Constructing Cubature Formulas of Degree 5 with Few Points

0.14 MB·English

by  Zhaoliang Meng

Constructing Cubature Formulas of Degree 5 with
Few Points
∗
1 Zhaoliang Meng a and Zhongxuan Luoa,b
1 †
0
a
2 School of Mathematical Sciences,Dalian University of Technology, Dalian, 116024, China

v bSchool of Software, Dalian University of Technology, Dalian, 116620, China
o
N
November 23, 2011
2
2

]
A
N Abstract
.
h
This paper will devote to construct a family of fifth degree cubature formulae
t
a for n-cube with symmetric measure and n-dimensional spherically symmetrical
m
region. The formula for n-cube contains at most n2 + 5n + 3 points and for
[ n-dimensional spherically symmetrical region contains only n2 +3n+3 points.

1 Moreover, the numbers can be reduced to n2 +3n+1 and n2 +n+1 if n = 7
v
respectively, the later of which is minimal.
4
9
Keywords: Fifth degree formula; Cubature formula; Product region;Spherically
0
5 symmetric region; Numerical integration
.
1
1
1 1 Introduction
1
:
v In this article, the fifth degree formula with respect to the following integral
i
X
r L(f) = ρ(X)f(X)dX (1)
a
ZΩ
will be considered where the region Ω Rn and ρ(X) 0 for X Ω. We study two
⊂ ≥ ∈
cases: one of which requires Ω = [ 1,1]n and ρ(X) = ρ (x )...ρ (x ) where ρ (x ) is
1 1 n n i i
−
symmetrical; and the other of which requires that L is a spherically symmetric integral
(see the definition of section 3 or refer to [9]).
Among all the cubature formulae, those with few points received most attention.
Without any special assumption, a general lower bound was given asfollows (see [2] and
the reference therein): for a cubature formula of degree 2k or 2k+1, the number of the
points N dimPk, where Pk denotes the polynomial space of dimension n with degree
≥ n n
∗This project is supported by NNSFC (Nos.109111440268,10801023) and “the Fundamental Re-
search Funds for the Central Universities”.
†Corresponding author: mzhl@ dlut.edu.cn
1
no more than k. For even degrees, no greater lower bound was given until now, see
Cools’ survey paper [2]. However, for odd degrees 2k+1, the lower bound for centrally
symmetrical integral can be improved as follows (see M¨oller [11]):
0 if s is odd,
N 2dimPk (2)
≥ n −(1 if s is even,
or explicitly,
n 1
n+k − s+k
+ 2s n if k is odd,
−
n s

N (cid:18) (cid:19) s=1 (cid:18) (cid:19) (3)
≥  n+k Xn−1 s+k 1
 + (1 2s n) − if k is even.
−
n − s
(cid:18) (cid:19) s=1 (cid:18) (cid:19)
 X



Particularly, for the fifth degree case,
N n2 +n+1.
≥
However, there is a big gap between the theoretical lower bound and the numbers of
points in the known cubature formulae, especially for the higher dimensional case (see
Cools’ survey papers [3, 4, 5, 6]). So far, only for special dimension, the minimal cuba-
ture rules can be obtained. Noskov and Schmid [13] derived a necessary and sufficient
condition of existence of the minimal cubature rules with respect to the integrals over
the ball.
For the product region, a good choice to construct cubature formula is to employ
Smolyak’smethod[1,7,14,15]. However, thenumber ofthepointsismuch greaterthan
the theoretical lower bound. Very recently, starting from Smolyak method, Hinrichs
and Novak [8] presented a kind of formula with n2 + 7n + 1 points for fifth degree
case and (n3 +21n2 +20n+3)/3 points for seventh degree case. Before them, except
Lu’s formula [10], the best upper bounds were of the form “ 2n2” and “ 4n3/3”
≈ ≈
respectively, where “ ” denotes the strong equivalence of sequences. Lu’s formula uses
≈
n2 +3n+3 points just for entire n-dimensional space with Gaussian weight function,
which is the minimal point formula for the general dimension among all the known
formulae until now.
In this paper we shall present a method to construct fifth degree formulae with
at most n2 + 5n+ 3 points for the integrals with the product form and with at most
n2+3n+3 points for the spherically symmetrical integrals, which employ fewer points
than Hinrichs and Novak’s formulae [8] and are more close to the theoretical lower
bound. This paper proceeds as follows. Section 2 will present a method to construct
fifth formula for n-cube with symmetrical product measure. And section 3 will devote
to the construction of cubature formulae for spherically symmetrical integrals. Finally,
section 4 will give some numerical integration rules by our method.
2
2 Fifth degree formula for symmetrical product mea-
sure
We begin with a well-known formula with respect to the surface of sphere:
U = X Rn : x2 +x2 +...+x2 = 1 .
n { ∈ 1 2 n }
Due to Stroud [16], the monomial integrals for ρ(x ,x ,...,x ) = 1 are
1 2 n
2πn/2
V = ,
Γ(n/2)
2Γ[(α +1)/2]...Γ[(α +1)/2]
xα1xα2 ...xαndX = 1 n ,
1 2 n Γ[(n+α +α +...+α )/2]
ZUn 1 2 n
provided all the α are even; otherwise the integral is zero.
i
Mysovskikh [12] derives a cubature formula of degree 5 (n 4) as follows:
≥
n+1 n(n+1)/2
L (f) = f(X)dX Q(f) = A f(a(j))+f( a(j)) +B f(b(j))+f( b(j))
1
≈ − −
ZUn j=1 j=1
X(cid:2) (cid:3) X (cid:2) (cid:3)
(4)
where
a(r) = (a(r),a(r),...,a(r)), r = 1,2,...,n+1, (5)
1 2 n
n+1 , i < r,
− n(n i+2)(n i+1)
− −
a(ir) =  q(n+1)(n−r+1), i = r, (6)
 n(n r+2)
 −
q0, i > r,

and 
n
b(j) = (b(j),b(j),...,b(j)) (a(k) +a(l)) : k < l, l = 1,2,...,n+1 . (7)
{ 1 2 n } ≡ 2(n 1)
nr − o
The corresponding coefficients are
A = n(7 n)V/[2(n+1)2(n+2)],
−
B = 2(n 1)2V/[n(n+1)2(n+2)].
−
We firstly construct a cubature formula of degree five for the integral with product
form:
L(f) = ρ (x )ρ (x )...ρ (x )f(X)dX, (8)
1 1 2 2 n n
Z[ 1,1]n
−
where
1 1
ρ (x )dx = 1 and ρ (x )x2k+1dx = 0.
i i i i i i i
Z 1 Z 1
− −
3
L(x2)
Define a diagonal matrix T = diag a ,a ,...,a , where a = i with
{ 11 22 nn} ii √γL1(x21x22)
r ·
a positive parameter γ. Let a˜(j) = Ta(j), b˜(k) = Tb(k), A˜ = γA,B˜ = γB and
n+1 n(n+1)/2
Q˜(f) = A˜ f(a˜(j))+f( a˜(j)) +B˜ f(b˜(j))+f( b˜(j)) . (9)
− −
j=1 j=1
X(cid:2) (cid:3) X (cid:2) (cid:3)
Then
˜
Q(x ) = 0
i
n+1 n(n+1)/2
Q˜(x x ) = A˜ 2a a(r) a a(r) +B˜ 2a b(r) a b(r)
i j ii i · jj j ii i · jj j
r=1 r=1
X(cid:2) (cid:3) X (cid:2) (cid:3)
n+1 n(n+1)/2
˜ (r) (r) ˜ (r) (r)
= a a A 2a a +B 2b b
ii jj i · j i · j
h Xr=1 (cid:2) (cid:3) Xr=1 (cid:2) (cid:3)i
L(x2)L(x2)
= i j γL (x x )
s γL1(x21x22) · 1 i j
√γL (x2)
1 i L(x2), for i = j;
= L (x2x2) · i
 1 1 2
0, for i = j;
p 6
n
√2π4(n+2) √γ L(x2), for i = j;
= 2√Γ(n2+2) · · i

0, for i = j;
 6

n+1
Q˜(x x x ) = A˜ a a(r) a a(r) a a(r) a a(r) a a(r) a a(r)
i k l ii i · kk k · ll l − ii i · kk k · ll l
r=1
X(cid:2) (cid:3)
n(n+1)/2
+B˜ a b(r) a b(r) a b(r) a b(r) a b(r) a b(r)
ii i · kk k · ll l − ii i · kk k · ll l
r=1
X (cid:2) (cid:3)
= 0
n+1
˜ ˜ (r) (r) (r) (r)
Q(x x x x ) = A 2a a a a a a a a
i j k l ii i · jj j · kk k · ll l
r=1
X(cid:2) (cid:3)
n(n+1)/2
+B˜ 2a b(r) a b(r) a b(r) a b(r)
ii i · jj j · kk k · ll l
r=1
X (cid:2) (cid:3)
n+1 n(n+1)/2
= a a a a A˜ 2a(r)a(r)a(r)a(r) +B˜ 2b(r)b(r)b(r)b(r)
ii jj kk ll i j k l i j k l
h Xr=1 (cid:2) (cid:3) Xr=1 (cid:2) (cid:3)i
L(x2)L(x2)L(x2)L(x2)
i j k l
= L (x x x x )
q L (x2x2) · 1 i j k l
1 1 2
4
L(x2x2), for i = j and k = l;
i k
= (L(x2))2 L1(x4i) , for i = j = k = l;
 i · L1(x21x22)

0, others
L(x2x2), for i = j and k = l;
 i k
= 3(L(x2))2, for i = j = k = l;
 i
0, others

and similarly Q˜[x xx x x ] = 0. Define L˜ = L Q˜ and compute the moments as
i j k l m
−
follows:
L˜(1) = 1 γ V,
− ·
L˜(x ) = 0,
i
0, for i = j,
L˜(x x ) = n 6
i j L(x2) 1 √2π4(n+2) √γ , for i = j,
 j · − 2√Γ(n2+2) ·
(cid:18) (cid:19)
L˜(x x x ) = 0,
i j k 
L(x4) 3 L(x2) 2, for i = j = k = l,
L˜(x x x x ) = j − j
i j k l
(0, (cid:0) (cid:1) others,
L˜(x x x x x ) = 0.
i j k l m
Remark 1. If R˜(f) is a cubature formula of degree five with respect to L˜, then L
Q˜ +R˜ is a cubature formula of degree five with respect to L. ≈
Now it is the time to state our main result.
Theorem 1. There exists a cubature formula C for L of degree five which uses at
n
most n2 +5n+3 points.
Proof. Suppose that an integration formula R˜ of degree five for L˜ is given by
v(i,j) = v e w , i = 1,2,...,n, j = 1,2
i,j i ij
(10)
(0,0...,0) C
where e = (0,...,0,1,0,...,0) is a vector of dimension n and the ith element one is
i
the only nonzero element. To enforce polynomial exactness of degree five, it suffices to
require (10) to satisfy the following equations:
2
w (v )m = L˜(xm), m = 1,2,3,4,5,i = 1,2,...,n,
ij i,j i
j=1
X (11)
n 2
w +C = L˜(1).
ij
i=1 j=1
XX
5
By a proper parameter γ, we can make L˜(x2) L˜(x4) > 0. Thus we get the required
˜ i · i
cubature formula R with
L˜(x4) L˜(x2) 2
v = i , v = v , w = w = i , i = 1,2,...,n,
i,1 sL˜(x2) i,2 − i,1 i,1 i,2 2L˜(x4)
i (cid:0) i(cid:1)
n
˜
C = L(1) (w +w ).
i,1 i,2
−
i=1
X
The corresponding cubature formula can be written as
n+1 n(n+1)/2
L(f) A˜ f(a˜(j))+f( a˜(j)) +B˜ f(b˜(j))+f( b˜(j))
≈ − −
j=1 j=1
X(cid:2) (cid:3) X (cid:2) (cid:3) (12)
n 2
+ w v(i,j) +Cf(0,0,...,0).
i,j
i=1 j=1
XX
The total number of the points is at most (n2 +3n+2)+(2n+1) = n2 +5n+3.
Remark 2. To ensure all the points inside the integration region, it is required that
L˜(x4)
a 1, and 0 < i 1. (13)
ii ≤ L˜(x2) ≤
i
If L(x4) < 3L(x2)2, then
j j
2Γ(n/2+2)
γ ≥ 2π−n2Γ(n/2+2) L(x2i) 2, √γ > (n+2)πn/4 ,
p (14)
(L(x2) L(x(cid:0)4)+3L(cid:1)(x2)2) 2Γ(n/2+2)
√γ i − i i
≥ L(x2) · (n+2)πn/4
i p
which always has a solution. Otherwise
L(x2) 2 2Γ(n +2)
i γ 2
(cid:0)L1(x21x22(cid:1)) ≤ ≤ πn2 (n+2)2
usually does not derive solution which means some points lie outside of the region.
Particularly, if ρ (x ) = (1 x2)αi(α > 1), then
i i − i i −
6
L(x4) 3L(x2)2 = < 0
j − j −(α /2+3)2(α /2+5)
i i
implies that the solution for γ always exsits.
Remark 3. Since A vanished when n = 7, the total number of the points is n2+3n+1
if n = 7.
6
Remark 4. If
√2πn4 (n+2) Γ(n +1)
1 √γ = 0, or γ = 2 ,
− 2 Γ(n +2) · πn/2(n+2)
2
then L˜(x2) = 0 and thups
j
γ Q(f)+(1 γ V)f(0,0,...,0)
· − ·
is a cubature formula of degree three. Furthermore, it is easy to check that
n ∂4f n
γ Q(f)+ γ f(0,0,...,0)+(1 γ γ V)f(0,0,...,0)
· i∂x4 − i − ·
i=1 i i=1
X X
is a cubature formula of degree five provided f is smooth enough, where
L(x4) 3(L(x2))2
γ = i − i .
i
4!
3 Fifth degree formula for spherically symmetrical
measure
L is a spherically symmetrical integral functional means the region is of the form
X Rn p X < q and the weight function W(X) is a function of X , where
{ ∈ | ≤ || || } || ||
< X,Y >= x y +...+x y and X = √< X,X > (see [9]). The classical examples
1 1 n n
|| ||
are the integrals over the ball and over the entire Rn space with Gaussian weight. In
this case, L(x2) = L(x2) and L(x2x2) = L(x2x2) for any i,j,k,l. Taking a =
i j i j k l ii
4 L(x21x22) and the same mathod explained, we have
γL1(x21x22)
·
q
˜
Q(x ) = 0
i
n
√2π4(n+2) γ L(x2x2), for i = j;
Q˜(xixj) = 2√Γ(n2+2) · · 1 2

0, p for i = j;
 6
Q˜(x x x ) = 0;
i j k 
L(x2x2), for i = j and k = l;
i k
Q˜(x x x x ) = 3(L(x2x2)), for i = j = k = l;
i j k l  1 2
0, others.
Q˜(xixjxkxlxm) = 0
and
L˜(1) = 1 γ V,
− ·
L˜(x ) = 0,
i
7
0, for i = j,
L˜(x x ) = n 6
i j L(x2) √2π4(n+2) γ L(x2x2), for i = j,
 1 − 2√Γ(n2+2) · · 1 2
L˜(x x x ) = 0, p
i j k 
L(x4) 3(L(x2x2)), for i = j = k = l,
L˜(x x x x ) = 1 − 1 2
i j k l
(0, others,
L˜(x x x x x ) = 0
i j k l m
where L˜ = L Q˜.
−
It is a slightly different from the product case, because we have the following
Theorem 2. Suppose that
L(f) = f(X)ρ(X)dX
ZΩ
is a spherically symmetric integral functional, then
L(x4) 3L(x2x2) = 0.
1 − 1 2
Proof. To calculate L(x4) and L(x2x2), we introduce the transformation which was
1 1 2
used in [16, Page 33]
x = rcosθ cosθ ...cosθ cosθ
1 n 1 n 2 2 1
− −
x = rcosθ cosθ ...cosθ sinθ
2 n 1 n 2 2 1
− −
x = rcosθ cosθ ...sinθ
3 n 1 n 2 2
− − (15)
............
x = rcosθ sinθ
n 1 n 1 n 2
− − −
x = rsinθ
n n 1
−
The Jacobian of transformation (15) is
J = rn 1(cosθ )n 2(cosθ )n 3...(cosθ )2(cosθ )
− n 1 − n 2 − 3 2
− −
and therefore
q p π/2
L(x4) = + − ρ(r) r n 1r4dr (cosθ )4dθ I
1 | | − 1 1 × 0
(cid:26)Zp Z q (cid:27) Z π/2
− −
q p π/2
L(x2x2) = + − ρ(r) r n 1r4dr (cosθ )2(sinθ )2dθ I
1 2 | | − 1 1 1 × 0
(cid:26)Zp Z q (cid:27) Z π/2
− −
where
π/2 π/2 π/2
I = (cosθ )5dθ (cosθ )6dθ ... (cosθ )n+2dθ .
0 2 2 3 3 n 1 n 1
− −
Z π/2 Z π/2 Z π/2
− − −
8
Hence
L(x41) = −ππ//22(cosθ1)4dθ1 = 3π/8 = 3
L(x21x22) ππ/2/R2(cosθ1)2(sinθ1)2dθ1 π/8
−
R
and therefore L(x4) 3L(x2x2) = 0. This completes the proof.
1 − 1 2
Let
√2πn4 (n+2)
L(x2) γ L(x2x2) = 0
1 − 2 Γ(n +2) · · 1 2
2 q
that is p
(L(x2))2 2Γ(n/2+2)
γ = 1 .
L(x2x2) · πn/2(n+2)2
1 2
In this case, we get a fifth degree formula with at most n2 +3n+3 points
n+1 n(n+1)/2
C (f) = A˜ f(a˜(j))+f( a˜(j)) +B˜ f(b˜(j))+f( b˜(j))
n
− − (16)
j=1 j=1
X(cid:2) (cid:3) X (cid:2) (cid:3)
+(1 γ V)f(0,0,...,0).
− ·
If n = 7, A˜ = 0 will yields a minimal cubature formula which only contains n2 +n+1
points.
4 Numerical Results
Assume n 4.
≥
Symmetric product region.
•
We consider the case of weight functions ρ = ρ = ... = ρ = 1/2.
1 2 n
In this case,
L˜(x4) = L(x4) 3(L(x2))2 = 1/5 1/3 = 2/15 < 0.
i j − j − −
To confirm all the points inside the region, it follows from (14) that
2Γ(n/2+2)
γ .
≥ 9πn/2
If we take γ = 2Γ(n/2+2), then
9πn/2
30(n 1)
a = 1, v = − ,
ii i,j
± 5(n 1)
p −
9
5 5 5 5 8 7
w = n2 + n , C = n3 n2 n+1.
i,j
−108 54 − 108 54 − 27 − 54
Due to (12), the corresponding cubature formula can be written as
1 5 8 7
f(X)dX ( n3 n2 n+1)f(0,0,...,0)
2n ≈ 54 − 27 − 54
Z[ 1,1]n
−
n
5 5 5 30(n 1) 30(n 1)
+( n2 + n ) f − e +f − e
i i
−108 54 − 108 5(n 1) − 5(n 1)
i=1 p − ! p − !!
X
(7 n)n2 n+1
+ − f( n/2+1 a(j))+f( n/2+1 a(j))
18(n+1)2 · − ·
j=1
X(cid:2) p p (cid:3)
n(n+1)/2
2(n 1)2
+ − f( n/2+1 b(j))+f( n/2+1 b(j))
9(n+1)2 · − ·
j=1
X (cid:2) p p (cid:3)
Spherically symmetric integral
•
(1). Entire n-dimensional space with Gaussian weight function.
Simple calculation yields
Γ(n/2+1)
γ = , and a = n/2+1.
πn/2(n+2) ii
p
Thus the fifth degree formula can be written as
2πn/2
e−x21−...−x2nf(x1,x2,...,xn)dx1...dxn f(0,0,...,0)
≈ n+2
Rn
Z
n2(7 n)πn/2 n+1
+ − f( n/2+1 a(j))+f( n/2+1 a(j))
2(n+1)2(n+2)2 · − ·
j=1
X(cid:2) p p (cid:3)
n(n+1)/2
2(n 1)2πn/2
+ − f( n/2+1 b(j))+f( n/2+1 b(j))
(n+1)2(n+2)2 · − ·
j=1
X (cid:2) p p (cid:3)
which is same with Lu’s formula [10].
(2). The n-dimensional unit ball.
Simple calculation yields
2(n+4)Γ(n/2+2) n+2
γ = , and a = .
(n+2)3πn/2 ii n+4
r
Thus the fifth degree formula can be written as
8πn/2
f(x ,x ,...,x )dx ...dx f(0,0,...,0)
1 2 n 1 n
≈ n(n+2)2Γ(n/2)
Bn
Z
(7 n)n(n+4)πn/2 n+1 n+2 n+2
+ − f( a(j))+f( a(j))
(n+1)2(n+2)3Γ(n/2) n+4 · − n+4 ·
j=1 r r
X(cid:2) (cid:3)
10