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A tensor NURBS patch exactly covers a quarter annulus

Statement

Assume Countable Choice. Fix 0<r0<1. Use a linear radial coordinate ρ(s)=r0+(1−r0)s and a rational quadratic angular curve with control points Q0=(1,0), Q1=(1,1), Q2=(0,1) and weights (1,1/2,1). Tensor their bases, use control points Pij=riQj for ri∈{r0,1}, and tensor weights wij=wj. The resulting patch maps [0,1]2 bijectively onto {(x,y):r0≤x2+y2≤1, x,y≥0} with strictly positive Jacobian, and its exact map satisfies the norm-equivalence hypotheses of Bi-Lipschitz NURBS maps preserve H1 and energy norms.

Facts & Assumptions

Given: The radial degree-one basis, the rational quadratic angular basis and controls above, and the positive parameter r0.

[A1]

The Axiom of Countable Choice (ACω) is used only for the pullback norm equivalence in [F3]; the explicit geometry calculations do not require choice.

[F1]

Tensor-product NURBS patch maps and rational spline spaces: tensor-product rational geometry uses the weighted denominator and positive-weight control points.

[F2]

Positive weights give a bounded partition-of-unity NURBS basis: positive weights and the tensor B-spline partition yield a positive denominator and a rational partition of unity.

[F3]

Bi-Lipschitz NURBS maps preserve H1 and energy norms under Countable Choice: a global bi-Lipschitz patch with positive bounded Jacobian and bounded derivative/inverse derivative has the stated H1 and energy norm equivalences.

Proof

1.1F1F2givenalgebra

Put a=2 and x0(t)=(1−t)2+at(1−t),y0(t)=t2+at(1−t),D(t)=(1−t)2+at(1−t)+t2. The weighted middle Bernstein coefficient is 2(1/2)t(1−t)=at(1−t), so the angular map is (X,Y)=(x0/D,y0/D). Direct expansion gives x02+y02−D2=−2t2(1−t)2+a2t2(1−t)2=0, so X2+Y2=1; also X,Y≥0 and D>0. The tensor-product weights and controls therefore give F(s,t)=(ρ(s)X(t),ρ(s)Y(t)).

2.1step 1.1algebra

Expanding the numerator derivatives gives x0y0′−y0x0′=a+2(1−a)t(1−t)=aD(t), because D(t)=1+(a−2)t(1−t) and a(a−2)=2(1−a). Consequently XY′−YX′=a/D(t)>0. Since X2+Y2=1, the angular coordinate θ(t)=arctan⁡(Y/X) has derivative θ′=XY′−YX′=a/D(t)>0 on the interior and extends monotonically from 0 to π/2. Thus t↦(X,Y) traverses the first-quadrant unit-circle arc exactly once. Since ρ increases bijectively from r0 to 1, F is injective and its image is exactly the stated quarter annulus.

3.1A1F1F3step 1.1step 2.1givenalgebra∎

The Jacobian is det⁡DF(s,t)=ρ′(s)ρ(s)(XY′−YX′)=(1−r0)ρ(s)2D(t)>0. On the closed square, ρ≥r0>0 and D is continuous and positive, so this determinant has positive minimum and finite maximum; DF is continuous and bounded, and the inverse matrix is bounded by its cofactor formula divided by the positive determinant. The physical inverse first recovers ρ=x2+y2 and the angle θ=atan2⁡(y,x). This angle is Lipschitz on the annular sector because ρ≥r0, and the inverse parameter t(θ) is Lipschitz because θ′(t)=2/D(t) has a positive minimum. Together with the linear inverse for s, this makes F−1 Lipschitz; bounded DF makes F Lipschitz. The quarter annulus has a finite piecewise smooth Lipschitz boundary, so all admissibility hypotheses in [F1] hold. Applying [F3] gives the announced pullback norm equivalence.

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