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On the dimension of bivariate periodic spline spaces type-1

triangulation

Citation for published version (APA):

Morsche, ter, H. G. (1988). On the dimension of bivariate periodic spline spaces type-1 triangulation. (RANA : reports on applied and numerical analysis; Vol. 8805). Technische Universiteit Eindhoven.

Document status and date: Published: 01/01/1988

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Department of Mathematics and Computing Science

RANA88-05 April 1988

ONTHEDIMENSION OF BIVARIATE PERIODIC SPLINE SPACES

TYPE-l TRIANGULATION

by

H.G. ter Morsche

Reports on Applied and Numerical Analysis

Department of Mathematics and Computing Science Eindhoven University of Technology

P.O. Box 513 5600 MB Eindhoven The Netherlands

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by H.G. ter Morsche

1.

Introduction

The extensive literature on spline functions shows that the problem of periodic spline interpola-tion is thoroughly studied in the univariate case, especially when the knots are equally spaced on the realline. Inthis case the number of knots in one period equals the dimension of the space of periodic spline functions, and the interpolating periodic spline function can be computed without matrix inversions. Moreover, the Fast Fourier transfonn is a powerful tool in these circumstances. The main reason that the theory of univariate spline functions with equidistant knots is so elegant is based on the fact that the corresponding spline-space is spanned by translates of a fixed com-pactly supported spline function, known as a B-spline function. The situation differs completely when the problem is lifted to more dimensions. Even the most straight forward generalization may cause difficulties. For instance, let us consider double-periodic bivariate cubic spline func-tions with global smoothness CIon a rectangular mesh in JR2•

By a rectangular mesh, we mean a subdivision of the x-y plane in squares bounded by the mesh-lines

x

=

111 ,Y

=

~2' which pass through the lattice points ~=(J.Ll ,~2)E ;;Z2 •

On each square oftherectangular mesh a cubic spline function coincides with a bivariate polyno-mial of degree at most three, and it is differentiable on 1R2 with continuous first order partial derivatives.

Itturns out that in this case the dimension of the space of(m n)- periodic cubic splines, i.e., cubic spline functions sfor which s (x

+

m , y)

=

s (x , y)

=

s (x , y

+

n) for all (x, y)E IR2 with integermand

n.

is equal to2m

+

2n

+

1. So,ifwe like to interpolate an(m , n)-periodic function at the lattice points J1E ;;Z2 then the number of interpolation conditions

m . n

differs generally from the dimension of the space of candidates for the interpolating function; the answer to the

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question ofunisolvence of the interpolation problem must unfortunately be negative.

The aim of this paper is to compute the dimension of various spaces of(m,n)-periodic spline functions with respect to a three direction mesh, also called a type-1 triangulation, as given below

. The mesh lines defines a uniform triangulation of JR2•

Evidently, the dimension we are looking for depends on the degree and the smoothness of the bivariate spline functions. Therefore, we introduce the spline spaceS~ as follows.

Definition 1.1. Letk and p be nonnegative integers. Then a spline functionsbelongs toS~ ifand only if

i) on each triangle of the type-I triangulation,

s

coincides with a bivariate polynomial of total degree at mostk.

ii) seep(JR2),i.e., all the partial derivates of S up to the order p exist and are continuous on

JR2.

Our interest is focussed on the(m , n)-periodic functions inS~.The set of these functions willbe denoted byS~!!, where!!

=

(n, m).

It appears that as in the one-dimensional situation (cf Schoenberg [3]) the so-called exponential or eigen splines play an important r6le to understand the spaceS~.The exponential splines may be interpreted as eigen functions of some shift operators; they can be defined as follows. If

(z1 ,z2)is a pair of complex numbers and

s

e S~ a nontrivial spline function satisfying

s (x

+

1 ,y)

=

Z 1 S(x ,y) ,

(1.1) s(x,y+1)= Z2 S(x,y), (x,y)e JR2

thens is called an exponential spline.

The importance of the exponential splinewillbe shown in the next section. There we shall prove thatS~!!is spanned by a basis consisting only of !!-periodic exponential splines. For the computa-tion of dimS~!!'the dimension of the spaceS~!!,we are thus interested in the quantity

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denoting the dimension of the subspaceSZt.%2 C S~ of exponential spline functions with (Zl ,zz)

fixed. This quantity will be computed in Section 3. Fmally, a complete answer to our problem of the computation of dimS~!!willbe given in the final section.

2. Exponential splines

The discrete Fourier transform, a basic tool in translation invariant periodic function spaces, can be applied in our problem with success. The bivariate discrete Fourier transform ofan!!-periodic sequence (!!=(n ,m» of (complex) numbers(aIL) ijJ.E % z)is an !!-periodic sequence(Fa)1Lgiven

by - -

-(2.1) (Fa) =

1-

~ a e-21ti(v.IL/~)

IL N kJ v - - ,

-

veR~-(2.3)

where the summation runs over the square.

(2.2) Rn

={

v

=

(VI,vz)E %z

I

O:s;VI:s;m-1 , O:s;vz:S; n -I } .

-

-VIllI vzllz

Moreover, (v,IIIn):= - -

+ - -

andN :=

nm .

- - - m

n

Itis well known that the inverse discrete Fourier transform can be written as follows

Now letSESt!!,and let the function(F s)1Lbe defined by

(Fs)lL(x,y)= N1

L

S(X-v1,y-vZ)e-Zlti~.~!!)

- veRa

Itcan easily be verified that(FS)ILE S~~.But the more important observation is based on the

fol-lowing relations:

-{

(Fs)~

(x

+

1,y)

=

e2ltillllm

(Fs)~

(x, y) , (Fs)1L (x,Y+1)

=

e2ltifJ2/n (Fs)1L (x,y) .

-

-In other words, if(Fs)1L is not the null function, then(Fs)1l is an !!-periodic exponential spline function withZl =eZltilL"'ilmandZ2=e2ltifJ2/n

-The function Scan be recovered from the functions (Fs)1L (IlERn) by inverse discrete Fourier

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-S(XI,X2)=

L

(FS)v(XI,X2)'

veRN

Infact, we have shown that the set of~-periodicexponential splines inS£ spans the spaceS£,~in the sense that everyS E

S£,

~ is a finite linear combination of~-periodicexponential splines. The

next question is to compute the number of elements in an independent set of~-periodic exponen-tial splines that spans the space st~. For this, we will investigate the subspace SZI •Z2 in S£ in more detail.

3. The spaceSZIoZ2

The contents of this section is rather technical; we do not explainallthe details. The basic idea is to represent an arbitrary polynomialp E ilk>i.e., the set of bivariate polynomials of total degree at mostk, by means of univariate (generalized) Euler-Frobenius polynomials or Bernoulli polyno-mials. Let zE Cbe given with z::#1, and leti be a positive integer. We introduce here the Euler-Frobenius polynomialx H Ei (z, x)of degreei as the uniquei-thdegree polynomial satisfying (3.1) Ei(z ,x+1) - zE i (z,x)= -.,Xi (l-z).

l.

The Bernoulli polynomialx H Bi(x)is defined here as the unique polynomial of degree i satis-fying

Bo(x) == 1,

(3.2)

(i~1) .

For notational purposes, it is sometimes convenientto writeEi (l ,x)

=

Bi(x).If(z1 ,Z2) is a pair of complex numbers andpe ilk,thenpcan be written as

(3.3) p(x,y)=

L

ai,j E i(ZI,X)Ej (Z2'Y)'

os;i+~k

Now letSbe an exponential spline function inS£ andpandqits polynomial pieces on the

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iy

1

o

1 ---+

x

Sinces is an exponential spline function in

cP

(R2),it is necessary that the following conditions

are satisfied.

(3.4)

(i) (d;p)(1,y)

=

Z1(d;q)(O,y) ,

(ii) (d~q)(x ,1) =Z2(d~p)(X,0) ,(1 =0, 1, ... ,p ,

(iii) q (x , y)=p (x , y)

+

r (x , y)

with r(x,y)=(x-y)p+1 r1(x,y)andrl E IIk -p-1

The partial derivatives toxand

yare

denoted by

d

x and

d

y,respectively.

On the other hand,ifp andqsatisfy (3.4)then due to (1.1)there is a unique exponential spline

s

which coincides with the polynomialponl\l and withqonl\2'

It follows from(3.4) and the representation(3.3)that

L

(J.j,j (Ef/) (ZI' 1)-ZI Efl> (ZI ,0» Ej (Z2' y)

=

Z1(d~r)(0,y) ,

QS;i+jSk

(3.5)

L

(J.j,j (EJ> (Z2' 1)-Z2EJ> (Z2'Ej(ZI ,x)=-(d~r)(x, 1) ,

QS;j+jSk

(1

=

0, 1, " ' ,p).

These relations may be simplified by using the properties of the polynomials Ej •To do so we

must distinguish the two cases:

A:Z1 ':t:-1 andZ2 ':t:-1, B: Z1= 1 orZ2=1 .

A:Z1 ':t:-1 andZ2':t:-1.

The polynomials Ej(Z1 ,x) and Ej(Z2'Y)

are

both Euler-Frobenius polynomials. So, by using

(3.1), we maywrite

(I) (I) { 0 (1;t:i),

(3.6) E , (z,I)-zE, (z,O)= l-z (I=i).

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k-l (l-Zl) Lal,j Ej (Z2' y)=Zl(d~,)(O,y) , j={) k-l . (l-z2)Lai,lEi(zl,x)=-(d~,)(x,1) (1=0,1, ... ,p), i={)

,(x,y)= (X_y)p+1 '1 (X ,y)"l E llk-p-1

Even a further simplification canbemade by using (3.6) again. We obtain

(3.7)

. 1 . ,

(1-z1)(I-Z2) a',j=Zl {d~dx ,(0,1) - Z2d~dx,(0, O)} ,

,

.

,

.

(1-z1)(I-z2) ai,'=zl

d

yd~,(O,I)-dyd~,(l,1),

1=0,1, ... ,p;j=O, 1,'" ,k-l;i=O, 1, ... ,k-I, ,(x,y)=(X-y)p+l 'l(X,y),

'rE

llk-p-1.

Let us fonnu1ate what we have reached up to now. If p and q are the polynomial pieces of an exponential spline in S~ based on the pair (Zl' Z2), then numbers ai,j and a polynomial , 1E llk-p-1 exist such that (3.7) is satisfied. On the other hand, if we can find numbersai,j and a polynomial '1 E llk-p-1 satisfying (3.7), then a polynomial p can be constructed by means of (3.3), and subsequently a polynomialqby means of (3.4)iii.These polynomials are used as a res-triction of a spline function

s

to the two trianglesL\1 andL\2'The function

s

can then be extended to the wholex-yplane on base of (1.1) giving an exponential spline inS~.

If'l is known then it follows immediately from (3.7) that the numbersai,j withi::;; p orj::;;pare detennined. The remaining numbers ai,j (i+j::;;k ,i~p+1, j~p+1) can be chosen freely. If

a+:=max(a ,0) then thetotalnumber offreeai,jis equal to t (k-2p-l)+ (k-2p).

In fact, we have computed the dimension of the space of exponential splines corresponding to the pair (z1 ,z2)for which the polynomial, vanishes. But, this is precisely the space of exponential splines corresponding to the reetangu1ar mesh, where the diagonal mesh lines are omitted. Apparently, a basis of this space is given by

Note that in any case

dimSZlo Z2S dimllk-p-1 +t (k-2p-1)+ (k-2p),

which is equal to zeroifp~k. This is not surprising;ifp~ kthenS~=llkand the space of poly-nomials contains no exponentials except the constants.

To each numberai,j there corresponds one or two equations in (3.7) involving ai,j' In order to avoid contradictions, we havetoimpose some conditions on the polynomial'1'The numbersai,j

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(3.8)

condition for the existence of the numbersCJ.i,imay be written in the following fonn

{

LiJ'=O,

i=0.1.2 • ... ,p;j=O,1.2,· .. ,p;i+j~k,

,(x.y)=(x-y)p+1 'l(X,y), r1 E Ilk-;>-l. where the linear functionalLiJisdefined by

(3.9) Li,ir =(a~ a~r)(1. 1) - zlz2(a~ a~r)(O,0) .

We wish to express LiJr in tenns of the polynomial

'1.

A straight forward computation will show that

(3.10)

(3.12)

For some values of(i,j), (3.10) is an empty sum; the corresponding condition Li,ir

=

0 may then be omitted. The values of(i,j)giving a nonempty condition belong to the set

(3.11) n~ = {(i ,j)E .7l2

I

p

+

1::;; i

+

j::;;k , i~

p,

j::;;

p }

-

-By settingL iJ ' 1=Li,i r, we d:fine linear functionalsLi,i on the polynomial space Ilk-p-1. Due

to (3.10) the linear functionalsLi,ican be expressed in tenns ofLi,i as follows.

- . i j .

Li,i=(-1Y (P+ 1)!

L

(i )(. )(-lY'Li.,il . i,+i,=i+i-p-1 1 J1

It is clear from the foregoing discussion that

(3.13) dimSZ,Z2=dim Ilk-p-1 - dim<Li'i I(i,j)E

n~

> +

t

(k - 2p-1)+(k - 2p)

H~re <LiJ 1(i.j)E n~

>

is the space of linear functionals on Ilk-;>-l spanned by the set

{Li,i J(i ,j)E

n£}.

-Since, every LiJ belongs to <.Li"iI I 0~ i1

+

it

~ k - P - 1

>,

we shall investigate the

depen-dence of the functionals Li,i first We have to distinguish two cases now to wit: z 1 z2

'*

1and

Zl Z2=1.

Lemma 3.1.

Ifz1Z2

'*

1then {Li,i J0~ i1

+it ::;;

k - P-1} is an independent set of functionals onTIk-;>-l' Proof. Note that dim TIk-;>-l=#{Li"i, I0::;; i1+

it ::;;

k - P-1 }, so it suffices to prove that if

q E TIk-;>-l and Li.,i, g

=

0 for all (i1,j1)' then g == O. So, let g E TIk-;>-l be such that

Li"i, g=O for all (i1,it). Then all the partial derivatives of the polynomial

g(x+1.y+l)-Zlz2g(.x,y) vanish at (0,0). Hence, g(x+1,y+1)=Zlz2g(X,y) for all

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In the previous proof, the assumption z1 Z2

*

1 is used only for the implication: g (x+1,y+ 1) = Z1 Z2 g(x, y) for all(x ,y)theng

=

O.IfZI Z2 = 1, then it can easily be shown that the property g (x+1, Y + 1) =g(x ,y) for all (x,y) implies g (x ,y)= h(x-y) where h is a univariate polynomial of degree at mostk - P-1. We denote byVthe space of all these polyno-mials ginIlk-p-1. Evidently, dim V=k-p.1fil +

it

=k-p-1 then

a~'

a;'

r1 is a constant. In this case the functional Lihi, is the null functional on llk-p-1' The set {Liloit IO~ i1+ it~ k-p-1} is dependent. However if we delete the null functionals, then according to the following lemma the remaining set will be independent.

Lemma 3.2.

IfZ1 Z2 = 1 then {Li.,i,

I

O~ i1 +

it

~k-p - 2} is an independent set of functionals onIlk-p-1 . Proof Let ge llk-p-1 be such that Lihi, g=0for all 0~ i1+

it

~ k - P- 2. Then all the partial derivatives of the polynomialg(x+ 1 ,Y+ 1) - g(x,y)vanish at (0, 0). Henceg e V. Therefore, dim <Lihi•I

O~

i1+

it~

k-p-2

>

= dimIlk-p-1 - dim V =

/7),

i.e., the number of elements in {Li.i ,

I

O~ i1 +

it

~ k-p-2}. This proves the Lemma.

0

-Now we return to the functionals LiJ'. They

are

described as linear combinations ofLi .J' with

, J' 1

i1+ it = i+j -p -1 (cf.3.12). So Li,i is the null functional for i+j =k, Z1 Z2 = 1. This means

that the conditions for rl e llk-p-l may be summed up by

{

Li,ir1 =0

«i

,j)e

n~

;zl z2*1), (3.14)

Li,ir1=0

«i

,j)e n~-1;ZI Z2*1).

We subdivide the collection of conditions into disjoints sets

Ac

(e=p + 1, ... ,k) given as fol-lows.

-Ac

= {Li,i I i

+

j

=

e ,i~ P,j~

p }

As a consequence of Lemma 3.1 and 3.2 linear functionals stemming from different nonempty sets

Ac

are

independent. Theref.?re, we havetoconsider linear functionals Li,i within one definite nonempty set

Ac.

A functional Li,i in

Ac

canbe written as

(3.15)

wherePj, is a polynomial of degree e-p-1;

(-1Y'

t(t-t) . " (t-i1

+

t)(e-t) ... (e- t -

it

+

1)

Pi, (t)

=

,

I ' I (te JR.),

11 .Jt . (i1+it=e-p-1).

c-p-l

The polynomialsPi.

are

independent, which can be shown as follows. Let

L

ail

Pi. (t) = 0 for i,=O

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all t E JR. Then by a successive substitution of t

=

c ,t

=

C - 1, .. , ,t

=

p + 2 we find 00

=

(II

= ... =

ac-p-2

=

0 , and thus (Ic-p-1

=

O.This proves the independence of the polynomi-alsPj ,.

Since every functional in

Ac

is a linear combination of the independent functionals {Li"it Iii +

h

=c - p - 1 }, we have that dim <Li,j I Li.jE

Ac

>

~c - p. On the other hand

there holds thatdim<Li,j I Li.jE

Ac

>

~ IAc I, where IAc I denotes the number of elements in

Ac.

We shall show now that indeed

-(3.16) dim

<

Li,j I Li•jE

Ac

>

=

min (I

Ac

I , c-p).

For this, it is sufficient to prove that any collection of no more than c - p functionals in

Ac

is

-

-independent. So let {Li,j E Ac liE I} be such that II I~ c - p, and assume

L

(IiLi.j , 1

=

0 for

iel

all'l E TIk-p-1. Then

L

(Li"it '1)

L

(IiPit (i)= 0 ('1 E TIk-p-1) .

i,+j,:c-p-1 iel

Because of the independence of the functionalsLj

"

J' one has

,

L(IjPj ,(i)=0 01=0,1, ... ,c-p-l).

iel

However, the polynomialsPj , are independent and II I~ c - p, which implies o'i = 0(iE I). So

we have proved relation (3.16).

Relation (3.16) combined with (3.13) gives the expression:

(3.17) k

L

min(JAc I ,c-p) (Zl Z2:;t1), c=p+1 k-1

L

min(IAcl,c-p) (Zl Z2=1). c=p+1

The last pan of our treatment of the case Z1:;tand 1 Z2:;t1 consists of an evaluation of the expres-sions in (3.17). It turns out that

k

(3.18)

L

min ( I A I, c -p)=t (k-p)+ (k-p+ 1)-(k-kp)+(kp+k-3p)+t (k-2p)+ (k-2p-I), c=p+1

where kp

=[

3P;

1

], i.e., the integer pan oft (3p + 1).

Lemma 3.3. For all k= 1 , 2, ... , ; p = 0, 1, 2, '" , and (z 1, z2) E (;2 with z 1:;t1 and z2:;tI, there holds

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The previous theorem is the ultimate result in the caseZ 1:¢:1,Z2:¢:1. Now we will continue with the situationZl:¢:1 orZ2=1.

B :Z 1=1orZ2= 1

First, we consider the case Zl Z2:¢:1. Hence Zl :¢:1 or Z2:¢:1. Without any restriction we may assumeZl :¢:1,Zz=1. The transformation T: JR2 ~ JR2 given by the matrix

[1-1]

o

-1

maps the type-I triangulation onto itself, moreover, ifs(x, y)is an exponential spline inS~ based on the pair (Zl,Z2), then s(f(x

,y»

is again an exponential spline in S~ based on (Zl ,(Zl Z2rl)=(Zl, zil). Since Zl:¢:1, zlzil = 1,it is clear that Lemma 3.3 may be applied. The result is

The situation Z 1

=

Z2

=

1 differs from the previous cases; it can not be obtained by a simple transformation. Instead of generalized Euler-Frobenius polynomials, we have to use the Bernoulli polynomialsBj (cf. (3.2». From (3.4) it follows that a sufficient and necessary condition for the

polynomial piecesp andq, p(x ,y)=

L

(J.j,jBj(x)Bj(y) ,

Q;,j+jSk q=p+r

to be a restriction of an exponential spline

s

withZ 1

=

z2

=

1 is given by

L

(J.i,j (B~l-l)(1) - B~l-l)(0» Bj (y)=

(ai-

1r) (0,y) , Q;,j+~k

L

(J.j,j (By-I)(1) - By-I)(0» Bj (x)= - (a~-l r) (x,1) ,

~j+~k

1=1, . o o

,p+l,

r(x,y)=(x-y)p+l rl (x,y).

By using the property that B~l-I)(1)- B~l-I)(O)=0(i:¢:l), Bg-1)(1)=Bg-1)(0) for all I, we may replace the system above by the equivalent system

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(3.19)

(3.21)

j=O

k-I

L

(Xi.1Bi(x)

=

_(a~-1r)(x, 1) ,

i=O

1=1,"',p+l, .

r(x ,y)=(x_y)p+l rl(x, y) .

A further simplification can be made, by using (3.3) again. We have

CJ.I.i=(a~-1

ai-I

r)(0, 1)_(a~-1 axl-Ir)(0, 0), (Xi.I =(a~-I a~-I r)(0, 1) - (a~-l a~-1r)(1 ,1) , i,j=l, "',k-I, 1 (X/.O=

J

(ai-1r)(O,y)dy,

a

1 Cl<J.I= -

J

(a~-Ir)(x, 1)dx , o

(ai-1

r)(0,y)

E

Ilk-l' (a~-lr)(x, 1)

E

Ilk-1

1=1, "',p+l,

r(x ,y)

=

(x - y)P+1r1(x ,y) .

Let us first consider the solutions for the rectangular mesh, i.e., solutions corresponding tor

=

O. Then

(X/.O=

ao.1

=0 (1=1 • ...•p+ 1),and,

(X/.j=(Ii.l= 0 (I= 1. . .. ,p + 1 ;i = 1, ... ,k -I ;j = 1, ... ,k -I)

The remaining(Xi,jcan be chosen freely, thetotal number of which is given by (3.20) 1

+

2(k-p-l)+

+t

(k-2p-3)+ (k-2p-2) ,

which is equal to the dimension of the space of the exponential splines (z1=z2 = 1) with respect to the rectangular mesh.

Note that the numbers (Xi.jfor i

=

1, ... , p+ 1 ,j

=

1, '" • p+ 1 and i+ jS; k occur twice in

(3.19). Therefore, a necessaryand sufficient condition for the polynomial r to be a solution of (3.19)is given by

r(x ,y)

=

(x-y)P+1 rl(x, y) ,

(a~r)(O,y)E nk-i - 1 • (a~r)(x, 1)E Ilk-i - 1,j,j =0,1, '" ,p.

a~ a~r (1. 1)

=

a~ a~r (0,0) , i=O,I, ..·.p;j=O.I, " ' , p

The next step is to convert the conditions(3.21) for the polynomial r into conditions for the poly-nomialrl. The computation is straight forward. We only present here the result It turns out that

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, 1must satisfy the conditions (3.22)

d

~ a~

,

1(0, 0)=0 , iS; Porj S; P , i

+

j

=

k - P - 1 , Lj,j'l=0 i,j=0, 1, .. , ,p ;i

+

j S;k - 2 .

-Here the functional Lj,jis defined in (3.12), where

L j),jJ'1

=d~ d~J'1

(1,

1)-d~J d~l'l

(0,0). Let the functional Li,jbedefined by Li,j'l=d~ d~'1 (0,0).

We prove now that the collection of functionals

{ Lj),jJ lOS;i1

+

h

s;k - P - 3 }U {L i,j liS; P orj S;p ,i

+

j =k - P - 1 }

is independent on ITk-P-l. As a consequence of Lemma 3.2 one has that the collection {LiJ>jJ lOS;i1 +

h

S;k-p-3} is independent on ITk-p-2. Note thatLi,jg=0 forallgE ITk-P-2.

So, we only have to prove that the collection

{Lj,j liS;Porj S;

p, i

+

j =k - P-1} is independent on ITk-p-l. However, this follows easily from the observation that {Li,j I i +j =k - P-1}is independent on ITk-p-l'

The total number of functionals in {Li,j liS;Porj S;

p,

i

+

j

=

k - P- 1} is equal to (k-p)+ - (k-3p-2)+. Hence

dim

<Li,j

liS; P orj S;p, i +j =k - P -1

>

=(k - p)+ - (k - 3p - 2)+ . Since

- k~

dim

<

Lj,j lOS;j , j S;p,

i

+

j S;k - 2

>

=

r,

min(I

Ac

I ,

c -

p) , c=p+l

one has that for z1=z2=1

k-2

- (k - p)+ +(k - 3p - 2)+ -

r,

min ( IAc I ,c - p ) . c=p

By using (3.18) this expression canbesimplified. We have

dimSZJ> Z1 =1 + 3 (k-p-l)+ + (k-3p-2)+ +(k-kp-2)(kp+k-3p-2).

This completes case B.

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Theorem 3.4. Forallk=I,2, ... ;p=O, 1,2, ... , and(ZltZ2)e Cthereholds

{

(k-kp)+ (k p+k-3p) (1E {Zl , Z2, Z3}),

dimSz"za= (k-p)+ +(k-l-k p)+ (k p+k-I-3p) (loccurs once in {Zl , Z2, Z3} ),

1+3(k-p-l)+ +(k-3p-2)+ +(k-k p-2)+(k p+k-3p-2) ({I} = {Zl • Z2, Z3} ) .

The comparison between smoothness and degree of a spline function is of importance with respect to the existence of spline functions inS~ having a compact support.

ITSis a compactly supported spline function then

L

S(X-VI ,Y-V2)Z~1

Z2

a (v"va)eZa

is a nontrivial spline function showing that dim Szlza

>

0 for all(Zl , Z2)e C2•As a consequence

of Theorem 3.4, k must then satisfy k

>

k p.This observation can also be derived from results published in a paper of de Boor and Hollig[1],where the number kpis fundamental for the study of the approximation order ofS~.

4. The dimension of periodic spline spaces

Itis now easy to compute the dimension of the spaceS~.!!. consisting of the

!!

=(n ,m)periodic function in S~. The space S~ has a basis of exponential splines (cf. Section 2). Therefore, in order to compute dim S~!!., we only have to add the dimensions of the spaces of exponential splines SZl,ia where

The total number of pairs(u ,v) corresponding to the various cases can be listed as follows

(m-l)(n-l)-(gcd(m.n)-I), (zl*I,Z2*I,ZlZ2*1), n-l m-l gcd (m, n)-1 1 (zl=I,Z2*1), (z2=1, Zl *1), (Zl=Z2=1).

Hereg cd(m,n) denotes the greatest common divisor ofmandn. Ourfinalresult is the contents of the next theorem.

Theorem 4.1.

Let

k

=

0, 1, ... and p

=

0, 1, .... Then the dimension of the space of

!!-periodic splines{[! =(n, m» inS~ is given by

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dimS£.~= (mn-n-m+2-g c d(m, n» (k-kp)+ (kp+k-3p) +

(m+n+g c d(m, n)-3)(k-p)+ + (k-l-kp)+ (kp+k-I-3p» +

1 + 3(k-p-l)+ + (k-3p-2)+ +(k-kp-2)+ (kp+k-3p-2).

The complicated formula for the dimension ofst~ can be simplified for the situation, when k = kp+1. This is the smallest degree k correspondingtoa given smoothness p such that S£

con-tains finitely supported splines. Examples of finitely supported splines in S£ are the so-called Box-splines (cf. de Boor, Hollig [1]).

Ifk=kp+ 1 then

{

m n + I(m + n + g c d (m , n» (p = 21) , dimS

£p+l.~

-- 2m n + 2 + l(m +n+g c d (m, n» (p=21 + 1).

For periodic cubic splines in S1 (p = 1,kp=2), this formula reduces to dim S1PI= 2

m n

+ 2 ,

'-which agrees with a result given in Ter Morsche [2], where the problem of periodic bivariate spline interpolation is studied.

References

[1] Boor, C de, Hollig, K.,Bivariate box splines and smooth pp functions on a three direction mesh. Journal of Computational and Applied Mathematics

2.

(1983), pp 13 - 28.

[2] Morsche, H.G., ter, Bivariate cubic periodic spline interpolation on athree direction mesh. EDT Report 86-Wsk-02 Eindhoven University of Technology, Eindhoven, 1986.

[3] Schoenberg, 1.1., Cardinal spline interpolation. Regional conference series in applied mathematics 12., SIAM, Philadelphia, 1973.

(17)

Number 87-16 87-17 87-18 88-01 88-02 88-03 88-04 88-05 Author(s) AA.F. van de Ven! Tani/Otomo/Shindo AJ.E.M. Janssen! S.J.L. van Eijndhoven

L.

Dortmans/

A Sauren!

AAF. van de Ven G.A Kluitenberg/

L.Restuccia RM.M. Mattheij

F.R de Hoog RM.M. Mattheij AF.M. ter Elst S.J.L. van Eijndhoven H.G. ter Morsche

Title

Magnetoelastic buckling of two nearby ferromagnetic rods in a magnetic field Spaces of type W, growth of Hennite coefficients, Wigner distribution and Bargmann transfonn

A note on the reduced creep function corresponding to the quasi-linear visco-elastic model proposed by Fung

On some generalizations of the Debye equation for dielectric relaxation

Direct Solution of Certain Sparse Linear Systems

On the conditioning of multipoint and integral boundary value problems

A Gevrey space characterization of cer-taingelfand-shilov spacesS~

On the dimension of bivariate periodic spline spaces type-I triangulation

Month Dec. '87 April '87 April '87 Feb. '88 Feb. '88 March '88 April '88 April '88

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