Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

300

we obtain

s

R g

d u p ( 1 − u 2 )( 1 − k 2 u 2 ) .

d t =

(5.284)

We have introduced here yet another substitution k 2 = sin 2 ϕ 2 < 1. By integrating equation (5.284) (with initial conditions: t 0 = 0, ϕ 0 = 0, u 0 = 0)we obtain s R g t = Z u 0 d u p ( 1 − u 2 )( 1 − k 2 u 2 ) . (5.285) The integral in this equation is an elliptic integral of the first kind, and thus for u we obtain u = sn s R g t ! which represents the required final solution.

Problem 215 Determine the final equations of the motion of a material point M along the surface of a smooth sphere - the spherical pendulum .

Figure 5.12: Spherical pendulum.

Solution It is convenient to observe this motion with respect to the spherical coordinate system, in which the equation of the sphere is f ( ρ , θ , z ) ≡ ρ 2 + z 2 − R 2 = 0 . (5.286) This relation represents the connecting equation! As in the case of mathematical pendulum, we start from the law on change in kinetic energy

A = T − h T 0 =

m 2

1 2 m ˙ ρ 2 + ρ 2 ˙ θ 2 + ˙ z 2 = mgz + mh

v 2 − mh ⇒

20 Using the relation cos 2 α

1 + cos α 2

2 =