Miodrag Iovanov. 165KB Jun 04 2011 12:09:14 AM

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

DETERMINING OF AN EXTREMAL DOMAIN FOR THE
FUNCTIONS FROM THE S-CLASS
by
Miodrag Iovanov
Abstract. Let S be the class of analytic functions of the form f(z) = z + a2 z2 +…, f(0)= 0,

f′(0)=1 defined on the unit disk z 0 given. For solving the problem we
shall use the variational method of Schiffer-Goluzin [1].
Key words: olomorf functions, variational method, extremal functions.

1. Let S the class of functions f(z) = z + a2 z2 +…, f(0)= 0, f′(0)=1 holomorf
and univalent in the unit disk z 0 existed.
Geometrically this is expressed like in the figure below:

In region (Ωe) any parallel (Re z >Re|ze| ) to Ox, is intersected f( z = r), in
one point. Because the class S is compact, exists this region. In this paper we will
resolve this problem with the variational method of Schiffer-Goluzin [1].

135

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class
2.Let z = r and let f ∈ S with Rez f′(z)=0, extremal function for exists the

maximum max Re f(z), f ∈ S. We consider a variation f∗(z) for the function f(z) given
by Schiffer-Goluzin formula [1],

ψ real, where
(1)

(2)

f * ( z ) = f ( z ) + λV ( z; ζ ;ψ ) + O(λ2 ), ζ < 1, λ > 0


V(z;ζ ;ψ ) = e iψ



 iψ zf ′( z )
⋅ζ
− e ⋅
z −ζ


 f (ζ ) 
f 2 ( z)
− e iψ f(z)
 −
f ( z ) − f (ζ )
 ζf ′(ζ ) 
2

 f (ζ ) 
z 2 f ′( z )  f (ζ ) 
− iψ
⋅ζ 
⋅
 .

 +e ⋅
1−ζ z
ζf ′(ζ ) 
ζf ′(ζ ) 
2

2

Is known that for λ sufficiently small, the function f∗(z) is in the class S. We
consider a variation z∗ for z:
z∗ =z + λ h + O( λ 2), h=

∂z ∗
∂λ

λ =0

where satisfy the conditions:
(3)

z ∗ = r şi Re z∗ f∗’ (z∗)=0

Observing that :

z∗

2

= z

2

+ 2λ Re( z h) +O( λ2 ) = r2.

Because z = r from relation (3) we obtain :
(4)

Re ( z h) = 0.

Replacing z with z∗ in f∗(z) we have : z∗f∗’(z∗) = A +B λ + O( λ2 ) where :

A = hf ' ( z )


 B = hf ' ( z ) + zhf ' ' ( z ) + zV ' ( z; ζ ;ψ )

The condition Re z∗f∗’(z∗) =0 from relation (3) become:

136

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class
Re {h( f ' ( z ) + zf " ( z )) + zV ' ( z; ζ ;ψ )} = 0 .

(5)

Because f(z) is extremes we have:

where is equivalent with :

Re f∗(z*) ≤ Re f(z)

Re { f ( z ) + λhf ′( z ) + L + λV ( z; ζ : ψ ) + L} ≤ Re f(z)
Re {hf ′( z ) + V ( z; ζ ;ψ )} ≤ 0.

or
(6)
From (5) ( h = −

z
h ) and (6) we obtain:
z
z
h( f ′( z ) + zf ′′( z )) + zV ′( z; ζ ;ψ ) − h (f ′(z ) + z ⋅ f ′′(z ) ) + z ⋅ V ′(z ; ξ ;ψ ) = 0
z

from where:

z z ⋅ V ( z; ζ ;ψ ) + z 2⋅V ′( z; ζ ;ψ )
/

(7)

h=

− zf ′( z ) − z 2 f ′′( z ) + z ⋅ f ′( z ) + z 2 ⋅ f ′′( z )

.

We will use the next denotations:
f= f(z),

w=f( ζ ) , l = f ′(z ), m = f ′′(z ) ,V = (z; ζ ;ψ ), V ′ = VZ′ ( z; ζ ;ψ ) .

With previous denotations, the relations (6) and (7) can be writhed as follows:
(8)

where p =

zl − z ⋅ l

− zl − z 2 m + z ⋅ l + z 2 ⋅ m

Re {pz V ′ +V } ≤ 0

(p real).

Ι .We suppose that Im (zl + z2m) ≠ 0 (-zl + z2m + z ⋅ l + z 2 ⋅ m ≠ 0 ). From
the relation (2) obtain:

137

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

 w 
f2
zl
 − e iψ ⋅
V =e ⋅

⋅ζ
− e iψ ⋅ f 
f −w
z −ζ
 ζ ⋅ w′ 
2

iψ

 w 
 w 
z 2l

 + e −iψ ⋅
⋅ ξ ⋅ 
⋅ 
/ 
1−ζ ⋅ z
 ζ ⋅ w′ 
ζ ⋅ w 

2

and

V′ = e

iψ

 w 
z( z − ζ ) ⋅ m − ζ ⋅ l
fl ( f − 2 w)
 − e iψ
⋅ζ
− e iψ ⋅ l ⋅ 
⋅
2
( f − w)
(z − ζ )2
 ζ ⋅ w′ 
2

+e

−iψ

z 2 (1 − ζ ⋅ z )m + zl ( 2 − ζ z )
⋅ζ
(1 − ζ z ) 2

[

 w 
 +
⋅ 
/ 
ζ ⋅w 
2

 w 
 .
⋅.
/ 
ζ ⋅w 
2

]

Replacing in relation (8) the expression of V and V / we obtain:
Re e iψ ( E − GF ) ≤ 0,

(9)
where:

[

]


f (− f − 2 pzl ) w + f 2 + pzlf
=
E

( f − w) 2

2

⋅l
zl
z
G = f +
ζ−
ζ + pzl +
z −ζ

(1 − zζ ) 2

2
+ pz[z ( z − ζ )m − ζ ⋅ l ] ⋅ ζ − pz z (1 − ζ ⋅ z ) ⋅ m + z ⋅ l ⋅ (2 − ζ ⋅ z ) ⋅ ζ

(z − ζ )2
(1 − z⋅) 2

2

 w 
 .
 F = 

 ζ w′ 

[

]

Because ψ is arbitrary, from relation (8) is result that the function w = f (ξ )
has to satisfy the differential equation :

138

2

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

[

∑t ζ

]

 ζ ⋅ w′  f (− f − 2 pzl ) w + f + pzlf
k =0
=

 ⋅
( f − w) 2
( z − ζ ) 2 (1 − z ⋅ ζ ) 2
 w 
2

(10)

2

4

k

k

where
t 0 = z 2 ( f + pzl ) + f

2
t1 = −2 zf (1 + r 2 ) + z 2 l − z l − 2 plz 2 (1 − r 2 ) + pz 3 m − pr 2 ( m z + 2l ),

4
2
2
2
4
4
2
t 2 = f (r + 4r + 1) − zl ( 2r + 1) − z ⋅ l (2r + r ) + pzl ( r + 4r + 1) −

2
2
4
− pz (2r ⋅ zm + mz + l ) + pz 2r (m ⋅ z + 2 ⋅ l ) + r ( m ⋅ z + l )
2

2
2
2
2
2
2
2
t 3 = −2 f z (1 − 2r ) + r zl ( z + 2) − z l (1 + 2r ) + pr ( −2r + mr z + 2mz − m z − 2l − 2r z m − 2r l )
2
4
2

4
t 4 = f z + p m( z − z ) + z (lz − l ⋅ z ) ..

[

[

]

]

The extremal function transforms the unit disk in the domain without external
points. To justify this thing is sufficient to suppose that the transformed domain by an
external function w = f (ζ ) has an external point w0 and to consider the function the
variation:

f * ( z ) = f ( z ) + λe i ψ

f 2 (z )
, λ >0, ψ real, f * ∈ S
f (z ) − w 0

3.Is known that the extrema function w = f (ζ ) transform the unit disk ζ <

1, in whole plane, cutted lengthwise of a finite number of analytically arc. Let q = e iθ ,
the point of the circle ζ =1 where corresponding the extremity of this kind of

section in which w / (q ) = 0 and ζ = q is double root for the polynom
Because ζ = q is double root for this polynom , we can write :

∑t ζ
4

k =0

k

k

(

= 1 − q ⋅ζ

) (a
2

0

139

)

+ a1ζ + a 2ζ 2 .

∑t ζ
4

k =0

k

k

.

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class
From the relation about the coefficients tk , k = 0,4 results that we can take

a0 = t 0 , a1 = −2kq , a2 = q 2 ⋅ t 4 .

The differential equation (10) can be write:

2

(11)

] (

[

)

2
1 − qζ (t 0 − 2kqζ + q 2 t 4ζ 2 )
 ζw′  f (− f − 2 pzl )w + f + pzlf
.
=

 ⋅
2
( f − w )2
 w 
(z − ζ )2 1 − zζ

(

2

)

4.After radical extraction in (11) we obtain:

[

]

(1 − qζ ) t 0 − 2kqζ + q 2 t 4ζ 2
f (− f − 2 pzl )w + f 2 + pzlf
dw =
.
w( f − w)
ζ ( z − ζ )(1 − zζ )

[

]

From double integration:

∫

w

(12)

0

ζ
(1 − qζ ) t0 − 2kqζ + q 2t4ζ 2
f (− f − 2 pzl )w + f 2 + pzlf
.
dw = ∫
w( f − w)
ζ ( z − ζ )(1 − zζ )
0

[

For calculation the integral from left side of (12) we denote:

I1 =

∫

]

f (− f − 2 pzl )w + f 2 + pzlf
dw .
w( f − w)

We observe that :

f (− f − 2 pzl ) ∫

w + a2
f 2 + pzlf
= a2.
dw where we denoted
− f − 2 pzl
w( f − w)
2
2
For the calculation of I 1 we make the substitution : w = u − a , dw = 2udu. We
I1 =

obtain I 1 =

f (− f − 2 pzl ) ∫

− 2u 2 du
,b2 = a + f .
(u 2 − a 2 )(u 2 − b 2 )

Observe that :

140

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

− 2u 2
2a 2
1
2b 2
1
=
⋅
−
⋅ 2
.
2
2
2
2
2
2
2
2
2
2
(u − a )(u − b ) b − a u − a
b − a u − b2
So:

I1 =

u−a
b
u − b
 a
f (− f − 2 pzl ) ⋅  2
ln
ln
− 2
2
2
u+a b −a
u + b 
b − a

or

f − (− f − 2 pzl )

I1 =

(13)

b2 − a2

where :

 u − a  a  u + b  b 
⋅ ln 
 ⋅
 
 u + a   u − b  

u = w + a2

.

For calculation the integral from right side of (12) we denote:

I2 = ∫

(1 − qζ ) t 0 − 2kqζ + q 2 t 4ζ 2

ζ ( z − ζ )(1 − zζ )

dζ .

We have: q 2 t 4ζ 2 − 2kqζ + t 0 = q 2 t 4 ⋅ (ζ − ζ 1 )(ζ − ζ 2 )

where ζ 1, 2 =

ζ1 =

δ

t4

k ± k 2 − t0t4

q and ζ 2 =

t4
t0

δ

with this denotations,

q. If we denotation k − k 2 − t 0 t 4 = δ observe that

q.
q 2 t 4 − 2kqζ + t 0 = q 2 t 4 (ζ −

δ
t4

q )(ζ −

calculate the integral I 2 make the substitution :
(14)

(ζ −

δ
t4

q )(ζ −

δ

t0

q ) = v(ζ −

From (14) obtain :

141

δ
t4

q) .

δ

t0

q ) . For the

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

ζ =σ ⋅

(15)

tt
v2 −α 2
δ
with σ = q and α 2 = 0 24 .
2
t4
δ
v −1

By an elementary calculation from (15) obtained successively:

(16)


σα 2 − z
v2 − β 2
2vσ (α 2 − 1)
2
ζ
ζ
σ
β
dv
z
z
d
=
−
=
−
⋅
=
,
(
)
cu
,

2
2
2
σ
z
−
v
v
−
−
(
1
)
1


v2 − δ 2
v2 − γ 2
1 − zσα 2
2
ζ
σ
ζ
σ
γ
q
q
cu
z
z
−
=
−
=
−
=
−
⋅
,
1
(
1
)
1
(
1
)

v2 −1
v2 −1
1 − zσ


t
δ
σ (1 − α 2 )v
1 − qσα 2
şi (ζ − q)(ζ − 0 q ) =
.
cu δ 2 =
δ
t4
v2 −1

1 − qσ

By using previous relations we obtain:

(17)

I2 =

2σq (1 − qσ )(1 − α 2 ) 2 t 4
(σ − z )(1 − zσ )

v 2 (v 2 − δ 2 )
∫ (v 2 − 1)(v 2 − α 2 )(v 2 − β 2 )(v 2 − γ 2 ) dv .

Let:

F (v ) =

v 2 (v 2 − δ 2 )
;
(v 2 − 1)(v 2 − α 2 )(v 2 − β 2 )(v 2 − γ 2 )

we are looking for a decomposition

(18)

F (v ) =

A
A
A
A
A
A1
A
A
+ 2 + 3 + 4 + 5 + 6 + 7 + 8 .
v −1 v +1 v −α v + α v − β v + β v − γ v + γ

From (1) we obtain the next values for the coefficients:

142

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class
not

1−δ 2
A
A
= τ1
=
=
2
 1
2
2
2
α
β
γ
2
(
1
)(
1
)(
1
)
−
−
−


α (α 2 − δ 2 )
A
A
=τ2
=
−
=
 3
4
2(α 2 − 1)(α 2 − β 2 )(α 2 − γ 2 )


β (β 2 − δ 2 )
A = −A =
= τ3
6
 5
2( β 2 − 1)( β 2 − α 2 )( β 2 − γ 2 )

γ (γ 2 − δ 2 )

A
A
=
−
=
=τ4 ⋅
8
2
2
2
2
2
 7
γ
γ
α
γ
β
2
(
1
)(
)(
)
−
−
−


(19)

We denote, µ =

2σq (1 − qσ )(1 − α 2 ) 2 ⋅ t 4
(σ − z )(1 − zσ )

; from (18) and (19) we obtain

for I 2 the expression:
(20)


v −1
v −α
v−β
v −γ 
+ τ 4 ln
+ τ 3 ln
+ τ 2 ln
I 2 = µ τ 1 ln
.
v +1
v +α
v+β
v + γ 


From the relation (14) observe that:

v(ζ ) =

(21)

ζ−

ζ−

δ
δ

t0

t4

q
,

v(0) = ±α

q

and from w = u 2 − a 2 we obtain:

(22)

u (ζ ) = w(ζ ) −

f 2 + 2 pzlf
, u (0) = ± a .
f + 2 pzl

With the relations (13) and (20) the relation (12) becomes:

143

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

I 1 0w = I 2

(12 / )

ζ
0

For ζ = 0 from (13), (20) and (12 / ) obtained the constant (which is obtained
from that two members of relations (13) and (20) corresponding to
(from left)): ln(−1)

a

f ( − f − 2 pzl )
b2 −a 2

+ µτ 2

u−a
v −α
from
u+a
v +α

, obtained in left member of equality (12 / ). Thus,

(12 / ) can be writhed:
f (− f − 2 pzl )
b2 − a2

+ ln (− 1)

 u (ζ ) − a  2

ln 
 u (ζ ) + a 

a f ( − f − 2 pzl )
b2 −a 2

+ µτ 2

=

 u (ζ ) + b 

⋅ 
 u (ζ ) − b 

2


+


τ2
  v(ζ ) − 1 τ 1
 v(ζ ) − β
 v(ζ ) − α 
 + ln
 + ln
= µ ln
  v(ζ ) + 1 
 v(ζ ) + β
 v(ζ ) + α 

 α − 1 τ 1
− µ ln 

 α + 1 

α − β
⋅ 
α + β





τ3

τ4
α −γ  
 .

 α + γ  

f ( − f − 2 pzl )
b2 − a2





τ3

a+b
ln

a−b

 v(ζ ) − γ
+ ln
 v(ζ ) + γ





τ4

b


−


With calculus the previsious equality can be writhed:
f ( − f − 2 pzl )

a
b
b2 −a 2


  a − u (ζ )   u (ζ ) + b a + b 
 ⋅
 
⋅
=

a + u (ζ )   u (ζ ) − b a − b  





(23) 
τ1
τ2
τ3
τ4 µ
 









−
−
−
+
−
−
ζ
α
ζ
ζ
β
α
β
ζ
γ
α
γ
v
v
v
v
(
)
1
(
)
(
)
(
)
= 







⋅
⋅
⋅
⋅
 .
  v(ζ ) + 1   α + v(ζ )   v(ζ ) + β α − β   v(ζ ) + γ α + γ  

 

where u (ζ ) and v(ζ ) is obtained by the relations (22) and (21).

144

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class
The relation (23) represent under the implicitly equation where is verify the
extremal function w = w(ζ ) , where realized max Re f ( z ).
f ∈S

II. We suppose that Im( zl + z m) = 0. In this case the expression of p, have
2

to zl − z l = 0 . How zl + z l = 0 implies zl = 0 . If l = 0( z = r > 0) then l = 0.

From (6) results that w(ζ ) have to verify the condition:

  f 2
Ree iψ 
−
  f − w

(24)

 w
f  /
 ζw

 
  ≤ 0 .
 

How ψ is arbitrary, real, results that w(ζ ) verify next differential equation:

(25)

or

f dw

w f −w

=

dζ

ζ


f
 =
.
f −w


 w
 /
 ζw

2

. From double integration:

∫w
w

(26)

f dw

0

f −w

=∫
ζ

0

dζ

ζ

.

With denotation
f − w = t , obtained w = f − t 2 , dw = −2tdt. The relation
(26) becomes (without limits of integration):

f (−2t )dt

∫ (f −t
ln

t−

t+

2

f
f

)⋅t

=∫

dζ

ζ

or

∫t

2 f

2

− ( f )2

= ln ζ or

145

dt = lnζ , from where

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

ln

(27)

f −w−

f −w+

f
f

w
0

= ln ζ .
ζ

0

After the calculus we obtained successively:


f
ln
f


and

ln f

f


−w − f

−w + f

− ln

f −w − f

f −w + f

0

= ln ζ − ln ζ

ζ =0

,

 ζ( f − w + f ) 2 
 ζ =0 = ln ζ
+ ln

−w
−w + f


−w − f


ln



from where:

and
(28)

f −

f +


⋅ 4 f  = ln ζ

f −w


f −w

f −

f +

f −w

f −w

=

ζ
4f

.

From (28) we obtained:
(29)

16 zf 2 ( z )
.
w( z ) =
(4 f ( z ) + z ) 2

How f(z) is considerate the extremal from (29) obtained after some calculus that

w( z ) =

z
. Observe that:
4

146

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class

Fig 2
and the condition Re zw / ( z ) = 0 implies x = 0( z = x + iy ) .

Then z = max Re w( z ) = 0 , so for z e = 0 and ∀ z > z e , any parallel to Ox

intersected f ( z = r ) in one point.

We exclude this ordinary case, it showing that the problem is true.
From I and II, result that the extremal function which is corresponding to the
extremal region Ω e from fig 1, has the implicitly form in equation (23).

5.Still remain to show how to determine the θ . For this we make in (11)

ζ → z; and after simplifications and by multiplication of equality from z l we
3

(

obtained:

)

r 3 (r 2 − 1)l l ⋅ p = (1 − qz ) 2 t 0 − 2kqz + q 2 ⋅ t 4 ⋅ z 2 ⋅ z ⋅ l
or :
(30)

3

r 3 (r 2 − 1)l l ⋅ p = ( z − qr 2 ) 2 (t 0 z l − 2kqlr 2 + q 2 t 4 l z ⋅ r 2 ) .

Because p is real, r 3 (r 2 − 1)l l p, is real from the relation (30) we obtain a
system with two equation and determinate θ and k :

(31)

[(

)

r 3 (r 2 − 1)l l p = Re z − qr 2 2 (t z l − 2kqlr 2 + q 2 t l zr 2 )
0
4



2
2
2
2
Im z − qr (t 0 z l − 2kqlr + q t 4 l zr ) = 0

[(

)

147

]

]

Miodrag Iovanov-Determining of an extremal domain for the functions
from the S-class
With θ and k determinated in this way the extremal function from equation
(27) is well determine and with its assisting we find z e = max Re f ( z ) with the

{

}

f ∈w

geometrical property enounced: in domain Ω e = z z > z e , any parallel to the Ox
axis intersect f ( z = r ) in one point (eventually in maximum one point).
References

[1].G.M. Goluzin, Geometriceskaia teoria complecsnogo peremenogo, MoscvaLeningrad, 1952.
[2].Petru T. Mocanu, O problemă extremală în clasa funcţiilor univalente,
Universitatea Babeş Bolyai-Cluj Napoca, Facultatea de Matematică, 1998.

Author:
Miodrag Iovanov, “Constantin Brâncuşi” University of Târgu-Jiu, Romania

148