Harmonic oscillator perturbed by a decreasing scalar potential

Abstract

In this paper we study the perturbation $L=H+V$, where $H=-\frac{{{d}^{2m}}}{d{{x}^{2m}}}+{{x}^{2m}}$ on $\mathbb {R}$, $m\in {{\mathbb {N}}^{*}}$ and V is a decreasing scalar potential. Let ${{\lambda }_{k}}$ be the $k^{th}$ eigenvalue of H. We suppose that the eigenvalues of L around ${{\lambda }_{k}}$ can be written in the form ${{\lambda }_{k}}+{{\mu }_{k}}$. The main result of the paper is an asymptotic formula for fluctuation $\{ {{\mu }_{k}} \}$ which is given by a transformation of V. In the case $m=1$ we recover a result on the harmonic oscillator.

1 Introduction and main results

We consider in $\mathbb {R}$ the operator H defined by

$$\begin{aligned} H=-\dfrac{d^{2m}}{dx^{2m}}+x^{2m},\quad m\in \mathbb {N}^* \end{aligned}$$

(1.1)

We recall that H [1] is essentially self-adjoint in $C_{0}^{\infty }(\mathbb {R})$ with compact resolvent. Its spectrum is the increasing sequence ${{\{ {{\lambda }_{k}} \}}_{k\ge 0}}$ of eigenvalues of finite multiplicity, such as there exists a positive integer ${{k}_{0}}$, for $k\ge {{k}_{0}}$, ${{\lambda }_{k}}$ is simple and has the following asymptotic expansion

$$\begin{aligned} \lambda _{k}^{\frac{1}{m}}=\frac{2\pi }{T}\left( k+\frac{1}{2} \right) +O(k^{-1})\quad k\rightarrow +\infty \end{aligned}$$

(1.2)

with

$$\begin{aligned} \displaystyle T=\int _{-1}^{1}{(1-{{u}^{2m}}}{{)}^{^{\frac{1}{2m}-1}}}du=\frac{1}{m}B\left( \frac{1}{2m},\frac{1}{2m}\right) \end{aligned}$$

(1.3)

where B is the beta function. Let $V\in {{\mathbb {C}}^{\infty }}(\mathbb {R},\mathbb {R})$ which satisfies the following estimate

$$\begin{aligned} | {{V}^{(n)}}(x) | \le {{C}_{n}}{{(1+{{x}^{2}})}^{-\frac{s}{2}}},\,x\in \mathbb {R},\, n\in \mathbb {N},\,s\in \mathbb {R}_{+}^{*}-\{ 1 \} \end{aligned}$$

(1.4)

Along this article we set

$$\begin{aligned} \delta =\left\{ \begin{array}{ll} s&{} \quad if\ 0<s<1\\ 1&{} \quad if\ s>1\\ \end{array}\right. \end{aligned}$$

(1.5)

Remark 1.1

So as not to burden our work, the case “$s =1$” will be treated in a different way, we will treat this case in another work.

The operator $L= H+V$ is essentially self-adjoint with compact resolvent [2]. The Min-Max theorem [3] shows that the spectrum of L around ${{\lambda }_{k}}$ can be written in the form ${{\lambda }_{k}}+{{\mu }_{k}}$. Our goal is to study the asymptotic behavior of the fluctuation ${{\mu }_{k}}$ when $k\rightarrow +\infty$, by expressing it using a transformation of V. Our main result is

Theorem 1.2

(Main Theorem) ${{\mu }_{k}}$ has the asymptotic expansion

$$\begin{aligned} \displaystyle {{\mu }_{k}}=\displaystyle \frac{1}{T}\int _{-1}^{1}{\frac{V(y\lambda _{k}^{\frac{1}{2m}})}{(1-{{y}^{2m}}){}^{1-\frac{1}{2m}}}}dy+\text {O}(\lambda _{k}^{\frac{-\delta -1}{2m}}), \quad \forall m\ge 2 \end{aligned}$$

For $m=1$, the case of the Harmonic Oscillator, the asymptotic behavior of $\mu _{k}$ is given by

Theorem 1.3

$\displaystyle {{\mu }_{k}}=\frac{1}{\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}{V\left( \sqrt{{{\lambda }_{k}}}\sin t \right) }dt+O\left( \lambda _{k}^{-\delta +\eta } \right)$

where $\eta \in ] 0,\frac{\delta }{2} [$.

Many authors interested in this kind of problems, especially the case of the Harmonic Oscillator [4, 5]. The case $m>1$ seems to us as not to have been treated yet. Briefly recall the content of [4], the author studies the perturbation $L=A+B$, where

$$\begin{aligned} A=\frac{1}{2}\left( -\frac{{{d}^{2}}}{d{{x}^{2}}}+{{x}^{2}}-1 \right) , \qquad B(x)\sim {{\left| x \right| }^{-\alpha }}\sum \limits _{m}{{{a}_{m}}}\cos {{\omega }_{m}}x \end{aligned}$$

and he proved that ${\mu }_{k}$ has the following asymptotic expansion

$$\begin{aligned} {{\mu }_{k}}\sim {{k}^{-(\frac{\alpha }{2}+\frac{1}{4})}}\widetilde{V}(\sqrt{2k})+\frac{C}{\sqrt{2k}}\quad k\rightarrow +\infty \end{aligned}$$

$\widetilde{V}$ represents the “Radon Transform” of V. In recent works, we find in [6] a study of

$$\begin{aligned} D=-\frac{{{d}^{2}}}{d{{x}^{2}}}+{{x}^{2}}+q(x) \end{aligned}$$

where real functions $q, q'$ and $x\rightarrow \int _{0}^{x}{q(s)ds}$ are bounded. ${\mu }_{k}$ has the asymptotic expansion

$$\begin{aligned} {{\mu }_{k}}=\frac{1}{2\pi }\displaystyle \int _{-\pi }^{\pi } q(\sqrt{\lambda }_{k}sin\theta ) \, \mathrm {d}\theta + O(k^{\frac{-1}{3}}) \quad as \ k\rightarrow +\infty \end{aligned}$$

(1.6)

We notice that, in the case $s<1$, the expansion (1.6) has the same main part as shown in Theorem 1.3, even though we don’t need to suppose that $x\rightarrow \int _{0}^{x}{q(s)ds}$ is bounded. In addition, if $s\in ] \frac{2}{3},1 [$, we get a better estimate and the same goes for $s>1$ because $\eta \in ] 0,\frac{1}{2} [$. We can also mention Pushnitski [7], who studied the case where $q\in C_{0}^{\infty }(\mathbb {R})$. He proved that ${\mu }_{k}$ admits the next development in series

$$\begin{aligned} {{\mu }_{k}}=\sum \limits _{j=1}^{+\infty }{{{c}_{j}}}\lambda _{k}^{\tfrac{-j}{2}} \quad \lambda _{k} \rightarrow +\infty \end{aligned}$$

(1.7)

with some coefficients ${{c}_{j}}\in \mathbb {R}$, in particular ${{c}_{1}}=\frac{1}{\pi }\int _{-\infty }^{+\infty }{q(x)} \, \mathrm {d}x$, and $c_{2}=0$.

Remark 1.4

We want to go further in studying the operator $H_{k,l}=-\frac{{{d}^{2k}}}{d{{x}^{2k}}}+{{x}^{2l}}$, where $k,l\in {{\mathbb {N}}^{*}}$, then by giving k the value of “1”, we’ll reach important results that have a lot off applications in the field of physics, especially the quartic oscillator.

Our main tool is the averaging Method of Weinstein [8, 9], whose origins go back to the classical work on celestial mechanics [10]. Note that for $m\in {{\mathbb {N}}^{*}}-\{ 1\}$, this method cannot be used directly in this case because H, viewed as a Pseudo-differential operator (OPD), doesn’t have a periodic flow, but the operator ${{H}^{\frac{1}{m}}}$ does have this property, so we start reducing ourselves to a perturbation (1.8) of the operator ${{H}^{\frac{1}{m}}}$

$$\begin{aligned} {{L}_{m}}=H^{\frac{1}{m}}+B ,\quad m\in {{\mathbb {N}}^{*}} \end{aligned}$$

(1.8)

where B is an operator to be determined. We apply the Averaging Method, firstly we replace B in perturbation (1.8) by the average

$$\begin{aligned} \displaystyle \overline{B}=\frac{1}{T}\int _{0}^{T}{{e}^{-itH^{\frac{1}{m}}}}B{{e}^{itH^{\frac{1}{m}}}dt} \end{aligned}$$

(1.9)

where T is the period of the flow of $H^{\frac{1}{m}}$, T is given by (1.3) [11]. The main advantage of this method is that $\overline{B}$ is a compact operator and $L_{m}$, ${{\overline{L}}_{m}}=H^{\frac{1}{m}}+\overline{B}$ are almost unitary equivalent, that means it exists a unitary operator U such as $U{{L}_{m}}{{U}^{-1}}-{{\overline{L}}_{m}}$ is compact. Note that $L_{m}$ and ${{\overline{L}}_{m}}$ are also with compact resolvent. Using min-max theorem, the parts of their spectrum around $\lambda _{k}^{\frac{1}{m}}$ are respectively of the form $\lambda _{k}^{{\frac{1}{m}}}+{{\upsilon }_{k}}$ and $\lambda _{k}^{\frac{1}{m}}+{{\overline{\upsilon }}_{k}}$. Then we study ${{\overline{\upsilon }}_{k}}$ by using a functional calculus of the operator H. We begin by establishing the link between ${{\upsilon }_{k}}$ and ${{\mu }_{k}}$.

Proposition 1.5

$$\begin{aligned} {{\mu }_{k}}=m\lambda _{k}^{1-\frac{1}{m}}{{\upsilon }_{k}}+O(\lambda _{k}^{-1}),\quad m\in {{\mathbb {N}}^{*}} \end{aligned}$$

Using a functional calculus for H, we obtain

Proposition 1.6

$$\begin{aligned} \displaystyle m\lambda _{k}^{1-\frac{1}{m}}{{\overline{\upsilon }}_{k}}=\frac{1}{T}\int _{-1}^{1}{\frac{V(\lambda _{k}^{\frac{1}{2m}}y)}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy+O(\lambda _{k}^{\frac{-\delta -1}{2m}})} \end{aligned}$$

The following proposition gives the relation between ${{\upsilon }_{k}}$ and ${{\overline{\upsilon }}_{k}}$

Proposition 1.7

$$\begin{aligned} {{\upsilon }_{k}}={{\overline{\upsilon }}_{k}}+\text {O}(\lambda _{k}^{\frac{-\delta +\eta }{m}+\frac{2}{m}-2}) \end{aligned}$$

where $\eta \in ] 0, \min (2,m-1+\tfrac{{\delta }}{2}) [$

Remark 1.8

Note that for $m=1$, H has a periodic flow of period $\pi$, so we can directly apply the Averaging Method to H.

This paper is organized as follows. The next section contains auxiliary fact concerning the proprieties of Weyl pseudo-differential operators which are the main tool in this article. In Sect. 3, we study the operator $L_{m}$ and show the relation between $\mu _{k}$ and $\upsilon _{k}$ by proving Proposition 1.5. The Sect. 4 is devoted to the functional calculus for the operator H, and we establish Proposition 1.6. In the last section, we study the relation between the spectrum of ${{L}_{m}}$ and ${\overline{L}_{m}}$ and we prove Proposition 1.7. Finally, we justify the asymptotic expansion given by Theorems 1.2 and 1.3

2 Weyl pseudo-differential operator

Let $\rho \in [ 0,1]$,$\;q\in \mathbb {R}$. $\Gamma _{\rho }^{q}(\mathbb {R}\times \mathbb {R})$ denote the space symbols associated with the temperate weight function ${{\mathbb {R}}^{2}}: (x,\xi )\rightarrow {{(1+{{x}^{2}}+{{\xi }^{2} })}^{\frac{q}{2}}}$ [15]. Precisely the space of function $a\in {{C}^{\infty }}({{\mathbb {R}}^{2}})$ satisfies $\forall \alpha ,\beta \in \mathbb {N},\exists \, {{C}_{\alpha ,\beta }}>0$

$$\begin{aligned} \left| \partial _{x}^{\alpha }\partial _{\xi }^{\beta }a(x,\xi ) \right| \le {{C}_{\alpha ,\beta }}{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{q-\rho (\alpha +\beta )}{2}}} \end{aligned}$$

(2.1)

We will use the standard Weyl quantization of the symbols. To be precise, if $a\in \Gamma _{\rho }^{q}$, then for $u\in S(\mathbb {R})$ the operator associated is defined by :

$$\begin{aligned} o{{p}^{w}}(a)u(x)=\frac{1}{{{(2\pi )}^{2}}}\int _{{{\mathbb {R}}\times \mathbb {R}}}{{e}^{i\langle x-y,\xi \rangle }}a \left( \tfrac{x+y}{2},\xi \right) u(y)dyd\xi \end{aligned}$$

(2.2)

Let us now introduce the notion of asymptotic expansion.

Definition 2.1

Let ${{a}_{j}}\in \Gamma _{\rho }^{{{q}_{j}}} (j\in {{\mathbb {N}}^{*}})$, we suppose that $q_{j}$ is a decreasing sequence tending towards $-\infty$. We say that $a\in {{C}^{\infty }}(\mathbb {R}\times \mathbb {R})$ has an asymptotic expansion and we write

$$\begin{aligned} a\sim \sum \limits _{j=1}^{\infty }{{{a}_{j}}} \end{aligned}$$

if

$$\begin{aligned} a-\sum \limits _{j=1}^{r-1}{{{a}_{j}}}\in \Gamma _{\rho }^{{{q}_{r}}} \quad \forall r\ge 2 \end{aligned}$$

We use the notation $G_{\rho }^{q}$ for the set of operators $op^{w}(a)$ if $a\in \Gamma _{\rho }^{q}$. In order to prove our main results, we shall recall some well-known results [12,13,14].

Theorem 2.2

(Calderon-Vailloncourt theorem) If $a\in \Gamma _{0}^{0}$ then the operator $op^{w}(a)$ is bounded.

Proposition 2.3

(Compactness) If $a\in \Gamma _{\rho }^{q}$ and $q<0,\,\rho \in [0,1 ]$, then the operator $op^{w}(a)$ is compact.

We will need the following proposition for the composition of pseudo-differential operators.

Proposition 2.4

Let $A\in G_{\rho }^{p}$, $B\in G_{\rho }^{q}$, $\rho \in ] 0,1 ]$, p and q$\in \mathbb {R}$. Then the operator $AB\in G_{\rho }^{p+q}$. Its Weyl symbol admits the following asymptotic behavior

$$\begin{aligned} c\sim \sum \limits _{j\ge 0}{{{c}_{j}}} \end{aligned}$$

In particular

$$\begin{aligned} c(x,\xi )-a(x,\xi ).b(x,\xi )\in \Gamma _{\rho }^{p+q-2\rho } \end{aligned}$$

where

$$\begin{aligned} {{c}_{j}}=\frac{1}{{{2}^{j}}}\sum \limits _{ \alpha +\beta =j}{\frac{{{(-\,1)}^{\left| \beta \right| }}}{\alpha !\beta !}}(\partial _{\xi }^{\alpha }\partial _{x}^{\beta }a)(\partial _{x}^{\alpha }\partial _{\xi }^{\beta }b) \end{aligned}$$

(2.3)

a and b are respectively the Weyl symbol of A and B

In the next proposition we are giving an extension where the case “$\rho =0$”, we have the following result

Proposition 2.5

If $A\in G_{1}^{m}$ and ${{({{B}_{i}})}_{i\in \{ 1,\ldots p \}}}$ is the set of operators such as ${{B}_{i}}\in G_{0}^{{{m}_{i}}}$ then:

• (i) The operator $A{{B}_{1}}\in G_{0}^{m+{{m}_{1}}}$, its Weyl symbol is giving by (2.3)

  where ${{c}_{j}}\in \Gamma _{0}^{m+{{m}_{1}}-j}$

• (ii) The commutator $[ A,{{B}_{1}} ]\in G_{0}^{m+{{m}_{1}}-1}$

• (iii) If A is elliptic and $m>0$, then ${{B}_{1}}\ldots {{B}_{P}}{{A}^{-\tfrac{{{m}_{1}}+\cdots +{{m}_{p}}}{m}}}$ is bounded.

Proposition 2.6

(Functional calculus) Let A be an elliptic operator included in $G_{1}^{m}$, its Weyl symbol admits the development $a\sim \sum \nolimits _{j\ge 0}{{{a}_{j}}}$. Then for any real number q we have ${{A}^{q}}\in G_{1}^{mq}$. Moreover, its weyl symbol admits the following asymptotic behavior

$$\begin{aligned} {{\sigma }_{{{A}^{q}}}}\sim \sum \limits _{j=0}^{+\infty }{{{\sigma }_{{{A}^{q}},j}}}, \quad {{\sigma }_{{{A}^{q}},j}}\in \Gamma _{1}^{mq-j} \end{aligned}$$

where ${{\sigma }_{{{A}^{q}},0}}={a_{0}}^{q}$, ${{\sigma }_{{{A}^{q}},1}}=qa_{1}.a_{0}^{q-1}$.

3 Reduction to a perturbation of ${H}^{{\frac{1}{m}}}$

If we translate H by a strictly positive constant, we can always assume that L is positive and $\left\| {{H}^{-1}}V \right\| <1$. We can reduce ourselves to a perturbation of ${{H}^{\frac{1}{m}}}$ by writing

$$\begin{aligned} {{(H+V)}^{\frac{1}{m}}}={{H}^{\frac{1}{m}}}+W \end{aligned}$$

consequently

$$\begin{aligned} W=B+{{H}^{\frac{1}{m}}}{{({{H}^{-1}}V)}^{2}}\sum \limits _{k=0}^{+\infty }{{{\alpha }_{k+2}}}{{({{H}^{-1}}V)}^{k}} \end{aligned}$$

where

$$\begin{aligned} B=\frac{1}{m}{{H}^{-(1-\frac{1}{m})}}V , \quad {{\alpha }_{k}}=\frac{\frac{1}{m}(\frac{1}{m}-1)\ldots (\frac{1}{m}-k+1)}{k!} \end{aligned}$$

(3.1)

We can write

$$\begin{aligned} {{L}^{\frac{1}{m}}}-{{L}_{m}}={{H}^{\frac{1}{m}}}{{({{H}^{-1}}V)}^{2}}\sum \limits _{k=0}^{+\infty }{{{\alpha }_{k+2}}}{{({{H}^{-1}}V)}^{k}} \end{aligned}$$

(3.2)

Since $\left\| {{H}^{-1}}V \right\| <1$, the operator $\sum \nolimits _{k=0}^{+\infty }{{{\alpha }_{k+2}}}{{({{H}^{-1}}V)}^{k}}$ is bounded in ${{L}^{2}}(\mathbb {R})$. Using the fact that, ${{H}^{-1}}\in G_{1}^{-2m}$, $V\in G_{0}^{0}$ and (iii) Proposition 2.5, we obtain that the operator $({{L}^{\frac{1}{m}}}-{{L}_{m}}){{H}^{2-\frac{1}{m}}}$ is bounded. We deduce that there exists a constant $c>0$ such that

$$\begin{aligned} -c{{H}^{-2+\frac{1}{m}}}\le {{L}^{\frac{1}{m}}}-{{L}_{m}}\le c{{H}^{-2+\frac{1}{m}}} \end{aligned}$$

(3.3)

According to the min-max theorem, we get

$$\begin{aligned} {{({{\lambda }_{k}}+{{\mu }_{k}})}^{\frac{1}{m}}}=\lambda _{k}^{\frac{1}{m}}+{{\upsilon }_{k}}+O\left( \lambda _{k}^{-2+\frac{1}{m}}\right) \end{aligned}$$

(3.4)

Using the fact that $\{ {{\mu }_{k}} \}$ is bounded and the Taylor’s formula for the function $t\rightarrow {{\left( 1+\frac{{{\mu }_{k}}}{t} \right) }^{\frac{1}{m}}}$, we obtain the estimate

$$\begin{aligned} {{\mu }_{k}}=m\lambda _{k}^{1-\frac{1}{m}}{{\upsilon }_{k}}+O(\lambda _{k}^{-1}) \end{aligned}$$

(3.5)

This completes the proof of Proposition 1.5.

4 Functional calculus for the operator H and the asymptotic behavior of ${{\overline{\upsilon }}_{k}}$

Recall that ${{\overline{L}}_{m}}$ is obtained by replacing B in ${{L}_{m}}$ by $\overline{B}$. Putting

$$\begin{aligned} \displaystyle \overline{V}=\frac{1}{T}\int _{0}^{T}{W(t)dt} ,\quad W(t)={{e}^{-it{{H}^{\frac{1}{m}}}}}V{{e}^{it{{H}^{\frac{1}{m}}}}} \end{aligned}$$

(4.1)

and From (3.1)

$$\begin{aligned} \overline{B}=\frac{1}{m}{{H}^{-\left( 1-\frac{1}{m}\right) }}\overline{V} \end{aligned}$$

(4.2)

We have the following proposition

Proposition 4.1

$\overline{V}\in G_{0}^{-\delta }$, and its Weyl symbol checks

$$\begin{aligned} {{\sigma }_{\overline{V}}}-{{\sigma }_{\overline{V},0}}\in \Gamma _{0}^{-\delta -1} \end{aligned}$$

where

$$\begin{aligned} \displaystyle {{\sigma }_{\overline{V},0}}(x,\xi )=\frac{1}{T}\int _{-1}^{1}{\frac{V(y.\sigma _{H}^{\frac{1}{2m}})}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}}dy \end{aligned}$$

and $\sigma _{H}$ is the Weyl symbol of H.

Proof

First we will study the Weyl symbol of the operator W(t). In order to apply the Egorov’s theorem (Theorem IV-10 [15]) checks his assumptions. Using Proposition 2.6, the operator $H^{\frac{1}{m}}\in G_{1}^{2}$ and ${{\sigma }_{{{H}^{\frac{1}{m}}}}}$ admits the following development

$$\begin{aligned} {{\sigma }_{{{H}^{\frac{1}{m}}}}}\sim \sum \limits _{j=0}^{+\infty }{{{\sigma }_{{{H}^{\frac{1}{m}}},j}}} \quad {{\sigma }_{{{H}^{\frac{1}{m}}},j}}\in \Gamma _{1}^{2-j} \end{aligned}$$

(4.3)

where ${{\sigma }_{{{H}^{\frac{1}{m}}},0}}={{({{\sigma }_{H}})}^{\frac{1}{m}}}$, ${{\sigma }_{{{H}^{\frac{1}{m}}},1}}={0}$

An elementary calculation shows that

$$\begin{aligned} \partial _{x}^{\alpha }\partial _{\xi }^{\beta }{{\sigma }_{{{H}^{\frac{1}{m}}},j}}\in {{L}^{\infty }}(\mathbb {R}\times \mathbb {R}),\quad \alpha ,\beta ,j\in \mathbb {N},\ \alpha +\beta +j>2 \end{aligned}$$

and

$$\begin{aligned} \partial _{x}^{\alpha }\partial _{\xi }^{\beta }\varphi (t)\in {{L}^{\infty }}(\mathbb {R}\times \mathbb {R}),\quad \alpha +\beta \ge 1 \end{aligned}$$

uniformly with respect to t, where $\varphi (t)$ denotes the Hamiltonian flow of $\sigma _{H}^{\frac{1}{m}}$. We recall that $\varphi (t)=(x(t),\xi (t))$ is a solution of the system

$$\begin{aligned} (S)\quad \left\{ \begin{array}{rl} &{} \frac{dx(t)}{dt}=\frac{\partial \sigma _{H}^{\frac{1}{m}}}{\partial \xi }=2{{E}^{\frac{1}{m}-1}}{{\xi }^{2m-1}}(t)\\ &{}\frac{d\xi (t)}{dt}=\frac{-\partial \sigma _{H}^{\frac{1}{m}}}{\partial x}=-2{{E}^{\frac{1}{m}-1}}{{x}^{2m-1}}(t)\\ &{}x(0)=x,\quad \xi (0)=\xi \\ &{} x^{2m}(t)+\xi ^{2m}(t)=x^{2m}+\xi ^{2m}=E \qquad (*)\\ \end{array}\right. \end{aligned}$$

$(*)$ is the first integral of this system

Now we apply the Egorov’s theorem, this yields, for $t\in \mathbb {R}$, W(t) is an (OPD), its Weyl symbol admits the following development ${{\sigma }_{W(t)}}\sim \sum \nolimits _{j\ge 0}{{{\sigma }_{W(t),j}}}$

$$\begin{aligned} \displaystyle {{\sigma }_{W(t),j}}=\int _{0}^{t}{{{i}^{-1}}}\sum \limits _{\underset{0\le l\le j-1}{\mathop {\alpha +\beta +l+k=j+1}}}C_{\alpha ,\beta }(\partial _{\xi }^{\alpha }\partial _{x}^{\beta }{{\sigma }_{{{H}^{1/m}},k}})(\partial _{\xi }^{\beta }\partial _{x}^{\alpha }{\sigma _{_{W,l}}}(\tau )| {{\varphi }^{t-\tau }} d\tau \end{aligned}$$

(4.4)

where

$$\begin{aligned} C_{\alpha ,\beta }=(1-(-1)^{\alpha +\beta })\Gamma (\alpha ,\beta ) \end{aligned}$$

in particular

$$\begin{aligned} {{\sigma }_{W(t),0}}(x,\xi )=Vox(t),\qquad {{\sigma }_{W(t),1}}(x,\xi )=0 \end{aligned}$$

(4.5)

We start by determining the class of ${{\sigma }_{\overline{V},0}}=\frac{1}{T}\int _{0}^{T}{V(x(t))dt}$. We can suppose that $x(0)>0$, $\frac{dx(0)}{dt}>0$ as initial conditions, other cases are treated in the same way, for now we must study the properties of the function x(t) on [0, T]. From the system (S) we get

$$\begin{aligned} dt=\pm \frac{dx}{ 2{{E}^{\frac{1}{m}-1}}{{(E-{{x}^{2m}})}^{1-\frac{1}{2m}}}} \end{aligned}$$

(4.6)

By combining the fact that x(t) is a smooth periodic function of period T with (4.6), we can ensure that the function x(t) reaches its maximum in $t_{0}$, $x(t_{0})={{E}^{\frac{1}{2m}}}$ and its minimum on $t_{1}$, $x(t_{1})=-{{E}^{\frac{1}{2m}}}$.

For now we have

$$\begin{aligned} \displaystyle {{\sigma }_{\bar{V},0}}=\frac{1}{T}\left[ \int _{0}^{{{t}_{0}}}{V(x(t))dt+\int _{{{t}_{0}}}^{{{t}_{1}}}{V(x(t))dt+\int _{{{t}_{1}}}^{T}{V(x(t))dt}}} \right] \end{aligned}$$

We make the change of variable $x(t)=u$, obviously x(t) is increasing on $[ 0,{{t}_{0}}]$, we get

$$\begin{aligned} \displaystyle \int _{0}^{{{t}_{0}}}{V(x(t))dt=}\frac{{{E}^{1-\frac{1}{m}}}}{2}\int _{x}^{{{E}^{\frac{1}{2m}}}}{\frac{V(u)}{{{(E-{{u}^{2m}})}^{1-\frac{1}{2m}}}}du} \end{aligned}$$

(4.7)

After the same calculation on $[ {{t}_{0}},{{t}_{1}}]$ and $[ {{t}_{1}},T]$ we obtain

$$\begin{aligned} \displaystyle {{\sigma }_{\overline{V},0}}=\frac{{{E}^{1-\frac{1}{m}}}}{T}\int _{-{{E}^{\frac{1}{2m}}}}^{{{E}^{\frac{1}{2m}}}}{\frac{V(u)}{{{(E-{{u}^{2m}})}^{1-\frac{1}{2m}}}}}du \end{aligned}$$

(4.8)

Now we apply the change of variable $y=u{{E}^{\frac{-1}{2m}}}$, we have

$$\begin{aligned} \displaystyle {{\sigma }_{\overline{V},0}}=\frac{1}{T}\int _{-1}^{1}{\frac{V(y\sigma _{H}^{\frac{1}{2m}})}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}}dy \end{aligned}$$

(4.9)

Let’s now determine the class to which $\displaystyle {{\sigma }_{\overline{V},0}}$ belongs. Using (1.4) we get

$$\begin{aligned} \displaystyle \left| {{\sigma }_{\overline{V},0}} \right| \le c\int _{0}^{1}{\frac{1}{{{(1+y\sigma _{H}^{\frac{1}{2m}})}^{s}}{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}}dy \end{aligned}$$

(4.10)

We have 2 cases

1st case$0<s<1$

$$\begin{aligned} \left| {{\sigma }_{\bar{V},0}} \right|&\le c{{(1+\sigma _{H}^{\frac{1}{2m}})}^{-s}}\int _{0}^{1}{\frac{1}{{{y}^{s}}{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy} \end{aligned}$$

(4.11)

$$\begin{aligned}&\le c{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-s}{2}}} \end{aligned}$$

(4.12)

2nd case$1<s$

We set $\sigma _{H}^{\frac{1}{2m}}=a$ and we split the integral into two parts,

$$\begin{aligned} {{\sigma }_{\bar{V},0}}&\le c{{\int \limits _{0}^{\frac{\sqrt{2}}{2}}{\left( \frac{1}{1+{{a}^{2}}{{y}^{2}}} \right) }}^{\frac{s}{2}}}\frac{1}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy\nonumber \\&\quad +\,c{{\int \limits _{\frac{\sqrt{2}}{2}}^{1}{\left( \frac{1}{1+{{a}^{2}}{{y}^{2}}} \right) }}^{\frac{s}{2}}}\frac{1}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy \end{aligned}$$

(4.13)

Since $s>1$ we have

$$\begin{aligned} \displaystyle {{\int _{0}^{\frac{\sqrt{2}}{2}}{\left( \frac{1}{1+{{a}^{2}}{{y}^{2}}} \right) }}^{\frac{s}{2}}}\frac{1}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy\le c\int _{0}^{\frac{\sqrt{2}}{2}}{\frac{1}{{{(1+{{a}^{2}}{{y}^{2}})}^{\frac{s}{2}}}}dy} \end{aligned}$$

After applying change both of variables $y=\frac{u}{1+u}$, and $v=u\sqrt{1+a^2}$,

we get

$$\begin{aligned} \displaystyle {{\int _{0}^{\frac{\sqrt{2}}{2}}{\left( \frac{1}{1+{{a}^{2}}{{y}^{2}}} \right) }}^{\frac{s}{2}}}\frac{1}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy\le \frac{c}{1+a} \end{aligned}$$

On the other side we have

$$\begin{aligned} \displaystyle {{\int _{\frac{\sqrt{2}}{2}}^{1}{\left( \frac{1}{1+{{a}^{2}}{{y}^{2}}} \right) }}^{\frac{s}{2}}}\frac{1}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy&\le \frac{c}{{{(1+a)}^{s}}}\int _{\frac{\sqrt{2}}{2}}^{1}{\frac{1}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy}\\ {}&\le \frac{c}{{{(1+a)}^{s}}} \end{aligned}$$

We obtain the following estimate

$$\begin{aligned} {{\sigma }_{\overline{V},0}}\le \frac{c}{\big (1+\sigma _{H}^{\frac{1}{2m}}\big )}\le c{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-1}{2}}}. \end{aligned}$$

(4.14)

The same estimates hold for $\partial _{x}^{\alpha }\partial _{\xi }^{\beta }\sigma _{\overline{V},0}(x,\xi )$, $\alpha , \beta \in {\mathbb { N}}$.

From (4.12) and (4.14) we have

$$\begin{aligned} {{\sigma }_{\bar{V},0}}\in \Gamma _{0}^{-\delta },\quad \hbox {for}\ s\in \mathbb {R}_{+}^{*}-\{ 1\} \end{aligned}$$

Combining (1.4), (4.4) and (4.5) we deduce that there exist $c>0$ such that: for all $t\in [0,T]$

$$\begin{aligned} |{\sigma }_{W(t)}-Vox(t)|\le c\,{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-1}{2}}}{{(1+{{x}^{2}(t)})}^{\frac{-s}{2}}} \end{aligned}$$

(4.15)

by integrating (4.15) along the interval [0, T] and following the same previous calculation we have

$$\begin{aligned} |{{\sigma }_{\overline{V}}}-\frac{1}{T}\int _{0}^{T}Vox(t)\,dt| \le \,c {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-1-\delta }{2}}} \end{aligned}$$

(4.16)

The same estimates hold for $\partial _{x}^{\alpha }\partial _{\xi }^{\beta }\sigma _{\overline{V}}(x,\xi )$, $\alpha , \beta \in {\mathbb { N}}$. Finally we conclude

$$\begin{aligned} {{\sigma }_{\overline{V}}}-\frac{1}{T}\int _{0}^{T}Vox(t)\,dt\in \Gamma _{0}^{-\delta -1} \end{aligned}$$

(4.17)

$\square$

In the following we will use a functional calculus for the operator H, this allows us to give the asymptotic behavior of ${{\overline{\upsilon }}_{k}}$. The functional calculus on (OPD) was studied in the case where the functions are in the Hörmander class $S_{1}^{r}$$(r \in \mathbb {R})$ see [15, 16]. In our work we are dealing with the case of the operator H plus a function in the class $S_{1-\frac{1}{2m}}^{r}$. More precisely, the set of functions $f\in {{C}^{\infty }}(\mathbb {R})$ such that for all $k\in \mathbb {N}$, there exist ${{C}_{k}}\ge 0$ such that

$$\begin{aligned} \left| {{f}^{(k)}}(x) \right| \le {{C}_{k}}{{(1+\left| x \right| )}^{r-(1-\frac{1}{2m})k}} \end{aligned}$$

(4.18)

We recall that the main symbol of $\overline{V}$ is written

$$\begin{aligned} {{\sigma }_{\overline{V},0}}=f({{\sigma }_{H}}) \end{aligned}$$

(4.19)

where

$$\begin{aligned} f(x)=\displaystyle \frac{1}{T}\int _{-1}^{1}{\frac{V(y{{x}^{\frac{1}{2m}}})}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy} \end{aligned}$$

a direct calculation shows that $f\in S_{1-\frac{1}{2m}}^{-\frac{\delta }{2m}}$ for $s\in \mathbb {R}_{+}^{*}-\{ 1\}$.

The operator f(H) is defined by a functional calculus of self-adjoint operators, then the spectrum of f(H) is the sequence ${{\{ f({{\lambda }_{k}})\}}_{k}}$. We have the following proposition

Proposition 4.2

f(H) is an OPD included in $G_{0}^{-\delta }$ and its Weyl symbol admits the following development

$$\begin{aligned}&{{\sigma }_{f(H)}}\sim \sum \limits _{j\ge 0}{{{\sigma }_{f(H),2j}}}\\&{{\sigma }_{f(H),2j}}=\sum \limits _{k=2}^{3j}{\frac{{{d}_{j,k}}}{k!}{{f}^{(k)}}({{\sigma }_{H}})}\quad \forall j\ge 1 \end{aligned}$$

where ${{d}_{j,k}}\in \Gamma _{1}^{2mk-4j}$ and ${{\sigma }_{f(H),2j}}\in \Gamma _{0}^{-\delta -j}$,

in particular

$$\begin{aligned} {{\sigma }_{f(H),0}}=f({{\sigma }_{H}})={{\sigma }_{\overline{V},0}}, \quad {{\sigma }_{f(H),1}}=0 \end{aligned}$$

Proof

For studying f(H) We follow the same strategy in [16] , we will use the Mellin transformation, this later consist of

• (1) To Study the operator ${{( H-\lambda )}^{-1}}$

• (2) To study the operator ${{H}^{-s}}$ using its Cauchy’s integral formula

  $$\begin{aligned} \displaystyle {{H}^{-s}}=\frac{1}{2\pi i}\int _{\Delta }{{{\lambda }^{-s}}}{{(H-\lambda )}^{-1}}d\lambda \end{aligned}$$

• (3) Studying f(H) using the representation formula

  $$\begin{aligned} f(H)=\displaystyle \frac{1}{2\pi i}\int _{\rho -i\infty }^{\rho +i\infty }{M[f](s)}{{H}^{-s}}ds \end{aligned}$$

where $r<0$ and $\rho <-r$

we only change the construction of the $(H-\lambda )^{-1}$-parametrix. We prove by induction that the $(H-\lambda )^{-1}$ is an OPD and its Weyl symbol admits the development ${{b}_{\lambda }}\sim \sum {{{b}_{j,\lambda }}}$ where

$$\begin{aligned} \left\{ \begin{array}{rl} &{} {{b}_{0,\lambda }}={{( {{\sigma }_{H}}-\lambda )}^{-1}} \\ &{}{{b}_{2j+1,\lambda }}=0\\ &{}{{b}_{2j,\lambda }}=\sum \limits _{k=2}^{3j}{{{(-1)}^{k}}{{d}_{j,k}}}.b_{0,\lambda }^{k+1},\quad {{d}_{j,k}}\in \Gamma _{1}^{2mk-4j} \\ \end{array}\right. \end{aligned}$$

$\square$

Proof of Proposition 1.6

using Proposition 4.2 and (4.19) formula we conclude

$$\begin{aligned} {{\sigma }_{f(H)}}-{{\sigma }_{\overline{V},0}}\in \Gamma _{0}^{-\delta -1} \end{aligned}$$

(4.20)

by combining (4.20) and Proposition 4.1 we get

$$\begin{aligned} {{\sigma }_{\overline{V}}}-{{\sigma }_{f(H)}}\in \Gamma _{0}^{-\delta -1} \end{aligned}$$

(4.21)

From (4.21) and Proposition 2.5 (iii) we deduce that the operator $(\overline{V}-f(H)){{H}^{\frac{\delta +1}{2m}}}$ is bounded. We can write

$$\begin{aligned} \frac{1}{m}{{H}^{\frac{1}{m}-1}}(\overline{V}-f(H))=\left[ {{\overline{L}}_{m}}-({{H}^{\frac{1}{m}}}+\tfrac{1}{m}{{H}^{\frac{1}{m}-1}}f(H)) \right] \end{aligned}$$

Finally we get that the operator $[ {{{\bar{L}}}_{m}}-({{H}^{\frac{1}{m}}}+\tfrac{1}{m}{{H}^{\frac{1}{m}-1}}f(H)) ]{{H}^{1+\tfrac{\delta }{2m}-\tfrac{1}{2m}}}$ is also bounded. According to min-max theorem we have

$$\begin{aligned} {{\overline{\upsilon }}_{k}}=\frac{1}{m}\lambda _{k}^{\tfrac{1}{m}-1}f({{\lambda }_{k}})+O\big (\lambda _{k}^{-\left( 1+\frac{\delta }{2m}-\frac{1}{2m}\right) }\big ) \end{aligned}$$

(4.22)

Then

$$\begin{aligned} \displaystyle m\lambda _{k}^{1-\frac{1}{m}}\overline{{{\upsilon }_{k}}}=\frac{1}{T}\int _{-1}^{1}{\frac{V\big (\lambda _{k}^{\frac{1}{2m}}y\big )}{{{(1-{{y}^{2m}})}^{1-\frac{1}{2m}}}}dy+O(\lambda _{k}^{\frac{-\delta -1}{2m}})} \end{aligned}$$

(4.23)

$\square$

Remark 4.3

We note that from (1.4) we have the following estimate

$$\begin{aligned} \lambda _{k}^{\tfrac{1}{m}-1}f({{\lambda }_{k}})=O(\lambda _{k}^{\tfrac{-\delta }{2m}+\tfrac{1}{m}-1}) \end{aligned}$$

5 The relation between the spectrum of ${{L}_{m}}$ and ${\overline{L}_{m}}$

Proof of Proposition 1.7

To establish Proposition 1.7, we need to prove the next result

Proposition 5.1

There exists a skew-symmetric operator $Q\in G_{0}^{-(2m-2+\delta )}$ such as the operator $({{e}^{Q}}{{L}_{m}}{{e}^{-Q}}-{{\overline{L}}_{m}}){{H}^{\frac{\delta -\eta }{m}+2-\frac{2}{m}}}$ is bounded, where $\eta \in ] 0,2 [$.

Proof

The Q operator is built using the $Q_{1}$ and $Q_{2}$ operators as follows

$$\begin{aligned} Q=Q_{1}+Q_{2} \end{aligned}$$

(5.1)

where

$$\begin{aligned} \displaystyle Q_{1}=\frac{i}{mT}{{H}^{\frac{1}{m}-1}}\int _{0}^{T}{(T-t)W(t)dt} \end{aligned}$$

and

$$\begin{aligned} \displaystyle {{Q}_{2}}=-\frac{1}{2T}\int _{0}^{T}{\left( T-t \right) }\int _{0}^{t}{\left[ \frac{1}{m}{{H}^{\frac{1}{m}-1}}W(t),\frac{1}{m}{{H}^{\frac{1}{m}-1}}W(r) \right] }drdt \end{aligned}$$

Before starting the proof we could make sure that

$$\begin{aligned} \left[ {{Q}_{1}},{{H}^{\frac{1}{m}}} \right] =\frac{1}{m}{{H}^{\frac{1}{m}-1}}(\overline{V}-V)\quad \left[ {{Q}_{2}},{{H}^{\frac{1}{m}}} \right] =-\frac{1}{2}\left[ {{Q}_{1}},\frac{1}{m}{{H}^{\frac{1}{m}-1}}V \right] -\overline{\overline{V}} \end{aligned}$$

(5.2)

where $\overline{\overline{V}}=\frac{1}{2Ti}\int _{0}^{T}{\int _{0}^{t}{\left[ \frac{1}{m}{{H}^{\frac{1}{m}-1}}W(t),\frac{1}{m}{{H}^{\frac{1}{m}-1}}W(r) \right] }}drdt$

We notice $AdQ.{{L}_{m}}=\left[ Q,{{L}_{m}} \right]$. The differential equation

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{dX}{dt}=\left[ Q,X \right] \\ X(0)={{L}_{m}}\\ \end{array} \right. \end{aligned}$$

(5.3)

has a unique solution

$$\begin{aligned} X(t)={{e}^{tADQ}}.{{L}_{m}}={{e}^{tQ}}{{L}_{m}}{{e}^{-tQ}} \end{aligned}$$

From (5.2) and (5.3) we get

$$\begin{aligned} {{e}^{Q}}{{L}_{m}}{{e}^{-Q}}-{{\overline{L}}_{m}}= & {} \left\{ -\overline{\overline{V}}+\frac{1}{2m}\left[ {{Q}_{2}},{{H}^{\frac{1}{m}-1}}V \right] \right\} \\&\quad +\,\frac{1}{2m}\left\{ \left[ Q,{{H}^{\frac{1}{m}-1}}\overline{V} \right] +\frac{1}{2}\left[ Q,\left[ {{Q}_{1,}}{{H}^{\frac{1}{m}-1}}V \right] \right] \right\} \\&\quad +\, \frac{1}{2}\left\{ \left[ Q,\left[ {{Q}_{2,}}\frac{1}{m}{{H}^{\frac{1}{m}-1}}V \right] \right] -\left[ Q,\overline{\overline{V}} \right] \right\} \\&\quad +\,\sum \limits _{n\ge 2}{\frac{{{(AdQ)}^{n}}}{(n+1)!}\left[ Q,{{H}^{\frac{1}{m}}} \right] }+\sum \limits _{n\ge 2}{\frac{{{(AdQ)}^{n}}}{(n+1)!}\left[ Q,\frac{1}{m}{{H}^{\frac{1}{m}-1}}V \right] } \end{aligned}$$

To complete the proof of proposition we need the following lemma

Lemma 5.2

$$\begin{aligned} {{Q}_{1}}\in G_{0}^{-(2m+2-\delta )} \quad and \quad \overline{\overline{V}}, Q_{2} \in G_{0}^{-(4m-4+2\delta -2\eta )} \end{aligned}$$

where $\eta \in ]0,2 [$

Proof

Using Proposition 2.5 (i) we can prove in analog way of Proposition 4.1 that ${{Q}_{1}}\in G_{0}^{-(2m+2-\delta )}$. Now let’s determine the class of $\overline{\overline{V}}$, we can write

$$\begin{aligned} \displaystyle \overline{\overline{V}}=\frac{1}{2Ti}\int _{0}^{T}{\left[ S(t),B(t) \right] }dt \end{aligned}$$

where

$$\begin{aligned} S(t)=\frac{1}{m}{{H}^{\tfrac{1}{m}-1}}W(t), \qquad B(t)=\int _{0}^{t}{S(r)dr} \end{aligned}$$

let’s start by clarifying the class of the operator $\int _{0}^{T}{S(t)B(t)dt}$. For now we are interested in the operator S(t)B(t), its Weyl symbol $c_{t}$ is given in [15] by

$$\begin{aligned} \displaystyle {{c}_{t}}(x,\xi )=\frac{1}{{{\pi }^{2}}}\int {{{e}^{-2i(r\rho -\omega \tau )}}}{{\sigma }_{S(t)}}(x+\omega ,\xi +\rho ){{\sigma }_{B(t)}}(x+r,\xi +\tau )d\rho d\omega d\tau dr. \end{aligned}$$

(5.4)

We split the oscillator integral $c_{t}$ into two parts $c_{t}^{(1)}$ and $c_{t}^{(2)}$, then we use the cutoff functions

$$\begin{aligned} {{\omega }_{1,\varepsilon }}(x,\xi ,\omega ,\tau ,r,\rho )=\chi \left[ \frac{{{\omega }^{2}}+{{\rho }^{2}}+{{r}^{2}}+{{\tau }^{2}}}{\varepsilon {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{\eta }{2}}}} \right] \quad \hbox {and} \quad {{\omega }_{2,\varepsilon }}=1-{{\omega }_{1,\varepsilon }} \end{aligned}$$

where $\chi \in C_{0}^{\infty }(\mathbb {R})$, $\chi \equiv 1$ in $\left[ -1,1 \right]$, $\chi \equiv 0$ in $\mathbb {R}\backslash ] -2,2 [$, $R={{\omega }^{2}}+{{\rho }^{2}}+{{r}^{2}}+{{\tau }^{2}}$, $\varepsilon >0$ and $\eta >0$. Let’s consider

$$\begin{aligned} {{d}_{j}}(x,\xi ,\omega ,\tau ,r,\rho )={{\omega }_{j,\varepsilon }}(x,\xi ,\omega ,\tau ,r,\rho ){{\sigma }_{S(t)}}(x+\omega ,\rho +\xi ){{\sigma }_{B(t)}}(x+r,\rho +\xi ) \end{aligned}$$

(5.5)

$c_{t}^{(1)}$ (resp $c_{t}^{(2)}$) the integral obtained in (5.4) by replacing the amplitude by $d_{1}$ (resp $d_{_{2}}$)

Study of $c_{t}^{(2)}$

On the support of $d_{2}$ we have $R\ge \varepsilon {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\tfrac{\eta }{2}}}$. We make an integration by parts using the operator

$$\begin{aligned} M=\frac{1}{2iR}(-\rho {{\partial }_{r}}-r{{\partial }_{\rho }}+\tau {{\partial }_{\omega }}+\omega {{\partial }_{\tau }}) \end{aligned}$$

We have for all $k\in \mathbb {N}$

$$\begin{aligned} c_{t}^{(2)}=\frac{1}{{{\pi }^{2}}}\int {{{e}^{-2i(r\rho -\omega \tau )}}{{({}^{t}M)}^{k}}}{{d}_{2}}\,d\rho \, d\omega \, d\tau \, dr \end{aligned}$$

then we obtain for all $k>0$

$$\begin{aligned} \left| c_{t}^{(2)} \right| \le {{C}_{k}}{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-\eta k}{4}}} \end{aligned}$$

Uniformly with respect to $t\in [ 0,T]$

Study of $c_{t}^{(1)}$

On the support of $d_{1}$ we have

$$\begin{aligned} c_{t}^{(1)}(x,\xi )= & {} \frac{1}{{{\pi }^{2}}}\int _{R\le 2\varepsilon {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{\eta }{2}}}}{{{e}^{-2i(r\rho -\omega \tau )}}}{{\sigma }_{S(t)}}(x+\omega ,\xi +\rho )\nonumber \\&\times \,{{\sigma }_{B(t)}}(x+r,\xi +\tau ){{\omega }_{1,\varepsilon }}d\rho d\omega d\tau dr \end{aligned}$$

(5.6)

$$\begin{aligned} \int _{0}^{T}{\left| c_{t}^{(1)} \right| }dt\le & {} c\int _{R\le 2\varepsilon {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{\eta }{2}}}}{d\rho d\omega d\tau dr}\int _{0}^{T}{\left| {{\sigma }_{S(t)}}(x+\omega ,\xi +\rho ) \right| }dt\nonumber \\&\times \,\int _{0}^{T}{\left| {{\sigma }_{S(t)}}(x+r ,\xi +\tau ) \right| }dt \end{aligned}$$

(5.7)

By using (4.4) and (1.4) we can deduce for all $\alpha ,\beta \in \mathbb {N}$

$$\begin{aligned} \left| \partial _{x}^{\alpha }\partial _{\xi }^{\beta }{{\sigma }_{S(t)}}(x,\xi ) \right| \le {{c}_{\alpha ,\beta }}{{(1+{{x}^{2}}+{{\xi }^{2}})}^{-(m-1)}}{{(1+{{x}^{2}}(t))}^{\frac{-s}{2}}} \end{aligned}$$

by integrating along the interval [0, T] and following the same reasoning in Proposition 4.1 we get

$$\begin{aligned} \left| \partial _{x}^{\alpha }\partial _{\xi }^{\beta }\int _{0}^{T}{{{\sigma }_{S(t)}}(x,\xi )dt} \right| \le {{c}_{\alpha ,\beta }}{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-(2m-2+\delta )}{2}}} \end{aligned}$$

(5.8)

From (5.7) and (5.8) we have

$$\begin{aligned} \int _{0}^{T}{\left| c_{t}^{(1)} \right| }dt\le & {} c\int _{R\le 2\varepsilon {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{\eta }{2}}}}{(1+{{(x+\omega )}^{2}}}+{{(\xi +\rho )}^{2}}{{)}^{\frac{-(2m-2+\delta )}{2}}}\nonumber \\&\times \,{{(1+{{(x+r)}^{2}}+{{(\xi +\tau )}^{2}})}^{\frac{-(2m-2+\delta )}{2}}}d\rho d\omega d\tau dr \end{aligned}$$

On the support of $d_{1}$, for $\varepsilon$ small enough and since $\eta \in ] 0,2 [$, there are positive constants $c, c ', C, C'$ such that

$$\begin{aligned} \left\{ \begin{array}{ll} &{}c{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\tfrac{1}{2}}}\le {{(1+{{(x+\omega )}^{2}}+{{(\rho +\xi )}^{2}})}^{\tfrac{1}{2}}}\le C{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\tfrac{1}{2}}} \\ &{}c'{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\tfrac{1}{2}}}\le {{(1+{{(x+r)}^{2}}+{{(\tau +\xi )}^{2}})}^{\tfrac{1}{2}}}\le C'{{(1+{{x}^{2}}+{{\xi }^{2}})}^{\tfrac{1}{2}}}\\ \end{array}\right. \end{aligned}$$

It follows that

$$\begin{aligned} \displaystyle \int _{0}^{T}{c_{t}^{(1)}dt\le C(1+{{x}^{2}}}+{{\xi }^{2}}{{)}^{-(2m-2+\delta )}}\int _{R\le 2\varepsilon {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{\eta }{2}}}}{d\rho d\omega d\tau dr} \end{aligned}$$

(5.9)

Finally

$$\begin{aligned} \displaystyle \int _{0}^{T}{c_{t}^{(1)}dt\le c(1+{{x}^{2}}}+{{\xi }^{2}}{{)}^{-(2m-2+\delta )+\eta }} \end{aligned}$$

(5.10)

At the end by denoting $\sigma$ the Weyl symbol of the operator $\displaystyle \int _{0}^{T}{S(t)B(t)dt}$,

we have

$$\begin{aligned} \left| \sigma \right|&\le \int _{0}^{T}{\left| c_{t}^{(1)} \right| }dt+\int _{0}^{T}{\left| c_{t}^{(2)} \right| }dt&\\ {}&\le C\left[ {{(1+{{x}^{2}}+{{\xi }^{2}})}^{\frac{-\eta k}{4}}}+{{(1+{{x}^{2}}+{{\xi }^{2}})}^{-(2m-2+\delta -\eta )}} \right] \\&\le C{{(1+{{x}^{2}}+{{\xi }^{2}})}^{-(2m-2+\delta -\eta )}} \end{aligned}$$

We obtain the same estimates for $\partial _{x}^{\alpha }\partial _{\xi }^{\beta }\sigma (x,\xi )$, we prove by the same way that $Q_{2}$$\in G_{0}^{-(4m-4+\delta -2\eta )}$$\square$

We return to the proof of Proposition 5.1, since $V\in G_{0}^{0},\overline{V}\in G_{0}^{-\delta }$, ${{Q}_{1}},Q\in G_{0}^{-(2m-2+\delta )}$ and $\overline{\overline{V}}$, $Q_{2}$$\in G_{0}^{-(4m-4+2\delta -2\eta )}$, by using Proposition 2.5 we have

$$\begin{aligned} \left\{ \begin{array}{ll} &{}\left\| \left\{ -\overline{\overline{V}}+\frac{1}{2m}\left[ {{Q}_{2}},{{H}^{\frac{1}{m}-1}}V \right] \right\} {{H}^{2-\frac{2}{m}+\frac{\delta -\eta }{m}}} \right\| \le C\\ &{}\left\| \left\{ \left[ Q,{{H}^{\tfrac{1}{m}-1}}\overline{V} \right] +\frac{1}{2}\left[ Q,\left[ {{Q}_{1}},{{H}^{\frac{1}{m}-1}}V \right] \right] \right\} {{H}^{2-\frac{2}{m}+\tfrac{\delta }{m}}} \right\| \le C\\ &{}\left\| \left\{ \frac{1}{2}\left[ Q,\left[ {{Q}_{2}},{{H}^{\frac{1}{m}-1}}V \right] \right] -\left[ Q,\overline{\overline{V}} \right] \right\} {{H}^{4-\frac{4}{m}+\frac{2\delta -2\eta }{m}}} \right\| \le C\\ &{}\left\| {{(ADQ)}^{n}}\left[ Q,{{H}^{\tfrac{1}{m}}} \right] {{H}^{2-\frac{2}{m}+\frac{\delta }{m}}} \right\| \le C{{\left\| Q \right\| }^{n-2}}\quad (n\ge 2)\\ &{}\left\| {{(ADQ)}^{n}}\left[ Q,{{H}^{\frac{1}{m}-1}}V \right] {{H}^{2-\frac{2}{m}+\tfrac{\delta }{m}}} \right\| \le C{{\left\| Q \right\| }^{n-2}}\\ \end{array}\right. \end{aligned}$$

From what precedes we deduce that $({{e}^{Q}}{{L}_{m}}{{e}^{-Q}}-{{\overline{L}}_{m}}){{H}^{\frac{\delta -\eta }{m}+2-\frac{2}{m}}}$ is bounded. $\square$

Come back to the proof of Proposition 1.7. We deduce from Proposition 5.1 that there exists a constant $c> 0$ such that

$$\begin{aligned} -c{{H}^{\frac{-\delta +\eta }{m}+\frac{2}{m}-2}}\le {{e}^{Q}}{{L}_{m}}{{e}^{-Q}}-{{\overline{L}}_{m}}\le c{{H}^{\frac{-\delta +\eta }{m}+\frac{2}{m}-2}} \end{aligned}$$

According to the min-max theorem

$$\begin{aligned} \upsilon _{k}={{\overline{\upsilon }}_{k}}+{\mathrm O}(\lambda _{k}^{\frac{-\delta +\eta }{m}+\tfrac{2}{m}-2}) \end{aligned}$$

(5.11)

To have a good estimate, let us specify the best choice of $\eta$. Combining Remark 4.3, (5.11) and Proposition 5.1, we choose

$$\begin{aligned} \eta \in ] 0, \min (2,m-1+\tfrac{{\delta }}{2}) [ \end{aligned}$$

(5.12)

$\square$

Now we prove Theorems 1.2 and 1.3. It is enough to combine Propositions 1.5, 1.6 and 1.7 we deduce

$$\begin{aligned} \displaystyle {{\mu }_{k}}=\displaystyle \frac{1}{T}\int _{-1}^{1}{\frac{V(y\lambda _{k}^{\frac{1}{2m}})}{(1-{{y}^{2m}}){}^{1-\frac{1}{2m}}}}dy+\text {O}(\lambda _{k}^{\frac{-\delta -1}{2m}}), \quad \forall m\ge 2 \end{aligned}$$

and for $m=1$ (harmonic oscillator case)

$$\begin{aligned} \displaystyle {{\mu }_{k}}=\frac{1}{\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}{V\left( \sqrt{{{\lambda }_{k}}}\sin t \right) }dt+O( \lambda _{k}^{-\delta +\eta }), \end{aligned}$$

where $\eta \in ] 0,\frac{\delta }{2} [$.