Let $ u:R \times [0, \infty) \rightarrow R $ be a solution of the initial value problem
$ \begin{cases} u_{tt}-u_{xx}=0, for (x,t) \in R \times (0,\infty) \\ u(x,0)=f(x), x\in R \\ u_t(x,0)=g(x), x\in R \end{cases} $
Suppose $f(x)=g(x)=0$ for $ x \notin [0,1], $
then we always have
$\\ 1. u(x,t)=0 \;for\; all\; (x,t) \in (-\infty,0) \times (0, \infty) \\ 2.u(x,t)=0 \;for\; all\; (x,t) \in (1,\infty) \times (0, \infty) \\ 3. u(x,t)=0 \;for \;all \;(x,t)\; satisfying\; x+t\lt 0. \\ 4.u(x,t)=0 \;for\; all\; (x,t)\; satisfying\; x-t\gt 1. $
I tried as follows: D'Alembert's solution of the wave equation $ \frac{\partial ^2u}{\partial t^2}=\frac{\partial ^2u}{\partial x^2}$................(1) let us introduce the new independent variables r=x+t and s=x-t so that u became a function of r and s. then $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial u}{\partial s} \\ and \; \frac{\partial ^2u}{\partial x^2}=\frac{\partial}{\partial x}(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial s})=\frac{\partial}{\partial r}(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial s})+\frac{\partial}{\partial s}(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial s})=\frac{\partial ^2u}{\partial r^2}+2\frac{\partial ^2u}{\partial r \partial s}+\frac{\partial ^2u}{\partial s^2}\\ similarly \;\;\frac{\partial ^2u}{\partial t^2}=\frac{\partial ^2u}{\partial r^2}-2\frac{\partial ^2u}{\partial r \partial s}+\frac{\partial ^2u}{\partial s^2}$
substituting in (1) we get $\frac{\partial ^2u}{\partial r \partial s}=0$...........(2)
Integrating (2) w.r.t. r we get $\frac{\partial u}{\partial r}=f(r)..........(3)$
Integrating (3) w.r.t.s we get $u=\int f(r)dr +\xi(s) \\ thus \; u=\phi(r)+\xi(s)...........(4) \\where \phi(r)=\int f(r)dr \\ u(x,t)=\phi(x+t)+\xi(x-t)$
is the general solution of (1) $u_t(x,t)=\phi'(x+t)-\xi'(x-t)$
Now, given that $u(x,0)=\phi(x)+\xi(x)=f(x)\;and \;u_t(x,o)=\phi'(x)-\xi'(x)=g(x)$
Suppose f(x)=g(x)=0 for $x\notin[0,1]$
we have $\phi(x)+\xi(x)=0 \Rightarrow \phi(x)=-\xi(x) \;........(5)\\ and \; \phi'(x)-\xi'(x)=0 \\ on\; integrating\; we\; get \\ \phi(x)=\xi(x)+k .......(6)$
from (5) and (6) we get $\phi(x)=k/2 \; and \; \xi(x)=-k/2\\ (4) \Rightarrow u(x,t)=0 $
Now am not getting how to check the conditions provided in the options, since u is independent of x and t in the final solution
Here the characteristic curves are $$x+t=constant,~~x-t=constant$$ and Range of influence of $[0,1]$ is the region $x+t\geq 0, x-t\leq 1 , t\geq 0$(to get this just draw the characteristics at $0$ & $1$). So, Domain of dependence of any point outside this region is a subset of $\mathbb{R}\setminus [0,1]$.
So option $3,4$ are correct.