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CH11_SHOCKS

# CH11_SHOCKS - Chapter 11 Shock Waves Here we shall follow...

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Unformatted text preview: Chapter 11 Shock Waves Here we shall follow closely the pellucid discussion in chapter 2 of the book by G. Whitham, beginning with the simplest possible PDE, ρ t + c ρ x = 0 . (11.1) The solution to this equation is an arbitrary right-moving wave (assuming c > 0), with profile ρ ( x,t ) = f ( x − c t ) , (11.2) where the initial conditions on eqn. 11.1 are ρ ( x,t = 0) = f ( x ). Nothing to see here, so move along. 11.1 Nonlinear Continuity Equation The simplest nonlinear PDE is a generalization of eqn. 11.1, ρ t + c ( ρ ) ρ x = 0 . (11.3) This equation arises in a number of contexts. One example comes from the theory of vehicular traffic flow along a single lane roadway. Starting from the continuity equation, ρ t + j x = 0 , (11.4) one posits a constitutive relation j = j ( ρ ), in which case c ( ρ ) = j ′ ( ρ ). If the individual vehicles move with a velocity v = v ( ρ ), then j ( ρ ) = ρv ( ρ ) ⇒ c ( ρ ) = v ( ρ ) + ρv ′ ( ρ ) . (11.5) It is natural to assume a form v ( ρ ) = c (1 − aρ ), so that at low densities one has v ≈ c , with v ( ρ ) decreasing monotonically to v = 0 at a critical density ρ = a − 1 , presumably corresponding to bumper-to-bumper traffic. The current j ( ρ ) then takes the form of an 1 2 CHAPTER 11. SHOCK WAVES inverted parabola. Note the difference between the individual vehicle velocity v ( ρ ) and what turns out to be the group velocity of a traffic wave, c ( ρ ). For v ( ρ ) = c (1 − aρ ), one has c ( ρ ) = c (1 − 2 aρ ), which is negative for ρ ∈ bracketleftbig 1 2 a − 1 ,a − 1 bracketrightbig . For vehicular traffic, we have c ′ ( ρ ) = j ′′ ( ρ ) < 0 but in general j ( ρ ) and thus c ( ρ ) can be taken to be arbitrary. Another example comes from the study of chromatography, which refers to the spatial separation of components in a mixture which is forced to flow through an immobile absorbing ‘bed’. Let ρ ( x,t ) denote the density of the desired component in the fluid phase and n ( x,t ) be its density in the solid phase. Then continuity requires n t + ρ t + V ρ x = 0 , (11.6) where V is the velocity of the flow, which is assumed constant. The net rate at which the component is deposited from the fluid onto the solid is given by an equation of the form n t = F ( n,ρ ) . (11.7) In equilibrium, we then have F ( n,ρ ) = 0, which may in principle be inverted to yield n = n eq ( ρ ). If we assume that the local deposition processes run to equilibrium on fast time scales, then we may substitute n ( x,t ) ≈ n eq ( ρ ( x,t ) ) into eqn. 11.6 and obtain ρ t + c ( ρ ) ρ x = 0 , c ( ρ ) = V 1 + n ′ eq ( ρ ) . (11.8) We solve eqn. 11.3 using the method of characteristics . Suppose we have the solution ρ = ρ ( x,t ). Consider then the family of curves obeying the ODE dx dt = c ( ρ ( x,t ) ) . (11.9) This is a family of curves, rather than a single curve, because it is parameterized by the initial condition x (0) ≡ ζ . Now along any one of these curves we must have....
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CH11_SHOCKS - Chapter 11 Shock Waves Here we shall follow...

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