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: $\Delta f=\frac{1}{\sqrt{|g|}} \partial_{i} (\sqrt{|g|}g^{ij} \partial_{j} f)$.


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?


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.

  • $\begingroup$ Thank you very much dear Professor Bryant for the wonderful explanation. Does your "YES" at the beginning of the message mean that it has a solution? because seeing your equation (1), I think my system has no solution. $\endgroup$
    – exxxit8
    Jan 12 at 12:16
  • 1
    $\begingroup$ @exxxit8: "Yes" is my answer to your question about whether there is a technique for answering the existence question without actually calculating the solution. $\endgroup$ Jan 12 at 14:25
  • $\begingroup$ Thank you dear Professor Bryant. Does this technique work also for $\Delta f=0$? $\endgroup$
    – exxxit8
    Jan 12 at 18:49
  • $\begingroup$ Dear Professor Bryant, Why you wrote $a(f)b(f)=1/2$, shouldn't it be $a(f)b(f)=f/2$? $\endgroup$
    – exxxit8
    Jan 12 at 18:59
  • 1
    $\begingroup$ @exxxit8: Yes to both your question comments: Taking $b(f) = 0$ works fine and I had a typo of $1/2$ that was supposed to be $f/2$ at the end. $\endgroup$ Jan 12 at 19:31

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