Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

260

5.8.3 Complete elliptic integrals of the first and second kind

π / 2 Z 0

F

;1; k 2 ,

d ϕ q 1 − k 2 sin 2 ϕ

π 2

1 2

1 2

K = K ( k )= F ( π / 2 , k )=

=

,

π / 2 Z 0

d ϕ q 1 − sin 2 α sin 2 ϕ ,

K ( α )=

(5.201)

π / 2 Z 0 q

F −

;1; k 2 ,

π 2

1 2

1 2

1 − k 2 sin 2 ϕ d ϕ ==

E ( k )= E ( π / 2 , k )=

,

π / 2 Z 0 q

1 − sin 2 α sin 2 ϕ d ϕ

E ( α )=

where F ( ± 1 2 ,

1 2 ;1; k

2 ) is the Gaussian hypergeometric function .

Complementary integrals By introducing the substitution k ′ = √ 1

− k 2 , where k ′ is the complementary modulus , and

k the modulus of the elliptic integral, we obtain the following values

π / 2 Z 0

d ϕ q 1 − ( 1 − k 2 ) sin 2 ϕ

K ′ ( k )= K ( k ′ )=

= F ( π / 2 , k ′ ) ,

(5.202)

π / 2 Z 0 q

E ′ ( k )= E ( k ′ )=

1 − ( 1 − k 2 ) sin 2 ϕ d ϕ = E ( π / 2 , k ′ )

It should be emphasized that in this case () ′ does not denote a derivative!!! Legendre relation K · E ′ + EK ′ − KK ′ = π 2

5.8.4 Jacobi elliptic functions

Jacobi elliptic functions are inverse elliptic integrals. If u ( ϕ , k ) (elliptic integral of the first kind) is given by

ϕ Z 0

sin ϕ Z 0

d ϕ q 1 − k 2 sin 2 ϕ

d t ( 10 t 2 ) √ 1

u =

(5.203)

=

− k 2 t 2

for k 2 < 1, then the inverse function is

ϕ = am u (amplitude of u ) .

(5.204)

sn u = sn ( u , k )= sin φ = sin ( am u ) , cn u = cn ( u , k )= cos φ = cos ( am u )= p 1 − sn 2 u , dn u = dn ( u , k )= q 1 − k 2 sin 2 φ = q 1 − k 2 sn 2 ( u )

(5.205)