On the Cauchy problem of Boltzmann equation with a very soft potential

This work proves the global existence to Boltzmann equation in the whole space with very soft potential \(\gamma \in [0,d)\) and angular cutoff, in the framework of small perturbation of equilibrium state. In this article, we generalize the estimate on linearized collision operator L to the case of very soft potential and obtain the spectrum structure of the linearized Boltzmann operator correspondingly. The global classic solution can be derived by the method of strongly continuous semigroup. For soft potential, the linearized Boltzmann operator could not give spectral gap; hence, we have to consider a weighted velocity space in order to obtain algebraic decay in time.

Authors and Affiliations

1. Department of Mathematics, City University of Hong Kong, Kowloon Tong, Hong Kong

   Dingqun Deng

2. Beijing Institute of Mathematical Sciences and Applications and Yau Mathematical Science Center, Tsinghua University, Beijing, People’s Republic of China

   Dingqun Deng

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Semigroup theory

Here, we give some useful semigroup theory, one may refer to [5, 7] for more details.

Definition A.1

Let (A, D(A)) be a linear unbounded densely defined operator from Hilbert space \(H_1\) into Hilbert space \(H_2\) with domain D(A). Define

Take \(y\in D(A^*)\), since \(\overline{D(A)}=H_1\), the functional \(F_y(x):=\langle Ax,y\rangle \) has a unique bounded extension on \(H_1\). Hence, Riesz representation theorem ensures the existence of a unique \(z\in H_1\) such that \(F_y(x) = \langle x,z\rangle \). Define \(A^*y:=z\), then

The operator \(A^*(H_2\rightarrow H_1)\) is linear and is called the adjoint of A.

Theorem A.2

Let \(A(X\rightarrow X)\) be the generator of the strongly continuous semigroup \(T(\cdot )\). Suppose that \(f: [0,\infty ) \rightarrow X\) is continuously differentiable on \([0,\infty )\). Then, for each \(x\in D(A)\), there exists a unique solution to

This solution is continuously differentiable and is given by

Theorem A.3

Let (A, D(A)) be a densely defined linear operator on a Banach space X. If both A and its adjoint \(A^*\) are dissipative, then the closure \(\overline{A}\) of A generates a contraction semigroup on X.

Theorem A.4

Let (A, D(A)) be the generator of a strongly continuous semigroup \((T(t))_{t\ge 0}\) on a Banach space X satisfying \(\Vert T(t)\Vert \le Me^{\omega t}\) for all \(t \ge 0\) and some \(\omega \in \mathbb {R}\), \(M\ge 1\). If \(B\in \mathscr {L}(X)\), then \(C := A + B\) with \(D(C) := D(A)\) generates a strongly continuous semigroup \((S(t))_{t\ge 0}\) satisfying \(\Vert S(t)\Vert \le Me^{(\omega +M\Vert B\Vert )t}\) and

for all \(t \ge 0\), \(x\in X\).

Hilbert–Schmidt operator

Let \(H_1\) be a separable Hilbert space, \(H_2\) be a Hilbert space.

Definition B.1

\(T\in \mathscr {L}(H_1,H_2)\) is called a Hilbert–Schmidt operator if there exists an orthonormal basis \(\{e_n\}^\infty _{n=1}\) of \(H_1\) such that \( \sum ^\infty _{n=1}\Vert Te_n\Vert ^2_{H_2}<\infty \).

Theorem B.2

If \(T\in \mathscr {L}(H_1,H_2)\) is a Hilbert–Schmidt operator, then T is compact.

Proof

Let \(\{e_n\}^\infty _{n=0}\) be an orthonormal basis of \(H_1\) such that \(\sum ^\infty _{n=1}\Vert Te_n\Vert ^2_{H_2}<\infty \). Then, for \(x\in H_1\),

since the right-hand side is absolutely convergent. Define \(T_n:H_1\rightarrow H_2\) by \( T_n(x) = \sum ^n_{k=1}(x,e_k)Te_k\). Then, \(T_n\) has finite rank and hence is compact.

On the other hand, for \(x\in H_1\) such that \(\Vert x\Vert _{H_1}\le 1\), we have

as \(n\rightarrow \infty \). Thus, T is compact since it is the limit of a sequence of compact operators.

Theorem B.3

Suppose \(L^2(X,\mu )\) and \(L^2(Y,\lambda )\) are two separable Hilbert space. If \(k\in L^2(X\times Y,\mu \otimes \lambda )\), then

is a Hilbert–Schmidt operator.

Proof

Let \(\{e_n\}^\infty _{n=0}\) be an orthonormal basis of \(L^2(Y,\lambda )\), then so is \(\{\overline{e_n}\}^\infty _{n=0}\). Since \(k\in L^2(X\times Y,\mu \otimes \lambda )\), we have \(k(x,\cdot )\in L^2(Y,\lambda )\), for almost all \(x\in X\). Thus, \(Ke_n\) is well defined for almost all \(x\in X\) and

Interpolation

Define \(\Vert f\Vert _{L^p_\beta } = \Vert (1+|\xi |)^\beta f\Vert _{L^p({\mathbb {R}^d})}\), \(p\in [1,\infty ]\).

Theorem C.1

Let \(\beta \in \mathbb {R}\), \(\gamma \in \mathbb {R}\setminus \{0\}\), \(p\in [1,\infty ]\). Suppose T is a linear operator defined on \(L^p_{\beta +n\cdot \gamma }\) such that

for some constants \(A_n>0\), for all \(n\in \{0,1,2,\dots \}\). Let \(\theta \in (0,1)\), \(n,m\in \{0,1,2,\dots \}\), pick \(\alpha =n\cdot \theta +m\cdot (1-\theta )\). Then, T is bounded linear operator from \(L^p_{\beta +\alpha \cdot \gamma }\) to \(L^p_\beta \), with

To prove this theorem, we will need the Hadamard three-lines theorem.

Theorem C.2

Let F be an analytic function in the open strip \(S=\{z\in \mathbb {C}:0<\text {Re}\lambda <1\}\). Suppose F is continuous and bounded on \(\overline{S}\) with

for some positive constants \(A_0,A_1\). Then, for any \(\theta \in [0,1]\), if \(\text {Re}z=\theta \), we have

Proof of Theorem C.1

1. 1.

   With loss of generality, assume \(m>n\). Fix \(\theta \in (0,1)\) and let

   $$\begin{aligned} \alpha = n\cdot \theta +m\cdot (1-\theta ). \end{aligned}$$

   Notice T is well defined on \(L^p_{\beta +\alpha \gamma }\), since \(L^p_{\beta +\alpha \gamma }\subset L^p_{\beta +m\gamma }\).

2. 2.

   Let \(f(\xi )=\sum ^K_{k=1}a_ke^{i\alpha _k}\chi _{A_k}(\xi )\) be any simple complex function on \({\mathbb {R}^d}\), with \(a_k>0\), \(\alpha _k\in \mathbb {R}\), \(\{A_k\}\) are pairwise disjoint bounded measurable subsets of \({\mathbb {R}^d}\). We would like to control

   $$\begin{aligned} \Vert Tf\Vert _{L^p_\beta } = \sup _{\Vert g\Vert _{L^{p'}}\le 1, g\text { is simple}} \left| \int _{\mathbb {R}^d}Tf(\xi )g(\xi )(1+|\xi |)^\beta \,\mathrm{d}\xi \right| . \end{aligned}$$

   Write \(g=\sum ^J_{j=1}b_je^{i\beta _j}\chi _{B_j}(\xi )\), with \(b_j>0\), \(\beta _j\in \mathbb {R}\), \(\{B_j\}\) are pairwise disjoint bounded measurable subsets of \({\mathbb {R}^d}\). Suppose \(z\in \{z\in \mathbb {C}:0\le \text {Re}z\le 1\}\). Define

   $$\begin{aligned} f_z(\xi ):=\sum ^K_{k=1}a_ke^{i\alpha _k}\chi _{A_k}(\xi )\cdot (1+|\xi |)^{(\alpha -nz-m(1-z))\gamma }. \end{aligned}$$

   Then, \(f_\theta (\xi ) = f(\xi )\), \(f_0(\xi ) = f(\xi )(1+|\xi |)^{(\alpha -m)\gamma }\), \(f_1(\xi ) = f(\xi )(1+|\xi |)^{(\alpha -n)\gamma }\). Define

   $$\begin{aligned} F(z) :&= \int _{\mathbb {R}^d}T(f_z)g(\xi )(1+|\xi |)^\beta \,\mathrm{d}\xi \\&=\sum ^K_{k=1}\sum ^J_{j=1}a_kb_je^{i\alpha _k}e^{i\beta _j} \int _{\mathbb {R}^d}T(\chi _{A_k}(\xi )(1+|\xi |)^{(\alpha -nz-m(1-z))\gamma })\chi _{B_j}(\xi )(1+|\xi |)^\beta \,\mathrm{d}\xi . \end{aligned}$$

   Now, we need to check F satisfies the assumptions in three-lines Theorem C.2.

(i). Claim: F(z) is continuous and bounded in \(\{0\le \text {Re}z\le 1\}\).

Indeed,

Notice \(A_k\) is bounded, there exists an open ball B(0, R) with radius \(R>0\) such that \(A_k\subset B(0,R)\), for all \(1\le k\le K\). Then, for \(z_1,z_2\in \{0\le \text {Re}z\le 1\}\), by Hölder’s inequality and the boundedness of T, we have

as \(|z_1-z_2|\rightarrow 0\). This proves the claim.

(ii). Claim: F(z) is analytic in \(\{0<\text {Re}z< 1\}\).

Indeed, similarly, for \(z,z_0\in \{0<\text {Re}z< 1\}\), by Hölder’s inequality and the boundedness of T and noticing that \(\{A_k\}^K_{k=1}\) is uniformly bounded, we have

for some \(s,t\in (0,1)\). Notice \(z\rightarrow z_0\) implies \(t,s\rightarrow 0\) and \(A_k\) are uniformly bounded; thus, the limit \(\lim _{z\rightarrow z_0}\frac{F(z)-F(z_0)}{z-z_0}\) exists and hence F(z) is analytic.

(iii). If \(\text {Re}z = 0\), by using Hölder’s inequality and the boundedness of T, we have

If \(\text {Re}z =1\), similarly we have

Therefore, we can apply the Hadamard’s three-lines theorem. When \(\text {Re}z=\theta \),

Thus, for any simple function f in \(L^p_{\beta +\alpha \gamma }\),

Since simple functions are dense in \(L^p_{\beta +\alpha \gamma }\), we proved the theorem.

Preliminary Lemmas and Properties of L

Lemma D.1

For \(A_1,A_2>0\), \(d\ge 3\). Denote \( b=\left( \frac{\xi +\xi _*}{2}\cdot \frac{\xi _*-\xi }{|\xi _*-\xi |}\right) \frac{\xi _*-\xi }{|\xi _*-\xi |}\). Then,

1. (1)

   For \(\alpha \in (-\infty ,d)\),

   $$\begin{aligned} \int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha }e^{-A_1|\xi _*-\xi |^2-A_2|b|^2}\,\mathrm{d}\xi _* \le C\frac{1}{1+|\xi |}. \end{aligned}$$

   (72)
2. (2)

   For \(\alpha \in [0,d)\),

   $$\begin{aligned} \int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha }e^{-A_1|\xi _*|^2}\,\mathrm{d}\xi _* \le C\frac{1}{(1+|\xi |)^\alpha }. \end{aligned}$$

   (73)
3. (3)

   For \(\alpha \in [0,d)\), \(\beta \in \mathbb {R}\),

   $$\begin{aligned} \int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha (1+|\xi _*|)^\beta }e^{-A_1|\xi _*-\xi |^2-A_2|b|^2}\,\mathrm{d}\xi _* \le C\frac{1}{(1+|\xi |)^{\beta +1}} \end{aligned}$$

   (74)

Proof

1. 1.

   If \(\alpha <d\),

   $$\begin{aligned}&\int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*-\xi |^2-A_2|b|^2)\,\mathrm{d}\xi _*\\&\quad \,= \int _{\mathbb {R}^d}\frac{1}{|\xi _*|^\alpha } \exp (-A_1|\xi _*|^2-\frac{A_2(|\xi _*|^2+2\xi _*\cdot \xi )^2}{4|\xi _*|^2})\,\mathrm{d}\xi _*\\&\quad \,= \int ^\infty _0 \frac{1}{r^{\alpha -d+1}}\exp (-A_1r^2)\ \int _{\mathbb {S}^{d-1}} \exp (-\frac{A_2(r+2\xi _*\cdot \xi )^2}{4})\,d\sigma (\xi _*)\mathrm{d}r. \end{aligned}$$

   By integration over sphere,

   $$\begin{aligned}&\int _{\mathbb {S}^{d-1}} \exp (-\frac{A_2(r+2\xi _*\cdot \xi )^2}{4})\,d\sigma (\xi _*)\\&\quad = \frac{2\pi ^{\frac{d-1}{2}}}{\Gamma (\frac{d-1}{2})} \int ^1_{-1}\exp (-\frac{A_2(r+2s|\xi |)^2}{4})(1-s^2)^{\frac{d-3}{2}} ds\\&\le C \int ^1_{-1}\exp (-\frac{A_2(r+2s|\xi |)^2}{4}) ds. \end{aligned}$$

   If \(|\xi |\ge 1\), we have

   $$\begin{aligned} \int ^1_{-1}\exp (-\frac{A_2(r+2s|\xi |)^2}{4}) ds&= \frac{1}{2|\xi |}\int ^{r+2|\xi |}_{r-2|\xi |}\exp (-\frac{A_2s^2}{4})\,ds \le C\frac{1}{1+|\xi |}, \end{aligned}$$

   and if \(|\xi |\le 1\), we have

   $$\begin{aligned} \int ^1_{-1}\exp (-\frac{A_2(r-2s|\xi |)^2}{4}) ds&\le 2\le \frac{C}{1+|\xi |}. \end{aligned}$$

   These estimates are independent of r, and we obtain (72).

2. 2.

   Fix \(0\le \alpha <d\). If \(|\xi |\le 1\), we have

   $$\begin{aligned}&\int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _*\\&\quad \,=\int _{|\xi _*-\xi |>2}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _* +\int _{|\xi _*-\xi |\le 2}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _*\\&\quad \,\le \int _{|\xi _*|>1}\frac{1}{2^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _* +\int _{|\xi _*-\xi |\le 2}\frac{1}{|\xi _*-\xi |^\alpha }\,\mathrm{d}\xi _*\\&\quad \,\le C. \end{aligned}$$

   If \(|\xi |\ge 1\), notice \(|\xi _*-\xi |\le \frac{|\xi |}{2}\) implies \(|\xi _*|\ge |\xi |/2\ge |\xi _*-\xi |\), we have

   $$\begin{aligned}&\int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _*\\&\quad \,\le \Big (\int _{|\xi _*-\xi |>\frac{|\xi |}{2}} +\int _{|\xi _*-\xi |\le \frac{|\xi |}{2}}\Big )\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _*\\&\quad \,\le \int _{|\xi _*-\xi |>\frac{|\xi |}{2}}\frac{1}{(\frac{|\xi |}{2})^\alpha }\exp (-A_1|\xi _*|^2)\,\mathrm{d}\xi _*\\&\qquad \qquad +\int _{|\xi _*-\xi |\le \frac{|\xi |}{2}}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-\frac{A_1|\xi _*-\xi |^2}{2})\,\mathrm{d}\xi _*\ \exp (-\frac{A_1|\xi |^2}{8})\\&\quad \,\le C\left( \frac{1}{|\xi |^\alpha } + \exp (-\frac{A_1|\xi |^2}{8})\right) \\&\quad \,\le \frac{C}{(1+|\xi |)^\alpha }. \end{aligned}$$

   This proves (73).

3. 3.

   Notice that \(|\xi _*|\le |\xi |/2\) implies \(|\xi _*-\xi |\ge |\xi |/2\). Thus, by (72),

   $$\begin{aligned}&\int _{\mathbb {R}^d}\frac{1}{|\xi _*-\xi |^\alpha (1+|\xi _*|)^\beta }\exp (-A_1|\xi _*-\xi |^2-A_2|b|^2)\,\mathrm{d}\xi _*\\&\quad \,\le \left( \int _{|\xi _*|>\frac{|\xi |}{2}} + \int _{|\xi _*|\le \frac{|\xi |}{2}}\right) \frac{1}{|\xi _*-\xi |^\alpha (1+|\xi _*|)^\beta }\exp (-A_1|\xi _*-\xi |^2-A_2|b|^2)\,\mathrm{d}\xi _*\\&\quad \,\le \frac{1}{(1+|\xi |/2)^\beta }\int _{|\xi _*|>\frac{|\xi |}{2}}\frac{1}{|\xi _*-\xi |^\alpha }\exp (-A_1|\xi _*-\xi |^2-A_2|b|^2)\,\mathrm{d}\xi _*\\&\qquad + \int _{|\xi _*|\le \frac{|\xi |}{2}} \frac{1}{|\xi _*-\xi |^\alpha }\exp (-\frac{A_1|\xi _*-\xi |^2}{2}-A_2|b|^2)\,\mathrm{d}\xi _*\ \exp (-\frac{A_1|\xi |^2}{8})\\&\quad \,\le \frac{C}{(1+|\xi |)^\beta }\left( \frac{1}{1+|\xi |}+\exp (-\frac{A_1|\xi |^2}{8})\right) . \end{aligned}$$

   This gives (74).

Proof of Theorem 2.1

1. 1.

   Firstly using \(\mathbf {M}\mathbf {M}_* = \mathbf {M}'\mathbf {M}'_*\), we have

   $$\begin{aligned} Lf&= \int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}}\mathbf {M}^{\frac{1}{2}}_*\left( f'_*\left( \mathbf {M}^{\frac{1}{2}}\right) '+f'\left( \mathbf {M}^{\frac{1}{2}}\right) '\right) q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*\\&\qquad - \mathbf {M}^{\frac{1}{2}}\int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}}\mathbf {M}^{\frac{1}{2}}_*q(\xi -\xi _*,\theta )\,\mathrm{d}n\mathrm{d}\xi _* - f\int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}}\mathbf {M}_*q(\xi -\xi _*,\theta )\,\mathrm{d}n\mathrm{d}\xi _*. \end{aligned}$$

   It suffices to deal with the first term, which denoted by I. Consider the unit vector \(m\in \text {Span}\{\xi -\xi _*,\omega \}\) such that \(m\perp \omega \). Then, performing a changing variable from \(\omega \) to m (where one needs to use polar coordinate on \(\mathbb {S}^{d-1}\) to do the change of variable), we have

   $$\begin{aligned}&\int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}}\mathbf {M}^{\frac{1}{2}}_*\left( \mathbf {M}^{\frac{1}{2}}\right) 'f'_*q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*\\&\quad = \int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}}\mathbf {M}^{\frac{1}{2}}_*\left( \mathbf {M}^{\frac{1}{2}}\right) '_*f'q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*. \end{aligned}$$

   Thus, by Fubini’s theorem,

   $$\begin{aligned} I&= 2\int _{\mathbb {S}^{d-1}}\int _{\mathbb {R}^d}\mathbf {M}^{\frac{1}{2}}_*\left( \mathbf {M}^{\frac{1}{2}}\right) '_*f'q(\xi -\xi _*,\theta )\, \mathrm{d}\xi _*\mathrm{d}\omega \\&=2\int _{\mathbb {S}^{d-1}}\int _{\mathbb {R}^d}\mathbf {M}^{\frac{1}{2}}(\xi _*+\xi )\mathbf {M}^{\frac{1}{2}}(\xi _*+\xi -(\xi _*\cdot \omega )\omega )f(\xi +(\xi _*\cdot \omega )\omega )q(\xi _*,\theta )\, \mathrm{d}\xi _*\mathrm{d}\omega \\&=4\int _{\mathbb {S}^{d-1}}\int _{\mathbb {R}_+}\int _{\mathbb {P}_\omega }\mathbf {M}^{\frac{1}{2}}(x+r\omega +\xi )\mathbf {M}^{\frac{1}{2}}(x+\xi )f(\xi +r\omega )q(x+r\omega ,\theta )\, \mathrm{d}x\mathrm{d}r\mathrm{d}\omega , \end{aligned}$$

   where \(\mathbb {P}_\omega \) is the hyperplane in \({\mathbb {R}^d}\) that is orthogonal to \(\omega \) and contains the origin. Here, we remark that one actually needs the boundedness on K to make sure that Fubini’s theorem can be applied. Then, by change of variable \(\xi _*\mapsto \xi _*-\xi \), we have

   $$\begin{aligned} I&=4\int _{{\mathbb {R}^d}}\int _{\mathbb {P}_{\xi _*}}\mathbf {M}^{\frac{1}{2}}(x+\xi _*+\xi )\mathbf {M}^{\frac{1}{2}}(x+\xi )f(\xi +\xi _*)\frac{q(x+\xi _*,\theta )}{|\xi _*|^{d-1}}\,\mathrm{d}x\mathrm{d}\xi _*\\&=4\int _{{\mathbb {R}^d}}\int _{\mathbb {P}_{\xi _*-\xi }}\frac{1}{(2\pi )^{-\frac{d}{2}}} \exp \left( -\frac{|x+a|^2}{2}-\frac{|b|^2}{2}-\frac{|\xi _*-\xi |^2}{8}\right) \\&\quad \times \frac{f(\xi _*)q(x+\xi _*-\xi ,\theta )}{|\xi _*-\xi |^{d-1}}\,\mathrm{d}x\mathrm{d}\xi _*\\&=4\int _{{\mathbb {R}^d}}\int _{\mathbb {P}_{\xi _*-\xi }}\frac{e^{-\frac{|x|^2}{2}}}{(2\pi )^{-\frac{d}{2}}}q(x-a+\xi _*-\xi ,\theta )\,\mathrm{d}x\\&\quad \times \exp \left( -\frac{|b|^2}{2}-\frac{|\xi _*-\xi |^2}{8}\right) \frac{f(\xi _*)}{|\xi _*-\xi |^{d-1}}\,\mathrm{d}\xi _*, \end{aligned}$$

   where \(\theta \) is angle between \(x-a+\xi _*-\xi \) and \(\xi _*-\xi \). This gives the expression of \(k_1\).

2. 2.

   \(\nu \) is defined by \( \nu (\xi ) = \int _{\mathbb {R}^d}\int _{\mathbb {S}^{d-1}}\mathbf {M}^{\frac{1}{2}}_*q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*\). Recall the cutoff assumption (4), then we have

   $$\begin{aligned} \nu (\xi )&= q_0(2\pi )^{d/4}\int _{\mathbb {R}^d}\exp (-\frac{|\xi _*|^2}{4})|\xi -\xi _*|^{-\gamma }\mathrm{d}\xi _*. \end{aligned}$$

   On the one hand, by (73) in Lemma D.1, we have

   $$\begin{aligned} \nu (\xi )&= q_0(2\pi )^{d/4}\int _{\mathbb {R}^d}\exp (-\frac{|\xi _*|^2}{4})|\xi -\xi _*|^{-\gamma }\,\mathrm{d}\xi _* \le C(1+|\xi |)^{-\gamma }. \end{aligned}$$

   On the other hand, noticing \(|\xi _*-\xi |\le 1+|\xi _*|+|\xi |\le (1+|\xi _*|)(1+|\xi |)\), we have

   $$\begin{aligned} \nu (\xi )&\ge q_0(2\pi )^{d/4}\int _{\mathbb {R}^d}\exp (-\frac{|\xi _*|^2}{4})(1+|\xi _*|)^{-\gamma }\,\mathrm{d}\xi _*\,(1+|\xi |)^{-\gamma }\\&\ge C(1+|\xi |)^{-\gamma }. \end{aligned}$$

   This proves statement (i).

3. 3.

   For the part \(k_1\), we firstly estimate \(J := \int _{\mathbb {P}_{\xi _*-\xi }}e^{-\frac{|x|^2}{2}} q(x-a+\xi _*-\xi ,\theta )\,\mathrm{d}x\). Notice \(a\in \mathbb {P}_{\xi _*-\xi }\) and integral is taken in \(\mathbb {P}_{\xi _*-\xi }\), we have

   $$\begin{aligned} J&\le \int _{\mathbb {P}_{\xi _*-\xi }}e^{-\frac{|x|^2}{2}} |x-a+\xi _*-\xi |^{-\gamma } |\cos \theta |\,\mathrm{d}x\\&\le \int _{\mathbb {P}_{\xi _*-\xi }}e^{-\frac{|x|^2}{2}} |x-a+\xi _*-\xi |^{-\gamma } \frac{|(x-a+\xi _*-\xi )\cdot (\xi _*-\xi )|}{|x-a+\xi _*-\xi |\,|\xi _*-\xi |}\,\mathrm{d}x\\&= \int _{\mathbb {P}_{\xi _*-\xi }}e^{-\frac{|x|^2}{2}} \frac{|\xi _*-\xi |}{(|x-a|^2+|\xi _*-\xi |^2)^{(\gamma +1)/2}}\,\mathrm{d}x. \end{aligned}$$

   If \(|\xi _*-\xi |\le 1\), by (73) in Lemma D.1,

   $$\begin{aligned} \frac{J}{|\xi _*-\xi |}&\le \int _{\mathbb {P}_{\xi _*-\xi }}e^{-\frac{|x|^2}{2}}\frac{1}{|x-a|^{\gamma +1}}\,\mathrm{d}x \le C\frac{1}{(1+|a|)^{\gamma +1}}. \end{aligned}$$

   If \(|\xi _*-\xi |> 1\), we split \(\mathbb {P}_{\xi _*-\xi }\) as

   $$\begin{aligned} B_1=\big \{x\in \mathbb {P}_{\xi _*-\xi }\,:\,|x-a|\le \frac{|a|}{2}\big \}, \quad B_2=\big \{x\in \mathbb {P}_{\xi _*-\xi }\,:\,|x-a|> \frac{|a|}{2}\big \}. \end{aligned}$$

   Noticing \(|x-a|\le \frac{|a|}{2}\) implies \(|x|>\frac{|a|}{2}\), we have

   $$\begin{aligned} \,\frac{J}{|\xi _*-\xi |}&\le \left( \int _{B_1 }+\int _{B_2 }\right) e^{-\frac{|x|^2}{2}} \frac{1}{(|x-a|^2+|\xi _*-\xi |^2)^{(\gamma +1)/2}}\,\mathrm{d}x\\&\le C\left( e^{-\frac{|a|^2}{16}} \frac{1}{|\xi _*-\xi |^{\gamma +1}} +\frac{1}{(|a|+|\xi _*-\xi |)^{\gamma +1}}\right) \\&\le C\frac{1}{(|a|+|\xi _*-\xi |)^{\gamma +1}}, \end{aligned}$$

   by using the fact that

   $$\begin{aligned} \sup _{a\in {\mathbb {R}^d},|\xi _*-\xi |\ge 1}e^{-\frac{|a|^2}{16}} \frac{(|a|+|\xi _*-\xi |)^{\gamma +1}}{|\xi _*-\xi |^{\gamma +1}} \le {\left\{ \begin{array}{ll} e^{-\frac{|a|^2}{16}}(2|a|)^{\gamma +1}, &{}\text { if }|a|>|\xi _*-\xi |,\\ 2^{\gamma +1}, &{}\text { if }|a|\le |\xi _*-\xi |. \end{array}\right. } \end{aligned}$$

   Thus, for any \(\varepsilon \in (0,1)\),

   $$\begin{aligned} |k_1(\xi ,\xi _*)|&\le \frac{C}{|\xi _*-\xi |^{d-2}}\frac{1}{(1+|a|+|\xi _*-\xi |)^{\gamma +1}}\ \exp (-\frac{|b|^2}{2}-\frac{|\xi _*-\xi |^2}{8}),\\&\le \frac{C}{|\xi _*-\xi |^{d-2}}\frac{1}{(1+|a|+|b|+|\xi _*-\xi |)^{\gamma +1}}\ \exp (-(1-\varepsilon )(\frac{|b|^2}{2}-\frac{|\xi _*-\xi |^2}{8})),\\&\le \frac{C_{\varepsilon }}{|\xi _*-\xi |^{d-2}}\frac{1}{(1+|\xi |+|\xi _*|)^{\gamma +1}}\ \exp (-(1-\varepsilon )(\frac{|b|^2}{2}-\frac{|\xi _*-\xi |^2}{8})), \end{aligned}$$

   where we use the fact that \(|a|^2+|b|^2 = \frac{|\xi _*+\xi |^2}{4}\).

Proof of Theorem 2.5

Notice \(\nu \) is a bounded positive function and K is linear continuous integral operator with symmetric kernel \(k(\xi ,\xi _*)\), so, \(L=\nu +K\) is self-adjoint. For the non-positiveness, we can use a well-known fact that

which is valid whenever the integral is absolutely convergent. Then, for \(f\in L^2({\mathbb {R}^d})\), since K is bounded on \(L^2\) and \(\nu \in L^\infty \), we can apply this identity to get

Let \(f\in L^2\) such that \(Lf=0\). Then, \((Lf,f)_{L^2}=0\). Using (75) and the fact \(\mathbf {M}\mathbf {M}_*=\mathbf {M}'\mathbf {M}'_*\), we have for a.e. \(\xi ,\xi _*\in {\mathbb {R}^d}\), \(\omega \in \mathbb {S}^{d-1}\) that

Thus, by the theory of collision invariant, \(f\in \) Span\(\{\mathbf {M}^{\frac{1}{2}}, \xi _1\mathbf {M}^{\frac{1}{2}}, \dots , \xi _d\mathbf {M}^{\frac{1}{2}}, |\xi |^2\mathbf {M}^{\frac{1}{2}}\}\).

Proof of Lemma 3.8

1. 1.

   Recall (2) and \(\Gamma (g,h):=\mathbf {M}^{-1/2}Q(\mathbf {M}^{1/2}g,\mathbf {M}^{1/2}h)\). For \(f,g\in L^\infty _\beta ({\mathbb {R}^d})\),

   $$\begin{aligned} |\Gamma (f,g)|&\le \int _{\mathbb {R}^d}\int _{S^{d-1}} \mathbf {M}_*^{1/2}\big (|f'g'_*| + |f'_*g'|+ |fg_*|+|f_*g|\big )q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*\\&=: \Gamma _1(f,g)+\Gamma _2(f,g)+\Gamma _3(f,g)+\Gamma _4(f,g). \end{aligned}$$

   Apply conservation laws (3), we control the weight \((1+|\xi |)^{\beta }\) by \((1+|\xi |)^{\beta }(1+|\xi _*|)^{\beta }\), \((1+|\xi '|)^{\beta }(1+|\xi '_*|)^{\beta }\) correspondingly.

   $$\begin{aligned}&\Vert \Gamma (f,g)\Vert _{L^\infty _{\beta +\gamma }({\mathbb {R}^d})}\\&\quad \,\le \sup _{\xi \in {\mathbb {R}^d}}\int _{\mathbb {R}^d}\int _{S^{d-1}} \mathbf {M}_*^{1/2}|f'g'_*| + |f'_*g'|+ |fg_*| \\&\qquad +|f_*g|) (1+|\xi |)^{\beta +\gamma }q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*\\&\quad \,\le 4\Vert f\Vert _{L^\infty _\beta }\Vert g\Vert _{L^\infty _\beta } \sup _{\xi \in {\mathbb {R}^d}}(1+|\xi |)^\gamma \int _{\mathbb {R}^d}\int _{S^{d-1}} \mathbf {M}_*^{1/2}q(\xi -\xi _*,\theta )\,\mathrm{d}\omega \mathrm{d}\xi _*\\&\quad \,\le 4\nu _1\Vert f\Vert _{L^\infty _\beta }\Vert g\Vert _{L^\infty _\beta }. \end{aligned}$$

   On the other hand, \(l>\frac{d}{2}\) assures that the Sobolev space \(H^l\) is a Banach algebra. Thus, for \(f,g\in L^\infty _\beta (H^l)\), \(\Vert \Gamma _j(f,g)\Vert _{H^l}\le \Gamma _j(\Vert f\Vert _{H^l},\Vert g\Vert _{H^l})\). This proves the first assertion.

2. 2.

   For \(\beta _0>\frac{d}{2}\), we have

   $$\begin{aligned} \Vert f\Vert _{L^2_{\alpha \gamma }}&\le \Vert f\Vert _{L^\infty _{\beta _0+\alpha \gamma }}\Vert (1+|\xi |)^{-2\beta _0}\Vert _{L^2}\le C\Vert f\Vert _{L^\infty _{\beta _0+\alpha \gamma }}. \end{aligned}$$

   For \(f,g\in L^2_{\alpha \gamma }(L^1)\), \(\beta _0>\frac{d}{2}\),

   $$\begin{aligned}&\big \Vert \Gamma (f,g)\big \Vert _{L^2_{\alpha \gamma }(L^1)}\\&\quad \le C\Big \Vert \int _{\mathbb {R}^d}\int _{S^{d-1}} \mathbf {M}_*^{\frac{1}{2}}\big (\Vert f'g'_*\Vert _{L^1_x} + \Vert f'_*g'\Vert _{L^1_x}\\&\qquad + \Vert fg_*\Vert _{L^1_x} +\Vert f_*g\Vert _{L^1_x}\big ) q\,\mathrm{d}\omega \mathrm{d}\xi _*\Big \Vert _{L^\infty _{\beta _0+\alpha \gamma }}\\&\quad \le C\big \Vert \Gamma (\Vert f\Vert _{L^2_x},\Vert g\Vert _{L^2_x})\big \Vert _{L^\infty _{\beta _0+\alpha \gamma }}\\&\quad \le C \Vert f\Vert _{L^\infty _{\beta _0-\gamma +\alpha \gamma }(L^2_x)}\Vert g\Vert _{L^\infty _{\beta _0-\gamma +\alpha \gamma }(L^2_x)}. \end{aligned}$$

   If \(\beta >\frac{d}{2}-\gamma +\alpha \gamma \), then we can find \(\beta _0\in (\frac{d}{2},\beta +\gamma -\alpha \gamma )\) to assure the above inequality valid. Then,

   $$\begin{aligned} \Vert \Gamma (f,g)\Vert _{L^2_{\alpha \gamma }(L^1)}&\le C\Vert f\Vert _{L^\infty _{\beta }(L^2_x)}\Vert g\Vert _{L^\infty _{\beta }(L^2_x)} \le C\Vert f\Vert _{L^\infty _{\beta }(H^l)}\Vert g\Vert _{L^\infty _{\beta }(H^l)}. \end{aligned}$$

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