Hermite Interpolation Using Möbius Transformations of Planar Pythagorean-Hodograph Cubics

and Applied Analysis 3 plane curves with rational offets. These curves are represented in the dual form, in which curves are specified using line coordinates instead of point coordinates. Pottmann showed how to design rational PH curves segments by G1 and G2 Hermite interpolations 17, 18 . However, in our work, what we need is only a suitable PH cubic and a PH-preserving transformation which is algebraically simple as possible and which can generate an extraparameter, and the latter is completely settled by the classical Möbius transformation. Moreover, the transformation is biholomorphic. Thus it preserves the topology of the preimage curve PH cubic . Therefore, the interpolants obtained by our method should have no cusp, although cusps are a generic feature of rational PH curves. They are simple curves or else loops. Hence, to obtain these, even avoiding the easy shortcut, there is no need to follow up the lengthy path with a far starting point. We just use the classical Möbius transformation of PH cubics, that is all. The rest of this paper is organized as follows. In Sections 2 and 3, we review some basic properties of Möbius transformations and planar PH cubics. In Section 4, we solve the C1 Hermite interpolation problem using the Möbius transformations of planar PH cubics. In Section 5, we present the condition on a Hermite dataset, which determine whether the corresponding Hermite interpolant has a loop, we also compare these new interpolants with PH quintics and show that the former have improved stability. We conclude this paper in Section 6. 2. Möbius Transformations A Möbius transformation Φ z is a bijective linear fractional transformation in the extended complex plane C∞ C ∪ {∞}, that is,


Introduction
Farouki and Sakkalis 1 introduced Pythagorean-hodograph PH curves, which are a special class of polynomial curves with a polynomial speed function.These curves have many computationally attractive features: in particular, their arc lengths and offset curves can be determined exactly.Farouki 2 reviews the abundant results on these curves obtained by many researchers.Hermite interpolation with PH curves is one of the main topics in this research Farouki and Neff 3 , Albrecht and Farouki 4 , J üttler 5 , J üttler and Mäurer 6 , Farouki et al. 7 , Pelosi et al. 8 , and Šír et al. 9 .In this paper, we solve the C 1 Hermite interpolation problem using the M öbius transformations of polynomial PH cubics in the plane.The use of M öbius transformation has been demonstrated in recent publications 10, 11 .In 11 , Bartoň et al. used a general M öbius transformation in R 3 , which is defined as a composition of an arbitrary number of inversions with respect to spheres or planes.They showed that μ • x t is a rational PH curve for any general M öbius transformation μ x 1 , x 2 , x 3 , if x t is a polynomial PH space curve in R 3 .The preservation of PH properties under transformation is first studied by Ueda 12 .They also presented an algorithm for G 1 Hermite interpolation.In this work, we use the classical M öbius transformation, a bijective linear fractional transformation in the extended complex plane for some complex numbers a, b, c, and d for which ad − bc / 0 13 .Using this transformation, we can solve the C 1 Hermite interpolation problems with PH cubics.In general, with PH cubics, we cannot solve C 1 Hermite interpolation problems, since their lack of parameters makes the problems overdetermined.But we can show that, for a C 1 Hermite interpolation problem, we are always able to obtain four interplants which are constructed by PH cubics.The M öbius transformation makes this possible, since it permits the introduction of a new extra parameter into the problem, which is to be reduced to a simple problem to determine PH cubics as follows: here we adapt the complex representation method 14 to solve the C 1 Hermite interpolation problem.The original problem is, for a Hermite dataset p i , p f , v i , v f , to find a polynomial PH curve r t and a M öbius transformation Φ z , which satisfy Next, by an appropriate translation, rotation, and scaling of the dataset, we can arrange that p i 0 and p f 1 and take a Möbius transformation which fixes 0 and 1, for some nonzero complex number α.Then the inverse image of the C 1 Hermite dataset under a M öbius transformation Φ makes 1.2 into which are suitable forms for adapting the complex representation method for details, see Section 4 .Farouki and Neff 3 already solved the C 1 Hermite interpolation problem with PH quintics.According to 1.2 , this is exactly the case in which r t is a quintic and Φ z is the identity, that is, α 1.On the other hand, our work in this paper is the case just when Φ z is not the identity, that is, α / 1.At the end of this paper, we will compare our interpolants with PH quintic ones for the same C 1 Hermite dataset.The interpolants obtained by our method are specific rational curves represented by complex rational functions.For planar rational curves, there already exists a general theory, which were introduced by Pottmann 15 and Fiorot and Gensane 16 : they studied rational plane curves with rational offets.These curves are represented in the dual form, in which curves are specified using line coordinates instead of point coordinates.Pottmann showed how to design rational PH curves segments by G 1 and G 2 Hermite interpolations 17, 18 .However, in our work, what we need is only a suitable PH cubic and a PH-preserving transformation which is algebraically simple as possible and which can generate an extraparameter, and the latter is completely settled by the classical M öbius transformation.Moreover, the transformation is biholomorphic.Thus it preserves the topology of the preimage curve PH cubic .Therefore, the interpolants obtained by our method should have no cusp, although cusps are a generic feature of rational PH curves.They are simple curves or else loops.Hence, to obtain these, even avoiding the easy shortcut, there is no need to follow up the lengthy path with a far starting point.We just use the classical M öbius transformation of PH cubics, that is all.
The rest of this paper is organized as follows.In Sections 2 and 3, we review some basic properties of M öbius transformations and planar PH cubics.In Section 4, we solve the C 1 Hermite interpolation problem using the M öbius transformations of planar PH cubics.In Section 5, we present the condition on a Hermite dataset, which determine whether the corresponding Hermite interpolant has a loop, we also compare these new interpolants with PH quintics and show that the former have improved stability.We conclude this paper in Section 6.

M öbius Transformations
A Möbius transformation Φ z is a bijective linear fractional transformation in the extended complex plane for some complex numbers a, b, c, and d for which ad − bc / 0 13 .Then Φ z is a one-to-one correspondence on the extended complex plane C ∞ with its inverse The derivative of Φ z is For any M öbius transformations Φ z and Ψ z , Ψ • Φ z is also a M öbius transformation.is a rational curve, also of degree n, with coefficients in the complex plane C.However, if we associate the complex plane C with R 2 , and express Φ • r t as a rational curve with real coefficients in R 2 then the result is generally a rational PH curve of degree 2n or n.If we perform a further M öbius transformation Ψ z , the rational curve Then there exist a polynomial curve s t and a Möbius transformation Ψ z , such that s 0 0, s 1 1, and Φ • r t Ψ • s t .
Proof.We can find a M öbius transformation Φ 1 z az b for some complex constants a and b such that a / 0 and s t Φ 1 • r t is a polynomial curve with s 0 0 and s 1 1.Consequently, we can obtain the M öbius transformation Ψ z A Möbius transformation Ψ z of this sort also fixes 0 and 1. Lemma 2.3.Let Φ z be a Möbius transformation with Φ 0 0 and Φ 1 1.Then there exists a nonzero complex constant α such that Proof.Let Ψ z az b / cz d be a M öbius transformation with Ψ 0 0 and Ψ ∞ ∞.Then from Ψ 0 0 we get b 0, and from Ψ ∞ ∞ we get c 0. Thus Ψ z αz, where α a/d.Let τ |α| and η arg α .Then we obtain Ψ z Now let Φ z be a M öbius transformation with Φ 0 0 and Φ 1 1.Let S z z/ z − 1 , so that S 0 0 and S ∞ 1.Then, since Ψ z , where Ψ z αz for some nonzero α.Thus we obtain Moreover, in the same way, we can obtain

Planar Pythagorean-Hodograph Cubics
A planar polynomial curve r t x t √ −1y t is a PH curve 19 if and only if there exist polynomials h t , u t , and v t , which satisfy In this paper, we will assume that h t is monic, meaning that its leading coefficient is 1.
A polynomial curve r t is a PH curve 14 if and only if there exists a polynomial h t and a polynomial curve w t such that r t h t w t 2 .

3.2
Suppose that the PH cubic r t is a line.Then the hodograph r t can be expressed as h t x 0 √ −1y 0 , where x 0 √ −1y 0 is a nonzero point and h t is the quadratic monic polynomial and h 0 , h 1 , and h 2 are real constants such that h 0 h 2 / 2h 1 .Let r t be a PH cubic for which r t w t 2 .Since w t is linear, we can write w t in Bernstein form: where w 0 and w 1 are distinct complex constants.The hodograph r t can then be expressed as

3.5
If we represent the PH cubic r t in the Bernstein form then we obtain where p 0 can be chosen arbitrarily.

First-Order Hermite Interpolation
We will now solve the C 1 Hermite interpolation problem using M öbius transformations of PH cubics.
Let p i and p f be the initial and final points to be interpolated, where p i / p f .Let v i r i e √ −1θ i and v f r f e √ −1θ f , respectively, be the initial vector at p i and the final vector at p f , where r i > 0 and r f > 0. For this Hermite dateset p i , p f , v i , v f , we want to find planar PH cubics r t and M öbius transformations Φ z which satisfy 1.2 , which are equivalent to

4.1
By an appropriate translation, rotation, and scaling of the data-set, we can arrange that p i 0 and p f 1.Then, from Lemmas 2.2 and 2.3, we seek some nonzero constants α and PH cubics r t , which satisfy 1.4 .

Case of r t h t h
In this case, 4.1 become From the second and third of these equations, we can see that Hermite interpolants r t exist if and only if θ i θ f mπ for some integers m.In this case, for α τe , where τ is any positive number, we have Consequently, we can obtain the PH cubics and their M öbius transformations and also let a w 2 0 /3, b w 2 1 /3 and k w 0 w 1 /3, then second and third equations in 4.6 imply that k v or k −v, and so we have Now let

4.10
where k v or k −v.Note that r m,1 r m,2 if and only if k is −1 or 1/3.From the PH cubics r m,j t we can obtain the M öbius transformations of the PH cubics Φ m,j r m,j t m 1, −1, and j 1, 2 , where

Best Interpolant
In this section we consider how to choose the best interpolant for a given Hermite data-set We will begin by presenting a condition under which the M öbius transformation of a PH cubic Φ • r t has a loop, where r t w 0 1 − t w 1 t 2 for some distinct complex constants w 0 and w 1 .Since Φ z represents a one-to-one correspondence on the extended complex plane, the condition that r t has a loop is both necessary and sufficient to establish that Φ • r t has a loop.Under the conditions r 0 0 and r 1 1, the PH cubic r t is given by r t A t − B 3 C, where The condition that there exist constants t 1 and t 2 , such that 0 ≤ t 1 < t 2 ≤ 1 and r t 1 r t 2 , is necessary and sufficient to establish that r t has a loop.From r t 1 r t 2 , we can obtain This equation is equivalent to and hence Consequently, r t has a loop if and only if B ∈ Ω 1 ∪ Ω 2 , see Figure 1 where On the other hand, the PH cubic r t can be represented by

5.9
Therefore we conclude as following.
Theorem 5.1.Suppose that Φ • r t is a Möbius transformation of a planar PH cubic, such that r 0 0, r 1 1, and r t w 0 1 − t w 1 t 2 for some distinct complex constants w 0 and w 1 (see Figure 2).Then Φ • r t 0 ≤ t ≤ 1 is a simple curve if and only if w 0 w 1 /3 / ∈ D, where For a given Hermite data-set 0, 1, v i , v f , the term v in 4.7 belongs to D if and only if are simple curves.From these simple curves we can choose a best interpolant, which is that with the least bending energy where κ is the curvature of Φ • r t .
where k 1, 5, 10, 20.We construct C 1 Hermite interpolants that satisfy these data-sets using M öbius transformations of PH cubics, and also PH quintics, all shown in Figure 5.The Möbius transformations of the PH cubics always provide two S-shaped simple curves and two other curves; the latter are C-shaped simple curves when k 1 or 5 and have a single loop in the other cases.As the parametric speed of the initial Hermite condition increases, the C-shaped interpolants change from simple curves to single loops, while the simple S-shaped interpolants retain their original shape characteristics.We also observe that, unlike the S-shaped interpolants produced by M öbius transformations of PH cubics, the S-shaped PH quintic interpolants may be simple like the curve labeled 4 in Figure 5 , or have one or two loops some PH quintics labeled 2 in Figure 5 are S-shaped double loops .
We observed the behavior of these interpolants as the parametric speed at the endpoints changes.As this speed increases, the arc-lengths of PH quintics increase rapidly, but the arc-length, of M öbius transformations of PH cubics are generally less affected.In particular, the simple S-shaped interpolants, produced by M öbius transformations of PH cubics show little change in arc-length.Table 1 shows that these latter interpolants have both lower bending energies and shorter arc-lengths, than all the other interpolants we are considering.
If we look at Table 1 and identify the most shapely interpolants with the lowest bending energies, we find that the best M öbius transformation of a PH cubic is always S-shaped and   simple.However, the merit of the PH quintic interpolants depends on the parametric speeds at their end points.For example, in Figures 5 a and 5 b , the curves labeled 4 are best, while the curves labeled 1 are best in Figures 5 c and 5 d .Looking closely at the PH quintic interpolants, we see that the simple S-shaped curve with the best shape when k 1 becomes less and less acceptable as the parametric speeds at the end-points increase.But the interpolants labeled 1 in Figures 5 a , 5 b , 5 c , and 5 d exhibit the opposite behavior: initially these curves are C-shaped loops with high bending energies when k 1; but as the parametric speed increases, they become C-shaped simple curves with lower bending energies.When k reaches 20, it has the best shape but the greatest arc-length.This suggests that the best-shaped interpolants, produced by M öbius transformations of PH cubics are more stable than the corresponding PH quintics, in the sense that the former largely achieve a lower arclength and bending energy than the latter, except when the end-point speeds are significantly asymmetric, as we see when k 20 in this example.

Conclusions
M öbius transformations preserve Pythagorean-hodograph properties.For any C 1 Hermite data-set, we can generally obtain four C 1 Hermite interpolants as M öbius transformations of PH cubics.We have proved that these interpolants are always simple curves or single loops, and that at least two of them must be simple.We have also presented the condition that an interpolant must meet if it is to be a simple curve.We compared the shape characteristics of C 1 Hermite interpolants, produced by M öbius transformations of PH cubics, together with their response to changes of parametric speed at their end points, with the same data for PH quintic interpolants satisfying an identical C 1 Hermite dataset: we found that interpolants produced by M öbius transformations of PH cubics generally have lower bending energies and shorter arc-lengths than PH quintics.
One avenue for further research is to look for ways of predicting how the geometry of M öbius transformation of PH cubics will be determined by a particular C 1 Hermite data-set.Another avenue to explore would be the application of M öbius transformations to other interpolation problems involving PH or MPH curves, in both two and three dimensions.In particular, we might look to complete the geometric characterization of M öbius transformation of PH cubics in C 1 Hermite interpolation.

4 . 9 where m 1
if k v, and m −1 if k −v.Then we have a a m and b b m , or a b m and b a m .Consequently, we can obtain the four PH cubics

Figure 1 :
Figure 1: Areas of Ω 1 and Ω 2 : B belongs to Ω 1 ∪ Ω 2 if and only if r t has a loop.

Figure 2 :
Figure 2: Area of D: w 0 w 1 /3 / ∈ D if and only if Φ • r t in Theorem 5.1 is a simple curve.

Figure 3 :
Figure 3: For the Hermite dataset 0, 1, 2e − √ −1π/4 , 2e − √ −1π/8 , the graph on the left shows that −v ∈ D and v / ∈ D; the central graph shows the PH cubics r t with their control polygons; the graph on the right shows the four interpolants.

Figure 4 :
Figure 4: For the Hermite data-set 0, 1, e − √ −13π/5 , e − √ −1π/5 , the graph on the left shows that −v / ∈ D and v / ∈ D; the central graph shows the PH cubics r t with their control polygons; the graph on the right shows the four interpolants.

Lemma 2.1. Let
This completes the proof.Lemma 2.1 means that M öbius transformations preserve the PH property, which is a special case of the result of Barto ň et al. 11 .For a polynomial PH curve r t of degree n, a Möbius transformation of r t Φ z be a Möbius transformation and r t be a polynomial PH curve.Then s t Φ • r t is a rational PH curve.Proof.Since s t Φ r t r t , we have s t |ad − bc| |cr t d| 2 r t |ad − bc| Re cr t d 2 Im cr t d 2 r t .2.5 Hermite data-set such that r i > 0 and r f > 0.a Let v be the vector given by 4.7 , and let k be v or −v.Then all C 1 Hermite interpolants using Möbius transformations of planar PH cubics r t , such that r t w t 2 for some linear curve w t , are Φ m,j r m,j t m 1, −1, and j 1, 2 , from 4.10 and 4.11 , where a m and b m are given by 4.9 .b C 1 Hermite interpolants using Möbius transformations of planar PH cubics r t , such that r t h 0 1 − t 2 h 1 2 1 − t t h 2 t 2 for some real number h 0 , h 1 , and h 2 such that h 0 h 2 / h 1 , exist if and only if θ i θ f mπ for some integers m.In this case, the interpolants Φ • r t are given by 4.5 , where r t is given by 4.3 and 4.4 , where α τe 11If k is nonreal, then both r m,1 t and r m,2 t are nonlinear.But if k is a real number, then −1 ≤ k ≤ 1/3 if and only if both r m,1 t and r m,2 t are linear.We can summarize these results.√−1θ f be a C 1

Table 1 :
Comparison of arc-length and bending energy for the interpolants of Figure5.