Cubature Formulae and Orthogonal Polynomials

CUBATURE FORMULAE AND ORTHOGONAL POLYNOMIALS*

1. P.MYSOVSKIKH

Leningrad

(Received 20 June 196’7)

IN this paper we construct cubature formulae for an arbitrary region of integration and a positive weight function, which are exact for third-degree polynomials and

have the least possible number of nodes. It is true that there exist two orthogonal second-degree polynomials in the region and a weight function which simultaneously vanish at exactly four real points, which can be taken as the nodes of the required cubature formula. When a certain condition is observed, imposed on the region of

integration and the weight function. The number of nodes which is 4 is the least possible. In this case there is an infinite number of cubature formulae and their complete set can be obtained by finding all the pairs of orthogonal polynomials which vanish simultaneously at exactly four points. Violation of this condition is necessary and sufficient for the least number of nodes of the cubature formula to be three. Such a cubature formula is unique.

The possibility of generalizing the results obtained is discussed.

Let Cl be the set of points on a plane and p (x, y) a function which is positive

in n and such that the moments

.uiA = ss P(X, Y)Z'Y~ dir dy, i, k = 0, 1, 2, . . . , 0

exist and poo > 0. Henceforth these conditions will be considered to be satisfied everywhere. We shall call the set n the region (of integration) and p (x, y) the

weight function.

Theorem 1

Among the orthogonal second-degree polynomials of the region Q and the

*Zh. vj%hisl. Mat. mat. Fiz. 9. 2, 419-425, 1969.

218 I. P. Mysovskikh

weight function p (x, y) there are two which simultaneously vanish at exactly four real points.

Proof. We write the basic orthogonal second-degree polynomials of the

region R and the weight p Cx, y):

(1)

If Bi = Q and cc3 = o we can take Pzo and poz as the orthogonal polynomials

which we have been discussing in the theorem. In fact the second-order curve

& = o consists of two non-coincident straight lines, parallel to the y-axis,

and the curve PO2 = 0 of two lines parallel to the x-axis.

Here we use the fact that if an orthogonal polynomial of the region and the

weight depend on only one of the variables, e.g. pk, O=~h+uia+--i + a2zh-2 + . . . +ac,

it has k different real roots.

We can consider that fh = 0. In fact let l3i # 0 On rotating the coordinate

axes through an angle 71 the polynomial P,, becomes the polynomial x2 - air -

I.&v + V, so that with a change of the angle of rotation 4 from 0 to n the

coefficient l%(cp) of y changes from p1 to -&. From continuity a value cp,= ‘po,

0 < ‘PO < rc, can be found for which l3i (CPO) 10. We shall consider the coefficient

a3 to be non-zero.

We now carry out a parallel translation of the coordinate axes, moving the

origin to the point (+, -az). The orthogonal polynomials (1) can be written in the form

Pso = x2 + air + Cl, Pi1 = XY + cz, Paz = y2 + usx + bsy + cs. (2)

We have retained the former coordinate notation. In the parallel translation the

coefficient of y in pro and the coefficient of x in POZ remain unchanged so that a3 # 0.

We shall look for a linear combination PO2 + aPil, which can be split into

the product of two factors of the first degree: POZ + ~911 = f~ + ax + P) (Y + Y).

To determine a, p and y we obtain ay = as, j3 + y = 63, BY = cs + cm, and

consequently /3 and y are the roots of the quadratic equation

Cubature formulae and orthogonal polynomials

z* - bgz + CJ + acz = 0.

Since v = a3 / a, for the determination of a-we have

c2a3 + c3a2 - bsa3a + a.2 = 0.

219

(3)

(4)

If cz # 0, then (4) has a real root ag’ Since a3 # 0, then a0 # G.

If c.2 = 0, equation (4) is a quadratic. Since a3 + 0, the sign of its

discriminant is the same as that of

d = bz2 - 4~s. (3)

Ascribing the condition of orthogonality of poz to 1, x and y and using the

fact that uI1 = kiz = pzi = o because c, = 0, we find that

I*20 c3 = --po1po3 - po22),

Pi0 a3 = - -c2,

I420 (6)

where A = poopzopoz - pol+20 - p,02p02 > 0, is the Gram determinant of the set of

functions 1, x, y. The second of the equalities (6) implies that CQ # 0. If

c3 < 0, then d > 0, as is obvious from (51.

Let us assume that c, > 0. By virtue of (6) we have

po1po3 - poz2 = E > 0. (7)

The condition that PO, should be orthogonal to y takes the form

On multiplying this equality by poi ff 0 and substituting for POIPOS from (7) we

obtain

Hence it follows that d > II

Since equation (4) always has a real root a0 # 0, the curve po2 + aoPii = o

22Q I. P. Mysovskikh

decomposes into two straight lines

Y + a05 + B = 0, El+y=o, (5)

where p and y are the roots of the quadratic equation (3) with a = UO. In the

case where c, = 0 the discriminant of this equation d > 0 and consequently B Z Y.

The orthogonal polynomials PZO and poz + a.Pli are also required if the

point of intersection of the lines (8)

Zf = (Y - B) I ao, yi = -y (9)

does not fall on the lines Pzo = 0.

We shall assume that the point (9) lies on one of the lines Pzo = 0. We

obtain three points of intersection of the curves Pzo = 0 and POZ + uoPii = 0.

Let us form a new linear combination of PO, and PI1:

PO2 + @ii = (Y + uoz + B) (Y + y) + (a - ao) (XY +cz).

We shall study the points of intersection of the curve poz + apli = 0 and the line

x = (y - I)) /ao. The ordinates of the points of intersection satisfy the equation

[

Y-B ys+ 2y+(u--ao)--- Y+YZ+(a--o)ca=O. a0 1

The discriminant of this quadratic equation is

(a-ao)” +(Y-B)‘-.

uo2 w

If all the orthogonal polynomials of the second degree do not vanish at the point (9), then y (y - fi) / a0 .- Cz = --P,,(Zi, Yi) # 0 and the first term on the

right-hand side of (10) must be made positive by a suitable choice of CT. Thus

the curve POZ + aPi, = o intersects the line z = y - fi) / a0 in two distinct points. Making the difference a - a, sufficiently small (and retaining its sign)

ensures that the points of intersection with the second of the lines pzo = 0

will be different.

We now assume that all the orthogonal polynomials vanish at the point (9).

Cubature formulae and orthogonal polynomials 221

The discriminant (10) in this case is written in the form D = (y - #I)" (a - ao)2 I a$.

We now prove that y # B. If y = B, it follows that all the orthogonal polynomials

vanish at the point 5% = 0, yi = -y. This leads in particular to the equality c, = 0, which is impossible (if c, = 0, as we have seen y # B). In this case

for sufficiently small cz - no # 0 the curve Paz + aPl; = 0 intersects

Pzo = 0 in four points.

From the proof it follows that there exists an infinite set of pairs of orthogonal polynomials possessing the property given in Theorem 1.

Theorem 2

Let K, and K, be orthogonal polynomials of the second degree in the region

fz with weight p (x, y) such that the curves

Ki = 0, Kz = 0 (11)

intersect in exactly four real points MS, i = 1, 2, 3, 4. Then the point M, may

be taken as nodes of the cubature formula

which is exact for polynomials of the third degree.

Proof. No three of the points Mi lie on one line, e.g. L = 0. In the opposite

case by Bezu’s theorem we find that K, and K, have a common linear factor L. This contradicts the fact that the curves (11) intersect in exactly four points.

To the points MC, i = 1, 2, 3, 4, we add the points MC, 5 < i G 10, SO that

all ten points do not lie on a third-order curve. This can be done in the following

way. Through M, we draw a straight line which does not go through any one of

the points Mi, M,, Mu, and take two points Ms. Me, on the line which are

distinct from M,. We now draw a line, not going through any one of the points

M &, . . . , M,, and take four points Y7,. . . , Mie on it. The points MS,. . . , i%flo

possess the required property, as follows from Bezu’s theorem.

We now prove that any polynomial of the third degree P(x, y) such that P(M,) =: o, i = 1, 2, 3, 4, can be written in the form

222 I. P. Mysouskikh

P(x,~) = L,Ki + LzKz, (13)

where L, and L, are polynomials of the first degree.

On the left- and right-hand sides of (13) there are polynomials of the third degree whose values are zero at the points Mi, i = 1, 2, 3, 4. For these polynomials to be equal it is sufficient that their values should be the same at the

points Mi, 5 < i < 10. TO determine the six coefficients of L, and L, we

obtain a linear algebraic system of six equations. We shall consider the corres-

ponding homogeneous system. The latter is equivalent to the problem of determining polynomials of the first degree in L, and L, such that

L,K, + L& = 0. (14)

If Li + 0 and LZ + 0, they are not proportional and it follows from (14) that

Kt = i;Lz, Kz = -LL,. This is impossible and consequently (14) is valid only

if L, = L, = 0. Consequently the homogeneous system has only a zero solution which proves (13).

Since the points M,, . . . , ~4,~ do not lie on a third-order algebraic curve they must be taken as the nodes of interpolation of the cubature formula

10

\ S P(X, Y)ftxt YldXdY Y zCjf(“j), (15) n j=l

which is exact for polynomials of the third degree.

We shall prove that cs = cs = . . = ci,, = 0, so that (15) has 4 nodes and

is the required formula. Let us construct a polynomial of the third degree

Pk(z gj, k = 5, 6, . . . , IO: such that Pk (.Mk) = i, pk (illi) = u, i f k. ‘1 &IlCe

Pk(it!fi) = 0 for i = 1, 2, 3, 4, statement (13) is valid for Pk.

We write that formula (15) is exact when f = Pk and obtain

c,, = 1 1 P (5. y)Pk (5, y)dx dy = 1 5 P (5, Y)[.&& + LA1 dx a!/ = 0. R R

The integral is zero because K, and K, are orthogonal polynomials. This proves Theorem 2.


Let us consider the integral

z = ( ( p (.t? Y)[P,llP02 - PIi dx dY, .s n

(16)

where P?o, PII, PO2 are the basic orthogonal polynomials of Q and p(x, y). As

we see, I is an important characteristic of the region and the weight.

We now carry out the affine transformation of the coordinates

u = ax $ b!/. L’ = c.r + a!/. (17)

We shall denote the basic orthogonal polynomials of R and p (x, y) in the new

coo:dinates by Pz_<, il i = 0, 1, 2. and the integral (16) in the new coordinates by 1. We now prove that

I- = 16131, (IS)

where 6 is the transformation determinankof (17). The equality (18) is true for

the whole affine transformation because I does not change for a parallel translation of the coordinate axes.

For the proof of (18) we use the representation of the basic orthogonal

polynomials by means of determinants. For instance we write

PO0 CL10 i;Ol 1 I CL10 llzo ill u

p&J = y - -

A &a El1 koz v Pzo bo PZl u2

Here ii, are the moments of a and p CC, y) in the new coordinate system

jk.4 = 161 ss P (2, Y) (ax + by) i (cx + dy)’ dx dy (19) Q

is the Gram determinant of the system 1, U, U.

224 1. P. Mysovskikh

We now introduce the notation

PO0 ko Ho1 Pea CL00 ho lLo1 PI1

M= ho p20 crll ha ho ko Pll PaI -

1401 Iril pea PO3 PO1 C"l1 Pas ha

cl20 ho h 0 kl P21 ha 0

If we replace urk by i;ik, on t&e right-hand side of this equality, we obtain

a number which we will denote by M.

From the definition of 7 we have

-AT = M. (20)

By replacing the moments F ik in the expression for G by the integrals of formula

(18) and calculating the determinant we find that M = 88M. Since M = ZA,

where A is the Gram determinant of the set of monomials 1, x, y, we find from (20) that

-iT = PZA.

It remains to note that X = /615A, and the equality (18) is proved. From (18) in

particular it follows that the integral (16) is an invariant of the orthogonal transformations.

In [ll integrals were considered which are related to the basic orthogonal polynomials of the third degree in fi and p (x, y):

The property of the numbers A, i3 and C, that at least one of them is non-

zero, is invariant for the affine transformations. If A, B, C are ikegrals of

(Zl), corresponding to the coordinates U, u, then 82 - UC = b*(Bz -UC), so

that the sign of the number B2 - 4AC is an invariant of the affine transformations.

The proof of these statements will not be given.

The condition Z # 0 is necessary and sufficient for the number of nodes 4

in the cubature formula (12) to be the least possible. This follows from Theorem 3.


Theorem 3

For a cubature formula with three nodes, which is exact for polynomials of

the third degree to exist, it is necessary and sufficient for the equality I = 0 to

be satisfied.

Proof. It has been proved above that a cubature formula exists with four

nodes, which is exact for polynomials of the third degree, with no three nodes

lying on one straight line. We choose a coordinate system so that two nodes are

situated on each of the coordinate axes. We can consider the system of coordina-

tes to be rectangular (in case of necessity we can carry out an affine transforma-

tion). The cubature formula is written in the form

ss P(S, y)f(=, y)dz dy 2 W(G O)+ Czf(~, O)+ Cd(O, YS)+ Cd@, ~4). (27) *

From the formula it is obvious that pci, = pZ, = pi2 = o and consequently

P,, = xy.

We evaluate the integral (16):

I= ss ~(2, y) (~2Poz - ~2yz)da:dy = ~430 + cspzo.

P

From (61 we obtain

1 1 = - --(plopSo - p202) (I”OlPOS - cLoz2). (23)

Since, in view of (22)

pi0 = ClZl’ + c2zzi. poi = C3Yd +cry,i, i = 1, 2, 3,

(23) can be written in the form

___IA = C,C~C3C4six2Y3Y4(~i -x2) 2(Y3 - y4J2. (24)

The theorem follows from (24). Let I = 0. The product ~~52~3~4 is non-zero

since no three nodes lie in one straight line. It follows from (241 that the

product of the remaining factors on the right-hand side of this equality is zero.

If for instance C, = 0 this means that the cubature formula (221 has in practice

226 I. P. Mysovskikh

three nodes. We come to the same conclusion if (zl -Q) (us - y*) = 0.

We now assume that a cubature formula exists with three nodes, which is

exact for polynomials of the third degree. We can consider that it is written in

the form (221 with C, = 0. But then it follows from (24) that I = 0.

We note that the cubature formula with three nodes, which is exact for

polynomials of the third degree, is unique. This statement follows from the fact

that all the orthogonal polynomials of the second degree vanish at the nodes of

this formula. The coefficients of such a formula are positive.

If Z # 0 we obtain an infinite set of formulae (12) with four nodes, which

are exact for polynomials of the third degree. We can show that Theorem 2 gives

the whole set of such formulae. Suppose that there is a cubature formula with

four nodes which is exact for polynomials of the third degree. We can consider

that no three nodes lie on a straight line (in the opposite case I = 0) and,

consequently, the formula can be written in the from (22). Obviously, as the K,

and K, of Theorem 2 we can take

Ki = xy, Kz = (2 I xi + Y / YS - 1) (5 I xs + Y I YA - 1).

Let us consider two examples.

Example 1. Let s2 possess central symmetry, the centre of symmetry being

the origin. This means that p(x, y) = p (-5, -y), (x, y) E Q. Then

pzo = x2 - I”20 / woo, Pll = XY - p1i / poo, PO2 = Y2 - PO2 I poo;

I= - ;(PPopor - pd) < 0,

so that the number of nodes, which is 4, in formula (12) is the least possible.

Note that the case of a centrally symmetric region is considered in [ 21 in another

way.

Example 2. As Q let us take the triangle z > 0, y > 0, x + y < 1 and p (x, y) ,= 1. Then

Pm = x2 - 0.8s + 0.1, Pii = xy - 0.2 (x + y) + 0.05, Paz = y2 - 0.8~ + 0.1,

and we obtain Z = -1/2(oo. Theorem 2 enables us to obtain a whole set of

cubature formulae (12).


The question arise as to whether the condition I J 0 is a constraint. In other words do regions and weight functions exist for which I = O? A positive answer

is given to this question below.

We now consider the region fl which is the union of the two rectangles

-I< I < 1, O<y<l EI --a<x<a, -E<Y<o,

where E and u are positive numbers and the weight function p(x, y) = 1. The region fi is symmetrical with respect to the axis of y and therefore I.L~O = pii = pi2

3 1130 = 0 - We also require that pzl = 0. This leads us to the equality = &ir.

In the case under consideration from (23) 22

1 = f-+oipo3 - p.022).

We now find

n,,, = 1 - D-2. noz = 213 (1 + o-‘/z), PO3 = ‘/2 (1 - o-5).

We introduce a new parameter ‘c,= O-‘/Z. From (25) we find that for the equality

1 = 0 to be satisfied it is necessary and sufficient that

cp(T) = ‘I_? (I- T4) (I- 90) - 41s (1 + ,‘)2 = 0. W)

Since q (0) > 0 and cp (1) < 0, there is a root ‘CO > 0 of equation (26)

The approximate values of the parameters are %. = 0.540199, o. = 3.42683, co =

0.157601. The region 0, defined hy the parameters o. and E,,, is the one required, and for it I = 0. Note also that if 0 = 4 and E = I/S we obtain a region for which

I> 0 so that the integral (16) can be a quantity of any sign.

A similar question is posed in 111: do a region Cl and weight (in [ll it is assumed that p(x, y) ,= 1) such that 11 = B = c = o

positive solution to a similar question in relation to the equality problem also has a positive solution.

We now formulate the generalization of Theorems 1 and 2.

Theorem 1’

p(x, y) > 0 exist (see (21)). A I = 0 that this

Two orthogonal polynomails of degree k exist for the region Cl and weight

p( x, y) which simultaneously vanish at exactly kZ real and different points.

Theorem 2’

228 1. P. Mysovskikh

If the orthogonal polynomials K, and K, of degree k are such that the algebraic curves Ki = 0 and Ka = 0 intersect at exactly k, real and different points

Mi, i = 1, 2,..., k2, the points Mi can be taken as the nodes of a cubature formula which is exact for all polynomials of degree less than or equal to 2k - 1.

Theorem 1’ has not been proved. The proof of Theorem 2’ is the same as that

of Theorem 2 if one of the polynomials K, or K, is split up into k linear factors.

Double integrals have been considered in this paper. In this case the interpolation cubature formula which is exact for polynomials of degree 2k - 1,

has 2k2 + k nodes. The cubature formula of the same algebraic degree of

accuracy, of which we speak in Theorem 2’) has k2 nodes so that we obtain a double advantage with regard to the number of nodes.

The generalization of Theorem 2’ to the case of integrals of multiplicity n leads to cubature formula which are exact for polynomials of degree 2k - 1 and .

kave kn nodes. The interpolation cubature formula, which is exact for polynomials of degree 2k - 1 has M(2k - 1, n) G (I/ IZ!) 2k(2k + 1) . . . (2k -I- n - 1) = (2” I n!)kn

nodes. If n = 3 the advantage in the number of nodes is insignificant and if n > 3

the number of nodes of the interpolation cubature formulae is less than kn

(commencing with some k). For instance if n = 4

M(2k - 1, 4) < k” for k > 7.

Translated by H. F. Cleaves

REFERENCES

1. RADON, J. Zur mechanischen Kubatur. Monatshefte Math. 52, 4. 286300. 1948.

2. GEORGIEV, G. Mechanical cubature formulae of a minimum number of terms with multiple integrals. Dokl. Akad. Nauk SSSR. 83, 4. 521-524. 1952.

Documents

Cubature Formulae and Orthogonal Polynomials