Suppose you had the non-linear transport equation $$u_t+uu_x=0$$ with initial data $$u(x,0)=g(x)=\begin{cases} 1, & x>1 \\-x, & -1\leq x< 0 \\x, & 0\leq x\leq 1\end{cases}$$

Solving the characteristic ODEs, we have that $u(x,t)=f(x-ut)$ where $f$ is some arbitrary $C^1$ function. The classical solution before the shock waves is given by:

$$u(x,t)=\begin{cases} 1, & x-ut>1 \\-x, & -1\leq x-ut< 0 \\ x, & 0\leq x-ut\leq 1\end{cases}$$

which can be rewritten also as:

$$u(x,t)=\begin{cases} 1, & x>1+ut \\-x, & -1+ut\leq x< ut \\ x, & ut\leq x\leq 1+ut\end{cases}$$

for the case $-1\leq x<0$, I believe the solution should be $u(x,t)=\dfrac{-x}{1-t}$, and for $0\leq x\leq 1$, the solution should be $u(x,t)=\dfrac{x}{1+t}$.

Now there is a shock wave at $t=-1$, as $u(x,t)=\dfrac{-x}{1-t}$ for $x\in [-1,0)$ has a singularity at $t=-1$ but no such singularity occurs with $u(x,t)=\dfrac{x}{1+t}$ for $x\in[0,1]$. For $x>1$, do we simply have $u(x,t)=\dfrac{x}{t}$?

This is where I get stuck, and I am unsure how to proceed to get the full solution $u(x,t)$ for small enough $t$.

Assuming I had a solution, it will be piecewise $C^1$, I can use the condition $T_b=\dfrac{-1}{\min_{x}g'(x)}$ on each of the pieces to determine when the shocks occur. However I am not sure how to work out the details or correctly interpret the shock waves in a physical sense.


1 Answer 1


This transport equation is the inviscid Burgers' equation. The figure below represents the characteristic curves in the $x$-$t$ plane:


Here, the breaking time where characteristics intersect for the first time is $$ T_b = \frac{-1}{\min_x g'(x)} = 1\, . $$ Indeed, the singularity in $\frac{-x}{1-t}$ occurs at $t=1$.

Now, let us compute the full solution. At $t=1$, characteristics coming from $-1<x<0$ intersect, and should stop at $t=1$. If we consider the position $(x,t)$ in the area $x>0$, $t>1$, we have data coming from $0<x<1$, but no other data. There is no shock wave in this area.

However, if we add the data $u(x,0) = 1$ for $x<-1$, we get the following sketch of the $x$-$t$ plane:

characteristics 2

A shock wave occurs at $t=1$, which speed is determined by the the data $u_L(x,t)=1$ on the left, and the data $u_R(x,t) = \frac{x}{1+t}$ on the right. The Rankine-Hugoniot condition indicates that the shock wave propagates along the curve $(x_s(t),t)$ given by \begin{aligned} x_s'(t) &= \frac{1}{2} \left(u_L(x_s(t),t)+u_R(x_s(t),t)\right) \\ &= \frac{1}{2}\left( 1 + \frac{x_s(t)}{1+t} \right) , \end{aligned} with the initial condition $x_s(1)=0$. Therefore, the shock is located at $x_s(t) = 1+t - \sqrt{2}\sqrt{1+t}$ for $t>1$.

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    $\begingroup$ I am working on a similar problem, but the right hand side is not zero but $-a*u$. So I am searching for the characteristic curves when I have the equation $ u_t + f(u)*u_x = -c*u $. Then the characteristic curves are no longer straight lines. But I am stuck with how I can determine the curves. Can you help me with that? $\endgroup$ Oct 24, 2018 at 14:42
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    $\begingroup$ I now was able to solve the problem I had, but thanks a lot for your help, your answer above was very helpful and the pictures are very good. $\endgroup$ Oct 24, 2018 at 20:26
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    $\begingroup$ I was able to solve the transport equation but posted the problem I still have here $\endgroup$ Oct 24, 2018 at 20:59

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