A NUMERICAL SOLUTION OF A HELMHOLTZ EQUATION USING BOUNDARY ELEMENTS - repository civitas UGM
October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
A NUMERICAL SOLUTION OF A HELMHOLTZ EQUATION USING BOUNDARY ELEMENTS ∗
IMAM SOLEKHUDIN
Mathematics and Mathematics Education, National Institute of Education,
Nanyang Technological University, 1 Nanyang Walk, Singapore 637616
and Department of Mathematics, Faculty of Mathematics and Natural Sciences,
Gadjah Mada University, Sekip Utara, Yogyakarta, 55281 Indonesia
∗E-mail: imamsolahuddin@yahoo.com
www.nie.edu.sg, www.ugm.ac.id
KENG-CHENG ANG
Mathematics and Mathematics Education, National Institute of Education,
Nanyang Technological University, 1 Nanyang Walk, Singapore 637616
E-mail: kengcheng.ang@nie.edu.sg
Helmholtz equation is a well known differential equation. Most boundary value
problems involving this equation are either difficult or impossible to solve ana-
lytically. In this study, we employ a dual reciprocity boundary element method
(DRBEM) to solve these problems. On the boundary and in the region bounded
by the boundary, a set of collocation points is chosen for the DRBEM. The
computational algorithm requires setting up and solving a system of linear al-
gebraic equation of the form, AX = B, based on this set of collocation points.
The solution to the boundary value problem is therefore approximated by the
solution of the algebraic equations. Examples are presented to test this method,
and results obtained are compared with their corresponding analytic solutions.
Keywords : Helmholtz equation; Dual reciprocity boundary element method; linear algebraic equation.
1. Introduction In this paper we examine a type of Helmholtz equation
2
2
∂ φ ∂ φ = φ, (1) +
2
2
∂x ∂y which is employed to model infiltration problems.
2
3 Numerous researchers such as Azis et al, Lobo et al and Clements
4 October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
simple problems, including problems which are a transformation of those
1 proposed by Batu.
In the present study, we shall consider the solution of Equation (1) subject to the boundary conditions ∂φ 2π αL
= for and y = 0, (2) 0 ≤ x ≤ − φ
∂n αL
2 ∂φ αL α(L + D) for and y = 0, (3)
= −φ < x ≤ ∂n (
2
2 ∂φ x = 0 and 0 < z < ∞
= 0 for , (4)
α(L+D)
∂n x = and 0 < z < ∞
2
∂φ α(L + D) for 0 < x < (5) = −φ and z = ∞,
∂n
2 using a Dual Reciprocity Boundary Element Method (DRBEM).
2. DRBEM In this section we briefly describe DRBEM employed to obtain numerical solutions of Equation (1) subject to the boundary conditions (2) - (5).
To implement this method, the domain in the y-direction is chosen to be between y = 0 and y = c, where c is a positive real number. It is assumed that ∂φ/∂n = −φ on y = c. The boundary is divided into five non-intersecting segments, C , C , C , C , and C , where
1
2
3
4
5
αL C and y = 0, (6)
1 : 0 ≤ x ≤
2 αL α(L + D)
C : and y = 0, (7)
2 < x ≤
2
2 α(L + D)
C : x = (8)
3 and 0 ≤ z ≤ c,
2 α(L + D)
C and y = c, (9)
4 : 0 ≤ x ≤
2 C
5 : x = 0 and (10) 0 ≤ y ≤ c.
5 According to Ang, solution of Equation (1) is
Z Z λ(ξ, η)φ(ξ, η) = ϕ(x, y; ξ, η)φ(x, y)dxdy (11) Z · R ¸
∂ ∂ φ(x, y) (φ(x, y)) ds(x, y),
- ∂n ∂n
(ϕ(x, y; ξ, η)) − ϕ(x, y; ξ, η)
C
where C = C , R is the region bounded by C,
1 ∪ C 2 ∪ C 3 ∪ C 4 ∪ C
5
1
2
2
ϕ(x, y; ξ, η) = ] (12) ln[(x − ξ) + (y − η) October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
is the fundamental solution of the two-dimensional Laplace’s equation, and 0, (ξ, η) / ∈ R ∪ C
1
λ(ξ, η) = , (ξ, η) lies on smooth part of C . (13)
2
1, (ξ, η) ∈ R
From (2) - (5), Equation (11) can be written as Z Z λ(ξ, η)φ(ξ, η) = ϕ(x, y; ξ, η)φ(x, y)dxdy (14)
Z ¸
R· ∂
- (ϕ(x, y; ξ, η)) + ϕ(x, y; ξ, η) φ(x, y)ds(x, y) ∪ ∪
∂n Z C 1 C 2 C 4 ∂ ∪ ∂n φ(x, y) (ϕ(x, y; ξ, η))ds(x, y) + Z C 3 C 5
2π ϕ(x, y; ξ, η)ds(x, y). −
αL
C 1 To obtain numerical solutions using DRBEM, each of C , i = 1, 2, 3, 4, 5, i
is discretized into a number of line segments joined end to end, and a num- ber of interior points is chosen. These interior points are chosen such that they are well spaced in the domain. The total number of the line segments and the total number of the interior points are N and M respectively. Each
(i) (i)
line segment is denoted by C , i = 1, 2, ..., N . On C
(i)
, (15) φ ≈ φ
∂φ
(i)
and , (16) ≈ p
∂n
(i) (i)
where φ and p are the value of φ and ∂φ/∂n at the mid-point of segment
(i) C respectively.
In order to compute (14) approximately, M + N collocation points are chosen. The first N collocation points are the mid-points of the line segments, and the rest M points are the chosen interior points. The
(1) (1) (2) (2) (N ) (N )
collocation points are denoted by (a , b ), (a , b ), ..., (a , b ),
(N +1) (N +1) (N +2) (N +2) (N +M ) (N +M ) (a , b ), (a , b ), ..., (a , b ).
The value of φ(x, y) in (14) may be approximated by
N +M X (i) (i) (i)
δ ρ(x, y; a , b ), (17) φ(x, y) ≃
i=1 (i)
where δ are the coefficients to be determined and
(i) (i) (i) 2 (i)
2
ρ(x, y; a , b ) ) ) ) = 1 + ((x − a + (y − b
(i) 2 (i) 2 3/2 October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
2 .
is a radial function on R From (17), we obtain Z Z N +M X (i) (i) (i)
δ Υ(ξ, η; a , b ), (19) ϕ(x, y; ξ, η)φ(x, y)dxdy ≃
R i=1
where Z · ∂
(i) (i) (i) (i) (i) (i)
Υ(ξ, η; a , b ) = λ(ξ, η)χ(ξ, η; a , b ) + ϕ(x, y; ξ, η) (χ(x, y; a , b )) ∂n
C ¸
∂
(i) (i)
, b ) (ϕ(x, y; ξ, η)) ds(x, y), (20) − χ(x, y; a
∂n and
1
1
(i) (i) (i) 2 (i) 2 (i) 2 (i)
2
2
χ(x, y; a , b ) = ) ) ] + ) ) ] [(x − a + (y − b [(x − a + (y − b
4
16
1
(i) 2 (i) 2 5/2
) ) + ] . (21) [(x − a + (y − b
25 The line integral in (20) can be approximated by ¯ X N Z ¯ ∂
(i) (i) ¯
(j) (j)
(χ(x, y; a , b )) ϕ(x, y; ξ, η)ds(x, y) ¯ (x,y)=(a ,b ) (j) ∂n
C j=1 X N Z
∂
(j) (j) (i) (i)
- χ(a , b ; a , b )) ϕ(x, y; ξ, η)ds(x, y). (22) (j)
∂n
C j=1 (i)
To compute δ in (19), point (x, y) in (17) is substituted by the collo- cation points. From this, we obtain a system of linear equations
N +M X (k) (k) (i) (k) (k) (i) (i)
φ(a , b ) = δ ρ(a , b ; a , b ), k = 1, 2, .., N + M. (23)
i=1
System of linear equations (23) can be inverted, and we obtain
N +M X (i) (ik) (k) (k)
δ = ω φ(a , b ), i = 1, 2, ..., N + M, (24)
k=1
where −
(ik) (k) (k) (i) (i)
1
- N X m=1
- 8 ∞ X October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
2
(nk)
, and y
(k)
are defined as follows: a
(nk)
= ( F
(k)
(a
(nk)
(n)
, b
(n)
) + ν
(nk)
− γ
(nk)
, if
, b
, n = 1, 2, ..., N + M (32) where a
specified over C
0 , otherwise . (31)
3
∪ C
5
1 , otherwise , (30)
β
2 =
½ 1 , (x, y) ∈ C
1
Equation (26) yields a system of linear algebraic equations containing N + M equations with N + M unknowns, and can be written as
(nk)
N +M X k=1
a
(nk)
y
(k)
=
N +M X k=1
b
∂φ ∂n
(n)
1 =
(n)
(a
(n)
, b
(n)
), if (a
(n)
, b
) ∈ C
(k)
1
0, otherwise , y
(k)
= φ
(k)
, and γ
(nk)
= ½ 1/2, if n = k
1
= ( 2π/αLF
F
2
(k)
1
(a
(n)
, b
(n)
) + F
(k)
(a
(nk)
(n)
, b
(n)
) + ν
(nk)
− γ
(nk)
, otherwise , b
½ 0 , (x, y) ∈ C
))ds(x, y), (29) β
October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
(n)
(n)
, b
(n)
) + F
(m)
2
(a
, b
1
(n)
)] − β
2
2π αL
N X m=1
F
(m)
1
(a
1 F (m)
(n)
)φ
Now, (ξ, η) in equation (14) is taken to be (a
(n)
, b
(n)
). The value of φ can be approximated by λ(a
(n)
, b
(n)
(n)
[β
=
N +M X k=1
ν
(nk)
φ
(k)
φ
(k)
(a
, b
(n)
(a
C (m)
ϕ(x, y; a
(n)
, b
(n)
)ds(x, y), (28) F
(m)
2
(n)
(n)
, b
(n)
) = Z
C (m)
∂ ∂n
(ϕ(x, y; a
(n)
, b
) = Z
, b
(n)
; a
), n = 1, 2, ..., N + M. (26) where ν
(nk)
=
N +M X i=1
Υ(a
(n)
, b
(n)
(i)
(n)
, b
(i)
)ω
(ik)
, (27) F
(m)
1
(a
0, if n 6= k (33) October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
When n= N + 1, N + 2, ..., N + M , it follows that
(nk) (nk) (nk)
a = ν , − γ
½ 1, if n = k
(nk)
and γ = . (34) 0, if n 6= k
Due to the size of the matrices, equation AX = B, (35) where
(nk)
A = [a ], (36)
(1) (2) (N +M −1) (N +M ) t
X = [y y ... y y ] , (37)
N +M N +M N +M X X X (1k) (2k) ([N +M ]k) t
B = [ b b ... b ] , (38)
k=1 k=1 k=1
is solved numerically using the built-in solver in MATLAB. Solutions ob- tained are then substituted to Equation (26) to acquire value of φ(x, y), where (x, y) ∈ R ∪ C.
3. Results and discussion The method described in the preceding section is implemented by consid-
1
ering two typical problems. These problems are adapted from Batu. To obtain the numerical solutions, DRBEM is applied using two different sets of boundary elements and interior collocation points. In Set (A), we let N = 400, M = 625 and in Set (B) we let N = 200, M = 900. These solutions are then compared and discussed with the corresponding exact solutions. For the simplicity, numerical solutions obtained using Set (A) and Set (B) are labeled Solution A and Solution B respectively.
Problem 1 Given Equation (1) subject to boundary conditions (2) - (5), where α = 0.002, L = 100 and D = 400.
The exact solution of this problem is ∞ ¡ ¢
nπ X √ − − sin 2 z z 1+(2nπ)
5
φ = πe + 20 ³ ´ cos(2nπx)e . (39)
2
p1 + (2nπ)
n=1 n 1 +
This solution is obtained using the method of separation variables described
1 by Batu with some slight modification.
The test points chosen to compare exact and numerical solutions are (0.5,y), where y ranges from 0.1 through to 1.3. The result is summarized
A DRBEM has been employed to obtain numerical solutions of two problems which have been solved analytically. Comparisons between ana- lytic solutions and numerical solutions show that the results are in good
n ³ 1 + p1 + (2nπ)
4. Summary A type of Helmholtz equation has analytic solution in the form of bound- ary integral equation. Because of the difficulties in solving this equation analytically, a linear algebraic system is used to solve numerically, which is solved using the built-in solver in MATLAB due to the size of the system.
Exact and numerical solutions for selected values of φ(0.5, y) of this problem are summarized in Table 2 It can be seen from Table 1 and Table 2, the numerical solutions ob- tained using the proposed DRBEM agree well with the exact solutions.
. (40) Like before, this was obtained using method of separation variables (Batu, 1978) with some modification.
1+(2nπ) 2
√
2 ´ cos(2nπx)e − z
2
¢
October 25, 2010 11:6 WSPC - Proceedings Trim Size: 9in x 6in Imamspaper
Table 1. Numerical and Exact solutions of
Φ(0.5, y) at selected values of z.nπ
sin ¡
n=1
z
The exact solution of this problem is φ = πe −
Problem 2 Given Equation (1) subject to boundary conditions (2) - (5), where α = 0.002, L = 250 and D = 250.
(x,y) Exact Solution A Solution B
(0.50,0.10) 2.155766 2.160531 2.165620
(0.50,0.20) 2.174371 2.178369 2.182683
(0.50,0.30) 2.105464 2.109038 2.113086
(0.50,0.40) 1.984910 1.988240 1.992249
(0.50,0.50) 1.840400 1.843554 1.847641
(0.50,0.60) 1.689396 1.692415 1.696603
(0.50,0.70) 1.541590 1.544495 1.548771
(0.50,0.80) 1.401803 1.404599 1.408942
(0.50,0.90) 1.272079 1.274772 1.279156
(0.50,1.00) 1.152975 1.155566 1.159968
(0.50,1.10) 1.044288 1.046781 1.051183
(0.50,1.20) 0.945458 0.947856 0.952243
(0.50,1.30) 0.855775 0.858083 0.862442
Table 2. Numerical and Exact solutions of
Φ(0.5, y) at selected values of z.(x,y) Exact Solution A Solution B
(0.50,0.10) 2.285952 2.287063 2.289185
(0.50,0.20) 2.270697 2.271343 2.272932
(0.50,0.30) 2.166691 2.167144 2.168514
(0.50,0.40) 2.020664 2.021055 2.022337
(0.50,0.50) 1.860346 1.860715 1.861968
(0.50,0.60) 1.700254 1.700620 1.701862
(0.50,0.70) 1.547424 1.547795 1.549027
(0.50,0.80) 1.404916 1.405293 1.406514
(0.50,0.90) 1.273734 1.274117 1.275324
(0.50,1.00) 1.153852 1.154241 1.155432
(0.50,1.10) 1.044753 1.045145 1.046319
(0.50,1.20) 0.945704 0.946098 0.947255
(0.50,1.30) 0.855906 0.856301 0.857439
From the description presented in Section 2, it is clear that this method can be applied to problems with boundaries of arbitrary shape. This makes the method very useful and powerful for problems which cannot be solved analytically.
References 1. V. Batu, S. S. S. A. J 42, 545 (1978).
2. M. I. Azis, D. L. Clements and M. Lobo, ANZIAM J. 44(E), C61 (2003).
3. M. Lobo, D. L. Clements, and N. Widana, ANZIAM J. 46(E), C1055 (2005).
4. D. L. Clements, M. Lobo, and N. Widana, Electronic J. of Boundary Elements
5, 1 (2007).
5. W. T. Ang, A Beginer’s Course in Boundary Element Methods (Universal
Publishers Boca Raton, Florida, 2007).