Mathematical Physics Vol 1

Chapter 7. Partial differential equations

380

Theorem27 In order for the solution of the Neumann problem to exist, it is necessary that the integral of the function f is annulled on the contour ℓ .

Proof Given that u is a harmonic function, it follows that x S ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 d x · d y = 0 .

(7.288)

Further, according to the Green formula (7.229), for v = 1, we obtain, for the case in the plane x S ∆ u d S = Z ℓ ∂ u ∂ n d l = Z ℓ f d l = 0 . (7.289)

Theorem28 Two solutions of the Neumann problem can differ only by an arbitrary constant. Proof Let us prove this Theorem for the case when f ( s )= 0. Assume that f ( s )= 0. Then u = 0 is a solution of the respective Neumann problem. Let v be some other solution of the same problem. Given that v is a harmonic function, we obtain x S v ∂ 2 v ∂ x 2 + ∂ 2 v ∂ y 2 d x d y = 0 . (7.290)

However, as

S S

∂ v ∂ x ∂ v ∂ y

d x d y = Z ℓ d x d y = Z ℓ

2

x S

d y − x

∂ 2 v ∂ x 2

∂ v ∂ x

v

v

d x d y

(7.291)

and

2

x S

d x − x

∂ 2 v ∂ y 2

∂ v ∂ y

v

v

d x d y ,

(7.292)

by adding the previous relations we obtain x S v ∂ 2 v ∂ x 2 + ∂ 2 v ∂ y 2 d x d y = Z ℓ v

∂ v ∂ n

d s − ∂ v ∂ x

(7.293)

S "

2 #

+

∂ v ∂ y

2

− x

d x d y .

Given that, according to the assumption, ∆ v = 0 and ∂ v ∂ n

= 0, it follows that

S "

2 #

∂ v ∂ x

+

∂ v ∂ y

2

x

d x d y = 0 ,

(7.294)