# Traffic flow with Dirac-$\delta$ source (on ramp)

I have been trying to solve the traffic flow equation with a singular source ($$D>0$$ large): $$\rho_t + f(\rho)_x = D\delta(x)$$ with the flux $$f(\rho)=\rho(1-\rho)$$ and the initial data $$\rho(x,0)=0.4$$.

I understand that the jump condition for small $$D$$ is $$f(\rho_r)-f(\rho_l)=D$$, where $$\rho_l=0.4$$. This gives me a real value for $$\rho_r$$. But when $$D$$ is large (eg:0.012), it does not give me a real value for $$\rho_r$$, so there should be something non trivial happening, like a shock (which makes sense physically too). However, I am not able to work this out explicitly. I tried to incorporate a shock term to the jump condition, but that gave me two unknowns in one equation.

Does anyone have any ideas on how I could proceed?

• Just thinking out loud, here. Looks like $\rho_r$ goes imaginary when $D>1/100,$ based on your value for $\rho_l.$ Can you edit your question to include what you had in mind for a shock term? Maybe we can work around the lack of information problem you mentioned. – Adrian Keister Dec 31 '18 at 16:24
• For $x\not=0,$ your pde reduces down to $\rho_t+\rho_x-2\rho\rho_x=0.$ The IC, of course, makes the trivial solution not work. Again, still just thinking out loud, here. – Adrian Keister Dec 31 '18 at 16:29
• This is a hyperbolic conservation law, which doesn't necessarily have a unique solution. The setup you have is a generalised Riemann problem, which may include shocks and rarefaction fans. – Eddy Dec 31 '18 at 16:35
• However, the shock condition and jump condition give you 2 equations, and there are 2 unknowns, so you should only have finitely many solutions. – Eddy Dec 31 '18 at 16:37
• The constant solution $\rho=2/5$ satisfies the pde and IC when $x\not=0$. – Adrian Keister Dec 31 '18 at 16:42

Another approach consists in tackling the problem directly. We write the corresponding Rankine-Hugoniot condition. Consider that $$\rho$$ has a single discontinuity with jump $$\rho_r-\rho_l$$ at $$x=\gamma(t)$$ between $$x_1$$ and $$x_2$$. Since $$\rho$$ is nonsmooth, we go back to the integral definition of the conservation law \frac{\text d}{\text d t} \int_{x_1}^{x_2} \rho\,\text d x = \left\lbrace\begin{aligned} & f(\rho)|_{x=x_1} - f(\rho)|_{x=x_2} & &\text{if}\quad 0 \notin [x_1,x_2] \\ & f(\rho)|_{x=x_1} - f(\rho)|_{x=x_2} + D & &\text{if}\quad 0 \in [x_1,x_2] \end{aligned}\right. where $$f(\rho) = \rho(1-\rho)$$. Then, the identity $$\frac{\text d}{\text d t} \int_{x_1}^{x_2} \rho\,\text d x = \int_{x_1}^{\gamma(t)} \rho_t\,\text d x + \int_{\gamma(t)}^{x_2} \rho_t\,\text d x + \gamma'(t)\left(\rho_l-\rho_r\right)$$ and the PDE yield the modified Rankine-Hugoniot condition \left\lbrace\begin{aligned} \gamma'(t) &= \frac{f(\rho_r)-f(\rho_l)}{\rho_r-\rho_l} & &\text{if}\quad \gamma\neq 0 \\ D &= {f(\rho_r)-f(\rho_l)} & &\text{if}\quad \gamma = 0 \end{aligned}\right. as $$x_2 \to x_1$$. The solution to the initial-value problem is made of two elementary wave solutions:
• At $$x=0$$, there is a static shock. If $$D<1/100$$, then the jump has left state $$\rho_l = 2/5$$ and right state $$\rho_r = \frac12 - \sqrt{\frac{1}{100} - D}$$ deduced from the Rankine-Hugoniot condition. Thus, only downstream car densities are impacted by the on-ramp. If $$D>1/100$$, then the jump has right state $$\rho_r = 2/5$$ and left state $$\rho_l = \frac12 + \sqrt{\frac{1}{100} + D}$$. In this case, upstream car densities are impacted by the on-ramp.
• There is an admissible rarefaction wave if $$D<1/100$$. The jump has left state $$\rho_l = \frac12 - \sqrt{\frac{1}{100} - D}$$ and right state $$\rho_r = \frac25$$. If $$D>1/100$$, then there is an admissible shock wave, with left state $$\rho_l = 2/5$$ and right state $$\rho_r = \frac12 + \sqrt{\frac{1}{100} + D}$$.