Boundary Element Method for Heat Transfer in a Buried Pipe ...

University of Reading School of Mathematics, Meteorology & Physics

Boundary Element Method for Heat Transfer in a Buried Pipe by Elena Panti August 18, 2008

.. This dissertation is submitted to the Department of Mathematics in partial fulfilment of the requirements for the degree of Master of Science

1

Acknowledgements

Firstly, I would like to thank my supervisors Dr. Peter K .Sweby and Dr. Steve Langdon for their much appreciated help and their endless generosity and patience. Also, I would like to acknowledge Mr.Chuk Ovuworie from Schlumberger Company who gave me the opportunity to study this subject. Secondly, a big thank you to all the teaching staff in the Maths Department and Mrs. Sue Davis for their support all over the year. I also want to thank all my friends and my family who have supported me during this year.

Declaration

I confirm that this is my own work and the use of all material from other sources has been properly and fully acknowledged.

Elena Panti.

2

Abstract

In this dissertation we explore the Boundary Element Method for heat transfer in a buried pipe. We are interested in modelling the steady-state heat transfer from buried pipes. We are studying the temperature through Laplace’s equation. First, we consider the interior and the exterior problem and then we move on to the full pipe problem. In the interior problem we solve the problem inside a circle. In the exterior problem we solve the problem outside the bounded domain and because the domain is a circle therefore we solve the problem outside the circle. For the full pipe problem we solve the problem outside the circle in a half plane. The boundary integral method tells us the value of the temperature on the pipe and on the ground surface. From there we can deduce the temperature anywhere below the ground surface. The theoretical results are supported by our numerical results.

3

Contents

1 Introduction 4 2 Boundary Element Method 10 2.1 Reformulation of a Partial Differential Equation as a Boundary Integral Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1.1 Bounded Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.1.2 Unbounded Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17 2.1.3 Full pipe Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19 2.1.4 Neumann Green’s function for a half plane Problem. . .. . . . .20 3 Methods for solving a single integral equation 25 3.1 Nyström Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..25 3.1.1 Example of a single periodic integral equation . . . . . . . . . . . . 28 3.1.2 Example of a single non-periodic integral equation . . . . . . . . .31 3.2 Collocation Method. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . ..33 4 Interior Problem for Laplace’s equation 35 4.1 Separation of variables in polar coordinates. . . . . . . . . . . . . . . . . . 43 5 Exterior Problem for Laplace’s equation 49 6 Full pipe flow Problem 53 7 Summary and Conclusions 61 Bibliography 62

4

Chapter 1

Introduction

This dissertation explores the Boundary Element Method for the Heat Transfer in a buried pipe. Heat transfer occurs due to temperature difference between the pipeline fluid and the ambient fluid which is air or water, overcoming thermal resistances offered by the pipe, coatings and ground. The state of the fluid (oil, liquid or gas) i.e. the density and the viscosity of the fluid, is defined by the temperature.

Figure1: Buried pipe (source: Partially Buried Pipe Heat Transfer (Powerpoint), Chuk Ovuworie from Schlumberger Company)

pir =radius of the pipe inside

por =radius of the pipe outside

cor = radius of the coating outside

aT =temperature of the ambient fluid

fT =temperature of the pipeline fluid

pR =resistance of the pipe

cR =resistance of the coatings

gR =resistance of the ground

z=distance from the centre of the pipe to the ground surface

5

We can express, at steady state, the rate of heat flow between the pipeline and the ambient fluid as:

)(2 afpopo TTUrQ −−= π

where

poU = overall heat transfer coefficient

porπ2 = pipe outside surface area

af TT − = temperature difference between the pipeline fluid and

the ambient fluid

At steady state the rate of heat flow (Q ) is the same through each of the

thermal layers.

We can also write Q across each thermal layer as:

Q =R

T∆

where

T∆ =temperature difference across the layer R= thermal resistance offered by the layer

The temperature difference between the pipeline fluid and inside wall is:

pipi

pifhr

QTT

π2−=− (1.1)

where

pih = the pipe inside film coefficient

pirπ2 = pipe inside surface area

fT =temperature of the pipeline fluid

piT =temperature of the pipe inside

6

The temperature difference between the pipe inside and outside walls is given by the following equation:

p

pipo

popik

rrTT

π2

)ln(=−

where

=pir radius of the pipe inside

=por radius of the pipe outside

pk =the pipe thermal conductivity

piT = the temperature of the pipe inside

=poT the temperature of the pipe outside

The easiest way to approach this problem is to assume radial symmetry. In this case, simplifying the problem within the pipe to one-dimension, dependent only on r we can easily solve the boundary value problem using traditional techniques. Below are the assumptions made in this analysis: Assumptions:

• Heat flow, denoted by Q (radial heat flow per length of pipe), is radially symmetric within the pipe and the coatings such that T=T(r).

• Heat flow is in a steady state (dQ/dt = 0)

• Conservation of heat energy reduces down to Laplace’s equation

01

2

22 =

∂∂

+∂∂

=∇r

T

rr

TT It is Laplace’s equation, but now

the θ∂

∂Tterm has disappeared.

(where r, θ are polar coordinates on the centre of the pipe).

7

Solving Laplace’s equation without the θ∂

∂T term, we obtain the general

solution for temperature:

BrArT += ln)( (1.2)

where A ,B are constants.

We do not need to worry that temperature relies on rln since we know the

temperature in the centre of the pipe already ( fT ) and thus we will never have

to compute the temperature at r equal to zero, where the solution breaks down.

For the pipe layer, the boundary conditions are:

• ( )

p

pifpi

rr k

TTh

r

T

pi

−−=

∂∂

=

• popo TrT =)(

We know that BrArT += ln)( .Then by using the boundary

conditionpopo TrT =)( we have :

popopo TBrArT =+= ln)( (1.3)

Now we want to find the constants A and B, using the boundary conditions.

r

A

r

TBrArT =

∂∂

⇒+= ln)(

pirr r

A

r

T

pi

=∂∂

⇒=

and by using the 1st boundary condition

( )p

pifpi

rr k

TTh

r

T

pi

−−=

∂∂

=

8

we have:

p

pifpi

pi k

TTh

r

A )( −−=

p

pifpipi

k

TThrA

)( −−=⇒ (1.4)

From (1.1) we have:

π2

)(Q

hrTT pipipif =−− (1.5)

Substitute (1.5) into (1.4) we have:

pk

QA

π2=

From (1.3) po

p

popopo rk

QTrATB ln

2ln

π−=−=

Finally, substitute A and B in (1.2) ,we have the following solution for the pipe layer:

po

pop

Tr

r

k

QrT +

= ln

2)(

π

The advantage of this method is that it relies on simple and easy to solve Ordinary Differential Equations. The problem with it, is that it over simplifies the situation, and does not take into account what is happening exterior to the pipe coating. In particular if we consider the fact that the pipe is buried at a finite depth, or even only partially buried, then the solution will certainly not be radially symmetric. In this dissertation we consider this more complicated case. The temperature of the ground with increasing depth (discounting the presence of the pipe) is given by:

( ) gTgyyT += .This is an asymptotic condition as y ±∞→∞→ x, . (1.6)

9

In this dissertation we begin by introducing the Boundary Element Method and we separate our problem in three stages: (1) Interior problem (bounded problem)

(2) Exterior problem (unbounded problem) (3) Full problem.

In the interior problem we will solve the problem inside a circle. In the exterior problem we will solve the problem outside a circle. For the full pipe problem we will solve the problem outside the circle in a half plane. In this dissertation, we will approach analytical and numerical solutions for each problem.

10

Chapter 2 Boundary Element Method

The Boundary Element Method (BEM) has been applied to a variety of heat transfer problems in the last thirty years. Initial applications of the method were for steady heat conduction problems described by Laplace’s equation [9]. The BEM is a method for solving Partial Differential Equations by reformulating as boundary integral equations and then solving them. Moreover, the boundary element method is derived through the discretisation of an integral equation that is mathematically equivalent to the original partial differential equation. The essential re-formulation of the PDE that underlies the BEM consists of an integral equation that is defined on the boundary of the domain and an integral that relates the boundary solution to the solution at points in the domain. The Boundary Element Method is often referred to as the Boundary Integral Method (BIM) or Boundary Integral Equation Method.

The advantages in the BEM arise from the fact that only the boundary (or boundaries) of the domain of the PDE requires sub-division. Thus, the dimension of the problem is effectively reduced by one, for example an equation governing a three-dimensional region is transformed into one over its surface. In cases where the domain is exterior to the boundary the extent of the domain is infinite and hence the advantages of the BEM are even more remarkable; the equation governing the infinite domain is reduced to an equation over the finite boundary.

2.1 Reformulation of a Partial Differential Equation as a Boundary Integral Equation

The main properties of potential functions ( 02 =∇ φ ) can be derived from

Gauss’ Theorem (divergence theorem) and its corollaries (Green’s identities).

(The partial differential operator, 2∇ or ∆ is called the Laplace operator, or just the

Laplacian).

11

Gauss Theorem [9]:

Let V be a region in space bounded by a closed surface S and F be a vector field acting on this region. The divergence theorem establishes that the total flux of the vector field F across the closed surface S must be equal to the volume integral of the divergence of this vector:

dVx

FdSnF

V i

i

S

ii ∫∫ ∂∂

=

where in =components of the unit vector normal to the surface S

Green’s first identity:

By substituting

i

ix

F∂∂

=ψ

φ into Gauss’ Theorem, we have:

dVxx

dSnx iV iS

i

i

∂∂

∂∂

=∂∂

∫∫ψ

φψ

φ (2.1)

Then we use the chain rule which give us:

ψφψφψ

φ 2∇+∂∂

∂∂

=

∂∂

∂∂

iiii xxxx (2.2)

Substitute (2.2) into the right hand side of (2.1) we get:

∫∫∫ ∇+∂∂

∂∂

=∂∂

VV iiS

dVdVxx

dSn

ψφψφψ

φ 2 (2.3)

This is called the Green’s first identity.

12

Green’s second identity:

The Green’s first identity is also valid when interchanging φ and ψ :

∫∫∫ ∇+∂∂

∂∂

=∂∂

VV iiS

dVdVxx

dSn

φψφψφ

ψ 2 (2.4)

Subtracting equation (2.4) from (2.3) gives Green’s second identity:

∫∫ ∇−∇=∂∂

−∂∂

VS

dVdSnn

)()( 22 φψψφφ

ψψ

φ

n the expressions of Green’s identities, the functions φ and ψ must be

differentiable at least to the orders that appear in the integrands.

2.1.1 Bounded Problem

Before we consider the pipe flow problem, we consider a simpler problem in a bounded domain in order to understand the main ideas.

We consider 02 =∇ T in D and n

T

∂∂

is known on C.

where D is a bounded (interior) domain with boundary C.

To begin with we have the following Green second identity:

dcnn

dyCD

∫∫ ∂∂

−∂∂

=∇−∇ )()( 22 φψ

ψφφψψφ

The Green’s function is designated as the fundamental solution.

G (||

1ln

2

1),

−−−− −

=yx

yxπ

where ),( 21 xxx =−

),( 21 yyy =−

13

then )(2

−−−=∇ yxGx δ for any fixed

−y or )(2

−−−=∇ yxGy δ for any fixed

−x ,

where δ is the Dirac Delta function.

Strictly speaking it is defined through the integral δ(x − y) f (x)dx = f (y)∫

We let T≡φ and G≡ψ

Where T=temperature, G=Green’s function. Thus we have:

dcn

TG

n

GTdxdxTGGT

DC

xx )()( 21

22

∂∂

−∂∂

=∇−∇∫∫ ∫ .

But 02 =∇ T so we have :

dcn

TG

n

GTdxdxyxT

DC

)()( 21 ∂∂

−∂∂

=−∫∫ ∫−−

δ

∫ ∂∂

−∂∂

=⇒C

dcn

TG

n

GTyyT )(),( 21 (2.5)

in the case that ( Dyy ∈), 21

or : ∫ ∂∂

−∂∂

=C

dcn

TG

n

GT )(0 if ( Dyy ∉), 21

14

D

D∂

εΩ∂

y ε

Suppose ∆G=0 inside a domain D

Suppose Dy∈

We choose G to be the solution of 0=∆G in D/ εΩ, hence we have:

0)( =∂∂

−∂∂

−∂∫ xd

n

TG

n

GT

Dε

−

Ω∂−

∂ ∂∂

−∂∂

+∂∂

−∂∂

⇒ ∫∫ xdn

TG

n

GTxd

n

TG

n

GT

D

)()(ε

=0 (2.6)

(from previous section)

We know G (||

1ln

2

1),

−−−− −

=yx

yxπ

so |||| Ryx =−−−

Therefore,

G (R

yx1

ln2

1),

π=

−− πεπ 2

1

2

1−=−=

∂∂

=∂∂

⇒RR

G

n

G

ε=radius of the circle y=centre of the circle

εΩ∂ =boundary of

the circle

15

Let 0→ε

Then the 2nd part of (2.6) which is −

Ω∂ ∂∂

−∂∂

∫ xdn

TG

n

GT )(

ε

can be separated in

two parts:

(i) ∫∫Ω∂

−−Ω∂

−=∂∂

εεπε

xdxTxdxn

GxT )(

2

1)()(

= ∫Ω∂

−+′+−

ε

εεπε

xdOyTyT ))()()((2

1 2

= ∫Ω∂

−

+

′−−

ε

εππε

xdOyT

yT )(2

)()(

2

1

= πεεππε

2)(2

)()(

2

1

+

′−− O

yTyT

= )()()( 2εε OyTyT +′−− )(yT− as 0→ε

(ii) −

Ω∂−

Ω∂ ∂∂

=∂∂

∫∫ xdn

Txd

n

TG

επεε

1ln

2

1

= πεεεπ

2)()(ln2

1

+∂∂

− Oyn

T

= 0)()(ln →

+∂∂

− εεε Oyn

T as 0→ε

16

Hence,

(2.6) )()]([)()( yTyTxdn

TG

n

GTxd

n

TG

n

GT

D

=−−=∂∂

−∂∂

−=∂∂

−∂∂

−Ω∂

−∂

∫∫ε

Dy∈ as 0→ε .

which is the same result as (2.5) but is defined in a slightly more rigorous answer.

Suppose now, Dy ∂∈ (y is on the boundary).In this case, the same

procedure as before can be applied with the difference that now we have a semicircle instead of a circle. Therefore the length of the boundary is π

instead of 2π in the derivations above.

Hence,

)(2

1)()( yTxd

n

TG

n

GTxd

n

TG

n

GT

D

=∂∂

−∂∂

−=∂∂

−∂∂

−Ω∂

−∂

∫∫ε

Dy ∂∈ (2.7)

D ε=radius of the circle y=centre of the circle

y

17

2.1.2 Unbounded Problem Suppose ∆T=0 and ∆G=0 are outside the domain D (exterior problem)

Suppose Dy∉

The following equation is equal to zero because y is outside the domain as we

have mentioned in equation (2.5) when ( Dyy ∉), 2.1 .

0)()()()()()( =∂∂

−∂∂

+∂∂

−∂∂

+∂∂

−∂∂

∫∫ ∫Ω∂∂ Ω∂

xdsn

TG

n

GTxds

n

TG

n

GTxds

n

TG

n

GT

RD ε

(2.8)

0)()()()()()( =∂∂

−∂∂

−∂∂

−∂∂

−=∂∂

−∂∂

⇒ ∫∫ ∫Ω∂∂ Ω∂

xdsn

TG

n

GTxds

n

TG

n

GTxds

n

TG

n

GT

RD ε

Same as before = - T(y)

D

R

RΩ∂

εΩ∂

ε=radius of the small

circle y=centre of the small

circle

εΩ∂ =boundary of the

small circle

RΩ∂ =boundary of the

big circle

R=radius of the big

circle

(2.9)

18

Now, we are going to find (2.9) in the limit as R ∞→

)(ln2

1)(

2

1)()()( xds

n

TRxds

RxTxds

n

TG

n

GT

RRR

∫∫∫Ω∂Ω∂Ω∂ ∂

∂−−=

∂∂

−∂∂

ππ

=0 (If ∆T=0) (Corollary of Green’s 2)

∫Ω∂

−−=R

xdsxTR

0)()(2

1

π

So overall, (2.8)

)()(2

1)()()( xdsxT

RyTxds

n

TG

n

GT

D R

∫ ∫∂ Ω∂

−=∂∂

−∂∂

π (2.8)

19

2.1.3 Full pipe Problem In the following plot the pipe is buried in the ground.

where 1Γ= ground surface

2Γ =pipe

D=exterior environment (below 1Γ and outside 2Γ )

We have the following assumptions: ∆T=0 in D

n

T

∂∂

=known on 1Γ =C1 (a constant)

n

T

∂∂

=known on 2Γ =C2 (a constant)

Τhe asymptotic condition as ±∞→∞→ xy ,

T gTgy +→

Now, to solve the full pipe problem we will make a rectangular domain as shown below:

D

2Γ

1Γ

x

y

20

2.1.4 Neumann Green’s function for a half plane Problem

where 2Γ =pipe

εΓ =small circle

ε=radius of small circle

RΩ =domain

We want to find the integral equation of the domain RΩ .

In the Domain : Known:

0),(

0

2

2

=∇

=∇

−−yxG

T

x

.

Γε

Γ2

R

RΩ

(0 ,0 )

y = R

y

x

x=R x= - R Γ1 Ground surface

21

We choose G such that 0ˆ=

∂∂n

G on Γ1. (*)

−−−−−− ′−

+−

=||

1ln

2

1

||

1ln

2

1),(ˆ

yxyxyxG

ππ

where y′ is the reflection of y in the line y = - z and where ),( 21 xxx =−

),( 21 yyy =−

),(ˆ−−yxG satisfies )()(ˆ2 ′−+−=∇

−−−−yxyxGx δδ .

To find the integral equation of the domain we add the integral equations of

the pipe, the circle, the ground surface and the lines RyRxRx =−== ,, .

∫∫∫∫∫∫ ∫Γ=−==−∩ΓΩ∂ Γ

+++++=∂∂

−∂∂

εRyRxRxRRR

dsn

TG

n

GT

],[12

)ˆˆ

( =0 (2.10)

The equation (2.10) is equal to zero from the earlier Green’s function. Consider the lim (as R ∞→ )

The 1st integral is : ∫∫ΓΓ

−∂∂

=∂∂

−∂∂

22

)()ˆˆ

()()ˆˆ

( 2 xdsCGn

GTxds

n

TG

n

GT

The 2nd integral is : ∫∫∞=

−∞=−∩Γ

−∂∂

=∂∂

−∂∂ x

xRR

xdsCGn

GTxds

n

TG

n

GT )()ˆ

ˆ()()ˆ

ˆ( 1

],[1

=0 from (*) The 3rd and 4th integrals are:

)()ˆˆ

()()ˆˆ

(],0[,],0[,

xdsn

TG

n

GTxds

n

TG

n

GT

RyRxRyRx ∂∂

−∂∂

+∂∂

−∂∂

∫∫∈−=∈=

22

We consider the asymptotic condition as ∞±∞→ xy ,

gTgyT +→ . (2.11)

Therefore,

0=∂∂x

T and g

y

T=

∂∂

We solve for gTgyTu −−= as ∞±∞→ xy , . (2.12)

Hence if we substitute (2.11) into (2.12) we have :

0=−−+=−−= ggg TgyTgyTgyTu .Thus 0→u as ∞±∞→ xy ,

0, →∂∂

∂∂

y

u

x

u

Thus the 3rd and 4th integrals are:

)()ˆˆ

()()ˆˆ

(],0[,],0[,

xdsn

TG

n

GTxds

n

TG

n

GT

RyRxRyRx ∂∂

−∂∂

+∂∂

−∂∂

∫∫∈−=∈=

=0

=0 =0 =0 =0

Therefore, the result of the addition of the 3rd and 4th integral is zero as R ∞→ .

In the 5th integral we consider:

−−−−−− ′−

+−

=||

1ln

2

1

||

1ln

2

1),(ˆ

yxyxyxG

ππ

where

−−−− −

=Φ||

1ln

2

1),(

yxyx

π,

−−−− ′−

=′Φ||

1ln

2

1),(

yxyx

π

23

))()(

1ln

2

1(),(

)( 2

22

2

1122 yxyxyyyx

yn −+−∂

∂−=

∂

Φ∂−=

∂

Φ∂

−− π

)(2])()[(2

1)()(

2

1),(

)(22

2

3

2

22

2

11

2

22

2

11 yxyxyxyxyxyxyn

−−+−

−−+−−=

∂

Φ∂ −

−− π

])()[(

)(

2

12

22

2

11

22

yxyx

yx

−+−−

=π

and in the same way

),()( −−

′∂

Φ∂yx

yn ])()[(

)(

2

12

22

2

11

22

yxyx

yx

++−+

−=π

Therefore, if we add ),,(−−∂

Φ∂yx

n),(

−−′

∂Φ∂

yxn

and 02 =y then the result is zero:

0),(),(02

=

′∂

Φ∂+

∂

Φ∂

=−−−−y

yxn

yxn

For that reason , the 5th integral is zero. So from (2.10) -

+−+−∂∂

= ∫∫ΓΓ

)()ˆ()()ˆˆ

(012

12 xdsCGxdsCGn

GT )()ˆ

ˆ( xds

n

TG

n

GT∫

Γ ∂∂

−∂∂

ε

),()()ˆ()()ˆˆ

(0 2112

12

yyTxdsCGxdsCGn

GT +−+−

∂∂

= ∫∫ΓΓ

Therefore,

=− ),( 21 yyT )()ˆ()()ˆˆ

(12

12 xdsCGxdsCGn

GT ∫∫

ΓΓ

−+−∂∂

24

∫∫∫ΓΓΓ

−+−=∂∂

−−122

)()ˆ()()ˆ()()(ˆ

),( 1221 xdsCGxdsCGxdsxTn

GyyT

We set

∫∫ΓΓ

−+−12

)()ˆ()()ˆ( 12 xdsCGxdsCG = )(yF

Hence,

)()()(ˆ

),(2

21 yFxdsxTn

GyyT =

∂∂

−− ∫Γ

We are going to solve this integral equation numerically. The general integral equation is of the following form:

∫Γ

=Κ+ )()(),()( xfdyyuyxxu

25

Chapter 3

Methods for solving a single integral equation We study certain Fredholm integral equations of the 2nd kind,

∫Γ

=Κ+ )()(),()( xfdyyuyxxu (3.1)

Where :

Kernel= ),( yxΚ is known

Γ is some closed boundary , Γ= ))(),(( 21 ςγςγ where [ ]πς 2,0∈

and 21,( γγ ) are periodic functions

)(xf is known

u is not known u is what we have to find

We can solve equation (3.1) by three methods: (i) Galerkin method (ii) Collocation method (iii) Nyström method The Galerkin method used for analysis, but the other two methods are easier for programming.

In this project we will use the Nyström and the Collocation method.

3.1 Nyström Method

To solve

∫ =+π2

0

)()(),()( xfdyyuyxKxu

We replace the integral by a quadrature rule. The easiest way is to use the trapezium rule.

26

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

)(2

0

ππ

ghnghghggh

dyyg +−++++≈∫

= )])1((......)()0([ hnghggh −+++ [2π periodic function

)2()0( πgg =⇒ ].

Where n

hπ2

= (n =quadrature parameter)

In our case we replace u bynu .

Therefore, we have:

)()])1(())1((,(....)(),()0()0,([)( xfhnuhnxKhuhxKuxKhxu nnnn =−−++++

x∀

We take hnhhhx )1(,.....,3,2,,0 −=

1st equation when 0=x :

)0()])1(())1((,0(....)(),0()0()0,0([)0( fhnuhnKhuhKuKhu nnnn =−−++++

2nd equation when hx = :

)()])1(())1((,(....)(),()0()0,([)( hfhnuhnhKhuhhKuhKhhu nnnn =−−++++

3rd equation when hx 2= :

)2()])1(())1((,2(....)(),2()0()0,2([)2( hfhnuhnhKhuhhKuhKhhu nnnn =−−++++

Last equation when hnx )1( −= :

))1(()])1(())1((,)1((

....)(),)1(()0()0,)1(([))1((

hnfhnuhnhnK

huhhnKuhnKhhnu

n

nnn

−=−−−+

++−+−+−

27

We have n equations with n unknowns:

))1((),......,3(),2(),(),0( hnuhuhuhuu nnnnn −

We write it as a matrix bAx =

−−+−−

−+

−+

−+

=

])1(,)1([1....................],)1([)0,)1([

.............................................................

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

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

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

hnhnKhhhnKhhnKh

hnhhKhhhKhhhkhhK

hnhhKhhhKhhhKhhK

hnhKhhKhhKhK

A

−

=

))1((

......

)3(

)2(

)(

)0(

hnu

hu

hu

hu

u

x

n

n

n

n

n

−

=

))1((

.......

)3(

)2(

)(

)0(

hnf

hf

hf

hf

f

b

28

• The conventional Nyström method is a simple and efficient mechanism for discretizing integral equations with non-singular kernels (K(x,y)).

• With a high-order quadrature rule, the solution one obtains by this method is a high-order approximation to the exact solution.

In the Nyström method we could use midpoint rule, Gaussian quadrature,Simpson’s rule and trapezoidal rule. Question: For a general integral equation in [a,b] of a general function which is the best quadrature? Answer:

If you have [0,2π] and a periodic function and if the function is ∞C ,then the

trapezoidal rule is exponentially accurate and also equivalent to replacing u

by its trigonometric interpolating polynomial and collocating at mesh points.

3.1.1 Example of a single periodic integral equation

In order to test our method we derive analytical solution:

Analytical solution:

Our example is of the form:

∫Γ

=Κ+ )()(),()( xfdyyuyxxu

Where the Kernel=K(x, y) = )2sin(2

1yx +

π

Γ is a closed boundary from 0 to π2

u(x)=cos( )x , u(y)=cos( y)

29

Thus,

∫ =++π

π

2

0

)()cos()2sin(2

1)cos( xfdyyyxx

)()cos(])2cos()sin()cos()2[(sin(2

1)cos(

2

0

xfdyyxyyxx =++⇒ ∫π

π

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

1)cos(

2

0

2xfdyyxyyxx =++⇒ ∫

π

π

)()2

)2sin()2cos(

2

1)2cos()2(sin(

2

1)cos(

2

0

xfdyy

xy

xx =++

+⇒ ∫π

π

)(4

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

24

)2sin()2sin(

2

1)cos(

2

0

xfyx

xyy

xx =

−++⇒

π

π

)())2sin(0(2

1)cos( xfxx =++⇒ π

π

Hence,

)2sin(2

1)cos()( xxxf +=

We program this example of a single periodic integral equation in Matlab and

we have the following results for the 2Luexactu −

where exactu =cos(x) and u is computed with the Nyström Method.

(Note: where 2Lexactuu −

2

12

0

2)(

−= ∫

π

exactuu )

30

TABLE 1

n 2Lexactuu −

2 3.9356e-016

4 2.8353e-016

8 3.6854e-016

16 5.2234e-016

32 7.5510e-016

We just use mesh points to evaluate the 2Lexactuu − . We always get zero

to machine precision.

We have used 2Lexactuu −

2

12

0

2)(

−= ∫

π

exactuu ) (3.2)

=2

1

1

1

^2

^^

)]()([

−∑

−

=

n

jexact hjuhjuh

with hh =^

and then we had zero. However this is not an accurate approximation to the error.

Instead, we need to work out integral (3.2) in a better way. We can work out

exactu everywhere because we have exact formula to work out an

approximation to u which valid anywhere. So we can use the following formula

where )(xuPn is called the trigonometric interpolating polynomial [10].

∑ ∑−

=

−

=

−+−+=

12

0

1

1

))(cos())(cos(212

1)()(

m

j

m

kn jhxmjhxk

mjhuxuP

Where 2

nm =

31

3.1.2 Example of a single non-periodic integral equation

Analytical solution:

∫Γ

=Κ+ )()(),()( xfdyyuyxxu

Where

the Kernel=K(x, y) =22 yx

Γ is a closed boundary from 0 to 1

u(x)=12

51

2x+ , u(y)=

12

51

2y+

So,

∫ =++

+

1

0

2222

)()12

51(

12

51 xfdyyx

yx

∫ =++

+⇒

1

0

4222

2

)()12

5(

12

51 xfdy

yxyx

x

)(12

1

3

1

12

51

1

0

52322

xfyxyxx

=

++

+⇒

)(12

1

3

1

12

51

1

0

5322

xfyyxx

=

++

+⇒

)(12

1

3

1

12

51 2

2

xfxx

=

++

+⇒

)(12

1

3

1

12

51 2 xfx =

+++⇒

32

Hence,

2

6

51)( xxf +=

We program this example of a single periodic integral equation in Matlab and

we have the following results for the 2Luexactu − .

We have used 2Lexactuu −

2

12

0

2)(

−= ∫

π

exactuu ) (3.2)

=2

1

1

1

^2

^^

)]()([

−∑

−

=

n

jexact hjuhjuh

TABLE 2

n 2Lexactuu −

2 0.5000 4 0.1221 8 0.0443 16 0.0191 32 0.0089

We computed the error at the mesh points. The error appears to half as we double the value of n .

If we compare our two examples, the periodic function with the non-periodic function, we can see that the periodic function is faster.

33

3.2 Collocation Method

The idea is to choose a finite-dimensional space of candidate solutions and a

number of points in the domain (called collocation points), and to select that

solution which satisfies the given equation at the collocation points.

To solve

∫ =+π2

0

)()(),()( xfdyyuyxKxu (3.3)

We seek an approximation nu of the form:

∑−

=

=1

0

)()(n

j

jjn xuxu φ where =jφ basis functions

Substitute u into (3.3)

∑ ∫ ∑−

=

−

=

=+1

0

2

0

1

0

)()()(),()()(n

j

n

jjjjj xfdyxuyyxKxux

π

φφ

∑ ∫−

=

=+⇒1

0

2

0

)()(])(),()([n

jjjj xfxudyyyxKx

π

φφ (3.4)

So we have one equation with n unknowns → the values of )( jxu .

To get n equations we fix (3.4) to hold at n points i.e. take ^^

1,....., nxxx = and

then we will have n equations with n unknowns.

If we choose ^^

1,....., nxx to be the same points as jx

∑ ∫−

=

=+⇒1

0

2

0

)()(])(),()([n

jmjjmmj xfxudyyyxKx

π

φφ

34

We replace the integral by a quadrature rule as in Nyström Method. The easiest way is to use the trapezium rule.

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

)(2

0

ππ

ghnghghggh

dyyg +−++++≈∫

= )])1((......)()0([ hnghggh −+++ (periodic function)

Where n

hπ2

= (n =quadrature parameter)

In our case we replace φ bynφ .

Therefore, we have:

)()()])1(())1((,(....)(),()0()0,([)( xfxuhnhnxKhhxKxKhx jnnnn =⋅−−++++ φφφφ

x∀

We take hnhhhx )1(,.....,3,2,,0 −=

We define nj uxu =)(

We have n equations with n unknowns:

))1((),......,3(),2(),(),0( hnuhuhuhuu nnnnn −

We use the trapezoidal rule in this method and we have exactly the same matrix as for the Nyström Method. Hence the Nyström Method is exact at mesh points.

35

Chapter 4

Interior Problem for Laplace’s equation

Consider 0=∆u ,

Ω is a circle with radius R

gn

u=

∂∂

We already know that the Fundamental solution G(||

1ln

2

1),

−−−− −

=yx

yxπ

and

that the form of the general integral equation is:

∫Γ

=Κ+ )()(),()( xfdyyuyxxu .

Thus, we set up the interior problem as an integral equation of the above form and we will solve it using a code in Matlab.

R

gn

u=

∂∂

Ω

36

From (2.7) of the bounded problem we have:

)(2

1)( yTxd

n

TG

n

GT

D

=∂∂

−∂∂

−∂∫ Dy ∂∈

−

∂−

∂∫∫ ∂

∂=

∂∂

+−⇒ xdn

TGxd

n

GTyT

DD

)(2

1 Dy ∂∈

Where in this case K(x,y)=n

G

∂∂

∫∫∂

−∂

−=

∂∂

=DD

xdGgxdn

TGxf )()()(

So we have to solve :

−

∂−

∂∫∫ ∂

∂=

∂∂

+− xdn

TGxd

n

GTyT

DD

22)(

Firstly, we have to find n

G

∂∂

with respect to x.

We know G(||

1ln

2

1),

−−−− −

=yx

yxπ

where ),( 21 xxx =−

),( 21 yyy =−

We substitute −x and

−y in G and then we have the following expression for G:

G(

( ) 2

22

2

11 )(

1ln

2

1),

yxyxyx

−+−=

−− π

)(xn

G

∂

∂ =

∂∂∂∂

⋅

=∇

−−

2

1

2

1

)(

)().(

x

G

x

G

xn

xnGxn x =

2

2

1

1 )()(x

Gxn

x

Gxn

∂∂

+∂∂

(4.1)

37

[ ] )(2)()(2

1)()(

2

111

2

32

22

2

11

2

22

2

11

1

yxyxyxyxyxx

G−−+−

−−+−=

∂∂ −

π

)(

])()[(

])()[(

2

111

2

3

2

22

2

11

2

1

2

22

2

11

1

yx

yxyx

yxyx

x

G−⋅

−+−

−+−−=

∂∂

π

])()[(

)(

2

12

22

2

11

11

1 yxyx

yx

x

G

−+−−

−=∂∂

π and in the same way

])()[(

)(

2

12

22

2

11

22

2 yxyx

yx

x

G

−+−−

−=∂∂

π

We set ςςςς sin,cos))sin(),(cos( 21 ==⇒= nnx

))sin(),(cos( tty =

After that we replace

1x

G

∂∂

,

2x

G

∂∂

, 1n and 2n into equation (4.1)

And finally we have:

)sin(])()[(

)(

2

1)cos(

])()[(

)(

2

12

22

2

11

22

2

22

2

11

11 ςπ

ςπ yxyx

yx

yxyx

yx

n

G

−+−−

−−+−

−−=

∂∂

[ ]))(sin())(cos(])()[(2

122112

22

2

11

yxyxyxyxn

G−+−

−+−−=

∂∂

⇒ ςςπ

We also replace )cos(1 ς=x )cos(1 ty =

)sin(2 ς=x )sin(2 ty =

38

[ ]]))sin()(sin())cos()[(cos(2

)sin(sinsin)cos(coscos22 tt

tt

n

G

−+−−+−

−=∂∂

⇒ςςπ

ςςςς

(Note: )sinsincos(cos22])sin(sin)cos[(cos 22 tttt ςςςς +−=−+−

)cos(22 t−−= ς

−

−=2

2cos22tς

= ]2

sin21[22 2

−

−−tς

=

−

+−2

sin422 2 tς

=

−

2sin4 2 tς

).

)2

(sin42

sinsinsincoscoscos

2

22

t

tt

n

G

−⋅

+−+−=

∂∂

⇒ς

π

ςςςς

)2

(sin8

sinsincoscos)sin(cos

2

22

t

tt

n

G

−+++−

=∂∂

⇒ς

π

ςςςς

)2

(sin8

sinsincoscos)sin(cos

2

22

t

tt

n

G

−+++−

=∂∂

⇒ς

π

ςςςς

(Note: Trigonometric identity: 1sincos 22 =+ ςς )

)2

(sin8

sinsincoscos1

2 t

tt

n

G

−++−

=∂∂

⇒ς

π

ςς

39

)2

(sin

))sinsincos(cos1(

8

1

2 t

tt

n

G

−+−

−=∂∂

⇒ς

ςςπ

Now we are going to simplify the numerator of this fraction.

)cos(1)sinsincos(cos1 ttt −−=+− ςςς

))2

(sin21(1 2 t−−−=

ς

)2

(sin2 2 t−=

ς

(Note: Trigonometric identities: ttt sinsincoscos)cos( ςςς +=−

ςςς 22 sincos2cos −=

ςςς 22 sinsin12cos −−=⇒

ςς 2sin212cos −=⇒ ).

Finally,

)2

(sin

)2

(sin2

8

1

2

2

t

t

n

G

−

−

−=∂∂

⇒ς

ς

π

π4

1−=

∂∂

⇒n

G (4.2)

Hence our integral equation is:

−

∂−

∂∫∫ =−− xdGgxdTyTDD

24

12)(

π

)(yf

We know G(||

1ln

2

1),

−−−− −

=yx

yxπ

therefore

40

dxxg

yx

dxxg

yx

Ggdxyf )(1

ln1

)(1

ln22

12)(

2

0

2

0

2

0

∫∫ ∫−−−−

−=

−==

ππ π

ππ

Lets substitute yxxyxx +=⇒−= ˆˆ

)ˆ()(

ˆ

yxgxg

xddx

+=

=

dxyxgx

xdyxgx

yfy

y

)(1

lnˆ)ˆ(ˆ

1ln

1)(

2

0

2

+=+= ∫∫−

−

ππ

π

because is periodic

To find )(yf which is equal to −

∂∫ xdGgD

we have to apply some quadrature

rule. The composite midpoint rule is appropriate for this equation, since the integrand is singular at x=0.

Composite midpoint rule:

For any function F and for any N 1≥ we have:

∫ ∑−

=

+≈

π2

0

1

0 2

)12()(

&

J

j hfhdxxf where

&h

π2=

Therefore, our )(yf is:

+

+

+≈+= ∑∫

−

=

yh

gh

hdxyxgx

yfj

&

j j 2

)12(

)12(

2ln)(

1ln)(

1

0

2

0

π

where 2

)12( hx

j += .

Finally, we know the kernel (K(x,y))and the right hand side of the integral equation f(y). Thus we will solve it numerically in Matlab to find T(y).

41

Below are the graphs of the numerical solution of the integral equation:

∫∫∂

−∂

−=

∂∂

+−DD

xGgdxdn

GTyT 22)( for different values of n.

Where

π4

1−=

∂∂n

G

dxyxgx

xgdGyfD

)(1

ln22)(2

0

+== ∫∫ −∂

π

)12sin()10cos()( θθθ +=g

0 1 2 3 4 5 6-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

n=4

n=8

n=16

Figure 4.1: Numerical solution of the interior problem for Laplace’s equation for n=4,8,16.

42

0 1 2 3 4 5 6 7-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

n=32

n=64

n=128

Figure 4.2: Numerical solution of the interior problem for Laplace’s equation for n=32,64,128. In the above diagrams the solution looks like converging. By increasing the value of n the solution becomes more accurate.

43

4.1 Separation of variables in polar coordinates We will use polar coordinates and separation of variables to solve analytically the interior problem:

011

2

2

2

2 =∂∂

+

∂∂

∂∂

=∇θu

rr

ur

rru

(4.3)

= 011

2

2

22

2

=∂∂

+∂∂

+∂∂

θu

rr

u

rr

u

We seek a solution of the form:

u=R(r). Θ( )θ

Τherefore,

( ) 0)()(1

)()((1

2

2

2=Θ

∂∂

+

Θ

∂∂

∂∂

θθ

θ rRr

rRr

rrr

01

)(1

)(2

2

2=

∂Θ∂

+

∂∂

∂∂

Θ⇒θ

θr

rRr

Rr

rr

Multiplying by r2/RΘ gives us:

01

2

2

=∂

Θ∂Θ

+

∂∂

∂∂

⇒θr

Rr

rR

r

Let us now separate the variables: i.e. let us collect all of the r -dependent terms on one side of the equation, and all of the θ-dependent terms on the other side. Thus,

2

21

θ∂Θ∂

Θ−=

∂∂

∂∂

⇒r

Rr

rR

r

44

The above equation has the form:

f(r)=f(θ)

where f(r) is a function of r and f(θ) is a function of θ. The only way in which the above equation can be satisfied, for general r and θ, is if both sides are equal to the same constant. Thus,

cr

Rr

rR

r=

∂Θ∂

Θ−=

∂∂

∂∂

⇒2

21

θ(constant)

The ordinary differential equations we get are then:

(a) 0=−

∂∂

∂∂

cRr

Rr

rr

(b) 02

2

=Θ+∂

Θ∂c

θ

We take (b) 02

2

=Θ+∂

Θ∂c

θ

Try Θ=λθe so cccee −±=⇒=+⇒=+ λλλ λθλθ

0022

cc BeAe −−− +=Θ⇒ θθ

We know that Θ(θ ) must be 2π periodic because is around a periodic boundary. If c<0 then is not periodic If c=0 then is not periodic

45

Thus when c>0 cici BeAe θθ −+=Θ⇒

= A(

cos( c θ)+B(

sin( c θ)

So write c= ⇒2ν )sin()cos( νθνθ BA +=Θ (4.4)

We take (a) 02 =−

∂∂

∂∂

Rr

Rr

rr ν (where c=

2ν )

02

2

2

=−

∂∂

+∂∂

⇒ Rr

R

r

Rrr ν

02

2

22 =−

∂∂

+∂∂

⇒ Rr

Rr

r

Rr ν

This is an Euler differential equation. The general solution to this simple case of Euler’s ordinary differential equation is given as:

νν −+= rCrCrR 21)( (4.5)

Combining equations (4.4) and (4.5) we have:

)()(),( θθ Θ= rRru

⇒ =),( θru )( 21

νν −+ rCrC . ))sin()cos(( νθνθ BA + (4.6)

46

As r→0 the term involving ν−r is unbounded. The only way to fix this is to

take 02 =C .

Therefore,

=),( θru νr ))sin()cos(( νθνθ BA + for any ν .

We know that equation (b) 02

2

=Θ+∂

Θ∂c

θ which its general solution is :

)sin()cos( νθνθ BA +=Θ

We use the boundary conditions Θ (0) =Θ (2π ) (i.e. periodic)

Θ (0) =A ⇔ Θ (0) =Θ (2π )

Θ (2π )=Acos(2 )πν

Thus,

cos (2 )πν ) =1 ,....2,1,0 ±±=⇒ v

Therefore the general solution of the problem is:

=),( θru ( )∑∞

=

+0

)sin()cos(ν

ννν νθνθ BAr (4.7)

In our problem gn

u=

∂∂

for a circle then gn

u

r

u=

∂∂

=∂∂

on r=a

( ) )()sin()cos(0

1 θθθ gmBmAmar

umm

m

m =+=∂∂

∑∞

=

− (4.8)

47

The interior Neumann problem is solvable if and only if:

∫ =π

θθ2

0

0)( dg

but there is no existence of a unique solution. (Theorem 6.26, Kress ‘Linear Integral Equations’) .

The coefficients A m and B m may be determined by a Fourier expansion on

πθ 20 ≤≤ . The important observation is that sine and cosine functions of different frequency are orthogonal. This means that, when multiplied and integrated, give zero result:

π if 0≠= nm

∫ =π

θθθ2

0

)cos()cos( dnm π2 if 0== nm

0 if nm ≠

π if 0≠= nm

∫ =π

θθθ2

0

)sin()sin( dnm 0 if 0== nm

0 if nm ≠

0)sin()cos(2

0

∫ =π

θθθ dnm , all nm,

(Fourier Series and Applications, Beatrice Pelloni,2006)

48

Fix an integer value 0≠n and multiply (4.8) by cos( )θn and sin( )θn , respectively.

Then perform the integration term by term:

∑ ∫ ∫∫∞

=

− +=0

2

0

2

0

12

0

])sin()cos()cos()cos([)cos()(m

mm

mdmnBdnmAmadng

π ππ

θθθθθθθθθ

n

n Ana π1−= =0

Hence,

∫−=π

θθθπ

2

01

)cos()(1

dngna

Ann 0≠n

While

∑ ∫ ∫∫∞

=

− +=0

2

0

2

0

12

0

])sin()sin()sin()cos([)sin()(m

mm

m dmnBdnmAmadngπ ππ

θθθθθθθθθ

=0

n

n Bna π1−=

Hence,

∫−=π

θθθπ

2

01

)sin()(1

dngna

Bnn 0≠n

The solution of the interior problem is not unique because If 0=∆u 0)( =+∆ cu

⇒

0=∂

∂

n

u 0

)(=

∂

+∂

n

cu for all constants c.

49

Chapter 5

Exterior Problem for Laplace’s equation For the exterior problem (unbounded problem) as we have seen in chapter 2 (2.8) the general integral is:

)()(2

1)()()( xdsxT

RyTxds

n

TG

n

GT

D R

∫ ∫∂ Ω∂

−=∂∂

−∂∂

π (5.1)

What happens as ∞→R ?

For the half plane problem we have the asymptotic condition as y→∞, x →±∞

ggg TgyyTT +=→ )( so gy

T=

∂∂

We solve for u=T- ggT and we consider u=T- gTgy −

As ∞→R then 0→u .

But because we have exterior problem and as ∞→R we need T(R)=o(1)

i.e. 0)( →RT

Therefore, the part ∫Ω∂ R

xdsxTR

)()(2

1

π of equation (5.1) disappears and then

we only have :

)()()(∫∂

=∂∂

−∂∂

D

yTxdsn

TG

n

GT (5.2)

50

The above equation (5.2) is the same equation as (2.6) . Thus, we are going to solve it numerically with the same way as the interior problem. For the exterior problem the analytical solution is the same as (4.6) :

=),( θru )( 21

νν −+ rCrC . ))sin()cos(( νθνθ BA +

except as r ∞→ the solution tends to zero. Thus, we take C1=0 and C2≠0 .

=),( θru ν−r . ))sin()cos(( νθνθ BA +

The exterior Neumann problem is uniquely solvable if and only if:

∫ =π

θθ2

0

0)( dg

(Theorem 6.28, Kress ‘Linear Integral Equations’) . Below are the graphs of the numerical solution integral equation:

51

∫∫∂

−∂

−=

∂∂

+−DD

xGgdxdn

GTyT 22)( for different values of n.

Where

π4

1=

∂∂n

G which is the same with the

n

G

∂∂

of the interior problem

but with different sign.

dxyxgx

xgdGyfD

)(1

ln22)(2

0

+== ∫∫ −∂

π

θθ cos)( =g

0 1 2 3 4 5 6-3

-2

-1

0

1

2

3

n=4

n=8

n=16

Figure 5.1: The numerical solution of the exterior problem for Laplace’s equation for n=4,8,16.

52

0 1 2 3 4 5 6 7-3

-2

-1

0

1

2

3

n=32

n=64

n=128

Figure 5.2: The numerical solution of the exterior problem for Laplace’s equation for n=32,64,128. Again we can see a convergence in both diagrams and as the value of n increases the solution becomes more accurate.

53

Chapter 6

Full pipe flow Problem To find the general integral equation for the full pipe flow problem we assume the following plot.

where

1C , 2C are constants.

We consider the asymptotic condition as ∞±∞→ xy ,

gTgyT +→ . where gT is a constant.

We solve for gTgyTu −−= as ∞±∞→ xy , and we end up :

0→u as ∞±∞→ xy ,

gCgy

T

y

u−=−

∂∂

=∂∂

1

31 CCgy

u

n

u=−=

∂∂

−=∂∂

⇒

∆Τ=0

2Cn

T

r

T=

∂∂

−=∂∂

1Cn

T

y

T=

∂∂

−=∂∂

x

y

n

Y= -z

n

54

The following is an integral equation approach:

0)(2

1)()()()( 3 =+−+

∂∂

−∂∂

∫∫∞

∞−

xuxdsGCxdsn

uG

n

Gu if x is on the

boundary of the pipe. We assume that C3 is zero .Therefore, we have the following integral equation:

∫Γ

+∂∂

−∂∂

= 0)()()(2

1xds

n

Gu

n

uGxu

)()()()()(2

1xds

n

uGxds

n

Guxu ∫∫

ΓΓ ∂∂

=∂∂

+⇒

This integral equation is the same as the integral equation of interior and exterior problem except that here the Green’s function is:

−−−−−− ′−

+−

=||

1ln

2

1

||

1ln

2

1),(

^

yxyxyxG

ππ (6.1)

We want to find dn

G∂.

We separate ),(^

−−yxG into two parts.

The 1st part is G(||

1ln

2

1),

−−−− −

=yx

yxπ

and the

2nd part we call it P(

||

1ln

2

1), ′−=′

−−

−−yx

yxπ

We already know from (4.2) π4

1−=

∂∂n

G

55

Thus, now we want to find n

P

∂∂

.

We know P(

||

1ln

2

1), ′−=′

−−

−−yx

yxπ

where ),( 21 xxx =−

)2,( 21 yzyy −−=′−

We substitute −x and

′−y in P and then we have the following expression for P:

P(

( ) 2

22

2

11 )2(

1ln

2

1),

yzxyxyx

+++−=′

−− π

n

P

∂∂

=

∂∂∂∂

⋅

=∇

−−

2

1

2

1

)(

)().(

x

P

x

P

xn

xnPxn x =

2

2

1

1 )()(x

Pxn

x

Pxn

∂∂

+∂∂

(6.2)

[ ] )(2)2()(2

1)2()(

2

111

2

32

22

2

11

2

22

2

11

1

yxyzxyxyzxyxx

P−+++−

−+++−=∂∂ −

π

)(

])2()[(

])2()[(

2

111

2

3

2

22

2

11

2

1

2

22

2

11

1

yx

yzxyx

yzxyx

x

P−⋅

+++−

+++−−=

∂∂

π

])2()[(

)(

2

12

22

2

11

11

1 yzxyx

yx

x

P

+++−−

−=∂∂

π and in the same way

])2()[(

)2(

2

12

22

2

11

22

2 yzxyx

yzx

x

P

+++−++

−=∂∂

π

We set ςςςς sin,cos))sin(),(cos( 21 ==⇒= nnx

))sin(),(cos( tty =

56

After that we replace

1x

P

∂∂

,

2x

P

∂∂

, 1n and 2n into equation (6.2)

And finally we have:

)sin(])2()[(

)2(

2

1)cos(

])2()[(

)(

2

12

22

2

11

22

2

22

2

11

11 ςπ

ςπ yzxyx

yzx

yzxyx

yx

n

P

+++−++

−+++−

−−=

∂∂

[ ])2)(sin())(cos(])2()[(2

122112

22

2

11

yzxyxyzxyxn

P+++−

+++−−=

∂∂

⇒ ςςπ

We also replace )cos(1 ς=x )cos(1 ty =

)sin(2 ς=x )sin(2 ty =

[ ]]))sin(2)(sin())cos()[(cos(2

)sin2(sinsin)cos(coscos22 tzt

tzt

n

P

+++−+++−

−=∂∂

⇒ςςπ

ςςςς

(Note: )sinsincos(cos22])sin2(sin)cos[(cos 22 tttzt ςςςς −−=+++−

tzzz sin4sin44 2 +++ ς

tzz

zt

sin4sin4

4)cos(22 2

++

++−=

ςς

tzz

zt

sin4sin4

42

2cos22 2

++

+

+

−=

ς

ς

tzzz

t

sin4sin44

]2

sin21[22

2

2

+++

+

−−=

ς

ς

57

tzz

zt

sin4sin4

42

sin4 22

+

++

+

=

ς

ς

).

]sin4sin44)2

(sin4[2

sin2sinsinsincoscoscos

22

22

tzzzt

ztt

n

P

++++

⋅

−−−+−=

∂∂

⇒ς

ςπ

ςςςςς

]sin4sin44)2

(sin4[2

sin2sinsincoscos1

22 tzzzt

ztt

n

P

++++

⋅

−−+−=

∂∂

⇒ς

ςπ

ςςς

tzzzt

ztt

n

P

sin4sin44)2

(sin4

sin2)sinsincos(cos1(

2

1

22 ++++

+−−−=

∂∂

⇒ς

ςςςς

π

(Note: Trigonometric identity: 1sincos 22 =+ ςς )

Now we are going to simplify the numerator of this fraction.

)cos(1)sinsincos(cos1 ttt +−=−− ςςς

))2

(sin21(1 2 t+−−=

ς

)2

(sin2 2 t+=

ς

(Note: Trigonometric identities: ttt sinsincoscos)cos( ςςς −=+

ςςς 22 sincos2cos −=

ςςς 22 sinsin12cos −−=⇒

ςς 2sin212cos −=⇒ ).

tzzzt

zt

n

P

sin4sin44)2

(sin4

sin22

sin2

2

1

22

2

++++

+

+

−=∂∂

⇒ς

ς

ςς

π

58

Finally,

dn

G∂=

n

P

n

G

∂∂

+∂∂

= +−π4

1

tzzzt

zt

sin4sin44)2

(sin4

sin22

sin2

2

1

22

2

++++

+

+

−ς

ς

ςς

π

Therefore, the integral equation of the full pipe flow problem is solved with the same way as the exterior and interior problem. The only difference is the Kernel. Below are the graphs of the numerical solution of the integral equation:

∫∫∂

−∂

−=

∂∂

+−DD

xGgdxdn

GTyT 22)( for different values of n.

Where

=∂∂n

G+−

π4

1

tzzzt

zt

sin4sin44)2

(sin4

sin22

sin2

2

1

22

2

++++

+

+

−ς

ς

ςς

π

dxyxgx

xgdGyfD

)(1

ln22)(2

0

+== ∫∫ −∂

π

)10cos()( θθ =g

59

0 1 2 3 4 5 6-5

-4

-3

-2

-1

0

1

2

3

n=4

n=8

n=16

Figure 6.1: The numerical solution of the full pipe flow problem for n=4,8,16.

60

0 1 2 3 4 5 6 7-5

-4

-3

-2

-1

0

1

2

3

4

n=32

n=64

Figure 6.2: The numerical solution of the full pipe flow problem for n=32,64. As we can see from the diagrams the solution converges and by increasing the value of n , the solution becomes more accurate.

61

Chapter 7 Summary and Conclusions This dissertation has used the Boundary Element Method to approximate the heat transfer in a buried pipe. In chapter 1 we started by becoming familiar with the pipe flow problem. We explained what the heat transfer is and simplified our problem within the pipe to one-dimension and made some assumptions for the problem. In chapter 2 we introduce the Boundary Element Method. The Boundary Element Method is a numerical method for solving Partial Differential Equations which have been formulated as integral equations. The advantages in the BEM arise from the fact that only the boundary of the domain of the PDE requires sub-division. So, the dimension of the problem is effectively reduced by one. We also reformulate the PDE as a Boundary Integral Equation. We looked at the bounded problem when y is on the domain and y is on the boundary. We also looked at the unbounded problem where y is not on the domain. Moreover, we looked at the full pipe problem. In chapter 3 we talked about the methods that we have used to solve a Fredholm integral equation of the 2nd kind. We had two examples of a single integral equation. One example of a periodic function and one of the non-periodic function. We end up that the periodic function is faster than the non-periodic function. In chapter 4 we saw the numerical solution of the interior boundary integral equation arising from the Laplace’s equation. We also looked at the analytical solution of the problem using polar coordinates and separation of variables. We came to the conclusion that the numerical solution of the interior problem of Laplace’s equation is converging and that by increasing the value of n the solution becomes more accurate. Furthermore, the solution of the interior problem is not unique. In chapter 5 we saw the numerical solution of the exterior boundary integral equation arising from the Laplace’s equation. We end up that the numerical solution of the exterior problem of Laplace’s equation is converging. Moreover, the solution of the exterior problem is unique. Finally, in chapter 6 we have approached an integral equation for the full pipe flow problem and we saw that the numerical solution of the full pipe flow problem is converging.

62

Bibliography [1] Bau, H.H. and Sadhal,S.S. (1982) Heat losses from fluid flowing in a Buried pipe,International Journal of Heat and Mass Transfer, 25 (11): 1621-1629 [2] Bau, H.H. (1984) Convective heat losses from a pipe buried in a Semi-infinite porous medium, International Journal of Heat and Mass Transfer, Vol 27,No.||, pp.2047-2056 [3] Beatrice Pelloni (2006) Fourier Series and Applications [4] Carslaw, H.S. and Jaeger, J.C. (1986) Conduction of Heat in Solids, 2nd Ed.,Oxford Science Publications [5] Chuk Ovuworie (2008) Partially Buried Pipe Heat Transfer, Powerpoint [6] Kendall E.Atkinson (1997) The Numerical Solution of Integral Equations of the Second Kind, Published by Cambridge University Press [7] Kreith, F. and Bohn, M. (1997) Principles of Heat Transfer,5th Ed., PWS Publishing Company [8] Kress, R. (1989) Linear Integral Equations. Berlin:Springer [9] Luiz C. Wrobel, M. H. Aliabadi, Wrobel. (2002) The Boundary Element Method: Applications in Thermo-Fluids and Acoustics, Published by John Wiley and Sons [10] S.Langdon and I.G.Graham. (1998) Boundary integral methods for singularly perturbed boundary value problems, IMA Journal of Numerical Analysis (2001) 21,217-237