Unformatted text preview: 2. The set of ODEs is dρ
dx
= ρ2
=0
dt
dt
The solution of the ﬁrst ODE requires the yet unknown ρ thus we tackle the second ODE
dρ = 0
Now integrate this
ρ(x, t) = K
At t = 0 the solution is
ρ(x0 , 0) = K = 4 x<0
3 x>0 Thus
ρ(x, t) = ρ(x0 , 0)
Now substitute this solution in the characteristic equation (ﬁrst ODE)
dx
= ρ2 (x0 , 0)
dt
Integrate x = ρ2 (x0 , 0) t + C For t = 0 we get
x0 = 0 + C
Thus
C = x0
and the characteristics are x = ρ2 (x0 , 0) t + x0 For x0 < 0 then ρ(x0 , 0) = 4 and the characteristic is then given by x = x0 + 16t
Therefore for x0 = x − 16t < 0 the solution is ρ = 4.
For x0 > 0 then ρ(x0 , 0) = 3 and the characteristic is then given by x = x0 + 9t
Therefore for x0 = x − 9t < 0 the solution is ρ = 3.
Notice that there is a shock (since the value of ρ is decreasing with increasing x). The
shock characteristc is given by
dxs
=
dt 1
3 · 43 − 1 · 33
3
=
4−3 The solution of this ODE is 1
(64
3 − 27)
37
=
1
3 37
t + xs (0)
3
xs (0) is where the shock starts, i.e. the discontinuity at time t = 0.
xs = 266 ...
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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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