About anaytic & harmonic function, CR equation

1. Sub:-COMPLEXVARIABLES AND NUMERICAL METHODS
SUBJECT CODE:-2141905
Topic:-ANALYTIC FUNCTION, Cauchy-riemann equations, Harmonic Conjugate
functions.
Branch:-Mechanical Engineering( 4th Sem.)
Prepared by:-Malaysinh Borasiya(130990119004)
Chintan Charola(130990119005)
Mayurrajsinh Chauhan(130990119006)

2. ANALYTIC FUNCTION
 A single valued function which is defined and differentiable
at each point of a domain D is said to be analytic in that
domain.
 A function is said to be analytic at a point if its derivative
exists not only at that point but in some neighbourhood of
that point.
 The point where the function is not analytic is called a
singular point of the function.

3. Notes
 Let f(z) and g(z) be analytic functions in some domain D;
(i) and f(z).g(z) are analytic functions in D.
(ii) is analytic in D except where g(z)=0.
(iii) f[g(z)] and g[f(z)] are analytic functions in D.
)()( zgzf 
)(
)(
zg
zf

4. Example -1 : State where the function is not analytic?
Solution:- f(z) is not analytic at as the function is not defined at these
points. Hence, the points are singular points.
Example-2 : State where the function is not
analytic?
Solution:- The function is not analytic at the points
and
and
1
2
)( 2


z
z
zf
iz 
iz 
)22)(2(
1
)( 2
2



zzz
z
zf
02 z 0222
 zz
2z iz 

 1
2
842

5. CAUCHY-RIEMANN PARTIAL DIFFERENTIAL
EQUATIONS
Theorem : If a function f(z) = u(x,y) + iv(x,y) is analytic at any point z = x+iy,
then the partial derivatives ux , uy , vx , vy should exist and satisfy the
equations ux =vy , uy = -vx or
Proof : Let f(z) be an analytic function at any point z.
exist
Now
…..(1)
x
v
y
u
y
v
x
u










,
z
zfzzf
z
zf



)()(
0
lim
)('



z
ivuvviuu
z
zf



)()()(
0
lim
)('



z
viu
z 





0
lim

6. CAUCHY-RIEMANN PARTIAL DIFFERENTIAL
EQUATIONS
For equation (1) limit exist along any path as Consider the path along real
axis(z=x).
from (1),
….(2)
Now consider the path along y axis i.e. z=iy
from (1),
….(3)
.0z

 00 xz 
x
viu
x
zf






0
lim
)('
x
v
i
x
u







 00 yz 
y
v
y
u
i
yi
viu
y
zf











 0
lim
)('

7. CAUCHY-RIEMANN PARTIAL DIFFERENTIAL
EQUATIONS
It is obvious that the limits (2) and (3) are same.
and
y
v
y
u
i
x
v
i
x
u












yx vu  xy vu 

8. Example 3 : Use C-R equation concept to find derivative of
Solution:- We have
and
So, C-R equations are satisfied everywhere in z-plane.
exists everywhere. Thus we get
Example 4 : Show that neither nor is an analytic function.
Solution:- We have
So, C-R equation is not satisfied.
is not analytic.
.)( 2
zzf 
xyiyxzzf 2)( 222

xyyxvyxyxu 2),(,),( 22

yx vxu  2 xy vyu  2
)(' zf
zyixiuvivuzf yyxx 222)(' 
zzf )( zzf )(
iyxzzf )(
yyxvxyxu  ),(,),(
1,0,0,1  yxyx vvuu yx vu 
zzf  )(

9. Continue…
Now
So, C-R equation is not satisfied.
is not analytic.
Example 5 : Show that is an analytic function, find
Solution:- We have,
and . So, w is analytic.
22
)( yxzzf 
0),(,),( 22
 yxvyxyxu
0,
22


 yx v
yx
x
u yx vu 
zzf  )(
2222
yx
iy
yx
x
w



 .
dz
dw
2222
,
yx
y
v
yx
x
u




,
)( 222
22
yx
xy
ux


 222
)(
2
yx
xy
vx


222
22
)( yx
xy
vy


,
)(
2
222
yx
xy
uy


yx vu  xy vu 

10. Continue…
Now,
Example 6 : Check whether the following functions are analytic or not at any point:
Solution:-
(a) we have
is not analytic anywhere.
(b)We have
is not
analytic anywhere.
xx ivu
dz
dw

222222
22
)(
2
)( yx
xy
i
yx
xy





.2)()()()( 2
ixyxzfbezfa z

)sin(cos)( yiyeeezf xiyxz
 
,cos yeu x
 ,sin yev x

,cos yeu x
x  yev x
y cos
yx vu 
z
ezf  )(
2
2)( ixyxzf 
,2
,2


xu
xu
xyv
xyv
y 2
2


yx vu  2
2)( ixyxzf 

11. Example 7 : Examine the analyticity of sinh z.
Solution:- Let z0 be any point in the domain.
exist at any point z.
is analytic.
0
0
00
0
0
sinhsinhlim)()(lim
zz
zz
zzzz
zfzf
zz 







0
00
0
2
sinh
2
cosh2
lim
zz
zzzz
zz 





 





 


0
0
0
0
0
0
cosh
2
2
sinh
lim
2
cosh
lim
z
zz
zz
zz
zz
zz






 





 






 


)(' zf
)(zf

12. Continue….
OR
We have
and
So, C-R equation is satisfied for any point.
So, f(z) is analytic function.
)sinh(sinh)( iyxzzf 
xyixy coshsinsinhcos 
,sinhcos),( xyyxu  xyyxv coshsin),( 
,coshcos xyux  xyvy coshcos
,sinhsin xyuy  ,sinhsin xyvx 
yx vu  xy vu 

13. POLAR FORM OF C.R. EQUATIONS
We have
are C-R equations in polar form.
and
 sin,cos ryrx 






 
x
y
yxr 122
tan,
r
vu
r
v
rr
u











1
,
1











 
r
v
i
r
u
ezf i
)('

14. Harmonic Function & Conjugate harmonic
function
Harmonic Function:- A function is said to be harmonic in a
domain D if
(1) Satisfy Laplace’s equation and
(2) are continuous functions of x and y in D.
Conjugate harmonic function:- If f(z)=u+iv is an analytic function of
z, then v is called a conjugate harmonic function of u and u in its turn is
termed a conjugate harmonic function of v. Or u and v are called conjugate
harmonic functions.
),( yx
),( yx 0 yyxx 
yyxyxx  ,,

15. Example 8 : Is the function u=x sin x cosh y - cos x sinh y harmonic?
Solution:- We have
And
also are continuous functions.
So, u is a harmonic function.
yxyyxxyxyxu
yxyyxxyxu
yxyyxxu
xx
x
sinhcoscoshsincoshcoscoshcos
sinhsincoshcoscoshsin
sinhcoscoshsin



yxyyxxyx sinhcoscoshsincoshcos2 
yxyyxyxyxxu
yxyyxyxxu
yy
y
sinhcoscoshcoscoshcoscoshsin
coshcossinhcossinhsin


yxyyxyxx sinhcoscoshcos2coshsin 
0 yyxx uu
yyxyxx uuu ,,

16. Example 9 : Show that is harmonic.
Solution:- We have
….(1)
22
yx
x
u


22
yx
x
u


222
22
222
22
)()(
)2()1)((
yx
xy
yx
xxyx
ux






422
2222222
)(
)2)((2)()2()(
yx
xyxxyxyx
uxx



322
23
322
3223
422
222222
)(
62
)(
4422
)(
)](4)2)()[((
yx
xyx
yx
xxyxyx
yx
xyxxyxyx










17. Continue…
…(2)
So, From equation (1) & (2),
Also are continuous functions.
So, u is a harmonic function.











422
22222
222
)(
2)(2)2()2()(
)(
2
yx
yyxxyxyx
u
yx
xy
u
yy
y
322
23
322
223
422
22222222
)(
62
)(
]822[
)(
)](8)2)([()(
yx
xyx
yx
xyxyx
yx
yxxyxyxyx









0 yyxx uu
yyxyxx uuu ,,

18. Example 10 : Determine a and b such that is harmonic and find
its conjugate harmonic.
Solution:- We have
since u is harmonic function,
and b assumes any value
Now,
bxyaxu  3
bxyaxu  3
0
63 2


yyy
xxx
ubxu
axubyaxu
0 yyxx uu
006  aax
bxubyubxyu yx  ,
dyvdxvdv yx 
bydybxdx
dyudxu xy



19. Continue… c
y
b
x
bv 
22
22






 c
bybx
ibxyivuzf
22
)(
22

20. Method of constructing a regular function
If only the real part of an analytic function f(z) is given then
where c is a real constant.
Replace x by and y by to find and put x=y=0 to find u(0,0)
in u(x,y)
ciu
i
zz
uzf 





 )0,0(
2
,
2
2)(
2
z
i
z
2






i
zz
u
2
,
2

21. Example 11 : Find an analytic function if
Solution:- We have
and u(0,0)=0
.33
xyxu ivuzf )(
xyxu 33

iz
z
i
zzz
i
zz
u 2
33
4
3
822
3
22
,
2

























ciu
i
zz
uzf 





 )0,0(
2
,
2
2)(
ciiz
z
zf  2
3
2
3
4
)(

22. Example 12 : Show that the function is harmonic and find the
corresponding analytic function.
Solution:- We have
So, u is harmonic.
Now,
xyxu  22
xyxu  22
22
212


yyy
xxx
uyu
uxu
0 yyxx uu
0)0,0(
222442
,
2
222







u
zzzzz
i
zz
u
cizzciu
i
zz
uzf 





 2
)0,0(
2
,
2
2)(