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Unformatted text preview: 3 2.5 2 1.5 1
u=2
0.5 u=0
u=1 0
−3 −2 −1 0 1 2 3 4 Figure 54: Solution for 7
[u] = 2 − 0 = 2
Thus dxs
=1
dt and the characteristic is then
xs = t + C.
To ﬁnd C , we substitute the point of intersection t = 1 and xs = 3
. Thus
2 3
=1+C
2
or
C= 1
2 The third shock characteristic is then
xs = t + 1
.
2 The shock characteristics and the solutions in each domain are given in the ﬁgure above. 274 ...
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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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