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Problem. Suppose we have

$$u_t + uu_x = 0, ~~~~~~~~u(x,0) = \begin{cases} x + 1, & x < 0, \\ x + 2, & x > 0. \end{cases} $$

What I have: We have the usual solution by method of characteristics to start: $u(x,t) = \phi(x - ut)$, and characteristics

$$ x = \phi(r)t + r, ~~~~~~~~z = \phi(r).$$

Using the initial data we get the following implicit solution $$ u(x,t) = \begin{cases} \frac{x + 1}{1 + t}, &x < t \tag{1} \\ \frac{x + 2}{1 + t}, &x > 2t. \end{cases} $$

We can plot the characteristics more easily using our parameterization with $r$ such that $$x(t,r) = \begin{cases} (r + 1)t + r, &r < 0 \\ (r + 2)t + r, &r > 0, \\ \end{cases} $$

Hence, we have a region where there are no characteristics emulating from the origin (0,0).

For these, I recall that rarefaction solution is given as $u(x,t) = f\left(\frac{x-r}{t}\right)$, for some $f$. Hence, we have $$\begin{align} u_t &= f'\left(\frac{x - r}{t}\right)\left(\frac{r - x}{t^2}\right) \\ u_x &= f'\left(\frac{x - r}{t}\right)\left(\frac{1}{t}\right), \end{align} $$

and we need to find some $f$ that satisfies $$ f'\left(\frac{x - r}{t}\right)\left(\frac{r - x}{t^2}\right) + f\left(\frac{x-r}{t}\right)f'\left(\frac{x - r}{t}\right)\left(\frac{1}{t}\right) = 0. \tag{2}$$

So, $f\left(\frac{x-r}{t}\right) = \frac{x - r}{t}$, and $(2)$ is satisfied.

I'm not sure on how to proceed from here though?

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    $\begingroup$ I have fixed some algebra mistakes in your rarefaction derivation. The rest is done in my answer below. $\endgroup$
    – EditPiAf
    Sep 21, 2020 at 9:21
  • $\begingroup$ Thank you. Please ignore my comment/question below if you saw it before I deleted. The edit to my algebra answered my question. $\endgroup$
    – EzioBosso
    Sep 21, 2020 at 9:34

1 Answer 1

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This Cauchy initial-value problem has increasing piecewise linear initial data, with a jump discontinuity at $x=0$. It isn't a Riemann problem, since the initial data isn't piecewise constant. The method of characteristics yields the implicit equation $u = \phi(x-ut)$ with $\phi = u(\cdot,0)$. Thus, \begin{equation} u = \left\lbrace\begin{aligned} &x-ut+1, \quad x-ut<0 \\ &x-ut+2, \quad x-ut>0 \end{aligned}\right. \quad = \quad\left\lbrace\begin{aligned} &\tfrac{x+1}{1+t}, \quad x<t ,\\ &\tfrac{x+2}{1+t}, \quad x>2t . \end{aligned}\right. \end{equation} Thus there is no characteristic curve between $x=t$ and $x=2t$. The data at $x>2t$ is moving faster than the data at $x<t$. A rarefaction wave starting at the origin $(x,t) = (0,0)$ is generated in between those lines. Its shape is governed by $u = x/t$, and we have \begin{equation} u(x,t) = \left\lbrace\begin{aligned} &\tfrac{x+1}{1+t}, && x<t ,\\ &\tfrac{x}{t} , && t<x<2t , \\ &\tfrac{x+2}{1+t}, && x>2t . \end{aligned}\right. \end{equation} Note that the solution is continuous for positive times $t>0$.

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