# Show the measure of supp($u$) is greater than $\frac{4}{3}$ where $-1 < \nabla \cdot v$

Let $x \in \mathbb{R}^3,$ and let $v(x) : \mathbb{R}^3 \to \mathbb{R}^3$ be smooth with the property that $-1 < \nabla \cdot v.$ Suppose $u$ solves $$u_t + \nabla u \cdot v = 0,$$ with $$u(x,0) = \chi_{|x| \leq 1}(x) = \begin{cases} 1 & |x| \leq 1 \\ 0 & \text{else} \end{cases}$$ Show that $\{x \in \mathbb{R}^3 \, : \, u(x,1) > 0\}$ has measure greater than $4/3.$

Here, we know the solution is given by $u(x,t) = u(x_0, 0),$ and the characteristics for $x_i, \; i=1,2,3$ are given by $$\frac{dx_i(t)}{dt} = v_1(x_1,x_2,x_3).$$

I've been struggling with this, so to start, I decided to convince myself that the problem is true for the easiest case of $v$ I could think of. Let $v(x) = v(x_1,x_2,x_3) = (-\frac{x_1}{3},-\frac{x_2}{3},-\frac{x_3}{3}).$ Sure, $\nabla \cdot v = -1,$ but let's see what happens.
The characteristic equations are $$\frac{dx_i(t)}{dt} = -\frac{x_i}{3},$$ so $x_i(t) = x_{i_0}e^{-t/3}.$ What this says is that at $t = 1,$ the radius of the initial ball has shrunk from $1 \to e^{-1/3},$ and therefore the volume of the support is $\frac{4\pi}{3} (e^{-1/3})^3 = \frac{4\pi}{3e} > \frac{4}{3}.$

However, this hasn't done me any good for showing it for arbitrary $v(x).$ Does anyone see a more intelligent method to go about this problem?

• I suppose that it is the support of $x \mapsto u(x-v(x))$ that is intended. Aug 14 '17 at 21:16
• Yes thank you, I will fix it Aug 14 '17 at 21:47
• Have you got a solution to this? Sep 2 '17 at 18:04
• Nope not yet :/ Sep 3 '17 at 18:54
• Sorry, I had to change the problem statement slightly. I think I had it factually incorrect as stated before. It is fixed now, sorry for any time wasted. Sep 11 '17 at 18:28

A sketch of an idea:

Set $U(t) = \int_{-\infty}^{\infty} u(x, t) \, dx.$ Then $$U(0) = \int_{-\infty}^{\infty} \chi_{|x|<1}(x) \, dx = \frac{4\pi}{3}$$ and $$U'(t) = \int_{-\infty}^{\infty} \partial_t u(x, t) \, dx = - \int_{-\infty}^{\infty} \nabla u(x, t) \cdot v \, dx \\ = \int_{-\infty}^{\infty} u(x, t) \, \nabla \cdot v \, dx > - \int_{-\infty}^{\infty} u(x, t) \, dx = - U(t).$$ where I have used partial integration and that $\nabla \cdot v > -1$.

The first question is now whether $U'(t) > -U(t)$ and $U(0)>0$ implies that $U(t) > U(0) \, e^{-t}$ for all $t>0$. If so, we have $$U(1) > \frac{4\pi}{3} \, e^{-1} = \frac43 \frac\pi e > \frac43.$$

The second question is whether we must have $u(x,t) \in [0, 1]$ for $t>0$, i.e. $u$ cannot increase and be larger than $1$. If so, then for $U(1) > \frac43$ to be true, we must have $m(\operatorname{supp}_x u(x, 1)) > \frac43.$

Since you know the solution is just the composition of $u$ with the flow of the vector field $v$, the support of $u$ at time $t$ is just the time-$t$ flow of the support at time zero; i.e.

$$\Omega_t := \mathrm{supp}(u(\cdot,t))= \Phi^v_t(\Omega_0)$$ where $\Omega_0 = \{|x| \le 1\}$ is the initial support and and $\Phi^v_t : \mathbb R^3 \to \mathbb R^3$ is the flow map, which is defined pointwise as the solution to the transport ODE $$\frac{d \Phi^v_t(x)}{dt}=v(\Phi_t^v(x))$$ with initial condition $\Phi_0^v(x) = x$. (This is related to your $x(t)$ simply by $\Phi_t^v(x_0) = x(t).$)

Thus we can forget about the function $u$ and just study the evolution of the volume of $\Omega_t$. Since motion of a small piece of the boundary (with outwards normal $\nu$ and area $dA$) at velocity $v$ produces an increase of volume at rate $v \cdot \nu\; dA$, we can compute $$\frac d{dt} V(\Omega_t) = \int_{\partial \Omega_t} v \cdot \nu \; dA = \int_{\Omega_t}\nabla \cdot v\; dV > -V(\Omega_t)$$ where we used the divergence theorem and the given bound on the divergence. Comparing to the ODE $dV/dt = -V$ (using e.g. Gronwall's inequality) we conclude that $V(\Omega_t) > V(\Omega_0) e^{-t}$; so $$V(\Omega_1) > \frac 1 e V(\Omega_0) =\frac 4 3 \frac \pi e>\frac 4 3.$$

• Have you used $u_t + \nabla u \cdot v = 0$ or taken the support of $x \mapsto u(x-v(x))$? Are these equivalent? Oct 9 '17 at 7:49
• @md2perpe: The former - as the OP wrote, we know the solution of this transport PDE is given by $u(x,t) = u \circ \Phi^v_{-t}(x).$ They are certainly not equivalent - not sure where your $u(x-v(x))$ is coming from, but this isn't even time-dependent. (If you're implying that the solution is $u(x,t)=u(x-tv(x),0)$, this is only true when $v$ is constant.) Oct 9 '17 at 7:59
• Upon review, this is essentially the same as your answer - the answer to your second question is yes, $u$ must maintain values in $[0,1]$ (in fact in $\{0,1\}$. Oct 9 '17 at 8:11
• As you can see from the comments and the history, the post has been edited a few times. My $u(x-v(x))$ was a comment to the original text that said "the support of $u(x-v)$. Oct 9 '17 at 8:39
• Ah, that explains it. Oct 9 '17 at 9:17