Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

304

The differential equation (5.288) now has the form R 2 · 4 ( z 1 − z 2 ) 2 u 2 ˙ u 2 = 2 g ( z 1 − z 2 ) u 2 ( z 1 − z 2 )( 1 − u 4 R 2 ˙ u 2 = 2 g ( 1 − u 2 ) z 1 − z 3 − ( z 1 − z 2 ) u g ( z 1 − z 2 ) ˙ u 2 =

2 ) z

2 ⇒

1 − z 3 − ( z 1 − z 2 ) u

2 ⇒ u 2

z 1 − z 2 z 1 − z 3

2 R 2

1 − u 2 1 −

If we introduce the following substitutions

= k 2 i λ = p

2 g ( z 1 − z 2 ) 2 R ,

z 1 − z 2 z 1 − z 3

we obtain the separable differential equation ˙ u = λ q ( 1 − u 2 )( 1 − k 2 u 2 ) ⇒

d u p ( 1 − u 2 )( 1 − k 2 u 2 )

= λ d t .

Bearing in mind the initial conditions: t 0 = 0 , u 0 = 0 we finally obtain

u Z 0

t Z 0

d u p ( 1 − u 2 )( 1 − k 2 u 2 )

λ d t = λ t .

I =

(5.289)

=

Thus integral I , as a function of the upper bound, cannot be expressed by a finite number of elementary functions, and it is called the normal elliptic integral of the first kind .

R Note that the function u must satisfy the condition 1 − u 2 ≥ 0 (the expression under the square root must be non-negative!), that is, u 2 ≤ 1, and thus this integral can obtain another form by substitution u = sin ϕ ⇒ d u = cos ϕ d ϕ d ϕ q 1 − k 2 sin 2 ϕ . The constant k is called the modulus of the elliptic integral , where 0 ≤ k 2 ≤ 1. It can be proved that, for boundary values k = 0 and k = 1, this integral is reduced to elementary functions. F ( ϕ , k )= ϕ Z 0

21 We have used here the expression for velocity in polar coordinates v 2 = ˙ ρ 2 + ρ 2 ˙ θ 2 + ˙ z 2 .