# Cannot understand method of characteristics

$u_t+uu_x =0$

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

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

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

What have I misinterpreted here?

*Edit: On the Wikipedia article http://en.wikipedia.org/wiki/Method_of_characteristics#Example 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

$t_s=1$, $x_s=u \left(\equiv u(x(s),t(s))\right)$ 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 $t=s$.

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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

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.

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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!

$\dfrac{dt}{ds}=1$ , 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. $u=F(x-ut)$

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

$F(x)=H(x)$

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

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

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