We'll consider the following equation defined on $x\in\mathbb{R}$, and $t>0$: $$y_t+y.y_x=0$$ with the initial condition: $$y(x,0)=\mathbb{1_{(-\infty,0)}(x)}$$ I tired to solve this equation using Method of characterestics:
let us define : $a=1$ and $b=y$ and $c=0.$
then we have $dy/dx=dy/dt=0$ and $dx/dt=y$
since the derivative of $y$ are equals to $0$. then $y$ is constant.
then we'll have $x(t)=y(x(t),t).t+A$ ($A$, is a constant of integration)
then, $y(x(t),t)=(x(t)-A)/t$
I did not know how to use initial condition.
This is the inviscous Burger's equation.
You got the characteristic curves $x(t)=y_0t+x_0$ and $y(t)=y(x(t),t)=y_0$ along these curves. Given a point $(x,t)$, the characteristic curve it lies on satisfies $$y_0=y(x,t)=y(x-y_0\Delta t, t-\Delta t)=const.=y(x-y_0t,0)=f(x-y_0t),$$ where $y(x,0)=f(x)$ is the initial condition.
This is a non-linear equation that may have multiple solutions. In the given case, the possible values for $y_0$ are $0$ and $1$. So if $x-t<0$ there is a characteristic with slope $y_0=1$ going through $(x,t)$, and if $x\ge 0$ then a characteristic of slope $y_0=0$. As there is an overlap for $t>0$, there will be a shock front. By Rankine-Hugoniot the curve of this front has slope $1/2$, so that $$ y(x,t)=\begin{cases}1,&x<t/2,\\0,&x\ge t/2.\end{cases} $$