Scattering Structure and Landau Damping for Linearized Vlasov Equations with Inhomogeneous Boltzmannian States

Abstract

We study the linearized Vlasov–Poisson–Ampère equation for non-constant Boltzmannian states with one region of trapped particles in dimension one and construct the eigenstructure in the context of the scattering theory. This is based on the use of semi-discrete variables (moments in velocity), and it leads to a new Lippmann–Schwinger variational equation. The continuity in quadratic norm of the operator is proved, and the well posedness is proved for a small value of the scaling parameter. It gives a proof of Linear Landau damping for inhomogeneous Boltzmannian states. The linear HMF model is an example.

Appendices

Appendix A. Monotony of the Time of Travel

1.1 A.1. First Branch: $e\in (-\varphi _0^-,\infty )$

We distinguish the behavior for large $e\gg 1$ in Lemma A.1, near the lower bound $e=-\varphi _0^-+s$ for small $s>0$ in Lemma 6.2 and the monotony in between in Lemma 6.3. It allows to define properly the inverse function in Proposition 3.8.

Lemma A.1

(Evident). One has ${\widehat{t}}_e= \frac{1}{\sqrt{2e}} + O\left( \frac{1}{ \text {e}^{\frac{3}{2}} } \right)$ for large $e\gg 1$.

Lemma 6.2

There exists a constant $C\in {\mathbb {R}}$ such that

$$\begin{aligned} {\widehat{t}}_{-\varphi _0^-+{\overline{s}}}= - \frac{1}{\sqrt{\varphi _0''(0)}} \log {\overline{s}} +C + O({\overline{s}}), \quad {\overline{s}}>0. \end{aligned}$$

(A.1)

Proof

One has ${\widehat{t}}_{-\varphi _0^-+{\overline{s}}}= \int _0^1 \frac{\text {d}z}{\sqrt{2({\overline{s}}+\varphi _0(z)-\varphi _0^-)}}$. Consider $A=\int _0^\frac{1}{2} \frac{\text {d}z}{\sqrt{2({\overline{s}}+\varphi _0(z)-\varphi _0^-)}}$ which is decomposed as $A=\underbrace{ \int _0^\frac{1}{2} \frac{\text {d}z}{\sqrt{2\left( {\overline{s}}+\frac{1}{2} \varphi _0''(0)z^2\right) }} }_{=A_1}$$+ \underbrace{\int _0^\frac{1}{2} \left( \frac{1}{\sqrt{2({\overline{s}}+\varphi _0(z)-\varphi _0^-)}} - \frac{1}{\sqrt{2({\overline{s}}+\frac{1}{2} \varphi _0''(0)z^2}} \right) \text {d}z}_{=A_2}$. Make the change of variable $z=\sqrt{\frac{2{\overline{s}} }{\varphi _0''(0)}}y$ and denote $a_{{\overline{s}}}= 1/(2\sqrt{2{\overline{s}}/\varphi _0''(0)})$, so $A_1= \frac{1}{\sqrt{\varphi _0''(0)}}\int _0^{a_{{\overline{s}}} }\frac{\text {d}y}{\sqrt{1+y^2}}= \frac{1}{\sqrt{\varphi _0''(0)}} \log \left( a_{{\overline{s}}} + \sqrt{1+a_{{\overline{s}}} ^2} \right)$. So $A_1= - \frac{1}{2 \sqrt{\varphi _0''(0)}} \log {\overline{s}} +d_{\varphi _0} + O({\overline{s}})$ for some constant $d_{\varphi _0}$. So $A= - \frac{1}{2 \sqrt{\varphi _0''(0)}} \log {\overline{s}} +e_{\varphi _0} + O({\overline{s}})$ for some constant $e_{\varphi _0}$. The integral $B=\int _\frac{1}{2}^1 \frac{\text {d}z}{\sqrt{2({\overline{s}}+\varphi _0(z)-\varphi _0^-)}}$ has the same asymptotic behavior. It yields the claim after summation of A and B. $\square$

Lemma 6.3

(Evident). In interval $(-\varphi _0^-, \infty )$, $e\mapsto {\widehat{t}}_e$ is monotone decreasing from $+\infty$ to 0.

Proof of Proposition 3.8

The first point is evident. Second point: let us shift for convenience the functions ${\overline{s}}(\lambda )=s(\lambda )+\varphi _0^-$ and $\psi (t)=\varphi (t)-\varphi _0^-$. The expansion of $s(\lambda )$ is immediate from Lemma A.1. By successive derivations of the first equation of (3.14), one obtains $- \frac{\sqrt{2}}{\lambda ^2}= - \frac{1}{2} \left( \int _0^1 \frac{\text {d}t}{({\overline{s}}(\lambda ) +\psi (t))^\frac{3}{2}} \right) {\overline{s}}'(\lambda )$, $\frac{2 \sqrt{2}}{\lambda ^3}= - \frac{1}{2} \left( \int _0^1 \frac{\text {d}t}{({\overline{s}}(\lambda ) +\psi (t))^\frac{3}{2}} \right) {\overline{s}}''(\lambda ) + \frac{3}{4} \left( \int _0^1 \frac{\text {d}t}{({\overline{s}}(\lambda ) +\psi (t))^\frac{5}{2}} \right) {\overline{s}}'(\lambda )^2$ and $- \frac{6 \sqrt{2}}{\lambda ^4} = - \frac{1}{2} \left( \int _0^1 \frac{\text {d}t}{({\overline{s}}(\lambda ) +\psi (t))^\frac{3}{2}} \right) {\overline{s}}'''(\lambda )$$+ \frac{9}{4} \left( \int _0^1 \frac{\text {d}t}{(s(\lambda ) +\psi (t))^\frac{5}{2}} \right) s'(\lambda ){\overline{s}}''(\lambda ) - \frac{15}{8} \left( \int _0^1 \frac{\text {d}t}{({\overline{s}}(\lambda ) +\psi (t))^\frac{7}{2}} \right) {\overline{s}}'(\lambda )^3$. For large $\lambda$, the estimates s on ${\overline{s}}'=s'$, ${\overline{s}}''=s''$ and ${\overline{s}}'''=s'''$ are obtained one after the other, and so on for increasing order of derivation. Third point: for small $0<\lambda$, the expansion ${\overline{s}}(\lambda )=\alpha _{\varphi _0} \text {e}^{-\frac{\sqrt{\varphi _0''(0)}}{2\lambda } }(1+ \sigma (\lambda ))$ is obtained from (A.1). Moreover, we note that there exists $\alpha >0$ and $\beta$ such that $\alpha t^2\le \psi (t)\le \beta t^2$ on the interval $t\in [0,1/2]$. Therefore, for small $\tau >0$ and $p>1$

$$\begin{aligned} \frac{\beta _p}{\tau ^{(p-1)/2}} \le 2 \int _0^{\frac{1}{2}} \frac{\text {d}t}{(\tau +\beta t^2)^\frac{p}{2}} \le \int _0^1 \frac{\text {d}t}{(\tau +\psi (t))^\frac{p}{2}} \le 2 \int _0^{\frac{1}{2}} \frac{\text {d}t}{(\tau +\alpha t^2)^\frac{p}{2}} \le \frac{\alpha _p}{\tau ^{(p-1)/2}} \end{aligned}$$

for some constants $\alpha _p$ and $\beta _p>0$. It yields the first bound $|{\overline{s}}'(\lambda )| \le c_1 \frac{{\overline{s}}(\lambda )}{\lambda ^2}$, and it is sufficient to insert this expression in the other identities to obtain the results for the first derivatives. $\square$

1.2 A.2. Second Branch: $e\in (-\varphi _0^+, -\varphi _0^-)$

We note $e= -\varphi _0^--s \Longleftrightarrow {\overline{s}} = -e -\varphi _0^-$ for small $s>0$.

Lemma 6.4

There exists $C\in {\mathbb {R}}$ such that ${\widehat{t}}_{-\varphi _0^- -{\overline{s}} }= - \frac{1 }{\sqrt{\varphi _0''(0)}} \log {\overline{s}} +C + \nu ({ {\overline{s}} })$ for $\varphi _0^+-\varphi _0^-> {\overline{s}} >0$, where $\lim _{{\overline{s}} \rightarrow 0^+}\nu ({ {\overline{s}} })=0$.

Proof

The proof is similar to the one of Lemma 6.2 in (3.14). One cuts the second integral (3.14) in two pieces and study the first piece $A= 2 \int _{{\widehat{a}}_e}^{x_0} \frac{ \text {d}z}{\sqrt{2(\psi (z) -{\overline{s}} )}}$ where $\psi (z)=\varphi _0(z)-\varphi _0^-$ and $\psi ''(0)=\varphi _0''(0)$. One has

$$\begin{aligned} A=2 \sqrt{\frac{2}{\varphi ''_0(0)}} \underbrace{ \int _{{\widehat{a}}_e}^{x_0 }\frac{ \frac{\psi '(z)}{2\sqrt{\psi (z)}} }{\sqrt{2(\psi (z)-{\overline{s}} )}} \text {d}z }_{=A_1}\,+\, 2 \sqrt{\frac{2}{\varphi ''_0(0)}} \underbrace{ \int _{{\widehat{a}}_e}^{x_0 } \frac{ \sqrt{\frac{\varphi ''_0(0)}{2}}- \frac{\psi '(z)}{2\sqrt{\psi (z)}} }{\sqrt{2(\psi (z)-{\overline{s}} )}} \text {d}z }_{=A_2}. \end{aligned}$$

Note that $\psi ({\widehat{a}}_e)=\varphi _0({\widehat{a}}_e)-\varphi _0^-=-e-\varphi _0^-={\overline{s}}$. A change of variable $\psi (z)={\overline{s}} u^2$ (the differential is $\frac{\psi '(z)}{2\sqrt{\psi (z)}} \text {d}z =\sqrt{ {\overline{s}} }\text {d}u$) in the first integral yields $A_1=\frac{1}{\sqrt{2}} \int _1^{c } \frac{\text {d}u}{\sqrt{u^2-1}} = \frac{1}{\sqrt{2}} \log \left( c+\sqrt{c^2-1} \right)$ with $c= \sqrt{\psi (x_0)/{\overline{s}} }$. So there exists a constant $k_{\varphi _0}$ such that $A_1= -\frac{1}{2 \sqrt{2}} \log {\overline{s}} + k_{\varphi _0} +o({\overline{s}} )$. Concerning $A_2$, we notice that $D(z):=\frac{ \sqrt{\frac{\varphi ''_0(0)}{2}}- \frac{\psi '(z)}{2\sqrt{\psi (z)}} }{\sqrt{\psi (z)}}= \frac{ \sqrt{2\psi ''(0)\psi (z)}- \psi '(z) }{2\psi (z)}$. But $\psi (0)=\psi '(0)=0$, $\psi ''(0)>0$ and $\psi$ is of class $W^{3,\infty }$. Therefore, $D\in L^\infty (I)$ and $A_2$ admits a limit as $s\rightarrow 0^+$: $A_2= \int _{{\widehat{a}}_e}^{x_0 } D(z) \frac{ \sqrt{ \psi (z) } }{\sqrt{2(\psi (z)-{\overline{s}} )}} \text {d}z = {{\widetilde{C}}}+o( {{\overline{s}}} )$. One obtains the asymptotic expansion of $A=A_1+A_2=-\frac{1}{2 \sqrt{2}} \log {\overline{s}} + {\widehat{C}} +o({ {\overline{s}}} )$. Adding a similar contribution for the second part of the integral $B=2 \sqrt{\frac{\varphi ''_0(0)}{2}}\int _{x_0}^{ {\widehat{b}}_e} \frac{ \text {d}z}{\sqrt{2(\psi (z) -{\overline{s}} )}}$ ends the proof. $\square$

Lemma 6.5

Under assumption 1.4, one has

$$\begin{aligned}&{\widehat{t}}_{-\varphi _0^++{\overline{e}}}= \frac{\sqrt{2} \pi }{ \sqrt{\psi ''_0(x_0)} }+ \frac{4\sqrt{2} \pi ( - (\psi ^\frac{1}{2})''(x_0) )}{ (\psi '')(x_0)^\frac{3}{2}} \sqrt{ {\overline{e}} } + o( { {\overline{e}}}),\nonumber \\&\quad \text{ where } \lim _{{{\overline{e}}\rightarrow 0}}o( { {\overline{e}}})/{\overline{e}} = 0\text{. } \end{aligned}$$

(A.2)

Proof

With the notations of Lemma 3.9, consider (3.15) and make a local expansion of $ug_u({\overline{e}})$ for small ${\overline{e}}$. One has ${\overline{e}} u^2= \psi (z)=\frac{1}{2} \psi ''(x_0)(z-x_0)^2+O(z-x_0)^3$ and $\psi '(z)= \psi ''(x_0)(z-x_0)+O(z-x_0)^2$. So one has $\psi '(z)= \sqrt{2 {\overline{e}} \psi ''(x_0)} u + O({\overline{e}})$. Plugging in (3.16), one gets

$$\begin{aligned}&\lim _{ {\overline{e}} \rightarrow 0^+} u g_u({\overline{e}}) = \lim _{ {\overline{e}} \rightarrow 0^+} u \frac{\sqrt{ {\overline{e}} }}{\psi '(z)}= \frac{1}{ \sqrt{2 \psi ''(x_0)} } . \end{aligned}$$

(A.3)

One has (3.17), so one obtains also

$$\begin{aligned} \sqrt{{\overline{e}}}\frac{\text {d}}{\text {d}{\overline{e}}} g_u({\overline{e}}) = \frac{2}{(2\psi ''(x_0))^\frac{3}{2}}(- \psi ^\frac{1}{2}(x_0))'' + O(\sqrt{{\overline{e}}}). \end{aligned}$$

(A.4)

Use (A.3) and a primitive of (A.4), so $ug_u({\overline{e}})= \frac{1}{ \sqrt{2 \psi ''(x_0)} } + \frac{4 u}{(2\psi ''(x_0))^\frac{3}{2}}(- \psi ^\frac{1}{2}(x_0))'' \sqrt{{\overline{e}}} + O({\overline{e}})$. Plug this Ansatz in (3.15) one gets $H({\overline{e}})=\int _0^1 \frac{2 }{\sqrt{1-u^2}} \left( \frac{1}{ \sqrt{2 \psi ''(x_0)} } + \frac{4 u}{(2\psi ''(x_0))^\frac{3}{2}}(- \psi ^\frac{1}{2}(x_0))'' \sqrt{{\overline{e}}} + O({\overline{e}}) \right) \text {d}u$$= \frac{\pi }{ \sqrt{2 \psi ''(x_0)} }+ \frac{8 }{(2\psi ''(x_0))^\frac{3}{2}}(- \psi ^\frac{1}{2}(x_0))'' \sqrt{{\overline{e}}} +O({\overline{e}})$. Adding the contribution which corresponds to $\int _{{\widehat{a}}_e}^{x_0}\ldots$ yields the claim. $\square$

Proof of Proposition 3.10

The first point is just a rephrasing of the previous results. The second point is the reciprocal expansion to the one of Lemma 6.5. Indeed, $\frac{1}{\lambda }- \frac{1}{{\mathcal {B}}}= \alpha \sqrt{{\overline{e}}}+\cdots$ yields ${\overline{e}}= \frac{(\lambda -{\mathcal {B}})^2}{\alpha ^2 \lambda ^2 {\mathcal {B}}^2}+\cdots = \frac{(\lambda -{\mathcal {B}})^2}{\alpha ^2 {\mathcal {B}}^4}+\cdots$. The coefficient in front of $(\lambda -{\mathcal {B}})^2$ in the proposition is precisely $\frac{1}{\alpha ^2 {\mathcal {B}}^4}$. The estimate for the derivative can be obtain (as before) by differentiation $-\frac{1}{\lambda ^2}= s_c'(\lambda ) \frac{\text {d}}{\text {d}e} \left( 2 \int _{{\widehat{a}}_e}^{{\widehat{b}}_e} \frac{\text {d}x}{\sqrt{2(e+\varphi _0(x))}}\right) _ {e=s_c(\lambda )}$. The derivative is $O(1/\sqrt{s_c(\lambda )+\varphi _0^+})$ by means of (3.18)–(A.4). Therefore with $\lambda \approx {\mathcal {B}}$, one gets $s_c'(\lambda )= O(\sqrt{s_c(\lambda )+\varphi _0^+})= O\left( \lambda - {\mathcal {B}} \right)$. The proof of the last point is similar to Proposition 3.8. $\square$

Appendix B. A Technical Result

The technical result below is an essential step to characterize, in Proposition 5.6, of the regularity of the $n_{a,b,p}^\varepsilon$ which are the bilinear forms in the region of trapped particles. The proof is elementary; however, it is reasonable to think that it solves a fundamental issue. Indeed, trapped particles, also called electrons hole, generate Abel type integrals [31][Eqs. (5) and (6)] which need specific treatment to control their singularity.

We use the notations of Sect. 3.1.4. For $0<\lambda < {\mathcal {B}}$ and ${\widehat{a}}_{s_c(\lambda )}\le x \le {\widehat{b}}_{s_c(\lambda )}$, let us define ${\widehat{f}}(\lambda )=\int _{{\widehat{a}}_{s_c(\lambda )} }^{{\widehat{b}}_{s_c(\lambda )}} \sin \left( 2 \pi p \lambda y_\lambda (x) \right) a(x) \text {d}x$ and $y_\lambda (x)=\int _{{\widehat{a}}_{s_c(\lambda )}}^x \frac{\text {d}t}{\sqrt{2(s_c (\lambda )+ \varphi _0(t))}}$. Our goal is to investigate the regularity of ${\widehat{f}}$, with the constraint that the regularity constant must be expressed in terms of the quadratic norm of a. The key issue is that it is not possible to have such a bound for the full derivative. Indeed, dropping non-essential terms, one has the local expansion $y_\lambda (x)= \frac{\sqrt{2}}{\varphi _0'({\widehat{a}}_{s_c(\lambda )})}\sqrt{x-{\widehat{a}}_{s_c(\lambda )}} +\cdots$ from which one gets $\frac{\text {d}}{\text {d}\lambda }y_\lambda (x)=\frac{C(\lambda )}{\sqrt{x-{\widehat{a}}_{s_c(\lambda )}}}+\cdots$. The problem is that $\frac{1}{\sqrt{x-{\widehat{a}}_{s_c(\lambda )}}}\not \in L^2(I)$ so is not convenient to obtain good bounds with respect to $\Vert a\Vert _{L^2(I)}$. One is forced to characterize in a weaker space, typically a Hölder space. In the following, we focus on 1 / 2-Hölder continuity with a weight which depends on $\lambda$.

Proposition B.1

There exists a constant $C>0$ independent of p, $\lambda ,\mu \in [0,{\mathcal {B}}]$ and $a\in L^2(I)$ such that

$$\begin{aligned} \left| {\widehat{f}}(\mu ) - {\widehat{f}}(\lambda )\right| \le C |p| \Vert a\Vert _{L^2(I)} \sqrt{\frac{{|\lambda - \mu | }}{ \min \left( \varphi _0'({\widehat{a}}_{s_c(\lambda )}) , \varphi _0'({\widehat{b}}_{s_c(\lambda )}) \right) }} , \quad \lambda <\mu . \end{aligned}$$

The proof is a corollary of Proposition B.3 and Lemma B.4 below. The starting point is the following simple inequality. By definition (3.14), one has $\frac{1}{2\lambda }= y_\lambda ({\widehat{b}}_{s_c(\lambda )})$. For convenience, we extend by continuity the function $y_\lambda$ in the entire interval [0, 1] setting $y_\lambda (x)=0$ for $0\le x \le {\widehat{a}}_{s_c(\lambda )}$ and $y_\lambda (x)=1/(2\lambda )$ for ${\widehat{b}}_{s_c(\lambda )}\le x \le 1$. Using the inequality $\left| \sin \alpha - \sin \beta \right| \le \left| \alpha -\beta \right|$, one gets from the definition $\left| {\widehat{f}}(\mu ) - {\widehat{f}}(\lambda )\right| \le 2\pi |p| \Vert a\Vert _{L^2(I)} \Vert \lambda y_\lambda -\mu y_\mu \Vert _{L^\infty (I)}$.

Lemma B.2

For $0<\lambda<\mu <{\mathcal {B}}$, one has $\Vert \lambda y_\lambda -\mu y_\mu \Vert _{L^\infty (I)}= \lambda \max \left( y_\lambda ({\widehat{a}}_{s_c(\mu )}), \frac{1}{2}-y_\lambda ({\widehat{b}}_{s_c(\mu )}) \right)$.

Proof

Split the interval $[0,1]=[0,{\widehat{a}}_{s_c(\lambda )}]\cup [{\widehat{a}}_{s_c(\lambda )},{\widehat{a}}_{s_c(\mu )}]\cup |{\widehat{a}}_{s_c(\mu )},{\widehat{b}}_{s_c(\mu )}] \cup [{\widehat{b}}_{s_c(\mu )},{\widehat{b}}_{s_c(\lambda )}] \cup [{\widehat{b}}_{s_c(\lambda )},1]$ for $0<\lambda<\mu <1$. By construction, the derivative of $g=\lambda y_\lambda -\mu y_\mu$ is zero in $(0,{\widehat{a}}_{s_c(\lambda )})$, positive in $({\widehat{a}}_{s_c(\lambda )}, {\widehat{a}}_{s_c(\mu )})$, positive in $({\widehat{b}}_{s_c(\mu )}, {\widehat{b}}_{s_c(\lambda )})$ and zero in $({\widehat{b}}_{s_c(\lambda )},1)$. In the interval $( {\widehat{a}}_{s_c(\mu )}, {\widehat{b}}_{s_c(\mu )})$ one gets that $g'(x)= \frac{\lambda }{\sqrt{2(s_c (\lambda )+ \varphi _0(x))}} - \frac{\mu }{\sqrt{2(s_c (\mu )+ \varphi _0(x))}}$. Proposition 3.10 yields that $s_c$ is decreasing function. Since $\lambda <\mu$, $s_c(\lambda )>s_c(\mu )$. So $g'(x)<0$ in this central interval. Therefore, $\Vert g\Vert _{L^\infty (I)}= \max \left( g( {\widehat{a}}_{s_c(\mu )}), - g( {\widehat{b}}_{s_c(\mu )}) \right)$. So the claim. $\square$

Proposition B.3

There exists a constant $C>0$ independent of p, $\lambda ,\mu \in [0,{\mathcal {B}}]$ and $a\in L^2(I)$ such that for $\lambda <\mu$

$$\begin{aligned} \left| {\widehat{f}}(\mu ) - {\widehat{f}}(\lambda )\right| \le C |p| \Vert a\Vert _{L^2(I)} \max \left( \sqrt{\frac{ |{\widehat{a}}_{s_c(\lambda )} - {\widehat{a}}_{s_c(\mu )}| }{\varphi _0'({\widehat{a}}_{s_c(\lambda )}) }}, \sqrt{\frac{ |{\widehat{b}}_{s_c(\lambda )} - {\widehat{b}}_{s_c(\mu )}| }{\varphi _0'({\widehat{b}}_{s_c(\lambda )})}} \right) .\nonumber \\ \end{aligned}$$

(B.1)

Proof

One studies $y_\lambda ({\widehat{a}}_{s_c(\mu )})$. By symmetry, it will be sufficient to get similar estimates for the other term $\frac{1}{2}-y_\lambda ({\widehat{b}}_{s_c(\mu )})$. A first estimate is $\lambda y_\lambda ({\widehat{a}}_{s_c(\mu )}) \le \frac{1}{2}$. Next $x\in \left[ {\widehat{a}}_{s_c(\lambda )}, x_0\right]$, one can rewrite $y_\lambda (x)= \int _{{\widehat{a}}_{s_c(\lambda )}}^x \frac{\text {d}t}{\sqrt{2( \varphi _0(t) - \varphi _0({\widehat{a}}_{s_c(\lambda )}) )}}$. Let ${\widehat{a}}_{s_c(\lambda )}\le d \le t\le {\widehat{a}}_{s_c(\mu )}$, one has the expansion $\varphi _0(t) - \varphi _0({\widehat{a}}_{s_c(\lambda )}) = \varphi _0'(d)(t- {\widehat{a}}_{s_c(\lambda )})$. So one has for ${\widehat{a}}_{s_c(\lambda )} \le t\le {\widehat{a}}_{s_c(\mu )}$

$$\begin{aligned} \varphi _0(t) \ge \left( \varphi _0'({\widehat{a}}_{s_c(\lambda )})- C\left( {\widehat{a}}_{s_c(\mu )} - {\widehat{a}}_{s_c(\lambda )} \right) \right) (t- {\widehat{a}}_{s_c(\lambda )}) \end{aligned}$$

where we used the boundedness of the second derivative to lower bound $\varphi _0'(d)\ge \varphi _0'({\widehat{a}}_{s_c(\lambda )})- C\left( {\widehat{a}}_{s_c(\mu )} - {\widehat{a}}_{s_c(\lambda )} \right)$ with $C>0$. Assuming that $\mu$ is sufficiently close to $\lambda$, one gets another estimate $y_\lambda (x)\le \frac{1}{ \sqrt{ \varphi _0'(a_{s_c(\lambda )})- C\left( {\widehat{a}}_{s_c(\lambda )} - {\widehat{a}}_{s_c(\mu )} \right) }} \int _{{\widehat{a}}_{s_c(\lambda )}}^x \frac{\text {d}t}{\sqrt{2( t - {\widehat{a}}_{s_c(\lambda )} )}} = \frac{ \sqrt{2( t - {\widehat{a}}_{s_c(\lambda )} )} }{ \sqrt{ \varphi _0'({\widehat{a}}_{s_c(\lambda )})- C\left( {\widehat{a}}_{s_c(\lambda )} - {\widehat{a}}_{s_c(\mu )} \right) } }.$ Combine the two estimates and use the notation $g(u,v)=\min \left( 1, \sqrt{\frac{ {2 u } }{v- Cu }} \right)$: so $y_\lambda ({\widehat{a}}_{s_c(\mu )}) \le g\left( {\widehat{a}}_{s_c(\mu )} - {\widehat{a}}_{s_c(\lambda )}, \varphi _0'({\widehat{a}}_{s_c(\lambda )})\right)$. The function g is homogenous of degree 0 and $g(w,1)\le C\sqrt{w}$ for some constant C. It yields $\lambda y_\lambda ({\widehat{a}}_{s_c(\mu )}) \le C \sqrt{\frac{ {\widehat{a}}_{s_c(\mu )} - {\widehat{a}}_{s_c(\lambda )} }{\varphi _0'({\widehat{a}}_{s_c(\lambda )}) }}$. A similar bound for the other term ends the proof of the claim of Proposition B.3. $\square$

Lemma B.4

The functions $\lambda \mapsto {\widehat{a}}_{s_c(\lambda )}$ and $\lambda \mapsto {\widehat{b}}_{s_c(\lambda )}$ are uniformly differentiable for $\lambda \in [0,{\mathcal {B}}]$.

Proof

The proof is provided for the first function. One has $s_c(\lambda )+\varphi _0({\widehat{a}}_{s_c(\lambda )})=0$ so $\frac{\text {d}}{\text {d}\lambda } {\widehat{a}}_{s_c(\lambda )}= - \frac{s_c'(\lambda )}{ \varphi _0'(\widehat{a}_{s_c(\lambda )})}$ for $\lambda \in (0,{\mathcal {B}})$ where $s_c(\lambda )\in C^1[0,{\mathcal {B}}]$. Assumption 1.4 yields that $\varphi _0'(\lambda )>0$ for $0<\lambda < {\mathcal {B}}$. The issue is the limit of the above expression for $0^+$ and ${\mathcal {B}}^-$.

• For small $\lambda$, Proposition 3.8 yields $s_c'(\lambda )= O\left( { \exp \left( {-\frac{\sqrt{\varphi _0''(0)}}{2\lambda } }\right) }/{ \lambda ^{2}} \right)$. For small $x\approx 0$, one has that $\varphi _0'(x)= O(\varphi _0(x)+\varphi _0^-)^\frac{1}{2}$ so

  $$\begin{aligned} \varphi _0'({\widehat{a}} _{s_c(\lambda )})= O(\varphi _0({\widehat{a}} _{s_c(\lambda )})+\varphi _0^-)^\frac{1}{2} =O\left( \exp \left( {-\frac{\sqrt{\varphi _0''(0)}}{2\lambda } }\right) \right) ^\frac{1}{2} \end{aligned}$$

  (B.2)

  and $\frac{\text {d}}{\text {d}\lambda } {\widehat{a}}_{s_c(\lambda )}= O\left( \frac{ \exp \left( {-\frac{\sqrt{\varphi _0''(0)}}{4\lambda } }\right) }{ \lambda ^{2}} \right) \text{ for } \text{ small } \lambda$. So $\lim _{\lambda \rightarrow 0^+} \frac{\text {d}}{\text {d}\lambda } {\widehat{a}}_{s_c(\lambda )}=0$.

• For $\lambda \approx {\mathcal {B}}$, Proposition 3.8 yields $s_c'(\lambda )= O\left( \lambda -{\mathcal {B}} \right)$. For $x\approx x_0$, one has that $\varphi _0'(x)= O(\varphi _0(x)+\varphi _0^+)^\frac{1}{2}$ so

  $$\begin{aligned} \varphi _0'({\widehat{a}}_{s_c(\lambda )})= O\left( \varphi _0({\widehat{a}}_{s_c(\lambda )})+\varphi _0^+ \right) ^\frac{1}{2}=O(\lambda -{\mathcal {B}}). \end{aligned}$$

  (B.3)

  Therefore, one has a limit $\lim _{\lambda \rightarrow {\mathcal {B}}^-} \frac{\text {d}}{\text {d}\lambda } {\widehat{a}}_{s_c(\lambda )}<\infty$.

• Since $\frac{\text {d}}{\text {d}\lambda } {\widehat{a}}_{s_c(\lambda )}\in C^0(0,{\mathcal {B}})$, the proof is ended. $\square$

Appendix C. A Useful Inequality

For $p,q\in {\mathbb {R}}^*$, one has the estimate $\left\| \int _{x\in I} f(\lambda ,x) \left( \frac{1}{\pi }P.V. \int _{\mu \in {\mathbb {R}}} \frac{g(\mu ,x)}{p\lambda -q\mu }\text {d}\mu \right) \text {d}x \right\| _{L^2_\lambda }^2$$\le \frac{1}{|pq|} \int _{x\in I} \Vert f \Vert _{L^2_\lambda }^2\text {d}x$$\times \int _{x\in I}\Vert g \Vert _{L^2_\lambda }^2\text {d}x$ where $L^2_\lambda =L^2_\lambda ({\mathbb {R}})$ is the quadratic norm is with respect to the variable $\lambda \in {\mathbb {R}}$. The proof which uses the fact that the Hilbert transform is an isometry in $L^2$ is left to the reader.

Appendix D. Verification that $V_{e,k,z}^\varepsilon$ is a Generalized Eigenvector

We provide a separate algebraic proof that $V_{e,k,z}^\varepsilon$ is a generalized eigenvector of ${\mathbf {i}} H^\varepsilon$. The proof is in the weak sense, assuming convergence of the sums and integrals (refer to Remark 5.8). It yields a direct verification of conditions (4.5).

Lemma D.1

One has $\left( V_{e,k,z}^\varepsilon , A W'\right) + \left( V_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) + \left( V_{e,k,z}^\varepsilon , D W \right) = \lambda _{e,k}^\varepsilon \left( V_{e,k,z}^\varepsilon , W\right)$ for all $W \in X_0$.

Let us start from (5.11). One has for all $W\in X_0$

$$\begin{aligned} \begin{array}{lllll} \left( V_{e,k,z}^\varepsilon , A W'\right) = \left( U_{e,k,z}^\varepsilon , A W'\right) + \left( a_{e,k,z}^\varepsilon (x) e_0, A W'\right) \\ \quad +\, \displaystyle \varepsilon \sum _{z'} \sum _{p\ne 0} P.V. \int _{s\in I_{z'}^\varepsilon } \frac{\left( \alpha a_{e,k,z}^\varepsilon e_2 ,\exp ({\varepsilon ^2 \varphi _0/2}) U_{s,p,z'}^\varepsilon \right) }{ \lambda _{s,p} ^\varepsilon - \lambda _{e,k}^\varepsilon } \left( U_{s,p,z'}^\varepsilon , A W' \right) { t_s^\varepsilon }\text {d}s \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{lllll} \left( V_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) = \left( U_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W\right) + \left( a_{e,k,z}^\varepsilon (x) e_0, \varepsilon ^2 E_0 B W\right) \\ \quad +\, \displaystyle \varepsilon \sum _{z'} \sum _{p\ne 0} P.V. \int _{s\in I_{z'}^\varepsilon } \frac{\left( \alpha a_{e,k,z}^\varepsilon e_2 ,\exp ({\varepsilon ^2 \varphi _0/2}) \qquad U_{s,p,z'}^\varepsilon \right) }{ \lambda _{s,p} ^\varepsilon - \lambda _{e,k}^\varepsilon } \left( U_{s,p,z'}^\varepsilon , \varepsilon ^2 E_0 B W\right) { t_s^\varepsilon }\text {d}s. \end{array}\nonumber \\ \end{aligned}$$

(D.1)

Summation of (D.1–D.3) and the use of (3.12) which is satisfied for all e, k, z yields

$$\begin{aligned}&\left( V_{e,k,z}^\varepsilon , A W'\right) + \left( V_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) \\&\quad =\left( U_{e,k,z}^\varepsilon , A W'\right) + \left( U_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) \\&\qquad +\left( a_{e,k,z}^\varepsilon (x) e_0, A W'\right) + \left( a_{e,k,z}^\varepsilon (x) e_0, \varepsilon ^2 E_0 B W \right) \\&\qquad +\, \displaystyle \varepsilon \sum _{z'} \sum _{p\ne 0} P.V. \int _{s\in I_{z'}^\varepsilon } \frac{\left( \alpha a_{e,k,z}^\varepsilon e_2 ,\exp ({\varepsilon ^2 \varphi _0/2}) U_{s,p,z'}^\varepsilon \right) }{ \lambda _{s,p} ^\varepsilon - \lambda _{e,k}^\varepsilon } \lambda _{s,p} ^\varepsilon \left( U_{s,p,z'}^\varepsilon , W\right) { t_s^\varepsilon }\text {d}s. \end{aligned}$$

So

$$\begin{aligned}&\left( V_{e,k,z}^\varepsilon , A W'\right) + \left( V_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) - \lambda _{e,k}^\varepsilon \left( V_{e,k,z}^\varepsilon , W\right) \\&\quad = \left( U_{e,k,z}^\varepsilon , A W'\right) + \left( U_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) - \lambda _{e,k}^\varepsilon \left( U_{e,k,z}^\varepsilon , W\right) \\&\qquad +\left( a_{e,k,z}^\varepsilon (x) e_0, A W'\right) + \left( a_{e,k,z}^\varepsilon (x) e_0, \varepsilon ^2 E_0 B W \right) - \lambda _{e,k}^\varepsilon \left( a_{e,k,z}^\varepsilon (x) e_0, W\right) \\&\qquad +\, \displaystyle \varepsilon \sum _{z'} \sum _{p\ne 0} \int _{s\in I_{z'}^\varepsilon } {\left( \alpha a_{e,k,z}^\varepsilon e_2 ,\exp ({\varepsilon ^2 \varphi _0/2}) U_{s,p,z'}^\varepsilon \right) } \left( U_{s,p,z'}^\varepsilon , W\right) { t_s^\varepsilon }\text {d}s\\&\quad = 0 - \lambda _{e,k}^\varepsilon \left( a_{e,k,z}^\varepsilon e_0, W\right) + \displaystyle \varepsilon \left( \exp ({\varepsilon ^2 \varphi _0/2}) \alpha a_{e,k,z}^\varepsilon e_2 , W\right) (\text{ use } \text{ the } \text{ identity } 3.6)]. \end{aligned}$$

Let $a\in L^2_0(I)$ and $c\in L^2(I)$ denote the first and third component of the infinite vector W. One recasts this as

$$\begin{aligned}&\left( V_{e,k,z}^\varepsilon , A W'\right) + \left( V_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) - \lambda _{e,k}^\varepsilon \left( V_{e,k,z}^\varepsilon , W\right) \nonumber \\&\quad = - \lambda _{e,k}^\varepsilon \left( a_{e,k,z}^\varepsilon , a\right) + \varepsilon \alpha \left( \exp ({\varepsilon ^2 \varphi _0/2}) a_{e,k,z}^\varepsilon , c\right) . \end{aligned}$$

(D.2)

On the other hand, another use of (5.11) yields

$$\begin{aligned} \begin{array}{lllll} \left( V_{e,k,z}^\varepsilon , D W \right) = \left( U_{e,k,z}^\varepsilon , D W\right) + \left( a_{e,k,z}^\varepsilon e_0, D W\right) \\ \quad +\, \displaystyle \varepsilon \sum _{z'} \sum _{p\ne 0} P.V. \int _{s\in I_{z'}^\varepsilon } \frac{\left( \alpha a_{e,k,z}^\varepsilon e_2 ,\exp ({\varepsilon ^2 \varphi _0/2}) U_{s,p,z'}^\varepsilon \right) }{ \lambda _{s,p} ^\varepsilon - \lambda _{e,k}^\varepsilon } \left( U_{s,p,z'}^\varepsilon , D W\right) { t_s^\varepsilon }\text {d}s. \end{array} \end{aligned}$$

(D.3)

One can write $DW= \alpha \exp (\varphi _0/2) a e_2 - \alpha 1^* (\exp (\varphi _0/2) c )e_0$. So

$$\begin{aligned} \begin{array}{lllll} \left( V_{e,k,z}^\varepsilon , D W \right) = \left( U_{e,k,z}^\varepsilon \cdot e_2 , \alpha \exp (\varepsilon ^2 \varphi _0/2) a\right) - \left( a_{e,k,z}^\varepsilon e_0, \alpha 1^*( \exp (\varepsilon ^2 \varphi _0/2) c )e_0\right) \\ \quad +\,\displaystyle \varepsilon \alpha \sum _{z'} \sum _{p\ne 0} P.V. \int _{s\in I_{z'}^\varepsilon } \frac{\left( \alpha a_{e,k,z}^\varepsilon e_2 ,\exp (\varepsilon ^2 \varphi _0/2) U_{s,p,z'}^\varepsilon \right) }{ \lambda _{s,p} ^\varepsilon - \lambda _{e,k}^\varepsilon } \left( U_{s,p,z'}^\varepsilon \cdot e_2 , \exp (\varepsilon ^2 \varphi _0/2) a\right) { t_s^\varepsilon }\text {d}s . \end{array} \end{aligned}$$

Since $a_{e,k,z}^\varepsilon \in L^2_0(I)$ is the unique solution to (4.6)–(5.3)–(5.7), one obtains

$$\begin{aligned} \left( V_{e,k,z}^\varepsilon , D W \right)= & {} \alpha \left( U_{e,k,z}^\varepsilon \cdot e_2 , \exp (\varepsilon ^2 \varphi _0/2) a\right) - \alpha \left( a_{e,k,z}^\varepsilon , \exp (\varepsilon ^2 \varphi _0/2) c \right) \\&+\,\frac{1}{\varepsilon }\lambda _{e,k}^\varepsilon (a_{e,k,z}^\varepsilon ,a) - \alpha \left( \exp (\varepsilon ^2 \varphi _0/2) U_{e,k,z}^\varepsilon \cdot e_2,a\right) . \end{aligned}$$

One adds to (D.2). So $\left( V_{e,k,z}^\varepsilon , A W'\right) + \left( V_{e,k,z}^\varepsilon , \varepsilon ^2 E_0 B W \right) - \lambda _{e,k}^\varepsilon \left( V_{e,k,z}^\varepsilon , W\right) +\varepsilon \left( V_{e,k,z}^\varepsilon , D W\right) =0$.