Harnack theorem

Harnack theorem

Introduction.

It is frequent to make use of

Cauchy integral formula

to represent analytically some

functions

that are useful in mathematical physics applications. However, it must be noted that such representation is not unique, so that the same function can be represented by different

integrals

of Cauchy’s type. An important case has to do with the equality of the two Cauchy integrals

1

2

⁢

π

⁢

i

⁢

∮

C

ψ

1

⁢

(

ζ

)

ζ

-

z

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

i

⁢

∮

C

ψ

2

⁢

(

ζ

)

ζ

-

z

⁢

𝑑

ζ

for all values of

z

in the interior of

C

. In general no conclusion can be drawn concerning the equality of the

density functions

ψ

1

⁢

(

ζ

)

and

ψ

2

⁢

(

ζ

)

. We shall see, however, that if some additional restriction are imposed on the density functions and on the

contour

C

, then the equality will occur. That is the matter of

Harnack theorem

. In considering the applications of the theory of functions of a complex variable to problems in continuum mechanics, for instance, we shall most frequently deal with the region

bounded

by the

unit circle

, i.e. the compact disc

|

z

|

≤

1

that we shall draw in the

z

-plane, its boundary will be denoted by

γ

and the points of

γ

by

ζ

=

e

i

⁢

θ

. All density functions of the

argument

θ

will be assumed to be

2

⁢

π

-

periodic

.

Theorem 1

.

1

1

A less restrictive form of Harnack theorem is discussed in

[

1

]

.

Let

f

⁢

(

θ

)

and

g

⁢

(

θ

)

be

continuous

real functions

of the argument

θ

defined on the boundary

γ

; if

1

2

⁢

π

⁢

i

⁢

∮

γ

f

⁢

(

θ

)

⁢

d

⁢

ζ

ζ

-

z

=

1

2

⁢

π

⁢

i

⁢

∮

γ

g

⁢

(

θ

)

⁢

d

⁢

ζ

ζ

-

z

(1)

then

f

⁢

(

θ

)

≡

g

⁢

(

θ

)

i

⁢

f

|

z

|

<

1

f

(

θ

)

=

g

(

θ

)

+

c

o

n

s

t

.

i

f

|

z

|

>

1

.

Proof.

1.

|

z

|

<

1

. From equality (1) we obtain

1

2

⁢

π

⁢

i

⁢

∮

γ

f

⁢

(

θ

)

-

g

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

≡

1

2

⁢

π

⁢

i

⁢

∮

γ

h

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

≡

0

,

where

h

⁢

(

θ

)

≡

f

⁢

(

θ

)

-

g

⁢

(

θ

)

. We shall prove that

h

⁢

(

θ

)

≡

0

. Since

|

z

|

<

1

we can write,

1

ζ

-

z

=

∑

n

=

0

∞

z

n

ζ

n

+

1

,

then

1

2

⁢

π

⁢

i

⁢

∮

γ

h

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

i

⁢

∮

γ

∑

n

=

0

∞

h

⁢

(

θ

)

ζ

n

+

1

⁢

z

n

⁢

d

⁢

ζ

=

1

2

⁢

π

⁢

i

⁢

∑

n

=

0

∞

(

a

n

-

i

⁢

b

n

)

⁢

z

n

,

(2)

the

complex form

of

Fourier series

2

2

The restrictions imposed upon the

expanded

function are known as the

Dirichlet conditions

, but it is sufficient to demand that it be a

function of bounded variation

.

where the coefficients are given by (we are using the Euler’s formula)

a

n

-

i

⁢

b

n

=

1

2

⁢

π

⁢

i

⁢

∮

γ

h

⁢

(

θ

)

ζ

n

+

1

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

∫

0

2

⁢

π

h

⁢

(

θ

)

⁢

e

-

i

⁢

n

⁢

θ

⁢

𝑑

θ

.

But (2)

vanishes

for all values of

z

, therefore

a

n

=

b

n

=

0

,

n

∈

ℕ

and a reference to Fourier complex expansion

h

⁢

(

θ

)

=

∑

n

=

-

∞

∞

c

n

⁢

e

i

⁢

n

⁢

θ

,

c

n

=

1

2

⁢

π

⁢

∫

0

2

⁢

π

h

⁢

(

t

)

⁢

e

-

i

⁢

n

⁢

t

⁢

𝑑

t

,

n

∈

ℤ

,

shows that all Fourier coefficients of the function

h

⁢

(

θ

)

vanish, and hence

h

⁢

(

θ

)

≡

0

.

2.

|

z

|

>

1

. By

analytic continuation

, we have

1

ζ

-

z

=

∑

n

=

0

∞

-

ζ

n

z

n

+

1

,

so that

1

2

⁢

π

⁢

i

⁢

∮

γ

h

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

=

-

1

2

⁢

π

⁢

i

⁢

∮

γ

∑

n

=

1

∞

ζ

n

-

1

⁢

h

⁢

(

θ

)

z

n

⁢

d

⁢

ζ

=

-

∑

n

=

1

∞

a

n

+

i

⁢

b

n

z

n

,

(3)

where

a

n

+

i

⁢

b

n

=

1

2

⁢

π

⁢

i

⁢

∮

γ

ζ

n

-

1

⁢

h

⁢

(

θ

)

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

∫

0

2

⁢

π

h

⁢

(

θ

)

⁢

e

i

⁢

n

⁢

θ

⁢

𝑑

θ

,

n

∈

ℕ

\

{

0

}

.

Since (3) vanishes for all values of

|

z

|

>

1

,

a

n

=

b

n

=

0

,

n

∈

ℕ

\

{

0

}

. Thus, all Fourier coefficients of

h

⁢

(

θ

)

, with the possible exception of

a

0

, vanish and hence

g

⁢

(

θ

)

=

f

⁢

(

θ

)

+

c

⁢

o

⁢

n

⁢

s

⁢

t

.

∎

Moreover, from this theorem if

|

z

|

>

1

and in addition to (1) we have the equality

1

2

⁢

π

⁢

i

⁢

∮

γ

f

⁢

(

θ

)

ζ

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

i

⁢

∮

γ

g

⁢

(

θ

)

ζ

⁢

𝑑

ζ

,

then

f

⁢

(

θ

)

=

g

⁢

(

θ

)

.

Corollary 1

.

Given the continuous real functions

f

1

,

f

2

,

g

1

,

g

2

and the following simultaneous equalities for all values of

z

.

1

2

⁢

π

⁢

i

⁢

∮

γ

f

1

+

i

⁢

f

2

ζ

-

z

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

i

⁢

∮

γ

g

1

+

i

⁢

g

2

ζ

-

z

⁢

𝑑

ζ

,

1

2

⁢

π

⁢

i

⁢

∮

γ

f

1

-

i

⁢

f

2

ζ

-

z

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

i

⁢

∮

γ

g

1

-

i

⁢

g

2

ζ

-

z

⁢

𝑑

ζ

,

then

g

1

=

f

1

,

g

2

=

f

2

,

i

⁢

f

|

z

|

<

1

,

and

g

1

=

f

1

+

c

o

n

s

t

.

,

g

2

=

f

2

+

c

o

n

s

t

.

,

i

f

|

z

|

>

1

.

By adding and substracting those equalities, this corollary follows from Harnack theorem.

References

1

N. I. Muskhelishvili’s,

Singular Integral Equations,

p.64, 1953.

2

E.C. Titchmarsh,

The Theory of Functions,

Oxford University Press, New York, 2d ed., pp. 64-101, 399-428.

3

W.F. Osgood,

Lehrbuch der Funktionentheorie

, Teubner Verlagsgesellschaft, Leipzig, vol. 1.

4

Ãâ°. Goursat,

Course d’analyse,

Gauthiers-Villars & Cie, Paris, vol. 2.

5

E. Picard,

Leçons sur quelques types simples d’ÃÂ©quations aux dÃÂ©rivÃÂ©es partielles,

Gauthiers-Villars & Cie, Paris.

Title

Harnack theorem

Canonical name

HarnackTheorem

Date of creation

2013-03-22 16:02:55

Last modified on

2013-03-22 16:02:55

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

8

Author

perucho (2192)

Entry type

Theorem

Classification

msc 30D10

Generated on Fri Feb 9 20:17:37 2018 by

LaTeXML