PDE with strange Auxiliary Conditions I am currently trying to find an $\textit{explicit solution}$ to this question but am getting stuck. So the question reads:

Find an explicit solution for the PDE:
  $$u_t+u\cdot u_x=0, \>\> u(x,0)=g(x)=\begin{cases}
1 &= \text{ if } x\leq 0\\
1-x &= \text{ if } 0\leq x \leq 1\\
0 &= \text{ if } x\geq 1
\end{cases}$$ 
  For all $x$ and $0\leq t\leq 1$. 

Now, I found the Characteristics:
$$\frac{\partial x}{\partial t}=z(t), \>\>\> \frac{\partial z}{\partial t}=0$$
Where $z(t)$ is the term on the RHS of the PDE. Now, setting the initial conditions:
$$x(0)=x_0, \>\> z(0)=g(x_0)$$
And the solutions are given by:
$$z(t)=g(x_0), \>\>\>\> x(t)=g(x_0)\cdot t +x_0$$
Can I conclude that $u(x,t)=g(x_0)$? This seems a bit too easy, however I know that I can set $u=\frac{x-x_0}{t}$, so would I have to rearrange the solution in terms of $x_0$ for each case?
 A: $$u_t+uu_x=0 \tag 1$$
The system of characteristic ODEs is :
$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$$
A first family of characteristic curves comes from $\quad du=0 \quad\to\quad u=c_1$
A second family of characteristic curves comes from $\quad \frac{dt}{1}=\frac{dx}{c_1} \quad\to\quad x-c_1t=c_2$
The general solution expressed on the form of implicit equation is :
$$u=F(x-ut) \tag 2$$
where $F$ is an arbitrary function, to be determined according to the boundary condition.
For $t=0 \qquad u(x,0)=F(x)=\begin{cases}
1 & \text{ if}\quad  x\leq 0\\
(1-x) & \text{ if}\quad  0\leq x \leq 1\\
0 & \text{ if}\quad  x\geq 1
\end{cases}$ 
Now, $F$ is known and must be put into the general solution $(2)$ in order to find the particular solution which satisfies the boundary condition.
$u(x,t)=F(x-ut) =\begin{cases}
1 & \text{ if} \quad (x-ut)\leq 0\\
1-(x-ut) & \text{ if}\quad  0\leq (x-ut) \leq 1\\
0 & \text{ if}\quad  (x-ut)\geq 1
\end{cases}$ 
In case of $0\leq (x-ut) \leq 1 \quad\to\quad u=1-(x-ut) \quad\to\quad u=\frac{1-x}{1-t}$
and $\quad (x-ut)=x-\frac{1-x}{1-t}t=\frac{x-t}{1-t}$ 
$$u(x,t) =\begin{cases}
1 & \text{ if} \quad (x-t)\leq 0\\
\frac{1-x}{1-t} & \text{ if}\quad  0\leq \frac{x-t}{1-t} \leq 1\\
0 & \text{ if}\quad  x\geq 1
\end{cases}$$
