Complex Analysis- Lecture 18 - Harmonic Functions

EX(1):- Find a harmonic conjugate v(x,y) of function u(x,y)=e^x cos?y
and f(Z).
SOL: - by C.R. equation u_x=v_y=e^x cos?y??v/?y=e^x cos?y
??v=e^x cos?y ?y?????v=???e^x cos?y ?y??
?v=e^x ???cos?y ?y+? (x) ? ?v=e^x sin?y+?(x) *
v_x=e^x sin?y+ ? (x) But by C.R. u_y=-v(x)
?u_y=-e^x sin?y?v_x=e^x sin?y
e^x sin?y+?^( ) (x)=e^x sin??y?? ?^( ) (x)=0??(x)=c
put in * we get
v=e^x sin?y+c_1 thus f(Z)=e^x cos??y+i(e^x sin??y+c_1 ? )?
=e^x cos??y+ie^x sin?y+ic_1 ?
f(Z)=e^x (cos?y+i sin?y )+c where c=ic_1
=e^x e^iy+c=e^(x+iy)+c=e^Z+c
Theorem:- The equation u_xx+u_yy=0 is an equivalent equation r^2 u_rr+ru_r+u_??=0 and u_rr+v_yy=0 is an equivalent equation
r^2 v_rr+rv_r+u_??=0 .
Proof:-To prove ? u?_xx+u_yy=0 is equivalent
r^2 u_rr+ru_r+u_??=0 By C-R equation
ru_r=v_? & u_?=-?rv?_r
We derivative first equation for r .

ru_rr+u_r (1)=v_?r?ru_rr+u_r=v_?r --------*
We derivative second equation for ? .
u_??=-rv_r??-1/r u_??=v_r? **
Since v_?r=v_r? , then
ru_rr+u_r=-1/r u_???ru_rr+u_r-1/r u_??=0
To prove v_xx+v_yy=0 is an equivalent
r^2 v_rr+rv_r+v_??=0 By C-R equation
ru_r=v_? & u_?=-rv_r
We derivative first equation for ? .
ru_r?=v_???u_r?=1/2 v_??
We derivative second equation for r .
u_?r=-ru_rr+v_r (-1)=-ru_rr-u_r?u_?v-ru_rr-v_r *
REMARK: - Since Z=x+iy & f(Z) then
f(Z)=u(x,y)+iv(x,y)
?f/?x=df/dZ??Z/?x=u_x (x,y)+iv_x (x,y)
df/dZ (1)=u_x (x,y)+iv_x (x,y)
Proof:- Since Z=re^i? & f(Z)=u(r,?)+iv(r,?)
Then
?f/?r=df/( dZ )??Z/?r=u_r+iv_r

?Z/?r=e^i? & f^ (Z)=df/dZ
f^ (Z)=e^i?=u_r+iv_r?f^ (Z)=e^(-i?) (u_r+iv_r ) *
?f/??=df/( dZ )??Z/??=u_?+iv_?
?Z/??=ire^i? & f^ (Z)=df/dZ
f^ (Z)ire^i?=u_?+iv_??f^ (Z)=e^(-i?)/ir (u_?+iv_? ) **
From * and ** we get
e^(-i?)=(u_r+iv_r )=e^(-i?) (1/ir u_?+1/r v_? )
?u_r+iv_r=(-i)/r u_?+1/r v_??u_r+iv_r=1/r v_?-i 1/r u_?
?u_r=1/r v_? & u_r=-1/r u_?