On a Boltzmann Equation for Compton Scattering from Non relativistic Electrons at Low Density

Abstract

A Boltzmann equation, used to describe the evolution of the density function of a gas of photons interacting by Compton scattering with electrons at low density and non relativistic equilibrium, is considered. A truncation of the very singular redistribution function is introduced and justified. The existence of weak solutions is proved for a large set of initial data. A simplified equation, where only the quadratic terms are kept and that appears at very low temperature of the electron gas, for small values of the photon’s energies, is also studied. The existence of weak solutions, and also of more regular solutions that are very flat near the origin, is proved. The long time asymptotic behavior of weak solutions of the simplified equation is described.

Appendices

Appendix A: Some Useful Estimates

Lemma A.1

If B satisfies (2.5)–(2.10), and $\varphi$ is L-Lipschitz on $[0,\infty )$, then for all $(x,y)\in \Gamma$:

$$\begin{aligned}&\displaystyle |k_{\varphi }(x,y)|\le LC_*Ae^{\frac{|x-y|}{2}},\quad A=\max \left\{ \frac{(1-\theta )^2}{\theta \delta (1+\theta )},\rho _*\right\} , \end{aligned}$$

(A.1)

$$\begin{aligned}&\displaystyle |\ell _{\varphi }(x,y)|\le \frac{LC_*(1-\theta )}{\theta ^2(1+\theta )}e^{\frac{x-y}{2}}. \end{aligned}$$

(A.2)

Moreover, the function $\mathcal {L}_{\varphi }$ given in (2.31) is continuous on $[0,\infty )$ and for all $x\in [0,\infty )$,

$$\begin{aligned} |\mathcal {L}_{\varphi }(x)|\le \frac{LC_*(1-\theta )}{\theta ^2(1+\theta )}\big (e^{\frac{x-\gamma _1(x)}{2}}-e^{\frac{x-\gamma _2(x)}{2}}\big ). \end{aligned}$$

(A.3)

In particular, $\mathcal {L}_{\varphi }(0)=0$.

Proof

We first prove (A.1). Let $(x,y)\in {\text {supp}}(B)=\Gamma$, and assume, by the symmetry of $k_{\varphi }$, that $0\le y\le x$. By the mean value theo, $|e^{-x}-e^{-y}|\le e^{-y}(x-y)$, and from (2.11) and the Lipschitz condition,

$$\begin{aligned} |k_{\varphi }(x,y)|&\le LC_*e^{\frac{x-y}{2}}\frac{(x-y)^2}{(x+y)xy}. \end{aligned}$$

Then by (2.8)–(2.10)

$$\begin{aligned} \frac{(x-y)^2}{(x+y)xy}\le \left\{ \begin{array}{ll} \frac{(1-\theta )^2}{\theta \delta _*(1+\theta )}&{} \quad \text {if }(x,y)\in \Gamma _1,\\ \rho _*&{} \quad \text {if }(x,y)\in \Gamma _2, \end{array} \right. \end{aligned}$$

and (A.1) follows.

In order to prove (A.2) we use (2.11) and the Lipschitz condition to have, for all $(x,y)\in \Gamma$,

$$\begin{aligned} |\ell _{\varphi }(x,y)|&\le LC_* e^{\frac{x-y}{2}}\frac{y|x-y|}{x(x+y)}. \end{aligned}$$

Using that $\Gamma \subset \{(x,y)\in [0,\infty )^2:\theta x\le y\le \theta ^{-1}x\}$, then

$$\begin{aligned} \frac{y|x-y|}{x(x+y)}\le \frac{(1-\theta )}{\theta ^2(1+\theta )}, \end{aligned}$$

and (A.2) follows. We obtain (A.3) directly from (A.2) and Remark 2.1.

We finally prove the continuity of $\mathcal {L}_{\varphi }$ on $[0,\infty )$. By (2.7)–(2.10), $\mathcal {L}_{\varphi }(x)$ is continuous for all $x>0$, so we only need to prove $\mathcal {L}_{\varphi }(x)\rightarrow 0$ as $x\rightarrow 0$. This follows from (A.3) and the mean value theo, using $\gamma _2(x)-\gamma _1(x)\le (\theta ^{-1}-\theta )x.$$\square$

Remark A.2

Under the hypothesis of Lemma A.1, the function $k_{\varphi }$ could not be continuous at the origin $(x,y)=(0,0)$, since we do not know if $\lim _{(x,y)\rightarrow (0,0)}k_{\varphi }(x,y)=0$. However we have the following.

Lemma A.3

If B satisfies (2.5)–(2.10), then $k_{\varphi }\in C([0,\infty )^2)$ for all $\varphi \in C^1([0,\infty ))$ with $\varphi '(0)=0$, and $k_{\varphi }(0,0)=0$.

Proof

By definition and (2.7), it is clear that $k_{\varphi }\in C([0,\infty )^2\setminus \{0\})$. If we prove that $\lim _{(x,y)\rightarrow (0,0)}k_{\varphi }(x,y)=0$, the continuity at the origin follows. To this end we mimic the proof of (A.1) using $\varphi (x)-\varphi (y)=\varphi '(\xi )(x-y)$ for some $\xi \in (\min \{x,y\},\max \{x,y\})$ instead of the Lipschitz condition, and we obtain

$$\begin{aligned} |k_{\varphi }(x,y)|&\le \max \left\{ \frac{(1-\theta )^2}{\theta \delta (1+\theta )},\rho \right\} |\varphi '(\xi )|e^{\frac{|x-y|}{2}} \end{aligned}$$

(A.4)

for all $(x,y)\in \Gamma$ and all $\varphi \in C^1([0,\infty )).$ If $\varphi '(0)=0$, it follows from (A.4) that $\lim _{(x,y)\rightarrow (0,0)}k_{\varphi }(x,y)=0$. $\square$

Proposition A.4

Suppose that B satisfies (2.5)–(2.10), $\varphi$ is L-Lipschitz on $[0,\infty )$, and $u\in \mathscr {M}_+([0,\infty ))$. Then

$$\begin{aligned} |K_{\varphi }(u,u)|&\le LC_*A\bigg (\int _{[0,\infty )}e^{\frac{x-\gamma _1(x)}{2}}u(x)dx\bigg )\bigg (\int _{[0,\infty )}u(y)dy\bigg ), \end{aligned}$$

(A.5)

$$\begin{aligned} |L_{\varphi }(u)|&\le \frac{LC_*(1-\theta )}{2\theta ^2(1+\theta )}\int _{[0,\infty )}\big (e^{\frac{x-\gamma _1(x)}{2}}-e^{\frac{x-\gamma _2(x)}{2}}\big )u(x)dx, \end{aligned}$$

(A.6)

where A is given in (A.1).

Proof

In order to prove (A.5), we use Remarks 2.1, 2.3 and (A.1):

$$\begin{aligned} |K_{\varphi }(u,u)|&\le LC_*A\int _0^{\infty }e^{\frac{x}{2}}u(x)\int _{\gamma _1(x)}^x e^{-\frac{y}{2}}u(y)dydx\\&\le LC_*A\int _0^{\infty } e^{\frac{x-\gamma _1(x)}{2}}u(x)\int _{\gamma _1(x)}^x u(y)dydx, \end{aligned}$$

from where (A.5) follows. The estimate (A.6) follows directly from (A.3). $\square$

Let us define now

$$\begin{aligned} K_{\varphi ,n}(u,u)&=\frac{1}{2}\iint _{[0,\infty )^2}k_{\varphi ,n}(x,y)u(t,x)u(t,y)dydx, \end{aligned}$$

(A.7)

$$\begin{aligned} k_{\varphi ,n}(x,y)&=b_n(x,y)(e^{-x}-e^{-y})(\varphi (x)-\varphi (y)), \end{aligned}$$

(A.8)

$$\begin{aligned} L_{\varphi ,n}(u_n)&=\frac{1}{2}\int _{[0,\infty )}\mathcal {L}_{\varphi ,n}(x)u(t,x)dx, \end{aligned}$$

(A.9)

$$\begin{aligned} \mathcal {L}_{\varphi ,n}(x)&=\int _0^{\infty }\ell _{\varphi ,n}(x,y)dy \end{aligned}$$

(A.10)

$$\begin{aligned} \ell _{\varphi ,n}(x,y)&=b_n(x,y)y^2e^{-y}(\varphi (x)-\varphi (y)). \end{aligned}$$

(A.11)

Remark A.5

Since $\phi _n\le x^{-1}$, the estimates (A.1)–(A.3) in Lemma A.1 hold for $k_{\varphi ,n}$, $\ell _{\varphi ,n}$ and $\mathcal {L}_{\varphi ,n}$ respectively, and estimates (A.5) and (A.6) in Lemma A.4 hold for $K_{\varphi ,n}(u,u)$ and $L_{\varphi ,n}(u)$ respectively, for all $n\in \mathbb {N}$.

Lemma A.6

$\mathcal {L}_{\varphi ,n}\rightarrow \mathcal {L}_{\varphi }$ as $n\rightarrow \infty$ uniformly on the compact sets of $[0,\infty )$ for all $\varphi$L-Lipschitz on $[0,\infty )$.

Proof

Let $R>0$ and $x\in [0,R]$. On the one hand, if $x\in [0,1/n]$, we have $|\mathcal {L}_{\varphi }(x)-\mathcal {L}_{\varphi ,n}(x)|\le 2|\mathcal {L}_{\varphi }(x)|\rightarrow 0$ as $n\rightarrow \infty$, since $\mathcal {L}_{\varphi }(0)=0$ (cf. Lemma A.1). On the other hand, if $x\in [1/n,R]$ and $y\in [1/n,n]$, by definition $\phi _n(x)\phi _n(y)=(xy)^{-1}$, and then

$$\begin{aligned} |\mathcal {L}_{\varphi }(x)-\mathcal {L}_{\varphi ,n}(x)|&\le \int _0^{\frac{1}{n}}| \ell _{\varphi }(x,y)-\ell _{\varphi ,n}(x,y)|dy\nonumber \\&\quad +\int _n^{\infty }|\ell _{\varphi }(x,y)-\ell _{\varphi ,n}(x,y)|dy. \end{aligned}$$

The two integrals in the right hand side above are treated in the same way. Using $|\ell _{\varphi }(x,y)-\ell _{\varphi ,n}(x,y)|\le |\ell _{\varphi }(x,y)|$ and (A.2),

$$\begin{aligned} \int _0^{\frac{1}{n}}|\ell _{\varphi }(x,y)|dy&\le \frac{LC_*(1-\theta )}{\theta ^2(1+\theta )}e^{\frac{R}{2}}\int _0^{\frac{1}{n}}e^{-\frac{y}{2}}dy\xrightarrow [n\rightarrow \infty ]{}0,\\ \int _n^{\infty }|\ell _{\varphi }(x,y)|dy&\le \frac{LC_*(1-\theta )}{\theta ^2(1+\theta )}e^{\frac{R}{2}}\int _n^{\infty }e^{-\frac{y}{2}}dy\xrightarrow [n\rightarrow \infty ]{}0, \end{aligned}$$

and the result follows. $\square$

Appendix B: The Function $\mathcal {B} _{ \beta }$, Properties and Scalings

In this Section, we describe several properties of the function $\mathcal {B}_\beta$. First, the parameter $\beta$ is used to scale the variables, in such a way that the total mass of the solution is conserved. Then, for each $\beta >0$ fixed, the behavior of $\mathcal {B}_\beta (k, k')$ is studied when k and $k'$ are varying on $(0, \infty )$.

1.1 B.1 $\beta$-Scalings of $\mathcal {B}_\beta$

It looks natural from (1.2) to introduce the scaled variable

$$\begin{aligned} \mathbf x =\beta \, \mathbf k , \end{aligned}$$

(B.1)

and define

$$\begin{aligned} F(\tau ,x)=f(t,k),\qquad \tau =\beta ^3 t,\quad x=\beta k. \end{aligned}$$

(B.2)

The scaling (B.2) preserves the total number of particles:

$$\begin{aligned} \int _0^{\infty } x^2 F(\tau , x)dx=\int _0^{\infty }k^2f(t,k)dk=\int _0^{\infty }k^2 f(0,k)dk\quad \forall \tau >0. \end{aligned}$$

In terms of F,

$$\begin{aligned} k^2\frac{\partial f}{\partial t}(t,k)= & {} \beta ^4x^2\frac{\partial F}{\partial \tau }(\tau ,x),\\ \tilde{q} (f, f')= & {} \beta ^6FF'\big (e^{-x}-e^{-x'}\big )+\beta ^3\big (F'e^{-x}-Fe^{-x'}\big ), \end{aligned}$$

and if we define

$$\begin{aligned} B_{\beta }(x,x')=\beta ^{-1}\mathcal {B}_{\beta }(k,k'), \end{aligned}$$

(B.3)

that is,

$$\begin{aligned} B_\beta (x, x')=\sqrt{\beta } e^{\frac{(x'+x )}{2}}\int _0^\pi \frac{(1+\cos ^2\theta )}{ |\mathbf x '-\mathbf x | }e^{-\beta \frac{m(x-x')^2+\frac{ |\mathbf x '-\mathbf x |^4}{4m\beta ^2}}{2 |\mathbf x '-\mathbf x |^2}} d\cos \theta , \end{aligned}$$

(B.4)

the Eq. (1.2) then reads

$$\begin{aligned} x^2\frac{\partial F}{\partial \tau }(\tau ,x)=&\int _0^\infty B_\beta (x, x')FF'\big (e^{-x}-e^{-x'}\big )xx'dx'+\nonumber \\&+\beta ^{-3}\int _0^\infty B_\beta (x, x')\big (F'e^{-x}-Fe^{-x'}\big )xx'dx'. \end{aligned}$$

(B.5)

If we now define

$$\begin{aligned} u(\tau , x)=x^2F(\tau , x) \end{aligned}$$

(B.6)

then from (B.5) we finally obtain

$$\begin{aligned} \frac{\partial u}{\partial \tau }(\tau ,x)&=\int _0^\infty \frac{B_{\beta }(x,x')}{xx'}\big (e^{-x}-e^{-x'}\big )uu'dx'+\nonumber \\&\quad +\beta ^{-3}\int _0^\infty \frac{B_{\beta }(x,x')}{xx'}\big (u'x^2e^{-x}-ux'^2e^{-x'}\big )dx', \end{aligned}$$

(B.7)

The second term in the right hand side of (B.7) seems then negligible when $\beta$ tends to $\infty$, but no rigorous result on that direction is known.

1.2 B.2 The Function $B_\beta (x, x')$ for $\beta$ Fixed.

In this Section we show some properties of the kernel $B_{\beta }$ defined in (B.4).

Proposition B.1

For all $\beta >0$, $x>0$ and $x'>0$,

$$\begin{aligned} B_\beta (x, x')\le \sqrt{\beta }\;\frac{4\big (10\max ^2\{x, x'\}+\min ^2\{x, x'\}\big )}{15 \max ^3\{x, x'\}}e^{\frac{(x'+x )}{2}}, \end{aligned}$$

(B.8)

and for all $x>0,\;x'>0$ with $x\ne x'$,

$$\begin{aligned} \lim _{ \beta \rightarrow \infty }B_\beta (x, x')=0. \end{aligned}$$

(B.9)

Proof

For all $x>0$ and $x'>0$,

$$\begin{aligned} \frac{e^{-\frac{(x+x')}{2}}}{\sqrt{\beta }}B_\beta (x, x')&\le \int _0^\pi \frac{(1+\cos ^2\theta )}{ |\mathbf x '-\mathbf x | }d\cos \theta =\int _{ -1 }^1 \frac{(1+t^2)}{ \sqrt{x^2+x'^2-2xx't} }dt \nonumber \\&=\frac{4\big (10\max ^2\{x, x'\}+\min ^2\{x, x'\}\big )}{15 \max ^3\{x, x'\}}, \end{aligned}$$

(B.10)

and then (B.8) holds. If $x'\not =x$, we have first

$$\begin{aligned} \lim _{ \beta \rightarrow \infty }e^{-\beta \frac{m(x-x')^2+\frac{ |\mathbf x '-\mathbf x |^4}{4m\beta ^2}}{2 |\mathbf x '-\mathbf x |^2}} =0\qquad \forall \theta \in [0,\pi ], \end{aligned}$$

and since

$$\begin{aligned} \frac{(1+\cos ^2\theta )}{ |\mathbf x '-\mathbf x | } e^{-\beta \frac{(x-x')^2+\frac{ |\mathbf x '-\mathbf x |^4}{\beta ^2}}{2m |\mathbf x '-\mathbf x |^2}} \le \frac{(1+\cos ^2\theta )}{ |\mathbf x '-\mathbf x |} \in L^1(d\cos \theta ) \quad \forall \beta >0, \end{aligned}$$

then (B.9) follows from Lebesgue’s convergence Theorem. $\square$

Proposition B.2

$$\begin{aligned} B_\beta (x, x)&=\sqrt{\beta }\left( \frac{2\sqrt{2\pi m\beta }}{x^2}+\mathcal {O}\left( \frac{1}{x}\right) ^3\right) e^x\quad \text {as}\quad x\rightarrow \infty , \end{aligned}$$

(B.11)

$$\begin{aligned} B_\beta (x, x)&=\sqrt{\beta }\;\frac{44}{15 }\left( \frac{1}{x}+1\right) +\mathcal {O}(x)\quad \text {as}\quad x\rightarrow 0. \end{aligned}$$

(B.12)

Proof

By definition, for all $x>0$,

$$\begin{aligned} B_\beta (x, x)&=\sqrt{\frac{\beta }{2}}\frac{e^{x}}{x}\int _0^\pi \frac{(1+\cos ^2\theta )}{\sqrt{1-\cos \theta } }e^{- \frac{x^2(1-\cos \theta )}{4m \beta }} d\cos \theta \\&=\sqrt{\beta }\frac{2e^x\sqrt{\beta m}}{x^6}\Big (\sqrt{2\pi }\big (6\beta ^2m^2-2\beta m x^2+x^4\big )Erf\left( \frac{x}{\sqrt{2\beta m}}\right) \\&\quad -12e^{-\frac{x^2}{2\beta m}}(\beta m)^{3/2}x\Big ), \end{aligned}$$

and the result follows. $\square$

The function $B_\beta$ is exponentially decreasing in the direction orthogonal to the first diagonal, as shown in the next two Propositions.

Proposition B.3

For all $\beta >0$,

$$\begin{aligned}&\nabla B_\beta (x, x')\cdot (1, -1)> 0 \quad \hbox {if}\quad x'>x>0, \end{aligned}$$

(B.13)

$$\begin{aligned}&\nabla B_\beta (x, x')\cdot (1, -1) < 0 \quad \hbox {if}\quad x>x'>0. \end{aligned}$$

(B.14)

Proof

It is only a straightforward calculation. With the help of Mathematica, using the change of variables $t=\cos \theta$,

$$\begin{aligned} \frac{\partial B_\beta }{\partial x}(x,x')&=\frac{e^{\frac{(x'+x )}{2}}}{4m\sqrt{\beta }}\int _{-1}^{1} \frac{(1+t^2)}{|\mathbf x '-\mathbf x |^{5}}e^{-\beta \frac{m(x-x')^2+\frac{ |\mathbf x '-\mathbf x |^4}{4m\beta ^2}}{2 |\mathbf x '-\mathbf x |^2}}\Theta (x, x', t) dt, \end{aligned}$$

(B.15)

$$\begin{aligned} \Theta (x, x', t)&=4(\beta m) ^2(t-1)x'(x-x')(x+x')-(x-tx') |\mathbf x '-\mathbf x |^4\nonumber \\&\quad +2\beta m \left( x'^2-2tx'(x-1)+(x-2)x\right) |\mathbf x '-\mathbf x |^2. \end{aligned}$$

(B.16)

The expression of $\frac{\partial B_\beta }{\partial x'}$ is obtained from (B.15) and (B.16) using the permutation $x\leftrightarrow x'$. Then,

$$\begin{aligned} \nabla B_\beta (x, x')\cdot (1, -1)&=\frac{e^{\frac{(x'+x )}{2}}}{4m\sqrt{\beta }}\int _{-1}^{1} \frac{(1+t^2)}{|\mathbf x '-\mathbf x |^{5} }e^{-\beta \frac{m(x-x')^2+\frac{ |\mathbf x '-\mathbf x |^4}{4m\beta ^2}}{2 |\mathbf x '-\mathbf x |^2}}\nonumber \\&\quad \times \big (\Theta (x, x', t) -\Theta (x',x,t)\big )dt,\nonumber \\ \Theta (x, x', t) -\Theta (x',x,t)&=(x'-x)\Big [4(\beta m) ^2(1-t)(x+x')^2\nonumber \\&\quad +4\beta m (1+t)|\mathbf x '-\mathbf x |^2+(1+t)|\mathbf x '-\mathbf x |^4 \Big ], \end{aligned}$$

(B.17)

and the result follows. $\square$

Proposition B.4

For all $\beta >0$, $x>0$ and $x'>0$,

$$\begin{aligned} B _{ \beta }(x, x')\le \mathscr {B} _{ \beta }(x, x'), \end{aligned}$$

(B.18)

where

$$\begin{aligned} \mathscr {B}_{\beta }(x, x')&=\sqrt{\beta } e^{-\beta \frac{m(x-x')^2+\frac{ (x-x')^4}{4m\beta ^2}}{2 (x+x')^2}}N (x+x', |x-x'|), \end{aligned}$$

(B.19)

$$\begin{aligned} N (p, q)&= \frac{ 8\, e^{\frac{p}{2}}\big (10(p+q)^2+(p-q)^2\big )}{15 (p+q)^3},\quad \forall p>0,\;\forall q>0. \end{aligned}$$

(B.20)

Proof

For all $\mathbf x \in \mathbb {R}^3$ and $\mathbf x '\in \mathbb {R}^3$ such that $|\mathbf x |=x$, $|\mathbf x '|=x'$,

$$\begin{aligned} |x-x'|\le |\mathbf x -\mathbf x '|\le x+x'. \end{aligned}$$

Therefore,

$$\begin{aligned} B _{ \beta }(x, x')\le \sqrt{\beta } e^{\frac{(x'+x )}{2}} e^{-\beta \frac{m(x-x')^2+\frac{ (x-x')^4}{4m\beta ^2}}{2(x+x')^2}}\int _0^\pi \frac{(1+\cos ^2\theta )}{ |\mathbf x '-\mathbf x | } d\cos \theta , \end{aligned}$$

and the result follows using (B.10). $\square$

Corollary B.5

$$\begin{aligned} \forall x>0, x'>0:\,\, B _{ \beta } (x, x')\le & {} B _{ \beta }\left( \frac{x+x'}{2}, \frac{x+x'}{2}\right) ,\nonumber \\ B_\beta \left( \frac{x+x'}{2}, \frac{x+x'}{2}\right)= & {} \sqrt{\beta }\,\left( \frac{2\sqrt{2\pi m\beta }}{(x+x')^2}+\mathcal {O}\left( \frac{1}{x+x'}\right) ^3\right) e^{\frac{x+x'}{2}} \end{aligned}$$

(B.21)

as $x+x'\rightarrow \infty$, and

$$\begin{aligned}&B_\beta \left( \frac{x+x'}{2}, \frac{x+x'}{2}\right) =\frac{44 \sqrt{\beta }\,}{15 }\left( \frac{1}{x+x'}+1\right) +\mathcal {O}(x+x'),\,\,\,x+x'\rightarrow 0. \end{aligned}$$

If $x+x'\rightarrow \infty$, and $|x-x'|\le \theta x$:

$$\begin{aligned} |e^{-x}-e^{-x'}|B_\beta (x, x')\le 2 \sqrt{\beta }\,\left( \frac{2\sqrt{2\pi m\beta }}{(x+x')^2}+\mathcal {O}\left( \frac{1}{x+x'}\right) ^3\right) \left| \, \sinh \left( \frac{\theta x}{2} \right) \right| . \end{aligned}$$

(B.22)

For all $\rho >0$ fixed and $x>0$, $x'>0$ such that $x+x'=\rho$,

$$\begin{aligned} B _{ \beta }(x, x')\le \sqrt{\beta }\,e^{-\beta \frac{(x-x')^2}{2m \rho ^2}}\Phi (\rho , |x-x'|). \end{aligned}$$

(B.23)

Proof

By Proposition B.3, the function $B_\beta$ is strictly decreasing in the direction orthogonal to the first diagonal, and then property (B.21) follows. In order to prove (B.22) we have first, when $x+x'\rightarrow \infty$,

$$\begin{aligned} |e^{-x}-e^{-x'}|B_\beta (x, x')\le 2 \left( \frac{2\sqrt{2\pi m\beta }}{(x+x')^2}+\mathcal {O}\left( \frac{1}{x+x'}\right) ^3\right) \left| \,\sinh \left( \frac{x'-x}{2} \right) \right| \end{aligned}$$

If moreover, $0\le x'-x \le \theta x$ then

$$\begin{aligned} 0\le (e^{-x}-e^{-x'})B_\beta (x, x')\le 2 \left( \frac{2\sqrt{2\pi m\beta }}{(x+x')^2}+\mathcal {O}\left( \frac{1}{x+x'}\right) ^3\right) \sinh \left( \frac{\theta x}{2} \right) \end{aligned}$$

If $-\theta x \le x'-x\le 0$ then,

$$\begin{aligned} 0\le -\sinh \left( \frac{x'-x}{2} \right) =\sinh \left( \frac{x-x'}{2} \right) \le \sinh \left( \frac{\theta x}{2} \right) , \end{aligned}$$

and (B.22) follows. $\square$

Proposition B.6

For all $\varphi \in C_c((0, \infty )\times (0, \infty ))$:

$$\begin{aligned}&\lim _{ \beta \rightarrow \infty }\iint _{ (0, \infty )^2 }\varphi (x, y) \Phi _\beta (x, y) \mathscr {B} _{ \beta }(x, y)dxdy\nonumber \\&\quad = \frac{88}{15}\sqrt{\frac{{m \pi }}{2}} erf(1)\int _{ (0, \infty ) }\varphi \left( \frac{z}{2}, \frac{z}{2} \right) e^{\frac{z}{2}}dz \end{aligned}$$

(B.24)

Proof

Define the new variables

$$\begin{aligned} \xi =x-y,\,\,\zeta =x+y,\,\,\,\psi (\xi , \zeta )=\varphi \left( \frac{\xi +\zeta }{2}, \frac{\zeta -\xi }{2} \right) \end{aligned}$$

and denote $\Psi _{\beta }(\xi ,\zeta )=\Phi _{\beta }(x,y)$. Then,

$$\begin{aligned} I=&\iint _{ (0, \infty )^2 }\varphi (x, y)\Phi _\beta (x, y) \mathscr {B} _{ \beta }(x, y)dxdy\\ =&\iint _{ D }e^{- \frac{\beta ^2\xi ^2+\xi ^4}{2m \beta \zeta ^2}}\Psi _{ \beta } (\xi , \zeta ) \mathscr {B}_\beta \left( \frac{\xi +\zeta }{2}, \frac{\zeta -\xi }{2} \right) \psi (\xi , \zeta )d\xi d\zeta \end{aligned}$$

where $D=\{(\zeta ,\xi )\in \mathbb {R}^2:\zeta >0,\;-\zeta<\xi <\zeta \}$. We write now,

$$\begin{aligned} \frac{\beta ^2\xi ^2+\xi ^4}{2m \beta \zeta ^2}=\frac{\beta \xi ^2}{2m\zeta ^2}\left( 1+ \frac{ \xi ^2 }{\beta ^2} \right) \end{aligned}$$

and the change of variables:

$$\begin{aligned} \sqrt{\frac{\beta }{2m}}\frac{\xi }{\zeta }=z_1,\,\,\zeta =z_2;\,\,\, \xi =\sqrt{\frac{2m}{\beta }}z_1z_2, \, \zeta =z_2 \end{aligned}$$

whose Jacobian is $\sqrt{2m/\beta }\, z_2$ and,

$$\begin{aligned} I=&\iint _{ \Omega } e^{-z_1^2\left( 1+\frac{2m z_1^2z_2^2}{\beta }\right) } \mathscr {B}_\beta \left( Z_1, Z_2 \right) \\&\times \Psi \left( \beta ^{-1} \sqrt{\frac{2m}{\beta }}z_1z_2, \beta ^{-1} z_2, \right) \psi \left( \sqrt{\frac{2m}{\beta }}z_1z_2,z_2\right) \sqrt{2m/\beta }\, z_2dz_1dz_2\\ Z_1=&\frac{1}{2}\left( z_2+\sqrt{\frac{2m}{\beta }} z_1z_2 \right) ,\,\,Z_2=\frac{1}{2}\left( z_2-\sqrt{\frac{2m}{\beta }} z_1z_2 \right) \end{aligned}$$

Due to the cut off function $\Phi _\beta (x, y)$, the actual domain of integration $\Omega _\beta$ is:

$$\begin{aligned} \Omega _\beta =\left\{ (z_1,z_2)\in \mathbb {R}\times \mathbb {R}^+;\,\, \sqrt{2m}|\,z_1|\le \theta z_2^{1/2}\left( 1-\frac{2m}{\beta }z_1^2 \right) ^{1/2} \right\} \end{aligned}$$

where $\Omega$ is the domain where $z_2>0$, $z_1\in (-1, 1)$. As $\beta \rightarrow \infty$,

$$\begin{aligned} \lim _{ \beta \rightarrow \infty }e^{-z_1^2\left( 1+\frac{2m z_1^2z_2^2}{\beta }\right) } \psi \left( \sqrt{\frac{2m}{\beta }}z_1z_2, z_2\right) = e^{-z_1^2} \psi \left( 0, z_2\right) . \end{aligned}$$

On the other hand, using (B.12), for all $z_1, z_2$,

$$\begin{aligned} \lim _{ \beta \rightarrow \infty } \frac{\mathscr {B}_\beta \left( Z_1,Z_2\right) }{\beta } = \frac{44\, e^{\frac{z_2}{2}} }{15 z_2} \end{aligned}$$

(B.25)

By definition of $\Psi$, for all $z_1\in \mathbb {R}$ and $z_2>0$ fixed, if $\beta$ is sufficiently large,

$$\begin{aligned} \Psi \left( \beta ^{-1} \sqrt{\frac{2m}{\beta }}z_1z_2, \beta ^{-1} z_2, \right) =1 \end{aligned}$$

Then,

$$\begin{aligned} \lim _{\beta \rightarrow \infty }I=&\frac{44}{15}\sqrt{2m} \iint _{\Omega } e^{-z_1^2} e^{\frac{z_2}{2}} \psi \left( 0, z_2\right) dz_1dz_2\\ =&\frac{44}{15}\sqrt{\frac{{m\pi }}{2}} erf (1)\int _{ (0, \infty ) }\varphi \left( \frac{z_2}{2}, \frac{z_2}{2} \right) e^{\frac{z_2}{2}}dz_2 \end{aligned}$$

$\square$

The function $B_\beta (x, y)\ge 0$ coincides with $\mathscr {B}_\beta (x, y)$ for $x=y$ and is below that function, that tends to a Dirac measure along the first diagonal as $\beta \rightarrow \infty$. From properties (B.9) and (B.24), the truncation of $\mathcal {B}_\beta$ may then be seen as reasonable.