Page 366 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 22: Solution of Laplace Differential Equation

Notes
2
1 d d
sin ( n n 1) 2 = 0
sin d d sin
1 d 2 d 2
= sin ( n n 1) 2 0
sin d dµ 1 µ

2sin cos d d d 2
= sin ( n n 1) 2 0
sin dµ d dµ 1 µ

2 d 2 d 2
= sin 2 2µ ( n n 1) 2 0
dµ dµ (1 µ )

2 d 2 d 2
or (1 µ ) 2 2µ ( n n 1) 2 0 ...(9)
dµ dµ (1 µ )

It is clear that will be a function of µ i.e.
z
( ) or (cos )
Hence the solution of Laplace equation is

V = (A r n B r (n 1) ) (cos ) A e i B e i ...(10)

where the solution of (µ) is

= A e a B e i ...(11)
For 2 m 2 , integer m, the solution is satisfied by associated Legendre polynomial P n m ( ) as
x
shown below:
Consider the Legendre equation

2
2 d y dy
(1 x ) 2 2x ( n n 1)y = 0 ...(12)
dx dx
Differentiating it m times and putting

m
d y
= m ...(13)
dx
We have

m
2
d m (1 x 2 d y 2 d m x dy ( n n 1) d y = 0
)
dx m dx 2 dx m dx dx m
or
m
m
2 d m 2 y d m 1 dmy d m 1 y d y d y
(1 x ) m 2 2.mx m y m (m 1)x m 2x m 1 2 m ( ) n (n 1) m 0
m
dx dx dx dx dx dx

LOVELY PROFESSIONAL UNIVERSITY 359