# Burgers' Equation with Initial and Boundary Conditions

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!

• 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. Commented Feb 15, 2019 at 10:17

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 $$x and $$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, & x0.$$
• 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. Commented Feb 7, 2019 at 3:58