Local Well-Posedness of Vlasov–Poisson–Boltzmann Equation with Generalized Diffuse Boundary Condition

Abstract

The Vlasov–Poisson–Boltzmann equation is a classical equation governing the dynamics of charged particles with the electric force being self-imposed. We consider the system in a convex domain with the Cercignani–Lampis boundary condition. We construct a uniqueness local-in-time solution based on an $L^\infty$-estimate and $W^{1,p}$-estimate. In particular, we develop a new iteration scheme along the characteristic with the Cercignani–Lampis boundary for the $L^\infty$-estimate, and an intrinsic decomposition of boundary integral for $W^{1,p}$-estimate.

Appendix

Lemma 15

For $R(u\rightarrow v;x,t)$ given by (1.11), given any u such that $u\cdot n(x)>0$,

$$\begin{aligned} \int _{n(x)\cdot v<0}R(u\rightarrow v;x,t)dv=1. \end{aligned}$$

(6.1)

Proof

We can transform the basis from $\{n,\tau _1,\tau _2\}$ to the standard bases $\{e_1,e_2,e_3\}$. For the sake of simplicity, we assume $T_w(x)=1$. The integration over ${\mathcal {V}}_\parallel$, after the orthonormal transformation, becomes integration over ${\mathbb {R}}^2$. We have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \frac{1}{r_\parallel (2-r_\parallel )} \exp \Big (\frac{|v_\parallel -(1-r_\parallel )u_\parallel |^2}{r_\parallel (2-r_\parallel )} \Big )dv_\parallel , \end{aligned}$$

which is obviously normalized.

Then we consider the integration over ${\mathcal {V}}_\perp$, which is $e_3<0$ after the transformation. We want to show

$$\begin{aligned} \frac{2}{r_\perp }\int _{-\infty }^0 -v_\perp e^{-\frac{|v_\perp |^2}{r_\perp }}e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}I_0(\frac{2(1-r_\perp )^{1/2}v_\perp u_\perp }{r_\perp })dv_\perp =1. \end{aligned}$$

(6.2)

The Bessel function reads

$$\begin{aligned} J_0(y)=\frac{1}{\pi }\int _0^{\pi } e^{iy\cos \theta }d\theta= & {} \sum _{k=0}^\infty \frac{1}{\pi }\int _0^\pi \frac{(iy\cos \theta )^k}{k!} d\theta =\sum _{k=0}^\infty \int _0^\pi \frac{(iy\cos \theta )^{2k}}{(2k)!} d\theta \\ \sum _{k=0}^\infty \int _0^\pi \frac{(-1)^k (y)^{2k} (\cos \theta )^{2k}}{(2k)!}d\theta= & {} \sum _{k=0}^\infty (-1)^k \frac{(\frac{1}{4}y^2)^k}{(k!)^2}, \end{aligned}$$

where we use the Fubini’s theorem and the fact that

$$\begin{aligned} \int _0^\pi \cos ^{2k}\theta =\frac{\pi }{2^{2k}}\left( \begin{array}{c} 2k \\ k \\ \end{array} \right) . \end{aligned}$$

Hence

$$\begin{aligned} I_0(y)=\frac{1}{\pi }\int _0^\pi e^{i(-iy)\cos \theta }d\theta =J_0(-iy)=\sum _{k=0}^\infty \frac{(\frac{1}{4}y^2)^k}{(k!)^2},\quad I_0(y)=I_0(-y). \end{aligned}$$

(6.3)

By taking the change of variable $v_\perp \rightarrow -v_\perp$, the LHS of (6.2) can be written as

$$\begin{aligned} \frac{2}{r_\perp }\int _{0}^\infty v_\perp e^{-\frac{|v_\perp |^2}{r_\perp }}e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}I_0(\frac{2(1-r_\perp )^{1/2}v_\perp u_\perp }{r_\perp })dv_\perp . \end{aligned}$$

Using (6.3) we rewrite the above term as

$$\begin{aligned} \sum _{k=0}^\infty \frac{2}{r_\perp }\int _0^\infty v_\perp e^{\frac{-|v_\perp |^2}{r_\perp }} e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\frac{(1-r_\perp )^k v_\perp ^{2k}u_\perp ^{2k}}{(k!)^2r_\perp ^{2k}}dv, \end{aligned}$$

(6.4)

where we use the Tonelli theorem. Rescale $v_\perp =\sqrt{r_\perp }v_\perp$ we have

$$\begin{aligned}&\frac{2}{r_\perp }\int _0^\infty v_\perp e^{\frac{-|v_\perp |^2}{r_\perp }} e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\frac{(1-r_\perp )^k v_\perp ^{2k}u_\perp ^{2k}}{(k!)^2r_\perp ^{2k}}dv\nonumber \\&\quad =2\int _0^\infty v_\perp e^{-|v_\perp |^2} e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\frac{(1-r_\perp )^k v_\perp ^{2k}u_\perp ^{2k}}{(k!)^2r_\perp ^{k}}dv \nonumber \\&\quad =2\int _0^\infty v_\perp ^{2k+1}e^{-|v_\perp |^2}dv e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\frac{(1-r_\perp )^k u_\perp ^{2k}}{(k!)^2r_\perp ^{k}} \nonumber \\&\quad =2\frac{k!}{2}e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\frac{(1-r_\perp )^k u_\perp ^{2k}}{(k!)^2r_\perp ^{k}}=e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\frac{(1-r_\perp )^k u_\perp ^{2k}}{k!r_\perp ^{k}}. \end{aligned}$$

(6.5)

Therefore, the LHS of (6.2) can be written as

$$\begin{aligned} e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}\sum _{k=0}^\infty \frac{(1-r_\perp )^k u_\perp ^{2k}}{k!r_\perp ^{k}} =e^{\frac{-(1-r_\perp )|u_\perp |^2}{r_\perp }}e^{\frac{(1-r_\perp )|u_\perp |^2}{r_\perp }}=1. \end{aligned}$$

$\square$

Lemma 16

For any $a>0,b>0,\varepsilon >0$ with $a+\varepsilon <b$,

$$\begin{aligned} \frac{b}{\pi }\int _{{\mathbb {R}}^2} e^{\varepsilon |v|^2} e^{a|v|^2}e^{-b|v-w|^2}dv=\frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2}. \end{aligned}$$

(6.6)

And when $\delta \ll 1$,

$$\begin{aligned} \frac{b}{\pi }\int _{|v-\frac{b}{b-a-\varepsilon }w|>\delta ^{-1}} e^{\varepsilon |v|^2} e^{a|v|^2}e^{-b|v-w|^2}dv\le & {} e^{-(b-a-\varepsilon )\delta ^{-2}} \frac{b}{b-a-\varepsilon } e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2} \end{aligned}$$

(6.7)

$$\begin{aligned}\le & {} \delta \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2} . \end{aligned}$$

(6.8)

Proof

$$\begin{aligned}&\frac{b}{\pi }\int _{{\mathbb {R}}^2} e^{\varepsilon |v|^2} e^{a|v|^2}e^{-b|v-w|^2}dv = \frac{b}{\pi }\int _{{\mathbb {R}}^2} e^{(a+\varepsilon -b)|v|^2} e^{2bv\cdot w} e^{-b|w|^2}dv \\&\quad =\frac{b}{\pi }\int _{{\mathbb {R}}^2} e^{(a+\varepsilon -b)|v+\frac{b}{a+\varepsilon -b}w|^2} e^{\frac{-b^2}{a+\varepsilon -b}|w|^2} e^{-b|w|^2}dv\\&\quad =\frac{b}{\pi }\int _{{\mathbb {R}}^2} e^{(a+\varepsilon -b)|v|^2}dv e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2}=\frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2}, \end{aligned}$$

where we apply change of variable $v+\frac{b}{a+\varepsilon -b}w\rightarrow v$ in the first step of the last line, then we obtain (6.6).

Following the same derivation

$$\begin{aligned}&\frac{b}{\pi }\int _{|v-\frac{b}{b-a-\varepsilon }w|>\delta ^{-1}} e^{\varepsilon |v|^2} e^{a|v|^2}e^{-b|v-w|^2}dv \\&\quad =\frac{b}{\pi }\int _{|v-\frac{b}{b-a-\varepsilon }w|>\delta ^{-1}} e^{(a+\varepsilon -b)|v-\frac{b}{b-a-\varepsilon }w|^2}dv e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2}\\&\quad \le e^{-(b-a-\varepsilon )\delta ^{-2}} \frac{b}{b-a-\varepsilon } e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2} \le \delta \frac{b}{b-a-\varepsilon } e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }|w|^2}, \end{aligned}$$

thus we obtain (6.8). $\square$

Lemma 17

For any $a>0,b>0,\varepsilon >0$ with $a+\varepsilon <b$,

$$\begin{aligned} 2b\int _{{\mathbb {R}}^+}v e^{\varepsilon v^2}e^{av^2} e^{-bv^2}e^{-bw^2}I_0(2bv w)dv=\frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}. \end{aligned}$$

(6.9)

And when $\delta \ll 1$,

$$\begin{aligned} 2b\int _{0< v<\delta }v e^{\varepsilon v^2}e^{av^2} e^{-bv^2}e^{-bw^2}I_0(2bv w)dv\le \delta \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}. \end{aligned}$$

(6.10)

Proof

$$\begin{aligned}&2b\int _{{\mathbb {R}}^+}v e^{\varepsilon v^2}e^{av^2} e^{-bv^2}e^{-bw^2}I_0(2bv w)dv \\&\quad =2b\int _{{\mathbb {R}}^+} v e^{(a+\varepsilon -b)v^2}I_0(2bv w) e^{\frac{b^2}{a+\varepsilon -b}w^2} e^{\frac{b^2}{b-a-\varepsilon }w^2} dv e^{-bw^2}\\&\quad =2(b-a-\varepsilon )\int _{{\mathbb {R}}^+}v e^{(a+\varepsilon -b)v^2}I_0(2bv w)e^{\frac{(bw)^2}{a+\varepsilon -b}}dv \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}\\&\quad =\frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}, \end{aligned}$$

where we use (6.2) in Lemma 15 in the last line, then we obtain (6.9).

Following the same derivation we have

$$\begin{aligned}&2b\int _{0< v< \delta }v e^{\varepsilon v^2}e^{av^2} e^{-bv^2}e^{-bw^2}I_0(2bv w)dv \\&\quad =2(b-a-\varepsilon )\int _{0<v<\delta } v e^{(a+\varepsilon -b)v^2}I_0(2bv w)e^{\frac{(bw)^2}{a+\varepsilon -b}}dv \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}. \end{aligned}$$

Using the definition of $I_0$ we have

$$\begin{aligned} I_0(y)=\frac{1}{\pi }\int _{0}^{\pi } e^{y\cos \phi }d\phi \le e^{y}. \end{aligned}$$

Thus when $a-b+\varepsilon <0$,

$$\begin{aligned}&2(b-a-\varepsilon )\int _{0<v<\delta } v e^{(a+\varepsilon -b)v^2}I_0(2bv w)e^{\frac{(bw)^2}{a+\varepsilon -b}}dv \\&\quad \le 2(b-a-\varepsilon )\int _{0<v<\delta } v e^{(a-b+\varepsilon )v^2}e^{2v b w}e^{\frac{(bw)^2}{a-b+\varepsilon }}\\&\quad =2(b-a-\varepsilon )\int _{0<v<\delta } v e^{(a-b+\varepsilon )(v+\frac{bw}{a-b+\varepsilon })^2}dv\\&\quad \le 2(b-a-\varepsilon )\int _{0<v<\delta }vdv<\delta , \end{aligned}$$

where we use $\delta \ll 1$ in the last step, then we obtain (6.10). Then we derive (6.13). $\square$

Lemma 18

For any $m,n>0$, when $\delta \ll 1$, we have

$$\begin{aligned} 2m^2\int _{\frac{n}{m}u_\perp +\delta ^{-1}}^\infty v_\perp e^{-m^2v_\perp ^2}I_0(2mnv_\perp u_\perp )e^{-n^2u_\perp ^2}dv_\perp \lesssim e^{-\frac{m^2}{4\delta ^{2}}}. \end{aligned}$$

(6.11)

In consequence, for any $a>0,b>0,\varepsilon >0$ with $a+\varepsilon <b$,

$$\begin{aligned} 2b\int _{\frac{b}{b-a-\varepsilon }w+\delta ^{-1}}^\infty v e^{\varepsilon v^2}e^{av^2} e^{-bv^2}e^{-bw^2}I_0(2bv w)dv\le & {} e^{\frac{-(b-a-\varepsilon )}{4\delta ^2}}\frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2} \end{aligned}$$

(6.12)

$$\begin{aligned}\le & {} \delta \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}. \end{aligned}$$

(6.13)

Proof

We discuss two cases. The first case is $v_\perp >2\frac{n}{m}u_\perp$. We bound $I_0$ as

$$\begin{aligned} I_0(2mnv_\perp u_\perp )\le \frac{1}{\pi }\int _0^\pi \exp \Big ( 2mnv_\perp u_\perp \Big ) d\theta =\exp \Big (2mnv_\perp u_\perp \Big ). \end{aligned}$$

The LHS of (6.11) is bounded by

$$\begin{aligned} 2m^2\int _{\max \{2\frac{n}{m}u_\perp ,\frac{n}{m}u_\perp +\delta ^{-1}\}}^\infty ve^{-m^2(v_\perp -\frac{n}{m}u_\perp )^2}dv. \end{aligned}$$

Using $v_\perp >2\frac{n}{m}u_\perp$ we have

$$\begin{aligned} (v_\perp -\frac{n}{m}u_\perp )^2\ge (\frac{v_\perp }{2}+\frac{v_\perp }{2}-\frac{n}{m}u_\perp )^2 \ge \frac{v_\perp ^2}{4}. \end{aligned}$$

Thus we can further bound LHS of (6.11) by

$$\begin{aligned} 2m^2\int _{\max \{2\frac{n}{m}u_\perp ,\frac{n}{m}u_\perp +\delta ^{-1}\}}^\infty v_\perp e^{-\frac{m^2 v_\perp ^2}{4}} dv_\perp \lesssim e^{-\frac{m^2}{4\delta ^2}}. \end{aligned}$$

The second case is $0\le v_\perp \le 2\frac{n}{m}u_\perp$. Since $\frac{n}{m}u_\perp +\delta ^{-1}<v_\perp$, without loss of generality, we can assume $u_\perp >\delta ^{-1}$. We compare the Taylor series of $v_\perp I_0(2mnv_\perp u_\perp )$ and $\exp \Big (2mnv_\perp u_\perp \Big )$. We have

$$\begin{aligned} v_\perp I_0(2mnv_\perp u_\perp )=\sum _{k=0}^\infty \frac{m^{2k}n^{2k}v_\perp ^{2k+1}u_\perp ^{2k}}{(k!)^2}, \end{aligned}$$

(6.14)

and

$$\begin{aligned} \exp \Big (2mnv_\perp u_\perp \Big )=\sum _{k=0}^\infty \frac{2^k m^k n^k v_\perp ^k u_\perp ^k}{k!}. \end{aligned}$$

(6.15)

We choose $k_1$ such that when $k>k_1$, we can apply the Sterling formula such that

$$\begin{aligned} \frac{1}{2}\le |\frac{k!}{k^ke^{-k}\sqrt{2\pi k}}|\le 2. \end{aligned}$$

Then we observe the quotient of the k-th term of (6.14) and the $2k+1$-th term of (6.15),

$$\begin{aligned}&\frac{m^{2k}n^{2k}v_\perp ^{2k+1}u_\perp ^{2k}}{(k!)^2}/\Big (\frac{2^{2k+1} m^{2k+1}n^{2k+1}v_\perp ^{2k+1} u_\perp ^{2k+1}}{(2k+1)!} \Big ) \\&\quad \le \frac{4}{k^{2k}e^{-2k}2\pi k}/\Big (\frac{2^{2k+1} mn u_\perp }{(2k+1)^{2k+1}e^{-(2k+1)}\sqrt{2\pi (2k+1)}} \Big )\\&\quad = \frac{4e}{2\pi mn}\Big (\frac{k+1/2}{k} \Big )^{2k+1} \frac{\sqrt{2\pi (2k+1)}}{u_\perp }\\&\quad = \frac{4e}{2\pi mn}\Big (\frac{2k+1}{2k} \Big )^{2k+1} \frac{\sqrt{2\pi (2k+1)}}{u_\perp }\le \frac{4e^2}{\sqrt{\pi } mn} \frac{\sqrt{k}}{u_\perp }. \end{aligned}$$

Thus we can take $k_u=u_\perp ^2$ such that when $k\le k_u$,

$$\begin{aligned} \sum _{k=k_1}^{k_u} \frac{m^{2k}n^{2k}v_\perp ^{2k+1}u_\perp ^{2k}}{(k!)^2}\le \frac{4e^2}{\sqrt{\pi }mn}\sum _{k=k_1}^{k_u} \frac{2^{2k+1} m^{2k+1}n^{2k+1}v_\perp ^{2k+1} u_\perp ^{2k+1}}{(2k+1)!}. \end{aligned}$$

(6.16)

Similarly we observe the quotient of the k-th term of (6.14) and the 2k-th term of (6.15),

$$\begin{aligned}&\frac{m^{2k}n^{2k}v_\perp ^{2k+1}u_\perp ^{2k}}{(k!)^2}/\Big (\frac{2^{2k} m^{2k}n^{2k}v_\perp ^{2k} u_\perp ^{2k}}{(2k)!} \Big )\\&\quad \le \frac{4v_\perp }{k^{2k}e^{-2k}2\pi k}/\Big (\frac{2^{2k}}{(2k)^{2k}e^{-2k}\sqrt{4\pi k}} \Big )=\frac{4v_\perp }{\sqrt{\pi } \sqrt{k}}. \end{aligned}$$

When $k>k_u=u_\perp ^2$, by $u_\perp >\delta ^{-1}$ and $v_\perp <2\frac{n}{m}u_\perp$ we have

$$\begin{aligned} \frac{4v_\perp }{\sqrt{\pi } \sqrt{k}}\le \frac{4v_\perp }{\sqrt{\pi }u_\perp }\le \frac{8n}{m\sqrt{\pi }}. \end{aligned}$$

Thus we have

$$\begin{aligned} \sum _{k=k_u}^\infty \frac{m^{2k}n^{2k}v_\perp ^{2k+1}u_\perp ^{2k}}{(k!)^2}\le \frac{8n}{m\sqrt{\pi }}\sum _{k=k_u}^\infty \frac{2^{2k} m^{2k}n^{2k}v_\perp ^{2k} u_\perp ^{2k}}{(2k)!}. \end{aligned}$$

(6.17)

Collecting (6.17) (6.16), when $v_\perp <2\frac{n}{m}u_\perp$, we obtain

$$\begin{aligned} v_\perp I_0(2mnv_\perp u_\perp )\lesssim \exp \Big (\frac{2(1-r_\perp )^{1/2} v_\perp u_\perp }{r_\perp } \Big ). \end{aligned}$$

(6.18)

By (6.18), we have

$$\begin{aligned}&\int _{\frac{n}{m}u_\perp +\delta ^{-1}}^{2\frac{n}{m}u_\perp }v_\perp I_0(2mnv_\perp u_\perp )) e^{-m^2v_\perp ^2}e^{n^2 v_\perp ^2}dv \nonumber \\&\quad \lesssim \int _{\frac{n}{m}u_\perp +\delta ^{-1}}^{2\frac{n}{m}u_\perp } e^{-m^2(v_\perp -\frac{n}{m}u_\perp )^2} dv\le e^{-m^2\delta ^{-2}}. \end{aligned}$$

(6.19)

Collecting (6.15) and (6.19) we prove (6.11).

Then following the same derivation as (6.9),

$$\begin{aligned}&2b\int _{\frac{b}{b-a-\varepsilon }w+\delta ^{-1}}^\infty v e^{\varepsilon v^2}e^{av^2} e^{-bv^2}e^{-bw^2}I_0(2bv w)dv\\&\quad =2(b-a-\varepsilon )\int _{\frac{b}{b-a-\varepsilon }w+\delta ^{-1}}^\infty v e^{(a+\varepsilon -b)v^2}I_0(2bv w)e^{\frac{(bw)^2}{a+\varepsilon -b}}dv \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}\\&\quad \le e^{\frac{-(b-a-\varepsilon )}{4\delta ^2}}\frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2} \le \delta \frac{b}{b-a-\varepsilon }e^{\frac{(a+\varepsilon )b}{b-a-\varepsilon }w^2}, \end{aligned}$$

where we apply (6.11) in the first step in the third line and take $\delta \ll 1$ in the last step of the third line. $\square$

Lemma 19

If $0<\frac{\theta }{4}<\rho$, if $0<{\tilde{\rho }}< \rho - \frac{\theta }{4}$, $0\le \lambda t< \theta$,

$$\begin{aligned} {\mathbf {k}}_{ \varrho }(v,u) \frac{e^{{\theta } |v|^2}}{e^{\mathcal {\theta } |u|^2}}\frac{e^{\lambda t\langle u\rangle }}{e^{\lambda t\langle v\rangle }} \lesssim {\mathbf {k}}_{{\tilde{\varrho }}}(v,u) . \end{aligned}$$

(6.20)

Proof

When $\langle u \rangle -\langle v\rangle \le 1$,

$$\begin{aligned} \frac{e^{\lambda s\langle u\rangle }}{e^{\lambda s\langle v\rangle }}\le e^{\lambda s}. \end{aligned}$$

When $\langle u\rangle -\langle v\rangle \ge 1$,

$$\begin{aligned} \langle u\rangle ^2-\langle v\rangle ^2=(\langle u\rangle -\langle v\rangle )(\langle u\rangle +\langle v\rangle )\ge \langle u\rangle -\langle v\rangle . \end{aligned}$$

Thus by $\langle u\rangle ^2=|u|^2+1$,

$$\begin{aligned} \frac{e^{\lambda s\langle u\rangle }}{e^{\lambda s\langle v\rangle }}\lesssim 1+\frac{e^{\lambda s | u|^2 }}{e^{\lambda s | v|^2}}. \end{aligned}$$

Note

$$\begin{aligned} \begin{aligned} {\mathbf {k}}_{ \varrho }(v,u) \frac{e^{\vartheta |v|^2}}{e^{\vartheta |u|^2}} = \frac{1}{|v-u| } \exp \left\{ - {\varrho } |v-u|^{2} - {\varrho } \frac{ ||v|^2-|u|^2 |^2}{|v-u|^2} + \vartheta |v|^2 - \vartheta |u|^2 \right\} . \end{aligned} \end{aligned}$$

Let $v-u=\eta$ and $u=v-\eta$. Then the exponent equals

$$\begin{aligned}&- \varrho |\eta |^{2}-\varrho \frac{||\eta |^{2}-2v\cdot \eta |^{2}}{ |\eta |^{2}}-\vartheta \{|v-\eta |^{2}-|v|^{2}\}-\lambda t \{|v|-|v-\eta |\} \\&\quad =-2 \varrho |\eta |^{2}+ 4 \varrho v\cdot \eta - 4 \varrho \frac{|v\cdot \eta |^{2}}{|\eta |^{2}}-\vartheta \{|\eta |^{2}-2v\cdot \eta \} \\&\quad =(-2 \varrho -\vartheta )|\eta |^{2}+(4 \varrho +2\vartheta )v\cdot \eta - 4 \varrho \frac{\{v\cdot \eta \}^{2}}{|\eta |^{2}}. \end{aligned}$$

If $0<\vartheta <4 \varrho$ then the discriminant of the above quadratic form of $|\eta |$ and $\frac{v\cdot \eta }{|\eta |}$ is

$$\begin{aligned} (4 \varrho +2\vartheta )^{2}-4 (-2 \varrho -\vartheta )(- 4 \varrho ) =4\vartheta ^{2}- 16 \varrho \vartheta <0. \end{aligned}$$

Hence, the quadratic form is negative definite. We thus have, for $0<{\tilde{\varrho }}< \varrho - \frac{\vartheta }{4}$, the following perturbed quadratic form is still negative definite

$$\begin{aligned} -(\varrho - {\tilde{\varrho }})|\eta |^{2}-(\varrho - {\tilde{\varrho }})\frac{||\eta |^{2}-2v\cdot \eta |^{2}}{|\eta |^{2}}-\vartheta \{|\eta |^{2}-2v\cdot \eta \} \le 0. \end{aligned}$$

For

$$\begin{aligned}&{\mathbf {k}}_{ \varrho }(v,u) \frac{e^{\vartheta |v|^2}}{e^{\vartheta |u|^2}}\frac{e^{\lambda t\langle u\rangle ^2}}{e^{\lambda t |v|^2}} \\&\quad = \frac{1}{|v-u| } \exp \left\{ - {\varrho } |v-u|^{2} - {\varrho } \frac{ ||v|^2-|u|^2 |^2}{|v-u|^2} + (\theta -\lambda t) |v|^2 - (\theta -\lambda t) |u|^2 \right\} . \end{aligned}$$

We just need to replace $\theta$ by $\theta -\lambda t$ in the previous computation. By $\lambda t\ll \theta$,

$$\begin{aligned} -(\varrho - {\tilde{\varrho }})|\eta |^{2}-(\varrho - {\tilde{\varrho }})\frac{||\eta |^{2}-2v\cdot \eta |^{2}}{|\eta |^{2}}-(\theta -\lambda t) \{|\eta |^{2}-2v\cdot \eta \} \le 0. \end{aligned}$$

Therefore, we conclude the lemma. $\square$