Suppose $u$ solves \begin{cases} (\partial_t^2-\Delta)u = 0 & \text{on } U\\ u(T,x) = 0 \\ u(t,x) = 0 & \text{on } \partial U. \end{cases} Since the energy $$ E(t) = \int_U |\nabla u|^2 + |\partial_t u|^2\,dx. $$ is constant, $E(t) = E(T) = 0$, it follows that $u \equiv 0$. From linearity of the wave equation, the backwards uniqueness result follows. In general, solving the wave equation backwards in time is not really different from solving it forwards in time.

Suppose $u$ solves \begin{cases} (\partial_t^2-\Delta)u = 0 & \text{on } U\\ u(T,x) = 0 \\ \partial_t u(T,x) = 0\\ u(t,x) = 0 & \text{on } \partial U. \end{cases} Since the energy $$ E(t) = \int_U |\nabla u|^2 + |\partial_t u|^2\,dx. $$ is constant, $E(t) = E(T) = 0$, it follows that $u \equiv 0$. From linearity of the wave equation, the backwards uniqueness result follows. In general, solving the wave equation backwards in time is not really different from solving it forwards in time.