Civil Engineering Reference
In-Depth Information
develop some criterion to determine accuracy. In general, for the method of
weighted residuals, the procedure is to continue obtaining solutions while
increasing the number of trial functions and note the behavior of the solution. If
the solution changes very little as we increase the number of trial functions, we
can say that the solution converges. Whether the solution converges to the cor-
rect solution is yet another question. While beyond the scope of this topic, a large
body of theoretical mathematics addresses the questions of convergence and
whether the convergence is to the correct solution. In the context of this work, we
assume that a converging solution converges to the correct solution. Certain
checks, external to the solution procedure, can be made to determine the “reason-
ableness” of a numerical solution in the case of physical problems. These checks
include equilibrium, energy balance, heat and fluid flow balance, and others dis-
cussed in following chapters.
In the previous examples, we used trial functions “concocted” to satisfy
boundary conditions automatically but not based on a systematic procedure.
While absolutely nothing is wrong with this approach, we now present a proce-
dure, based on polynomial trial functions, that gives a method for increasing the
number of trial functions systematically and, hence, aids in examining conver-
gence. The procedure is illustrated in the context of the following example.
EXAMPLE 5.4
Solve the problem of Examples 5.1 and 5.2 by assuming a general polynomial form for
the solution as
y
*(
x
)
=
c
0
+
c
1
x
+
c
2
x
2
+···
■
Solution
For a first trial, we take only the quadratic form
y
*(
x
)
=
c
0
+
c
1
x
+
c
2
x
2
and apply the boundary conditions to obtain
y
*(0)
=
0
=
c
0
y
*(1)
=
0
=
c
1
+
c
2
The second boundary condition equations show that
c
1
and
c
2
are not independent if the
homogeneous boundary condition is to be satisfied exactly. Instead, we obtain the
con-
straint
relation
c
2
=−
c
1
. The trial solution becomes
y
*(
x
)
=
c
1
x
+
c
2
x
2
=
c
1
x
−
c
1
x
2
=
c
1
x
(1
−
x
)
and is the same as the solution obtained in Example 5.1.
Next we add the cubic term and write the trial solution as
y
*(
x
)
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
