Module 4 : Macroscopic And Mesoscopic Traffic Flow Modeling

Lecture 17 : Traffic Flow Modeling Analogies

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	Method of Characteristics Consider $k(x,t)$ at each point of $x$ and $t$, and $\frac{\partial k}{\partial t} + \frac{\partial k}{\partial x}[f(k) + \frac{df}{dk}k] = 0$ in the total derivative of $k$ along a curve which has slope $\frac{\partial x}{\partial t} = f(k) + \frac{df}{dk}k$. ie., Along any curve in $(x,t)$, consider $x,k$ as function of $t$. file=qfMacroCharacteristics.eps,width=4cm Total derivative of $k$ will be $$\frac{dk}{dt} = \frac{\partial k}{\partial t} + \frac{\partial k}{\partial x} . \frac{dx}{dt}$$ $$= \frac{\partial k}{\partial t} + \frac{\partial k}{\partial x}\left[f(k) + \frac{df}{dk}k\right]$$ At the solution, $\frac{dk}{dt} =0$, $k$ is constant along the curve, $f(k) + k.\frac{df}{dk}$ is constant along the curve. That is, $$x(t) = x_0+\left[\frac{dx}{dt}\right]~t$$ $$= x_0+\left[f(k) + k.\frac{dt}{dk}\right]t$$ Note that the solution is to construct some curve $e$ so that: (a) $k_t + c(k) .k_x$ is the total derivative of $k$ along the curve (ie., directional derivative) and (b) slope of the curve $\frac{dx}{dt} = c(k)$. We know $k(x,t)$. Therefore directional derivative $k(x,t)$ along $t$ $$\frac{dk(x,t)}{dt} = \frac{\partial k}{\partial t} + \frac{dx}{dt}. \frac{\partial k}{\partial x}$$ $$= \frac{\partial k}{\partial t} + \left[f(k) + k. \frac{df}{dk}\right] \frac{\partial k }{\partial x}$$ $$=$$ 0 $$\mathrm{ie.,} \, \frac{dk}{dt} =$$ 0 That is $k$ is constant along the curve $e$ or $\frac{dx}{dt} = f(k) + k \frac{df}{dk}$ is constant along curve $e$. Therefore $e$ must be straight line. $$x(t) = x_0 + \left[f(k) + k.\frac{df}{dk}\right]t$$ If $k(x,0) =k_0$ is initial condition $$x(t) = x_0 + \left[f(k_0) + k_0. \frac{df}{dk} \bigg\vert _{k=k_0}\right]t$$ This function is plotted below along with a fundamental q-k diagram.