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BEM 3/8/09
1
University of Hertfordshire
Department of Mathematics
Study on the Boundary Element Method
by Using a Spreadsheet
Wattana Toutip
Technical Report 1 May 1998
Abstract
In recent years the Boundary Element Method (BEM) has been
developed in various ways to solve problems in applied science and engineering.
Using spreadsheet method is one of a variety of those method offered the possibility
of implementations without the need of high-level programming language. The
purpose of this report is to explain the basic theory for the readers who have
mathematical background at undergraduate level. In addition it illustrate step by step
to implement a spreadsheet for solving two-dimensional potential problems.
BEM 3/8/09
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Introduction
The boundary Element Method (BEM) has become an important tool for solving
problems in applied science and engineering. It is new-coming compared with those
of Finite Different Method (FDM) and Finite Element Method (FEM). Although
FEM currently plays the famous roles, many users of FEM do not understand the
method itself, using it as a set of rules which lead to approximate results (Paris and
Canas,1997) and it is clear that from a user point of view BEM will always lead to a
much simpler set of rules than FEM.
In recent years an increasing amount of research work has been carried out on the
application of BEM. Consequently it is necessary to simplify this method in various
ways. Using spreadsheet method is one of the most powerful methods for solving
two-dimensional potential problems and need not high-level programming language.
Furthermore the spreadsheet facilities are very convenient for investigation of the
properties of the solutions such as convergence and for changing the geometry or
boundary conditions (Davies and Crann, 1996)
This paper is an attempt to explain the basic theory of BEM and intended mainly for
users who want to understand the mathematical foundation of the method by using
only undergraduate mathematical background. In addition, to make it clear we
illustrate step by step how to implement spreadsheet based on the basic theory for
solving a few two-dimensional potential problems.
Finally we assume that the reader familiar with using “ Excel “, especially relative
and absolute address, replication, named variables and the use of options such as
Calculation, Solver, etc. We shall use the notation R and R to denote replication
across and replicate down respectively.
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1. Preliminary Concept
In this section we briefly discuss some mathematical background which directly use
in the Boundary Element Method (BEM). The first part is concerned with area of a
triangle in rectangular co-ordinate. The second is on the Gaussian Integration which
is one of the most powerful tool in numerical integration. The rest of this section
contain a number of necessary basic theory used in BEM.
1.1 Area of a triangle
Suppose that P1 ),( 11 yx , P2 ),( 22 yx and P3 ),( 33 yx be the end points of a triangle as
shown in Figure 1
We see that the area of this triangle is obtained by vu2
1
By using the cross product in 3-dimension space and rearrange the result we obtain
Area x y x y x y x y x y x y 1 2 2 3 3 1 2 1 3 2 1 3 (1.1.1)
1.2 Gaussian Integration
To approximate the definite integral f x dxa
b
( ) we can change variables by using
linear transformation as shown in Figure 2.
P2
P1 P3
u
v
Figure 1: Triangle comprised from vector u and v
Figure 2: Transformation from the domain bxa to 11 t
transform
x
)(xf )(xf
a b -1 1 t
)(tf
)(tf
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Suppose BAtx then BAa )1( and BAb )1( .
By solving these equations we obtain the transformation
xb a b a
t
( ) ( )2 2
(1.2.1)
Hence
I f x dxb a
f b a b a t dta
b
( )( )
( ( ) ( ) )2
12
1
1
and we can approximate the integral I by
I b a w f b a b a tii
N
i
1
2 1
12( ) ( ( ( ) )) (1.2.2)
where wi and t i we can take from the Gauss - Legendre N-point Method as
shown in Table 1.2.1 which one can see the details in [2] or in general numerical
method textbooks.
Table 1.2.1 Values for Gaussian Quadrature
Number of
Points
Value of t Value of w Valid up to
degree
1 0 2 1
2 -0.57735027
0.57735027
1.0
1.0
3
3
-0.77459667
0.0
0.77459667
0.55555555
0.88888889
0.55555555
5
4
-0.86113631
-0.33998104
0.33998104
0.86113631
7
1.3 Potential Problem
Let us consider a scalar function , defined on a region D bounded by the closed
curve C , is called a “harmonic function “ if it is a solution of the Laplace equation
2 0( )x , x D
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Suppose that u satisfies Laplace equation so that
2 0u in D (1.3.1)
subject to Dirichlet condition
u u s 0 ( ) on C0
and the Neumann condition
u
nq q s 1( ) on C1
where D is a two - dimensional region bounded by the closed curve C = C0 + C1 as
shown in Figure 3
u u s 0 ( )
u
nq q s 1( )
2 0u
The combination with these boundary conditions is called a “ Potential Problem “. It
has become traditional in the boundary element literature to write the flux term,
u
n ,
as q. Problems governed by the Laplace equation appear in different fields of applied
science and engineering.
1.4 Green’s Theorem
Suppose u and v be two scalar functions defined in D bounded by the curve C in
two - dimensional space which are continuous and admit continuous partial
derivatives.
We have
( )u v u v u v2
and ( )v u v u u v2
C0
C
1
D
n
Figure 3: Boundary condition of the potential problem
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Taking the difference between these two equations and integrating over D then
DD
dAuvvudAuvvu 22 (1.4.1)
Applying the Gauss Theorem to the left hand side then
CD
uvvudAuvvu .)()( nds
Hence
dsn
uv
n
vudAuvvu
CD
(1.4.2)
Equation (1.4.1) after (1.4.2) we obtain
dsn
uv
n
vudAuvvu
CD
22
(1.4.3)
This is the expression of the well known Second Green Theorem.
1.5 Fundamental Solution of the Laplace Equation
Suppose that R is the position vector of a point, Q, relative to point, P, inside D.
Surround P by a small disc, center P radius , as shown in Figure 4.
Let 0rrR , R = R and u*( R ) is the fundamental solution of the Laplace
equation which satisfies
1qq
0uu
C0
Q R
D P
D
Figure 4: Neighborhood of the point p in the domain D
n
C1
r
0r
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(*2 u )0r
where )(r is Dirac delta function which property
0)( r except at 0rr
and D
f ( r )() 0r dA = f )( 0r
We see that
0*2 u
everywhere except at 0r
To find the fundamental solution we first find the solution of the above homogeneous
equation. In cylindrical co-ordinates , by the fact that
22
2 2
2
2
1 1u R
u
R R
u
R R
u** *
( )
and that the solution depends only on the variable R we obtain
01 **2
dR
du
RdR
ud
This ordinary differential equation can be solved easily and then we get
u C R D* ln
Since D is an arbitrary constant we can choose D = 0 and then we obtain
u C R* ln (1.5.1)
To find the constant C we apply the property of Dirac delta function and then
2 1u dA r dA
DD
* ( )
Applying the Divergence Theorem we obtain
C
u*1 .nds dsn
u
C
*
Since
u
n
u
R
R
n
* *
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then
u
n
C
R
*
, where n is outward normal
Thus 1 20
2C
RRd C
Finally we have C 1
2
Substituting in (1.5.1) we obtain
u R R*( ) ln 1
2 (1.5.2)
which is the fundamental solution of Laplace equation.
1.6 Integral Equation ( Internal points )
From Figure 4, we apply the Second Green‟s Theorem (1.4.3) with v = u* to the
region D - D then
( ) ( )* **
*u u u u dA uu
nuu
nds
D D C C
2 2
Since both u and u* are harmonic in D - D the integral on the left hand side is zero
therefore
( ) ( )*
**
*uu
nu
u
nds u
u
nu
u
nds
C C
(1.6.1)
Substituting (1.5.2) in (1.6.1) and the fact that
u
nq we have
1
2
1
2
1
20 0
((ln )
ln ) lim ( ln ) lim ( ln )uR
nq R ds u
nR ds q R ds
C C C
In the limit as 0 , since u is continuous with continuous partial derivative then
u up and q qp
Hence
1
2
1
2
1
20 0
((ln )
ln ) lim ( ) lim ( ln )uR
nq R ds u ds q ds
C
p p
C C
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Computing the right hand side and using L‟Hospital rule at the last term we obtain
u uR
nq R dsp
C
1
2
(
(ln )ln ) (1.6.2)
or dsnRR
uqRuC
p
.
1()ln(
2
12
(1.6.3)
i.e. u u q uq dsp
C
( )* * , where qu
n
**
(1.6.4)
1.7 Integral Equation ( Boundary Points )
Suppose that P itself is a point on the boundary at which there is a kink with angle
as shown in Figure 5 . If the boundary is smooth at P then = .
This section will illustrate the property of the boundary point . In the similar analysis
as shown in the previous section . We obtain from Figure 6 as
dsn
uu
n
uuds
n
uu
n
uudAuuuu
CCCDD
*
)()()( **
**
2**2
(1.7.1)
Again the left hand side is zero therefore
( ) ( )*
**
*
*
uu
nuu
nds u
u
nuu
nds
C C C
C
P
D
Figure 5
P
1
2
n D
D-D
C-
C Figure 6
C*
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Taking limit as 0 then
dsn
uuds
n
uuds
n
uu
n
uu
CCC
**
)(lim)(lim)( *
0
*
0
**
Therefore
1
2
1
2
1
20 01
2
1
2
((ln )
ln ) lim lim ( ln )uR
nq R ds u d q d
C
p p
Computing the right hand side and using L‟Hospital‟s rule at the last term and we
have 2 1 .
So that
u u
R
nq R dsp
C
((ln )
ln ) (1.7.2)
or u R q uRR n dsp
C
( ln ) (
^12
(1.7.3)
i.e. cu u q uq dsp
C
( )* * , where c
2 (1.7.4)
It is clear that if P is an interior point then 2 and hence c 1 . If P is a
boundary point and curve C smooth then therefore c 12 .
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2. The Boundary Element Method
This section contains two main objects of this report . The first is concerned with the
fundamental concept of the Boundary Element Method . The second we explain the
spreadsheet implementation and summarize how to implement it. We follow the
notation of Davies and Crann [1].
2.1 Fundamental Concept
We have obtained the integral equation (1.7.2) from the previous section which is the
important tool . This section will apply it to approximate u and q by the Boundary
Element Method as follows.
Suppose that w s j Nj ( ): ,2, ,..., 1 3 is a set of linearly independent function of
position, s, around CN, where, if node j is at the point sj then w si j ij( ) , with the
Kronecker delta function given by
ji
jiij
,0
,1
Then the boundary element approximation to u and q are given by
~( ) ( )
~( ) ( )
u s w s U
q s w s Q
j j
j
N
j j
j
N
1
1
(2.1.1)
We shall choose the nodes to be at the midpoint of the elements and w sj ( ) to be
piecewise constant, yielding the so called “constant element”, with w sj ( ) given by
N 1
2
3 ...
...
C
CN
[N]
[1]
[2]
[3]
[...]
[...]
Figure 7: Partition of the boundary into N elements
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otherwise
jssw j
,0
][,1)(
By substituting (2.1.1) into (1.7.2) we have
dsRQswn
RUswU
C
i
N
j
jj
iN
j
jjii
ln))(()(ln
))((11
Hence
N
j
j
C
ij
N
j
j
i
C
jii QdsRswUdsn
RswU
NN11
ln)()(ln
)(
(2.1.2)
where iR |R| and Ri is the position vector of a boundary point relative to node i.
Since w sj ( ) is non-zero only in element [j] , in which it takes value 1 , we can
write equation (2.1.2) as
j
N
j
ijj
N
j
ijii QGUHU
11
ˆ (2.1.3)
where (ln )[ ]
Hn
R dsij
j
ij
(2.1.4)
and G R dsij ij
j
ln[ ]
(2.1.5)
as shown in Figure 8
We see that
target element
base node
[j]
i
dij
jn
Figure 8: Distance from the node i to the element [j]
ijR
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i j
ijjijjijR
RnRnR1
.1.cosˆˆ. ijij dd
where dij is the perpendicular distance from node i to target element [j].
Hence
).(ln)(ln ijij RgradRn
jn
ijR
1jij nR ˆ.ˆ
= 12Rd
ij
ij
Let jijij ldA2
1 be the area of the triangle which has element [j] as its base and node
i as a vertex. To perform the integrals in (2.1.4) and (2.1.5) we use a N-point
Gaussian quadrature ( 1.2.2 ) over element [j] , with length l j , so that
( )
Hd l
Rdt A w
Rij
ij j
ij
ij g
ij gg
p
2
1 12
1
1
21
(2.1.6)
also Gl
R dtl
w Rij
j
ij
j
g
g
p
ij g
4 41
1
1
ln ln( ) (2.1.7)
If, in equation (2.1.3), we write
H Hij ij i ij (2.1.8)
then
j
N
j
ijj
N
j
N
j
ijijijii QGUUHU
11 1
Hence
H U G Qij j
j
N
ij
j
N
j
1 1
0 (2.1.9)
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Let H = Hij N N
U = U j N1
G = Gij N N
Q = Qj N1
We can write (2.1.9) in the matrix form as
HU + GQ = 0 (2.1.10)
If i j then Rii and ni are orthogonal , so that R nii i 0
Hence
NiH ii ,...,3,2,1,0ˆ (2.1.11)
For Gii we can integrate analytically to obtain
G lii i
li 1 2ln( ) (2.1.12)
Since u 1 is harmonic in D with 1u and 0q on C , equation (2.1.3)
yields
i ij
j
N
H
1
therefore we obtain
H Hii ij
jj i
N
1
(2.1.13)
When equation (2.1.10) has been solved we know the values of Ui and Qi at all
nodal points. We now use equation (1.7.2) to obtain the solution at k internal
points by the followings.
In a similar manner to which we approximate Hij and Gij for nodal points we can
obtain kjH and kjG for internal points as follows:
H A wR
kj kj g
kj gg
p
12
1 ( ) (2.1.14)
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Gl
w Rkj
j
g
g
p
kj g
4 1
ln( ) (2.1.15)
Finally, we obtain the solution for internal points as
U H U G Qk kj j
j
N
kj
j
N
j
1
2 1 1( ) (2.1.16)
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2.2 Spreadsheet Implementation
We begin this section with discussing some features in geometry then we summarize
the algorithm of the Boundary Element Method applied in a spreadsheet . Finally we
finish this section by implementing the spreadsheet to solve some potential problems.
Consider the potential problem on the region D with the boundary C in rectangular
co-ordinate as shown in Figure 9
It is clear that if P(x , y) is an arbitrary point then the relationships between co-
ordinate and angle, , are as follows:
x = |OP|cos (2.2.1)
y = |OP|sin (2.2.2)
Let us discuss only on each of those elements, boundary points and internal points
because we can use the facilities of Excel to replicate them easily and we will
describe later.
Suppose that (xi , yi) be the co-ordinate of node i
(xj , yj) be the co-ordinate of node j of element [j]
(xi ,yi)
(xk ,yk) *
* *
*
(Xj ,Yj)
(Xj+1,Yj+1)
I II
II
I (xj ,yj)
O X
Y
P(x , y)
Figure 9: Geometry of the boundary element method
Rkj
Rij
(xg ,yg)
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(Xj ,Yj) be the co-ordinate of end point of element [j]
(xk , yk) be the co-ordinate of internal point k
and (xg ,yg) be the co-ordinate of Gauss quadrature point
as shown in Figure 9.
We can easily compute the areas Aij and Akj by using ( 1.1.1) as the following .
A x y X Y X y X y X Y x Yij i j j j j i j i j j i j 1 1 1 1 (2.2.3)
and A x y X Y X y X y X Y x Ykj k j j j j k j k j j k j 1 1 1 1 (2.2.4)
The integration over each element is obtained by using the Gauss-Legendre N-point
method. We have chosen a 3-point rule as shown in Figure 9. By using (1.2.1) we
obtain
tXXxx jjj )(2
11
and tYYyy jjj )(2
11
where the value of t and w we take from Table 1.2.1 as the following
1321
321
,88888889.0,55555555.0:
77459667.0,0.0,77459667.0:
wwwww
tttt
g
g (2.2.5)
So that in element [j] we obtain the Guassian quadrature points as
gjjjg tXXxx )(2
11 (2.2.6)
and gjjjg tYYyy )(2
11 (2.2.7)
Hence
( ) ( ) ( )R x x y yij g i g i g
2 2 2 (2.2.8)
and ( ) ( ) ( )R x x y ykj g k g k g
2 2 2 (2.2.9)
Now we are ready to construct the whole system of linear equation of all elements.
First we obtain Hij by using (2.1.6) and (2.1.11) as
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3
1
22 ,)()/(
,0
ˆ
g
igiggij
ij jiyyxxwA
ji
H (2.2.10)
Then we construct Hij by using (2.1.8) and (2.1.13) as
jiH
jiHH
ij
N
i
ij
ij
,ˆ
,ˆ
1 (2.2.11)
and construct Gij by using (2.1.7) and (2.1.12) as
3
1
22 ),)()ln((4
)),2/ln(1(
g
igigg
j
jj
ijjiyyxxw
l
jill
G (2.2.12)
Next we add the Dirichlet and Neumann conditions of the boundary points so that we
have U j and Q j in hand. Consequently we obtain the system of linear equation as
the following.
H U G Q bij N N j N ij N N j N j N
1 1 1 (2.2.13)
By setting bj 0 for j = 1,2,3,...,N by (2.2.10) then using the “ Solver option “ in
Excel we obtain the solutions U j and Q j
Next we construct Hkj by using (2.1.6) as
H A w x x y ykj kj g g k g k
g
/ (( ) ( ) )2 2
1
3
(2.2.14)
and obtain Gkj by using (2.1.7) as
Gl
w x x y ykj
j
g g k g k
g
4
2 2
1
3
ln(( ) ( ) ) (2.2.15)
Finally by using (2.1.16) the interior solution Uk is obtained by
U H U G Qk kj j kj j
j
N
j
N
1
2 11 k = 1,2,3,...,M (2.2.16)
We finish this section by summarizing the process described above in the following
flow chart.
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START
Data input
Internal points
(xk , yk)
Boundary points
(xi ,yi)
Computing areas
Aij
End points
(Xj , Yj)
Computing length
lj
Quadrature points
(xg , yg)
Quadrature value
tg and wg
Computing areas
Akj
Constructing matrix
Hij
Constructing matrix
Hkj
Constructing matrix
Hij
Constructing matrix
Gij Constructing matrix
Gij
Dirichlet conditions
U j
Neumann conditions
Q j
Solving equations
HU + GQ = 0
by „ Solver „ option
Boundary solutions
U j and Q j
Internal Solutions
Uk
STOP
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Example
This example concerns the following problem in a circle:
2 0u in x y2 2 4
subject to the boundary conditions
u x y on x y x2 2 4 0 ,
q x y 12 ( ) on x y x2 2 4 0 ,
The analytic solution is u x y x y( , )
Solution ( with 8 , 16 , 32 and 64 elements)
Firstly we shall use just 8 elements with 7 internal points as shown in Figure 10.
In Table 1 we show the calculation of the element geometry about nodal points ,
internal points ,end points and lengths
The nodal points (xi , yi) are obtained by equation (2..2.1), (2.2.2) and are entered
in cell
: Boundary
node
: Internal
point
1
7
6
5
4
3
2
8
yxu )(2
1yxq
y
x
02 u
2
-2
2
-2
Figure 10: Partition the boundary into 8 elements and 7 internal points
BEM 3/8/09
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B7 (R) as =2*COS((A7-1)*2*PI()/8)
and C7(R) as =2*SIN((A7-1)*2*PI()/8)
The end points (Xj , Yj ) are obtained by the similar way and are entered in cell
H7(R) as =(2/COS(2*PI()/16))*COS((2*A7-3)*(2*PI()/16))
and I7(R) as =(2/COS(2*PI()/16))*SIN((2*A7-3)*(2*PI()/16))
The calculation of co-ordinates of the endpoints and the nodal points are shown in
Table 1..
Table 1: Co-ordinate of nodal points, internal points and endpoints of elements
The length lj of each element we can compute directly by distance formula and are
entered in cell
K7(R) as =SQRT((H7-H8)^2+(I7-I8)^2)
The internal points we firstly choose 7 points as show in Table1.
In Table 2 we show the calculation of areas of triangles from nodal points and
internal points.
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The area Aij are obtained by using equation (2.2.3) and are entered in cell
B31(RR) as
=ABS(1/2*(-$B7*B$20-B$19*C$20-
C$19*$C7+B$19*$C7+C$19*B$20+$B7*C$20))
The area Akj are obtained by using equation (2.2.4) and are entered in cell
B42(RR) as
=ABS(1/2*(-$E7*B$20-B$19*C$20-
C$19*$F7+B$19*$F7+C$19*B$20+$E7*C$20))
Table 2: Area of triangles which the vertices are at the nodal points and internal
points
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In Table 3 we show weighted value and the calculation of Gauss points .
The Gauss points are given by equations (2.2.6) and (2.2.7) and are enter in cell
C54 .. C56 (R) and C59 .. C61 (R) as follows:
=B24+((C19-B19)/2)*$C$65
and =B25+((C20-B20)/2)*$C$65
Table3 : The weighted values and co-ordinates of the Gauss points
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The coefficient matrix Aij are shown in Table 4.
Table 4: The coefficient matrix ijH
We construct the matrix ijH by using equation (2.2.10) and are entered in cell B71
as the following.
=IF($A71=B$70,0,B31*($I$65*((C$54-$B7)^2+(C$59-$C7)^2)^(-
1)+$J$65*((C$55-$B7)^2+(C$60-$C7)^2)^(-1)+$K$65*((C$56-$B7)^2+(C$61-
$C7)^2)^(-1)))
The coefficient matrices Hij , Gij and the boundary solutions are shown in Table 5.
Table 5: The coefficient matrices Hij and Gij
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The matrix H is given by equation (2.2.11) and is entered in cell B82 (R R) as
=IF($A82=B$81,-SUM($B71:$I71),B71)
The matrix G is given by equation (2.2.12) and is entered in cell B93 (R R) as
=IF($A93=B$92,$K7*(1-LN($K7/2)),(-$K7/4)*($I$65*LN((C$54-
$B7)^2+(C$59-$C7)^2)+$J$65*LN((C$55-$B7)^2+(C$60-
$C7)^2)+$K$65*LN((C$56-$B7)^2+(C$61-$C7)^2)))
The fixed boundary value from boundary condition are indicated by *** in column J
and the rows of the products HU and GQ are entered in cells L82 (R) and L93
(R) as
=MMULT(B82:I82,K$82:K$89)
and
=MMULT(B93:I93,K$93:K$100)
respectively.
The solution is invoked using “ Solver “ option . The right hand side of equation
(2.2.13) is accumulated in cell N93 (R) as
=L82+L93
and then cells N93 .. N100 are set to zero by equation (2.2.10) in the “ Solver “
parameter box as shown in Figure 10
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26
The result of “ Solver “ process is shown in Figure 11.
Figure 10: The Solver active box
Figure 11: The status of the Solver result
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Table 6: The boundary solutions and the matrices kjH and Gkj
The matrix kjH is given by equation (2.1.14) and is entered in cell B116 (R R)
as
=B105*($I$65*((C$54-$E7)^2+(C$59-$F7)^2)^(-1)+$J$65*((C$55-
$E7)^2+(C$60-$F7)^2)^(-1)+$K$65*((C$56-$E7)^2+(C$61-$F7)^2)^(-1))
The matrix Gkj is given by equation (2.2.15) and is entered in cell B126 (R R)
as
=(-$K7/4)*($I$65*LN((C$54-$E7)^2+(C$59-$F7)^2)+$J$65*LN((C$55-
$E7)^2+(C$60-$F7)^2)+$K$65*LN((C$56-$E7)^2+(C$61-$F7)^2))
Finally the internal solution Uk is given by equation (2.2.16) and is entered in cell
G143 (R)
=(L117+L128)/(2*PI())
as shown in Table 7.
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Table 7: The internal solution of the problem
The spreadsheet of the previous case ( 8 points) is easily expanded to larger sizes
and we consider solutions with 16 , 32 and 64 equal elements together with 40
internal points. The solutions are shown in Table 8.
The spreadsheet provides an environment in which it is very convenient to investigate
properties of the solutions as shown in Figure 12. We also see from Figure 12 that the
errors at points very near the boundary are worse than those at the other interior
points. This is behaviour of boundary element solutions.
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Table 8: The internal solutions of 8, 16, 32 and 64 elements with 40 internal points.
Solutions by N-nodal points Exact Errors in each case
Solution
Internal points 8 16 32 64 u = x + y 8 16 32 64
x(k) y(k) U(k) U(k) U(k) U(k) u(k)
1 1.95 0 5.494 3.325 2.352 2.013 1.95 3.544 1.375 0.402 0.063
2 1.85 0 2.5 1.975 1.854 1.848 1.85 0.65 0.125 0.004 0.002
3 1.75 0 1.906 1.743 1.741 1.747 1.75 0.156 0.007 0.009 0.003
4 1.65 0 1.645 1.62 1.64 1.648 1.65 0.005 0.03 0.01 0.002
5 1.55 0 1.482 1.515 1.54 1.548 1.55 0.068 0.035 0.01 0.002
6 1.45 0 1.358 1.415 1.441 1.448 1.45 0.092 0.035 0.009 0.002
7 1.35 0 1.249 1.316 1.341 1.348 1.35 0.101 0.034 0.009 0.002
8 1.25 0 1.147 1.218 1.242 1.248 1.25 0.103 0.032 0.008 0.002
9 1.15 0 1.05 1.12 1.142 1.148 1.15 0.1 0.03 0.008 0.002
10 1.05 0 0.954 1.022 1.042 1.048 1.05 0.096 0.028 0.008 0.002
11 0.95 0 0.859 0.923 0.943 0.948 0.95 0.091 0.027 0.007 0.002
12 0.85 0 0.765 0.825 0.843 0.848 0.85 0.085 0.025 0.007 0.002
13 0.75 0 0.671 0.727 0.744 0.748 0.75 0.079 0.023 0.006 0.002
14 0.65 0 0.577 0.628 0.644 0.649 0.65 0.073 0.022 0.006 0.001
15 0.55 0 0.483 0.53 0.545 0.549 0.55 0.067 0.02 0.005 0.001
16 0.45 0 0.389 0.432 0.445 0.449 0.45 0.061 0.018 0.005 0.001
17 0.35 0 0.295 0.333 0.346 0.349 0.35 0.055 0.017 0.004 0.001
18 0.25 0 0.2 0.235 0.246 0.249 0.25 0.05 0.015 0.004 0.001
19 0.15 0 0.106 0.137 0.146 0.149 0.15 0.044 0.013 0.004 1E-03
20 0.05 0 0.011 0.038 0.047 0.049 0.05 0.039 0.012 0.003 9E-04
21 -0.05 0 -0.08 -0.06 -0.05 -0.05 -0.05 0.033 0.01 0.003 8E-04
22 -0.15 0 -0.18 -0.16 -0.15 -0.15 -0.15 0.028 0.009 0.002 7E-04
23 -0.25 0 -0.27 -0.26 -0.25 -0.25 -0.25 0.023 0.007 0.002 6E-04
24 -0.35 0 -0.37 -0.36 -0.35 -0.35 -0.35 0.018 0.006 0.002 5E-04
25 -0.45 0 -0.46 -0.45 -0.45 -0.45 -0.45 0.013 0.005 0.001 4E-04
26 -0.55 0 -0.56 -0.55 -0.55 -0.55 -0.55 0.009 0.003 0.001 3E-04
27 -0.65 0 -0.65 -0.65 -0.65 -0.65 -0.65 0.004 0.002 7E-04 2E-04
28 -0.75 0 -0.75 -0.75 -0.75 -0.75 -0.75 5E-05 9E-04 3E-04 1E-04
29 -0.85 0 -0.85 -0.85 -0.85 -0.85 -0.85 0.004 3E-04 2E-05 4E-05
30 -0.95 0 -0.94 -0.95 -0.95 -0.95 -0.95 0.007 0.002 3E-04 4E-05
31 -1.05 0 -1.04 -1.05 -1.05 -1.05 -1.05 0.01 0.003 6E-04 1E-04
32 -1.15 0 -1.14 -1.15 -1.15 -1.15 -1.15 0.012 0.004 9E-04 2E-04
33 -1.25 0 -1.24 -1.25 -1.25 -1.25 -1.25 0.011 0.005 0.001 3E-04
34 -1.35 0 -1.34 -1.34 -1.35 -1.35 -1.35 0.006 0.006 0.001 3E-04
35 -1.45 0 -1.46 -1.44 -1.45 -1.45 -1.45 0.006 0.006 0.002 4E-04
36 -1.55 0 -1.59 -1.54 -1.55 -1.55 -1.55 0.036 0.005 0.002 5E-04
37 -1.65 0 -1.76 -1.65 -1.65 -1.65 -1.65 0.106 4E-04 0.002 5E-04
38 -1.75 0 -2.03 -1.78 -1.75 -1.75 -1.75 0.282 0.025 0.001 6E-04
39 -1.85 0 -2.66 -2.01 -1.86 -1.85 -1.85 0.81 0.16 0.012 2E-04
40 -1.95 0 -5.84 -3.38 -2.36 -2.02 -1.95 3.888 1.432 0.412 0.065
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8 points & Exact Solutions
-6
-4
-2
0
2
4
6
1.9
5
1.6
5
1.2
5
0.8
5
0.4
5
0.0
5
-0.3
5
-0.7
5
-1.1
5
-1.5
5
-1.9
5
16 points & Exact Solutions
-4
-3
-2
-1
0
1
2
3
4
1.9
5
1.6
5
1.2
5
0.8
5
0.4
5
0.0
5
-0.3
5
-0.7
5
-1.1
5
-1.5
5
-1.9
5
32 points & Exact Solutions
-3
-2
-1
0
1
2
3
1.9
5
1.6
5
1.2
5
0.8
5
0.4
5
0.0
5
-0.3
5
-0.7
5
-1.1
5
-1.5
5
-1.9
5
64 points & Exact Solutions
-3
-2
-1
0
1
2
3
1.9
5
1.6
5
1.2
5
0.8
5
0.4
5
0.0
5
-0.3
5
-0.7
5
-1.1
5
-1.5
5
-1.9
5
8 , 16 , 32 , 64 points Solutions
-6
-4
-2
0
2
4
6
1.951.651.250.850.450.05-0.35-0.75-1.15-1.55-1.95
Series1
Series2
Series3
Series4
8 , 16 , 32 , 64 points Errors
0
0.5
1
1.5
2
2.5
3
3.5
4
1.9
5
1.6
5
1.2
5
0.8
5
0.4
5
0.0
5
-0.3
5
-0.7
5
-1.1
5
-1.5
5
-1.9
5
Series1
Series2
Series3
Series4
Figure 12: The errors of the internal solution for each partition
31
References
Brebbia, C.A. (1978) The boundary element method for engineers, London: Pentech
Press.
Chiskawski, R.D. and Brebbia, C.A. (1991) Boundary element method in acoustics,
Addison-Wesley.
Colley, S.J. (1998) Vector Calculus, Prentice-Hall.
Davies, A.J. (1980) The finite element method: A first approach, Oxford University
Press.
Davies, A.J. and Crann, D. (1996) Using a spreadsheet for the boundary element
method: Interior and exterior problems. 13, Mathematics Department:
University of Hertfordshire.
Gerald, C.F. and Wheatley, P.O. (1994) Applied numerical analysis, Addison-
Wesley.
Humi, M. and Miller, W.B. (1992) Boundary value problems and patial differential
equation, PWS-KENT.
Paris, F. and Canas, J. (1997) Boundary element method: Foundations and
Application, Oxford: Oxford University Press.
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