# Burgers Equation with Rarefaction

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?

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

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, 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 x2t . \end{aligned}\right. Thus there is no characteristic curve between $$x=t$$ and $$x=2t$$. The data at $$x>2t$$ is moving faster than the data at $$x. A rarefaction wave starting at the origin $$(x,t) = (0,0)$$ is generated in between those lines. Its shape is governed by $$u = x/t$$, and we have u(x,t) = \left\lbrace\begin{aligned} &\tfrac{x+1}{1+t}, && x2t . \end{aligned}\right. Note that the solution is continuous for positive times $$t>0$$.