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Proof.
Once the fundamental solution of the d'Alembert-Euler equations is based on the function space
of homogeneous polynomials
P r ( q,p )=
α + β = r
C αβ q α p β (0
α
r ) ,
(D.17)
the vectorial Laplace-Beltrami equation has to be fulfilled. ( r
1) constraints are given for ( r +1)
coecients such that for any r
2 two linear independent harmonic polynomials exist which we
are going to construct.
P r ( q,p )= c r, 0 q r + c r− 1 , 1 q r− 1 p + c r− 2 , 2 q r− 2 p 2 + c r− 3 , 3 q r− 3 p 3 + c r− 4 , 4 q r− 4 p 4 + ···
... + c 3 ,r− 3 q 3 p r− 3 + c 2 ,r− 2 q 2 p r− 2 + c 1 ,r− 1 qp r− 1 + c 0 ,r p r ,
(D.18)
ΔP r ( q,p )=( 2
∂q 2 + 2
∂p 2 ) P r ( q,p )=
1) c r, 0 q r− 2 +( r
3) c r− 2 , 2 q r− 4 p 2 +( r
5) c r− 4 , 4 q r− 6 p 4 +
= r ( r
2)( r
4)( r
···
... +2 c r− 2 , 2 q r− 2 +4 · 3 c r− 4 , 4 q r− 4 p 4 + ···
(D.19)
... +( r − 1)( r − 2) c r− 1 , 1 q r− 3 p +( r − 3)( r − 4) c r− 3 , 3 q r− 5 p 3 + ···
... +3
2 c r− 3 , 3 q r− 3 p +
·
···
.
Obviously, the recurrence relation ( D.20 ) connects the coecients of even second index with each
other, similarly the coecients of odd second index according to the following set of coecient
pairs: c r, 0
|
c r− 2 , 2 ,c r− 1 , 1
|
c r− 3 , 3 ,c r− 2 , 2
|
c r− 4 , 4 ,c r− 3 , 3
|
c r− 5 , 5 etc.
c k,r−k k ( k − 1) + c k− 2 ,r−k +2 ( r − k +2)( r − k +1)=0
(D.20)
c k,r−k k ( k − 1)
c k− 2 ,r−k +2 =
k +1) .
( r
k +2)( r
Those harmonic polynomials with coecients of even second index are denoted by P r, 1 , while
alternatively those with coecients of odd second index are denoted by P r, 2 . Once we chose
c r, 0 = 1, the recurrence relation leads to ( D.21 )or( D.22 ). In contrast, once we choose c r− 1 , 1 = r
we are led to ( D.23 )or( D.24 ).
r ( r 1)
2
q r− 2 p 2 + ( r 3)( r 2)( r 1)
4
P r, 1 ( q,p )= q r
q r− 4 p 4 + ··· ,
(D.21)
3
2
1) s r
q r− 2 s p 2 s ,
[ 2
]
P r, 1 ( q,p )=
(
(D.22)
2 s
s=0
 
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