Traffic Waves - Revision history

Levi: Some fixes

2025-08-05T06:52:42Z

Some fixes

← Older revision		Revision as of 06:52, 5 August 2025
Line 98:		Line 98:
	\| [[Image:Velocity.jpg\|thumb\|350px\|<math>v(\rho)</math> versus <math>\rho</math>]]		\| [[Image:Velocity.jpg\|thumb\|350px\|<math>v(\rho)</math> versus <math>\rho</math>]]
	\| [[Image:Q_flux.jpg\|thumb\|350px\|<math>q(\rho)</math> versus <math>\rho</math>]]		\| [[Image:Q_flux.jpg\|thumb\|350px\|<math>q(\rho)</math> versus <math>\rho</math>]]
	\| [[Image:C_speed.jpg\|thumb\|350px\|<math>~~\rho~~(\rho)</math> versus <math>\rho</math>]]		\| [[Image:C_speed.jpg\|thumb\|350px\|<math>c(\rho)</math> versus <math>\rho</math>]]
	\|}		\|}

Line 166:		Line 166:
	</math></center>		</math></center>
	which allows us to calculate the solution stepping forward in time, but not to determine the solution given		which allows us to calculate the solution stepping forward in time, but not to determine the solution given
	a value of <math>(x,t)</math> (because we ~~have no way of knowing what~~ <math>c(\rho_0(x_0))</math> is.		a value of <math>(x,t)</math> (because we generally cannot solve the equation <math>x_0 + c(\rho_0(x_0))t = x</math> for <math>x_0</math>).

	The characteristics are a family of straight lines which will all have different slopes. If two characteristics		The characteristics are a family of straight lines which will all have different slopes. If two characteristics
Line 250:		Line 250:
	</math></center>		</math></center>
	then we know that on the lines of the expansion fan (which all start at <math>x=0</math>) we have		then we know that on the lines of the expansion fan (which all start at <math>x=0</math>) we have
	<math>c(\rho) = x/t</math>. We can rearrange this and solve for <math>x</math> and obtain		<math>c(\rho) = x/t</math> (as the characteristic lines themselves are <math>x = c(\rho(x, t))t</math>). We can rearrange this and solve for <math>x</math> and obtain
	<math>\rho =\frac{ 1}{2} \rho_{\max} (1-x/v_0 t)</math>		<math>\rho(x, t) =\frac{ 1}{2} \rho_{\max} \left(1-\dfrac x{v_0 t}\right)</math>

	The solution is then given by		The solution is then given by
Line 259:		Line 259:
	\begin{matrix}		\begin{matrix}
	\rho_{L},& x < c(\rho_L) t\\		\rho_{L},& x < c(\rho_L) t\\
	\frac{ \rho_{\max}}{2} (1-x/v_0 t),& c(\rho_L) t \leq x \leq c(\rho_R) t\\		\frac{ \rho_{\max}}{2} \left(1-\dfrac x{v_0 t}\right),& c(\rho_L) t \leq x \leq c(\rho_R) t\\
	\rho_{R},& x > c(\rho_R) t		\rho_{R},& x > c(\rho_R) t
	\end{matrix}		\end{matrix}
Line 300:		Line 300:
	=== Shocks ===		=== Shocks ===

	So far we have only considered the case when <math>c(x_0)</math> is monotonically increasing so that		So far we have only considered the case when <math>g(x_0)=c(\rho_0(x_0))</math> is monotonically increasing so that
	two characteristics never cross. We now consider the case when characteristics can meet.		two characteristics never cross. We now consider the case when characteristics can meet.
	A movie of this case is shown below.		A movie of this case is shown below.
Line 310:		Line 310:
	with		with
	<center><math>		<center><math>
	X_1(t) = x_0 + c(x_0)t\,		X_1(t) = x_0 + g(x_0)t\,
	</math></center>		</math></center>
	<center><math>		<center><math>
	X_2(t) = x_0 + \delta x + c(x_0+\delta x)t\,		X_2(t) = x_0 + \delta x + g(x_0+\delta x)t\,
	</math></center>		</math></center>
	Then these curves will meet at time <math>T</math> where		Then these curves will meet at time <math>T</math> where
	<center><math>		<center><math>
	x_0 + c(x_0)T = x_0 + \delta x + c(x_0+\delta x)T\,		x_0 + g(x_0)T = x_0 + \delta x + g(x_0+\delta x)T\,
	</math></center>		</math></center>
	which implies that		which implies that
	<center><math>		<center><math>
	T = -\frac{1}{c^{\prime}(x_0)}		T = -\frac{1}{g^{\prime}(x_0)}
	</math></center>		</math></center>
	Note the following		Note the following
	* If <math>c^{\prime}(x) > 0 </math> then no shock will form.		* If <math>g^{\prime}(x) > 0 </math> then no shock will form.
	* The shock first forms at the minimum positive value of		* The shock first forms at the minimum positive value of <math> - \frac{1}{g^{\prime}(x)} </math> for <math> -\infty < x <\infty </math>.
	<math> - \frac{1}{c^{\prime}(x)} </math> for <math> -\infty < x <\infty </math>.

	==== Shock Fitting ====		==== Shock Fitting ====

Meylan: /* Shocks */

2023-07-27T05:04:42Z

Shocks

← Older revision		Revision as of 05:04, 27 July 2023
Line 351:		Line 351:
	this into the governing integral equation we obtain		this into the governing integral equation we obtain
	<center><math>		<center><math>
	\frac{\partial}{\partial t} \left( \int_{x_1}^{s(t)} + \int_{s(t)}^{x_2}		\frac{\partial}{\partial t} \left( \int_{x_1}^{s(t)} \rho(x,t)\mathrm{d}x + \int_{s(t)}^{x_2}
	\rho(x,t)\mathrm{d}x \right) = q(x_1,t) - q(x_2,t)		\rho(x,t)\mathrm{d}x \right) = q(x_1,t) - q(x_2,t)
	</math></center>		</math></center>

Meylan: /* Part 1 */

2020-08-11T05:19:57Z

Part 1

← Older revision		Revision as of 05:19, 11 August 2020
Line 456:		Line 456:
	=== Part 1 ===		=== Part 1 ===

	{{#ev:youtube\|~~zXEvCJyHQSg~~}}		{{#ev:youtube\|SWIXL97H0fw}}

	=== Part 2 ===		=== Part 2 ===

Meylan: /* Part 2 */

2020-08-11T05:19:37Z

Part 2

← Older revision		Revision as of 05:19, 11 August 2020
Line 460:		Line 460:
	=== Part 2 ===		=== Part 2 ===

	{{#ev:youtube\|~~vRRVU4VrxV0~~}}		{{#ev:youtube\|j652tvrgm4c}}

	=== Part 3 ===		=== Part 3 ===

Meylan: /* Part 3 */

2020-08-11T05:19:17Z

Part 3

← Older revision		Revision as of 05:19, 11 August 2020
Line 464:		Line 464:
	=== Part 3 ===		=== Part 3 ===

	{{#ev:youtube\|~~xkGdvFGit_c~~}}		{{#ev:youtube\|QVtYsUGtb7I}}


	=== Part 4 ===		=== Part 4 ===

Meylan: /* Part 4 */

2020-08-11T05:18:53Z

Part 4

← Older revision		Revision as of 05:18, 11 August 2020
Line 469:		Line 469:
	=== Part 4 ===		=== Part 4 ===

	{{#ev:youtube\|~~rW-voMoG0KI~~}}		{{#ev:youtube\|Ik2Q7H_aIV8}}

	=== Part 5 ===		=== Part 5 ===

Meylan: /* Part 5 */

2020-08-11T05:18:31Z

Part 5

← Older revision		Revision as of 05:18, 11 August 2020
Line 473:		Line 473:
	=== Part 5 ===		=== Part 5 ===

	{{#ev:youtube\|~~zXEvCJyHQSg~~}}		{{#ev:youtube\|lNeFNHAqgEA}}

	=== Part 6 ===		=== Part 6 ===

Meylan: /* Part 6 */

2020-08-11T05:18:12Z

Part 6

← Older revision		Revision as of 05:18, 11 August 2020
Line 477:		Line 477:
	=== Part 6 ===		=== Part 6 ===

	{{#ev:youtube\|~~vRRVU4VrxV0~~}}		{{#ev:youtube\|Yo0sW3Nh2FI}}

	=== Part 7 ===		=== Part 7 ===

Meylan: /* Part 7 */

2020-08-11T05:17:51Z

Part 7

← Older revision		Revision as of 05:17, 11 August 2020
Line 481:		Line 481:
	=== Part 7 ===		=== Part 7 ===

	{{#ev:youtube\|~~xkGdvFGit_c~~}}		{{#ev:youtube\|gOEMfOQZAAM}}


	=== Part 8 ===		=== Part 8 ===

Meylan: /* Lecture Videos */

2020-08-11T05:17:28Z

Lecture Videos

← Older revision		Revision as of 05:17, 11 August 2020
Line 470:		Line 470:

	{{#ev:youtube\|rW-voMoG0KI}}		{{#ev:youtube\|rW-voMoG0KI}}

			=== Part 5 ===

			{{#ev:youtube\|zXEvCJyHQSg}}

			=== Part 6 ===

			{{#ev:youtube\|vRRVU4VrxV0}}

			=== Part 7 ===

			{{#ev:youtube\|xkGdvFGit_c}}


			=== Part 8 ===

			{{#ev:youtube\|ZwvTZcWLZKM}}

	[[Category:Simple Nonlinear Waves]]		[[Category:Simple Nonlinear Waves]]