V 024 Ox 0y2az0z Let V2d=4 P aza So we can transform the function VO=O =0 (a) 022022 as y=V2(v2y)=V2P=0 for V2P=0 11
11 x y zz = + = 2 2 2 2 2 2 4 P z z = = 2 2 Let 4 So we can transform the function 0 4 = 0 ( ) 2 2 4 a z z = as ( ) 0 4 2 2 2 = = P = For 0 2 P =
V 024 Ox 0y2az0z 令 V2d=4 P aza 于是可将方程式 VO=O =0 (a) 022022 变换成为 y=V2(v2y)=V2P=0 由 V2P=0 12
12 x y zz = + = 2 2 2 2 2 2 4 P z z = = 2 2 令 4 于是可将方程式 0 4 = 0 ( ) 2 2 4 a z z = 变换成为 ( ) 0 4 2 2 2 = = P = 由 0 2 P =
dextrane Methods for Pane bestial It is obvious known p is harmonic function which can be obtained by real part of analytical function. Supposefz)as analytical function and let P=(f(z)+f(=) For V=4 P aza yields ee24′=42(()+r() Let f(=)=4q)(=) thi lUs ((2)+((z ) azaz 2 13
13 It is obvious known P is harmonic function which can be obtained by real part of analytical function. Suppose f(z) as analytical function and let ( ( ) ( )) 2 1 P = f z + f z P z z = = 2 2 For 4 ( ( ) ( ))] 2 1 [ 4 1 4 1 2 P f z f z z z = = + f (z) = 4'(z) ( '( ) '( )) 2 1 2 z z z z = + Let yields thus
可知,是调和函数可由解析函数的实部得到。设 f(z)为解析函数,可令 P 2 (∫(=)+f(z) 由 Vφ=4 P aza 得 ea24=42(()+( 令 f(=)=4(z) 则 ((2)+((z ) azaz 2 14
14 可知，P是调和函数可由解析函数的实部得到。设 f(z)为解析函数，可令 ( ( ) ( )) 2 1 P = f z + f z P z z = = 2 2 由 4 ( ( ) ( ))] 2 1 [ 4 1 4 1 2 P f z f z z z = = + f (z) = 4'(z) ( '( ) '( )) 2 1 2 z z z z = + 令 得 则
Integrating the above equation with respect to z yields p az 2 (2'(=)+z()+v(二) Then integrating with respect to z yields =(0()+0(-)+|v(二)dz+8(E) 2 Let (=)dz=(=) l e (z)=x(=) thus p=(E0(二)+20(-)+(=)+g(2) 15
15 ( '( ) '( ) ( )) 2 1 z z z z z z = + + Then integrating with respect to z yields ( ( ) ( ) ( )d ( )) 2 1 z z z z z z g z = + + + (z)dz (z) Let = i.e. (z) = '(z) ( ( ) ( ) ( ) ( )) 2 1 = z z + z z + z + g z Integrating the above equation with respect to yields z thus