Consider the continuity equation $u_t + \nabla \cdot (u v) = 0$ where $u : \mathbb{R}^3 \times (0,\infty) \to \mathbb{R},$ with smooth $v(x) : \mathbb{R}^3 \to \mathbb{R}^3$ and initial data $u_0.$
Suppose $-1 < \nabla \cdot v$ and $$u_0(x) = \begin{cases} 1 & -1 \leq |x| \leq 1 \\ 0 & \text{else} \end{cases}.$$ Show that $\Omega := \{x : u(x,1) > 0\}$ has volume greater than $\frac{4}{3}.$
Any ideas on how one might go about showing this, any suggestions? I've tried to solve it, but I'm stumped on how the given assumption on the divergence of $v$ would help even if I did solve it. Thanks!
To start we could look at the characteristic equations, where I will let $u_t := q,$ $u := z,$ $\nabla u = p,$ and $x = (x,y,z)$ (excluding the time variable). \begin{align} \frac{dz}{dt} &= q + v \cdot p = -z(\nabla \cdot v)\\ \frac{dx}{dt} &= v.\\ \end{align} We immediately get \begin{align} x - vt &= x_0\\ u(x,t) &= z(x_0(x,t),0) e^{-(\nabla \cdot v) t}\\ &= z(x - vt,0) e^{-(\nabla \cdot v) t}\\ &= \begin{cases} e^{-(\nabla \cdot v) t} & -1 \leq |x - vt| \leq 1\\ 0 & \text{else} \end{cases} \end{align}
We have that $u(x,1) = \chi_{[-1,1]}(|x-v(x)|) e^{-(\nabla \cdot v(x))},$ but from here I am not sure where to go since the power of the exponential is a number smaller than 1.