On the High-Level Error Bound for Multiquadric and Inverse Multiquadric Interpolations

0.14 MB·English

by  Lin-Tian Luh

On the High-Level Error Bound for
Multiquadric and Inverse Multiquadric
6 Interpolations
0
0
2
LIN-TIAN LUH
n
a
J
February 2, 2008

9

]
A
Abstract It’s well-known that there is a so-called high-level error bound
N
for multiquadric and inverse multiquadric interpolations, which was put for-
.
h 1
wardbyMadych andNelsonin1992. It’softheform f(x) s(x) λd f
t | − | ≤ k kh
a where 0 < λ < 1 is a constant, d is the fill distance which roughly speaking
m
measures the spacing of the data points, s(x) is the interpolating function
[
of f(x), and h denotes the multiquadric or inverse multiquadric. The error

1
boundconverges very fastasd 0.The constant λis very sensitive. Aslight
v
→
8 change of it will result in a huge change of the error bound. Unfortunately
5
λ can not be calculated, or even approximated. This is a famous question in
1
1 the theory of radial basis functions. The purpose of this paper is to answer
0 the question.
6
0
/ Key words. radial basis function, conditionally positive definite function,
h
t interpolation, multiquadric, inverse multiquadric
a
m
: AMS subject classification. 41A05, 41A15, 41A25, 41A30, 41A63
v
i
X
r 1 Introduction
a
Let h be a continuous function on Rn which is conditionally positive def-
inite of order m. Given data (x , f ), j = 1,..., N, where X = x ,...,x
j j 1 N
is a subset of points in Rn and the f ′s are real or complex n{umbers, th}e
j
so-called h spline interpolant of these data is the function s defined by
N
(1.1) s(x) = p(x)+ c h(x x ),
j j
−
j=1
X
1
′
where p(x) is a polynomial in and the c s are chosen so that
m−1 j
P
N
(1.2) c q(x ) = 0
j j
j=1
X
for all polynomials q in and
m−1
P
N
(1.3) p(x )+ c h(x x ) = f , i = 1,..., N.
i j i j i
−
j=1
X
Here denotes the class of those polynomials of Rn of degree m 1.
m−1
P ≤ −
It is well known that the system of equations (1.2)and (1.3) has a unique
solution when X is a determining set for and h is strictly conditionally
m−1
P
positive definite. For more details please see [5]. Thus, in this case, the
interpolant s(x) is well defined.
We remind the reader that X is said to be a determining set for if
m−1
P
p is in and p vanishes on X imply that p is identically zero.
m−1
P
In this paper h is defined by formula
β
(1.4) h(x) := Γ( )(c2 + x 2)β2, β R 2N≥0, c > 0,
−2 | | ∈ \
where x is the Euclidean norm of x, Γ is the classical gamma function
| |
and β, c are constants. The function h is called multiquadric or inverse
multiquadric, respectively, depending on β > 0, or β < 0.
In [6] Madych and Nelson obtained bounds on the pointwise difference
between a function f and the h spline which agrees with f on a subset X
of Rn. These estimates involve a parameter that measures the spacing of
the points in X and are O dℓ as d 0 where l depends on h. Later in
→
[7] they found that for multiquadrics and inverse multiquadrics, the estimate
(cid:0) (cid:1)
1
can be improved to O λd as d 0, where λ is a constant which satisfies
→
0 < λ < 1. The conditi(cid:16)ons(cid:17)on f are the same as those in [6].
1.1 A Bound for Multivariate Polynomials
A key ingredient in the development of our estimates is the following
lemma which gives a bound on the size of a polynomial on a cube in Rn in
terms of its values on a discrete subset which is scattered in a sufficiently
uniform manner. We cite it directly from [7] and omit its proof.
Lemma 1.1 For n = 1,2,..., define γ by the formulas γ = 2 and, if n > 1,
n 1
γ = 2n(1+γ ). Let Q be a cube in Rn that is subdivided into qn identical
n n−1
2
subcubes. Let Y be a set of qn points obtained by selecting a point from each
of those subcubes. If q γ (k +1), then for all p in
n k
≥ P
sup p(x) e2nγn(k+1)sup p(y) .
| | ≤ | |
x∈Q y∈Y
1.2 A Variational Framework for Interpolation
The precise statement of our estimates concerning h splines requires a
certain amount of technical notation and terminology which is identical to
that used in [6].For the convenience of the reader we recall several basic
notions.
The space of complex-valued functions on Rn that are compactly sup-
ported and infinitely differentiable is denoted by . The Fourier transform
D
of a function φ in is
D
φˆ(ξ) = e−i<x,ξ>φ(x)dx.
Z
A continuous function h is conditionally positive definite of order m if
h(x)φ(x)⋆φ˜(x)dx 0
≥
Z
holds whenever φ = p(D)ψ with ψ in and p(D) a linear homogeneous
constant coefficient differential operator oDf order m. Here φ˜(x) = φ( x) and
−
⋆ denotes the convolution product
φ ⋆φ (t) = φ (x)φ (t x)dx.
1 2 1 2
−
Z
As pointed out in [6] , this definition of conditional positive definiteness is
equivalent to that of [5] which is generally used.
If h is a continuous conditionally positive definite function of order m,
the Fourier transform of h uniquely determines a positive Borel measure µ
on Rn 0 and constants a , r =2m as follows: For all ψ
r
\{ } | | ∈ D
ξr
(1.5) h(x)ψ(x)dx = ψˆ(ξ) χˆ(ξ) Drψˆ(0) dµ(ξ)
 − r!
Z Z  |γX|<2m 
a
+  Drψˆ(0) r, 
r!
|γ|≤2m
X
3
where for every choice of complex numbers c , α = m,
α
| |
a c c 0.
α+β α β
≥
|α|=m|β|=m
X X
Here χ is a function in such that 1 χˆ(ξ) has a zero of order 2m+1 at
ξ = 0; both of the inteDgral ξ−2mdµ(ξ), dµ(ξ) are finite. The
0<|ξ|<1| | |ξ|≥1
choice of χ affects the value of the coefficients a for γ < 2m.
R γR | |
Our variational framework for interpolation is supplied by a space we
denote by . If
h,m
C
= φ : xαφ(x)dx = 0 for all α < m ,
m
D ∈ D | |
(cid:26) Z (cid:27)
then is the class of those continuous functions f which satisfy
h,m
C
1
2
(1.6) f(x)φ(x)dx c(f) h(x y)φ(x)φ(y)dxdy
≤ −
(cid:12)Z (cid:12) (cid:26)Z (cid:27)
(cid:12) (cid:12)
(cid:12) (cid:12)
for some constant c(f) and all φ in . If f , let f denote the
(cid:12) (cid:12) Dm ∈ Ch,m k kh
smallest constant c(f) for which (1.6) is true. Recall that f is a semi-
k kh
norm and is a semi-Hilbert space; in the case m = 0 it is a norm and a
h,m
C
Hilbert space respectively.
2 Main Results
We first recall that the function h defined in (1.4) is conditionally positive
definite of order m = 0 if β < 0, and m = β if β > 0. This can be found in
2
[8] and many relevant papers. Then we have the following lemma.
(cid:6) (cid:7)
Lemma 2.1 Let h be as in (1.4) and m be its order of conditional positive
definiteness. There exists a positive constant ρ such that
n+β+1
(2.1) ξ kdµ(ξ) √2 √π n+1 nα cβ−k ρk k!
n 0
| | ≤ · · · ·△ · ·
ZRn (cid:16) (cid:17)
(cid:0) (cid:1)
for all integer k 2m+2 where µ is defined in (1.5), α denotes the volume
n
≥
of the unit ball in Rn, c is as in (1.4), and is a positive constant.
0
△
4
Proof. Let K denote a modified Bessel function of the second kind. Then
ν
ξ kdµ(ξ)
| |
ZRn
ξ −n+2β
= ξ k 2πn2 | | Kn+β (c ξ )dξ
ZRn | | · ·(cid:18)2c(cid:19) · 2 | |
−n+β
= 2πn2 1 2 ξ k−n+2β Kn+β (c ξ )dξ
(cid:18)2c(cid:19) ·ZRn | | · 2 | |
√π 2πn2 1 −n+2β ξ k−n+2β 1 dξ
∼ √2 · · 2c | | · c ξ ec|ξ|
(cid:18) (cid:19) ZRn | |·
= √π 2πn2 1 −n+2β n αn ∞rk−pn+2β rn−1 dr
√2 · · 2c · · · c r ec|r|
(cid:18) (cid:19) Z0 | |·
√π n n+β 1 ∞ rk+n−2β−p3
= √2 ·2π2 ·(2c) 2 · √c ·n·αn ecr dr
Z0
√π n n+β 1 1 ∞ rk+n−2β−3
= √2 ·2π2 ·(2c) 2 · √c ·n·αn · ck+n−2β−1 Z0 er dr
′
∞ rk n β 3
= 2n+2β+1 ·πn+21 ·n·αn ·cβ−k er dr where k′ = k + −2 − .
Z0
′
Note that if β < 0, then m = 0 and k 2m+2 = 2. This implies k > 0. If
β > 0, then m = β and k 2m+2≥= 2 β +2. This implies k′ > 0. In
2 ≥ 2
′
any case k > 0.
(cid:6) (cid:7) (cid:6) (cid:7)
′′ ′
Now we divide the proof into three cases. Let k = k which is the
′
smallest integer greater than or equal to k .
(cid:6) (cid:7)
′′ ′′
Case1. Assume k > k. Let k = k +s. Then
′ ′′
∞ rk ∞ rk
′′
dr dr = k ! = (k +s)(k +s 1) (k +1)k!
er ≤ er − ···
Z0 Z0
and
′ ′′
∞ rk +1 ∞ rk +1
′′
dr dr = (k +1)! = (k+s+1)(k+s) (k+2)(k+1)k!.
er ≤ er ···
Z0 Z0
Note that
(k +s+1)(k +s) (k +2) k +s+1
··· = .
(k +s)(k +s 1) (k +1) k +1
− ···
5
(i)Assume β < 0. Then m = 0 and k 2. This gives
≥
k +s+1 3+s
.
k +1 ≤ 3
Let ρ = 3+s. Then
3
′′
∞ rk +1
dr ρk+1 (k +1)!.
er ≤ △0 · ·
Z0
′′
if ∞ rk dr ρk k!. The smallest k′′ is k′′ = 2+s. Now,
0 er ≤ △0 · · 0
R ∞ rk0′′ ′′
dr = k ! = (2+s)(2+s 1) (3) k! where k = 2
er 0 − ··· ·
Z0
(2+s)(2+s 1) (3)
= − ··· ρkk!(k = 2)
ρ2 ·
(2+s)(2+s 1) (3)
= ρ2 2! where = − ··· .
△0 · · △0 ρ2
′
It follows that ∞ rk dr ρk k!. for all k 2.
0 er ≤ △0 · · ≥
R
(ii)Assume β > 0, Then m= β and k 2m+2. This gives
2 ≥
k +s+1 2m(cid:6) +(cid:7) 3+s s
= 1+ .
k +1 ≤ 2m+3 2m+3
Let ρ = 1+ s . Then
2m+3
′′
∞ rk +1
dr ρk+1 (k +1)!
er ≤ △0 · ·
Z0
′′
if ∞ rk dr ρk k!. The smallest k′′ is k′′=2m+2+s when k=2m+2.
0 er ≤ △0 · · 0
Now,
R ′′
∞ rk0 dr
0 er
R ′′
= k ! = (2m+2+s)(2m+1+s) (2m+3)(2m+2)!
0 ···
(2m+2+s)(2m+1+s) (2m+3)
= ··· ρ2m+2 (2m+2)!
ρ2m+2 · ·
(2m+2+s)(2m+1+s) (2m+3)
= ρ2m+2 (2m+2)! where = ··· .
△0 · · △0 ρ2m+2
6
′
It follows that ∞ rk dr ρkk! for all k 2m+2.
0 er ≤ △0 ≥
R ′′ ′′
Case2. Assume k < k. Let k = k s where s > 0. Then
−
′ ′′
∞ rk ∞ rk 1
′′
dr dr = k ! = (k s)! = k!
er ≤ er − k(k 1) (k s+1) ·
Z0 Z0 − ··· −
and
′ ′′
∞ rk +1 ∞ rk +1
dr dr
er ≤ er
Z0 Z0
1
′′
= (k +1)! = (k s+1)! = (k +1)!.
− (k +1)k (k s+2) ·
··· −
Note that
1 1 k(k 1) (k s+1) k s+1
= − ··· − = − .
(k +1)k (k s+2) k(k 1) (k s+1) (k +1)k (k s+2) k +1
(cid:26) ··· − (cid:30) − ··· − (cid:27) ··· −
′′
(i) Assume β < 0, then m = 0 and k 2. Since k = k s 1 holds for
≥ − ≥
all k 2, it must be that s = 1. Thus
≥
k s+1 s 1
− = 1 = 1 1 for all k 2.
k +1 − k +1 − k +1 ≤ ≥
Let ρ = 1. Then
′′ ′′
∞ rk +1 ∞ rk
dr ρk+1(k +1)! if dr ρkk!.
er ≤ △0 er ≤ △0
Z0 Z0
′′ ′′
The smallest k is k = k s = 2 s. Now,
0 0 − −
′′
∞ rk0 ′′
dr = k ! = (2 s)! = 1! = 1
er 0 −
Z0
1
= k! where k = 2
2
1
= ρkk!
2
1
= ρkk! where = .
0 0
△ △ 2
′
It follows that ∞ rk dr ρkk! for all k 2.
0 er ≤ △0 ≥
(ii)Assume β > 0 Then m= β and k 2m+2. This gives
R 2 ≥
k (cid:6)s+(cid:7) 1 s
− = 1 1.
k +1 − k +1 ≤
7
′′ ′′
Let ρ = 1 Then ∞ rk +1dr ρk+1 (k +1)! if ∞ rk dr ρk k!.
0 er ≤ △0 · · 0 er ≤ △0 · ·
′′ ′′
The smallestk is k = 2m+2. Hence the smallestk isk =k s = 2m+2 s.
R 0 R 0 0− −
Now,
′′
∞ rk0 ′′
dr = k ! = (2m+2 s) = (k s)!
er 0 − 0 −
Z0
1
= (k )!
k (k 1) (k s+1) · 0
0 0 0
− ··· −
1
= ρk0k ! where = .
△0 · 0 △0 (2m+2)(2m+1) (2m s+3)
··· −
′
It follows that ∞ rk dr ρkk! for all k 2m+2.
0 er ≤ △0 ≥
R ′′
Case3. Assume k = k. Then
′ ′′ ′
∞ rk ∞ rk ∞ rk +1
dr dr = k! and dr (k +1)!.
er ≤ er er ≤
Z0 Z0 Z0
Let ρ = 1. Then
′
∞ rk
dr ρkk! for all k where = 1.
er ≤ △0 △0
Z0
(cid:3)
Thelemmaisnowanimmediateresultofthethreecases.
Remark. For the convenience of the reader we should express the constants
and ρ in a clear form. It’s easily shown that
0
△
′′
(a) k > k if and only if n β > 3;
−
′′
(b) k < k if and only if n β 1;
− ≤
′′
(c) k = k if and only if 1 < n β 3
′′ − ≤
where k and k are as in the proof of the lemma.
We thus have the following situations.
(a) n-β > 3. Let s = n−β−3 . Then
2
(i)if β < 0, ρ = 3+s and = (2+s)(1+s)···3;
(cid:6)3 (cid:7)△0 ρ2
(ii)if β > 0,ρ = 1 + s and = (2m+2+s)(2m+1+s)···(2m+3)
2 β +3 △0 ρ2m+2
⌈2⌉
where m = β .
2
(cid:6) (cid:7)
8
(b) n-β 1. Let s = n−β−3 . Then
≤ − 2
(i)if β < 0,ρ = 1 and = 1;
(cid:6) (cid:7) △0 2
(ii)if β > 0,ρ = 1 and = 1 ; where
△0 (2m+2)(2m+1)···(2m−s+3)
m = β .
2
(cid:6) (cid:7)
(c) 1 < n β 3. Then ρ = 1 and = 1.
0
− ≤ △
Beforeintroducing our main theorem, we need the following lemma which
is cited directly form [7].
Lemma 2.2 Let Q,Y, and γ be as in Lemma1.1. Then, given a point x in
n
Q ,there is a measure σ supported on Y such that
p(y)dσ(y) = p(x)
ZRn
for all p in , and
k
P
d σ (y) e2nγn(k+1).
| | ≤
ZRn
Now we need another lemma.
Lemma 2.3 For any positive integer k,
(2k)!
2k.
k! ≤
p
Proof. This inequality holds fork = 1 obviously. Weproceed by induction.
[2(k +1)]! (2k +2)! (2k)! (2k +2) (2k +1)
= = ·
(k +1)! k!(k +1) k! · k +1
p p p p
(2k)! (2k +2)2 (2k +2)
2k = 2k+1.
≤ k! · k +1 ≤ · k +1
p p
(cid:3)
Because of the local nature of the result, we first restrict our attention to
the case where x lies in a cube.
Theorem 2.4. Suppose h is defined as in (1.4) and m is its order of con-
ditional positive definiteness. Let µ be its corresponding measure as in (1.5)
.Then, given a positive number b , there are positive constants δ and λ,0 <
0 0
λ < 1, which depend on b for which the following is true:
0
9
If f and s is the h spline that interpolates f on a subset X of Rn,
h,m
∈ C
then
n+β+1 n+1 β 1
(2.2) |f(x)−s(x)| ≤ 2 4 ·π 4 ·√nαn ·c2 · △0 ·λδ ·kfkh
holds for all x in a cube E provided that (i)E has sidpe b and b b ,(ii)0 <
0
≥
δ δ and (iii)every subcube E of side δ contains a point of X. Here, α
0 n
≤
denotes the volume of the unit ball in Rn and c, are as in (2.1).
0
△
The numbers δ and λ can be expressed specifically as
0
1
1 2 3Cγn
δ = , λ =
0
3Cγ (m+1) 3
n (cid:18) (cid:19)
where
2 ρ
C = max 2ρ′√ne2nγn, , ρ′ = .
3b c
(cid:26) 0(cid:27)
The number ρ can be found in the remark immediately following Lemma2.1,
and γ is defined in Lemma1.1.
n
Proof. First,letρ,γ ,andb betheconstantsappearingin(2.1),Lemma1.1.
n 0
and Theorem2.4, respectively. Let
2 ρ
B = 2ρ′√ne2nγn and C = max B, where ρ′ = .
3b c
(cid:26) 0(cid:27)
Let
1
δ = ,
0 3Cγ (m+1)
n
where m is the order of conditional positive definiteness of h.
Now, let x be any point of the cube E and recall that Theorem4.2 of [6]
implies that
(2.3) f(x) s(x) c f y x kd σ (y)
| − | ≤ kk kh | − | | |
ZRn
whenever k > m, where σ is any measure supported on X such that
(2.4) p(y)dσ(y) = p(x)
ZRn
for all polynomials p in . Here
k−1
P
1
ξ 2k 2
c = | | dµ(ξ)
k
(k!)2
(ZRn )
whenever k > m. By (2.1), for all 2k 2m+2,
≥
10