I am trying to find the solution of linear advection equation of the form:
$\frac {\partial c}{\partial t}+u\frac {\partial c}{\partial x}=0$
$c(x,0)=0$
$c(0,t)=\{c_0 \ \text{for}\ t \leq t_1 \text{and}\ 0 \ \text{for}\ t>t_1 $
Can anybody help me?
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
The transport equation simply moves your solution to the right with speed $u$ (assuming $u > 0$ and that $u$ is independent of $t$ and $x$).
Until time $t_1$, you will move a constant solution equal to $c_0$ into the domain at a speed $u$. The "jump" from $c_0$ to $0$ (your initial value), originally located at $x=0$, will thus travel rightward with speed $u$ and is thus located at $x = u*t$ (because it starts at $x=0$). So until time $t_1$, your solution will be
\begin{equation} c(x,t) = c_0 \end{equation}
for $0 \leq x \leq u t$ and
\begin{equation} c(x,t) = 0 \end{equation} for $x > u t$.
At time $t_1$, your boundary condition changes and you now start to move a second jump from $c_0$ to $0$ into the domain, again with speed $u$. Therefore, you now have a "sawtooth" shaped solution with the right edge at $x = u*t$ and the left edge at $x = u(t - t_1)$.
Thus, for $t > t_1$, your solution is
\begin{equation} c(x,t) = c_0 \end{equation}
if $u(t - t_1) \leq x \leq u t$ and
\begin{equation} c(x,t) = 0 \end{equation} for all other values of $x$.
