5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

characteristics

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 $t>1$.

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Infinite_28 Oct 24 '18 at 14:42
  • 1
    $\begingroup$ 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. $\endgroup$ – Infinite_28 Oct 24 '18 at 20:26
  • 1
    $\begingroup$ I was able to solve the transport equation but posted the problem I still have here $\endgroup$ – Infinite_28 Oct 24 '18 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.