Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

302

- the connecting equation

z · ˙ z

ρ 2 + z 2 − R 2 = 0 ⇒ ˙ ρ = −

(3)

√ R 2

.

− z 2

From (2) i (1) we obtain

2

2 C

˙ ρ 2 +

+ ˙ z 2 = 2 gz + h

ρ

(4)

ρ 42

From (3) and (4) it follows that z 2 ˙ z 2 R 2 − z 2 + C 2 R 2 − z 2 + ˙ z 2 = 2 gz + h ⇒ ˙ z 2 1 + = 2 gz + h ⇒ R 2 ˙ z 2 = R 2 − z 2 ( 2 gz + h ) − C 2 = − 2 gz 3 − z 2 h + 2 gRz + hR 2 − C 2 | {z } P ( z ) . Analysis. P ( z ) is a third degree polynomial. As the left hand side is the square of real functions, only solutions for which P ( z ) ≥ 0 make sense. This polynomial has three zeroes, where lim z →± ∞ P ( z )= ∓ ∞ , P ( ± R )= − C 2 . In the initial moment the point was at position z 0 and had initial velocity ˙ z 0 . Observe the interval − R ≤ z 0 ≤ R inwhich P ( z 0 )= R 2 − z 2 0 ( 2 gz 0 + h ) − C 2 = R 2 ˙ z 2 0 > 0. In order for the point to start moving, it is necessary that the initial velocity is P ( z 0 ) > 0 (see Figure 5.13). z 2 R 2 − z 2 + C 2 R 2 − z 2

Figure 5.13: Analysis of the task 215 - sketch of the polynomial P ( z ) .

As the polynomial has three zeroes, its sign changes three times (at points: z 1 , z 2 i z 3 ), and we will thus observe the following intervals z 0 < z 1 < R , − R < z 2 < z 0 , − ∞ < z 3 < − R . However, as the point moves along a hemisphere, the z coordinate is limited to the interval − R ≤ z ≤ R , and due to the condition P ( z ) > 0, to the interval z 2 ≤ z ≤ z 1 . As the polynomial zeroes are z 1 , z 2 and z 3 , it can be represented in the form P ( z )= − g ( z − z 1 )( z − z 2 )( z − z 3 ) .