Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 390

1

It is clear that Δ g = 0and g | ∂Ω = by Eq. (7.112). Finally, the Green function

4 π r M 0 M

is

1
r M 0 M −

1
4

R
r 0

1
r M 1 M

G ( M , M 0 )= . (7.113)

π

Let ( r sin θ cos ϕ , r sin θ sin ϕ , r cos θ ) be the spherical coordinates of point M in Ω .

Here r < R ,0 ≤ θ ≤ π , 0 ≤ ϕ ≤ 2 π . Note that

r 0 + r 1 +

R 2

r M 0 M = r 2 − 2 r 0 r cos ψ , r M 1 M = r 2 − 2 r 1 r cos ψ , r 0 r 1 = ,

where r = r OM , ψ is the angle between OM 0 (or OM 1 )and OM . Thus

⎡ ⎤

1
4

1 R

⎣ ⎦ ,

G ( M , M 0 )= r 0 + − r 0 r 2

π

r 2 − 2 r 0 r cos ψ − 2 R 2 r 0 r cos ψ + R 4

=

and, by noting that the external normal of ∂Ω is along r r OM ,

∂Ω = ∂ r = R = −

∂ G G

∂

1
4

r − r 0 cos ψ

3
/ 2

∂ n r π

r 0 + r 2 ( − 2 r 0 r cos ψ )

r = R

r 0 r R 2 r 0 cos ( − ψ ) R

−

3
/ 2

r 0 r 2 2 R 2 r 0 r cos R 4

( − ψ + )

R 2 r 0

1 −

= −

/ 2 .

3

4 π R

r 0 + ( R 2 − 2 r 0 R cos ψ )

Hence the solution of

Δ u ( r , θ , ϕ )= F ( r , θ , ϕ ) , 0 < r < R , 0 ≤ θ ≤ π , 0 ≤ ϕ ≤ 2 π ,

(7.114)

u | r = R = f ( R , θ , ϕ ) .

is

2π π

R 2 r 0

R
4

−

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

3
/ 2

π

r 0 + ( R 2 − 2 r 0 R cos ψ )

0 0

⎡

R 2π π

1
4

1

⎣

r 0 +

−

π 0 0 0

r 2 − 2 r 0 r cos ψ