Metastability Results for a Class of Linear Boltzmann Equations

Abstract

We consider a semiclassical linear Boltzmann model with a non-local collision operator. We provide sharp spectral asymptotics for the small spectrum in the low temperature regime from which we deduce the rate of return to equilibrium as well as a metastability result. The main ingredients are resolvent estimates obtained via hypocoercive techniques and the construction of sharp Gaussian quasimodes through an adaptation of the WKB method.

Appendices

Appendix A: Proof of Lemma 1.4

Let us begin by showing that there exists a self-adjoint operator A such that

$$\begin{aligned} \varrho (H_0)=b_h^*\circ A \circ b_h. \end{aligned}$$

(A.1)

Since $\varrho (0)=0$, there exists an analytic function $\tilde{\varrho }$ such that $\varrho (z)=z\tilde{\varrho }(z)$ and $|\tilde{\varrho }(z)|\le C \langle z \rangle ^{-1}$. Using Cauchy’s formula, one easily gets that for all $z_0\in \{\textrm{Re}\, z > -\frac{1}{2C}\}$ and f an analytic function on $\{\textrm{Re}\, z > -\frac{1}{C}\}$ satisfying $f(z)=O(\langle z \rangle ^{-\beta })$ for some $\beta >0$, we have that

$$\begin{aligned} f(z_0)=\frac{-1}{2i\pi }\int _{\{\textrm{Re}\, z = -\frac{1}{2C}\}} f(z) (z_0-z)^{-1} \textrm{d}z. \end{aligned}$$

(A.2)

Working with a Hilbert basis of eigenfunctions of $H_0$, this identity yields

$$\begin{aligned} f(H_0)=\frac{-1}{2i\pi }\int _{\{\textrm{Re}\, z = -\frac{1}{2C}\}} f(z) (H_0-z)^{-1} \textrm{d}z. \end{aligned}$$

(A.3)

Besides, denoting

$$\begin{aligned} b_h=\begin{pmatrix} b_h^1\\ \vdots \\ b_h^d \end{pmatrix}, \end{aligned}$$

we have $b_h H_0=(b_h^j H_0)_{1 \le j\le d}$ and using the identity $b_h^j H_0=b_h^* b_hb_h^j+hb _h^j$, we get $b_h H_0=H_1 b_h$ where

$$\begin{aligned} H_1=\begin{pmatrix} H_0+h &{} &{} \\ &{} \ddots &{} \\ &{} &{} H_0+h \end{pmatrix}. \end{aligned}$$

(A.4)

In particular, if u is an eigenfunction of $H_0$ associated with a positive eigenvalue, the function $b_h u$ is an eigenfunction of $H_1$ associated with the same eigenvalue and therefore

$$\begin{aligned} H_0 (H_0-z)^{-1}=b_h^* (H_1-z)^{-1} b_h. \end{aligned}$$

(A.5)

It follows using (A.3) with $f=\tilde{\varrho }$ that (A.1) holds with $A=\tilde{\varrho }(H_0+h)\otimes \textrm{Id}$:

$$\begin{aligned} \varrho (H_0)=H_0 \tilde{\varrho }(H_0)=b_h^* \circ \tilde{\varrho }(H_0+h)\otimes \textrm{Id} \circ b_h. \end{aligned}$$

We can improve the integrability in the integral representation of $\tilde{\varrho }(H_0+h)$ by writing

$$\begin{aligned} \tilde{\varrho }(z)=\frac{\tilde{\varrho }(z)}{1+z}+\frac{\varrho (z)-\varrho _\infty }{1+z}+\frac{\varrho _\infty }{1+z} \end{aligned}$$

which yields always thanks to (A.3)

$$\begin{aligned} \tilde{\varrho }(H_0+h)\otimes \textrm{Id}&=\frac{-1}{2i\pi }\int _{\{\textrm{Re}\, z = -\frac{1}{2C}\}} \frac{\tilde{\varrho }(z)}{1+z} (H_1-z)^{-1} \textrm{d}z\nonumber \\&\quad +\frac{-1}{2i\pi }\int _{\{\textrm{Re}\, z = -\frac{1}{2C}\}} \frac{\varrho (z)-\varrho _\infty }{1+z} (H_1-z)^{-1} \textrm{d}z+\varrho _\infty (H_1+1)^{-1}. \nonumber \\ \end{aligned}$$

(A.6)

Besides, it is well known (see, for instance, [4]) that the resolvent $(H_1-z)^{-1}$ is a pseudo-differential operator and we denote its symbol $R_z(v,\eta )$. Thanks to [3], we even have the explicit expression $R_z(v,\eta )=G_z(v^2/2+2\eta ^2)\,\textrm{Id}$ where $G_z$ is an entire function defined by

$$\begin{aligned} G_z(\mu )&=\,2h^{-1}\int _0^1(1-s)^{-\frac{z}{h}}(1+s)^{\frac{z}{h}+d-2}\textrm{e}^{-\frac{s}{h}\mu } \textrm{d}s\\&=\,2\int _0^{h^{-1}}(1-h\sigma )^{-\frac{z}{h}}(1+h\sigma )^{\frac{z}{h}+d-2}\textrm{e}^{-\sigma \mu } \textrm{d}\sigma . \end{aligned}$$

Let us then set in view of (A.6)

$$\begin{aligned} M^h(v,\eta )&=\frac{-1}{2i\pi }\int _{\{\textrm{Re}\, z = -\frac{1}{2C}\}} \frac{\tilde{\varrho }(z)}{1+z} R_z(v,\eta ) \textrm{d}z +\frac{-1}{2i\pi }\int _{\{\textrm{Re}\, z = -\frac{1}{2C}\}}\nonumber \\&\frac{\varrho (z)-\varrho _\infty }{1+z} R_z(v,\eta ) \textrm{d}z +\varrho _\infty R_{-1}(v,\eta ) \end{aligned}$$

(A.7)

and we now want to show that $M^h$ is a matrix of symbols matching the properties listed in Hypothesis 1.3. To this purpose, we need to study more carefully the function $R_z$ for z fixed such that $\textrm{Re}\,z\le -1/2C$. We already saw that it is analytic in both variables v and $\eta$. Now, if we take $(v,\eta )\in \mathbb R^d\times \Sigma _\tau$ and put $\mu =v^2/2+2\eta ^2$, we get that $\mu$ belongs to the sector

$$\begin{aligned} D_\tau =\{\mu \in \mathbb C;\, |\textrm{Im}\,\mu |\le \textrm{Re}\,\mu +4d\tau ^2\}. \end{aligned}$$

One can then easily adapt Theorem 10 from [3] to show that for $n\in \mathbb N$ and $\mu \in D_\tau$, we have

$$\begin{aligned} |\partial _\mu ^n G_z(\mu )|&\le C \int _0^{h^{-1}} \sigma ^n (1-h\sigma )^{-\textrm{Re}\,z/h}(1+h\sigma )^{\textrm{Re}\,z/h}\textrm{e}^{-\textrm{Re}\,\mu \sigma } \textrm{d}\sigma \nonumber \\&\le C \int _0^{+\infty } \sigma ^n \textrm{e}^{-(\textrm{Re}\,\mu -2\textrm{Re}\,z) \sigma } \textrm{d}\sigma \le C_n\langle \mu \rangle ^{-(n+1)} \end{aligned}$$

(A.8)

since $\textrm{Re}\,\mu -2\textrm{Re}\,z>0$ for $\tau$ small enough. From (A.8) we can already conclude that $M^h\in \mathcal M_d\big (S^0_\tau (\langle (v,\eta ) \rangle ^{-2})\big )$. Thus, $\tilde{\varrho }(H_0+h)\otimes \textrm{Id}=\textrm{Op}_h(M^h)$ with $M^h$ sending $\mathbb R^{2d}$ in $\mathcal M_d(\mathbb R)$ as $H_0$ is self-adjoint. Moreover, since $R_z$ is diagonal and even in the variable $\eta$, it is also the case of $M^h$. It only remains to prove that $M^h$ satisfies items b) and d) from Hypothesis 1.3. In order to avoid some tedious computations, instead of proving the whole expansion from item b), we only show that $M^h$ admits a principal term $M_0$ in $\mathcal M_d\big (S^0_\tau (\langle (v,\eta ) \rangle ^{-2})\big )$ from which we will deduce that item d) is satisfied. One easily gets for $\textrm{Re} \, z\le -1/2C$ and $\mu \in D_\tau$ fixed by dominated convergence that

$$\begin{aligned} \lim _{h\rightarrow 0}G_z(\mu )=2\int _0^\infty \textrm{e}^{\sigma (2z-\mu )}\textrm{d}\sigma =\frac{1}{\mu /2-z}=:G_z^0(\mu ). \end{aligned}$$

(A.9)

We would like to get some estimates of the derivatives $\partial _\mu ^n (G_z-G_z^0)$ in $O(h\langle \mu \rangle ^{-n-1})$ on $D_\tau$ uniformly in $z\in \{\textrm{Re}\, z\le -1/2C\}$ in order to apply the formula (A.7) to those. We have

$$\begin{aligned}&\partial _\mu ^n (G_z-G_z^0)(\mu )=2\int _0^{h^{-1}} \nonumber \\&\quad \bigg [ \exp \bigg (z\,\Big [\frac{1}{h} \ln \Big (\frac{1+h\sigma }{1-h\sigma }\Big )-2\sigma \Big ]+(d-2)\ln (1+h\sigma )\bigg )-1\bigg ](-\sigma )^n\textrm{e}^{\sigma (2z-\mu )}\textrm{d}\sigma \nonumber \\&\qquad - 2\int _{h^{-1}}^{\infty } (-\sigma )^n e^{\sigma (2z-\mu )}\textrm{d}\sigma \nonumber \\&\quad =2\int _0^{h^{-1}/2} \bigg [ \exp \bigg (z\,\Big [\frac{1}{h} \ln \Big (\frac{1+h\sigma }{1-h\sigma }\Big )-2\sigma \Big ]+(d-2)\ln (1+h\sigma )\bigg )-1\bigg ]\\&\qquad (-\sigma )^n\textrm{e}^{\sigma (2z-\mu )}\textrm{d}\sigma +O\Big (\textrm{e}^{\frac{\textrm{Re}\,(2z-\mu )}{Ch}}\Big )\nonumber . \end{aligned}$$

(A.10)

Let us denote

$$\begin{aligned} g_{z,h}(\sigma )=\bigg [ \exp \bigg (z\,\Big [\frac{1}{h} \ln \Big (\frac{1+h\sigma }{1-h\sigma }\Big )-2\sigma \Big ]+(d-2)\ln (1+h\sigma )\bigg )-1\bigg ](-\sigma )^n \end{aligned}$$

and observe that for all $0\le k \le n$, one has

$$\begin{aligned} \partial _\sigma ^k g_{z,h}(0)=0 \qquad \text {and }\qquad \partial _\sigma ^k g_{z,h}(h^{-1}/2)=O(h^{-n}\langle z \rangle ^k). \end{aligned}$$

(A.11)

Besides, on $\sigma \in [0,h^{-1}/2]$, it holds

$$\begin{aligned} \partial _\sigma ^{n+1} g_{z,h}(\sigma )=\sum _{j=1}^{n+1}O\big (h\langle z \rangle ^{j} \langle \sigma \rangle ^{j}\sigma ^{j-1}\big ). \end{aligned}$$

(A.12)

Now, let us do $n+1$ integrations by parts in the first term from (A.10). By (A.11), each boundary term is $O(h^{-n}\langle z \rangle ^{k} \langle 2z-\mu \rangle ^{-(k+1)}\textrm{e}^{\textrm{Re}\,(2z-\mu )/Ch})$, while the remaining integral term satisfies

$$\begin{aligned}&\bigg |\frac{2}{(\mu -2z)^{n+1}}\int _0^{h^{-1}/2}\partial _\sigma ^{n+1} g_{z,h}(\sigma ) \textrm{e}^{\sigma (2z-\mu )}\textrm{d}\sigma \bigg |\\&\quad \le C_nh\sum _{j=1}^{n+1}\frac{\langle z\rangle ^j}{|2z-\mu |^{n+1}}\int _0^\infty \sigma ^{j-1}\langle \sigma \rangle ^{j} \textrm{e}^{\sigma \, \textrm{Re}(2z-\mu )}\textrm{d}\sigma \\&\quad \le C_n h \langle \mu \rangle ^{-(n+1)} \end{aligned}$$

thanks to (A.12). Thus, we have shown that for $n\in \mathbb N$, $\mu \in D_\tau$ and $\textrm{Re}\,z\le -1/2C$,

$$\begin{aligned} |\partial _\mu ^n (G_z-G_z^0)(\mu )|\le C_n h \langle \mu \rangle ^{-(n+1)}. \end{aligned}$$

Putting $R_z^0(v,\eta )=G_z^0(v^2/2+2\eta ^2)\,\textrm{Id}$ and defining $M_0(v,\eta )$ as in (A.7) with $R_z$ replaced by $R_z^0$, we deduce that

$$\begin{aligned} |\partial ^\alpha (M^h-M_0)(v,\eta )|\le C_\alpha h\langle (v,\eta ) \rangle ^{-2}\qquad \text {on }\mathbb R^d\times \Sigma _\tau \end{aligned}$$

so item b) from Hypothesis 1.3 holds true. Finally, by definition of $M_0$ and thanks to (A.9) and (A.2), we have

$$\begin{aligned} M_0(v,\eta )=\tilde{\varrho }\big (v^2/4+\eta ^2\big )\,\textrm{Id} \ge \frac{1}{C} \langle (v,\eta ) \rangle ^{-2}\,\textrm{Id} \end{aligned}$$

(A.13)

by assumption on $\varrho$. Therefore, item d) from Hypothesis 1.3 holds true and the proof is complete.

Appendix B: Linear Algebra Lemma

We use the following lemma which is inspired by [1], Lemma 2.6.

Lemma B.1

Let $M\in \mathcal M_{d'}(\mathbb C)$ such that $M=S(A+T)$ with S Hermitian and invertible, A skew-Hermitian and T Hermitian positive semidefinite. Suppose moreover that

$$\begin{aligned} M(\textrm{Ker}\,T) \cap \textrm{Ker}\,T=\textrm{Ker}\,M\cap \textrm{Ker}\,T=\{0\}. \end{aligned}$$

Then, M has no spectrum in $i\mathbb R$.

Proof

Let $\lambda \in \mathbb R$ and $X\in \textrm{Ker}\, [M-i\lambda ]$, we first show that $X \in$ Ker T. Since T is Hermitian positive semidefinite, it is sufficient to show that $\langle TX,X\rangle =0$. Using the properties of S, A and T, we have

$$\begin{aligned} \langle TX,X\rangle&=\textrm{Re}\,\big \langle (A+T)X,X\big \rangle \\&=\textrm{Re}\,\big \langle S^{-1}S(A+T)X,X\big \rangle \\&=\textrm{Re}\,\big (i\lambda \big \langle S^{-1}X,X\big \rangle \big )\\&=0 \end{aligned}$$

so $X \in$ Ker T. Thanks to the assumption, it only remains to prove that $X \in$ Ker M. This can be done easily by noticing that

$$\begin{aligned} MX=i\lambda X \in M(\textrm{Ker}\,T) \cap \textrm{Ker}\,T \end{aligned}$$

so $MX=0$ by assumption.

Appendix C: Asymptotic Expansions

Let $d'\in \mathbb N^*$. Here, we use the convention $\sum _{j=0}^{-1}a_j=0$ for any sequence $(a_j)_{j\ge 0}$ in a vector space. For $K\subseteq \mathbb R^{d'}$, the notation $a=O_{\mathcal C^{\infty }(K)}(h^N)$ (respectively, $a=O_{L^{\infty }(K)}(h^N)$) means that for all $\beta \in \mathbb N^{d'}$, there exists $C_{\beta , N}$ such that $\Vert \partial ^\beta a\Vert _{\infty ,K}\le C_{\beta , N} h^N$ (resp. there exists $C_{ N}$ such that $\Vert a\Vert _{\infty ,K}\le C_{ N} h^N$). We will also use the notations from Definition 1.2 and (1.6).

Proposition C.1

Let $m\in \mathbb N^*$; $d_1,\dots , d_m \in \mathbb N^*$ and for $1 \le j \le m$, $K_j\subset \mathbb R^{d_j}$ some compact sets. Let a smooth function

$$\begin{aligned} \phi _h:\prod _{j=1}^m K_j\rightarrow K\subset \Sigma _\tau \end{aligned}$$

such that $\phi _h=O_{\mathcal C^{\infty }(\prod _{j=1}^m K_j)}(1)$. Consider $g^h \sim _h \sum _{n \ge 0}h^ng_n$ in $S^0_\tau (1)$ or in $\mathcal C^{\infty }(K)$ if $\phi _h$ actually takes values in $\mathbb R^d$. Then,

$$\begin{aligned} g^h\circ \phi _h \sim _h \sum _{n \ge 0}h^n(g_n\circ \phi _h) \end{aligned}$$

in $\mathcal C^{\infty }(\prod _{j=1}^m K_j)$.

Proof

Let $N \in \mathbb N$ and denote $r_N=g^h-\sum _{n=0}^{N-1}h^n g_n=O_{S^0_\tau (1)}(h^N)$.

$$\begin{aligned} g^h\circ \phi _h&=\left( \sum _{n=0}^{N-1}h^n g_n+r_N\right) \circ \phi _h\\&=\sum _{n=0}^{N-1}h^n (g_n\circ \phi _h) +r_N\circ \phi _h. \end{aligned}$$

But since all the derivatives of $\phi _h$ are bounded uniformly in h, and the ones of $r_N$ are $O_{L^{\infty }(\Sigma _\tau )}(h^N)$, we see that $r_N\circ \phi _h$ is $O_{\mathcal C^{\infty }(\prod _{j=1}^m K_j)}(h^N)$ so we have the announced result.

Proposition C.2

Since the matrix $M^h$ from Hypothesis 1.3 satisfies $M^h \sim \sum _{n \ge 0}h^nM_n$ in $\mathcal M_{d}\big (S^0_\tau (\langle (v,\eta ) \rangle ^{-2})\big )$, the vector of symbols $g^h$ defined in Remark 3.12 also admits a classical expansion $g^h \sim \sum _{n \ge 0}h^ng_n$ in $\mathcal M_{1,d}\big (S^0_\tau (\langle (v,\eta ) \rangle ^{-1})\big )$, where the $(g_n)$ are given by

$$\begin{aligned} g_0(x,v,\eta )=\Big (-i\, {}^t\eta + \, \frac{{}^tv}{2}\Big )M_0(x,v, \eta ) \end{aligned}$$

and

$$\begin{aligned} g_n(x,v,\eta )=\left( -i\, {}^t\eta + \, \frac{{}^tv}{2}\right) M_n(x,v,\eta )-\frac{1}{2} \left( {}^t\nabla _v-\frac{i}{2} {}^t\nabla _\eta \right) M_{n-1}(x,v,\eta ) \end{aligned}$$

for $n\ge 1.$

Proof

We have

$$\begin{aligned} g^h=(-i\, {}^t\eta +\, {}^tv/2)M^h-\frac{h}{2}\left( {}^t\nabla _v-\frac{i}{2} {}^t\nabla _\eta \right) M^h \end{aligned}$$

, and the last term clearly admits the expansion

$$\begin{aligned} -\sum _{n\ge 1}h^n\frac{1}{2} \left( {}^t\nabla _v-\frac{i}{2} {}^t\nabla _\eta \right) M_{n-1} \end{aligned}$$

in $S^0_\tau (\langle (v,\eta ) \rangle ^{-2})$. For the first term of $g^h$, it suffices to notice that for any $N\in \mathbb N$,

$$\begin{aligned} \Big (-i\, {}^t\eta + \, \frac{{}^tv}{2}\Big )\;O_{\mathcal M_{d}\big (S^0_\tau (\langle (v,\eta ) \rangle ^{-2})\big )}(h^N)=O_{\mathcal M_{1,d}\big (S^0_\tau (\langle (v,\eta ) \rangle ^{-1})\big )}(h^N). \end{aligned}$$

Proposition C.3

Let K a compact set in $\mathbb R^{d'}$ and $a\sim _h \sum _{n\ge 0}h^n a_n$ in $\mathcal C^{\infty }(K)$ such that for all $n \ge 0$, $a_n\sim _h \sum _{j\ge 0}h^j a_{n,j}$ in $\mathcal C^{\infty }(K)$. Then,

$$\begin{aligned} a \sim _h \sum _{n \ge 0}h^n \sum _{j=0}^n a_{j,n-j} \quad \text {in } \mathcal C^{\infty }(K). \end{aligned}$$

Proof

It suffices to write for $N \in \mathbb N$

$$\begin{aligned} a&=\sum _{n=0}^{N-1} h^n \left( \sum _{j=0}^{N-1-n} h^j a_{n,j}+O_{\mathcal C^{\infty }(K)}(h^{N-n}) \right) +O_{\mathcal C^{\infty }(K)}(h^N)\\&=\sum _{n = 0}^{N-1}h^n \sum _{j=0}^n a_{j,n-j} +O_{\mathcal C^{\infty }(K)}(h^N). \end{aligned}$$

Proposition C.4

Let K a compact set in $\mathbb R^{d'}$ and $a \in \mathcal C^\infty (K)$ such that for all $\beta \in \mathbb N^{d'}$, there exists $a_{\beta ,j}\in \mathcal C^\infty (K)$ such that $\partial ^\beta a \sim \sum _{j\ge 0}h^ja_{\beta ,j}$ in $L^\infty (K)$. Then, $a_{\beta ,j}=\partial ^\beta a_{0,j}$, i.e.,

$$\begin{aligned} a \sim \sum _{j\ge 0}h^ja_{0,j} \quad \text {in } \mathcal C^\infty (K). \end{aligned}$$

Proof

For simplicity, we take $d'=1$. Let us denote $a_j=a_{0,j}$. By induction, it is sufficient to prove the result for $\beta =1$, i.e., prove that $a_{1,j}=a_j'$. Here again, it suffices to prove the case $j=0$ which we can then apply to the function $h^{-1}(a-a_0)$ and so on. Let x in the interior of K and $t\in \mathbb R^*$ in a neighborhood of 0. We look at the differential fraction

$$\begin{aligned} \frac{a_0(x+t)-a_0(x)}{t}&=\frac{a(x+t)-a(x)}{t}+\frac{O(h)}{t}\\&=a'(x)+t\int _0^1 (1-s)a''(x+st) \textrm{d}s+\frac{O(h)}{t}\\&=a_{1,0}(x)+O(h)+t\int _0^1 (1-s)a''(x+st) \textrm{d}s+\frac{O(h)}{t}\\&\xrightarrow [h \rightarrow 0]{} a_{1,0}(x)+t\int _0^1 (1-s)a_{2,0}(x+st) \textrm{d}s. \end{aligned}$$

Taking now the limit $t \rightarrow 0$, we get $a_0'(x)=a_{1,0}(x)$ which was the desired result.

Proposition C.5

Recall the notation (3.11) and let $K\subset \mathbb R^{d'}$ a compact set, $\Psi :K\rightarrow D(0, \tau )^d$ a smooth function such that $\Psi \sim \sum _{j\ge 0}h^j \Psi _j$ in $\mathcal C^\infty (K)$ and b an analytic function on $\Sigma _\tau$. Then,

$$\begin{aligned} b \circ \Psi \sim \sum _{j\ge 0}h^j b_j \end{aligned}$$

(C.1)

in $\mathcal C^\infty (K)$, with

$$\begin{aligned} b_0=b \circ \Psi _0 \quad \; \text {and for }j \ge 1, \;\quad b_j=\sum _{|\beta |=1}^j \frac{\partial ^\beta b \circ \Psi _0}{\beta !}\sum _{s\in S_{\beta ,j}} \prod _{k \in K_\beta } \left( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \big (\Psi _{a_l}\big )_k \right) , \end{aligned}$$

where $K_\beta =$ supp $\beta =\{k \in \llbracket 1,d \rrbracket \,;\,\beta _k\ne 0\}$, $S_{\beta ,j}=\{s \in \mathbb N^d \,;\, \textrm{supp}\, s=K_\beta , \, |s|=j \text { and }s\ge \beta \}$ and $A_{\beta , s, k}=\{a \in (\mathbb N^*)^{\beta _k} \,;\, |a|=s_k\}$.

Proof

We first prove that (C.1) holds in $L^\infty (K)$. Doing a Taylor expansion of b, we have for $N \in \mathbb N^*$ that

$$\begin{aligned} b\circ \Psi&= b \circ \Psi _0 + \sum _{|\beta |=1}^{N-1} \frac{\partial ^\beta b \circ \Psi _0 }{\beta !}(\Psi - \Psi _0)^\beta + O\Big ((\Psi - \Psi _0)^N\Big ) \nonumber \\&= b \circ \Psi _0 + \sum _{|\beta |=1}^{N-1} \frac{\partial ^\beta b \circ \Psi _0 }{\beta !}(\Psi - \Psi _0)^\beta +O_{L^\infty (K)}(h^N) \end{aligned}$$

(C.2)

since $\Psi -\Psi _0=O_{\mathcal C^\infty (K)}(h)$. Now, one can see that

$$\begin{aligned} (\Psi -\Psi _0)^\beta \sim \sum _{j\ge |\beta |}h^j\sum _{s\in S_{\beta ,j}} \prod _{k \in K_\beta } \Bigg ( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \big (\Psi _{a_l}\big )_k \Bigg ) \end{aligned}$$

so (C.2) gives

$$\begin{aligned} b\circ \Psi&= b \circ \Psi _0+\sum _{|\beta |=1}^{N-1} \frac{\partial ^\beta b \circ \Psi _0 }{\beta !} \Bigg [\sum _{j= |\beta |}^{N-1} h^j\sum _{s\in S_{\beta ,j}} \prod _{k \in K_\beta } \left( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \left( \Psi _{a_l}\right) _k \right) +O_{\mathcal C^\infty (K)}(h^N) \Bigg ]\\&\quad +O_{L^\infty (K)}(h^N) \\&=b \circ \Psi _0+\sum _{j=1}^{N-1} h^j \sum _{|\beta |=1}^{j} \frac{\partial ^\beta b \circ \Psi _0 }{\beta !} \sum _{s\in S_{\beta ,j}} \prod _{k \in K_\beta } \Big ( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \big (\Psi _{a_l}\big )_k \Big ) +O_{L^\infty (K)}(h^N) \end{aligned}$$

which proves that (C.1) holds in $L^\infty (K)$.

Besides, the derivatives of $b \circ \Psi$ are linear combinations of products of some derivatives of $\Psi$ with some $\partial ^\gamma b \,\circ \Psi$ where $\gamma$ is a integer multi-index. Hence, the expansion of $\Psi$ in $\mathcal C^\infty (K)$ and the result that we just proved applied to $\partial ^\gamma b \circ \Psi$ instead of $b\circ \Psi$ yield that for all $\beta \in \mathbb N^{d'}$, $\partial ^\beta (b\circ \Psi )$ admits a classical expansion in $L^\infty (K)$ whose coefficients are smooth. Therefore, Proposition C.4 enables us to conclude that (C.1) holds in $\mathcal C^\infty (K)$.

Corollary C.6

Using the notations from the proof of Lemma 4.1, we have

$$\begin{aligned} g_n\Big (x,\frac{v+v'}{2},\eta +i\psi (x,v,v')\Big )\sim \sum _{j\ge 0} h^j g_{n,j}(x,v,v',\eta ) \quad \text {on } B_0(\textbf{s},2r)\times B_\infty (0,2r) \end{aligned}$$

with

$$\begin{aligned} g_{n,0}(x,v,v',\eta )=g_n\Big (x,\frac{v+v'}{2},\eta +i\psi _0(x,v,v')\Big ) \end{aligned}$$

and for $j \ge 1$

$$\begin{aligned} g_{n,j}(x,v,v',\eta )=iD_\eta g_n\Big (x,\frac{v+v'}{2},\eta +i\psi _0(x,v,v')\Big ) \big (\psi _j(x,v,v')\big )+R^1_j(\ell _0, \dots , \ell _{j-1}) \end{aligned}$$

where $R^1_j: \big ( \mathcal C^\infty (B_0(\textbf{s},2r) )\big )^j \rightarrow \mathcal C^\infty (B_0(\textbf{s},2r) ).$

Proof

Since $\psi (\textbf{s},0, 0)=O(h)$, we can suppose that r was chosen small enough so that $(x,v,v',\eta )\mapsto \eta +i\psi (x,v,v')$ sends $B_0(\textbf{s},2r) \times B_\infty (0,2r)$ in $D(0, \tau )^d$ Hence, we can use Proposition C.5 to get that

$$\begin{aligned} g_n\Big (x,\frac{v+v'}{2},\eta +i\psi (x,v,v')\Big )\sim \sum _{j\ge 0} h^j g_{n,j}(x,v,v',\eta ) \quad \text {on } B_0(\textbf{s},2r)\times B_\infty (0,2r) \end{aligned}$$

with

$$\begin{aligned} g_{n,0}(x,v,v',\eta )=g_n\Big (x,\frac{v+v'}{2},\eta +i\psi _0(x,v,v')\Big ) \end{aligned}$$

and for $j \ge 1$

$$\begin{aligned} g_{n,j}(x,v,v',\eta )&=\sum _{|\beta |=1}^j \frac{i^{|\beta |}}{\beta !}\partial ^\beta _\eta g_n \Big (x,\frac{v+v'}{2},\eta +i\psi _0(x,v,v')\Big )\sum _{s\in S_{\beta ,j}} \nonumber \\&\quad \prod _{k \in K_\beta } \bigg ( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \big (\psi _{a_l}\big )_k \bigg ) \end{aligned}$$

(C.3)

where $K_\beta =$ supp $\beta =\{k \in \llbracket 1,d \rrbracket \,;\,\beta _k\ne 0\}$, $S_{\beta ,j}=\{s \in \mathbb N^d \,;\, \textrm{supp}\, s=K_\beta , \, |s|=j \text { and }s\ge \beta \}$ and $A_{\beta , s, k}=\{a \in (\mathbb N^*)^{\beta _k} \,;\, |a|=s_k\}$. Now, we see thanks to (C.3) that the terms of $g_{n,j}(x,v,v',\eta )$ for which $|\beta |=1$ yield

$$\begin{aligned} iD_\eta g_n\Big (x,\frac{v+v'}{2},\eta +i\psi _0(x,v,v')\Big ) \big (\psi _j(x,v,v')\big ), \end{aligned}$$

while the terms for which $|\beta |>1$ only feature the functions $\ell _0, \dots , \ell _{j-1}$.

Finally, we state the version of Laplace’s method for integral approximation that we use in this paper.

Proposition C.7

Let $x_0 \in \mathbb R^{d'}$, K be a compact neighborhood of $x_0$ and $\varphi \in \mathcal C^\infty (K)$ such that $x_0$ is a non-degenerate minimum of $\varphi$ and its only global minimum on K. Let also $a_h \sim \sum _{j\ge 0}h^j a_j$ in $\mathcal C^\infty (K)$ and denote $H\in \mathcal M_{d'}(\mathbb R)$ the Hessian of $\varphi$ at $x_0$. The integral

$$\begin{aligned} \frac{\det (H)^{1/2}}{(2\pi h)^{d'/2}} \int _{K}a_h(x)\textrm{e}^{-\frac{\varphi (x)-\varphi (x_0)}{h}}\textrm{d}x \end{aligned}$$

admits a classical expansion whose first term is given by $a_0(x_0)$.

Appendix D: Proof of Lemma 4.3

According to the proof of Corollary C.6 and the end of the proof of Lemma 4.1 from which we keep the notations, we have the following expression for $R_j$:

$$\begin{aligned}&R_j(\ell _0, \dots , \ell _{j-1})(x,v) \nonumber \\&\quad =\mathop {\sum _{n_1+n_2+n_3+n_4=j}}_{\begin{array}{c} n_3, n_4 \ne j \end{array}}\frac{1}{i^{n_1}n_1!}\big (\partial _{v'} \cdot \partial _\eta \big )^{n_1} \Big ( g_{n_2,n_3}(x,v,v',\eta ) \partial _v \ell _{n_4}(x,v') \Big ) \Bigg | \mathop {}_{\begin{array}{c} v'=v \\ \eta =0 \end{array}} \\&\qquad +\sum _{|\beta |=2}^j \frac{i^{|\beta |}}{\beta !}\partial ^\beta _\eta g_0 \Big (x,\frac{v+v'}{2},i\big (\frac{v}{2} +\ell _0 (x,v)\, \partial _v \ell _0(x,v)\big )\Big )\nonumber \\&\qquad \times \sum _{s\in S_{\beta ,j}} \prod _{k \in K_\beta } \bigg ( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \big (\psi _{a_l}(x,v,v)\big )_k \bigg ) \partial _v \ell _0(x,v) \nonumber \\&\qquad +iD_\eta g_0\big (x,v,i(v/2+\ell _0 (x,v)\, \partial _v \ell _0(x,v))\big )\sum _{k=1}^{j-1}\big (\ell _k\partial _v \ell _{j-k}\big )(x,v)\; \partial _v \ell _0(x,v).\nonumber \end{aligned}$$

(D.1)

Using Lemma 4.2 and (C.3), it is clear that the last two terms of $R_j(\ell _0, \cdots , \ell _{j-1})$ given by (D.1) and the terms of the first sum for which $n_1=0$ are real valued. For the rest of the first term, we start by noticing that one can establish by induction that for $n_1 \ge 1$,

$$\begin{aligned} \big (\partial _{v'}\cdot \partial _\eta \big )^{n_1}=\sum _{p\in \llbracket 1,d\rrbracket ^{n_1}}\partial _{v'}^{\gamma (p)}\partial _{\eta }^{\gamma (p)} \end{aligned}$$

(D.2)

where using the notation (3.22), we define $\gamma (p)=\sum _{k=1}^{n_1}e_{p_k}$ (note that $|\gamma (p)|=n_1$). Besides, we have for $0 \le n_2 \le j$ and $p\in \llbracket 1,d\rrbracket ^{n_1}$

$$\begin{aligned} \partial _{\eta }^{\gamma (p)}g_{n_2,0}(x,v,v',0)=\partial _{\eta }^{\gamma (p)}g_{n_2}\Big (x,\frac{v+v'}{2},i\psi _0(x,v,v')\Big )\in i^{n_1}\mathbb R^d \end{aligned}$$

(D.3)

according to Lemma 4.2 and in the case $j\ge 2$, for $1 \le n_3 \le j-1$

$$\begin{aligned} \qquad \;\,\partial _{\eta }^{\gamma (p)}&g_{n_2,n_3}(x,v,v',0) \\&=\sum _{|\beta |=1}^{n_3} \frac{i^{|\beta |}}{\beta !}\partial ^{\beta +\gamma (p)}_\eta g_{n_2} \Big (x,\frac{v+v'}{2},i\psi _0(x,v,v')\Big ) \sum _{s\in S_{\beta ,n_3}} \prod _{k \in K_\beta } \nonumber \\&\quad \bigg ( \sum _{a \in A_{\beta , s, k}} \prod _{l=1}^{\beta _k} \big (\psi _{a_l}\big )_k \bigg )\in i^{n_1}\mathbb R^d \nonumber \end{aligned}$$

(D.4)

where we used (C.3) and Lemma 4.2 once again. The combination of (D.2), (D.3) and (D.4) enables us to conclude that the term

$$\begin{aligned} \mathop {\sum _{n_1+n_2+n_3+n_4=j}}_{\begin{array}{c} n_1\ne 0;\, n_3, n_4 \ne j \end{array}}\frac{1}{i^{n_1}n_1!}\big (\partial _{v'} \cdot \partial _\eta \big )^{n_1} \Big ( g_{n_2,n_3}(x,v,v',\eta ) \partial _v \ell _{n_4}(x,v') \Big ) \Bigg | \mathop {}_{\begin{array}{c} v'=v \\ \eta =0 \end{array}} \end{aligned}$$

from (D.1) is also real so $R_j(\ell _0^{\textbf{s}}, \dots , \ell _{j-1}^{\textbf{s}})$ is real valued. For the last statement, it suffices to use the formula (D.1) after noticing that $\psi$ (and hence the $(g_{n_2,n_3})$) remain unchanged when $\ell$ is replaced by $-\ell$.