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


$f$ must satisfy the following PDEs:


$\begin{cases} 
\Delta f=f/2 \\ |\nabla f|^2=\frac{(f^2+3f)}{2}+1
\end{cases}$


Considering that $\nabla f$ is the gradient of $f$, where $f$ is a smooth positive function on manifold $M$, where $M$ is the Poincaré half plan, so the gradient is referred to the metric: $g=\frac{dx^2+dy^2}{y^2}$; (The gradient of a smooth function on a manifold is $\nabla f=g^{ij}\frac{\partial f}{\partial x^j} \frac{\partial}{\partial x\_i}$), and $\Delta f$ is the Laplace-Beltrami operator for $f$ on manifold $M$ (so referred again to the metric $g=\frac{dx^2+dy^2}{y^2}$), and the Laplace-Beltrami of a smooth function on a manifold is determined by: $\Delta f=\frac{1}{\sqrt{|g|}} \partial\_{i} (\sqrt{|g|}g^{ij} \partial\_{j} f)$.


QUESTIONS:


Is there a possible solution of that pde system? and if "yes",


How can it be shown that that pde system admits at least one solution without having to calculate it?


Is there a technique to understand if a solution exists even without calculating it?


**Sorry if I reactivate this question:**


How does equation (1) vary if, instead of considering a manifold with constant Gaussian curvature -1, we consider in a more general way a manifold with constant negative Gaussian curvature?



Solution existence in a pde system
Yes. Here is a general approach to this problem: Suppose that one has two functions a>0 and b on some interval $I\subset\mathbb{R}$ and one wants to know whether there is a solution $f$ to the system
$$
|\nabla f|^2 = a(f)^2,\qquad \Delta f = a(f)b(f)
$$
on some (nonempty) open set in the Poincaré upper half plane (i.e., a Riemannian surface with Gauss curvature identically equal to $-1$).


Then I claim that there is such a solution $f$ if and only if the functions $a$ and $b$ satisfy the differential equation
$$
a(t)a''(t)-a'(t)^2 + 2b(t)a'(t)-a(t)b'(t)-b(t)^2 + 1 = 0.\tag1
$$


To see this, note that, if such an $f$ exists, then the metric $g$ can be written in the form $g = {\omega\_1}^2 + {\omega\_2}^2$ where $\omega\_1 = (\mathrm{d}f)/a(f)$
and where $\ast\mathrm{d}f = a(f)\,\omega\_2$. Since $\mathrm{d}(\ast\mathrm{d}f) = \Delta f\,\omega\_1\wedge\omega\_2$, it follows that $\mathrm{d}(a(f)\,\omega\_2) = a(f)b(f)\,\omega\_1\wedge\omega\_2$. This implies, since $\mathrm{d}(a(f)) = a'(f)\,\mathrm{d}f=a(f)a'(f)\,\omega\_1$, that we must have $\mathrm{d}\omega\_2 = \bigl(b(f)-a'(f)\bigr)\,\omega\_1\wedge\omega\_2$. Next, since $\mathrm{d}\omega\_1 = -\omega\_{12}\wedge\omega\_2$ and $\mathrm{d}\omega\_2 =\omega\_{12}\wedge\omega\_1$, it follows that $\omega\_{12} = \bigl(a'(f)-b(f)\bigr)\,\omega\_2$. Finally, the equation $\mathrm{d}\omega\_{12} = K\,\omega\_1\wedge\omega\_2 = -\omega\_1\wedge\omega\_2$ expands to yield the equation (1).


Conversely, if $a$ and $b$ satisfy (1), consider the equations
$$
\omega\_1 = \mathrm{d}f/a(f),\quad 
\mathrm{d}\omega\_2 = \bigl(b(f)-a'(f)\bigr)\,\omega\_1\wedge\omega\_2
= \bigl(b(f)-a'(f)\bigr)/a(f)\,\mathrm{d}f\wedge\omega\_2\tag2
$$
By linear ODE, there will exist a function $c>0$ on the interval $I$ (unique up to a constant multiple) such that
$$
c'(f) = c(f) \bigl(b(f)-a'(f)\bigr)/a(f).\tag3
$$
Then the above equations (2) and (3) imply that $\mathrm{d}\bigl(\omega\_2/c(f)\bigr)=0$. Consequently, assuming that the domain is simply-connected, $\omega\_2 = c(f)\,\mathrm{d}h$ for some function $h$. Now, the equations (1) and (3) imply that the metric
$$
g = \left(\frac{\mathrm{d}f}{a(f)}\right)^2 + \left(c(f)\,\mathrm{d}h\right)^2
$$
on $I\times\mathbb{R}$ (with coordinates $f$ and $h$) has constant Gauss curvature -1, and hence is isometrically immersed onto a domain in the Poincaré upper half plane.


In the OP's particular case, it suffices to solve the equations
$$
a(f)^2 =\frac{f^2+3f}2 + 1\qquad \text{and}\quad a(f)b(f) = f/2
$$
for $a$ and $b$ and then check whether (1) is satisfied.


**The more general case of constant negative curvature:** If the givn metric $g$ has constant Gauss curvature $K<0$, then consider the rescaled metric $\bar g = (-K)\, g$, which has curvature $\bar K = -1$. We also have
$$
|\nabla f|^2\_{\bar g} = (-K)^{-1}\,|\nabla f|^2\_{g}
\quad\text{and}\quad \Delta\_{\bar g} f = (-K)^{-1}\,\Delta\_g f\,,
$$
so it's now easy to figure out what $\bar a$ and $\bar b$ are. Thus, the correct condition in the more general case is to apply (1) to $(\bar a,\bar b)$.


