Introduction
The equations
The main results
Proof of main theorems
Conclusions
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
1 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
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.
2 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Introduction: the interpolation problem
For given data points
T 0 , T 1 , . . . , T 5 ∈ R2 ,
T i 6= T i+1 ,
find a cubic curve P 3 : [0, 1] → R2 that interpolates these
points,
P 3 (ti ) = T i , i = 0, 1, . . . , 5,
(1)
where the parameters are ordered as
0 =: t0 < t1 < · · · < t4 < t5 := 1.
(2)
The admissible parameters ti are lying in the open simplex
D := t := (ti )5i=0 ; 0 =: t0 < t1 < · · · < t4 < t5 := 1 ,
with the boundary ∂D where at least two different ti
coincide.
3 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
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).
4 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Introduction: motivation
T3
T2
T2
T3
T1
T0
T1
T4
T4
T0 T5
T5
T4
T5
T5
T3
T2
T1
T4
T2
T3
T0
T1
T0
5 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
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
∆T
determinants:
T i , ∆T
Tj ,
Di,j := det ∆T
D0,2
D0,1
, λ2 :=
,
λ1 :=
D1,2
D1,2
D2,4
D3,4
λ3 :=
, λ4 :=
,
D2,3
D2,3
D1,3
D2,3
δ :=
, µ :=
.
D1,2
D1,2
6 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The equations: notation
DT4
DT2
DT3
DT1
∆<0
∆<0
∆>0
DT2
DT4
Λ1 >0
Λ2 >0
DT0
Λ3 >0
Λ4 >0
DT3
∆>0
DT1
DT0
Λ1 >0
Λ2 >0
Convex data
Nonconvex data
Convex data: D1,2 D2,3 6= 0, λi > 0 and µ > 0.
Nonconvex data: D1,2 D2,3 6= 0, λi > 0 and µ < 0.
7 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The equations
Equations: P 3 (ti ) = 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` , t`+1 , . . . , t`+4 ]:
P 3 = [t` , t`+1 , . . . , t`+4 ] (T
T i )5i=0 ,
[t` , t`+1 , . . . , t`+4 ]P
0=
`+4
X
i=`
1
T i,
ω̇` (ti )
` = 0, 1,
where
ω` (t) := (t − t` )(t − t`+1 ) . . . (t − t`+4 ),
` = 0, 1.
8 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The equations
After some transformations the system for the unknown
parameters ti becomes
1
1
1
(1 + λ2 ) +
+
µ= 0,
ω̇0 (t0 )
ω̇0 (t1 ) ω̇0 (t4 )
1
1
1
λ1 +
+
(1 + δ)= 0,
ω̇0 (t0 )
ω̇0 (t3 ) ω̇0 (t4 )
1
δ
1
1
1+
+
+
λ4 = 0,
ω̇1 (t1 )
µ
ω̇1 (t2 ) ω̇1 (t5 )
1
1
1 1
+
+
(1 + λ3 )= 0.
ω̇1 (t1 ) µ ω̇1 (t4 ) ω̇1 (t5 )
(3)
9 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The main results: notation
Unfortunately the existence conditions do not depend on
the signs of λ , µ and δ only.
Let us define
γ1 :=
λ2 (1 + λ2 )
p
,
λ1 (1 + λ2 ) + λ1 (1 + λ2 )(λ1 + λ2 )
γ2 :=
λ3 (1 + λ3 )
p
,
λ4 (1 + λ3 ) + λ4 (1 + λ3 )(λ4 + λ3 )
λ := (λi )4i=1 .
10 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The main results: notation
The following functions will give boundary conditions on
the constants
p
2µ − γ1 + γ12 + 4µ(1 + γ1 )
λ , µ):=
ϑ1 (λ
,
2γ1
p
2 − µγ2 + µ2 γ22 + 4µ(1 + γ2 )
λ , µ):=
ϑ2 (λ
,
2γ2
s
λ1 µ λ4
µ λ1 (λ1 + λ2 )
λ , µ):=
ϑ3 (λ
+
+
,
λ2
λ3 λ 2
1 + λ2
s
λ1 µ λ4
1 λ4 (λ3 + λ4 )
λ , µ):=
ϑ4 (λ
+
+
,
λ2
λ3 λ 3
1 + λ3
11 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The main results: existence theorem
Theorem
Suppose that the data are convex, i.e., D1,2 D2,3 6= 0, µ > 0 and λi > 0, i = 1, 2, 3, 4.
λ , µ) = ϑ2 (λ
λ , µ) or ϑ1 (λ
λ , µ) 6= ϑ2 (λ
λ , µ) and
If ϑ1 (λ
λ , µ)}
δ < min {ϑ` (λ
`=1,2
or
λ , µ)} ,
δ > max {ϑ` (λ
`=1,2
then the interpolating curve P 3 that interpolates data points T i , i = 0, 1, . . . , 5, exists.
Example:
T3
T2
T1
T0
T4
T5
12 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
The main results: existence theorem
Theorem
Suppose that the data imply an inflection point, i.e., D1,2 D2,3 6= 0, µ < 0 and λi > 0
for all i. If
λ , µ), ϑ4 (λ
λ , µ)) ,
δ ∈ (ϑ3 (λ
then the interpolating curve P 3 that interpolates data points T i , i = 0, 1, . . . , 5, exists.
Example:
T4
T5
T3
T2
T0
T1
13 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
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.
14 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Step 1: a particular case
Data points:
1 − 2c
2
1
,
T ∗3 =
0
T ∗0 =
The constants:
∗
λ
ϑ1 (λ
∗
, µ∗ )
λ
= ϑ2 (λ
λ∗ = 1,
, µ∗ )
−1 − c
,
1
1+c
T ∗4 =
,
(−1)s
T ∗1 =
,
= 4,
δ∗ =
∗
λ
ϑ3 (λ
1
2
(1 + (−1)s ) c,
, µ∗ )
= −1,
T ∗2 =
−1
,
0
T ∗5 =
−1 + 2c
2(−1)s
.
µ∗ = (−1)s ,
∗
λ , µ∗ ) = 1
ϑ4 (λ
Data 1: s = 0, c = 1:
Data 2: s = 0, c = 5,
Data 3: s = 1, c = 1,
δ∗ = 1
δ∗ = 5
δ∗ = 0
T0
T5
T0
T0
T5
T1
T4
T1
T1
T4
T2
T3
T4
T2
T3
T2 T3
T5
15 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Step 1: a particular case
The equations:
1
1
2
+
+ (−1)s
= 0,
ω̇0 (t0 )
ω̇0 (t1 )
ω̇0 (t4 )
1
1
c
c
1
+
+ 1 + + (−1)s
= 0,
ω̇0 (t0 )
ω̇0 (t3 )
2
2 ω̇0 (t4 )
2
1
1
+
+ (−1)s
= 0,
ω̇1 (t5 )
ω̇1 (t4 )
ω̇1 (t1 )
1
1
c
c
1
+
+ 1 + + (−1)s
= 0.
ω̇1 (t5 )
ω̇1 (t2 )
2
2 ω̇1 (t1 )
Symmetric solution: ∆t0 = ∆t4 , ∆t1 = ∆t2 .
Nonsymmetric solutions come in pairs.
16 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Step 1: a particular case
Admissible symmetric solutions:
T0
s = 0, c = 1
s = 0, c = 5
s = 1, c ∈ R
t3
0.61053
0.56382
3
5
t4
0.86209
0.85642
9
10
T5
T0
T0
T5
T1
T4
T1
T1
T4
T2
T3
T4
T2
T3
T2 T3
T5
17 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Step 2: solution region
We need to show that t 6∈ ∂D, i.e.,
∆ti := ti+1 − ti ≥ const > 0, i = 0, 1, . . . , 4.
Lemma 1: Suppose that λ > 0, µ > 0 and
λ , µ) 6= ϑ2 (λ
λ , µ). Then ∆ti → 0, i = 0, 1, 2, 3,
ϑ1 (λ
λ , µ).
implies δ → ϑ1 (λ
Similarly, from ∆ti → 0, i = 1, 2, 3, 4, it follows
λ , µ).
δ → ϑ2 (λ
Lemma 2: Suppose that λ > 0, µ < 0. Then
λ , µ), and
∆ti → 0, i = 0, 1, 2, 4, implies δ → ϑ3 (λ
λ , µ).
similarly, ∆ti → 0, i = 0, 2, 3, 4, implies δ → ϑ4 (λ
The rest of the cases can be proved just by assuming
λ > 0 and µ > 0 or µ < 0.
18 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
Step 3: Homotopy and Brouwer’s degree theorem
Denote the nonlinear system (3) by F (tt ; λ , δ, µ) = 0 .
λ ∗ , δ ∗ , µ∗ ) as described.
Choose particular data (λ
Define a convex homotopy H (tt ; α) so that
H (tt ; 0) = F (tt ; λ ∗ , δ ∗ , µ∗ ),
H (tt ; 1) = F (tt ; λ , δ, µ),
and that the set of solutions
S := {tt ∈ D;
H (tt , α) = 0 , α ∈ [0, 1]}
is bounded away from the boundary ∂D.
Homotopy invariance of the Brouwer’s mapping degree
concludes the proof.
19 / 20
Introduction
The equations
The main results
Proof of main theorems
Conclusions
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.
20 / 20