Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 371

Theorem 2 .Let u ( M ) be a harmonic function in Ω .Forany M 0 ∈ Ω , u ( M 0 ) must

be

R 2
S M 0
R

1

u ( M 0 )= u ( M ) d S , (7.80)

4 π

where the S M 0
R

stands for the spherical surface of a sphere (inside Ω )ofcenter M 0

and radius R .

Proof. Take in the fundamental integral formula of harmonic functions

(Eq. (7.69)) to be S M R . Equation (7.69) and theorem 1 thus yield

∂Ω

1 1

r

d S

1
4 R ∂

u

u ∂
∂

( )=

n −

u M 0

π ∂ r

S M 0
R

R 2
S M 0
R

1
4

u
R 2 d S

1

= = u d S .

π 4 π

S M 0
R

Remark 1 . For a two-dimensional harmonic function u = u ( x , y ) , its value at point

M 0 ( x 0 , y 0 ) is

2π

1
2

u ( M 0 )= u ( x 0 + R cos θ , y 0 + R sin θ ) d θ ,

π

0

which is its mean value over the circumference of a circle of center M 0 and radius
R .

Remark 2 .If u is an harmonic function in a open domain Ω ,wehave

1

V M 0
R

u ( M 0 )= u ( M ) d Ω , ⊂ Ω , (7.81)

4

3 π

R 3

V M 0
R

where V M 0
R

denotes a sphere inside Ω of center M 0 and radius R . Note that for any

S M r with 0

< r ≤ R ,wehave

r 2 u ( M 0 ) 4 π = u ( M ) d S .

S M 0
r

Thus an integration from r = 0to r = R will arrive at Eq. (7.81).

Theorem 3 (Liouville Theorem). Let u be a harmonic function in the entire domain
R 3 .If u is bounded from below or above, it must be a constant.

−

Proof. If u is a harmonic function, u is also a harmonic function. Therefore, we

need to prove the theorem only for the case u ≥ c (bounded from below).