Other numerical methods 259
Then, in order to apply the weighted residual method, a trial function
is selected as

f
x
x
=
+
+
α α
α
0
1
2
2
(9.19)
Since the trial function needs to satisfy the aforementioned boundary conditions, the unknown constants are estimated as α o = 1 and α 2  =
1 – α 1 , which leads to

f
x
x
x x
x
= +
− +
= − +
−
1
1
1
1
1
1
2
2
1
α
α
α
(
)
(
)
(9.20)
Substitution of Equation 9.20 into Equation 9.13 results in the residual
as follows:
R = −2m + (c − 2m − 2cx)α 1 + kx − 2cx
= (12x − 10) α 1  + 16x − 4 ≠ 0
(9.21)
Collocation method: The collocation method requires the selection
of one point within the solution domain 0 ≤ x ≤ 1 since there is only
one unknown (α 1 ). After selecting the point x 1 = 0.5, then the residual
is set equal to zero according to Equations 9.6 and 9.21:
R = (12·0.5 − 10)α 1 + 16·0.5 − 4 = −4α 1 + 4 = 0
(9.22)
From Equation 9.22 α 1 = 1 and the approximate solution is

f
x x
= + −
1
2
2
(9.23)
Sub-domain method: For the sub-domain method, the solution
domain is discretized into a single sub-domain D (0 to 1). In this
method the solution is obtained by solving the following equation
resulting from Equation 9.7:
R
m c m cx
kx cx dx
x
= −
+ −
−
+ −
=
−
+
∫
[
(
)
]
[(
)
2
2
2
2
12 10
16
1
0
1
1
α
α
x x
dx
−
=
∫
4
0
0
1
]
(9.24)
After integration, the constant parameter is found as α 1 = 1, so that the
approximate solution is the same as the one obtained by the collocation method (Equation 9.23).
Least squares method: For the least squares method, first, the derivative of the residual with respect to the unknown constant is calculated as
∂
∂
= −
−
=
−
R c m cx
x
α 1
2
2
12 10
(9.25)