1 Introduction↩︎

In this paper, we study well-posedness of the Vlasov–Poisson equation in \(L^{2}\)-based Sobolev spaces with relatively low regularity. The equation describes evolution of a distribution \(f(t,x,v) : \mathbb{R}\times \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}\): \[\label{eq:VP}\left\{ \begin{align} \partial_{t} f + v \cdot \nabla_{x} f \pm \nabla_{x} U \cdot \nabla_{v} f = 0, & \\ \Delta_{x} U = \rho = \int_{\mathbb{R}^{n}} f dv. & \end{align} \right.\tag{1}\] We shall take \(n \ge 3\), and then \(U(t,x)\) is explicitly given by \[\begin{align} U(t,x) = \int_{\mathbb{R}^{n}} \Gamma_{n}(x-y) \rho(t,y) dy, \qquad \Gamma_{n}(x) = \frac{1}{n(2-n)\omega_{n}} |x|^{2-n} \end{align}\] where \(\omega_{n}\) is the volume of the unit \(n\)-ball. Moreover, we consider both signs \(\pm\) in 1 : the positive sign corresponds to collision-less plasma dynamics and the negative to astrophysical situations, most notably galaxy dynamics ([1]–[3]).

Our main result gives local well-posedness of solutions to 1 in \(H^{s}\) spaces.

1.1 Motivation for studying low regularity problems↩︎

One motivation for having well-posedness of 1 in low regularity spaces is that it provides a framework for studying the evolution of “singular structures” present in the initial data. For instance, Theorem 1 allows for data having algebraic singularities \(f_{0} \sim |x|^{-\alpha}\) for some \(\alpha>0\). This point can be illustrated by the extreme case when \(f\) is given as a time-dependent measure on the \((x,v)\) space. This includes the case of electron sheets, which model physical situations where the plasma density is highly concentrated along a hypersurface; see [4] for a nice exposition. Even in the simplest case of one spatial dimension, such singular solutions show very interesting dynamical behavior, including singularity formation and non-uniqueness of weak solutions ([4]–[7]).

Recently, there has been a lot of progress on well/ill-posedness of compressible and incompressible Euler equations in low regularity Sobolev spaces. In the incompressible case, ill-posedness results (norm inflation and non-existence) are available once the initial velocity belongs to \(W^{s,p}(\mathbb{R}^n)\) with \(s \le \frac{n}{p} + 1\) and not better, which basically shows that there are no regularizing effects in time ([8]–[13]). The situation for the compressible case is much more complicated; not only different Sobolev regularity for the velocity, vorticity, and density can be prescribed, there are dispersive smoothing effects which give well-posedness at relatively low regularity ([14]–[17]). In several cases, instantaneous shock formation can be used to show ill-posedness sharply below such well-posedness results ([18]–[20]). Although the Vlasov-Poisson equation is closely related to these fluid systems, the corresponding low regularity problem in Sobolev spaces seems to be much less explored. There does not seem to be any ill-posedness results for the Vlasov–Poisson equation in Sobolev spaces, and it is not clear whether the threshold \(s> n/2 - 1/4\) in Theorem 1 is sharp. This is because there are no examples in the literature showing the sharpness of the gain of a \(1/4\)-order derivative in Lemma 1 in the time-dependent case, to the best of our knowledge.

1.2 Previous works on well-posedness↩︎

Let us now review various works on well-posedness of 1 , focusing on the whole space case \(\mathbb{R}^n \times \mathbb{R}^n\). A more extensive list of references, as well as history, can be found in the textbooks [1], [21], [22].

The classical paper of Horst [2] proves local well-posedness for \(C^1\) data in the whole space with any \(n\ge 1\), with global well-posedness for dimensions 1 and 2. Global well-posedness for classical solutions in \(n = 3\) was obtained in [23]–[25], see [26], [27] for the case of bounded domains. For \(n \ge 4\), classical solutions can blow up in finite time ([3]). Recent works by Chen and He ([28]–[30]) give local well-posedness in (weighted in \(v\)) Sobolev, Bessel potential, and Besov spaces. In the case of \(L^{2}\)-based spaces, they require \(s> \frac{n}{2}+1\). Their Besov local well-posedness works in \(B^{s}_{p, r}(\mathbb{R}^n)\), where \(s>\frac{n}{p}+1\), \(1<p<\frac{n}{2}\), \(n\geq 3\), and \(1<r<\infty\). They point out that this well-posedness result covers some unbounded data.

On the other hand, it has been well-known since [31] that global existence of weak solutions holds with little regularity assumptions on the data, see also [24], [32], [33] for the theory of weak solutions. Regarding the question of uniqueness for weak solutions, Loeper [34] gave a very nice criterion which states that \(\rho \in L^{\infty}([0,T);L^\infty(\mathbb{R}^{n}))\) ensures uniqueness up to time \(T\). In this result, regarding \(f\), it is sufficient to assume that \(f_{0}\) is a non-negative bounded measure on \(\mathbb{R}^3\times\mathbb{R}^3\). This uniqueness criterion was strengthened in several directions. Among those, [35] showed that for \(n = 1\), BV regularity is sufficient for well-posedness. Moreover, [36], [37] extended the uniqueness criterion to \(\rho\) belonging to the so-called Yudovich spaces, which allows the \(L^p\) norm of \(\rho\) to grow at some rate as \(p\to\infty\).

Lastly, we remark that the proof of uniqueness in our Theorem 1 is also based on the idea of Loeper. The main difference is that in our case, \(\rho_{0}\) does not necessarily belong to \(L^{\infty}(\mathbb{R}^n)\) (although it does for \(t>0\) by velocity averaging). For clarity, we provide a proof of this extension in the Appendix.

1.3 Organization of the paper↩︎

The rest of the paper is organized as follows. In §2, we obtain the necessary \(H^{s}\) a priori estimate for the solution. Then we prove the main result by proving existence and uniqueness in §3 and §4, respectively. In the Appendix, two technical results are proved, namely the velocity averaging lemma ([38], [39]) and a lemma on the existence and uniqueness of flows generated by \(L^2([0,T];C^{1+\alpha})\) vector fields.

Acknowledgments. We thank Seok-Bae Yun for helpful discussions. IJ and ST have been supported by the Samsung Science and Technology Foundation under Project No. SSTF-BA2002-04.

2 A Priori Estimate↩︎

In this section, we assume the existence of a solution to 1 with sufficient regularity and an \(H^{s}\) a priori estimate for the solution. We shall only consider \(t \ge 0\), since \(t<0\) can be treated in the same way. Rigorous proof of existence and uniqueness will be given in the next section. Moreover, we shall restrict ourselves to the case \(s\in(\frac{n}{2}-\frac{1}{4}, \frac{n}{2}+1]\), since the range \(s>\frac{n}{2}+1\) was already covered by [30] and is simpler.

The rest of this section is devoted to the proof of this proposition. To begin with, we observe that \(L^1\) and \(L^2\) estimates follow immediately since the characteristic vector field \((v, -\nabla_x U(x))\) is divergence-free in \(\mathbb{R}^n\times\mathbb{R}^n\) and thus \(L^1\) and \(L^2\) norms are conserved: \[\Vert f \Vert_{L^1_{x, v}}=\Vert f_0 \Vert_{L^1_{x, v}}\quad \text{and} \quad\Vert f \Vert_{L^2_{x, v}}=\Vert f_0 \Vert_{L^2_{x, v}}.\] Hence, it suffices to estimate the homogeneous Sobolev norm \(\dot{H}^{s}\). To this end, we take the Fourier transform in the \((x, v)\) variables. The dual variables for \(v\) and \(x\) are denoted by \(\mu\) and \(\xi\), respectively. This gives \[\begin{align} \partial_t \widehat{f}(t, \mu, \xi)+\nabla_{\mu}\cdot\xi\widehat{f}(t, \mu, \xi)-\int \widehat{\nabla U}(\xi-\eta)\cdot i\mu\widehat{f}(t, \mu, \eta)d\eta=0, \end{align}\] and we obtain \[\begin{align} &\frac{1}{2}\frac{d}{dt}\iint\left( \vert \mu \vert^{2s}+\vert \xi \vert^{2s} \right)\left\vert \widehat{f}(t, \mu, \xi) \right\vert^2 d\xi d\mu =-\text{Re}\iint \left(\vert \mu \vert^{2s}+\vert \xi \vert^{2s}\right) \overline{\widehat{f}}\xi\cdot\nabla_{\mu}\widehat{f}d\xi d\mu\\&\qquad +\text{Re} \iint \left(\vert \mu \vert^{2s}+\vert \xi \vert^{2s}\right) \overline{\widehat{f}}\int \widehat{\nabla U}(\xi-\eta)\cdot i\mu\widehat{f}(t, \mu, \eta)d\eta d\xi d\mu =:I+II \end{align}\] We estimate the term \(I\) by using integration by parts: \[\begin{align} \left\vert I \right\vert&=\left\vert \frac{1}{2}\iint \left( \vert \mu \vert^{2s}+\vert \xi \vert^{2s} \right) \xi\cdot \left(\overline{\widehat{f}}\nabla_{\mu}\widehat{f}+ \widehat{f}\nabla_{\mu}\overline{\widehat{f}} \right) d\xi d\mu \right\vert =\left\vert \frac{1}{2}\iint \left( \vert \mu \vert^{2s}+\vert \xi \vert^{2s} \right) \xi\cdot \nabla_\mu \vert \widehat{f} \vert^2 d\xi d\mu \right\vert\\&= \frac{1}{2}\left\vert \iint \vert\widehat{f}\vert^2\xi\cdot\nabla_\mu \left( \vert \mu \vert^{2s}+\vert \xi \vert^{2s} \right) d\xi d\mu \right\vert\leq C\iint \left( \vert \mu \vert^{2s}+\vert \xi \vert^{2s} \right) \vert\widehat{f}\vert^2 d\xi d\mu \leq C\left\Vert f \right\Vert_{\dot{H}^s_{x, v}}^2. \end{align}\] To estimate the term \(II\), we need to use a symmetry to exploit some cancellations. First, note that \[\begin{align} \operatorname{Re}\iint\vert\mu\vert^{2s}\overline{\widehat{f}(t, \mu, \xi)}\int\widehat{\nabla U}(\xi-\eta)\cdot i\mu\widehat{f}(t, \mu, \eta)d\eta d\mu d\xi=0 \end{align}\] since \[\begin{align} \iint\vert\mu\vert^{2s}\overline{\widehat{f}(t, \mu, \xi)}\int\widehat{\nabla U}(\xi-\eta)\cdot i\mu\widehat{f}(t, \mu, \eta)d\eta d\mu d\xi &=\iint\vert\mu\vert^{2s}\overline{\widehat{f}(t, \mu, \eta)}\int\widehat{\nabla U}(\eta-\xi)\cdot i\mu\widehat{f}(t, \mu, \xi)d\xi d\mu d\eta\\ & =-\overline{\iint\vert\mu\vert^{2s}\overline{\widehat{f}(t, \mu, \xi)}\int\widehat{\nabla U}(\xi-\eta)\cdot i\mu\widehat{f}(t, \mu, \eta)d\eta d\mu d\xi} \end{align}\] by interchanging \(\xi\) and \(\eta\). Furthermore, we have \[\begin{align} \operatorname{Re}\iiint\vert\xi\vert^{s}\overline{\widehat{f}(t, \mu, \xi)}\widehat{\nabla U}(\xi-\eta)\cdot i\mu\vert \eta \vert^{s}\widehat{f}(t, \mu, \eta)d\eta d\mu d\xi=0 \end{align}\] in a similar way. Thus, we have that \[\begin{align} II=\operatorname{Re}\iiint\vert\xi\vert^{s}\overline{\widehat{f}(t, \mu, \xi)}(\vert \xi \vert^s-\vert \eta \vert^{s})\widehat{\nabla U}(\xi-\eta)\cdot i\mu\widehat{f}(t, \mu, \eta)d\eta d\mu d\xi. \end{align}\] Using \(\vert \vert\xi\vert^s-\vert \eta \vert^s \vert\leq C\vert\xi-\eta\vert(\vert\xi-\eta\vert^{s-1}+\vert\eta\vert^{s-1})\), we write \(\vert II\vert\leq C(II_1+II_2)\), where \[\begin{align} II_1 := \iiint\vert\xi\vert^{s}\vert\overline{\widehat{f}(t, \mu, \xi)}\vert \xi -\eta\vert^{s}\vert\widehat{\nabla U}(\xi-\eta)\vert\vert\mu\vert\vert\widehat{f}(t, \mu, \eta)\vert d\eta d\mu d\xi, \end{align}\] \[\begin{align} II_2 := \iiint\vert\xi\vert^{s}\vert\overline{\widehat{f}(t, \mu, \xi)}\vert \xi -\eta\vert\vert\widehat{\nabla U}(\xi-\eta)\vert\vert\eta\vert^{s-1}\vert\mu\vert\vert\widehat{f}(t, \mu, \eta)\vert d\eta d\mu d\xi . \end{align}\] We first estimate \(II_2\): we have \[\begin{align} II_2&\leq \Vert \vert\xi\vert^s\widehat{f}(t) \Vert_{L^2_{\mu, \xi}}\left\Vert \int\vert \xi-\eta \vert\vert \widehat{\nabla U}(\xi-\eta) \vert\vert\eta\vert^{s-1}\vert\mu \widehat{f}(t, \mu, \eta)\vert d\eta\right\Vert_{L^2_{\mu, \xi}}\\&\leq C\Vert \vert\xi\vert^s\widehat{f}(t) \Vert_{L^2_{\mu, \xi}}\Vert\vert\xi\vert\widehat{\nabla U}(\xi)\Vert_{L^1_{\xi}}\Vert\vert\xi\vert^{s-1}\vert\mu\vert\vert\widehat{f}(t)\vert\Vert_{L^2_{\mu, \xi}} \leq C\Vert \nabla_x^2U\Vert_{H^{\frac{n}{2}+\epsilon}_x}\Vert f \Vert^2_{H^s_{x, v}} \end{align}\] for any \(\epsilon>0\), where we have used the Cauchy–Schwarz inequality, Young’s convolution inequality, and the Sobolev embedding in the first, second, and third step, respectively. For the estimate for \(II_1\), we proceed as follows: for some \(\delta>0\) sufficiently small, we define \[\begin{align} \frac{1}{q}:=\frac{1}{2}+\frac{s-1}{n}-\delta \quadand \quad \frac{1}{p}:=\frac{3}{2}-\frac{1}{q} \end{align}\] so that Young’s convolution inequality in \(\xi\) gives \[\begin{align} \left\Vert \int\vert\xi-\eta\vert^s\vert\widehat{\nabla U}(\xi-\eta)\vert\vert \mu\widehat{f}(t, \mu, \eta) \vert d\eta \right\Vert_{L^2_{\xi, \mu}}&\leq C\Vert \vert\xi\vert^s\vert\widehat{\nabla U}(\xi)\vert \Vert_{L^p_\xi}\Vert \vert\mu\vert\vert\widehat{f}(\mu, \xi)\vert \Vert_{L^q_\xi L^2_\mu}\\&\leq C\Vert (1+\vert\xi\vert^2)^{\frac{\frac{n}{2}+1+\epsilon}{2}}\vert\widehat{\nabla U}(\xi)\vert \Vert_{L^2_{\xi}}\Vert(1+\vert\xi\vert^2)^{\frac{s-1}{2}}\vert\mu\vert\vert\widehat{f}(\mu, \xi)\vert\Vert_{L^2_{\xi, \mu}}\\& \leq C\Vert \nabla_x U \Vert_{H^{\frac{n}{2}+1+\epsilon}_{x}}\Vert f \Vert_{H^s_{x, v}} \end{align}\] for some \(\epsilon>0\). Thus, we finally get the following estimate: for any \(\epsilon>0\), for some \(C\) depending on \(\epsilon\), \[\begin{align} \label{eq:apriori-1} \frac{d}{dt}\Vert f(t) \Vert_{H^s_{x, v}}\leq C(1+\Vert \nabla_xU(t)\Vert_{H^{\frac{n}{2}+1+\epsilon}_{x}})\Vert f(t) \Vert_{H^s_{x, v}} \end{align}\tag{2}\] We have to estimate \(\Vert \nabla_xU(t) \Vert_{H^{\frac{n}{2}+1+\epsilon}_x}\) in order to close the a priori estimate. We use the following version of the velocity averaging lemma:

We shall apply the lemma with \(h = \Lambda^s_x f\), where \(\Lambda^s_x\) is the Fourier multiplier operator corresponding to the function \(\vert \xi \vert^s\). Applying \(\Lambda^s_x\) to both sides of the equation for \(f\), we get \[\begin{align} \partial_t(\Lambda^s_x f)+v\cdot\nabla_x(\Lambda^s_x f)=\nabla_v\cdot\Lambda^s_x (\nabla_x U\,f). \end{align}\] Clearly, we have \[\begin{align} \Vert \Lambda^s_x f\Vert_{L^2([0, T]; L^2_{x, v})}\leq CT^{\frac{1}{2}}\Vert f\Vert_{L^\infty([0, T]; H^s_{x, v})}, \quad \Vert \Lambda^s_x (\nabla_x U f) \Vert_{L^2([0, T]; L^2_{x, v})}&\leq CT^{\frac{1}{2}}\Vert \nabla_x U f \Vert_{L^\infty([0, T];L^2_v H^s_x)}. \end{align}\] At this stage, we recall the following Kato–Ponce estimate ([40], [41]).

Using this lemma, we obtain the following estimate: \[\begin{align} \Vert\Lambda^s_x(\nabla_x U\,f(t, \cdot, v))\Vert_{L^2(\mathbb{R}^n)}\leq C(\Vert \nabla_x U \Vert_{L^\infty(\mathbb{R}^n)}\Vert \Lambda^s_x f(t, \cdot, v) \Vert_{L^2(\mathbb{R}^n)}+\Vert f(t, \cdot, v) \Vert_{L^p(\mathbb{R}^n)}\Vert \Lambda^{s+1}_xU \Vert_{L^q(\mathbb{R}^n)}), \end{align}\] where \(\frac{1}{2}=\frac{1}{p}+\frac{1}{q}\) and \(p, q\) will be chosen as follows: first, choose \(p\) such that the following embedding holds: \[H^s(\mathbb{R}^n)\hookrightarrow L^p(\mathbb{R}^n)\] This is satisfied whenever \(-\frac{n}{p}\leq s-\frac{n}{2}\), that is, \(\frac{1}{p}\geq \frac{1}{2}-\frac{s}{n}\). Furthermore, we want the following embedding to hold: \[H^{s+1}(\mathbb{R}^n)\hookrightarrow W^{s, q}(\mathbb{R}^n)\] which is satisfied if \(s-\frac{n}{q}\leq s+1-\frac{n}{2}\). Using the relationship between \(p\) and \(q\), this is equivalent to \(p\geq n\). Choosing \(p:=2s+2\), all the conditions are satisfied since \[2s+2>n+\frac{3}{2}>n \quad and \quad \frac{1}{2}-\frac{s}{n}<\frac{1}{2}-\frac{s}{2s+\frac{1}{2}}<\frac{1}{2}-\frac{s}{2s+2}=\frac{1}{2s+2}=\frac{1}{p}.\] Using the embedding relations, we obtain \[\begin{align} \Vert\Lambda^s_x(\nabla_x U\,f(t, \cdot, v))\Vert_{L^2_x}&\leq C(\Vert \nabla_x U \Vert_{L^\infty_x}\Vert \Lambda^s_x f \Vert_{L^2_x}+\Vert f \Vert_{L^p_x}\Vert \Lambda^{s+1}_xU \Vert_{L^q_x}) \leq C\Vert \nabla_xU \Vert_{H^{s+1}_x}\Vert f(t, \cdot, v) \Vert_{H^s_x}. \end{align}\] By elliptic regularity, we have \(\Vert \nabla_x^2 U \Vert_{H^{s}_x}\leq C\Vert \rho \Vert_{H^{s}_x}.\) Moreover, since \(\rho\in L^1\cap L^2\subset L^\frac{2n}{n+2}\), we have \[\nabla_x U\in \dot{W}^{1, \frac{2n}{n+2}}\hookrightarrow L^2.\] Thus, we have the bound \(\Vert\nabla_xU\Vert_{L^2}\leq C\Vert \rho \Vert_{L^{\frac{2n}{n+2}}}\leq C(\Vert\rho\Vert_{L^1_x}+\Vert\rho\Vert_{H^s_x})\leq C(\Vert f_0\Vert_{L^1_{x, v}}+\Vert\rho\Vert_{H^s_x})\) and hence \[\Vert\nabla_xU\Vert_{H^{s+1}_x}\leq C(\Vert f_0\Vert_{L^1_{x, v}}+\Vert\rho\Vert_{H^s_x}).\] Now, we show the estimate for \(\rho\) as follows. \[\begin{align} \Vert\rho\Vert^2_{H^s_x}&=\left\Vert\int_{\vert v\vert\leq Q} f(t, \cdot, v)dv\right\Vert^2_{H^s_x}\leq CQ^n \int_{\vert v \vert\leq Q}\Vert f(t, \cdot, v)\Vert^2_{H^s_x}dv=CQ^n \int_{\vert v\vert\leq Q}\int \vert \mathcal{F}_xf(t, \xi, v) \vert^2 (1+\vert\xi\vert^2)^{\frac{s}{2}} d\xi dv \\&\leq CQ^n \iint \vert \mathcal{F}_{x, v}f(t, \xi, \mu) \vert^2 (1+\vert\xi\vert^2)^\frac{s}{2} d\xi d\mu \leq CQ^n \iint \vert \mathcal{F}_{x, v}f(t, \xi, \mu) \vert^2 (1+\vert\xi\vert^2+\vert\mu\vert^2)^\frac{s}{2} d\xi d\mu\\& \leq CQ^n\Vert f(t, \cdot, \cdot)\Vert_{H^s_{x, v}}^2 \end{align}\] Thus, we conclude that \[\begin{align} \Vert\Lambda^s_x(\nabla_x U\,f(t, \cdot, v))\Vert_{L^2_x}\leq C\left(\Vert f_0 \Vert_{L^1_{x, v}}+Q^\frac{n}{2}\Vert f(t, \cdot, v)\Vert_{H^s_x}\right)\Vert f(t\cdot, \cdot) \Vert_{H^s_{x, v}}, \end{align}\] and by integrating with respect to \(v\) and taking the supremum with respect to \(t\) gives \[\begin{align} \Vert\Lambda^s_x(\nabla_x U\,f)\Vert_{L^\infty([0, T];L^2_{x, v})}\leq CQ^\frac{n}{2}T^\frac{1}{2}\left(\Vert f_0 \Vert_{L^1_{x, v}}+\Vert f\Vert_{L^\infty([0, T];H^s_{x, v})}\right)\Vert f\Vert_{L^\infty([0, T];H^s_{x, v})}. \end{align}\] We now use Lemma 1 to conclude that (by choosing \(\epsilon=s+\frac{1}{4}-\frac{n}{2}\)) \[\Vert\nabla_x^2 U\Vert_{L^2([0, T];H^{s+\frac{1}{4}}_x)} \leq C\Vert\rho\Vert_{L^2([0, T];H^{s+\frac{1}{4}}_x)}\leq CT^\frac{1}{2}(1+\sup_{t\in [0, T]}Q(t))^n\Vert f\Vert_{L^\infty([0, T];H^s_{x, v})}(1+\Vert f\Vert_{L^\infty([0, T];H^s_{x, v})}).\] Together with the fact that \(\Vert\nabla_xU\Vert_{L^{2}_x}\leq C(\Vert f_0\Vert_{L^1_{x, v}}+\Vert\rho\Vert_{H^s_x})\leq C(\Vert f_0\Vert_{L^1_{x, v}}+\Vert\rho\Vert_{H^{s+\frac{1}{4}}_x})\) gives that \[\Vert\nabla_x U\Vert_{L^2([0, T];H^{s+\frac{5}{4}}_x)} \leq CT^\frac{1}{2}(1+\sup_{t\in [0, T]}Q(t))^n(1+\Vert f\Vert_{L^\infty([0, T];H^s_{x, v})})^2.\] Now, recall the equation 2 \[\begin{align} \frac{d}{dt}\Vert f(t) \Vert_{H^s_{x, v}}\leq C(1+\Vert \nabla_x U(t)\Vert_{H^{\frac{n}{2}+1+\epsilon}_{x}})\Vert f(t) \Vert_{H^s_{x, v}}. \end{align}\] Using the Gronwall inequality, one obtains with \(R(T):=1+\sup_{t\in [0, T]}Q(t)\) that \[\begin{align} \Vert f(t) \Vert_{H^s_{x, v}}&\leq \Vert f_0 \Vert_{H^s_{x, v}} \exp\left(CT+C\int_{0}^{t} \Vert\rho(s)\Vert_{H^{\frac{n}{2}+\epsilon}} ds\right) \leq \Vert f_0 \Vert_{H^s_{x, v}} \exp\left(CT R^{n}(T) (1+\Vert f\Vert_{L^\infty([0, T];H^s_{x, v})})^2\right). \end{align}\] Defining \(F(T):=1+\sup_{t\in [0, T]}\Vert f(t)\Vert_{H^s_{x, v}}\), we have the following inequality: \[F(T)\leq (1+\Vert f_0 \Vert_{H^s_{x, v}})\exp\left(CTR^n(T)F^2(T)\right).\] On the other hand, the estimate for \(R(t)\) simply follows from the fact that \(f\) is conserved along the flow: \[\begin{align} \frac{d}{dt}R(t) &\leq \left\vert \frac{d}{dt}Q(t) \right\vert\leq C\Vert \nabla_x U(t) \Vert_{L^\infty_x}\leq C \Vert \nabla_xU(t) \Vert_{H^{\frac{n}{2}+\epsilon}_x} \\ &\leq C \Vert \nabla_xU(t) \Vert_{H^{s+1}_x} \leq CQ^\frac{n}{2}(t)\Vert f(t) \Vert_{H^s_{x, v}} \leq CR^{\frac{n}{2}}(t)F(t). \end{align}\] We will use a bootstrapping argument by assuming the continuity of \(F(T)\) and \(R(T)\) in \(T\): assume that \(F(t)\leq 4F(0)\) and \(R(t)\leq 4 R(0)\) for \(0\leq t\leq T\). Then, this implies that for \(t\in [0, T]\), \[\begin{align} F(t)\leq F(0)\exp\left( CT(4R(0))^n (4F(0))^2 \right) \leq 2F(0) \quadand\quad R(t)\leq R(0)+CT(4F(0))(4R(0))^\frac{n}{2}\leq 2R(0), \end{align}\] if we pick \(T>0\) sufficiently small depending only on \(F(0)\) and \(R(0)\). Thus, the standard bootstrapping argument works so that there exists \(T>0\) depending only on the initial data, such that \(\Vert f(t)\Vert_{H^s_{x, v}}\leq 2\Vert f_0\Vert_{H^s_{x, v}}\) and \(R(t)\leq 2R(0)\) for \(0\leq t\leq T\). This finishes the proof of a priori estimate in Proposition 3.
For later purpose, we also derive a Lipschitz-in-time estimate of \(f\). We have\[\label{eq:apriori-Lip} \begin{align} \Vert\partial_tf\Vert_{H^{-1}_{x, v}}&\leq \Vert v\cdot\nabla_xf \Vert_{ H^{-1}_{x, v}}+\Vert\nabla_xU\cdot\nabla_vf\Vert_{H^{-1}_{x, v}}\leq \Vert v\cdot\nabla_xf \Vert_{L^2_v H^{-1}_{x}}+\Vert\nabla_xU\cdot\nabla_vf\Vert_{L^2_x H^{-1}_{v}}\\& \leq Q \Vert \nabla_xf \Vert_{L^2_vH^{-1}_x}+\Vert\nabla_x U\Vert_{L^\infty_x}\Vert\nabla_vf\Vert_{L^2_xH^{-1}_v}\\& \leq \Vert f\Vert_{L^2_{x, v}}(Q+C(\Vert \rho \Vert_{H^s_x}+\Vert f_0 \Vert_{L^1_{x, v}}))\leq \Vert f\Vert_{L^2_{x, v}}(Q+C\Vert f_0 \Vert_{L^1_{x, v}}+CQ^\frac{n}{2}\Vert f \Vert_{H^{s}_{x, v}}). \end{align}\tag{3}\]

3 Construction of Solution↩︎

In this section, we justify the above a priori estimate rigorously. To do this, we recall that ([1]–[3]) the Vlasov–Poisson equation has a unique local in time smooth solution for smooth and compactly supported in \(v\) initial data. Furthermore, the solution can be continued to be smooth if the solution is compactly supported in \(v\), uniformly in \(t\). This means that we can obtain a unique solution for regularized initial data, and we shall obtain the desired solution by an approximation argument.
Therefore, pick an initial data \(f_0\in H^s_{x,v}\) compactly supported in \(v\) for \(s\in (\frac{n}{2}-\frac{1}{4}, \frac{n}{2}+1]\) and let \(\{f_{k0}\}_{k=1}^{\infty}\) be a smooth function with compact support such that \(f_{k0}\to f_0\) in \(H^s_{x,v}\) and in \(L^1\) as \(k\to\infty\). Let \(\{f_k\}_{k=1}^{\infty}\) be solutions to the Vlasov–Poisson equation with initial data \(f_{k0}\). Since \(f_{k0}\) are approximations of \(f_0\), we can assume that \(\Vert f_{k0}\Vert_{H^s_{x, v}}\leq 2\Vert f_0 \Vert_{H^s_{x, v}}\), \(\Vert f_{k0} \Vert_{L^1_{x, v}}\leq 2\Vert f_0\Vert_{L^1_{x, v}}\), and \(Q_k(0)\leq 2Q(0)\), where \(Q_{k}(0)\) are such that \(f_{k0}(x, v, 0)=0\) for \(\vert v\vert\geq Q_k(0)\) and analogous for \(Q(0)\). Thus, by the a priori estimates given above, there exists a time \(T>0\), depending only on the initial data (actually, \(\Vert f_0\Vert_{L^1_{x, v}}\), \(\Vert f_0\Vert_{H^s_{x, v}}\), and \(Q(0)\)) such that \(f_k(t)\) is defined for \(t\in[0, T]\), \(\Vert f_k(t)\Vert_{H^s_{x, v}}\leq 2\Vert f_{k0}\Vert_{H^s_{x, v}}\leq 4\Vert f_0\Vert_{H^s_{x, v}}\), and \(Q_k(t)\leq 2Q_k(0)\leq 4Q(0)\) for such \(t\). Thus, we have that \(f_k\) is uniformly bounded in \(L^\infty_t H^s_{x, v}\). By the Banach–Alaoglu theorem, we can extract a subsequence of \(\{f_k\}_{k=1}^\infty\), which we again denote by \(\{f_k\}_{k=1}^\infty\), such that it converges to \(f\) in \(L^\infty([0, T];\,H^s_{x, v})\) in the weak-* topology.
Next, by the a priori Lipschitz estimate for \(f\) in \(H^{-1}_{x, v}\) norm given in 3 , we see that \(\Vert \partial_tf_k \Vert_{L^\infty([0, T];\,H^{-1}_{x,v})}\) is uniformly bounded, whose bound only depends on the initial conditions. Hence, passing through a subsequence, we see that \(\partial_t f_k\) converges to \(\partial_t f\) in \(L^\infty([0, T];\,H^{-1}_{x, v} )\) in its weak-* topology.
Moreover, we have from the a priori estimate that \[\begin{align} \Vert\rho_k\Vert_{L^2([0, T];H^{s+\frac{1}{4}}_x)}\leq CT^\frac{1}{2}(1+\sup_{t\in [0, T]}Q(t))^n(1+\Vert f_k\Vert_{L^\infty([0, T];H^s_{x, v})})^2, \end{align}\] where \(\rho_k(t, x)=\int f_k(t, x, v)dv\). Therefore, for \(U_k\) satisfying \(\Delta_x U_k(t, x)=\rho_k(t, x)\), we have that \[\Vert\nabla_x U_k\Vert_{L^2_t H^{s+\frac{5}{4}}_x}\leq CT^\frac{1}{2}(1+Q(0))^n\Vert f_0\Vert_{H^s_{x,v}}(1+\Vert f_0\Vert_{H^s_{x,v}})\] and therefore, \(\nabla_x U_k\) is bounded in \(L^2([0, T];\, H^{s+\frac{5}{4}}(\mathbb{R}^n))\). Also, integrating both sides of the equation \[\begin{align} \partial_t f_k+ v \cdot \nabla_x f_k - \nabla_x U_k\cdot \nabla_v f_k=0 \end{align}\] in \(v\), we have \[\begin{align} \partial_t \rho_k(t, x) + \int_{\vert v \vert\leq Q_k(t)} v \cdot \nabla_x f_k(t, x, v) \,dv=0 \end{align}\] and thus \[\begin{align} \partial_t \nabla_x U_k(t, x) + \int_{\vert v \vert\leq Q_k(t)} v \cdot \nabla_x\Delta_x^{-1}\nabla_x f_k(t, x, v) \,dv=0. \end{align}\] Since \(\nabla_x\Delta_x^{-1}\nabla_x\) is a Calderon-Zygmund operator, it has \(L^2\) boundedness. Thus, by squaring both sides of the equation and integrating in \(t\) and \(x\) gives \[\begin{align} \Vert \partial_t \nabla_x U_k \Vert^2_{L^2([0, T];\, L^2(\mathbb{R}^n))}&\leq C \sup_{t\in[0, T]} Q^{2n+2}_k(t) \int_0^T\int_{\vert v\vert\leq Q_k(t)}\int \vert \nabla_x\Delta_x^{-1}\nabla_x f_k(t, x, v)\vert^2 dxdvdt \\&\leq C \sup_{t\in[0, T]} Q^{2n+2}_k(t) \int_{0}^{T}\iint \vert f_k(t, x, v)\vert^2 dxdvdt \leq CT Q^{2n+2}(0) \Vert f_0 \Vert_{L^2_{x, v}}^2, \end{align}\] and hence \(\partial_t\nabla_x U_k\) is bounded in \(L^2([0, T];\, L^2(\mathbb{R}^n))\).
Now, we are in a position to apply the Aubin–Lions Lemma. However, since the space is not compact, it needs a little care. First, we pick a sequence of balls \(B_M\) in \(\mathbb{R}^n\) centered at the origin with radius \(M\in\mathbb{N}\). Then, we have the inclusion: \[H^{s+\frac{5}{4}}(B_M)\hookrightarrow H^{s+\frac{1}{4}}(B_M)\hookrightarrow L^2(B_M)\] where the inclusions are compact. Thus, the Aubin–Lions Lemma implies that for every subsequence of \(\{ \nabla_x U_k \}_{k=1}^\infty\), there exists a further subsequence such that it converges strongly in \(L^2([0, T]: H^{s+\frac{1}{4}}(B_M))\) and hence in \(L^2([0, T]: C(B_M))\) by Sobolev embedding. Thus, a diagonal argument shows that there exists a subsequence of \(\{ \nabla_x U_k \}_{k=1}^\infty\) so that it converges strongly in \(L^2([0, T]: C(B_M))\) for all \(M\in\mathbb{N}\). By the weak-* convergence of \(f_k\) to \(f\), we also see that the limit is \(\nabla_x U\), where \(\Delta_x U=\rho\).
We are ready to prove the existence of a distributional solution. Pick a smooth function \(\varphi(t, x, v)\) supported in \([0, T]\times\mathbb{R}^n\times\mathbb{R}^n\). Then, we have \[\begin{align} \iiint_{[0, T]\times \mathbb{R}^n\times\mathbb{R}^n} \left[\partial_t f_k+ v \cdot \nabla_x f_k - \nabla_x U_k\cdot \nabla_v f_k\right]\varphi(t, x, v)dxdvdt =0 . \end{align}\] Integrating by parts gives \[\begin{align} \iiint [-\partial_t\varphi(t, x, v)-v\cdot\nabla_x\varphi(t, x, v)+\nabla_x U_k(t, x, v)\cdot \nabla_v\varphi(t, x, v) ]f_k(t, x, v)dxdvdt=0. \end{align}\] Now, by weak-* convergence of \(f_k\) and strong convergence of \(\nabla_x U_k\), we can take the limit \(k\to\infty\) to obtain \[\begin{align} \iiint [-\partial_t\varphi(t, x, v)-v\cdot\nabla_x\varphi(t, x, v)+\nabla_x U(t, x, v)\cdot \nabla_v\varphi(t, x, v) ]f(t, x, v)dxdvdt=0, \end{align}\] and thus \(f\) is the desired distributional solution. Finally, since we have \[\begin{align} \iiint_{[0, T]\times\mathbb{R}^n\times \mathbb{R}^n} \psi(x, v, t)f_k(x, v, t)dxdvdt=0 \end{align}\] for all \(\psi\) with compact support in \([0, T]\times\mathbb{R}^n\times \mathbb{R}^n\) whose support is disjoint from \(\{(x, v, t):\,\vert v \vert\leq 4Q(0)\}\), by weak convergence of \(f_k\) to \(f\), we have \[\begin{align} \iiint_{[0, T]\times\mathbb{R}^n\times \mathbb{R}^n} \psi(x, v, t)f(x, v, t)dxdvdt=0 \end{align}\] by setting \(\psi(x, v, t)\) to be \(\eta(t)\varphi(x, v)\) and by choosing \(\eta\) as an approximation of the Dirac delta, we see that \[\begin{align} \iint_{\mathbb{R}^n\times \mathbb{R}^n} \varphi(x, v)f(x, v, t)dxdv=0 \end{align}\] for almost every \(t\in[0, T]\), where the exceptional points may depend on \(\varphi\). However, since we know that \(L^2\) is separable, we see that this dependency can be removed by taking the countable union of exceptional points, which also has measure zero. Thus, we see that \(f\) is compactly supported in \(v\).
Finally, we show that the solution we have just constructed actually belongs to \(C([0, T];\, H^s_{x, v})\). To show this, we need to show the weak continuity in time and the continuity of the norm. The weak continuity is simply obtained from the Lipschitz estimates. Pick a test function \(\psi\). We have \[\begin{align} \left\vert\left< f(u')-f(u), \psi \right>\right\vert=\left\vert\left< \int_{u}^{u'} \partial_tf(t)dt, \psi \right>\right\vert=\int_{u}^{u'}\left\vert\left< \partial_tf(t), \psi \right>\right\vert dt \leq \vert u'-u\vert \Vert\partial_tf\Vert_{L^\infty([0, T];\, H^{-1}_{x, v})} \Vert \psi\Vert_{H^1_{x, v}} \end{align}\] where the bracket denotes the dual pairing. Thus, the weak continuity is proven. To prove the continuity of the norm, we prove the following chain of inequalities: \[\begin{align} \Vert f(t) \Vert_{H^s_{x, v}}\leq \liminf_{t'\to t} \Vert f(t') \Vert_{H^s_{x, v}}\leq \limsup_{t'\to t} \Vert f(t') \Vert_{H^s_{x, v}}\leq \Vert f(t) \Vert_{H^s_{x, v}} \end{align}\] Inequality on the left is just the consequence of weak continuity in time. To prove inequality on the right, we recall the following a priori estimate: \[\begin{align} F(T)\leq F(0)\exp\left(CtR^n(T)F^2(T)\right) \end{align}\] where \(F(t)\) and \(Q\) are given as before. Actually, with a slight care, one can prove the following:² \[\begin{align} 1+\sup_{t'\in[t-\delta, t+\delta]} \Vert f(t') \Vert_{H^s_{x, v}}\leq (1+\Vert f(t) \Vert_{H^s_{x, v}})\exp(\sqrt{\delta T} C(1+Q(0))^n(1+\Vert f_0 \Vert_{H^s_{x, v}})^2 ) \end{align}\] Sending \(\delta\to 0\) gives \[\begin{align} \limsup_{t'\to t} \Vert f(t') \Vert_{H^s_{x, v}}\leq \Vert f(t) \Vert_{H^s_{x, v}} \end{align}\] which is the desired result. Therefore, \(f\) is contained in \(C([0, T];\, H^s_{x, v})\).

4 Uniqueness of Solution↩︎

In this section, we prove the uniqueness of solutions. To do this, we first establish the fact that the solution \(f\) is transported along the characteristic flow (the flow generated by the vector field \((v, -\nabla_x U(x, t))\)). We first establish the existence and uniqueness of the flow, given the solution \(f\). We first state the following lemma. The proof is given in the appendix.

We apply this Lemma to the vector field \((v, -\nabla_x U)\in\mathbb{R}^n\times\mathbb{R}^n\). We check the conditions for the Lemma. First, \((v, -\nabla_xU)\) is continuous in \(x\) since by the a priori estimates, \(\Vert \nabla_x U \Vert_{H^{s+1}_x}\leq C \Vert \rho \Vert_{H^s_x}\leq CQ(0)^\frac{n}{2}\Vert f_0 \Vert_{H^s_{x, v}}\) and \(s+1>\frac{n}{2}+\frac{3}{4}\), thus \(\nabla_x U\) is in \(C^\frac{3}{4}(\mathbb{R}^n)\) by Sobolev embedding. The rest of the conditions are already verified in the a priori estimates, just note that \(s>\frac{n}{2}-\frac{1}{4}\).
We now show that \(f\) is transported along the characteristic flow. Indeed, let \(\Phi_t\) be the characteristic flow. Fix a fixed test function \(\varphi(t, x, v)\). We first have \[\begin{align} \iiint f(t, \Phi_t(x, v))\partial_t\varphi(t, x, v)dtdvdx=\iiint f(t, x, v)(\partial_t\varphi)(t, \Phi_t^{-1}(x, v))dtdvdx \end{align}\] where we have used the fact that \(\Phi_t\) is a measure-preserving diffeomorphism. Let \(\psi(t, x, v):=\varphi(t, \Phi_t^{-1}(x, v))\). Then, \(\varphi(t, x, v)=\psi(t, \Phi_t(x,v))\) and thus by using the chain rule, we have \[\begin{align} \partial_t\varphi(t, x, v)=(\partial\psi)(t, \Phi_t(x, v))+v\cdot\nabla_x\psi(t, \Phi_t(x, v))-\nabla_xU(t, \Phi_t(x,v))\cdot\nabla_v\psi(t, \Phi_t(x,v)), \end{align}\] and thus \[(\partial_t\varphi)(t, \Phi_t^{-1}(x, v))=(\partial_t+v\cdot\nabla_x-\nabla_xU(x, t)\cdot\nabla_v)\psi(t, x, v)\] Combining the results give \[\begin{align} \iiint f(t, \Phi_t(x, v))\partial_t\varphi(t, x, v)dtdvdx&=-\iiint f(t, x, v)(\partial_t+v\cdot\nabla_x-\nabla_xU(x, t)\cdot\nabla_v)\psi(t, x, v)dtdvdx =0 \end{align}\] since \(\psi\) is a suitable test function (\(\Phi\) is a \(C^1\) diffeomorphism) and \(f\) is the distributional solution of the Vlasov–Poisson equation \[\partial_tf+v\cdot\nabla_xf-\nabla_xU\cdot\nabla_vf=0\] This implies that \(f(t, \Phi_t(x,v))=f_0(x,v)\) for almost every \(t\), and therefore \(f\) is transported along the characteristic flow.
We now prove uniqueness. It follows the proof of Theorem 2.1 in the paper [34], with minor modifications. Suppose that there are two solutions, \(f_1\) and \(f_2\), with initial data \(f_0\). For simplicity, we assume from now on that \(f_{0} \ge 0\). It is clarified in [34] that the non-negativity assumption is not necessary at all. Let \[\Phi_i(t, x, v)=(X_i(t, x, v),\, \Xi_i(t, x, v))\] be the corresponding characteristic flow for \(i=1,2\). We aim to estimate the following quantity: \[\begin{align} P(t):=\frac{1}{2}\iint f_0(x, v) \vert \Phi_1(t, x, v)-\Phi_2(t, x, v) \vert^2 dvdx. \end{align}\] Differentiating in time, we have \[\begin{align} \frac{d}{dt}P(t)&=\iint f_0(x, v)(\Phi_1(t, x, v)-\Phi_2(t, x, v))\cdot(\partial_t \Phi_1(t, x, v)-\partial_t\Phi_2(t, x, v))\\& =\iint f_0(x, v) (X_1(t, x, v)-X_2(t, x, v))\cdot(\Xi_1(t, x, v)-\Xi_2(t, x, v))dvdx \\&\quad -\iint f_0(x, v)(\Xi_1(t, x, v)-\Xi_2(t, x, v))\cdot(\nabla_x U_1(t, X_1(t, x, v))-\nabla_xU_2(t, X_2(t, x, v))) dvdx \\& =: A_1+A_2. \end{align}\] By the Cauchy–Schwarz inequality, \(\vert A_1\vert\leq P(t)\). On the other hand, \(A_2\) is bounded by \[\begin{align} &\left[ \iint f_0(x, v)\vert \Xi_1(t, x, v)-\Xi_2(t, x, v) \vert^2 dvdx \right]^\frac{1}{2} \left[\iint f_0(x, v)\vert \nabla_x U_1(t, X_1(t, x, v))-\nabla_xU_2(t, X_2(t, x, v)) \vert^2 dvdx \right]^\frac{1}{2}\\&\leq (2P(t))^\frac{1}{2}\left[ \iint f_0(x, v)\vert \nabla_x U_1(t, X_1(t, x, v))-\nabla_xU_2(t, X_2(t, x, v)) \vert^2 dvdx \right]^\frac{1}{2}\\& \leq (2P(t))^\frac{1}{2}\left[ \iint f_0(x, v)\vert \nabla_x U_1(t, X_1(t, x, v))-\nabla_xU_2(t, X_1(t, x, v)) \vert^2 dvdx \right]^\frac{1}{2}\\& \quad + (2P(t))^\frac{1}{2}\left[ \iint f_0(x, v)\vert \nabla_x U_2(t, X_1(t, x, v))-\nabla_xU_2(t, X_2(t, x, v)) \vert^2 dvdx \right]^\frac{1}{2} \\ & =: (2P(t))^\frac{1}{2}\left([T_1(t)]^\frac{1}{2}+[T_2(t)]^\frac{1}{2}\right). \end{align}\] From [34], we can estimate \(T_1(t)\) as \[\begin{align} T_1(t)\leq 2\max\{\Vert \rho_1(t) \Vert_{L^\infty}, \Vert\rho_2(t)\Vert_{L^\infty}\}^2 P(t)\leq 4(\Vert \rho_1(t) \Vert^2_{H^{s+\frac{1}{4}}}+\Vert \rho_2(t) \Vert^2_{H^{s+\frac{1}{4}}})P(t). \end{align}\] The estimate of \(T_2(t)\) needs to be done differently from the case of [34]: we estimate \[\begin{align} T_2(t)&\leq \iint f_0(x, v) \Vert \nabla_x U_2(t) \Vert_{C^1(\mathbb{R}^n)}^2\vert X_1(t, x, v)-X_2(t, x, v) \vert^2dvdx\\& \leq \Vert\nabla_x U_2(t)\Vert_{H^{s+\frac{5}{4}}(\mathbb{R}^n)}^2 \iint f_0(x, v)\vert X_1(t, x, v)-X_2(t, x, v) \vert^2dvdx \leq C\Vert \rho_2(t) \Vert^2_{H^{s+\frac{1}{4}}(\mathbb{R}^n)}P(t). \end{align}\] Combining these estimates, we have the following inequality: \[\begin{align} \frac{d}{dt}P(t)\leq CP(t)\left(1+\Vert \rho_1(t) \Vert^2_{H^{s+\frac{1}{4}}}+\Vert \rho_2(t) \Vert^2_{H^{s+\frac{1}{4}}}\right)^\frac{1}{2}. \end{align}\] The Gronwall’s inequality gives \[\begin{align} P(t)&\leq P(0)\exp\left( CT+C\int_0^T\left( \Vert\rho_1(t')\Vert_{H^{s+\frac{1}{4}}}+\Vert\rho_2(t')\Vert_{H^{s+\frac{1}{4}}} \right)dt' \right)\\&\leq P(0)\exp\left(CT+C\sqrt{T}\left(\Vert\rho_1\Vert_{L^2([0, T];\,H^{s+\frac{1}{4}})}+\Vert\rho_2\Vert_{L^2([0, T];\,H^{s+\frac{1}{4}})}\right)\right). \end{align}\] Since \(P(0)=0\), we conclude that \(P(t)=0\). This means that \(\Phi_1(x, v, t)=\Phi_2(x, v, t)\) for all points \((x, v)\) where \(f_0(x, v)\neq 0\). Since \(f_1\) and \(f_2\) are transported along the flow, as we have seen earlier, \(f_1=f_2\), and the proof is complete.

5 Proof of the Averaging Lemma↩︎

In this section, we prove Lemma 1. To do this, we first extend \(f\) and \(g\) to be 0 outside \([0, T]\times\mathbb{R}^n\times\mathbb{R}^n\). Taking the Fourier transform with respect to \((t, v)\), we have \[\begin{align} \left\Vert \int f(\cdot, \cdot, v)dv \right\Vert^2_{L^2_tH^\frac{1}{4}_x}&\leq C\iint(1+\vert \xi \vert^\frac{1}{2})\left\vert \int\widehat{f}(\tau, \xi, v)dv \right\vert^2 d\xi d\tau \\& \leq CQ^n\Vert f \Vert^2_{L^2_{t, x, v}}+C\iint_{\vert\xi\vert\geq 1} \vert\xi\vert^\frac{1}{2} \left\vert \int\widehat{f}(\tau, \xi, v) dv \right\vert^2 d\xi d\tau. \end{align}\]

Thus, all we have to do is to estimate \(\int\widehat{f}(\tau, \xi, v) dv\) for \(\vert\xi\vert\geq 1\). To do this, we split the domain of integration into two parts by using the cutoff function \(\zeta\left(\frac{\tau+\xi\cdot v}{\vert\xi\vert^\frac{1}{2}}\right)\). Here, \(\zeta\) is a smooth function such that \(\zeta(z)=1\) for \(\vert z\vert\leq 1\), \(\zeta(z)=0\) for \(\vert z\vert\geq 2\), and \(\vert\zeta^{'}(z)\vert\leq 2\) for all \(z\in\mathbb{R}^n\). Then, \[\begin{align} \int\widehat{f}(\tau, \xi, v) dv &= \int\widehat{f}(\tau, \xi, v)\zeta\left(\frac{\tau+\xi\cdot v}{\vert\xi\vert^\frac{1}{2}}\right)dv + \int\widehat{f}(\tau, \xi, v)\left(1-\zeta\left(\frac{\tau+\xi\cdot v}{\vert\xi\vert^\frac{1}{2}}\right)\right) dv =: I_1+I_2. \end{align}\]

For \(I_1\), we use the Cauchy–Schwartz inequality: \[\begin{align} \vert I_1 \vert \leq C\Vert \widehat{f}(\tau, \xi, \cdot)\Vert_{L^2_v}\left( \int_{\vert v\vert\leq Q}\zeta^2\left( \frac{ \tau+\xi\cdot v }{\vert \xi\vert^\frac{1}{2}} \right)dv \right)^\frac{1}{2}\leq CQ^\frac{n-1}{2} \Vert\widehat{f}(\tau, \xi, \cdot)\Vert_{L^2_v}\frac{1}{\vert\xi\vert^\frac{1}{4}}. \end{align}\] This is obtained as follows: we first decompose \(v=v_{\parallel}+v_\perp\), where \(v_{\parallel}\) denotes the component of \(v\) parallel to \(\xi\) and \(v_\perp\) the component orthogonal to \(\xi\). Then, we have \[\begin{align} \int_{\vert v\vert\leq Q}\zeta^2\left( \frac{ \tau+\xi\cdot v }{\vert \xi\vert^\frac{1}{2}} \right)dv&\leq \int_{\vert\tau+\xi\cdot v_\parallel\vert\leq 2\vert\xi\vert^\frac{1}{2}}\int_{\vert v_\perp \vert\leq (Q^2-v_\parallel^2)^\frac{1}{2}}dv_\perp dv_\parallel \leq C\int_{\vert\tau+\xi\cdot v_\parallel\vert\leq 2\vert\xi\vert^\frac{1}{2}} (Q^2-v_\parallel^2)^\frac{n-1}{2} dv_\parallel\\& \leq C\int_{\vert\tau+\xi\cdot v_\parallel\vert\leq 2\vert\xi\vert^\frac{1}{2}} Q^{n-1} dv_\parallel = CQ^{n-1}\frac{1}{\vert\xi\vert^\frac{1}{2}}. \end{align}\] To estimate \(I_2\), we use the following equation, which is obtained by taking the Fourier transform to the equation \(\partial_t f+v\cdot \nabla_x f=\nabla_v\cdot g\): \[\begin{align} i(\tau+\xi\cdot v) \widehat{f}(\tau, \xi, v)=\nabla_v\cdot \widehat{g}(\tau, \xi, v) \end{align}\] Then, since the function \(\frac{1}{\tau+\xi\cdot v}\left( 1-\zeta\left( \frac{\tau+\xi\cdot v}{\vert \xi \vert^\frac{1}{2}} \right)\right)\) is smooth due to the assumption that \(\vert\xi\vert\geq 1\), we can multiply it to both sides of the equation to get: \[\begin{align} \widehat{f}(\tau, \xi, v)\left(1-\zeta\left(\frac{\tau+\xi\cdot v}{\vert\xi\vert^\frac{1}{2}}\right)\right) = \frac{-i}{\tau+\xi\cdot v}\left( 1-\zeta\left( \frac{\tau+\xi\cdot v}{\vert \xi \vert^\frac{1}{2}} \right)\right) \nabla_v\cdot\widehat{g}(\tau, \xi, v). \end{align}\] We integrate in \(v\) and then integrate by parts in the right-hand side, which is permitted by the definition of distributional derivatives. This gives \[\begin{align} \int\widehat{f}(\tau, \xi, v)\left(1-\zeta\left(\frac{\tau+\xi\cdot v}{\vert\xi\vert^\frac{1}{2}}\right)\right)dv = \int \widehat{g}(\tau, \xi, v)\cdot\nabla_v\left(\frac{i}{\tau+\xi\cdot v}\left( 1-\zeta\left( \frac{\tau+\xi\cdot v}{\vert \xi \vert^\frac{1}{2}} \right)\right)\right)dv. \end{align}\] The left-hand side is just \(I_2\), so we have by the Cauchy–Schwarz inequality that \[\begin{align} \vert I_2\vert\leq \Vert \widehat{g}(\tau, \xi., \cdot)\Vert_{L^2_v} \left\Vert \nabla_v\left(\frac{i}{\tau+\xi\cdot v}\left( 1-\zeta\left( \frac{\tau+\xi\cdot v}{\vert \xi \vert^\frac{1}{2}} \right)\right)\right) \right\Vert_{L^2(\{ \vert v\vert\leq Q \})} . \end{align}\] Thus, we need to estimate the following: \[\begin{align} \int_{\vert v\vert\leq Q} \left\vert \nabla_v\left(\frac{1}{\tau+\xi\cdot v}\left( 1-\zeta\left( \frac{\tau+\xi\cdot v}{\vert \xi \vert^\frac{1}{2}} \right)\right)\right)\right\vert^2 dv. \end{align}\] Again, we decompose \(v=v_\parallel+v_\perp\) as before and note that the integrand is independent of \(v_\perp\). Thus, we write the above expression as \[\begin{align} &\int_{\vert\tau+\xi\cdot v_\parallel\vert\leq 2\vert\xi\vert^\frac{1}{2}}\int_{\vert v_\perp \vert\leq (Q^2-v_\parallel^2)^\frac{1}{2}} \left\vert \frac{d}{dv_\parallel} \left(\frac{1}{\tau+\vert\xi\vert v_\parallel}\left( 1-\zeta\left( \frac{\tau+\vert\xi\vert v_\parallel}{\vert \xi \vert^\frac{1}{2}} \right)\right)\right) \right\vert^2 dv_\perp dv_\parallel \\&\leq CQ^{n-1} \int_{\vert\tau+\xi\cdot v_\parallel\vert\leq 2\vert\xi\vert^\frac{1}{2}} \left\vert \frac{d}{dv_\parallel} \left(\frac{1}{\tau+\vert\xi\vert v_\parallel}\left( 1-\zeta\left( \frac{\tau+\vert\xi\vert v_\parallel}{\vert \xi \vert^\frac{1}{2}} \right)\right)\right) \right\vert^2 dv_\parallel. \end{align}\] By a reparameterization, we see that the integrand is a bounded quantity, and the above expression is bounded by \(CQ^{n-1}\vert \xi \vert^{-\frac{1}{2}}\). Thus, we obtain \[\begin{align} \vert I_2\vert\leq CQ^\frac{n-1}{2}\Vert \widehat{g}(\tau, \xi., \cdot)\Vert_{L^2_v}\vert \xi \vert^{-\frac{1}{4}}. \end{align}\] Combining all together gives \[\begin{align} \left\Vert \int f(\cdot, \cdot, v)dv \right\Vert_{L^2_tH^\frac{1}{4}_x}\leq C(1+Q)^{\frac{n}{2}}\left(\Vert f\Vert_{L^2([0, T]\times\mathbb{R}^n\times\mathbb{R}^n)}+\Vert g \Vert_{L^2_{t, x, v}([0, T]\times\mathbb{R}^n\times\mathbb{R}^n)}, \right), \end{align}\] and this was what we wanted to show. This ends the proof.

6 Proof of Lemma 3↩︎

In this section, we prove Lemma 3.

References↩︎