On a flat disk, I can solve the Laplace's $\nabla^2 u = 0$ equation with different boundary conditions on semi circles like shown here 
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The answer to this is:
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Here $a$ is the disk radius, and sum is over positive odd integers.
This is an unsolved problem in ML Boas's book (pg 644), but I'm interested in a more general case where the disk is hyperbolic and the boundary conditions on upper and lower semi circles are $u_1$ and $u_2$.
If the disk was flat, it's easy to adjust the constants to get the answer $$ u = \frac{u_1 + u_2}{2} + \frac{2}{\pi} (u_1 - u_2) \sum_{odd\; n} \left(\frac{r}{a}\right)^n \frac{\sin n\theta}{n} $$
Now I'm trying to see what should happen when disk is hyperbolic, and I'm getting confused. I know that in general for 2d spaces one can write the metric as $$ ds^2 = f(z, \overline{z}) dz d\overline{z} $$ and therefore Laplace's equation on hyperbolic disk will reduce to Laplace's equation on the flat disk (only in the $z, \overline{z}$ coordinates?) and therefore the above solution should hold, provided I identify what is my $r$ and $a$ correctly.
The correct mapping between coordinate systems I suppose then should be polar to hyperbolic polar for which $$ ds^2 = d\rho^2 + \sinh^2\rho d\phi^2 $$
Then that would mean $r \to \rho$ and $\theta \to \phi$. So
$$ u = \frac{u_1 + u_2}{2} + \frac{2}{\pi} (u_1 - u_2) \sum_{odd\; n} \left(\frac{\rho}{a}\right)^n \frac{\sin n\phi}{n} $$
should be the answer?
Does this make sense, or am I making some stupid mistake? How would you solve this problem?