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7.3 Are the approximate and exact solutions identical at the collocation points for the
collocation method? Explain.
Hint:
See the results of Problem 7.2.
x
u
y
u
7.4 Solve the partial differential equation
∂
+
∂
=−
1 with the boundary conditions
u
=
0on
x
=±
1 and
y
=±
1
.
Use a boundary residual method with collocation.
−
(
x
2
+
y
2
)/
.
Hint:
A particular solution is
4
Two complementary functions are
1
,x
4
−
6
x
2
y
2
+
y
4
.
A trial solution, which satisfies the differential equation, would
be
u
=−
(
x
2
+
y
2
)/
4
+
u
1
+
u
2
(
x
4
−
6
x
2
y
2
+
y
4
).
Choose
u
1
,
u
2
to satisfy
u
=
0on
the square
x
=±
1
,y
=±
1
.
For collocation points
(
x
1
,y
1
)
=
(
1
,
0
)
and
(
x
2
,y
2
)
=
(
1
,
1
)
,
u
1
=
3
/
10
,
u
2
=−
1
/
20
.
This gives
u
=
0
.
3at
x
=
y
=
0
,
as compared to the
exact solution of
u
=
0
.
2947
.
7.5 Use the subdomain interior method to solve the problem posed in Problem 7.2.
Hint:
For the trial solution, use (Problem 7.2)
u
=
u
1
x
(
1
−
x
)
,R
=−
2
u
1
+
1
,
1
0
u
1
x
(
1
−
x
)
+
x
.
For a single subdomain 0
≤
x
≤
R
(
x
)
dx
=
0
=−
11
/
6
u
1
+
1
/
2
=
0 gives
u
1
=
3
/
11
,
so that
u
=
3
x
(
1
−
x
)/
11
.
Results:
Solution
u
x
Exact
Subdomain
1/4
0.044
0.051
1/2
0.070
0.068
3/4
0.060
0.051
7.6 Use Galerkin's method to solve the differential equation of Problem 7.2.
Hint:
For the trial solution
u
=
u
1
x
(
1
−
x
)
,
1
1
x
(
1
−
x
)
R
(
x
)
dx
=
x
(
1
−
x
)
[
−
2
u
1
+
u
1
x
(
1
−
x
)
+
x
]
dx
=
0
0
0
gives
u
1
=
5
/
18 and
u
=
5
x
(
1
−
x
)/
18
.
Results:
Solution
u
x
Exact
Galerkin
1/4
0.044
0.052
1/2
0.070
0.069
3/4
0.060
0.052
7.7 Study the solution of the boundary value problem (with a partial differential equation)
2
2
∂
x
u
+
∂
y
u
=
0
,
0
<
x
<
1
,
0
<
y
<
∞
with boundary conditions
u
(
0
,y
)
=
u
(
1
,y
)
=
0 for
y
>
0
,u
(
x,
0
)
=
x
(
1
−
x
)
,
and
u
(
x, y
→∞
)
=
0 for 0
≤
x
≤
1.
=
(a) Use the trial solution
u
u
1
(
y
)
x
(
1
−
x
).
Find the boundary conditions on
u
1
and
the residual function.
(b) Apply collocation along
x
=
1
/
3
.
=
e
−
3
y
.
=
(
).
Answer:
u
1
Note
u
1
u
1
y
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