For $u_t = \Delta u + f$ in $\Omega \times (0,T)$ with $u(x,0) = u_0$, $u = g$ on $\partial\Omega \times (0,T)$:

If $u_0 \in L^2(\Omega)$ and $f \in L^2(0,T; L^2(\Omega))$, there exists unique weak solution $u \in L^2(0,T; H^1(\Omega))$ with $u_t \in L^2(0,T; H^{-1}(\Omega))$. Moreover, $u \in C([0,T]; L^2(\Omega))$ and satisfies energy estimate: $\|u(t)\|_{L^2}^2 + \int_0^t\|\nabla u\|_{L^2}^2 \leq C(\|u_0\|_{L^2}^2 + \|f\|_{L^2(0,t; L^2)}^2)$

For $u_{tt} = c^2\Delta u + f$ with $u(x,0) = u_0$, $u_t(x,0) = u_1$:

If $u_0 \in H^1(\Omega)$, $u_1 \in L^2(\Omega)$, $f \in L^1(0,T; L^2(\Omega))$, there exists unique weak solution with $u \in C([0,T]; H^1)$, $u_t \in C([0,T]; L^2)$, and energy conservation (for $f = 0$): $E(t) = \frac{1}{2}\int(u_t^2 + c^2|\nabla u|^2) = E(0)$

For hyperbolic equation $u_{tt} = c^2\Delta u$, if $u_0, u_1$ have support in ball $B_R(0)$, then $u(x,t) = 0$ for $|x| > R + ct$. This sharp result (domain of dependence) fails completely for parabolic equations.