MAE294B/SIO203B: Methods in Applied Mechanics
Winter Quarter 2010
http://maecourses.ucsd.edu/mae294b
Solution IV
1
The leadingorder solution
y
0
satisfies
y
0
+
y
0
x

1
xy
0
=
0
.
This can be solved by separation of variables to yield
y
0
=
±
√
1
+
Ax

2
. For the two boundary
conditions here, the solutions are
y
0
=
1 and
y
0
=
√
1
+
3
x

2
, so in both cases the limit for large
x
is 1. For small
x
, from the equation one cannot neglect
ε
with respect to
x
when
x
=
O
(
ε
)
. However
the magnitude of
y
may also change, as it clearly does in case 2, so care is needed. The general
rescaling
x
=
ε
α
X
,
y
=
ε
β
Y
gives
Y
X
+
Y
X
+
ε
1

α
X

X
(
X
+
ε
1

α
)
2
ε

2
β
Y
=
0
.
To find the appropriate exponents, we will need the
O
(
ε
)
equation for
y
=
O
(
1
)
:
y
1
+
1
+
1
y
2
0
y
1
x
=
y
0
x
2

2
x
2
y
0
,
y
1
(
1
) =
0
.
For the first case, we find
y
1
=
x

2

x

1
. Hence
y
=
1
+
ε
(
x

2

x

1
)+
···
. The expansion becomes
disordered when
ξ
=
ε

1
/
2
x
=
O
(
1
)
, at which point
y
=
O
(
1
)
. That is, we need to solve the case
with
α
=
1
/
2 and
β
=
0, which to leading order gives
y
ξ
+
y
ξ

1
ξ
y
=
0
.
We have solved this before and the answer is
y
=
1
+
B
ξ

2
. Van Dyke’s rule gives
H
0
E
1
y
=
H
0
[
1
+
ε
(
x

2

x

1
)] =
H
0
[
1
+
ξ

2

ε
1
/
2
ξ

1
] =
1
+
ξ

2
,
E
1
H
0
y
=
E
1
1
+
B
ε
x

2
=
1
+
1
2
B
ε
x

2
.
Hence
B
=
2 and
y
=
1
+
2
ξ

2
. This is still not enough because this solution is no longer valid
for
X
=
ε

1
x
=
O
(
1
)
. Then, from what we have just found,
y
=
O
(
ε

1
/
2
)
and we have
α
=
1 and
β
=

1
/
2. The governing equation becomes, to leading order and using
Y
because the scaling has
changed,
Y
X
+
Y
X
+
1
=
0
,
1
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which has the solution
Y
=
C
(
X
+
1
)

1
.
Match with the solution in the
ξ
region (noting that
ξ
=
ε
1
/
2
X
):
H
0
E
0
y
=
H
0
[
1
+
2
ε

1
X

2
] =
√
2
ε

1
/
2
X

1
,
E
0
H
0
y
=
E
0
[
ε

1
/
2
C
(
ε

1
/
2
ξ
+
1
)

1
] =
C
ξ

1
,
so
C
=
√
2 and the solution becomes
y
=
2
/
ε
(
X
+
1
)

1
. Hence
y
(
0
) =
2
/
ε
.
For the second case, the governing equation for
y
1
is less tractable, but shows that
y
1
∼
x

2
again
for small
x
.
Hence the variable
ξ
is again relevant and
α
=
1
/
2, but this time
y
=
O
(
x

1
)
for
small
x
and so
β
=

α
=

1
/
2. Then the governing equation becomes, in terms of
Y
=
ε
1
/
2
y
and
ξ
=
ε

1
/
2
x
,
Y
ξ
+
Y
ξ
=
0
with solution
Y
=
D
ξ

1
. Van Dyke’s rule gives
H
0
E
0
y
=
H
0
[
1
+
3
ε

1
ξ

2
] =
√
3
ε

1
/
2
ξ

1
,
E
0
H
0
y
=
E
0
[
ε

1
/
2
D
ε
1
/
2
x

1
] =
Dx

1
so
D
=
√
3. Now once again we need to work in the region
X
=
O
(
1
)
, i.e.
α
=
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 Winter '08
 Young,W
 Boundary value problem, Boundary conditions, Van Dyke

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