Given $u_t=(u_x)^2$ and an initial condition $u(x,0)=\begin{cases} 1-x & \text{ if } x \in [0,1) \\ 0 & \text{ if } x \in [1,\infty) \end{cases} $
Using Charpit's equations by defining $F(x,y,u,p=u_x,q=u_t)= u_t-(u_x)^2 = 0$, and parameterising characteristic projections by $(s,\tau)$, we obtain
$q=p^2 \implies F=q-p^2$ and so characteristic projections are defined by $ t'(s)=1,x'(s)=-2p$ and on the projections, $p$ and $q$ vary according to;$ \dot{p}=\dot{q}=0$ and the solution is given by $u'(s)=q-2p^2=p^2-2p^2=-p^2$
Now the problem is defining conditions on s=0, I have the solutions(given below), but I have no intuitive idea of how they are yielded, from the initial conditions given.
The aformentioned conditions are given by (on $s=0$) : $t=0,x=\tau,u=1-\tau,p=\mu, q=\mu^2$
An obvious solution for the PDE $\quad u_t=u_x^2\quad$ is : $$u(x,t)=\alpha x +\alpha^2 t+\beta$$ With the conditions the solution is : $$u(x,t)=\begin{cases} -x+t+1 & \text{ if } x \in [0,1) \qquad\qquad \alpha=-1\quad;\quad \beta=1 \\ 0 & \text{ if } x \in [1,\infty) \qquad\qquad \alpha=0\quad;\quad \beta=0 \end{cases} $$ Or with the Heaviside step function $$u(x,y)=(-x+t+1)\text{H}(1-x)$$