Specifications

University of Pretoria etd – Combrinck, M (2006)
where c can take on any value (implying an infinite number of solutions). An easy solution
is to use definite integrals instead, so that
(4.15)).()()( afbfdxxf
b
a
−=
′
∫
From equation 4.12, the system of equations that has to be solved now reduces to a general
form as shown in equation 4.16.
(4.16).
)
1
()(
.
.
.
)()(
)(
.
.
.
)
1
(
...
1
.....
.....
....
21
1
..
1211
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
=
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
′
′
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
n
xf
n
xf
a
xf
b
xf
n
xf
xf
nn
a
n
a
a
n
aaa
The elements of
A (a
ii
) must now be defined so as to produce numerical integration of
the
. Simpson’s rule is a standard and very effective method suitable for polynomial
functions. For equally spaced points it is given by
)(xf
i
′
[]
()()
.
11
,)
1
()(4)
1
(
3
)
1
()
1
(
1
1
)(
−
−=−
+
=
+
′
+
′
⋅+
−
′
≈
−
−
+
=
∫
+
−
′
i
x
i
x
i
x
i
xh
i
xf
i
xf
i
xf
h
i
xf
i
xf
i
x
i
x
dxxf
where
(4.17)
This formula will generate n-2 equations. In order to have a square matrix and unique
solutions we need two more independent equations. This is done by deriving two end-
point Simpson’s rule formulas. The midpoint Simpson’s rule is derived by fitting a second
order polynomial through three consecutive points and integrating this polynomial from
the first to the third point. The same strategy is followed for the end points except that we
integrate from the first to the second point at the start of the sequence and from the
second to the third point at the end. This results in the following two equations.
() .
,)(
4
5
)(2)(
4
1
3
)()()(
,)(
4
1
)(2)(
4
5
3
)()()(
121
22112
1
2
1
i
nnnnn
x
x
x
x
xxh
xfxfxf
h
xfxfdxxf
xfxfxf
h
xfxfdxxf
n
n
−=
⎥
⎦
⎤
⎢
⎣
⎡
′
+
′
⋅+
′
−≈−=
′
⎥
⎦
⎤
⎢
⎣
⎡
′
−
′
⋅+
′
≈−=
′
+
−−−
∫
∫
−
1i
where
(4.19)
and
(4.18)
For equally spaced points, equation 4.16 can thus be rewritten as
43