Kinematic Waves. These are waves which result from the conservation equation. t + I = 0. (2)

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1 Introduction Kinemtic Wves These re wves which result from the conservtion eqution E t + I = 0 (1) where E represents sclr density field nd I, its outer flux. The one-dimensionl form of (1) is E t + I = 0. (2) As n exmple, we consider gs of density ρ streming with velocity v. In this cse, the conservtion of mss cn be expressed by (1), if we tke E = ρ nd I = ρv, ρ t For one-dimensionl problem, we hve + (ρv) = 0. (3) ρ t + (ρv) = 0. (4) We note tht neither (3) nor (4) re sufficient to determine both ρ nd v, since we hve one eqution nd more thn one vrible. We now confine our ttention to the one-dimensionl eqution (2) nd further ssume tht the flux depends only on the density E., i.e., I = I(E). We cn the write E t + di E = 0. (5) de It is customry to tke the following nottions: E = u, c(u) = di, nd to write (5) de s t + c(u) = 0 (6) Eqution (6) is first order qusi-liner prtil differentil eqution. Generl Solution: The chrcteristics equtions for (6) re given by dt 1 = dx c(u) = du 0. (7) 1

2 The lst eqution shows tht long chrcteristic curve, The first eqution of (7) then reduces to which cn be integrted immeditely to give Therefore the generl solution is given in n implicit form, The eqution u = k 2. (8) dt 1 = dx c(k 2 ), (9) x c(k 2 )t = k 1. (10) u = f[x c(u)t]. (11) dx dt = c(k 2) (12) represents the projection of the chrcteristic on the x-t plne. Let C be this projection which is here stright line tht intersects the x xis t ξ nd long which u is constnt. t C dx/dt = c(k2) o ξ x Figure 1: The projection of the chrcteristics on the t x plne. Initil Vlue Problem: To find solution to (6), we need to specify the vlues of the function u t some time t = 0, for exmple. Let u(x, 0) = f(x), < x < +. (13) 2

3 The eqution for C is given by C intersects the x-xis t ξ, i.e., x = ξ + c(u)t (14) t t = 0, x = ξ, u = f(ξ), (15) Since u is constnt long C nd depends only on ξ, the slope of C is c(u) = c[f(ξ)] is known function of ξ. Thus we hve the following prmetric representtion of the solution: x = ξ + c[(f(ξ)]t (16) u = f(ξ) (17) It is esy to verify tht (16 nd 17) represent solution to the initil vlue problem defined by (6 nd 13). First we note tht t = f (ξ) ξ t, = f (ξ) ξ. To evlute ξ t ξ nd, we first differentite (16) with respect to t 0 = ξ t + dc(u) du du dξ ξ t + c[f(ξ)], (18) t which gives ξ t = c[f(ξ)] 1 + dc(u) f du (ξ)t. (19) We then differentite (16) with respect to x, 1 = ξ + dc(u) du ξ t, (20) du dξ which gives We finlly get, ξ = dc(u) f du (ξ)t. (21) t = c[f(ξ)]f (ξ) 1 + dc(u) f du (ξ)t, = which substituted in (6) gives n identity. f (ξ) 1 + dc(u) f du (ξ)t, (22) 3

4 Vlidity of Solution We now exmine whether the solution (16 nd 17) to the initil vlue problem (6, 13) is vlid for ll time t 0 or whether there is limit t b where the solution will brek down if t > t b. The expressions for the derivtives nd given by equtions (22) t suggest tht those derivtives my become singulr for t = 1/[(dc/du)f (ξ)]. On the other hnd, let us recll tht the theorem for the existence nd uniqueness of the Cuchy boundry-vlue problem sttes tht chrcteristic C must intersect the x xis only once. So, since (6) is nonliner, the chrcteristics C depend on the initil conditions. If, for exmple, two chrcteristics intersect this will violte the conditions stted in the existence nd uniqueness theorem. Let us then exmine wht hppens when two chrcteristics intersect ech other. At the intersection both chrcteristics hve the sme x nd t s shown in Figure 2. We consider two chrcteristics one intersecting the x xis t ξ, the other t ξ + δξ. The equtions for these chrcteristics re given below. t Cξ Cξ+δξ O ξ ξ+δξ x Figure 2: Intersection of two chrcteristics. x = ξ + c[f(ξ)]t, (23) x = ξ + δξ + c[f(ξ + δξ)]t = ξ + δξ + c[f(ξ)]t + dc du f (ξ)δξt, (24) where we hve expnded (24) with respect to the smll prmeter δξ. Subtrcting (23) from (24), we get t = 1 dc du f (ξ). (25) 4

5 This is exctly the time t which both nd become singulr. Since t > 0, there t is no intersection if dc f (ξ) > 0. On the other hnd, if du dc du f (ξ) < 0, (26) the time t first breking will occur t the mximum vlue of [ dc du f (ξ)], t b = 1 dc f. (27) du (ξ) mx These results cn lso be directly derived from the generl form of the solution (11). Differentiting u with respect to t, we get nd When t = 1 f dc(u)/du, t = [c(u) + tdc(u)/du t ]f [x c(u)t], (28). t t = c(u)f 1 + tf dc(u)/du. (29) Evolution of the solution in time: If dc(u)/du > 0, then (22) shows tht s time increses steepening of the slope of the initil distribution for f < 0 nd flttening for f > 0. On the other hnd, if dc(u)/du < 0, then (22) shows tht s time increses flttening of the slope of the initil distribution for f > 0 nd steepening for f > 0. As n exmple, consider the eqution t + u = 0, (30) with the initil condition t t = 0, u = exp( x 2 /2). Here, c = u nd dc/du = 1. The breking occurs t t = Figure 3 shows the evolution of the solution in time. As time increses the slope of the wve steepens to the right nd flttens to the left. For t > , the solution becomes multivlued nd hence physiclly uncceptble. Trffic Flow: Let ρ(x, t) be density per unit length. The crs re moving t speed v(x, t). The cr flux, the number of crs crossing point per unit time, is : q(x, t) = ρv. The 5

6 1 0.8 u(x,t) t=0 t=0.5 t=1 t= t= x Figure 3: Evolution of the solution in time. Breking occurs t t= Observe how nonliner effects fltten the wve to the left nd steepen it to the right. conservtion reltion between ρ nd q is summrized s follows d dt x2 ρdx = crs in - crs out (31) x 1 = q(x 1, t) q(x 2, t) (32) x2 q = dx (33) x 1 So, The wve velocity ρ t + q = 0. (34) c(ρ) = dq dρ = v + ρ v ρ. (35) 6

7 Since v < 0, it follows tht c < v. Hence for driver the wve will be moving ρ with the negtive velocity c v < 0, implying the wve propgtes bckwrd. As n exmple we consider the trffic through the Lincoln tunnel in New York. Dt were gthered, nd it ws found tht q = ρ(1 ρ ρ j ) (36) = 17.2 mph (37) ρ j = 250 vpm (38) c(ρ) = dq dρ = 2ρ (1 ) ρ j (39) dc dρ = 2 ρ j (40) Cr flux: vehicles per hour Cr density: vehicles per mile Figure 4: Cr flux versus cr density in the Lincoln tunnel. We ssume the following initil distribution for the crs. ρ(x, 0) = ρe αx2 (41) 7

8 20 15 Wve velocity: mile per hour Cr density: vehicles per mile Figure 5: Wve velocity versus cr density. The wve velocity c(ρ) = dq/dρ, hence Let, c[ρ(ξ)] = (1 2 ρ ρ j e αξ2 ) (42) dc dξ = 4α ρ ξe αξ2 ρ j t b = [ 1 ] c min = + ρ j u 4α ρ (43) 1 (44) ξe αξ2 mx z = ξe αξ2 (45) dz dξ = (1 2αξ 2 )e αξ2. (46) 1 The mximum occurs t ξ = 2α t b = ρ j e 2 ρ 2α. (47) 8

9 Dmped Wves Consider the eqution t + u + u = 0, (48) where is positive constnt. The chrcteristic equtions for re (48) dt 1 = dx u = du u (49) Integrting (49), we get Applying the initil condition, du = dt u (50) u = k 1 e t (51) x = k 1 e t + k 2. (52) u(x, 0) = f(x), u = f(ξ) (53) ξ = k 2 k 1 k 2 = ξ + f(ξ) t = 0, u = k 1 = f(k 2 ) (54) (55) (56) x = ξ + 1 e t f(ξ) (57) u = f(ξ)e t (58) let us exmine the breking of the solution. Two chrcteristic curves interct x = ξ e t f(ξ 1 ) (59) x = ξ e t f(ξ 2 ) (60) 0 = (ξ 2 ξ 1 ) + 1 e t (f(ξ 2 ) f(ξ 1 )) (61) 0 = e t (f(ξ 2 ) f(ξ 1 )) ξ 2 ξ 1 (62) 0 = e t f (ξ). (63) 9

10 This implies tht f (ξ) < 0, nd f (ξ) <. (64) The breking will occur only if the initil curve hs enough negtive slope. The breking time is t b = { ln(1 + /f (ξ)) } min. (65) 10

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