Mathematical Physics Vol 1

7.8 Examples

427

Given that left hand side is a function of r only, and the right hand side of t , we can conclude that these expressions have to be constant, i.e. ∆ P P = 1 c 2 ¨ T T = − k 2 . (7.426) Thus, the initial partial differential equation (7.424), by applying this method, is decomposed on two differential equations (7.426): ¨ T + ω 2 T = 0 , ( ω = kc ) , (7.427) ∆ P + k 2 P = 0 . (7.428) The solution of the first of equation (7.427) (second order differential equation with constant coefficients) is of the form T = C 1 sin ω t + C 2 cos ω t . (7.429) Observe now the equation (7.428). As we have assumed that r is a function of spherical coordinates, this equation can be represented in the form (see the form of Delta operator in spherical coordinates) 1 r 2 ∂ ∂ r r 2 ∂ P ∂ r + 1 r 2 sin θ ∂ ∂θ sin θ ∂ P ∂θ + 1 r 2 sin 2 θ ∂ 2 P ∂ϕ 2 + k 2 P = 0 . (7.430) For solving this partial differential equation we shall, once again, use the method of separation of variables, i.e. assume that the function P is a function of the form P = R ( r ) · Θ ( θ ) · Φ ( ϕ ) . (7.431) Substituting (7.431) into (7.430) and dividing by R ΘΦ we obtain 1 r 2 R d d r r 2 d R d r + 1 r 2 Θ sin θ d d θ sin θ d Θ d θ + 1 r 2 Φ sin 2 θ d 2 Φ d ϕ 2 + k 2 = 0 . (7.432) If we multiply this equation by r 2 sin 2 θ , we can notice that the member 1 Φ d 2 Φ d ϕ 2 depends on ϕ only, while the remaining members depend on r and θ . Based on this, as in the previous similar case, we conclude that 1 Φ d 2 Φ d ϕ 2 = − m 2 , (7.433)

d d r

d R d r −

d d θ

d Θ d θ −

sin 2 θ R

sin θ Θ

ρ 2

k 2 r 2 sin 2 θ = − m 2 ,

sin θ

(7.434)

−

where m is a constant. From (7.433) it follows that the function Φ is of the form Φ = C 3 sin m ϕ + C 4 cos m ϕ . Substituting (7.435) into (7.432) we obtain 1 r 2 R d d r r 2 d R d r + 1 r 2 Θ sin θ d d θ sin θ d Θ d θ + 1 r 2 sin 2 θ

(7.435)

( − m 2 )+ k 2 = 0 .