Differential Equations Lecture Work Solutions 266

# Differential Equations Lecture Work Solutions 266 - 2 The...

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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.

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