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Shock formation by characteristic intersection
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Past exam of the mathematics course of the University of Cambridge
/
2008
/
iii
/
Paper 69
/
3
/
Solution
2026-10-03
0
Past exam of the mathematics course of the University of Cambridge
/
2024
/
ii
/
Paper 4
/
39D
/
b
/
ii
/
Solution
Created
2026-09-24
Updated
2026-09-25
View more
Put
A
=
(
γ
+
1
)
/2
.
A
shock formation by characteristic intersection
first
occurs when the
map
from emission
time
to
position
loses monotonicity:
0
=
∂
τ
∂
x
=
(
1
−
A
)
X
˙
−
c
0
+
A
X
¨
(
t
−
τ
)
.
(1)
Thus
t
=
τ
+
A
X
¨
(
τ
)
c
0
+
(
A
−
1
)
X
˙
(
τ
)
.
(2)
For
X
=
c
0
t
3
/
T
2
, write
s
=
τ
/
T
to obtain
T
t
=
2
A
3
A
−
1
s
+
6
A
s
1
.
(3)
Minimizing over
s
>
0
gives
s
2
=
[
3
(
3
A
−
1
)
]
−
1
and hence
T
t
shock
=
f
(
γ
)
=
3
(
γ
+
1
)
2
(
3
γ
+
1
)
.
(4)
Total
articles
:
2