Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$u_t+uu_x =0$

$u(x,0)\equiv u_0(x)=\begin{cases} 0 & x<0 \\ 1 & x>0 \end{cases}$

I want to parametrise $u(x(s),t(s))$. This is the first thing that is conceptually quite difficult to picture, but I get the idea that we are parameterising.

$$\frac{du}{ds}=\frac{\partial u}{\partial x}\frac{dx}{ds} +\frac{\partial u}{\partial t}\frac{dt}{ds}$$

In this case

$$\frac{du}{ds} = 0$$

So I understand this is just a trick to use and so $\frac{dx}{ds}=u$, $\frac{dt}{ds}=1$.

Next $\frac{dx}{dt}=u$, do we always do this, i.e. do we always divide those derivatives to eliminate the parameter? If I do then I lose it forever, and it seems that the constant I get should actually be the parameter, $s$, that is why it is used strangely below, I cannot seem to work out how to do this properly.

$x=ut+\text{const.}\equiv ut+c$

If $t=0$ then $x=c$ and $u(x,0)=u_0(c)=\begin{cases} 0 & c<0 \\ 1 & c>0 \end{cases}$

Is $x=u_0t+c$ equivalent? *

Then $x=\begin{cases} c & c<0 \\ t+c & c>0 \end{cases}$

This should read

$x=\begin{cases} s & s<0 \\ t+s & s>0 \end{cases}$

What have I misinterpreted here?

*Edit: On the Wikipedia article it uses the fact that $u_s=0$; the solution is constant along the characteristic - then that $(x_s,t_s)$ and $(x_0,0)$ are on the same characteristic (How?) to deduce $u(x_s,t_s)=u(x_0,0)$ where $(x_s,t_s) = (a,1) $.

This would seem to be a solution to taking $u=u_0$ in my (above) attempt

** In another attempt, after reading the Wikipedia page,

$t_s=1$, $x_s=u \left(\equiv u(x(s),t(s))\right)$ and $u_s=0$ as before,

$\int_{t_0}^t dt=\int_{0}^s ds \Rightarrow t=s+t_0$ and since $t_0=0$ (Why?) $t=s$.

$x_s=u$ I am guessing that all characteristics must originate from some initial point in the the initial value problem. So for this reason, $u(x,0)=u(x(s),t(s))$ since $u$ is constant on the characteristic.

This leaves me with $\int_{x_0}^x dx=\int_0^s u_0 ds$ which leads me to $x=u_0 t+x_0$ since $u$ is constant along the path of $s$ and is therefore a constant in that integral, and I replace $s$ with $t$. The only way to get to the answer from here is with $x_0=s$ - I cannot see how you would get here, since $t=s$.

share|cite|improve this question
Note that you are required to find $u$ in terms of $x$ and $t$ rather than just required to find $x$ in terms of $t$ , so the answer you accepted has problems! – doraemonpaul Apr 20 '13 at 17:56
I am prescribed some curve $\Gamma$ initially. $u$ does not vary along $u(x(t),t)$, so given that I know $u(x(0),0)$ all of the characteristics coming from $u(x(0),0)$ described by $x(t)$ give a solution surface for all $t>0$. – shilov Apr 20 '13 at 19:16
The help I needed was mainly with characteristics - I did not care too much for solving the equation once I can do the characteristic slopes. Thanks for the help though. – shilov Apr 20 '13 at 19:21
up vote 1 down vote accepted

There is no particular reason to parametrize $t$, and it is simpler not to. From the chain rule, along a curve $(x(t),t)$, you have $$ \frac{d}{dt}(u(x(t),t) = u_t+x'(t)u_x, $$ and set this equal to $u_t+uu_x=0$, from which $u$ is constant on lines where $x'=u$. In your case, all vertical lines starting from $(x,0)$ when $x$ is negative, so $u=0$ in the whole left half plane. Also all lines of slope 1 starting from $(x,0)$ when $x$ is positive, so $u=1$ in that whole region. That leaves a wedge shape where $u$ has not been defined yet. The only way a line can fit into that is if it is $x=mt$ with $m$ between 0 and 1. That gives you a solution $u=x/t$ in that wedge, and together with the first two regions, now $u$ is defined and continuous in the whole half space $t>0$.

You might parametrize both $t$ and $x$ for an equation like $a(x,t,u)u_t+b(x,t,u)u_x=0$, but not needed here.

share|cite|improve this answer
I can see that you parametrise such that $u$ is constant on the characteristic. The condition for this is $x'=u$. Your picture of all straight lines for $x<0$ and slope 1 lines from $(x,0)$ is clear but where do you get the vertical and slope 1 lines from? – shilov Mar 20 '13 at 1:59
I'm having trouble with the comments section, but I wanted to add a question: since $x'=u$ shouldn't the slope $x<0$ be horizontal not vertical? – shilov Mar 20 '13 at 2:12
The 0 and 1 come from your initial values, $u(x,0)$. Draw $x$ horizontal, $t$ vertical, so that the evolution goes upward. – Bob Terrell Mar 20 '13 at 11:57
Yes that's how I drew the axes but I completely missed the fact that $x'=0$ would be vertical with t on the vertical axis. I understand that now – shilov Mar 20 '13 at 12:11
Also, it definitely makes sense $u=\frac{x}{t}$ now you have explained it - but I doubt I would be able to have seen that beforehand - method for seeing that? – shilov Mar 20 '13 at 12:27

I think you are thinking this question too complicated.

Just follow the following procedure is OK!

Follow the method in

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$

$u(x,0)=\begin{cases}0&x<0\\1&x>0\end{cases}=H(x)$ :


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

Hence $u(x,t)=\begin{cases}0&x<0\\1&x>t\\c&\text{neither}~x<0~\text{nor}~x>t\end{cases}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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