Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

258

These types of integrals can be reduced to three basic elliptic integrals Z d z p ( 1 − z 2 )( 1 − k 2 z 2 ) , Z z 2 d z p ( 1 − z 2 )( 1 − k 2 z 2 ) , Z d z (5.192) where k and h are constants, and h can be an imaginary value, which Legendre named elliptic integrals of the first, second and third kind. The constant k is called the elliptic modulus , and takes its values in the interval 0 < k < 1. The substitution z = sin ϕ yields the so called Legendre elliptic integrals of the first, second and third kind F ( ϕ , k )= Z d ϕ q 1 − k 2 sin 2 ϕ , E ( ϕ , k )= Z q 1 − k 2 sin 2 ϕ d ϕ , Π ( n , ϕ , k )= Z d ϕ ( 1 + h sin 2 ϕ ) q 1 − k 2 sin 2 ϕ . (5.193) These integrals are called normal (incomplete), if they are functions of the upper boundary ϕ , or complete , if the upper boundary ϕ = π / 2. ( 1 + hz 2 ) p ( 1 − z 2 )( 1 − k 2 z 2 ) ,

ϕ Z 0

x Z 0

d ϕ q 1 − k 2 sin 2 ϕ

d t p ( 1 − t 2 )( 1 − k 2 t 2 ) ,

x = sin ϕ , k 2 < 1 ,

F ( ϕ , k )=

=

√ 1

ϕ Z 0 q

x Z 0

− k 2 t 2

d t , x = sin ϕ , k 2 < 1 ,

1 − k 2 sin 2 ϕ d ϕ , =

E ( ϕ , k )=

(5.194)

√ 1

− t 2

ϕ Z 0

x Z 0

d ϕ ( 1 + h sin 2 ϕ ) q 1 − k 2 sin 2 ϕ

1 1 + ht 2

d t p ( 1 − t 2 )( 1 − k 2 t 2 ) .

Π ( n , ϕ , k )=

=

For the first two integrals Legendre even produced tables with two arguments: independent variable ϕ and parameter k , that is, ϕ and θ , if the substitution k = sin θ is introduced. The angles are expressed in degrees. Solutions of many practical problems lead to these integrals. Besides the already mentioned determining of the length of the arc of an ellipse, let us also mention the task of movement of mathematical and spherical pendulum (see Examples 214 and 215, p. 297 and 300).

5.8.1 Some properties of the integral F ( ϕ , k )

1. F ( 0 , k )= 0, which is a direct consequence of the definition ( R 0 0 = 0) 2. F ( ϕ + m π , k )= F ( ϕ , k )+ mF ( π , k ) .

(5.195)

Proof (for m = 1)

ϕ + π Z 0

ϕ + π Z π

π Z 0

d ϕ q 1 − k 2 sin 2 ϕ

d ϕ q 1 − k 2 sin 2 ϕ

d ϕ q 1 − k 2 sin 2 ϕ

F ( ϕ + π , k )=

=

+

=

π Z 0

θ Z 0

= d ϕ q 1 − k 2 sin 2 ϕ = F ( ϕ , k )+ F ( π , k ) .

d θ p 1 − k 2 sin 2 θ

+

=