Mathematical Physics Vol 1

Chapter 7. Partial differential equations

374

Let u and v be two harmonic functions in region V . According to the second Green formula (7.229) and Theorem 25 (p. 371) (7.233), we obtain x S ∪ σ u ∂ v ∂ n d S − x S ∪ σ v ∂ u ∂ n d S = 0 . (7.244)

We have seem that the function 1 r =

1 p ( x − a ) 2 +( y − b ) 2 +( z − c ) 2

(7.245)

is harmonic, so let us assume that v = 1 / r . For such a choice of v the relation (7.244) becomes x S ∪ σ − u 1 r 2 − 1 r ∂ u ∂ n d S = 0 , (7.246) that is x S − u 1 r 2 − 1 r ∂ u ∂ n d S + x σ − u 1 r 2 − 1 r ∂ u ∂ n d S = 0 . (7.247) Given that σ is a sphere, with boundary equation r = R , we obtain x S − u 1 r 2 − 1 r ∂ u ∂ n d S = 1 R 2 x σ u d S + 1 R x σ ∂ u ∂ n d S . (7.248) Integrals at the right hand side of the previous equation can be represented as follows 1 R 2 x σ u d S = 1 R 2 4 π R 2 u ∗ = 4 π u ∗ , (7.249) where u ∗ is the mean value of function u on surface σ , (it is known, does not depend of S and can be moved in front of the integral), (4 π R 2 is the area of σ , i.e, the area of the sphere) and 1 R x σ ∂ u ∂ n d S = 1 R 4 π R 2 ∂ u ∂ n ∗ , (7.250)

where

∂ u ∂ n

∗ is the mean value of the derivative on the sphere.

Further, given that

4 π R

∂ u ∂ n

∗

u ∗ = u ( A ) , lim R → 0

lim R → 0

= 0 ,

(7.251)

" x σ

d S = 4 π R 2 ,

lim ( x , y , z ) → A

u ( x , y , z )= u ( A ) ,

d S = u ( A ) 4 π R 2 , #

x σ

u ( x , y , z ) d S = u ( A ) x σ

lim ( x , y , z ) → A

and that on the surface S : u | S = f , we obtain, from (7.248) 4 π u ( A )= − x S f r 2 + 1 r ∂ u ∂ n d S .

(7.252)