PDE with Laplacian and squared of the gradient

Question

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.

Answer 1

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.

Answer 2

No, every global solution is constant.

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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.

Answer 3

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.

Answer 4

If singularity is allowed ,a simpler counterexample may be ln(|x1| + 1)