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Consider a first-order PDE: $$u_t + (1 + 2u)u_x = 0$$

valid on $$0 \leq x \leq \infty$$ $$0 \leq t \leq \infty$$

with Initial condition: $$u(x, 0) = 0$$ and boundary condition:

$$u(0,t) = \begin{cases} 1, & 0\leq t \leq 1 \\ t, & 1\leq t \leq \infty \end{cases} $$

I haven't encountered 1st-order PDEs with both an initial condition and boundary condition before, so I'm a bit confused on how to analytically solve. I generally understand the method of characteristics, and that an implicit general solution of $u = f(x - (1+2u)t)$. However when pluggin in initial conditions, I seem to get contradictory statements that I'm not sure how to interpret.

Plugging in $u(x,0)$ results in $f(x) = 0$. Plugging in $u(0,t)$ results in $f(-t^2-2t) = t$ and $f(-3t) = 1$ respectively. Not sure how to combine all of these, so some insight would be greatly appreciated!

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  • $\begingroup$ This is the Lighthill-Whitham-Richards (LWR) traffic flow model (Burgers' equation $u_t+ u u_x = 0$ is indeed quite similar). However, the physical interpretation as a flow of cars on a one-way road is lost since the car density $u$ becomes larger than one. $\endgroup$ – Harry49 Feb 15 at 10:17
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I believe that the best way to solve this is to draw the $x$-$t$ plane and use the characteristic lines and the fact that the solution to this equation is constant along the characteristic lines.

The first section of the $x$-$t$ plane is when $x>t$ (right of the line with slope 1), which corresponds to information from the initial condition, so $u=0$ in this region.

The second region is where information is carried from the BC $u=1$ from $t\in(0,1)$. In this region, the characteristics have slope $1/3$ in the $x$-$t$ plane, so the characteristics will collide with $x=t$, forming a shock. Nevertheless, we have $u=1$ when $x<t$ and $x>3(t-1)$.

For the third region, which corresponds to $x<3(t-1)$, the slopes of the characteristics vary depending on where they start on the $t$-axis, the value of which we will call $s$. The characteristic lines then have the form $t=x/(1+2s)+s$ and along this characteristic, $u=s$. We can solve this first equation for $s$, which yields a quadratic equation with 2 real roots. One of them is not feasible (remember we need $s>1$) and the other is, so we can write $u(x,t)=s=(-1 + 2 t + \sqrt{(2t+1)^2 - 8 x})/4$ $$ u(x,t) = \begin{cases} 0, & x\geq t \\ 1, & x<t \text{ and }x\geq 3(t-1) \\ \frac{-1 + 2 t + \sqrt{(2t+1)^2 - 8 x}}{4}, & x<3(t-1) \end{cases}, \ \forall x,t>0. $$

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  • $\begingroup$ Thanks for the very thorough response! It makes sense once you can see it visually with the characteristics sketch. I'm wondering however what's your strategy in sketching the characteristics in the x-t plane. I often have difficulty in doing that, mostly from not knowing which equations to look at. $\endgroup$ – albell26 Feb 6 at 18:36
  • $\begingroup$ If we have $u_t+f(x,t,u)u_x=0$, then $f$ determines the characteristic line passing through a point in the $x$-$t$ plane. In this case, $f=1+2u$, so it does not depend explicitly on $x$ or $t$, which indicates that the characteristics will be straight lines. $f$ can be interpreted as the speed at which $x$ changes with respect to $t$ along the characteristic. In this light, you need to make sure that the slopes are correct in the $x$-$t$ plane. If $1+2u$ is large, then the slopes will be small, since we are looking at $x$ vs. $t$ instead of the other way around. $\endgroup$ – whpowell96 Feb 7 at 3:58

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