Does a conformal map between two bounded convex surfaces exist generally?

Suppose we have two bounded surfaces, which are both flat or convex, is there a conformal map that links the two surfaces in general, such as $x^2+y^2<a^2, z=0$ and $z=\sqrt{a^2-x^2-y^2}$?

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The two surfaces should be the same topological type, and if that topological type is disk or sphere than the answer is yes, otherwise, not so much.

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If they both are topological spheres, then they are conformally equivalent, but not always if they are both topological disks. E.g., the punctured sphere and the hemisphere are not conformally equivalent, even though they are both topological disks. – Lukas Geyer Dec 5 '13 at 19:36
@LukasGeyer he did say bounded... – Igor Rivin Dec 5 '13 at 20:01
Yes, but the punctured sphere is bounded. Or do you think he meant "surface with boundary"? – Lukas Geyer Dec 5 '13 at 20:57
@LukasGeyer I meant surface with boundary indeed, such as a bump, partial cylindrical surface etc. Is there a general method to find the conformal map between the two surfaces, if it exists? I'm wondering if we can find a solution by solving a Laplace equation with certain boundary conditions. – Tony Dong Dec 6 '13 at 2:21
That's better. The best-known guy in the numerical analysis side is Nick Trefethen (google Trefethen/Driscoll, it's a very nice book). Start there. – Igor Rivin Dec 6 '13 at 2:38

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to either the unit disk, the whole plane, or the sphere. The question whether it is the sphere is easy to decide by topology, the question whether it is the disk or the plane is harder. E.g., a punctured sphere is conformally equivalent to the plane, a hemisphere is conformally a disk. (In these two examples it is easy to check that the uniformizing map is stereographic projection.)

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