Bardos-Degond’s solutions to Vlasov-Poisson

This is part of my lecture notes on Kinetic Theory of Gases, taught at Penn State last semester, Fall 2017. In this part, I’d like to introduce this nice Bardos-Degond 1985’s global solutions to the Vlasov-Poisson system. Of course, the global smooth solutions are already constructed, without any restriction on size of initial data (e.g., Pfaffelmoser, Schaeffer ’91; see also the previous lecture), however they give no information on their asymptotic behavior at large time. Now, for initial data that are sufficiently small near zero, Bardos and Degond were able to construct global smooth solutions that decay in large time. To my knowledge, this was the first result where dispersion is rigorously shown for kinetic equations (they appear to be motivated by similar results for nonlinear wave equations where dispersion was (still is) the key to deduce the global behavior at the large time; e.g., Klainerman, Ponce, Shatah, among others, in the early 80s).

Precisely, we consider the Vlasov-Poisson system:

$\displaystyle \begin{aligned} \partial_t f + v \cdot \nabla_x f + E \cdot \nabla_v f & =0, \\ E = -\nabla \phi, \qquad -\Delta \phi &=\rho, \end{aligned} \ \ \ \ \ (1)$

on ${\mathbb{R}^3 \times \mathbb{R}^3}$ , with ${\rho = \int_{\mathbb{R}^3} f\; dv}$ denoting the charge density. The starting point in constructing Bardos-Degond’s solutions is the dispersive estimate for the free transport

$\displaystyle \partial_t f + v\cdot \nabla_x f =0$

posed on ${\mathbb{R}^3\times \mathbb{R}^3}$ . Indeed, the solution reads ${f(x,v,t) = f_0(x-vt, v)}$ and hence

$\displaystyle \begin{aligned} \rho(x,t) & = \int_{\mathbb{R}^3} f_0(x-vt,v) \; dv = \frac{1}{t^3}\int_{\mathbb{R}^3} f_0 (y, \frac{x-y}{t})\; dy \\&\le \frac{1}{t^3} \int_{\mathbb{R}^3} \sup_v f_0(x,v) \; dx . \end{aligned}\ \ \ \ \ (2)$

This implies that as long as ${\sup_v f_0(x,v)}$ is integrable, the density satisfies ${\| \rho (t)\|_{L^\infty} \lesssim t^{-3}}$ . Here and in what follows, we consider the time ${t\ge 1}$ , for otherwise the density is already bounded by ${\|f_0\|_{L^\infty_xL^1_v}}$ . In addition, a direct computation also yields ${\|\partial_x^k \rho(t)\|_{L^\infty} \lesssim t^{-3-k}}$ , for ${k\ge 0}$ , under some extra regularity assumption on initial data.

Going back to the Vlasov-Poisson system (1), once the density is shown to decay at order ${t^{-3}}$ , the standard elliptic estimates yield decaying for the electric field; namely,

$\displaystyle \|E(t)\|_{L^\infty} \lesssim \|\rho(t)\|_{L^\infty}^{2/3}\|\rho(t)\|_{L^1}^{1/3} \lesssim t^{-2}.\ \ \ \ \ (3)$

On the other hand, the standard elliptic estimates yield that ${\|\nabla_xE(t)\|_{L^\infty}}$ decays essentially as fast as ${\|\rho(t)\|_{L^\infty}}$ , possibly up to some logarithmic growth in time: precisely,

$\displaystyle \| \nabla E \|_{L^\infty} \lesssim \|\rho\|_{L^\infty} \log (2+ \|\rho\|_{L^1}^{1/3}+ \|\nabla \rho\|_{L^\infty}) \lesssim t^{-3}.\ \ \ \ \ (4)$

The decay will be sufficient to apply the standard nonlinear iteration, yielding the global solution to the Vlasov-Poisson system.

Theorem 1 (Bardos-Degond ’85) There exists a positive constant ${\epsilon_0}$ so that for any initial data ${f_0}$ satisfying

$\displaystyle \|f_0\|_{L^1_{x,v}} + \sup_x\| D_{x,v}^kf_0(x,\cdot)\|_{L^1_v}+ \sup_v\| D_{x,v}^kf_0(\cdot, v)\|_{L^1_x} \le \epsilon_0$

for ${k=0,1}$ , smooth solutions ${(f(t), E(t))}$ to the Vlasov-Poisson system (1) exist globally in time. In addition, there is some universal constant ${C_0}$ so that

$\displaystyle (1+t)^3\| \rho(t)\|_{W^{1,\infty}_x} + (1+t)^{2}\|E(t)\|_{L^\infty} + (1+t)^3\|\nabla_xE (t)\|_{L^\infty} \le C_0 \epsilon_0,$

for all ${t\ge 0}$ .

In what follows, I give a proof of the Bardos-Degond’s theorem.

1.1. Estimates on characteristics

Recall the particle trajectories ${(X(s;t,x,v),V(s;t,x,v))}$ solving

$\displaystyle \dot X = V , \qquad \dot V = E(X(s),s),$

with ${(X(t;t,x,v),V(t;t,x,v)) = (x,v)}$ . Assuming a priori that ${E}$ is sufficiently small and decays as in (3), we show that the particle trajectories are not far from those from the free transport. To keep track of the iteration, let us introduce

$\displaystyle \epsilon(t): = \sup_{0\le s\le t}(1+s)^{3}\| \nabla_xE(s)\|_{L^\infty} .\ \ \ \ \ (5)$

Lemma 2 There is a universal constant ${C_0}$ so that for any ${s,t,x,v}$ , as long as ${ \epsilon(t)}$ remains sufficiently small, there hold

$\displaystyle \begin{aligned} | \nabla_x X(s;t,x,v) - \mathrm{Id} | + | \nabla_x V(s;t,x,v)| &\le C_0 \epsilon(t) , \\ | \nabla_v X(s;t,x,v) - (s-t)\mathrm{Id} | + | \nabla_v V(s;t,x,v)| &\le C_0 (t-s)\epsilon(t) . \end{aligned}$

In particular, there holds

$\displaystyle | \mathrm{det} (\nabla_v X(s;t,x,v) )|\ge (t-s)^3(1 - C_0 \epsilon(t)),\ \ \ \ \ (6)$

for any ${0\le s\le t}$ .

Proof: Fix ${(t,x,v)}$ . For any ${0\le s\le t}$ , we compute

$\displaystyle \frac{d^2}{ds} \nabla_x X(s) = \nabla_x E(X(s),s) \nabla_x X(s)$

with ${ \nabla_x X(t) = \mathrm{Id}}$ and ${\frac{d}{ds} \nabla_x X(t) = \nabla_x V(t) = 0}$ . Integrating with respect to time yields

$\displaystyle \nabla_x X(s) = \mathrm{Id} + \int_s^t \int_\tau^t \nabla_x E(X(\theta),\theta) \nabla_x X (\theta) d\theta d\tau .$

Let ${\mathcal{T}(\nabla_x X)}$ be the right hand side. Using (5), we check that

$\displaystyle | \mathcal{T} (A) |\le 1 + \epsilon(t) |A| \int_s^t \int_\tau^t (1+\theta)^{-3}d\theta d\tau \le 1 + C_0 \epsilon(t) |A|$

for some universal constant ${C_0}$ . That is, ${\mathcal{T}}$ maps from ${L^\infty(\mathbb{R}_+)}$ to itself and it is contractive for ${\epsilon}$ sufficiently small. This proves the claimed estimate for ${ \nabla_x X(s)}$ . The estimates for the other quantities follow similarly. The lower bound (6)follows from the estimate on ${\nabla_v X}$ . $\Box$

1.2. Density etimates

We obtain the following a priori estimates on the density ${\rho(x,t)}$ .

Lemma 3 Let ${\epsilon(t)}$ be defined as in (5). Assume that ${f_0}$ and its derivatives satisfy

$\displaystyle \sup_x\| D_{x,v}^kf_0(x,\cdot)\|_{L^1_v}+ \sup_v\| D_{x,v}^kf_0(\cdot, v)\|_{L^1_x} \le \epsilon_0$

for ${k=0,1}$ and for some finite ${\epsilon_0}$ . Then, as long as ${\epsilon(t)}$ remains sufficiently small, there holds

$\displaystyle \begin{aligned} (1+t)^3\| \rho(t)\|_{W^{1,\infty}_x} &\le C_0 \epsilon_0(1+\epsilon(t)) \end{aligned}$

for ${t\ge 0}$ .

Proof: Recall that solutions to the Vlasov-Poisson system satisfy

$\displaystyle f(t,x,v) = f_0(X(0;t,x,v), V(0; t,x, v)).$

Thus, similar to (2), we compute

$\displaystyle \begin{aligned} \rho(x,t) &= \int_{\mathbb{R}^3} f_0(X(0;t,x,v), V(0; t,x, v))\; dv \\ &= \int_{\mathbb{R}^3} f_0(y, V(0; t,x, v))\; \frac{dy}{|\mathrm{det} (\nabla_v X (0;t,x,v))|} . \end{aligned}$

Using (6), we obtain

$\displaystyle | \rho(x,t)|\le C_0 t^{-3} (1 + \epsilon(t))\sup_v \| f_0(\cdot,v) \|_{L^1_x}.$

which proves the estimate for ${\rho}$ , when ${t\ge 1}$ . For ${t\le 1}$ , we have ${|\rho(x,t)|\le \sup_x\| f_0(x,\cdot)\|_{L^1_v}}$ . As for derivatives, we compute

$\displaystyle \begin{aligned} \partial_x \rho(x,t) &= \int_{\mathbb{R}^3} \Big[ \partial_x X(0) \cdot \nabla_xf_0 + \partial_x V(0) \cdot \nabla_vf_0\Big](X(0;t,x,v), V(0; t,x, v))\; dv . \end{aligned}$

Since both ${\nabla_x X}$ and ${\nabla_xV}$ are uniformly bounded from Lemma 2, the estimate for ${\partial_x \rho}$ follows similarly. $\Box$

1.3. Nonlinear iteration

Let us now introduce the nonlinear iteration. Set

$\displaystyle \zeta(t) = \sup_{0\le s\le t} \Big[ M(1+s)^3\| \rho(s)\|_{W^{1,\infty}_x} + (1+s)^{2}\|E(s)\|_{L^\infty} + (1+s)^{3}\|\nabla_xE (s)\|_{L^\infty}\Big].$

Our goal is to prove that for sufficiently large ${M}$ and sufficiently small ${\epsilon_0}$ , there holds

$\displaystyle \zeta(t) \le C_0 ( \epsilon_0 + \zeta^2(t) ),\ \ \ \ \ (7)$

which would end the proof of Theorem 1, upon taking ${\epsilon_0}$ sufficiently small. Note that the local existence theory shows that ${\zeta(t)}$ exists for some short time and is continuous. The estimate (7) yields a global solution, upon using the standard continuous induction.

We need to check each term in the definition of ${\zeta(t)}$ . First, using Lemma 3, we have

$\displaystyle (1+s)^3\| \rho(s)\|_{W^{1,\infty}_x} \le C_0 \epsilon_0 (1 + \epsilon(t)) .$

Note that by definition (5), ${\epsilon(t) \le \zeta(t)}$ . As for the fields, the elliptic estimates yield

$\displaystyle \begin{aligned} \|E(s)\|_{L^\infty} &\le C_0 \|\rho(s)\|_{L^\infty}^{2/3}\|\rho(s)\|_{L^1}^{1/3} \le C_0 M^{-2/3}(1+s)^{-2} \| f_0\|_{L^1_{x,v}}^{1/3}\zeta^{2/3}(t) \\ &\le (1+s)^{-2} \Big[ \frac14 \zeta(t) + C_0M^{-2} \| f_0\|_{L^1_{x,v}}\Big] \\ &\le (1+s)^{-2} \Big[ \frac14 \zeta(t) + C_0M^{-2} \epsilon_0\Big] . \end{aligned}$

Next, using the elliptic estimate 4, we estimate

$\displaystyle \begin{aligned} \| \nabla_x E(s) \|_{L^\infty} &\le C_0 \|\rho(s)\|_{L^\infty} \log (2+ \|\rho(s)\|_{L^1}^{1/3}+ \|\nabla_x \rho(s)\|_{L^\infty}) \\ &\le C_0M^{-1} (1+s)^{-3} \zeta(t) \log (2 + \zeta(t)) \\ &\le C_0 M^{-1} (1+s)^{-3} \zeta(t) (1 + \zeta(t)) . \end{aligned}$

Thus, taking ${M}$ large so that ${C_0M^{-1}\le \frac14}$ and ${\epsilon_0}$ small so that ${C_0 \epsilon_0 \le \frac14}$ , we obtain (7) at once. The Bardos-Degond’s global solutions follow.

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Landau damping for screened Vlasov-Poisson on the whole space | Snapshots in Mathematics ! says:

September 9, 2019 at 10:34 am

[…] Introducing the change of variable and recalling that the map is a diffeomorphism, one essentially reduces to the similar calculation as done for the free transport. I also discuss in somewhat details the proof here on this blog post. […]

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Bardos-Degond’s solutions to Vlasov-Poisson

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