# Numerical Investigation on Convergence Rate of Singular Boundary Method

Authors: Junpu Li; Wen Chen; Zhuojia Fu
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

The singular boundary method (SBM) is a recent boundary-type collocation scheme with the merits of being free of mesh and integration, mathematically simple, and easy-to-program. Its essential technique is to introduce the concept of the source intensity factors to eliminate the singularities of fundamental solutions upon the coincidence of source and collocation points in a strong-form formulation. In recent years, several numerical and semianalytical techniques have been proposed to determine source intensity factors. With the help of these latest techniques, this short communication makes an extensive investigation on numerical efficiency and convergence rates of the SBM to an extensive variety of benchmark problems in comparison with the BEM. We find that in most cases the SBM and BEM have similar convergence rates, while the SBM has slightly better accuracy than the direct BEM. And the condition number of SBM is lower than BEM. Without mesh and numerical integration, the SBM is computationally more efficient than the BEM.


---

## Body

## 1. Introduction

The boundary element method (BEM) [1–4] and the method of fundamental solutions (MFS) [5–8] are two important numerical methods for science and engineering applications. However, it is a sophisticated mathematical and time-consuming issue for numerical integration over the singularities in the BEM. As for the MFS, the location of fictitious boundary is vital for its numerical accuracy and reliability and remains an open issue to optimally determine especially for complex-shaped or multiconnected domain problems.To remedy these drawbacks in the BEM and MFS, several numerical schemes have been proposed, such as the boundary collocation method (BCM) [9], the modified collocation Trefftz method (MCTM) [10–12], the regularized meshless method (RMM) [13–16], the modified method of fundamental solutions (MMFS) [17], the boundary distributed source (BDS) method [18–20], and the nonsingular method of fundamental solutions [21].The singular boundary method (SBM) [22] was proposed by the authors in 2009 and is a strong-form boundary discretization technique. In order to regularize the singularities of fundamental solutions upon the coincidence of source and collocation points, the concept of the source intensity factors (SIFs) was first introduced, which is also called the origin intensity factors (OIFs) in some literatures [22]. Under the extensive studies, four techniques have been proposed to determine the source intensity factors of both the fundamental solution and its derivative, namely, inverse interpolation technique (IIT), semianalytical technique with subtracting and adding-back desingularization (SAT1), semianalytical technique with integral mean value approach (SAT2), and semianalytical technique with the empirical formula (SAT3). Numerical investigation shows that the SBM can provide accurate solutions in potential [22], Helmholtz [23], acoustic and elastic waves [24, 25], and water wave problems [26] with arbitrarily complex-shaped computational geometries.This short communication will make a comparison on numerical efficiency and convergence rate under the extensive benchmark testing with the best approach to determining the source intensity factors. A brief outline of the paper is as follows: Section2 will describe the techniques to determine SIFs which will be used in Section 3 for solving 2D and 3D Laplace, Helmholtz, and modified Helmholtz equations. Section 3 will make a comparison on numerical efficiency and convergence rate under the extensive benchmark testing in comparison with the direct BEM. At last, Section 4 will make a conclusion.


## 2. The Latest Techniques to Determine Source Intensity Factors

This section will describe the techniques to determine source intensity factors for solving 2D and 3D Laplace, Helmholtz, and modified Helmholtz equations.


### 2.1. SBM for Laplace Equations

This section describes the SBM for Laplace equations. Consider the Laplace equations:(1)
Δ
u
x
=
0
,
x
∈
Ω
,
u
x
=
u
-
x
,
x
∈
Γ
D
,
∂
u
x
∂
n
=
∂
u
-
x
∂
n
,
x
∈
Γ
N
,where Δ denotes the Laplacian operator, the solutions u
(
x
) are the potentials in domain Ω, u
-
(
x
) is the known function, and n is the unit outward normal on physical boundary. Γ
D and Γ
N represent the essential boundary (Dirichlet) and the natural boundary (Neumann) conditions.By adopting the fundamental solutionG
0 of Laplace equation, the solution u
(
x
) is approximated by a linear combination of fundamental solutions with respect to different source points s
j as below:(2)
u
x
=
∑
j
=
1
N
α
j
G
0
x
,
s
j
,where N is the number of source points and α
j is the jth unknown coefficient. The fundamental solution G
0 will have singularities when x
m
=
s
j. To solve this problem, we introduce the concept of source intensity factors in SBM. We place all computing nodes on the same physical boundary. So the source points {
s
j
} and the collocation points {
x
m
} are the same set of boundary nodes. When x
m
=
s
j, we use source intensity factors replacing the singular terms in formulation (2). Thus the SBM formulation can be expressed as(3)
u
x
m
=
∑
j
=
1
N
α
j
G
0
x
m
,
s
j
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
G
0
x
m
,
s
j
+
α
m
U
0
j
j
,
x
m
∈
Γ
D
,
q
x
m
=
∂
u
x
m
∂
n
x
=
∑
j
=
1
N
α
j
∂
G
0
x
m
,
s
j
∂
n
x
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
∂
G
0
x
m
,
s
j
∂
n
x
+
α
m
Q
0
j
j
,
x
m
∈
Γ
N
,where U
0
j
j and Q
0
j
j are defined as the source intensity factors corresponding to the fundamental solutions and the unit outward normal of fundamental solutions, namely, the diagonal elements of the SBM interpolation matrix.Therefore, to solve all kinds of physical and mechanical problems with formulations (3), the key issue is to determine the source intensity factors.In recent years, four techniques have been proposed to determine the abovementioned source intensity factors. And the merits and demerits of these techniques have already been extensively investigated in some literatures [22–26]. In this study, we only list the best formulations of these techniques to investigate the numerical efficiency and convergence rates of the SBM in comparison with the BEM under several typical benchmark problems.In this study, we use formulas (4) and (5) to determine the source intensity factors for 2D and 3D Laplace problems with Neumann boundary condition.The formulation can be expressed as(4)
Q
0
j
j
=
-
V
m
-
∑
j
=
1
j
≠
m
N
Π
j
∂
G
0
x
m
,
s
j
∂
n
s
,
(5)
U
0
j
j
=
u
I
x
m
-
∑
j
=
1
,
j
≠
m
N
α
j
G
0
x
m
,
s
j
-
u
I
x
I
-
∑
j
=
1
N
α
j
G
0
x
I
,
s
j
α
m
,where x
I is one inner point inside the domain Ω, u
I
=
x
+
y is the sample solution in 2D problems, and u
I
=
x
+
y
+
z is the sample solution in 3D problems. G
0 is the fundamental solution of Laplace equation, Π
j
=
L
j
/
L
m, and Π
m
=
1. L
j is the corresponding influence area of source point s
j. For 2D problem, L
j is half length of the curve s
j
-
1
s
j
+
1 between source points s
j
-
1 and s
j
+
1 as shown in Figure 1(a). And, for 3D problem, it is the corresponding infinitesimal area as shown in Figure 1(b).Figure 1
The schematic configuration of (a) the source pointss
j and the curve s
j
-
1
s
j
+
1
⌢ on 2D problems and (b) the source points s
j and the corresponding infinitesimal area L
j on 3D problems.
(a)
(b)Then, we use the following formula to determine the source intensity factors for 2D Laplace equations with Dirichlet boundary condition:(6)
U
0
j
j
=
ln
⁡
L
j
/
2
π
2
π
.And we use formula (7) to determine the source intensity factors for 3D Laplace equations with Dirichlet boundary condition:(7)
U
0
j
j
=
Θ
x
j
-
∑
i
=
1
,
i
≠
j
N
G
0
x
j
,
s
i
L
i
L
j
,where Θ
x
j
=
∫
Γ
G
0
x
j
,
s
d
Γ
s
,
G
0
x
j
,
s
=
1
/
4
π
r
x
j
,
s.Formulas (4) and (5) are also called the semianalytical technique with subtracting and adding-back desingularization (SAT1). Formula (6) is also called the semianalytical technique with the empirical formula (SAT3). Formula (7) is also called the semianalytical technique with integral mean value approach (SAT2). And one can find more details about these techniques in the literatures [22–26].


### 2.2. SBM for Helmholtz Equations and Modified Helmholtz Equations

In analogy to the SBM for Laplace equations, the corresponding SBM approximate solution for Helmholtz and modified Helmholtz equations can be represented as(8)
u
x
m
=
∑
j
=
1
N
α
j
G
i
i
x
m
,
s
j
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
G
i
i
x
m
,
s
j
+
α
m
U
j
j
,
x
m
∈
Γ
D
,
q
x
m
=
∂
u
x
m
∂
n
x
=
∑
j
=
1
N
α
j
∂
G
i
i
x
m
,
s
j
∂
n
x
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
∂
G
i
i
x
m
,
s
j
∂
n
x
+
α
m
Q
j
j
,
x
m
∈
Γ
N
,where U
j
j and Q
j
j are the source intensity factors corresponding to the fundamental solution and the unit outward normal of fundamental solution. N is the number of source point. α
j is the jth unknown coefficient, i
i
=
1 or 2. G
1
x
m
,
s
j
=
i
/
4
H
0
1
(
k
r
m
j
) is the fundamental solution for 2D Helmholtz equations, G
2
x
m
,
s
j
=
1
/
2
π
K
0
(
k
r
m
j
) for 2D modified Helmholtz equations, G
1
x
m
,
s
j
=
e
x
p
(
i
k
r
m
j
)
/
4
π
r
m
j for 3D Helmholtz problems, and G
2
x
m
,
s
j
=
c
o
s
h
(
k
r
m
j
)
/
4
π
r
m
j for 3D modified Helmholtz problems. H
0
(
1
) represents the zero-order Hankel function of the first kind, K
0 represents the zero-order modified Bessel function of the second kind, and r
m
j
=
x
m
-
s
j
2 is the Euclidean distance between collocation points x
m and source points s
j.Due to the property of the same order of the singularities between the fundamental solutions of Helmholtz-type equations and Laplace equation, the source intensity factors of Helmholtz equation and modified Helmholtz equation can be expressed as(9)
U
j
j
x
m
,
s
j
=
U
0
j
j
x
m
,
s
j
+
B
,
r
m
j
⟶
0
,
(10)
Q
j
j
x
m
,
s
j
=
Q
0
j
j
x
m
,
s
j
,
r
m
j
⟶
0
,where B
=
-
1
/
2
π
l
n
k
/
2
+
γ
-
i
π
/
2 in 2D Helmholtz problem, B
=
-
1
/
2
π
l
n
k
/
2
+
γ in 2D modified Helmholtz problem, B
=
i
k
/
4
π in 3D Helmholtz problem, and B
=
0 in 3D modified Helmholtz problem, γ
=
0.57721566490153286
…. Therefore, we can get U
j
j and Q
j
j through U
0
j
j and Q
0
j
j with (9) and (10).


## 2.1. SBM for Laplace Equations

This section describes the SBM for Laplace equations. Consider the Laplace equations:(1)
Δ
u
x
=
0
,
x
∈
Ω
,
u
x
=
u
-
x
,
x
∈
Γ
D
,
∂
u
x
∂
n
=
∂
u
-
x
∂
n
,
x
∈
Γ
N
,where Δ denotes the Laplacian operator, the solutions u
(
x
) are the potentials in domain Ω, u
-
(
x
) is the known function, and n is the unit outward normal on physical boundary. Γ
D and Γ
N represent the essential boundary (Dirichlet) and the natural boundary (Neumann) conditions.By adopting the fundamental solutionG
0 of Laplace equation, the solution u
(
x
) is approximated by a linear combination of fundamental solutions with respect to different source points s
j as below:(2)
u
x
=
∑
j
=
1
N
α
j
G
0
x
,
s
j
,where N is the number of source points and α
j is the jth unknown coefficient. The fundamental solution G
0 will have singularities when x
m
=
s
j. To solve this problem, we introduce the concept of source intensity factors in SBM. We place all computing nodes on the same physical boundary. So the source points {
s
j
} and the collocation points {
x
m
} are the same set of boundary nodes. When x
m
=
s
j, we use source intensity factors replacing the singular terms in formulation (2). Thus the SBM formulation can be expressed as(3)
u
x
m
=
∑
j
=
1
N
α
j
G
0
x
m
,
s
j
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
G
0
x
m
,
s
j
+
α
m
U
0
j
j
,
x
m
∈
Γ
D
,
q
x
m
=
∂
u
x
m
∂
n
x
=
∑
j
=
1
N
α
j
∂
G
0
x
m
,
s
j
∂
n
x
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
∂
G
0
x
m
,
s
j
∂
n
x
+
α
m
Q
0
j
j
,
x
m
∈
Γ
N
,where U
0
j
j and Q
0
j
j are defined as the source intensity factors corresponding to the fundamental solutions and the unit outward normal of fundamental solutions, namely, the diagonal elements of the SBM interpolation matrix.Therefore, to solve all kinds of physical and mechanical problems with formulations (3), the key issue is to determine the source intensity factors.In recent years, four techniques have been proposed to determine the abovementioned source intensity factors. And the merits and demerits of these techniques have already been extensively investigated in some literatures [22–26]. In this study, we only list the best formulations of these techniques to investigate the numerical efficiency and convergence rates of the SBM in comparison with the BEM under several typical benchmark problems.In this study, we use formulas (4) and (5) to determine the source intensity factors for 2D and 3D Laplace problems with Neumann boundary condition.The formulation can be expressed as(4)
Q
0
j
j
=
-
V
m
-
∑
j
=
1
j
≠
m
N
Π
j
∂
G
0
x
m
,
s
j
∂
n
s
,
(5)
U
0
j
j
=
u
I
x
m
-
∑
j
=
1
,
j
≠
m
N
α
j
G
0
x
m
,
s
j
-
u
I
x
I
-
∑
j
=
1
N
α
j
G
0
x
I
,
s
j
α
m
,where x
I is one inner point inside the domain Ω, u
I
=
x
+
y is the sample solution in 2D problems, and u
I
=
x
+
y
+
z is the sample solution in 3D problems. G
0 is the fundamental solution of Laplace equation, Π
j
=
L
j
/
L
m, and Π
m
=
1. L
j is the corresponding influence area of source point s
j. For 2D problem, L
j is half length of the curve s
j
-
1
s
j
+
1 between source points s
j
-
1 and s
j
+
1 as shown in Figure 1(a). And, for 3D problem, it is the corresponding infinitesimal area as shown in Figure 1(b).Figure 1
The schematic configuration of (a) the source pointss
j and the curve s
j
-
1
s
j
+
1
⌢ on 2D problems and (b) the source points s
j and the corresponding infinitesimal area L
j on 3D problems.
(a)
(b)Then, we use the following formula to determine the source intensity factors for 2D Laplace equations with Dirichlet boundary condition:(6)
U
0
j
j
=
ln
⁡
L
j
/
2
π
2
π
.And we use formula (7) to determine the source intensity factors for 3D Laplace equations with Dirichlet boundary condition:(7)
U
0
j
j
=
Θ
x
j
-
∑
i
=
1
,
i
≠
j
N
G
0
x
j
,
s
i
L
i
L
j
,where Θ
x
j
=
∫
Γ
G
0
x
j
,
s
d
Γ
s
,
G
0
x
j
,
s
=
1
/
4
π
r
x
j
,
s.Formulas (4) and (5) are also called the semianalytical technique with subtracting and adding-back desingularization (SAT1). Formula (6) is also called the semianalytical technique with the empirical formula (SAT3). Formula (7) is also called the semianalytical technique with integral mean value approach (SAT2). And one can find more details about these techniques in the literatures [22–26].


## 2.2. SBM for Helmholtz Equations and Modified Helmholtz Equations

In analogy to the SBM for Laplace equations, the corresponding SBM approximate solution for Helmholtz and modified Helmholtz equations can be represented as(8)
u
x
m
=
∑
j
=
1
N
α
j
G
i
i
x
m
,
s
j
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
G
i
i
x
m
,
s
j
+
α
m
U
j
j
,
x
m
∈
Γ
D
,
q
x
m
=
∂
u
x
m
∂
n
x
=
∑
j
=
1
N
α
j
∂
G
i
i
x
m
,
s
j
∂
n
x
,
x
m
∈
Ω
∖
∂
Ω
,
∑
j
=
1
j
≠
m
N
α
j
∂
G
i
i
x
m
,
s
j
∂
n
x
+
α
m
Q
j
j
,
x
m
∈
Γ
N
,where U
j
j and Q
j
j are the source intensity factors corresponding to the fundamental solution and the unit outward normal of fundamental solution. N is the number of source point. α
j is the jth unknown coefficient, i
i
=
1 or 2. G
1
x
m
,
s
j
=
i
/
4
H
0
1
(
k
r
m
j
) is the fundamental solution for 2D Helmholtz equations, G
2
x
m
,
s
j
=
1
/
2
π
K
0
(
k
r
m
j
) for 2D modified Helmholtz equations, G
1
x
m
,
s
j
=
e
x
p
(
i
k
r
m
j
)
/
4
π
r
m
j for 3D Helmholtz problems, and G
2
x
m
,
s
j
=
c
o
s
h
(
k
r
m
j
)
/
4
π
r
m
j for 3D modified Helmholtz problems. H
0
(
1
) represents the zero-order Hankel function of the first kind, K
0 represents the zero-order modified Bessel function of the second kind, and r
m
j
=
x
m
-
s
j
2 is the Euclidean distance between collocation points x
m and source points s
j.Due to the property of the same order of the singularities between the fundamental solutions of Helmholtz-type equations and Laplace equation, the source intensity factors of Helmholtz equation and modified Helmholtz equation can be expressed as(9)
U
j
j
x
m
,
s
j
=
U
0
j
j
x
m
,
s
j
+
B
,
r
m
j
⟶
0
,
(10)
Q
j
j
x
m
,
s
j
=
Q
0
j
j
x
m
,
s
j
,
r
m
j
⟶
0
,where B
=
-
1
/
2
π
l
n
k
/
2
+
γ
-
i
π
/
2 in 2D Helmholtz problem, B
=
-
1
/
2
π
l
n
k
/
2
+
γ in 2D modified Helmholtz problem, B
=
i
k
/
4
π in 3D Helmholtz problem, and B
=
0 in 3D modified Helmholtz problem, γ
=
0.57721566490153286
…. Therefore, we can get U
j
j and Q
j
j through U
0
j
j and Q
0
j
j with (9) and (10).


## 3. Numerical Results and Discussions

In this section, the efficiency, accuracy, and convergence of the SBM are tested to 2D and 3D Laplace, Helmholtz, and modified Helmholtz problems. The numerical accuracy is calculated by the relative root mean square errors (Rerr)L
e
r
r
u which is defined as follows:(11)
Lerr
u
=
1
/
T
N
∑
k
=
1
T
N
u
k
-
u
-
k
2
1
/
T
N
∑
k
=
1
T
N
u
-
k
2
,where u
-
k and u
k are the analytical and numerical solutions at x
i, respectively, and T
N is the total number of test points in the interest domain. To demonstrate the convergence rate of the SBM and BEM, the following formulations are introduced:(12a)
C
2
D
=
-
ln
⁡
E
r
r
o
r
N
1
-
ln
⁡
E
r
r
o
r
N
2
ln
⁡
N
1
-
ln
⁡
N
2
,
f
o
r
2
D
p
r
o
b
l
e
m
s
,
(12b)
C
3
D
=
-
2
ln
⁡
E
r
r
o
r
N
1
-
ln
⁡
E
r
r
o
r
N
2
ln
⁡
N
1
-
ln
⁡
N
2
,
f
o
r
3
D
p
r
o
b
l
e
m
s
,where E
r
r
o
r
N
1 and E
r
r
o
r
N
2 represent the errors L
e
r
r
u of the SBM and BEM with N
1 and N
2 boundary nodes, respectively. In the BEM, the direct formulation introduced in [1] is adopted and the integrations of the fundamental solution and its normal gradient over each element are calculated through 8-point Gauss quadrature for 2D problems and 9-point Gauss quadrature for 3D problems.


### 3.1. Error Analysis

#### 3.1.1. 2D Laplace Problems

Example 1.
Consider 2D Laplace problems in a circular domain with radius 1. And the exact solution is given by(13)
u
x
,
y
=
cos
⁡
x
exp
⁡
y
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (6) to determine the SIFs. And the numerical results are shown in Figures 2 and 3.Figure 2
Relative errors of Example1 by using the SBM and the BEM. C
S
B
M
=
3.0; C
B
E
M
=
2.0.Figure 3
Condition numbers of the SBM and the BEM in Example1.Example 2.
Consider 2D Laplace problems in a square domain with length 1. And the exact solution is given by(14)
u
x
,
y
=
cos
⁡
x
exp
⁡
y
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (6) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 4.Figure 4
Relative errors of Example2 by using the SBM and the BEM. C
S
B
M
=
1.9; C
B
E
M
=
1.9.


#### 3.1.2. 2D Helmholtz Problems

Example 3.
Consider 2D Helmholtz problems in a circular domain with radius 1. And the exact solution is given by(15)
u
x
,
y
=
cos
⁡
k
x
+
cos
⁡
k
y
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (6) and (9) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 5. Table 1 lists the numerical results of the SBM and the BEM with different wavenumbers (k
=
1, k
=
10, and k
=
20).Table 1
Numerical results for Helmholtz problem in circular domain.
N
k
=
1
k
=
10
k
=
20
SBM
BEM
SBM
BEM
SBM
BEM
40
2.70
E
-
4
2.87
E
-
4
1.63
E
-
2
2.31
E
-
2
6.15
E
-
1
2.48
E
-
1
80
2.50
E
-
6
7.31
E
-
5
1.78
E
-
3
5.95
E
-
3
2.54
E
-
2
1.86
E
-
2
160
1.92
E
-
7
1.84
E
-
5
2.18
E
-
4
1.53
E
-
3
2.92
E
-
3
5.33
E
-
3
320
2.40
E
-
8
4.62
E
-
6
2.71
E
-
5
3.86
E
-
4
3.58
E
-
4
1.44
E
-
3
640
2.99
E
-
9
1.16
E
-
6
3.38
E
-
6
9.72
E
-
5
4.45
E
-
5
3.76
E
-
4
1280
3.74
E
-
10
2.89
E
-
7
4.23
E
-
7
2.43
E
-
5
5.56
E
-
6
9.59
E
-
5Figure 5
Relative errors of Example3 by using the SBM and the BEM. C
S
B
M
=
3.1; C
B
E
M
=
1.9.Example 4.
Consider 2D Helmholtz problems in a square domain with length 1. And the exact solution is given by(16)
u
x
,
y
=
cos
⁡
k
x
+
cos
⁡
k
y
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4), (6), (9), and (10) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 6.Figure 6
Relative errors of Example4 by using the SBM and the BEM. C
S
B
M
=
1.6; C
B
E
M
=
1.6.


#### 3.1.3. 2D Modified Helmholtz Problems

Example 5.
Consider 2D modified Helmholtz problems in a circular domain with radius 1. And the exact solution is given by(17)
u
x
,
y
=
exp
⁡
k
x
+
y
2
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (6) and (9) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 7.Figure 7
Relative errors of Example5 by using the SBM and the BEM. C
S
B
M
=
3.1; C
B
E
M
=
2.0.Example 6.
Consider 2D modified Helmholtz problems in a square domain with length 1. And the exact solution is given by(18)
u
x
,
y
=
e
x
p
k
x
+
y
2
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4), (6), (9), and (10) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 8. Table 2 lists the numerical results of the SBM and the BEM with different wavenumbers (k
=
1, k
=
10, and k
=
20).Table 2
Numerical results for Helmholtz problem in square domain.
N
k
=
1
k
=
10
k
=
20
SBM
BEM
SBM
BEM
SBM
BEM
60
4.51
E
-
3
3.31
E
-
4
5.01
E
-
2
6.81
E
-
3
1.33
E
-
1
2.51
E
-
2
120
1.10
E
-
5
1.14
E
-
5
1.27
E
-
4
2.28
E
-
3
9.99
E
-
4
7.71
E
-
3
240
4.44
E
-
6
3.86
E
-
5
3.99
E
-
5
7.28
E
-
4
1.93
E
-
4
2.28
E
-
3
480
1.39
E
-
6
1.28
E
-
5
1.10
E
-
5
2.30
E
-
4
3.70
E
-
5
6.90
E
-
4
960
4.27
E
-
7
4.18
E
-
6
3.40
E
-
6
7.24
E
-
5
9.00
E
-
6
2.06
E
-
4
1920
1.17
E
-
7
1.36
E
-
6
9.34
E
-
7
2.27
E
-
5
2.25
E
-
6
6.18
E
-
5Figure 8
Relative errors of Example6 by using the SBM and the BEM. C
S
B
M
=
1.6; C
B
E
M
=
1.6.It can be observed that the present SBM performs more accurate solutions than the BEM with the same number of boundary nodes, and the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
2
D
=
2) in 2D problems. And, in some cases, the convergence rate can reach 3. In addition, the condition number of SBM is lower than BEM. Without mesh and integration, the SBM is computationally more efficient than the BEM.


#### 3.1.4. 3D Laplace Problems

Example 7.
Consider 3D Laplace problems in a sphere domain with radius 1. And the exact solution is given by(19)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (7) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 9.Figure 9
Relative errors of Example7 by using the SBM and the BEM. C
S
B
M
=
2.3; C
B
E
M
=
2.0.Example 8.
Consider 3D Laplace problems in a cube domain with length 1. And the exact solution is given by(20)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: x
=
1, y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (5) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 10.Figure 10
Relative errors of Example8 by using the SBM and the BEM. C
S
B
M
=
2.7; C
B
E
M
=
2.6.Example 9.
Consider 3D Laplace problems in a tire domain. The tire surface is defined by the following equation:(21)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
y
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
z
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5. The distribution of source points on tire surface is shown in Figure 11. And the exact solution is given by (22)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (7) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 12.Figure 11
The distribution of source points on tire surface.Figure 12
Relative errors of Example9 by using the SBM and the BEM. C
S
B
M
=
2.3; C
B
E
M
=
2.2.


#### 3.1.5. 3D Helmholtz Problems

Example 10.
Consider 3D Helmholtz problems in a sphere domain with radius 1. And the exact solution is given by(23)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 13 and 14. Table 3 lists the numerical results of the SBM and the BEM in case (
1
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 3
Numerical results for Helmholtz problem in sphere domain in case (1).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
100
1.51
E
-
3
1.14
E
-
2
1.74
E
-
1
9.47
E
-
2
8.09
E
-
1
9.33
E
-
1
400
3.84
E
-
3
2.80
E
-
3
2.70
E
-
2
2.09
E
-
2
1.27
E
-
1
4.90
E
-
2
900
1.58
E
-
4
1.23
E
-
3
9.66
E
-
3
8.71
E
-
3
3.16
E
-
2
1.32
E
-
2
1600
8.73
E
-
5
6.94
E
-
4
4.85
E
-
3
4.79
E
-
3
1.29
E
-
2
7.42
E
-
3
2500
5.53
E
-
5
4.23
E
-
4
2.90
E
-
3
3.02
E
-
3
6.57
E
-
3
4.78
E
-
3Figure 13
Relative errors of Example10 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Figure 14
Relative errors of Example10 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Example 11.
Consider 3D Helmholtz problems in a cube domain with length 1 and the exact solution is given by(24)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions: (1)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n,x
m
∈
Γ
N.
(2)
x
=
1
,
y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m
,
x
m
∈
Γ
D.
In this example, we use formulas (4) and (10) to determine the SIFs in case (
1
) and formulas (4), (10), (5), and (9) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 15–17. Table 4 lists the numerical results of the SBM and the BEM in case (
2
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 4
Numerical results for Helmholtz problems in cube domain in case (2).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
96
1.71
E
-
3
5.11
E
-
3
2.06
E
-
2
1.03
E
-
1
2.41
E
-
1
1.83
E
-
1
384
1.24
E
-
4
1.59
E
-
3
3.87
E
-
3
3.81
E
-
2
2.35
E
-
2
3.19
E
-
2
864
5.72
E
-
5
9.03
E
-
4
2.02
E
-
3
1.90
E
-
2
7.35
E
-
3
1.44
E
-
2
1536
2.70
E
-
5
6.26
E
-
4
1.48
E
-
3
1.11
E
-
2
3.38
E
-
3
8.31
E
-
3
2400
2.13
E
-
5
4.78
E
-
4
8.57
E
-
4
7.08
E
-
3
1.81
E
-
3
5.37
E
-
3Figure 15
Relative errors of Example11 when k
=1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.0; C
B
E
M
=
1.1.Figure 16
Relative errors of Example11 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.3; C
B
E
M
=
1.5.Figure 17
Condition numbers of Example11 by using the SBM and the BEM in case (
2
).Example 12.
Consider 3D Helmholtz problems in a tire domain. The tire surface is defined by the following equation:(25)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
x
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
y
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5, and the exact solution is given by(26)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 18 and 19.Figure 18
Relative errors of Example12 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
1.9; C
B
E
M
=
1.7.Figure 19
Relative errors of Example12 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.0; C
B
E
M
=
1.7.


#### 3.1.6. 3D Modified Helmholtz Problems

Example 13.
Consider 3D modified Helmholtz problems in a sphere domain with radius 1. And the exact solution is given by(27)
u
x
,
y
,
z
=
exp
⁡
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 20 and 21.Figure 20
Relative errors of Example13 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Figure 21
Relative errors of Example13 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.2; C
B
E
M
=
2.0.Example 14.
Consider 3D modified Helmholtz problems in a cube domain with length 1 and the exact solution is given by(28)
u
x
,
y
,
z
=
e
x
p
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
(2)
x
=
1, y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (10) to determine the SIFs in case (
1
) and formulas (4), (10), (5), and (9) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
5 in case (
1
) and k
=
1 in case (
2
) are shown in Figures 22 and 23. Table 5 lists the numerical results of the SBM and the BEM in case (
1
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 5
Numerical results for modified Helmholtz problems in cube domain in case (1).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
96
7.01
E
-
3
1.03
E
-
2
3.63
E
-
2
8.68
E
-
2
7.07
E
-
2
2.18
E
-
1
384
1.58
E
-
3
3.01
E
-
3
9.53
E
-
3
2.45
E
-
2
2.25
E
-
2
4.84
E
-
2
864
7.35
E
-
4
3.31
E
-
3
4.47
E
-
3
1.20
E
-
2
1.09
E
-
2
2.19
E
-
2
1536
4.25
E
-
4
3.29
E
-
3
2.61
E
-
3
7.21
E
-
3
6.41
E
-
3
1.29
E
-
2
2400
2.76
E
-
4
3.21
E
-
3
1.68
E
-
3
4.78
E
-
3
4.18
E
-
3
8.61
E
-
3Figure 22
Relative errors of Example14 when k
=
5 by using the SBM and the BEM in case (
1
). C
S
B
M
=
1.9; C
B
E
M
=
1.8.Figure 23
Relative errors of Example14 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.4; C
B
E
M
=
2.4.Example 15.
Consider 3D modified Helmholtz problems in a tire domain. The tire surface is defined by the following equation:(29)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
x
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
y
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5, and the exact solution is given by(30)
u
x
,
y
,
z
=
e
x
p
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 24 and 25.
It can be observed that the present SBM performs more accurate solutions than the BEM with the same number of boundary nodes, and the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
3
D
=
2) in all these 3D examples.Figure 24
Relative errors of Example15 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
3.1; C
B
E
M
=
1.5.Figure 25
Relative errors of Example15 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.0; C
B
E
M
=
1.5.


## 3.1. Error Analysis

### 3.1.1. 2D Laplace Problems

Example 1.
Consider 2D Laplace problems in a circular domain with radius 1. And the exact solution is given by(13)
u
x
,
y
=
cos
⁡
x
exp
⁡
y
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (6) to determine the SIFs. And the numerical results are shown in Figures 2 and 3.Figure 2
Relative errors of Example1 by using the SBM and the BEM. C
S
B
M
=
3.0; C
B
E
M
=
2.0.Figure 3
Condition numbers of the SBM and the BEM in Example1.Example 2.
Consider 2D Laplace problems in a square domain with length 1. And the exact solution is given by(14)
u
x
,
y
=
cos
⁡
x
exp
⁡
y
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (6) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 4.Figure 4
Relative errors of Example2 by using the SBM and the BEM. C
S
B
M
=
1.9; C
B
E
M
=
1.9.


### 3.1.2. 2D Helmholtz Problems

Example 3.
Consider 2D Helmholtz problems in a circular domain with radius 1. And the exact solution is given by(15)
u
x
,
y
=
cos
⁡
k
x
+
cos
⁡
k
y
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (6) and (9) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 5. Table 1 lists the numerical results of the SBM and the BEM with different wavenumbers (k
=
1, k
=
10, and k
=
20).Table 1
Numerical results for Helmholtz problem in circular domain.
N
k
=
1
k
=
10
k
=
20
SBM
BEM
SBM
BEM
SBM
BEM
40
2.70
E
-
4
2.87
E
-
4
1.63
E
-
2
2.31
E
-
2
6.15
E
-
1
2.48
E
-
1
80
2.50
E
-
6
7.31
E
-
5
1.78
E
-
3
5.95
E
-
3
2.54
E
-
2
1.86
E
-
2
160
1.92
E
-
7
1.84
E
-
5
2.18
E
-
4
1.53
E
-
3
2.92
E
-
3
5.33
E
-
3
320
2.40
E
-
8
4.62
E
-
6
2.71
E
-
5
3.86
E
-
4
3.58
E
-
4
1.44
E
-
3
640
2.99
E
-
9
1.16
E
-
6
3.38
E
-
6
9.72
E
-
5
4.45
E
-
5
3.76
E
-
4
1280
3.74
E
-
10
2.89
E
-
7
4.23
E
-
7
2.43
E
-
5
5.56
E
-
6
9.59
E
-
5Figure 5
Relative errors of Example3 by using the SBM and the BEM. C
S
B
M
=
3.1; C
B
E
M
=
1.9.Example 4.
Consider 2D Helmholtz problems in a square domain with length 1. And the exact solution is given by(16)
u
x
,
y
=
cos
⁡
k
x
+
cos
⁡
k
y
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4), (6), (9), and (10) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 6.Figure 6
Relative errors of Example4 by using the SBM and the BEM. C
S
B
M
=
1.6; C
B
E
M
=
1.6.


### 3.1.3. 2D Modified Helmholtz Problems

Example 5.
Consider 2D modified Helmholtz problems in a circular domain with radius 1. And the exact solution is given by(17)
u
x
,
y
=
exp
⁡
k
x
+
y
2
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (6) and (9) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 7.Figure 7
Relative errors of Example5 by using the SBM and the BEM. C
S
B
M
=
3.1; C
B
E
M
=
2.0.Example 6.
Consider 2D modified Helmholtz problems in a square domain with length 1. And the exact solution is given by(18)
u
x
,
y
=
e
x
p
k
x
+
y
2
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4), (6), (9), and (10) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 8. Table 2 lists the numerical results of the SBM and the BEM with different wavenumbers (k
=
1, k
=
10, and k
=
20).Table 2
Numerical results for Helmholtz problem in square domain.
N
k
=
1
k
=
10
k
=
20
SBM
BEM
SBM
BEM
SBM
BEM
60
4.51
E
-
3
3.31
E
-
4
5.01
E
-
2
6.81
E
-
3
1.33
E
-
1
2.51
E
-
2
120
1.10
E
-
5
1.14
E
-
5
1.27
E
-
4
2.28
E
-
3
9.99
E
-
4
7.71
E
-
3
240
4.44
E
-
6
3.86
E
-
5
3.99
E
-
5
7.28
E
-
4
1.93
E
-
4
2.28
E
-
3
480
1.39
E
-
6
1.28
E
-
5
1.10
E
-
5
2.30
E
-
4
3.70
E
-
5
6.90
E
-
4
960
4.27
E
-
7
4.18
E
-
6
3.40
E
-
6
7.24
E
-
5
9.00
E
-
6
2.06
E
-
4
1920
1.17
E
-
7
1.36
E
-
6
9.34
E
-
7
2.27
E
-
5
2.25
E
-
6
6.18
E
-
5Figure 8
Relative errors of Example6 by using the SBM and the BEM. C
S
B
M
=
1.6; C
B
E
M
=
1.6.It can be observed that the present SBM performs more accurate solutions than the BEM with the same number of boundary nodes, and the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
2
D
=
2) in 2D problems. And, in some cases, the convergence rate can reach 3. In addition, the condition number of SBM is lower than BEM. Without mesh and integration, the SBM is computationally more efficient than the BEM.


### 3.1.4. 3D Laplace Problems

Example 7.
Consider 3D Laplace problems in a sphere domain with radius 1. And the exact solution is given by(19)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (7) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 9.Figure 9
Relative errors of Example7 by using the SBM and the BEM. C
S
B
M
=
2.3; C
B
E
M
=
2.0.Example 8.
Consider 3D Laplace problems in a cube domain with length 1. And the exact solution is given by(20)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: x
=
1, y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (5) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 10.Figure 10
Relative errors of Example8 by using the SBM and the BEM. C
S
B
M
=
2.7; C
B
E
M
=
2.6.Example 9.
Consider 3D Laplace problems in a tire domain. The tire surface is defined by the following equation:(21)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
y
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
z
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5. The distribution of source points on tire surface is shown in Figure 11. And the exact solution is given by (22)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (7) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 12.Figure 11
The distribution of source points on tire surface.Figure 12
Relative errors of Example9 by using the SBM and the BEM. C
S
B
M
=
2.3; C
B
E
M
=
2.2.


### 3.1.5. 3D Helmholtz Problems

Example 10.
Consider 3D Helmholtz problems in a sphere domain with radius 1. And the exact solution is given by(23)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 13 and 14. Table 3 lists the numerical results of the SBM and the BEM in case (
1
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 3
Numerical results for Helmholtz problem in sphere domain in case (1).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
100
1.51
E
-
3
1.14
E
-
2
1.74
E
-
1
9.47
E
-
2
8.09
E
-
1
9.33
E
-
1
400
3.84
E
-
3
2.80
E
-
3
2.70
E
-
2
2.09
E
-
2
1.27
E
-
1
4.90
E
-
2
900
1.58
E
-
4
1.23
E
-
3
9.66
E
-
3
8.71
E
-
3
3.16
E
-
2
1.32
E
-
2
1600
8.73
E
-
5
6.94
E
-
4
4.85
E
-
3
4.79
E
-
3
1.29
E
-
2
7.42
E
-
3
2500
5.53
E
-
5
4.23
E
-
4
2.90
E
-
3
3.02
E
-
3
6.57
E
-
3
4.78
E
-
3Figure 13
Relative errors of Example10 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Figure 14
Relative errors of Example10 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Example 11.
Consider 3D Helmholtz problems in a cube domain with length 1 and the exact solution is given by(24)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions: (1)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n,x
m
∈
Γ
N.
(2)
x
=
1
,
y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m
,
x
m
∈
Γ
D.
In this example, we use formulas (4) and (10) to determine the SIFs in case (
1
) and formulas (4), (10), (5), and (9) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 15–17. Table 4 lists the numerical results of the SBM and the BEM in case (
2
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 4
Numerical results for Helmholtz problems in cube domain in case (2).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
96
1.71
E
-
3
5.11
E
-
3
2.06
E
-
2
1.03
E
-
1
2.41
E
-
1
1.83
E
-
1
384
1.24
E
-
4
1.59
E
-
3
3.87
E
-
3
3.81
E
-
2
2.35
E
-
2
3.19
E
-
2
864
5.72
E
-
5
9.03
E
-
4
2.02
E
-
3
1.90
E
-
2
7.35
E
-
3
1.44
E
-
2
1536
2.70
E
-
5
6.26
E
-
4
1.48
E
-
3
1.11
E
-
2
3.38
E
-
3
8.31
E
-
3
2400
2.13
E
-
5
4.78
E
-
4
8.57
E
-
4
7.08
E
-
3
1.81
E
-
3
5.37
E
-
3Figure 15
Relative errors of Example11 when k
=1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.0; C
B
E
M
=
1.1.Figure 16
Relative errors of Example11 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.3; C
B
E
M
=
1.5.Figure 17
Condition numbers of Example11 by using the SBM and the BEM in case (
2
).Example 12.
Consider 3D Helmholtz problems in a tire domain. The tire surface is defined by the following equation:(25)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
x
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
y
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5, and the exact solution is given by(26)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 18 and 19.Figure 18
Relative errors of Example12 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
1.9; C
B
E
M
=
1.7.Figure 19
Relative errors of Example12 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.0; C
B
E
M
=
1.7.


### 3.1.6. 3D Modified Helmholtz Problems

Example 13.
Consider 3D modified Helmholtz problems in a sphere domain with radius 1. And the exact solution is given by(27)
u
x
,
y
,
z
=
exp
⁡
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 20 and 21.Figure 20
Relative errors of Example13 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Figure 21
Relative errors of Example13 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.2; C
B
E
M
=
2.0.Example 14.
Consider 3D modified Helmholtz problems in a cube domain with length 1 and the exact solution is given by(28)
u
x
,
y
,
z
=
e
x
p
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
(2)
x
=
1, y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (10) to determine the SIFs in case (
1
) and formulas (4), (10), (5), and (9) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
5 in case (
1
) and k
=
1 in case (
2
) are shown in Figures 22 and 23. Table 5 lists the numerical results of the SBM and the BEM in case (
1
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 5
Numerical results for modified Helmholtz problems in cube domain in case (1).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
96
7.01
E
-
3
1.03
E
-
2
3.63
E
-
2
8.68
E
-
2
7.07
E
-
2
2.18
E
-
1
384
1.58
E
-
3
3.01
E
-
3
9.53
E
-
3
2.45
E
-
2
2.25
E
-
2
4.84
E
-
2
864
7.35
E
-
4
3.31
E
-
3
4.47
E
-
3
1.20
E
-
2
1.09
E
-
2
2.19
E
-
2
1536
4.25
E
-
4
3.29
E
-
3
2.61
E
-
3
7.21
E
-
3
6.41
E
-
3
1.29
E
-
2
2400
2.76
E
-
4
3.21
E
-
3
1.68
E
-
3
4.78
E
-
3
4.18
E
-
3
8.61
E
-
3Figure 22
Relative errors of Example14 when k
=
5 by using the SBM and the BEM in case (
1
). C
S
B
M
=
1.9; C
B
E
M
=
1.8.Figure 23
Relative errors of Example14 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.4; C
B
E
M
=
2.4.Example 15.
Consider 3D modified Helmholtz problems in a tire domain. The tire surface is defined by the following equation:(29)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
x
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
y
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5, and the exact solution is given by(30)
u
x
,
y
,
z
=
e
x
p
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 24 and 25.
It can be observed that the present SBM performs more accurate solutions than the BEM with the same number of boundary nodes, and the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
3
D
=
2) in all these 3D examples.Figure 24
Relative errors of Example15 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
3.1; C
B
E
M
=
1.5.Figure 25
Relative errors of Example15 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.0; C
B
E
M
=
1.5.


## 3.1.1. 2D Laplace Problems

Example 1.
Consider 2D Laplace problems in a circular domain with radius 1. And the exact solution is given by(13)
u
x
,
y
=
cos
⁡
x
exp
⁡
y
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (6) to determine the SIFs. And the numerical results are shown in Figures 2 and 3.Figure 2
Relative errors of Example1 by using the SBM and the BEM. C
S
B
M
=
3.0; C
B
E
M
=
2.0.Figure 3
Condition numbers of the SBM and the BEM in Example1.Example 2.
Consider 2D Laplace problems in a square domain with length 1. And the exact solution is given by(14)
u
x
,
y
=
cos
⁡
x
exp
⁡
y
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (6) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 4.Figure 4
Relative errors of Example2 by using the SBM and the BEM. C
S
B
M
=
1.9; C
B
E
M
=
1.9.


## 3.1.2. 2D Helmholtz Problems

Example 3.
Consider 2D Helmholtz problems in a circular domain with radius 1. And the exact solution is given by(15)
u
x
,
y
=
cos
⁡
k
x
+
cos
⁡
k
y
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (6) and (9) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 5. Table 1 lists the numerical results of the SBM and the BEM with different wavenumbers (k
=
1, k
=
10, and k
=
20).Table 1
Numerical results for Helmholtz problem in circular domain.
N
k
=
1
k
=
10
k
=
20
SBM
BEM
SBM
BEM
SBM
BEM
40
2.70
E
-
4
2.87
E
-
4
1.63
E
-
2
2.31
E
-
2
6.15
E
-
1
2.48
E
-
1
80
2.50
E
-
6
7.31
E
-
5
1.78
E
-
3
5.95
E
-
3
2.54
E
-
2
1.86
E
-
2
160
1.92
E
-
7
1.84
E
-
5
2.18
E
-
4
1.53
E
-
3
2.92
E
-
3
5.33
E
-
3
320
2.40
E
-
8
4.62
E
-
6
2.71
E
-
5
3.86
E
-
4
3.58
E
-
4
1.44
E
-
3
640
2.99
E
-
9
1.16
E
-
6
3.38
E
-
6
9.72
E
-
5
4.45
E
-
5
3.76
E
-
4
1280
3.74
E
-
10
2.89
E
-
7
4.23
E
-
7
2.43
E
-
5
5.56
E
-
6
9.59
E
-
5Figure 5
Relative errors of Example3 by using the SBM and the BEM. C
S
B
M
=
3.1; C
B
E
M
=
1.9.Example 4.
Consider 2D Helmholtz problems in a square domain with length 1. And the exact solution is given by(16)
u
x
,
y
=
cos
⁡
k
x
+
cos
⁡
k
y
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4), (6), (9), and (10) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 6.Figure 6
Relative errors of Example4 by using the SBM and the BEM. C
S
B
M
=
1.6; C
B
E
M
=
1.6.


## 3.1.3. 2D Modified Helmholtz Problems

Example 5.
Consider 2D modified Helmholtz problems in a circular domain with radius 1. And the exact solution is given by(17)
u
x
,
y
=
exp
⁡
k
x
+
y
2
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (6) and (9) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 7.Figure 7
Relative errors of Example5 by using the SBM and the BEM. C
S
B
M
=
3.1; C
B
E
M
=
2.0.Example 6.
Consider 2D modified Helmholtz problems in a square domain with length 1. And the exact solution is given by(18)
u
x
,
y
=
e
x
p
k
x
+
y
2
.Boundary conditions: y
=
0 and x
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0 and y
=
1 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4), (6), (9), and (10) to determine the SIFs. The numerical results of SBM and BEM when k
=
1 are shown in Figure 8. Table 2 lists the numerical results of the SBM and the BEM with different wavenumbers (k
=
1, k
=
10, and k
=
20).Table 2
Numerical results for Helmholtz problem in square domain.
N
k
=
1
k
=
10
k
=
20
SBM
BEM
SBM
BEM
SBM
BEM
60
4.51
E
-
3
3.31
E
-
4
5.01
E
-
2
6.81
E
-
3
1.33
E
-
1
2.51
E
-
2
120
1.10
E
-
5
1.14
E
-
5
1.27
E
-
4
2.28
E
-
3
9.99
E
-
4
7.71
E
-
3
240
4.44
E
-
6
3.86
E
-
5
3.99
E
-
5
7.28
E
-
4
1.93
E
-
4
2.28
E
-
3
480
1.39
E
-
6
1.28
E
-
5
1.10
E
-
5
2.30
E
-
4
3.70
E
-
5
6.90
E
-
4
960
4.27
E
-
7
4.18
E
-
6
3.40
E
-
6
7.24
E
-
5
9.00
E
-
6
2.06
E
-
4
1920
1.17
E
-
7
1.36
E
-
6
9.34
E
-
7
2.27
E
-
5
2.25
E
-
6
6.18
E
-
5Figure 8
Relative errors of Example6 by using the SBM and the BEM. C
S
B
M
=
1.6; C
B
E
M
=
1.6.It can be observed that the present SBM performs more accurate solutions than the BEM with the same number of boundary nodes, and the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
2
D
=
2) in 2D problems. And, in some cases, the convergence rate can reach 3. In addition, the condition number of SBM is lower than BEM. Without mesh and integration, the SBM is computationally more efficient than the BEM.


## 3.1.4. 3D Laplace Problems

Example 7.
Consider 3D Laplace problems in a sphere domain with radius 1. And the exact solution is given by(19)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (7) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 9.Figure 9
Relative errors of Example7 by using the SBM and the BEM. C
S
B
M
=
2.3; C
B
E
M
=
2.0.Example 8.
Consider 3D Laplace problems in a cube domain with length 1. And the exact solution is given by(20)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: x
=
1, y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (5) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 10.Figure 10
Relative errors of Example8 by using the SBM and the BEM. C
S
B
M
=
2.7; C
B
E
M
=
2.6.Example 9.
Consider 3D Laplace problems in a tire domain. The tire surface is defined by the following equation:(21)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
y
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
z
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5. The distribution of source points on tire surface is shown in Figure 11. And the exact solution is given by (22)
u
x
,
y
,
z
=
exp
⁡
2
z
cos
⁡
x
sin
⁡
y
+
1
.Boundary conditions: all Dirichlet boundary conditions: u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formula (7) to determine the SIFs. The numerical results of SBM and BEM are shown in Figure 12.Figure 11
The distribution of source points on tire surface.Figure 12
Relative errors of Example9 by using the SBM and the BEM. C
S
B
M
=
2.3; C
B
E
M
=
2.2.


## 3.1.5. 3D Helmholtz Problems

Example 10.
Consider 3D Helmholtz problems in a sphere domain with radius 1. And the exact solution is given by(23)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 13 and 14. Table 3 lists the numerical results of the SBM and the BEM in case (
1
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 3
Numerical results for Helmholtz problem in sphere domain in case (1).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
100
1.51
E
-
3
1.14
E
-
2
1.74
E
-
1
9.47
E
-
2
8.09
E
-
1
9.33
E
-
1
400
3.84
E
-
3
2.80
E
-
3
2.70
E
-
2
2.09
E
-
2
1.27
E
-
1
4.90
E
-
2
900
1.58
E
-
4
1.23
E
-
3
9.66
E
-
3
8.71
E
-
3
3.16
E
-
2
1.32
E
-
2
1600
8.73
E
-
5
6.94
E
-
4
4.85
E
-
3
4.79
E
-
3
1.29
E
-
2
7.42
E
-
3
2500
5.53
E
-
5
4.23
E
-
4
2.90
E
-
3
3.02
E
-
3
6.57
E
-
3
4.78
E
-
3Figure 13
Relative errors of Example10 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Figure 14
Relative errors of Example10 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Example 11.
Consider 3D Helmholtz problems in a cube domain with length 1 and the exact solution is given by(24)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions: (1)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n,x
m
∈
Γ
N.
(2)
x
=
1
,
y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m
,
x
m
∈
Γ
D.
In this example, we use formulas (4) and (10) to determine the SIFs in case (
1
) and formulas (4), (10), (5), and (9) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 15–17. Table 4 lists the numerical results of the SBM and the BEM in case (
2
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 4
Numerical results for Helmholtz problems in cube domain in case (2).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
96
1.71
E
-
3
5.11
E
-
3
2.06
E
-
2
1.03
E
-
1
2.41
E
-
1
1.83
E
-
1
384
1.24
E
-
4
1.59
E
-
3
3.87
E
-
3
3.81
E
-
2
2.35
E
-
2
3.19
E
-
2
864
5.72
E
-
5
9.03
E
-
4
2.02
E
-
3
1.90
E
-
2
7.35
E
-
3
1.44
E
-
2
1536
2.70
E
-
5
6.26
E
-
4
1.48
E
-
3
1.11
E
-
2
3.38
E
-
3
8.31
E
-
3
2400
2.13
E
-
5
4.78
E
-
4
8.57
E
-
4
7.08
E
-
3
1.81
E
-
3
5.37
E
-
3Figure 15
Relative errors of Example11 when k
=1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.0; C
B
E
M
=
1.1.Figure 16
Relative errors of Example11 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.3; C
B
E
M
=
1.5.Figure 17
Condition numbers of Example11 by using the SBM and the BEM in case (
2
).Example 12.
Consider 3D Helmholtz problems in a tire domain. The tire surface is defined by the following equation:(25)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
x
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
y
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5, and the exact solution is given by(26)
u
x
,
y
,
z
=
cos
⁡
k
x
+
cos
⁡
k
y
+
cos
⁡
k
z
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 18 and 19.Figure 18
Relative errors of Example12 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
1.9; C
B
E
M
=
1.7.Figure 19
Relative errors of Example12 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.0; C
B
E
M
=
1.7.


## 3.1.6. 3D Modified Helmholtz Problems

Example 13.
Consider 3D modified Helmholtz problems in a sphere domain with radius 1. And the exact solution is given by(27)
u
x
,
y
,
z
=
exp
⁡
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 20 and 21.Figure 20
Relative errors of Example13 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
2.1; C
B
E
M
=
2.0.Figure 21
Relative errors of Example13 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.2; C
B
E
M
=
2.0.Example 14.
Consider 3D modified Helmholtz problems in a cube domain with length 1 and the exact solution is given by(28)
u
x
,
y
,
z
=
e
x
p
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
(2)
x
=
1, y
=
1, and z
=
1 are Neumann boundary conditions; ∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N. x
=
0, y
=
0, and z
=
0 are Dirichlet boundary conditions; u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
In this example, we use formulas (4) and (10) to determine the SIFs in case (
1
) and formulas (4), (10), (5), and (9) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
5 in case (
1
) and k
=
1 in case (
2
) are shown in Figures 22 and 23. Table 5 lists the numerical results of the SBM and the BEM in case (
1
) with different wavenumbers (k
=
1, k
=
5, and k
=
10).Table 5
Numerical results for modified Helmholtz problems in cube domain in case (1).
N
k
=
1
k
=
5
k
=
10
SBM
BEM
SBM
BEM
SBM
BEM
96
7.01
E
-
3
1.03
E
-
2
3.63
E
-
2
8.68
E
-
2
7.07
E
-
2
2.18
E
-
1
384
1.58
E
-
3
3.01
E
-
3
9.53
E
-
3
2.45
E
-
2
2.25
E
-
2
4.84
E
-
2
864
7.35
E
-
4
3.31
E
-
3
4.47
E
-
3
1.20
E
-
2
1.09
E
-
2
2.19
E
-
2
1536
4.25
E
-
4
3.29
E
-
3
2.61
E
-
3
7.21
E
-
3
6.41
E
-
3
1.29
E
-
2
2400
2.76
E
-
4
3.21
E
-
3
1.68
E
-
3
4.78
E
-
3
4.18
E
-
3
8.61
E
-
3Figure 22
Relative errors of Example14 when k
=
5 by using the SBM and the BEM in case (
1
). C
S
B
M
=
1.9; C
B
E
M
=
1.8.Figure 23
Relative errors of Example14 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.4; C
B
E
M
=
2.4.Example 15.
Consider 3D modified Helmholtz problems in a tire domain. The tire surface is defined by the following equation:(29)
x
,
y
,
z
∣
x
=
R
+
r
cos
⁡
φ
cos
⁡
θ
,
x
=
R
+
r
cos
⁡
φ
sin
⁡
θ
,
y
=
r
sin
⁡
φ
,
0
≤
θ
,
φ
<
2
π
,where R
=
2, r
=
0.5, and the exact solution is given by(30)
u
x
,
y
,
z
=
e
x
p
k
x
+
y
+
z
3
.Boundary conditions:(1)
all Dirichlet boundary conditions:u
x
m
=
u
-
x
m, x
m
∈
Γ
D.
(2)
all Neumann boundary conditions:∂
u
x
m
/
∂
n
=
∂
u
-
x
m
/
∂
n, x
m
∈
Γ
N.
In this example, we use formulas (7) and (9) to determine the SIFs in case (
1
) and formulas (4) and (10) to determine the SIFs in case (
2
). The numerical results of SBM and BEM when k
=
1 are shown in Figures 24 and 25.
It can be observed that the present SBM performs more accurate solutions than the BEM with the same number of boundary nodes, and the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
3
D
=
2) in all these 3D examples.Figure 24
Relative errors of Example15 when k
=
1 by using the SBM and the BEM in case (
1
). C
S
B
M
=
3.1; C
B
E
M
=
1.5.Figure 25
Relative errors of Example15 when k
=
1 by using the SBM and the BEM in case (
2
). C
S
B
M
=
2.0; C
B
E
M
=
1.5.


## 4. Conclusion

This short communication makes an extensive investigation on numerical efficiency and convergence rates of the SBM to several 2D and 3D benchmark examples about Laplace, Helmholtz, and modified Helmholtz equations in comparison with the direct BEM.Through numerical experiments, we find that the present SBM results are in good agreement with the exact solutions in both 2D and 3D problems. And the SBM converge remarkably with the increasing boundary node numberN. The error analysis shows that the present SBM have rapid convergence rate (C
2
D
=
C
3
D
=
2) in both 2D and 3D problems for Laplace, Helmholtz, and modified Helmholtz equations. And, in some cases, the convergence rate of SBM can reach 3. In addition, we find from the tables that the present SBM results are still in good agreement with the exact solutions in Helmholtz and modified Helmholtz problems with different wavenumber.Furthermore, we find that in most cases the SBM and BEM have similar convergence rates, while the SBM has slightly better accuracy than the direct BEM. In addition, the condition number of SBM is lower than BEM in most cases. Without mesh and integration, the SBM is computationally more efficient than the BEM.



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