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Past exam of the mathematics course of the University of Cambridge
/
2025
/
iii
/
Paper 321
/
2
/
c
/
ii
/
Solution
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2025
iii
Paper 321
2
c
ii
Created
2026-09-24
Updated
2026-09-25
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For
m
=
4
,
n
=
2
, the
equation
is the one-dimensional
porous medium equation
S
τ
=
(
S
3
)
xx
. Use
a
similarity solution
S
=
τ
−
1/4
f
(
ξ
)
and
ξ
=
x
τ
−
1/4
. Then
−
4
1
(
f
+
ξ
f
′
)
=
(
f
3
)
′′
.
(1)
The no-
mass-flux
condition at
x
=
0
sets
the integration constant to zero:
(
f
3
)
′
=
−
4
1
ξ
f
.
(2)
With
f
(
0
)
=
1
,
f
(
ξ
)
=
(
1
−
12
ξ
2
)
+
1/2
.
(3)
Consequently
Σ
(
r
,
t
)
=
Σ
0
(
r
0
r
)
−
3/2
τ
−
1/4
[
1
−
12
τ
1/2
r
/
r
0
]
+
1/2
.
(4)
Its edge is
R
=
12
r
0
τ
1/2
∝
t
1/2
. Moreover,
M
=
4
π
Σ
0
r
0
2
∫
0
∞
S
d
x
=
4
π
Σ
0
r
0
2
∫
0
12
f
(
ξ
)
d
ξ
(5)
is
time
-independent. This agrees with part (
b)
, since
2
−
m
+
2
n
=
2
.
Ancestors
(12)
ii
c
2
Paper 321
iii
2025
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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