Weak solution of Burgers equation (Riemann problem)

Question

Problem

Show that, for every $\alpha > 1$ the function $$ u_{\alpha}(x,t) = \begin{align}\begin{cases}-1 & x< (-1-\alpha)\frac{t}{2} \\ -\alpha & 0 > x > (-1-\alpha)\frac{t}{2} \\ \alpha & 0 < x< (1+\alpha)\frac{t}{2} \\ 1 & x >(1+\alpha)\frac{t}{2} \end{cases} \end{align} \tag{1}$$ is a weak solution to the problem $$ \begin{align}\begin{cases} u_{t} + u u_{x} =0 & t > 0 , x \in \mathbb{R} \\ u(x,0) = \phi(x) \end{cases} \end{align} \tag{2}$$ where $\phi(x)$ is given by $$ \phi(x) = \begin{align}\begin{cases} -1 & x < 0 \\ 1 & x > 0 \end{cases} \end{align} \tag{3}$$ Is it also an entropy solution? At least for some $\alpha$?

My question is mostly about how to approach this problem. I'm looking in the book. There is another problem like it where it says they connect a function.

$$ g(x)= \begin{align}\begin{cases} 0 & x < 0 \\ 1 & x > 0 \end{cases} \end{align} \tag{4}$$

through rarefaction waves to construct a weak solution. I do not understand what the entropy solution is.

Answer 1

I guess that there are sign mistakes in the solution $u_\alpha$. Indeed, according to the Rankine-Hugoniot condition, $u_\alpha$ isn't even a weak solution of Burgers' equation! If $u_\alpha$ were a weak solution, then the first discontinuity would have speed $-\frac{1}{2}(1+\alpha)$ (instead of $\frac{1}{2}(1-\alpha)$), and the second discontinuity would have speed $\frac{1}{2}(1+\alpha)$ (instead of $\frac{1}{2}(\alpha-1)$). To prove that a Rankine-Hugoniot satisfying version of $u_\alpha$ is a weak solution, see this post for methodology.

Let us imagine that this problem is fixed by setting the appropriate signs in the expression of $u_\alpha$. Thus, one can notice that $u_\alpha$ is not an entropy solution of this Riemann problem, since the middle discontinuity with speed zero does not satisfy the Lax entropy condition.