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?

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?


1 Answer 1


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 $(\bar a,\bar b)$.

  • $\begingroup$ 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, 2021 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, 2021 at 14:25
  • $\begingroup$ Thank you dear Professor Bryant. Does this technique work also for $\Delta f=0$? $\endgroup$
    – exxxit8
    Jan 12, 2021 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, 2021 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, 2021 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.