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Positive NURBS weights do not imply an admissible patch

Statement

On [0,1]2, take degree-one tensor Bernstein functions in each coordinate, all four weights equal to one, and control net P00=(0,0),P10=(1,0),P01=(0,1),P11=(1,−1). The resulting positive-weight bilinear NURBS map is F(ξ,η)=(ξ,η(1−2ξ)). Its Jacobian vanishes at ξ=1/2, changes sign across that line, and the map is not injective; hence positive weights alone do not give an admissible patch.

Facts & Assumptions

Given: The degree-one tensor Bernstein basis, the four control points and unit weights in the Statement.

[F1]

Tensor-product NURBS patch maps and rational spline spaces defines F as the rational weighted sum of its control points and requires global injectivity and positive cell Jacobian for admissibility.

[F2]

Positive weights give a bounded partition-of-unity NURBS basis: unit weights yield denominator one and nonnegative basis functions summing to one.

Proof

1.1F1F2givenalgebra

The degree-one Bernstein functions are 1−ξ,ξ and 1−η,η. With all weights one, [F2] gives W=1, so the tensor rational basis is the ordinary bilinear basis. Its first component is ξ and its second component is (1−ξ)η+ξ(−η)=η(1−2ξ).

2.1F1step 1.1algebra∎

Differentiation gives DF=(10−2η1−2ξ),det⁡DF=1−2ξ. Thus the Jacobian is zero on ξ=1/2, positive for ξ<1/2, and negative for ξ>1/2. Moreover, F(1/2,0)=F(1/2,1)=(1/2,0) with distinct parameter points. These violate two separate admissibility requirements in [F1] despite all weights being positive.

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