Civil Engineering Reference
In-Depth Information
1.2
Exact
Galerkin
1.0
0.8
y
(
x
)
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1.0
Figure 5.3
Solutions to Example 5.3.
Substitution into the differential equation results in the residual
d
2
y
*
d
x
2
=
2
c
1
+
c
1
x
2
=
c
1
x
2
R
(
x
;
c
1
)
=
+
y
*
−
4
x
−
c
1
x
+
x
−
4
x
−
c
1
x
+
2
c
1
−
3
x
and the weighted residual integral becomes
1
1
x
(
x
−
1)(
c
1
x
2
N
1
(
x
)
R
(
x
;
c
1
)d
x
=
+
c
1
x
−
2
c
1
−
3
x
)d
x
=
0
0
0
While algebraically tedious, the integration is straightforward and yields
c
1
=
5
/
6
so the approximate solution is
5
6
x
(
x
−
1)
+
x
5
6
x
2
1
6
x
y
*(
x
)
=
=
+
As in the previous example, we have the luxury of comparing the approximate solution
to the exact solution, which is
y
(
x
)
=
4
x
−
3
.
565 sin
x
The approximate solution and the exact solution are shown in Figure 5.3 for comparison.
Again, the agreement is observed to be reasonable but could be improved by adding a
second trial function.
How does one know when the MWR solution is accurate enough? That is,
how do we determine whether the solution is close to the exact solution? This
question of
convergence
must be addressed in all approximate solution tech-
niques. If we do not know the exact solution, and we seldom do, we must
