15-2 A fundamental solution for the Korn–Lichtenstein equations 323
While (15.47), (15.50), and (15.51) represent the polynomial solution of (15.42), namely for x(l, b),
a corresponding solution for (15.43) could be found as soon as we replace x and y, namely for the
polynomial solution y(l, b). Let us write down the n − 1 constraints for n + 1 polynomials given by
the zero identiy of the sum of the three terms of (15.47) (sx ll , first term), (15.50) (r b x b , second term),
and (15.51) (rx bb , third term).
Corollary 15.1 (Laplace–Beltrami equations solved in the function space of bivariate polynomials).
If a polynomial (15.29)–(15.32) of degree n fulfills the Laplace–Beltrami equations (15.42) and (15.43),
then there are n − 1 coefficient constraints, namely
n = 2 :
2s 0 x 20 + 2r 0 x 02 + r 1 x 01 = 0 ,
(15.52)
2s 0 y 20 + 2r 0 y 02 + r 1 y 01 = 0 ;
(15.53)
n = 3 :
6s 0 x 30 + 2r 0 x 12 + r 1 x 11 = 0 ,
(15.54)
6s 0 y 30 + 2r 0 y 12 + r 1 y 11 = 0 ,
(15.55)
s 0 x 21 + s 1 x 20 + 3r 0 x 03 + 2r 1 x 02 + r 2 x 01 = 0 ,
(15.56)
s 0 y 21 + s 1 y 20 + 3r 0 y 03 + 2r 1 y 02 + r 2 y 01 = 0 ;
(15.57)
n = 4 :
12s 0 x 40 + 2r 0 x 22 + r 1 x 21 = 0 ,
(15.58)
12s 0 y 40 + 2r 0 y 22 + r 1 y 21 = 0 ,
(15.59)
3s 0 x 31 + 3s 1 x 30 + 3r 0 x 13 + 2r 1 x 12 + r 2 x 11 = 0 ,
(15.60)
3s 0 y 31 + 3s 1 y 30 + 3r 0 y 13 + 2r 1 y 12 + r 2 y 11 = 0 ,
(15.61)
2s 0 x 22 + 2s 1 x 21 + 2s 2 x 20 + 12r 0 x 04 + 9r 1 x 03 + 6r 2 x 02 + 3r 3 x 01 = 0 ,
(15.62)
2s 0 y 22 + 2s 1 y 21 + 2s 2 y 20 + 12r 0 y 04 + 9r 1 y 03 + 6r 2 y 02 + 3r 3 y 01 = 0 ;
(15.63)
and in general
sx ll + (rx b ) b =
∞
n=2
n−2
i=0
i
j=0
(j + 1)[(i − j + 1)r j+1 x n−i−2,i−j+1 +
+(j + 2)r i−j x n−i−2,j+2 ] + (n − i)(n − i − 1)s i x n−i,i−j
l
n−i−2 b
i = 0 ,
sy ll + (ry b ) b =
∞
n=2
n−2
i=0
i
j=0
(j + 1)[(i − j + 1)r j+1 y n−i−2,i−j+1 +
+(j + 2)r i−j y n−i−2,j+2 ] + (n − i)(n − i − 1)s i y n−i,i−j
l
n−i−2 b
i = 0 .
(15.64)
End of Corollary.
While (15.47), (15.50), and (15.51) represent the polynomial solution of (15.42), namely for x(l, b),
a corresponding solution for (15.43) could be found as soon as we replace x and y, namely for the
polynomial solution y(l, b). Let us write down the n − 1 constraints for n + 1 polynomials given by
the zero identiy of the sum of the three terms of (15.47) (sx ll , first term), (15.50) (r b x b , second term),
and (15.51) (rx bb , third term).
Corollary 15.1 (Laplace–Beltrami equations solved in the function space of bivariate polynomials).
If a polynomial (15.29)–(15.32) of degree n fulfills the Laplace–Beltrami equations (15.42) and (15.43),
then there are n − 1 coefficient constraints, namely
n = 2 :
2s 0 x 20 + 2r 0 x 02 + r 1 x 01 = 0 ,
(15.52)
2s 0 y 20 + 2r 0 y 02 + r 1 y 01 = 0 ;
(15.53)
n = 3 :
6s 0 x 30 + 2r 0 x 12 + r 1 x 11 = 0 ,
(15.54)
6s 0 y 30 + 2r 0 y 12 + r 1 y 11 = 0 ,
(15.55)
s 0 x 21 + s 1 x 20 + 3r 0 x 03 + 2r 1 x 02 + r 2 x 01 = 0 ,
(15.56)
s 0 y 21 + s 1 y 20 + 3r 0 y 03 + 2r 1 y 02 + r 2 y 01 = 0 ;
(15.57)
n = 4 :
12s 0 x 40 + 2r 0 x 22 + r 1 x 21 = 0 ,
(15.58)
12s 0 y 40 + 2r 0 y 22 + r 1 y 21 = 0 ,
(15.59)
3s 0 x 31 + 3s 1 x 30 + 3r 0 x 13 + 2r 1 x 12 + r 2 x 11 = 0 ,
(15.60)
3s 0 y 31 + 3s 1 y 30 + 3r 0 y 13 + 2r 1 y 12 + r 2 y 11 = 0 ,
(15.61)
2s 0 x 22 + 2s 1 x 21 + 2s 2 x 20 + 12r 0 x 04 + 9r 1 x 03 + 6r 2 x 02 + 3r 3 x 01 = 0 ,
(15.62)
2s 0 y 22 + 2s 1 y 21 + 2s 2 y 20 + 12r 0 y 04 + 9r 1 y 03 + 6r 2 y 02 + 3r 3 y 01 = 0 ;
(15.63)
and in general
sx ll + (rx b ) b =
∞
n=2
n−2
i=0
i
j=0
(j + 1)[(i − j + 1)r j+1 x n−i−2,i−j+1 +
+(j + 2)r i−j x n−i−2,j+2 ] + (n − i)(n − i − 1)s i x n−i,i−j
l
n−i−2 b
i = 0 ,
sy ll + (ry b ) b =
∞
n=2
n−2
i=0
i
j=0
(j + 1)[(i − j + 1)r j+1 y n−i−2,i−j+1 +
+(j + 2)r i−j y n−i−2,j+2 ] + (n − i)(n − i − 1)s i y n−i,i−j
l
n−i−2 b
i = 0 .
(15.64)
End of Corollary.