Pde system problem

Question

I have a big problem to solve this system $\Delta f-hf^2=0$

$|\nabla f|^2+hf^3=0$

where $h$ is a constant, $f$ is a 2-dimensional smooth function, $\Delta f$ is Laplacian of $f$ (i.e. $\Delta f=f_{xx}+f_{yy}$) and $\nabla f$ is the gradient of $f$.

ADD In first case $f$ is defined on $R^2$ and in second case $f$ is defined on surface $S$ ($f:S \rightarrow (0, \infty)$). is there a solution? Thank you for help

MODIFICATION after Igor Khavkine answer: and if the system is

$\Delta f-hf^2+cf=0$

$|\nabla f|^2+hf^3=0$

(c is another constant)

Answer 1

I assume that, in the surface case, the OP wants to interpret $S$ as a surface endowed with a Riemannian metric and wants to understand the solutions to the equations $\Delta f - hf^2 = 0$ and $|\nabla f|^2 + hf^3 = 0$ for a given constant $h$.

Clearly, if $h=0$, the only solutions are to have $f$ be constant, so one can assume that $h\not=0$. Then, setting $u = -hf$, the above equations are equivalent to $\Delta u + u^2 = 0$ and $|\nabla u|^2 - u^3 = 0$, so it suffices to solve these latter equations.

Let $g$ be the metric on $S$ and assume that $u$ is a nonzero (and, hence, necessarily positive) solution to the above equations on a simply-connected open subset $S'\subset S$. The second equation implies that $\omega_1 = u^{-3/2}\,\mathrm{d}u$ is a $1$-form with $g$-norm $1$ on $S'$, and hence $g$ can be written in the form $g = {\omega_1}^2 + {\omega_2}^2$ on $S'$ for some $\omega_2$, which is also a unit 1-form.

Fix an orientation by requiring that $\omega_1\wedge\omega_2$ be the $g$-area form on $S'$. Then $\star \mathrm{d}u = u^{3/2}\,\omega_2$, and since $\mathrm{d}(\star \mathrm{d}u) = \Delta u\, \omega_1\wedge\omega_2$, it follows that $$ \tfrac32\,u^{1/2}\mathrm{d}u\wedge\omega_2 + u^{3/2}\,\mathrm{d}\omega_2 =\mathrm{d}(u^{3/2}\,\omega_2) = -u^2\,\omega_1\wedge\omega_2 = -u^{1/2}\,\mathrm{d}u\wedge\omega_2\,, $$ or $$ \mathrm{d}\omega_2 = -\tfrac52\,u^{-1}\,\mathrm{d}u\wedge\omega_2\,, $$ which can be written in the form $$ \mathrm{d}\bigl(u^{5/2}\,\omega_2\bigr) = 0. $$ Since $S'$ is simply connected, it follows that there exists a function $v$ on $S'$ such that $u^{5/2}\,\omega_2 = \mathrm{d}v$. Consequently, the metric $g$ has the form $$ g = {\omega_1}^2 + {\omega_2}^2 = u^{-3}\,\mathrm{d}u^2 + u^{-5}\,\mathrm{d}v^2. $$ This metric can be placed in 'polar form' by setting $u = 4r^{-2}$ and $v = 32\theta$, in which case, it becomes $$ g = \mathrm{d}r^2 + r^{10}\,\mathrm{d}\theta^2, $$ This is a singular, incomplete metric at $r=0$ (where $u$ goes to $\infty$), though it is complete at $r=\infty$. Its Gauss curvature is $K = -(r^5)''/r^5 = -20 r^{-2} = -5u < 0$.

The original $f$ then takes the form $f = -u/h = -4/(hr^2)$.

Thus, up to isometry, there is essentially only one solution to the OP's original problem. (In particular, as Igor showed in his answer, there is no nonconstant solution when the background metric $g$ is flat.) The modified problem (with the additional constant $c$) can be solved using essentially the same techniques.

Answer 2

I will assume that your system is simply defined on $\mathbb{R}^2$. I don't know what you mean for this system to be defined on a surface $S$, since you haven't told us anything about $S$. Note that your equations are in the form $E_2 := f_{xx} + f_{yy} -hf^2 = 0$ and $E_1 := f_x^2+f_y^2 + hf^3=0$. I will assume also that $h$ is a non-zero constant.

If you differentiate $E_2$ you can algebraically solve $(E_2,(E_1)_x, (E_1)_y) = 0$ for the highest derivatives and get \begin{align*} F_1 &:= f_{xy} - 4 f_x f_y/f = 0, \\ F_2 &:= f_{xx} + 4 f_y^2/f + 3hf^2/2 = 0, \\ F_3 &:= f_{yy} + 4 f_x^2/f + 3hf^2/2 = 0. \end{align*}

If you cross-differentiate the $F$-equations you can algebraically eliminate all second derivatives from $((F_1)_x - (F_2)_y, (F_1)_y - (F_3)_x) = 0$ and end up with two first order equations. Using $E_1$ to simplify them, unless I made some calculation mistake, the result is equivalent to $f_x = 0$ and $f_y = 0$. Finally, eliminating all derivatives from $E_1$, we end up with the condition $f=0$. Thus $f(x,y) \equiv 0$ is the only solution to your system of equations.

If $h=0$, then after the first step, you end up with the equations \begin{align*} F_1 &:= f_{xy} = 0, \\ F_2 &:= f_{xx} = 0, \\ F_3 &:= f_{yy} = 0. \end{align*} The only solutions are $f(x,y) = Ax+By+C$ with constants $A$, $B$, $C$. Equation $E_1=0$ forces the constraint $A^2+B^2=0$. So, if you only want positive solutions, the only possibility is $f(x,y) \equiv C$, for some positive constant $C>0$.