Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol99.Issue1-2.1998:
Journal of Computational and Applied Mathematics 99 (1998) 239–253
Partial di erential equations having orthogonal
polynomial solutions
Y.J. Kim a , K.H. Kwon a; ∗ , J.K. Lee b
b
a
Department of Mathematics, KAIST, Taejon 305-701, South Korea
Department of Mathematics, Sunmoon university, Cheonan, South Korea
Received 30 October 1997; received in revised form 19 May 1998
Abstract
We show that if a second order partial di erential equation:
L[u] := Auxx + 2Buxy + Cuyy + Dux + Euy = n u
has orthogonal polynomial solutions, then the di erential operator L[·] must be symmetrizable and can not be parabolic
in any nonempty open subset of the plane. We also nd Rodrigues type formula for orthogonal polynomial solutions of
c 1998 Elsevier Science B.V. All rights reserved.
such di erential equations.
AMS classi cation: 33C50; 35P99
Keywords: Orthogonal polynomials in two variables; Partial di erential equations; Symmetrizability; Rodrigues type
formula
1. Introduction
As a natural generalization of classical orthogonal polynomials of one variable, Krall and She er
[5] considered orthogonal polynomials in two variables satisfying a second order partial di erential
equation with polynomial coecients:
L[u] = Auxx + 2Buxy + Cuyy + Dux + Euy = n u:
(1.1)
Krall and She er rst nd necessary and sucient conditions in order for orthogonal polynomials
to satisfy the di erential equation (1.1) and then classify such di erential equations, up to a complex
linear change of variables, which have at least weak orthogonal polynomials as solutions.
∗
Corresponding author. Fax: 42 869 2710; e-mail: khkwon@jacobi.kaist.ac.kr.
c 1998 Elsevier Science B.V. All rights reserved.
0377-0427/98/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 6 0 - 5
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Later, Littlejohn [8] noted that all di erential equations found in [5] are elliptic in the region
of orthogonality in case the orthogonalizing weight is known. However, the type of di erential
operators cannot be determined properly from the speci c forms of the equations given in [5] since
the type of a di erential operator is not preserved under a complex change of variables. In fact,
Krall and She er found at least one di erential equation (see Example 3.3), which is hyperbolic
everywhere and has orthogonal polynomials as solutions.
In this work, using the functional calculus on moment functionals developed in [6, 8], we rst
prove that if the di erential equation (1.1) has orthogonal polynomials as solutions, then L[·] cannot
be parabolic in any nonempty open subset of the plane and must be symmetrizable. We also establish
Rodrigues type formula for orthogonal polynomial solutions to the di erential equation (1.1).
2. Preliminaries
For any integer nS¿ 0, let Pn be the space of real polynomials in two variables of (total) degree 6 n and P = n¿0 Pn . By a polynomial system (PS), we mean a sequence of polynomials
n
n
{ n−j; j }∞
n=0; j=0 such that deg( n−j; j ) = n; 0 6 j 6 n and { n−j; j }j=0 are linearly independent modulo
Pn−1 for n ¿ 0 (P−1 = {0}).
n
∞
∞
We let n :=[ n; 0 ; n−1; 1 ; : : : ; 0; n ]T and denote the PS { n−j; j }∞
n=0; j=0 by {n }n=0 . A PS {Pn }n=0 ;
T
Pn = [Pn; 0 ; Pn−1; 1 ; : : : ; P0; n ] , where
Pn−k; k (x; y) = xn−k yk modulo Pn−1 ;
0 6 k 6 n;
is called a monic PS. For any PS {n }∞
n=0 , where
n−j; j (x; y) =
n
X
anjk xn−k yk modulo Pn−1 ;
0 6 j 6 n;
k=0
−1
the matrix An = [anjk ]nj; k=0 ; n ¿ 0; is nonsingular. Then, the monic PS {Pn }∞
n=0 de ned by Pn = An n
is called the normalization of {n }∞
n=0 :
For a moment functional on P, we call
ij := h; xi y j i;
i; j ¿ 0
the moments of and
00
10
n := .
..
0n
10
01
···
20
..
.
11
..
.
···
1n
0; n+1
···
..
.
0n
;
0;2n
1n
..
.
n¿0
the Hankel determinants of . We call to be quasi-de nite if n 6= 0; n ¿ 0.
n
For a matrix = [ ij (x; y)]mi=0; j=0
of polynomials we let
h; i = [h;
m
n
ij i]i=0; j=0 :
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
241
For any PS {n }∞
n=0 , there is a unique moment functional , called the canonical moment functional of {n }∞
,
de ned
by the conditions
n=0
h; 1i = 1
and
h; n i = 0; n ¿ 1:
De nition 2.1. A PS {n }∞
n=0 is called a weak orthogonal polynomial system (WOPS) if there is
a nonzero moment functional such that h; m nT i = 0; m 6= n. If moreover Hn := h; n nT i; n ¿ 0,
is a nonsingular diagonal matrix, we call {n }∞
n=0 an orthogonal polynomial system(OPS). In this
case, we say that {n }∞
is
a
WOPS
or
an
OPS
relative to :
n=0
Note that if {n }∞
n=0 is a WOPS relative to , then must be a nonzero constant multiple of the
∞
canonical moment functional of {n }∞
n=0 . If {n }n=0 is an OPS relative to , then the normalization
∞
{Pn }∞
n=0 of {n }n=0 is a WOPS, but not necessarily an OPS relative to :
The following is proved in [5] (see also [11]).
Proposition 2.2. For a moment functional ; the following statements are equivalent.
(i) is quasi-de nite;
(ii) there is an OPS {n }∞
n=0 relative to ;
(iii) there is a unique monic WOPS {Pn }∞
n=0 relative to ;
(iv) there is a monic WOPS {Pn }∞
relative
to such that
n=0
T
Hn := h; Pn Pn i; n ¿ 0; is nonsingular.
T
If { n }∞
n=0 is a WOPS relative to such that h; n n i; n ¿ 0, are nonsingular, then there are
orthogonal matrices On : (n + 1) × (n + 1); n ¿ 0, such that
On h; n nT iOnT = h; (On n )(On n )T i;
n¿0
are nonsingular diagonal matrices. Hence {n := On n }∞
n=0 , is an OPS relative to . In fact, for a
quasi-de nite moment functional , let Vn ; n ¿ 0, be the vector space of polynomials of degree n,
which are orthogonal to any polynomial of degree ¡ n relative to . Then, {n }∞
n=0 , where n is a
basis (respectively, an orthogonal basis) of Vn , forms a WOPS (respectively, an OPS) relative to
and Hn := h; n nT i; n ¿ 0, are nonsingular (respectively, nonsingular diagonal).
For any moment functional and any polynomial (x; y), we let
h@x ; i = −h; x i;
h ; i = h; i;
h@y ; i = −h; y i
∈ P:
Then we have
Lemma 2.3. Let and be moment functionals and R(x; y) a polynomial. Then
(i) = 0 if and only if @x = 0 or @y = 0:
Assume that is quasi-de nite and let {n }∞
n=0 be an OPS relative to . Then
(ii) R(x; y) = 0 if and only if R(x; y) = 0.
(iii) h ; n i = 0 for n ¿ k (k ¿ 0 an integer) if and only if = (x; y) for some polynomial (x; y)
of degree 6 k .
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Proof. See Lemma 2.3 in [6].
3. Di erential operator L[·]
We now consider the di erential equation (1.1). Krall and She er [5] showed that if the di erential
equation (1.1) has a PS as solutions, then it must be of the form
L[u] = Auxx + 2Buxy + Cuyy + Dux + Euy
= (ax2 + d1 x + e1 y + f1 )uxx + (2axy + d2 x + e2 y + f2 )uxy
+ (ay2 + d3 x + e3 y + f3 )uyy + (gx + h1 )ux + (gy + h2 )uy
= n u; n = an(n − 1) + gn:
(3.1)
Following Krall and She er, we say that the di erential operator L[·] is admissible if m 6= n for
m 6= n or equivalently n 6= 0 for n ¿ 1:
It is shown in [5] that the di erential equation (3.1) has a unique monic PS as solutions if and
only if L[·] is admissible.
Theorem 3.1. If the di erential equation (3.1) has an OPS as solutions; then B2 − AC 6≡ 0 in
any nonempty open subset of the plane. In other words; the di erential operator L[·] cannot be
parabolic in any nonempty open subset of the plane.
In order to prove Theorem 3.1, we rst need the following fact:
Lemma 3.2. If the di erential equation (3.1) has a PS {n }∞
n=0 as solutions; then the canonical
moment functional of {n }∞
must
satisfy
n=0
∗
L [] = 0;
(3.2)
h; Di = h; Ei = h; A + xDi = h; C + yEi = h; B + yDi = h; B + xEi = 0;
where L∗ [·] is the formal Lagrange adjoint of L[·] given by
L∗ [u] = (Au)xx + 2(Bu)xy + (Cu)yy − (Du)x − (Eu)y :
(3.3)
(3.4)
Proof. See Lemma 3.1 in [6].
Proof of Theorem 3.1. Assume that the di erential equation (3.1) has an OPS {n }∞
n=0 as solutions.
∞
Let be the canonical moment functional of {n }n=0 . We assume that
B2 − AC = −a[d3 x3 + (e3 − d2 )x2 y + (d1 − e2 )xy2 + e1 y3 ]
+( 41 d22 − af3 − d1 d3 )x2 + ( 12 d2 e2 + af2 − d1 e3 − d3 e1 )xy
+( 41 e22 − e1 e3 − af1 )y2 + ( 12 d2 f2 − d1 f3 − f1 d3 )x
+( 21 e2 f2 − f1 e3 − e1 f3 )y + 41 f22 − f1 f3
≡0
(3.5)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
in some nonempty open subset of the plane. Then, all the coecients of B2 − AC must be 0. We
now consider the following three cases separately.
Case 1: a + g =
6 0 and g 6= 0: Then, we may assume that g = 1 and h1 = h2 = 0 so that (3.3) gives
10 = 01 = 0
and
(a + 1)20 + f1 = (a + 1)02 + f3 = 2(a + 1)11 + f2 = 0:
Therefore
2
=
1 = 20 02 − 11
1
[f1 f3 − 14 f22 ] = 0;
(a + 1)2
which is a contradiction since 1 6= 0 when is quasi-de nite.
Case 2: a + g 6= 0 and g = 0: Then we may assume that a = 1 so that (3.5) gives
d3 = e 1 = 0 ;
d 2 = e3 ;
d1 = e2 ;
f1 = 41 d21 ;
f2 = 12 d1 d2 ;
f3 = 41 d22 :
Hence the di erential equation (3.1) becomes, after replacing x + 12 d1 by x and y + 12 d2 by y,
x2 uxx + 2xyuxy + y2 uyy = n(n − 1)u:
(3.6)
Then we have by (3.3), 20 = 02 = 11 = 0 so that 1 = 0, which is a contradiction.
Case 3: a + g = 0 and g 6= 0: Then we may assume that a = − 1; g = 1, and h1 = h2 = 0. In this case,
it is easy to see that if the di erential equation (3.1) has a PS as solutions, then f1 = f2 = f3 = 0. We
then have from (3.5) d1 = e1 = d2 = e2 = d3 = e3 = 0 so that the di erential equation (3.1) becomes
x2 uxx + 2xyuxy + y2 uyy − xux − yuy = n(n − 2)u:
Then we have from (3.2)
Amn := hL∗ []; xm yn i = h; L[xm yn ]i = (m + n)(m + n − 2)mn = 0;
m; n ¿ 0:
Hence mn = 0 for m + n ¿ 3 so that 2 = 0, which is a contradiction.
While Krall and She er considered only admissible di erential equations, we do not assume the
admissibility of L[·] in Theorem 3.1 since we do not know whether L[·] must be admissible or not
when the di erential equation (3.1) has an OPS as solutions.
When L[·] is not admissible, the di erential equation (3.1) may have either no PS as solutions
or in nitely many distinct monic PS’s as solutions. For example, the di erential equation (3.6) has
in nitely many monic PS’s {Pn }∞
n=0 as solutions;
Pmn (x; y) = xm yn
for m + n =
6 1
and
P10 = x + ; P01 = y + ;
where and are arbitrary constants. Note that for any choice of and ; {Pn }∞
n=0 can not be an
OPS but for = = 0; {xm yn }∞
is
a
WOPS
relative
to
(
x;
y
)
:
m; n=0
Example 3.3. Krall and She er [5] showed that the di erential equation
L[u] = 3yuxx + 2uxy − xux − yuy = −nu
(3.7)
has an OPS as solutions. Since B2 − AC = 1 ¿ 0, the di erential operator L[·] in (3.7) is hyperbolic
everywhere.
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Littlejohn [8] observed that all di erential equations found by Krall and She er [5], that have
orthogonal polynomial solutions whose weight functions are known, are elliptic in the region of
orthogonality.
However, since Krall and She er used a complex linear change of variables in their classi cation,
the type of the di erential equation cannot be properly determined from the speci c forms of the
equations given in [5].
For example, Krall and She er obtained the following di erential equation (see the equation
(5.13) in [5]) by a complex linear change of variables
uxx + uyy − xux − yuy = −nu
(3.8)
from the equation (see Eq. (5.12) in [5])
f2 uxy + gxux + gyuy = gnu:
(3.9)
Note that since f2 (6= 0) is real, the di erential equation (3.9) is hyperbolic everywhere while
Eq. (3.8) is elliptic everywhere.
∞
Proposition 3.4. Let {n }∞
n=0 be a PS satisfying the di erential equation (3:1). If {n }n=0 is at
∞
∗
least a WOPS; then the canonical moment functional of {n }n=0 satis es L [] = 0 and
M1 [] := (A)x + (B)y − D = 0;
(3.10)
M2 [] := (B)x + (C)y − E = 0:
(3.11)
Proof. See Corollary 3.3 in [6].
Lemma 3.5. For the di erential operator L[·] in (3.1) and any moment functional ; the following
statements are equivalent.
(i) M1 [] = M2 [] = 0;
(ii) L[·] is formally symmetric on polynomials; that is;
hL[P ]; Qi = hL[Q]; Pi;
P and Q in P:
Furthermore; if L[P ] = P and L[Q] = Q for 6= ; then h; PQi = 0 for any moment functional
satisfying M1 [] = M2 [] = 0.
Proof. See Lemma 3.6 in [6].
We call the functional equations M1 [] = M2 [] = 0 in (3.10) and (3.11) the moment equations
for the di erential equation (3.1).
Lemma 3.6. If the di erential equation (3.1) is admissible and has a WOPS as solutions; then
the moment equations M1 [] = M2 [] = 0 can have only one solution up to a nonzero constant
multiple.
Proof. See Corollary 3.5 in [6].
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
245
As a consequence of Theorem 3.1 and Lemma 3.5, we can now prove:
Corollary 3.7. If the di erential equation (3.1) has an OPS {n }∞
n=0 as solutions; then
(i) the moment equations M1 [] = M2 [] = 0 can have only one solution up to a nonzero constant
multiple; which must be quasi-de nite;
(ii) the di erential equation (3.1) can have only one monic WOPS as solutions (but may have in nitely many other monic PS’s as solutions; which are not WOPS’s; unless L[·] is admissible).
∞
Proof. Let be the canonical moment functional of {n }∞
n=0 and {Pn }n=0 the normalization of
∞
{n }n=0 . Then is quasi-de nite and M1 [] = M2 [] = 0 by Proposition 3.4 and {Pn }∞
n=0 is a monic
WOPS satisfying the di erential equation (3.1).
(i) If L[·] is admissible, then it follows from Lemma 3.6. We now assume that L[·] is not
admissible, that is, N = 0 for some N ¿ 1. Let be any moment functional satisfying the moment
equations M1 [ ] = M2 [ ] = 0: Then Lemma 3.5 implies h ; n i = 0; n 6= 0; N since n 6= 0 for n 6= 0; N .
Therefore we have, by Lemma 2.3(iii), = (x; y) for some polynomial (x; y) of degree 6 N:
Then
M1 [ ] = M1 [ ] = ( x A + y B) + M1 [] = ( x A + y B) = 0;
M2 [ ] = M2 [ ] = ( x B + y C ) + M2 [] = ( x B + y C ) = 0:
Hence we have by Lemma 2.3(ii)
x A + y B = 0
and
x B + y C = 0;
which implies that x = y = 0, that is, (x; y) must be a constant since B2 −AC 6≡ 0 by Theorem 3.1.
(ii) Let {Qn }∞
n=0 be any monic WOPS relative to satisfying the di erential equation (3.1). Then
satis es the moment equations M1 [ ] = M2 [ ] = 0 by Proposition 3.4. Hence by (i), = c for some
∞
∞
nonzero constant c. Therefore {Qn }∞
n=0 is also a monic WOPS relative to so that {Qn }n=0 = {Pn }n=0
by Proposition 2.2 since is quasi-de nite.
Corollary 3.7 provides a unique existence of orthogonal polynomial solutions to the di erential
equation (3.1) in the monic level whether L[·] is admissible or not. However, the di erential equation
(3.1) may have in nitely many distinct OPS’s, which are not monic, as solutions if it has at least
one.
De nition 3.8 (Littlejohn [8]). The di erential operator L[·] in (3.1) is symmetric if L[·] = L∗ [·].
We say that L[·] is symmetrizable if there is a non-trivial function s(x; y) such that s(x; y) is C 2 in
some open set and sL[·] is symmetric.
In this case, we call s(x; y) a symmetry factor of L[·]:
It is easy to see (cf. [8]) that a function s(x; y) is a symmetry factor of L[·] if and only if s(x; y)
satis es
M1 [s] = (As)x + (Bs)y − Ds = 0;
(3.12)
M2 [s] = (Bs)x + (Cs)y − Es = 0:
(3.13)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
We call the simultaneous equations (3.12) and (3.13) the symmetry equations for L[·]. We rst see
by an example that not every di erential operator L[·] is symmetrizable.
Example 3.9 (Littlejohn [8]). Consider the di erential equation
L[u] = x2 uxx + 2xyuxy + y2 uyy + (gx + h1 )ux + (gy + h2 )uy = n(n − 1 + g)u:
(3.14)
Note rst that the di erential (3.14) is parabolic everywhere so that it cannot have an OPS as
solutions by Theorem 3.1. The corresponding symmetry equations are
x2 sx + xysy + ((3 − g)x − h1 )s = 0;
xysx + y2 sy + ((3 − g)y − h2 )s = 0
so that (h2 x − h1 y)s = 0: Hence, if h21 + h22 6= 0, then s(x; y) = 0 on {(x; y) ∈ R2 | h2 x − h1 y 6= 0} and
so L[·] cannot be symmetrizable. However, if g = 3 and h1 = h2 = 0, then L[·] is symmetrizable with
s(x; y) = arctan(y=x) as a symmetry factor.
Proposition 3.10. Assume that B2 − AC 6≡ 0. Then the di erential operator L[·] is symmetrizable
if and only if
@ A(Bx + Cy − E ) − B(Ax + By − D)
@ C (Ax + By − D) − B(Bx + Cy − E )
=
:
@y
B2 − AC
@x
B2 − AC
(3.15)
We call (3.15) the compatibility condition for the symmetrizability of L[·]:
Proof. (⇒): We may solve the symmetry equations (3.12) and (3.13) for sx and sy as
(B2 − AC )sx = [C (Ax + By − D) − B(Bx + Cy − E )]s;
(3.16)
(B2 − AC )sy = [A(Bx + Cy − E ) − B(Ax + By − D)]s;
(3.17)
from which (3.15) follows.
(⇐): Assume that the compatibility condition (3.15) is satis ed. Choose any open rectangle
R = (x0 − r; x0 + r ) × (y0 − r; y0 + r ) (r ¿ 0) in the plane such that (B2 − AC )(x; y) 6= 0 for all (x; y)
in R. If we set
Z
x
s(x; y) = exp
x0
f(t; y) d t +
Z
y
y0
g(x0 ; t ) d t ;
(x; y) ∈ R;
where f(x; y) = [C (Ax + By − D) − B(Bx + Cy − E )](B2 − AC )−1 and g(x; y) = [A(Bx + Cy − E ) −
B(Ax + By − D)](B2 − AC )−1 , then s(x; y) satis es (3.12) and (3.13). Hence L[·] is symmetrizable
with s(x; y) as a symmetry factor.
Example 3.11. Consider the following di erential equation
L[u] = x2 uxx + 2(xy + 1)uxy + y2 uyy + ux + uy = n(n − 1)u:
(3.18)
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
247
Then B2 − AC = 2xy + 1 6≡ 0 but L[·] is not symmetrizable since the compatibility condition (3.15)
is not satis ed.
Theorem 3.12. If the di erential equation (3.1) has an OPS {n }∞
n=0 as solutions; then L[·] must
be symmetrizable.
Proof. Let be the canonical moment functional of {n }∞
n=0 . Then satis es M1 [] = M2 [] = 0
by Proposition 3.4. Solving the moment equations M1 [] = M2 [] = 0 for x and y , we obtain
x =
and
y =
;
(3.19)
where = B2 − AC; = C (Ax + By − D) − B(Bx + Cy − E );
= A(Bx + Cy − E ) − B(Ax + By − D): Then
(x; y) 6≡ 0 by Theorem 3.1 and we have from (3.19) ( x )y − ( y )x = ( )y − (
)x and so
( y − x
) = ( y −
x ): Hence, we have by Lemma 2.3(ii), y − y = x
−
x since is
quasi-de nite. Therefore the compatibility condition (3.15) is satis ed so that L[·] is symmetrizable
by Proposition 3.10.
The symmetrizability of ordinary di erential equations of arbitrary order having OPS’s in one
variable as solutions is recently proved in [7].
4. Rodrigues type formula
From now on, we may and shall assume (see Theorem 3.12) that the di erential operator L[·]
in (3.1) is symmetrizable and let w(x; y)(6≡ 0) be a symmetry factor of L[·]. That is, w(x; y) is any
nonzero solution of the symmetry equations
M1 [w] = (Aw)x + (Bw)y − Dw = 0;
M2 [w] = (Bw)x + (Cw)y − Ew = 0:
(4.1)
Solving M1 [w] = M2 [w] = 0 for wx and wy yields (cf. (3.16) and (3.17))
wx = w
and
wy =
w:
(4.2)
Note that deg( ) 6 3; deg( ) 6 2 and deg(
) 6 2: Decompose A; B, and C into
A = A1 A2 ;
B = A1 B1 C1 ;
C = C1 C2 ;
(4.3)
where A1 6≡ 0; C1 6≡ 0: Then
= A1 C1 0 ;
0 = A1 B12 C1 − A2 C2 ;
= C1 0 ;
0 = A1 B1 (E − Bx − Cy ) − C2 (D − Ax − By );
= A1
0 ;
0 = B1 C1 (D − Ax − By ) − A2 (E − Bx − Cy )
so that Eq. (4.2) become
pwx = 0 w
and
qwy =
0 w;
where p = A1 0 and q = C1 0 :
(4.4)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Note that p 6≡ 0; q 6≡ 0 and deg(p) 6 3; deg(q) 6 3; deg( 0 ) 6 2; deg(
0 ) 6 2:
Lemma 4.1 (cf. Suetin [9], p. 183). If
(A1 )y = (C1 )x = 0;
(4.5)
then for any integers m and n ¿ 0
1 m n m n
@ @ (p qw ) :=
w x y
mn (x; y )
(4.6)
is a polynomial of degree 6 2(m + n): If moreover;
deg(p); deg(q) 6 2
then deg(
mn ) 6 m
and
deg( 0 ); deg(
0 ) 6 1;
(4.7)
+ n:
Proof. Assume that the condition (4.5) holds. Then for any polynomial (x; y) and any integers m
and n, we have
@x (pm qn w) = (pm−1 qn 1 )w
and
@y (pm qn w) = (pm qn−1 2 )w;
where
1 = mpx + nA1 ( 0 )x + px + 0 ;
2 = mC1 ( 0 )y + nqy + qy +
0 :
Since deg(1 )6 deg() + max{deg(p) − 1; deg( 0 )} and deg(2 )6 deg() + max{deg(q) − 1;
deg(
0 )}, the conclusions follow easily by induction on m and n.
However, in general, mn need not satisfy the di erential equation (3.1) and we cannot say
anything on orthogonality of { mn }, unless w(x; y) is an orthogonalizing weight function for an
OPS satisfying the di erential equation (3.1).
Hence we now formalize Lemma 4.1 by using, instead of a symmetry factor w(x), a moment
functional solution of the moment equations M1 [] = M2 [] = 0:
Theorem 4.2. Assume that the di erential equation (3.1) has a WOPS {n }∞
n=0 as solutions and
let be the canonical moment functional of {n }∞
.
Assume
that
satis es
n=0
px = 0
and
qy =
0 :
(4.8)
(i) If the condition (4.5) holds; then for any integers m and n ¿ 0
@xm @yn (pm qn ) =
where
h; xk yl
mn (x; y )
(4.9)
mn ;
is a polynomial of degree 6 2(m + n); and
mn (x; y )i = 0;
0 6 k + l 6 m + n and (k; l) 6= (m; n):
(ii) If the conditions (4.5) and (4.7) hold; then deg(
mn ) 6 m
+ n.
(4.10)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
(iii) If is quasi-de nite and the conditions (4.5) and (4.7) hold; then { n }∞
n=0 ; where n = {
n−1;1 ; : : : ; 0n }; is a WOPS relative to and satis es the di erential equation (3.1).
n0 ;
Before proving Theorem 4.2, we rst note that under the assumptions as in Theorem 4.2,
satis es the moment equations M1 [] = M2 [] = 0 by Proposition 3.4, which, however, do not imply
(4.8) in general.
Proof of Theorem 4.2. The proof of (i) and (ii) except (4.10) is essentially the same as that of
Lemma 4.1. Now,
h; xk y‘
mn (x; y )i = h mn ; x
k
y‘ i = (−1)m+n hpm qn ; @xm @yn (xk y‘ )i
from which (4.10) follows. Finally, we further assume that is quasi-de nite. We rst claim that
deg( mn ) = m + n and h; xm yn mn i =
6 0 for m and n ¿ 0. Assume that deg( mn ) 6 m + n − 1 for some
m + n ¿ 1: Then mn ≡ 0 since h; mn i = 0 for any (x; y) in Pm+n−1 by (4.10) and is quaside nite. Hence @xm @yn (pm qn ) = 0 so that either p(x; y) ≡ 0 or q(x; y) ≡ 0, which is a contradiction.
Hence deg( mn ) = m + n and h; xm yn mn i =
6 0 for all m and n ¿ 0 by (4.10).
We now claim that for each n ¿ 0; { n−j; j }nj=0 are linearly independent modulo Pn−1 P
so that
n
{ n }∞
is
a
PS.
For
any
integer
n
¿
1,
let
C
;
C
;
:
:
:
;
C
be
constants
such
that
(
x;
y
)
=
0
1
n
n=0
j=0 Cj
n−j; j is of degree 6 n− 1. Then (x; y ) ≡ 0 since h; (x; y )(x; y )i = 0 for any ∈ Pn−1 by (4.10).
Then
h; (x; y)xn−k yk i = Ck h;
n−k; k x
n−k
yk i = 0;
06k 6n
so that C0 = C1 = · · · = Cn = 0 by the rst claim. Hence { n−j; j }nj=0 are linearly independent modulo
∞
Pn−1 and so { n }∞
n=0 is a WOPS relative to by (4.10). Finally, in order to see that { n }n=0 satisfy
∞
∞
the di erential equation (3.1), we let {Pn }n=0 and {Qn }0 be the normalizations of {n }∞
n=0 and
∞
∞
∞
{ n }n=0 respectively. Then {Pn }n=0 and {Qn }0 are monic WOPS’s relative to so that Pn = Qn ,
∞
∞
n ¿ 0, by Proposition 2.2. Since {n }∞
n=0 satisfy the di erential equation (3.1), {Pn }n=0 and {Qn }n=0
also satisfy the di erential equation (3.1).
Combining Lemma 4.1 and Theorem 4.2, we now have:
Theorem 4.3. Assume that the di erential equation (3.1) has an OPS {n }∞
n=0 relative to as
solutions and let w(x; y) be a symmetry factor of L[·]: If the conditions (4.5), (4.7), and (4.8)
hold; then the PS { mn } de ned by (4.6) is a WOPS relative to and satis es the di erential
equation (3.1).
However, h; n nT i, with { n }∞
n=0 as in Theorem 4.3, is nonsingular but need not be diagonal,
that is, { n }∞
is
a
WOPS
but
need
not be an OPS in general (see Example 4.5 below).
n=0
We may call (4.6) and (4.9) Rodrigues type formulas for orthogonal polynomial solutions of the
di erential equations (3.1).
Example 4.4. Assume that Ay = B = Cx = 0 so that the di erential equation (3.1) is of the form
L[u] = A(x)uxx + C (y)uyy + D(x)ux + E (y)uy = n u;
(4.11)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
where A(x) = d1 x + f1 ; C (y) = e3 y + f3 ; D(x) = gx + h1 ; E (y) = gy + h2 and g 6= 0: Then it is shown
in [6] that
• the di erential equation (4.11) has a unique monic PS {Pn }∞
n=0 as solutions;
• Pn0 (x; y) = Pn0 (x); P0n (x; y) = P0n (y), and Pmn (x; y) = Pm0 (x)P0n (y) for m and n ¿ 0;
• {Pn }∞
n=0 is a WOPS;
• Pn0 (x) and P0n (y); n ¿ 0, satisfy
A(x)Pn0′′ (x) + D(x)Pn0′ (x) = n Pn0 (x)
(4.12)
C (y)P0n′′ (y) + E (y)P0n′ (y) = n P0n (y):
(4.13)
and
In decomposition (4.3), we take A2 = C2 = −1 and B1 = 0. Then
A1 = −A;
p = A;
C1 = −C;
0 = −1;
0 = D − A ′ ;
0 = E − C ′ ;
q=C
so that the conditions (4.5) and (4.7) hold. On the other hand, the canonical moment func(x)
tional of {Pn }∞
⊗ (y) ; where (x) and (y) are the canonical moment funcn=0 is equal to =
∞
∞
tionals of {Pn0 (x)}0 and {P0n (y)}n=0 , respectively. Since A(x) (d = d x) (x) = (D(x) − A′ (x))(x) and
C (y) (d = d y) (y) = (E (y) − C ′ (y))(y) ; = (x) ⊗ (y) satis es the condition (4.8). Hence, by Theorem 4.2,
@xm @yn (A(x)m C (y)n ) =
where {
mn }
mn ;
m; n ¿ 0;
(4.14)
is a WOPS relative to . In fact, we have
mn (x; y ) = g
m+n
Pm0 (x)P0n (y);
m; n ¿ 0;
so that the Rodrigues type formula (4.14) is nothing but the tensor product of one dimensional
∞
Rodrigues formulas for {Pn0 (x)}∞
n=0 and {P0n (y )}n=0 (see [1, 2, 10] for Rodrigues formula for classical
orthogonal polynomials of one variable).
We may, of course, replace by a symmetry factor w(x; y) = w1 (x)w2 (y) of the di erential
equation (4.11), where w1 (x) and w2 (y) are symmetry factors of the di erential equations (4.12)
and (4.13), respectively.
Example 4.5. Consider the di erential equation for circle polynomials:
L[u] = (x2 − 1)uxx + 2xyuxy + (y2 − 1)uyy + gxux + gyuy = n u:
(4.15)
In decomposition (4.3), we take A1 = C1 = 1 so that B1 = B = xy and
= p = q = x 2 + y 2 − 1;
= 0 = (g − 3)x;
=
0 = (g − 3)y:
Then by Lemma 4.1,
1 m n 2
@ @ [(x + y2 − 1)m+n w] =
w x y
mn (x; y );
m; n ¿ 0
is a polynomial of degree 6 m + n, where w(x; y) = (x2 + y2 − 1)
di erential equation (4.15). In general, deg( mn ) 6= m + n.
g−3
2
is a symmetry factor of the
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
For example, we have
00 (x; y ) = 1
10 (x; y ) = (g
− 1)x;
20 (x; y ) = (g
+ 1)(gx2 + y2 − 1);
02 (x; y ) = (g
and
01 (x; y ) = (g
2
251
− 1)y;
11 (x; y ) = (g
+ 1)(g − 1)xy;
2
− 1)(x + gy − 1):
However, if g 6= 1; 0; −1; : : :, then the di erential equation (4.15) has an OPS {n }∞
n=0 (called the
∞
circle polynomials) as solutions. Then the canonical moment functional of {n }n=0 satis es the
condition (4.8) so that by Theorem 4.3, { mn } is a WOPS. But, even in this case, { mn } is not an
OPS. For if we let be the canonical moment functional of { mn }, then satis es L∗ [] = 0 (cf.
Lemma 3.2), that is,
(m + n)(m + n − 1 + g)mn − m(m − 1)m−2; n − n(n − 1)m; n−2 = 0;
m; n ¿ 0
so that
00 = 1;
10 = 01 = 11 = 30 = 21 = 12 = 03 = 31 = 13 = 0;
3
1
1
20 = 02 =
;
40 = 04 =
;
22 =
:
g+1
(g + 1)(g + 3)
(g + 1)(g + 3)
Hence, we have
H2 = h; 2 2T i =
2g
(g + 1)(g − 1)
0
g+3
2
0
2
g−1
0
0 ;
2g
which is nonsingular but not diagonal.
Example 4.6. Consider the di erential equation for triangle polynomials:
L[u] = (x2 − x)uxx + 2xyuxy + (y2 − y)uyy + [(a + b + c + 3)x − (1 + a)]ux
+ [(a + b + c + 3)y − (1 + b)]uy = n u:
(4.16)
In decomposition (4.3), we take A1 = x; C1 = y so that
0 = x + y − 1;
p = x(x + y − 1);
0 = (a + c)x + ay − a;
0 = bx + (b + c)y − b;
q = y(x + y − 1):
Then by Lemma 4.1,
1 m n m n
@ @ [x y (x + y − 1)m+n w] =
w x y
mn (x; y );
m; n ¿ 0
(4.17)
is a polynomial of degree 6 m + n, where w(x; y) = xa yb (x + y − 1)c is a symmetry factor of
the di erential equation (4.16). We now assume that a; b; c ¿ −1 so that the di erential equation
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
(4.16) has an OPS {n }∞
n=0 (called the triangle polynomials) as solutions and let be the canonical moment functional of {n }∞
n=0 . Then satis es the moment equations M1 [] = M2 [] = 0 or
equivalently
xy(x + y − 1)x = y((a + c)x + ay − a);
xy(x + y − 1)y = x(bx + (b + c)y − b):
(4.18)
Let
u := x(x + y − 1)x − [(a + c)x + ay − a] = px − 0 ;
v := y(x + y − 1)y − [bx + (b + c)y − b] = qy −
0 :
Then, (4.18) becomes
yu = xv = 0:
(4.19)
On the other hand, we can see by using M1 [] = M2 [] = 0 that
u = −v = xyx − xyy + (bx − ay):
(4.20)
Hence, we have from (4.19) and (4.20)
hu; xm yn i = hv; xm yn i = 0;
m + n ¿ 1:
(4.21)
We also have from (4.20)
hu; 1i = −hv; 1i = hxyx − xyy + (bx − ay); 1i = (b + 1)10 − (a + 1)01
and from (3.3)
h; Di = (a + b + c + 3)10 − (1 + a) = 0;
h; Ei = (a + b + c + 3)01 − (1 + b) = 0
so that hu; 1i = hv; 1i = 0. Therefore, together with (4.21), we have u = v = 0, that is, satis es the
condition (4.8). Hence, by Theorem 4.3, { mn } given by (4.17) is a WOPS relative to and satis es
the di erential equation (4.16).
Rodrigues type formulas for circle and triangle polynomials have been known before (see
[3, chapter 12] and [4, Section 4]).
We nally give a negative example for which Rodrigues type formula (4.6) does not hold.
Example 4.7. Krall and She er [5] showed that the di erential equation
L[u] = 3yuxx + 2uxy − xux − yuy = −nu
(4.22)
has an OPS as solutions. In decomposition (4.3), we must choose A1 = 1 and C1 = 1 in order for
the condition (4.5) to be satis ed. Then
= 0 = 1;
= 0 = −y;
=
0 = 3y2 − 3x;
p=q=1
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
253
so that by Lemma 4.1,
1 m nw
@ @ =
w x y
mn ;
m; n ¿ 0
is a polynomial of degree 6 2(m + n), where w(x; y) = exp(y3 − xy) is a symmetry factor of the
di erential equation (4.22). For example,
10 (x; y ) = −y
and
01 (x; y ) = 3y
2
−x
and 10 satis es the di erential equation (4.22) but
(4.22).
01
does not satisfy the di erential equation
Acknowledgements
This work is partially supported by KOSEF(95-0701-02-01-3), GARC, and the Korean Ministry
of Education (BSRI 97-1420). Authors thank referees for giving many valuable suggestions and
directing attention to Refs. [3, 4].
References
[1] W.C. Brenke, On polynomial solutions of a class of linear di erential equations of the second order, Bull. Amer.
Math. Soc. 36 (1930) 77– 84.
[2] C.W. Cryer, Rodrigues formulas and the classical orthogonal polynomials, Boll. Unione. Mat. Ital. 25 (1970) 1–11.
[3] A. Erdelyi, W. Magnus, F. Oberhettinger, F.G. Tricomi, Higher Transcendental Functions, vol. II, McGraw-Hill,
New York, 1953.
[4] T.H. Koornwinder, Two-variable analogues of the classical orthogonal polynomials, in: R.A. Askey (Ed.), Theory
and Applications of Special functions, Academic Press, New York, 1975, pp. 435– 495.
[5] H.L. Krall, I.M. She er, Orthogonal polynomials in two variables, Ann. Mat. Pura Appl. Serie 4 76 (1967) 325–376.
[6] Y.J. Kim, K.H. Kwon, J.K. Lee, Orthogonal polynomials in two variables and second order partial di erential
equations, J. Comput. Appl. Math. 82 (1997) 239 – 260.
[7] K.H. Kwon, G.J. Yoon, Symmetrizability of di erential equations having orthogonal polynomial solutions,
J. Comput. Appl. Math. 83 (1997) 257– 268.
[8] L.L. Littlejohn, Orthogonal polynomial solutions to ordinary and partial di erential equations, in: M. Alfaro et al.
(Eds.), Proc. 2nd Internat. Symp. Orthogonal Polynomials and their Applications, Segovia (Spain), 1986, Lect. Notes
Math., vol. 1329, Springer, Berlin, 1988, pp. 98 –124.
[9] P.K. Suetin, Orthogonal Polynomials in Two Variables, Nauka, Moscow, 1988 (in Russian).
[10] F. Marcellan, A. Branquinho, J. Petronilho, Classical orthogonal polynomials: a functional approach, Acta Appl.
Math. 34 (1994) 283–303.
[11] Y. Xu, On multivariate orthogonal polynomials, SIAM J. Math. Anal. 24 (1993) 783–794.
Partial di erential equations having orthogonal
polynomial solutions
Y.J. Kim a , K.H. Kwon a; ∗ , J.K. Lee b
b
a
Department of Mathematics, KAIST, Taejon 305-701, South Korea
Department of Mathematics, Sunmoon university, Cheonan, South Korea
Received 30 October 1997; received in revised form 19 May 1998
Abstract
We show that if a second order partial di erential equation:
L[u] := Auxx + 2Buxy + Cuyy + Dux + Euy = n u
has orthogonal polynomial solutions, then the di erential operator L[·] must be symmetrizable and can not be parabolic
in any nonempty open subset of the plane. We also nd Rodrigues type formula for orthogonal polynomial solutions of
c 1998 Elsevier Science B.V. All rights reserved.
such di erential equations.
AMS classi cation: 33C50; 35P99
Keywords: Orthogonal polynomials in two variables; Partial di erential equations; Symmetrizability; Rodrigues type
formula
1. Introduction
As a natural generalization of classical orthogonal polynomials of one variable, Krall and She er
[5] considered orthogonal polynomials in two variables satisfying a second order partial di erential
equation with polynomial coecients:
L[u] = Auxx + 2Buxy + Cuyy + Dux + Euy = n u:
(1.1)
Krall and She er rst nd necessary and sucient conditions in order for orthogonal polynomials
to satisfy the di erential equation (1.1) and then classify such di erential equations, up to a complex
linear change of variables, which have at least weak orthogonal polynomials as solutions.
∗
Corresponding author. Fax: 42 869 2710; e-mail: khkwon@jacobi.kaist.ac.kr.
c 1998 Elsevier Science B.V. All rights reserved.
0377-0427/98/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 6 0 - 5
240
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Later, Littlejohn [8] noted that all di erential equations found in [5] are elliptic in the region
of orthogonality in case the orthogonalizing weight is known. However, the type of di erential
operators cannot be determined properly from the speci c forms of the equations given in [5] since
the type of a di erential operator is not preserved under a complex change of variables. In fact,
Krall and She er found at least one di erential equation (see Example 3.3), which is hyperbolic
everywhere and has orthogonal polynomials as solutions.
In this work, using the functional calculus on moment functionals developed in [6, 8], we rst
prove that if the di erential equation (1.1) has orthogonal polynomials as solutions, then L[·] cannot
be parabolic in any nonempty open subset of the plane and must be symmetrizable. We also establish
Rodrigues type formula for orthogonal polynomial solutions to the di erential equation (1.1).
2. Preliminaries
For any integer nS¿ 0, let Pn be the space of real polynomials in two variables of (total) degree 6 n and P = n¿0 Pn . By a polynomial system (PS), we mean a sequence of polynomials
n
n
{ n−j; j }∞
n=0; j=0 such that deg( n−j; j ) = n; 0 6 j 6 n and { n−j; j }j=0 are linearly independent modulo
Pn−1 for n ¿ 0 (P−1 = {0}).
n
∞
∞
We let n :=[ n; 0 ; n−1; 1 ; : : : ; 0; n ]T and denote the PS { n−j; j }∞
n=0; j=0 by {n }n=0 . A PS {Pn }n=0 ;
T
Pn = [Pn; 0 ; Pn−1; 1 ; : : : ; P0; n ] , where
Pn−k; k (x; y) = xn−k yk modulo Pn−1 ;
0 6 k 6 n;
is called a monic PS. For any PS {n }∞
n=0 , where
n−j; j (x; y) =
n
X
anjk xn−k yk modulo Pn−1 ;
0 6 j 6 n;
k=0
−1
the matrix An = [anjk ]nj; k=0 ; n ¿ 0; is nonsingular. Then, the monic PS {Pn }∞
n=0 de ned by Pn = An n
is called the normalization of {n }∞
n=0 :
For a moment functional on P, we call
ij := h; xi y j i;
i; j ¿ 0
the moments of and
00
10
n := .
..
0n
10
01
···
20
..
.
11
..
.
···
1n
0; n+1
···
..
.
0n
;
0;2n
1n
..
.
n¿0
the Hankel determinants of . We call to be quasi-de nite if n 6= 0; n ¿ 0.
n
For a matrix = [ ij (x; y)]mi=0; j=0
of polynomials we let
h; i = [h;
m
n
ij i]i=0; j=0 :
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
241
For any PS {n }∞
n=0 , there is a unique moment functional , called the canonical moment functional of {n }∞
,
de ned
by the conditions
n=0
h; 1i = 1
and
h; n i = 0; n ¿ 1:
De nition 2.1. A PS {n }∞
n=0 is called a weak orthogonal polynomial system (WOPS) if there is
a nonzero moment functional such that h; m nT i = 0; m 6= n. If moreover Hn := h; n nT i; n ¿ 0,
is a nonsingular diagonal matrix, we call {n }∞
n=0 an orthogonal polynomial system(OPS). In this
case, we say that {n }∞
is
a
WOPS
or
an
OPS
relative to :
n=0
Note that if {n }∞
n=0 is a WOPS relative to , then must be a nonzero constant multiple of the
∞
canonical moment functional of {n }∞
n=0 . If {n }n=0 is an OPS relative to , then the normalization
∞
{Pn }∞
n=0 of {n }n=0 is a WOPS, but not necessarily an OPS relative to :
The following is proved in [5] (see also [11]).
Proposition 2.2. For a moment functional ; the following statements are equivalent.
(i) is quasi-de nite;
(ii) there is an OPS {n }∞
n=0 relative to ;
(iii) there is a unique monic WOPS {Pn }∞
n=0 relative to ;
(iv) there is a monic WOPS {Pn }∞
relative
to such that
n=0
T
Hn := h; Pn Pn i; n ¿ 0; is nonsingular.
T
If { n }∞
n=0 is a WOPS relative to such that h; n n i; n ¿ 0, are nonsingular, then there are
orthogonal matrices On : (n + 1) × (n + 1); n ¿ 0, such that
On h; n nT iOnT = h; (On n )(On n )T i;
n¿0
are nonsingular diagonal matrices. Hence {n := On n }∞
n=0 , is an OPS relative to . In fact, for a
quasi-de nite moment functional , let Vn ; n ¿ 0, be the vector space of polynomials of degree n,
which are orthogonal to any polynomial of degree ¡ n relative to . Then, {n }∞
n=0 , where n is a
basis (respectively, an orthogonal basis) of Vn , forms a WOPS (respectively, an OPS) relative to
and Hn := h; n nT i; n ¿ 0, are nonsingular (respectively, nonsingular diagonal).
For any moment functional and any polynomial (x; y), we let
h@x ; i = −h; x i;
h ; i = h; i;
h@y ; i = −h; y i
∈ P:
Then we have
Lemma 2.3. Let and be moment functionals and R(x; y) a polynomial. Then
(i) = 0 if and only if @x = 0 or @y = 0:
Assume that is quasi-de nite and let {n }∞
n=0 be an OPS relative to . Then
(ii) R(x; y) = 0 if and only if R(x; y) = 0.
(iii) h ; n i = 0 for n ¿ k (k ¿ 0 an integer) if and only if = (x; y) for some polynomial (x; y)
of degree 6 k .
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Proof. See Lemma 2.3 in [6].
3. Di erential operator L[·]
We now consider the di erential equation (1.1). Krall and She er [5] showed that if the di erential
equation (1.1) has a PS as solutions, then it must be of the form
L[u] = Auxx + 2Buxy + Cuyy + Dux + Euy
= (ax2 + d1 x + e1 y + f1 )uxx + (2axy + d2 x + e2 y + f2 )uxy
+ (ay2 + d3 x + e3 y + f3 )uyy + (gx + h1 )ux + (gy + h2 )uy
= n u; n = an(n − 1) + gn:
(3.1)
Following Krall and She er, we say that the di erential operator L[·] is admissible if m 6= n for
m 6= n or equivalently n 6= 0 for n ¿ 1:
It is shown in [5] that the di erential equation (3.1) has a unique monic PS as solutions if and
only if L[·] is admissible.
Theorem 3.1. If the di erential equation (3.1) has an OPS as solutions; then B2 − AC 6≡ 0 in
any nonempty open subset of the plane. In other words; the di erential operator L[·] cannot be
parabolic in any nonempty open subset of the plane.
In order to prove Theorem 3.1, we rst need the following fact:
Lemma 3.2. If the di erential equation (3.1) has a PS {n }∞
n=0 as solutions; then the canonical
moment functional of {n }∞
must
satisfy
n=0
∗
L [] = 0;
(3.2)
h; Di = h; Ei = h; A + xDi = h; C + yEi = h; B + yDi = h; B + xEi = 0;
where L∗ [·] is the formal Lagrange adjoint of L[·] given by
L∗ [u] = (Au)xx + 2(Bu)xy + (Cu)yy − (Du)x − (Eu)y :
(3.3)
(3.4)
Proof. See Lemma 3.1 in [6].
Proof of Theorem 3.1. Assume that the di erential equation (3.1) has an OPS {n }∞
n=0 as solutions.
∞
Let be the canonical moment functional of {n }n=0 . We assume that
B2 − AC = −a[d3 x3 + (e3 − d2 )x2 y + (d1 − e2 )xy2 + e1 y3 ]
+( 41 d22 − af3 − d1 d3 )x2 + ( 12 d2 e2 + af2 − d1 e3 − d3 e1 )xy
+( 41 e22 − e1 e3 − af1 )y2 + ( 12 d2 f2 − d1 f3 − f1 d3 )x
+( 21 e2 f2 − f1 e3 − e1 f3 )y + 41 f22 − f1 f3
≡0
(3.5)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
in some nonempty open subset of the plane. Then, all the coecients of B2 − AC must be 0. We
now consider the following three cases separately.
Case 1: a + g =
6 0 and g 6= 0: Then, we may assume that g = 1 and h1 = h2 = 0 so that (3.3) gives
10 = 01 = 0
and
(a + 1)20 + f1 = (a + 1)02 + f3 = 2(a + 1)11 + f2 = 0:
Therefore
2
=
1 = 20 02 − 11
1
[f1 f3 − 14 f22 ] = 0;
(a + 1)2
which is a contradiction since 1 6= 0 when is quasi-de nite.
Case 2: a + g 6= 0 and g = 0: Then we may assume that a = 1 so that (3.5) gives
d3 = e 1 = 0 ;
d 2 = e3 ;
d1 = e2 ;
f1 = 41 d21 ;
f2 = 12 d1 d2 ;
f3 = 41 d22 :
Hence the di erential equation (3.1) becomes, after replacing x + 12 d1 by x and y + 12 d2 by y,
x2 uxx + 2xyuxy + y2 uyy = n(n − 1)u:
(3.6)
Then we have by (3.3), 20 = 02 = 11 = 0 so that 1 = 0, which is a contradiction.
Case 3: a + g = 0 and g 6= 0: Then we may assume that a = − 1; g = 1, and h1 = h2 = 0. In this case,
it is easy to see that if the di erential equation (3.1) has a PS as solutions, then f1 = f2 = f3 = 0. We
then have from (3.5) d1 = e1 = d2 = e2 = d3 = e3 = 0 so that the di erential equation (3.1) becomes
x2 uxx + 2xyuxy + y2 uyy − xux − yuy = n(n − 2)u:
Then we have from (3.2)
Amn := hL∗ []; xm yn i = h; L[xm yn ]i = (m + n)(m + n − 2)mn = 0;
m; n ¿ 0:
Hence mn = 0 for m + n ¿ 3 so that 2 = 0, which is a contradiction.
While Krall and She er considered only admissible di erential equations, we do not assume the
admissibility of L[·] in Theorem 3.1 since we do not know whether L[·] must be admissible or not
when the di erential equation (3.1) has an OPS as solutions.
When L[·] is not admissible, the di erential equation (3.1) may have either no PS as solutions
or in nitely many distinct monic PS’s as solutions. For example, the di erential equation (3.6) has
in nitely many monic PS’s {Pn }∞
n=0 as solutions;
Pmn (x; y) = xm yn
for m + n =
6 1
and
P10 = x + ; P01 = y + ;
where and are arbitrary constants. Note that for any choice of and ; {Pn }∞
n=0 can not be an
OPS but for = = 0; {xm yn }∞
is
a
WOPS
relative
to
(
x;
y
)
:
m; n=0
Example 3.3. Krall and She er [5] showed that the di erential equation
L[u] = 3yuxx + 2uxy − xux − yuy = −nu
(3.7)
has an OPS as solutions. Since B2 − AC = 1 ¿ 0, the di erential operator L[·] in (3.7) is hyperbolic
everywhere.
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Littlejohn [8] observed that all di erential equations found by Krall and She er [5], that have
orthogonal polynomial solutions whose weight functions are known, are elliptic in the region of
orthogonality.
However, since Krall and She er used a complex linear change of variables in their classi cation,
the type of the di erential equation cannot be properly determined from the speci c forms of the
equations given in [5].
For example, Krall and She er obtained the following di erential equation (see the equation
(5.13) in [5]) by a complex linear change of variables
uxx + uyy − xux − yuy = −nu
(3.8)
from the equation (see Eq. (5.12) in [5])
f2 uxy + gxux + gyuy = gnu:
(3.9)
Note that since f2 (6= 0) is real, the di erential equation (3.9) is hyperbolic everywhere while
Eq. (3.8) is elliptic everywhere.
∞
Proposition 3.4. Let {n }∞
n=0 be a PS satisfying the di erential equation (3:1). If {n }n=0 is at
∞
∗
least a WOPS; then the canonical moment functional of {n }n=0 satis es L [] = 0 and
M1 [] := (A)x + (B)y − D = 0;
(3.10)
M2 [] := (B)x + (C)y − E = 0:
(3.11)
Proof. See Corollary 3.3 in [6].
Lemma 3.5. For the di erential operator L[·] in (3.1) and any moment functional ; the following
statements are equivalent.
(i) M1 [] = M2 [] = 0;
(ii) L[·] is formally symmetric on polynomials; that is;
hL[P ]; Qi = hL[Q]; Pi;
P and Q in P:
Furthermore; if L[P ] = P and L[Q] = Q for 6= ; then h; PQi = 0 for any moment functional
satisfying M1 [] = M2 [] = 0.
Proof. See Lemma 3.6 in [6].
We call the functional equations M1 [] = M2 [] = 0 in (3.10) and (3.11) the moment equations
for the di erential equation (3.1).
Lemma 3.6. If the di erential equation (3.1) is admissible and has a WOPS as solutions; then
the moment equations M1 [] = M2 [] = 0 can have only one solution up to a nonzero constant
multiple.
Proof. See Corollary 3.5 in [6].
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
245
As a consequence of Theorem 3.1 and Lemma 3.5, we can now prove:
Corollary 3.7. If the di erential equation (3.1) has an OPS {n }∞
n=0 as solutions; then
(i) the moment equations M1 [] = M2 [] = 0 can have only one solution up to a nonzero constant
multiple; which must be quasi-de nite;
(ii) the di erential equation (3.1) can have only one monic WOPS as solutions (but may have in nitely many other monic PS’s as solutions; which are not WOPS’s; unless L[·] is admissible).
∞
Proof. Let be the canonical moment functional of {n }∞
n=0 and {Pn }n=0 the normalization of
∞
{n }n=0 . Then is quasi-de nite and M1 [] = M2 [] = 0 by Proposition 3.4 and {Pn }∞
n=0 is a monic
WOPS satisfying the di erential equation (3.1).
(i) If L[·] is admissible, then it follows from Lemma 3.6. We now assume that L[·] is not
admissible, that is, N = 0 for some N ¿ 1. Let be any moment functional satisfying the moment
equations M1 [ ] = M2 [ ] = 0: Then Lemma 3.5 implies h ; n i = 0; n 6= 0; N since n 6= 0 for n 6= 0; N .
Therefore we have, by Lemma 2.3(iii), = (x; y) for some polynomial (x; y) of degree 6 N:
Then
M1 [ ] = M1 [ ] = ( x A + y B) + M1 [] = ( x A + y B) = 0;
M2 [ ] = M2 [ ] = ( x B + y C ) + M2 [] = ( x B + y C ) = 0:
Hence we have by Lemma 2.3(ii)
x A + y B = 0
and
x B + y C = 0;
which implies that x = y = 0, that is, (x; y) must be a constant since B2 −AC 6≡ 0 by Theorem 3.1.
(ii) Let {Qn }∞
n=0 be any monic WOPS relative to satisfying the di erential equation (3.1). Then
satis es the moment equations M1 [ ] = M2 [ ] = 0 by Proposition 3.4. Hence by (i), = c for some
∞
∞
nonzero constant c. Therefore {Qn }∞
n=0 is also a monic WOPS relative to so that {Qn }n=0 = {Pn }n=0
by Proposition 2.2 since is quasi-de nite.
Corollary 3.7 provides a unique existence of orthogonal polynomial solutions to the di erential
equation (3.1) in the monic level whether L[·] is admissible or not. However, the di erential equation
(3.1) may have in nitely many distinct OPS’s, which are not monic, as solutions if it has at least
one.
De nition 3.8 (Littlejohn [8]). The di erential operator L[·] in (3.1) is symmetric if L[·] = L∗ [·].
We say that L[·] is symmetrizable if there is a non-trivial function s(x; y) such that s(x; y) is C 2 in
some open set and sL[·] is symmetric.
In this case, we call s(x; y) a symmetry factor of L[·]:
It is easy to see (cf. [8]) that a function s(x; y) is a symmetry factor of L[·] if and only if s(x; y)
satis es
M1 [s] = (As)x + (Bs)y − Ds = 0;
(3.12)
M2 [s] = (Bs)x + (Cs)y − Es = 0:
(3.13)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
We call the simultaneous equations (3.12) and (3.13) the symmetry equations for L[·]. We rst see
by an example that not every di erential operator L[·] is symmetrizable.
Example 3.9 (Littlejohn [8]). Consider the di erential equation
L[u] = x2 uxx + 2xyuxy + y2 uyy + (gx + h1 )ux + (gy + h2 )uy = n(n − 1 + g)u:
(3.14)
Note rst that the di erential (3.14) is parabolic everywhere so that it cannot have an OPS as
solutions by Theorem 3.1. The corresponding symmetry equations are
x2 sx + xysy + ((3 − g)x − h1 )s = 0;
xysx + y2 sy + ((3 − g)y − h2 )s = 0
so that (h2 x − h1 y)s = 0: Hence, if h21 + h22 6= 0, then s(x; y) = 0 on {(x; y) ∈ R2 | h2 x − h1 y 6= 0} and
so L[·] cannot be symmetrizable. However, if g = 3 and h1 = h2 = 0, then L[·] is symmetrizable with
s(x; y) = arctan(y=x) as a symmetry factor.
Proposition 3.10. Assume that B2 − AC 6≡ 0. Then the di erential operator L[·] is symmetrizable
if and only if
@ A(Bx + Cy − E ) − B(Ax + By − D)
@ C (Ax + By − D) − B(Bx + Cy − E )
=
:
@y
B2 − AC
@x
B2 − AC
(3.15)
We call (3.15) the compatibility condition for the symmetrizability of L[·]:
Proof. (⇒): We may solve the symmetry equations (3.12) and (3.13) for sx and sy as
(B2 − AC )sx = [C (Ax + By − D) − B(Bx + Cy − E )]s;
(3.16)
(B2 − AC )sy = [A(Bx + Cy − E ) − B(Ax + By − D)]s;
(3.17)
from which (3.15) follows.
(⇐): Assume that the compatibility condition (3.15) is satis ed. Choose any open rectangle
R = (x0 − r; x0 + r ) × (y0 − r; y0 + r ) (r ¿ 0) in the plane such that (B2 − AC )(x; y) 6= 0 for all (x; y)
in R. If we set
Z
x
s(x; y) = exp
x0
f(t; y) d t +
Z
y
y0
g(x0 ; t ) d t ;
(x; y) ∈ R;
where f(x; y) = [C (Ax + By − D) − B(Bx + Cy − E )](B2 − AC )−1 and g(x; y) = [A(Bx + Cy − E ) −
B(Ax + By − D)](B2 − AC )−1 , then s(x; y) satis es (3.12) and (3.13). Hence L[·] is symmetrizable
with s(x; y) as a symmetry factor.
Example 3.11. Consider the following di erential equation
L[u] = x2 uxx + 2(xy + 1)uxy + y2 uyy + ux + uy = n(n − 1)u:
(3.18)
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
247
Then B2 − AC = 2xy + 1 6≡ 0 but L[·] is not symmetrizable since the compatibility condition (3.15)
is not satis ed.
Theorem 3.12. If the di erential equation (3.1) has an OPS {n }∞
n=0 as solutions; then L[·] must
be symmetrizable.
Proof. Let be the canonical moment functional of {n }∞
n=0 . Then satis es M1 [] = M2 [] = 0
by Proposition 3.4. Solving the moment equations M1 [] = M2 [] = 0 for x and y , we obtain
x =
and
y =
;
(3.19)
where = B2 − AC; = C (Ax + By − D) − B(Bx + Cy − E );
= A(Bx + Cy − E ) − B(Ax + By − D): Then
(x; y) 6≡ 0 by Theorem 3.1 and we have from (3.19) ( x )y − ( y )x = ( )y − (
)x and so
( y − x
) = ( y −
x ): Hence, we have by Lemma 2.3(ii), y − y = x
−
x since is
quasi-de nite. Therefore the compatibility condition (3.15) is satis ed so that L[·] is symmetrizable
by Proposition 3.10.
The symmetrizability of ordinary di erential equations of arbitrary order having OPS’s in one
variable as solutions is recently proved in [7].
4. Rodrigues type formula
From now on, we may and shall assume (see Theorem 3.12) that the di erential operator L[·]
in (3.1) is symmetrizable and let w(x; y)(6≡ 0) be a symmetry factor of L[·]. That is, w(x; y) is any
nonzero solution of the symmetry equations
M1 [w] = (Aw)x + (Bw)y − Dw = 0;
M2 [w] = (Bw)x + (Cw)y − Ew = 0:
(4.1)
Solving M1 [w] = M2 [w] = 0 for wx and wy yields (cf. (3.16) and (3.17))
wx = w
and
wy =
w:
(4.2)
Note that deg( ) 6 3; deg( ) 6 2 and deg(
) 6 2: Decompose A; B, and C into
A = A1 A2 ;
B = A1 B1 C1 ;
C = C1 C2 ;
(4.3)
where A1 6≡ 0; C1 6≡ 0: Then
= A1 C1 0 ;
0 = A1 B12 C1 − A2 C2 ;
= C1 0 ;
0 = A1 B1 (E − Bx − Cy ) − C2 (D − Ax − By );
= A1
0 ;
0 = B1 C1 (D − Ax − By ) − A2 (E − Bx − Cy )
so that Eq. (4.2) become
pwx = 0 w
and
qwy =
0 w;
where p = A1 0 and q = C1 0 :
(4.4)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
Note that p 6≡ 0; q 6≡ 0 and deg(p) 6 3; deg(q) 6 3; deg( 0 ) 6 2; deg(
0 ) 6 2:
Lemma 4.1 (cf. Suetin [9], p. 183). If
(A1 )y = (C1 )x = 0;
(4.5)
then for any integers m and n ¿ 0
1 m n m n
@ @ (p qw ) :=
w x y
mn (x; y )
(4.6)
is a polynomial of degree 6 2(m + n): If moreover;
deg(p); deg(q) 6 2
then deg(
mn ) 6 m
and
deg( 0 ); deg(
0 ) 6 1;
(4.7)
+ n:
Proof. Assume that the condition (4.5) holds. Then for any polynomial (x; y) and any integers m
and n, we have
@x (pm qn w) = (pm−1 qn 1 )w
and
@y (pm qn w) = (pm qn−1 2 )w;
where
1 = mpx + nA1 ( 0 )x + px + 0 ;
2 = mC1 ( 0 )y + nqy + qy +
0 :
Since deg(1 )6 deg() + max{deg(p) − 1; deg( 0 )} and deg(2 )6 deg() + max{deg(q) − 1;
deg(
0 )}, the conclusions follow easily by induction on m and n.
However, in general, mn need not satisfy the di erential equation (3.1) and we cannot say
anything on orthogonality of { mn }, unless w(x; y) is an orthogonalizing weight function for an
OPS satisfying the di erential equation (3.1).
Hence we now formalize Lemma 4.1 by using, instead of a symmetry factor w(x), a moment
functional solution of the moment equations M1 [] = M2 [] = 0:
Theorem 4.2. Assume that the di erential equation (3.1) has a WOPS {n }∞
n=0 as solutions and
let be the canonical moment functional of {n }∞
.
Assume
that
satis es
n=0
px = 0
and
qy =
0 :
(4.8)
(i) If the condition (4.5) holds; then for any integers m and n ¿ 0
@xm @yn (pm qn ) =
where
h; xk yl
mn (x; y )
(4.9)
mn ;
is a polynomial of degree 6 2(m + n); and
mn (x; y )i = 0;
0 6 k + l 6 m + n and (k; l) 6= (m; n):
(ii) If the conditions (4.5) and (4.7) hold; then deg(
mn ) 6 m
+ n.
(4.10)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
(iii) If is quasi-de nite and the conditions (4.5) and (4.7) hold; then { n }∞
n=0 ; where n = {
n−1;1 ; : : : ; 0n }; is a WOPS relative to and satis es the di erential equation (3.1).
n0 ;
Before proving Theorem 4.2, we rst note that under the assumptions as in Theorem 4.2,
satis es the moment equations M1 [] = M2 [] = 0 by Proposition 3.4, which, however, do not imply
(4.8) in general.
Proof of Theorem 4.2. The proof of (i) and (ii) except (4.10) is essentially the same as that of
Lemma 4.1. Now,
h; xk y‘
mn (x; y )i = h mn ; x
k
y‘ i = (−1)m+n hpm qn ; @xm @yn (xk y‘ )i
from which (4.10) follows. Finally, we further assume that is quasi-de nite. We rst claim that
deg( mn ) = m + n and h; xm yn mn i =
6 0 for m and n ¿ 0. Assume that deg( mn ) 6 m + n − 1 for some
m + n ¿ 1: Then mn ≡ 0 since h; mn i = 0 for any (x; y) in Pm+n−1 by (4.10) and is quaside nite. Hence @xm @yn (pm qn ) = 0 so that either p(x; y) ≡ 0 or q(x; y) ≡ 0, which is a contradiction.
Hence deg( mn ) = m + n and h; xm yn mn i =
6 0 for all m and n ¿ 0 by (4.10).
We now claim that for each n ¿ 0; { n−j; j }nj=0 are linearly independent modulo Pn−1 P
so that
n
{ n }∞
is
a
PS.
For
any
integer
n
¿
1,
let
C
;
C
;
:
:
:
;
C
be
constants
such
that
(
x;
y
)
=
0
1
n
n=0
j=0 Cj
n−j; j is of degree 6 n− 1. Then (x; y ) ≡ 0 since h; (x; y )(x; y )i = 0 for any ∈ Pn−1 by (4.10).
Then
h; (x; y)xn−k yk i = Ck h;
n−k; k x
n−k
yk i = 0;
06k 6n
so that C0 = C1 = · · · = Cn = 0 by the rst claim. Hence { n−j; j }nj=0 are linearly independent modulo
∞
Pn−1 and so { n }∞
n=0 is a WOPS relative to by (4.10). Finally, in order to see that { n }n=0 satisfy
∞
∞
the di erential equation (3.1), we let {Pn }n=0 and {Qn }0 be the normalizations of {n }∞
n=0 and
∞
∞
∞
{ n }n=0 respectively. Then {Pn }n=0 and {Qn }0 are monic WOPS’s relative to so that Pn = Qn ,
∞
∞
n ¿ 0, by Proposition 2.2. Since {n }∞
n=0 satisfy the di erential equation (3.1), {Pn }n=0 and {Qn }n=0
also satisfy the di erential equation (3.1).
Combining Lemma 4.1 and Theorem 4.2, we now have:
Theorem 4.3. Assume that the di erential equation (3.1) has an OPS {n }∞
n=0 relative to as
solutions and let w(x; y) be a symmetry factor of L[·]: If the conditions (4.5), (4.7), and (4.8)
hold; then the PS { mn } de ned by (4.6) is a WOPS relative to and satis es the di erential
equation (3.1).
However, h; n nT i, with { n }∞
n=0 as in Theorem 4.3, is nonsingular but need not be diagonal,
that is, { n }∞
is
a
WOPS
but
need
not be an OPS in general (see Example 4.5 below).
n=0
We may call (4.6) and (4.9) Rodrigues type formulas for orthogonal polynomial solutions of the
di erential equations (3.1).
Example 4.4. Assume that Ay = B = Cx = 0 so that the di erential equation (3.1) is of the form
L[u] = A(x)uxx + C (y)uyy + D(x)ux + E (y)uy = n u;
(4.11)
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Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
where A(x) = d1 x + f1 ; C (y) = e3 y + f3 ; D(x) = gx + h1 ; E (y) = gy + h2 and g 6= 0: Then it is shown
in [6] that
• the di erential equation (4.11) has a unique monic PS {Pn }∞
n=0 as solutions;
• Pn0 (x; y) = Pn0 (x); P0n (x; y) = P0n (y), and Pmn (x; y) = Pm0 (x)P0n (y) for m and n ¿ 0;
• {Pn }∞
n=0 is a WOPS;
• Pn0 (x) and P0n (y); n ¿ 0, satisfy
A(x)Pn0′′ (x) + D(x)Pn0′ (x) = n Pn0 (x)
(4.12)
C (y)P0n′′ (y) + E (y)P0n′ (y) = n P0n (y):
(4.13)
and
In decomposition (4.3), we take A2 = C2 = −1 and B1 = 0. Then
A1 = −A;
p = A;
C1 = −C;
0 = −1;
0 = D − A ′ ;
0 = E − C ′ ;
q=C
so that the conditions (4.5) and (4.7) hold. On the other hand, the canonical moment func(x)
tional of {Pn }∞
⊗ (y) ; where (x) and (y) are the canonical moment funcn=0 is equal to =
∞
∞
tionals of {Pn0 (x)}0 and {P0n (y)}n=0 , respectively. Since A(x) (d = d x) (x) = (D(x) − A′ (x))(x) and
C (y) (d = d y) (y) = (E (y) − C ′ (y))(y) ; = (x) ⊗ (y) satis es the condition (4.8). Hence, by Theorem 4.2,
@xm @yn (A(x)m C (y)n ) =
where {
mn }
mn ;
m; n ¿ 0;
(4.14)
is a WOPS relative to . In fact, we have
mn (x; y ) = g
m+n
Pm0 (x)P0n (y);
m; n ¿ 0;
so that the Rodrigues type formula (4.14) is nothing but the tensor product of one dimensional
∞
Rodrigues formulas for {Pn0 (x)}∞
n=0 and {P0n (y )}n=0 (see [1, 2, 10] for Rodrigues formula for classical
orthogonal polynomials of one variable).
We may, of course, replace by a symmetry factor w(x; y) = w1 (x)w2 (y) of the di erential
equation (4.11), where w1 (x) and w2 (y) are symmetry factors of the di erential equations (4.12)
and (4.13), respectively.
Example 4.5. Consider the di erential equation for circle polynomials:
L[u] = (x2 − 1)uxx + 2xyuxy + (y2 − 1)uyy + gxux + gyuy = n u:
(4.15)
In decomposition (4.3), we take A1 = C1 = 1 so that B1 = B = xy and
= p = q = x 2 + y 2 − 1;
= 0 = (g − 3)x;
=
0 = (g − 3)y:
Then by Lemma 4.1,
1 m n 2
@ @ [(x + y2 − 1)m+n w] =
w x y
mn (x; y );
m; n ¿ 0
is a polynomial of degree 6 m + n, where w(x; y) = (x2 + y2 − 1)
di erential equation (4.15). In general, deg( mn ) 6= m + n.
g−3
2
is a symmetry factor of the
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
For example, we have
00 (x; y ) = 1
10 (x; y ) = (g
− 1)x;
20 (x; y ) = (g
+ 1)(gx2 + y2 − 1);
02 (x; y ) = (g
and
01 (x; y ) = (g
2
251
− 1)y;
11 (x; y ) = (g
+ 1)(g − 1)xy;
2
− 1)(x + gy − 1):
However, if g 6= 1; 0; −1; : : :, then the di erential equation (4.15) has an OPS {n }∞
n=0 (called the
∞
circle polynomials) as solutions. Then the canonical moment functional of {n }n=0 satis es the
condition (4.8) so that by Theorem 4.3, { mn } is a WOPS. But, even in this case, { mn } is not an
OPS. For if we let be the canonical moment functional of { mn }, then satis es L∗ [] = 0 (cf.
Lemma 3.2), that is,
(m + n)(m + n − 1 + g)mn − m(m − 1)m−2; n − n(n − 1)m; n−2 = 0;
m; n ¿ 0
so that
00 = 1;
10 = 01 = 11 = 30 = 21 = 12 = 03 = 31 = 13 = 0;
3
1
1
20 = 02 =
;
40 = 04 =
;
22 =
:
g+1
(g + 1)(g + 3)
(g + 1)(g + 3)
Hence, we have
H2 = h; 2 2T i =
2g
(g + 1)(g − 1)
0
g+3
2
0
2
g−1
0
0 ;
2g
which is nonsingular but not diagonal.
Example 4.6. Consider the di erential equation for triangle polynomials:
L[u] = (x2 − x)uxx + 2xyuxy + (y2 − y)uyy + [(a + b + c + 3)x − (1 + a)]ux
+ [(a + b + c + 3)y − (1 + b)]uy = n u:
(4.16)
In decomposition (4.3), we take A1 = x; C1 = y so that
0 = x + y − 1;
p = x(x + y − 1);
0 = (a + c)x + ay − a;
0 = bx + (b + c)y − b;
q = y(x + y − 1):
Then by Lemma 4.1,
1 m n m n
@ @ [x y (x + y − 1)m+n w] =
w x y
mn (x; y );
m; n ¿ 0
(4.17)
is a polynomial of degree 6 m + n, where w(x; y) = xa yb (x + y − 1)c is a symmetry factor of
the di erential equation (4.16). We now assume that a; b; c ¿ −1 so that the di erential equation
252
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
(4.16) has an OPS {n }∞
n=0 (called the triangle polynomials) as solutions and let be the canonical moment functional of {n }∞
n=0 . Then satis es the moment equations M1 [] = M2 [] = 0 or
equivalently
xy(x + y − 1)x = y((a + c)x + ay − a);
xy(x + y − 1)y = x(bx + (b + c)y − b):
(4.18)
Let
u := x(x + y − 1)x − [(a + c)x + ay − a] = px − 0 ;
v := y(x + y − 1)y − [bx + (b + c)y − b] = qy −
0 :
Then, (4.18) becomes
yu = xv = 0:
(4.19)
On the other hand, we can see by using M1 [] = M2 [] = 0 that
u = −v = xyx − xyy + (bx − ay):
(4.20)
Hence, we have from (4.19) and (4.20)
hu; xm yn i = hv; xm yn i = 0;
m + n ¿ 1:
(4.21)
We also have from (4.20)
hu; 1i = −hv; 1i = hxyx − xyy + (bx − ay); 1i = (b + 1)10 − (a + 1)01
and from (3.3)
h; Di = (a + b + c + 3)10 − (1 + a) = 0;
h; Ei = (a + b + c + 3)01 − (1 + b) = 0
so that hu; 1i = hv; 1i = 0. Therefore, together with (4.21), we have u = v = 0, that is, satis es the
condition (4.8). Hence, by Theorem 4.3, { mn } given by (4.17) is a WOPS relative to and satis es
the di erential equation (4.16).
Rodrigues type formulas for circle and triangle polynomials have been known before (see
[3, chapter 12] and [4, Section 4]).
We nally give a negative example for which Rodrigues type formula (4.6) does not hold.
Example 4.7. Krall and She er [5] showed that the di erential equation
L[u] = 3yuxx + 2uxy − xux − yuy = −nu
(4.22)
has an OPS as solutions. In decomposition (4.3), we must choose A1 = 1 and C1 = 1 in order for
the condition (4.5) to be satis ed. Then
= 0 = 1;
= 0 = −y;
=
0 = 3y2 − 3x;
p=q=1
Y.J. Kim et al. / Journal of Computational and Applied Mathematics 99 (1998) 239–253
253
so that by Lemma 4.1,
1 m nw
@ @ =
w x y
mn ;
m; n ¿ 0
is a polynomial of degree 6 2(m + n), where w(x; y) = exp(y3 − xy) is a symmetry factor of the
di erential equation (4.22). For example,
10 (x; y ) = −y
and
01 (x; y ) = 3y
2
−x
and 10 satis es the di erential equation (4.22) but
(4.22).
01
does not satisfy the di erential equation
Acknowledgements
This work is partially supported by KOSEF(95-0701-02-01-3), GARC, and the Korean Ministry
of Education (BSRI 97-1420). Authors thank referees for giving many valuable suggestions and
directing attention to Refs. [3, 4].
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