Civil Engineering Reference
In-Depth Information
A graphical comparison of the two solutions is depicted in Figure 5.1, which shows that
the approximate solution is in reasonable agreement with the exact solution. However,
note that the one-term approximate solution is symmetric over the interval of interest.
That this is not correct can be seen by examining the differential equation. The prime
driving “force” is the quadratic term in
x
; therefore, it is unlikely that the solution is
symmetric. The following example expands the solution and shows how the method
approaches the exact solution.
EXAMPLE 5.2
Obtain a two-term Galerkin solution for the problem of Example 5.1 using the trial
functions
=
x
2
(
x
−
N
1
(
x
)
=
x
(
x
−
1)
N
2
(
x
)
1)
■
Solution
The two-term approximate solution is
+
c
2
x
2
(
x
−
y
*
=
c
1
x
(
x
−
1)
1)
and the second derivative is
d
2
y
*
d
x
2
=
2
c
1
+
2
c
2
(3
x
−
1)
Substituting into the differential equation, we obtain the residual
R
(
x
;
c
1
,
c
2
)
=
2
c
1
+
2
c
2
(3
x
−
1)
−
10
x
2
−
5
Using the trial functions as the weighting functions per Galerkin's method, the residual
equations become
1
x
(
x
−
1)[2
c
1
+
2
c
2
(3
x
−
1)
−
10
x
2
−
5] d
x
=
0
0
1
x
2
(
x
−
1)[2
c
1
+
2
c
2
(3
x
−
1)
−
10
x
2
−
5] d
x
=
0
0
After integration and simplification, we obtain the algebraic equations
c
1
3
−
c
2
6
+
4
3
=
0
−
c
1
6
−
2
c
2
15
+
3
4
=
0
−
Simultaneous solution results in
19
6
5
3
c
1
=
c
2
=
