Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 374

region D between ∂ K R and ∂ K R / 2 ,

1
r +

r
2 R 2 −

3
2 R ,

v ( r )=

x 2

y 2 z 2 , M

where r = + + ( x , y , z ) ∈ D .In D ,wehave

∂ K R = ∂ r = R = −

∂ v v 1

2 R 2 <

v ( r ) | ∂ K R = v ( R )= 0 , 0 ,

∂ r ∂ r

r 2 ∂ r 2

2 R 2 −

1

r 2 ∂

1 v

r 2 ∂

1 1

n 2 r >

Δ v = = = 0 , R / 2 < r < R .

∂ r ∂ r ∂ r

To analyze the normal derivative of u ( M ) at M 0 , consider another new function

w ( M )= u ( M ) − ε v ( M ) , ε > 0 , M ∈ D ,

we attempt to prove that w ( M ) takes its minimum value at M 0 .For w ( M )= u ( M ) −

ε v ( M ) ,wehave

w ( M )= u ( M ) , M ∈ ∂ K R ,

w ( M ) − w ( M 0 )= u ( M ) − ε v ( M ) − u ( M 0 )

= u ( M ) − u ( M 0 ) − ε v ( R / 2 ) , M ∈ ∂ K R / 2 .

Define d = min

r 2 [

u ( M ) − u ( M 0 )] > 0 and consider 0 < ε < d / v ( R / 2 ) ,wehave

= R /

w ( M ) ≥ w ( M 0 ) , M ∈ ∂ K R / 2 .

Thus

w ( M ) ≥ w ( M 0 ) , M ∈ ∂ K R / 2 ∪ 2 K r .

Also, w ( M ) cannot take its minimumvalue in D . Otherwise, at the point of minimum

value,

w xx ≥ 0 , w yy ≥ 0 , w zz ≥ 0 ,

so Δ w ≥ 0. Or, alternately,

Δ w = Δ u − εΔ v = − εΔ v < 0 .

( ) ∈

Therefore, w M , M D , must take its minimum value at M 0 ,so

∂ w

n = ∂

u

n − ε ∂

v

0 r ∂

u

n ≤ ε ∂

v

n ≤ n <

0 .

∂ ∂ ∂ ∂ ∂

Harmonic Functions and Analytical Functions

Let f be an analytical function in a plane domain D .Bythe

theory of complex functions, u

( z )= u ( x , y )+ i v ( x , y )

( x , y ) and v ( x , y ) must be differentiable at any point