Consider the 1D wave equation($x \in \mathbb R$ and $t>0$): $$u_{tt} - c^2 u_{xx} = q(x,t)$$ where $q(x,t) = (1-x^2)sin(t)$ when $|x|\leq 1$ and $q(x,t)=0$ if $|x|>1$. $u(x,0)=0$ and $u_t(x,0)=0$. Here $c$ is a positive constant. Prove: $u(x,t)=0$ for $|x|>ct+1$.
By D'Alembert formula, we have $u(x,t)=\frac{1}{2c} \int_{0}^{t} \int_{x-c(t-s)}^{x+c(t-s)} q(y,s)dyds$. My understanding is if we want $u=0$, we need the absolute values of end points to be greater than 1. However I tried, both $t$ and $s$ are always included. Besides, in the question, the expression of $q$ when $|x|<1$ doesn't really matter, right?