I'm trying to prove this. First I tried to multiply the equation by $\phi(x,t)$ and use the Gronwall Lemma, but it didn't work.
Can anyone help?
Here's the problem:
Given a smooth field $u:\; \mathbb{R^n}\;\to\;\mathbb{R^n}$ consider the PDE
$$\phi_t+\langle u(x),\nabla_x\phi\rangle=0,\;t>0, \; x \in \mathbb{R^n} $$
a) Suppose $div\;u(x)=0$ for all $x \in \mathbb{R^n}$. Show that
$$\int_\mathbb{R^n}{\mid\phi(x,t)\mid^2} =\; \int_\mathbb{R^n}{f(x)^2} $$
b) Suppose $div\;u(x) \ge 1$ for all $x \in \mathbb{R^n}$. Show that
$$\int_\mathbb{R^n}{\mid\phi(x,t)\mid^2} \ge e^t \int_\mathbb{R^n}{f(x)^2} $$
I guess the initial condition is $\phi(x,0) = f(x)$.
Gronwall isn't really needed here. You should be able to work out that $$\frac{d}{dt} \int_{\mathbb R^n} |\phi(x,t)|^2 \, dx = \int_{\mathbb R^n} |\phi(x,t)|^2 \mathrm{div\, }u \, dx,\quad t > 0.$$
Let $\Phi(t) = \displaystyle \int_{\mathbb R^n} |\phi(x,t)|^2 \, dt$.
If $\mathrm{div\, }u = 0$ then $\Phi'(t) = 0$ for $t > 0$ so that $\Phi$ is constant. Your answer to (a) is incorrect.
If $\mathrm{div\, } u \ge 1$ then $\Phi'(t) \ge \Phi(t)$. Then $$[e^{-t} \Phi(t)]' = e^{-t} \Phi'(t) - e^{-t} \Phi(t) = e^{-t}( \Phi'(t) - \Phi(t)) \ge 0.$$ It follows that $e^{-t} \Phi(t) \ge \Phi(0)$ which is your stated inequality.