1 / 20 Geometric Interpolation by Planar Cubic Polynomials Jernej Kozak, Marjeta Krajnc Faculty of Mathematics and Physics University of Ljubljana Institute of Mathematics, Physics and Mechanics Avignon, 30.6.2006
2 / 20 Introduction: geometric interpolation Geometric interpolation schemes are an important tool for approximation of curves. The basic idea: parameter values are not prescribed in advance. The freedom of choosing a parametrization is used to increase an approximation order and to improve the shape of the interpolant. Nonlinear equations. Existence? Implementation? Asymptotic analysis. Geometric conditions for the existence of the interpolant.
Introduction: the interpolation problem For given data points T 0, T 1,..., T 5 R 2, T i T i+1, find a cubic curve P 3 : [0, 1] R 2 that interpolates these points, P 3 (t i ) = T i, i = 0, 1,..., 5, (1) where the parameters are ordered as 0 =: t 0 < t 1 < < t 4 < t 5 := 1. (2) The admissible parameters t i are lying in the open simplex D := { t := (t i ) 5 i=0 ; 0 =: t 0 < t 1 < < t 4 < t 5 := 1 }, with the boundary D where at least two different t i coincide. 3 / 20
4 / 20 Introduction: the interpolation problem The nonlinear part of the problem is to determine the parameters t D. The coefficients of P 3 are then obtained by using any standard interpolation scheme (Newton, Lagrange,...) componentwise. The main motivation is the shape of the interpolant P 3. A comparison: Cubic geometric scheme quintic interpolation scheme (uniform, chord length parameterization).
5 / 20 Introduction: motivation T 3 T 2 T 0 T 1 T 2 T3 T 4 T 1 T 4 T 5 T 0 T 5 T 5 T 3 T 4 T 5 T 1 T 2 T 3 T 4 T 2 T 0 T 0 T 1
6 / 20 The equations: notation The existence conditions will be based on data differences T i := T i+1 T i and on the signs and ratios of determinants: D i,j := det ( T i, T j ), λ 1 := D 0,1 D 1,2, λ 2 := D 0,2 D 1,2, λ 3 := D 2,4 D 2,3, λ 4 := D 3,4 D 2,3, δ := D 1,3 D 1,2, µ := D 2,3 D 1,2.
7 / 20 The equations: notation T 4 T 2 T 3 T 1 0 0 0 T 3 T 2 0 Λ 1 0 Λ 2 0 T 0 Convex data T 4 Λ 3 0 Λ 4 0 Λ 1 0 Λ 2 0 T 1 T 0 Nonconvex data Convex data: D 1,2 D 2,3 0, λ i > 0 and µ > 0. Nonconvex data: D 1,2 D 2,3 0, λ i > 0 and µ < 0.
8 / 20 The equations Equations: P 3 (t i ) = T i, i = 0, 1,..., 5. Separate the equations for unknown parameters t from the equations for unknown coefficients of P 3. How? Apply the divided differences [t l, t l+1,..., t l+4 ]: where [t l, t l+1,..., t l+4 ]P 3 = [t l, t l+1,..., t l+4 ] (T i ) 5 i=0, 0 = l+4 i=l 1 ω l (t i ) T i, l = 0, 1, ω l (t) := (t t l )(t t l+1 )... (t t l+4 ), l = 0, 1.
9 / 20 The equations After some transformations the system for the unknown parameters t i becomes 1 ω 0 (t 0 ) (1 + λ 2) + 1 ω 0 (t 1 ) + 1 µ= 0, ω 0 (t 4 ) 1 ω 0 (t 0 ) λ 1 + 1 ω 0 (t 3 ) + 1 (1 + δ)= 0, (3) ω 0 (t 4 ) ( 1 1 + δ ) + 1 ω 1 (t 1 ) µ ω 1 (t 2 ) + 1 ω 1 (t 5 ) λ 4= 0, 1 1 ω 1 (t 1 ) µ + 1 ω 1 (t 4 ) + 1 ω 1 (t 5 ) (1 + λ 3)= 0.
10 / 20 The main results: notation Unfortunately the existence conditions do not depend on the signs of λ, µ and δ only. Let us define γ 1 := γ 2 := λ 2 (1 + λ 2 ) λ 1 (1 + λ 2 ) + λ 1 (1 + λ 2 )(λ 1 + λ 2 ), λ 3 (1 + λ 3 ) λ 4 (1 + λ 3 ) + λ 4 (1 + λ 3 )(λ 4 + λ 3 ), λ:= (λ i ) 4 i=1.
11 / 20 The main results: notation The following functions will give boundary conditions on the constants ϑ 1 (λ, µ):= 2µ γ 1 + γ 2 1 + 4µ(1 + γ 1) 2γ 1, ϑ 2 (λ, µ):= 2 µγ 2 + µ 2 γ2 2 + 4µ(1 + γ 2), 2γ 2 ϑ 3 (λ, µ):= λ 1µ + λ 4 λ 1 (λ 1 + λ 2 ), λ 2 1 + λ 2 λ 3 + µ λ 2 ϑ 4 (λ, µ):= λ 1µ λ 2 + λ 4 λ 3 + 1 λ 3 λ 4 (λ 3 + λ 4 ) 1 + λ 3,
12 / 20 The main results: existence theorem Theorem Suppose that the data are convex, i.e., D 1,2 D 2,3 0, µ > 0 and λ i > 0, i = 1, 2, 3, 4. If ϑ 1 (λ, µ) = ϑ 2 (λ, µ) or ϑ 1 (λ, µ) ϑ 2 (λ, µ) and δ < min l=1,2 {ϑ l (λ, µ)} or δ > max l=1,2 {ϑ l (λ, µ)}, then the interpolating curve P 3 that interpolates data points T i, i = 0, 1,..., 5, exists. Example: T3 T2 T1 T0 T4 T5
13 / 20 The main results: existence theorem Theorem Suppose that the data imply an inflection point, i.e., D 1,2 D 2,3 0, µ < 0 and λ i > 0 for all i. If δ (ϑ 3 (λ, µ), ϑ 4 (λ, µ)), then the interpolating curve P 3 that interpolates data points T i, i = 0, 1,..., 5, exists. Example: T4 T5 T3 T2 T0 T1
14 / 20 Proof of main theorems One needs to show that the nonlinear system (3) has at least one solution t D. The sketch of the proof: 1 For particular data the system (3) has an odd number of admissible solutions t D. 2 Under the conditions of Theorem 1 and Theorem 2 the system (3) cannot have a solution arbitrary close to the boundary D. 3 The convex homotopy and Brouwer s degree theorem carry the conclusions from a particular to the general case.
T 2 T 3 T 4 15 / 20 Step 1: a particular case Data points: ( T 1 2c 0 = 2 ( ) T 1 3 = 0 ) ( ) ( ), T 1 c 1 =, T 1 1 2 =, 0 ( ) ( ), T 1 + c 4 = ( 1) s, T 1 + 2c 5 = 2( 1) s. The constants: λ = 1, δ = 1 2 (1 + ( 1)s ) c, µ = ( 1) s, ϑ 1 (λ, µ ) = ϑ 2 (λ, µ ) = 4, ϑ 3 (λ, µ ) = 1, ϑ 4 (λ, µ ) = 1 Data 1: s = 0, c = 1: δ = 1 Data 2: s = 0, c = 5, δ = 5 Data 3: s = 1, c = 1, δ = 0 T 0 T 5 T 0 T 5 T 0 T 1 T 1 T 4 T 1 T 4 T 2 T 3 T 2 T 3 T 5
16 / 20 Step 1: a particular case The equations: 2 ω 0 (t 0 ) + 1 ω 0 (t 1 ) + 1 ( 1)s ω 0 (t 4 ) = 0, ) 1 ω 0 (t 0 ) + 1 (1 ω 0 (t 3 ) + + c 2 + c ( 1)s 2 1 ω 0 (t 4 ) = 0, 2 ω 1 (t 5 ) + 1 ω 1 (t 4 ) + 1 ( 1)s ω 1 (t 1 ) = 0, ) 1 ω 1 (t 5 ) + 1 (1 ω 1 (t 2 ) + + c 2 + c ( 1)s 2 1 ω 1 (t 1 ) = 0. Symmetric solution: t 0 = t 4, t 1 = t 2. Nonsymmetric solutions come in pairs.
T 2 T 3 T 4 17 / 20 Step 1: a particular case Admissible symmetric solutions: s = 0, c = 1 s = 0, c = 5 s = 1, c R t 3 0.61053 0.56382 t 4 0.86209 0.85642 3 5 9 10 T 0 T 5 T 0 T 5 T 0 T 1 T 1 T 4 T 1 T 4 T 2 T 3 T 2 T 3 T 5
18 / 20 Step 2: solution region We need to show that t D, i.e., t i := t i+1 t i const > 0, i = 0, 1,..., 4. Lemma 1: Suppose that λ > 0, µ > 0 and ϑ 1 (λ, µ) ϑ 2 (λ, µ). Then t i 0, i = 0, 1, 2, 3, implies δ ϑ 1 (λ, µ). Similarly, from t i 0, i = 1, 2, 3, 4, it follows δ ϑ 2 (λ, µ). Lemma 2: Suppose that λ > 0, µ < 0. Then t i 0, i = 0, 1, 2, 4, implies δ ϑ 3 (λ, µ), and similarly, t i 0, i = 0, 2, 3, 4, implies δ ϑ 4 (λ, µ). The rest of the cases can be proved just by assuming λ > 0 and µ > 0 or µ < 0.
19 / 20 Step 3: Homotopy and Brouwer s degree theorem Denote the nonlinear system (3) by F (t; λ, δ, µ) = 0. Choose particular data (λ, δ, µ ) as described. Define a convex homotopy H(t; α) so that H(t; 0) = F (t; λ, δ, µ ), H(t; 1) = F (t; λ, δ, µ), and that the set of solutions S := {t D; H(t, α) = 0, α [0, 1]} is bounded away from the boundary D. Homotopy invariance of the Brouwer s mapping degree concludes the proof.
20 / 20 Conclusions and future work The sufficient, entirely geometric conditions on data points that imply the existence of the interpolant were given. The conditions where the interpolant cannot exist can be found. Hermite interpolation, G 1 splines, G 2 splines.