Flow Around A Cylinder



A Traffic Flow Model

Frank Massey

1. Introduction.

These notes look at some equations that are used to model the flow of traffic along a highway. We are interested in the following physical quantities

x = position along a highway (miles),

t = time (hours),

( = ((x,t) = density of vehicles at position x and time t (vehicles per mile),

v = v(x,t) = velocity of the vehicles at position x and time t (miles per hour).

( = ((x,t)

= ((x,t)v(x,t)

= rate at which vehicles are passing point x at time t (vehicles per hour)

= flux (or flow rate) at position x and time t

c = c(x,t)

= wave velocity at position x and time t (miles per hour)

= velocity at which disturbances in ( and v propagate (usually less than v)

Traffic flow is often modeled by three equations. The first, called the continuity equation, is

(1.1) = -

or

(1.2) = -

As we shall see in section 4, the continuity equation expresses conservation of vehicles in regions of the x-t plane in which ((x,t) and v(x,t) are smooth functions of x and t.

The second equation, called the Rankine-Hugoniot (RH) equation, replaces the continuity equation along curves x = ((t) in the x-t plane in which ((x,t) and v(x,t) have a jump discontinuity. It is

(1.3) = .

Here ((x+,t) = (((,t) and ((x-,t) = (((,t) and similarly for (. Places where ((x,t) and v(x,t) have a discontinuity are called shocks and the curve x = ((t) is called a shock wave. The RH equation describes how shock waves propagate. A derivation of (1.3) is in section 4.

The third equation expresses a car's velocity in terms of the traffic density. We shall suppose that

(1.4) v = V(()

where V(() is some given function. Then ( is also a function of the density, i.e.

(1.5) ( = ((()

where (((), the flux function, is given by

(1.6) ((() = (V(().

Then (1.2) and (1.3) become

(1.7) = - c(()

(1.8) =

where

(1.9) c(() = ('(() = V(() + (V'(()

is the wave velocity, as we shall see in section 5. Since V(() is usually a decreasing function of (, the wave velocity is usually less than the velocity of the vehicles. In fact the wave velocity will be negative for traffic densities for which the flux is decreasing with (. This usually occurs at higher traffic densities.

Note that we can write the RH equation (1.8) as

= C[ ((((t)-,t)), ((((t)+,t)) ]

where

C[ p, q ] = =

is the average of c(() over the interval p ( ( ( q. If c(() is monotone then C[ p, q ] lies between the values of c(p) and c(q).

We are interested in finding ((x,t) in some region in the x-t plane by solving (1.7) and (1.8) along with initial and boundary conditions reflecting given information about the density on some part of the boundary. One possibility is the highway is infinite in both directions in which case we might prescribe the density and velocity at a starting time t = 0, i.e. the initial conditions are

(1.10) ((x,0) = (o(x) for all x,

where (o(x) is a given function.

Another possibility is that the highway is half infinite occupying the positive x axis. In that case we might prescribe the density at a starting time t = 0, i.e. the initial conditions are

(1.11) ((x,0) = (o(x) for x > 0.

At the end of the highway we also prescribe the density, so the boundary conditions are

(1.12) ((0,t) = (b(x) for t > 0.

2. The case where the velocity is a linear function of the density.

A simple example that we shall consider is when the velocity is a piecewise linear function of the density, i.e.

(2.1) v = V(() = ) if ( ( (s, 0 if ( > (s)) .

where

(s = saturation density of the vehicles

= density at which the velocity drops to zero,

vo = velocity when the density is zero.

Typical values are vo = 60 miles/hour and (s = 150 vehicles/mile/lane. The value of (s = 150 corresponds to about one vehicle every 35 feet.

If we combine (2.1) with (1.6) – (1.9) and assume ( ( (s we get

(2.2) ( = ((() = vo( ( 1 - ) = [ ,4) - (( - )2 ]

c = c(() = - ( ( - ) = 2v - vo

(2.3) = ( ( - )

(2.4) = ( (s - ((((t)+,t) - ((((t)-,t) )

Note that the wave velocity is also a linear function of the traffic density and vehicle velocity. Since v < vo for ( > 0, the wave velocity is less than the vehicle velocity. Moreover, the wave velocity is negative when the density is greater than (s/2.

These equations simplify somewhat if we make the change of variables

(2.5) u = ( ( - )

Then

(2.6) ( = ( u + vo )

Note that ( = 0 corresponds to u = - vo and ( = (s corresponds to u = vo and (2.3) and (2.4) become

(2.7) = u

(2.8) = -

In terms of u the wave velocity c is just – u. The last equation says that a shock wave propagates at a rate equal to the of the average of the wave velocity on the two sides of the shock. Note that (2.7) and (2.8) are of the form (1.7) and (1.8) with ((u) = - u2/2.

3. Car paths and conservation of vehicles.

In order to precisely state the conservation of vehicles assumption, we need to consider the paths followed by the vehicles. We let

x(t;() = path followed by a car that starts out at position ( at time t = s. In particular, x(s;() = (.

It is assumed that the velocity of a car is equal to v evaluated at the position of the car, i.e.

(3.1) = v(x(t;(),t).

If W = [a,b] is some interval we let

(3.2) n(W,t) = number of vehicles in W at time t

= .

(3.3) Wt = [x(t;a), x(t;b)]

= interval occupied at time t by the vehicles that start out in W at time s.

The conservation of vehicles assumption says that if we focus on a collection of moving vehicles, then the total number of vehicles doesn't change with time, i.e.

(3.4) n(Wt,t) = n(W,s)

for all t for any region W. We shall show that this gives rise to the continuity equation (1.1).

4. Derivation of the continuity and Rankine-Hugoniot equations.

First we derive the continuity equation (1.1) in regions where ((x,t) and v(x,t) are smooth functions of x and t. Let ( be an open set in the x-t plane in which ((x,t) and v(x,t) are smooth. Let s be a value of t and suppose a ( x ( b is an interval such that (x,s) is in ( for a ( x ( b. Then (x,t) is in ( for x(t;a)( x ( x(t;b) and t near s. Note that (3.4) is equivalent to

= 0. Using (3.2) and (3.3), one can see that this implies

(x,t) dx) + ((x(t;b),t) - ((x(t;a),t) = 0.

Using (3.1) this becomes

(x,t) dx) + ((x(t;b),t) v(x(t;b),t) - ((x(t;a),t) v(x(t;a),t) = 0

or

(x,t) + (x,t) ] dx) = 0.

Putting t = s one obtains

(x,s) + (x,s) ] dx) = 0.

Since the interval [a, b] can be any interval, this implies (1.1) holds in the region where ((x,t) and v(x,t) are smooth.

Next we derive the RH equation along curves x = ((t) in the x-t plane in which ((x,t) and v(x,t) have a jump discontinuity. To be more precise, suppose that x = ((t) defines a curve C in the x-t plane and ( is an open set in the x-t plane that contains C. Suppose that ((x,t) and v(x,t) are smooth in ( \ C. Furthermore, suppose ((x,t) and v(x,t) and their first partial derivatives have limits as x( ((t)- and x( ((t)+ for each fixed t. We shall show that (1.3) holds on C.

Let s be a value of t and suppose a ( x ( b is an interval such that (x,s) is in ( for a ( x ( b and such that a < ((s) < b. Then for t near s one has (x,t) in ( and x(t;a)  ½ , the function gt(() increases, then decreases, and then increases again. Therefore, there is more than one solution ( to (5.4) for some x corresponding to more than one characteristic going though (x,t). The solution will have developed a discontinuity after t = ½, and the propagation of this discontinuity is governed by the RH equation which we look at in the next section. [pic]

[pic]

6. Solving the Rankine-Hugoniot equation.

In this section we consider how to combine the Rankine-Hugoniot equation with the solution derived in section 5 using the characteristics in order to obtain the shock waves. We shall assume we are solving (2.7) along with the initial condition u(x,0) = uo(x) for all x. For simplicity we shall assume

(6.1) uo is bounded.

(6.2) Either uo is uniformly Lipschitz continuous or there exists a point b such that uo is uniformly Lipschitz continuous on -( < x ( b-( and b+( ( x < ( for every ( > 0.piecewise C1. We assume uo(x) approaches a limit as x approaches b from the right and left.

(6.3) There is a, c such that the following is true.

i. a ( b ( c

ii. uo(x) is non increasing for x ( a and x ( c and non decreasing for a ( x ( c

iii. u is non decreasing on a ( x ( b and u is non increasing on b ( x < (. Thus u has a maximum at x = b so

q = max{ u(x) : -( < x < (} = u(b) > 0.

Using the notation of section 5, let gt(() = ( - t uo((). One has = 1 - tu((). For fixed t ( 0, one has 1 – tq = min{ : -( < ( < (} = . Let tc be defined by (5.6).

If 0 ( t < tc, then > 0 for all (. So gt(() is increasing. Thus gt-1(x) is a function and (5.5) defines u(x,t).

If t = tc then = 0. Therefore, either gt-1(x) doesn't exist or its derivative with respect to x is infinite at x = gt(b).

If t > tc, then there is an interval ((t) < ( < ((t) where is negative. Note that a ( ((t) ( b ( ((t) ( c. So gt(() is increasing for ( ( ((t) and ( ( ((t) and decreasing for ((t) ( ( ( ((t). Let ((t) = gt(((t)) and ((t) = gt(((t)). Then ((t) < ((t) and for ((t) < x < ((t) there is more than one ( to such that gt(() = x corresponding to more than one characteristic going though (x,t). Let g(x) be the inverse function to the restriction of gt to - ( < ( ( ((t) and g(x) be the inverse function to the restriction of gt to ((t) ( ( < (. Then g(x) is defined for x < ((t) and g(x) is defined for ((t) < x. The functions u-(x,t) and u+(x,t) defined by

u-(x,t)  = (x) - x,t) u+(x,t)  = (x) - x,t)

are solutions to (2.7) and they are defined for x < ((t) and ((t) < x respectively. In the Rankine-Hugoniot equation (2.8) we have u(((t)-,t) = u-(((t),t) and u(((t)+,t) = u+(((t),t). Therefore, the Rankine-Hugoniot equation becomes

(6.4) = -

where x = ((t). This is a first order differential equation and the initial condition is

x(tc) = gtc(() = b - tc uo(b).

This can be solved, usually numerically, to find ((t). The solution u(x,t) is equal to u-(x,t) for x < ((t) and u+(x,t) for x > ((t).

Example 6.1. Suppose uo(x) = )

[pic]

Note that q = max{ u(x) : -( < x < (} = u(0) = (. So tc = 0. If t > 0 the function gt(() increases for ( < 0, has a jump discontinuity downward at ( = 0 and then increases again for ( > 0.

gt(() = )

[pic]

[pic]

Therefore ((t) = ((t) = 0 and ((t) = t and ((t) = -2t. Also g(x) = x - t and g(x) = x + 2t. So u-(x,t) = -1 and u+(x,t) = 2 and the RH equation (6.4) is

=

or

= - ½

Therefore x = ((t) = - t/2 and the shock wave travels in the negative direction with speed ½. So

u(x,t) = )

[pic]

[pic]

Example 6.2. Suppose uo(x) is as in example 5.2. In this case b = 0 and as noted tc = ½. For t > ½ the equation = 0 has solutions ( = ((t) = -1 + and ( = ((t) = 1 - . So gt(() is increasing for t ( ((t) = -1 + and t ( ((t) = 1 - and decreasing for -1 + ( t ( 1 - . A calculation shows that ((t) = - and ((t) = . The same calculation that lead to (5.7) shows that

u-(x,t) = ,2t2) - if -1 + t ( x ( ((t)))

u+(x,t) = ,2t2) - if ((t) ( x ( 1 - t, 2 if 1 - t ( x))

So the RH equation (6.4) is

(6.5) = + - ,4t2)

as long as –1 + t ( x(t) ( 1 – t. The initial condition is x(½) = 0. Normally it would be difficult to solve (6.5) analytically, but with the initial condition x(½) = 0, the solution is just x(t) = 0, at least for ½ ( t ( 1. Therefore, for ½ ( t ( 1 the formula (5.7) continues to hold.

[pic]

[pic]

When t = 1, the equation (6.5) becomes = 0, so x(t) = 0 for t ( 1 also. For t ( 1 one has

u(x,t) = )

[pic]

[pic]

Example 6.3. Suppose uo(x) is as in example 5.3. In this case b = 0 and as noted tc = ½. For t > ½ the equation = 0 has solutions ( = ((t) = -1 + and ( = ((t) = 1 - , and gt(() is increasing for t ( ((t) and t ( ((t) and decreasing for ((t) ( t ( ((t). A calculation shows that ((t) = - and ((t) = . The same calculation that lead to (5.7) shows that

u-(x,t) = ,2t2) - if -1 + t ( x ( ((t)))

u+(x,t) = ,t2) - if ((t) ( x ( 2 - 2t, 2 if 2 - 2t ( x))

For –1 + t ( x(t) ( 2 – 2t the RH equation (6.4) becomes

(6.6) = + - 2,4t2)

The initial condition is x(½) = 0. It does not seem possible to solve (6.5) analytically, so we obtain a numerical solution.

[pic]

[pic]

As we see the shock wave travels in the negative direction. When x = ((t) hits the curve x = -1 + t the right side of (6.6) changes. This corresponds to the region where u = -1 catching up with the shock wave.

[pic]

[pic]

Therefore for ½ ( t ( 0.886963 one has

u(x,t) = ,2t2) - if -1 + t ( x ( ((t), ,t2) - if ((t) ( x ( 2 - 2t, 2 if 2 - 2t ( x))

[pic]

[pic]

[pic]

[pic]

After t = 0.886963 the right side of (6.6) changes so that (6.4) becomes

= + + ,2t2)

as long as x(t) ( 2 – 2t. When x = ((t) hits the curve x = 2 - 2t the right side of (6.6) again changes. This corresponds to shock wave catching up to the region where u = 1.

[pic]

[pic]

Therefore for 0.886963 ( t ( 1.11111 one has

u(x,t) = ,t2) - if ((t) ( x ( 2 - 2t, 2 if 2 - 2t ( x))

[pic]

[pic]

After t = 1.11111 the right side of (6.6) changes again so that (6.4) becomes = - ½. The initial condition is x(1.11111) = ((1.11111) The solution is x(t) = - (t - 1.11111)/2 + ((1.11111). For t ( 1.11111 one has

u(x,t) = )

[pic]

[pic]

References.

1. Chester, Clive R., Techniques in Partial Differential Equations, McGraw-Hill, New York, 1971.

2. Chorin, Alexandre, J. and Jerrold E. Marsden, A Mathematical Introduction to Fluid Mechanics, Third Edition, Springer, New York, 1992.

3. Knobel, Roger, An Introduction to the Mathematical Theory of Waves, American Mathematical Society, Providence, 2000.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download