Once differentiable, piecewise degree three polynomials on triangulated planar domains

2011-08-05T16:21:22Z

Here is an easily described, but very difficult, problem that I (and a number of other people) really would like to see solved during our life times. The basic problem is to compute the dimension of a certain vector space. Suppose we are given a triangulation (a tessellation by triangles where any two triangles share at most a common edge or a common vertex) of a polygonal domain in the plane. Let $S$ be the space of once differentiable functions on that triangulation that on each triangle can be represented as a bivariate polynomial of degree 3. $S$ is clearly a vector space. It is known that the dimension of that space is greater than or equal to $3V_B + 2V_I + 1 + \sigma$ where $V_B$ is the number of boundary vertices, $V_I$ is the number of interior vertices, and $\sigma$ is the number of singular vertices of the underlying triangulation. A singular vertex is an interior vertex that has exactly four edges attached where those edges form two parallel pairs. (In other words, a singular vertex is the intersection of the diagonals of a convex quadrilateral.) It is known that generically the dimension of $S$ equals the given expression, and there is no case known where the dimension is larger than that expression. Many people conjecture that the lower bound equals the dimension for all triangulations. Prove, or disprove, that conjecture.

Some background: triangulations are the natural generalization of a partition of an interval to two variables, and $S$ is a spline space with potential for a wide range of practical problems, such as data fitting or solving partial differential equations. The problem has been known among approximation theorists since the early 1970s, and despite efforts by a number of people the problem is still unsolved. For more information on spaces like $S$ see the recent book by Lai and Schumaker [Spline Functions on Triangulations, Cambridge University Press] in particular section 9.9. The basic issue with the kind of spline space considered here is that the dimension depends not just on the topology of the triangulation, but also on its geometry. An arbitrarily small change of the location of the vertices can change the dimension. If the polynomial degree is four (instead of three) it is known that the dimension can change only when a vertex switches between being singular and non-singular. On the other hand, if the polynomial degree is two (instead of three) many configurations other than singular vertices are known where the dimension changes with the geometry. So the polynomial degree three straddles the boundary, and we would like to know to which camp it belongs.

Once differentiable, piecewise degree three polynomials on triangulated planar domains - MathOverflow

Once differentiable, piecewise degree three polynomials on triangulated planar domains