Given the hyperbolic Vlasov equation $$ \frac{\partial f }{\partial t} +v\nabla_x f + F(t,x)\nabla_vf =0$$ where $f=f(t,x,v)$ and $(t,x,v)\in \mathbb{ R}\times\mathbb{R}^{n}\times \mathbb{R}^{n} $. I wonder how can be proved that $$ \Vert f(t,x,v)\Vert_{L^p(\mathbb{R}^{2n})} = \Vert f(0,x,v)\Vert_{L^p(\mathbb{R}^{2n})}, \quad p\in [1,\infty] $$ Any hint is welcome. Thank in advance.

  • $\begingroup$ I see you've posted a bunch of PDE questions; I'd suggest including the arXiv tag ap.analysis-of-pdes to them if you do so in the future. $\endgroup$ Oct 28 '19 at 19:36
  • $\begingroup$ Thanks for that suggestion, I will add that tag for future questions. $\endgroup$ Oct 28 '19 at 19:42

The "one phrase answer" is "divergence theorem".

Slightly wordier but a bit formally (for ease of typing I write $dz = dx~dv$ for the volume on phase space)

$$ \partial_t \int f^p dz = \int \partial_t f^p dz $$


$$ 0 = \int \nabla_x \cdot (vf^p) dz $$

assuming $f$ decays suitably fast at infinity, and similarly

$$ 0 = \int \nabla_v \cdot (Ff^p) dz $$


$$ \partial_t f^p + \nabla_x \cdot (vf^p) + \nabla_v \cdot (F f^p) = p f^{p-1} \left[ \partial_t + v\cdot \nabla_x + F \cdot \nabla_v \right]f = 0 $$

($\nabla_x$ trivially acts on $v$ and $\nabla_v$ trivially acts on $F(t,x)$) and the result follows.

To be precise, one has to interpret $|f|^p = \lim_{\epsilon\to 0} (\sqrt{\epsilon^2 + |f|^2}- \epsilon)^{p}$, and approximate your $f\in L^p$ with $f \in \mathcal{S}\cap L^p$ or similar.

For the case $p = \infty$ it suffices to notice that $f$ is constant on the integral curves of the vector field $(1, v, F(t,x))$ on $\mathbb{R}\times\mathbb{R}^n \times \mathbb{R}^n$.

Finally, something stronger is true: let $G:\mathbb{R}\to\mathbb{R}$ be smooth, then $\int G\circ f ~dz$ is invariant in time if $f$ solves Vlasov.

  • $\begingroup$ Wow! Thank you. I was really stuck... +1 $\endgroup$ Oct 28 '19 at 19:42

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