}^2$. Does anybody know that the following PDE
$$\Delta  u+|\nabla  u|^2=0$$
 has any non-constant general solution or not? It would be appreciated if any one can give a reference for such PDE's.
 A: There are two helpful ways to rewrite this PDE: $e^{-u}\Delta e^u = 0$ and $\partial \bar{\partial} u + (\partial u) (\bar{\partial} u) = 0$, where $\partial = \frac{\partial}{\partial z}$, $\bar{\partial} = \frac{\partial}{\partial \bar{z}}$ are the usual complex differentials for $z=x+iy$.
Thus, it is easy to generate solutions by solving either the Laplace $\Delta e^u = 0$ or Cauchy-Riemann $\bar{\partial} u = 0$ equations. Although, since you want $u$ to be real, the latter doesn't help so much.
If you want $u$ to be singularity free, then $e^u$ must also be singularity free and in addition to be a positive harmonic function. All such functions are constants if they are globally defined on $\mathbb{R}^2$ (Liouville's theorem). But they do exist on smaller domains, like the unit disk.
A: No, every global solution is constant.

Suppose $u$ is a global solution to $\Delta u + |\nabla u|^2 = 0$, and define
$$ f(r) = -\int_{\partial B(x_0, r)} \partial_n u $$
(here $\partial_n$ is the outward normal derivative, and the integral is taken with respect to the arc-length measure). Then
$$ f(r) = -\int_{B(x_0, r)} \Delta u = \int_{B(x_0, r)} |\nabla u|^2 . $$
If $u$ is non-constant, we may choose $x_0$ so that $f(r) > 0$ for all $r > 0$. Differentiating with respect to $r$, we find that
$$ f'(r) = \int_{\partial B(x_0, r)} |\nabla u|^2 \geqslant \int_{\partial B(x_0, r)} |\partial_n u|^2 \geqslant \frac{1}{2 \pi r} \biggl(\int_{\partial B(x_0, r)} \partial_n u\biggr)^2 = \frac{(f(r))^2}{2 \pi r} \, .$$
Therefore,
$$ (-1 / f)'(r) \geqslant \frac{1}{2 \pi r} \, , $$
and thus
$$ \frac{1}{f(1)} - \frac{1}{f(R)} \geqslant \log R . $$
When $\log R > 1 / f(1)$, this is a contradiction.
A: There are separation-of-variables solutions of the form
$$\eqalign{u(x,y) &= X(x) + Y(y)\cr
\text{where}\cr
          X'' &= c - (X')^2 \cr
          Y'' &= -c - (Y')^2\cr}$$
There are radially symmetric solutions of the form
$$ u(x,y) = \ln(a \ln(x^2 + y^2) + b) $$
These will generally have singularities.  I don't know if there are any global solutions.
A: If singularity is allowed ,a simpler counterexample may be ln(|x1| + 1) 
