Show that a solution of $u_t+(|u|^\alpha)_x=0$ violates entropy condition 
Consider $$u_t+(|u|^\alpha)_x=0, \quad\alpha>1$$ Given the initial condition
  $$u(x,0)=\begin{cases} 0, x<0\\1,x>0\end{cases}$$
a) Find a solution for $u(x, t)$ that is continuous for all $t > 0$ and satisﬁes the initial condition.
b) Find a solution for $u(x, t)$ that includes a shock obeying the
  appropriate jump condition.
(c) Show that one of these solutions does not satisfy the entropy
  condition.

My attempt:
We can rewrite the equation $$u_t+(|u|^\alpha)_x=u_t+\alpha u |u|^{\alpha-2}u_x$$ then we parametrize \begin{align} x(0,r) &= r \\ t(0,r) &= 1 \\ u(0,r) &= u_0 \end{align}
The characteristics satisfying IVP are
\begin{align} 
x_s &= \alpha u |u|^{\alpha-2} \\ 
\implies x &= \alpha u |u|^{\alpha-2}t+r \\
t_s &= 1 \\ 
\implies t &= s \\
u_s &=0 \\ 
\implies u &= u_0 \end{align}
The projection on $(x,t)$-plane is given by $$t = \frac{x-r}{\alpha u |u|^{\alpha-2}}$$ So if $x<0$, the denominator is $0$. How do I find the solution?
 A: The characteristics are the curves $x=x_0+\alpha u|u|^{\alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant.
With the given initial data, we have
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 &&\text{if}\quad x <0\\
&U (x/t) &&\text{if}\quad 0 \leq x \leq \alpha t\\
&1 &&\text{if}\quad \alpha t <x
\end{aligned}
\right.
$$
where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (\xi)$ with $\xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains
$$
U (x/t) = \left.\left (\xi\to\alpha \xi|\xi|^{\alpha-2}\right)^{-1}\right|_{\xi=x/t} = \left(\frac{x}{\alpha t}\right)^\frac {1}{\alpha-1} .
$$
Concerning the shock wave solution
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 &&\text{if}\quad x <st\\
&1 &&\text{if}\quad st <x
\end{aligned}
\right.
$$
with
$$
s = \frac{|1|^\alpha - |0|^\alpha}{1 - 0} = 1
$$
following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since
$$
\alpha\, 0\, |0|^{\alpha-2} < s < \alpha\, 1\, |1|^{\alpha-2} .
$$
Indeed, the Lax entropy condition requires opposite inequalities.
Finally, the entropy solution is the rarefaction wave.
