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The Beltrami Pseudosphere

$$[x = a \sin p \cos t , y= - a ( \cos p + \log \tan p/2 ) , z= b+ a \sin p \sin t \; ], (.1 <p<\pi/2), (0< t< 2 \pi), \; (b>a) $$

is bent to a non-axisymmetric surface so that its straight axis of symmetry along $z=b$ before bending goes to circle $ y^2+z^2 = b^2 $ to radius $b$ in an isometric mapping preserving its Gauss curvature $ K=-1/a^2 $. The plot is drawn for $a=1,b=2$ units.

EDIT1:

Bending is such that 1) Opposites points of a diameter on the meridian in $yz$ plane before deformation remain in the $yz$ plane and 2) Normals intersecting on pseudosphere straight axis before deformation intersect on circle $ y^2+z^2= b^2 $ after deformation as sketched.

How is a parametrization of bent surface found ?

It is expected that the cuspidal circle to distort and smaller tubular part getting split as positive $K$ cannot develop in a closed toroidal configuration.

BentPseudosphere

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    $\begingroup$ I don't think such a thing is possible if the surface is to be at least $C^2$. While you haven't described exactly what you want, it appears that such a surface would have to have a point where $z$ reaches a local minimum (down near the bottom of the circle you have drawn). However, a surface of negative curvature cannot have a point at which $z$ is a local minimum since a negative curvature surface, being locally saddle-shaped, must always cut across its tangent planes. $\endgroup$
    – Robert Bryant
    Commented Jul 6, 2016 at 11:35
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    $\begingroup$ Even with this final remark added (clarifying that you don't want to maintain the entire tubular structure, just the patch $0<t<2\pi$), you haven't described exactly what you want the surface to be. The 'axis of symmetry' is not part of the original surface, so you are only roughly describing where you want the 'bent' surface to go, so there probably no uniqueness and hence no way to determine an explicit parametrization. Presumably, you want the curve $t=\pi/2$ to spiral in to the circle and lie in the plane $x=0$, but you will need to say exactly how you want that to happen. $\endgroup$
    – Robert Bryant
    Commented Jul 6, 2016 at 13:41
  • $\begingroup$ Thanks for giving attention for the question. My question certainly needs a revision, and sorry about that ... the axis of symmetry $ z=b$ is not on the surface but in the air, unconnected I shall collect my thoughts and pose it properly. Regards $\endgroup$
    – Narasimham
    Commented Jul 6, 2016 at 15:10
  • $\begingroup$ Is the problem now better posed? $\endgroup$
    – Narasimham
    Commented Jul 7, 2016 at 14:33
  • $\begingroup$ Better posed, but not well-posed. Basically, all you are requiring is that the the bent surface should intersect the plane $x=0$ at right angles in two curves that are asymptotic (somehow) to the circle $y^2+z^2=b^2$ in the $x=0$ plane. There is still a lot of choice for these two curves, but the two curves are not independent. Basically, once you choose the intersection curve outside the circle, the intersection curve inside the circle with be uniquely determined, if it exists at all. To determine a formula, you would need to at least specify the outer curve completely. $\endgroup$
    – Robert Bryant
    Commented Jul 7, 2016 at 15:06

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