Классические методы математической физики. Алексеев Г.В. - 52 стр.

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[−1, 1] λ = λ
n
d
m
P
n
/dx
m
P
n
P (x) ≡ P
m
n
(x)
λ = λ
n
= n(n + 1)
P
m
n
(x) = (1 − x
2
)
m/2
d
m
P
n
(x)
dx
m
.
L
2
(−1, 1)
m m − 1
d
dx
(1 − x
2
)
m
d
m
P
n
(x)
dx
m
+[λ − m(m − 1)] (1−x
2
)
m−1
d
m−1
P
n
(x)
dx
m−1
= 0.
L
m
n,k
=
1
Z
−1
P
m
n
(x)P
m
k
(x)dx =
1
Z
−1
(1 − x
2
)
m
d
m
P
n
dx
m
d
m
P
k
dx
m
dx.
L
m
n,k
=
(1 − x
2
)
m
d
m
P
n
dx
m
d
m−1
P
k
dx
m−1
1
−1
−
1
Z
−1
d
m−1
P
k
dx
m−1
d
dx
(1 − x
2
)
m
d
m
P
n
dx
m
dx =
= [n(n + 1) − m(m − 1)]
1
Z
−1
(1 − x
2
)
m−1
d
m−1
P
k
dx
m−1
d
m−1
P
n
dx
m−1
dx =
= [n(n + 1) − m(m − 1)]L
m−1
n,k
= (n + m)(n − m + 1)L
m−1
n,k
.
L
m
n,k
= (n + m)(n − m + 1)(n + m − 1)(n −m + 2)L
m−2
n,k
= ... =
= (n + m)(n + m − 1)...(n + 1)n(n − 1)...(n − m + 1)L
0
n,k
=
=
(n + m)!(n − m)!
n!(n − m + 1)!
L
0
n,k
=
(n + m)!
(n − m)!
L
0
n,k
.
L
0
n,k
≡
1
Z
−1
P
n
(x)P
k
(x)dx =
0, k 6= n,
2
2n+1
, k = n.
(4.29), èìååò íåòðèâèàëüíûå ãëàäêèå íà [−1, 1] ðåøåíèÿ òîëüêî ïðè λ = λn ,
ïðè÷åì ýòèìè ðåøåíèÿìè ÿâëÿþòñÿ ïðîèçâîäíûå dm Pn /dxm îò ïîëèíîìîâ
Ëåæàíäðà Pn . Âåðíóâøèñü ê èñõîäíîìó óðàâíåíèþ (4.23), ïðèõîäèì ê âû-
âîäó î òîì, ÷òî ïðèñîåäèíåííûå óíêöèè Ëåæàíäðà P (x) ≡ Pnm (x) ñóùå-
ñòâóþò òîëüêî ïðè λ = λn = n(n + 1) è îïðåäåëÿþòñÿ îðìóëàìè
                                                                  m
                                                      2 m/2 d     Pn (x)
                                Pnm (x)   = (1 − x )                     .                 (4.31)
                                                                  dxm
   Âû÷èñëèì íîðìû â L2 (−1, 1) ïðèñîåäèíåííûõ óíêöèé Ëåæàíäðà è îä-
íîâðåìåííî äîêàæåì èõ îðòîãîíàëüíîñòü. Ñ ýòîé öåëüþ çàìåíèì â óðàâíå-
íèè (4.30) m íà m − 1. Ïîëó÷èì óðàâíåíèå
                m                                      m−1
                       
  d        2 m d Pn (x)                        2 m−1 d     Pn (x)
     (1 − x )             +[λ − m(m − 1)] (1−x  )                 = 0. (4.32)
 dx              dxm                                    dxm−1
Ïîëîæèì
                       Z1                               Z1                    m
                                                                        Pn dmPk
                                                                       2 md
            Lm
             n,k   =        Pnm (x)Pkm(x)dx       =          (1 − x )           dx.
                                                                      dxm dxm
                       −1                             −1

Èíòåãðèðóÿ îäèí ðàç ïî ÷àñòÿì, èìååì ñ ó÷åòîì (4.32), ÷òî

                    m     m−1                1        Z1
                                                             dm−1Pk d             m
                                                                                   
               2 m d Pn d     Pk                                             2 m d Pn
Lm
 n,k   = (1 − x )                                 −                    (1 − x )         dx =
                    dxm dxm−1                −1              dxm−1 dx             dxm
                                                      −1

                                             Z1                         m−1
                                                             2 m−1 d      Pk dm−1Pn
         = [n(n + 1) − m(m − 1)]                  (1 − x )                          dx =
                                                                       dxm−1 dxm−1
                                            −1
                                  m−1                      m−1
          = [n(n + 1) − m(m − 1)]Ln,k = (n + m)(n − m + 1)Ln,k .
Ïîñëåäîâàòåëüíî ïðèìåíÿÿ ïîñëåäíþþ ðåêóððåíòíóþ îðìóëó, èìååì
       Lm                                              m−2
        n,k = (n + m)(n − m + 1)(n + m − 1)(n − m + 2)Ln,k = ... =

         = (n + m)(n + m − 1)...(n + 1)n(n − 1)...(n − m + 1)L0n,k =
                   (n + m)!(n − m)! 0     (n + m)! 0
                       =           Ln,k =         L .
                    n!(n − m + 1)!        (n − m)! n,k
Íî â ñèëó ñâîéñòâ ïîëèíîìîâ Ëåæàíäðà
                                Z1                            
                                                                  0,       k 6= n,
                   L0n,k    ≡        Pn (x)Pk (x)dx =               2
                                                                  2n+1
                                                                       ,   k = n.
                                −1

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