Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 391

⎤

R

⎦ F

r 2 sin

− r 0 r 2 ( , θ , ϕ ) .

r θ d θ d ϕ d r

2 R 2 r 0 r cos R 4

− ψ +

Here the coordinates of M 0 are ( r 0 , θ 0 , ϕ 0 ) and the area element d S on surface r = R

is

R 2 sin d S = θ d θ d ϕ .

Remark 1 .When F ( M )= 0, the solution reduces to

2π π

R 2 r 0 R

4

−

u ( r 0 , θ 0 , ϕ 0 )=

3
/ 2

π

r 0 + R 2

( − 2 r 0 R cos ψ )

0 0

· f ( R , θ , ϕ ) sin θ d θ d ϕ (7.115)

which is called the Poisson formula for problems in a sphere .

Remark 2 .Let e 0 and e be the unit vectors of OM 0 and OM , respectively.
Since

e 0 =( sin θ 0 cos ϕ 0 , sin θ 0 sin ϕ 0 , cos θ 0 ) ,

e =( sin θ cos ϕ , sin θ sin ϕ , cos θ ) ,

we have

cos ψ = e 0 · e = sin θ sin θ 0 ( cos ϕ cos ϕ 0 + sin ϕ sin ϕ 0 )+ cos θ cos θ 0

= sin θ sin θ 0 cos ( ϕ − ϕ 0 )+ cos θ cos θ 0 .

Remark 3 . For the external problems in the domain x 2 y 2 z 2 R 2 ,

+ + >

∂ G

n = − ∂

G

r . ∂ ∂

The solution of external problems thus differs from that in Eq. (7.115) only by a
sign, i.e.

2π π

r 0 − R 2

R
4

u ( r 0 , θ 0 , ϕ 0 )= f ( R , θ , ϕ ) sin θ d θ d ϕ ,

π

3
/ 2

r 0 + ( R 2 − 2 r 0 R cos ψ )

0 0

: x 2 y 2 z 2 R 2 .

where M 0 ( r 0 , θ 0 , ϕ 0 ) is a point outside the sphere Ω + + <