I have to do the following assignment:
Consider the wave equation on a semi-infinite interval $$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2u}{\partial x^2} \hspace{1cm} 0<x<\infty$$ with the free boundary condition $$\frac{\partial u}{\partial x}(0,t)=0$$ and the initial conditions $$u(x,0)= f(x)= \cases{0 \hspace{0.5cm} 0<x<2 \\ 1 \hspace{0.5cm} 2<x<3 \\ 0 \hspace{0.5cm} x>3} $$ $$\frac{\partial u}{\partial t}(x,0) =g(x)= 0$$
I have to determine the solution using D'Alembert formula / general solution of a one-dimensional wave equation. I don't know how to handle the different cases of $f(x)$. I know I should extend $f(x)$ as an even function. I thought this looks like this: $$f_{e}(x)= \cases{0 \hspace{1cm} x<-3 \\ 1 \hspace{0.5cm} -3<x<-2 \\ 0 \hspace{0.5cm} -2<x<2 \\ 1 \hspace{1cm} 2<x<3 \\ 0 \hspace{1cm} x>3}$$ The solution should now be $$u(x,t) = \frac{1}{2}(f_{e}(x-ct)+f_{e}(x+ct))$$ Then you usually consider the cases where $x<ct$ and $x>ct$. If $x>ct$, $u(x,t)$ is just $u(x,t)=\frac{1}{2}(f(x-ct)+f(x+ct))$. But here's what I don't get. How should I quantify this? Since we have that $$f(x-ct) = \cases{0 \hspace{1cm} ct<x<2+ct \\ 1 \hspace{1cm} 2+ct<x<3+ct \\ 0 \hspace{1cm} x>3+ct}$$ and $$f(x+ct) = \cases{0 \hspace{1cm} -ct<x<2-ct \\ 1 \hspace{1cm} 2-ct<x<3-ct \\ 0 \hspace{1cm} x>3-ct}$$
How should I combine these different cases for $f(x-ct)$ and $f(x+ct)$ and so find $u(x,t)$ for $x>ct$? The same problem I have for when $x<ct$.