If I have a smooth positive scalar function $h$ defined on a 2-dimensional manifold $M$, then $h:M\rightarrow (0, \infty)$, where the metric of $M$ is $g=\frac{dx^2+dy^2}{y^2}$.


$h$ must satisfy the following $|\nabla h|^2=\frac{(h+1)^2}{2}$.


Considering that the gradient of a smooth function on manifold is $\nabla h=g^{ij}\frac{dh}{dx^j}e\_i$, which is one possible solution of that pde? I can not find one.


Or I just need to understand if it admits a solution.


EDIT:


I think that $\nabla h= y^2h\_xe\_1+y^2h\_ye\_2$,


where $h\_x$ and $h\_y$ are partial derivarives, and if I set $e\_1=(1,0)$ and $e\_2=(0,1)$,


I obtain: $y^2(h\_x, h\_y)$.


Now if I want to find a possible solution for $h$ such that $h\_x=0$, I obtain:


$y^2(0, h\_y)$ and my initial equation becomes:


$y^4(h\_y)^2=(h+1)^2/2$, or
$y^2(h\_y)=(h+1)/\sqrt{2}$.


This is a ODE... So, is it correct say that Its solution is a possible solution for $h$ such that $h\_x=0$?


EDIT 2 (after Bryant's answer):


if $h$ is such that must be satisfy:
$|\nabla h|^2=\frac{(h+1)^2}{2}$ and $\Delta h=0$, what change for the solution?


(with $\Delta h$ I mean the Palacian with positive sign, i e. for example if $M$ were $R^n$ it would be expressed as $\nabla h=h\_{xx}+h\_{yy}$).



Gradient of a function defined on a Riemannian-manifold
The requirement that $h$ be positive coupled with the assumption that the metric on $M$ be complete implies that there is no solution. It doesn't really matter what the metric is as long as it's complete. Here is why:


The equation $|\nabla h|^2 = \tfrac12 (h+1)^2$ implies that, if we set $f = \log (h+1)$, then we have $|\nabla f|^2 = 1/2$. Because the gradient of $f$ has constant length, it follows that the gradient lines of $f$ are geodesics and that, if $\gamma:(-\infty,\infty)\to M$ is a unit speed geodesic $f$-gradient line, oriented so that $f(\gamma(t))$ is increasing with $t$, then we have $f(\gamma(t)) = f(\gamma(0)) + t/\sqrt2$. It follows that
$$
h(\gamma(t)) = \mathrm{e}^{f(\gamma(0)) + t/\sqrt2} - 1,
$$
so that, as $t\to-\infty$, we will have $h(\gamma(t))\to -1 < 0$, contradicting the assumption that $h$ is positive on $M$.


**Addition (10/21/20):** Since the OP has added to the question and has asked for an answer to the additional question, here it is: Consider a more general situation, where a simply-connected Riemannian surface $(M^2,g)$ supports a function $h:M\to\mathbb{R}$ that satisfies $|\nabla h|^2 = F(h)^2$ and $\Delta h = 0$ where $F:\mathbb{R}\to\mathbb{R}$ is smooth and $h$ satisfies $F(h)>0$.


One can assume that the Riemannian surface is oriented, so that the Hodge star is well-defined. Then it follows that one can write $g = {\omega\_1}^2 + {\omega\_2}^2$ for some oriented coframing $(\omega\_1,\omega\_2)$ such that $\mathrm{d}h = F(h)\,\omega\_1$ (using the first equation). Then the second equation $\Delta h = 0$ is equivalent to $\mathrm{d}(\*\mathrm{d}h) = 0$, and, since $\*\mathrm{d}h = F(h)\,\omega\_2$, it follows that there is a function $y$ on $M$ so that $F(h)\,\omega\_2 = \mathrm{d}y$. Consequently,
$$
g = {\omega\_1}^2 + {\omega\_2}^2 = \frac{\mathrm{d}h^2 + \mathrm{d}y^2}{F(h)^2}.
$$
Thus, $(h,y):M\to \mathbb{R}^2$ defines a conformal submersion of $M$ into $\mathbb{R}^2$. If the metric on $M$ is to be complete, then $h$ must stay in an interval $I\subset\mathbb{R}$ on which $F(h)>0$ and the integration of the $1$-form $\eta = \frac{\mathrm{d}h}{F(h)}$ on $I$ must diverge to infinity at both ends of the interal. Conversely, these conditions are sufficient to guarantee that there is a solution to the above equations on a complete Riemannian surface, namely, take $M= I\times \mathbb{R}$ and let $g$ be as above (with coordinate $h:I\to\mathbb{R}$ in the first factor and $y$ on the second factor).


