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In-Depth Information
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Problems
dx 2 EI d 2
d 2
7.1 Show that the differential operator for an Euler-Bernoulli beam L
=
dx 2 is both
self-adjoint and positive definite.
Hint:
L
0 v
L
0 v
dx 2 EI d 2
d 2
w
dx 2
L
w
dV
=
dx
L
d 2
EI d 2
w
dx 2
v
dx 2 dx
L
0
L
0
=− v
V
|
+ θ
M
|
+
0
L
0 w
L
0 w
dx 2 EI d 2
d 2
v
dx 2 dx
=
=
L
v
dV
Therefore, L is self-adjoint.
Since 0
= 0
dx 2 dx, 0 w
= 0
EI d 2
w
dx 2
d 2
v
d 2
w
dx 2
2 dx so that L is
v
L
w
dV
L
w
dV
EI
(
)
positive definite.
7.2 Solve the differential equation u +
u
=−
x for 0
≤
x
≤
1, with u
(
0
) =
u
(
1
) =
0
.
Use collocation and compare your answer with the exact solution, which is u
(
x
) =
/
−
.
sin x
sin 1
x
Hint:
Choose a simple trial solution such as
u
=
u 1 x
(
1
−
x
)
which satisfies the
boundary conditions, as would x 2
(
1
−
x
)
,x 3
(
1
−
x
)
, etc.
d x +
R
=
L
u
−
f
=−
2
u 1 +
u 1 x
(
1
−
x
) +
x,
where L
=
1 ,
f
=−
x
1
For a single collocation point at x
=
1
/
2 ,R x = 1 / 2 =−
2
u 1 +
4
u 1 +
1
/
2
.
Results:
Solution u
x
Exact
Collocation
1/4
0.044
0.054
1/2
0.070
0.071
3/4
0.060
0.054
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