1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ I suppose that it is the support of $x \mapsto u(x-v(x))$ that is intended. $\endgroup$
    – md2perpe
    Aug 14 '17 at 21:16
  • $\begingroup$ Yes thank you, I will fix it $\endgroup$
    – Merkh
    Aug 14 '17 at 21:47
  • $\begingroup$ Have you got a solution to this? $\endgroup$
    – md2perpe
    Sep 2 '17 at 18:04
  • $\begingroup$ Nope not yet :/ $\endgroup$
    – Merkh
    Sep 3 '17 at 18:54
  • $\begingroup$ 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. $\endgroup$
    – Merkh
    Sep 11 '17 at 18:28
1
$\begingroup$

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 $m(\operatorname{supp}_x u(x, 1)) > \frac43.$

$\endgroup$
0
$\begingroup$

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 $$V(\Omega_1) > \frac 1 e V(\Omega_0) =\frac 4 3 \frac \pi e>\frac 4 3.$$

$\endgroup$
5
  • $\begingroup$ Have you used $u_t + \nabla u \cdot v = 0$ or taken the support of $x \mapsto u(x-v(x))$? Are these equivalent? $\endgroup$
    – md2perpe
    Oct 9 '17 at 7:49
  • $\begingroup$ @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.) $\endgroup$ Oct 9 '17 at 7:59
  • $\begingroup$ 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 $\{0,1\}$. $\endgroup$ Oct 9 '17 at 8:11
  • $\begingroup$ 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 $u(x-v)$. $\endgroup$
    – md2perpe
    Oct 9 '17 at 8:39
  • $\begingroup$ Ah, that explains it. $\endgroup$ Oct 9 '17 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.