I am currently trying to solve a problem, but am having a bit of difficulty. The question is:
Consider the 1D Wave Equation for $u(x,t)$ for $x\in\mathbb R$ and $t>0$: $$\begin{cases} u_{tt}&=u_{xx}\\u(x,0) &= 0\\ u_t(x,0) &= \psi(x)\end{cases}$$ Where $$\psi(x) = \begin{cases} 1 & \text{ if } |x-3|\leq 1 \text{ or } |x+3|\leq 1\\ 0 & \text{ otherwise}\end{cases}$$ $\boldsymbol{A}$: At $t=1$ and $t=4$, what are the positions on the string where the displacement $u$ is nonzero? $\boldsymbol{B}$: When at $t=10$, what are the positions on the string where the displacement is maximal? $\boldsymbol{C}$: What is $u(0,t)$ as a function of $t$?
By D'Alembert, I know the solution is given by: $$u(x,t) = \frac{1}{2}\int_{x-t}^{x+t}\psi(\tau)d\tau$$
If $t=1$, we have:
$$u(x,1) =\frac{1}{2}\int_{x-1}^{x+1}\psi(\tau) d\tau$$ So we need $\tau$ to be either in $[2,4]$ or $[-4,-2]$. So thus $x=3,-3$?
If $t=4$, we have: $$u(x,4) =\frac{1}{2}\int_{x-4}^{x+4}\psi(\tau) d\tau$$ So we still need $\tau$ to be in $[-4,-2]$ and $[2,4]$. So then is $x\in[-8,8]$?
If $t=10$, then we have: $$u(x,10) =\frac{1}{2}\int_{x-10}^{x+10}\psi(\tau) d\tau$$ Wouldn't the maximal position be at $x=0$? Since it gives the biggest interval of integration?