Transfinite mean value interpolation
Christopher Dyken and Michael Floater
Centre of Mathematics for Applications,
Department of Informatics,
University of Oslo
Id: maiatalk.tex,v 1.9 2007/08/23 07:35:58 dyken Exp
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Transfinite interpolation
Given Ω ⊂ R2 , a convex or non-convex set, possibly with holes.
Lagrange transfinite interpolation
We are given f : ∂Ω → R.
Find g : Ω → R that interpolates f on ∂Ω.
Hermite transfinite interpolation
∂f
We are given f : ∂Ω → R and ∂n
: ∂Ω → R.
Find g : Ω → R that interpolates f and ∂g
∂n matches
∂f
∂n
on ∂Ω.
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Lagrange can be solved by solving the harmonic equation
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Hermite can be solved by solving the biharmonic equation
=⇒ But we want something simpler. . .
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Pseudo-harmonic interpolation [Gordon, Wixom 1974]
Let
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x be a point inside the
convex set Ω;
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Q(x, θ) be the infinite line
through x in the direction θ.
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Let p1 (x, θ) and p2 (x, θ) be
the two intersections
between Q(x, θ) and ∂Ω,
p1(x , θ )
Q (x , θ )
θ
Ω
x
dΩ
then we define
Z 2π
1
lerp f (p1 (x, θ)), f (p2 (x, θ)),
gGW (x) =
2π 0
p2(x , θ )
kp1 (x,θ)−xk
kp1 (x,θ)−p2 (x,θ)k
dθ.
Works only for convex sets.
I Evaluation requires numerical integration
=⇒ must find intersections for each integration point!
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A mean value approach
Let
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L(x, θ) be the semi-infinite line
originating at x in the direction θ.
p(x, θ) be the intersection of
L(x, θ) and ∂Ω.
p (x , θ )
L (x , θ )
θ
Ω
x
dΩ
and define the “radially linear” function F as
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F (x + r (cos θ, sin θ)) = lerp g (x), f (p(x, θ)),
r
kp(x,θ)−xk
.
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We want F to satisfy the Mean Value property at x.
Let Γ be any circle at x with
radius r , then
Γ
Z
Ω
1
F (x) =
F (z)dz,
2πr Γ
whose unique solution is
1
g (x) =
φ(x)
Z
0
2π
f (p(x, θ))
dθ,
kp(x, θ) − xk
p (x , θ )
r
θ
x
dΩ
Z
φ(x) =
0
2π
1
dθ,
kp(x, θ) − xk
which is the “angle integral” formula for the MV interpolant g .
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Generalizes to non-convex sets
Evaluation still requires numerical integration.
Still must find an intersection for each integration point!
How do we differentiate this thing?
Luckily, we have two other formulas. . .
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The boundary integral formula [Ju, Schaefer, Warren 2005]
Suppose c : [a, b] → R2 is an anticlockwise representation of ∂Ω.
Then
dθ
(c(t) − x) × c0 (t)
=
dt
kc(t) − xk2
c (t )
which gives
Z
φ(x) =
a
b
c0 (t)
(c(t) − x) ×
kc(t) − xk3
dt,
and
g (x) =
θ
Ω
x
dΩ
1
φ(x)
Z
a
b
(c(t) − x) × c0 (t)
f (c(t)dt.
kc(t) − xk3
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The polygonal formula
Suppose Ω is a polygon with vertices p1 , . . . , pn .
Then
p i+1
1 X
wi (x)f (pi ),
φ(x)
g (x) =
Ω
αi
i
α i−1
where
φ(x) =
pi
x
X
wi (x),
i
p i−1
dΩ
and
wi (x) =
tan(αi−1 (x)/2) + tan(αi (x)/2)
.
kpi − xk
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We have three formulations for the MV interpolant:
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The polygonal formula:
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The boundary integral formula
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closed form
easy to find expressions for derivatives.
needs adaptive numerical quadrature for evaluation.
easy to find expressions for derivatives.
The angle integral
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describes the interpolant along a particular direction.
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The MV weight function
A lot of properties can be deduced from the “weight function” ψ
Z 2π
1
1
=1
dθ,
ψ(x) =
φ(x)
kp(x, θ) − xk
0
Z b
(c(t) − x) × c0 (t)
=1
dt,
kc(t) − xk3
a
X
tan(αi−1 (x)/2) + tan(αi (x)/2)
=1
.
kpi − xk
i
ψ
∂ψ
∂x
∂ψ
∂y
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Minimum principle for ψ
For arbitrary Ω, we have that
Z
∆φ(x) = 3
0
2π n(x,θ)
X
j=1
(−1)j−1
dθ
kpj (x, θ) − xk3
from which follows that
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2
2
∆ψ = ∂ ψ
+ ∂ ψ
∂x 2
∂y 2
ψ has no local minima in Ω.
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Bounds on ψ
For all x ∈ Ω we have that
1
dist(x, ∂Ω) ≤ ψ(x) ≤ c dist(x, ∂Ω),
2π
0.5
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c depends on dist(ME , ∂Ω),
the distance between ∂Ω
and its exterior medial axis
If Ω is convex, then c = 12 .
=⇒ For all x ∈ Ω, ψ > 0.
0.4
0.3
0.2
0.1
0
−1
0
1
The plot shows the upper and lower bounds and ψ along a crosssection when Ω is the unit disc.
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Proof of interpolation for the Lagrange MV interpolant
If
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f is continuous,
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∂Ω and any line intersects a bounded number of times,
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and dist(ME , ∂Ω) > 0
then
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g interpolates f .
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Normal derivatives of ψ and g
If
dist(ME , ∂Ω) > 0
and
dist(MI , ∂Ω) > 0,
then, for all y ∈ ∂Ω,
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the inward normal derivative for ψ is
∂ψ
1
(y) =
∂n
2
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the inward normal derivate for the Lagrange interpolant g is
∂g
(y) =
∂n
1
2
Z
a
b
(c(t) − y) × c0 (t)
f (c(t)) − f (y) dt.
3
kc(t) − xk
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Hermite mean value interpolation
In one variable, we have the problem
p(xi ) = f (xi )
and
p 0 (xi ) = f 0 (xi ),
i = 0, 1.
One approach of expressing p is
p(x) = g0 (x) + ψ(x)g1 (x),
where
I g0 and g1 are Lagrange interpolants,
I ψ vanishes at x0 and x1 and ψ 0 is nonzero at x0 and x1 .
Which gives the conditions
g0 (xi ) = f (xi )
and
g1 (xi ) =
f 0 (xi ) − g00 (xi )
.
ψ 0 (xi )
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In two variables, we can generalize a similar problem,
p(y) = f (y)
and
∂p
∂f
(y) =
(y),
∂n
∂n
y ∈ ∂Ω.
and let p be on the form
p(x) = g0 (x) + ψ(x)g1 (x).
We can use the MV-ψ since
ψ(y) = 0
and
∂ψ
1
(y) = ,
∂n
2
y ∈ ∂Ω,
and let g0 and g1 be MV Lagrange interpolants.
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Then, for y ∈ ∂Ω we get the conditions
g0 (y) = f (y)
,
∂f
∂g0
∂ψ
g1 (y) =
(y) −
(y)
(y).
∂n
∂n
∂n
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Application: Smooth mappings
Reference shape
Computational domain
(extended Gordon & Hall)
MV-Lagrange
MV-Hermite
Conjecture: Lagrange interpolation from convex sets to convex
sets is always injective.
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Application: WEB-splines [Höllig, Reif, Wipper 2001]
Idea: Use ψ as a weight function for WEB-splines
Parametric circle
Two nested ellipses
Polygon
Piecwise cubic Bézier curve
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Solution to Poisson’s equation using bicubic web-splines
Using implicit weight function
Grid
10 × 8
20 × 16
40 × 32
80 × 64
L2 error
7.3e-02
2.9e-02
1.6e-03
4.4e-05
order
1.31
4.21
5.17
Using MV weight function
Grid
10 × 8
20 × 16
40 × 32
80 × 64
L2 error
9.5e-02
4.1e-02
2.4e-03
1.4e-04
order
1.21
4.12
4.01
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Inhomogenenous Poisson’s equation
True solution
MV Lagrange interpolant
Homogeneous solution
Inhomogeneous solution
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Conclusions
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The Lagrange mean value interpolant does in fact
interpolate.
Constructed a Hermite mean value interpolant.
The mean value weight function has nice properties:
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positive;
C ∞ -smooth;
bounded by the distance function:
=⇒ a very smooth distance-like function without ridges along the
inner medial axis!
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constant normal derivate;
has no local minima in Ω;
The mean value constructions are relatively easy to compute:
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The polygonal case has a closed form.
The boundary integral must be calculated numerically, but:
Strong influence of the boundary region closest to the point of
evaluation.
=⇒ Adaptivity pays off.
I Simpler than solving a PDE.
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Thank you for listening!
Questions?
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