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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!

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  • $\begingroup$ Isn't $u_0$ also a function on $\mathbb{R}^n$? $\endgroup$ – Chappers Aug 3 '17 at 17:46
  • $\begingroup$ Good catch, thanks $\endgroup$ – Indefinite Aug 3 '17 at 19:31
  • $\begingroup$ So $u: \Bbb R^n \times (0, \infty) \to \Bbb R$, right? $\endgroup$ – Robert Lewis Aug 3 '17 at 19:34
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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.

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    $\begingroup$ I have a question: how can you solve $dx/dt = v$ to get $x - vt = x_0$ when $v$ is a non-constant vector field? $\endgroup$ – Robert Lewis Aug 3 '17 at 20:09
  • $\begingroup$ $v$ is a constant with respect to time (it maps $\mathbb{R}^3 \to \mathbb{R}^3$), which you are integrating the equation $\frac{dx}{dt}$ with respect to time. $\endgroup$ – Merkh Aug 3 '17 at 21:42

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