Mathematical Physics Vol 1

Chapter 7. Partial differential equations

378

From the similarity of triangles (Fig. 5.5) follows the proportionality of their sides r r 1 = l R = R l 1 ⇒ 1 r 1 = l R 1 r . Given that 1 / r is a harmonic function in the region bounded by the sphere S , we can apply the equation (7.254) u ( a , b , c )= 1 4 π x S f ∂ ∂ n 1 r − 1 r ∂ u ∂ n d S . (7.278) According to (7.277) and the second Green formula (7.229) we get x S u ∂ ∂ n 1 r 1 − 1 r 1 ∂ u ∂ n d S = 0 . (7.279) Further on, according to the assumptions of this problem, we have that u = f for M ( x , y , z ) ∈ S , and thus we obtain x S f ∂ ∂ n 1 r 1 − 1 r 1 ∂ u ∂ n d S = 0 · R 4 π l ⇒ R 4 π l x S f ∂ ∂ n 1 r 1 − 1 r 1 ∂ u ∂ n d S = 0 . (7.280) (7.277)

R Note that we could not apply this relation to function 1 / r , as it is not defined at point A ( a , b , c ) , inside the region S , while the function 1 / r 1 is not defined at point A 1 ( a 1 , b 1 , c 1 ) , which is outside the observed region.

Finally, from (7.280) and (7.278), and using (7.277), we obtain

S

∂ ∂ n 1 4 π x S

1 r −

∂ ∂ n

1 r 1

1 4 π x

R l

u ( a , b , c )=

f

f

d S ⇒

(7.281)

∂ ∂ n

1 r 1

1 r −

R l

u ( a , b , c )=

f

d S

the solution for Dirichlet problem for a sphere. Let us now define the function G by the relation

G = G ( x , y , z , a , b , c )= 1

R l

1 r 1

(7.282)

r −

=

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

1

p ( x − a 1 )

=

2 +( y − b

2 +( z − c

2

1 )

1 )

with the following properties: • G is a harmonic function with respect to point M ( x , y , z ) , within the sphere S , except at point A ( a , b , c ) ; • G is a harmonic function with respect to point A ( a , b , c ) , within the sphere S , except at point M ( x , y , z ) ; • function G − 1 / r is a harmonic function at all points points inside the region S ; • G is annulled on the sphere S .