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+
x
2
This gives
u
1
=−
0
.
333
.
As a second approximation, use
u
=
x
u
1
(
−
x
)
+
The integrals
1
0
0
,
1
0
x
3
u
2
(
−
x
).
Rdx
=
xR dx
=
0 give two nonlinear algebraic
3
x
2
3
x
2
x
3
equations for
u
1
and
u
2
. Thus,
u
(
x
)
=
/
2
−
/
4
+
/
4
.
Results:
Solution
u
First Approx.
Second Approx.
x
Exact
Method of Moments
Method of Moments
1/4
0.323
0.313
0.332
1/2
0.581
0.583
0.594
3/4
0.803
0.813
0.809
7.10 Consider the problem posed in Problem 7.9. This system represents the steady state
heat conduction across a slab with a conductivity of 1
+
au,
where
u
is a nondimen-
sional temperature differential.
(a) Solve the problem with the method of collocation for
a
=
1
.
Compare your result
with the exact solution.
(b) Same as step (a), but use Galerkin's method. Be careful as this is a nonlinear
system.
7.11 Use a least squares approach to solve the differential equation of Problem 7.2.
=
Hint:
With the trial function
u
u
1
x
(
1
−
x
)
,
which satisfies the boundary condi-
+
/∂
tions,
R
=−
2
u
1
u
1
x
(
1
−
x
)
+
x,
∂
R
u
1
=−
2
+
x
(
1
−
x
).
With
1
R
∂
R
∂
dV
=
[
−
2
u
1
+
u
1
x
(
1
−
x
)
+
x
][
−
2
+
x
(
1
−
x
)
]
dx
=
0
u
1
V
0
This gives
u
1
=
55
/
202 and
u
(
x
)
=
55
x
(
1
−
x
)/
202
Results:
Solution
u
x
Exact
Last Squares
1/4
0.044
0.051
1/2
0.070
0.068
3/4
0.060
0.051
7.12 Show that under certain conditions a self-adjoint operator Galerkin's method leads
to symmetric equations in the undetermined coefficients
u
.
7.13 Show that equations for the coefficients
u
,
as obtained by the least squares method,
are both symmetric and positive definite.
7.14
Su
ppose a simply supported beam of length
L
is subjected to the distributed load
p
z
=
Calculate the distribution of bending moment using three weighted-
residual methods, e.g., collocation, least squares, and Galerkin. Compare the results
with the exact solution. Let
L
sin
(π
x
/
L
).
=
.
1
dx
2
Hint:
Recall that
d
2
M
/
=−
p
z
.
7.15 Suppose
a
beam
of
unit
length
with
unit
distributed
load
(
p
z
=
1
force
/
length
is fixed on both ends and rests on an elastic foundation. The governing
equation would be
EId
4
)
dx
4
w/
+
w
=−
k
p
z
,
where
k
is the foundation modulus.
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