What curvature conditions make a surface rigid? Consider a compact surface $S$, possibly with boundary, embedded in $\mathbb{R}^3$, with the induced Riemannian metric.  I believe that if $S$ has constant positive Gaussian curvature (that is, $S$ is a piece of a sphere), it is rigid.  Here rigid means that all isometric embeddings are related by rigid transformations of the ambient space.
Are surfaces with nonconstant positive curvature also rigid?  Conversely, if $S$ has negative curvature, is it always locally nonrigid (a sufficient small neighborhood of any point is nonrigid)?
 A: Rigidity of closed surfaces of positive curvature in $R^3$ is a theorem by S. Cohn-Vossen from 1927. See for instance 
"Isometric Embedding of Riemannian Manifolds in Euclidean Spaces" by Q.Han, theorem 8.1.2. 
As for $C^2$-smooth closed surfaces of negative curvature in $R^3$, they do not exist; hence, one can say that they all are rigid. 
A: Can we then generalize with two statements in the following way?



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*All closed compact smooth surfaces of arbitrary Gaussian curvature and $ 4 \pi$ integral curvature  are rigid. Example surface shown above composed of continuous constant positive and negative regions is rigid..(please ignore all the lines drawn on the surface).

*All arbitrary compact smooth surfaces with boundary condition fixed as in eigenvalue problems or as in mechanics of materials finite element structural deformation problems in which all six degrees of freedom are annulled, are rigid. 
EDIT: I shall delete above image after sometime, in view of error (that no one pointed out ) I find subsequent to posting -- As second order derivatives along meridian are discontinuous, it is particularly not a good example. For further discussion, please consider the following surface formed by displaced revolved cosine curve $ y = 2 + cos(x) $  where edges $x= 0$ and $ x =  \pi$ are fixed in the manner of Eigenfunction solutions or FEA.
Can it then be mathematically considered rigid by satisfaction of above two conditions?

