Considering a Jordan curve $\Gamma\subset \mathbb{R}^3$, we know from Douglas and Rado that there exists a unique minimal disk which bounds $\Gamma$. We also know that there exists exterior solutions, for example graph over $\mathbb{R}^2\setminus \Omega$ where $\Omega$ is the interior of the projection of $\Gamma$ over $\mathbb{R}^2\times\{0\}$ when the projection is one to one. Moreover we know that these surfaces are asymptotic to a plane or a catenoid. My question is about condition over Γ for existence of such minimal surface asymptotic to a plane? and do we know some curve for which there exits several solutions?
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2$\begingroup$ This was crossposted at math.SE: math.stackexchange.com/questions/35170 . Raphael, in general, posting the same question in multiple places at once is discouraged - you don't want to make someone work hard to giving you a great answer just to hear "thanks, but I've already got it from another forum". I'm unable to judge whether this question is appropriate for MathOverflow (since this isn't my area), but you should probably close one of the two versions you've posted. $\endgroup$– Zev ChonolesCommented Apr 26, 2011 at 11:55
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$\begingroup$ Zev, I ask to math.SE for deleting the other post since I think this is more appropriate for MO. $\endgroup$– RaphaelCommented Apr 26, 2011 at 14:09
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2$\begingroup$ The question has been removed from math.stackexchange. I agree that this is not, at first sight, an unreasonable question for MO. $\endgroup$– Willie WongCommented Apr 26, 2011 at 14:54
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Evidently what you want is Friedrich Tomi and Rugang Ye, "The Exterior Plateau Problem," Mathematische Zeitschrift, vol. 205 (1990), pages 233-245. I have only the first page available, but it appears that, once a normal direction to the surface at infinity is specified, so is the surface they construct. So there are, at least, infinitely many solutions. I should think their surfaces are planar at infinity, since a sufficently symmetric curve $$ x = \cos \theta, y = \sin \theta, z = \sin 4 \theta $$ with a catenoidal exterior minimal surface that faces "up" would have a second catenoidal surface that faces "down." I imagine they discuss uniqueness for planar solutions.
Actually, that shows me how to give a $\Gamma$ with three answers. Draw some very non-symmetric curve around the neck of the catenoid. This curve has two existing catenoidal exterior minimal surfaces, the upper end of the catenoid and the lower end. If Tomi and Ye's solution with the same normal at infinity is planar, that gives a third solution. Of course, this is not very different from saying that the catenoid and the plane give three solutions exterior to the unit circle in the $xy$ plane.
Note that, by the Meeks-Hoffman strong half space theorem, a properly immersed minimal surface in $R^3$ that lies between two parallel planes is itself a plane. This is a comment only on the situation when an exterior solution and an interior solution match up to give a complete surface.
answered Apr 26, 2011 at 19:36
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$\begingroup$ I will read attentively the paper of Ye and Tomi. But my question was about the uniqueness of solution with planar ends. Of course if $\Gamma$ is planar then the only solution should be the exterior of the plane. But if $\Gamma$ is not planar does their exists exterior solutions with different planar ends and how is characterize the normal direction at infinity? $\endgroup$– RaphaelCommented Apr 27, 2011 at 7:31
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$\begingroup$ ...Anyway, Thank you for your answer $\endgroup$– RaphaelCommented Apr 27, 2011 at 7:34
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$\begingroup$ You might also want to have a look to this other paper (specially section 3) of Tomi & Jorge: Tomi, F., and L. P. Jorge. "The exterior Plateau problem in higher codimension." Journal of Mathematical Sciences 149.6 (2008): 1741-1754. $\endgroup$– CoffeeCommented Feb 11, 2014 at 1:50