MIXED BOUNDARY VALUE PROBLEM FOR A QUARTER-PLANE WITH A ROBIN CONDITION

MIXED BOUNDARY VALUE PROBLEM FOR A

QUARTER-PLANE WITH A ROBIN CONDITION

A. Aghili

*

Department of Mathematics, Faculty of Sciences, Gilan University, Rasht, P.O. Box 1841, Islamic Republic of Iran

Abstract

We consider a mixed boundary value problem for a quarter-plane with a Robin

condition on one edge. We have developed two procedures, one based on the

advanced theory of dual integral equations and the other, in our opinion simpler

technique, relying on conformal mapping. Both of the procedures are of interest,

because the former may be easier to adapt to other boundary value problems.

Keywords: Mixed boundary value problems; Dual integral equations; Fredholm’s integral

equation of second kind

* _{E-mail: [email protected]}

Introduction

Problem. Find the solution of the following mixed

boundary value problem.

P.D.E: ∆Φ = 0, x > 0, y > 0.

B.C. 1) Φx(0,y) + h(y) Φ(0,y) = q(y), y > 0.

2) Φy(x,0) = f(x), 0 < x < 1.

3) Φ(x,0) = 0, x > 1.

Note. Our strategy is to create a sequence of tractable

boundary value problems leading to standard integral equations which may be solved numerically.

Solution. We first find the solution to our auxiliary

problem by replacing the Robin condition by Φx(0,y) =

) (y

θ , where θ(y) is an admissible boundary function. Let Φ(x,y) be the desired function, then Φ(x,y)

has the following integral representation:

. ) cos( ) exp( ) (

) sin( ) exp( ) ( ) , (

0 0

∫

∫

∞ ∞

− +

− =

Φ

ξ ξ ξ ξ

ξ ξ ξ ξ

d x y B

d y x A

y x

(1.1)

Let

Φ ∂

∂

x (0,y) =

∫

. (1.2)

∞

= −

0

) ( ) sin( )

( y d y

Aξ ξ ξ θ

From Fourier’s inversion-formula for the sine transform

∫

∞ − = 0 ) sin( ) ( 2 )

( s s ds

A θ ξ

π ξ

ξ . (1.3)

From (1.3) we obtain the following equation:

ds x s s s d x A

∫

∫

∞ ∞ + − = − 0 2 2 0 ) ( 2 ) exp( ) ( θ π ξ ξ ξ

ξ . (1.4)

The conditions on y = 0 lead us to the pair of dual integral equations

∫

∫

∞ ∞ = − − 0 0 ) ( ) cos( ) ( ) exp( ) ( x f d x B d x A ξ ξ ξ ξ ξ ξ ξ ξ

, 0<x<1, (1.5)

∫

∞ = 0 0 ) cos( )

(ξ xξ dξ

B , x>1, (1.6)

The equation (1.6) is satisfied by [1,3,4]:

∫

=

1

0

0( )

) ( 2 )

( tg t J t dt

B ξ

π

ξ ,

where g (t) is determined to satisfy (1.5). Thus (1.5) may be rewritten as

) ( ) ( 2 ) sin( ) ( 0 2 2 0 x f ds x s s s d x B dx d = + − −

∫

∫

∞ ∞ θ π ξ ξ ξ

, 0<x<1. (1.7)

Following Sneddon’s elementary solution [4], we have

∫

∫

−∞ ₊ − − =

0 2 2

0 2 2

) ( ) ( ) ( x s ds s x t dx x f t g t _θ , (1.8)

∫

∫

∞ ∞ − − = Φ 0 0 ) exp( ) ( ) sin( ) ( ) , 0

( y Aξ yξ dξ Bξ yξ dξ, (1.9)

from (1.3) and [2]

. ) ( 2 ln ) ( 1 ) , 0 ( 1

0 2 2

0

∫

∫

+ + − + − = Φ ∞ dt y t t tg ds s y s y s y π θ π (1.10)

Now we turn to the Robin condition, which gives us the equation . ] ) ( 2 ln ) ( 1 )[ ( ) ( ) ( 1

0 2 2 0

∫

∫

+ + − + − − = ∞ dt y t t tg ds s y s y s y h y q y π θ π θ (1.11)

Further simplification is possible. We have

. ] 1 1 )[ln ( ] ) )( ( [ ) ( ) ( 0 2 2 1 2 2 2 2 1 0 1

0 2 2

ds s y s y s dx y t t s tdt x f dt y t t tg x

∫

∫

∫

∫

∞ + + + + − + + − = + θ

Thus, θ(y) is determined by the Fredholm’s integral equation of the second kind:

, 0 ) ( 1 1 ln ) ( ) ( ) (

0 2 2

where

dx y x

y x x

f y h y q y

F

∫

+ + −

+ −

=

1

0 2 2

2 2

1 1

ln ) ( 2 ) ( ) ( ) (

π .

Putting the pieces together, θ(y) determines g(t) which in turn allows us to compute

) , (x y

Φ .

Note. In the above integral equation we used the

following Integrals:

a)

, } ) )( (

{ ) (

) (

1

0

1

2 2 2 2

0 2 2

1

0 2 2

∫

∫

∫

∫

+ −

= − +

x t

y t x t

tdt dx

x f

x t

dx x f y t

tdt

but the inner integral of right side can be written after changing in variables:

α α 2 2

2

2 _{=x cos +sin}

t

and

α α

α d

x

dt₌(1₋ 2)sin cos

. sin ) 1 ( ) (

cos 1

) )( (

2

0 2 2 2 2

2 1

2 2 2 2

α α α π

d x

y x

x y t x t

tdt x

∫

∫

− + +

−

= + −

Using another change of variable of the form

u

x =

− sinα

1 2 _{one gets,}

. 1 1

ln

) (

) )( (

2 2

2 2 1

0 2 2 2

1

2 2 2 2

2

y x

y x

u y x

du y

t x t

tdt x

x

+ + −

+ −

= + + =

+

−

∫

∫

−

b)

= + +

∫

1

0 (t2 y2)(t2 s2) tdt

∫

₋

− + +

1

0 2 2 2 2 2 2₎2

2 ( ) 2

(t y s y s

tdt

.

Let

u s y s y t

2 2

2 2 2 2

2₊ + ₌ −

. 1 1

ln

) 2 ( ) 2 (

2 2

1

0 2 2 2 2 2 2 2

s y

s y

s y s

y t

tdt

+ + + +

= − − + +

∫

Next we show how the conformal mapping w = 2sin-1_{z may be gainfully used to solve the problem.}

Let

w = 2sin-1_{z, x + iy =}

2 sinu+iv

implies that,

x =

2 cosh 2

sinu v, y =

2 sinh 2 cosu v

Let

) , (x y

The boundary conditions change to:

0 ) ,

( =

ΔΨu v

1- )

2 (sinh )

, 0 ( ) 2 (sinh sec

) , 0

( v hv h v v q v

u + Ψ =

Ψ ,

2-

2 cos ) 2 (sinh sec

) 0 ,

(u hv f u u

v =

Ψ , 0 < u < π,

3- Ψ(π,v)=0, v > 0,

We formulate an auxiliary problem for Laplace’s equation

0 ) ,

( =

ΔΨu v

1- Ψu(0,v)=θ(v), v > 0,

2-

2 cos ) 2 (sinh )

0 ,

(u f u u

v =

Ψ , 0 < u < π,

3- Ψ(π,v)=0, v > 0.

Since we have a non-homogenous boundary condition, has the following integral representation:

) , (u v

Ψ

. ) cos( ) sinh( ) (

) 2 1 cos( ] ) 2 1 ( exp[ )

, (

0 1

∫

∑

∞ ∞

=

− +

− −

− =

Ψ

ξ ξ ξ π

ξ u v d

c

u n v n a

v u

n

n (1.12)

From boundary conditions (1) and (2) we obtain

) ( )

cos( ) sinh( ) (

0

v d

v u

cξ π− ξ ξ ξ =−θ

∫

∞

, (1.13)

2 cos ) 2 (sinh )

2 1 cos( ) 2 1 (

1

u u f u n a n

n

n − =

−

∑

∞

=

, (1.14) or,

udu n

u u f a

n _n )

2 1 cos( 2 cos ) 2 (sin 2 )

2 1 (

0

− −

=

−

∫

π

π . (1.15)

From the above relation one gets

udu n

u u f n

a_n )

2 1 cos( 2 cos ) 2 (sin ) 1 2 (

4

0

− −

−

=

∫

π

π . (1.16)

Then

∫

∑

∞ ∞

=

+

− − =

Ψ

0 1

) cos( sinh ) (

] ) 2 1 ( exp[ )

, 0 (

ξ ξ πξ ξ

ν

d v c

v n a

n n u

, (1.17)

and,

=

Ψ_u(0,v) ( )cosh cos( ) ( )

0

v d v

cξ πξ ξ ξ =θ

−

∫

∞

. (1.18)

We need to express Ψ(0,v) in terms of θ and f. Therefore from (1.18) we get

ds s s

c

∫

∞

− =

0

) cos( ) ( 2 cosh

)

( θ ξ

π πξ ξ

ξ . (1.19)

The above relation yields:

. cosh

sinh cos cos ) ( 2

) cos( sinh ) (

0 0

0

∫

∫

∫

∞ ∞

∞

−

=

ξ πξ

ξ

πξ ξ

ξ θ

π

ξ νξ πξ ξ

d v

s ds s

d c

(1.20)

follows [2]:

)] 4 coth( ) 4 log[coth( 1

)] ( cos ) ( [cos tanh 1

0

s v s v

d s v s

v

− +

−

= − +

+

∫

∞

π

ξ ξ

ξ ξ

πξ π

(1.21)

As in previous case, the Robin condition will give a Fredholm’s integral equation of the second kind with a complicated kernel that can be solved

) 2 (sinh )

, 0 ( ) 2 (sinh sec

)

(v hv+h v Ψ v =q v

θ (1.22)

with,

, ) cos( sinh ) (

] ) 2 1 ( exp[ )

, 0 (

0 1

∫

∑

∞ ∞

=

+

− − =

Ψ

ξ ξ πξ ξ

ν

d v c

v n a

n n

=

∫

∞

0

) cos( sinh )

(ξ πξ vξ dξ c

ds s v s v

s ]

4 coth 4 coth )[log ( 2

0

− +

∫

∞

θ ,

udu n

u u f n

a_n )

2 1 cos( 2 cos ) 2 (sin ) 1 2 (

4

0

− −

−

=

∫

π

π .

References

1. Aghili A. Doctoral Dissertation, State University of New York, Stony Brook, New York (1999).

2. Erdelyi A., Magnus W., Oberhettinger F. and Tricomi

F.G. Tables of Integral Transforms. Vol. I, McGraw Hill Book Company Inc. (1954).

3. Melrose G. Triple trigonometric series and their

application to Mixed Boundary value problems. Ph.D. Thesis, Old Dominion University, Norfolk, Virginia (1984).