# Solution existence in a pde system

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?

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.

• 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. Jan 12 at 12:16
• @exxxit8: "Yes" is my answer to your question about whether there is a technique for answering the existence question without actually calculating the solution. Jan 12 at 14:25
• Thank you dear Professor Bryant. Does this technique work also for $\Delta f=0$? Jan 12 at 18:49
• Dear Professor Bryant, Why you wrote $a(f)b(f)=1/2$, shouldn't it be $a(f)b(f)=f/2$? Jan 12 at 18:59
• @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. Jan 12 at 19:31