Unformatted text preview: 7.
ut + uux = 0 u (x, 0) = 2 for x < 0
1 for 0 < x < 1
0 for x > 1 First ﬁnd the shock characteristic for those with speed u = 2 and u = 1
[q ] = 1 22
1
3
u
= (22 − 12 ) =
21
2
2
[u] = 2 − 1 = 1 Thus dxs
3
=
dt
2
and the characteristic through x = 0 is then
xs = 3
t
2 Similarly for the shock characteristic for those with speed u = 1 and u = 0
[q ] = 12
u
2 1
0 = 12
1
(1 − 02 ) =
2
2 [u] = 1 − 0 = 1
Thus 1
dxs
=
dt
2
and the characteristic through x = 1 is then
xs = 1
t+1
2 Now these two shock charateristic going to intersect. The point of intersection is found
by equating xs in both, i.e.
1
3
t+1= t
2
2
3
The solution is t = 1 and xs = . Now the speeds are u = 2 and u = 0
2
[q ] = 12
u
2 2
0 = 12
(2 − 02 ) = 2
2 273 ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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