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
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.
1
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.
2
Contents
1 Introduction
2 Boundary Element Method
4
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
3
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)
rpi =radius of the pipe inside
rpo =radius of the pipe outside
rco = radius of the coating outside
Ta =temperature of the ambient fluid
T f =temperature of the pipeline fluid
R p =resistance of the pipe
Rc =resistance of the coatings
Rg =resistance of the ground
z=distance from the centre of the pipe to the ground surface
4
We can express, at steady state, the rate of heat flow between the pipeline and
the ambient fluid as:
Q = −2πrpoU po (T f − Ta )
where
U po = overall heat transfer coefficient
2πrpo = pipe outside surface area
T f − Ta = 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=
∆T
R
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:
T f − T pi = −
Q
2πrpi h pi
(1.1)
where
h pi = the pipe inside film coefficient
2πrpi = pipe inside surface area
T f =temperature of the pipeline fluid
T pi =temperature of the pipe inside
5
The temperature difference between the pipe inside and outside walls is given by
the following equation:
T pi − T po =
ln(r po r pi )
2πk p
where
rpi = radius of the pipe inside
rpo = radius of the pipe outside
k p =the pipe thermal conductivity
T pi = the temperature of the pipe inside
T po = 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
∂ 2T 1 ∂T
=0
∇T= 2 +
∂r
r ∂r
2
It is Laplace’s equation, but now
the
∂T
term has disappeared.
∂θ
(where r, θ are polar coordinates on the centre of the pipe).
6
Solving Laplace’s equation without the
∂T
∂θ
term, we obtain the general
solution for temperature:
T ( r ) = A ln r + B
(1.2)
where A ,B are constants.
We do not need to worry that temperature relies on ln r since we know the
temperature in the centre of the pipe already ( T f ) 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:
•
•
∂T
∂r
=−
h pi (T f − T pi )
r = r pi
kp
T (r po ) = T po
We know that T ( r ) = A ln r +
condition T (r po ) = T po we have :
B .Then by using the boundary
T ( rpo ) = A ln rpo + B = T po
(1.3)
Now we want to find the constants A and B, using the boundary conditions.
T (r ) = A ln r + B ⇒
⇒
st
∂T A
=
∂r r
∂T
∂r
=
r = rpi
and by using the 1 boundary condition
A
rpi
∂T
∂r
=−
r = r pi
h pi (T f − T pi )
kp
7
we have:
h pi (T f − T pi )
A
=−
rpi
kp
⇒ A=−
rpi h pi (T f − T pi )
(1.4)
kp
From (1.1) we have:
− (T f − T pi )rpi h pi =
Q
2π
(1.5)
Substitute (1.5) into (1.4) we have:
A=
From (1.3)
Q
2πk p
B = T po − A ln rpo = T po −
Q
ln rpo
2πk p
Finally, substitute A and B in (1.2) ,we have the following solution for the pipe
layer:
T (r ) =
 r 
Q
ln  + T po
2πk p  rpo 
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:
T ( y ) = gy + Tg .This is an asymptotic condition as y → ∞, x → ±∞ .
(1.6)
8
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.
9
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 ( ∇ φ = 0 ) can be derived from
Gauss’ Theorem (divergence theorem) and its corollaries (Green’s identities).
2
(The partial differential operator, ∇ or ∆ is called the Laplace operator, or just the
2
Laplacian).
10
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:
∂Fi
dV
V ∂xi
∫ Fi ni dS = ∫
S
where
ni =components of the unit vector normal to the surface S
Green’s first identity:
∂ψ
into Gauss’ Theorem, we have:
∂xi
By substituting Fi = φ
∂ψ
∫ φ ∂x
S
∂  ∂ψ
 φ
x
∂
∂xi
V
i
ni dS = ∫
i

dV

(2.1)
Then we use the chain rule which give us:
∂  ∂ψ
φ
∂xi  ∂xi
 ∂φ ∂ψ
 =
+ φ∇ 2ψ
 ∂xi ∂xi
(2.2)
Substitute (2.2) into the right hand side of (2.1) we get:
∫φ
S
∂ψ
∂φ ∂ψ
dS = ∫
dV + ∫ φ∇ 2ψdV
∂n
V ∂xi ∂xi
V
(2.3)
This is called the Green’s first identity.
11
Green’s second identity:
The Green’s first identity is also valid when interchanging
∫ψ
S
φ and ψ :
∂φ
∂ψ ∂φ
dS = ∫
dV + ∫ψ∇ 2φdV
∂n
V ∂xi ∂xi
V
(2.4)
Subtracting equation (2.4) from (2.3) gives Green’s second identity:
∫ (φ
S
∂ψ
∂φ
−ψ
)dS = ∫ (φ∇ 2ψ − ψ∇ 2φ )dV
∂n
∂n
V
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 ∇ T = 0 in D and
2
∂T
is known on C.
∂n
where D is a bounded (interior) domain with boundary C.
To begin with we have the following Green second identity:
∫ (φ∇ ψ −ψ∇ φ )dy = ∫ (φ
2
2
D
C
∂ψ
∂φ
−ψ
)dc
∂n
∂n
The Green’s function is designated as the fundamental solution.
G ( x, y ) =
− −
1
1
ln
2π | x − y |
−
where
x = ( x1 , x2 )
−
−
y = ( y1 , y 2 )
−
12
then ∇ x G = δ ( x − y ) for any fixed y
2
−
−
δ is the Dirac Delta function.
where
or ∇ y G = δ ( x − y ) for any fixed x ,
2
−
−
Strictly speaking it is defined through the integral
We let
−
−
∫ δ (x − y) f (x)dx = f (y)
φ ≡ T and ψ ≡ G
Where T=temperature, G=Green’s function.
Thus we have:
∫∫D (T∇
2
x
G − G∇ 2 xT )dx1dx2 = ∫ (T
C
∂G
∂T
− G )dc .
∂n
∂n
But ∇ T = 0 so we have :
2
∫∫D Tδ ( x− − y− )dx1dx2 = ∫ (T
C
⇒ T ( y1 , y 2 ) = ∫ (T
C
∂G
∂T
− G )dc
∂n
∂n
∂G
∂T
− G )dc
∂n
∂n
(2.5)
in the case that ( y1 , y 2 ) ∈ D
or :
0 = ∫ (T
C
∂G
∂T
− G )dc
∂n
∂n
if
( y1 , y 2 ) ∉ D
13
Suppose ∆G=0 inside a domain D
Suppose y ∈ D
D
∂D
ε=radius of the circle
y=centre of the circle
∂Ω ε =boundary of
the circle
ε
y
∂Ω ε
We choose G to be the solution of ∆G = 0 in D/ Ω ε , hence we have:
∫
∂Dε
(T
∂G
∂T
∂G
∂T
∂G
∂T
− G ) d x = 0 ⇒ ∫ (T
− G )d x + ∫ (T
− G )d x =0 (2.6)
−
∂n
∂n
∂n
∂n − ∂Ω
∂n
∂n −
∂D
ε
(from previous section)
We know G ( x, y )
−
=
−
1
1
ln
2π | x − y |
−
so | x − y |=| R |
−
−
−
Therefore,
G ( x, y ) =
−
−
∂G ∂G
1
1
1
1
=
=−
=−
ln ⇒
∂n ∂R
2πR
2πε
2π R
14
Let
ε →0
Then the 2nd part of (2.6) which is
∫
∂Ω ε
(T
∂G
∂T
− G )d x can be separated in
∂n −
∂n
two parts:
(i)
∂G
1
∫ T ( x) ∂n ( x)d x− = − 2πε ∫ T ( x)d x−
∂Ωε
∂Ωε
= −
1
∫
2πε ∂Ω
(T ( y ) + εT ′( y ) + O (ε 2 ))d x
−
ε
T ′( y )
 1

T ( y) −
+ O (ε ) ∫ d x
2π
 2πε
 ∂Ω −
= −
ε
T ′( y )
 1

+ O(ε ) 2πε
T ( y) −
2π
 2πε

= −
= − T ( y ) − εT ′( y ) + O (ε )
2
(ii)
∫
∂Ωε
G
− T ( y ) as ε → 0
∂T
1 1 ∂T
ln
dx
dx= ∫
∂n − ∂Ω 2π ε ∂n −
ε
=−
1
 ∂T

ln ε  ( y ) + O(ε ) 2πε
2π
 ∂n

 ∂T

( y ) + O(ε ) → 0 as ε → 0
 ∂n

= − ε ln ε 
15
Hence,
(2.6)
∫ (T
∂D
∂G
∂T
∂G
∂T
− G )d x = − ∫ (T
− G )d x = −[−T ( y )] = T ( y )
∂n
∂n −
∂n
∂n −
∂Ω
ε
y ∈ D as ε → 0 .
which is the same result as (2.5) but is defined in a slightly more rigorous
answer.
Suppose now, y ∈ ∂D (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 2 π in the derivations above.
D
ε=radius of the circle
y=centre of the circle
y
Hence,
∫ (T
∂D
∂G
∂T
∂G
∂T
1
− G )d x = − ∫ (T
− G )d x = T ( y )
∂n
∂n −
∂n
∂n − 2
∂Ω
y ∈ ∂D
(2.7)
ε
16
2.1.2 Unbounded Problem
Suppose ∆T=0 and ∆G=0 are outside the domain D (exterior problem)
Suppose y ∉ D
∂Ω
R
ε=radius of the small
circle
y=centre of the small
circle
∂Ω ε =boundary of the
small circle
R
D
∂Ω R =boundary of the
big circle
R=radius of the big
circle
∂Ω ε
The following equation is equal to zero because y is outside the domain as we
have mentioned in equation (2.5) when ( y1. , y 2 ) ∉ D .
∫ (T
∂D
∂G
∂T
∂G
∂T
∂G
∂T
− G ) ds ( x) + ∫ (T
− G )ds ( x) + ∫ (T
− G ) ds ( x) = 0
∂n
∂n
∂n
∂n
∂n
∂n
∂Ω
∂Ω
ε
R
(2.8)
⇒ ∫ (T
∂D
∂G
∂T
∂G
∂T
∂G
∂T
− G )ds ( x) = − ∫ (T
− G )ds ( x) − ∫ (T
− G )ds ( x) = 0
∂n
∂n
∂n
∂n
∂n
∂n
∂Ω
∂Ω
ε
R
Same as before
= - T(y)
(2.9)
17
Now, we are going to find (2.9) in the limit as R → ∞
∫ (T
∂Ω R
∂G
∂T
∂T
1
1
− G )ds ( x) = − ∫ T ( x)
ln R ds( x)
ds ( x) − ∫
∂n
∂n
∂n
2πR
∂Ω
∂Ω 2π
R
R
=0 (If ∆T=0)
(Corollary of Green’s 2)
=−
1
∫ T ( x)ds( x) − 0
2πR ∂Ω
R
So overall, (2.8)
∫ (T
∂D
∂G
∂T
1
− G )ds ( x) = T ( y ) −
∫ T ( x)ds( x)
∂n
∂n
2πR ∂Ω
(2.8)
R
18
2.1.3 Full pipe Problem
In the following plot the pipe is buried in the ground.
Γ1
x
D
y
Γ2
where
Γ1= ground surface
Γ2 =pipe
D=exterior environment (below Γ1 and outside Γ2 )
We have the following assumptions:
∆T=0 in D
∂T
=known on Γ1 =C1 (a constant)
∂n
∂T
=known on Γ2 =C2 (a constant)
∂n
Τhe asymptotic condition as y → ∞, x → ±∞
T → gy + Tg
Now, to solve the full pipe problem we will make a rectangular domain as
shown below:
19
2.1.4 Neumann Green’s function for a half plane Problem
(0 ,0 )
x= - R
Γ1 Ground
surface
x=R
R
Γε
ΩR
Γ2
x
y= R
y
where Γ2 =pipe
Γε =small circle
ε=radius of small circle
Ω R =domain
We want to find the integral equation of the domain Ω R .
In the Domain :
Known:
∇ 2T = 0
∇ 2 x G ( x, y ) = 0
− −
.
20
We choose Ĝ such that
∂Gˆ
= 0 on Γ1. (*)
∂n
1
1
1
1
Gˆ ( x, y ) =
+
ln
ln
− −
2π | x − y | 2π | x − y ′ |
−
−
−
−
where y ′ is the reflection of y in the line y = - z and where
x = ( x1 , x2 )
−
y = ( y1 , y 2 )
−
Gˆ ( x, y ) satisfies ∇ 2 xGˆ = δ ( x− y) + δ ( x − y′ ) .
−
− −
−
−
−
To find the integral equation of the domain we add the integral equations of
the pipe, the circle, the ground surface and the lines x = R, x = − R, y = R .
∫
∂Ω R
∂Gˆ ˆ ∂T
− G )ds = ∫ + ∫ + ∫ + ∫ + ∫ + ∫
(T
∂n
∂n
Γ
Γ ∩[ − R , R ]
x= R
x=− R
y=R
Γε
2
=0
(2.10)
1
The equation (2.10) is equal to zero from the earlier Green’s function.
Consider the lim (as R → ∞ )
The 1st integral is :
∫ (T
Γ2
The 2nd integral is :
∂Gˆ ˆ ∂T
∂Gˆ ˆ
− G )ds ( x) = ∫ (T
− GC 2 )ds ( x )
∂n
∂n
∂
n
Γ
2
∫
(T
Γ1 ∩[ − R , R ]
x =∞
∂Gˆ ˆ ∂T
∂Gˆ ˆ
− G ) ds ( x ) = ∫ (T
− GC1 )ds ( x )
∂n
∂n
∂n
x = −∞
=0 from (*)
The 3rd and 4th integrals are:
∫
x = R , y∈[ 0 , R ]
(T
∂Gˆ ˆ ∂T
∂Gˆ ˆ ∂T
(
T
− G )ds ( x)
− G )ds ( x) +
∫
∂n
∂n
∂n
∂n
x =− R , y∈[ 0 , R ]
21
We consider the asymptotic condition as y → ∞, x ± ∞
T → gy + Tg .
(2.11)
Therefore,
∂T
∂T
= 0 and
=g
∂y
∂x
We solve for
u = T − gy − Tg as y → ∞, x ± ∞ .
(2.12)
Hence if we substitute (2.11) into (2.12) we have :
u = T − gy − Tg = gy + Tg − gy − Tg = 0 .Thus u → 0 as y → ∞, x ± ∞
∂u ∂u
, →0
∂x ∂y
Thus the 3rd and 4th integrals are:
∫
x = R , y∈[ 0 , R ]
(T
∂Gˆ ˆ ∂T
∂Gˆ ˆ ∂T
− G )ds ( x ) +
− G ) ds ( x) =0
T
(
∫
∂n
∂n
∂n
∂n
x = − R , y∈[ 0 , R ]
=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:
1
1
1
1
+
Gˆ ( x, y ) =
ln
ln
− −
2π | x − y | 2π | x − y ′ |
−
where Φ ( x, y ) =
− −
−
−
−
1
1
1
1
′
ln
ln
, Φ ( x, y ) =
− −
2π | x − y |
2π | x − y ′ |
−
−
−
−
22
1
∂Φ
∂Φ
∂ 1
( x, y ) = −
( ln
=−
)
∂n( y ) − −
∂y2
∂y2 2π
( x1 − y1 ) 2 + ( x2 − y2 ) 2
3
−
1
∂Φ
 1
( x, y ) = −
( x1 − y1 ) 2 + ( x2 − y2 ) 2  − [( x1 − y1) 2 + ( x2 − y2 ) 2 ] 2 2( x2 − y2 )
2π
∂n( y ) − −
 2
=
∂Φ
( x, y′)
∂n( y ) − −
( x2 − y 2 )
1
2π [( x1 − y1 ) 2 + ( x2 − y 2 ) 2 ]
=−
and in the same way
( x2 + y 2 )
1
2π [( x1 − y1 ) 2 + ( x2 + y 2 ) 2 ]
∂Φ
Therefore, if we add ∂Φ ( x, y ),
( x, y ′) and y 2 = 0 then the result is zero:
∂n − − ∂n − −
∂Φ
 ∂Φ
′ 
=0
 ∂n ( x− , y− ) + ∂n ( x− , y− )

 y2 =0
For that reason , the 5th integral is zero.
So from (2.10)
-
0 = ∫ (T
Γ2
0 = ∫ (T
Γ2
∂Gˆ ˆ
∂Gˆ ˆ ∂T
− GC 2 )ds ( x) + ∫ ( −Gˆ C1 ) ds ( x ) + ∫ (T
− G )ds ( x)
∂n
n
∂n
∂
Γ
Γε
1
∂Gˆ ˆ
− GC 2 ) ds ( x ) + ∫ ( −Gˆ C1 ) ds ( x ) + T ( y1 , y 2 )
∂n
Γ
1
Therefore,
− T ( y1 , y 2 ) =
∫ (T
Γ2
∂Gˆ ˆ
− GC 2 )ds ( x) + ∫ (−Gˆ C1 ) ds ( x)
∂n
Γ
1
23
− T ( y1 , y 2 ) −
∂Gˆ
∫ ∂n T ( x)ds( x) = ∫ (−Gˆ C2 )ds( x) + ∫ (−Gˆ C1 )ds( x)
Γ
Γ
Γ
2
2
1
We set
∫ (−Gˆ C 2 )ds( x) + ∫ (−Gˆ C1 )ds( x) = F ( y)
Γ2
Γ1
Hence,
− T ( y1 , y 2 ) −
∂Gˆ
∫ ∂n T ( x)ds( x) = F ( y )
Γ
2
We are going to solve this integral equation numerically. The general integral
equation is of the following form:
u ( x) + ∫ Κ ( x, y )u ( y )dy = f ( x)
Γ
24
Chapter 3
Methods for solving a single integral equation
We study certain Fredholm integral equations of the 2nd kind,
u ( x) + ∫ Κ ( x, y )u ( y )dy = f ( x)
(3.1)
Γ
Where :
Kernel= Κ ( x, y ) is known
Γ is some closed boundary ,
Γ = (γ 1 (ς ), γ 2 (ς ))
where ς ∈ [0,2π ]
and (γ 1 , γ 2 ) are periodic functions
f (x) 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
2π
u ( x) + ∫ K ( x, y )u ( y )dy = f ( x)
0
We replace the integral by a quadrature rule. The easiest way is to use the
trapezium rule.
25
2π
h
∫ g ( y)dy ≈ 2 [ g (0) + 2 g (h) + 2 g (2h) + ...... + 2 g ((n − 1)h) + g (2π )]
0
Where
= h[ g (0) + g ( h) + ...... + g (( n − 1) h)]
h=
2π
n
[2π periodic function
⇒ g (0) = g (2π ) ].
( n =quadrature parameter)
In our case we replace u
by u n .
Therefore, we have:
u n ( x ) + h[ K ( x,0)u n (0) + K ( x, h)u n ( h) + .... + K ( x, (( n − 1) h)u n (( n − 1) h)] = f ( x )
∀x
We take x = 0, h,2h,3h,....., ( n − 1) h
1st equation when x = 0 :
u n (0) + h[ K (0,0)u n (0) + K (0, h)u n ( h) + .... + K (0, ((n − 1)h)u n ((n − 1) h)] = f (0)
2nd equation when x = h :
u n (h) + h[ K (h,0)u n (0) + K (h, h)u n ( h) + .... + K ( h, ((n − 1)h)u n ((n − 1) h)] = f (h)
3rd equation when x = 2h :
u n (2h) + h[ K (2h,0)u n (0) + K (2h, h)u n (h) + .... + K (2h, ((n − 1)h)u n ((n − 1) h)] = f (2h)
Last equation when x = ( n − 1)h :
u n ((n − 1)h) + h[ K ((n − 1)h,0)u n (0) + K ((n − 1)h, h)u n ( h) + .... +
+ K ((n − 1)h, ((n − 1)h)u n ((n − 1)h)] = f ((n − 1)h)
26
We have n equations with n unknowns:
u n (0), u n (h), u n (2h), u n (3h),......, u n ((n − 1)h)
We write it as a matrix
Ax = b
.......
hK (0, h)
hK (0,2h)
hK (0, (n − 1)h)
 1 + hK (0,0)

 hK (h,0)

1 + hK (h, h)
.........
hK (h,2h)
hK ( h, (n − 1)h)



A =  hK (2h,0)
1 + hK ( 2h,2h) ..........
hk (2h, h)
hK (2h, (n − 1)h)


........
............
............
.............
................


h[ K (n − 1) h,0) h[ K ( n − 1)h, h]
.........
........... 1 + h[ K (n − 1)h, ( n − 1)h]
 u n (0) 


 u n ( h) 
 u ( 2 h) 
n

x=
 u n (3h) 


......


 u ((n − 1)h) 

 n
f (0) 



f (h) 

 f (2h) 

b=
 f (3h) 
 ....... 


 f ((n − 1)h) 


27
•
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:
u ( x) + ∫ Κ ( x, y )u ( y )dy = f ( x)
Γ
Where the Kernel=K(x, y) =
1
sin( 2 x + y )
2π
Γ is a closed boundary from 0 to 2π
u(x)=cos( x ) ,
u(y)=cos( y)
28
Thus,
cos( x) +
⇒ cos( x) +
1 2π
∫ sin(2 x + y ) cos( y)dy = f ( x)
2π 0
1 2π
∫ [(sin(2 x) cos( y ) + sin( y ) cos(2 x)] cos( y)dy = f ( x)
2π 0
1 2π
⇒ cos( x) +
(sin(2 x) cos 2 ( y ) + sin( y ) cos(2 x) cos( y ))dy = f ( x)
∫
2π 0
sin(2 y )
cos(2 y ) + 1
1 2π
⇒ cos( x) +
x
+
x
)dy = f ( x)
cos(
2
)
(sin(
2
)
∫
2
2
2π 0
1
⇒ cos( x) +
2π
⇒ cos( x) +
2π
cos(2 x) sin( 2 y ) 
sin( 2 y ) y

+
−
x
x
sin(
2
)
sin(
2
)

 = f ( x)
4
4
2
0
1
(0 + π sin(2 x)) = f ( x)
2π
Hence,
1
f ( x) = cos( x) + sin( 2 x)
2
We program this example of a single periodic integral equation in Matlab and
we have the following results for the u − uexact L 2
where u exact =cos(x) and u is computed with the Nyström Method.
1
(Note: where
u − u exact
2
 2π
2
(
u
u
)
=
−
exact
 )
∫
L2

0
29
TABLE 1
u − u exact
n
L2
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 u − u exact
L2
. We always get zero
to machine precision.
1
We have used
u − u exact
2
 2π
2
(
u
u
)
=
−
exact
 )
∫
L2

0

^ n −1

j =1
^
(3.2)
^

= h ∑ [u ( j h) − u exact ( j h)] 
2
1
2

^
with h = h 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
u exact everywhere because we have exact formula to work out an
approximation to u which valid anywhere. So we can use the following formula
where Pn u (x) is called the trigonometric interpolating polynomial [10].
Pn u ( x) =
Where m =
m −1
 1 

+
u
jh
1
2
cos(k ( x − jh)) + cos(m( x − jh)) 
(
)

∑
∑


j =0
k =1
 2m 
2 m −1
n
2
30
3.1.2 Example of a single non-periodic integral equation
Analytical solution:
u ( x) + ∫ Κ ( x, y )u ( y )dy = f ( x)
Γ
Where
2
the Kernel=K(x, y) = x y
2
Γ is a closed boundary from 0 to 1
u(x)=1 +
5x 2
,
12
u(y)= 1 +
5y2
12
So,
 5x 2  1
5y2 2 2
1 +
 + ∫ (1 +
) x y dy = f ( x)
12
12  0

 5x 2  1 2 2 5x 2 y 4
 + ∫ ( x y +
⇒ 1 +
)dy = f ( x)
12
12

 0
1
 5x 2  1 2 3 1 2 5 
 + x y + x y  = f ( x)
⇒ 1 +
12   3
12
0

1
 5x 2 
1 
1
 + x 2  y 3 + y 5  = f ( x)
⇒ 1 +
12 
12  0
3

 5x 2 
1 1 
 + x 2  +  = f ( x)
⇒ 1 +
12 
 3 12 

5 1 1
⇒ 1 + x 2  + +  = f ( x)
 12 3 12 
31
Hence,
f ( x) = 1 +
5 2
x
6
We program this example of a single periodic integral equation in Matlab and
we have the following results for the u − uexact L 2 .
1
We have used
u − u exact
2
 2π
2
(
u
u
)
=
−
exact
 )
∫
L2

0

^ n −1

j =1
^
(3.2)
^

= h ∑ [u ( j h) − u exact ( j h)] 
2
1
2

TABLE 2
n
u − u exact
2
0.5000
4
0.1221
8
0.0443
16
0.0191
32
0.0089
L2
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.
32
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
2π
u ( x) + ∫ K ( x, y )u ( y )dy = f ( x)
(3.3)
0
We seek an approximation un of the form:
n −1
un = ∑ φ j ( x)u ( x j )
where
j =0
φ j = basis functions
Substitute u into (3.3)
2π
n −1
n −1
∑φ j ( x)u ( x j ) + ∫ K ( x, y )∑φ j ( y )u ( x j )dy =
j =0
j =0
f ( x)
0
n −1
2π
j =0
0
⇒ ∑ [φ j ( x) + ∫ K ( x, y )φ j ( y )dy ]u ( x j ) = f ( x)
(3.4)
So we have one equation with n unknowns → the values of u ( x j ) .
^
^
To get n equations we fix (3.4) to hold at n points i.e. take x = x1 ,....., xn and
then we will have n equations with n unknowns.
^
^
If we choose x1 ,....., xn to be the same points as x j
n −1
2π
j =0
0
⇒ ∑ [φ j ( xm ) + ∫ K ( xm , y )φ j ( y )dy ]u ( x j ) = f ( xm )
33
We replace the integral by a quadrature rule as in Nyström Method. The
easiest way is to use the trapezium rule.
2π
h
∫ g ( y)dy ≈ 2 [ g (0) + 2 g (h) + 2 g (2h) + ...... + 2 g ((n − 1)h) + g (2π )]
0
= h[ g (0) + g ( h) + ...... + g (( n − 1) h)]
Where
h=
2π
n
(periodic function)
( n =quadrature parameter)
In our case we replace
φ by φ n .
Therefore, we have:
φn ( x) + h[ K ( x,0)φn (0) + K ( x, h)φn (h) + .... + K ( x, ((n − 1)h)φn ((n − 1)h)] ⋅ u ( x j ) = f ( x)
∀x
We take x = 0, h,2h,3h,....., ( n − 1) h
We define u ( x j ) = u n
We have n equations with n unknowns:
u n (0), u n (h), u n (2h), u n (3h),......, u n ((n − 1)h)
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.
34
Chapter 4
Interior Problem for Laplace’s equation
Consider
∆u = 0 ,
Ω is a circle with radius R
∂u
=g
∂n
Ω
R
∂u
= g
∂n
We already know that the Fundamental solution G( x, y ) =
− −
1
1
and
ln
2π | x − y |
−
−
that the form of the general integral equation is:
u ( x) + ∫ Κ ( x, y )u ( y )dy = f ( x) .
Γ
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.
35
From (2.7) of the bounded problem we have:
∫ (T
∂D
∂G
∂T
1
− G )d x = T ( y )
∂n
∂n − 2
1
∂G
∂T
⇒ − T ( y) + ∫ T
d x = ∫G d x
2
∂n − ∂D ∂n −
∂D
Where in this case K(x,y)=
f ( x) = ∫ (G
∂D
y ∈ ∂D
y ∈ ∂D
∂G
∂n
∂T
)d x = ∫ (Gg )d x
−
∂n − ∂D
So we have to solve :
− T ( y) + 2 ∫ T
∂D
Firstly, we have to find
We know G( x, y ) =
− −
∂G
∂T
d x = 2 ∫G d x
∂n −
∂n −
∂D
∂G
with respect to x.
∂n
1
1
ln
where x = ( x1 , x2 )
−
2π | x − y |
−
−
y = ( y1 , y 2 )
−
We substitute x and y in G and then we have the following expression for G:
−
G( x, y ) =
− −
−
1
ln
2π
1
( x1 − y1 )
2
+ ( x2 − y 2 ) 2
 ∂G 


 n1 ( x)   ∂x1 
∂G
∂G
∂G
 ⋅
= n( x ).∇ x G = 
= n1 ( x)
+ n2 ( x )
∂n(x) − −
∂x1
∂x2
 n2 ( x)   ∂G 
 ∂x 
 2
(4.1)
36
∂G 1
 1
( x1 − y1 ) 2 + ( x2 − y 2 ) 2  −  ( x1 − y1 ) 2 + ( x2 − y 2 ) 2
=
∂x1 2π
 2
[
]
−
3
2
2( x1 − y1 )
1
2 2
1 [( x1 − y1 ) + ( x2 − y 2 ) ]
∂G
⋅ ( x1 − y1 )
=−
3
2π
∂x1
2
2 2
[( x1 − y1 ) + ( x2 − y 2 ) ]
2
( x1 − y1 )
1
∂G
=−
∂x1
2π [( x1 − y1 ) 2 + ( x2 − y 2 ) 2 ]
and in the same way
( x2 − y 2 )
1
∂G
=−
∂x2
2π [( x1 − y1 ) 2 + ( x2 − y 2 ) 2 ]
We set x = (cos(ς ), sin(ς )) ⇒ n1 = cos ς , n2 = sin ς
y = (cos(t ), sin(t ))
After that we replace
∂G ∂G
,
, n1 and n2 into equation (4.1)
∂x1 ∂x2
And finally we have:
( x2 − y 2 )
( x1 − y1 )
∂G
1
1
=−
sin(ς )
cos(ς ) −
2
2
∂n
2π [( x1 − y1 ) 2 + ( x2 − y 2 ) 2 ]
2π [( x1 − y1 ) + ( x2 − y 2 ) ]
⇒
∂G
1
[cos(ς )( x1 − y1 ) + sin(ς )( x2 − y 2 )]
=−
2
2π [( x1 − y1 ) + ( x2 − y 2 ) 2 ]
∂n
We also replace
x1 = cos(ς )
y1 = cos(t )
x2 = sin(ς )
y 2 = sin(t )
37
⇒
[cos ς (cos ς − cos t ) + sin ς (sin ς − sin t )]
∂G
=−
∂n
2π [(cos(ς ) − cos(t )) 2 + (sin(ς ) − sin(t )) 2 ]
(Note: [(cos ς − cos t ) + (sin ς − sin t ) ] = 2 − 2(cos ς cos t + sin ς sin t )
2
2
= 2 − 2 cos(ς − t )
ς −t 
= 2 − 2 cos 2

 2 
ς −t 
= 2 − 2[1 − 2 sin 2 
]
 2 
2ς −t 
= 2 − 2 + 4 sin 

 2 
ς − t 
= 4 sin 2 
 ).
 2 
⇒
∂G − cos 2 ς + cos ς cos t − sin 2 ς + sin ς sin t
=
ς −t
∂n
2π ⋅ 4 sin 2 (
)
2
∂G − (cos 2 ς + sin 2 ς ) + cos ς cos t + sin ς sin t
=
⇒
ς −t
∂n
8π sin 2 (
)
2
∂G − (cos 2 ς + sin 2 ς ) + cos ς cos t + sin ς sin t
=
⇒
ς −t
∂n
8π sin 2 (
)
2
(Note: Trigonometric identity: cos
⇒
2
ς + sin 2 ς = 1)
∂G − 1 + cos ς cos t + sin ς sin t
=
ς −t
∂n
8π sin 2 (
)
2
38
⇒
∂G
1 (1 − (cos ς cos t + sin ς sin t ))
=−
ς −t
∂n
8π
sin 2 (
)
2
Now we are going to simplify the numerator of this fraction.
1 − (cos ς cos t + sin ς sin t ) = 1 − cos(ς − t )
= 1 − (1 − 2 sin 2 (
= 2 sin 2 (
ς −t
2
ς −t
2
))
)
(Note: Trigonometric identities: cos(ς − t ) = cos ς cos t + sin ς sin t
cos 2ς = cos 2 ς − sin 2 ς
⇒ cos 2ς = 1 − sin 2 ς − sin 2 ς
⇒ cos 2ς = 1 − 2 sin 2 ς ).
Finally,
2 sin 2 (
)
2
ς −t
sin 2 (
)
2
∂G
1
=−
⇒
∂n
8π
⇒
ς −t
∂G
1
=−
4π
∂n
(4.2)
Hence our integral equation is:
− T ( y) − 2 ∫ T
∂D
1
d x = 2 ∫ Ggd x
−
4π −
∂D
f ( y)
We know G( x, y ) =
− −
1
1
therefore
ln
2π | x − y |
−
−
39
2π
f ( y ) = ∫ 2Ggdx =
0
1
1 2π
1
1 2π
2
ln
g
(
x
)
dx
ln
g ( x)dx
=
∫
∫
2π 0
π
0
x− y
x− y
−
−
−
−
Lets substitute xˆ = x − y ⇒ x = xˆ + y
dx = dxˆ
g ( x) = g ( xˆ + y )
f ( y) =
1 2π − y
π
∫
−y
2π
1
1
ln g ( xˆ + y )dxˆ = ∫ ln g ( x + y )dx
xˆ
x
0
because is periodic
To find f ( y ) which is equal to
∫ Ggd x−
we have to apply some quadrature
∂D
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:
& −1  ( 2 + 1) h 
j

f ( x)dx ≈ h ∑ f 
2
J =0 

where h =
2π
&
& −1 
  (2 j + 1)h
1
2
g 
+
g ( x + y )dx ≈ h ∑ ln
 (2 + 1)h  
x
2
j =0
 
 j

y 

2π
∫
0
Therefore, our f ( y ) is:
2π
f ( y ) = ∫ ln
0
where
x=
(2 j + 1)h
2
.
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).
40
Below are the graphs of the numerical solution of the integral equation:
− T ( y) + 2 ∫ T
∂D
∂G
d x = 2 ∫ Ggd x for different values of n.
−
∂n − ∂D
Where
∂G
1
=−
∂n
4π
2π
f ( y ) = 2 ∫ 2G gd x = ∫ ln
∂D
−
0
1
g ( x + y )dx
x
g (θ ) = cos(10θ ) + sin(12θ )
0.6
n=4
n=8
n=16
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
1
2
3
4
5
6
Figure 4.1: Numerical solution of the interior problem for Laplace’s equation
for n=4,8,16.
41
1
n=32
n=64
n=128
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
1
2
3
4
5
6
7
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.
42
4.1 Separation of variables in polar coordinates
We will use polar coordinates and separation of variables to solve analytically
the interior problem:
1 ∂  ∂u  1 ∂ 2 u
=0
∇ u=
r  +
r ∂r  ∂r  r 2 ∂θ 2
2
(4.3)
∂ u 1 ∂u 1 ∂ u
=0
+
+
∂r 2 r ∂r r 2 ∂θ 2
2
=
2
We seek a solution of the form:
u=R(r). Θ( θ )
Τherefore,
2
1 ∂ ∂
 1 ∂
(R(r )Θ(θ ) ) = 0
r
(
R
(
r
)
(
θ
)
Θ
+


2
2
θ
r ∂r  ∂r
r
∂

1 ∂  ∂R 
1 ∂ 2Θ
=0
⇒ Θ(θ )
 r  + R(r ) 2
r ∂θ 2
r ∂r  ∂r 
2
Multiplying by r /RΘ gives us:
r ∂  ∂R  1 ∂ 2 Θ
=0
⇒
r  +
R ∂r  ∂r  Θ ∂θ 2
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,
1 ∂ 2Θ
r ∂  ∂R 
⇒
r  = −
Θ ∂θ 2
R ∂r  ∂r 
43
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,
1 ∂ 2Θ
r ∂  ∂R 
= c (constant)
⇒
r  = −
R ∂r  ∂r 
Θ ∂θ 2
The ordinary differential equations we get are then:
∂  ∂R 
 r  − cR = 0
∂r  ∂r 
(a)
r
(b)
∂ 2Θ
+ cΘ = 0
∂θ 2
We take (b)
Try
Θ= e
λθ
∂ 2Θ
+ cΘ = 0
∂θ 2
so
⇒ Θ = Aeθ
λ2 e λθ + ce λθ = 0 ⇒ λ2 + c = 0 ⇒ λ = ± − c
−c
+ Be −θ
−c
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
44
Thus when c>0 ⇒ Θ = Ae
iθ c
+ Be −iθ
(
c
(
= A cos( c θ)+ B sin( c θ)
2
So write c=ν ⇒ Θ = A cos(νθ ) + B sin(νθ )
We take (a) r
∂  ∂R 
2
 r  −ν R = 0
∂r  ∂r 
(4.4)
(where c=ν )
2
 ∂ 2 R ∂R 
 − ν 2 R = 0
⇒ r  r 2 +
∂r 
 ∂r
∂R
∂2R
−ν 2 R = 0
+r
⇒r
2
∂r
∂r
2
This is an Euler differential equation. The general solution to this simple case
of Euler’s ordinary differential equation is given as:
R(r ) = C1r ν + C 2 r −ν
(4.5)
Combining equations (4.4) and (4.5) we have:
u (r ,θ ) = R(r )Θ(θ )
⇒ u (r ,θ ) = (C1r ν + C2 r −ν ) . ( A cos(νθ ) + B sin(νθ ))
(4.6)
45
As r → 0 the term involving r
take C 2 = 0 .
−ν
is unbounded. The only way to fix this is to
Therefore,
u (r ,θ ) = rν ( A cos(νθ ) + B sin(νθ ))
We know that equation (b)
for any ν .
∂ 2Θ
+ cΘ = 0 which its general solution is :
∂θ 2
Θ = A cos(νθ ) + B sin(νθ )
We use the boundary conditions Θ (0) =Θ (2 π )
Θ (0) =A
Θ (2 π )=Acos(2 πν )
(i.e. periodic)
⇔ Θ (0) =Θ (2 π )
Thus,
cos (2 πν ) ) =1 ⇒ v = 0,±1,±2,....
Therefore the general solution of the problem is:
∞
u (r ,θ ) = ∑ rν ( Aν cos(νθ ) + Bν sin(νθ ) )
(4.7)
ν =0
In our problem
∂u
∂u ∂u
= g for a circle then
=
= g on r=a
∂n
∂r ∂n
∂u ∞
= ∑ ma m−1 ( Am cos(mθ ) + Bm sin(mθ ) ) = g (θ )
∂r m=0
(4.8)
46
The interior Neumann problem is solvable if and only if:
2π
∫ g (θ )dθ = 0
0
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
0 ≤ θ ≤ 2π .
The important observation is that sine and cosine functions of different
frequency are orthogonal. This means that, when multiplied and integrated,
give zero result:
2π
∫ cos(mθ ) cos(nθ )dθ =
π
if
m=n≠0
2π
if
m=n=0
0
if
m≠n
π
if
0
if
m=n=0
0
if
m≠n
0
2π
∫ sin(mθ ) sin(nθ )dθ =
0
m=n≠0
2π
∫ cos(mθ ) sin(nθ )dθ =0
,
all m, n
0
(Fourier Series and Applications, Beatrice Pelloni,2006)
47
Fix an integer value n ≠ 0 and multiply (4.8) by cos( nθ ) and sin( nθ ) , respectively.
Then perform the integration term by term:
2π
∞
∫ g (θ ) cos(nθ )dθ =m∑=0[ma
m −1
2π
2π
0
0
Am ∫ cos(mθ ) cos(nθ )dθ + Bm ∫ cos(nθ ) sin(mθ )dθ ]
0
= na n−1πAn
=0
Hence,
1 2π
An =
g (θ ) cos(nθ )dθ
πna n−1 ∫0
n≠0
While
2π
∞
2π
2π
0
m=0
0
0
m −1
∫ g (θ ) sin(nθ )dθ =ma ∑ [ Am ∫ cos(mθ ) sin(nθ )dθ + Bm ∫ sin(nθ ) sin(mθ )dθ ]
=0
= na πBn
n −1
Hence,
Bn =
1 2π
g (θ ) sin( nθ )dθ
πna n −1 ∫0
n≠0
The solution of the interior problem is not unique because
If
∆(u + c) = 0
∆u = 0
⇒
∂u
=0
∂n
∂ (u + c)
=0
∂n
for all constants c.
48
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:
∫ (T
∂D
1
∂G
∂T
− G )ds ( x) = T ( y ) −
∫ T ( x)ds ( x)
∂n
∂n
2πR ∂Ω
(5.1)
R
What happens as R → ∞ ?
For the half plane problem we have the asymptotic condition as y→∞, x →±∞
T → Tgg ( y ) = gy + Tg
We solve for
so ∂T = g
∂y
u=T- Tgg and we consider
u=T- gy − Tg
As R → ∞ then u → 0 .
But because we have exterior problem and as R → ∞ we need T(R)=o(1)
i.e. T ( R ) → 0
Therefore, the part
1
∫ T ( x)ds( x) of equation (5.1) disappears and then
2πR ∂Ω
R
we only have :
∫ (T
∂D
∂G
∂T
− G )ds ( x) = T ( y )
∂n
∂n
(5.2)
49
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) :
u (r ,θ ) = (C1r ν + C2 r −ν ) . ( A cos(νθ ) + B sin(νθ ))
except
as r → ∞ the solution tends to zero. Thus, we take C1=0 and C2≠0 .
u (r ,θ ) = r −ν . ( A cos(νθ ) + B sin(νθ ))
The exterior Neumann problem is uniquely solvable if and only if:
2π
∫ g (θ )dθ = 0
0
(Theorem 6.28, Kress ‘Linear Integral Equations’) .
Below are the graphs of the numerical solution integral equation:
50
− T ( y) + 2 ∫ T
∂D
∂G
d x = 2 ∫ Ggd x for different values of n.
−
∂n − ∂D
Where
∂G 1
∂G
which is the same with the
of the interior problem
=
∂n
∂n 4π
but with different sign.
2π
f ( y ) = 2 ∫ 2G gd x = ∫ ln
∂D
−
0
1
g ( x + y )dx
x
g (θ ) = cos θ
3
n=4
n=8
n=16
2
1
0
-1
-2
-3
0
1
2
3
4
5
6
Figure 5.1: The numerical solution of the exterior problem for Laplace’s
equation for n=4,8,16.
51
3
n=32
n=64
n=128
2
1
0
-1
-2
-3
0
1
2
3
4
5
6
7
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.
52
Chapter 6
Full pipe flow Problem
To find the general integral equation for the full pipe flow problem we assume
the following plot.
n
∂T
∂T
=−
= C1
∂y
∂n
Y= -z
∆Τ=0
n
x
∂T
∂T
= C2
=−
∂n
∂r
y
where
C1 , C 2 are constants.
We consider the asymptotic condition as y → ∞, x ± ∞
T → gy + Tg .
where
Tg is a constant.
u = T − gy − Tg as y → ∞, x ± ∞ and we end up :
u → 0 as y → ∞, x ± ∞
We solve for
∂u ∂T
− g = C1 − g
=
∂y ∂y
⇒
∂u
∂u
= − = g − C1 = C3
∂y
∂n
53
The following is an integral equation approach:
∫ (u
∞
∂G
∂u
1
− G )ds( x) + ∫ (−GC3 )ds( x) + u ( x) = 0 if x is on the
∂n
∂n
2
−∞
boundary of the pipe.
We assume that C3 is zero .Therefore, we have the following integral
equation:
1
∂u
∂G
u ( x) = ∫ (G − u )ds( x) + 0
2
∂n
∂n
Γ
1
∂G
∂u
⇒ u ( x) + ∫ (u )ds( x) = ∫ (G )ds( x)
2
∂n
∂n
Γ
Γ
This integral equation is the same as the integral equation of interior and
exterior problem except that here the Green’s function is:
^
G ( x, y ) =
− −
We want to find
1
1
1
1
+
ln
ln
2π | x − y | 2π | x − y ′ |
−
−
−
(6.1)
−
∂Ĝ
.
dn
^
We separate G ( x, y ) into two parts.
− −
The 1st part is G( x, y ) =
− −
1
1
and the
ln
2π | x − y |
−
2nd part we call it P( x, y ′) =
−
−
−
1
1
ln
2π | x − y ′ |
−
We already know from (4.2)
−
∂G
1
=−
∂n
4π
54
Thus, now we want to find
We know P( x, y ′) =
−
−
∂P
.
∂n
1
1
where x = ( x1 , x2 )
ln
−
2π | x − y ′ |
−
−
′
y = ( y1 ,−2 z − y 2 )
−
′
We substitute x and y in P and then we have the following expression for P:
−
′
P( x, y ) =
− −
−
1
ln
2π
1
(x1 − y1 )2 + ( x2 + 2 z + y2 ) 2
 ∂P 


 n1 ( x)   ∂x1 
∂P
∂P
∂P
 ⋅
= n( x ).∇ x P = 
= n1 ( x)
+ n2 ( x )
∂n − −
∂x1
∂x2
 n2 ( x)   ∂P 
 ∂x 
 2
(6.2)
1
∂P
 1
( x1 − y1 ) 2 + ( x2 + 2 z + y 2 ) 2  −  ( x1 − y1 ) 2 + ( x2 + 2 z + y 2 ) 2
=
∂x1 2π
 2
[
]
−
3
2
1
2 2
∂P
1 [( x1 − y1 ) + ( x2 + 2 z + y 2 ) ]
=−
⋅ ( x1 − y1 )
3
∂x1
2π
[( x1 − y1 ) 2 + ( x2 + 2 z + y 2 ) 2 ] 2
2
( x1 − y1 )
1
∂P
=−
2
∂x1
2π [( x1 − y1 ) + ( x2 + 2 z + y 2 ) 2 ]
and in the same way
( x2 + 2 z + y 2 )
∂P
1
=−
∂x2
2π [( x1 − y1 ) 2 + ( x2 + 2 z + y 2 ) 2 ]
We set x = (cos(ς ), sin(ς )) ⇒ n1 = cos ς , n2 = sin ς
y = (cos(t ), sin(t ))
55
2( x1 − y1 )
After that we replace
∂P ∂P
,
, n1 and n2 into equation (6.2)
∂x1 ∂x2
And finally we have:
( x2 + 2 z + y 2 )
( x1 − y1 )
1
1
∂P
=−
sin(ς )
cos(ς ) −
2
2
∂n
2π [( x1 − y1 ) 2 + ( x2 + 2 z + y 2 ) 2 ]
2π [( x1 − y1 ) + ( x2 + 2 z + y 2 ) ]
⇒
∂P
1
[cos(ς )( x1 − y1 ) + sin(ς )( x2 + 2 z + y 2 )]
=−
2
2π [( x1 − y1 ) + ( x2 + 2 z + y 2 ) 2 ]
∂n
We also replace
⇒
x1 = cos(ς )
y1 = cos(t )
x2 = sin(ς )
y 2 = sin(t )
[cos ς (cos ς − cos t ) + sin ς (sin ς + 2 z + sin t )]
∂P
=−
2π [(cos(ς ) − cos(t )) 2 + (sin(ς ) + 2 z + sin(t )) 2 ]
∂n
(Note:
[(cos ς − cos t ) 2 + (sin ς + 2 z + sin t ) 2 ] = 2 − 2(cos ς cos t − sin ς sin t )
+ 4 z 2 + 4 z sin ς + 4 z sin t
= 2 − 2 cos(ς + t ) + 4 z 2
+ 4 z sin ς + 4 z sin t
ς +t 
2
= 2 − 2 cos 2
 + 4z
 2 
+ 4 z sin ς + 4 z sin t
ς + t 
= 2 − 2[1 − 2 sin 2 
]
 2 
+ 4 z 2 + 4 z sin ς + 4 z sin t
56
ς +t 
2
= 4 sin 2 
 + 4z +
 2 
4 z sin ς + 4 z sin t
).
⇒
∂P − cos 2 ς + cos ς cos t − sin 2 ς − sin ς sin t − 2 z sin ς
=
ς +t
∂n
2π ⋅ [4 sin 2 (
) + 4 z 2 + 4 z sin ς + 4 z sin t ]
2
⇒
⇒
∂P
=
∂n
− 1 + cos ς cos t − sin ς sin t − 2 z sin ς
ς +t
2π ⋅ [ 4 sin 2 (
) + 4 z 2 + 4 z sin ς + 4 z sin t ]
2
1 (1 − (cos ς cos t − sin ς sin t ) + 2 z sin ς
∂P
=−
ς +t
∂n
2π
) + 4 z 2 + 4 z sin ς + 4 z sin t
4 sin 2 (
2
(Note: Trigonometric identity: cos
2
ς + sin 2 ς = 1)
Now we are going to simplify the numerator of this fraction.
1 − (cos ς cos t − sin ς sin t ) = 1 − cos(ς + t )
= 1 − (1 − 2 sin 2 (
= 2 sin 2 (
ς +t
2
ς +t
2
))
)
(Note: Trigonometric identities: cos(ς + t ) = cos ς cos t − sin ς sin t
cos 2ς = cos 2 ς − sin 2 ς
⇒ cos 2ς = 1 − sin 2 ς − sin 2 ς
⇒ cos 2ς = 1 − 2 sin 2 ς ).
ς +t 
2 sin 2 
 + 2 z sin ς
1
∂P
2 

=−
⇒
2π 4 sin 2 ( ς + t ) + 4 z 2 + 4 z sin ς + 4 z sin t
∂n
2
57
Finally,
ς +t 
2 sin 2 
 + 2 z sin ς
∂Ĝ ∂G ∂P
1
1
2 

=
=−
+
+−
dn ∂n ∂n
4π
2π 4 sin 2 ( ς + t ) + 4 z 2 + 4 z sin ς + 4 z sin t
2
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:
− T ( y) + 2 ∫ T
∂D
∂G
d x = 2 ∫ Ggd x for different values of n.
−
∂n − ∂D
Where
ς +t 
2 sin 2 
 + 2 z sin ς
∂G
1
1
2 

=−
+−
∂n
4π
2π 4 sin 2 ( ς + t ) + 4 z 2 + 4 z sin ς + 4 z sin t
2
2π
f ( y ) = 2 ∫ 2G gd x = ∫ ln
∂D
−
0
1
g ( x + y )dx
x
g (θ ) = cos(10θ )
58
3
n=4
n=8
n=16
2
1
0
-1
-2
-3
-4
-5
0
1
2
3
4
5
6
Figure 6.1: The numerical solution of the full pipe flow problem for n=4,8,16.
59
4
n=32
n=64
3
2
1
0
-1
-2
-3
-4
-5
0
1
2
3
4
5
6
7
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.
60
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 nonperiodic function. We end up that the periodic function is faster than the nonperiodic 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.
61
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
62