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Legendre Functions
SOLO HERMELIN
Updated: 20.02.13
1
http://www.solohermelin.com
Table of Content
SOLO
2
Legendre Functions
Introduction
Legendre Polynomials History
Second Order Linear Ordinary Differential Equation (ODE)
Laplace’s Homogeneous Differential Equation
Legendre Polynomials
The Generating Function of Legendre Polynomials
Rodrigues' Formula
Series Solutions – Frobenius’ Method
Recursive Relations for Legendre Polynomial Computation
Orthogonality of Legendre Polynomials
Expansion of Functions, Legendre Series
Schlaefli Integral
Laplace’s Integral Representation
Neumann Integral
Table of Content (continue)
SOLO
3
Legendre Functions
Associate Lagrange Differential Equation
Laplace Differential Equation in Spherical Coordinates
Analysis of Associate Lagrange Differential Equation
Associate Lagrange Differential Equation (2nd
Way)
Generating Function for Pn
m
(t)
Recurrence Relations for Pn
m
(t)
Orthogonality of Associated Legendre Functions
Recurrence Relations for Θn
m
Functions
Spherical Harmonics
References
Boundary Value Problems and Sturm–Liouville Theory
SOLO
4
Legendre Polynomials
Adrien-Marie Legendre
(1752 –1833(
In mathematics, Legendre functions are solutions to Legendre's
differential equation:
( ) ( ) ( ) ( ) 011 2
=++





− xPnnxP
xd
d
x
xd
d
nn
They are named after Adrien-Marie Legendre. This
ordinary differential equation is frequently encountered in
physics and other technical fields. In particular, it occurs
when solving Laplace's equation (and related partial
differential equations) in spherical coordinates.
The Legendre polynomials were first introduced in 1785 by Adrien-Marie
Legendre, in “Recherches sur l’attraction des sphéroides homogènes”, as the
coefficients in the expansion of the Newtonian potential
( )∑
∞
=






=






−





+
=
−+
=
−
0
222
cos
'1
cos
'
2
'
1
11
cos'2'
1
'
1
n
n
n
P
r
r
r
r
r
r
rrrrrrrr
γ
γ
γ

Return to Table of Content
Second Order Linear Ordinary Differential Equation (ODE)
Consider a Second Order Linear Ordinary Differential Equation (ODE) define by
the Operator
SOLO
( ) ( ) ( ) ( ) ( )xuxp
xd
ud
xp
xd
ud
xpxu 212
2
0: ++=L
defined in the interval a ≤ x ≤ b, with the coefficients p0 (x), p1 (x), p2 (x) real in
this region and with the first 2 – i derivatives of pi (x) continuous . Also we
require that p0 (x) is nonzero in the interval.
Define the Quadratic Form of the Operator L as:
( ) ( ) ( ) dxuup
xd
ud
p
xd
ud
pdxxuxuxu
b
a
b
a
∫∫ 





++== 212
2
0: LLu,
( ) ( ) ( )
( ) ( ) ( ) ( )∫ ∫∫
∫ ∫∫
∫∫∫
+−++





−





=
+−+−





=
++
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
xdupuxdup
xd
d
uupudxup
xd
d
uup
xd
d
u
xd
ud
up
xdupuxdup
xd
d
uupudx
xd
ud
up
xd
d
xd
ud
up
xdupxdu
xd
ud
pxdu
xd
ud
p
21102
2
00
21100
2
212
2
0
5
SOLO
( ) ( ) ( ) dxuup
xd
ud
p
xd
ud
pdxxuxuxu
b
a
b
a
∫∫ 





++== 212
2
0: LLu,
( ) ( )
( ) ( )
b
a
b
a
b
a
b
a
b
a
xu
xd
pd
pxu
upu
xd
ud
upu
xd
pd
u
xd
ud
upupuup
xd
d
u
xd
ud
up












−=






+−−=+





−





0
1
10
0
0100
Therefore:
( ) ( ) ( ) ( )∫ 





+−+











−=
b
a
b
a
dxupup
xd
d
up
xd
d
uxu
xd
pd
pxu 2102
2
0
1
Define the Adjoint Operator:
( ) ( ) ( ) upup
xd
d
up
xd
d
xu 2102
2
: +−=L
6
Second Order Linear Ordinary Differential Equation (ODE)
SOLO
( ) dx
u
up
xd
ud
p
xd
ud
puxu
b
a
∫ 





++=
  
L
Lu, 212
2
0
The two Integrals are equal if:
( ) ( ) ( ) ( )∫ 





+−+











−=
b
a
b
a
dxupup
xd
d
up
xd
d
uxu
xd
pd
pxu
  
uL
2102
2
0
1
or:
( ) ( ) ( ) ( ) ( ) ( )xuxq
xd
xud
xp
xd
d
xuxu +





== LL
( ) ( )
02 1
01
2
0
2
2102
2
212
2
0
=





−+





−=






+−−





++
xd
ud
p
xd
pd
uu
xd
pd
xd
pd
u
upup
xd
d
up
xd
d
uup
xd
ud
p
xd
ud
pu
( ) ( )
xd
xpd
xp 0
1 =
If this condition is satisfied, then the terms at the boundary x = a and x = b also
vanish, and we have by defining p (x): = p0 (x) and q (x) := p2 (x)
7
Second Order Linear Ordinary Differential Equation (ODE)
SOLO
( ) ( ) ( ) ( ) ( ) ( )xuxq
xd
xud
xp
xd
d
xuxu +





== LL
This Operator is called Self-Adjoint
( )
( )
( )
( )
( )
( )
( )
( ) ( ) ( ) ( )





++





=





∫∫ xuxp
xd
ud
xp
xd
ud
xptd
tp
tp
xp
xutd
tp
tp
xp xx
212
2
0
0
1
00
1
0
expexp
1
L
1
( )
( )
( )
( )
( )
( )
( )
( )
( )xutd
tp
tp
xp
xp
xd
ud
td
tp
tp
xd
d
xq
x
xp
x
    






+


















= ∫∫ 0
1
0
2
0
1
expexp This is clearly Self-Adjoint.
We can see that p0 (x) is in the denominator. This is the reason that we requested the
p0 (x) to be nonzero in the interval a ≤ x ≤ b. p0 (x1) = 0 means that the Differential
Equation is not Second Order at that point.
We can always transform the Non-Self-Adjoint Operator to a Self-Adjoint one
by multipling L by
( )
( )
( ) 





∫x
td
tp
tp
xp 0
1
0
exp
1
8
Second Order Linear Ordinary Differential Equation (ODE)
Consider a Second Order Linear Ordinary Differential Equation (ODE)
SOLO
( ) ( ) ( ) ( ) ( ) ( ) ( ) 2
2
210 :'':'0''':
xd
ud
u
xd
ud
uxuxpxuxpxuxpxu ===++=L
If we know one Solution u1 (x) we can find a second u2 (x).
Proof
If we have two Solutions u1 (x) and u2 (x), then
0'''
0'''
222120
121110
=++
=++
upupup
upupup
Multiply first equation by u2 (x) and second by u1 (x) and subtract:
( ) ( ) 0'''''' 1221112210 =−+− uuuupuuuup
The Wronskian W is defined as:
1221
21
21
''
''
: uuuu
uu
uu
W −==
0' 10 =+ WpWp
9
Theorem: “Method of Reduction of Order”
Second Order Linear Ordinary Differential Equation (ODE)
Consider a Second Order Linear Ordinary Differential Equation (ODE)
SOLO
( ) ( ) ( ) ( ) ( ) ( ) ( ) 2
2
210 :'':'0''':
xd
ud
u
xd
ud
uxuxpxuxpxuxpxu ===++=L
Theorem: “Method of Reduction of Order”
If we know one Solution u1 (x) we can find a second u2 (x).
Proof (continue -1)
0' 10 =+ WpWp xd
p
p
W
Wd
0
1
−= ( ) ( ) ( )
( ) 







−= ∫
x
x
d
p
p
xWxW
0 0
1
0 exp τ
τ
τ
From: ( ) ( ) ( ) ( ) ( )xuxuxuxuxW 1221 '' −=
q.e.d.
Therefore: ( ) ( ) ( )
( )∫=
x
x
d
u
W
xuxu
0
2
1
12 τ
τ
τ
( )
( )
( )
( )
( )
( )
( ) 





=−=
1
2
22
1
1
1
2
2
1
''
u
u
xd
d
xu
xu
xu
xu
xu
xu
xW
Divide by u1
2
(x)
x0 and W (x0) are given (or not).
10
Second Order Linear Ordinary Differential Equation (ODE)
If the Second Order Linear Ordinary Differential Equation (ODE) is in the
Self-Adjoint Mode:
SOLO
Then:
The Wronskian is given by
( ) ( ) ( ) ( ) ( ) ( ) 0=+





== xuxq
xd
xud
xp
xd
d
xuxu LL
( )
( ) ( ) ( ) ( ) ( ) 0
1
0
2
2
=++ xuxq
xd
xud
xd
xpd
xd
xud
xp
p
p 
( ) ( ) ( )
( )
( ) ( )
( )
( )xp
xp
xW
p
pd
xWd
p
p
xWxW
x
x
x
x
0
00
0
0
0
0
0
1
0
00
expexp =








−=








−= ∫∫ τ
τ
τ
11
Return to Table of Content
Second Order Linear Ordinary Differential Equation (ODE)
12
SOLO
The Laplace’s Homogeneous Differential Equation is:
We want to find the Potential Φ at the point F (field) due to all the sources (S) in the
volume V, including its boundaries .
n
i
iSS
1=
=
iS
nS

n
i
iSS
1=
=dV
dSn
→
1
V
Fr

Sr

F
0r
SF rrr

−= iS
nS
dV
dSn
→
1
V
Fr

Sr

F
0r SF rrr

−=
F inside V F on the boundary of V
Therefore is the vector from S to F.SF rrr

−=
→→→
++= zzyyxxr 111

→→→
++= zzyyxxr SSSS 111

→→→
++= zzyyxxr FFFF 111

Let define the operator
that acts only on the
source coordinate.
→→→
∂
∂
+
∂
∂
+
∂
∂
=∇ z
z
y
y
x
x SSS
S 111
Sr

Laplace’s Homogeneous Differential Equation
( ) 02
=Φ∇ r
Pierre-Simon,
marquis de Laplace
(1749 1827)
13
SOLO
Laplace’s Homogeneous Differential Equation
Laplace Differential Equation in Spherical Coordinates
( ) 02
=Φ∇ r
Spherical Coordinates:
θ
ϕθ
ϕθ
cos
sinsin
cossin
rz
ry
rx
=
=
=
A Solution in Spherical Coordinates is: ( ) 0
1
≠=Φ r
r
r
( ) 0
111
22
≠−=∇−=∇=Φ∇ r
r
r
r
r
rr
r

( ) ( ) 00
33111
35333
2
≠=−⋅=⋅∇−⋅





−∇=





⋅−∇=Φ∇⋅∇=Φ∇ r
r
rr
r
r
r
r
r
r
r
rr

r
r
r
z
r
y
r
x
zyx
zyx
r zyxzyx

=++=++





∂
∂
+
∂
∂
+
∂
∂
=∇ 111111 222
( ) 3111111 =++⋅





∂
∂
+
∂
∂
+
∂
∂
=⋅∇ zyx
zyx
r zyxzyx

2222
zyxr ++=
14
SOLO
Laplace’s Homogeneous Differential Equation
Laplace Differential Equation in Spherical Coordinates
Spherical Coordinates:
θ
ϕθ
ϕθ
cos
sinsin
cossin
rz
ry
rx
=
=
=
θcos=
∂
∂
z
r
2222
zyxr ++=
0
cos11
22
≠−=
∂
∂
−=





∂
∂
=
∂
Φ∂
r
rz
r
rrzz
θ
0
1cos3331
3
2
5
22
4332
2
≠
−
=
−
=
∂
∂
+−=





−
∂
∂
=
∂
Φ∂
r
rr
rz
z
r
r
z
rr
z
zz
θ
0
cos9cos159153
5
26
3
4
23
7
23
6
22
55
22
3
3
≠
−
=
−
−=
∂
∂−
−
∂
∂
−
=




 −
∂
∂
=
∂
Φ∂
r
r
r
r
rzz
z
r
r
rz
r
z
r
rz
r
rz
zz
θθ
We note that ∂n
Φ/∂zn
is a n-degree polynomial in cos θ divided by rn+1
.
( ) ( ) ( ) ( ) ( ) ( )zyx
zn
hzyx
z
hzyx
z
hzyxhzyx n
nn
n
,,
!
1
,,
!2
1
,,,,,, 2
2
2
∂
Φ∂−
++
∂
Φ∂
+
∂
Φ∂
−Φ=−Φ 
Using Taylor’s Series development we obtain
15
SOLO
Laplace’s Homogeneous Differential Equation
Laplace Differential Equation in Spherical Coordinates
Spherical Coordinates:
θ
ϕθ
ϕθ
cos
sinsin
cossin
rz
ry
rx
=
=
=
2222
zyxr ++=
( )
( )
( )






∂
∂−
++





∂
∂
+





∂
∂
−=
−++
=−Φ
rzn
h
rz
h
rz
h
rhzyx
hzyx n
nn
n 1
!
11
!2
1111
,, 2
2
2
222

( )
∑
∞
=






∂
∂−
=
+− 0
22
1
!
1
cos2
1
n
n
nn
n
rzn
h
hhrr θ
or:
Let define:
( ) ( )






∂
∂−
= +
rzn
rP n
nn
n
n
1
!
1
:cos 1
θ
( ) rhP
r
h
rhhrr n
n
n
≤





=
+−
∑
∞
=0
22
cos
1
cos2
1
θ
θ
Therefore:
16
SOLO
Laplace’s Homogeneous Differential Equation
Laplace Differential Equation in Spherical Coordinates
Spherical Coordinates:
θ
ϕθ
ϕθ
cos
sinsin
cossin
rz
ry
rx
=
=
=
2222
zyxr ++=
Let define w:=Pn (cos θ) and substitute w/rn+1
in the
Laplace Equations in Spherical Coordinates instead of Φ
Therefore, since w:=Pn (cos θ) (Partial Derivative becomes Total Derivative):
0
sin
1
sin
sin
11
0
12
2
22121
2
2
=





∂
∂
+











∂
∂
∂
∂
+











∂
∂
∂
∂
+++

nnn
r
w
rr
w
rr
w
r
r
rr φθθ
θ
θθ
0sin
sin
111
1
=





∂
∂
∂
∂
+


 +
−
∂
∂
+
θ
θ
θθ
w
rr
n
r
w nn
or:
( ) 0sin
sin
1
1 =





++
θ
θ
θθ d
wd
d
d
wnn
Substitute t=cos θ (dt = - sinθ dθ):

( ) ( ) 





−=








−−=










=





−−
td
wd
t
td
d
td
wd
t
td
d
d
td
td
wd
d
td
d
d
d
wd
d
d 2
sin
2
1sin
11
sin
1
sin
sin
1
sin
sin
1
2

 θ
θ
θθθ
θ
θθθ
θ
θθ
Finally we obtain: ( ) ( ) 1011 2
≤=++





− twnn
td
wd
t
td
d
Return to Table of Content
17
SOLO
This is the Legendre Differential Equation and Pn (t) the Legendre Polynomials
are one of the two solutions of the ODE.
( ) ( ) ( ) ( ) 1011 2
≤=++





− ttPnntP
td
d
t
td
d
nn
( )∑
∞
=






=






+−
0
2
cos
cos21
1
n
n
n
P
r
h
r
h
r
h
θ
θ
We found
( ) 1cos
cos21
1
0
2
≤=
+−
∑
∞
=
uPu
uu n
n
n
θ
θFor u:=h/r
The left-hand side is called “The Generating Function of Legendre
Polynomials”
Legendre Polynomials
The Generating Function of Legendre Polynomials
18
SOLO
Let use Taylor expansion
for the function:
( ) ( )
( )
( ) ( )
( ) ( )
( )  +++++= n
n
x
n
f
x
f
x
f
fxf
!
0
!2
0
1
0
0 2
21
( )
( ) 1,
21
1
0
2
≤=
−+
∑
∞
=
tutPu
tuu n
n
n
( ) ( ) ( ) 101 2
1
=−=
−
fxxf
( )
( ) ( ) ( )
( )
2
1
01
2
1 1
2
3
1
=−=
−
fxxf
( )
( ) ( ) ( )
( )
2
3
2
1
01
2
3
2
1 2
2
5
2
⋅=−⋅=
−
fxxf
( )
( ) ( ) ( )
( ) ( ) ( )
!2
!2
!2
!2
2
1231
2
12
2
3
2
1
01
2
12
2
3
2
1
2
2
12
n
n
n
nnn
fx
n
xf nn
n
n
n
n
n
=⋅
−⋅
=
−
⋅=−
−
⋅=
+
− 

Tacking x = 2 u t - u2
we obtain
( )
( )
( )
( ) 12
!2
!2
21
1
0
2
222
≤−=
−−
∑
∞
=
uutu
n
n
utu n
n
n
Let prove that Pn (t) are indeed Polynomials.
Legendre Polynomials
The Generating Function of Legendre Polynomials
19
SOLO
Using Taylor expansion we obtained:
( )
( ) 1,
21
1
0
2
≤=
−+
∑
∞
=
tutPu
tuu n
n
n
Take the binomial expansion of (2 u t - u2
)n
we obtain
( )
( )
( )
( ) 12
!2
!2
21
1
0
2
222
≤−=
−−
∑
∞
=
uutu
n
n
utu n
n
n
( )
( )
( )
( )
( )
( ) ( ) ( )
( )
( ) 12
!!!2
!2
12
!!
!
1
!2
!2
21
1
0 0
2
0 0
222
≤
−
−=
−
−=
−−
∑∑∑ ∑
∞
= =
+−
∞
= =
−
uut
knkn
n
ut
knk
n
n
n
utu n
n
k
knkn
n
k
n
n
k
kknk
n
Change Variables in the second sum from n+k to n
( )
( ) ( )
( ) ( )
( )
[ ]
( )


−
=





≤
−−
−
−=
−−
∑ ∑
∞
= =
−
−
evennifn
oddnifnn
uut
knknk
kn
utu n
n
k
nkn
kn
k
2/1
2/
2
12
!2!!2
!22
1
21
1
0
2/
0
2
222
Equating in the two power series the un
coefficients, we obtain
( ) ( ) ( )
( ) ( )
[ ]
∑=
−
≤
−−
−
−=
2/
0
2
1
!2!!2
!22
1
n
k
kn
n
k
n tt
knknk
kn
tP
Polynomial
of order n in t
Return to Rodrtgues Formula
Return to Frobenius Series
We can see that for n odd the polynomial Pn (t) has only odd powers of t and
for n even only even powers of t.
Legendre Polynomials
The Generating Function of Legendre Polynomials
20
SOLO
Let use Taylor expansion
for the function:
( ) ( )
( )
( ) ( )
( ) ( )
( )  +++++= n
n
x
n
f
x
f
x
f
fxf
!
0
!2
0
1
0
0 2
21
( )
( ) 1,
21
1
0
2
≤=
−+
∑
∞
=
tutPu
tuu n
n
n
Tacking x = u2
-2 u t we obtain
( ) ( ) ( ) 101 2
1
=+=
−
fxxf
( )
( ) ( ) ( )
( )
2
1
01
2
1 1
2
3
1
−=+−=
−
fxxf
( )
( ) ( ) ( )
( )
2
3
2
1
01
2
3
2
1 2
2
5
2
⋅=+⋅=
−
fxxf
( )
( ) ( ) ( ) ( )
( ) ( )
2
12
2
3
2
1
101
2
12
2
3
2
1
1 2
12 −
⋅−=+
−
⋅−=
+
− n
fx
n
xf
nn
n
nn

( )
( ) ( ) ( ) ( )
( )∑
∞
=
=




 +−
+




 −
+




 −
++=
−+−−−+−−=
−+
0
24
4
3
3
2
20
4232222
2
8
33035
2
35
2
13
1
2
128
35
2
16
5
2
8
3
2
1
1
21
1
n
n
n
tPu
tt
u
tt
u
t
utuu
tuutuutuutuu
ttuu


Legendre Polynomials
The Generating Function of Legendre Polynomials
SOLO
21
Legendre Polynomials
The first few Legendre polynomials are:
( )
( )
( )
( )
( )
( )
( )
( )
( )
( ) 256/63346530030900901093954618910
128/31546201801825740121559
128/35126069301201264358
16/353156934297
16/51053152316
8/1570635
8/330354
2/353
2/132
1
10
246810
3579
2468
357
246
35
24
3
2
−+++−
+−+−
+−+−
−+−
−+−
+−
+−
−
−
xxxxx
xxxxx
xxxx
xxxx
xxx
xxx
xx
xx
x
x
xPn n
22
SOLO
( ) 1,
21
1
0
2
≤=
+−
∑
∞
=
tutPu
utu n
n
n
Substitute u by – u in this equation:
( ) ( ) ( ) 11
21
1
00
2
≤−=−=
++
∑∑
∞
=
∞
=
utPutPu
utu n
n
n
n
n
nn
which results in the following identity: ( ) ( ) ( )tPtP n
n
n 1−=−
For t =1 we have
( ) 11
1
1
21
1
0
2
≤=
−
=
+−
∑
∞
=
uPu
uuu n
n
n
But 1
1
1
0
≤=
−
∑
∞
=
uu
u n
n
By equalizing the coefficients of un
in the two sums, we obtain:
( ) nPn ∀=11
and
( ) ( ) ( ) ( )n
n
n
n PP 1111 −=−=−
Legendre Polynomials
The Generating Function of Legendre Polynomials
23
SOLO
( ) 1,
21
1
0
2
≤=
+−
∑
∞
=
tutPu
utu n
n
n
For t = 0 we obtain:
Legendre Polynomials
( ) 10
1
1
0
2
≤=
+
∑
∞
=
uPu
u n
n
n
( ) ( ) ( ) ( ) ( ) ( ) ( ){ } ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )∑∑∑
∞
=
∞
=
∞
=
−
−=
⋅−⋅⋅−⋅⋅
−=
−⋅⋅
−=
+
−−−−
++
−−
+





−+=+=
+
0
22
2
0
1
2
0
2
22222/12
2
!2
!2
1
!2
222642
!2
12531
1
!2
12531
1
!
2/123/12/1
!2
3/12/1
2
1
11
1
1
n
n
n
n
n
nn
n
n
n
n
n
n
n
n
un
n
nn
n
un
n
un
t
n
n
ttu
u
  




Therefore we have:
( ) ( )
( )
( ) 10
!2
!2
1
1
1
00
22
2
2
≤=−=
+
∑∑
∞
=
∞
=
uPu
n
un
u n
n
n
n
n
n
n
By equating coefficients of un
on both sides we obtain:
( ) ( ) ( )
( )
( ) 00
!2
!2
10
12
222
=
−=
+n
n
n
n
P
n
n
P
Return to Table of Content
The Generating Function of Legendre Polynomials
Derivation of Legendre Polynomials via Rodrigues’ Formula
SOLO
24
Legendre Polynomials
Benjamin Olinde Rodrigues
(1794-1851)
In mathematics, Rodrigues' Formula (formerly called the Ivory–
Jacobi formula) is a formula for Legendre polynomials
independently introduced by Olinde Rodrigues (1816), Sir James
Ivory (1824) and Carl Gustav Jacobi (1827). The name
"Rodrigues’ formula" was introduced by Heine in 1878, after
Hermite pointed out in 1865 that Rodrigues was the first to
discover it, and is also used for generalizations to other orthogonal
polynomials. Askey (2005) describes the history of the Rodrigues’
Formula in detail.
Rodrigues stated his formula for Legendre polynomials Pn
Carl Gustav Jacob Jacobi
(1804 –1851)
( ) ( )[ ]n
n
n
nn x
xd
d
n
xP 1
!2
1 2
−=
SOLO
25
Legendre Polynomials
Olinde Rodrigues
(1794-1851)
Start from the function: ( ) .12
constkxky
n
=−=
( ) 12
12:'
−
−==
n
xxkn
xd
yd
y
( ) ( ) ( ) 22212
2
2
11412:''
−−
−−+−==
nn
xxnknxkn
xd
yd
y
Let compute:
( ) ( ) ( ) ( ) ( ) '12211412''1
12222
yxnynxxnknxknyx
nn
−+=−−+−=−
−
or: ( ) ( ) 02'12''12
=−−+− ynyxnyx
Let differentiate the last equation n times with respect to x:
( )[ ] ( ) ( ) ( ) ( )
( ) ( ) ''1''2''1
00''1
3
''1
2
''1
1
''1''1
2
2
1
1
2
3
3
0
2
3
3
2
2
2
2
2
1
1
222
y
xd
d
nny
xd
d
xny
xd
d
x
y
xd
d
x
xd
dn
y
xd
d
x
xd
dn
y
xd
d
x
xd
dn
y
xd
d
xyx
xd
d
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
−
−
−
−
−
−
−
−
−
−
−++−=
+−





+−





+−





+−=− 

( ) ( ) ( ) ( ) 





+−=











+−=− −
−
−
−
''12'
1
'12'12 1
1
1
1
y
xd
d
ny
xd
d
xny
xd
d
x
xd
dn
y
xd
d
xnyx
xd
d
n n
n
n
n
n
n
n
n
n
n
Derivation of Legendre Polynomials via Rodrigues’ Formula
SOLO
26
Legendre Polynomials
Olinde Rodrigues
(1794-1851)
Start from the function: ( ) .12
constkxky
n
=−=
( ) ( ) 02'12''12
=−−+− ynyxnyxDifferentiate n times with
respect to x:
( ) ( ) ''1''2''1 2
2
1
1
2
y
xd
d
nny
xd
d
xny
xd
d
x n
n
n
n
n
n
−
−
−
−
−++−
( ) 02''12 1
1
=−





+−+ −
−
y
xd
d
ny
xd
d
ny
xd
d
xn n
n
n
n
n
n
Define: a Polynomial( ) ( )[ ]n
n
n
n
n
x
xd
d
k
xd
yd
xw 1: 2
−==
( ) ( ) ( )[ ] 02'121'2''12
=−+−+−++− wnwnwxnwnnwxnwx
( ) ( ) ( ) ( )[ ] 02121'2''12
=−−+−+−++− wnnnnnwxnxxnwx
( ) ( ) 01'2''12
=+−+− wnnwxwx
This is Legendre’s Differential Equation. We proved that one of the solutions
are Polynomials. We can rewrite this equation in a Sturm-Liouville Form:
( ) ( ) 0112
=+−





− wnnw
xd
d
x
xd
d
Derivation of Legendre Polynomials via Rodrigues’ Formula
SOLO
27
Legendre Polynomials
Olinde Rodrigues
(1794-1851)
Let find k such that:
by using the fact that Pn (1) = 1
( ) ( )[ ]n
n
n
n
n
n x
xd
d
k
xd
yd
xP 12
−==
( ) ( )[ ] ( ) ( ) ( ) ( ) 





−+=








+−=−= ∑>0
22
1!2111
i
inn
v
n
u
n
n
n
n
n
n
n xxaxnkxx
xd
d
kxk
xd
d
xP

!2
1
n
k n
=
We recover the Rodrigues Formula:
( ) ( )[ ]n
n
n
nn x
xd
d
n
xP 1
!2
1 2
−=
Let use Leibnitz’s Rule (Binomial Expansion for the n Derivative
of a Product - with u:=(x-1)n
and v:=(x+1)n
):
( )
( )
( ) udvudvdnvddu
nn
vddunvdu
vdud
mnm
n
vud
nnnnn
n
m
mnmn
+++
−
++=
−
=⋅
−−−
=
−
∑
1221
0
!2
1
!!
!

We have:
( )  
( )
   1!2
!2
1
1
1!20
12
0
21
00
==







+++
−
++==
=
−−−
nkudvudvdnvddu
nn
vddunvdukxP n
xn
nnnnn
n
n

We can see from this Formula that Pn (x) is indeed a Polynomial of Order n in t.
Derivation of Legendre Polynomials via Rodrigues’ Formula
SOLO
28
Legendre Polynomials
Olinde Rodrigues
(1794-1851)
Let find an explicit expression for Pn (x) from Rodrigues’ Formula:
( ) ( )[ ]n
n
n
nn
n
n x
xd
d
nxd
yd
xP 1
!2
1 2
−==
Start with:
( ) ( )
( )∑=
−
−
−
=−
n
m
mnmn
x
mnm
n
x
0
22
1
!!
!
1
( ) ( )[ ] ( )
( )
( ) ( )
( )
( )
( )
( )
( ) ( )
( ) ( )
( )


+
=
−−
−
−=
−−
−=
+−−
−
−=−=
∑
∑
∑
=
−
−=
≥
−−
≥
−−
oddnn
evennn
p
x
knknk
kn
x
nm
m
mnm
xnmmm
mnm
n
n
x
xd
d
n
xP
p
k
knk
n
knm
n
pm
nmmn
n
n
pm
nmmn
n
n
n
n
nn
2/12
2/
!2!!
!22
1
2
1
!2
!2
!!
1
1
2
1
12122
!!
!
1
!2
1
1
!2
1
0
2
2
22

( ) ( )



<
≥+−−
=
2/0
2/121222
nm
nmnmmm
x
xd
d m
n
n

We recover the result obtained by the Generating Function of Legendre Polynomials
Return to Frobenius Series
Return to Table of Content
Derivation of Legendre Polynomials via Rodrigues’ Formula
SOLO
Series Solutions – Frobenius’ Method
Ferdinand Georg
Frobenius
(1849 –1917)
( ) ( ) .0121 2
2
22
constrealryrr
xd
yd
yx
xd
yd
x =++−−
Example: General Legendre Equation
∑
∞
=
+
=
0λ
λ
λ
k
xay
Let check the Frobenius’s expansion:
We have:
( ) ( ) ( )∑∑
∞
=
−+
∞
=
−+
−++=+=
0
2
2
2
0
1
1&
λ
λ
λ
λ
λ
λ λλλ kk
xkka
xd
yd
xka
xd
yd
Substitute in the General Legendre Equation:
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )[ ] 0111
121
00
2
000
2
=+++−++−++=
+++−−−++
∑∑
∑∑∑
∞
=
+
∞
=
−+
∞
=
+
∞
=
+
∞
=
+−+
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λλ
λ
λλλλ
λλλ
kk
kkkk
xkkrraxkka
xrraxkaxxkka
29
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 1)
( ) ( ) ( ) ( ) ( )[ ] 0111
00
2
=+++−++−++ ∑∑
∞
=
+
∞
=
−+
λ
λ
λ
λ
λ
λ λλλλ kk
xkkrraxkka
Denote, in the first sum λ = j +2 and in the second sum λ = j, to obtain:
( ) ( )
( ) ( ) ( ) ( ) ( )[ ]{ } 01112
11
0
2
1
1
2
0
=+++−+++++++
++−
∑
∞
=
+
+
−−
j
jk
jj
kk
xjkjkrrajkjka
xkkaxkka
All the coefficients of xk+j
must be zero, therefore
( ) 001 00 ≠=− akka
( ) 011 =+kka
( ) ( ) ( ) ( ) ( )[ ] 011122 =+++−+++++++ jkjkrrajkjka jj
30
Ferdinand Georg
Frobenius
1849 - 1917
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 2)
( ) ( ) ( )
( ) ( )
( )[ ] ( )
( ) ( )
,2,1,0
12
1
12
11
2 =
++++
++++−
−=
++++
+++−+
−=+ j
jkjk
jkrjkr
a
jkjk
jkjkrr
aa jjj
( ) 01 =−kk
( ) 011 =+kka
( ) ( ) ( ) ( ) ( )[ ] 011122 =+++−+++++++ jkjkrrajkjka jj
0&1.3
0&0.2
0&0.1
1
1
1
==
≠=
==
ak
ak
akThree possible
solutions
The equation that, k (k+1) = 0, comes from the coefficient of the lowest power
of x, and is called Indicial Equation. It has two solutions for k
k = 0 and k = 1
The equation
gives the recursive relation
31
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 3)
( ) 1001: ==⇒=− korkkkEquationIndicial
1. Using k = 0 and a1 = 0 we obtain a series of even powers of x
( ) ( )
( ) ( )
( ) ( )
( ) ( )
,2,1,0
21
1
21
11
2 =
++
++−
−=
++
+−+
−=+ j
jj
jrjr
a
jj
jjrr
aa jjj
( ) ( ) ( ) ( ) ( ) ( )






+
++−
+
+
−== 42
0
!4
312
!2
1
1: x
rrrr
x
rr
axpxy reven
( ) 01231 ==== +naaa 
The recurrence relation results in the following expression for the coefficients
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )


,3,2,1
!2
123124222
1 02 =
−+++⋅−+−+−
−= ma
m
mrrrrrmrmr
a
m
m
32
( ) ( ) ( )
( ) ( )
( )[ ] ( )
( ) ( )
,2,1,0
12
1
12
11
2 =
++++
++++−
−=
++++
+++−+
−=+ j
jkjk
jkrjkr
a
jkjk
jkjkrr
aa jjj
( ) ∑
∞
=
= 0
2
2m
m
meven xaxy
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 4)
( ) 1001: ==⇒=− korkkkEquationIndicial
3. Using k = 1 and a1 = 0 we obtain a series of odd powers of x
( ) ( )( )
( ) ( )
( )( )
( ) ( )
.2,1,0
32
21
32
211
2 =
++
++−−
−=
++
++−+
−=+ j
jj
jrjr
a
jj
jjrr
aa jjj
( ) ( ) ( )( ) ( ) ( )( ) ( )






+
++−−
+
+−
−== 53
0
!4
4213
!3
21
: x
rrrr
x
rr
xaxqxy rodd
( ) 01231 ==== +naaa 
The recurrence relation results in the following expression for the coefficients
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )


,3,2,1
!12
24213212
1 02 =
+
+++−+−+−
−= ma
m
mrrrrmrmr
a
m
m
33Since pr (x) and qr (x) are two linearly independent solutions of the 2nd
Order
Linear Lagrange ODE, the final solution is y = c1 pn (x) + c2 qn (x)
( ) ( ) ( )
( ) ( )
( )[ ] ( )
( ) ( )
,2,1,0
12
1
12
11
2 =
++++
++++−
−=
++++
+++−+
−=+ j
jkjk
jkrjkr
a
jkjk
jkjkrr
aa jjj
( ) ∑
∞
=
+
= 0
12
2m
m
modd xaxy
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 5)
Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have
Convergence Test:
( ) ( ) ( )
( ) ( ) 



≥≥
<<
=
++++
+++−+
=
∞→
+
+
∞→ divergex
convergex
xx
jkjk
jkjkrr
xa
xa
jj
j
j
j
j 11
11
12
11
limlim 22
2
2
The even series stops at j = n. The expansion is a Polynomial of order n (even).
1. If n is even, using k = 0 and a1 = 0 we obtain a series of even powers of x .
( ) ( )
( ) ( )
,..1,0
21
11
2 =
++
+−+
−=+ j
jj
jjnn
aa jj
For the case that r = n, a positive integer:
,...2,1,,02 ++==+ nnnjaj
3. If n is odd, using k = 1 and a1 = 0 we obtain a series of odd powers of x .
( ) ( )( )
( ) ( )
,..2,1
32
211
2 =
++
++−+
−=+ j
jj
jjnn
aa jj
,...2,1,,1,02 ++−==+ nnnnjaj
The odd series stops at j = n-1. The expansion is a Polynomial of order n (odd).34
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 6)
1. If n is even, using k = 0 and a1 = 0 we obtain a series of even powers of x
( ) ( ) ( )( ) ( )






+
++−
+
+
−= 42
0
!4
312
!2
1
1 x
nnnn
x
nn
axyeven
3. If n is odd using k = 1 and a1 = 0 we obtain a series of odd powers of x
( ) ( )( ) ( ) ( )( ) ( )






+
++−−
+
+−
−= 53
0
!4
4231
!3
21
x
nnnn
x
nn
xaxyodd
( ) ( )
( ) ( )( ) ( )






+−=




 ++−
+
+
−==
−=




 +
−==
==
42
0
42
0
2
0
2
0
0
6
70
101
!4
3414244
!2
144
14
31
!2
122
12
0
xxaxxayn
xaxayn
ayn
even
even
even
( )( )






−=




 +−
−==
==
3
0
3
0
0
3
5
!3
2313
3
1
xxaxxayn
xayn
odd
odd
We obtain the Legendre Polynomials Solutions for a0 = 1. 35
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 7)
1. The recurrence relation for even x powers results in the following expression for the
coefficients
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
2/,,3,2,1
!2
123124222
1 02 nma
m
mnnnnnmnmn
a
m
m 

=
−+++⋅−+−+−
−=
( ) ( ) ( ) ( ) ( ) ( )
( )!
!
2121224222 11
2
mp
p
ppmpmpnnmnmn mm
pn
−
=−+−+−=−+−+− −−
=

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )
( )( ) ( )
( )
( ) ( )( ) ( )
( )
( ) ( )!!22
!!22
212
1
!2
!22
242
242
1231
!
!
1231
22
mpp
pmp
mpppp
mp
mnnn
mnnn
mnnn
n
n
mnnn
m
pn
m
pn
+
+
=
+++
+
=
+++
+++
−+++=−+++
==




( )
( )
( )
( ) ( ) ( )
( )
( )
( ) ( )
( ) ( )
( ) ( )
2/,,3,2,1
!!!22
!221
!!!2!2
!
2
!22
1
!!2!2
!!22
!
!
2
1
1
0
0
2
2
02 nm
snssn
sn
a
snssnn
n
sn
a
mpmp
pmp
mp
p
a n
sthatsucha
choose
s
nmps
m
m =
−−
−−
=
−−












−
−=
+
+
−
−=
−
−
−=
( ) ( )
( ) ( )∑=
−−
<
−−
−
−=
2/
0
2
1
!2!!2
!22
1
n
s
sn
n
s
even xx
snsns
sn
y 36
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 8)
3. The recurrence relation for odd x powers results in the following expression for the
coefficients
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) 2
1
,,3,2,1
!12
24213212
1 02
−
=
+
+++−+−+−
−=
n
ma
m
mnnnnmnmn
a
m
m 

( ) ( ) ( ) ( )( ) ( )
( )!
!
222242222213212 1
12
mp
p
ppmpmpnmnmn m
pn
−
=−+−+−=−+−+− −
+=

( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )( )
( )
( ) ( )!!122
!!122
212
224222
1225232
!12
!12
242
1
1
12
mpp
pmp
mppp
mppp
mppp
p
p
mnnn
m
m
pn
++
++
=
+++
+++
++++
+
+
=+++
−
−
+=


( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( )
( ) ( ) 2
1
,,3,2,1
!2!1!!
!
2
1
!122
1
!12!!!12
!!122
1 0
2
0
2
2
−
=
−−−











 −
−−
−=
++−+
++
−=
−
−=
n
ma
snsnsn
n
sn
a
mmpmpp
pmp
a
sp
mps
m
m 
37
( )
( ) ( )
( )[ ] ( ) ( )
( )
( )
( ) ( ) 2
1
,...,1,0
!2!!!2
!
2
1
!22
1
!2!12!!
!
2
1
!12222
1 0
2
0
2
−
=
−−











 −
−
−=
−−−−











 −
−−−
−=
−
−=
−
−=
n
sa
snsnsn
n
sn
a
snsnsnsn
n
snsn
sp
mps
sp
mps
Series Solutions – Frobenius’ Method
Legendre Polynomials
Example: Legendre Equation (continue – 9)
3. The recurrence relation for odd x powers results in the following expression for the
coefficients
( ) ( )
( ) ( )
( ) ( )
( ) ( ) 2
1
,...,1,0
!2!!2
!22
1
!2!!
!22
!
2
1
!2
1
1
0
0
2
2
1
2
−
=
−−
−
−=
−−
−











 −
−=
−−
− n
s
snsns
sn
a
snsns
snn
n
a n
s
thatsucha
Choose
s
n
m
( ) ( ) ( )
( ) ( )
( )
1&
!2!!
!22
1
2
1 2/1
0
2
<
−−
−
−= ∑
−
=
−−
xoddnx
snsns
sn
xy
n
s
sns
nodd
We also found that the solution for k = 0 and a1 = 0 is
( ) ( ) ( )
( ) ( )
1&
!2!!2
!22
1
2/
0
2
<
−−
−
−= ∑=
−−
xevennx
snsns
sn
xy
n
s
sn
n
s
even
We recover the result obtained by the Generating Function of Legendre Polynomials
and by Rodrigues’ Formula 38
Therefore we can unify those two relations to obtain:
( ) ( ) ( )
( ) ( )
[ ]
1
!2!!2
!22
1
2/
0
2
<
−−
−
−= ∑=
−−
xx
snsns
sn
xP
n
s
sn
n
s
n
Series Solutions – Frobenius’ Method
SOLO
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 10)
2. Using k = 0 and a1 ≠ 0 we obtain an infinite series and not a polynomial.
This is the Second Solution of the Legendre Differential Equation.
The solution is the sum of the two infinite series, one with even powers of x
and the other with odd powers of x. The series solution, in this case,
diverges
at x = ± 1.
Those are Legendre Functions of the Second Kind.
The Polynomial solutions are Legendre Functions of the First Kind.
39
( ) ( )
( ) ( )
( ) ( )
( ) ( )
,2,1,0
21
1
21
11
2 =
++
++−
−=
++
+−+
−=+ j
jj
jnjn
a
jj
jjnn
aa jjj
In this case we have
The recurrence relation results in the following expression for the coefficients
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )


,3,2,1
!2
123124222
1 02 =
−+++⋅−+−+−
−= ma
m
mnnnnnmnmn
a
m
m
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )


,3,2,1
!12
123124222
1 112 =
+
−+++⋅−+−+−
−=+ ma
m
mnnnnnmnmn
a
m
m
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 11)
40
We want to find a series that converges for |x| > 1.
Let return to the conditions to have a series solution for Legendre ODE
For k = 0 and a0 = 0
By substituting j = m – 2 we obtain
( ) ( )
( ) ( )
,2,1,0
1
21
2 ==
++−
++
−= + ja
jnjn
jj
a jj
( )
( ) ( ) mm a
mnmn
mm
a
12
1
2
−++−
−
−=−
( ) ( ) .0121 2
2
22
constrealryrr
xd
yd
yx
xd
yd
x =++−−
( ) ( )
( ) ( )
( ) ( )
( ) ( )
,2,1,0
21
1
21
11
2 =
++
++−
−=
++
+−+
−=+ j
jj
jrjr
a
jj
jjrr
aa jjj
We can make am+2, am+4, am+6,…. To vanish for m = r = n or m = -n – 1.
If we start for am ≠ 0 we can obtain the following recursive formula
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 12)
41
( )
( ) ( ) mm a
mnmn
mm
a
12
1
2
−++−
−
−=−
Tacking m = n we obtain
( )
( )
( ) ( )
( )
( )( ) ( )
( ) ( ) nnn
nn
a
nn
nnnn
a
n
nn
a
a
n
nn
a
321242
321
324
32
122
1
24
2
−⋅−⋅⋅
−−−
+=
−
−−
−=
−
−
−=
−−
−
The first solution can be written as
( ) ( )
( )
( )( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) 





+
+−−⋅−⋅⋅
−−−−
++
−⋅−⋅⋅
−−−
+
−
−
−= −−−



 innnn
n x
innni
ininnn
x
nn
nnnn
x
n
nn
xaxy 242
1
1223212242
1221
321242
321
122
1
Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have
( )( )
( ) 



≤≥
><
=
+−
−−−
= −−
∞→+−
+−
−
−
∞→ divergex
convergex
xx
ini
inin
xa
xa
iin
in
in
in
j 11
11
1222
122
limlim 22
22
22
2
2
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 13)
42
( )
( ) ( ) mm a
mnmn
mm
a
12
1
2
−++−
−
−=−
Tacking m =- n - 1 we obtain
( ) ( )
( )
( ) ( )
( )
( ) ( )( ) ( )
( ) ( ) 135
13
523242
4321
524
43
322
21
−−−−−−
−−−−
+⋅+⋅⋅
++++
+=
+
++
=
+
++
+=
nnn
nn
a
nn
nnnn
a
n
nn
a
a
n
nn
a
The second solution can be written as
( ) ( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) 





+
+++⋅+⋅⋅
+−+++
++
+⋅+⋅⋅
++++
+
+
++
+= −−−−−−−−−
−− 


 12531
12
1225232242
21221
523242
4321
322
21 innnn
n x
innni
ininnn
x
nn
nnnn
x
n
nn
xaxy
Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have
( ) ( )
( ) 



≤≥
><
=
++
+−+
= −−
∞→+−
+−
−−
−−−
∞→ divergex
convergex
xx
ini
inin
xa
xa
iin
in
in
in
j 11
11
1222
212
limlim 22
12
12
12
12
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 14)
43
( )( )
!
1353212
n
nn
an
⋅⋅−−
=
By choosing
we have
( ) ( ) ( )( ) ( )
( )
( )( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1]
1223212242
1221
1
321242
321
122
1
[
!
1353212
2
42
1
>+
+−−⋅−⋅⋅
−−−−
−+
+
−⋅−⋅⋅
−−−
+
−
−
−
⋅⋅−−
==
−
−−
xx
innni
ininnn
x
nn
nnnn
x
n
nn
x
n
nn
xyxP
ini
nnn
n






Return to Table of Content
Finally ( ) ( ) ( )
( ) ( )
1
!2!!2
!22
1
0
2
>
−−
−
−= ∑
∞
=
−
xx
inini
in
xP
i
in
n
i
n
( )( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( )
( )
( )
( )
( )
( ) ( )
( ) ( )!2!!2
!22
1
!2
!22
!2!2
1
1
!2!2
135122
1
1223212242
1221
1
!
1353212
inini
in
in
in
iiniin
in
innni
ininnn
n
nn
n
i
ini
i
i
i
i
−−
−
−=
−
−
−
−=
−
⋅⋅−−
−=
+−−⋅−⋅⋅
−−−−
−
⋅⋅−−
−



Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 14)
44
( )( ) 1351212
!
1
⋅⋅−+
== −−
nn
n
aa nn
By choosing
we have
( ) ( )
( )( )
( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
1]
1225232242
21221
523242
4321
322
21
[
1351212
!
12
531
2
>+
+++⋅+⋅⋅
+−+++
+
+
+⋅+⋅⋅
++++
+
+
++
+
⋅⋅−+
==
−−−
−−−−−−
xx
innni
ininnn
x
nn
nnnn
x
n
nn
x
nn
n
xyxQ
in
nnn
n






( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
( )!122!
!2!
!
!12
!
!2
!122
!12
!
!2
2
1
!122
2222!12
!
!2
!2
1
1225232242
21221
2
++
+++
=
+
++
++
=
++
++++
=
+++⋅+⋅⋅
+−+++
ini
inin
n
n
n
in
in
n
n
in
in
innn
n
in
iinnni
ininnn
i
i
i



( ) ( ) ( )
( )
( )
1
!122!
!2!
2
0
12
>
++
++
= ∑
∞
=
++−
xx
ini
inin
xQ
i
inn
n
Go to Neumann Integral
( )( ) ( )!12
!2
!
1351212
!
+
=
⋅⋅−+ n
n
n
nn
n n

Finally
Series Solutions – Frobenius’ Method
Legendre Polynomials
SOLO
Example: Legendre Equation (continue – 15)
45
Summarize
( ) ( ) ( )
( )
( )
1
!122!
!2!
2
0
12
>
++
++
= ∑
∞
=
++−
xx
ini
inin
xQ
i
inn
n
Return to Table of Content
( ) ( ) ( )
( ) ( )
1
!2!!
!22
1
2
1
0
2
>
−−
−
−= ∑
∞
=
−
xx
inini
in
xP
i
ini
nn
( ) ( ) ( )
( ) ( )
[ ]
1
!2!!
!22
1
2
1 2/
0
2
<
−−
−
−= ∑=
−−
xx
inini
in
xP
n
i
ini
nn
Return to
Similar to Rodrigues Formula
Using Frobenius’ Method we found that solutions of Legendre ODE
are
( ) ( ) integerpositive0121 2
2
22
nynn
xd
yd
yx
xd
yd
x =++−−
Series Solutions – Frobenius’ Method
Legendre Polynomials
( )
4,3,2,1
1
=
>
l
xxPl
( )
4,3,2,1
1
=
>
l
xxQl
46
SOLO
( ) 1
21
1
0
2
≤=
+−
∑
∞
=
utPu
utu n
n
n
For u=0 we obtain ( ) 10 =tP
For t = 1 we obtain ( ) ( ) 111
1
1
21
1
00
2
=⇒==
−
=
+−
∑∑
∞
=
∞
=
n
n
n
n
n
n
PPuu
uuu
For t = -1 we obtain ( ) ( ) ( ) ( )n
n
n
n
n
n
nn
PPuu
uuu
1111
1
1
21
1
00
2
−=⇒=−=
+
=
++
∑∑
∞
=
∞
=
Let find a Recursive Relation for Legendre Polynomial computation
Start with:
( ) ( )
( ) ( )
( )
( )
( )






∂
∂
+
−
=





∂
∂−
+
−
=





∂
∂
+
−
= ++
+
+
++
+
+
11
1
1
11
2
1 cos
1
11
!
1
1
11
!1
1cos
n
n
n
nn
n
nn
n
n
r
P
znrznnrznr
P θθ
( )
( )
( ) ( ) ( )( ) ( )






∂
∂
+
∂
∂+
−
+
−
= +++
+
zd
Pd
rz
r
r
Pn
nr
P n
nn
n
n
n θ
θ
θθθ cos
cos
cos1cos1
1
1cos
122
1
Recursive Relations for Legendre Polynomial Computation
First Recursive Relation
Legendre Polynomials
47
SOLO
Recursive Relations for Legendre Polynomial Computation
( )
( )
( ) ( ) ( )( ) ( )






∂
∂
+
∂
∂+
−
+
−
= +++
+
zd
Pd
rz
r
r
Pn
nr
P n
nn
n
n
n θ
θ
θθθ cos
cos
cos1cos1
1
1cos
122
1
θcosrz =
θcos=
∂
∂
z
r
 z
r
z
r
z
z
∂
∂
+
∂
∂
=
∂
∂
=
θ
θ
θ
cos
cos1
cos
rz
θθ 2
cos1cos −
=
∂
∂
( )
( )
( ) ( ) ( )( )





 −
+
+
−
+
−
= +++
+
rd
Pd
rr
Pn
nr
P n
nn
n
n
n θ
θ
θ
θ
θθ 2
122
1 cos1
cos
cos1
cos
cos1
1
1cos
( ) ( ) ( )( )
θ
θθ
θθθ
cos
cos
1
cos1
coscoscos
2
1
d
Pd
n
PP n
nn
+
−
−=+
Substituting t = cos θ we obtain
( ) ( ) ( )
td
tPd
n
t
tPttP n
nn
1
1 2
1
+
−
−=+
First Recursive Relation (continue – 1)
Legendre Polynomials
48
SOLO
Recursive Relations for Legendre Polynomial Computation
( ) ( ) ( ) 0,1
1
1 2
1 ≥≤
+
−
−=+ nt
td
tPd
n
t
tPttP n
nn
Use to start ( ) ( ) 01 0
0 =⇒=
td
tPd
tP
( )
( )
( )
( )
( )
8
157063
8
33035
2
35
2
13
35
5
24
4
3
3
2
2
1
ttt
tP
tt
tP
tt
tP
t
tP
ttP
+−
=
+−
=
−
=
−
=
=
First Recursive Relation (continue – 2)
Legendre Polynomials
49
SOLO
Recursive Relations for Legendre Polynomial Computation
Start from ( ) ( ) 1
21
1
:,
0
2
≤=
+−
= ∑
∞
=
utPu
utu
tug
n
n
n
Let differentiate both sides with respect to u and rearranging
( ) ( )∑
∞
=
−
+−=
+−
−
=
∂
∂
1
12
2
21
21 n
n
n
tPunutu
utu
ut
u
g
( ) ( ) ( ) ( )∑∑
∞
=
−
∞
=
+−=−
1
12
0
21
n
n
n
n
n
n
tPunututPuut
( ) ( ) ( ) ( )∑∑
∞
=
+−
∞
=
+
+−=−
1
11
0
1
2
n
n
nnn
n
n
nn
tPunutnuntPuut
( ) ( ) ( ) ( ) ( ) ( ) ( )∑∑∑∑∑
∞
=
−
∞
=
∞
=
+
∞
=
−
∞
=
−+−+=−
2
1
10
1
1
1
0
121
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
tPuntPutntPuntPutPut
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )




≥+=−+
=−=−
==
+− 2112
122
0
11
1201
10
ntPuntPuntPutn
ntPtutPutPutuPt
ntPtPt
n
n
n
n
n
n
We can see that the last relation agrees also with the previous
relations, for n = 0 and n = 1.
Second Recursive Relation
Legendre Polynomials
50
SOLO
Recursive Relations for Legendre Polynomial Computation
We find the Recursive Relation:
( ) ( ) ( ) 1
11
12
11 ≥
+
−
+
+
= −+ ntP
n
n
tPt
n
n
tP nnn
( ) 10 =tP
This is called the Bonnet’s Recursion Relation. It starts with:
Examples:
( ) ( ) ( )
2
3
1 01
2
tPtPt
tPn
−
=→=
( )
2
1
2
3 2
2 −= ttP
( )
( )
( )

tP
tP
ttttPn
1
2
3
2
2
1
2
3
3
5
2 2
3 −





−=→=

( )
2
35 3
3
tt
tP
−
=
( )
( ) ( )

tPtP
tttttPn
23
2
1
2
3
4
3
2
3
2
5
4
7
3 23
4 





−−





−=→= ( )
8
33035 24
4
+−
=
tt
tP
( ) ttP =1
Second Recursive Relation (continue)
Legendre Polynomials
51
SOLO
Recursive Relations for Legendre Polynomial Computation
Start from ( ) ( ) 1
21
1
:,
0
2
≤=
+−
= ∑
∞
=
utPu
utu
tug
n
n
n
Let differentiate both sides with respect to t and rearranging
( )
( )
∑
∞
=
=
+−
=
∂
∂
0
2/32
21 n
nn
td
tPd
u
utu
u
t
g
( ) ( ) ( )
∑∑
∞
=
∞
=
+−=
0
2
0
21
n
nn
n
n
n
td
tPd
uututPuuor
Equaling coefficients of each power of u gives
( ) ( ) ( ) ( )
td
tPd
td
tPd
t
td
tPd
tP nnn
n
11
2 −+
+−=
( ) ( ) ( ) ( ) ( )tPntPntPtn nnn 11112 −+ ++=+
Differentiate the Second Recursive Relation
with respect to t and rearranging
( ) ( ) ( )
( )
( )
( )
( )
td
tPd
n
n
td
tPd
n
n
td
tPd
ttP nnn
n
11
1212
1 −+
+
+
+
+
=+
Third Recursive Relation
Legendre Polynomials
52
SOLO
Recursive Relations for Legendre Polynomial Computation
We found ( ) ( ) ( ) ( )
td
tPd
ttP
td
tPd
td
tPd n
n
nn
211
+=+ −+
( )
( )
( )
( )
( ) ( ) ( )
td
tPd
ttP
td
tPd
n
n
td
tPd
n
n n
n
nn
+=
+
+
+
+ −+ 11
1212
1
Third Recursive Relation (continue)
Let solve for and in terms of and( )tPn
( )
td
tPd n( )
td
tPd n 1+( )
td
tPd n 1−
( ) ( ) ( )
td
tPd
ttPn
td
tPd n
n
n
+−=−1
( ) ( ) ( ) ( )
td
tPd
ttPn
td
tPd n
n
n
++=+
11
Subtracting the first relation from the second gives the Third Recursive Relation
( ) ( ) ( ) ( )
td
tPd
td
tPd
tPn nn
n
11
12 −+
−=+
Legendre Polynomials
53
SOLO
Recursive Relations for Legendre Polynomial Computation
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }yPxPyPxP
yx
n
yPxPk
tPtntPn
td
tPd
t
tPntPtn
td
tPd
t
tPn
td
tPd
t
td
tPd
tPn
td
tPd
td
tPd
t
tPn
td
tPd
td
tPd
tPntPtntPn
tP
n
n
tP
n
n
tPt
tPin
td
tPd
nnnn
n
k
kk
nn
n
nn
n
n
nn
n
nn
n
nn
nnn
nnn
l
i
in
n
11
0
1
2
1
2
1
1
1
11
11
11
1
2
1
0
12
1
129
1118
17
6
5
124
01213
1212
1
2
1421
++
=
+
−
−
−
−
−+
−+
−+






−
=
−−
−
−
+
=+
+−+=−
−=−
=−
=−
+=−
=++−+
+
+
+
+
=
−−=
∑
∑
Recursive Relation between Legendre Polynomials and their Derivatives)
Legendre Polynomials
Return to Table of Content
54
SOLO
Orthogonality of Legendre Polynomials
Define ( ) ( )tPwtPv nm == :&:
We use Legendre’s Differential Equations:
( ) ( ) 011 2
=++





− vmm
td
vd
t
td
d
( ) ( ) 011 2
=++





− wnn
td
wd
t
td
d
Multiply first equation by w and integrate from t = -1 to t = +1.
( ) ( ) 011
1
1
1
1
2
=++





− ∫∫
+
−
+
−
dtwvmmdtw
td
vd
t
td
d
Integrate the first integral by parts we get
( ) ( ) ( ) 0111
1
1
1
1
2
0
1
1
2
=++−−− ∫∫
+
−
+
−
+=
−=
dtwvmmdt
td
wd
td
vd
tw
td
vd
t
t
t
  
In the same way, multiply second equation by v and integrate from t = -1 to t = +1.
( ) ( ) 011
1
1
1
1
2
=++−− ∫∫
+
−
+
−
dtwvnndt
td
wd
td
vd
t
Legendre Polynomials
55
SOLO
Orthogonality of Legendre Polynomials
( ) ( ) 011
1
1
1
1
2
=++−− ∫∫
+
−
+
−
dtwvmmdt
td
wd
td
vd
t
Subtracting those two equations we obtain
( ) ( ) 011
1
1
1
1
2
=++−− ∫∫
+
−
+
−
dtwvnndt
td
wd
td
vd
t
( ) ( )[ ] ( ) ( )[ ] ( ) ( ) 01111
1
1
1
1
=+−+=+−+ ∫∫
+
−
+
−
dttPtPnnmmdtwvnnmm nm
This gives the Orthogonality Condition for m ≠ n
( ) ( ) nmdttPtP nm ≠=∫
+
−
0
1
1
To find let square the relation and integrate
between t = -1 to t = +1. Due to orthogonality only the integrals of terms
having Pn
2
(t) survive on the right-hand side. So we get
( )∫
+
−
1
1
2
dttPn
( )∑
∞
=
=
+− 0
2
21
1
n
n
n
tPu
utu
( )∑ ∫∫
∞
=
+
−
+
−
=
+− 0
1
1
22
1
1 2
21
1
n
n
n
dttPudt
utu
Legendre Polynomials
56
SOLO
Orthogonality of Legendre Polynomials
( )∑ ∫∫
∞
=
+
−
+
−
=
+− 0
1
1
22
1
1 2
21
1
n
n
n
dttPudt
utu
( ) ( )
( )
1
1
1
ln
1
1
1
ln
2
1
21ln
2
1
21
1
2
21
1
2
1
1 2
<
−
+
=
+
−
−
=−+
−
=
−+
+=
−=
+
−∫ u
u
u
uu
u
u
tuu
u
dt
tuu
t
t
( ) ( ) ( ) ( ) ( ) ( ) ( )
∑∑∑
∞ ++∞ +∞ +
+
−−
−=
+
−
−−
+
−=−−+
0
11
0
1
0
1
1
1
1
1
1
1
1
1
1
1ln
1
1ln
1
n
uu
un
u
un
u
u
u
u
u
u
nn
n
n
n
n
n
( ) ( ) ( )
( )
( ) ( )
( ) ∑∑∑∑
∞∞ +∞ ++
+
∞ ++
+
=
+
=
+
−−
−+
+
−−
−=
0
2
0
12
0
0
1212
12
0
1212
2
12
2
12
1
2
12
1
1
12
1
1 n
nnn
n
nn
n
u
nn
u
un
uu
un
uu
u
  
Let compute first
Therefore
( )∑ ∫∑∫
∞ +
−
∞+
−
=
+
=
+− 0
1
1
22
0
2
1
1 2
12
2
21
1
dttPuu
n
dt
utu
n
nn
Comparing the coefficients of u2n
we get ( )
12
21
1
2
+
=∫
+
− n
dttPn
Legendre Polynomials
( ) ( ) nmmn
n
dttPtP δ
12
21
1 +
=∫
+
−
Hence
57
SOLO
Using Rodrigues’ Formula
let calculate
( ) ( ) ( ) ( )[ ]∫∫
+
−
+
−
−
=
1
1
2
1
1
1
!2
1
dt
td
td
tP
n
dttPtP n
nn
knnk
Legendre Polynomials
( ) ( )[ ]n
nn
nn
td
td
n
tP
1
!2
1 2
−
=
Orthogonality of Legendre Polynomials (Second Method)
Assume, without loss of generality, that n > k, and integrate by parts
( ) ( ) ( ) ( )[ ]
( ) ( )[ ] ( ) ( )[ ] ( ) ( ) ( )
∫∫
∫∫
+
−
+
− −
−
+
−
−
−
+
−
+
−
−−==
−
−
−
=
−
=
1
1
2
1
1 1
21
0
1
1
1
21
1
1
2
1
1
1
!2
1
1
1
!2
11
!2
1
1
!2
1
dt
td
tPd
t
n
dt
td
td
td
tPd
ntd
td
tP
n
dt
td
td
tP
n
dttPtP
n
k
n
n
n
n
n
nn
k
nn
nn
kn
n
nn
knnk

  
Since Pk (t) is a Polynomial of Order k and we assume that n > k, we have
( ) 0=n
k
n
td
tPd
Therefore
( ) ( ) nkdttPtP nk ≠=∫
+
−
0
1
1
58
SOLO
For k = n we have
Legendre Polynomials
Orthogonality of Legendre Polynomials (Second Method) (continue – 1)
( )[ ] ( ) ( ) ( )
∫∫
+
−
+
−
−−=
1
1
2
1
1
2
1
!2
1
1 dt
td
tPd
t
n
dttP n
n
n
n
n
n
n
But we found that Pn (t) is given by:
( ) ( ) ( )
( ) ( )
[ ]
∑=
−
−−
−
−=
2/
0
2
!2!!2
!22
1
n
k
kn
n
k
n t
knknk
kn
tP
Therefore to compute it is sufficient to consider only the highest power of t
in the series, i.e. for k = 0, and we obtain
( )
n
n
n
td
tPd
( ) ( )
( )
( )
!2
!2
!
!2
!2
2
n
n
n
n
n
td
tPd
nnn
n
n
==
( )[ ] ( ) ( ) ( ) ( )
( )
( )∫∫∫
+
−
+
−
+
−
−=−−=
1
1
2
22
1
1
2
1
1
2
1
!2
!2
!2
!2
1
!2
1
1 dtt
n
n
dt
n
n
t
n
dttP
n
nn
n
n
n
n
SOLO
Legendre Polynomials
Orthogonality of Legendre Polynomials (Second Method) (continue – 2)
( )[ ] ( ) ( ) ( ) ( )
( )
( )∫∫∫
+
−
+
−
+
−
−=−−=
1
1
2
22
1
1
2
1
1
2
1
!2
!2
!2
!2
1
!2
1
1 dtt
n
n
dt
n
n
t
n
dttP
n
nn
n
n
n
n
( ) ∫∫∫
++
=+
−
=−=−
π
π
θ
θθθθ
0
12
0
12
cos1
1
2
sinsin1 dddtt nn
tn
Therefore
( )
( ) ( ) ( ) ( ) ( )
( )
( )!12
!2
cos
2222
!2
11212
!2
sin
31212
2222
sin
12
2
sin
212
2
0
1
00
12
0
12
+
=
−⋅
⋅
−⋅+
=
−⋅+
−⋅
=
+
=
+
−+
∫∫∫ n
n
nn
n
nn
n
d
nn
nn
d
n
n
d
nnn
nn

  


π
πππ
θθθθθθθ
( ) ( ) ( )∫∫∫∫
+−−+
−=+−==
ππ
π
ππ
θθθθθθθθθθθθ
0
1212
0
212
0
0
2
0
2
0
12
sinsin2cossin2cossincossinsin dndndd nnnnnn
  
( )[ ] ( )
( )
( ) ( )
( )
( )
( )
,2,1,0
12
2
!12
!2
!2
!2
1
!2
!2
212
22
1
1
2
22
1
1
2
=
+
=
+
=−=
+
+
−
+
− ∫∫ n
nn
n
n
n
dtt
n
n
dttP
n
n
n
nn
Return to Table of Content
59
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Using Sturm-Liouville Theory it can be seen that the Legendre
Polynomials that are Solution of the Legendre ODE, form an orthogonal
and “Complete” Set, meaning that we can expand any function f (t) ,
Piecewise Continuous in the interval -1 ≤ t ≤+1. Therefore we can define a
series of Legendre Polynomials that converges in the mean to the function
f (t)
( ) ( ) 11
0
≤≤−= ∑
∞
=
ttPatf
n
nn
The coefficients an can be defined using the
Orthogonality Property of Legendre Polynomials
( ) ( ) ( ) ( ) m
n
m
mnnm a
m
tdtPtPatdtPtf
mn
12
2
0
2
12
1
1
1
1
+
== ∑ ∫∫
∞
=
+
+
−
+
−
  
δ
( ) ( )∫
+
−
+
=
1
1
2
12
tdtPtf
m
a mm
( ) ( ) ( ) ( ) 11
2
12
0
1
1
≤≤−






+
= ∑ ∫
∞
=
+
−
ttPtdtPtf
n
tf
n
nn
60
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
At any discontinuous point t0 ( f(t0- ) ≠ f(t0+) ) we have
( ) ( )[ ] ( ) ( ) ( ) 11
2
12
2
1
0
0
0
1
1
00 ≤≤−






+
=+ ∑ ∫
∞
=
+
−
+− ttPtdtPtf
n
tftf
n
nn
If f (t) is defined in the interval –a ≤ t ≤ +a then
( ) ( ) ataatPatf
n
nn ≤≤−= ∑
∞
=0
/ ( ) ( )∫
+
−
+
=
a
a
nn tdatPtf
a
n
a /
2
12
61
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
If f (t) is an odd function ( f(-t ) =- f(t) ) we have
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
( )
( ) ( )
( ) ( )
( ) ( )



=+−−=
+==
+
∫
∫∫
∫∫∫
+
++
−−
+
−→
−
+
−
1
0
1
0
1
0
1
1
0
0
1
1
1
2
0
12
2
oddndttPtf
evenn
dttPtfdttPtf
dttPtfdttPtfdttPtfa
n
n
n
tP
n
tf
n
tt
nnn
n
n

  
If f (t) is an even function ( f(-t ) = f(t) ) we have
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
( )
( ) ( )
( ) ( )
( ) ( )




=+−−=
+==
+
∫∫∫
∫∫∫
+
++
−
+
−→
−
+
−
oddn
evenndttPtf
dttPtfdttPtf
dttPtfdttPtfdttPtfa
n
n
n
tP
n
tf
n
tt
nnn
n
n 0
2
12
2
1
0
1
0
1
0
1
1
0
0
1
1
1

  
62
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Using Rodrigues’ Formula
let calculate
( ) ( )[ ]n
nn
nn
td
td
n
tP
1
!2
1 2
−
=
( ) ( ) ( )[ ] ( )
( ) ( )[ ] ( )[ ] ( ) ( ) ( ) ( )
∫∫
∫∫
+
−
+
−
−
−
+
−
−
−
+
−
+
−
−−==
−
−
−
=
−
=
1
1
2
1
1
1
21
0
1
1
1
21
1
1
21
1
1
!2
1
1
1
!2
11
!2
1
1
!2
1
td
td
tfd
t
n
td
td
tfd
td
td
ntd
td
tf
n
tdtf
td
td
n
tdtftP
n
n
n
n
n
n
nn
nn
nn
n
n
nn
nn

  
( ) ( ) ( ) ( ) ( ) ( ) ( )
∫∫∫
+
−
+
−
+
−
−=−−=
1
1
2
1
1
2
1
1
1
!2
1
1
!2
1
1 td
td
tfd
t
n
td
td
tfd
t
n
tdtPtf n
n
n
nn
n
n
n
n
n
or
63
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Example f (t) = tk
,|t| < 1
( ) ( ) ( ) ( ) ( ) ( ) ( )




≥−+−−
+
<
=−
+
=
+
=
∫
∫∫ +
−
−
+
+
−
+
+
−
nktdttnkkk
n
n
nk
td
td
td
t
n
n
tdtPt
n
a nkn
n
n
kn
n
nn
k
n
1
1
2
1
1
1
2
1
1
1
111
!2
12
0
1
!2
12
2
12

We have
( )
( ) ( )
( ) ( )
( )
( )( )
( ) ( )
( )
( )( ) ( )[ ]
( ) ( ) ( )
( ) ( )
( )
( )
( )
( )
( )∫
∫
∫
∫∫
+
−
+
−−
+
−
−−−+
+
−
−−+
+
−
−−+
+
−
−−+=
−=
+
−
−
−
−
−
+
=
−
+++
−−−−−−−−
=
=−
++
−−−−
=
−
+
−−
+−
+
−=−
−−
1
1
2
1
1
22222
1
1
422222
1
1
222122
0
1
1
122122
1
2
1
1
1
2222
1
!2
!22
!2
!2
1
2222122
12222322122
1
222122
322122
1
122
122
1
122
1
1
122
222
tdt
nm
nm
mn
n
tdtt
mnnn
nmnmnmnm
tdtt
nn
nmnm
tdtt
n
nm
tt
n
tdtt
mn
nmnm
nmnmmn
nmn
nmnnmntu
tdtdv
nmn
nm
n




  
Take k = 2m and since t2m
is even, they are only even coefficients nonzero so
we take 2 n instead of n, and 2n ≤ 2m
64
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Example f (t) = tk
, |t| < 1 (continue – 1)
( )
( )
( )
( )
( )




≥−
−
+
<
=
∫
+
−
−
+
nmtdtt
nm
m
n
n
nm
a nmn
n
n
1
1
2222
12
2
1
!22
!2
!22
14
0
We have for f (t) = t2m
( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )( )
( )!122
!2
!2
!22
!2
!2
1
!2
!22
!2
!2
1
2122
1
1
2
1
1
2222
++
+
−
−
+
=
−
−
−
+
=−
++
−−
+
−
+
−−
+
−
−
∫∫
nm
mn
nm
nm
mn
n
tdt
nm
nm
mn
n
tdtt
nm
nmnm
mn
nmnm
nmn
( )
( )
( )
( )
( )
( )
( )
( )
( )( )
( )
( ) ( ) ( )
( ) ( )!122!
!!2142
!122
!2
!2
!22
!2
!2
!22
!2
!22
14 22122
122
++−
++
=
++
+
−
−
+−
+
=
++
−−+
nmnm
nmmn
nm
nm
nm
nm
nm
n
nm
m
n
n
a
nnm
nmnmnn
( ) ( )
( )!12
!2
1
212
1
1
2
+
=−
+
+
−∫ n
n
dtt
n
n
Where we used the previous result
Therefore
( ) ( ) ( )
( ) ( )
( )∑= ++−
++
=
m
n
n
n
m
tP
nmnm
nmmn
t
0
2
2
2
!122!
!!2142
65
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
( ) ( ) ( ) ( ) ( ) ( ) ( )




≥−+−−
+
<
=−
+
=
+
=
∫
∫∫ +
−
−
+
+
−
+
+
−
nktdttnkkk
n
n
nk
td
td
td
t
n
n
tdtPt
n
a nkn
n
n
kn
n
nn
k
n
1
1
2
1
1
1
2
1
1
1
111
!2
12
0
1
!2
12
2
12

We have
( )
( ) ( )
( ) ( )
( )
( )( )
( ) ( )
( )
( )( ) ( )[ ]
( ) ( ) ( )
( ) ( )
( )
( )
( )
( )
( )( )
( )
( )∫
∫
∫
∫∫
+
−
++
++
−−
+
−
−−−++
+
−
−−+
+
−
−−+
+
−
−−+=
−=
+
−
−+
−
++
+
−
−
++
+
=
−
++++
−−−−−−−−
=
=−
++
−−−−
=
−
+
−−
+−
+
−=−
−−
+
1
1
12
2122
1
1
222212
1
1
422322
1
1
222222
0
1
1
122222
1
2
1
1
1
22122
1
!122
!2
!2
!22
!12
!12
1
12322222
12222322122
1
322222
322122
1
222
122
1
222
1
1
122
2122
tdt
nm
mn
nm
nm
mn
n
tdtt
mnnn
nmnmnmnm
tdtt
nn
nmnm
tdtt
n
nm
tt
n
tdtt
mn
nm
nmnm
nmnmmn
nmn
nmnnmntu
tdtdv
nmn
nm
n




  
Take k = 2m+1 and since t2m+1
is odd, they are only odd coefficients nonzero so
we take 2 n+1 instead of n, and 2n+1 ≤ 2m+1
66
Example f (t) = tk
, |t| < 1 (continue – 2)
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
( )
( )
( )
( )
( )




≥−
−+
+
<
=
∫
+
−
−+
+
+
nmtdtt
nm
m
n
n
nm
a nmn
n
n
1
1
22122
22
12
1
!22
!2
!122
34
0
We have for f (t) = t2m+1
( ) ( )
( )
( )
( )
( )( )
( )
( )
( )
( )
( )
( )
( )( )
( )!322
!12
!2
!22
!12
!12
1
!122
!2
!2
!22
!12
!12
1
2322
1
1
12
21221
1
22122
++
++
−
−
++
+
=
−
++
+
−
−
++
+
=−
++
−−
+
−
++
++
−−
+
−
−+
∫∫
nm
mn
nm
nm
mn
n
tdt
nm
mn
nm
nm
mn
n
tdtt
nm
nmnm
mn
nm
nmnm
nmn
( )
( )
( )
( )
( )
( )
( )
( )
( )( )
( )
( ) ( ) ( )
( ) ( )!322!
!1!12342
!322
!12
!2
!22
!12
!12
!22
!2
!122
34 122322
2212
++−
++++
=
++
++
−
−
++
+
−+
+
=
+++
−−++
nmnm
nmmn
nm
nm
nm
nm
nm
n
nm
m
n
n
a
nnm
nmnmnn
( ) ( )
( )!12
!2
1
212
1
1
2
+
=−
+
+
−∫ n
n
dtt
n
n
Where we used the previous result
Therefore
( ) ( ) ( )
( ) ( )
( )∑=
+
+
+
++−
++++
=
m
n
n
n
m
tP
nmnm
nmmn
t
0
12
12
12
!322!
!1!12342
Return to Neumann Integral
67
Example f (t) = tk
, |t| < 1 (continue – 3)
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Neumann Integral
1
1
1
111
0
2
12
0
12
2
0
1
0
<+==





=
−
=
−
∑∑∑∑
∞
=
−∞
=
+
∞
=
+
∞
= x
t
x
t
x
t
x
t
x
t
x
x
txtx m
m
m
m
m
m
m
m
m
m
m
Start from
Use
( ) ( ) ( )
( ) ( )
( )∑=
+
+
+
++−
++++
=
m
n
n
n
m
tP
nmnm
nmmn
t
0
12
12
12
!322!
!1!12342
( ) ( ) ( )
( ) ( )
( )∑= ++−
++
=
m
n
n
n
m
tP
nmnm
nmmn
t
0
2
2
2
!122!
!!2142
( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )
( )
( ) ( ) ( )
( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( )
( )
( ) ( ) ( )
( )
( )
( ) ( ) ( ) ( )
( )
( )∑ ∑∑ ∑
∑ ∑∑ ∑
∑ ∑∑ ∑
∞
=
∞
=
+
−
+∞
=
∞
=
++−
+=
∞
=
∞
=
+
−
+∞
=
∞
=
+−
∞
= =
−
+
+∞
=
+−
=
++
+++++
+
++
+++
=
++−
++++
+
++−
++
=
++−
++++
+
++−
++
=
−
0 0
12
2
12
0 0
2
212
2
0
12
2
12
0
2
12
2
0 0
2
12
12
0
12
0
2
2
!322!
!12!122342
!124!
!2!22142
!322!
!1!12342
!122!
!!2142
!322!
!1!12342
!122!
!!21421
n i
n
m
n
n i
n
in
ninm
n nm
n
m
n
n nm
n
m
nOrder
Summation
Change
m
m
n
m
n
n
m
m
m
n
n
n
tPx
nmi
ininn
tPx
ini
ininn
tPx
nmnm
nmmn
tPx
nmnm
nmmn
xtP
nmnm
nmmn
xtP
nmnm
nmmn
tx
68
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Neumann Integral (continue – 1)
( ) ( ) ( )
( )
( )
( ) ( ) ( ) ( )
( )
( )∑ ∑∑ ∑
∞
=
+
∞
=
−
+∞
=
∞
=
++−






++
+++++
+





++
+++
=
− 0
12
0
2
12
0
2
0
212
2
!322!
!12!122342
!124!
!2!221421
n
n
i
m
n
n
n
i
in
n
tPx
nmi
ininn
tPx
ini
ininn
tx
We found
( ) ( ) ( )
( )
( )
1
!122!
!2!
2
0
12
>
++
++
= ∑
∞
=
++−
xx
ini
inin
xQ
i
inn
n
Use the Frobenius Series development of Legendre Functions of the
Second Kind Qn (x)
We have ( ) ( ) ( ) ( ) ( ) ( )∑∑
∞
=
++
∞
=
+++=
− 0
1212
0
22 3414
1
n
nn
n
nn tPxQntPxQn
tx
( ) ( ) ( )∑
∞
=
<+=
− 0
12
1
n
nn xttPxQn
tx
Return to Qn
Frobenius Series
69
SOLO
Legendre Polynomials
Expansion of Functions, Legendre Series
Neumann Integral (continue – 2)
We found
( ) ( ) ( )∑
∞
=
+=
− 0
12
1
n
nn tPxQn
tx
Multiply both sides by Pm (t) and integrate between -1 to +1
( )
( ) ( ) ( ) ( ) ( )xQtdtPtPxQntd
tx
tP
m
n
n
mnn
m
nm
212
0
12
2
1
1
1
1
=+=
−
∑ ∫∫
∞
=
+
+
−
+
−
  
δ
We obtain
( )
( )
∫
+
−
−
=
1
1
2
1
td
tx
tP
xQ
n
n
Franz Neumann's Integral of 1848
Franz Ernst Neumann
(1798 –1895)
Return to Table of Content
70
SOLO
71
Legendre Polynomials
Schlaefli Integral
Start with
Using Rodrigues's Formula we obtain( ) ( )[ ]n
n
n
nn x
xd
d
n
xP 1
!2
1 2
−=
( ) ( )
∫ −
= td
zt
tf
j
zf
π2
1
Cauchy's Integral
with ( ) ( )n
zzf 12
−=
( ) ( )
∫ −
−
=− td
zt
t
j
z
n
n 1
2
1
1
2
2
π
Differentiate n times this equation with respect to z and multiply by 1/ (2n
n!)
( ) ( )
( )∫ +
−
−
−
=− td
zt
t
j
z
zd
d
n n
nn
n
n
n
n 1
2
2 1
2
2
1
!2
1
π
with the contour enclosing the point t = z.
Schlaefli Integral ( ) ( )
( )∫ +
−
−
−
= td
zt
t
j
zP n
nn
n 1
2
1
2
2
π
Return to Table of Content
SOLO
72
Legendre Polynomials
Laplace’s Integral Representation
Start with
( )
( )
( )[ ]
( )[ ] ( )[ ]
( )
[ ] [ ] ( ) ( )
22
0
1
2
0
22
0
2
0
22
2/tan
0
22
2
0
2
2
0
121
2
1
1
tan
1
2
1
1
1
1
1
1
2
11
2
11
2
2/tan12/tan1
2/tan1
2/tan1
2/tan1
1
cos1
λ
ππ
λλ
λ
λ
λ
λ
λ
λ
λ
λλλφλφ
φφ
φ
φ
λ
φ
φλ
φ φπππ
−
=
−
=







+
−
−
=








+
−
+
+
−
−
=
−++
=
−++
=
−++
+
=
+
−
+
=
+
∞
−
∞
∞∞=
∫
∫∫∫∫∫
t
t
td
t
td
tt
tdddd t
( )
( )
( )
( )
( ) { }[ ] ( ) { }∑
∞
=
−−
−
±=
−±−=−±−−=
−±−
−
=
−
−
±
=
+ 0
2
1
2
22
1
1
cos11cos111
cos11
1
cos
1
1
1
1
cos1
1
2
n
n
n
xu
xu
xxuxuxxuxu
xuxu
xu
xu
xu
φφ
φ
φ
φλ
λ
Let write
where we used ( ) ∑
∞
=
−
=−
0
1
1
n
n
aa
( )
( )
( )
( )
( )
( ) ( )
( )
2222
2
22
1
1
2
21
1
11
1
1
1
1
1
1
1
2
uxu
xu
xuxu
xu
xu
xu
xu
xu
+−
−
=
−−−
−
=
−
−
−
=
−
−
−
±=λ
λ
( ) { } ( ) { } ( )
22
0 0
2
0 0
2
0 21
1
1
cos11cos11
cos1 uxu
xu
dxxuxudxxuxu
d
n
n
n
n
n
n
+−
−
=
−
=−±−=−±−=
+
∑ ∫∫ ∑∫
∞
=
∞
=
π
λ
π
φφφφ
φλ
φ πππ
SOLO
73
Legendre Polynomials
Laplace’s Integral Representation (continue -
1)
{ } ( )∑∑ ∫
∞
=
∞
=
=
+−
=−±
0
2
0 0
2
21
cos1
n
n
n
n
n
n
xPu
uxu
dxxu π
π
φφ
π
We obtained
Equating un
coefficients we obtain :
( ) { }∫ −±=
π
φφ
π 0
2
cos1
1
dxxxP
n
n
( ) ( ) 0112
=+−





− ynny
xd
d
x
xd
d
If we replace in the Legendre ODE n by –n – 1
the equation does not change. Therefore , and( ) ( )xPxP nn 1−−=
( ) { }∫
−−
−±=
π
φφ
π 0
1
2
cos1
1
dxxxP
n
n
Substitute x = cosθ
( ) { }∫ ±=
π
φφθθ
π
θ
0
cossincos
1
cos djP
n
n
Laplace’s First Integral
Laplace’s Second Integral
SOLO
74
Legendre Polynomials
Laplace’s Integral Representation (continue -
2)
Use the Generating Function
[ ] ( )∑
∞
=
=
+− 0
2/12
21
1
n
n
n
tPu
uut
Substitute t = cosθ and u = ejφ
[ ] ( )∑
∞
=
=
+− 0
2/12
cos
cos21
1
n
n
nj
jj
Pe
ee
θ
θ
ϕ
ϕϕ
[ ] [ ] [ ] [ ] [ ] 2/12/12/12/12/12
coscos2cos2cos21 θϕθθ ϕϕϕϕϕϕ
−=+−=+− − jjjjjj
eeeeeeBut
Therefore
( )
( )
( )
( )






>
−
<
−
=
−
∞
=
∑
θϕ
θϕ
θϕ
θϕ
θ
πϕ
ϕ
ϕ
2/12/
2/12/
0
coscos2
1
coscos2
1
cos
j
j
n
n
nj
e
e
Pe
Equating the real and
imaginary parts, we obtain
( ) ( )
( )
( )
( )
( )






>
−
<
−
=∑
∞
=
θϕ
θϕ
ϕ
θϕ
θϕ
ϕ
θϕ
2/1
2/1
0
coscos2
2/sin2
coscos2
2/cos2
coscos
n
nPn
( ) ( )
( )
( )
( )
( )






>
−
<
−
−
=∑
∞
=
θϕ
θϕ
ϕ
θϕ
θϕ
ϕ
θϕ
2/1
2/1
0
coscos2
2/cos2
coscos2
2/sin2
cossin
n
nPn
SOLO
75
Legendre Polynomials
Laplace’s Integral Representation (continue -
3)
Let multiply first relation by cos (nφ) and the second by sin (nφ) and integrate
over φ on (0,π), we obtain two integrals
( ) ( )
( )
( )
( )
( )






>
−
<
−
=∑
∞
=
θϕ
θϕ
ϕ
θϕ
θϕ
ϕ
θϕ
2/1
2/1
0
coscos2
2/sin2
coscos2
2/cos2
coscos
i
iPi ( ) ( )
( )
( )
( )
( )






>
−
<
−
−
=∑
∞
=
θϕ
θϕ
ϕ
θϕ
θϕ
ϕ
θϕ
2/1
2/1
0
coscos2
2/cos2
coscos2
2/sin2
cossin
i
iPi
( ) ( ) ( ) ( ) ( ) ( )
( )
( ) ( )
( )∫ ∫∑∫ −
+
−
==
∞
=
θ π
θ
δ
π
π
ϕ
θϕ
ϕϕ
ϕ
θϕ
ϕϕ
θ
π
θϕϕϕ
0
2/12/1
0
2
0 coscos2
cos2/sin2
coscos2
cos2/cos2
cos
2
coscoscos d
n
d
n
PPdni n
i
i
in
  
( ) ( ) ( ) ( ) ( ) ( )
( )
( ) ( )
( )∫ ∫∑∫ −
+
−
−==
∞
=
θ π
θ
δ
π
π
ϕ
θϕ
ϕϕ
ϕ
θϕ
ϕϕ
θ
π
θϕϕϕ
0
2/12/1
0
2
0 coscos2
sin2/cos2
coscos2
sin2/sin2
cos
2
cossinsin d
n
d
n
PPdni n
i
i
in
  
( ) ( ) ( )
( )
( ) ( )
( ) 





−
+
−
= ∫ ∫
θ π
θ
ϕ
θϕ
ϕϕ
ϕ
θϕ
ϕϕ
π
θ
0
2/12/1
coscos
cos2/sin
coscos
cos2/cos2
cos d
n
d
n
Pn
( ) ( ) ( )
( )
( ) ( )
( ) 





−
+
−
−= ∫ ∫
θ π
θ
ϕ
θϕ
ϕϕ
ϕ
θϕ
ϕϕ
π
θ
0
2/12/1
coscos
sin2/cos
coscos
sin2/sin2
cos d
n
d
n
Pn
Dirichlet Integrals
Johann Peter Gustav
Lejeune Dirichlet
(1805 –1859)
SOLO
76
Legendre Polynomials
Add and subtract those two equations
( ) ( ) ( )
( )
( ) ( )
( ) 





−
+
−
= ∫ ∫
θ π
θ
ϕ
θϕ
ϕϕ
ϕ
θϕ
ϕϕ
π
θ
0
2/12/1
coscos
cos2/sin
coscos
cos2/cos2
cos d
n
d
n
Pn
( ) ( ) ( )
( )
( ) ( )
( ) 





−
+
−
−= ∫ ∫
θ π
θ
ϕ
θϕ
ϕϕ
ϕ
θϕ
ϕϕ
π
θ
0
2/12/1
coscos
sin2/cos
coscos
sin2/sin2
cos d
n
d
n
Pn
Dirichlet Integrals
( ) ( )[ ]
( )
( )[ ]
( ) 





−
+
+
−
+
= ∫ ∫
θ π
θ
ϕ
θϕ
ϕ
ϕ
θϕ
ϕ
π
θ
0
2/12/1
coscos
2/1sin
coscos
2/1cos
2
1
cos d
n
d
n
Pn
( )[ ]
( )
( )[ ]
( )∫ ∫ −
−
−
−
−
=
θ π
θ
ϕ
θϕ
ϕ
ϕ
θϕ
ϕ
0
2/12/1
coscos
2/1sin
coscos
2/1cos
0 d
n
d
n
Replace n by n + 1 in the last equation and substitute in the previous
( ) ( )[ ]
( )
( )[ ]
( )∫ ∫ −
+
=
−
+
=
θ π
θ
ϕ
θϕ
ϕ
π
ϕ
θϕ
ϕ
π
θ
0
2/12/1
coscos
2/1sin2
coscos
2/1cos2
cos d
n
d
n
Pn
Mehler Integrals
Gustav Ferdinand
Mehler
(1835 - 1895)
Return to Table of Content
Laplace’s Integral Representation (continue -
4)
SOLO
77
Legendre Polynomials
We found
Return to Table of Content
Integrals in terms of sin(iθ) and cos(iθ)
( ) ( ) mnnk
n
dttPtP δ
12
21
1 +
=∫
+
−
( ) ( ) mnnk
n
dPP δθθθθ
π
12
2
sincoscos
0 +
=∫
θcos=t
( ) ( ) ( )
( ) ( )




≥
++−
+
<
=
+
= −
+
−
∫ nm
nmnm
nmm
nm
a
n
tdtPt n
nn
m
!122!
!!22
0
14
2 12
2
1
1
2
2
( ) ( ) ( )
( ) ( )




≥
++−
+++
<
=
+
= +
+
+
−
+
+
∫ nm
nmnm
nmm
nm
a
n
tdtPt n
nn
m
!322!
!1!122
0
34
2 22
12
1
1
12
12
( ) ( ) ( )
( ) ( )




≥
++−
+
<
= −
∫ nm
nmnm
nmm
nm
dP n
n
m
!122!
!!22
0
sincoscos 12
0
2
2
π
θθθθ
( ) ( ) ( )
( ) ( )




≥
++−
+++
<
= +
+
+
∫ nm
nmnm
nmm
nm
dP n
n
m
!322!
!1!122
0
sincoscos 22
0
12
12
π
θθθθ
Ordinary Differential EquationsSOLO
Second Order Linear Ordinary Differential Equation (ODE)
78
Legendre Functions of the Second Kind Qn (x)
( ) [ ]
( ) ( )∑
∫
∞
=
−−
∞ −−
=
−
−
+−
−+=
0
2
12/12
0
1
2
1
cosh21
cosh1
n
n
n
n
n
txQ
x
xt
ttx
dxxxQ θθ
( ) ( )n
n
n
nn x
xd
d
n
xP 1
!2
1 2
−=
( ) ( ) 1<+= xxQBxPAy nn
( ) ( ) ( )xP
xd
d
xxP nm
m
mm
n
2/2
1−=
( ) ( ) ( )xQ
xd
d
xxQ nm
m
mm
n
2/2
1−=
( ) ( ) ( )xW
x
x
xPxQ nnn 1
1
1
ln
2
1
−−
−
+
=
( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
3
2
2
5
2
3
1
1
1
ln
2
1
2
033
022
011
0
+−=
−=
−=
−
+
=
xxQxPxQ
xxQxPxQ
xQxPxQ
x
x
xQ
( ) ( ) ( )∑=
−−− =
n
m
mnmn xPxP
m
xW
1
11
1 ( ) ( ) ( )
( ) ( )
( )
( ) ( ) ( )
[ ]
( )
( )


−
−
=


 −





 −
−
+−
−
−
+
=
−+
−−
−
−
+
=
∑
∑
=
−
=+





 −
=
−−
evennifn
oddnifnn
xP
m
nm
mn
x
x
xP
xP
rnr
rn
x
x
xPxQ
n
m
mnn
mr
n
r
rnnn
2/2
2/1
2
1
2
1
122
1
1
ln
2
1
12
142
1
1
ln
2
1
1
12
2
1
0
12
Legendre Ordinary Differential EquationSOLO
79
Legendre Functions of the Second Kind Qn (x)
( ) ( ) ( )xPxuxy nnn =
With Pn (x) being a solution of the Legendre Differential Equation
we look for the second solution having the form
( ) ( ) 0121 2
2
2
=++−− wnn
xd
wd
x
xd
wd
x
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2
2
2
2
2
2
2
xd
xPd
xu
xd
xPd
xd
xud
xP
xd
xud
xd
xyd
xd
xPd
xuxP
xd
xud
xd
xyd
n
n
nn
n
nn
n
nn
nn
++=
+=
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 01221121 2
2
22
2
2
2
=++−−−+−+− xPxunn
xd
xPd
xuxxP
xd
xud
x
xd
xPd
xxu
xd
xPd
xd
xud
xxP
xd
xud
x nn
n
nn
nn
n
nn
n
n
Substituting in the Legendre ODE we obtain
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 01212121
0
2
2
22
2
2
2
=





++−−+−−+− xuxPnn
xd
xPd
x
xd
xPd
xxP
xd
xud
x
xd
xPd
xd
xud
xxP
xd
xud
x nn
nn
n
nnn
n
n
  
or
SOLO
80
Legendre Functions of the Second Kind Qn (x) (continue – 1)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 02121 2
2
2
2
=−−+− xP
xd
xud
x
xd
xPd
xd
xud
xxP
xd
xud
x n
nnn
n
n
Equivalent to
( )
( )
( )
( )
0
1
2/
2
/
/
2
22
=
−
−+
x
x
xP
xdxPd
xdxud
xdxud
n
n
n
n
( ) ( ) ( ) 01lnln2ln 2
=−++ x
xd
d
xP
xd
d
xd
xud
xd
d
n
n
or
Integrating we obtain
( ) ( )[ ] ( ) .1lnlnln 22
constxxP
xd
xud
n
n
=−++
Therefore ( ) ( )[ ] ( ) AconstxxP
xd
xud
n
n
==−⋅⋅ .1 22
( )
( )[ ] ( )∫ −⋅
= 22
1 xxP
xd
Axu
n
n
This means that the second solution has the form
( ) ( ) ( ) ( )
( )[ ] ( )∫ −⋅
== 22
1 xxP
xd
xPxuxPxQ
n
nnnn
Legendre Ordinary Differential Equation
SOLO
81
Legendre Functions of the Second Kind Qn (x) (continue – 2)
We obtained
( ) ( ) ( ) ( )
( )[ ] ( )
1
1 22
<
−⋅
== ∫ x
xxP
xd
xPxuxPxQ
n
nnnn
( ) ( ) ( )[ ] ( )
( ) ( )[ ]
x
x
xxxd
xxxxP
xd
xPxQ
−
+
=++−−=





+
+
−
=
−⋅
= ∫∫ 1
1
ln
2
1
1ln1ln
2
1
1
1
1
1
2
1
1 2
1
2
01
00

Let calculate Q0 (x), Q1 (x)
( ) ( ) ( )[ ] ( ) ( ) 1
1
1
ln
2
1
1
2/1
1
2/1
1
1
1
22222
1
11
2
−
−
+
=





+
+
+
−
=
−
=
−⋅
= ∫∫∫ x
xx
xd
xxx
xxd
xx
x
xxP
xd
xPxQ
x
x 
Legendre Ordinary Differential Equation
SOLO
82
Legendre Functions of the Second Kind Qn (x) (continue – 2)
( )
( ) ( )
( ) ( )
( ) ( )
3
2
2
5
1
1
ln
2
1
3
2
2
5
1
1
ln
4
35
2
3
1
1
ln
2
1
2
3
1
1
ln
4
43
1
1
1
ln
2
1
1
1
1
ln
2
1
1
ln
2
1
2
3
23
3
2
2
2
11
0
+−





−
+
=+−





−
+−
=
−





−
+
=−





−
+−
=
−





−
+
=−





−
+
=






−
+
=
x
x
x
xP
x
x
xxx
xQ
x
x
x
xP
x
x
xx
xQ
x
x
xP
x
xx
xQ
x
x
xQ
Legendre Ordinary Differential Equation
SOLO
83
Legendre Functions of the Second Kind Qn (x) (continue – 3)
To obtain a general formula for Qn (x), start from the Polynomial Pn (x) that has
n zeros αi, i=1,2,…,n
( ) ( ) ( ) ( )nnn xxxkxP ααα −−−= 21
( )[ ] ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )∑=






−
+
−
+
+
+
−
=
+⋅−⋅−−−
=
−⋅
n
i i
i
i
i
nnn
x
d
x
c
x
b
x
a
xxxxxkxxP
1
2
00
22
2
2
1
222
11
11
1
1
1
αα
ααα 
( ) ( )[ ] ( ) ( )[ ] ( )[ ] ( ) ( ) ( )∑=






−
+
−
−⋅+−++=
n
i i
i
i
i
nnn
x
d
x
c
xxPxPxbxPxa
1
2
222
0
2
0 1111
αα
If we put x=1 and x = -1, and remembering that Pn (1) = 1 and Pn(-1)=(-1)n
, we obtain
2/100 == ba
Let prove that
( )
( )[ ] ( )
0
1
1
22
2
=














−⋅
−=
= ixn
ii
xxP
x
xd
d
c
α
α
Legendre Ordinary Differential Equation
SOLO
84
Legendre Functions of the Second Kind Qn (x) (continue – 4)
Let prove that ( )
( )[ ] ( )
0
1
1
22
2
=














−⋅
−=
= ixn
ii
xxP
x
xd
d
c
α
α
Start with
( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) i
x
iixi xatfinitexfprovided
xd
xfd
xxfxxfx
xd
d
i
i
αααα
α
α ==





−+−=−
=
=
02
22
The only terms that are not finite in
at x = αi are the terms ci/(x-αi) and di/(x-αi)2
, therefore
( )[ ] ( ) ( ) ( )∑=






−
+
−
+
+
+
−
=
−⋅
n
i i
i
i
i
n x
d
x
c
x
b
x
a
xxP 1
2
00
22
111
1
αα
( )
( )[ ] ( )
( )
( ) ( )
( )[ ] i
x
iii
xi
i
i
i
i
xn
i cdxc
xd
d
x
d
x
c
x
xd
d
xxP
x
xd
d
iii
=






+−=














−
+
−
−=






−⋅
−
=== ααα
α
αα
αα 2
2
22
2
1
1
Therefore if we write Pn (x) = (x-αi) Li (x), we have
( )[ ] ( ) ( )[ ] ( )
( )
( )[ ] ( )
( ) ( ) ( )
( )[ ] ( )
( ) ( ) ( )
( )[ ] ( )22
2
223
2
2322222
1
/1
2
1
/122
1
/2
1
2
1
1
iii
iiiiii
xi
ii
xi
i
ixi
i
L
xdxLdL
xxL
xdxLdxxLx
xxL
xdxLd
xxL
x
xxLxd
d
c
iii
αα
αααα
ααα
−⋅
=−−
=








−⋅
−⋅−
=








−⋅
−
−⋅
=














−⋅
=
===
Legendre Ordinary Differential Equation
SOLO
85
Legendre Functions of the Second Kind Qn (x) (continue – 5)
We proved that ( )
( )[ ] ( ) ixn
ii
xxP
x
xd
d
c
α
α
=














−⋅
−= 22
2
1
1
( ) ( ) ( )
( )[ ] ( )22
2
1
/1
2
iii
iiiiii
i
L
xdxLdL
c
αα
αααα
−⋅
=−−
=Therefore if we write Pn (x) = (x-αi) Li (x), we have
Since Pn (x) = (x-αi) Li (x) satisfies the Legendre ODE, we have
( ) ( ) ( ){ } ( ) ( ){ } ( ) ( ) ( ){ } 0121 2
2
2
=−++−−−− xLxnnxLx
xd
d
xxLx
xd
d
x iiiiii ααα
Performing the calculation and substituting x = αi, we have
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 01221 2
2
2
=






−++





+−−





+−−
= ix
iii
i
i
ii
i xLxnnxL
xd
xLd
xx
xd
xLd
xd
xLd
xx
α
ααα
( ) ( ) ( ) 0212
2
=−
=
− iii
ii
i L
xd
xLd
αα
α
α
Substituting in ci equation we get
( ) ( ) ( )
( )[ ] ( ) 0
1
/1
2 22
2
=
−⋅
=−−
=
iii
iiiiii
i
L
xdxLdL
c
αα
αααα
Therefore ( )[ ] ( ) ( )∑= −
+





+
+
−
=
−⋅
n
i i
i
n x
d
xxxxP 1
222
1
1
1
1
2
1
1
1
α
Legendre Ordinary Differential Equation
SOLO
86
Legendre Functions of the Second Kind Qn (x) (continue – 6)
We can prove that but the exact value is not important, as
we shall see
( )
( )[ ] ( ) ixn
i
i
xxP
x
d
α
α
=





−⋅
−
= 22
2
1
Therefore
( )[ ] ( ) ( )∑= −
+





+
+
−
=
−⋅
n
i i
i
n x
d
xxxxP 1
222
1
1
1
1
2
1
1
1
α
( )[ ] ( ) ( ) ∑∑∫∫∫ == −
−
−
+
=
−
+





+
+
−
=
−⋅
n
i i
i
n
i i
i
n
x
d
x
x
dx
x
d
dx
xxxxP
dx
11
222
1
1
ln
2
1
1
1
1
1
2
1
1 αα
( ) ( ) ( )
∑= −
−
−
+
=
n
i i
n
inn
x
xP
d
x
x
xPxQ
11
1
ln
2
1
α
Since Pn (x)/(x-αi) is a polynomial of order (n-1) the sum above is also a
polynomial of order (n-1), and we define it as
( ) ( )
∑=
−
−
=
n
i i
n
in
x
xP
dxW
1
1 :
α
so
( ) ( ) ( ) 1
1
1
ln
2
1
1 <−
−
+
= − xxW
x
x
xPxQ nnn
Legendre Ordinary Differential Equation
SOLO
87
Legendre Functions of the Second Kind Qn (x) (continue – 7)
To find Wn-1 (x) let use the fact that Qn (x) is a solution of Legendre’s ODE
or
( ) ( ) ( ) ( ) ( ) 0121 2
2
2
=++−− xQnn
xd
xQd
x
xd
xQd
x n
nn
( ) ( ) ( ) ( )
xd
xWd
xP
xx
x
xd
xPd
xd
xQd n
n
nn 1
2
1
1
1
1
ln
2
1 −
−
−
+
−
+
=
( ) ( ) ( )
( )
( ) ( )
2
1
2
2222
2
2
2
1
2
1
2
1
1
ln
2
1
xd
xWd
xP
x
x
xd
xPd
xx
x
xd
xPd
xd
xQd n
n
nnn −
−
−
+
−
+
−
+
=
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0121
1
2
2
1
2
1
1
ln
2
1
121
1
1
2
1
2
2
22
0
2
2
2
=+−+−−
−
−+
−
+
−
+






++−−
−
−−
xWnn
xd
xWd
x
xd
xWd
xxP
x
x
xd
xPd
xP
x
x
x
x
xPnn
xd
xPd
x
xd
xPd
x
n
nn
n
n
n
n
nn
  
( ) ( ) ( ) ( ) ( ) ( )
xd
xPd
xWnn
xd
xWd
x
xd
xWd
x n
n
nn
2121 1
1
2
1
2
2
=++−− −
−−
Legendre Ordinary Differential Equation
( ) ( ) ( ) 1
1
1
ln
2
1
1 <−
−
+
= − xxW
x
x
xPxQ nnn
SOLO
88
Legendre Functions of the Second Kind Qn (x) (continue – 8)
Use the Recursive Formula
( ) ( ) ( ) ( ) ( )
xd
xPd
xWnn
xd
xWd
x
xd
d n
n
n
211 1
12
=++






− −
−
( ) ( ) ( ) ( )
( )
∑






−
=
−−−−=
1
2
1
0
121421
l
i
in
n
xPin
xd
xPd
Since Wn-1(x) is a polynomial of order n – 1, let write
( ) ( ) ( ) ( )
( )
∑






−
=
−−−−− =++=
1
2
1
0
1231101
n
i
ininnn xPaxPaxPaxW 
Substitute those two equation in the Wn-1(x) O.D.E. to obtain
( ) ( ){ }
( )
( ) ( )
( )
( ) ( )
( )
∑∑∑






−
=
−−






−
=
−−






−
=
−− −−=++−
1
2
1
0
12
1
2
1
0
12
1
2
1
0
12
2
142211
l
i
in
n
i
ini
n
i
ini xPinxPannxPx
xd
d
a
But by Legendre O.D.E.: ( ) ( ){ } ( ) ( ) ( ) 02121 1212
2
=−−−+− −−−− xPininxPx
xd
d
inin
( ) ( ) ( ){ } ( )
( )
( ) ( )
( )
∑∑






−
=
−−






−
=
−− −−=++−−−−
1
2
1
0
12
1
2
1
0
12 14221212
l
i
in
n
i
ini xPinxPnninina
Let solve
Legendre Ordinary Differential Equation
SOLO
89
Legendre Functions of the Second Kind Qn (x) (continue – 9)
The coefficients of the same polynomial in both sides must be equal
( ) ( ) ( ){ } ( )
( )
( ) ( )
( )
∑∑






−
=
−−






−
=
−− −−=++−−−−
1
2
1
0
12
1
2
1
0
12 14221212
l
i
in
n
i
ini xPinxPnninina
( ) ( ) ( ){ } ( )14221212 −−=++−−−− innnininai
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )( )12224244
1221212
222
2
+−=−+−=++−+−+−=
++−+−−=++−−−−
iininininniniinn
nnininnninin
But
which gives
( ) ( )12
142
+−
−−
=
iin
in
ai
( )
( ) ( )
( )
( )
( ) ( ) ( ) ( )
[ ][ ]
∑ ∑∑ = =
−−
=+






−
=
−−−
+−
+−
=





 −
−
+−
=
+−
−−
=
n
m
n
m
mnmn
mi
n
i
inn xP
mn
mn
m
xP
m
n
mn
m
xP
iin
in
xW
1 1
12
1
2
1
0
121
12
12221
2
1
1221
12
142
and
( ) ( ) ( )∑=
−−− =
n
m
mnmn xPxP
m
xW
1
11
1
?????
Legendre Ordinary Differential Equation
SOLO
90
Legendre Functions of the Second Kind Qn (x) (continue -10)
( )
3,2,1,0
10
=
<≤
n
xxQn
Legendre Ordinary Differential Equation
( )
( ) ( )
( ) ( )
( ) ( )
3
2
2
5
1
1
ln
2
1
3
2
2
5
1
1
ln
4
35
2
3
1
1
ln
2
1
2
3
1
1
ln
4
43
1
1
1
ln
2
1
1
1
1
ln
2
1
1
ln
2
1
2
3
23
3
2
2
2
11
0
+−





−
+
=+−





−
+−
=
−





−
+
=−





−
+−
=
−





−
+
=−





−
+
=






−
+
=
x
x
x
xP
x
x
xxx
xQ
x
x
x
xP
x
x
xx
xQ
x
x
xP
x
xx
xQ
x
x
xQ
SOLO
91
Legendre Functions of the Second Kind Qn (x) (continue -11)
Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x)
Start with
( )
( ) ( )12
2212
1
1
1
1 +−−
++
−=
−
n
nn
x
xx
( ) ( ) ( ) ( )
( ) ( )( ) ( ) ( )
( ) ( )( )( ) ( )
( )
( ) ( ) ( )( ) ( )1
32211
11,
,121,11,1:
++−
+−+−+−
−++=
−++=−+=−=
ini
nnn
uinnuf
unnufunufuuf


( )
( ) ( ) ( )
1
!!
!
!!
!1
1
1 12
0
2
0
2212
>
+
=
+
=
−
++−
∞
=
−
∞
=
++ ∑∑ xx
in
in
x
in
in
xx
in
i
i
i
nn
( )
( )
( ) ( )( ) ( ) 1
!
21
!
0
00
<
+++
== ∑∑
∞
=
∞
=
uu
i
innn
u
i
f
uf
i
i
i
i
i

Taylor expansion around u = 0
Use u = x-2
Legendre Ordinary Differential Equation
SOLO
92
Legendre Functions of the Second Kind Qn (x) (continue -12)
Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x)
Integrate relative to x
( )
( )
( )
( ) ( )
( )
( )
( )
( )
( ) ( )
( )
( )
∑∑
∫∑∫∑∫
∞
=
++−
∞
=
∞
++−
∞
++−
∞
=
∞
++−
∞
=
∞
+
−+
+
=
++
+
=
+
=
+
=
−
0
122
0
122
12
0
2
12
0
212
122!!
!
122!!
!
!
!
!
!
1
i
in
i
xin
x
in
ix
in
ix
n
x
inni
in
inni
in
d
n
in
d
n
ind
η
ηηηη
η
η
Integrate this n more times
( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
( )
( ) ( )
( )
( )
∑
∑
∑ ∫ ∫∫∫ ∫
∞
=
++−
∞
=
++−
∞
=
∞ ∞
++−
∞ ∞ ∞
+
+
++
++
=
+++++++
+
=
++
+
=
−
0
12
0
12
0
122
12
1
!122!!
!2!
122222122!!
!
122!!
!
1
i
in
i
in
i x
nin
x
n
n
x
inni
inin
x
ininininni
in
d
inni
ind


ηη η
ηη
η
η
Legendre Ordinary Differential Equation
SOLO
93
Legendre Functions of the Second Kind Qn (x) (continue -13)
Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x)
(continue)
( )
( ) ( )
( )
( )
1
!122!
!2!
!
1
1 0
12
12
1
>
++
++
=
−
∑∫∫ ∫
∞
=
++−
∞ ∞ ∞
+
+
xx
ini
inin
n
d
i
in
x
n
n
η η η
η

We found, using Frobenius Series
By comparing those two relations we obtain
( )
( )
1
1
!2 12
1
>
−
= ∫∫ ∫
∞∞ ∞
+
+
x
d
nxQ
x
n
n
n
n
η η η
η

Return to
Frobenius Series
( ) ( ) ( )
( )
( )
1
!122!
!2!
2
0
12
>
++
++
= ∑
∞
=
++−
xx
ini
inin
xQ
i
inn
n
Legendre Ordinary Differential Equation
SOLO
94
Legendre Functions of the Second Kind Qn (x) (continue -14)
Another Expression for Legendre Functions of the Second Kind Qn (x)
Start with the following Differential Equation
( ) ( ) 02121 2
2
2
=+−+− un
xd
ud
xn
xd
ud
x
One of the Solutions is . Check:( ) ( )n
xxu 12
1 −=
( ) ( ) ( ) ( ) 22212
2
1
2
121
11412&12
−−−
−−+−=−=
nnn
xxnnxn
xd
ud
xxn
xd
ud
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) 01211411412
2121
21221222
1
1
2
1
2
2
=−−−−+−−−−−=
+−+−
−− nnnn
xnxnxnxxnnxn
un
xd
ud
xn
xd
ud
x
The Second Solution can be find using u1 (x) by finding a function v (x) that satisfies:
( ) ( ) ( )n
xxvxu 12
2 −=
Legendre Ordinary Differential Equation
SOLO
95
Legendre Functions of the Second Kind Qn (x) (continue -15)
Another Expression for Legendre Functions of the Second Kind Qn (x)
(continue – 1)
( ) ( ) ( ) ( )
( ) ( ) 01221
21212121
112
2
1
12
0
1
1
2
1
2
2
2
2
2
2
2
2
=−+





+−+






+−+−=+−+−
xd
vd
uxnu
xd
vd
u
xd
vd
xd
ud
x
un
xd
ud
xn
xd
ud
xvun
xd
ud
xn
xd
ud
x
  
The Second Solution can be find using u1 (x) by finding a function v (x) that satisfies:
( ) ( ) ( )n
xxvxu 12
2 −=
2
1
2
1
12
2
2
2
2
1
1
2
2&
xd
ud
v
xd
ud
xd
vd
u
xd
vd
xd
ud
xd
ud
vu
xd
vd
xd
ud
++=+=
( )
( )
( )
( ) ( ) ( )
( )1
121
2
2
1
12
2
1
12
/
/
2
1
2
1
2
1
1
1
2
1
22
−
+
−=
−
−
−
−
=
−
−
−
−
=
x
xn
u
x
uxn
x
uxn
u
xd
ud
x
uxn
xdvd
xdvd
Legendre Ordinary Differential Equation
SOLO
96
Legendre Functions of the Second Kind Qn (x) (continue -16)
Another Expression for Legendre Functions of the Second Kind Qn (x)
(continue – 2)
Therefore
( )
( )1
12
/
/
2
22
−
+
−=
x
xn
xdvd
xdvd
( ) ( )
( )
( )
( )∫
∞
++
−
=⇒
−
=⇒−+−=
x
nn
d
xv
xxd
vd
xn
xd
vd
1212
2
11
1
1ln1
η
η
( ) ( ) ( ) ( )
( )∫
∞
+
−
−==
x
n
n d
xxuxvxu 12
2
12
1
1
η
η
Differentiate the Differential Equation ( ) ( ) 02121 2
2
2
=+−+− un
xd
ud
xn
xd
ud
x
( ) ( ) ( )
( ) ( ) ( ) 0122221
2121221
2
2
3
3
2
2
2
2
2
2
=−+−+−=
+−+−+−−
xd
ud
n
xd
ud
xn
xd
ud
x
xd
ud
n
xd
ud
xd
d
xn
xd
ud
n
xd
ud
x
xd
ud
xd
d
x
Legendre Ordinary Differential Equation
SOLO
97
Legendre Functions of the Second Kind Qn (x) (continue -17)
Another Expression for Legendre Functions of the Second Kind Qn (x)
(continue – 3)
Differentiate relative to x the Ordinary Differential Equation n times
( ) ( ) ( ) 0122221: 2
2
3
3
2
=−+−+−
xd
ud
n
xd
ud
xn
xd
ud
xODE
xd
d
( ) ( ) ( ) 0223321: 2
2
4
4
2
2
2
=−+−+−
xd
ud
n
xd
ud
xn
xd
ud
xODE
xd
d
( ) ( ) 02121: 2
2
2
=+−+− un
xd
ud
xn
xd
ud
xODE
( ) ( ) 0121: 1
1
2
2
2
=++−− +
+
+
+
n
n
n
n
n
n
n
n
xd
ud
nn
xd
ud
x
xd
ud
xODE
xd
d
Derive ( ) ( )[ ] ( )( ) 021121: 1
1
2
2
2
=−+++−+− +
+
+
+
i
i
i
i
i
i
i
i
xd
ud
ini
xd
ud
xin
xd
ud
xODE
xd
d
( ) ( )[ ]{ } ( )[ ] ( )( ){ }
( ) ( )[ ] ( )( ) 0122221
21121221:
1
1
1
1
3
3
2
1
1
1
1
3
3
2
=−−+++−+−
−+++−++−+−+−
+
+
+
+
+
+
+
+
+
+
+
+
i
i
i
i
i
i
i
i
i
i
i
i
i
i
xd
ud
ini
xd
ud
xin
xd
ud
x
xd
ud
iniin
xd
ud
xin
xd
ud
xODE
xd
d
xd
d
Proof
by
Induction q.e.d.
i = n
i → i+1
Legendre Ordinary Differential Equation
SOLO
98
Legendre Functions of the Second Kind Qn (x) (continue -18)
Another Expression for Legendre Functions of the Second Kind Qn (x) (continue – 4)
This is the Legendre Ordinary Differential Equation for dn
u/dxn
, having the solution
Pn (x) and Qn (x). Thus, the solution Pn (x) and Qn (x) can be written in the following
form:
( ) ( ) 0121: 1
1
2
2
2
=++−− +
+
+
+
n
n
n
n
n
n
n
n
xd
ud
nn
xd
ud
x
xd
ud
xODE
xd
d
We obtain
( ) ( )[ ]n
n
n
nn
n
nn x
xd
d
nxd
ud
n
xP 1
!2
1
!2
1 21
−==
( ) ( )
( )
( )
( )
( )
( )
1
1
1
!2
!21
!2
!21
12
22
>








−
−
−
=
−
= ∫
∞
+
x
d
x
xd
d
n
n
xd
ud
n
n
xQ
x
n
n
n
nnn
n
nnn
n
η
η
Rodrigues Formula
An integral for Qn (x) valid in |x| < 1 can be obtained from the previous result
( ) ( )
( )
( )
( )
1
1
1
!2
!21
0
12
2
<








−
−
−
= ∫ +
x
d
x
xd
d
n
n
xQ
x
n
n
n
nnn
n
η
η
Return to Table of Content
Legendre Ordinary Differential Equation
99
SOLO
Laplace Differential Equation in Spherical Coordinates
0
sin
1
sin
sin
11
2
2
222
2
2
2
=
∂
Φ∂
+





∂
Φ∂
∂
∂
+





∂
Φ∂
∂
∂
=Φ∇
φθθ
θ
θθ rrr
r
rr
Let solve this equation by the method of Separation of Variables,
by assuming a solution of the form :
( ) ( )ϕθ,SrR=Φ
Spherical Coordinates:
θ
ϕθ
ϕθ
cos
sinsin
cossin
rz
ry
rx
=
=
=
In Spherical Coordinates the Laplace equation becomes:
Substituting in the Laplace Equation and dividing by Φ gives:
0sinsin
sin
11
2
2
22
2
2
=





∂
∂
+





∂
∂
∂
∂
+





φθ
θ
θ
θ
θ
SS
Srrd
Rd
r
rd
d
Rr
The first term is a function of r only, and the second of angular
coordinates. For the sum to be zero each must be a constant, therefore:
λ
φθ
θ
θ
θ
θ
λ
−=





∂
∂
+





∂
∂
∂
∂
=





2
2
2
2
sinsin
sin
1
1
SS
S
rd
Rd
r
rd
d
R
Associate Legendre Differential Equation
100
SOLO
λ
φθ
θ
θ
θ
θ
−=





∂
∂
+





∂
∂
∂
∂
2
2
2
sinsin
sin
1 SS
S
( ) ( ) ( )φθφθ ,,, SrRr =Φ
We obtain:
Multiply this by S sin2
θ and put to get:( ) ( ) ( )φθφθ ΦΘ=,S
0
1
sinsinsin
1
2
2
2
=
Φ
Φ
+





+




 Θ
∂
∂
Θ φ
θλ
θ
θ
θ
θ
d
d
d
d
Again, the first term, in the square bracket, and the last term must
be equal and opposite constants, which we write m2
, -m2
. Thus:
Φ−=
Φ
=Θ





−+




 Θ
2
2
2
2
2
0
sin
sin
sin
1
m
d
d
m
d
d
d
d
φ
θ
λ
θ
θ
θθ
The Φ ( ) must be periodical in (a period of 2 π) and becauseϕ ϕ
this we choose the constant m2
, with m an integer. Thus:
( ) φφφ mbma sincos +=Φ
Laplace Differential Equation in Spherical Coordinates
( ) φφ
φ mjmj
ee +−
=Φ ,or
With m integer, we have the Orthogonality Condition
21
21
,
2
0
2 mm
mjmj
dee δπφ
π
φφ
=⋅∫
+−
Return to Table of Content
Associate Legendre Differential Equation
101
SOLO
( ) ( ) ( )φθ ΦΘ=Φ rR
We obtain:
or:
Θ−=Θ−
Θ
+
Θ
λ
θθ
θ
θ 2
2
2
2
sin
cot
m
d
d
d
d
0
sin
sin
sin
1
2
2
=Θ





−+




 Θ
∂
∂
θ
λ
θ
θ
θθ
m
d
d
Analysis of Associate Lagrange Differential Equation
Laplace Differential Equation in Spherical Coordinates
Change of variables: t = cos θ
θθ dtd sin−=
td
d
d
d Θ
−=
Θ
θ
θ
sin

( ) td
d
t
td
d
t
td
d
d
d
td
d
td
d
td
d
td
d
d
d
d
d
d
d Θ
−
Θ
−=
Θ
+
Θ
=




 Θ
−−=




 Θ
=
Θ
−
2
2
2
cossin/1
2
2
2
2
2
1
sin
sinsinsinsin

θθ
θ
θθ
θθθθ
θθθ
Θ





−
−+
Θ
−
Θ
−
Θ
=Θ





−+
Θ
+
Θ
=
=
2
2
2
2cos
2
2
2
2
1sin
cot0
t
m
td
d
t
td
d
t
td
dm
d
d
d
d t
λ
θ
λ
θ
θ
θ
θ
We obtain:
1cos
0
1
2 2
2
2
2
≤⇒=
=Θ





−
−+
Θ
−
Θ
tt
t
m
td
d
t
td
d
θ
λ
Associate Legendre Differential Equation
102
SOLO
( ) ( ) ( )ϕθ gfrR=Φ
We obtain:
or:
ff
m
d
fd
d
fd
λ
θθ
θ
θ
−=−+ 2
2
2
2
sin
cot
Let try to factorize the left-hand side, second order differential
equation into two first-order operators:
0
sin
sin
sin
1
2
2
=





−+





∂
∂
f
m
d
fd
θ
λ
θ
θ
θθ
The two equations are identical if the coefficient are equal. This
is obtained by choosing α and β as follows:
( ) integer1
1
2
==−
=+
mmαβ
βα
( )
( ) ( ) ff
d
fd
d
fd
ff
d
fd
d
fd
f
d
fd
d
d
f
d
d
d
d
βα
θ
αβ
θ
θβα
θ
θβα
θ
β
θ
θβα
θ
θβ
θ
θα
θ
θβ
θ
θα
θ
θ
−−−++=
+−++=






+





+=





+





+
−
22
2
1
sin
1
2
22
2
sin
1
1cot
cot
sin
cot
cotcotcotcot
2

Laplace Differential Equation in Spherical Coordinates
Analysis of Associate Lagrange Differential Equation (continue – 1)
Associate Legendre Differential Equation
103
SOLO
( ) ( ) ( )ϕθ gfrR=Φ
We obtain:
and:
ff
m
d
fd
d
fd
λ
θθ
θ
θ
−=−+ 2
2
2
2
sin
cot
We have two solutions for α and β as follows:
1.β1 = m, α1 = 1-m
2.β2 = -m, α2 = 1+m
Since m is an integer α, β are also integers.
( ) integer1
1
2
==−
=+
mmαβ
βα
( ) ( ) ( ) fff
d
fd
d
fd

λ
βαλβα
θ
αβ
θ
θβα
θ
+−=−−−++ '
sin
1
1cot 22
2
βα −=1
integer22
== mmβ
Let define the two operators:
( )
( )
( ) θ
θ
θ
θ
cot1
cot
++=
−=
−
+
m
d
d
M
m
d
d
M
m
m
Laplace Differential Equation in Spherical Coordinates
Analysis of Associate Lagrange Differential Equation (continue – 2)
Associate Legendre Differential Equation
104
SOLO
( ) ( ) ( )ϕθ gfrR=Φ
We obtain:
We have two solutions for α and β as follows:
1.β1 = m, α1 = 1-m
2.β2 = -m, α2 = 1+m
Since m is an integer α, β are also integers.
( )
( )
( ) θ
θ
θ
θ
cot1
cot
++=
−=
−
+
m
d
d
M
m
d
d
M
m
m
ff
d
d
d
d
λθβ
θ
θα
θ
−=





+





+ cotcot
( )

( )

( )

( ) ( )11
11
11
11
1 mmmm fmmfMM








−−−=
−
−
−
+
−
αβ
βα
λ
We have:
( )

( )

( )

( ) ( )22
22
22
1 mmmm fmmfMM








+−−=
−
+−
αβ
βα
λ
fm
(1)
– the solution for α1, β1
fm
(2)
– the solution for α2, β2
Laplace Differential Equation in Spherical Coordinates
Analysis of Associate Lagrange Differential Equation (continue – 3)
Associate Legendre Differential Equation
105
SOLO
( ) ( ) ( )ϕθ gfrR=Φ
We obtain:
( ) ( ) ( )
( )[ ] ( )11
11 1 mmmm fmmfMM −−−=−
−
+
− λ
( ) ( ) ( )
( )[ ] ( )22
1 mmmm fmmfMM +−−=+−
λ
Let operate on first of those equations with
( )−
−1mM
( ) ( ) ( ) ( )
[ ] ( )[ ] ( ) ( )
[ ]1
1
1
111 1 mmmmmm fMmmfMMM −
−
−
−
+
−
−
− −−−= λ
In the Second Equation replace m by m-1
( ) ( ) ( )
( )[ ] ( )2
1
2
111 1 −−
+
−
−
− −−−= mmmm fmmfMM λ
Let operate on second of those equations with
( )+
mM
1
2
( ) ( ) ( ) ( )
[ ] ( )[ ] ( ) ( )
[ ]22
1 mmmmmm fMmmfMMM ++−+
+−−= λ
In the First Equation replace m by m+1
( ) ( ) ( )
( )[ ] ( )1
1
1
1 1 ++
−+
+−−= mmmm fmmfMM λ
( ) ( ) ( )21
1 mmmm fpfM =+
−
( ) ( ) ( )1
1
2
+
+
= mmmm fqfM
Laplace Differential Equation in Spherical Coordinates
pm is a constant
qm is a constant
mm →−1
mm →−1
Analysis of Associate Lagrange Differential Equation (continue – 4)
Associate Legendre Differential Equation
106
SOLO
We
obtained:( ) ( ) ( )21
1 mmmm fpfM =+
−
( ) ( ) ( )1
1
2
+
+
= mmmm fqfM
Laplace Differential Equation in Spherical Coordinates
( )
( )
( )
θ
θ
θ
θ
cot
cot1
m
d
d
M
m
d
d
M
m
m
−=
++=
+
−
where:
Theorem:
( ) ( )
( ) ( ) ( )
( )∫∫
+−
−=
ππ
θθθθθθθθ
00
sinsin dgMfdfMg mm
where f and g are arbitrary bounded function of θ.
Proof:
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )
( )∫∫
∫
∫∫
+
−
−=





−−=






+−−=






++=
ππ
π
π
ππ
θθθθθθθ
θ
θ
θθ
θ
θθ
θθθθθθ
θ
θθθθ
θθθ
θ
θθθθθθ
00
0
0
0
00
sinsin
sin
cos
sin
cos1sinsin
cot1sinsin
dgMfdgfmg
d
d
f
dgfmg
d
d
ffg
dfm
d
d
gdfMg
m
m
  
Those are Recursive Relations in m.
Analysis of Associate Lagrange Differential Equation (continue – 5)
Associate Legendre Differential Equation
107
SOLO
Laplace Differential Equation in Spherical Coordinates
( ) ( )
( )( ) ( ) ( )
( )( )∫∫
+−
−=
ππ
θθθθθθθθ
00
sinsin dgMfdfMg mm
( )
( ) ( )
( ) ( ) ( )
( )∫∫ +
−+
++
−
+
−
−=
ππ
θθθθ
0
11
0
11 sinsin dfMMfdfMfM mmmmmmmm
( )
( ) ( )
( ) ( ) ( )
∫
∫∫
=
=+
−
+
−
π
ππ
θθ
θθθθ
0
22
00
11
sin
sinsin
dfp
dfpfpdfMfM
mm
mmmmmmmm
Choose f := fm+1 and g := Mm
(-)
fm+1:
( ) ( )
( ) ( )[ ]{ }
( )[ ] ∫
∫∫
+
+++
−+
+
+−=
+−−−=−
π
ππ
θθλ
θλθθθ
0
2
1
0
11
0
11
sin1
1sinsin
dfmm
dfmmfdfMMf
m
mmmmmm
Assuming: we have1sinsin
0
2
1
0
2
== ∫∫ +
ππ
θθθθ dfdf mm
( ) ( ) λλ ≤+→+−= 11 mmmmpm
Analysis of Associate Lagrange Differential Equation (continue – 6)
Associate Legendre Differential Equation
108
SOLO
Laplace Differential Equation in Spherical Coordinates
( ) ( )
( )( ) ( ) ( )
( )( )∫∫
+−
−=
ππ
θθθθθθθθ
00
sinsin dgMfdfMg mm
Choose now f := Mm
(+)
fm and g := fm, to obtain as before
We obtained
( ) ( ) λλ ≤+→+−= 11 mmmmqm
( ) ( ) λλ ≤+→+−= 11 mmmmpm
We see that m, an integer, is bounded by λ, therefore we must choose λ as
( ) 0integer1 >=+= lllλ
In this case we have mMAX = l, mmin = -(l+1), or
( ) integer0,integer1 =>=≤≤+− mllml
Therefore
( ) ( ) ( ) lmlmmllqp mm ≤≤+−→+−+== 111
Analysis of Associate Lagrange Differential Equation (continue – 7)
Associate Legendre Differential Equation
109
SOLO
Laplace Differential Equation in Spherical Coordinates
We obtained
( ) ( ) ( ) lmlmmllqp mm ≤≤+−→+−+== 111
( )
( )
( )
θ
θ
θ
θ
cot
cot1
m
d
d
M
m
d
d
M
m
m
−=
++=
+
−
( ) ( ) ( )21
1 mmmm fpfM =+
−
( ) ( ) ( )1
1
2
+
+
= mmmm fqfM
( ) ( )
( ) ( ) ( )
( )
( ) ( )
( ) lml
mmllfm
d
d
fmmllfm
d
d
m
mm
≤≤+−







+−+=





−
+−+=





++ +
1
11cot
11cot1
2
21
1
θ
θ
θ
θ
Substituting m = - (l+1) in the First Equation and m = l in the Second we obtain:
( )
( )







=





−
=





− −
0cot
0cot
2
1
l
l
fl
d
d
fl
d
d
θ
θ
θ
θ
( )
( )
( )
( )
( )
( )
θ
θ
θ
θ
sin
sin
sin
sin
2
2
1
1
d
l
f
fd
d
l
f
fd
l
l
l
l
=
=
+
+
−
−
( ) ( )
θl
lll Cff sin
21
== +−
If we can find a Solution for a particular m, we can use the Recursive Relations
above to find the others.
Analysis of Associate Lagrange Differential Equation (continue – 8)
Associate Legendre Differential Equation
110
SOLO
Laplace Differential Equation in Spherical Coordinates
We obtained ( ) ( )
θl
lll Cff sin
21
== +−
Cl must be chosen to normalize fi and f-l:
1sinsin
0
122
0
2
== ∫∫
+
±
ππ
θθθθ dCdf l
ll
( )
( )!12
!21
212
2
+
=
+
n
n
C
n
l
( )
( )212
!2
!12
n
n
C nl +
+
=
( ) ( ) ( )∫∫∫∫
+−−+
−=+−=−=
ππ
π
ππ
θθθθθθθθθθθθ
0
1212
0
212
0
0
2
0
2
0
12
sinsin2cossin2cossincossinsin dndndd nnnnnn
  
Therefore
( )
( ) ( ) ( ) ( ) ( )
( )
( )!12
!2
cos
2222
!2
11212
!2
sin
31212
2222
sin
12
2
sin
212
2
0
1
00
12
0
12
+
=
−⋅
⋅
−⋅+
=
−⋅+
−⋅
=
+
=
+
−+
∫∫∫ n
n
nn
n
nn
n
d
nn
nn
d
n
n
d
nnn
nn

  


π
πππ
θθθθθθθ
Analysis of Associate Lagrange Differential Equation (continue – 9)
Associate Legendre Differential Equation
111
SOLO
Laplace Differential Equation in Spherical Coordinates
The Normalized Solution for m = l is defined as
( )
( )
θl
n
lm
l
lm
l
n
n
sin
!2
!12
212 +
−== +
=Θ=Θ
The Solutions for m < l can be found using the Recursive Relation:
( )
( ) ( )
( )1,,,0,,1,cot
11
11
1
1
1
+−−−=Θ





+
−−+
=Θ=Θ −
−
−
−
llllmm
d
d
mmll
M
p
m
l
m
lm
m
m
l θ
θ
( ) ( ) ( )21
1 mmmm fpfM =+
−
From which the Normalized Solutions for m < l are given by
Or we can find the Solutions for m >- l by using the Recursive Relation:
( ) ( ) ( )1
1
2
+
+
= mmmm fqfM
( )
( ) ( )
llllmm
d
d
mmll
M
q
m
l
m
lm
m
m
l ,1,,0,,,1cot
11
111
−−−−=Θ





−
+−+
=Θ=Θ ++
θ
θ
Analysis of Associate Lagrange Differential Equation (continue – 10)
Associate Legendre Differential Equation
112
SOLO
Laplace Differential Equation in Spherical Coordinates
( )
( )
θl
n
lm
l
lm
l
n
n
sin
!2
!12
212 +
−== +
=Θ=Θ
( )
( ) ( )
( )1,,,0,,1,cot
11
11
1
1
1
+−−−=Θ





+
−−+
=Θ=Θ −
−
−
−
llllmm
d
d
mmll
M
p
m
l
m
lm
m
m
l θ
θ
Examples:
( )
( ) ( )
llllmm
d
d
mmll
M
q
m
l
m
lm
m
m
l ,1,,0,,,1cot
11
111
−−−−=Θ





−
+−+
=Θ=Θ ++
θ
θ
Analysis of Associate Lagrange Differential Equation (continue – 11)
2
cos
1
1
cos
0
1
2
cos
1
1
0
0
1
2
3
sin
2
3
2
3
cos
2
3
1
2
3
sin
2
3
1
2
1
0
x
x
xl
l
x
x
x
−−=−=Θ
==Θ
−==Θ=
=Θ=
=
−
=
=
θ
θ
θ
θ
θ
θ
( )
( ) ( )
( )2
cos
22
2
2
cos
1
2
2
cos
20
2
2
cos
1
2
2
cos
22
2
1
4
15
sin
4
15
1
2
15
cossin
2
15
13
22
5
1cos3
22
5
1
2
15
cossin
2
15
1
4
15
sin
4
15
2
x
xx
x
xx
xl
x
x
x
x
x
−−=−=Θ
−−=−=Θ
−=−=Θ
−==Θ
−==Θ=
=
−
=
−
=
=
=
θ
θ
θ
θ
θ
θ
θθ
θ
θθ
θ
Return to Table of Content
Associate Legendre Differential Equation
SOLO
113
Associated Legendre Functions
Let Differentiate this equation m times with respect to t, and use
Leibnitz Rule of
Product Differentiation: ( ) ( )[ ]
( )
( ) ( )
im
im
i
im
i
m
m
td
tgd
td
tsd
imi
m
tgts
td
d
−
−
=
∑ −
=⋅
0 !!
!
Start with: ( ) ( ) ( ) ( ) 1011 2
≤=++





− ttwnntw
td
d
t
td
d
nn
or: ( ) ( ) ( ) ( ) ( ) 10121 2
2
2
≤=++−− ttwnntw
td
d
ttw
td
d
t nnn
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( )
( )twmmtwtmtwttw
td
d
t
td
d m
n
m
n
m
nnm
m
1211
122
2
2
2
−−−−=





−
++
( ) ( )
( ) ( )
( )twmtwttw
td
d
t
td
d m
n
m
nnm
m
+=




 +1
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
( )twnntwmtwttwmmtwtmtwt
m
n
m
n
m
n
m
n
m
n
m
n 122121
1122
++−−−−−−
+++
( ) ( )
( ) ( ) ( )
( ) ( ) ( )[ ] ( )
( ) 011121
122
=+−+++−−=
++
twmmnntwtmtwt
m
n
m
n
m
n
2nd
Way
SOLO
114
Associated Legendre Functions
( ) ( )
( ) ( ) ( )
( ) ( ) ( )[ ] ( )
( ) 011121
122
=+−+++−−
++
twmmnntwtmtwt
m
n
m
n
m
n
Define: ( ) ( )
( )twty
m
n=:
( ) ( )
( ) ( ) ( )
( ) ( ) ( )[ ] ( ) 011121 122
=+−+++−− tymmnntytmtyt
Now define: ( ) ( ) ( )tyttu
m
22
1: −=
Let compute:
( ) ( ) ( )1221
22
11 ytyttm
td
ud mm
−+−−=
−
( ) ( ) ( ) ( )11
22222
111 ytyttm
td
ud
t
mm
+
−+−−=−
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( )21
221221221
2222222
1121111 ytyttmyttmyttmytm
td
ud
t
td
d mmmmm
+−
−+−+−−−−+−−=





−
( ) ( ) ( )
( ) ( )
( ) ( )[ ]{ } ( ) ( ) y
t
tm
mnnmmtymmnnytmytt
mm






−
+−+−+−++−+++−−−= 2
22
222
0
12222
1
11111211
  
We get: ( ) ( ) 0
1
11 2
2
2
=





−
−++





− u
t
m
nn
td
ud
t
td
d
2nd
Way
SOLO
115
Associated Legendre Functions
Define: ( ) ( ) ( )tw
td
d
ttu nm
mm
22
1: −=
We get: ( ) ( ) 0
1
11 2
2
2
=





−
−++





− u
t
m
nn
td
ud
t
td
d
Start with Legendre Differential Equation:
( ) ( ) ( ) ( ) 1011 2
≤=++





− ttwnntw
td
d
t
td
d
nn
Summarize
But this is the Differential Equation of f (θ) obtained by solving Laplace’s Equation
by Separation of Variables in Spherical Coordinates .
02
=Φ∇
( ) ( ) ( ) ( )φθφθ ΦΘ=Φ rRr ,,
The Solutions of this Differential Equation are called Associated Legendre
Functions, because they are derived from the Legendre Polynomials, and
Legendre Functions of the Second Kind Qn (x) and are denoted:
( ) ( ) ( )tP
td
d
ttP nm
mm
m
n
22
1: −=
2nd
Way
( ) ( ) ( )tQ
td
d
ttQ nm
mm
m
n
22
1: −=
SOLO
116
Associated Legendre Functions
Let use Rodrigues Formula for Pn (t):
We see that we can define the Associated Legendre Function even for negative
m (In the Differential equation m2
appears):
( ) ( ) ( )tP
td
d
ttP nm
mm
m
n
22
1: −=
( ) ( )[ ]n
n
n
nn t
td
d
n
tP 1
!2
1 2
−=
we obtain:
Associated Legendre Functions
( ) ( ) ( )[ ]n
mn
mnm
n
m
n t
td
d
t
n
tP 11
!2
1
: 222
−−= +
+
From this equation we obtain:
( ) ( )tPtP nn =
0
2nd
Way
SOLO
117
Associated Legendre Functions
Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are
( ) ( ) ( )[ ]n
mn
mnm
n
m
n t
td
d
t
n
tP 11
!2
1
: 222
−−= +
+
Let Differentiate this equation n+m times with respect to t, and use
Leibnitz Rule of
Product Differentiation: ( ) ( )[ ]
( )
( ) ( )
im
im
i
im
i
m
m
td
tgd
td
tsd
imi
m
tgts
td
d
−
−
=
∑ −
=⋅
0 !!
!
( )[ ] ( ) ( )[ ]nn
mn
mn
n
mn
mn
tt
td
d
t
td
d
1112
+⋅−=− +
+
+
+
Since we differentiate (t-1)n and (t+1)n, all the differentials greater then n will vanish,
therefore we get
( ) ( )[ ] ( ) ( )
( )[ ] ( )[ ]
( ) ( )
( )[ ] ( )[ ]
( )
( ) ( )
( )[ ] ( )[ ]∑
−
=
+
+
−
−
−+
−+
+
+
+−
+−
+
=
++−
+
+
++−
+
+=+⋅−
mn
i
n
im
im
n
in
in
n
n
n
n
m
nmn
n
m
m
n
n
mmn
nn
mn
mn
t
td
d
t
td
d
imin
mn
t
td
d
t
td
d
mn
mn
t
td
d
t
td
d
mn
mn
tt
td
d
0
11
!!
!
0011
!!
!
11
!!
!
0011


2nd
Way
SOLO
118
Associated Legendre Functions
Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are
( ) ( ) ( )[ ]n
mn
mnm
n
m
n t
td
d
t
n
tP 11
!2
1
: 222
−−= +
+
( ) ( )[ ] ( )
( ) ( )
( )[ ] ( )[ ]
( )
( ) ( )
( ) ( )( ) ( ) ( )( )∑
∑
−
=
−−
−
=
+
+
−
−
+
+
++−−−−+−
+−
+
=
+−
+−
+
=+⋅−
mn
i
imni
mn
i
n
im
im
n
in
in
nn
mn
mn
timlnntinn
imin
mn
t
td
d
t
td
d
imin
mn
tt
td
d
0
0
111111
!!
!
11
!!
!
11

( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( )( ) ( ) ( )( ) ( ) ( )!11!11
1111
!!
1111
imnimnininiiimim
imnnninn
inim
imnnninn
−−+−−−−−+−++
+−−−⋅+−
=
−+
+−−−⋅+−


( ) ( ) ( ) ( )
( )!!
1111
imni
innnimnn
−−
+−−⋅++−
=

( ) ( ) ( )[ ]
( ) ( )
( )
( ) ( )( ) ( ) ( )( )∑
−
=
−−+
+
+
++−−−++−
−−
+−
=
−−=
mn
i
minmi
n
m
n
mn
mnm
n
m
n
tinnntimnn
imni
mn
n
t
td
d
t
n
tP
0
2/2/
2/
222
111111
!!
!
!2
1
11
!2
1
:

2nd
Way
SOLO
119
Associated Legendre Functions
Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are
( ) ( ) ( )[ ]
( ) ( )
( )
( ) ( )( ) ( ) ( )( )∑
−
=
−−+
+
+
++−−−++−
−−
+−
=
−−=
mn
i
minmi
n
m
n
mn
mnm
n
m
n
tinnntimnn
imni
mn
n
t
td
d
t
n
tP
0
2/2/
2/
222
111111
!!
!
!2
1
11
!2
1
:

Using this formula we can write the Associated Legendre Functions for – m as
( ) ( ) ( )[ ]n
mn
mnm
n
m
n t
td
d
t
n
tP 11
!2
1
: 222
−−= −
−
−−
Using Leibnitz Rule of Product Differentiation we obtain
( ) ( )[ ] ( )
( )
( )[ ] ( )[ ]∑
−
=
−−
−−
−
−
+−
−−
−
=+−
mn
i
n
i
i
n
imn
imn
nn
mn
mn
t
td
d
t
td
d
imni
mn
tt
td
d
0
11
!!
!
11
( )
( )
( ) ( ) ( ) ( ) ( ) ( )∑
−
=
−+
++−−⋅−++−
−−
−
=
mn
i
inim
tinnntimnn
imni
mn
0
111111
!!
!

( ) ( ) ( )[ ]
( ) ( )
( )
( ) ( ) ( ) ( ) ( ) ( )∑
−
=
−−+
−
−
−
−−
++−−⋅−++−
−−
−−
=
−−=
mn
i
minmi
n
m
n
mn
mnm
n
m
n
tinnntimnn
imni
mn
n
t
td
d
t
n
tP
0
2/2/
2/
222
111111
!!
!
!2
1
11
!2
1
:

2nd
Way
SOLO
120
Associated Legendre Functions
We obtained
( ) ( ) ( )[ ]
( ) ( )
( )
( ) ( )( ) ( ) ( )( )∑
−
=
−−+
+
+
++−−−++−
−−
+−
=
−−=
mn
i
minmi
n
m
n
mn
mnm
n
m
n
tinnntimnn
imni
mn
n
t
td
d
t
n
tP
0
2/2/
2/
222
111111
!!
!
!2
1
11
!2
1
:

( ) ( ) ( )[ ]
( ) ( )
( )
( ) ( ) ( ) ( ) ( ) ( )∑
−
=
−−+
−
−
−
−−
++−−⋅−++−
−−
−−
=
−−=
mn
i
minmi
n
m
n
mn
mnm
n
m
n
tinnntimnn
imni
mn
n
t
td
d
t
n
tP
0
2/2/
2/
222
111111
!!
!
!2
1
11
!2
1
:

( ) ( ) ( )
( )
( )tP
mn
mn
tP
m
n
mm
n
!
!
1
+
−
−=
−
Therefore
2nd
Way
SOLO
121
Associated Legendre Functions
Examples
( ) ( ) ( )tP
td
d
ttP nm
mm
m
n
22
1: −=
( ) ( ) 10 0
0
0 === tPtPn
( ) ( )
( )
 ( )
( ) ( )
( ) ( ) ( ) θ
θ
θ
θ
θ
θ
sin1
cos:
sin111
cos
2
1
21
1
1
1
cos
1
0
1
cos
2
1
22
1
21
1
1
−=−−=−=
===
=−=−==
=
−
=
=
t
t
t
tP
ttPtP
ttPtP
tt
td
d
ttPn
( ) ( ) ( )
( )
( )tP
mn
mn
tP
m
n
mm
n
!
!
1
+
−
−=
−
( ) ( )
( )
( )
( ) ( )
( )
( )
( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( ) ( ) θ
θθ
θ
θθ
θ
θ
θ
θ
θ
θ
2
cos
22
2
22
2
cos
2
1
211
2
2cos2
2
0
2
cos
2
1
2
2
2
1
21
2
2
cos
2
2
2
2
2
2
22
2
sin
8
1
1
8
1
!4
!0
1
2
cossin
13
!3
1
1
2
1cos3
2
13
cossin313
2
13
1
sin313
2
13
12
1
2
2
1
=
−
=
−
=
=
=
=−=−=
−=−−=
−
=
−
==
=−=




 −
−=
=−=




 −
−==
t
t
tP
t
t
tP
t
tP
ttPtP
tttP
t
tPtP
tt
t
td
d
ttP
t
t
td
d
ttPn



( ) 10 =tP ( )
2
1
2
3 2
2 −= ttP( ) ttP =1
2nd
Way
SOLO
122
Associated Legendre Functions
Examples
( ) ( ) ( )tP
td
d
ttP nm
mm
m
n
22
1: −= ( ) ( ) ( )
( )
( )tP
mn
mn
tP
m
n
mm
n
!
!
1
+
−
−=
−
2nd
Way
Return to Table of Content
SOLO
123
Associated Legendre Functions
Generating Function for Pn
m
(t)
Start with ( ) ( ) 1
21
1
:,
0
2
≤=
+−
= ∑
∞
=
utPu
utu
tug
n
n
n
Let derivate this function relative to t
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( )
( )
m
m
mm
m
m
m
m
u
m
m
utuu
m
m
mutu
t
tug
uutu
t
tug
uutu
t
tug
uutu
t
tug
!12
!12
21
!12
2242
123121
,
2
2
5
2
3
2
1
21
,
2
2
3
2
1
21
,
2
2
1
21
,
1
2/12
1
1
2/12
32/72
3
3
22/52
2
2
2/32
−
−
+−=
−
−⋅
−⋅+−=
∂
∂
−





−





−





−+−=
∂
∂
−





−





−+−=
∂
∂
−





−+−=
∂
∂
−
−−
−
−−
−
−
−
  



( ) ( ) ( ) ( )
( )
( )
( )
( ) ( ) ( )∑∑
∞
=
∞
=
+−
=−=
+−
−
−
−
=
∂
∂
−=
00
22
2/12
2/2
1
2/2
1
21
1
!12
!12,
1:,
n
m
n
n
n
nm
mm
n
m
mm
mm
m
m
m tPutP
td
d
tu
utu
ut
m
m
t
tug
ttug
( ) ( ) ( )tP
td
d
ttP nm
mm
m
n
22
1: −=Use
We obtain
SOLO
124
Associated Legendre Functions
( ) ( ) ( ) ( )
( )
( )
( )
( )∑
∞
=
+−
=
+−
−
−
−
=
∂
∂
−=
0
2/12
2/2
1
2/2
21
1
!12
!12,
1,
n
m
n
n
m
mm
mm
m
m
m tPu
utu
ut
m
m
t
tug
ttug
The Generating Function for Pn
m
(t) is
The Generating Function for Pn (t) is
( ) ( ) 1
21
1
:,
0
2
≤=
+−
= ∑
∞
=
utPu
utu
tug
n
n
n
Return to Table of Content
Generating Function for Pn
m
(t)
SOLO
125
Associated Legendre Functions
Recurrence Relations for Pn
m
(t)
Start with
Because of the existence of two indices in Pn
m
(t), we have a wide variety of
recurrence relations.
( ) ( )
( )
( )
( )
( )∑
∞
=
+−
=
+−
−
−
−
=
0
2/12
2/2
1
21
1
1
!12
!12
,
n
m
n
n
m
m
c
m
mm tPu
utu
ut
m
m
tug
m
  
or
( )
( )∑
∞
=
−
+
=
+− 0
2/12
21
1
n
m
n
mn
mm tPu
utu
c
Differentiate with respect to u
( ) ( )
( )
( ) ( )
( )( )
( )
( ) ( ) ( )
( )( ) ( ) ( )( ) ( )∑∑
∑
∑
∞
=
−
∞
=
∞
=
−
+
∞
=
−−
+
−+−=−+⇒
−+−=
+−
−+⇒
−=
+−
−+
⇒
0
12
0
0
12
2/12
0
1
2/32
2112
21
21
12
21
222/1
n
m
n
n
n
m
n
m
n
m
n
n
m
m
m
n
m
n
mn
mm
tPumnututPuutm
tPumnutu
utu
uc
utm
tPumn
utu
utm
c
Arrange so all the terms, in the last relation, have the same power of u, by
shifting indexes
( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ }∑∑
∞
−+
∞
− −−+−−−+=+−+
n
m
n
m
n
m
n
n
n
m
n
m
n
m
tPmntPmnttPmnutPmttPmu 111 1211212
First Recurrence Relation
SOLO
126
Associated Legendre Functions
Equating the terms in um
, and rearranging gives the First Recurrence Equation:
( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ }∑∑
∞
−+
∞
− −−+−−−+=+−+
n
m
n
m
n
m
n
n
n
m
n
m
n
m
tPmntPmnttPmnutPmttPmu 111 1211212
First Recurrence Relation (continue)
( ) ( ) ( ) ( ) ( ) ( )tPmntPmnttPn
m
n
m
n
m
n 11 112 +− +−++=+
Second Recurrence Relation
Start with ( )
( )
( ) ( )
( )
( )∑
∞
−+
−
−
−
==
+−
=
n
m
mm
m
n
n
m
m
m
m t
m
m
ctPu
utu
uc
tug
2/2
12/12
1
!12
!12
21
,
or
( ) ( ) ( )
( )
( ) 2/2
1
2/12
1
!12
!12
21
m
n
mm
m
n
nmm
m t
m
m
ctPuutuuc −
−
−
=+−= ∑
∞
−
+
Differentiate with respect to u
( ) ( ) ( ) ( ){ } ( )∑
∞
−+−−
+−++−−+=
n
m
n
nmnmm
m tPunutuuututumumc 12/122/121
2121222/1
( )
( ) ( ) ( ){ } ( )∑
∞
−
−
−
+−+−+=
+− n
m
n
nn
m
m
m
tPunutuutum
utu
umc 12
2/12
1
21222/1
21
ud
d
Recurrence Relations for Pn
m
(t)
SOLO
127
Associated Legendre Functions
Second Recurrence Relation (continue - 1)
or
( )
( )
( ) ( ) ( )
( )
( ) ( )
( )
( ) ( ) ( ) 1
2/1212/2
2
2/122/2
1
1121
!22
!32
1
12
2212
1
!12
!12
−
−
−−
−−=−
−
−
−
−
−−
=−
−
−
= m
m
m
m
mm ctmt
m
m
t
m
mm
t
m
m
c
( )( )
( )
( ) ( ) ( ){ } ( )∑
∞
−
−
−
−
+−+−+=
+−
−−
n
m
n
nn
m
m
m
tPunutuutum
utu
uc
mtm 12
2/12
1
12/12
21222/1
21
112
( )( ) ( ) ( ) ( ) ( ){ } ( )
    
RHS
n
m
n
nn
LHS
n
m
n
n
tPunutuutumtPutmm ∑∑
∞
−−
+−+−+=−− 1212/12
21222/1112
Rearrange the RHS , by changing indexes, to have powers of un
( ) ( ) ( ){ } ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }
( ) ( ) ( ) ( ) ( ) ( ){ }∑
∑
∑
∞
+−
∞
−+−
∞
−
++++−+=
−+−+++−+=
+−+−+
n
nm
n
m
n
m
n
n
nm
n
m
n
m
n
m
n
m
n
n
m
n
nn
utPntPtnmtPnm
utPntPtntPntPtmtPm
tPunutuutum
11
111
12
11222
1211212
21222/1
Equating the terms in un
of LHS and RHS we obtain:
( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tPntPtnmtPnmtPtmm
m
n
m
n
m
n
m
n 11
12/12
11222112 +−
−
++++−+=−−
Recurrence Relations for Pn
m
(t)
SOLO
128
Associated Legendre Functions
Second Recurrence Relation (continue - 2)
Use the First Recurrence Formula in RHS2:
( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )    
2
11
2
12/12
11222112
RHS
m
n
m
n
m
n
LHS
m
n tPntPtnmtPnmtPtmm +−
−
++++−+=−−
( ) ( ) ( ) ( ) ( ) ( )tPmntPmnttPn
m
n
m
n
m
n 11 112 +− +−++=+
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]tPmntPmn
n
nmtPntPnmRHS
m
n
m
n
m
n
m
n 1111 1
12
1
122122 +−+− +−++
+
++−+++=
( ) ( )( ) ( )( )[ ] ( ) ( ) ( ) ( ) ( )[ ] ( )
( ) ( ) ( ) ( )tPmmtPmm
tPmnnmnntPmnnmnmnRHSn
m
n
m
n
m
n
m
n
11
11
1212
1122112122212212
+−
+−
−+−−=
+−++−++++++−++=⋅+
Equating (2 n + 1) LHS2 with (2 n + 1) RHS2 we obtain the Second Recurrence
Formula:
( ) ( ) ( ) ( )tPtPtPtn
m
n
m
n
m
n 11
12
112 −+
−
−=−+
Recurrence Relations for Pn
m
(t)
SOLO
129
Associated Legendre Functions
( ) ( ) ( ) ( ) ( ){ } ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ){ }
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }tPmnmntPmnmn
n
tPt
tPtP
n
tPt
tPmntPmntPtn
tPmmnntP
t
tm
tP
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
1
1
1
1
2
1
1
1
1
2
11
12
1
2111
12
1
14
12
1
13
1122
011
1
2
1
−
+
−
−
+
−
+
+
+−
+
+
+−+−−−++
+
=−
−
+
=−
+−++=+
=+−++
−
−
Return to Table of Content
Recurrence Relations for Pn
m
(t)
SOLO
130
Associated Legendre Functions
Orthogonality of Associated Legendre Functions
( ) ( ) ( )[ ]n
mn
mnm
n
m
n t
td
d
t
n
tP 11
!2
1
: 222
−−= +
+
Let Compute
( ) ( ) ( ) ( )[ ] ( )[ ]∫∫
+
−
+
+
+
+
+
+
−
−−−=
1
1
222
1
1
111
!!2
1
dtt
td
d
t
td
d
t
qp
dttPtP
q
mq
mq
p
mp
mp
m
qp
m
q
m
p
Define X := x2
-1
( ) ( ) ( ) [ ] [ ]∫∫
+
−
+
+
+
+
+
+
−
−
=
1
1
1
1
!!2
1
dtX
td
d
X
td
d
X
qp
dttPtP q
mq
mq
p
mp
mp
m
qp
m
m
q
m
p
If p ≠ q, assume q > p and integrate by parts q + m times
[ ] [ ] mqidtX
td
d
vdX
td
d
X
td
d
u q
imq
imq
p
mp
mp
m
i
i
+==





= −+
−+
+
+
,,1,0 
All the integrated parts will vanish at the boundaries t = ± 1 as long as there is a factor
X = x2
-1. We have, after integrating m + q times
( ) ( ) ( ) ( ) [ ]∫∫
+
−
+
+
+
+
+
++
−





−−
=
1
1
1
1
!!2
11
dtX
td
d
X
td
d
X
qp
dttPtP p
mp
mp
m
mq
mq
q
qp
mqm
m
q
m
p
SOLO
131
Associated Legendre Functions
( ) ( ) ( ) ( ) [ ] 1:
!!2
11 2
1
1
1
1
−=




−−
= ∫∫
+
−
+
+
+
+
+
++
−
xXdtX
td
d
X
td
d
X
qp
dttPtP p
mp
mp
m
mq
mq
q
qp
mqm
m
q
m
p
Because the term Xm
contains no power greater than x2m
, we must have
q + m – i ≤ 2 m
or the derivative will vanish. Similarly,
p + m + i ≤ 2 p
Adding both inequalities yields
q ≤ p
which contradicts the assumption that q > p, therefore is no solution for i and the
integral vanishes.
Let expand the integrand on the right-side using Leibniz’s formula
( )
( )∑
+
=
++
++
−+
−+
+
+
+
+
−+
+
=




 mq
i
p
imp
imp
m
imq
imq
qp
mp
mp
m
mq
mq
q
X
td
d
X
td
d
imqi
mq
XX
td
d
X
td
d
X
0 !!
!
( ) ( ) qpdttPtP m
q
m
p ≠=∫
+
−
0
1
1
This proves that the Associated Legendre Functions are Orthogonal (for the same m).
Orthogonality of Associated Legendre Functions
SOLO
132
Associated Legendre Functions
( )[ ] ( ) ( ) 1:
!!2
11 2
1
1
2
21
1
2
−=




−−
= ∫∫
+
−
+
+
+
++
−
xXdtX
td
d
X
td
d
X
pp
dttP p
mp
mp
m
mp
mp
p
p
mp
m
p
Let expand the integrand on the right-side using Leibniz’s formula
( )
( )∑
+
=
++
++
−+
−+
+
+
+
+
−+
+
=




 mp
i
p
imp
imp
m
imp
imp
pp
mp
mp
m
mp
mp
p
X
td
d
X
td
d
impi
mp
XX
td
d
X
td
d
X
0 !!
!
For the case p = q we have
Because X = x2
– 1 the only non-zero term is for i = p - m
( )[ ] ( ) ( )
( ) ( ) ( )
( ) ( )
1:
!2!!2
!1 2
1
1
!2
2
2
!2
2
2
22
1
1
2
−=⋅
−
+−
= ∫∫
+
−
+
−
xXdtX
td
d
X
td
d
X
mmpp
mp
dttP
p
p
p
p
m
m
m
m
p
p
p
m
p

( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( )
( ) ( )
( )
( )
( )!
!
12
2
sin1
!!2
!2!1
1
!!2
!2!1
!12
!2
1
0
12
22
cos1
1
2
22
212
mp
mp
p
d
mpp
pmp
dtx
mpp
pmp
p
p
pp
p
ptp
X
p
p
p
p
−
+
⋅
+
=
−
−
+−
=−
−
+−
=
+
−
+
=+
−
+
∫∫
  

θθ
πθ
Orthogonality of Associated Legendre Functions
SOLO
133
Associated Legendre Functions
( ) ( ) ( ) ( ) ( )
( ) qp
m
q
m
p
t
m
q
m
p
mp
mp
p
dPPdttPtP ,
0
cos1
1
!
!
12
2
sincoscos δθθθθ
πθ
−
+
+
== ∫∫
=+
−
Therefore the Orthonormal Associated Legendre Functions is
( ) ( )
( )
( ) mnmP
mn
mnn
Θ
m
n
m
n ≤≤−
+
−+
= θθ cos
!
!
2
12
cos
Orthogonality of Associated Legendre Functions
SOLO
134
Associated Legendre Functions
Examples
( ) ( ) 10 0
0
0 === tPtPn
( ) ( )
( )
( ) ( ) θ
θ
θ
θ
θ
θ
sin1
cos
sin11
cos
2
1
21
1
cos
0
1
cos
2
1
21
1
−=−−=
==
=−==
=
−
=
=
t
t
t
ttP
ttP
ttPn
( ) ( )
( ) ( )
( )
( ) ( )
( ) θ
θθ
θ
θθ
θ
θ
θ
θ
θ
θ
2
cos
22
2
cos
2
1
21
2
2cos2
0
2
cos
2
1
21
2
2
cos
22
2
sin
8
1
1
8
1
2
cossin
1
2
1
2
1cos3
2
13
cossin313
sin3132
=
−
=
−
=
=
=
=−=
−=−−=
−
=
−
=
=−=
=−==
t
t
t
t
t
tP
tttP
t
tP
tttP
ttPn
2nd
Way
( ) ( )
( )
( ) mnmP
mn
mnn
Θ
m
n
m
n ≤≤−
+
−+
= θθ cos
!
!
2
12
cos
2
cos
1
1
cos
0
1
2
cos
1
1
0
0
1
2
3
sin
2
3
2
3
cos
2
3
1
2
3
sin
2
3
1
2
1
0
t
t
tl
l
t
t
t
−−=−=Θ
==Θ
−==Θ=
=Θ=
=
−
=
=
θ
θ
θ
θ
θ
θ
( )
( ) ( )
( )2
cos
22
2
2
cos
1
2
2
cos
20
2
2
cos
1
2
2
cos
22
2
1
4
15
sin
4
15
1
2
15
cossin
2
15
13
22
5
1cos3
22
5
1
2
15
cossin
2
15
1
4
15
sin
4
15
2
t
tt
t
tx
tl
t
t
t
t
t
−−=−=Θ
−−=−=Θ
−=−=Θ
−==Θ
−==Θ=
=
−
=
−
=
=
=
θ
θ
θ
θ
θ
θ
θθ
θ
θθ
θ
We recovered the previous results
SOLO
135
Associated Legendre Functions
Recurrence Relations for Θn
m
Functions
2nd
Way
Use the First Recurrence Formula for Pn
m
Functions:
( ) ( )
( )
( ) mnmtΘ
mn
mn
n
tP
m
n
m
n ≤≤−
−
+
+
=
!
!
12
2
( ) ( ) ( ) ( ) ( ) ( )tPmntPmnttPn
m
n
m
n
m
n 11 112 +− +−++=+
( ) ( )
( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( )tΘ
mn
mn
n
mntΘ
mn
mn
n
mntΘ
mn
mn
n
tn
m
n
m
n
m
n 11
!1
!1
32
2
1
!1
!1
12
2
!
!
12
2
12 +−
+−
++
+
+−+
−−
−+
−
+=
−
+
+
+
We have
to obtain
Cancellation of common terms leads to
( ) ( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
( )tΘ
nn
mnmn
tΘ
nn
mnmn
tΘt
m
n
m
n
m
n 11
12123212
11
−+
−+
−+
+
++
+−++
=
First Recurrence Relations for Θn
m
Functions
SOLO
136
Associated Legendre Functions 2nd
Way
Use the Second Recurrence Formula for Pn
m
Functions:
( ) ( )
( )
( ) mnmtΘ
mn
mn
n
tP
m
n
m
n ≤≤−
−
+
+
=
!
!
12
2
We have
to obtain
Cancellation of common terms leads to
Second Recurrence Relations for Θn
m
Functions
( ) ( ) ( ) ( )tPtPtPtn
m
n
m
n
m
n 11
12
112 −+
−
−=−+
( ) ( )
( )
( ) ( )
( )
( ) ( )
( )
( )tΘ
mn
mn
n
tΘ
mn
mn
n
tΘ
mn
mn
n
tn
m
n
m
n
m
n 11
12
!1
!1
12
2
!1
!1
32
2
!1
!1
12
2
112 −+
−
−−
−+
−
−
+−
++
+
=
−−
−+
+
−+
( ) ( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
( )tΘ
nn
mnmn
tΘ
nn
mnmn
tΘt
m
n
m
n
m
n 11
12
1212
1
3212
1
1 −+
−
+−
+−−
−
++
+++
=−
Return to Table of Content
Recurrence Relations for Θn
m
Functions
SOLO
137
Spherical Harmonics
The Spherical Harmonics that forms an orthogonal system,
were first introduced by Pierre Simon de Laplace in 1782
Pierre-Simon,
marquis de Laplace
(1749-1827)
Spherical Harmonics were first investigated in connection with
the Newtonian potential of Newton's law of universal
gravitation in three dimensions. In 1782, Pierre-Simon de
Laplace had, in his “Mécanique Céleste”, determined that the
gravitational potential at a point x associated to a set of point
masses mi located at points xi was given by
In 1867, William Thomson (Lord Kelvin) and Peter
Guthrie Tait introduced the solid spherical harmonics
in their “Treatise on Natural Philosophy”, and also
first introduced the name of "spherical harmonics"
for these functions. The solid harmonics were
homogeneous solutions of Laplace's equation
William Thomson,
1st Baron Kelvin
(1824 –1907)
Peter Guthrie Tait
(1831 –1901)
SOLO
138
Spherical Harmonics
( ) ( )
( )
( ) mnmP
mn
mnn
Θ
m
n
m
n ≤≤−
+
−+
= θθ cos
!
!
2
12
cos
0
sin
1
sin
sin
11
2
2
222
2
2
2
=
∂
Φ∂
+





∂
Φ∂
∂
∂
+





∂
Φ∂
∂
∂
=Φ∇
φθθ
θ
θθ rrr
r
rr
By Separation of Variables, we assume a solution of the form
( ) ( ) ( ) ( )φθφθ ΦΘ=Φ rRr ,,
Laplace Equation in Spherical Coordinates is:
f (θ) and g (φ) are described by the Differential Equations:
( )
Φ−=
Φ
=Θ





−++




 Θ
∂
∂
2
2
2
2
2
0
sin
1sin
sin
1
m
d
d
m
nn
d
d
φ
θθ
θ
θθ
The Orthonormal Solutions, as functions of the parameter m are:
( ) ϕ
π
ϕ mjm
eG
2
1
=
The functions Θn
m
are Orthonormal with respect to the angle θ and the
functions Gm
are Orthonormal with respect to angle φ.
SOLO
139
Spherical Harmonics
( ) ( ) ( ) ( ) ( ) ( )
( )
( ) mnmeP
mn
mnn
GΘY mjm
n
mmm
n
mm
n ≤≤−
+
−+
−=−= ϕ
θ
π
ϕθϕθ cos
!
!
4
12
1cos1:,
For r = constant (Spherical Surfaces) we define the function:
We can see that:
The Functions Yn
m
are Orthonormal with respect to both angles θ and φ.
( ) ( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
( )
( )
( )
( )
( ) ( ) 2121
21
2,1
212
2
1
1
21
2
2
1
1
,
02
22
1
11
2
0022
222
11
111
2
0 0
sincoscos
!
!
2
12
!
!
2
12
2
1
sincoscos1
!
!
2
12
!
!
2
12
sin,,
nn
m
n
m
n
mmm
mmjm
n
m
n
mm
m
n
m
n
dPP
mn
mnn
mn
mnn
dedPP
mn
mnn
mn
mnn
ddYY
mm
δθθθθ
ϕ
π
θθθθ
ϕθθϕθϕθ
π
θ
δ
π
ϕ
ϕ
π
θ
π
ϕ
π
θ
=
+
−+
+
−+
=
−
+
−+
+
−+
=
∫
∫∫
∫ ∫
=
==
=
−
=
+
= =
  
We can write ( ) ( ) 2121
2
2
1
1 ,,
4
,, mmnn
m
n
m
n dYY δδϕθϕθ
π
=Ω∫
Where d Ω = sin θ dθ dφ is the element of solid angle.
SOLO
140
Spherical Harmonics
One of the most important properties of Spherical Harmonics lies in the
Completeness Property, a consequence of the Sturm-Liouville form of the
Laplace Equation. This property means that any function f (θ,φ) (with
sufficient continuity properties) can be extended in a uniformly Convergent
Double Series of Spherical Harmonics
This expansion holds in the sense of mean-square convergence, which is to say
that
( ) ( )∑ ∑
∞
= −=
=
0
,,
n
n
nm
m
n
m
n Yff ϕθϕθ
( ) ( ) 0sin,,
2
0 0
2
0
=−∫ ∫ ∑ ∑
= =
∞
= −=
π
ϕ
π
θ
ϕθθϕθϕθ ddYff
n
n
nm
m
n
m
n
The expansion coefficients are the analogs of Fourier coefficients, and can be obtained
by multiplying the above equation by the complex conjugate of a spherical harmonic,
integrating over the solid angle Ω, and utilizing the above orthogonality relationships.
This is justified rigorously by basic Hilbert space theory. For the case of
orthonormalized harmonics, this gives:
( ) ( ) ( ) ( )∫ ∫∫ = =Ω
=Ω=
π
ϕ
π
θ
θθϕθϕθϕϕθϕθ
2
0 0
sin,,,, dYfddYff m
n
m
n
m
n
SOLO
141
Spherical Harmonics
To prove the fn
m
relation let compute
( ) ( )∑ ∑
∞
= −=
=
0
,,
n
n
nm
m
n
m
n Yff ϕθϕθ
( ) ( ) ( ) ( ) ( ) ( )
∑ ∑
∑ ∑ ∫∫ ∑ ∑∫
∞
= −=
∞
= −= ΩΩ
∞
= −=Ω
==
Ω=Ω=Ω
0
,,
00
1
1
1
1
11
,,,,,,
l
m
n
l
lm
mmln
m
l
l
l
lm
m
l
m
n
m
l
l
l
lm
m
l
m
l
m
n
m
n
ff
dYYfdYfYdYf
δδ
ϕθϕθϕθϕθϕθϕθ
Let choose f (θ,φ) = Pn
m
(θ,φ)
( ) ( )∫ ∫= =
=
π
ϕ
π
θ
ϕθθϕθθ
2
0 0
sin, ddYPf k
l
m
n
k
l
( ) ( ) ( )
( )
( ) klkeP
kl
kln
Y kjk
l
kk
l ≤≤−
+
−+
−= ϕ
θ
π
ϕθ cos
!
!
4
12
1:,
( ) ( )
( )
( ) ( )
( )
( )
∫∫ =
−
+
+
+
=
+
−+
−=
π
ϕ
ϕ
δ
π
θ
ϕθθθθ
π
2
0
!
!
12
2
0
,
sin
!
!
4
12
1 dedPP
ml
mln
f mj
mn
mn
n
m
l
m
n
mm
l
ln
  
( ) ( ) ( )
( ) qp
m
q
m
p
mp
mp
p
dPP ,
0
!
!
12
2
sincoscos δθθθθ
π
−
+
+
=∫
SOLO
142
Spherical Harmonics
Proof
( ) ( ) ( )ϕθϕθ ,1, m
n
mm
n YY −
−=
( ) ( ) ( )
( )
( ) ϕ
θ
π
ϕθ mjm
n
mm
n eP
mn
mnn
Y −
+
−+
−= cos
!
!
4
12
1,
( ) ( ) ( ) ( ) ( ) ( )
( )
( ) mnmeP
mn
mnn
GΘY mjm
n
mmm
n
mm
n ≤≤−
+
−+
−=−= ϕ
θ
π
ϕθϕθ cos
!
!
4
12
1cos1:,
( ) ( )
( )
( ) ( )
( )
( )
( ) ( ) ( )
( )
( ) ( ) ( )ϕθθ
π
θ
π
ϕ
ϕ
,1cos
!
!
4
12
11
cos
!
!
1
!
!
4
12
1
m
n
mmjm
n
mm
mjm
n
mm
YeP
mn
mnn
eP
mn
mn
mn
mnn
−−−−
−−−
−=
−
++
−−=
−
+
−
+
−+
−=
q.e.d.
SOLO
143
Spherical Harmonics
Visualization of Spherical Harmonics
The Laplace spherical harmonics Yl
m
can be visualized by
considering their “nodal lines", that is, the set of points on the
sphere where Re[Yl
m
], or alternatively where Im[Yl
m
]. Nodal
lines of are composed of circles: some are latitudes and
others are longitudes. One can determine the number of nodal
lines of each type by counting the number of zeros of Yl
m
in the
latitudinal and longitudinal directions independently. For the
latitudinal direction, the real and imaginary components of
the associated Legendre polynomials each possess ℓ−|m| zeros,
whereas for the longitudinal direction, the trigonometric sin
and cos functions possess 2|m| zeros.
Schematic representation of Yl
m
on
the unit sphere and its nodal lines.
Re[Yl
m
] is equal to 0 along m great
circles passing through the poles, and
along n-m circles of equal latitude.
The function changes sign each time
it crosses one of these lines.
When the spherical harmonic order m is zero (upper-left in
the figure), the spherical harmonic functions do not depend
upon longitude, and are referred to as zonal. Such spherical
harmonics are a special case of zonal spherical functions.
When ℓ = |m| (bottom-right in the figure), there are no zero
crossings in latitude, and the functions are referred to as
sectoral. For the other cases, the functions checker the
sphere, and they are referred to as tesseral.
SOLO
144
Spherical Harmonics
( )
( )
( )
( )
( )
( )
( ) ( )
( )
( ) ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
θ
π
ϕθ
θθ
π
ϕθ
θ
π
ϕθ
θθ
π
ϕθ
θ
π
ϕθ
θ
π
ϕθ
θ
π
ϕθ
θ
π
ϕθ
π
ϕθ
j
j
j
j
j
j
eY
eY
Y
eY
eY
eY
Y
eY
Y
222
2
1
2
20
2
1
2
222
2
1
1
0
1
1
1
0
0
sin
2
15
4
1
,
cossin
2
15
2
1
,
1cos3
5
4
1
,
cossin
2
15
2
1
,
sin
2
15
4
1
,
sin
2
3
2
1
,
cos
3
2
1
,
sin
2
3
2
1
,
1
2
1
,
=
−
=
−=
=
=
−
=
=
=
=
−−
−−
−−
Return to Table of Content
145
SOLO
We obtain:
This is a Sturm-Liouville Differential Equation:
( )

( )
( )

22
0sin
sin
sin
2
π
θ
π
θλ
θθ
θ
θ θ
θ
θ
≤≤−=+−







∂
∂
ff
m
d
fd
r
q
p 
Laplace’s Homogeneous Differential Equation
Analysis of f (θ) Equation
( ) ( ) ( ) bxaconstyxryxq
xd
yd
xp
xd
d
<<==++





λλ 0
Laplace Differential Equation in Spherical Coordinates
Ordinary Differential Equations
Boundary Value Problems and Sturm–Liouville Theory
Jacques Charles François
Sturm
(1855–1803)
Joseph Liouville
(1809–1882),
( ) ( ) ( ) bxaconstyxryxq
xd
yd
xp
xd
d
<<==++





λλ 0
Many problems of mathematical physics lead to differential
equations of the form
( ) ( ) ( )
( ) ( ) ( )



>+=+
>+=+
00'
00'
2
2
2
121
2
2
2
121
bbbybbyb
aaayaaya
with the Boundary Conditions B.C. [y] = 0:
When p (x), q (x), r (x) are continuous on [a,b] and p (x)>0
and r (x)>0 on [a,b], we refer to this system as a
Regular Sturm-Liouville Boundary Value Problem.
The Sturm-Liouville Problem is a two-point boundary
second-order linear differential equation problem, in which
we want to define the parameter λ such that the equation has
non trivial solutions y(t)≠0. Such problems are called
Eigenvalues Problems and the corresponding number λ an
Eigenvalue.
SOLO
146
Return to Rodrigues’ Formula
Ordinary Differential Equations
Jacques Charles François
Sturm
(1855–1803)
Joseph Liouville
(1809–1882),
( ) ( ) ( ) bxaconstyxryxq
xd
yd
xp
xd
d
<<==++





λλ 0
Many problems of mathematical physics lead to differential
equations of the form
with the Boundary Conditions B.C. [y] = 0:
When p (x), q (x), r (x) are continuous on (a,b) and p (x)>0
and r (x)>0 on (a,b). we refer to this system as a
Singular Sturm-Liouville Boundary Value Problem if one
of the following conditions occurs:
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) unboundedisbac
xrorxqorxporxrorxqorxpb
xporxpa
bx
bx
ax
ax
bx
bx
ax
ax
,
limlim
0lim0lim
∞→∞→
==
<
→
>
→
<
→
>
→
SOLO
( ) ( ) ( )
( ) ( ) ( )



>+=+
>+=+
00'
00'
2
2
2
121
2
2
2
121
bbbybbyb
aaayaaya
147
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Define the Operator: [ ]( ) ( ) ( ) yxq
xd
yd
xp
xd
d
xyL +





=:
The Sturm–Liouville Equation can be rewritten as: [ ]( ) ( ) ( ) 0=+ xyxrxyL λ
Lagrange's Identity (Boundary Value Problem)
Proof:
Joseph-Louis Lagrange
(1736 –1813),
[ ] [ ] ( ) [ ]( )vuWxp
xd
d
uLvvLu ,=−
where is the Wronskian of u and v.[ ]
xd
ud
v
xd
vd
uvuW −=:,
Theorem 1: Let u and v be functions having continuous
second derivatives on the interval [a,b]. Then
[ ] [ ] ( ) ( ) ( ) ( ) uvxq
xd
ud
xp
xd
d
vvuxq
xd
vd
xp
xd
d
uuLvvLu −





−+





=−
( ) ( ) ( ) ( ) 





−





−





+





=
xd
ud
xp
xd
d
v
xd
ud
xp
xd
vd
xd
vd
xp
xd
ud
xd
vd
xp
xd
d
u
( ) ( ) ( ) 











−=





−





=
xd
ud
v
xd
vd
uxp
xd
d
xd
ud
xpv
xd
d
xd
vd
xpu
xd
d
q.e.d.
SOLO
148
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Proof:
[ ] [ ]( )( ) ( ) [ ]( ) ( ) [ ]( )
( ) ( )( ) ( ) ( ) ( ) ( )[ ] ( ) ( )( ) ( ) ( ) ( ) ( )[ ] 0////
,,
21212121 =−−−−−−−=
−=−∫
avauaaavaaauapbvbubbbvbbbubp
avuWapbvuWbpxdxuLvvLu
b
a
q.e.d.
Green’s Formula (Boundary Value Problem)
When a2≠0 and b2≠0: ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )bvbbbvavaaav
bubbbuauaaau
2121
2121
/',/'
/',/'
−=−=
−=−=
SOLO
149
Boundary Value Problems and Sturm–Liouville Theory
Corollary 1: Let u and v be functions having continuous second derivatives on the
interval [a,b]. Then
George Green
1793-1841
tomb stone
If u and v satisfies the Boundary Conditions (y=u,v):
then:
[ ] [ ]( )( ) ( ) [ ]( ) b
a
b
a
vuWxpxdxuLvvLu ,=−∫
[ ] [ ]( )( ) 0=−∫
b
a
xdxuLvvLu
( ) ( )
( ) ( )


===+
===+
00'
00'
:..
2121
2121
bbnotaybayb
aanotayaaya
CB
Ordinary Differential Equations
Is the Inner Product on the set of Continuous
Real-valued functions on [a,b].
Green’s Formula (Boundary Value Problem)
( ) ( )∫=
b
a
xdxgxfgf :,
Takes a particularly useful and elegant form when expressed in form of a Inner
Product:
Equation [ ] [ ]( )( ) 0=−∫
b
a
xdxuLvvLu
Using this notation we can rewrite the Green’s Formula:
[ ] [ ] vuLvLu ,, =
A Linear Differential Operator that satisfies this relation for all u and v in its
domain is called a Selfadjoint Operator. We have shown that the Sturm-Luville
Operator with the domain of functions that have
continuous second derivatives on [a,b] ans satisfy the Boundary Conditions, then
L is a Selfadjoint Operator. Selfadjoint Operators are like symmetric matrices in
that that their Eigenvalues are Real-valued. We will prove this for Regular
Sturm-Liouville Boundary Value Problems.
[ ]( ) ( ) ( ) yxq
xd
yd
xp
xd
d
xyL +





=:
SOLO
150
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Proof:
Theorem 2: The eigenvalues λ for the Regular Sturm-Liouville Boundary
Value Problem
are Real and have Real-valued Eigenvalues
[ ]( ) ( ) ( ) 0=+ xyxrxyL λ
Assume λ, possible a complex number, is an Eigenvalue for the Sturm-Liouville
Problem with eigenfunction φ (x)
[ ]( ) ( ) ( ) 0=+ xxrxL ϕλϕ
and φ (x) satisfies the Boundary Conditions:
( ) ( )
( ) ( ) 00'
00'
2121
2121
===+
===+
bbnotbbbb
aanotaaaa
ϕϕ
ϕϕ
Take the Complex Conjugate and use the fact that p, q and r are Real-valued:
[ ]( ) ( ) ( ) [ ]( ) ( ) ( ) 0=+=+ xxrxLxxrxL ϕλϕϕλϕ
Since a1, a2, b1, b2 are Real-valued, also satisfies the Boundary Conditions,
hence is an Eigenvalue with the Eigenfunction .
ϕ
ϕλ
[ ]( ) ( ) ( ) ( ) ( )∫∫ −=
b
a
b
a
xdxxxrxdxxL ϕϕλϕϕ
SOLO
151
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Proof (continue):
q.e.d.
[ ]( ) ( ) ( ) ( ) ( )∫∫ −=
b
a
b
a
xdxxxrxdxxL ϕϕλϕϕ
Using Green’s Formula for Boundary
Value:
[ ] [ ]( )( ) 0=−∫
b
a
xdxuLvvLu
[ ]( ) ( ) ( ) [ ]( ) ( ) ( ) ( )∫∫∫ −==
b
a
b
a
b
a
xdxxxrxdxLxxdxxL ϕϕλϕϕϕϕ
Therefore: ( ) ( ) ( ) ( ) ( ) ( )∫∫ −=−
b
a
b
a
xdxxxrxdxxxr ϕϕλϕϕλ
( ) ( ) ( ) ( ) λλϕϕλλ =⇒=−
>
>
∫ 0
0
0
  
b
a
xdxxxr
If φ (x) is not real-valued, since λ is real, we can see that Real {φ (x)} and Im
{φ (x)} are also Eigenfunctions, and we can use Real {φ (x)} as the Real-
valued Eigenfunction.
λ is Real.
SOLO
152
Boundary Value Problems and Sturm–Liouville Theory
Theorem 2: The eigenvalues λ for the Regular Sturm-Liouville Boundary Value
Problem
are Real and have Real-valued Eigenvalues
[ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
Ordinary Differential Equations
Proof :
q.e.d.
Theorem 3: All the eigenvalues λ of the Regular Sturm-Liouville Boundary Value
Problem
are simple (for each Eigenvalue λ there is only one Linearly Independent
Eigenfunction)
Assume φ (x) and ψ (x) are two Linearly Independent Eigenfunctions
corresponding to the same Eigenvalue λ (thus φ (x) and ψ (x) ) satisfy the
differential equation for the same λ and the Boundary Condition):
Assume a2 ≠ 0 (the same argument if a1≠ 0):
If the Wronskian of two solutions of a second-order linear homogeneous
equation is zero at a point in the interval [a,b] then the solutions are linearly
dependent, and this is a contradiction.
( ) ( ) ( ) ( ) 00'&0' 212121 ===+=+ aanotaaaaaaaa ψψϕϕ
( ) ( ) ( ) ( ) ( ) ( )aaaaaaaa ψψϕϕ 2121 /'&/' −=−=
Compute the Wronskian W [φ,ψ] at x = a
[ ] ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 0//'', 2121 =−−−=−= aaaaaaaaaaaaW ψϕψϕψϕψϕψϕ
[ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
SOLO
153
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Proof :
q.e.d.
Theorem 4: Eigenfunctions of the Regular Sturm-Liouville Boundary Value
Problem
are Orthogonal with respect to the weight function r (x) on [a,b].
Let λ and μ be distinct (λ≠μ) and Real Eigenvalues with corresponding Real-
valued Eigenfunctions φ (x) and ψ (x) respectively
Use Green’s Formula:
[ ]( ) ( ) ( ) [ ]( ) ( ) ( ) 0&0 =+=+ xxrxLxxrxL ψµψϕλϕ
( ) [ ]( ) ( ) [ ]( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ }
( ) ( ) ( ) ( )∫
∫∫
−=
+−=−=
b
a
b
a
b
a
dxxxxr
dxxxrxxxrxdxxLxxLx
ψϕµλ
ϕλψψµϕϕψψϕ0
( ) ( ) ( ) 0=∫
b
a
dxxxxr ψϕ
[ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
SOLO
154
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Uniform Convergence of Eigenfunction Extension
SOLO
155
Boundary Value Problems and Sturm–Liouville Theory
Theorem 5: The Eigenvalues of the Regular Sturm-Liouville Boundary Value
Problem
form a countable, increasing sequence λ1<λ2<λ3<…
with lim n→∞ λn=+∞
[ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
Ordinary Differential Equations
Uniform Convergence of Eigenfunction Expension
Theorem 6: Let be an Orthonormal Sequence of
Eigenfunctions for the Regular Sturm-Liouville Boundary Value Problem
Let f (x) be a continuous function on [a,b] with f’(x) piecewise continuous in [a,b].
If f satisfies the Boundary Conditions then
{ }∞
=1nnφ ( )nkkn δφφ =,
[ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
( ) ( ) bxaxcxf
n
nn <<= ∑
∞
=
,
1
φ
where
( ) ( ) ( )∫=
b
a
nn dxxfxxrc φ
The Eigenfunction expension converges uniformly in [a,b]. Since this is true
for all continuous f (x) and piecewise continuous f’(x) this Functional Space
is Completely Covered by the Eigenfunctions.
Eigenfunction Expension
The series of functions is uniformly convergent to f (x)
if for all ε>0 and for all x∈(a,b) we can find a number N (ε)
such that ( ) ( ) ( )εε Nnxfxfn >∀<−
( ) ( )∑=
=
n
i
iin xcxf
1
φ
SOLO
156
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential Equations
Pointwise Convergence of Eigenfunction Expension
SOLO
157
Boundary Value Problems and Sturm–Liouville Theory
Theorem 7: Let be an Orthonormal Sequence of
Eigenfunctions for the Regular Sturm-Liouville Boundary Value Problem
Let f (x) and f’(x) piecewise continuous in [a,b]. There for any x in (a,b)
where
[ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
{ }∞
=1nnφ ( )nkkn δφφ =,
( ) ( ) ( )∫=
b
a
nn dxxfxxrc φ
( ) ( ) ( ) ( )n
xx
xx
n
xx
xx
xfxfxfxf
n
n
n
n
>
→
+
<
→
−
== lim&lim
( ) ( ) ( )[ ] bxaxfxfxc
n
nn <<+= +−
∞
=
∑ ,
2
1
1
φ
Ordinary Differential EquationsSOLO
Equation Regular
singularity
x =
Iregular
singularity
x =
1. Hypergeometric
x (x-1) y”+[(1+a+b) x –c] + aby = 0
0, 1, ∞ -
2. Legendre
(1-x2
) y” – 2 x y’ + l (l+1) = 0
-1, 1, ∞ -
3. Chebyshev
(1-x2
) y”–x y’ + n2
y= 0
-1, 1, ∞ -
4. Confluent Hypergeometric
x y” +(c –x) y’ –a y = 0
0 ∞
5. Bessel
x2
y”+x y’ +(x2
- n2
)y= 0
0 ∞
6. Laguerre
x y” +(1 –x) y’ +a y = 0
0 ∞
7. Simple Harmonic Oscillator
y” + ω2
y = 0
- ∞
8. Hermite
y”-2 x y’ 2 α y= 0
- ∞
158
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential EquationsSOLO
Equation p (x) q (x) λ r (x)
Legendre 1-x2
0 l (l+1) 1
Shifted Legendre x (1-x) 0 l (l+1) 1
Associated Legendre 1-x2
-m2
/(1-x2
) l (l+1) 1
Chebyshev I (1-x2
)1/2
0 n2
(1-x2
)-1/2
Shifted Chebyshev I [x (1-x)]1/2
0 n2
[x (1-x)]-1/2
Chebyshev II (1-x2
)3/2
0 n (n+2) (1-x2
)1/2
Ultraspherical
(Gegenbauer)
(1-x2
)α+1/2
0 n (n+2 α) (1-x2
)α-1/2
Bessel x -n2
/x a2
x
Laguerre x e-x
0 α e-x
Associated Laguerre xk+1
e-x
0 α-k xk
e-x
Hermite e-x2
0 2 α e-x2
Simple Harmonic
Oscillator
1 0 n2
1
Self-Adjoint ODE
159
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential EquationsSOLO
Equation Interval r (x) Standard Normalization
Legendre -1 ≤ x ≤1 1
Shifted Legendre 0 ≤ x ≤1 1
Chebyshev I -1 ≤ x ≤1 (1-x2
)-1/2
Shifted Chebyshev I 0 ≤ x ≤1 [x (1-x)]-1/2
Chebyshev II -1 ≤ x ≤1 (1-x2
)1/2
Laguerre 0 ≤ x < ∞ e-x
Associated Laguerre 0 ≤ x < ∞ xk
e-x
Hermite -∞ ≤ x < ∞ e-x2
Orthogonal Polynomials Generated by Gram-Schmidt Orthogonalization
( )[ ]
12
2
1
1
2
+
=∫
+
−
n
dxxPn
( )[ ]
12
1
*
1
0
2
+
=∫
+
n
dxxPn
( )[ ]
( ) 


=
≠
=
−
∫
+
−
0
02/
1
1
1
2/12
2
n
n
dx
x
xTn
π
π
( )[ ]
( ) 


=
≠
=
−
∫
+
0
02/
1
*
1
0
2/12
2
n
n
dx
x
xTn
π
π
( )[ ] ( ) 2
1
1
1
2/122 π
=−∫
+
−
dxxxUn
( )[ ] 1
0
2
=∫
∞
−
dxexL x
n
( )[ ] ( )
!
!
0
2
n
kn
dxexxL xkk
n
+
=∫
∞
−
( )[ ] !2 2/1
0
2 2
ndxexH nx
n π=∫
∞
−
160
Boundary Value Problems and Sturm–Liouville Theory
Ordinary Differential EquationsSOLO
Equation a b r (x)
Legendre -1 1 1
Shifted Legendre 0 1 1
Associated Legendre -1 1 1
Chebyshev I -1 1 (1-x2
)-1/2
Shifted Chebyshev I 0 1 [x (1-x)]-1/2
Chebyshev II -1 1 (1-x2
)1/2
Laguerre 0 ∞ e-x
Associated Laguerre 0 ∞ xk
e-x
Hermite -∞ ∞ e-x2
Simple Harmonic
Oscillator
0
-π
2π
π
1
1
The Weighting Function r (x) is established by putting the ODE in Self-Adjoint form.
Return to Expansion of Functions, Legendre Series
Return to Table of Content Return to Spherical Harmonics
161
Boundary Value Problems and Sturm–Liouville Theory
References
SOLO
162
Legendre Functions
http://en.wikipedia.org/wiki/
Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary
Value Problems“, 4th
Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue
Problems and Sturm-Liouville Equations”
Byron Jr., F.W., Fuller, R.W., “Mathematics of Classical and Quantum
Physics”, Dover, 1969, 1970
G.M. Wysin, “Associated Legendre Functions and Dipole Transition Matrix
Elements”, http://www.phys.ksu.edu/personal/wysin
Wallace, P.R., “Mathematical Analysis of Physical Problems”, Dover 1972, 1984
Afken, Weber, “Mathematical Methods for Physicists”, Harcourt Academic Press,
5th
Ed., 2001
W. W. Bell, “Special Functions for Scientists and Engineers”, D. van
Nostrand Company ltd., 1968
S.I. Hayek, “Advanced Mathematical Methods in Science and Engineering”,
Marcel Dekker Inc., 2001
L.C. Andrews, “Special Functions for Engineers and Applied Mathematics”,
MacMillan Publishing Companie, 1985
Return to Table of Content
163
SOLO
Technion
Israeli Institute of Technology
1964 – 1968 BSc EE
1968 – 1971 MSc EE
Israeli Air Force
1970 – 1974
RAFAEL
Israeli Armament Development Authority
1974 – 2013
Stanford University
1983 – 1986 PhD AA

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Legendre functions

  • 1. Legendre Functions SOLO HERMELIN Updated: 20.02.13 1 http://www.solohermelin.com
  • 2. Table of Content SOLO 2 Legendre Functions Introduction Legendre Polynomials History Second Order Linear Ordinary Differential Equation (ODE) Laplace’s Homogeneous Differential Equation Legendre Polynomials The Generating Function of Legendre Polynomials Rodrigues' Formula Series Solutions – Frobenius’ Method Recursive Relations for Legendre Polynomial Computation Orthogonality of Legendre Polynomials Expansion of Functions, Legendre Series Schlaefli Integral Laplace’s Integral Representation Neumann Integral
  • 3. Table of Content (continue) SOLO 3 Legendre Functions Associate Lagrange Differential Equation Laplace Differential Equation in Spherical Coordinates Analysis of Associate Lagrange Differential Equation Associate Lagrange Differential Equation (2nd Way) Generating Function for Pn m (t) Recurrence Relations for Pn m (t) Orthogonality of Associated Legendre Functions Recurrence Relations for Θn m Functions Spherical Harmonics References Boundary Value Problems and Sturm–Liouville Theory
  • 4. SOLO 4 Legendre Polynomials Adrien-Marie Legendre (1752 –1833( In mathematics, Legendre functions are solutions to Legendre's differential equation: ( ) ( ) ( ) ( ) 011 2 =++      − xPnnxP xd d x xd d nn They are named after Adrien-Marie Legendre. This ordinary differential equation is frequently encountered in physics and other technical fields. In particular, it occurs when solving Laplace's equation (and related partial differential equations) in spherical coordinates. The Legendre polynomials were first introduced in 1785 by Adrien-Marie Legendre, in “Recherches sur l’attraction des sphéroides homogènes”, as the coefficients in the expansion of the Newtonian potential ( )∑ ∞ =       =       −      + = −+ = − 0 222 cos '1 cos ' 2 ' 1 11 cos'2' 1 ' 1 n n n P r r r r r r rrrrrrrr γ γ γ  Return to Table of Content
  • 5. Second Order Linear Ordinary Differential Equation (ODE) Consider a Second Order Linear Ordinary Differential Equation (ODE) define by the Operator SOLO ( ) ( ) ( ) ( ) ( )xuxp xd ud xp xd ud xpxu 212 2 0: ++=L defined in the interval a ≤ x ≤ b, with the coefficients p0 (x), p1 (x), p2 (x) real in this region and with the first 2 – i derivatives of pi (x) continuous . Also we require that p0 (x) is nonzero in the interval. Define the Quadratic Form of the Operator L as: ( ) ( ) ( ) dxuup xd ud p xd ud pdxxuxuxu b a b a ∫∫       ++== 212 2 0: LLu, ( ) ( ) ( ) ( ) ( ) ( ) ( )∫ ∫∫ ∫ ∫∫ ∫∫∫ +−++      −      = +−+−      = ++ b a b a b a b a b a b a b a b a b a b a b a b a b a b a xdupuxdup xd d uupudxup xd d uup xd d u xd ud up xdupuxdup xd d uupudx xd ud up xd d xd ud up xdupxdu xd ud pxdu xd ud p 21102 2 00 21100 2 212 2 0 5
  • 6. SOLO ( ) ( ) ( ) dxuup xd ud p xd ud pdxxuxuxu b a b a ∫∫       ++== 212 2 0: LLu, ( ) ( ) ( ) ( ) b a b a b a b a b a xu xd pd pxu upu xd ud upu xd pd u xd ud upupuup xd d u xd ud up             −=       +−−=+      −      0 1 10 0 0100 Therefore: ( ) ( ) ( ) ( )∫       +−+            −= b a b a dxupup xd d up xd d uxu xd pd pxu 2102 2 0 1 Define the Adjoint Operator: ( ) ( ) ( ) upup xd d up xd d xu 2102 2 : +−=L 6 Second Order Linear Ordinary Differential Equation (ODE)
  • 7. SOLO ( ) dx u up xd ud p xd ud puxu b a ∫       ++=    L Lu, 212 2 0 The two Integrals are equal if: ( ) ( ) ( ) ( )∫       +−+            −= b a b a dxupup xd d up xd d uxu xd pd pxu    uL 2102 2 0 1 or: ( ) ( ) ( ) ( ) ( ) ( )xuxq xd xud xp xd d xuxu +      == LL ( ) ( ) 02 1 01 2 0 2 2102 2 212 2 0 =      −+      −=       +−−      ++ xd ud p xd pd uu xd pd xd pd u upup xd d up xd d uup xd ud p xd ud pu ( ) ( ) xd xpd xp 0 1 = If this condition is satisfied, then the terms at the boundary x = a and x = b also vanish, and we have by defining p (x): = p0 (x) and q (x) := p2 (x) 7 Second Order Linear Ordinary Differential Equation (ODE)
  • 8. SOLO ( ) ( ) ( ) ( ) ( ) ( )xuxq xd xud xp xd d xuxu +      == LL This Operator is called Self-Adjoint ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )      ++      =      ∫∫ xuxp xd ud xp xd ud xptd tp tp xp xutd tp tp xp xx 212 2 0 0 1 00 1 0 expexp 1 L 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )xutd tp tp xp xp xd ud td tp tp xd d xq x xp x            +                   = ∫∫ 0 1 0 2 0 1 expexp This is clearly Self-Adjoint. We can see that p0 (x) is in the denominator. This is the reason that we requested the p0 (x) to be nonzero in the interval a ≤ x ≤ b. p0 (x1) = 0 means that the Differential Equation is not Second Order at that point. We can always transform the Non-Self-Adjoint Operator to a Self-Adjoint one by multipling L by ( ) ( ) ( )       ∫x td tp tp xp 0 1 0 exp 1 8 Second Order Linear Ordinary Differential Equation (ODE)
  • 9. Consider a Second Order Linear Ordinary Differential Equation (ODE) SOLO ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 210 :'':'0''': xd ud u xd ud uxuxpxuxpxuxpxu ===++=L If we know one Solution u1 (x) we can find a second u2 (x). Proof If we have two Solutions u1 (x) and u2 (x), then 0''' 0''' 222120 121110 =++ =++ upupup upupup Multiply first equation by u2 (x) and second by u1 (x) and subtract: ( ) ( ) 0'''''' 1221112210 =−+− uuuupuuuup The Wronskian W is defined as: 1221 21 21 '' '' : uuuu uu uu W −== 0' 10 =+ WpWp 9 Theorem: “Method of Reduction of Order” Second Order Linear Ordinary Differential Equation (ODE)
  • 10. Consider a Second Order Linear Ordinary Differential Equation (ODE) SOLO ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 210 :'':'0''': xd ud u xd ud uxuxpxuxpxuxpxu ===++=L Theorem: “Method of Reduction of Order” If we know one Solution u1 (x) we can find a second u2 (x). Proof (continue -1) 0' 10 =+ WpWp xd p p W Wd 0 1 −= ( ) ( ) ( ) ( )         −= ∫ x x d p p xWxW 0 0 1 0 exp τ τ τ From: ( ) ( ) ( ) ( ) ( )xuxuxuxuxW 1221 '' −= q.e.d. Therefore: ( ) ( ) ( ) ( )∫= x x d u W xuxu 0 2 1 12 τ τ τ ( ) ( ) ( ) ( ) ( ) ( ) ( )       =−= 1 2 22 1 1 1 2 2 1 '' u u xd d xu xu xu xu xu xu xW Divide by u1 2 (x) x0 and W (x0) are given (or not). 10 Second Order Linear Ordinary Differential Equation (ODE)
  • 11. If the Second Order Linear Ordinary Differential Equation (ODE) is in the Self-Adjoint Mode: SOLO Then: The Wronskian is given by ( ) ( ) ( ) ( ) ( ) ( ) 0=+      == xuxq xd xud xp xd d xuxu LL ( ) ( ) ( ) ( ) ( ) ( ) 0 1 0 2 2 =++ xuxq xd xud xd xpd xd xud xp p p  ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )xp xp xW p pd xWd p p xWxW x x x x 0 00 0 0 0 0 0 1 0 00 expexp =         −=         −= ∫∫ τ τ τ 11 Return to Table of Content Second Order Linear Ordinary Differential Equation (ODE)
  • 12. 12 SOLO The Laplace’s Homogeneous Differential Equation is: We want to find the Potential Φ at the point F (field) due to all the sources (S) in the volume V, including its boundaries . n i iSS 1= = iS nS  n i iSS 1= =dV dSn → 1 V Fr  Sr  F 0r SF rrr  −= iS nS dV dSn → 1 V Fr  Sr  F 0r SF rrr  −= F inside V F on the boundary of V Therefore is the vector from S to F.SF rrr  −= →→→ ++= zzyyxxr 111  →→→ ++= zzyyxxr SSSS 111  →→→ ++= zzyyxxr FFFF 111  Let define the operator that acts only on the source coordinate. →→→ ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ z z y y x x SSS S 111 Sr  Laplace’s Homogeneous Differential Equation ( ) 02 =Φ∇ r Pierre-Simon, marquis de Laplace (1749 1827)
  • 13. 13 SOLO Laplace’s Homogeneous Differential Equation Laplace Differential Equation in Spherical Coordinates ( ) 02 =Φ∇ r Spherical Coordinates: θ ϕθ ϕθ cos sinsin cossin rz ry rx = = = A Solution in Spherical Coordinates is: ( ) 0 1 ≠=Φ r r r ( ) 0 111 22 ≠−=∇−=∇=Φ∇ r r r r r rr r  ( ) ( ) 00 33111 35333 2 ≠=−⋅=⋅∇−⋅      −∇=      ⋅−∇=Φ∇⋅∇=Φ∇ r r rr r r r r r r r rr  r r r z r y r x zyx zyx r zyxzyx  =++=++      ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ 111111 222 ( ) 3111111 =++⋅      ∂ ∂ + ∂ ∂ + ∂ ∂ =⋅∇ zyx zyx r zyxzyx  2222 zyxr ++=
  • 14. 14 SOLO Laplace’s Homogeneous Differential Equation Laplace Differential Equation in Spherical Coordinates Spherical Coordinates: θ ϕθ ϕθ cos sinsin cossin rz ry rx = = = θcos= ∂ ∂ z r 2222 zyxr ++= 0 cos11 22 ≠−= ∂ ∂ −=      ∂ ∂ = ∂ Φ∂ r rz r rrzz θ 0 1cos3331 3 2 5 22 4332 2 ≠ − = − = ∂ ∂ +−=      − ∂ ∂ = ∂ Φ∂ r rr rz z r r z rr z zz θ 0 cos9cos159153 5 26 3 4 23 7 23 6 22 55 22 3 3 ≠ − = − −= ∂ ∂− − ∂ ∂ − =      − ∂ ∂ = ∂ Φ∂ r r r r rzz z r r rz r z r rz r rz zz θθ We note that ∂n Φ/∂zn is a n-degree polynomial in cos θ divided by rn+1 . ( ) ( ) ( ) ( ) ( ) ( )zyx zn hzyx z hzyx z hzyxhzyx n nn n ,, ! 1 ,, !2 1 ,,,,,, 2 2 2 ∂ Φ∂− ++ ∂ Φ∂ + ∂ Φ∂ −Φ=−Φ  Using Taylor’s Series development we obtain
  • 15. 15 SOLO Laplace’s Homogeneous Differential Equation Laplace Differential Equation in Spherical Coordinates Spherical Coordinates: θ ϕθ ϕθ cos sinsin cossin rz ry rx = = = 2222 zyxr ++= ( ) ( ) ( )       ∂ ∂− ++      ∂ ∂ +      ∂ ∂ −= −++ =−Φ rzn h rz h rz h rhzyx hzyx n nn n 1 ! 11 !2 1111 ,, 2 2 2 222  ( ) ∑ ∞ =       ∂ ∂− = +− 0 22 1 ! 1 cos2 1 n n nn n rzn h hhrr θ or: Let define: ( ) ( )       ∂ ∂− = + rzn rP n nn n n 1 ! 1 :cos 1 θ ( ) rhP r h rhhrr n n n ≤      = +− ∑ ∞ =0 22 cos 1 cos2 1 θ θ Therefore:
  • 16. 16 SOLO Laplace’s Homogeneous Differential Equation Laplace Differential Equation in Spherical Coordinates Spherical Coordinates: θ ϕθ ϕθ cos sinsin cossin rz ry rx = = = 2222 zyxr ++= Let define w:=Pn (cos θ) and substitute w/rn+1 in the Laplace Equations in Spherical Coordinates instead of Φ Therefore, since w:=Pn (cos θ) (Partial Derivative becomes Total Derivative): 0 sin 1 sin sin 11 0 12 2 22121 2 2 =      ∂ ∂ +            ∂ ∂ ∂ ∂ +            ∂ ∂ ∂ ∂ +++  nnn r w rr w rr w r r rr φθθ θ θθ 0sin sin 111 1 =      ∂ ∂ ∂ ∂ +    + − ∂ ∂ + θ θ θθ w rr n r w nn or: ( ) 0sin sin 1 1 =      ++ θ θ θθ d wd d d wnn Substitute t=cos θ (dt = - sinθ dθ):  ( ) ( )       −=         −−=           =      −− td wd t td d td wd t td d d td td wd d td d d d wd d d 2 sin 2 1sin 11 sin 1 sin sin 1 sin sin 1 2   θ θ θθθ θ θθθ θ θθ Finally we obtain: ( ) ( ) 1011 2 ≤=++      − twnn td wd t td d Return to Table of Content
  • 17. 17 SOLO This is the Legendre Differential Equation and Pn (t) the Legendre Polynomials are one of the two solutions of the ODE. ( ) ( ) ( ) ( ) 1011 2 ≤=++      − ttPnntP td d t td d nn ( )∑ ∞ =       =       +− 0 2 cos cos21 1 n n n P r h r h r h θ θ We found ( ) 1cos cos21 1 0 2 ≤= +− ∑ ∞ = uPu uu n n n θ θFor u:=h/r The left-hand side is called “The Generating Function of Legendre Polynomials” Legendre Polynomials The Generating Function of Legendre Polynomials
  • 18. 18 SOLO Let use Taylor expansion for the function: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )  +++++= n n x n f x f x f fxf ! 0 !2 0 1 0 0 2 21 ( ) ( ) 1, 21 1 0 2 ≤= −+ ∑ ∞ = tutPu tuu n n n ( ) ( ) ( ) 101 2 1 =−= − fxxf ( ) ( ) ( ) ( ) ( ) 2 1 01 2 1 1 2 3 1 =−= − fxxf ( ) ( ) ( ) ( ) ( ) 2 3 2 1 01 2 3 2 1 2 2 5 2 ⋅=−⋅= − fxxf ( ) ( ) ( ) ( ) ( ) ( ) ( ) !2 !2 !2 !2 2 1231 2 12 2 3 2 1 01 2 12 2 3 2 1 2 2 12 n n n nnn fx n xf nn n n n n n =⋅ −⋅ = − ⋅=− − ⋅= + −   Tacking x = 2 u t - u2 we obtain ( ) ( ) ( ) ( ) 12 !2 !2 21 1 0 2 222 ≤−= −− ∑ ∞ = uutu n n utu n n n Let prove that Pn (t) are indeed Polynomials. Legendre Polynomials The Generating Function of Legendre Polynomials
  • 19. 19 SOLO Using Taylor expansion we obtained: ( ) ( ) 1, 21 1 0 2 ≤= −+ ∑ ∞ = tutPu tuu n n n Take the binomial expansion of (2 u t - u2 )n we obtain ( ) ( ) ( ) ( ) 12 !2 !2 21 1 0 2 222 ≤−= −− ∑ ∞ = uutu n n utu n n n ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 12 !!!2 !2 12 !! ! 1 !2 !2 21 1 0 0 2 0 0 222 ≤ − −= − −= −− ∑∑∑ ∑ ∞ = = +− ∞ = = − uut knkn n ut knk n n n utu n n k knkn n k n n k kknk n Change Variables in the second sum from n+k to n ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( )   − =      ≤ −− − −= −− ∑ ∑ ∞ = = − − evennifn oddnifnn uut knknk kn utu n n k nkn kn k 2/1 2/ 2 12 !2!!2 !22 1 21 1 0 2/ 0 2 222 Equating in the two power series the un coefficients, we obtain ( ) ( ) ( ) ( ) ( ) [ ] ∑= − ≤ −− − −= 2/ 0 2 1 !2!!2 !22 1 n k kn n k n tt knknk kn tP Polynomial of order n in t Return to Rodrtgues Formula Return to Frobenius Series We can see that for n odd the polynomial Pn (t) has only odd powers of t and for n even only even powers of t. Legendre Polynomials The Generating Function of Legendre Polynomials
  • 20. 20 SOLO Let use Taylor expansion for the function: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )  +++++= n n x n f x f x f fxf ! 0 !2 0 1 0 0 2 21 ( ) ( ) 1, 21 1 0 2 ≤= −+ ∑ ∞ = tutPu tuu n n n Tacking x = u2 -2 u t we obtain ( ) ( ) ( ) 101 2 1 =+= − fxxf ( ) ( ) ( ) ( ) ( ) 2 1 01 2 1 1 2 3 1 −=+−= − fxxf ( ) ( ) ( ) ( ) ( ) 2 3 2 1 01 2 3 2 1 2 2 5 2 ⋅=+⋅= − fxxf ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 12 2 3 2 1 101 2 12 2 3 2 1 1 2 12 − ⋅−=+ − ⋅−= + − n fx n xf nn n nn  ( ) ( ) ( ) ( ) ( ) ( )∑ ∞ = =      +− +      − +      − ++= −+−−−+−−= −+ 0 24 4 3 3 2 20 4232222 2 8 33035 2 35 2 13 1 2 128 35 2 16 5 2 8 3 2 1 1 21 1 n n n tPu tt u tt u t utuu tuutuutuutuu ttuu   Legendre Polynomials The Generating Function of Legendre Polynomials
  • 21. SOLO 21 Legendre Polynomials The first few Legendre polynomials are: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 256/63346530030900901093954618910 128/31546201801825740121559 128/35126069301201264358 16/353156934297 16/51053152316 8/1570635 8/330354 2/353 2/132 1 10 246810 3579 2468 357 246 35 24 3 2 −+++− +−+− +−+− −+− −+− +− +− − − xxxxx xxxxx xxxx xxxx xxx xxx xx xx x x xPn n
  • 22. 22 SOLO ( ) 1, 21 1 0 2 ≤= +− ∑ ∞ = tutPu utu n n n Substitute u by – u in this equation: ( ) ( ) ( ) 11 21 1 00 2 ≤−=−= ++ ∑∑ ∞ = ∞ = utPutPu utu n n n n n nn which results in the following identity: ( ) ( ) ( )tPtP n n n 1−=− For t =1 we have ( ) 11 1 1 21 1 0 2 ≤= − = +− ∑ ∞ = uPu uuu n n n But 1 1 1 0 ≤= − ∑ ∞ = uu u n n By equalizing the coefficients of un in the two sums, we obtain: ( ) nPn ∀=11 and ( ) ( ) ( ) ( )n n n n PP 1111 −=−=− Legendre Polynomials The Generating Function of Legendre Polynomials
  • 23. 23 SOLO ( ) 1, 21 1 0 2 ≤= +− ∑ ∞ = tutPu utu n n n For t = 0 we obtain: Legendre Polynomials ( ) 10 1 1 0 2 ≤= + ∑ ∞ = uPu u n n n ( ) ( ) ( ) ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑∑∑ ∞ = ∞ = ∞ = − −= ⋅−⋅⋅−⋅⋅ −= −⋅⋅ −= + −−−− ++ −− +      −+=+= + 0 22 2 0 1 2 0 2 22222/12 2 !2 !2 1 !2 222642 !2 12531 1 !2 12531 1 ! 2/123/12/1 !2 3/12/1 2 1 11 1 1 n n n n n nn n n n n n n n n un n nn n un n un t n n ttu u        Therefore we have: ( ) ( ) ( ) ( ) 10 !2 !2 1 1 1 00 22 2 2 ≤=−= + ∑∑ ∞ = ∞ = uPu n un u n n n n n n n By equating coefficients of un on both sides we obtain: ( ) ( ) ( ) ( ) ( ) 00 !2 !2 10 12 222 = −= +n n n n P n n P Return to Table of Content The Generating Function of Legendre Polynomials
  • 24. Derivation of Legendre Polynomials via Rodrigues’ Formula SOLO 24 Legendre Polynomials Benjamin Olinde Rodrigues (1794-1851) In mathematics, Rodrigues' Formula (formerly called the Ivory– Jacobi formula) is a formula for Legendre polynomials independently introduced by Olinde Rodrigues (1816), Sir James Ivory (1824) and Carl Gustav Jacobi (1827). The name "Rodrigues’ formula" was introduced by Heine in 1878, after Hermite pointed out in 1865 that Rodrigues was the first to discover it, and is also used for generalizations to other orthogonal polynomials. Askey (2005) describes the history of the Rodrigues’ Formula in detail. Rodrigues stated his formula for Legendre polynomials Pn Carl Gustav Jacob Jacobi (1804 –1851) ( ) ( )[ ]n n n nn x xd d n xP 1 !2 1 2 −=
  • 25. SOLO 25 Legendre Polynomials Olinde Rodrigues (1794-1851) Start from the function: ( ) .12 constkxky n =−= ( ) 12 12:' − −== n xxkn xd yd y ( ) ( ) ( ) 22212 2 2 11412:'' −− −−+−== nn xxnknxkn xd yd y Let compute: ( ) ( ) ( ) ( ) ( ) '12211412''1 12222 yxnynxxnknxknyx nn −+=−−+−=− − or: ( ) ( ) 02'12''12 =−−+− ynyxnyx Let differentiate the last equation n times with respect to x: ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ''1''2''1 00''1 3 ''1 2 ''1 1 ''1''1 2 2 1 1 2 3 3 0 2 3 3 2 2 2 2 2 1 1 222 y xd d nny xd d xny xd d x y xd d x xd dn y xd d x xd dn y xd d x xd dn y xd d xyx xd d n n n n n n n n n n n n n n n n − − − − − − − − − − −++−= +−      +−      +−      +−=−   ( ) ( ) ( ) ( )       +−=            +−=− − − − − ''12' 1 '12'12 1 1 1 1 y xd d ny xd d xny xd d x xd dn y xd d xnyx xd d n n n n n n n n n n n Derivation of Legendre Polynomials via Rodrigues’ Formula
  • 26. SOLO 26 Legendre Polynomials Olinde Rodrigues (1794-1851) Start from the function: ( ) .12 constkxky n =−= ( ) ( ) 02'12''12 =−−+− ynyxnyxDifferentiate n times with respect to x: ( ) ( ) ''1''2''1 2 2 1 1 2 y xd d nny xd d xny xd d x n n n n n n − − − − −++− ( ) 02''12 1 1 =−      +−+ − − y xd d ny xd d ny xd d xn n n n n n n Define: a Polynomial( ) ( )[ ]n n n n n x xd d k xd yd xw 1: 2 −== ( ) ( ) ( )[ ] 02'121'2''12 =−+−+−++− wnwnwxnwnnwxnwx ( ) ( ) ( ) ( )[ ] 02121'2''12 =−−+−+−++− wnnnnnwxnxxnwx ( ) ( ) 01'2''12 =+−+− wnnwxwx This is Legendre’s Differential Equation. We proved that one of the solutions are Polynomials. We can rewrite this equation in a Sturm-Liouville Form: ( ) ( ) 0112 =+−      − wnnw xd d x xd d Derivation of Legendre Polynomials via Rodrigues’ Formula
  • 27. SOLO 27 Legendre Polynomials Olinde Rodrigues (1794-1851) Let find k such that: by using the fact that Pn (1) = 1 ( ) ( )[ ]n n n n n n x xd d k xd yd xP 12 −== ( ) ( )[ ] ( ) ( ) ( ) ( )       −+=         +−=−= ∑>0 22 1!2111 i inn v n u n n n n n n n xxaxnkxx xd d kxk xd d xP  !2 1 n k n = We recover the Rodrigues Formula: ( ) ( )[ ]n n n nn x xd d n xP 1 !2 1 2 −= Let use Leibnitz’s Rule (Binomial Expansion for the n Derivative of a Product - with u:=(x-1)n and v:=(x+1)n ): ( ) ( ) ( ) udvudvdnvddu nn vddunvdu vdud mnm n vud nnnnn n m mnmn +++ − ++= − =⋅ −−− = − ∑ 1221 0 !2 1 !! !  We have: ( )   ( )    1!2 !2 1 1 1!20 12 0 21 00 ==        +++ − ++== = −−− nkudvudvdnvddu nn vddunvdukxP n xn nnnnn n n  We can see from this Formula that Pn (x) is indeed a Polynomial of Order n in t. Derivation of Legendre Polynomials via Rodrigues’ Formula
  • 28. SOLO 28 Legendre Polynomials Olinde Rodrigues (1794-1851) Let find an explicit expression for Pn (x) from Rodrigues’ Formula: ( ) ( )[ ]n n n nn n n x xd d nxd yd xP 1 !2 1 2 −== Start with: ( ) ( ) ( )∑= − − − =− n m mnmn x mnm n x 0 22 1 !! ! 1 ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )   + = −− − −= −− −= +−− − −=−= ∑ ∑ ∑ = − −= ≥ −− ≥ −− oddnn evennn p x knknk kn x nm m mnm xnmmm mnm n n x xd d n xP p k knk n knm n pm nmmn n n pm nmmn n n n n nn 2/12 2/ !2!! !22 1 2 1 !2 !2 !! 1 1 2 1 12122 !! ! 1 !2 1 1 !2 1 0 2 2 22  ( ) ( )    < ≥+−− = 2/0 2/121222 nm nmnmmm x xd d m n n  We recover the result obtained by the Generating Function of Legendre Polynomials Return to Frobenius Series Return to Table of Content Derivation of Legendre Polynomials via Rodrigues’ Formula
  • 29. SOLO Series Solutions – Frobenius’ Method Ferdinand Georg Frobenius (1849 –1917) ( ) ( ) .0121 2 2 22 constrealryrr xd yd yx xd yd x =++−− Example: General Legendre Equation ∑ ∞ = + = 0λ λ λ k xay Let check the Frobenius’s expansion: We have: ( ) ( ) ( )∑∑ ∞ = −+ ∞ = −+ −++=+= 0 2 2 2 0 1 1& λ λ λ λ λ λ λλλ kk xkka xd yd xka xd yd Substitute in the General Legendre Equation: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 0111 121 00 2 000 2 =+++−++−++= +++−−−++ ∑∑ ∑∑∑ ∞ = + ∞ = −+ ∞ = + ∞ = + ∞ = +−+ λ λ λ λ λ λ λ λ λ λ λ λ λ λλ λ λλλλ λλλ kk kkkk xkkrraxkka xrraxkaxxkka 29 Legendre Polynomials
  • 30. SOLO Example: Legendre Equation (continue – 1) ( ) ( ) ( ) ( ) ( )[ ] 0111 00 2 =+++−++−++ ∑∑ ∞ = + ∞ = −+ λ λ λ λ λ λ λλλλ kk xkkrraxkka Denote, in the first sum λ = j +2 and in the second sum λ = j, to obtain: ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]{ } 01112 11 0 2 1 1 2 0 =+++−+++++++ ++− ∑ ∞ = + + −− j jk jj kk xjkjkrrajkjka xkkaxkka All the coefficients of xk+j must be zero, therefore ( ) 001 00 ≠=− akka ( ) 011 =+kka ( ) ( ) ( ) ( ) ( )[ ] 011122 =+++−+++++++ jkjkrrajkjka jj 30 Ferdinand Georg Frobenius 1849 - 1917 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 31. SOLO Example: Legendre Equation (continue – 2) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ,2,1,0 12 1 12 11 2 = ++++ ++++− −= ++++ +++−+ −=+ j jkjk jkrjkr a jkjk jkjkrr aa jjj ( ) 01 =−kk ( ) 011 =+kka ( ) ( ) ( ) ( ) ( )[ ] 011122 =+++−+++++++ jkjkrrajkjka jj 0&1.3 0&0.2 0&0.1 1 1 1 == ≠= == ak ak akThree possible solutions The equation that, k (k+1) = 0, comes from the coefficient of the lowest power of x, and is called Indicial Equation. It has two solutions for k k = 0 and k = 1 The equation gives the recursive relation 31 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 32. SOLO Example: Legendre Equation (continue – 3) ( ) 1001: ==⇒=− korkkkEquationIndicial 1. Using k = 0 and a1 = 0 we obtain a series of even powers of x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,2,1,0 21 1 21 11 2 = ++ ++− −= ++ +−+ −=+ j jj jrjr a jj jjrr aa jjj ( ) ( ) ( ) ( ) ( ) ( )       + ++− + + −== 42 0 !4 312 !2 1 1: x rrrr x rr axpxy reven ( ) 01231 ==== +naaa  The recurrence relation results in the following expression for the coefficients ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )   ,3,2,1 !2 123124222 1 02 = −+++⋅−+−+− −= ma m mrrrrrmrmr a m m 32 ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ,2,1,0 12 1 12 11 2 = ++++ ++++− −= ++++ +++−+ −=+ j jkjk jkrjkr a jkjk jkjkrr aa jjj ( ) ∑ ∞ = = 0 2 2m m meven xaxy Series Solutions – Frobenius’ Method Legendre Polynomials
  • 33. SOLO Example: Legendre Equation (continue – 4) ( ) 1001: ==⇒=− korkkkEquationIndicial 3. Using k = 1 and a1 = 0 we obtain a series of odd powers of x ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) .2,1,0 32 21 32 211 2 = ++ ++−− −= ++ ++−+ −=+ j jj jrjr a jj jjrr aa jjj ( ) ( ) ( )( ) ( ) ( )( ) ( )       + ++−− + +− −== 53 0 !4 4213 !3 21 : x rrrr x rr xaxqxy rodd ( ) 01231 ==== +naaa  The recurrence relation results in the following expression for the coefficients ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )   ,3,2,1 !12 24213212 1 02 = + +++−+−+− −= ma m mrrrrmrmr a m m 33Since pr (x) and qr (x) are two linearly independent solutions of the 2nd Order Linear Lagrange ODE, the final solution is y = c1 pn (x) + c2 qn (x) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ,2,1,0 12 1 12 11 2 = ++++ ++++− −= ++++ +++−+ −=+ j jkjk jkrjkr a jkjk jkjkrr aa jjj ( ) ∑ ∞ = + = 0 12 2m m modd xaxy Series Solutions – Frobenius’ Method Legendre Polynomials
  • 34. SOLO Example: Legendre Equation (continue – 5) Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have Convergence Test: ( ) ( ) ( ) ( ) ( )     ≥≥ << = ++++ +++−+ = ∞→ + + ∞→ divergex convergex xx jkjk jkjkrr xa xa jj j j j j 11 11 12 11 limlim 22 2 2 The even series stops at j = n. The expansion is a Polynomial of order n (even). 1. If n is even, using k = 0 and a1 = 0 we obtain a series of even powers of x . ( ) ( ) ( ) ( ) ,..1,0 21 11 2 = ++ +−+ −=+ j jj jjnn aa jj For the case that r = n, a positive integer: ,...2,1,,02 ++==+ nnnjaj 3. If n is odd, using k = 1 and a1 = 0 we obtain a series of odd powers of x . ( ) ( )( ) ( ) ( ) ,..2,1 32 211 2 = ++ ++−+ −=+ j jj jjnn aa jj ,...2,1,,1,02 ++−==+ nnnnjaj The odd series stops at j = n-1. The expansion is a Polynomial of order n (odd).34 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 35. SOLO Example: Legendre Equation (continue – 6) 1. If n is even, using k = 0 and a1 = 0 we obtain a series of even powers of x ( ) ( ) ( )( ) ( )       + ++− + + −= 42 0 !4 312 !2 1 1 x nnnn x nn axyeven 3. If n is odd using k = 1 and a1 = 0 we obtain a series of odd powers of x ( ) ( )( ) ( ) ( )( ) ( )       + ++−− + +− −= 53 0 !4 4231 !3 21 x nnnn x nn xaxyodd ( ) ( ) ( ) ( )( ) ( )       +−=      ++− + + −== −=      + −== == 42 0 42 0 2 0 2 0 0 6 70 101 !4 3414244 !2 144 14 31 !2 122 12 0 xxaxxayn xaxayn ayn even even even ( )( )       −=      +− −== == 3 0 3 0 0 3 5 !3 2313 3 1 xxaxxayn xayn odd odd We obtain the Legendre Polynomials Solutions for a0 = 1. 35 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 36. SOLO Example: Legendre Equation (continue – 7) 1. The recurrence relation for even x powers results in the following expression for the coefficients ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2/,,3,2,1 !2 123124222 1 02 nma m mnnnnnmnmn a m m   = −+++⋅−+−+− −= ( ) ( ) ( ) ( ) ( ) ( ) ( )! ! 2121224222 11 2 mp p ppmpmpnnmnmn mm pn − =−+−+−=−+−+− −− =  ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )!!22 !!22 212 1 !2 !22 242 242 1231 ! ! 1231 22 mpp pmp mpppp mp mnnn mnnn mnnn n n mnnn m pn m pn + + = +++ + = +++ +++ −+++=−+++ ==     ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2/,,3,2,1 !!!22 !221 !!!2!2 ! 2 !22 1 !!2!2 !!22 ! ! 2 1 1 0 0 2 2 02 nm snssn sn a snssnn n sn a mpmp pmp mp p a n sthatsucha choose s nmps m m = −− −− = −−             − −= + + − −= − − −= ( ) ( ) ( ) ( )∑= −− < −− − −= 2/ 0 2 1 !2!!2 !22 1 n s sn n s even xx snsns sn y 36 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 37. SOLO Example: Legendre Equation (continue – 8) 3. The recurrence relation for odd x powers results in the following expression for the coefficients ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 ,,3,2,1 !12 24213212 1 02 − = + +++−+−+− −= n ma m mnnnnmnmn a m m   ( ) ( ) ( ) ( )( ) ( ) ( )! ! 222242222213212 1 12 mp p ppmpmpnmnmn m pn − =−+−+−=−+−+− − +=  ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )!!122 !!122 212 224222 1225232 !12 !12 242 1 1 12 mpp pmp mppp mppp mppp p p mnnn m m pn ++ ++ = +++ +++ ++++ + + =+++ − − +=   ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 ,,3,2,1 !2!1!! ! 2 1 !122 1 !12!!!12 !!122 1 0 2 0 2 2 − = −−−             − −− −= ++−+ ++ −= − −= n ma snsnsn n sn a mmpmpp pmp a sp mps m m  37 ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) 2 1 ,...,1,0 !2!!!2 ! 2 1 !22 1 !2!12!! ! 2 1 !12222 1 0 2 0 2 − = −−             − − −= −−−−             − −−− −= − −= − −= n sa snsnsn n sn a snsnsnsn n snsn sp mps sp mps Series Solutions – Frobenius’ Method Legendre Polynomials
  • 38. Example: Legendre Equation (continue – 9) 3. The recurrence relation for odd x powers results in the following expression for the coefficients ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 ,...,1,0 !2!!2 !22 1 !2!! !22 ! 2 1 !2 1 1 0 0 2 2 1 2 − = −− − −= −− −             − −= −− − n s snsns sn a snsns snn n a n s thatsucha Choose s n m ( ) ( ) ( ) ( ) ( ) ( ) 1& !2!! !22 1 2 1 2/1 0 2 < −− − −= ∑ − = −− xoddnx snsns sn xy n s sns nodd We also found that the solution for k = 0 and a1 = 0 is ( ) ( ) ( ) ( ) ( ) 1& !2!!2 !22 1 2/ 0 2 < −− − −= ∑= −− xevennx snsns sn xy n s sn n s even We recover the result obtained by the Generating Function of Legendre Polynomials and by Rodrigues’ Formula 38 Therefore we can unify those two relations to obtain: ( ) ( ) ( ) ( ) ( ) [ ] 1 !2!!2 !22 1 2/ 0 2 < −− − −= ∑= −− xx snsns sn xP n s sn n s n Series Solutions – Frobenius’ Method SOLO Legendre Polynomials
  • 39. SOLO Example: Legendre Equation (continue – 10) 2. Using k = 0 and a1 ≠ 0 we obtain an infinite series and not a polynomial. This is the Second Solution of the Legendre Differential Equation. The solution is the sum of the two infinite series, one with even powers of x and the other with odd powers of x. The series solution, in this case, diverges at x = ± 1. Those are Legendre Functions of the Second Kind. The Polynomial solutions are Legendre Functions of the First Kind. 39 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,2,1,0 21 1 21 11 2 = ++ ++− −= ++ +−+ −=+ j jj jnjn a jj jjnn aa jjj In this case we have The recurrence relation results in the following expression for the coefficients ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )   ,3,2,1 !2 123124222 1 02 = −+++⋅−+−+− −= ma m mnnnnnmnmn a m m ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )   ,3,2,1 !12 123124222 1 112 = + −+++⋅−+−+− −=+ ma m mnnnnnmnmn a m m Series Solutions – Frobenius’ Method Legendre Polynomials
  • 40. SOLO Example: Legendre Equation (continue – 11) 40 We want to find a series that converges for |x| > 1. Let return to the conditions to have a series solution for Legendre ODE For k = 0 and a0 = 0 By substituting j = m – 2 we obtain ( ) ( ) ( ) ( ) ,2,1,0 1 21 2 == ++− ++ −= + ja jnjn jj a jj ( ) ( ) ( ) mm a mnmn mm a 12 1 2 −++− − −=− ( ) ( ) .0121 2 2 22 constrealryrr xd yd yx xd yd x =++−− ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,2,1,0 21 1 21 11 2 = ++ ++− −= ++ +−+ −=+ j jj jrjr a jj jjrr aa jjj We can make am+2, am+4, am+6,…. To vanish for m = r = n or m = -n – 1. If we start for am ≠ 0 we can obtain the following recursive formula Series Solutions – Frobenius’ Method Legendre Polynomials
  • 41. SOLO Example: Legendre Equation (continue – 12) 41 ( ) ( ) ( ) mm a mnmn mm a 12 1 2 −++− − −=− Tacking m = n we obtain ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) nnn nn a nn nnnn a n nn a a n nn a 321242 321 324 32 122 1 24 2 −⋅−⋅⋅ −−− += − −− −= − − −= −− − The first solution can be written as ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )       + +−−⋅−⋅⋅ −−−− ++ −⋅−⋅⋅ −−− + − − −= −−−     innnn n x innni ininnn x nn nnnn x n nn xaxy 242 1 1223212242 1221 321242 321 122 1 Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have ( )( ) ( )     ≤≥ >< = +− −−− = −− ∞→+− +− − − ∞→ divergex convergex xx ini inin xa xa iin in in in j 11 11 1222 122 limlim 22 22 22 2 2 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 42. SOLO Example: Legendre Equation (continue – 13) 42 ( ) ( ) ( ) mm a mnmn mm a 12 1 2 −++− − −=− Tacking m =- n - 1 we obtain ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 135 13 523242 4321 524 43 322 21 −−−−−− −−−− +⋅+⋅⋅ ++++ += + ++ = + ++ += nnn nn a nn nnnn a n nn a a n nn a The second solution can be written as ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )       + +++⋅+⋅⋅ +−+++ ++ +⋅+⋅⋅ ++++ + + ++ += −−−−−−−−− −−     12531 12 1225232242 21221 523242 4321 322 21 innnn n x innni ininnn x nn nnnn x n nn xaxy Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have ( ) ( ) ( )     ≤≥ >< = ++ +−+ = −− ∞→+− +− −− −−− ∞→ divergex convergex xx ini inin xa xa iin in in in j 11 11 1222 212 limlim 22 12 12 12 12 Series Solutions – Frobenius’ Method Legendre Polynomials
  • 43. SOLO Example: Legendre Equation (continue – 14) 43 ( )( ) ! 1353212 n nn an ⋅⋅−− = By choosing we have ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1] 1223212242 1221 1 321242 321 122 1 [ ! 1353212 2 42 1 >+ +−−⋅−⋅⋅ −−−− −+ + −⋅−⋅⋅ −−− + − − − ⋅⋅−− == − −− xx innni ininnn x nn nnnn x n nn x n nn xyxP ini nnn n       Return to Table of Content Finally ( ) ( ) ( ) ( ) ( ) 1 !2!!2 !22 1 0 2 > −− − −= ∑ ∞ = − xx inini in xP i in n i n ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )!2!!2 !22 1 !2 !22 !2!2 1 1 !2!2 135122 1 1223212242 1221 1 ! 1353212 inini in in in iiniin in innni ininnn n nn n i ini i i i i −− − −= − − − −= − ⋅⋅−− −= +−−⋅−⋅⋅ −−−− − ⋅⋅−− −    Series Solutions – Frobenius’ Method Legendre Polynomials
  • 44. SOLO Example: Legendre Equation (continue – 14) 44 ( )( ) 1351212 ! 1 ⋅⋅−+ == −− nn n aa nn By choosing we have ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1] 1225232242 21221 523242 4321 322 21 [ 1351212 ! 12 531 2 >+ +++⋅+⋅⋅ +−+++ + + +⋅+⋅⋅ ++++ + + ++ + ⋅⋅−+ == −−− −−−−−− xx innni ininnn x nn nnnn x n nn x nn n xyxQ in nnn n       ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )!122! !2! ! !12 ! !2 !122 !12 ! !2 2 1 !122 2222!12 ! !2 !2 1 1225232242 21221 2 ++ +++ = + ++ ++ = ++ ++++ = +++⋅+⋅⋅ +−+++ ini inin n n n in in n n in in innn n in iinnni ininnn i i i    ( ) ( ) ( ) ( ) ( ) 1 !122! !2! 2 0 12 > ++ ++ = ∑ ∞ = ++− xx ini inin xQ i inn n Go to Neumann Integral ( )( ) ( )!12 !2 ! 1351212 ! + = ⋅⋅−+ n n n nn n n  Finally Series Solutions – Frobenius’ Method Legendre Polynomials
  • 45. SOLO Example: Legendre Equation (continue – 15) 45 Summarize ( ) ( ) ( ) ( ) ( ) 1 !122! !2! 2 0 12 > ++ ++ = ∑ ∞ = ++− xx ini inin xQ i inn n Return to Table of Content ( ) ( ) ( ) ( ) ( ) 1 !2!! !22 1 2 1 0 2 > −− − −= ∑ ∞ = − xx inini in xP i ini nn ( ) ( ) ( ) ( ) ( ) [ ] 1 !2!! !22 1 2 1 2/ 0 2 < −− − −= ∑= −− xx inini in xP n i ini nn Return to Similar to Rodrigues Formula Using Frobenius’ Method we found that solutions of Legendre ODE are ( ) ( ) integerpositive0121 2 2 22 nynn xd yd yx xd yd x =++−− Series Solutions – Frobenius’ Method Legendre Polynomials ( ) 4,3,2,1 1 = > l xxPl ( ) 4,3,2,1 1 = > l xxQl
  • 46. 46 SOLO ( ) 1 21 1 0 2 ≤= +− ∑ ∞ = utPu utu n n n For u=0 we obtain ( ) 10 =tP For t = 1 we obtain ( ) ( ) 111 1 1 21 1 00 2 =⇒== − = +− ∑∑ ∞ = ∞ = n n n n n n PPuu uuu For t = -1 we obtain ( ) ( ) ( ) ( )n n n n n n nn PPuu uuu 1111 1 1 21 1 00 2 −=⇒=−= + = ++ ∑∑ ∞ = ∞ = Let find a Recursive Relation for Legendre Polynomial computation Start with: ( ) ( ) ( ) ( ) ( ) ( ) ( )       ∂ ∂ + − =      ∂ ∂− + − =      ∂ ∂ + − = ++ + + ++ + + 11 1 1 11 2 1 cos 1 11 ! 1 1 11 !1 1cos n n n nn n nn n n r P znrznnrznr P θθ ( ) ( ) ( ) ( ) ( )( ) ( )       ∂ ∂ + ∂ ∂+ − + − = +++ + zd Pd rz r r Pn nr P n nn n n n θ θ θθθ cos cos cos1cos1 1 1cos 122 1 Recursive Relations for Legendre Polynomial Computation First Recursive Relation Legendre Polynomials
  • 47. 47 SOLO Recursive Relations for Legendre Polynomial Computation ( ) ( ) ( ) ( ) ( )( ) ( )       ∂ ∂ + ∂ ∂+ − + − = +++ + zd Pd rz r r Pn nr P n nn n n n θ θ θθθ cos cos cos1cos1 1 1cos 122 1 θcosrz = θcos= ∂ ∂ z r  z r z r z z ∂ ∂ + ∂ ∂ = ∂ ∂ = θ θ θ cos cos1 cos rz θθ 2 cos1cos − = ∂ ∂ ( ) ( ) ( ) ( ) ( )( )       − + + − + − = +++ + rd Pd rr Pn nr P n nn n n n θ θ θ θ θθ 2 122 1 cos1 cos cos1 cos cos1 1 1cos ( ) ( ) ( )( ) θ θθ θθθ cos cos 1 cos1 coscoscos 2 1 d Pd n PP n nn + − −=+ Substituting t = cos θ we obtain ( ) ( ) ( ) td tPd n t tPttP n nn 1 1 2 1 + − −=+ First Recursive Relation (continue – 1) Legendre Polynomials
  • 48. 48 SOLO Recursive Relations for Legendre Polynomial Computation ( ) ( ) ( ) 0,1 1 1 2 1 ≥≤ + − −=+ nt td tPd n t tPttP n nn Use to start ( ) ( ) 01 0 0 =⇒= td tPd tP ( ) ( ) ( ) ( ) ( ) 8 157063 8 33035 2 35 2 13 35 5 24 4 3 3 2 2 1 ttt tP tt tP tt tP t tP ttP +− = +− = − = − = = First Recursive Relation (continue – 2) Legendre Polynomials
  • 49. 49 SOLO Recursive Relations for Legendre Polynomial Computation Start from ( ) ( ) 1 21 1 :, 0 2 ≤= +− = ∑ ∞ = utPu utu tug n n n Let differentiate both sides with respect to u and rearranging ( ) ( )∑ ∞ = − +−= +− − = ∂ ∂ 1 12 2 21 21 n n n tPunutu utu ut u g ( ) ( ) ( ) ( )∑∑ ∞ = − ∞ = +−=− 1 12 0 21 n n n n n n tPunututPuut ( ) ( ) ( ) ( )∑∑ ∞ = +− ∞ = + +−=− 1 11 0 1 2 n n nnn n n nn tPunutnuntPuut ( ) ( ) ( ) ( ) ( ) ( ) ( )∑∑∑∑∑ ∞ = − ∞ = ∞ = + ∞ = − ∞ = −+−+=− 2 1 10 1 1 1 0 121 n n n n n n n n n n n n n n n tPuntPutntPuntPutPut ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )     ≥+=−+ =−=− == +− 2112 122 0 11 1201 10 ntPuntPuntPutn ntPtutPutPutuPt ntPtPt n n n n n n We can see that the last relation agrees also with the previous relations, for n = 0 and n = 1. Second Recursive Relation Legendre Polynomials
  • 50. 50 SOLO Recursive Relations for Legendre Polynomial Computation We find the Recursive Relation: ( ) ( ) ( ) 1 11 12 11 ≥ + − + + = −+ ntP n n tPt n n tP nnn ( ) 10 =tP This is called the Bonnet’s Recursion Relation. It starts with: Examples: ( ) ( ) ( ) 2 3 1 01 2 tPtPt tPn − =→= ( ) 2 1 2 3 2 2 −= ttP ( ) ( ) ( )  tP tP ttttPn 1 2 3 2 2 1 2 3 3 5 2 2 3 −      −=→=  ( ) 2 35 3 3 tt tP − = ( ) ( ) ( )  tPtP tttttPn 23 2 1 2 3 4 3 2 3 2 5 4 7 3 23 4       −−      −=→= ( ) 8 33035 24 4 +− = tt tP ( ) ttP =1 Second Recursive Relation (continue) Legendre Polynomials
  • 51. 51 SOLO Recursive Relations for Legendre Polynomial Computation Start from ( ) ( ) 1 21 1 :, 0 2 ≤= +− = ∑ ∞ = utPu utu tug n n n Let differentiate both sides with respect to t and rearranging ( ) ( ) ∑ ∞ = = +− = ∂ ∂ 0 2/32 21 n nn td tPd u utu u t g ( ) ( ) ( ) ∑∑ ∞ = ∞ = +−= 0 2 0 21 n nn n n n td tPd uututPuuor Equaling coefficients of each power of u gives ( ) ( ) ( ) ( ) td tPd td tPd t td tPd tP nnn n 11 2 −+ +−= ( ) ( ) ( ) ( ) ( )tPntPntPtn nnn 11112 −+ ++=+ Differentiate the Second Recursive Relation with respect to t and rearranging ( ) ( ) ( ) ( ) ( ) ( ) ( ) td tPd n n td tPd n n td tPd ttP nnn n 11 1212 1 −+ + + + + =+ Third Recursive Relation Legendre Polynomials
  • 52. 52 SOLO Recursive Relations for Legendre Polynomial Computation We found ( ) ( ) ( ) ( ) td tPd ttP td tPd td tPd n n nn 211 +=+ −+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) td tPd ttP td tPd n n td tPd n n n n nn += + + + + −+ 11 1212 1 Third Recursive Relation (continue) Let solve for and in terms of and( )tPn ( ) td tPd n( ) td tPd n 1+( ) td tPd n 1− ( ) ( ) ( ) td tPd ttPn td tPd n n n +−=−1 ( ) ( ) ( ) ( ) td tPd ttPn td tPd n n n ++=+ 11 Subtracting the first relation from the second gives the Third Recursive Relation ( ) ( ) ( ) ( ) td tPd td tPd tPn nn n 11 12 −+ −=+ Legendre Polynomials
  • 53. 53 SOLO Recursive Relations for Legendre Polynomial Computation ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }yPxPyPxP yx n yPxPk tPtntPn td tPd t tPntPtn td tPd t tPn td tPd t td tPd tPn td tPd td tPd t tPn td tPd td tPd tPntPtntPn tP n n tP n n tPt tPin td tPd nnnn n k kk nn n nn n n nn n nn n nn nnn nnn l i in n 11 0 1 2 1 2 1 1 1 11 11 11 1 2 1 0 12 1 129 1118 17 6 5 124 01213 1212 1 2 1421 ++ = + − − − − −+ −+ −+       − = −− − − + =+ +−+=− −=− =− =− +=− =++−+ + + + + = −−= ∑ ∑ Recursive Relation between Legendre Polynomials and their Derivatives) Legendre Polynomials Return to Table of Content
  • 54. 54 SOLO Orthogonality of Legendre Polynomials Define ( ) ( )tPwtPv nm == :&: We use Legendre’s Differential Equations: ( ) ( ) 011 2 =++      − vmm td vd t td d ( ) ( ) 011 2 =++      − wnn td wd t td d Multiply first equation by w and integrate from t = -1 to t = +1. ( ) ( ) 011 1 1 1 1 2 =++      − ∫∫ + − + − dtwvmmdtw td vd t td d Integrate the first integral by parts we get ( ) ( ) ( ) 0111 1 1 1 1 2 0 1 1 2 =++−−− ∫∫ + − + − += −= dtwvmmdt td wd td vd tw td vd t t t    In the same way, multiply second equation by v and integrate from t = -1 to t = +1. ( ) ( ) 011 1 1 1 1 2 =++−− ∫∫ + − + − dtwvnndt td wd td vd t Legendre Polynomials
  • 55. 55 SOLO Orthogonality of Legendre Polynomials ( ) ( ) 011 1 1 1 1 2 =++−− ∫∫ + − + − dtwvmmdt td wd td vd t Subtracting those two equations we obtain ( ) ( ) 011 1 1 1 1 2 =++−− ∫∫ + − + − dtwvnndt td wd td vd t ( ) ( )[ ] ( ) ( )[ ] ( ) ( ) 01111 1 1 1 1 =+−+=+−+ ∫∫ + − + − dttPtPnnmmdtwvnnmm nm This gives the Orthogonality Condition for m ≠ n ( ) ( ) nmdttPtP nm ≠=∫ + − 0 1 1 To find let square the relation and integrate between t = -1 to t = +1. Due to orthogonality only the integrals of terms having Pn 2 (t) survive on the right-hand side. So we get ( )∫ + − 1 1 2 dttPn ( )∑ ∞ = = +− 0 2 21 1 n n n tPu utu ( )∑ ∫∫ ∞ = + − + − = +− 0 1 1 22 1 1 2 21 1 n n n dttPudt utu Legendre Polynomials
  • 56. 56 SOLO Orthogonality of Legendre Polynomials ( )∑ ∫∫ ∞ = + − + − = +− 0 1 1 22 1 1 2 21 1 n n n dttPudt utu ( ) ( ) ( ) 1 1 1 ln 1 1 1 ln 2 1 21ln 2 1 21 1 2 21 1 2 1 1 2 < − + = + − − =−+ − = −+ += −= + −∫ u u u uu u u tuu u dt tuu t t ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∑∑∑ ∞ ++∞ +∞ + + −− −= + − −− + −=−−+ 0 11 0 1 0 1 1 1 1 1 1 1 1 1 1 1ln 1 1ln 1 n uu un u un u u u u u u nn n n n n n ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∑∑∑∑ ∞∞ +∞ ++ + ∞ ++ + = + = + −− −+ + −− −= 0 2 0 12 0 0 1212 12 0 1212 2 12 2 12 1 2 12 1 1 12 1 1 n nnn n nn n u nn u un uu un uu u    Let compute first Therefore ( )∑ ∫∑∫ ∞ + − ∞+ − = + = +− 0 1 1 22 0 2 1 1 2 12 2 21 1 dttPuu n dt utu n nn Comparing the coefficients of u2n we get ( ) 12 21 1 2 + =∫ + − n dttPn Legendre Polynomials ( ) ( ) nmmn n dttPtP δ 12 21 1 + =∫ + − Hence
  • 57. 57 SOLO Using Rodrigues’ Formula let calculate ( ) ( ) ( ) ( )[ ]∫∫ + − + − − = 1 1 2 1 1 1 !2 1 dt td td tP n dttPtP n nn knnk Legendre Polynomials ( ) ( )[ ]n nn nn td td n tP 1 !2 1 2 − = Orthogonality of Legendre Polynomials (Second Method) Assume, without loss of generality, that n > k, and integrate by parts ( ) ( ) ( ) ( )[ ] ( ) ( )[ ] ( ) ( )[ ] ( ) ( ) ( ) ∫∫ ∫∫ + − + − − − + − − − + − + − −−== − − − = − = 1 1 2 1 1 1 21 0 1 1 1 21 1 1 2 1 1 1 !2 1 1 1 !2 11 !2 1 1 !2 1 dt td tPd t n dt td td td tPd ntd td tP n dt td td tP n dttPtP n k n n n n n nn k nn nn kn n nn knnk     Since Pk (t) is a Polynomial of Order k and we assume that n > k, we have ( ) 0=n k n td tPd Therefore ( ) ( ) nkdttPtP nk ≠=∫ + − 0 1 1
  • 58. 58 SOLO For k = n we have Legendre Polynomials Orthogonality of Legendre Polynomials (Second Method) (continue – 1) ( )[ ] ( ) ( ) ( ) ∫∫ + − + − −−= 1 1 2 1 1 2 1 !2 1 1 dt td tPd t n dttP n n n n n n n But we found that Pn (t) is given by: ( ) ( ) ( ) ( ) ( ) [ ] ∑= − −− − −= 2/ 0 2 !2!!2 !22 1 n k kn n k n t knknk kn tP Therefore to compute it is sufficient to consider only the highest power of t in the series, i.e. for k = 0, and we obtain ( ) n n n td tPd ( ) ( ) ( ) ( ) !2 !2 ! !2 !2 2 n n n n n td tPd nnn n n == ( )[ ] ( ) ( ) ( ) ( ) ( ) ( )∫∫∫ + − + − + − −=−−= 1 1 2 22 1 1 2 1 1 2 1 !2 !2 !2 !2 1 !2 1 1 dtt n n dt n n t n dttP n nn n n n n
  • 59. SOLO Legendre Polynomials Orthogonality of Legendre Polynomials (Second Method) (continue – 2) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( )∫∫∫ + − + − + − −=−−= 1 1 2 22 1 1 2 1 1 2 1 !2 !2 !2 !2 1 !2 1 1 dtt n n dt n n t n dttP n nn n n n n ( ) ∫∫∫ ++ =+ − =−=− π π θ θθθθ 0 12 0 12 cos1 1 2 sinsin1 dddtt nn tn Therefore ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )!12 !2 cos 2222 !2 11212 !2 sin 31212 2222 sin 12 2 sin 212 2 0 1 00 12 0 12 + = −⋅ ⋅ −⋅+ = −⋅+ −⋅ = + = + −+ ∫∫∫ n n nn n nn n d nn nn d n n d nnn nn       π πππ θθθθθθθ ( ) ( ) ( )∫∫∫∫ +−−+ −=+−== ππ π ππ θθθθθθθθθθθθ 0 1212 0 212 0 0 2 0 2 0 12 sinsin2cossin2cossincossinsin dndndd nnnnnn    ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,2,1,0 12 2 !12 !2 !2 !2 1 !2 !2 212 22 1 1 2 22 1 1 2 = + = + =−= + + − + − ∫∫ n nn n n n dtt n n dttP n n n nn Return to Table of Content 59
  • 60. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Using Sturm-Liouville Theory it can be seen that the Legendre Polynomials that are Solution of the Legendre ODE, form an orthogonal and “Complete” Set, meaning that we can expand any function f (t) , Piecewise Continuous in the interval -1 ≤ t ≤+1. Therefore we can define a series of Legendre Polynomials that converges in the mean to the function f (t) ( ) ( ) 11 0 ≤≤−= ∑ ∞ = ttPatf n nn The coefficients an can be defined using the Orthogonality Property of Legendre Polynomials ( ) ( ) ( ) ( ) m n m mnnm a m tdtPtPatdtPtf mn 12 2 0 2 12 1 1 1 1 + == ∑ ∫∫ ∞ = + + − + −    δ ( ) ( )∫ + − + = 1 1 2 12 tdtPtf m a mm ( ) ( ) ( ) ( ) 11 2 12 0 1 1 ≤≤−       + = ∑ ∫ ∞ = + − ttPtdtPtf n tf n nn 60
  • 61. SOLO Legendre Polynomials Expansion of Functions, Legendre Series At any discontinuous point t0 ( f(t0- ) ≠ f(t0+) ) we have ( ) ( )[ ] ( ) ( ) ( ) 11 2 12 2 1 0 0 0 1 1 00 ≤≤−       + =+ ∑ ∫ ∞ = + − +− ttPtdtPtf n tftf n nn If f (t) is defined in the interval –a ≤ t ≤ +a then ( ) ( ) ataatPatf n nn ≤≤−= ∑ ∞ =0 / ( ) ( )∫ + − + = a a nn tdatPtf a n a / 2 12 61
  • 62. SOLO Legendre Polynomials Expansion of Functions, Legendre Series If f (t) is an odd function ( f(-t ) =- f(t) ) we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )    =+−−= +== + ∫ ∫∫ ∫∫∫ + ++ −− + −→ − + − 1 0 1 0 1 0 1 1 0 0 1 1 1 2 0 12 2 oddndttPtf evenn dttPtfdttPtf dttPtfdttPtfdttPtfa n n n tP n tf n tt nnn n n     If f (t) is an even function ( f(-t ) = f(t) ) we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )     =+−−= +== + ∫∫∫ ∫∫∫ + ++ − + −→ − + − oddn evenndttPtf dttPtfdttPtf dttPtfdttPtfdttPtfa n n n tP n tf n tt nnn n n 0 2 12 2 1 0 1 0 1 0 1 1 0 0 1 1 1     62
  • 63. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Using Rodrigues’ Formula let calculate ( ) ( )[ ]n nn nn td td n tP 1 !2 1 2 − = ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] ( )[ ] ( ) ( ) ( ) ( ) ∫∫ ∫∫ + − + − − − + − − − + − + − −−== − − − = − = 1 1 2 1 1 1 21 0 1 1 1 21 1 1 21 1 1 !2 1 1 1 !2 11 !2 1 1 !2 1 td td tfd t n td td tfd td td ntd td tf n tdtf td td n tdtftP n n n n n n nn nn nn n n nn nn     ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∫∫∫ + − + − + − −=−−= 1 1 2 1 1 2 1 1 1 !2 1 1 !2 1 1 td td tfd t n td td tfd t n tdtPtf n n n nn n n n n n or 63
  • 64. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Example f (t) = tk ,|t| < 1 ( ) ( ) ( ) ( ) ( ) ( ) ( )     ≥−+−− + < =− + = + = ∫ ∫∫ + − − + + − + + − nktdttnkkk n n nk td td td t n n tdtPt n a nkn n n kn n nn k n 1 1 2 1 1 1 2 1 1 1 111 !2 12 0 1 !2 12 2 12  We have ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫ ∫ ∫ ∫∫ + − + −− + − −−−+ + − −−+ + − −−+ + − −−+= −= + − − − − − + = − +++ −−−−−−−− = =− ++ −−−− = − + −− +− + −=− −− 1 1 2 1 1 22222 1 1 422222 1 1 222122 0 1 1 122122 1 2 1 1 1 2222 1 !2 !22 !2 !2 1 2222122 12222322122 1 222122 322122 1 122 122 1 122 1 1 122 222 tdt nm nm mn n tdtt mnnn nmnmnmnm tdtt nn nmnm tdtt n nm tt n tdtt mn nmnm nmnmmn nmn nmnnmntu tdtdv nmn nm n        Take k = 2m and since t2m is even, they are only even coefficients nonzero so we take 2 n instead of n, and 2n ≤ 2m 64
  • 65. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Example f (t) = tk , |t| < 1 (continue – 1) ( ) ( ) ( ) ( ) ( )     ≥− − + < = ∫ + − − + nmtdtt nm m n n nm a nmn n n 1 1 2222 12 2 1 !22 !2 !22 14 0 We have for f (t) = t2m ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )!122 !2 !2 !22 !2 !2 1 !2 !22 !2 !2 1 2122 1 1 2 1 1 2222 ++ + − − + = − − − + =− ++ −− + − + −− + − − ∫∫ nm mn nm nm mn n tdt nm nm mn n tdtt nm nmnm mn nmnm nmn ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )!122! !!2142 !122 !2 !2 !22 !2 !2 !22 !2 !22 14 22122 122 ++− ++ = ++ + − − +− + = ++ −−+ nmnm nmmn nm nm nm nm nm n nm m n n a nnm nmnmnn ( ) ( ) ( )!12 !2 1 212 1 1 2 + =− + + −∫ n n dtt n n Where we used the previous result Therefore ( ) ( ) ( ) ( ) ( ) ( )∑= ++− ++ = m n n n m tP nmnm nmmn t 0 2 2 2 !122! !!2142 65
  • 66. SOLO Legendre Polynomials Expansion of Functions, Legendre Series ( ) ( ) ( ) ( ) ( ) ( ) ( )     ≥−+−− + < =− + = + = ∫ ∫∫ + − − + + − + + − nktdttnkkk n n nk td td td t n n tdtPt n a nkn n n kn n nn k n 1 1 2 1 1 1 2 1 1 1 111 !2 12 0 1 !2 12 2 12  We have ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )∫ ∫ ∫ ∫∫ + − ++ ++ −− + − −−−++ + − −−+ + − −−+ + − −−+= −= + − −+ − ++ + − − ++ + = − ++++ −−−−−−−− = =− ++ −−−− = − + −− +− + −=− −− + 1 1 12 2122 1 1 222212 1 1 422322 1 1 222222 0 1 1 122222 1 2 1 1 1 22122 1 !122 !2 !2 !22 !12 !12 1 12322222 12222322122 1 322222 322122 1 222 122 1 222 1 1 122 2122 tdt nm mn nm nm mn n tdtt mnnn nmnmnmnm tdtt nn nmnm tdtt n nm tt n tdtt mn nm nmnm nmnmmn nmn nmnnmntu tdtdv nmn nm n        Take k = 2m+1 and since t2m+1 is odd, they are only odd coefficients nonzero so we take 2 n+1 instead of n, and 2n+1 ≤ 2m+1 66 Example f (t) = tk , |t| < 1 (continue – 2)
  • 67. SOLO Legendre Polynomials Expansion of Functions, Legendre Series ( ) ( ) ( ) ( ) ( )     ≥− −+ + < = ∫ + − −+ + + nmtdtt nm m n n nm a nmn n n 1 1 22122 22 12 1 !22 !2 !122 34 0 We have for f (t) = t2m+1 ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )!322 !12 !2 !22 !12 !12 1 !122 !2 !2 !22 !12 !12 1 2322 1 1 12 21221 1 22122 ++ ++ − − ++ + = − ++ + − − ++ + =− ++ −− + − ++ ++ −− + − −+ ∫∫ nm mn nm nm mn n tdt nm mn nm nm mn n tdtt nm nmnm mn nm nmnm nmn ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )!322! !1!12342 !322 !12 !2 !22 !12 !12 !22 !2 !122 34 122322 2212 ++− ++++ = ++ ++ − − ++ + −+ + = +++ −−++ nmnm nmmn nm nm nm nm nm n nm m n n a nnm nmnmnn ( ) ( ) ( )!12 !2 1 212 1 1 2 + =− + + −∫ n n dtt n n Where we used the previous result Therefore ( ) ( ) ( ) ( ) ( ) ( )∑= + + + ++− ++++ = m n n n m tP nmnm nmmn t 0 12 12 12 !322! !1!12342 Return to Neumann Integral 67 Example f (t) = tk , |t| < 1 (continue – 3)
  • 68. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Neumann Integral 1 1 1 111 0 2 12 0 12 2 0 1 0 <+==      = − = − ∑∑∑∑ ∞ = −∞ = + ∞ = + ∞ = x t x t x t x t x t x x txtx m m m m m m m m m m m Start from Use ( ) ( ) ( ) ( ) ( ) ( )∑= + + + ++− ++++ = m n n n m tP nmnm nmmn t 0 12 12 12 !322! !1!12342 ( ) ( ) ( ) ( ) ( ) ( )∑= ++− ++ = m n n n m tP nmnm nmmn t 0 2 2 2 !122! !!2142 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∞ = ∞ = + − +∞ = ∞ = ++− += ∞ = ∞ = + − +∞ = ∞ = +− ∞ = = − + +∞ = +− = ++ +++++ + ++ +++ = ++− ++++ + ++− ++ = ++− ++++ + ++− ++ = − 0 0 12 2 12 0 0 2 212 2 0 12 2 12 0 2 12 2 0 0 2 12 12 0 12 0 2 2 !322! !12!122342 !124! !2!22142 !322! !1!12342 !122! !!2142 !322! !1!12342 !122! !!21421 n i n m n n i n in ninm n nm n m n n nm n m nOrder Summation Change m m n m n n m m m n n n tPx nmi ininn tPx ini ininn tPx nmnm nmmn tPx nmnm nmmn xtP nmnm nmmn xtP nmnm nmmn tx 68
  • 69. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Neumann Integral (continue – 1) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑ ∑∑ ∑ ∞ = + ∞ = − +∞ = ∞ = ++−       ++ +++++ +      ++ +++ = − 0 12 0 2 12 0 2 0 212 2 !322! !12!122342 !124! !2!221421 n n i m n n n i in n tPx nmi ininn tPx ini ininn tx We found ( ) ( ) ( ) ( ) ( ) 1 !122! !2! 2 0 12 > ++ ++ = ∑ ∞ = ++− xx ini inin xQ i inn n Use the Frobenius Series development of Legendre Functions of the Second Kind Qn (x) We have ( ) ( ) ( ) ( ) ( ) ( )∑∑ ∞ = ++ ∞ = +++= − 0 1212 0 22 3414 1 n nn n nn tPxQntPxQn tx ( ) ( ) ( )∑ ∞ = <+= − 0 12 1 n nn xttPxQn tx Return to Qn Frobenius Series 69
  • 70. SOLO Legendre Polynomials Expansion of Functions, Legendre Series Neumann Integral (continue – 2) We found ( ) ( ) ( )∑ ∞ = += − 0 12 1 n nn tPxQn tx Multiply both sides by Pm (t) and integrate between -1 to +1 ( ) ( ) ( ) ( ) ( ) ( )xQtdtPtPxQntd tx tP m n n mnn m nm 212 0 12 2 1 1 1 1 =+= − ∑ ∫∫ ∞ = + + − + −    δ We obtain ( ) ( ) ∫ + − − = 1 1 2 1 td tx tP xQ n n Franz Neumann's Integral of 1848 Franz Ernst Neumann (1798 –1895) Return to Table of Content 70
  • 71. SOLO 71 Legendre Polynomials Schlaefli Integral Start with Using Rodrigues's Formula we obtain( ) ( )[ ]n n n nn x xd d n xP 1 !2 1 2 −= ( ) ( ) ∫ − = td zt tf j zf π2 1 Cauchy's Integral with ( ) ( )n zzf 12 −= ( ) ( ) ∫ − − =− td zt t j z n n 1 2 1 1 2 2 π Differentiate n times this equation with respect to z and multiply by 1/ (2n n!) ( ) ( ) ( )∫ + − − − =− td zt t j z zd d n n nn n n n n 1 2 2 1 2 2 1 !2 1 π with the contour enclosing the point t = z. Schlaefli Integral ( ) ( ) ( )∫ + − − − = td zt t j zP n nn n 1 2 1 2 2 π Return to Table of Content
  • 72. SOLO 72 Legendre Polynomials Laplace’s Integral Representation Start with ( ) ( ) ( )[ ] ( )[ ] ( )[ ] ( ) [ ] [ ] ( ) ( ) 22 0 1 2 0 22 0 2 0 22 2/tan 0 22 2 0 2 2 0 121 2 1 1 tan 1 2 1 1 1 1 1 1 2 11 2 11 2 2/tan12/tan1 2/tan1 2/tan1 2/tan1 1 cos1 λ ππ λλ λ λ λ λ λ λ λ λλλφλφ φφ φ φ λ φ φλ φ φπππ − = − =        + − − =         + − + + − − = −++ = −++ = −++ + = + − + = + ∞ − ∞ ∞∞= ∫ ∫∫∫∫∫ t t td t td tt tdddd t ( ) ( ) ( ) ( ) ( ) { }[ ] ( ) { }∑ ∞ = −− − ±= −±−=−±−−= −±− − = − − ± = + 0 2 1 2 22 1 1 cos11cos111 cos11 1 cos 1 1 1 1 cos1 1 2 n n n xu xu xxuxuxxuxu xuxu xu xu xu φφ φ φ φλ λ Let write where we used ( ) ∑ ∞ = − =− 0 1 1 n n aa ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2222 2 22 1 1 2 21 1 11 1 1 1 1 1 1 1 2 uxu xu xuxu xu xu xu xu xu +− − = −−− − = − − − = − − − ±=λ λ ( ) { } ( ) { } ( ) 22 0 0 2 0 0 2 0 21 1 1 cos11cos11 cos1 uxu xu dxxuxudxxuxu d n n n n n n +− − = − =−±−=−±−= + ∑ ∫∫ ∑∫ ∞ = ∞ = π λ π φφφφ φλ φ πππ
  • 73. SOLO 73 Legendre Polynomials Laplace’s Integral Representation (continue - 1) { } ( )∑∑ ∫ ∞ = ∞ = = +− =−± 0 2 0 0 2 21 cos1 n n n n n n xPu uxu dxxu π π φφ π We obtained Equating un coefficients we obtain : ( ) { }∫ −±= π φφ π 0 2 cos1 1 dxxxP n n ( ) ( ) 0112 =+−      − ynny xd d x xd d If we replace in the Legendre ODE n by –n – 1 the equation does not change. Therefore , and( ) ( )xPxP nn 1−−= ( ) { }∫ −− −±= π φφ π 0 1 2 cos1 1 dxxxP n n Substitute x = cosθ ( ) { }∫ ±= π φφθθ π θ 0 cossincos 1 cos djP n n Laplace’s First Integral Laplace’s Second Integral
  • 74. SOLO 74 Legendre Polynomials Laplace’s Integral Representation (continue - 2) Use the Generating Function [ ] ( )∑ ∞ = = +− 0 2/12 21 1 n n n tPu uut Substitute t = cosθ and u = ejφ [ ] ( )∑ ∞ = = +− 0 2/12 cos cos21 1 n n nj jj Pe ee θ θ ϕ ϕϕ [ ] [ ] [ ] [ ] [ ] 2/12/12/12/12/12 coscos2cos2cos21 θϕθθ ϕϕϕϕϕϕ −=+−=+− − jjjjjj eeeeeeBut Therefore ( ) ( ) ( ) ( )       > − < − = − ∞ = ∑ θϕ θϕ θϕ θϕ θ πϕ ϕ ϕ 2/12/ 2/12/ 0 coscos2 1 coscos2 1 cos j j n n nj e e Pe Equating the real and imaginary parts, we obtain ( ) ( ) ( ) ( ) ( ) ( )       > − < − =∑ ∞ = θϕ θϕ ϕ θϕ θϕ ϕ θϕ 2/1 2/1 0 coscos2 2/sin2 coscos2 2/cos2 coscos n nPn ( ) ( ) ( ) ( ) ( ) ( )       > − < − − =∑ ∞ = θϕ θϕ ϕ θϕ θϕ ϕ θϕ 2/1 2/1 0 coscos2 2/cos2 coscos2 2/sin2 cossin n nPn
  • 75. SOLO 75 Legendre Polynomials Laplace’s Integral Representation (continue - 3) Let multiply first relation by cos (nφ) and the second by sin (nφ) and integrate over φ on (0,π), we obtain two integrals ( ) ( ) ( ) ( ) ( ) ( )       > − < − =∑ ∞ = θϕ θϕ ϕ θϕ θϕ ϕ θϕ 2/1 2/1 0 coscos2 2/sin2 coscos2 2/cos2 coscos i iPi ( ) ( ) ( ) ( ) ( ) ( )       > − < − − =∑ ∞ = θϕ θϕ ϕ θϕ θϕ ϕ θϕ 2/1 2/1 0 coscos2 2/cos2 coscos2 2/sin2 cossin i iPi ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫ ∫∑∫ − + − == ∞ = θ π θ δ π π ϕ θϕ ϕϕ ϕ θϕ ϕϕ θ π θϕϕϕ 0 2/12/1 0 2 0 coscos2 cos2/sin2 coscos2 cos2/cos2 cos 2 coscoscos d n d n PPdni n i i in    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫ ∫∑∫ − + − −== ∞ = θ π θ δ π π ϕ θϕ ϕϕ ϕ θϕ ϕϕ θ π θϕϕϕ 0 2/12/1 0 2 0 coscos2 sin2/cos2 coscos2 sin2/sin2 cos 2 cossinsin d n d n PPdni n i i in    ( ) ( ) ( ) ( ) ( ) ( ) ( )       − + − = ∫ ∫ θ π θ ϕ θϕ ϕϕ ϕ θϕ ϕϕ π θ 0 2/12/1 coscos cos2/sin coscos cos2/cos2 cos d n d n Pn ( ) ( ) ( ) ( ) ( ) ( ) ( )       − + − −= ∫ ∫ θ π θ ϕ θϕ ϕϕ ϕ θϕ ϕϕ π θ 0 2/12/1 coscos sin2/cos coscos sin2/sin2 cos d n d n Pn Dirichlet Integrals Johann Peter Gustav Lejeune Dirichlet (1805 –1859)
  • 76. SOLO 76 Legendre Polynomials Add and subtract those two equations ( ) ( ) ( ) ( ) ( ) ( ) ( )       − + − = ∫ ∫ θ π θ ϕ θϕ ϕϕ ϕ θϕ ϕϕ π θ 0 2/12/1 coscos cos2/sin coscos cos2/cos2 cos d n d n Pn ( ) ( ) ( ) ( ) ( ) ( ) ( )       − + − −= ∫ ∫ θ π θ ϕ θϕ ϕϕ ϕ θϕ ϕϕ π θ 0 2/12/1 coscos sin2/cos coscos sin2/sin2 cos d n d n Pn Dirichlet Integrals ( ) ( )[ ] ( ) ( )[ ] ( )       − + + − + = ∫ ∫ θ π θ ϕ θϕ ϕ ϕ θϕ ϕ π θ 0 2/12/1 coscos 2/1sin coscos 2/1cos 2 1 cos d n d n Pn ( )[ ] ( ) ( )[ ] ( )∫ ∫ − − − − − = θ π θ ϕ θϕ ϕ ϕ θϕ ϕ 0 2/12/1 coscos 2/1sin coscos 2/1cos 0 d n d n Replace n by n + 1 in the last equation and substitute in the previous ( ) ( )[ ] ( ) ( )[ ] ( )∫ ∫ − + = − + = θ π θ ϕ θϕ ϕ π ϕ θϕ ϕ π θ 0 2/12/1 coscos 2/1sin2 coscos 2/1cos2 cos d n d n Pn Mehler Integrals Gustav Ferdinand Mehler (1835 - 1895) Return to Table of Content Laplace’s Integral Representation (continue - 4)
  • 77. SOLO 77 Legendre Polynomials We found Return to Table of Content Integrals in terms of sin(iθ) and cos(iθ) ( ) ( ) mnnk n dttPtP δ 12 21 1 + =∫ + − ( ) ( ) mnnk n dPP δθθθθ π 12 2 sincoscos 0 + =∫ θcos=t ( ) ( ) ( ) ( ) ( )     ≥ ++− + < = + = − + − ∫ nm nmnm nmm nm a n tdtPt n nn m !122! !!22 0 14 2 12 2 1 1 2 2 ( ) ( ) ( ) ( ) ( )     ≥ ++− +++ < = + = + + + − + + ∫ nm nmnm nmm nm a n tdtPt n nn m !322! !1!122 0 34 2 22 12 1 1 12 12 ( ) ( ) ( ) ( ) ( )     ≥ ++− + < = − ∫ nm nmnm nmm nm dP n n m !122! !!22 0 sincoscos 12 0 2 2 π θθθθ ( ) ( ) ( ) ( ) ( )     ≥ ++− +++ < = + + + ∫ nm nmnm nmm nm dP n n m !322! !1!122 0 sincoscos 22 0 12 12 π θθθθ
  • 78. Ordinary Differential EquationsSOLO Second Order Linear Ordinary Differential Equation (ODE) 78 Legendre Functions of the Second Kind Qn (x) ( ) [ ] ( ) ( )∑ ∫ ∞ = −− ∞ −− = − − +− −+= 0 2 12/12 0 1 2 1 cosh21 cosh1 n n n n n txQ x xt ttx dxxxQ θθ ( ) ( )n n n nn x xd d n xP 1 !2 1 2 −= ( ) ( ) 1<+= xxQBxPAy nn ( ) ( ) ( )xP xd d xxP nm m mm n 2/2 1−= ( ) ( ) ( )xQ xd d xxQ nm m mm n 2/2 1−= ( ) ( ) ( )xW x x xPxQ nnn 1 1 1 ln 2 1 −− − + = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 5 2 3 1 1 1 ln 2 1 2 033 022 011 0 +−= −= −= − + = xxQxPxQ xxQxPxQ xQxPxQ x x xQ ( ) ( ) ( )∑= −−− = n m mnmn xPxP m xW 1 11 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( )   − − =    −       − − +− − − + = −+ −− − − + = ∑ ∑ = − =+       − = −− evennifn oddnifnn xP m nm mn x x xP xP rnr rn x x xPxQ n m mnn mr n r rnnn 2/2 2/1 2 1 2 1 122 1 1 ln 2 1 12 142 1 1 ln 2 1 1 12 2 1 0 12
  • 79. Legendre Ordinary Differential EquationSOLO 79 Legendre Functions of the Second Kind Qn (x) ( ) ( ) ( )xPxuxy nnn = With Pn (x) being a solution of the Legendre Differential Equation we look for the second solution having the form ( ) ( ) 0121 2 2 2 =++−− wnn xd wd x xd wd x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 xd xPd xu xd xPd xd xud xP xd xud xd xyd xd xPd xuxP xd xud xd xyd n n nn n nn n nn nn ++= += ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 01221121 2 2 22 2 2 2 =++−−−+−+− xPxunn xd xPd xuxxP xd xud x xd xPd xxu xd xPd xd xud xxP xd xud x nn n nn nn n nn n n Substituting in the Legendre ODE we obtain ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 01212121 0 2 2 22 2 2 2 =      ++−−+−−+− xuxPnn xd xPd x xd xPd xxP xd xud x xd xPd xd xud xxP xd xud x nn nn n nnn n n    or
  • 80. SOLO 80 Legendre Functions of the Second Kind Qn (x) (continue – 1) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 02121 2 2 2 2 =−−+− xP xd xud x xd xPd xd xud xxP xd xud x n nnn n n Equivalent to ( ) ( ) ( ) ( ) 0 1 2/ 2 / / 2 22 = − −+ x x xP xdxPd xdxud xdxud n n n n ( ) ( ) ( ) 01lnln2ln 2 =−++ x xd d xP xd d xd xud xd d n n or Integrating we obtain ( ) ( )[ ] ( ) .1lnlnln 22 constxxP xd xud n n =−++ Therefore ( ) ( )[ ] ( ) AconstxxP xd xud n n ==−⋅⋅ .1 22 ( ) ( )[ ] ( )∫ −⋅ = 22 1 xxP xd Axu n n This means that the second solution has the form ( ) ( ) ( ) ( ) ( )[ ] ( )∫ −⋅ == 22 1 xxP xd xPxuxPxQ n nnnn Legendre Ordinary Differential Equation
  • 81. SOLO 81 Legendre Functions of the Second Kind Qn (x) (continue – 2) We obtained ( ) ( ) ( ) ( ) ( )[ ] ( ) 1 1 22 < −⋅ == ∫ x xxP xd xPxuxPxQ n nnnn ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] x x xxxd xxxxP xd xPxQ − + =++−−=      + + − = −⋅ = ∫∫ 1 1 ln 2 1 1ln1ln 2 1 1 1 1 1 2 1 1 2 1 2 01 00  Let calculate Q0 (x), Q1 (x) ( ) ( ) ( )[ ] ( ) ( ) 1 1 1 ln 2 1 1 2/1 1 2/1 1 1 1 22222 1 11 2 − − + =      + + + − = − = −⋅ = ∫∫∫ x xx xd xxx xxd xx x xxP xd xPxQ x x  Legendre Ordinary Differential Equation
  • 82. SOLO 82 Legendre Functions of the Second Kind Qn (x) (continue – 2) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 5 1 1 ln 2 1 3 2 2 5 1 1 ln 4 35 2 3 1 1 ln 2 1 2 3 1 1 ln 4 43 1 1 1 ln 2 1 1 1 1 ln 2 1 1 ln 2 1 2 3 23 3 2 2 2 11 0 +−      − + =+−      − +− = −      − + =−      − +− = −      − + =−      − + =       − + = x x x xP x x xxx xQ x x x xP x x xx xQ x x xP x xx xQ x x xQ Legendre Ordinary Differential Equation
  • 83. SOLO 83 Legendre Functions of the Second Kind Qn (x) (continue – 3) To obtain a general formula for Qn (x), start from the Polynomial Pn (x) that has n zeros αi, i=1,2,…,n ( ) ( ) ( ) ( )nnn xxxkxP ααα −−−= 21 ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑=       − + − + + + − = +⋅−⋅−−− = −⋅ n i i i i i nnn x d x c x b x a xxxxxkxxP 1 2 00 22 2 2 1 222 11 11 1 1 1 αα ααα  ( ) ( )[ ] ( ) ( )[ ] ( )[ ] ( ) ( ) ( )∑=       − + − −⋅+−++= n i i i i i nnn x d x c xxPxPxbxPxa 1 2 222 0 2 0 1111 αα If we put x=1 and x = -1, and remembering that Pn (1) = 1 and Pn(-1)=(-1)n , we obtain 2/100 == ba Let prove that ( ) ( )[ ] ( ) 0 1 1 22 2 =               −⋅ −= = ixn ii xxP x xd d c α α Legendre Ordinary Differential Equation
  • 84. SOLO 84 Legendre Functions of the Second Kind Qn (x) (continue – 4) Let prove that ( ) ( )[ ] ( ) 0 1 1 22 2 =               −⋅ −= = ixn ii xxP x xd d c α α Start with ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) i x iixi xatfinitexfprovided xd xfd xxfxxfx xd d i i αααα α α ==      −+−=− = = 02 22 The only terms that are not finite in at x = αi are the terms ci/(x-αi) and di/(x-αi)2 , therefore ( )[ ] ( ) ( ) ( )∑=       − + − + + + − = −⋅ n i i i i i n x d x c x b x a xxP 1 2 00 22 111 1 αα ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )[ ] i x iii xi i i i i xn i cdxc xd d x d x c x xd d xxP x xd d iii =       +−=               − + − −=       −⋅ − === ααα α αα αα 2 2 22 2 1 1 Therefore if we write Pn (x) = (x-αi) Li (x), we have ( )[ ] ( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )[ ] ( )22 2 223 2 2322222 1 /1 2 1 /122 1 /2 1 2 1 1 iii iiiiii xi ii xi i ixi i L xdxLdL xxL xdxLdxxLx xxL xdxLd xxL x xxLxd d c iii αα αααα ααα −⋅ =−− =         −⋅ −⋅− =         −⋅ − −⋅ =               −⋅ = === Legendre Ordinary Differential Equation
  • 85. SOLO 85 Legendre Functions of the Second Kind Qn (x) (continue – 5) We proved that ( ) ( )[ ] ( ) ixn ii xxP x xd d c α α =               −⋅ −= 22 2 1 1 ( ) ( ) ( ) ( )[ ] ( )22 2 1 /1 2 iii iiiiii i L xdxLdL c αα αααα −⋅ =−− =Therefore if we write Pn (x) = (x-αi) Li (x), we have Since Pn (x) = (x-αi) Li (x) satisfies the Legendre ODE, we have ( ) ( ) ( ){ } ( ) ( ){ } ( ) ( ) ( ){ } 0121 2 2 2 =−++−−−− xLxnnxLx xd d xxLx xd d x iiiiii ααα Performing the calculation and substituting x = αi, we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] 01221 2 2 2 =       −++      +−−      +−− = ix iii i i ii i xLxnnxL xd xLd xx xd xLd xd xLd xx α ααα ( ) ( ) ( ) 0212 2 =− = − iii ii i L xd xLd αα α α Substituting in ci equation we get ( ) ( ) ( ) ( )[ ] ( ) 0 1 /1 2 22 2 = −⋅ =−− = iii iiiiii i L xdxLdL c αα αααα Therefore ( )[ ] ( ) ( )∑= − +      + + − = −⋅ n i i i n x d xxxxP 1 222 1 1 1 1 2 1 1 1 α Legendre Ordinary Differential Equation
  • 86. SOLO 86 Legendre Functions of the Second Kind Qn (x) (continue – 6) We can prove that but the exact value is not important, as we shall see ( ) ( )[ ] ( ) ixn i i xxP x d α α =      −⋅ − = 22 2 1 Therefore ( )[ ] ( ) ( )∑= − +      + + − = −⋅ n i i i n x d xxxxP 1 222 1 1 1 1 2 1 1 1 α ( )[ ] ( ) ( ) ∑∑∫∫∫ == − − − + = − +      + + − = −⋅ n i i i n i i i n x d x x dx x d dx xxxxP dx 11 222 1 1 ln 2 1 1 1 1 1 2 1 1 αα ( ) ( ) ( ) ∑= − − − + = n i i n inn x xP d x x xPxQ 11 1 ln 2 1 α Since Pn (x)/(x-αi) is a polynomial of order (n-1) the sum above is also a polynomial of order (n-1), and we define it as ( ) ( ) ∑= − − = n i i n in x xP dxW 1 1 : α so ( ) ( ) ( ) 1 1 1 ln 2 1 1 <− − + = − xxW x x xPxQ nnn Legendre Ordinary Differential Equation
  • 87. SOLO 87 Legendre Functions of the Second Kind Qn (x) (continue – 7) To find Wn-1 (x) let use the fact that Qn (x) is a solution of Legendre’s ODE or ( ) ( ) ( ) ( ) ( ) 0121 2 2 2 =++−− xQnn xd xQd x xd xQd x n nn ( ) ( ) ( ) ( ) xd xWd xP xx x xd xPd xd xQd n n nn 1 2 1 1 1 1 ln 2 1 − − − + − + = ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2222 2 2 2 1 2 1 2 1 1 ln 2 1 xd xWd xP x x xd xPd xx x xd xPd xd xQd n n nnn − − − + − + − + = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0121 1 2 2 1 2 1 1 ln 2 1 121 1 1 2 1 2 2 22 0 2 2 2 =+−+−− − −+ − + − +       ++−− − −− xWnn xd xWd x xd xWd xxP x x xd xPd xP x x x x xPnn xd xPd x xd xPd x n nn n n n n nn    ( ) ( ) ( ) ( ) ( ) ( ) xd xPd xWnn xd xWd x xd xWd x n n nn 2121 1 1 2 1 2 2 =++−− − −− Legendre Ordinary Differential Equation ( ) ( ) ( ) 1 1 1 ln 2 1 1 <− − + = − xxW x x xPxQ nnn
  • 88. SOLO 88 Legendre Functions of the Second Kind Qn (x) (continue – 8) Use the Recursive Formula ( ) ( ) ( ) ( ) ( ) xd xPd xWnn xd xWd x xd d n n n 211 1 12 =++       − − − ( ) ( ) ( ) ( ) ( ) ∑       − = −−−−= 1 2 1 0 121421 l i in n xPin xd xPd Since Wn-1(x) is a polynomial of order n – 1, let write ( ) ( ) ( ) ( ) ( ) ∑       − = −−−−− =++= 1 2 1 0 1231101 n i ininnn xPaxPaxPaxW  Substitute those two equation in the Wn-1(x) O.D.E. to obtain ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∑∑∑       − = −−       − = −−       − = −− −−=++− 1 2 1 0 12 1 2 1 0 12 1 2 1 0 12 2 142211 l i in n i ini n i ini xPinxPannxPx xd d a But by Legendre O.D.E.: ( ) ( ){ } ( ) ( ) ( ) 02121 1212 2 =−−−+− −−−− xPininxPx xd d inin ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ∑∑       − = −−       − = −− −−=++−−−− 1 2 1 0 12 1 2 1 0 12 14221212 l i in n i ini xPinxPnninina Let solve Legendre Ordinary Differential Equation
  • 89. SOLO 89 Legendre Functions of the Second Kind Qn (x) (continue – 9) The coefficients of the same polynomial in both sides must be equal ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ∑∑       − = −−       − = −− −−=++−−−− 1 2 1 0 12 1 2 1 0 12 14221212 l i in n i ini xPinxPnninina ( ) ( ) ( ){ } ( )14221212 −−=++−−−− innnininai ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )12224244 1221212 222 2 +−=−+−=++−+−+−= ++−+−−=++−−−− iininininniniinn nnininnninin But which gives ( ) ( )12 142 +− −− = iin in ai ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ][ ] ∑ ∑∑ = = −− =+       − = −−− +− +− =       − − +− = +− −− = n m n m mnmn mi n i inn xP mn mn m xP m n mn m xP iin in xW 1 1 12 1 2 1 0 121 12 12221 2 1 1221 12 142 and ( ) ( ) ( )∑= −−− = n m mnmn xPxP m xW 1 11 1 ????? Legendre Ordinary Differential Equation
  • 90. SOLO 90 Legendre Functions of the Second Kind Qn (x) (continue -10) ( ) 3,2,1,0 10 = <≤ n xxQn Legendre Ordinary Differential Equation ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 5 1 1 ln 2 1 3 2 2 5 1 1 ln 4 35 2 3 1 1 ln 2 1 2 3 1 1 ln 4 43 1 1 1 ln 2 1 1 1 1 ln 2 1 1 ln 2 1 2 3 23 3 2 2 2 11 0 +−      − + =+−      − +− = −      − + =−      − +− = −      − + =−      − + =       − + = x x x xP x x xxx xQ x x x xP x x xx xQ x x xP x xx xQ x x xQ
  • 91. SOLO 91 Legendre Functions of the Second Kind Qn (x) (continue -11) Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x) Start with ( ) ( ) ( )12 2212 1 1 1 1 +−− ++ −= − n nn x xx ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )1 32211 11, ,121,11,1: ++− +−+−+− −++= −++=−+=−= ini nnn uinnuf unnufunufuuf   ( ) ( ) ( ) ( ) 1 !! ! !! !1 1 1 12 0 2 0 2212 > + = + = − ++− ∞ = − ∞ = ++ ∑∑ xx in in x in in xx in i i i nn ( ) ( ) ( ) ( )( ) ( ) 1 ! 21 ! 0 00 < +++ == ∑∑ ∞ = ∞ = uu i innn u i f uf i i i i i  Taylor expansion around u = 0 Use u = x-2 Legendre Ordinary Differential Equation
  • 92. SOLO 92 Legendre Functions of the Second Kind Qn (x) (continue -12) Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x) Integrate relative to x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∑∑ ∫∑∫∑∫ ∞ = ++− ∞ = ∞ ++− ∞ ++− ∞ = ∞ ++− ∞ = ∞ + −+ + = ++ + = + = + = − 0 122 0 122 12 0 2 12 0 212 122!! ! 122!! ! ! ! ! ! 1 i in i xin x in ix in ix n x inni in inni in d n in d n ind η ηηηη η η Integrate this n more times ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∑ ∑ ∑ ∫ ∫∫∫ ∫ ∞ = ++− ∞ = ++− ∞ = ∞ ∞ ++− ∞ ∞ ∞ + + ++ ++ = +++++++ + = ++ + = − 0 12 0 12 0 122 12 1 !122!! !2! 122222122!! ! 122!! ! 1 i in i in i x nin x n n x inni inin x ininininni in d inni ind   ηη η ηη η η Legendre Ordinary Differential Equation
  • 93. SOLO 93 Legendre Functions of the Second Kind Qn (x) (continue -13) Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x) (continue) ( ) ( ) ( ) ( ) ( ) 1 !122! !2! ! 1 1 0 12 12 1 > ++ ++ = − ∑∫∫ ∫ ∞ = ++− ∞ ∞ ∞ + + xx ini inin n d i in x n n η η η η  We found, using Frobenius Series By comparing those two relations we obtain ( ) ( ) 1 1 !2 12 1 > − = ∫∫ ∫ ∞∞ ∞ + + x d nxQ x n n n n η η η η  Return to Frobenius Series ( ) ( ) ( ) ( ) ( ) 1 !122! !2! 2 0 12 > ++ ++ = ∑ ∞ = ++− xx ini inin xQ i inn n Legendre Ordinary Differential Equation
  • 94. SOLO 94 Legendre Functions of the Second Kind Qn (x) (continue -14) Another Expression for Legendre Functions of the Second Kind Qn (x) Start with the following Differential Equation ( ) ( ) 02121 2 2 2 =+−+− un xd ud xn xd ud x One of the Solutions is . Check:( ) ( )n xxu 12 1 −= ( ) ( ) ( ) ( ) 22212 2 1 2 121 11412&12 −−− −−+−=−= nnn xxnnxn xd ud xxn xd ud ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 01211411412 2121 21221222 1 1 2 1 2 2 =−−−−+−−−−−= +−+− −− nnnn xnxnxnxxnnxn un xd ud xn xd ud x The Second Solution can be find using u1 (x) by finding a function v (x) that satisfies: ( ) ( ) ( )n xxvxu 12 2 −= Legendre Ordinary Differential Equation
  • 95. SOLO 95 Legendre Functions of the Second Kind Qn (x) (continue -15) Another Expression for Legendre Functions of the Second Kind Qn (x) (continue – 1) ( ) ( ) ( ) ( ) ( ) ( ) 01221 21212121 112 2 1 12 0 1 1 2 1 2 2 2 2 2 2 2 2 =−+      +−+       +−+−=+−+− xd vd uxnu xd vd u xd vd xd ud x un xd ud xn xd ud xvun xd ud xn xd ud x    The Second Solution can be find using u1 (x) by finding a function v (x) that satisfies: ( ) ( ) ( )n xxvxu 12 2 −= 2 1 2 1 12 2 2 2 2 1 1 2 2& xd ud v xd ud xd vd u xd vd xd ud xd ud vu xd vd xd ud ++=+= ( ) ( ) ( ) ( ) ( ) ( ) ( )1 121 2 2 1 12 2 1 12 / / 2 1 2 1 2 1 1 1 2 1 22 − + −= − − − − = − − − − = x xn u x uxn x uxn u xd ud x uxn xdvd xdvd Legendre Ordinary Differential Equation
  • 96. SOLO 96 Legendre Functions of the Second Kind Qn (x) (continue -16) Another Expression for Legendre Functions of the Second Kind Qn (x) (continue – 2) Therefore ( ) ( )1 12 / / 2 22 − + −= x xn xdvd xdvd ( ) ( ) ( ) ( ) ( )∫ ∞ ++ − =⇒ − =⇒−+−= x nn d xv xxd vd xn xd vd 1212 2 11 1 1ln1 η η ( ) ( ) ( ) ( ) ( )∫ ∞ + − −== x n n d xxuxvxu 12 2 12 1 1 η η Differentiate the Differential Equation ( ) ( ) 02121 2 2 2 =+−+− un xd ud xn xd ud x ( ) ( ) ( ) ( ) ( ) ( ) 0122221 2121221 2 2 3 3 2 2 2 2 2 2 =−+−+−= +−+−+−− xd ud n xd ud xn xd ud x xd ud n xd ud xd d xn xd ud n xd ud x xd ud xd d x Legendre Ordinary Differential Equation
  • 97. SOLO 97 Legendre Functions of the Second Kind Qn (x) (continue -17) Another Expression for Legendre Functions of the Second Kind Qn (x) (continue – 3) Differentiate relative to x the Ordinary Differential Equation n times ( ) ( ) ( ) 0122221: 2 2 3 3 2 =−+−+− xd ud n xd ud xn xd ud xODE xd d ( ) ( ) ( ) 0223321: 2 2 4 4 2 2 2 =−+−+− xd ud n xd ud xn xd ud xODE xd d ( ) ( ) 02121: 2 2 2 =+−+− un xd ud xn xd ud xODE ( ) ( ) 0121: 1 1 2 2 2 =++−− + + + + n n n n n n n n xd ud nn xd ud x xd ud xODE xd d Derive ( ) ( )[ ] ( )( ) 021121: 1 1 2 2 2 =−+++−+− + + + + i i i i i i i i xd ud ini xd ud xin xd ud xODE xd d ( ) ( )[ ]{ } ( )[ ] ( )( ){ } ( ) ( )[ ] ( )( ) 0122221 21121221: 1 1 1 1 3 3 2 1 1 1 1 3 3 2 =−−+++−+− −+++−++−+−+− + + + + + + + + + + + + i i i i i i i i i i i i i i xd ud ini xd ud xin xd ud x xd ud iniin xd ud xin xd ud xODE xd d xd d Proof by Induction q.e.d. i = n i → i+1 Legendre Ordinary Differential Equation
  • 98. SOLO 98 Legendre Functions of the Second Kind Qn (x) (continue -18) Another Expression for Legendre Functions of the Second Kind Qn (x) (continue – 4) This is the Legendre Ordinary Differential Equation for dn u/dxn , having the solution Pn (x) and Qn (x). Thus, the solution Pn (x) and Qn (x) can be written in the following form: ( ) ( ) 0121: 1 1 2 2 2 =++−− + + + + n n n n n n n n xd ud nn xd ud x xd ud xODE xd d We obtain ( ) ( )[ ]n n n nn n nn x xd d nxd ud n xP 1 !2 1 !2 1 21 −== ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 !2 !21 !2 !21 12 22 >         − − − = − = ∫ ∞ + x d x xd d n n xd ud n n xQ x n n n nnn n nnn n η η Rodrigues Formula An integral for Qn (x) valid in |x| < 1 can be obtained from the previous result ( ) ( ) ( ) ( ) ( ) 1 1 1 !2 !21 0 12 2 <         − − − = ∫ + x d x xd d n n xQ x n n n nnn n η η Return to Table of Content Legendre Ordinary Differential Equation
  • 99. 99 SOLO Laplace Differential Equation in Spherical Coordinates 0 sin 1 sin sin 11 2 2 222 2 2 2 = ∂ Φ∂ +      ∂ Φ∂ ∂ ∂ +      ∂ Φ∂ ∂ ∂ =Φ∇ φθθ θ θθ rrr r rr Let solve this equation by the method of Separation of Variables, by assuming a solution of the form : ( ) ( )ϕθ,SrR=Φ Spherical Coordinates: θ ϕθ ϕθ cos sinsin cossin rz ry rx = = = In Spherical Coordinates the Laplace equation becomes: Substituting in the Laplace Equation and dividing by Φ gives: 0sinsin sin 11 2 2 22 2 2 =      ∂ ∂ +      ∂ ∂ ∂ ∂ +      φθ θ θ θ θ SS Srrd Rd r rd d Rr The first term is a function of r only, and the second of angular coordinates. For the sum to be zero each must be a constant, therefore: λ φθ θ θ θ θ λ −=      ∂ ∂ +      ∂ ∂ ∂ ∂ =      2 2 2 2 sinsin sin 1 1 SS S rd Rd r rd d R Associate Legendre Differential Equation
  • 100. 100 SOLO λ φθ θ θ θ θ −=      ∂ ∂ +      ∂ ∂ ∂ ∂ 2 2 2 sinsin sin 1 SS S ( ) ( ) ( )φθφθ ,,, SrRr =Φ We obtain: Multiply this by S sin2 θ and put to get:( ) ( ) ( )φθφθ ΦΘ=,S 0 1 sinsinsin 1 2 2 2 = Φ Φ +      +      Θ ∂ ∂ Θ φ θλ θ θ θ θ d d d d Again, the first term, in the square bracket, and the last term must be equal and opposite constants, which we write m2 , -m2 . Thus: Φ−= Φ =Θ      −+      Θ 2 2 2 2 2 0 sin sin sin 1 m d d m d d d d φ θ λ θ θ θθ The Φ ( ) must be periodical in (a period of 2 π) and becauseϕ ϕ this we choose the constant m2 , with m an integer. Thus: ( ) φφφ mbma sincos +=Φ Laplace Differential Equation in Spherical Coordinates ( ) φφ φ mjmj ee +− =Φ ,or With m integer, we have the Orthogonality Condition 21 21 , 2 0 2 mm mjmj dee δπφ π φφ =⋅∫ +− Return to Table of Content Associate Legendre Differential Equation
  • 101. 101 SOLO ( ) ( ) ( )φθ ΦΘ=Φ rR We obtain: or: Θ−=Θ− Θ + Θ λ θθ θ θ 2 2 2 2 sin cot m d d d d 0 sin sin sin 1 2 2 =Θ      −+      Θ ∂ ∂ θ λ θ θ θθ m d d Analysis of Associate Lagrange Differential Equation Laplace Differential Equation in Spherical Coordinates Change of variables: t = cos θ θθ dtd sin−= td d d d Θ −= Θ θ θ sin  ( ) td d t td d t td d d d td d td d td d td d d d d d d d Θ − Θ −= Θ + Θ =      Θ −−=      Θ = Θ − 2 2 2 cossin/1 2 2 2 2 2 1 sin sinsinsinsin  θθ θ θθ θθθθ θθθ Θ      − −+ Θ − Θ − Θ =Θ      −+ Θ + Θ = = 2 2 2 2cos 2 2 2 2 1sin cot0 t m td d t td d t td dm d d d d t λ θ λ θ θ θ θ We obtain: 1cos 0 1 2 2 2 2 2 ≤⇒= =Θ      − −+ Θ − Θ tt t m td d t td d θ λ Associate Legendre Differential Equation
  • 102. 102 SOLO ( ) ( ) ( )ϕθ gfrR=Φ We obtain: or: ff m d fd d fd λ θθ θ θ −=−+ 2 2 2 2 sin cot Let try to factorize the left-hand side, second order differential equation into two first-order operators: 0 sin sin sin 1 2 2 =      −+      ∂ ∂ f m d fd θ λ θ θ θθ The two equations are identical if the coefficient are equal. This is obtained by choosing α and β as follows: ( ) integer1 1 2 ==− =+ mmαβ βα ( ) ( ) ( ) ff d fd d fd ff d fd d fd f d fd d d f d d d d βα θ αβ θ θβα θ θβα θ β θ θβα θ θβ θ θα θ θβ θ θα θ θ −−−++= +−++=       +      +=      +      + − 22 2 1 sin 1 2 22 2 sin 1 1cot cot sin cot cotcotcotcot 2  Laplace Differential Equation in Spherical Coordinates Analysis of Associate Lagrange Differential Equation (continue – 1) Associate Legendre Differential Equation
  • 103. 103 SOLO ( ) ( ) ( )ϕθ gfrR=Φ We obtain: and: ff m d fd d fd λ θθ θ θ −=−+ 2 2 2 2 sin cot We have two solutions for α and β as follows: 1.β1 = m, α1 = 1-m 2.β2 = -m, α2 = 1+m Since m is an integer α, β are also integers. ( ) integer1 1 2 ==− =+ mmαβ βα ( ) ( ) ( ) fff d fd d fd  λ βαλβα θ αβ θ θβα θ +−=−−−++ ' sin 1 1cot 22 2 βα −=1 integer22 == mmβ Let define the two operators: ( ) ( ) ( ) θ θ θ θ cot1 cot ++= −= − + m d d M m d d M m m Laplace Differential Equation in Spherical Coordinates Analysis of Associate Lagrange Differential Equation (continue – 2) Associate Legendre Differential Equation
  • 104. 104 SOLO ( ) ( ) ( )ϕθ gfrR=Φ We obtain: We have two solutions for α and β as follows: 1.β1 = m, α1 = 1-m 2.β2 = -m, α2 = 1+m Since m is an integer α, β are also integers. ( ) ( ) ( ) θ θ θ θ cot1 cot ++= −= − + m d d M m d d M m m ff d d d d λθβ θ θα θ −=      +      + cotcot ( )  ( )  ( )  ( ) ( )11 11 11 11 1 mmmm fmmfMM         −−−= − − − + − αβ βα λ We have: ( )  ( )  ( )  ( ) ( )22 22 22 1 mmmm fmmfMM         +−−= − +− αβ βα λ fm (1) – the solution for α1, β1 fm (2) – the solution for α2, β2 Laplace Differential Equation in Spherical Coordinates Analysis of Associate Lagrange Differential Equation (continue – 3) Associate Legendre Differential Equation
  • 105. 105 SOLO ( ) ( ) ( )ϕθ gfrR=Φ We obtain: ( ) ( ) ( ) ( )[ ] ( )11 11 1 mmmm fmmfMM −−−=− − + − λ ( ) ( ) ( ) ( )[ ] ( )22 1 mmmm fmmfMM +−−=+− λ Let operate on first of those equations with ( )− −1mM ( ) ( ) ( ) ( ) [ ] ( )[ ] ( ) ( ) [ ]1 1 1 111 1 mmmmmm fMmmfMMM − − − − + − − − −−−= λ In the Second Equation replace m by m-1 ( ) ( ) ( ) ( )[ ] ( )2 1 2 111 1 −− + − − − −−−= mmmm fmmfMM λ Let operate on second of those equations with ( )+ mM 1 2 ( ) ( ) ( ) ( ) [ ] ( )[ ] ( ) ( ) [ ]22 1 mmmmmm fMmmfMMM ++−+ +−−= λ In the First Equation replace m by m+1 ( ) ( ) ( ) ( )[ ] ( )1 1 1 1 1 ++ −+ +−−= mmmm fmmfMM λ ( ) ( ) ( )21 1 mmmm fpfM =+ − ( ) ( ) ( )1 1 2 + + = mmmm fqfM Laplace Differential Equation in Spherical Coordinates pm is a constant qm is a constant mm →−1 mm →−1 Analysis of Associate Lagrange Differential Equation (continue – 4) Associate Legendre Differential Equation
  • 106. 106 SOLO We obtained:( ) ( ) ( )21 1 mmmm fpfM =+ − ( ) ( ) ( )1 1 2 + + = mmmm fqfM Laplace Differential Equation in Spherical Coordinates ( ) ( ) ( ) θ θ θ θ cot cot1 m d d M m d d M m m −= ++= + − where: Theorem: ( ) ( ) ( ) ( ) ( ) ( )∫∫ +− −= ππ θθθθθθθθ 00 sinsin dgMfdfMg mm where f and g are arbitrary bounded function of θ. Proof: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )∫∫ ∫ ∫∫ + − −=      −−=       +−−=       ++= ππ π π ππ θθθθθθθ θ θ θθ θ θθ θθθθθθ θ θθθθ θθθ θ θθθθθθ 00 0 0 0 00 sinsin sin cos sin cos1sinsin cot1sinsin dgMfdgfmg d d f dgfmg d d ffg dfm d d gdfMg m m    Those are Recursive Relations in m. Analysis of Associate Lagrange Differential Equation (continue – 5) Associate Legendre Differential Equation
  • 107. 107 SOLO Laplace Differential Equation in Spherical Coordinates ( ) ( ) ( )( ) ( ) ( ) ( )( )∫∫ +− −= ππ θθθθθθθθ 00 sinsin dgMfdfMg mm ( ) ( ) ( ) ( ) ( ) ( ) ( )∫∫ + −+ ++ − + − −= ππ θθθθ 0 11 0 11 sinsin dfMMfdfMfM mmmmmmmm ( ) ( ) ( ) ( ) ( ) ( ) ∫ ∫∫ = =+ − + − π ππ θθ θθθθ 0 22 00 11 sin sinsin dfp dfpfpdfMfM mm mmmmmmmm Choose f := fm+1 and g := Mm (-) fm+1: ( ) ( ) ( ) ( )[ ]{ } ( )[ ] ∫ ∫∫ + +++ −+ + +−= +−−−=− π ππ θθλ θλθθθ 0 2 1 0 11 0 11 sin1 1sinsin dfmm dfmmfdfMMf m mmmmmm Assuming: we have1sinsin 0 2 1 0 2 == ∫∫ + ππ θθθθ dfdf mm ( ) ( ) λλ ≤+→+−= 11 mmmmpm Analysis of Associate Lagrange Differential Equation (continue – 6) Associate Legendre Differential Equation
  • 108. 108 SOLO Laplace Differential Equation in Spherical Coordinates ( ) ( ) ( )( ) ( ) ( ) ( )( )∫∫ +− −= ππ θθθθθθθθ 00 sinsin dgMfdfMg mm Choose now f := Mm (+) fm and g := fm, to obtain as before We obtained ( ) ( ) λλ ≤+→+−= 11 mmmmqm ( ) ( ) λλ ≤+→+−= 11 mmmmpm We see that m, an integer, is bounded by λ, therefore we must choose λ as ( ) 0integer1 >=+= lllλ In this case we have mMAX = l, mmin = -(l+1), or ( ) integer0,integer1 =>=≤≤+− mllml Therefore ( ) ( ) ( ) lmlmmllqp mm ≤≤+−→+−+== 111 Analysis of Associate Lagrange Differential Equation (continue – 7) Associate Legendre Differential Equation
  • 109. 109 SOLO Laplace Differential Equation in Spherical Coordinates We obtained ( ) ( ) ( ) lmlmmllqp mm ≤≤+−→+−+== 111 ( ) ( ) ( ) θ θ θ θ cot cot1 m d d M m d d M m m −= ++= + − ( ) ( ) ( )21 1 mmmm fpfM =+ − ( ) ( ) ( )1 1 2 + + = mmmm fqfM ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lml mmllfm d d fmmllfm d d m mm ≤≤+−        +−+=      − +−+=      ++ + 1 11cot 11cot1 2 21 1 θ θ θ θ Substituting m = - (l+1) in the First Equation and m = l in the Second we obtain: ( ) ( )        =      − =      − − 0cot 0cot 2 1 l l fl d d fl d d θ θ θ θ ( ) ( ) ( ) ( ) ( ) ( ) θ θ θ θ sin sin sin sin 2 2 1 1 d l f fd d l f fd l l l l = = + + − − ( ) ( ) θl lll Cff sin 21 == +− If we can find a Solution for a particular m, we can use the Recursive Relations above to find the others. Analysis of Associate Lagrange Differential Equation (continue – 8) Associate Legendre Differential Equation
  • 110. 110 SOLO Laplace Differential Equation in Spherical Coordinates We obtained ( ) ( ) θl lll Cff sin 21 == +− Cl must be chosen to normalize fi and f-l: 1sinsin 0 122 0 2 == ∫∫ + ± ππ θθθθ dCdf l ll ( ) ( )!12 !21 212 2 + = + n n C n l ( ) ( )212 !2 !12 n n C nl + + = ( ) ( ) ( )∫∫∫∫ +−−+ −=+−=−= ππ π ππ θθθθθθθθθθθθ 0 1212 0 212 0 0 2 0 2 0 12 sinsin2cossin2cossincossinsin dndndd nnnnnn    Therefore ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )!12 !2 cos 2222 !2 11212 !2 sin 31212 2222 sin 12 2 sin 212 2 0 1 00 12 0 12 + = −⋅ ⋅ −⋅+ = −⋅+ −⋅ = + = + −+ ∫∫∫ n n nn n nn n d nn nn d n n d nnn nn       π πππ θθθθθθθ Analysis of Associate Lagrange Differential Equation (continue – 9) Associate Legendre Differential Equation
  • 111. 111 SOLO Laplace Differential Equation in Spherical Coordinates The Normalized Solution for m = l is defined as ( ) ( ) θl n lm l lm l n n sin !2 !12 212 + −== + =Θ=Θ The Solutions for m < l can be found using the Recursive Relation: ( ) ( ) ( ) ( )1,,,0,,1,cot 11 11 1 1 1 +−−−=Θ      + −−+ =Θ=Θ − − − − llllmm d d mmll M p m l m lm m m l θ θ ( ) ( ) ( )21 1 mmmm fpfM =+ − From which the Normalized Solutions for m < l are given by Or we can find the Solutions for m >- l by using the Recursive Relation: ( ) ( ) ( )1 1 2 + + = mmmm fqfM ( ) ( ) ( ) llllmm d d mmll M q m l m lm m m l ,1,,0,,,1cot 11 111 −−−−=Θ      − +−+ =Θ=Θ ++ θ θ Analysis of Associate Lagrange Differential Equation (continue – 10) Associate Legendre Differential Equation
  • 112. 112 SOLO Laplace Differential Equation in Spherical Coordinates ( ) ( ) θl n lm l lm l n n sin !2 !12 212 + −== + =Θ=Θ ( ) ( ) ( ) ( )1,,,0,,1,cot 11 11 1 1 1 +−−−=Θ      + −−+ =Θ=Θ − − − − llllmm d d mmll M p m l m lm m m l θ θ Examples: ( ) ( ) ( ) llllmm d d mmll M q m l m lm m m l ,1,,0,,,1cot 11 111 −−−−=Θ      − +−+ =Θ=Θ ++ θ θ Analysis of Associate Lagrange Differential Equation (continue – 11) 2 cos 1 1 cos 0 1 2 cos 1 1 0 0 1 2 3 sin 2 3 2 3 cos 2 3 1 2 3 sin 2 3 1 2 1 0 x x xl l x x x −−=−=Θ ==Θ −==Θ= =Θ= = − = = θ θ θ θ θ θ ( ) ( ) ( ) ( )2 cos 22 2 2 cos 1 2 2 cos 20 2 2 cos 1 2 2 cos 22 2 1 4 15 sin 4 15 1 2 15 cossin 2 15 13 22 5 1cos3 22 5 1 2 15 cossin 2 15 1 4 15 sin 4 15 2 x xx x xx xl x x x x x −−=−=Θ −−=−=Θ −=−=Θ −==Θ −==Θ= = − = − = = = θ θ θ θ θ θ θθ θ θθ θ Return to Table of Content Associate Legendre Differential Equation
  • 113. SOLO 113 Associated Legendre Functions Let Differentiate this equation m times with respect to t, and use Leibnitz Rule of Product Differentiation: ( ) ( )[ ] ( ) ( ) ( ) im im i im i m m td tgd td tsd imi m tgts td d − − = ∑ − =⋅ 0 !! ! Start with: ( ) ( ) ( ) ( ) 1011 2 ≤=++      − ttwnntw td d t td d nn or: ( ) ( ) ( ) ( ) ( ) 10121 2 2 2 ≤=++−− ttwnntw td d ttw td d t nnn ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )twmmtwtmtwttw td d t td d m n m n m nnm m 1211 122 2 2 2 −−−−=      − ++ ( ) ( ) ( ) ( ) ( )twmtwttw td d t td d m n m nnm m +=      +1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )twnntwmtwttwmmtwtmtwt m n m n m n m n m n m n 122121 1122 ++−−−−−− +++ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) 011121 122 =+−+++−−= ++ twmmnntwtmtwt m n m n m n 2nd Way
  • 114. SOLO 114 Associated Legendre Functions ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) 011121 122 =+−+++−− ++ twmmnntwtmtwt m n m n m n Define: ( ) ( ) ( )twty m n=: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) 011121 122 =+−+++−− tymmnntytmtyt Now define: ( ) ( ) ( )tyttu m 22 1: −= Let compute: ( ) ( ) ( )1221 22 11 ytyttm td ud mm −+−−= − ( ) ( ) ( ) ( )11 22222 111 ytyttm td ud t mm + −+−−=− ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )21 221221221 2222222 1121111 ytyttmyttmyttmytm td ud t td d mmmmm +− −+−+−−−−+−−=      − ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]{ } ( ) ( ) y t tm mnnmmtymmnnytmytt mm       − +−+−+−++−+++−−−= 2 22 222 0 12222 1 11111211    We get: ( ) ( ) 0 1 11 2 2 2 =      − −++      − u t m nn td ud t td d 2nd Way
  • 115. SOLO 115 Associated Legendre Functions Define: ( ) ( ) ( )tw td d ttu nm mm 22 1: −= We get: ( ) ( ) 0 1 11 2 2 2 =      − −++      − u t m nn td ud t td d Start with Legendre Differential Equation: ( ) ( ) ( ) ( ) 1011 2 ≤=++      − ttwnntw td d t td d nn Summarize But this is the Differential Equation of f (θ) obtained by solving Laplace’s Equation by Separation of Variables in Spherical Coordinates . 02 =Φ∇ ( ) ( ) ( ) ( )φθφθ ΦΘ=Φ rRr ,, The Solutions of this Differential Equation are called Associated Legendre Functions, because they are derived from the Legendre Polynomials, and Legendre Functions of the Second Kind Qn (x) and are denoted: ( ) ( ) ( )tP td d ttP nm mm m n 22 1: −= 2nd Way ( ) ( ) ( )tQ td d ttQ nm mm m n 22 1: −=
  • 116. SOLO 116 Associated Legendre Functions Let use Rodrigues Formula for Pn (t): We see that we can define the Associated Legendre Function even for negative m (In the Differential equation m2 appears): ( ) ( ) ( )tP td d ttP nm mm m n 22 1: −= ( ) ( )[ ]n n n nn t td d n tP 1 !2 1 2 −= we obtain: Associated Legendre Functions ( ) ( ) ( )[ ]n mn mnm n m n t td d t n tP 11 !2 1 : 222 −−= + + From this equation we obtain: ( ) ( )tPtP nn = 0 2nd Way
  • 117. SOLO 117 Associated Legendre Functions Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are ( ) ( ) ( )[ ]n mn mnm n m n t td d t n tP 11 !2 1 : 222 −−= + + Let Differentiate this equation n+m times with respect to t, and use Leibnitz Rule of Product Differentiation: ( ) ( )[ ] ( ) ( ) ( ) im im i im i m m td tgd td tsd imi m tgts td d − − = ∑ − =⋅ 0 !! ! ( )[ ] ( ) ( )[ ]nn mn mn n mn mn tt td d t td d 1112 +⋅−=− + + + + Since we differentiate (t-1)n and (t+1)n, all the differentials greater then n will vanish, therefore we get ( ) ( )[ ] ( ) ( ) ( )[ ] ( )[ ] ( ) ( ) ( )[ ] ( )[ ] ( ) ( ) ( ) ( )[ ] ( )[ ]∑ − = + + − − −+ −+ + + +− +− + = ++− + + ++− + +=+⋅− mn i n im im n in in n n n n m nmn n m m n n mmn nn mn mn t td d t td d imin mn t td d t td d mn mn t td d t td d mn mn tt td d 0 11 !! ! 0011 !! ! 11 !! ! 0011   2nd Way
  • 118. SOLO 118 Associated Legendre Functions Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are ( ) ( ) ( )[ ]n mn mnm n m n t td d t n tP 11 !2 1 : 222 −−= + + ( ) ( )[ ] ( ) ( ) ( ) ( )[ ] ( )[ ] ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )∑ ∑ − = −− − = + + − − + + ++−−−−+− +− + = +− +− + =+⋅− mn i imni mn i n im im n in in nn mn mn timlnntinn imin mn t td d t td d imin mn tt td d 0 0 111111 !! ! 11 !! ! 11  ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )!11!11 1111 !! 1111 imnimnininiiimim imnnninn inim imnnninn −−+−−−−−+−++ +−−−⋅+− = −+ +−−−⋅+−   ( ) ( ) ( ) ( ) ( )!! 1111 imni innnimnn −− +−−⋅++− =  ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )∑ − = −−+ + + ++−−−++− −− +− = −−= mn i minmi n m n mn mnm n m n tinnntimnn imni mn n t td d t n tP 0 2/2/ 2/ 222 111111 !! ! !2 1 11 !2 1 :  2nd Way
  • 119. SOLO 119 Associated Legendre Functions Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )∑ − = −−+ + + ++−−−++− −− +− = −−= mn i minmi n m n mn mnm n m n tinnntimnn imni mn n t td d t n tP 0 2/2/ 2/ 222 111111 !! ! !2 1 11 !2 1 :  Using this formula we can write the Associated Legendre Functions for – m as ( ) ( ) ( )[ ]n mn mnm n m n t td d t n tP 11 !2 1 : 222 −−= − − −− Using Leibnitz Rule of Product Differentiation we obtain ( ) ( )[ ] ( ) ( ) ( )[ ] ( )[ ]∑ − = −− −− − − +− −− − =+− mn i n i i n imn imn nn mn mn t td d t td d imni mn tt td d 0 11 !! ! 11 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑ − = −+ ++−−⋅−++− −− − = mn i inim tinnntimnn imni mn 0 111111 !! !  ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑ − = −−+ − − − −− ++−−⋅−++− −− −− = −−= mn i minmi n m n mn mnm n m n tinnntimnn imni mn n t td d t n tP 0 2/2/ 2/ 222 111111 !! ! !2 1 11 !2 1 :  2nd Way
  • 120. SOLO 120 Associated Legendre Functions We obtained ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )∑ − = −−+ + + ++−−−++− −− +− = −−= mn i minmi n m n mn mnm n m n tinnntimnn imni mn n t td d t n tP 0 2/2/ 2/ 222 111111 !! ! !2 1 11 !2 1 :  ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑ − = −−+ − − − −− ++−−⋅−++− −− −− = −−= mn i minmi n m n mn mnm n m n tinnntimnn imni mn n t td d t n tP 0 2/2/ 2/ 222 111111 !! ! !2 1 11 !2 1 :  ( ) ( ) ( ) ( ) ( )tP mn mn tP m n mm n ! ! 1 + − −= − Therefore 2nd Way
  • 121. SOLO 121 Associated Legendre Functions Examples ( ) ( ) ( )tP td d ttP nm mm m n 22 1: −= ( ) ( ) 10 0 0 0 === tPtPn ( ) ( ) ( )  ( ) ( ) ( ) ( ) ( ) ( ) θ θ θ θ θ θ sin1 cos: sin111 cos 2 1 21 1 1 1 cos 1 0 1 cos 2 1 22 1 21 1 1 −=−−=−= === =−=−== = − = = t t t tP ttPtP ttPtP tt td d ttPn ( ) ( ) ( ) ( ) ( )tP mn mn tP m n mm n ! ! 1 + − −= − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) θ θθ θ θθ θ θ θ θ θ θ 2 cos 22 2 22 2 cos 2 1 211 2 2cos2 2 0 2 cos 2 1 2 2 2 1 21 2 2 cos 2 2 2 2 2 2 22 2 sin 8 1 1 8 1 !4 !0 1 2 cossin 13 !3 1 1 2 1cos3 2 13 cossin313 2 13 1 sin313 2 13 12 1 2 2 1 = − = − = = = =−=−= −=−−= − = − == =−=      − −= =−=      − −== t t tP t t tP t tP ttPtP tttP t tPtP tt t td d ttP t t td d ttPn    ( ) 10 =tP ( ) 2 1 2 3 2 2 −= ttP( ) ttP =1 2nd Way
  • 122. SOLO 122 Associated Legendre Functions Examples ( ) ( ) ( )tP td d ttP nm mm m n 22 1: −= ( ) ( ) ( ) ( ) ( )tP mn mn tP m n mm n ! ! 1 + − −= − 2nd Way Return to Table of Content
  • 123. SOLO 123 Associated Legendre Functions Generating Function for Pn m (t) Start with ( ) ( ) 1 21 1 :, 0 2 ≤= +− = ∑ ∞ = utPu utu tug n n n Let derivate this function relative to t ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) m m mm m m m m u m m utuu m m mutu t tug uutu t tug uutu t tug uutu t tug !12 !12 21 !12 2242 123121 , 2 2 5 2 3 2 1 21 , 2 2 3 2 1 21 , 2 2 1 21 , 1 2/12 1 1 2/12 32/72 3 3 22/52 2 2 2/32 − − +−= − −⋅ −⋅+−= ∂ ∂ −      −      −      −+−= ∂ ∂ −      −      −+−= ∂ ∂ −      −+−= ∂ ∂ − −− − −− − − −       ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑∑ ∞ = ∞ = +− =−= +− − − − = ∂ ∂ −= 00 22 2/12 2/2 1 2/2 1 21 1 !12 !12, 1:, n m n n n nm mm n m mm mm m m m tPutP td d tu utu ut m m t tug ttug ( ) ( ) ( )tP td d ttP nm mm m n 22 1: −=Use We obtain
  • 124. SOLO 124 Associated Legendre Functions ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∑ ∞ = +− = +− − − − = ∂ ∂ −= 0 2/12 2/2 1 2/2 21 1 !12 !12, 1, n m n n m mm mm m m m tPu utu ut m m t tug ttug The Generating Function for Pn m (t) is The Generating Function for Pn (t) is ( ) ( ) 1 21 1 :, 0 2 ≤= +− = ∑ ∞ = utPu utu tug n n n Return to Table of Content Generating Function for Pn m (t)
  • 125. SOLO 125 Associated Legendre Functions Recurrence Relations for Pn m (t) Start with Because of the existence of two indices in Pn m (t), we have a wide variety of recurrence relations. ( ) ( ) ( ) ( ) ( ) ( )∑ ∞ = +− = +− − − − = 0 2/12 2/2 1 21 1 1 !12 !12 , n m n n m m c m mm tPu utu ut m m tug m    or ( ) ( )∑ ∞ = − + = +− 0 2/12 21 1 n m n mn mm tPu utu c Differentiate with respect to u ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )∑∑ ∑ ∑ ∞ = − ∞ = ∞ = − + ∞ = −− + −+−=−+⇒ −+−= +− −+⇒ −= +− −+ ⇒ 0 12 0 0 12 2/12 0 1 2/32 2112 21 21 12 21 222/1 n m n n n m n m n m n n m m m n m n mn mm tPumnututPuutm tPumnutu utu uc utm tPumn utu utm c Arrange so all the terms, in the last relation, have the same power of u, by shifting indexes ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ }∑∑ ∞ −+ ∞ − −−+−−−+=+−+ n m n m n m n n n m n m n m tPmntPmnttPmnutPmttPmu 111 1211212 First Recurrence Relation
  • 126. SOLO 126 Associated Legendre Functions Equating the terms in um , and rearranging gives the First Recurrence Equation: ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ }∑∑ ∞ −+ ∞ − −−+−−−+=+−+ n m n m n m n n n m n m n m tPmntPmnttPmnutPmttPmu 111 1211212 First Recurrence Relation (continue) ( ) ( ) ( ) ( ) ( ) ( )tPmntPmnttPn m n m n m n 11 112 +− +−++=+ Second Recurrence Relation Start with ( ) ( ) ( ) ( ) ( ) ( )∑ ∞ −+ − − − == +− = n m mm m n n m m m m t m m ctPu utu uc tug 2/2 12/12 1 !12 !12 21 , or ( ) ( ) ( ) ( ) ( ) 2/2 1 2/12 1 !12 !12 21 m n mm m n nmm m t m m ctPuutuuc − − − =+−= ∑ ∞ − + Differentiate with respect to u ( ) ( ) ( ) ( ){ } ( )∑ ∞ −+−− +−++−−+= n m n nmnmm m tPunutuuututumumc 12/122/121 2121222/1 ( ) ( ) ( ) ( ){ } ( )∑ ∞ − − − +−+−+= +− n m n nn m m m tPunutuutum utu umc 12 2/12 1 21222/1 21 ud d Recurrence Relations for Pn m (t)
  • 127. SOLO 127 Associated Legendre Functions Second Recurrence Relation (continue - 1) or ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2/1212/2 2 2/122/2 1 1121 !22 !32 1 12 2212 1 !12 !12 − − −− −−=− − − − − −− =− − − = m m m m mm ctmt m m t m mm t m m c ( )( ) ( ) ( ) ( ) ( ){ } ( )∑ ∞ − − − − +−+−+= +− −− n m n nn m m m tPunutuutum utu uc mtm 12 2/12 1 12/12 21222/1 21 112 ( )( ) ( ) ( ) ( ) ( ){ } ( )      RHS n m n nn LHS n m n n tPunutuutumtPutmm ∑∑ ∞ −− +−+−+=−− 1212/12 21222/1112 Rearrange the RHS , by changing indexes, to have powers of un ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ }∑ ∑ ∑ ∞ +− ∞ −+− ∞ − ++++−+= −+−+++−+= +−+−+ n nm n m n m n n nm n m n m n m n m n n m n nn utPntPtnmtPnm utPntPtntPntPtmtPm tPunutuutum 11 111 12 11222 1211212 21222/1 Equating the terms in un of LHS and RHS we obtain: ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tPntPtnmtPnmtPtmm m n m n m n m n 11 12/12 11222112 +− − ++++−+=−− Recurrence Relations for Pn m (t)
  • 128. SOLO 128 Associated Legendre Functions Second Recurrence Relation (continue - 2) Use the First Recurrence Formula in RHS2: ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )     2 11 2 12/12 11222112 RHS m n m n m n LHS m n tPntPtnmtPnmtPtmm +− − ++++−+=−− ( ) ( ) ( ) ( ) ( ) ( )tPmntPmnttPn m n m n m n 11 112 +− +−++=+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]tPmntPmn n nmtPntPnmRHS m n m n m n m n 1111 1 12 1 122122 +−+− +−++ + ++−+++= ( ) ( )( ) ( )( )[ ] ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( )tPmmtPmm tPmnnmnntPmnnmnmnRHSn m n m n m n m n 11 11 1212 1122112122212212 +− +− −+−−= +−++−++++++−++=⋅+ Equating (2 n + 1) LHS2 with (2 n + 1) RHS2 we obtain the Second Recurrence Formula: ( ) ( ) ( ) ( )tPtPtPtn m n m n m n 11 12 112 −+ − −=−+ Recurrence Relations for Pn m (t)
  • 129. SOLO 129 Associated Legendre Functions ( ) ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }tPmnmntPmnmn n tPt tPtP n tPt tPmntPmntPtn tPmmnntP t tm tP m n m n m n m n m n m n m n m n m n m n m n m n 1 1 1 1 2 1 1 1 1 2 11 12 1 2111 12 1 14 12 1 13 1122 011 1 2 1 − + − − + − + + +− + + +−+−−−++ + =− − + =− +−++=+ =+−++ − − Return to Table of Content Recurrence Relations for Pn m (t)
  • 130. SOLO 130 Associated Legendre Functions Orthogonality of Associated Legendre Functions ( ) ( ) ( )[ ]n mn mnm n m n t td d t n tP 11 !2 1 : 222 −−= + + Let Compute ( ) ( ) ( ) ( )[ ] ( )[ ]∫∫ + − + + + + + + − −−−= 1 1 222 1 1 111 !!2 1 dtt td d t td d t qp dttPtP q mq mq p mp mp m qp m q m p Define X := x2 -1 ( ) ( ) ( ) [ ] [ ]∫∫ + − + + + + + + − − = 1 1 1 1 !!2 1 dtX td d X td d X qp dttPtP q mq mq p mp mp m qp m m q m p If p ≠ q, assume q > p and integrate by parts q + m times [ ] [ ] mqidtX td d vdX td d X td d u q imq imq p mp mp m i i +==      = −+ −+ + + ,,1,0  All the integrated parts will vanish at the boundaries t = ± 1 as long as there is a factor X = x2 -1. We have, after integrating m + q times ( ) ( ) ( ) ( ) [ ]∫∫ + − + + + + + ++ −      −− = 1 1 1 1 !!2 11 dtX td d X td d X qp dttPtP p mp mp m mq mq q qp mqm m q m p
  • 131. SOLO 131 Associated Legendre Functions ( ) ( ) ( ) ( ) [ ] 1: !!2 11 2 1 1 1 1 −=     −− = ∫∫ + − + + + + + ++ − xXdtX td d X td d X qp dttPtP p mp mp m mq mq q qp mqm m q m p Because the term Xm contains no power greater than x2m , we must have q + m – i ≤ 2 m or the derivative will vanish. Similarly, p + m + i ≤ 2 p Adding both inequalities yields q ≤ p which contradicts the assumption that q > p, therefore is no solution for i and the integral vanishes. Let expand the integrand on the right-side using Leibniz’s formula ( ) ( )∑ + = ++ ++ −+ −+ + + + + −+ + =      mq i p imp imp m imq imq qp mp mp m mq mq q X td d X td d imqi mq XX td d X td d X 0 !! ! ( ) ( ) qpdttPtP m q m p ≠=∫ + − 0 1 1 This proves that the Associated Legendre Functions are Orthogonal (for the same m). Orthogonality of Associated Legendre Functions
  • 132. SOLO 132 Associated Legendre Functions ( )[ ] ( ) ( ) 1: !!2 11 2 1 1 2 21 1 2 −=     −− = ∫∫ + − + + + ++ − xXdtX td d X td d X pp dttP p mp mp m mp mp p p mp m p Let expand the integrand on the right-side using Leibniz’s formula ( ) ( )∑ + = ++ ++ −+ −+ + + + + −+ + =      mp i p imp imp m imp imp pp mp mp m mp mp p X td d X td d impi mp XX td d X td d X 0 !! ! For the case p = q we have Because X = x2 – 1 the only non-zero term is for i = p - m ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1: !2!!2 !1 2 1 1 !2 2 2 !2 2 2 22 1 1 2 −=⋅ − +− = ∫∫ + − + − xXdtX td d X td d X mmpp mp dttP p p p p m m m m p p p m p  ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )! ! 12 2 sin1 !!2 !2!1 1 !!2 !2!1 !12 !2 1 0 12 22 cos1 1 2 22 212 mp mp p d mpp pmp dtx mpp pmp p p pp p ptp X p p p p − + ⋅ + = − − +− =− − +− = + − + =+ − + ∫∫     θθ πθ Orthogonality of Associated Legendre Functions
  • 133. SOLO 133 Associated Legendre Functions ( ) ( ) ( ) ( ) ( ) ( ) qp m q m p t m q m p mp mp p dPPdttPtP , 0 cos1 1 ! ! 12 2 sincoscos δθθθθ πθ − + + == ∫∫ =+ − Therefore the Orthonormal Associated Legendre Functions is ( ) ( ) ( ) ( ) mnmP mn mnn Θ m n m n ≤≤− + −+ = θθ cos ! ! 2 12 cos Orthogonality of Associated Legendre Functions
  • 134. SOLO 134 Associated Legendre Functions Examples ( ) ( ) 10 0 0 0 === tPtPn ( ) ( ) ( ) ( ) ( ) θ θ θ θ θ θ sin1 cos sin11 cos 2 1 21 1 cos 0 1 cos 2 1 21 1 −=−−= == =−== = − = = t t t ttP ttP ttPn ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) θ θθ θ θθ θ θ θ θ θ θ 2 cos 22 2 cos 2 1 21 2 2cos2 0 2 cos 2 1 21 2 2 cos 22 2 sin 8 1 1 8 1 2 cossin 1 2 1 2 1cos3 2 13 cossin313 sin3132 = − = − = = = =−= −=−−= − = − = =−= =−== t t t t t tP tttP t tP tttP ttPn 2nd Way ( ) ( ) ( ) ( ) mnmP mn mnn Θ m n m n ≤≤− + −+ = θθ cos ! ! 2 12 cos 2 cos 1 1 cos 0 1 2 cos 1 1 0 0 1 2 3 sin 2 3 2 3 cos 2 3 1 2 3 sin 2 3 1 2 1 0 t t tl l t t t −−=−=Θ ==Θ −==Θ= =Θ= = − = = θ θ θ θ θ θ ( ) ( ) ( ) ( )2 cos 22 2 2 cos 1 2 2 cos 20 2 2 cos 1 2 2 cos 22 2 1 4 15 sin 4 15 1 2 15 cossin 2 15 13 22 5 1cos3 22 5 1 2 15 cossin 2 15 1 4 15 sin 4 15 2 t tt t tx tl t t t t t −−=−=Θ −−=−=Θ −=−=Θ −==Θ −==Θ= = − = − = = = θ θ θ θ θ θ θθ θ θθ θ We recovered the previous results
  • 135. SOLO 135 Associated Legendre Functions Recurrence Relations for Θn m Functions 2nd Way Use the First Recurrence Formula for Pn m Functions: ( ) ( ) ( ) ( ) mnmtΘ mn mn n tP m n m n ≤≤− − + + = ! ! 12 2 ( ) ( ) ( ) ( ) ( ) ( )tPmntPmnttPn m n m n m n 11 112 +− +−++=+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tΘ mn mn n mntΘ mn mn n mntΘ mn mn n tn m n m n m n 11 !1 !1 32 2 1 !1 !1 12 2 ! ! 12 2 12 +− +− ++ + +−+ −− −+ − += − + + + We have to obtain Cancellation of common terms leads to ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tΘ nn mnmn tΘ nn mnmn tΘt m n m n m n 11 12123212 11 −+ −+ −+ + ++ +−++ = First Recurrence Relations for Θn m Functions
  • 136. SOLO 136 Associated Legendre Functions 2nd Way Use the Second Recurrence Formula for Pn m Functions: ( ) ( ) ( ) ( ) mnmtΘ mn mn n tP m n m n ≤≤− − + + = ! ! 12 2 We have to obtain Cancellation of common terms leads to Second Recurrence Relations for Θn m Functions ( ) ( ) ( ) ( )tPtPtPtn m n m n m n 11 12 112 −+ − −=−+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tΘ mn mn n tΘ mn mn n tΘ mn mn n tn m n m n m n 11 12 !1 !1 12 2 !1 !1 32 2 !1 !1 12 2 112 −+ − −− −+ − − +− ++ + = −− −+ + −+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )tΘ nn mnmn tΘ nn mnmn tΘt m n m n m n 11 12 1212 1 3212 1 1 −+ − +− +−− − ++ +++ =− Return to Table of Content Recurrence Relations for Θn m Functions
  • 137. SOLO 137 Spherical Harmonics The Spherical Harmonics that forms an orthogonal system, were first introduced by Pierre Simon de Laplace in 1782 Pierre-Simon, marquis de Laplace (1749-1827) Spherical Harmonics were first investigated in connection with the Newtonian potential of Newton's law of universal gravitation in three dimensions. In 1782, Pierre-Simon de Laplace had, in his “Mécanique Céleste”, determined that the gravitational potential at a point x associated to a set of point masses mi located at points xi was given by In 1867, William Thomson (Lord Kelvin) and Peter Guthrie Tait introduced the solid spherical harmonics in their “Treatise on Natural Philosophy”, and also first introduced the name of "spherical harmonics" for these functions. The solid harmonics were homogeneous solutions of Laplace's equation William Thomson, 1st Baron Kelvin (1824 –1907) Peter Guthrie Tait (1831 –1901)
  • 138. SOLO 138 Spherical Harmonics ( ) ( ) ( ) ( ) mnmP mn mnn Θ m n m n ≤≤− + −+ = θθ cos ! ! 2 12 cos 0 sin 1 sin sin 11 2 2 222 2 2 2 = ∂ Φ∂ +      ∂ Φ∂ ∂ ∂ +      ∂ Φ∂ ∂ ∂ =Φ∇ φθθ θ θθ rrr r rr By Separation of Variables, we assume a solution of the form ( ) ( ) ( ) ( )φθφθ ΦΘ=Φ rRr ,, Laplace Equation in Spherical Coordinates is: f (θ) and g (φ) are described by the Differential Equations: ( ) Φ−= Φ =Θ      −++      Θ ∂ ∂ 2 2 2 2 2 0 sin 1sin sin 1 m d d m nn d d φ θθ θ θθ The Orthonormal Solutions, as functions of the parameter m are: ( ) ϕ π ϕ mjm eG 2 1 = The functions Θn m are Orthonormal with respect to the angle θ and the functions Gm are Orthonormal with respect to angle φ.
  • 139. SOLO 139 Spherical Harmonics ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) mnmeP mn mnn GΘY mjm n mmm n mm n ≤≤− + −+ −=−= ϕ θ π ϕθϕθ cos ! ! 4 12 1cos1:, For r = constant (Spherical Surfaces) we define the function: We can see that: The Functions Yn m are Orthonormal with respect to both angles θ and φ. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2121 21 2,1 212 2 1 1 21 2 2 1 1 , 02 22 1 11 2 0022 222 11 111 2 0 0 sincoscos ! ! 2 12 ! ! 2 12 2 1 sincoscos1 ! ! 2 12 ! ! 2 12 sin,, nn m n m n mmm mmjm n m n mm m n m n dPP mn mnn mn mnn dedPP mn mnn mn mnn ddYY mm δθθθθ ϕ π θθθθ ϕθθϕθϕθ π θ δ π ϕ ϕ π θ π ϕ π θ = + −+ + −+ = − + −+ + −+ = ∫ ∫∫ ∫ ∫ = == = − = + = =    We can write ( ) ( ) 2121 2 2 1 1 ,, 4 ,, mmnn m n m n dYY δδϕθϕθ π =Ω∫ Where d Ω = sin θ dθ dφ is the element of solid angle.
  • 140. SOLO 140 Spherical Harmonics One of the most important properties of Spherical Harmonics lies in the Completeness Property, a consequence of the Sturm-Liouville form of the Laplace Equation. This property means that any function f (θ,φ) (with sufficient continuity properties) can be extended in a uniformly Convergent Double Series of Spherical Harmonics This expansion holds in the sense of mean-square convergence, which is to say that ( ) ( )∑ ∑ ∞ = −= = 0 ,, n n nm m n m n Yff ϕθϕθ ( ) ( ) 0sin,, 2 0 0 2 0 =−∫ ∫ ∑ ∑ = = ∞ = −= π ϕ π θ ϕθθϕθϕθ ddYff n n nm m n m n The expansion coefficients are the analogs of Fourier coefficients, and can be obtained by multiplying the above equation by the complex conjugate of a spherical harmonic, integrating over the solid angle Ω, and utilizing the above orthogonality relationships. This is justified rigorously by basic Hilbert space theory. For the case of orthonormalized harmonics, this gives: ( ) ( ) ( ) ( )∫ ∫∫ = =Ω =Ω= π ϕ π θ θθϕθϕθϕϕθϕθ 2 0 0 sin,,,, dYfddYff m n m n m n
  • 141. SOLO 141 Spherical Harmonics To prove the fn m relation let compute ( ) ( )∑ ∑ ∞ = −= = 0 ,, n n nm m n m n Yff ϕθϕθ ( ) ( ) ( ) ( ) ( ) ( ) ∑ ∑ ∑ ∑ ∫∫ ∑ ∑∫ ∞ = −= ∞ = −= ΩΩ ∞ = −=Ω == Ω=Ω=Ω 0 ,, 00 1 1 1 1 11 ,,,,,, l m n l lm mmln m l l l lm m l m n m l l l lm m l m l m n m n ff dYYfdYfYdYf δδ ϕθϕθϕθϕθϕθϕθ Let choose f (θ,φ) = Pn m (θ,φ) ( ) ( )∫ ∫= = = π ϕ π θ ϕθθϕθθ 2 0 0 sin, ddYPf k l m n k l ( ) ( ) ( ) ( ) ( ) klkeP kl kln Y kjk l kk l ≤≤− + −+ −= ϕ θ π ϕθ cos ! ! 4 12 1:, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∫∫ = − + + + = + −+ −= π ϕ ϕ δ π θ ϕθθθθ π 2 0 ! ! 12 2 0 , sin ! ! 4 12 1 dedPP ml mln f mj mn mn n m l m n mm l ln    ( ) ( ) ( ) ( ) qp m q m p mp mp p dPP , 0 ! ! 12 2 sincoscos δθθθθ π − + + =∫
  • 142. SOLO 142 Spherical Harmonics Proof ( ) ( ) ( )ϕθϕθ ,1, m n mm n YY − −= ( ) ( ) ( ) ( ) ( ) ϕ θ π ϕθ mjm n mm n eP mn mnn Y − + −+ −= cos ! ! 4 12 1, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) mnmeP mn mnn GΘY mjm n mmm n mm n ≤≤− + −+ −=−= ϕ θ π ϕθϕθ cos ! ! 4 12 1cos1:, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )ϕθθ π θ π ϕ ϕ ,1cos ! ! 4 12 11 cos ! ! 1 ! ! 4 12 1 m n mmjm n mm mjm n mm YeP mn mnn eP mn mn mn mnn −−−− −−− −= − ++ −−= − + − + −+ −= q.e.d.
  • 143. SOLO 143 Spherical Harmonics Visualization of Spherical Harmonics The Laplace spherical harmonics Yl m can be visualized by considering their “nodal lines", that is, the set of points on the sphere where Re[Yl m ], or alternatively where Im[Yl m ]. Nodal lines of are composed of circles: some are latitudes and others are longitudes. One can determine the number of nodal lines of each type by counting the number of zeros of Yl m in the latitudinal and longitudinal directions independently. For the latitudinal direction, the real and imaginary components of the associated Legendre polynomials each possess ℓ−|m| zeros, whereas for the longitudinal direction, the trigonometric sin and cos functions possess 2|m| zeros. Schematic representation of Yl m on the unit sphere and its nodal lines. Re[Yl m ] is equal to 0 along m great circles passing through the poles, and along n-m circles of equal latitude. The function changes sign each time it crosses one of these lines. When the spherical harmonic order m is zero (upper-left in the figure), the spherical harmonic functions do not depend upon longitude, and are referred to as zonal. Such spherical harmonics are a special case of zonal spherical functions. When ℓ = |m| (bottom-right in the figure), there are no zero crossings in latitude, and the functions are referred to as sectoral. For the other cases, the functions checker the sphere, and they are referred to as tesseral.
  • 144. SOLO 144 Spherical Harmonics ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ϕ ϕ ϕ ϕ ϕ ϕ θ π ϕθ θθ π ϕθ θ π ϕθ θθ π ϕθ θ π ϕθ θ π ϕθ θ π ϕθ θ π ϕθ π ϕθ j j j j j j eY eY Y eY eY eY Y eY Y 222 2 1 2 20 2 1 2 222 2 1 1 0 1 1 1 0 0 sin 2 15 4 1 , cossin 2 15 2 1 , 1cos3 5 4 1 , cossin 2 15 2 1 , sin 2 15 4 1 , sin 2 3 2 1 , cos 3 2 1 , sin 2 3 2 1 , 1 2 1 , = − = −= = = − = = = = −− −− −− Return to Table of Content
  • 145. 145 SOLO We obtain: This is a Sturm-Liouville Differential Equation: ( )  ( ) ( )  22 0sin sin sin 2 π θ π θλ θθ θ θ θ θ θ ≤≤−=+−        ∂ ∂ ff m d fd r q p  Laplace’s Homogeneous Differential Equation Analysis of f (θ) Equation ( ) ( ) ( ) bxaconstyxryxq xd yd xp xd d <<==++      λλ 0 Laplace Differential Equation in Spherical Coordinates
  • 146. Ordinary Differential Equations Boundary Value Problems and Sturm–Liouville Theory Jacques Charles François Sturm (1855–1803) Joseph Liouville (1809–1882), ( ) ( ) ( ) bxaconstyxryxq xd yd xp xd d <<==++      λλ 0 Many problems of mathematical physics lead to differential equations of the form ( ) ( ) ( ) ( ) ( ) ( )    >+=+ >+=+ 00' 00' 2 2 2 121 2 2 2 121 bbbybbyb aaayaaya with the Boundary Conditions B.C. [y] = 0: When p (x), q (x), r (x) are continuous on [a,b] and p (x)>0 and r (x)>0 on [a,b], we refer to this system as a Regular Sturm-Liouville Boundary Value Problem. The Sturm-Liouville Problem is a two-point boundary second-order linear differential equation problem, in which we want to define the parameter λ such that the equation has non trivial solutions y(t)≠0. Such problems are called Eigenvalues Problems and the corresponding number λ an Eigenvalue. SOLO 146 Return to Rodrigues’ Formula
  • 147. Ordinary Differential Equations Jacques Charles François Sturm (1855–1803) Joseph Liouville (1809–1882), ( ) ( ) ( ) bxaconstyxryxq xd yd xp xd d <<==++      λλ 0 Many problems of mathematical physics lead to differential equations of the form with the Boundary Conditions B.C. [y] = 0: When p (x), q (x), r (x) are continuous on (a,b) and p (x)>0 and r (x)>0 on (a,b). we refer to this system as a Singular Sturm-Liouville Boundary Value Problem if one of the following conditions occurs: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) unboundedisbac xrorxqorxporxrorxqorxpb xporxpa bx bx ax ax bx bx ax ax , limlim 0lim0lim ∞→∞→ == < → > → < → > → SOLO ( ) ( ) ( ) ( ) ( ) ( )    >+=+ >+=+ 00' 00' 2 2 2 121 2 2 2 121 bbbybbyb aaayaaya 147 Boundary Value Problems and Sturm–Liouville Theory
  • 148. Ordinary Differential Equations Define the Operator: [ ]( ) ( ) ( ) yxq xd yd xp xd d xyL +      =: The Sturm–Liouville Equation can be rewritten as: [ ]( ) ( ) ( ) 0=+ xyxrxyL λ Lagrange's Identity (Boundary Value Problem) Proof: Joseph-Louis Lagrange (1736 –1813), [ ] [ ] ( ) [ ]( )vuWxp xd d uLvvLu ,=− where is the Wronskian of u and v.[ ] xd ud v xd vd uvuW −=:, Theorem 1: Let u and v be functions having continuous second derivatives on the interval [a,b]. Then [ ] [ ] ( ) ( ) ( ) ( ) uvxq xd ud xp xd d vvuxq xd vd xp xd d uuLvvLu −      −+      =− ( ) ( ) ( ) ( )       −      −      +      = xd ud xp xd d v xd ud xp xd vd xd vd xp xd ud xd vd xp xd d u ( ) ( ) ( )             −=      −      = xd ud v xd vd uxp xd d xd ud xpv xd d xd vd xpu xd d q.e.d. SOLO 148 Boundary Value Problems and Sturm–Liouville Theory
  • 149. Ordinary Differential Equations Proof: [ ] [ ]( )( ) ( ) [ ]( ) ( ) [ ]( ) ( ) ( )( ) ( ) ( ) ( ) ( )[ ] ( ) ( )( ) ( ) ( ) ( ) ( )[ ] 0//// ,, 21212121 =−−−−−−−= −=−∫ avauaaavaaauapbvbubbbvbbbubp avuWapbvuWbpxdxuLvvLu b a q.e.d. Green’s Formula (Boundary Value Problem) When a2≠0 and b2≠0: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )bvbbbvavaaav bubbbuauaaau 2121 2121 /',/' /',/' −=−= −=−= SOLO 149 Boundary Value Problems and Sturm–Liouville Theory Corollary 1: Let u and v be functions having continuous second derivatives on the interval [a,b]. Then George Green 1793-1841 tomb stone If u and v satisfies the Boundary Conditions (y=u,v): then: [ ] [ ]( )( ) ( ) [ ]( ) b a b a vuWxpxdxuLvvLu ,=−∫ [ ] [ ]( )( ) 0=−∫ b a xdxuLvvLu ( ) ( ) ( ) ( )   ===+ ===+ 00' 00' :.. 2121 2121 bbnotaybayb aanotayaaya CB
  • 150. Ordinary Differential Equations Is the Inner Product on the set of Continuous Real-valued functions on [a,b]. Green’s Formula (Boundary Value Problem) ( ) ( )∫= b a xdxgxfgf :, Takes a particularly useful and elegant form when expressed in form of a Inner Product: Equation [ ] [ ]( )( ) 0=−∫ b a xdxuLvvLu Using this notation we can rewrite the Green’s Formula: [ ] [ ] vuLvLu ,, = A Linear Differential Operator that satisfies this relation for all u and v in its domain is called a Selfadjoint Operator. We have shown that the Sturm-Luville Operator with the domain of functions that have continuous second derivatives on [a,b] ans satisfy the Boundary Conditions, then L is a Selfadjoint Operator. Selfadjoint Operators are like symmetric matrices in that that their Eigenvalues are Real-valued. We will prove this for Regular Sturm-Liouville Boundary Value Problems. [ ]( ) ( ) ( ) yxq xd yd xp xd d xyL +      =: SOLO 150 Boundary Value Problems and Sturm–Liouville Theory
  • 151. Ordinary Differential Equations Proof: Theorem 2: The eigenvalues λ for the Regular Sturm-Liouville Boundary Value Problem are Real and have Real-valued Eigenvalues [ ]( ) ( ) ( ) 0=+ xyxrxyL λ Assume λ, possible a complex number, is an Eigenvalue for the Sturm-Liouville Problem with eigenfunction φ (x) [ ]( ) ( ) ( ) 0=+ xxrxL ϕλϕ and φ (x) satisfies the Boundary Conditions: ( ) ( ) ( ) ( ) 00' 00' 2121 2121 ===+ ===+ bbnotbbbb aanotaaaa ϕϕ ϕϕ Take the Complex Conjugate and use the fact that p, q and r are Real-valued: [ ]( ) ( ) ( ) [ ]( ) ( ) ( ) 0=+=+ xxrxLxxrxL ϕλϕϕλϕ Since a1, a2, b1, b2 are Real-valued, also satisfies the Boundary Conditions, hence is an Eigenvalue with the Eigenfunction . ϕ ϕλ [ ]( ) ( ) ( ) ( ) ( )∫∫ −= b a b a xdxxxrxdxxL ϕϕλϕϕ SOLO 151 Boundary Value Problems and Sturm–Liouville Theory
  • 152. Ordinary Differential Equations Proof (continue): q.e.d. [ ]( ) ( ) ( ) ( ) ( )∫∫ −= b a b a xdxxxrxdxxL ϕϕλϕϕ Using Green’s Formula for Boundary Value: [ ] [ ]( )( ) 0=−∫ b a xdxuLvvLu [ ]( ) ( ) ( ) [ ]( ) ( ) ( ) ( )∫∫∫ −== b a b a b a xdxxxrxdxLxxdxxL ϕϕλϕϕϕϕ Therefore: ( ) ( ) ( ) ( ) ( ) ( )∫∫ −=− b a b a xdxxxrxdxxxr ϕϕλϕϕλ ( ) ( ) ( ) ( ) λλϕϕλλ =⇒=− > > ∫ 0 0 0    b a xdxxxr If φ (x) is not real-valued, since λ is real, we can see that Real {φ (x)} and Im {φ (x)} are also Eigenfunctions, and we can use Real {φ (x)} as the Real- valued Eigenfunction. λ is Real. SOLO 152 Boundary Value Problems and Sturm–Liouville Theory Theorem 2: The eigenvalues λ for the Regular Sturm-Liouville Boundary Value Problem are Real and have Real-valued Eigenvalues [ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
  • 153. Ordinary Differential Equations Proof : q.e.d. Theorem 3: All the eigenvalues λ of the Regular Sturm-Liouville Boundary Value Problem are simple (for each Eigenvalue λ there is only one Linearly Independent Eigenfunction) Assume φ (x) and ψ (x) are two Linearly Independent Eigenfunctions corresponding to the same Eigenvalue λ (thus φ (x) and ψ (x) ) satisfy the differential equation for the same λ and the Boundary Condition): Assume a2 ≠ 0 (the same argument if a1≠ 0): If the Wronskian of two solutions of a second-order linear homogeneous equation is zero at a point in the interval [a,b] then the solutions are linearly dependent, and this is a contradiction. ( ) ( ) ( ) ( ) 00'&0' 212121 ===+=+ aanotaaaaaaaa ψψϕϕ ( ) ( ) ( ) ( ) ( ) ( )aaaaaaaa ψψϕϕ 2121 /'&/' −=−= Compute the Wronskian W [φ,ψ] at x = a [ ] ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 0//'', 2121 =−−−=−= aaaaaaaaaaaaW ψϕψϕψϕψϕψϕ [ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ SOLO 153 Boundary Value Problems and Sturm–Liouville Theory
  • 154. Ordinary Differential Equations Proof : q.e.d. Theorem 4: Eigenfunctions of the Regular Sturm-Liouville Boundary Value Problem are Orthogonal with respect to the weight function r (x) on [a,b]. Let λ and μ be distinct (λ≠μ) and Real Eigenvalues with corresponding Real- valued Eigenfunctions φ (x) and ψ (x) respectively Use Green’s Formula: [ ]( ) ( ) ( ) [ ]( ) ( ) ( ) 0&0 =+=+ xxrxLxxrxL ψµψϕλϕ ( ) [ ]( ) ( ) [ ]( ){ } ( ) ( ) ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( )∫ ∫∫ −= +−=−= b a b a b a dxxxxr dxxxrxxxrxdxxLxxLx ψϕµλ ϕλψψµϕϕψψϕ0 ( ) ( ) ( ) 0=∫ b a dxxxxr ψϕ [ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ SOLO 154 Boundary Value Problems and Sturm–Liouville Theory
  • 155. Ordinary Differential Equations Uniform Convergence of Eigenfunction Extension SOLO 155 Boundary Value Problems and Sturm–Liouville Theory Theorem 5: The Eigenvalues of the Regular Sturm-Liouville Boundary Value Problem form a countable, increasing sequence λ1<λ2<λ3<… with lim n→∞ λn=+∞ [ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ
  • 156. Ordinary Differential Equations Uniform Convergence of Eigenfunction Expension Theorem 6: Let be an Orthonormal Sequence of Eigenfunctions for the Regular Sturm-Liouville Boundary Value Problem Let f (x) be a continuous function on [a,b] with f’(x) piecewise continuous in [a,b]. If f satisfies the Boundary Conditions then { }∞ =1nnφ ( )nkkn δφφ =, [ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ ( ) ( ) bxaxcxf n nn <<= ∑ ∞ = , 1 φ where ( ) ( ) ( )∫= b a nn dxxfxxrc φ The Eigenfunction expension converges uniformly in [a,b]. Since this is true for all continuous f (x) and piecewise continuous f’(x) this Functional Space is Completely Covered by the Eigenfunctions. Eigenfunction Expension The series of functions is uniformly convergent to f (x) if for all ε>0 and for all x∈(a,b) we can find a number N (ε) such that ( ) ( ) ( )εε Nnxfxfn >∀<− ( ) ( )∑= = n i iin xcxf 1 φ SOLO 156 Boundary Value Problems and Sturm–Liouville Theory
  • 157. Ordinary Differential Equations Pointwise Convergence of Eigenfunction Expension SOLO 157 Boundary Value Problems and Sturm–Liouville Theory Theorem 7: Let be an Orthonormal Sequence of Eigenfunctions for the Regular Sturm-Liouville Boundary Value Problem Let f (x) and f’(x) piecewise continuous in [a,b]. There for any x in (a,b) where [ ]( ) ( ) ( ) ..&0 CBxyxrxyL =+ λ { }∞ =1nnφ ( )nkkn δφφ =, ( ) ( ) ( )∫= b a nn dxxfxxrc φ ( ) ( ) ( ) ( )n xx xx n xx xx xfxfxfxf n n n n > → + < → − == lim&lim ( ) ( ) ( )[ ] bxaxfxfxc n nn <<+= +− ∞ = ∑ , 2 1 1 φ
  • 158. Ordinary Differential EquationsSOLO Equation Regular singularity x = Iregular singularity x = 1. Hypergeometric x (x-1) y”+[(1+a+b) x –c] + aby = 0 0, 1, ∞ - 2. Legendre (1-x2 ) y” – 2 x y’ + l (l+1) = 0 -1, 1, ∞ - 3. Chebyshev (1-x2 ) y”–x y’ + n2 y= 0 -1, 1, ∞ - 4. Confluent Hypergeometric x y” +(c –x) y’ –a y = 0 0 ∞ 5. Bessel x2 y”+x y’ +(x2 - n2 )y= 0 0 ∞ 6. Laguerre x y” +(1 –x) y’ +a y = 0 0 ∞ 7. Simple Harmonic Oscillator y” + ω2 y = 0 - ∞ 8. Hermite y”-2 x y’ 2 α y= 0 - ∞ 158 Boundary Value Problems and Sturm–Liouville Theory
  • 159. Ordinary Differential EquationsSOLO Equation p (x) q (x) λ r (x) Legendre 1-x2 0 l (l+1) 1 Shifted Legendre x (1-x) 0 l (l+1) 1 Associated Legendre 1-x2 -m2 /(1-x2 ) l (l+1) 1 Chebyshev I (1-x2 )1/2 0 n2 (1-x2 )-1/2 Shifted Chebyshev I [x (1-x)]1/2 0 n2 [x (1-x)]-1/2 Chebyshev II (1-x2 )3/2 0 n (n+2) (1-x2 )1/2 Ultraspherical (Gegenbauer) (1-x2 )α+1/2 0 n (n+2 α) (1-x2 )α-1/2 Bessel x -n2 /x a2 x Laguerre x e-x 0 α e-x Associated Laguerre xk+1 e-x 0 α-k xk e-x Hermite e-x2 0 2 α e-x2 Simple Harmonic Oscillator 1 0 n2 1 Self-Adjoint ODE 159 Boundary Value Problems and Sturm–Liouville Theory
  • 160. Ordinary Differential EquationsSOLO Equation Interval r (x) Standard Normalization Legendre -1 ≤ x ≤1 1 Shifted Legendre 0 ≤ x ≤1 1 Chebyshev I -1 ≤ x ≤1 (1-x2 )-1/2 Shifted Chebyshev I 0 ≤ x ≤1 [x (1-x)]-1/2 Chebyshev II -1 ≤ x ≤1 (1-x2 )1/2 Laguerre 0 ≤ x < ∞ e-x Associated Laguerre 0 ≤ x < ∞ xk e-x Hermite -∞ ≤ x < ∞ e-x2 Orthogonal Polynomials Generated by Gram-Schmidt Orthogonalization ( )[ ] 12 2 1 1 2 + =∫ + − n dxxPn ( )[ ] 12 1 * 1 0 2 + =∫ + n dxxPn ( )[ ] ( )    = ≠ = − ∫ + − 0 02/ 1 1 1 2/12 2 n n dx x xTn π π ( )[ ] ( )    = ≠ = − ∫ + 0 02/ 1 * 1 0 2/12 2 n n dx x xTn π π ( )[ ] ( ) 2 1 1 1 2/122 π =−∫ + − dxxxUn ( )[ ] 1 0 2 =∫ ∞ − dxexL x n ( )[ ] ( ) ! ! 0 2 n kn dxexxL xkk n + =∫ ∞ − ( )[ ] !2 2/1 0 2 2 ndxexH nx n π=∫ ∞ − 160 Boundary Value Problems and Sturm–Liouville Theory
  • 161. Ordinary Differential EquationsSOLO Equation a b r (x) Legendre -1 1 1 Shifted Legendre 0 1 1 Associated Legendre -1 1 1 Chebyshev I -1 1 (1-x2 )-1/2 Shifted Chebyshev I 0 1 [x (1-x)]-1/2 Chebyshev II -1 1 (1-x2 )1/2 Laguerre 0 ∞ e-x Associated Laguerre 0 ∞ xk e-x Hermite -∞ ∞ e-x2 Simple Harmonic Oscillator 0 -π 2π π 1 1 The Weighting Function r (x) is established by putting the ODE in Self-Adjoint form. Return to Expansion of Functions, Legendre Series Return to Table of Content Return to Spherical Harmonics 161 Boundary Value Problems and Sturm–Liouville Theory
  • 162. References SOLO 162 Legendre Functions http://en.wikipedia.org/wiki/ Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations” Byron Jr., F.W., Fuller, R.W., “Mathematics of Classical and Quantum Physics”, Dover, 1969, 1970 G.M. Wysin, “Associated Legendre Functions and Dipole Transition Matrix Elements”, http://www.phys.ksu.edu/personal/wysin Wallace, P.R., “Mathematical Analysis of Physical Problems”, Dover 1972, 1984 Afken, Weber, “Mathematical Methods for Physicists”, Harcourt Academic Press, 5th Ed., 2001 W. W. Bell, “Special Functions for Scientists and Engineers”, D. van Nostrand Company ltd., 1968 S.I. Hayek, “Advanced Mathematical Methods in Science and Engineering”, Marcel Dekker Inc., 2001 L.C. Andrews, “Special Functions for Engineers and Applied Mathematics”, MacMillan Publishing Companie, 1985 Return to Table of Content
  • 163. 163 SOLO Technion Israeli Institute of Technology 1964 – 1968 BSc EE 1968 – 1971 MSc EE Israeli Air Force 1970 – 1974 RAFAEL Israeli Armament Development Authority 1974 – 2013 Stanford University 1983 – 1986 PhD AA

Editor's Notes

  • #6: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #7: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #8: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #9: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #10: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #11: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #12: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”
  • #22: http://en.wikipedia.org/wiki/Legendre_polynomials
  • #24: W. W. Bell, “Special Functions for Scientists and Engineers”, D. van Nostrand Company ltd., 1968
  • #25: http://en.wikipedia.org/wiki/Rodrigues%27_formula http://books.google.co.il/books?id=oTyJYUx8Jr4C&amp;pg=PA105&amp;redir_esc=y#v=onepage&amp;q&amp;f=false http://www-history.mcs.st-andrews.ac.uk/Biographies/Rodrigues.html
  • #26: http://en.wikipedia.org/wiki/Rodrigues%27_formula
  • #27: http://en.wikipedia.org/wiki/Rodrigues%27_formula
  • #28: http://en.wikipedia.org/wiki/Rodrigues%27_formula
  • #29: http://en.wikipedia.org/wiki/Rodrigues%27_formula
  • #30: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #31: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #32: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #33: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #34: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #35: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #36: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
  • #37: Arfken, Weber, “Mathematical Methods for the Physicists”, 5 Ed., Harcourt – Academic Press, 2001 K.T. Tang, “Mathematical Methods for Engineers and Scientists”, Vol 3, Springer-Verlag, 2007 http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius
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