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GOVERNMENT INSTITUTE OF SCIENCE
COLLEGE,NAGPUR
MATHEMATICS
B.Sc. Sem-5
PAPER-1:ANALYSIS
UNIT – III
ANALYTIC FUNCTIONS
SUBJECT: HARMONIC FUNCTIONS, ORTHOGONAL
FAMILIES,CONSTRUCTION OF ANALYTIC FUNCTION.
HARMONIC FUNCTIONS :
A real valued function u of two variables x and y is said to be
harmonic if it has a continuous partial derivatives and
satisfies the equation-
𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0
or
𝛻 2 u =0
This equation is also known as Laplace equation.
The functions u(x , y) and v(x , y) which satisfy Laplace’s
equation are called harmonic functions.
Theorem :-
If f(z)=u+iv is an analytic function of z=x+iy then prove
that u and v are harmonic functions.
Proof:- Let f(z)=u+iv be an analytic function of z=x+iy.
∴ By theorem,
C-R equations are satisfied.
𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑣
i.e. = … 1 𝑎𝑛𝑑 = − …(2)
𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥
To prove :1)u is harmonic.
i.e. 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0
Diff. eqn (1) w.r.t. x and eqn (2) w.r.t. y ,we get
𝜕 𝜕𝑢 𝜕 𝜕𝑣 𝜕 𝜕𝑢 𝜕 𝜕𝑣
= and = −
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑥
On adding, we get
𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0
To prove :2) v is harmonic.
i.e. 𝑣𝑥𝑥 + 𝑣𝑦𝑦 = 0
Diff. eqn (1) w.r.t. x and eqn (2) w.r.t. y ,we get
𝜕 𝜕𝑢 𝜕 𝜕𝑣 𝜕 𝜕𝑢 𝜕 𝜕𝑣
= and = −
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑥
On adding, we get
𝑣𝑥𝑥 + 𝑣𝑦𝑦 = 0
Hence proved.
ORTHOGONAL FAMILIES :
Two families of curves u(x ,y)=𝑐1 and v(x ,y )=𝑐2 are said
to be an orthogonal system if they intersect at right angles
at each of their points of intersection.
Theorem :-
If w= f(z)=u(x ,y )+i v(x ,y ) is an analytic function in
domain D, then the one parameter families of curves
u(x ,y)=𝑐1 and v(x ,y )=𝑐2 form two orthogonal families.
Proof :- Let f(z)=u(x ,y )+i v(x ,y ) is an analytic function in
domain D.
Hence Cauchy-Riemann equations are satisfied.
𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑣
i.e. . = 𝑎𝑛𝑑 =−
𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥
Taking differential of u(x ,y)=𝑐1 and v(x ,y )=𝑐2 , we get
du =0 and dv = 0
𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑣
⇒ 𝑑𝑥 + 𝑑𝑦 = 0 𝑎𝑛𝑑 𝑑𝑥 + 𝑑𝑦 = 0
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
⇒𝑚1 = slope of the tangent to the curve u(x ,y )=𝑐1 at any point
𝜕𝑦 𝑢
(x ,y )= = − 𝑥
𝜕𝑥 𝑢𝑦
and 𝑚2 = slope of the tangent to the curve v(x ,y )=𝑐2 at any
𝜕𝑦 𝑣
point (x ,y )= = − 𝑥
𝜕𝑥 𝑣𝑦
𝑢𝑥 𝑣𝑥 𝑢𝑥 𝑣𝑥 𝑢𝑥 𝑣𝑥
Now, 𝑚1 𝑚2 = − − = = = −1
𝑢𝑦 𝑣𝑦 𝑢𝑦 𝑣𝑦 −𝑣𝑥 𝑢𝑥
⇒The curves u(x ,y)=𝑐1 and v(x ,y )=𝑐2 form two orthogonal
families.
𝑦
Question 1:If 𝑢 = 𝑥2 − 𝑦2, 𝑣 = − 2 2 then show that both u
𝑥 +𝑦
and v satisfies Laplace’s equation but u+iv is not an analytic
function of z.
Proof:-To show that u+iv is not an analytic function of z,we
have to show that u and v do not satisfy the Cauchy-Riemann
equations 𝑢𝑥 = 𝑣𝑦 , 𝑢𝑦 = −𝑣𝑥 .
Now,
𝜕𝑢 𝜕𝑢
= 2𝑥 , = -2y
𝜕𝑥 𝜕𝑦
𝜕𝑣 −1 2𝑥 2𝑥𝑦 𝜕𝑣 − 𝑥 2 +𝑦 2 −2𝑦 −𝑦 𝑥 2 −𝑦 2
= −𝑦 = , = =−
𝜕𝑥 𝑥 2 +𝑦 2 2 𝑥 2 +𝑦 2 2 𝜕𝑦 𝑥 2 +𝑦 2 2 𝑥 2 +𝑦 2 2
𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑣
Clearly, ≠ , ≠
𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥
⇒ u+iv is not an analytic function of z.
To prove that u and v both satisfy Laplace’s equation, we have
to show that 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0 and 𝑣𝑥𝑥 + 𝑣𝑦𝑦 = 0
Now, 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 2-2=0
∴ 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0
And
2 2
𝑥 2 +𝑦 2 −2 𝑥 2 +𝑦 2 2𝑥𝑥 2𝑦 𝑥 2 +𝑦 2 − 𝑦 2 −𝑥 2 2 𝑥 2 +𝑦 2 2𝑦
𝑣𝑥𝑥 + 𝑣𝑦𝑦 = 2𝑦 +
𝑥 2 +𝑦 2 4 𝑥 2 +𝑦 2 4
2𝑦 𝑦 2 −3𝑥 2 3𝑥 2 −𝑦 2 2𝑦
= + =0
𝑥 2 +𝑦 2 3 𝑥 2 +𝑦 2 3
∴ 𝑣𝑥𝑥 + 𝑣𝑦𝑦 =0
Hence proved.
Contruction of analytic function :
Method-1 :Milne Thomson’s Method:
Case -1]:If the real part u of f(z) is given.
𝜕𝑢 𝜕𝑢
Step-1] Find and.
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑢
Step-2] Find f’(z)= +i = -i
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦
Step-3]Put x=z and y=0 in f’(z).
Step-4]Integrate f’(z) to obtain f(z).
Case -2] If imaginary part v of f(z)is given.
𝜕𝑣 𝜕𝑣
Step-1] Find and
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑣
Step-2] Find f’(z)= +i = +i
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥
Step-3] Put x=z and y=0 in f’(z).
Step-4] Integrate f’(z) to obtain f(z).
Method-2:By finding harmonic conjugate v and
construction of analytic function f(z) :
Given :-u is given.
𝜕𝑢 𝜕𝑢
Step-1] Find and .
𝜕𝑥 𝜕𝑦
𝜕𝑣 𝜕𝑣 𝜕𝑢 𝜕𝑢
Step-2] dv= dx+ dy=- dx+ dy=Mdx + Ndy …(1)
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥
𝜕𝑢 𝜕𝑢
Step-3] M=- , N=
𝜕𝑦 𝜕𝑥
𝜕𝑀 𝜕𝑁
Find and
𝜕𝑦 𝜕𝑦
𝜕𝑀 𝜕𝑁
If = then eqn (1) is exact differential.
𝜕𝑦 𝜕𝑦
Step-4]Find v by using,
v= 𝑀𝑑𝑥 + 𝑁𝑑𝑦 + 𝑐
(taking y constant) (terms of N which do
not contain x)
Step-5] f(z)=u +iv
Question 1:-Find the analytic function f(z)= u + iv of which the
real part is u=𝑒 𝑥 𝑥 cos 𝑦 − 𝑦 sin 𝑦 .
Solution:-We have u=𝑒 𝑥 𝑥 cos 𝑦 − 𝑦 sin 𝑦 .
𝜕𝑢 𝜕𝑢
=𝑒 (cos y)+𝑒 (x cosy –ysiny) , =𝑒 𝑥 (-xsiny-ycosy-siny)
𝑥 𝑥
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑢
Method-1]f’(z)= +i = -i
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦
⇒ f(z)= 𝑒 𝑥 𝑐𝑜𝑠𝑦 + 𝑒 𝑥 (xcosy-ysiny)+i𝑒 𝑥 (-xsiny-ycosy-siny)dz
Put x=z and y=0 in above equation,
⇒ f(z)= 𝑒 𝑧 𝑐𝑜𝑠0 + 𝑒 𝑧 (zcos0-0sin0)+i𝑒 𝑧 (-zsin0-0cos0-sin0)dz
= (𝑒 𝑧 + 𝑒 𝑧 z)dz +c
=𝑒 𝑧 - 𝑒 𝑧 +z𝑒 𝑧 +c
f(z) =z𝑒 𝑧 +c
𝜕𝑣 𝜕𝑣 𝜕𝑢 𝜕𝑢
Method :-2]dv= dx+ dy=- dx+ dy
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥
=-𝑒 𝑥 (-xsiny-ycosy-siny)dx +𝑒 𝑥 (xcosy-ysiny+cosy)dy
which is of the type dv=Mdx + Ndy,
where M =-𝑒 𝑥 (-xsiny-ycosy-siny) and N=𝑒 𝑥 (xcosy-ysiny+cosy)
𝜕𝑀
=-𝑒 𝑥 −xcosy + ysiny − cosy − cosy
𝜕𝑦
=𝑒 𝑥 (2cosy+xcosy-ysiny)
𝜕𝑁
=𝑒 𝑥 cosy+(xcosy-ysiny+cosy) 𝑒 𝑥 =𝑒 𝑥 (2cosy+xcosy-ysiny)
𝜕𝑦
Hence the given eqn is exact differential. Hence its solution is on
integrating v ,we get
v= 𝑀𝑑𝑥 + 𝑁𝑑𝑦 + 𝑐
(taking y constant) (terms of N which do not contain x)
v= −𝑒 𝑥 (−xsiny−ycosy−siny)𝑑𝑥 + 0𝑑𝑦 + 𝑐
=siny 𝑥𝑒 𝑥 𝑑𝑥 + (𝑦𝑐𝑜𝑠𝑦 + 𝑠𝑖𝑛𝑦) 𝑒 𝑥 𝑑𝑥 + 0 + 𝑐
=[(x-1)siny+ycosy+siny] 𝑒 𝑥 +c
=(xsiny+ycosy) 𝑒 𝑥 +c
∴f(z)=u + iv
=𝑒 𝑥 𝑥 cos 𝑦 − 𝑦 sin 𝑦 + 𝑖[(xsiny+ycosy) 𝑒 𝑥 +c]
=x𝑒 𝑥 (cosy +isiny)+iy𝑒 𝑥 (cosy +isiny)+ic
=(x+iy) 𝑒 𝑥 𝑒 𝑖𝑦 + 𝑐 ′
∴f(z)=z𝑒 𝑧 + 𝑐′
