I am trying to solve the following initial boundary-value problem for the constant-coefficient, first-order wave equation:
$2u_{x}+u_{t} = 0$, $x>0$, $t>0$
I.C.: $u(x,0)= \begin{cases} x-1, & 0<x<1 \\ 1, & x\geq 1 \end{cases}$
B.C.: $u(0,t)=0$, $t>0$
Since $a = 2$, $f(x) = \begin{cases} x-1, & 0 < x < 1, \\ 1, x \geq 1\end{cases}$, $g(t)=0$, the solution is given by
$\displaystyle u(x,t) = \begin{cases} f(x-2t), & x>2t \\ g\displaystyle \left(\frac{2t-x}{2} \right), & x < 2t \end{cases}$.
I figured out that the solution, then, must be $u(x,t) = \begin{cases} 0, & 0 < x < 2t\\ (x-2t)-1, & 2t < x < 2t+1 \\ 1,& x > 2t + 1 \end{cases}$
(Unless I am incorrect in the solution, in which case, please correct me!)
Now, what I am having trouble with is the second half of what I am asked to do for this problem, which is to graph $u(x,t)$ as a function of $x$ for $t =0$, $t = 1$, and $t = 2$.
Quite frankly, what is bothering me is that what I get for $u(x,0)$, $u(x,1)$, and $u(x,2)$ are not continuous, so I must be making a mistake somewhere:
$\mathbf{t = 0}: \, u(x,0) = \begin{cases} 0, & x=0 \\ x-1, & 0<x<1 \\ 1,& x>1 \end{cases}$ which is not continuous at $x=0$ or $x=1$.
$\mathbf{t=1}: \, u(x,1) = \begin{cases} 0, & 0 < x < 2 \\ x-3, & 2 < x < 3 \\ 1, &x>3 \end{cases}$ which is not continuous at $x = 2$ or $x=3$.
$\mathbf{t=2}: \, u(x,2) = \begin{cases} 0, & 0 < x < 4 \\ x-5, & 4<x< 5 \\ 1, & x> 5\end{cases}$ which is not continuous at $x=4$ or $x=5$.
Now, it is my understanding that the solution $u(x,t)$ is supposed to be continuous, and I don't understand what I am doing wrong!
Please help! And if possible, could you give me a straightforward solution without any hints or questions? I am really in a hurry and don't have time for that right now.
Thank you!