Let us consider the following problem
$$u_{tt}-c^2u_{xx} =0 \hspace{1cm} x,t\in\mathbb{R} \\ u(x,0)=f(x) \\ u_t(x,0)=g(x)$$
where $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$
Suppose that
$$g(x)=0 \hspace{1cm} \forall x\in \mathbb{R}$$ $$f(x)=0 \hspace{1cm} x\notin [-1,1]$$
The question is: where's the support of the solution to the problem u(x,t)?
I know that applying d'Alembert formula and the fact that $g(x)=0$ we obtain
$$u(x,t)= \frac{1}{2}[f(x+ct)+f(x-ct)]$$
but I don't know how do I use the fact that $f(x)=0$ for $x\notin [-1,1]$ to find the support.
And now, the question's the same but now suppose
$$f(x)=0 \hspace{1cm} \forall x\in \mathbb{R}$$ $$g(x)=0 \hspace{1cm} x\notin [-1,1]$$
Once again, applying d'Alembert formula and the fact that $f(x)=0$ we obtain
$$u(x,t)=\frac{1}{2c} \int_{x-ct}^{x+ct}g(s)ds$$
But I really don't know what to do next.
A helpful concept is the domain of influence. The initial position at $(x_0,0)$ influences the values of $u$ at $(x_0\pm ct, t)$. The initial velocity at $(x_0,0)$ influences the values of $u$ on the interval from $(x_0 - ct, t)$ to $(x_0 + ct, t)$. Both these statements are a consequence of d'Alembert's formula.
Case 1: $g\equiv 0$, $f$ is supported on $[-1,1]$
Draw the lines $x = x_0\pm ct $ for each $x_0\in [-1,1]$. They fill in two half-strips:
The support of $u$ is contained in the union of these two strips. You may or may not want to described it by inequalities; I think a sketch would convey the idea better.
Case 2: $f\equiv 0$, $g$ is supported on $[-1,1]$
Again, draw the lines $x = x_0\pm ct $ for each $x_0\in [-1,1]$, but this time also fill in the angle between them. The result is the shape bounded by $x=-1-ct$, $x=1+ct$, and the segment $[-1,1]$ itself. The support of $u$ is contained in this region.