reference_2[1]

# reference_2[1] - Reference Answer of Assignment 2 SCI 3510...

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Unformatted text preview: Reference Answer of Assignment 2 ( SCI 3510 Mathematical and Statistical Modelling ) The Greenshield’s model is adopted for the velocity-density relation without special notification in the following problems. (1) Suppose that the initial density is given by ρ ( x ) =                  3 4 ρ opt x ∈ (-∞ ,- 1] 2- x 4 ρ opt x ∈ [- 1 , 1 ] 1 4 ρ opt x ∈ [ 1 , + ∞ ) (a) Sketch the characteristic lines for the above initial density. (b) Give the density ρ ( x, t ) and its graph at time t = 1 and 2. Solution (a) The characteristic equation is given by x- φ ( s ) t = s, where φ ( s ) = c ( ρ ( s )) = U max parenleftbigg 1- ρ ( s ) ρ opt parenrightbigg . Then we have φ ( s ) =                  1 4 U max s ∈ (-∞ ,- 1 ] 2 + s 4 U max s ∈ [- 1 , 1 ] 3 4 U max s ∈ [ 1 ,-∞ ) Case 1: s ≤ - 1 The characteristic lines are given by x- 1 4 U max = s i.e. t = 4 U max ( x- s ) Case 2:- 1 ≤ s ≤ 1 The characteristic lines are given by x- 1 4 (2 + s ) U max = s i.e. t = 4 (2 + s ) U max ( x- s ) Case 3: s ≥ 1 The characteristic lines are given by x- 3 4 U max = s i.e. t = 4 3 U max ( x- s ) 2 The characteristic lines are sketched as below: a45 x a54 t a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a12 a12 a12 a12 a12 a12 a20 a20 a20 a20 a20 a20 a28 a28 a28 a28 a28 a28 a44 a44 a44 a44 a44 a44 a44 a44 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 •-1 θ 1 • 1 θ 2 tan ( θ 1 ) = 4 U max , tan ( θ 2 ) = 4 3 U max (b) From (a) we have φ prime ( s ) =          s ∈ (-∞ ,- 1) 1 4 U max s ∈ (- 1 , 1) s ∈ ( 1 ,-∞ ) Then φ prime ( s ) ≥ 0 for all s , i.e. Δ negationslash = 0. We can use the method of characteristics to solve the problem. The following is an analytical approach to find the density. a45 x a54 t a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a2 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 a17 •-1 θ 1 • 1 θ 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright L 1 L 2 It is easy to know that L 1 =...
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reference_2[1] - Reference Answer of Assignment 2 SCI 3510...

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