Schwarz and Poisson formulas

Schwarz and Poisson formulas

Introduction

Fundamental boundary-value problems of

potential theory

, i.e. (

http://planetmath.org/Ie

), the so-called Dirichlet and

Neumann problems

occur in

many of applied mathematics such as hydrodynamics, elasticity and electrodynamics. While solving the two-dimensional problem for special of boundaries is likely to present serious computational difficulties, it is possible to write down

formulas

for a circular (

http://planetmath.org/Circle

) boundary. We shall give

Schwarz and Poisson formulas

that solve the

Dirichlet problem

for a circular domain.

Schwarz formula

Without loss of generality, we shall consider the

compact

disc

D

¯

:

|

z

|

≤

1

in the

z

-

plane, its boundary will be denoted by

γ

and any point on this one by

ζ

=

e

i

⁢

θ

. Let it be required to determine a

harmonic function

u

⁢

(

x

,

y

)

, which on the boundary

γ

assumes the values

u

|

γ

=

f

⁢

(

θ

)

,

(1)

where

f

⁢

(

θ

)

is a

continuous

single-valued function

of

θ

. Let

v

⁢

(

x

,

y

)

be the

conjugate harmonic function

which is determined to within an arbitrary constant from the knowledge of the

function

u

.

1

1

Since

u

+

i

⁢

v

is an

analytic function

of

z

=

x

+

i

⁢

y

,it is clear from the

Cauchy-Riemann equations

that the function

v

⁢

(

x

,

y

)

is determined by

v

⁢

(

x

,

y

)

=

∫

z

0

z

∂

⁡

v

∂

⁡

x

⁢

𝑑

x

+

∂

⁡

v

∂

⁡

y

⁢

d

⁢

y

=

∫

z

0

z

-

∂

⁡

u

∂

⁡

y

⁢

d

⁢

x

+

∂

⁡

u

∂

⁡

x

⁢

d

⁢

y

,

where the

integral

is evaluated over an arbitrary path joining some point

z

0

with an arbitrary point

z

belonging to the

unitary

open disc

D

. We are concerned to a simply connected domain, so that the function

v

⁢

(

x

,

y

)

will be

single-valued

.

Then the function

w

⁢

(

z

)

=

u

⁢

(

x

,

y

)

+

i

⁢

v

⁢

(

x

,

y

)

is an analytic function for all values of

z

∈

D

. We shall suppose that

w

⁢

(

z

)

∈

C

⁢

(

D

¯

)

the class of

continuous functions

. Therefore, we may write the boundary condition (1) as

w

⁢

(

ζ

)

+

w

¯

⁢

(

ζ

¯

)

=

2

⁢

f

⁢

(

θ

)

o

⁢

n

⁢

γ

.

(2)

We define here

w

¯

⁢

(

ζ

)

=

w

⁢

(

ζ

¯

)

¯

and

w

¯

⁢

(

ζ

¯

)

=

w

⁢

(

ζ

)

¯

. Next, we multiply (2) by

1

2

⁢

π

⁢

i

⁢

d

⁢

ζ

ζ

-

z

and, by integrating over

γ

, we obtain

1

2

⁢

π

⁢

i

⁢

∫

γ

w

⁢

(

ζ

)

ζ

-

z

⁢

𝑑

ζ

+

1

2

⁢

π

⁢

i

⁢

∫

γ

w

¯

⁢

(

ζ

¯

)

ζ

-

z

⁢

𝑑

ζ

=

1

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

,

(3)

which, by Harnack’s

theorem

, is to (2). Notice that the first integral on the left is equal to

w

⁢

(

z

)

by Cauchy’s integral formula, and for the same reason

2

2

From Taylor’s formula

w

⁢

(

z

)

=

w

⁢

(

0

)

+

w

′

⁢

(

0

)

⁢

z

+

1

2

!

⁢

w

′′

⁢

(

0

)

⁢

z

2

+

O

⁢

(

z

3

)

.

But on

γ

,

z

¯

=

1

/

ζ

, so

w

¯

⁢

(

ζ

¯

)

=

w

¯

⁢

(

0

)

+

w

¯

′

⁢

(

0

)

⁢

1

ζ

+

1

2

!

⁢

w

¯

′′

⁢

(

0

)

⁢

1

ζ

2

+

O

⁢

(

1

ζ

3

)

and term-by-term integration gives the desired result recalling that

1

2

⁢

π

⁢

i

⁢

∫

γ

d

⁢

ζ

ζ

n

⁢

(

ζ

-

z

)

=

{

1

,

i

⁢

f

⁢

n

=

0

,

0

,

o

⁢

t

⁢

h

⁢

e

⁢

r

⁢

w

⁢

i

⁢

s

⁢

e

.

the second one is equal to

w

¯

⁢

(

0

)

. Let

w

¯

⁢

(

0

)

=

a

-

i

⁢

b

, thus (3) becomes

w

⁢

(

z

)

=

1

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

-

a

+

i

⁢

b

.

(4)

By setting

z

=

0

in (4), we get

a

+

i

⁢

b

=

1

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

ζ

⁢

𝑑

ζ

-

a

+

i

⁢

b

,

whence

2

⁢

a

=

1

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

ζ

⁢

𝑑

ζ

=

1

π

⁢

i

⁢

∫

0

2

⁢

π

f

⁢

(

θ

)

⁢

𝑑

θ

.

(5)

As one would expect,

b

is left undetermined because the conjugate harmonic function

v

⁢

(

x

,

y

)

is determined to within an arbitrary real constant. Finally we substitute

a

from (5) in (4),

w

⁢

(

z

)

=

1

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

ζ

-

z

⁢

𝑑

ζ

-

1

2

⁢

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

ζ

⁢

𝑑

ζ

+

i

⁢

b

=

1

2

⁢

π

⁢

i

⁢

∫

γ

f

⁢

(

θ

)

⁢

ζ

+

z

ζ

-

z

⁢

d

⁢

ζ

ζ

+

i

⁢

b

,

(6)

the aimed Schwarz formula.

3

3

It is possible to prove that, if

f

⁢

(

θ

)

satisfies

Hölder condition, then the function

w

⁢

(

z

)

given by (6) will be continuous in

D

¯

. Such a condition is less restrictive than the requirement of the

existence

of a

bounded

derivative

.

Poisson formula

If we substitute

z

=

ρ

⁢

e

i

⁢

ϕ

and

ζ

=

e

i

⁢

θ

in (6) and separate the real and

imaginary parts

, we find

ℜ

⁡

w

⁢

(

z

)

≡

u

⁢

(

ρ

,

ϕ

)

=

1

2

⁢

π

⁢

∫

0

2

⁢

π

(

1

-

ρ

2

)

⁢

f

⁢

(

θ

)

1

-

2

⁢

ρ

⁢

cos

⁡

(

θ

-

ϕ

)

+

ρ

2

⁢

𝑑

θ

.

(7)

This is the Poisson formula (so-called also

Poisson integral

), which gives the solution of Dirichlet problem. It is possible to prove that (7) also the solution under the

assumption

that

f

⁢

(

θ

)

is a

piecewise continuous

function.

4

4

See

[

1

]

.

It is also possible to generalize the formulas obtained above so as to make them apply to any simply connected region. This is done by introducing a

mapping function

and the idea of

conformal mapping

of simply connected domains.

5

5

For a discussion of Neumann problem, see

[

2

]

.

References

1

O. D. Kellog,

Foundations

of Potential Theory

, Dover, 1954.

2

G. C. Evans,

The Logarithmic

Potential

, Chap. IV, New York, 1927.

Title

Schwarz and Poisson formulas

Canonical name

SchwarzAndPoissonFormulas

Date of creation

2013-03-22 16:05:58

Last modified on

2013-03-22 16:05:58

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

12

Author

perucho (2192)

Entry type

Theorem

Classification

msc 30D10

Defines

Schwarz formula

Defines

Poisson formula

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