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Chapter 4
Handling corner problems in DRM
As mentioned in Example 3.7.1, the normal derivative at the corners using DRBEM 2
needs to be improved. This chapter aims to deal with such problems. The first section
considers a program using the gradient approach in the dual reciprocity method to
resolve corner conditions in a variety of linear problems. We discuss non-linear
problems in Chapter 5. The second section presents an application to the heat
conduction problem. The last section gives concluding remarks.
4.1 The GRADRM program
Some improvement is needed in results of the problems when the boundary contains
corners as mentioned in Example 3.7.1. Unfortunately, the multiple node method
cannot be used with DRM. Collocation at the same point forces the matrix F in
equation (3.38) to be singular. However, the gradient approach as discussed in section
2.3 is introduced to resolve the problem. DRBEM2 is modified using this gradient
approach. We call the new code GRADRM. To test on several problems, GRADRM
is implemented to solve all linear problems in the form
y
uyxp
x
uyxpuyxpyxpu
),(),(),(),( 4321
2. (4.1)
We apply both the radial basis function rf 1 and the augmented thin plate spline
cybxarrf log2 . The program is tested on seven cases:
Case 1: 04321 pppp
Case 2: 0432 ppp , 1),(1 yxp
Case 3: 0432 ppp , 2
1 ),( xyxp
Case 4: 0432 ppp , yxeyxp 2
1 5),(
Case 5: 0431 ppp , 1),(2 yxp
82
Case 6: 0421 ppp , 1),(3 yxp
Case 7: 021 pp , 1),(,1),( 43 yxpyxp
Case 1 04321 pppp
This is Laplace's equation and is used to test GRADRM and compare the result with
the previous programs.
Example 4.1.1 The square problem
Consider the potential problem
02 u (4.2)
with boundary conditions as used in the mixed boundary problem in Example 1.4.1.
The internal solutions are compared with those of using LINBEM, MULBEM and
DRBEM2 and shown in Table 4.1.1. We see that the solution obtained using
GRADRM is better than the other three methods. We note that there is no difference
between two radial basis functions because the Laplace solution does not use the
interpolation process
Table 4.1.1 The potential at the internal points
Normal derivatives on the boundary nodes are shown in Figure 4.1.1 and we can see
that the normal derivative obtained using GRADRM agrees well with the exact value.
It is much better compared with that using DRBEM2.
Point LINBEM MULBEM DRBEM2 GRADRM Exact x y Standard Multiple Linear ATPS Linear ATPS solution
2.00 2.00 200.862 200.044 200.393 200.393 200.000 200.000 200 4.00 4.00 99.909 99.957 99.607 99.607 100.000 100.000 100 2.00 4.00 200.980 200.044 200.393 200.393 200.000 200.000 200 3.00 3.00 150.382 150.000 150.000 150.000 150.000 150.000 150 4.00 2.00 99.752 99.897 99.608 99.608 100.000 100.000 100
83
Case 2 0432 ppp , 1),(1 yxp
In this case the test is the Poisson problem whose domain has corners on a square
plate. Again this problem is posed on the boundary in which N .
Example 4.1.2 Square plate
Consider the Poisson equation
12 u (4.3)
on a square ( , ) : 6 6, 6 6x y x y with homogeneous Dirichlet condition.
The symmetry of the region means we need consider only the second quadrant as
shown Figure 4.1.2, so that 0q on 0x and 0y , 0u on 6x and 6y .
The exact solution may be written in the form
(2 1) (2 1) 2112 12 2
1
( , ) cos cosh 18n n
n
n
u x y a x y x
(4.4)
Figure 4.1.1 Normal derivative along the boundary from DRBEM2 and GRADRM
-60.0
-40.0
-20.0
0.0
20.0
40.0
60.0
1 2 3 4 5 6 7 8 9 10 11 12 1
Exact DRBEM2 GRADRM
Normal derivative from DRBEM2 and GRADRM using TPS
q
Node
ATPS
84
where
32 1
2
2 12
sin1 12
3 (2 1) cosh
n
n na
n
(4.5)
We can see from Table 4.1.2 that GRADRM using Linear works almost as well as
MULDRM. The result is better than that of the other programs. The GRADRM
solution using ATPS is slightly worse than that using Linear. As we found in Toutip et
al. (2000), it should be better as the number of elements increases. To examine this
proposition we partition the boundary into 24 elements. The result is shown in Table
4.1.3.
Table 4.1.2 The potential at the internal points using 12 elements
12 u
(-6,0)
(-6,6) (0,6)
(0,0) 1 4
7 10
0q
0u
0u
Figure 4.1.2 Discretisation of the boundary into 12 elements with
5 internal points
0q
Internal MULDRM DRBEM2 GRADRM Exact point Standard Multiple Linear ATPS Linear ATPS solution
(-2,2) 8.418 8.690 8.421 8.483 8.691 8.755 8.690 (-4,2) 5.582 5.772 5.589 5.631 5.773 5.815 5.748 (-3,3) 6.358 6.547 6.365 6.410 6.548 6.594 6.522 (-2,4) 5.584 5.772 5.589 5.631 5.773 5.815 5.748 (-4,4) 3.889 3.987 3.898 3.928 3.989 4.018 3.928
85
Table 4.1.3 The potential at internal points using 24 elements
We can see from Table 4.1.3 that GRADRM using Linear works as well as
MULBEM using multiple nodes. We compute the RMS error of the GRADRM
solution and find that it is 0.016 for ATPS and 0.043 for Linear. Therefore, the
GRADRM result using ATPS improves more than Linear as the number of elements
increases. The normal derivative on the boundary using 24 elements is shown in
Figure 4.1.3
We can see from Figure 4.1.3 that the normal derivative from DRBEM2 is worse at
the corners such as (0,6) and (-6,0). On the other hand, the derivative from GRADRM
agrees well with the exact value.
Internal MULDRM GRADRM Exact point Standard Multiple Linear ATPS solution
(-2,2) 8.551 8.623 8.623 8.703 8.690 (-4,2) 5.653 5.710 5.711 5.763 5.748 (-3,3) 6.428 6.480 6.481 6.537 6.522 (-2,4) 5.654 5.710 5.711 5.763 5.748 (-4,4) 3.885 3.912 3.912 3.948 3.928
Figure 4.1.3 Normal derivative from DRBEM2 and GRADRM
Normal derivative from DRBEM2 and GRADRM
-4.5 -4.0
-3.5 -3.0
-2.5 -2.0
-1.5 -1.0
-0.5 0.0
(0,6) (-2,6) (-4,6) (-6,6) (-6,4) (-6,2) (-6,0)
Exact DRBEM2 (ATPS) GRADRM (ATPS)
q
Point
86
Case 3 0432 ppp , 2
1 ),( xyxp
This case we again examine the torsion problem in Example 3.4.1 in the case where
2( , )b x y x .
Example 4.1.3 The Poisson problem on elliptical domain
Consider the Poisson equation
22 xu (4.6)
on the elliptical domain as used in Example 3.4.1. The exact solution is given by
1
46.33850
246
1 22
22 yx
yxu (4.7)
which satisfies the boundary condition 0u on and produces
yyyyxx
xxyxq 2.833296246
1
22.839650
246
1 3223 (4.8)
The internal solution at the internal nodes along the line 0y is shown in Table
4.1.4.
Table 4.1.4 The internal solution on the line 0y .
Since the boundary is smooth, we can see from Table 4.1.4 that the solutions from
DRBEM2 and GRADRM are the same. We also see that the Linear result is slightly
Internal MULDRM DRBEM2 GRADRM Exact point Standard Multiple Linear ATPS Linear ATPS solution
(-1.5,0) 0.267 0.265 0.263 0.263 0.263 0.263 0.260 (-0.9,0) 0.234 0.237 0.237 0.258 0.237 0.258 0.240 (-0.3,0) 0.140 0.142 0.143 0.172 0.143 0.172 0.151
(0,0) 0.124 0.127 0.127 0.158 0.127 0.158 0.137 (0.3,0) 0.141 0.142 0.143 0.172 0.143 0.172 0.151 (0.9,0) 0.235 0.236 0.237 0.258 0.237 0.258 0.240 (1.5,0) 0.264 0.265 0.263 0.263 0.263 0.263 0.260
87
better than the ATPS result because of the small numbers of interpolation elements.
However the convergence of the ATPS solution is more rapid than that of Linear
(Toutip, 2000).
As we found in Example 3.4.1 the normal derivative from MULDRM using the
transformation to Laplace problem is again poor in this problem. However, the normal
derivatives from DRBEM2 and GRADRM agree well with the exact value except at
the end of the major axis of the elliptical boundary because of the big change of the
normal vectors at these points as shown in Figure 4.1.4
Case 4 0432 ppp , yxeyxp 2
1 5),(
This case we return to examine the problem in Example 3.6.3.
Example 4.1.4 The mixed problem
Consider the Poisson equation
yxeu 22 5 (4.9)
Figure 4.1.4 Normal derivative along the boundary
Normal derivative along the boundary
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Exact DRBEM2(ATPS) GRADRM(ATPS) MULDRM
q
Node
88
on the domain and boundary conditions as in Example 3.6.3.
We see from Table 4.1.5 that the internal solution using GRADRM is better than that
using DRBEM2. The RMS error of the GRADRM solution using ATPS is 0.011 and
is 0.016 using Linear. Hence, the GRADRM result using ATPS is the better of the two
methods.
Table 4.1.5 The internal solution from DRBEM2 and GRADRM
The normal derivative on the boundary is shown in Figure 4.1.5.
x y DRBEM2 GRADRM Exact Linear ATPS Linear ATPS solution
0.75 0.25 5.722 5.741 5.750 5.769 5.754 0.50 0.25 3.454 3.459 3.500 3.504 3.490 0.25 0.25 2.075 2.062 2.139 2.126 2.117 0.25 0.50 2.691 2.685 2.731 2.725 2.718 0.50 0.50 4.445 4.454 4.471 4.480 4.482 0.75 0.50 7.351 7.363 7.359 7.372 7.389 0.50 0.75 5.738 5.737 5.745 5.745 5.755
Figure 4.1.5 Normal derivative along the boundary from DRBEM2 and GRADRM
Normal Derivative from DRBEM2 and GRADRM
-10
-5
0
5
10
15
20
25
1 1 2 3 4 5 6 7 8 9 9 10 11 12 13 14 15 16 17 17 18 19 20 21 22 23 24
Exact DRBEM(ATPS) GRADRM(ATPS)
q
Node
89
We see from Figure 4.1.5 that the normal derivative agrees well with the exact value
except at the corners such as nodes 1 and 9. Furthermore, the normal derivative using
GRADRM at such nodes is better than those using DRBEM2.
Case 5 0431 ppp , 1),(2 yxp
This case we return to investigate the Poisson-type problem in Example 3.7.2.
Example 4.1.5 The case uu 2
Consider the Poisson-type equation
uu 2 (4.10)
with boundary condition as used in Example 3.7.2.The solution of the problem is also
compared with that of the method of Partridge et al.(1992) and the exact solution in
Table 4.1.6.
We see from Table 4.1.6 that the solutions obtained with GRADRM and DRBEM2
give the same result. This is because the domain of the problem has a smooth
boundary. However, it is slightly better than the results in the reference at almost
every point. The ATPS solution is better than the Linear solution as shown by the
absolute errors in Figure 4.1.6.
Table 4.1.6 The internal solutions of the problem in Example 4.1.5.
Point x y P and B DRBEM2 GRADRM Exact f = 1+r f=1+r+r^2 Linear ATPS Linear ATPS solution
1 1.50 0.00 0.994 0.995 0.996 0.997 0.996 0.997 0.997 2 1.20 -0.35 0.928 0.932 0.928 0.931 0.928 0.931 0.932 3 0.60 -0.45 0.562 0.566 0.562 0.564 0.562 0.564 0.565 4 0.00 -0.45 0.000 0.000 0.000 0.000 0.000 0.000 0.000 5 0.90 0.00 0.780 0.784 0.780 0.782 0.780 0.782 0.783 6 0.30 0.00 0.294 0.296 0.294 0.295 0.294 0.295 0.295 7 0.00 0.00 0.000 0.000 0.000 0.000 0.000 0.000 0.000
90
We can see from Figure 4.1.7 that the normal derivative from GRADRM is the same
as that from DRBEM2 in Example 3.7.2 (see Figure 3.7.3). This is because the
boundary is smooth.
Case 6 0421 ppp , 1),(3 yxp
Example 4.1.6 The case x
uu
2
Consider the Poisson equation
Figure 4.1.6 Absolute error of internal solutions from GRADRM
Absolute error of internal solution from GRADRM
0.0000
0.0010
0.0020
0.0030
0.0040
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
GRADRM(Linear) GRADRM(ATPS)
Absolute error
Node
Figure 4.1.7 Normal derivative along the boundary from GRADRM
-0.50 -0.40 -0.30 -0.20 -0.10 0.00 0.10 0.20 0.30 0.40 0.50
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Exact Linear ATPS
Normal derivative along the boundary from GRADRM
q
Node
91
x
uu
2 (4.11)
on the elliptical domain and discretise as in Example 3.4.1. We consider the Dirichlet
problem for which the exact solution is
xu e (4.12)
The solution of the problem is also compared with that of the method of Partridge et
al. (1992) and the exact solution in Table 4.1.7
Table 4.1.7 The internal solutions of the problem in Example 4.1.6
Point x y P and B DRBEM2 GRADRM Exact f = 1+r f=1+r+r^2 Linear ATPS Linear ATPS solution
1 1.50 0.00 0.229 0.214 0.229 0.225 0.229 0.225 0.223 2 1.20 -0.35 0.307 0.274 0.307 0.305 0.307 0.305 0.301 3 0.60 -0.45 0.555 0.523 0.555 0.553 0.555 0.553 0.549 4 0.00 -0.45 1.003 1.006 1.003 1.005 1.003 1.005 1.000 5 0.90 0.00 0.411 0.363 0.412 0.410 0.412 0.410 0.406 6 0.30 0.00 0.745 0.725 0.745 0.745 0.745 0.745 0.741 7 0.00 0.00 1.002 1.002 1.002 1.005 1.002 1.005 1.000
Figure 4.1.8 Normal derivative along the boundary from GRADRM
Normal derivative from GRADRM
-2.0
0.0
2.0
4.0
6.0
8.0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1
Exact Linear ATPS
q
Node
92
As for the previous example, we see from Table 4.1.7 that the solutions obtained
using GRADRM and the DRBEM2 give the same result. Once again, the ATPS result
is better than that using Linear. The normal derivative on the boundary is shown in
Figure 4.1.8 and we see that the normal derivative from both radial basis functions
agrees well with the exact value.
Case 7 021 pp , 1),(,1),( 43 yxpyxp
This case we return to investigate the problem in Example 3.7.3.
Example 4.1.7 The case y
u
x
uu
2
Consider the Poisson-type equation
y
u
x
uu
2
(4.13)
with the boundary conditions as in Example 3.7.3. The solution of the problem is also
compared with that of the method in Partridge et al. (1992) and the exact solutions in
Table 4.1.8.
As before we see that GRADRM and DRBEM2 give the same result for a problem
with a smooth boundary. We also can see from Table 4.1.8 that this statement is true
for this problem. Furthermore, the GRADRM result using ATPS is more accurate
compared with that using Linear.
Table 4.1.8 The internal solutions of the problem in Example 4.1.7
The normal derivative on the boundary is shown in Figure 4.1.9.
Point x y P and B DRBEM2 GRADRM Exact f = 1+r f=1+r+r^2 Linear ATPS Linear ATPS solution
1 1.50 0.00 1.231 1.214 1.231 1.225 1.231 1.225 1.223 2 1.20 -0.35 1.714 1.669 1.714 1.717 1.714 1.717 1.720 3 0.60 -0.45 2.107 2.057 2.107 2.109 2.107 2.109 2.117 4 0.00 -0.45 2.557 2.547 2.557 2.560 2.557 2.560 2.568 5 0.90 0.00 1.400 1.345 1.401 1.404 1.401 1.404 1.406 6 0.30 0.00 1.731 1.691 1.731 1.737 1.731 1.737 1.741 7 0.00 0.00 1.989 1.963 1.989 1.997 1.989 1.997 2.000
93
We can see from Figure 4.1.9 that the normal derivative from GRADRM agrees well
with the exact value. The ATPS result is more accurate than the Linear one.
4.2 A heat conduction problem
All linear problems we have tested so far involve constant coefficients in variable
terms on the right hand side. This section discusses a problem which contains non-
constant coefficients. The heat conduction problem is such a problem. We give the
formulation of the problem in section 4.2.1. We investigate three examples in section
4.2.2. A problem of this type will be coupled with another problem in Chapter 7.
Functionally graded materials are becoming of interest as processes for their
production are improving. In such materials, physical properties vary rapidly over
short distances thus allowing a smooth transition from one material to another without
the possible problems which can occur at the interface between materials of, say,
significantly differing thermal conductivity. Typically, thermal conductivities vary
exponentially in one variable only. The Green's function for exponentially varying
thermal conductivity has been developed by Gray et al. (2000) and a similar
formulation is developed by Li and Evans (1991) for convective-diffusion problems.
Figure 4.1.9 Normal derivative along the boundary from GRADRM
-2.0
0.0
2.0
4.0
6.0
8.0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1
Exact Linear TPS
Normal derivative along the boundary from GRADRM
q
Node
94
We shall consider in the following examples, the application of the dual reciprocity
method to heat conduction in functionally graded materials.
4.2.1 Formulation of the heat conduction problem
Steady state isotropic heat conduction in a solid is governed by the equation
( ) 0K u (4.14)
where ( , )u u x y is the temperature function and ( , , )K K u x y is the thermal
conductivity.
In two-dimension space, equation (4.14) may be written in the form
2 1 K u K uu
K x x y y
(4.15)
For exponential conductivity in functionally graded materials
( , ) ( ) xK x y K x Ae
We see that equation (4.15) can be solved using GRADRM as implemented in section
4.1. We discuss the computational results in the following subsection.
4.2.2 Computational results
We test the GRADRM program with the following three examples.
Example 4.2.1
Consider the heat conduction problem (4.15) on the unit square
( , ) : 0 1, 0 1x y x y with the heat conductivity
35 xK e (4.16)
95
The boundary condition is shown in Figure 4.2.1. The exact solution is given by Gray
et al. (2000) as
3
3
1100
1
xeu
e
. (4.17)
The boundary is partitioned in to 16, 32 and 64 elements with 9 internal points. The
temperature solutions along the line 0y from GRADRM are compared with those
from the exact solutions in Figure 4.2.2. The symbol 16/9 means that the method uses
16 elements with 9 internal points.
The normal derivative along the line 1x is shown is Figure 4.2.3.
Figure 4.2.1 Boundary conditions of the problem
x
y
(0,0) (1,0)
(1,1) (0,1)
0u 100u
0q
0q
13 9
5 1
Figure 4.2.2 Temperature along the line 0y
0.0
20.0
40.0
60.0
80.0
100.0
0.0 0.2 0.4 0.6 0.8 1.0
16/9 32/9 64/9 Exact
Temperature along the line y = 0u
x
96
We can see from Figure 4.2.2 that the temperature solutions from GRADRM agree
very well with those from the exact solution for all cases but the normal derivative
using Linear as in Figure 4.2.3 is not good enough. However, the normal derivative is
improved when using ATPS or increasing the number of internal nodes as shown in
Figure 4.2.4.
Example 4.2.2
Figure 4.2.3 Normal derivative along the line 1x
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
0 0.25 0.5 0.75 1
16/9 32/9 64/9 Exact
Normal derivative along the line x = 1
q
y
12.0
13.0
14.0
15.0
16.0
0 0.25 0.5 0.75 1
Linear(32/9) Linear(32/49) TPS(32/9)
TPS(32/49) Exact
Normal derivative along the line x = 1
q
y
Figure 4.2.4 Normal derivative from Linear and ATPS along 1x
12.0
13.0
14.0
15.0
16.0
0 0.25 0.5 0.75 1 Linear(32/9) Linear(32/49) ATPS(32/9)
ATPS(32/49) Exact
Normal derivative along the line x = 1 q
y
Figure 4.2.4 Normal derivative from Linear and ATPS along 1x
97
This example uses the heat conduction problem posed on the same domain in the
previous example. The discretisation is also the same as in Example 4.2.2. The
boundary conditions are given in Figure 4.2.5.
The exact solution is also given by Gray et al. (2000) as
31000 xu ye (4.18)
The temperature solution from GRADRM is shown in Figure 4.2.6.
We can see from Figure 4.2.6 that the temperature solutions from GRADRM also
agree very well with those from the exact solution for all cases. The behaviour of the
Figure 4.2.5 Boundary conditions for the problem
x
y
(0,0) (1,0)
(1,1) (0,1)
33000q ye
31000 xq e
13 9
5 1
31000 xq e
1000u y
0
200
400
600
800
1000
0.0 0.2 0.4 0.6 0.8 1.0
16/9 32/9 64/9 Exact
Temperature along the line y = 1u
x
Figure 4.2.6 Temperature along the line 1y
98
normal derivative acts the same as that in the previous example (see Figure 4.2.3 and
Figure 4.2.4).
These examples are rather simple and can be solved using the standard dual
reciprocity method. In practice, functionally graded materials themselves do often
offer relatively simple geometry, probably due to the fact that the mechanical process
involved in their production allows the coating of simple shapes only. Nevertheless,
we should like to test the GRADRM on a more challenging problem and it needs
ATPS and discretisation to improve the results as in the following example.
Example 4.2.3
We consider steady-state heat flow in a quarter annulus bounded by the circles
2 2 1x y , 2 2 2x y and the lines 0x and 0y . The mixed boundary
conditions and the discretisation into 40 elements are shown in Figure 4.2.7.
The exponential thermal conductivity is given by
1
2 2 23 3( )( , ) ( ) 5 5r x yK x y K r e e (4.19)
The boundary is discretised into 20, 40 and 80 elements with 16, 81 and 361 internal
points respectively. The domain problem is cylindrically symmetric and, although
there is no exact solution, we compare the results with an accurate finite difference
solution of the partial differential equation written in polar co-ordinates. The internal
100u
0q
0q
0u
O
21
1 31
11
Figure 4.2.7 Boundary conditions of the problem
r
99
solution along the line 5 with different r using GRADRM is compared with those
using the finite difference method (FDM) in Table 4.2.1. The solution along the curve
1.4r is shown is Table 4.2.2.
Table 4.2.1 The temperature along the line 5
We observe from Table 4.2.1 and Table 4.2.2 that the solution using GRADRM
approaches the reference solution as the number of nodes increases. The solution
using ATPS approaches the reference solution more rapidly than that using Linear, as
expected. Also, both solutions using 80 elements with 361 internal points agree very
well with the FDM solution using 200 nodes as shown in Figure 4.2.8.
Table 4.2.2 The solution along the curve 1.4r
r GRADRM(20/16) GRADRM(40/81) GRADRM(80/361) FDM
Linear ATPS Linear ATPS Linear ATPS 200 1.2 59.01 57.64 55.8 55.02 55.05 54.06 54.59 1.4 85.13 81.38 80.79 79.84 80.26 79.04 79.70 1.6 95.83 92.08 92.46 91.58 92.08 90.95 91.54 1.8 100.29 97.31 98.06 97.19 97.64 96.88 97.23
60 70 80 90
100 u
The solution from GRADRM (80/361)
GRADRM(20/16) GRADRM(40/81) GRADRM(80/361) FDM
Linear ATPS Linear ATPS Linear ATPS 200 91.90 87.86 83.56 82.09 81.11 78.91 79.70 86.17 82.90 81.20 80.37 80.40 78.84 79.70 85.13 81.39 80.79 79.84 80.26 79.04 79.70 85.13 81.39 80.79 79.85 80.26 78.84 79.70 86.17 82.90 81.20 80.39 80.41 78.28 79.70 91.90 87.86 83.56 82.12 81.11 78.16 79.70
0 10 5
3 10 2 5
2
100
We see from Table 4.2.3 that the normal derivative along the curve 1r agrees well
with the FDM values.
Table 4.2.3 The normal derivative along the curve 1r
For the normal derivative along the curve 2r , we can see from Table 4.2.4 that it is
quite poor using a small number of nodes. However it improves as the number of
nodes increases.
Table 4.2.4 The normal derivative along the curve 2r
10 5
3 10 2 5
GRADRM(20/16) GRADRM(40/81) GRADRM(80/361) FDM Linear ATPS Linear ATPS Linear ATPS 200
-479 -490 -387 -392 -395 -389 -392 -392 -401 -392 -397 -394 -390 -392 -392 -400 -392 -397 -394 -390 -392 -479 -490 -387 -392 -395 -390 -392
101
Although the 40/81 and 80/361 solutions using either Linear or ATPS give good flux
values along 1r , they are not so good along 2r , with ATPS only marginally
better than Linear. We notice that our discretisation has elements on 2r which are
twice the size of those on 1r . We consider now 50/72 and 100/342 discretisations
with twice as many nodes on 2r so that the meshes on the curved boundaries are
the same size and we consider the Linear case. We have removed the internal nodes
along the curve nearest to the curved boundary 1r so that the total number of
degrees of freedom is comparable ( 121 for the 40/81and 122 for the 50/72; 441 for
the 80/361 and 442 for the 100/342). The normal derivative along the curved
boundaries is shown in Table 4.2.5 and Table 4.2.6.
Table 4.2.5 The normal derivative along the curve 1r
Table 4.2.6 The normal derivative along the curve 2r
10 5
3 10 2 5
GRADRM(20/16) GRADRM(40/81) GRADRM(80/361) FDM Linear ATPS Linear ATPS Linear ATPS 200
-25.64 -8.58 0.46 5.06 5.60 11.78 9.44 -13.64 -3.76 1.89 5.56 6.10 11.00 9.44 -13.64 -3.76 1.89 5.55 6.18 11.12 9.44 -25.64 -8.59 0.44 5.03 5.50 12.32 9.44
10 5
3 10 2 5
GRADRM GRADRM FDM
40/81 50/72 80/361 100/342 200
0.46 6.18 5.60 7.62 9.44
1.89 6.69 6.10 7.66 9.44
1.89 6.39 6.18 7.78 9.44
0.44 6.17 5.50 7.63 9.44
GRADRM GRADRM FDM
40/81 50/72 80/361 100/342 200
-387 -363 -395 -386 -392
-392 -371 -394 -386 -392
-392 -371 -394 -386 -392
-387 -363 -395 -386 -392
10 5
3 10 2 5
102
We see from Table 4.2.6 that we do indeed see an improvement in the normal
derivative along 2r but we notice a degradation in flux along 1r as in Table
4.2.5. This is not unexpected since the removal of the internal nodes along the first
curve closest to the boundary 1r leads, in principal, to a worse approximation of
the form (3.8) which has an adverse effect on the boundary conditions close to that
boundary. According to Golberg et al. (1998) when using the dual reciprocity method
there is a rule of thumb which states that internal nodes should be placed a point
approximately one element length from the boundary to ensure a reasonable
approximation in the neighbourhood of the boundary.
To improve the normal derivative along the curve 1r from the previous
discretisation, we now remove the internal nodes along the middle curve. We call the
new discretisation 50/72(N) and 100/342(N). The normal derivative is compared with
that from the previous discretisations in Table 4.2.7 and Table 4.2.8.
Table 4.2.7 The normal derivative along 1r
Table 4.2.8 The normal derivative along 2r
10
5
3 10 2 5
GRADRM GRADRM FDM
50/72 50/72(N) 100/342 100/342(N) 200
6.18 5.90 7.62 7.49 9.44
6.69 6.39 7.66 7.64 9.44
6.39 6.13 7.78 7.53 9.44
6.17 5.85 7.63 7.56 9.44
10 5
3 10 2 5
GRADRM GRADRM FDM
50/72 50/72(N) 100/342 100/342(N) 200
-363 -378 -386 -392 -392
-371 -386 -386 -392 -392
-371 -386 -386 -392 -392
-363 -378 -386 -392 -392
103
We can see from Table 4.2.7 that the new discretisation gives the encouraging result
especially the 100/342(N). The normal derivative along 2r using the new
discretisation is almost the same as the previous one as shown in Table 4.2.8. The
solution, u, from both new discretisations is in good agreement with FDM solution as
shown in Table 4.2.9.
Table 4.2.9 The solution, u, along the curve 1.4r
4.3 Concluding remarks
The gradient approach is suitable for this method. GRABEM and DRBEM2 are
integrated into GRADRM. In this development, GRADRM can solve all linear
problems as mentioned before and can resolve corner and discontinuous boundary
condition problems. It is our attention to develop the GRADRM code to solve non-
linear problems in Chapter 5.
For radial basis functions we have shown that the augmented thin plate spline
cybxarrf log2 works, in general, better than the linear function rf 1 .
However, this is not always the case and often there is not a significant difference
between them.
We have tested the GRADRM program with the heat conduction problem. The
computational results from the program agree very well with those from the exact
solutions and the reference solutions. We shall discuss further applications in Chapter
6 and Chapter 7.
0 10
5
3 10 2 5
2
GRADRM GRADRM FDM
40/81 50/72 50/72(N) 80/361 100/342 100/342(N) 200
83.56 80.64 80.81 81.11 80.18 80.21 79.70
81.20 80.18 80.26 80.40 80.03 80.06 79.70
80.79 80.17 80.23 80.26 80.01 80.03 79.70
80.79 80.17 80.23 80.26 80.01 80.03 79.70
81.20 80.18 80.26 80.41 80.03 80.06 79.70
83.56 80.64 80.81 81.11 80.17 80.20 79.70
