Surjectivity of certain adjoint operators and applications

Abstract

This paper is an extension and generalization of some previous works, such as the study of M. Benalili and A. Lansari. Indeed, these authors, in their work about the finite co-dimension ideals of Lie algebras of vector fields, restricted their study to fields $X_0$ of the form $X_0=\sum _{i=1}^{n}( \alpha _i \cdot x_i+\beta _i\cdot x_i^{1+m_i}) \frac{\partial }{\partial x_i}$, where $\alpha _i, \beta _i$ are positive and $m_i$ are even natural integers. We will first study the sub-algebra U of the Lie-Fréchet space E, containing vector fields of the form $Y_0 = X_0^+ + X_0^- + Z_0$, such as $X_0\left( x,y\right) =A\left( x,y\right) =\left( A^{-}\left( x \right) ,A^{+}\left( y\right) \right)$, with $A^-$ (respectively, $A^+$) a symmetric matrix having eigenvalues $\lambda < 0$ (respectively, $\lambda >0$) and $Z_0$ are germs infinitely flat at the origin. This sub-algebra admits a hyperbolic structure for the diffeomorphism $\psi _{t*}=(exp\cdot tY_0)_*$. In a second step, we will show that the infinitesimal generator $ad_{-X}$ is an epimorphism of this admissible Lie sub-algebra U. We then deduce, by our fundamental lemma, that $U=E$.

1 Introduction

The ideals of finite codimension in Lie algebras of vector fields have recently received a lot of attention. Some authors such as Pursel and Shanks [9], by studying the invertibility of the Lie bracket $[X,Y]=ad_X(Y)$ which is an infinitesimal generator of an one-parameter group t, $\gamma _t=(\exp tX)^*$, in Lie algebras containing a germ of vector fields X do not vanish at the origin O, have treated the finite-codimensional ideals of these algebras. This result has been prolonged in the Banach-Lie algebras of vector fields infinitely flat at 0 containing germs which vanish at the origin of the form $X_0=\sum _{i=1}^{n}( \alpha _i \cdot x_i+Z_0(x))$, where $\alpha _i$ are of constant signs [2, 3, 8]. Among the motivations and possible applications of these results:

• The properties of the injectivity of the exponential function of a vector field have given rise to the existence of the Fourier series [11].

• The properties of the surjectivity of the directional derivative of the exponential function have given rise to the existence of the inversibility of the exponential function through the Nash–Moser theorem where positive results were obtained first in a Fréchet space and then in a hyperbolic type Fréchet space by integrating the diffeomorphisms in the smooth flows [12].

• Boris Kolev in [7] studied the particular case of a Lie–Poisson canonical structure.

• M. BENALILI in [1] has studied suitable spectral properties of the adjoint operators induced by appropriate perturbations of some hyperbolic linear vector fields of the form $Y_0 = X_0^+ + X_0^- + Z_0$, where $Z_0$ is k-flat in the unit ball.

  This paper is an extension on the basis of these works where we will first study the sub-algebra U of the Lie–Fréchet space E, containing vector fields of the form $Y_0 = X_0^+ + X_0^- + Z_0$, such as $X_0\left( x,y\right) =A\left( x,y\right) =\left( A^{-}\left( x \right) , A^{+}\left( y\right) \right)$, with $A^-$ (respectively, $A^+$) a symmetric matrix having eigenvalues $\lambda < 0$ (respectively, $\lambda >0$) and $Z_0$ are germs infinitely flat at the origin. This sub-algebra admits a hyperbolic structure for the diffeomorphism $\psi _{t*}=(exp \cdot tY_0)_*$. In a second step, we will show that the infinitesimal generator $ad_{-X}$ is an epimorphism of this admissible Lie sub-algebra U. We then deduce, by our fundamental lemma, that $U=E$.

Part I: Admissible hyperbolic-type algebra

2 Definitions

2.1 Fréchet space

Let ${\mathbb {R}}^n$ be the Euclidean space provided with the scalar product $\left\langle .,.\right\rangle$ and $\left\| . \right\|$ the norm induced by this scalar product.

(a) Let E be the space of vector fields X of class $C^\infty$ on ${\mathbb {R}}^n$, satisfying:

$\forall r \in {\mathbb {N}}, \exists M_r >0 \text{ such } \text{ that: }$

${\left\{ \begin{array}{ll} \forall x \in {\mathbb {R}}^n \\ \forall \alpha =(\alpha _1, \ldots , \alpha _n)\in {\mathbb {N}}^n\\ \forall k \in {\mathbb {N}} \end{array}\right. } \text { and } {\left\{ \begin{array}{ll} \left| \alpha \right| = \alpha _1 + \cdots + \alpha _n \\ \left| \alpha \right| + k \le r \end{array}\right. }.$

we have $\left\| D^\alpha X(x) \right\| \left( 1+\left\| x \right\| ^2 \right) ^{k/2} \leqslant M_r$.

We define on E a graduation of seminorms:

$$\begin{aligned} \left\| X \right\| _r= \underset{ x \in {\mathbb {R}}^n }{ {\text {sup}} } \, \underset{ k+\left| \alpha \right| \le r }{\max } \left\| D^\alpha X(x) \right\| \left( 1+\left\| x \right\| ^2 \right) ^{k/2} \end{aligned}$$

so $( E,\left\| \cdots \right\| _r)$ is called a Fréchet space [6].

(b) Let G the Schwartz space, which is the vector space of class $C^{\infty }$ functions on ${\mathbb {R}}^n$ satisfying:

$G=\{f\in C^{\infty }({\mathbb {R}}^n)/ \forall p\ge 0, \forall x\in {\mathbb {R}}^{n},\forall \alpha \in {\mathbb {N}}^{n}$ $\exists C_{p}>0$ satisfying $\Vert D^{\alpha }f(x)\Vert (1+\Vert x\Vert ^{2})^{p}\le C_{p}\}$.

G is a space where the Fourier transform exists as well as its inverse.

We define on G a graduation of seminorms as follows:

$$\begin{aligned} \left\| f \right\| _r= \underset{ x \in {\mathbb {R}}^n }{ {\text {sup}} } \, \underset{ k+\left| \alpha \right| \le r }{\max } \left\| D^\alpha f(x) \right\| \left( 1+\left\| x \right\| ^2 \right) ^{k/2} \end{aligned}$$

so $( G,\left\| \cdots \right\| _r)$ is called a Fréchet space [6].

2.2 Smooth flow

(a) Adjoint diffeomorphisms

Definition 2.1

For all $X,Y \in E$ such that $Y = \sum _{i=1}^{n} u_i(x)\frac{\partial }{\partial x_i}$, where $u_i \in C^\infty ({\mathbb {R}}^n)$, we define the adjoint diffeomorphisms $\phi _t^*$ and $(\phi _t)_*$ by

$$\begin{aligned} \left\{ \begin{matrix} (\phi _t)^* Y= (exp t X)^*\cdot Y =((D\phi _t)^{-1}\cdot Y)\circ \phi _t=\sum _{j=1}^{n}e^{-tX}\cdot u_{j}\left( e^{tX}\cdot x\right) \frac{\partial }{\partial x_{j}} \\ (\phi _t)_* Y= (exp t X)_*\cdot Y = (D\phi _t\cdot Y)\circ (\phi _t)^{-1}= \sum _{j=1}^{n} e^{tX}\cdot u_{j}\left( e^{-tX}\cdot x\right) \frac{\partial }{\partial x_{j}} \end{matrix}\right. \end{aligned}$$

Proprieties. For all $X \in E$, we associate the $X-$flow $\phi _t=exptX$, so

(i) $ad_X$ (respectively, $ad_{-X}$) is an infinitesimal generator of the one-parameter group $\phi _t^*$ (respectively, $(\phi _t)_*$ ) on E; that is, $\phi _t^*$ is a solution of the following dynamic system:

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{{\text {d}}}{{\text {d}}t}\left( \phi _{t}\right) ^{*}Y= \left( \phi _{t}\right) ^{*}\cdot ad_{X}\left( Y\right) = \phi _{t}^* \left[ X,Y \right] \\ \left( \phi _{0}\right) ^{*}Y=Y \end{array}\right. } \end{aligned}$$

and, respectively, $(\phi _t)_*$ is the solution of

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{{\text{ d }}}{{\text{ d }}t}\left( \phi _{t}\right) _{*}Y= \left( \phi _{t}\right) _{*}\cdot ad_{-X}\left( Y\right) = \left( \phi _{t}\right) _{*} \left[ Y,X \right] \\ \left( \phi _{0}\right) _{*}Y=Y \end{array}\right. } \end{aligned} \end{aligned}$$

(ii) $(\phi _t)_* = \phi _{-t}^*$; $\phi _t^* = (\phi _{-t})_*, \forall t > 0$.

Proof

(i) In fact,

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}\left( \left( \phi _{t}\right) _{*}Y\right) \left( x\right)= & {} \frac{{\text {d}}}{{\text {d}}t}D\left[ \phi _{t}\left( \phi _{-t}\left( x\right) \right) \right] \cdot Y\left( \phi _{-t}\left( x\right) \right) +D\phi _{t}\left( \phi _{-t}\left( x\right) \right) \cdot \frac{{\text {d}}}{{\text {d}}t}Y\left( \phi _{-t}\left( x\right) \right) \\= & {} D_{x}X\left( \phi _{t}\circ \phi _{-t}\left( x\right) \right) \cdot D\phi _{t}\left( \phi _{-t}\left( x\right) \right) \cdot Y\left( \phi _{-t}\left( x\right) \right) \\&-D\phi _{t}\left( \phi _{-t}\left( x\right) \right) \cdot D_y\left( Y\left( \phi _{-t}\left( x\right) \right) \right) \cdot X\circ \phi _{-t}\left( x\right) \quad \text {where } y=\phi _{-t}(x) \\= & {} D_{x}X\left( x\right) \left( D\phi _{t}\cdot Y\right) \left( \phi _{-t}\left( x\right) \right) -\left( D\phi _{t} \cdot X\right) \left( \phi _{-t}\left( x\right) \right) . D_{y} Y(y) \\= & {} D_xX\left( x\right) \left( \left( \phi _{t}\right) _{*}\cdot Y\right) \left( x\right) -\left( \left( \phi _{t}\right) _{*}\cdot X\right) \left( x\right) \cdot D_yY\left( y\right) \\= & {} \left( \phi _{t}\right) _{*}\left[ Y,X\right] \left( x\right) \\= & {} \left( \phi _{t}\right) _{*} ad_{-X}\left( Y\right) \left( x\right) . \end{aligned}$$

That is to say:

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}\left( \phi _{t}\right) _{*}Y= \left( \phi _{t}\right) _{*} \left[ Y,X\right] ; \end{aligned}$$

so,

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}((\phi _{t})_{*}Y)(x) \mid _{t=0^{+}}= & {} \ [Y,\ X](x)=ad_{-X}(Y)(x), \end{aligned}$$

i.e. that $ad_{-X}$ is an infinitesimal generator of the one-parameter group $(\phi _t)_*$ on E, and by the same reasoning, we will have

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}\left( \phi _{t}\right) ^{*}Y=\left( \phi _{t}\right) ^{*} \left[ X,Y\right] . \end{aligned}$$

(ii) Let us show that $(\phi _t)_*=\phi _{-t}^*$; we put $y=\phi _{-t}(x)$:

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}\left( \left( \phi _{t}\right) _{*}Y\right) \left( x\right)= & {} D_{x}X\left( \phi _{t}\circ \phi _{-t}\left( x\right) \right) \cdot D\phi _{t}\left( \phi _{-t}\left( x\right) \right) \cdot Y\left( \phi _{-t}\left( x\right) \right) \\&-D\phi _{t}\left( \phi _{-t}\left( x\right) \right) \cdot D\left( Y\left( \phi _{-t}\left( x\right) \right) \right) \cdot X\circ \phi _{-t}\left( x\right) \\= & {} D_{x}X\left( x\right) \left( (D\phi _{-t})^{-1} \cdot Y\right) \left( \phi _{-t}\left( x\right) \right) -\left( (D\phi _{-t})^{-1} \cdot X\right) \left( \phi _{-t}\left( x\right) \right) . D_{y} Y(y) \\= & {} D_xX\left( x\right) \left( \left( \phi _{-t}\right) ^{*}\cdot Y\right) \left( x\right) -\left( \left( \phi _{-t}\right) ^{*}\cdot X\right) \left( x\right) .D_yY\left( y\right) \\= & {} \left( \phi _{-t}\right) ^{*}\left[ Y,X\right] \left( x\right) \\= & {} \left( \phi _{-t}\right) ^{*} ad_{-X}\left( Y\right) \left( x\right) . \end{aligned}$$

Now, according to Property (i):

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}\left( \left( \phi _{t}\right) _{*}Y\right) \left( x\right) = \left( \phi _{t}\right) _{*} ad_{-X}\left( Y\right) \left( x\right) , \end{aligned}$$

then $\left( \phi _{t}\right) _{*} = \left( \phi _{-t}\right) ^{*}$.

- The same reasoning can be applied in the case: $\phi _t^* = (\phi _{-t})_*$ $\square$.

(b) Smooth flow Let $X \in E$ and the $X-$flow $\phi _t=exptX$. We say that an adjoint flow $\phi _t^*$ decays tamely on E of degree r and base b if

• (i) $\phi _t^*$ preserves E $\forall t\ge 0$,

• (ii) for any integer $k \ge b$ there is an integer $l_k = k+r$ and a strictly positive, continuous and decreasing function $C_k(t)$ defined on $[0,\infty )$ satisfying:

  $$\begin{aligned} \left\| \phi _t^* Z \right\| _k\leqslant C_k(t)\left\| Z \right\| _{l_k} \qquad (\forall t\geqslant 0; \forall Z \in E), \end{aligned}$$

  and the improper integral:

  $$\begin{aligned} \int _{0}^{\infty } C_k(t){\text {d}}t \qquad converges \quad \forall k\ge b. \end{aligned}$$

  Alternately, $\phi _t^*$ can be replaced by $(\phi _t)_* = \phi _{-t}^*$ according to the asymptotic behaviour of $\phi _t$ [12].

2.3 Fréchet space with hyperbolic structure

(a) Admissible algebra

Definition 2.2

(Admissible algebra) Let U be a Lie–Fréchet sub-algebra of E. U is said to be admissible with respect to the vector field $Y_0=X_0+Z_0$ if and only if it satisfies the following conditions:

(i) Let $X_0$ be a vector field on E such that

$$\begin{aligned} \begin{aligned} X_0.\left( \begin{array}{c}x\\ y\end{array}\right) = A.\left( \begin{array}{c}x\\ y\end{array}\right) = \begin{pmatrix} A^- &{}{} 0 \\ 0 &{}{} A^+ \end{pmatrix} \cdot \left( \begin{array}{c}x\\ y\end{array}\right) = \left( \begin{array}{c}A^-x\\ A^+y\end{array}\right) \\ \forall (x,y) \in K \subset {\mathbb {R}}^k \times {\mathbb {R}}^l ={\mathbb {R}}^n \end{aligned} \end{aligned}$$

$A^-$ (respectively, $A^+$) is a real symmetric matrix of type ($k \times k$) (respectively, $l \times l$) with $k+l=n$, having strictly negative eigenvalues (respectively, strictly positive).

The matrix $A^-$ satisfies: $\forall m\ge 2, \forall i=1,k: \exists a_R, a_L> 0, \rho _{1} >1$, such that

$$\begin{aligned} {\left\{ \begin{array}{ll} -a_{L}\le \lambda _i \le -a_R \ < 0 \\ m\cdot a_R-a_{L}\ge \rho _{1} > 1 \end{array}\right. } \quad (I). \end{aligned}$$

And, respectively, for $A^+$:

$\forall m\ge 2, \forall i=1,l: \exists b_R, b_L> 0, \rho _{2} >1$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} 0 < b_L \le \lambda _i \le b_R \\ m\cdot b_L-b_{R}\ge \rho _{2} > 1 \end{array}\right. } \quad \text {(II)}. \end{aligned}$$

We adopt the following notation for the rest of this paper:

$$\begin{aligned} X_0=X_0^- + X_0^+;\quad X_{0}\left( x,y\right) =\left( {X_0}^-\left( x\right) ,X_{0}^{+}\left( y\right) \right) = \left( {A}^- \cdot x ,A^{+}\cdot y \right) \in {\mathbb {R}} ^{k}\times {\mathbb {R}} ^{l}, \end{aligned}$$

from which we have

$$\begin{aligned} \phi _t(x,y)=\left( exp tX_0 \right) (x,y)= \left( \left( exp tX_0^- \right) (x), \left( exp tX_0^+ \right) (y) \right) ; \forall (x,y) \in {\mathbb {R}}^k \times {\mathbb {R}}^l. \end{aligned}$$

(ii) There exists $U_1$ (respectively, $U_2$) a Lie–Fréchet sub-algebra of E containing the vector field $Y_0^-=X_0^- + Z_0^-$ (respectively, $Y_0^+=X_0^+ + Z_0^+$) such that

$$\begin{aligned} U = U_1 \oplus U_2, \end{aligned}$$

where $Z_0=Z_0^- + Z_0^+$ is a perturbation of $X_0$ such that

$$\begin{aligned} Z_0(x,y)=\big (Z_0^-(x),Z_0^+(y) \big ) \in {\mathbb {R}}^k \times {\mathbb {R}}^l={\mathbb {R}}^n; \forall (x,y)\in K \subset {\mathbb {R}}^k \times {\mathbb {R}}^l \end{aligned}$$

$Z_0^-$ (respectively, $Z_0^+$ ) is an infinitely flat germ at the origin, i.e. satisfying the following estimate:

$$\begin{aligned} \forall x\in {\mathbb {R}}^{n}, \left\{ \begin{array}{l} \forall \alpha =(\alpha _{1},\ \ldots ,\ \alpha _{n})\in {\mathbb {N}} ^{n}\\ |\alpha |=\sum _{{{i}}=1}^{n}\alpha _{i}\ \end{array}\right. \end{aligned}$$

There exists $M_{\alpha } > 0$ such that

$$\begin{aligned} ||D^{\alpha }Z_0^-{(x)}|| \le M_{\alpha }\cdot ||x||^{k_{1}},\ \forall k_{1}>1 \quad \forall x \in {\mathbb {R}}^{k}, \end{aligned}$$

(resp)

$$\begin{aligned} ||D^{\alpha }Z_{0}^{+}(y)||\le M_{\alpha }\cdot ||y||^{k_{2}},\ \forall k_{2}>1 \quad \forall y \in {\mathbb {R}}^{l}. \end{aligned}$$

(iii) $\forall Y \in U;\exists ! Y^i \in U_i(i=1,2)$ such that $Y=Y^1+Y^2$ and $\forall (x,y) \in {\mathbb {R}}^{n} = \Omega _1 \cup \Omega _2$, we have

$$\begin{aligned}Y(x,y) = \left( Y^1(x),Y^2(y) \right) \in \Omega _1 \cup \Omega _2;\\(exp t Y_0)_* Y(x,y) = \left( (exp t Y_0^-)_* Y^1(x) ,(exp t Y_0^+)_* Y^2(y) \right) \in \Omega _1 \cup \Omega _2,\end{aligned}$$

where

$$\begin{aligned} {\left\{ \begin{array}{ll} \Omega _1= \left\{ (x,y) \in {\mathbb {R}}^{k} \times {\mathbb {R}}^{l} / Y(x,y) = (Y^1(x),0) \right\} \\ \Omega _2= \left\{ (x,y) \in {\mathbb {R}}^{k} \times {\mathbb {R}}^{l} / Y(x,y) = (0,Y^2(y)) \right\} . \end{array}\right. } \end{aligned}$$

(4i) There exists $\delta >0$ such that $U_i^\delta = \{ Y^i \in U_i / supp Y^i \subseteq \Omega _i^\delta \}$ with $\Omega _i^\delta = \Omega _i + B_{\delta }$ and $B_{\delta } \subset \Omega _j, i \ne j$.

(5i) U is locally closed, i.e. for any sequence $( X_m^i )_{m \ge 0}$ in $U_i$ such that $supp X_m^i \subset \Omega _i$ converges to $X^i$ in $\Omega _i$.

(b) Hyperbolic structure

Definition 2.3

Let E be a Fréchet space of vector fields, $X \in E$ and the $X-$flow $\psi _t=exptX$. E has a tame hyperbolic structure for $(\psi _t)_*$ if and only if $\exists \delta ' > 0$ /

(i):

$\psi _{t*}$ is invariant on $E; \forall t \in {\mathbb {R}}$,

(ii):

$(\psi _t)_*$ decays on $E_1^{\delta '}; \forall t \ge 0$,

(iii):

$(\psi _{-t} )_*$ decays on $E_2^{\delta '}; \forall t \ge 0$,

where $E_{i}^{\delta '}= \{ Z \in E / supp Z \subset \Omega _{i}^{\delta '} \}$ with $\Omega _{i}^{\delta '} = \Omega _{i} + B_{\delta '}$, $\text {such that } B_{\delta '} \subset \Omega _{j}, i \ne j \text { and } E=E_1 \oplus E_2 = E_1^{\delta '} + E_{2}^{\delta '}$ [12].

3 Estimations

To show that U has a hyperbolic structure, we need some estimates.

3.1 Estimation of $exp tY_0^+$ and $exp tY_0^-$

(a) Estimation of $exp t X_0^+$ and $exp t X_0^-$

Lemma 3.1

Let $X_{0}^{+} \in U_2$ (respectively, $X_{0}^{-} \in U_1$) Then $X_{0}^{+}$-flow has for estimate, for all $t\geqslant 0$ :

$$\begin{aligned} {\left\{ \begin{array}{ll} \parallel y \parallel e^{b_{L}t}\leqslant \parallel (\exp tX_{0}^{+})\left( y \right) \parallel \leqslant \parallel y \parallel e^{b_{R}t} \\ \parallel y \parallel e^{-b_{R}t}\leqslant \parallel (\exp -tX_{0}^{+})\left( y \right) \parallel \leqslant \parallel y \parallel e^{-b_{L}t} \end{array}\right. } (\forall y \in {\mathbb {R}}^l), \end{aligned}$$

respectively, for the $X_{0}^{-}$-flow:

$$\begin{aligned} \left\{ \begin{array}{c} \parallel x\parallel e^{-a_{L}t}\leqslant \parallel (\exp tX_{0}^{-})\left( x\right) \parallel \leqslant \parallel x\parallel e^{-a_{R}t} \\ \parallel x\parallel e^{a_{R}t}\leqslant \parallel (\exp -tX_{0}^{-})\left( x\right) \parallel \leqslant \parallel x\parallel e^{a_{L}t} \end{array} \right. (\forall x \in {\mathbb {R}}^k). \end{aligned}$$

Proof

We put $\phi _{t}^{+}\left( y \right) = (\exp tX_{0}^{+})\left( y \right)$, from which

$$\begin{aligned} \frac{1}{2}\frac{{\text {d}}}{{\text {d}}t}\parallel \phi _{t}^{+}\left( y \right) \parallel ^{2}= \langle \phi _{t}^{+}\left( y \right) ,X_{0}^+\circ \phi _{t}^{+}\left( y \right) \rangle . \end{aligned}$$

According to Wintner theorem [10], $\langle y, A^+y\rangle \le d \mid y \mid ^2$, $y \in {\mathbb {R}}^l$, where the constant d is the largest eigenvalue of the symmetric matrix $A^+$. We get

$$\begin{aligned} \inf _{i=1,\cdots ,l} \lambda _{i} \cdot \parallel \phi _{t}^{+}\left( y \right) \parallel ^{2}\leqslant \frac{1}{2}\frac{{\text {d}}}{{\text {d}}t} \parallel \phi _{t}^{+}\left( y \right) \parallel ^{2}\leqslant \sup _{i=1,..,l} \lambda _{i} \cdot \parallel \phi _{t}^{+}\left( y \right) \parallel ^{2}. \end{aligned}$$

We finally have

$$\begin{aligned} \parallel y \parallel e^{b_{L}t}\leqslant \parallel \phi _{t}^{+}\left( y \right) \parallel \leqslant \parallel y \parallel e^{b_{R}t}. \end{aligned}$$

The same should be applicable to $\phi _{-t}^{+}$ by replacing t by $(-t)$:

$$\begin{aligned} \parallel y \parallel e^{-b_{R}t}\leqslant \parallel \phi _{-t}^{+}\left( y \right) \parallel \leqslant \parallel y \parallel e^{-b_{L}t}. \end{aligned}$$

On the other hand, applying the same reasoning to $X_0^-$ shows that

$$\begin{aligned} \left\{ \begin{array}{l} \parallel x\parallel e^{-a_{L}t}\leqslant \parallel \phi _{t}^{-}\left( x\right) \parallel \leqslant \parallel x\parallel e^{-a_{R}t} \\ \parallel x\parallel e^{a_{R}t}\leqslant \parallel \phi _{-t}^{-}\left( x\right) \parallel \leqslant \parallel x\parallel e^{a_{L}t} \end{array} \right. \end{aligned}$$

$\square$

(b) Estimation of $exp t Y_0^+$ and $exp t Y_0^-$

We put $Y^-_0 = X^-_0 + Z^-_0$ (respectively, $Y^+_0 = X^+_0 + Z^+_0$) Then the vector form of $Y^-_0 - flow$ (respectively, $Y^+_0 - flow$) will be

$$\begin{aligned} \psi _{t}^{-}(x)=exp t (X_0^- + Z_0^-) (x)=\phi _t ^- (x)+ \int _0^t \phi _{t-s} ^- \circ Z_0^- (\psi _s^- (x)) {\text {d}}s \qquad \mathbf (1) , \end{aligned}$$

respectively,

$$\begin{aligned} \psi _{t}^{+}(y)=exp t (X_0^+ + Z_0^+) (y)=\phi _t ^+ (y)+ \int _0^t \phi _{t-s} ^+ \circ Z_0^+ (\psi _s^+ (y)) {\text {d}}s. \end{aligned}$$

Solution of the following dynamic system:

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\frac{{\text {d}}}{{\text {d}}t} \psi _{t}^{-}\left( x\right) = X_0^- \circ \psi _{t}^{-}\left( x\right) +Z_{0}^{-} \circ \psi _{t}^{-}\left( x\right) \\ &{}\psi _{0}^{-}(x)=x, \end{array}\right. } \end{aligned}$$

(respectively,

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\frac{{\text {d}}}{{\text {d}}t} \psi _{t}^{+}\left( y \right) = X_0^+ \circ \psi _{t}^{+}\left( y \right) +Z_{0}^{+} \circ \psi _{t}^{+}\left( y \right) \\ &{}\psi _{0}^{+}(y)=y. \end{array}\right. } \end{aligned}$$

Lemma 3.2

Let $Y_{0}^{-} \in U_1$ (respectively, $Y_{0}^{+} \in U_2$ ) Then the vector fields $Y_0^-$ (respectively, $Y_0^+$) is complete, and the flow $\psi _{t}^{-}$(respectively, $\psi _{t}^{+})$ would satisfy the following estimates:

$$\begin{aligned} \left\{ \begin{array}{c} \parallel x\parallel \cdot e^{-a_{L}t}\leqslant \parallel (\exp tY_0^{-})\left( x\right) \parallel \leqslant \parallel x\parallel \cdot e^{-a_{R}t} \\ \parallel x\parallel \cdot e^{a_{R}t}\leqslant \parallel (\exp -tY_0^{-})\left( x\right) \parallel \leqslant \parallel x\parallel \cdot e^{a_{L}t}, \end{array} ;\text { }\forall t\geqslant 0\right. \end{aligned}$$

(respectively,

$$\begin{aligned} \left. \left\{ \begin{array}{c} \parallel y \parallel \cdot e^{b_{L}t}\leqslant \parallel (\exp tY_0^{+})\left( y \right) \parallel \leqslant \parallel y \parallel \cdot e^{b_{R}t} \\ \parallel y \parallel \cdot e^{-b_{R}t}\leqslant \parallel (\exp -tY_0^{+})\left( y \right) \parallel \leqslant \parallel y \parallel \cdot e^{-b_{L}t} \end{array} ;\text { }\forall t\geqslant 0\right. \right) . \end{aligned}$$

Proof

Consider the equation below:

$$\begin{aligned} \frac{1}{2}\frac{{\text {d}}}{{\text {d}}t}\parallel \psi _{t}^{-}\left( x\right) \parallel ^{2}= \langle \psi _{t}^{-}\left( x\right) ,X_{0}^{-}\left( \psi _{t}^{-}\left( x\right) \right) +Z_{0}^{-}\left( \psi _{t}^{-}\left( x\right) \right) \rangle . \end{aligned}$$

After taking $\zeta =\parallel \psi _{t}^{-}\left( x\right) \parallel$, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} -a_{L}\zeta ^{2}-M_{0}\zeta ^{k_1+1}\leqslant \frac{1}{2}\frac{{\text {d}}}{{\text {d}}t}\zeta ^{2}\leqslant -a_{R}\zeta ^{2}+M_{0} \zeta ^{k_1+1} \\ \zeta \left( 0\right) =\parallel x\parallel . \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} -a_{L} \zeta ^{1-k_1}-M_{0}\leqslant \zeta ^{-k_1}\frac{{\text {d}}}{{\text {d}}t} \zeta \leqslant -a_{R} \zeta ^{1-k_1}+M_{0}. \end{aligned}$$

If we put $z=\zeta ^{1-k_1}$, we will have $dz=\left( 1-k_1\right) \zeta ^{-k_1}d\zeta$ and the system becomes

$$\begin{aligned} -a_{L}z-M_{0}\leqslant \frac{1}{1-k_1}\frac{{\text {d}}}{{\text {d}}t}z\leqslant -a_{R}z+M_{0} \end{aligned}$$

which has as solutions

$$\begin{aligned} b_{1}e^{-a_{R}\left( 1-k_1\right) t}+\frac{M_{0}}{a_{L}}\leqslant z\leqslant b_{2}e^{-a_{L}\left( 1-k_1\right) t}-\frac{M_{0}}{a_{R}} \end{aligned}$$

such as $b_{1}, b_{2}$, two functions of x and as $a_{L},a_{R}>0,$ then

$$\begin{aligned} b_{1}e^{-a_{R}\left( 1-k_1\right) t}\leqslant z\leqslant b_{2}e^{-a_{L}\left( 1-k_1\right) t}. \end{aligned}$$

As $1-k_1<0$, then

$$\begin{aligned} b_{3}e^{-a_{L}t}\leqslant \zeta \leqslant b_{4}e^{-a_{R}t}, \end{aligned}$$

where $b_3, b_4$ also two functions of x,

from which

$$\begin{aligned} \parallel x\parallel \cdot e^{-a_{L}t}\leqslant \parallel \psi _{t}^{-}\left( x\right) \parallel \leqslant \parallel x\parallel \cdot e^{-a_{R}t}. \end{aligned}$$

Similarly, we will have

$$\begin{aligned} \parallel x\parallel \cdot e^{a_{R}t}\leqslant \parallel \psi _{-t}^{-}\left( x\right) \parallel \leqslant \parallel x\parallel \cdot e^{a_{L}t}. \end{aligned}$$

And by a similar reasoning, we come to

$$\begin{aligned} \left\{ \begin{array}{l} \parallel y \parallel \cdot e^{b_{L}t}\leqslant \parallel \psi _{t}^{+}\left( y \right) \parallel \leqslant \parallel y \parallel \cdot e^{b_{R}t} \\ \parallel y \parallel \cdot e^{-b_{R}t}\leqslant \parallel \psi _{-t}^{+}\left( y \right) \parallel \leqslant \parallel y \parallel \cdot e^{-b_{L}t} \end{array} ;\text { }\forall t\geqslant 0\right. \end{aligned}$$

$\square$

(c) The final estimate of the $l'-{th}$ derivative of $Y_{0}-$ flow

(i) Estimation of the first derivative of the $Y_{0}-$ flow:

We denote $\eta ^{-}\left( t,x,\upsilon _1 \right) =D\psi _{t}^{-}\left( x\right) \upsilon _1$ and $\eta ^{+}\left( t,y,\upsilon _2 \right) =D\psi _{t}^{+}\left( y\right) \upsilon _2$, where $\upsilon _1, \upsilon _2 \in {\mathbb {R}}^{k}.$ The first derivative with respect to x of $Y_{0}^{-}$-flow (respectively, $Y_{0}^{+}$-flow ) is a solution of the dynamic system:

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{{\text {d}}}{{\text {d}}t}\eta ^{-}\left( t,x,\upsilon _1 \right) =\left( D_{\xi _1}X_{0}^{-}+D_{\xi _1}Z_{0}^{-}\right) \cdot \eta ^{-}\left( t,x,\upsilon _1 \right) \\ \eta ^{-}\left( 0,x,\upsilon _1 \right) =\upsilon _1 \end{array}\right. } \end{aligned}$$

with $\xi _1=\psi _{t}^{-}\left( x\right)$ and, respectively,

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{{\text {d}}}{{\text {d}}t}\eta ^{+}\left( t,y,\upsilon _2 \right) =\left( D_{\xi _2}X_{0}^{+}+D_{\xi _2}Z_{0}^{+}\right) \cdot \eta ^{+}\left( t,y,\upsilon _2 \right) \\ \eta ^{+}\left( 0,y,\upsilon _2 \right) =\upsilon _2 \end{array}\right. } \end{aligned}$$

with $\xi _2=\psi _{t}^{+}\left( y\right)$.

Lemma 3.3

The derivative of $Y_{0}^{-}$-flow (respectively, of $Y_{0}^{+}$-flow) has the following estimates for all $t\geqslant 0$:

$$\begin{aligned} \left\{ \begin{array}{l} e^{-a_{L}t}\leqslant \parallel (D\exp tY_0^-)\left( x\right) \parallel \leqslant e^{-a_{R}t} \\ e^{a_{R}t}\leqslant \parallel (D\exp -tY_0^-)\left( x\right) \parallel \leqslant e^{a_{L}t}, \end{array} \right. \end{aligned}$$

(respectively,

$$\begin{aligned} \left. \left\{ \begin{array}{l} e^{b_{L}t}\leqslant \parallel (D\exp tY_0^+)\left( y\right) \parallel \leqslant e^{b_{R}t} \\ e^{-b_{R}t}\leqslant \parallel (D\exp -tY_0^+)\left( y\right) \parallel \leqslant e^{-b_{L}t} \end{array} \right. \right) . \end{aligned}$$

Proof

Let $z_1=\parallel \eta ^{-}\left( t,x,v \right) \parallel$, and consider the following equation:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}\frac{{\text{ d }}}{{\text{ d }}t} z_1^{2} =\frac{1}{2}\frac{{\text{ d }}}{{\text{ d }}t}\parallel \eta ^{-}\left( t,x,v_1 \right) \parallel ^{2}\\ = \langle \eta ^{-}\left( t,x,v_1 \right) ,\left( D_{\xi _1}X_{0}^{-}+D_{\xi _1}Z^-_{0}\right) \cdot \eta ^{-}\left( t,x,v_1 \right) \rangle \\ z_1(0) =\parallel v_1 \parallel . \end{array}\right. } \end{aligned} \end{aligned}$$

We will have

$$\begin{aligned} {\left\{ \begin{array}{ll} z_1^2 \left( -a_L-M_1e^{-a_Lt k_1} \right) \leqslant \frac{ 1}{2}\frac{{\text {d}}}{{\text {d}}t} z_1^{2}\leqslant {z_1}^{2}.\left( -a_R+M_1e^{-a_Rt k_1} \right) \\ {z_1(0)} =\parallel v_1 \parallel . \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} \left( -a_L-M_1e^{-a_Lt k_1} \right) {\text {d}}t \leqslant \frac{dz_1}{z_1} \leqslant \left( -a_R+M_1e^{-a_Rt k_1} \right) {\text {d}}t. \end{aligned}$$

We will then have

$$\begin{aligned} \begin{array}{c} -a_Lt + \frac{M_1}{a_L k_1}e^{-a_Lt k_1} + b_6 \leqslant ln z_1 \leqslant -a_R.t- \frac{M_1}{a_R k_1}e^{-a_Rt k_1} + b_5, \\ \\ \end{array} \end{aligned}$$

where $b_5, b_6$ are two functions depending on x.

As $-a_Lt + b_6 \leqslant ln z_1 \leqslant -a_R.t + b_5$, then $\left\| v_1 \right\| \cdot e^{-a_L.t} \le z_1 \le \left\| v_1 \right\| \cdot e^{-a_R.t}$. Taking $\left\| v_1 \right\| = 1$, we get

$$\begin{aligned} e^{-a_L t} \leqslant \parallel D\psi _{t}^{-} (x) \parallel \leqslant e^{-a_R t} \qquad \forall x \in {\mathbb {R}}^k, \forall t>0. \end{aligned}$$

We can easily deduce

$$\begin{aligned} e^{a_R t} \leqslant \parallel D\psi _{-t}^{-} (x) \parallel \leqslant e^{a_L t}. \end{aligned}$$

And by a similar reasoning, we come to

$$\begin{aligned} \left. \left\{ \begin{array}{c} e^{b_{L}t}\leqslant \parallel D\psi _{t}^{+}\left( y \right) \parallel \leqslant e^{b_{R}t} \\ e^{-b_{R}t}\leqslant \parallel D\psi _{-t}^{+}\left( y \right) \parallel \leqslant e^{-b_{L}t} \end{array} \right. \right) . \end{aligned}$$

$\square$

(ii) The final estimate of the $l'-{th}$ derivative of $Y_{0}-$ flow

Lemma 3.4

The $l'-{th}$ derivative of $Y_{0}^-$ -flow ( resp of $Y_{0}^{+}$-flow ) has the following estimates for all $t\geqslant 0, \forall l' \in {\mathbb {N}}$ :

$$\begin{aligned} \left\{ \begin{array}{c} \parallel D^{l'}\psi _{t}^{-}\left( x\right) \parallel \leqslant M''_{l}e^{-a_{R}t}\\ \parallel D^{l'}\psi _{-t}^{-}\left( x\right) \parallel \leqslant M''_{l}e^{a_{L}t} \end{array} \quad \forall x \in {\mathbb {R}}^k \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{l} \parallel D^{l'}\psi _{t}^{+}\left( y\right) \parallel \leqslant M''_{l}e^{b_{R}t} \\ \parallel D^{l'}\psi _{-t}^{+}\left( y\right) \parallel \leqslant M''_{l}e^{-b_{L}t} \end{array} \right. \quad \forall y \in {\mathbb {R}}^l. \end{aligned}$$

Proof

We can generalize the previous result using a recurrence reasoning as follows:

• (1) For $j=0,1$, the result is verified.

• (2) So, let us suppose the statement true until $\left( j-1\right)$ order, i.e.

  $$\begin{aligned} \parallel D^{l'}\psi _{t}^{-}\left( x\right) \parallel \leqslant M'_{l'}e^{-a_{R}t}; \qquad \forall l' \le j-1. \end{aligned}$$

• (3) Let us show that this last property is true at the order j. Let the $j^{th}$ derivative of $Y_{0}^-$ -flow

  $$\begin{aligned} \eta _j ^{-}\left( t,x,v \right) =D^{j}\psi _{t}^{-}\left( x\right) v ^{j}\ \ , \forall v \in {\mathbb {R}}^{n} \end{aligned}$$

  solution of the following dynamic system:

  $$\begin{aligned} {\left\{ \begin{array}{ll} \frac{{\text {d}}}{{\text {d}}t}\eta _j ^{-}\left( t,x,v \right) =D_{\xi _1}Y_{0}^{-}\cdot \eta _{j}^{-}\left( t,x,v \right) +G_{j}^{-}\left( t,x,v \right) \\ \eta _j^{-}\left( 0,x,v,\ldots ,v \right) =v \end{array}\right. } \end{aligned}$$

  with

  $$\begin{aligned} {\left\{ \begin{array}{ll} G_{j}^{-}\left( t,x,v \right) =\sum _{k'=2}^{j} D_{\xi _1}^{k'}Z^-_{0}\left( \xi _1 \right) \sum \nolimits _{\begin{array}{c} i_{1}+i_{2}+\cdots +i_{k'}=j \\ i_{k'}>0 \end{array} } D_{x}^{i_{1}}\psi _{t}^{-}\left( x\right) v ^{i_{1}}\cdots D_{x}^{i_{k'}}\psi _{t}^{-}\left( x\right) v ^{i_{k'}} \\ \xi _1=\psi _{t}^{-}\left( x\right) . \end{array}\right. } \end{aligned}$$

Using the so-called resolvent transform method discussed, for example, in [4], Y. Domar studied the closed ideals in some Banach algebras [5]), we deduce that

$$\begin{aligned} \eta _{j}^{-}\left( t,x,v ,\ldots ,v \right) =D\psi _{t}^{-}\left( x\right) v +\int _{0}^{t}D\psi _{t-s}^{-}\left( \psi _{s}^{-}\left( x\right) \right) G_{j}^{-}\left( s,x,v \right) {\text {d}}s. \end{aligned}$$

The integral in the preceding expression is well defined at the point $s=0$ because

$$\begin{aligned} \underset{s\rightarrow 0^{+}}{\lim }D\psi _{t-s}^{-}\left( \psi _{s}^{-}\left( x\right) \right) =D\psi _{t}^{-}\left( x\right) \end{aligned}$$

and there are constants $A_{l}>0$ such that

$$\begin{aligned} \underset{s\rightarrow 0^{+}}{\lim }G_{j}^{-}\left( s,x,v \right) = \overset{m}{\underset{k''=2}{\sum }}A_{k'}D_{\xi _1}^{k'}Z_{0}^{-}\left( \xi _1 \right) v ^{k'}. \end{aligned}$$

Since v is arbitrary, we can choose $\parallel v \parallel =1$, and put

$$\begin{aligned} I_{j}=\int _{0}^{t}\parallel D\psi _{t-s}^{-}\left( \psi _{s}^{-}\left( x\right) \right) \parallel .\parallel G_{j}^{-}\left( s,x,v \right) \parallel ds. \end{aligned}$$

Let $K\subset {\mathbb {R}} ^{k}$ a compact set, so $I_{j}$ converges uniformly when t tends to $+\infty$ for all $x\in K.$

As

$$\begin{aligned} \parallel D^{k'}Z_{0}^-\left( x\right) \parallel ^K \leqslant M_{k'}{ \ \ , \ } \forall k'\geqslant 1 \ , \text {and }\forall x\in {\mathbb {R}} ^{k} \end{aligned}$$

and using the recurrence hypotheses, we arrive at

$$\begin{aligned} I_{j}\leqslant \overset{j}{\underset{k'=2}{\sum }}M_{k'} \overset{}{\underset{i_1+i_2+..+i_k'=j}{\sum }} M'_{i_1} \cdots M'_{i_k'} \int _{0}^{t}e^{-\left( t-s+s k'\right) a_R }{\text {d}}s\leqslant {M'}_{j}e^{-a_{R}t}. \end{aligned}$$

The integral $I_ {j}$ is then uniformly convergent with respect to x when t tends to $+\infty$. Consequently, there are constants $M_{j}^{''}=sup (1,M'_j)>0$ such tat

$$\begin{aligned} \parallel \eta _{j}^{-}\parallel\leqslant & {} \parallel D\psi _{t}^{-}\left( x\right) v \parallel +\int _{0}^{+\infty }\parallel D\psi _{t-s}^{-}\left( \psi _{s}^{-}\left( x\right) \right) \parallel .\parallel G_{j}^{-}\left( s,x,v \right) \parallel {\text {d}}s \\\leqslant & {} M_{j}^{''}e^{-a_{R}t}. \end{aligned}$$

(4) Conclusion:

$$\begin{aligned} \parallel D^{l'}\psi _{t}^{-}\left( x\right) \parallel \leqslant M''_{l}e^{-a_{R}t}, \forall l' \in {\mathbb {N}}; \quad \forall x \in {\mathbb {R}}^k. \end{aligned}$$

By similar reasoning, we shall have

$$\begin{aligned} \parallel D^{l'}\psi _{-t}^{-}\left( x\right) \parallel \leqslant M''_{l}e^{a_{L}t} \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{l} \parallel D^{l'}\psi _{t}^{+}\left( y\right) \parallel \leqslant M''_{l}e^{b_{R}t} \\ \parallel D^{l'}\psi _{-t}^{+}\left( y\right) \parallel \leqslant M''_{l}e^{-b_{L}t} \end{array} \right. \quad \forall y \in {\mathbb {R}}^l. \end{aligned}$$

$\square$

3.2 Estimation of $(exp tY_0^+)_*$ and $(exp tY_0^-)_*$

As $\psi _t = \phi _t \circ f_t$, the estimate of $\psi _{{t}_*}$ is deduced according to the estimates of $\phi _{{t}_*}$ and $f_{{t}_*}$.

(a) The absorbents $N_1^+$ and $N_2^+$

A closed subset $N_i^+$ is called positive absorbent for the flow $\phi _t$, if and only if for any compact ${K_i}$ in $\Omega _i$, there exists $t_{K_i} > 0$, such that $\phi _t(K_i) \subset N_i$.

Example: Let $K_i$ be a compact of $\Omega _i, (i=1,2)$ Then $\exists \delta >0$, such that $\delta = \underset{x \in K_i}{min } \left\| x \right\|$, we can then have the following absorbents:

$$\begin{aligned} {\left\{ \begin{array}{ll} N_1^+= \left\{ (x,y) \in {\mathbb {R}}^{k} \times {\mathbb {R}}^{l} / Y(x,y) = (Y^1(x),0) \text { and } \left\| x \right\| \ge \delta \right\} \subset \Omega _1 \\ N_2^+= \left\{ (x,y) \in {\mathbb {R}}^{k} \times {\mathbb {R}}^{l} / Y(x,y) = (0,Y^2(y)) \text { and } \left\| y \right\| \ge \delta \right\} \subset \Omega _2 \end{array}\right. } \end{aligned}$$

(b) Estimation of $(exp tX_0^+)_*$ and $(exp tX_0^-)_*$

Lemma 3.5

The diffeomorphism $(\phi _t^-)_*$ ( respectively, $(\phi _{-t}^+)_*$) decays tamely on $U_1$ (respectively, on $U_2$) of degree 0 and base $m_1$ (respectively, $m_2$), in other words:

(i) For every arbitrary positive constant $\rho '_1$, there is a constant $m_1= \left[ r. \frac{a_L}{a_R} \right] +\rho '_1 > 0$ and $C_1 > 0$ such that

$$\begin{aligned} ||(\exp {t}X_0^-)_{*}\cdot Y^1||_{r}^{\Omega _1} \le C_1\cdot e^{-t\cdot \rho '_1 a_R}. \left\| Y^1 \right\| _{r+m_1}^{\Omega _1} \qquad \forall Y^1 \in U_1. \end{aligned}$$

(ii) For every arbitrary positive constant $\rho '_2$, there is a constant $m_2= \left[ r\cdot \frac{b_R}{b_L} \right] +\rho '_2 > 0$ and $C_2 > 0$ such that

$$\begin{aligned} ||(exp{-t}X_0^+)_{*}\cdot Y^2||_{r}^{\Omega _2} \le C_2\cdot e^{-t \rho '_2\cdot b_{L}}\cdot \left\| Y^2 \right\| _{r+m_2}^{\Omega _2} \qquad \forall Y^2 \in U_2. \end{aligned}$$

Proof

(i) Consider the following diffeomorphism:

$$\begin{aligned}&(\phi _{t}^-)_{*}\cdot Y^1=(D\phi _{t}^{-}\cdot Y^1) \circ (\phi _{t}^{-})^{-1} \\&(\phi _{t}^-)_{*}\cdot Y^1=\sum _{j=1}^{k}e^{tX_{0}^-} \left[ u_{j} \left( e^{\mathrm {-t}X_{0}^{-}}\cdot x\right) \right] \frac{\partial }{\partial x_{j}} \end{aligned}$$

with

$$\begin{aligned} Y^1=\sum _{j=1}^{k}u_{{\dot{j}}}(x)\frac{\partial }{\partial x_{j}} \in U_1. \end{aligned}$$

For any integer $r \ge 0$ and any multi-index $\beta =(\beta _{1}\ldots .,\beta _{k})\in {\mathbb {N}}^{k}$ of module $|\beta |=\beta _{1}+\ldots +\beta _{k}$, we obtain for $x \in {\mathbb {R}}^k$:

$$\begin{aligned} D_{x}^{\beta }(\phi _{t}^{-})_{*}\cdot Y^1(x) =(D_{x}^{\beta }e^{tX_{0}^-}\cdot u_{1}(e^{-tX_0^-}\cdot x), \ldots , D_{x}^{\beta }e^{t X_{0}^-}\cdot u_{k}(e^{-tX_0^-}\cdot x) ). \end{aligned}$$

We will have

$$\begin{aligned} \begin{aligned} || (D_{x}^{\beta }(\phi _{t}^{-})_{*}\cdot Y^1( x)) ||\ \le e^{-a_{R}\mathrm {t}} \sup _{j=1\ldots k } \left\| D_{x}^{\beta }u_{j}(e^{-tX_0^-}\cdot x) \right\| . \end{aligned} \end{aligned}$$

Since $|| \phi _{-t}^{-} (x) ||\le ||x||e^{{a}_{L}t}$, and posing $z=e^{-tX_{0} ^-}x$, then

$$\begin{aligned} || (D_{x}^{\beta }(\phi _{t}^{-})_{*}\cdot Y^1( x)) || \le e^{-a_{R} t} \cdot e^{t |\beta | a_{L} } \cdot \sup _{j=1\ldots k } \left\| D_{z}^{\beta }\cdot u_{j}(z) \right\| , \end{aligned}$$

from which we obtain

$$\begin{aligned} || (\phi _{t}^{-})_{*}\cdot Y^1 ||_r^{\Omega _1} \le e^{-t (- r a_{L} +a_R)} \left\| Y^1 \right\| _r^{ \phi _{-t}^{-}(\Omega _1) }, \forall r \in {\mathbb {N}}. \end{aligned}$$

Let $\rho '_1$ be an arbitrary positive constant Then there is $m_1= \left[ r. \frac{a_L}{a_R} \right] +\rho '_1 > 0$, we will have

$$\begin{aligned} || (\phi _{t}^{-})_{*} \cdot Y^1||_{r}^{\Omega _1} \le e^{-t(a_{R}-ra_{L})}.\ ||Y^1||_{r+m_1}^{ \phi _{-t}^{-}(\Omega _1) } \cdot \sup _{x \in {\mathbb {R}}^k } \left( \frac{1}{1+\left\| \phi _{-t}^- (x) \right\| ^2} \right) ^{\frac{m_1}{2}}. \end{aligned}$$

From Lemma 3.1, $\parallel x\parallel e^{a_{R}t}\leqslant \parallel \phi _{-t}^{-}\left( x\right) \parallel$, there is a constant $C_1 > 0$ such that

$$\begin{aligned} ||(\phi _{t}^-)_{*}\cdot Y^1||_{r}^{\Omega _1} \le C_1\cdot e^{-t\cdot \rho '_1 a_R}. \left\| Y^1 \right\| _{r+m_1}^{N_1^+}. \end{aligned}$$

And as the integral $\int _{0}^{\infty } e^{-t \rho '_1 a_R} {\text {d}}t$ is convergent, $(\phi _{t}^-)_{*}$ decays tamely on $U_{1}.$

(ii) We have, respectively,

For any vector field $Y^2 \in U_2$, with $Y^2=\sum _{j=1}^{l}v_{{\dot{j}}}(y)\frac{\partial }{\partial y_{j}}$, we consider

$$\begin{aligned} (\phi _{-t}^+)_{*}\cdot Y^2=\sum _{j=1}^{l}e^{-tX_{0}^+} \left[ v_{j} \left( e^{\mathrm {t}X_{0}^{+}}\cdot y \right) \right] \frac{\partial }{\partial y_{j}}. \end{aligned}$$

For any integer $r\ge 0$ and any multi-index $\beta '=(\beta '_{1}\ldots .,\beta '_{l})\in {\mathbb {N}}^{l}$ of module $|\beta '|=\beta '_{1}+\ldots +\beta '_{l}$, we obtain

$$\begin{aligned} D_{y}^{\beta '}(\phi _{-t}^{+})_{*}\cdot Y^1( y) =(D_{y}^{\beta '}e^{-t X_{0}^+}.v_{1}(e^{tX_0^+}\cdot y), \ldots , D_{y}^{\beta '}e^{-t X_{0}^+}.v_{l}(e^{tX_0^+}\cdot y) ). \end{aligned}$$

We will have

$$\begin{aligned} || (D_{y}^{\beta '}(\phi _{-t}^{+})_{*}\cdot Y^2( y)) || \le e^{-b_{L}\mathrm {t}} \sup _{j=1\ldots l} \left\| D_{y}^{\beta '}v_{j}(e^{tX_0^+}\cdot y) \right\| . \end{aligned}$$

Since $|| \phi _{t}^{+} (y) ||\le ||y||e^{b_{R}t}$, and posing $z'=e^{tX_{0} ^+}\cdot y$,

$$\begin{aligned} || (D_{y}^{\beta '}(\phi _{-t}^{+})_{*}\cdot Y^2( y)) || \le e^{-b_{L} t} \cdot e^{t |\beta '| b_{R} } \cdot \sup _{j=1\ldots l } \left\| D_{z'}^{\beta '}.v_{j}(z') \right\| , \end{aligned}$$

from which we obtain

$$\begin{aligned} || (\phi _{-t}^{+})_{*}\cdot Y^2 ||_r^{\Omega _2} \le e^{-t (- r b_{R} +b_L)} \left\| Y^2 \right\| _r^{ \phi _{t}^{+}(\Omega _2) }. \end{aligned}$$

Let $\rho '_2$ be an arbitrary positive constant Then there is $m_2= \left[ r. \frac{b_R}{b_L} \right] +\rho '_2 > 0$, we will have

$$\begin{aligned} || (\phi _{-t}^{+})_{*} \cdot Y^2||_{r}^{\Omega _2} \le e^{-t(b_{L}-rb_{R})}. ||Y^2||_{r+m_2}^{ \phi _{t}^{+}(\Omega _2) }. \sup _{y \in {\mathbb {R}}^l } \left( \frac{1}{1+\left\| \phi _t^+ (y) \right\| ^2} \right) ^{\frac{m_2}{2}}. \end{aligned}$$

From Lemma 3.1, $\parallel y \parallel e^{b_{L}t}\leqslant \parallel \phi _{t}^{+}\left( y \right) \parallel$ , there is a constant $C_2 > 0$ such that

$$\begin{aligned} ||(\phi _{-t}^+)_*\cdot Y^2||_{r}^{\Omega _2} \le C_2\cdot e^{-t \rho '_2.b_{L}}. \left\| Y^2 \right\| _{r+m_2}^{N_2^+}. \end{aligned}$$

And as the integral $\int _{0}^{\infty } e^{-t(\rho '_2.b_{L})} {\text {d}}t$ is convergent, $(\phi _{-t}^+)_{*}$ operates smoothly on $U_{2}.$ $\square$

(b) Wave operators

Let

$$\begin{aligned} \phi _{t}^{-}= & {} \exp tX_{0}^{-} \ \ \ \text {and} \ \ \psi _{t}^{-}=\exp tY_{0}^{-}\text { } \\ \ \text {( resp }\phi _{t}^{+}= & {} \exp tX_{0}^{+} \ \ \ \text {and} \ \ \psi _{t}^{+}=\exp tY_{0}^{+}\text { )} \end{aligned}$$

and the diffeomorphism $f_{t}^{-}\left( x\right) =\left( \phi _{-t}^{-}\circ \psi _{t}^{-}\right) \left( x\right) \in G$. According to expression (1) (in Sect. 3.1) of $\psi _t^-(x)$, the diffeomorphism $f_{t}^{-}$ becomes

$$\begin{aligned} f_{t}^{-}\left( x\right) = x+\int _{0}^{t} \phi _{-s}^- (Z_{0}^{-} \circ \psi _{s}^{-}) (x) {\text {d}}s \end{aligned}$$

(respectively,

$$\begin{aligned} f_{-t}^{+}\left( y \right) = y-\int _{0}^{t} \phi _{s}^+ (Z_{0}^{+} \circ \psi _{-s}^{+}) (y) {\text {d}}s). \end{aligned}$$

The wave operator is defined by

$$\begin{aligned} f^{-}=\underset{t\rightarrow +\infty }{\lim }\phi _{-t}^{-}\circ \psi _{t}^{-}\ \ \ (resp\ f^{+}=\underset{t\rightarrow +\infty }{\lim }\phi _{-t}^{+}\circ \psi _{t}^{+}). \end{aligned}$$

Lemma 3.6

(1) $f_{t}^{-}$, $D^{r}f_{t}^{-}$ (respectively, $f_{-t}^{+}$, $D^{r}f_{-t}^{+}$), $f_{-t}^{-}$ and $D^{r}f_{-t}^{-}$ (respectively, $f_{t}^{+}$ and $D^{r}f_{t}^{+}$) are infinite globally uniformly bounded $\forall t>0, \forall r \in {\mathbb {N}}.$

(2) For all compact $K_1 \subseteq \Omega _1$ (respectively, $K_2 \subseteq \Omega _2$), there is $\varepsilon > 0$ such that $\forall t \in \mathbb {R^+}$:

$$\begin{aligned} \Vert f_{-t}^{-} - id \Vert ^{K_1}_r \le \varepsilon \text { and } \Vert f_{t}^{+} - id \Vert ^{K_2}_r \le \varepsilon . \end{aligned}$$

Proof

(1) Let us show that $f_{t}^{-}$ and $D^{r}f_{t}^{-}$ (respectively, $f_{-t}^{+}$ and $D^{r}f_{-t}^{+}$) are infinite globally uniformly bounded $\forall t>0, \forall r \in {\mathbb {N}}$.

Let

$$\begin{aligned} f_{t}^{-}\left( x\right) = x+\int _{0}^{t} \phi _{-s}^- (Z_{0}^{-} \circ \psi _{s}^{-}) (x) {\text {d}}s, \end{aligned}$$

then for all $r \in {\mathbb {N}}; \forall v \in {\mathbb {R}}^n$ such as $\left\| v \right\| = 1$, we have

$$\begin{aligned} D_x^{r} \left( f_{t}^{-} - id \right) \left( x\right) v^{r} = \int _{0}^{t} e^{-A^- s} D_x^{r} (Z_{0}^{-} \circ \psi _{s}^{-} (x)). v^{r} {\text {d}}s. \end{aligned}$$

As

$$\begin{aligned} D_x^{r} (Z_0^- \circ \psi _{s}^{-}(x)).v^{r}= \sum _{\gamma =1}^{r} D_{\xi }^\gamma Z_0^- (\xi ) \sum _{l_1+\cdots l_{\gamma }=r} D^{l_1} \psi _s^-(x).v^{l_1}\cdots D^{l_\gamma } \psi _s^-(x).v^{l_\gamma }, \end{aligned}$$

where $\xi =\psi _s^-(x)$.

According to Lemma 3.2, there exist constants $M_i > 0$ such that $\parallel \psi _s^-(x) \parallel ^{K_1} \le M_1 e^{-s a_R}$ with $K_1$ a compact included in $\Omega _1$ and $\forall k_1 > 1$; we will then have

$$\begin{aligned} \parallel D_{\xi }^\gamma Z_0^- (\xi ) \parallel \le M_{\gamma } \parallel {\xi } \parallel ^{k_1}= M_{\gamma } \Vert \psi _{s}^{-} (x) \Vert ^{k_1}, \end{aligned}$$

from which,

$$\begin{aligned} \Vert D_x^{r} \left( f_{t}^{-} - id \right) \left( x\right) \Vert ^{K_1}\le & {} M_2 \int _{0}^{t} e^{ s a_L} \Vert \psi _{s}^{-} (x) \Vert ^{k_1} {\text {d}}s \\\le & {} M_3 \int _{0}^{t} e^{ s a_L} \cdot e^{ -s a_R k_1} {\text {d}}s \\\le & {} M_3 \int _{0}^{t} e^{ -s[ k_1 a_R - a_L]} {\text {d}}s. \end{aligned}$$

By hypothesis (I) in Sect. 2.3, $\exists \rho _1> 1/ k_1 a_R - a_L \ge \rho _1 > 1$, then

$$\begin{aligned} \Vert D_x^{r} \left( f_{t}^{-} - id \right) \left( x\right) \Vert ^{K_1} \le M_3 \int _{0}^{t} e^{ -s \rho _1} {\text {d}}s \le \dfrac{M_3}{\rho _1}, \end{aligned}$$

from which,

$$\begin{aligned} \Vert \left( f_{t}^{-} - id \right) \left( x\right) \Vert ^{K_1}_r \le \dfrac{M_3}{\rho _1}; \qquad \forall t >0. \end{aligned}$$

It follows that $f_{t}^{-}$ and $Df_{t}^{-}$ are infinite-uniformly bounded $\forall t >0; \forall r \in {\mathbb {N}}$.

And by a similar reasoning, with $K_2$ a compact included in $\Omega _2$ and with the hypothesis (II) in Sect. 2.3$\exists \rho _2> 1/ k_2 b_L - b_R \ge \rho _2 > 1$, we can demonstrate that

$$\begin{aligned} \Vert \left( f_{-t}^{+} - id \right) \left( y\right) \Vert ^{K_2}_r \le \dfrac{M_4}{\rho _2}, \end{aligned}$$

from which $f_{-t}^{+}$ is infinite-globally bounded $\forall t\geqslant 0$, i.e.

$$\begin{aligned} \Vert D_y^{r} \left( f_{-t}^{+} - id \right) \left( y\right) \Vert ^{K_2} \leqslant M_4 \int _{0}^{t} e^{-s (k_2 b_L - b_R)} {\text {d}}s \le M_4 \int _{0}^{t} e^{ -s \rho _2} {\text {d}}s = M_4 \dfrac{1-e^{-t \rho _2}}{\rho _2} \le \dfrac{M_4}{\rho _2}, \end{aligned}$$

where $k_2.b_L-b_R \ge \rho _2> 1;k_2 > 1$.

Let us show that $f_{-t}^{-}$ and $D^{r}f_{-t}^{-}$ (respectively, $f_{t}^{+}$ and $D^{r}f_{t}^{+}$) are infinite globally uniformly bounded $\forall t>0, \forall l' \in {\mathbb {N}}$.

Let

$$\begin{aligned} f_{-t}^{-}\left( x\right) = x-\int _{0}^{t} \phi _{s}^- (Z_{0}^{-} \circ \psi _{-s}^{-}) (x) {\text {d}}s, \end{aligned}$$

then, for all $r \in {\mathbb {N}}; \forall v \in {\mathbb {R}}^n$ such as $\left\| v \right\| = 1$, we have

$$\begin{aligned} D_x^{r} \left( f_{-t}^{-} - id \right) \left( x\right) v^{r} = -\int _{0}^{t} e^{A^- s} D_x^{r} (Z_{0}^{-} \circ \psi _{-s}^{-} (x)). v^{r} {\text {d}}s. \end{aligned}$$

As

$$\begin{aligned} D_x^{r} (Z_0^- \circ \psi _{-s}^{-}(x)).v^{r}= \sum _{\gamma =1}^{r} D_{\xi '}^\gamma Z_0^- (\xi ') \sum _{l_1+\cdots l_{\gamma }=r} D^{l_1} \psi _{-s}^-(x).v^{l_1}\cdots D^{l_\gamma } \psi _{-s}^-(x).v^{l_\gamma } \qquad , \text { where } \xi '=\psi _{-s}^-(x). \end{aligned}$$

According to Lemma 3.2, $\parallel \psi _{-s}^-(x) \parallel ^{K_1} \le M e^{s a_L}$, we deduce that there is a constant $M' > 0$ such that

$$\begin{aligned} \Vert D_x^{r} \left( f_{-t}^{-} - id \right) \left( x\right) \Vert ^{K_1}\le & {} M' \int _{0}^{t} e^{ -s a_R} \Vert Z_0^- \Vert _{r+m'_1}^{\psi _{-s}^{-} (K_1)}\sup _{x \in K_1} \left( \frac{1}{1+\parallel \psi _{-s}^-(x)\parallel ^2} \right) ^{m'_1/2}e^{r sa_L}{\text {d}}s \\\le & {} M' \int _{0}^{t} e^{ -s[(1+m'_1)a_R - (k_1+r)a_L]} {\text {d}}s. \end{aligned}$$

We choose

$$\begin{aligned} m'_1=(r+k_1).\left[ \frac{a_L}{a_R}+1 \right] + \rho >0 \end{aligned}$$

with $\rho$ any positive constant.

$$\begin{aligned} \Vert D_x^{r} \left( f_{-t}^{-} - id \right) \left( x\right) \Vert ^{K_1} \le M' \int _{0}^{t} e^{ -s (\rho +1)a_R} {\text {d}}s = M' \dfrac{1-e^{-t (\rho +1)}}{(\rho +1) a_R} \end{aligned}$$

As $K_1$ is arbitrary, we can find $\varepsilon _1>0$ and $\rho >0$ such that

$$\begin{aligned} \Vert D_x^{r} \left( f_{-t}^{-} - id \right) \left( x\right) \Vert ^{K_1} \le \varepsilon _1; \qquad \forall t >0. \end{aligned}$$

(3.1)

It follows that $f_{-t}^{-}$ and $D^{r}f_{-t}^{-}$ are infinite-uniformly bounded $\forall t >0; \forall r \in {\mathbb {N}}$.

And by a similar reasoning, there is a constant $M''> 0$ such that

$$\begin{aligned} \Vert D_y^{r} \left( f_{t}^{+} - id \right) \left( y\right) \Vert ^{K_2} \leqslant M'' \int _{0}^{t} e^{-s \left( (m'_2+1)b_L - (r+k_2)b_R \right) } {\text {d}}s. \end{aligned}$$

By choosing

$$\begin{aligned} m'_2=(r+k_2).\left[ \frac{b_R}{b_L}+1 \right] + \rho ' >0 \end{aligned}$$

with $\rho '$ an arbitrary positive constant, we deduce that

$$\begin{aligned} \Vert D_y^{r} \left( f_{-t}^{+} - id \right) \left( y\right) \Vert ^{K_2} \le M'' \int _{0}^{t} e^{ -s (\rho '+1)b_R} {\text {d}}s. \end{aligned}$$

As $K_2$ is arbitrary, we can find $\varepsilon _2>0$ and $\rho ' >0$ such that

$$\begin{aligned} \Vert D_y^{r} \left( f_{t}^{+} - id \right) \left( y\right) \Vert ^{K_2} \le \varepsilon _2; \end{aligned}$$

(3.2)

from which $f_{t}^{+}$ and $D^{r}f_{t}^{+}$ are infinite-globally bounded $\forall t\geqslant 0$. $\square$

2) From Eqs.  (3.1) and  (3.2), $\forall t >0; \exists \varepsilon = max \{ \varepsilon _1, \varepsilon _2 \}$:

$$\begin{aligned} \Vert f_{-t}^{-} - id \Vert ^{K_1}_{r} \le \varepsilon ; \quad \text { and } \quad \Vert f_{t}^{+} - id \Vert ^{K_2}_r \le \varepsilon . \end{aligned}$$

c) Estimation of $(exp tY_0^+)_*$ and $(exp tY_0^-)_*$

Let $B_{\varepsilon } = B(0,\varepsilon )$ be the ball centered at the origin with radius a certain $\varepsilon > 0$.

Lemma 3.7

For all $Y^i \in U_i$ and $\forall t \ge 0$, there exists $C'$, $C''$ positive constants such that

$$\begin{aligned} ||(\exp {t}Y_0^-)_{*}\cdot Y^1||_{r}^{\Omega _1} \le C'\cdot e^{-t \rho '_1.a_{R}}. \Vert Y^1 \Vert _{r+m_1}^{\Omega _1^{\varepsilon } } \end{aligned}$$

and, respectively,

$$\begin{aligned} ||(\exp {-t} Y_0^+)_{*}\cdot Y^2||_{r}^{\Omega _2} \le C''\cdot e^{-t \rho '_2. b_L}.\Vert Y^2 \Vert _{r+m_2}^{ \Omega _2^{\varepsilon } }. \end{aligned}$$

Proof

Let $Y^1 \in U_1$ and $\forall t \ge 0$, we have

$$\begin{aligned} (\exp {t}Y_0^-)_{*}\cdot Y^1= & {} (\psi _{t}^-)_* Y^1 = (\phi _t^- \circ f_t^-)_* \cdot Y^1 \\= & {} (D(\phi _t^- \circ f_t^-) \cdot Y^1 ) \circ (\phi _t^- \circ f_t^-)^{-1} \\= & {} D \phi _t^- (f_t^- \circ (f_t^-)^{-1} \circ \phi _{-t}^- ) \cdot Y^1 ((f_t^-)^{-1} \circ \phi _{-t}^-).D f_t^- ((f_t^-)^{-1} \circ \phi _{-t}^-) \\= & {} D \phi _t^- (\phi _{-t}^-)\cdot Y^1 (\psi _{-t}^-). Df_t^-(\psi _{-t}^-) \\= & {} D \phi _t^- (\phi _{-t}^-)\cdot Y^1 (\phi _{-t}^- \circ f_{-t}^-).Df_{t}^- (\phi _{-t}^- \circ f_{-t}^-). \end{aligned}$$

Let $K_1$ be a compact of $\Omega _1$, we then deduce the following estimate:

$$\begin{aligned}\parallel (\exp {t}Y_0^-)_*\cdot Y^1 \parallel ^{K_1} \le \parallel D \phi _t^- (\phi _{-t}^-)\cdot Y^1 (\phi _{-t}^- \circ f_{-t}^-) \parallel ^{K_1} \cdot \parallel Df_{t}^- (\phi _{-t}^- \circ f_{-t}^-)\parallel ^{K_1}. \end{aligned}$$

According to Lemma 3.6, $Df_{t}^-$ and $f_{-t}^-$ are uniformly bounded, then there is a constant $C > 0$ such that

$$\begin{aligned} \begin{aligned}\parallel (\exp {t}Y_0^-)_*\cdot Y^1 \parallel ^{K_1}_r \le C \parallel (\phi _{t}^-)_*\cdot Y^1 \parallel ^{f^-_{-t}(K_1)}_r \quad \forall r \ge 0 \quad (*). \end{aligned} \end{aligned}$$

As

$$\begin{aligned} \left| \left\| f^-_{-t} \right\| ^{K_1} - \left\| id \right\| ^{K_1} \right| \leqslant \left\| f^-_{-t}-id \right\| ^{K_1} \leqslant \varepsilon , \end{aligned}$$

then

$$\begin{aligned} \left\| id \right\| ^{K_1}-\varepsilon \leqslant \left\| f^-_{-t}\right\| ^{K_1} \leqslant \left\| id \right\| ^{K_1}+\varepsilon ; \end{aligned}$$

hence,

$$\begin{aligned} K_1-B_{\varepsilon } \subseteq f^-_{-t}(K_1) \subseteq K_1 + B_{\varepsilon } \qquad (**). \end{aligned}$$

We will then have, from equations (*) and (**).

$$\begin{aligned} \parallel (\exp {t}Y_0^-)_*\cdot Y^1 \parallel ^{K_1}_r \le C' e^ {-a_R t \rho '_1} \parallel Y^1 \parallel _{r+m_1}^{\Phi _{-t}^-(f^-_{-t}(K_1))} \le C' e^ {-a_R t \rho '_1} \parallel Y^1 \parallel _{r+m_1}^{\Phi _{-t}^-(K_1+B_{\varepsilon })}, \end{aligned}$$

from which,

$$\begin{aligned} \Phi ^-_{-t} \left( f^-_{-t} \left( K_1 \right) \right) \subseteq \Omega _1^{\varepsilon }. \end{aligned}$$

We will then have

$$\begin{aligned} \parallel (\exp {t}Y_0^-)_*\cdot Y^1 \parallel ^{K_1}_r \le C' e^ {-a_R t \rho '_1} \parallel Y^1 \parallel ^{\Omega _1^{\varepsilon }}_{r+m_1}. \end{aligned}$$

As $K_1$ is arbitrary on $\Omega _1$, so

$$\begin{aligned} \parallel (\exp {t}Y_0^-)_*\cdot Y^1 \parallel ^{\Omega _1}_r \le C' e^ {-a_R t \rho '_1} \parallel Y^1 \parallel ^{\Omega _1^{\varepsilon }}_{r+m_1}. \end{aligned}$$

We proceed in the same way to demonstrate

$$\begin{aligned} ||(\exp {-t} Y_0^+)_{*}\cdot Y^2||_{r}^{\Omega _2} \le C''\cdot e^{-t \rho '_2. b_L}.\left\| Y^2 \right\| _{r+m_2}^{ \Omega _2^{\varepsilon } } \end{aligned}$$

$\square$

4 Admissible algebra with hyperbolic structure

4.1 $(expt X_0)_*$ decays tamely

Lemma 4.1

(i) $(\phi _{t})_*$ is invariant on $U; \forall t \in {\mathbb {R}}$.

(ii) $(\phi _t)_*$ decays tamely on $U_1$ and $(\phi _{-t})_*$ decays tamely on $U_2$ for all $t \ge 0$. That is to say, $\forall t \ge 0$, we have

$$\begin{aligned} ||( \exp tX_0)_* \cdot Y||_{r}^{\Omega _1} \le e^{-\omega t}. ||Y||_{r+m_1}^{\Omega _1} \quad \forall \omega \in {\mathbb {R}}^{*+}, \forall Y \in U \end{aligned}$$

and

$$\begin{aligned} || (\exp -tX_0)_* \cdot Y||_{r}^{\Omega _2} \le e^{-\omega ' t}. ||Y||_{r+m_2}^{\Omega _2} \quad \forall \omega ' \in {\mathbb {R}}^{*+}, \forall Y \in U. \end{aligned}$$

Proof

Let $Y \in U=U_1 \oplus U_2$; then $\exists ! Y^i \in U_i(i=1,2)$ such that $Y=Y^1+Y^2$;

and let $t \in {\mathbb {R}}, \forall (x,y) \in {\mathbb {R}}^n=\Omega _1 \cup \Omega _2$, we have

$$\begin{aligned} (\phi _{t} )_* Y(x,y) = \left( (\phi _{t}^- )_* Y^1(x) ,(\phi _{t}^+)_* Y^2(y) \right) . \end{aligned}$$

(i) We will show that $(\phi _{t})_* Y \in U, \forall t \in {\mathbb {R}}$.

- We project $(\phi _{t} )_* Y$ on $U_1,$ So

$$\begin{aligned} (\phi _{t} )_* Y(x,y) = \left( (\phi _{t}^- )_* Y^1(x) , 0 \right) , \end{aligned}$$

from which $(\phi _{t} )_* Y \in U$.

- Similarly, we project $(\phi _{t} )_* Y$ on $U_2,$ So

$$\begin{aligned} (\phi _{t} )_* Y(x,y) = \left( 0, (\phi _{t}^+ )_* Y^2(x) \right) , \end{aligned}$$

from which $(\phi _{t} )_* Y \in U$.

We deduce that $(\phi _t)_*$ is invariant on $U, \forall t \in {\mathbb {R}}$.

(ii) Let us first show that $(\phi _t)_*$ decays tamely on $U_1; \forall t >0$:

$$\begin{aligned} || (\phi _{t})_{*} \cdot Y||_{r}^{\Omega _1}= & {} || (\phi _{t}^-)_{*} \cdot Y^1||_{r}^{\Omega _1} \quad \text {and for all } \rho '_1 > 0, \text{ we } \text{ have } \\ || (\phi _{t})_{*} \cdot Y||_{r}^{\Omega _1}\le & {} C_1\cdot e^{-t \rho '_1 a_R}. ||Y^1||_{r+m_1}^{\Omega _1} \quad \text {(see Lemma 3.5)} \\\le & {} C_1\cdot e^{-\omega t} ||Y||_{r+m_1}^{\Omega _1} \text { with } \omega =\rho '_1.a_R. \end{aligned}$$

Then we show that $(\phi _{-t} )_*$ decays tamely on $U_2; \forall t >0$:

$$\begin{aligned} || (\phi _{-t})_{*} \cdot Y||_{r}^{\Omega _2}= & {} || (\phi _{-t}^+)_{*} \cdot Y^2||_{r}^{\Omega _2} \quad \text {and for all } \rho '_2 > 0, \text{ we } \text{ have }\\ || (\phi _{-t})_{*} \cdot Y||_{r}^{\Omega _2}\le & {} C_2\cdot e^{-t \rho '_2 b_L} \cdot ||Y^2||_{r+m_2}^{\Omega _2} \quad \text {(see Lemma 3.5)} \\\le & {} C_2\cdot e^{-\omega ' t} ||Y||_{r+m_2}^{\Omega _2} \qquad \text {with } \omega ' =\rho '_2.b_L. \end{aligned}$$

$\square$

4.2 $(expt Y_0)_*$ decays tamely

Lemma 4.2

(i) $(\psi _{t})_{*}$ is invariant on $U, \forall t \in {\mathbb {R}}$.

(ii) The diffeomorphism $(\psi _{t})_{*}$ decays tamely on $U_1^{\varepsilon }$ and $(\psi _{-t})_{*}$) decays tamely on $U_2^{\varepsilon }$, that is to say, $\forall t \ge 0$, there exist $\omega \in {\mathbb {R}}^{*+}$ and $\omega ' \in {\mathbb {R}}^{*+}$ such that

$$\begin{aligned} || (\psi _{t})_{*} \cdot Y||_{r}^{\Omega _1} \le e^{-\omega t}. ||Y||_{r+m_1}^{\Omega _1^{\varepsilon }}, \end{aligned}$$

respectively,

$$\begin{aligned} || (\psi _{-t})_{*} \cdot Y||_{r}^{\Omega _2} \le e^{-\omega ' t}. ||Y||_{r+m_2}^{\Omega _2^{\varepsilon }}. \end{aligned}$$

Proof

Let $Y \in U=U_1 \oplus U_2$ Then $\exists ! Y^i \in U_i(i=1,2)$ such that $Y=Y^1+Y^2$.

Let $t \in {\mathbb {R}}, \forall (x,y) \in {\mathbb {R}}^n = \Omega _1 \cup \Omega _2$

$$\begin{aligned} (\psi _{t} )_* Y(x,y) = \left( (\psi _{t}^- )_* Y^1(x) ,(\psi _{t}^+)_* Y^2(y) \right) . \end{aligned}$$

(i) We will prove that $(\psi _{t})_* Y \in U, \forall t \ge 0$.

- We project on $U_1 \Rightarrow Y(x,y)= (Y^1 (x),0)$.

Therefore,

$$\begin{aligned} (\psi _{t} )_* Y(x,y) = \left( (\psi _{t}^- )_* Y^1(x) , 0 \right) , \end{aligned}$$

from which $(\psi _{t} )_* Y \in U$

- Similarly, if we project on $U_2 \Rightarrow Y(x,y)= (0,Y^2 (x))$,

then

$$\begin{aligned} (\psi _{t} )_* Y(x,y) = \left( 0, (\psi _{t}^+ )_* Y^2(x) \right) , \end{aligned}$$

from which $(\psi _{t} )_* Y \in U$. We deduce that $(\psi _t)_*$ is invariant on $U, \forall t \in {\mathbb {R}}$.

(ii) We note that

$$\begin{aligned} || (\psi _{t})_{*} \cdot Y||_{r}^{\Omega _1}= & {} || (\psi _{t}^-)_{*} \cdot Y^1||_{r}^{\Omega _1} \\\le & {} C'\cdot e^{-t \rho '_1.a_R}.||Y^1||_{r+m_1}^{\Omega _1^ {\varepsilon } } \qquad \text {(cf Lemma 3.7)}\\\le & {} C'_1\cdot e^{- \omega t}||Y||_{r+m_1}^{ \Omega _1^{\varepsilon } } \qquad \text {with } \omega =\rho '_1.a_R. \end{aligned}$$

This means that $(\psi _{t})_{*}$ decays tamely on $U_1^{\varepsilon }$.

- Similarly,

$$\begin{aligned} || (\psi _{-t})_{*} \cdot Y||_{r}^{ \Omega _2}= & {} || (\psi _{-t}^+)_{*} \cdot Y^2||_{r}^{\Omega _2} \\\le & {} C''\cdot e^{-t \rho '_2. b_L}.||Y^2||_{r+m_2}^{\Omega _2^{\varepsilon }} \qquad \text {(cf Lemma 3.7)}\\\le & {} C''_2\cdot e^{- \omega ' t}.||Y||_{r+m_2}^{ \Omega _2^{\varepsilon } } \qquad \text {with } \omega '=\rho '_2.b_L. \end{aligned}$$

We deduce that $(\psi _{-t})_{*}$ decays tamely on $U_2^{\varepsilon }$. $\square$

Theorem 4.3

The admissible algebra U admits a hyperbolic structure for the flow $\psi _{t*}$.

Proof

According to Lemma 4.2, we deduce that U has a hyperbolic structure for $\psi _t$. $\square$

Part II: applications

We will show, as an application, that the ideal of finite codimension extends over all the hyperbolic-type admissible algebra, and this using the following lemma:

Fundamental lemma: ([8]) Let V be a finite codimension subspace of a ${\mathbb {R}}-$ vector space E and an endomorphism $\psi$ on E such that

• 1. $\psi (V) \subset V$;

• 2. $\psi +b.id$ is surjective on V for all $b \in {\mathbb {R}}$;

• 3. for all numbers $b,c \in {\mathbb {R}}$ such as $b^2-4c<0$, the operator $\psi ^2 + b \psi + c.id$ is surjective on V.

Then $V= E$.

5 Surjectivity of the operator $ad_{Y_0}+bI$

In this section, we study the surjectivity of some linear operators. Note by

$\varphi = (ad_{-Y_0^-},ad_{Y_0^+})$, where $ad_{-Y_0^-}=\varphi _1$ and $ad_{Y_0^+}=\varphi _2$ two adjoint endomorphisms and id the identity application.

Lemma 5.1

For all $b \in {\mathbb {R}}$, the operator $\varphi _1+b.id_{{\mathbb {R}}^k}$ ( respectively, $\varphi _2+b.id_{{\mathbb {R}}^l}$) is surjective on $U_1^{\varepsilon }$ (respectively, on $U_2^{\varepsilon }$ ).

Proof

Let $Y^i \in U_i (i=1,2)$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} Y^1=\overset{}{\underset{i=1}{\overset{k}{\sum }}f_{i}\left( x \right) \frac{ \partial }{\partial x_{i}}} \in U_1\\ Y^2= \overset{}{\underset{j=1}{\overset{l}{\sum }}g_{j}\left( y\right) \frac{ \partial }{\partial y_{j}}} \in U_2. \end{array}\right. } \end{aligned}$$

By putting

$$\begin{aligned} {\left\{ \begin{array}{ll} W_1=\int _{0}^{+\infty }R(t) (\psi _t ^{-})_* Y^1 {\text {d}}t \in U_1^{\varepsilon }\\ W_2=\int _{0}^{+\infty }R(t) (\psi _{-t}^{+})_* Y^2 {\text {d}}t \in U_2^{\varepsilon } \end{array}\right. } \end{aligned}$$

- First stage: Let us prove that $W_i$ is a solution of the equation:

$$\begin{aligned} (\varphi _i + b.id) (W_i)&=Y^i, (i=1,2), \forall b \in {\mathbb {R}} \\ ad_{-Y_0^-}(W_1)&=\lim _{s \rightarrow 0 } \frac{{\text {d}}}{{\text {d}}s} (\psi _{s}^{-})_* \left( \int _{0}^{+\infty }R(t) (\psi _{t} ^{-})_* Y^1 {\text {d}}t \right) \\&= \lim _{s \rightarrow 0 } \int _{0}^{+\infty }R(t) \frac{{\text {d}}}{{\text {d}}s} \left( \psi _{(t+s)} ^{-} \right) _* Y^1 {\text {d}}t\\&=\int _{0}^{+\infty }R(\tau ) \frac{{\text {d}}}{{\text {d}}\tau } (\psi _{\tau }^{-})_* Y^1 d\tau \text {, } \qquad \text{ where } \tau = t+s \\&=R(\tau ) (\psi _{\tau } ^{-})_*\cdot Y^1 |_0^{+\infty }-\int _{0}^{+\infty } (\psi _{\tau }^{-})_* Y^1 \frac{{\text {d}}}{{\text {d}}\tau } R(\tau ) d\tau \\ ad_{-Y_0^-}(W_1)+bW_1&=R(\tau ) (\psi _{\tau } ^{-})_* Y^1|_0^{+\infty }+ \int _{0}^{+\infty } (bR(\tau )-R'(\tau )) (\psi _{\tau }^{-})_* Y^1 d\tau . \end{aligned}$$

We put

$$\begin{aligned} {\left\{ \begin{array}{ll} R'(\tau )-b R(\tau ) =0\\ \qquad \qquad \qquad \qquad \qquad then \quad R(\tau ) = -e^{b\tau } \\ R(0)= -1 \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \lim _{\tau \rightarrow +\infty } \parallel R(\tau ) (\psi _{\tau }^-)_*\cdot Y^1\parallel _r ^{\Omega _1}&\le C' \lim _{\tau \rightarrow +\infty } e^{b \tau }. e^{-\tau \rho '_1 a_R}. \left\| Y^1 \right\| _{r+m_1}^{\Omega _1^{\varepsilon } } \\&\le C' \lim _{\tau \rightarrow +\infty } e^{-\tau [-b+\rho '_1 a_R]}. \left\| Y^1 \right\| _{r+m_1}^{\Omega _1^{\varepsilon }}. \end{aligned}$$

As $\rho '_1$ is arbitrary, then

$$\begin{aligned} \forall b \in {\mathbb {R}}, \exists \rho '_{1,b}> 0 / -b+ \rho '_{1,b}. a_R > 0, \end{aligned}$$

then

$$\begin{aligned} \lim _{\tau \rightarrow +\infty } \parallel R(\tau ) (\psi _{\tau }^-)_*. Y^1\parallel _r = 0. \end{aligned}$$

It results

$$\begin{aligned} ad_{-Y_0^-}(W_1)+bW_1&== R(\tau ) \left( \psi _{\tau } ^{-} \right) _* Y^1 |_0^{+\infty } = Y^1. \end{aligned}$$

Similarly,

$$\begin{aligned} ad_{Y_0^+}(W_2)&=\lim _{s \rightarrow 0 } \frac{{\text {d}}}{{\text {d}}s} (\psi _{-s}^{+})_* \left( \int _{0}^{+\infty }R(t) (\psi _{-t} ^{+})_* Y^2 {\text {d}}t \right) \\&= \lim _{s \rightarrow 0 } \int _{0}^{+\infty }R(t) \frac{{\text {d}}}{{\text {d}}s} \left( (\psi _{-s-t} ^{+})_* \right) Y^2 {\text {d}}t\\&=\int _{0}^{+\infty }R(\tau ) \frac{{\text {d}}}{{\text {d}}\tau } (\psi _{-\tau }^{+})_* Y^2 d\tau \text {, } \qquad \text{ where } \tau = s+t \\&=R(\tau ) (\psi _{-\tau } ^{+})_*\cdot Y^2 |_0^{+\infty }-\int _{0}^{+\infty } \left( (\psi _{-\tau }^{+})_*. Y^2 \right) \frac{d}{d\tau } R(\tau ) d\tau . \end{aligned}$$

By adding $bW_2$ in both members, we will have

$$\begin{aligned} ad_{Y_0^+}(W_2)+bW_2&=R(\tau ) (\psi _{-\tau } ^{+})_* Y^2|_0^{+\infty }+ \int _{0}^{+\infty } (-R'(\tau )+bR(\tau )) (\psi _{-\tau }^{*})_* \cdot Y^2 d\tau . \end{aligned}$$

We put

$$\begin{aligned} {\left\{ \begin{array}{ll} -R'(\tau )+b R(\tau ) =0\\ \qquad \qquad \qquad \qquad \qquad then \quad R(\tau )=-e^{b\tau }\\ R(0)= -1. \end{array}\right. } \end{aligned}$$

We have

$$\begin{aligned} \lim _{\tau \rightarrow +\infty } \parallel R(\tau ) (\psi _{-\tau }^+)_*\cdot Y^2\parallel _r ^{\Omega _2}&\le C'' \lim _{\tau \rightarrow +\infty } e^{b \tau }. e^{-\tau \rho '_2 b_L}. \left\| Y^2 \right\| _{r+m_2}^{\Omega _2^{\varepsilon }} \\&\le C'' \lim _{\tau \rightarrow +\infty } e^{-\tau [-b+\rho '_2 b_L]}. \left\| Y^2 \right\| _{r+m_2}^{\Omega _2^{\varepsilon }}. \end{aligned}$$

As $\rho '_2$ is arbitrary, then

$$\begin{aligned} \forall b \in {\mathbb {R}}, \exists \rho '_{2,b}> 0 /- b+ \rho '_{2,b}. b_L > 0, \end{aligned}$$

from which

$$\begin{aligned} \lim _{\tau \rightarrow +\infty } \parallel R(\tau ) (\psi _{-\tau }^+)_*. Y^2\parallel _r = 0. \end{aligned}$$

Therefore,

$$\begin{aligned} ad_{Y_0^+}(W_2)+bW_2&=R(\tau ) \left( (\psi _{-\tau } ^{+})_* \right) Y^2 |_0^{+\infty } = Y^2. \end{aligned}$$

- Second stage: Let us prove that $W_1$ is of class $C^\infty$ on any compact.

Let $K_1$ be a compact of $\Omega _1$, and as

$$\begin{aligned} W_1=\int _{0}^{+\infty }R(t) (\psi _t ^{-})_* Y^1 {\text {d}}t, \end{aligned}$$

according to Lemma 3.7, we will have

$$\begin{aligned} \parallel R (t) (exp t Y_0^-)_*\cdot Y^1 \parallel _r^{K_1} \le C' e^{-t( -b+\rho '_1.a_R)}\parallel Y^1\parallel _{r+m_1}^{\Omega _1^{\varepsilon }}. \end{aligned}$$

As $\rho '_1$ is arbitrary, then

$$\begin{aligned}&\forall b \in {\mathbb {R}}, \exists \rho '_{1,b}> 0 / -b+ \rho '_{1,b}. a_R > 0, \\&\int _{0}^{+\infty } e^{ -t (-b+\rho '_{1,b}.a_R) } {\text {d}}t \qquad \text {converges } \forall b \in {\mathbb {R}}. \end{aligned}$$

It follows that $W_1$ converges uniformly, $\forall x \in \Omega _1$, where ultimately $W_1$ is of class $C^\infty$ on any compact of $\Omega _1$. The same reasoning is valid to prove that $W_2$ is of class $C^\infty$ on any compact of $\Omega _2$. $\square$

Lemma 5.2

For all $b \in {\mathbb {R}}$, $\varphi +b.id$ is surjective on U.

Proof

Let $Y \in U$, seeking a $W \in U$ such that $\forall (x,y) \in {\mathbb {R}}^n; Y(x,y) = (\varphi +b.id) W(x,y)$, $\forall b \in {\mathbb {R}}$.

As $Y \in U= U_1 \oplus U_2$, $\exists ! Y^i \in U_i$ such that

$$\begin{aligned} Y(x,y) = (Y^1(x) , Y^2(y))=(\varphi _1 W_1(x)+b W_1(x), \varphi _2 W_2(y)+b W_2(y)). \end{aligned}$$

We project on $U_i$:

• On $U_1$, we have $Y(x,y) = (Y^1(x),0)=(\varphi _1 W_1(x)+b W_1(x), 0)$, and according to Lemma 5.1$\exists \varepsilon > 0$ and $W_1$, such that $W_1(x)=-\int _{0}^{+\infty } e^{bt} (\psi _t ^{-})_* Y^1 (x){\text {d}}t \in \Omega _1^{\varepsilon }$.

• On $U_2$, we have $Y(x,y)=(0,Y^2(y))=(0,\varphi _2 W_2(y)+b W_2(y))$, and according to Lemma 5.1$\exists \varepsilon > 0$ and $W_2$, such that $W_2(y)=-\int _{0}^{+\infty } e^{bt} (\psi _{-t} ^{+})_* Y^2 (y){\text {d}}t \in \Omega _2^{\varepsilon }$.

It follows that

$$\begin{aligned} W_i \in U_i^{\varepsilon } \end{aligned}$$

.

We put $W(x,y)=(W_1(x),W_2(y)) \in {\mathbb {R}}^k \times {\mathbb {R}}^l={\mathbb {R}}^n$.

As $W=W_1 + W_2 \in U_1^\varepsilon + U_2^\varepsilon = U$, it follows that $\varphi + b.id$ is surjective on U, for all $b \in {\mathbb {R}}$. $\square$

6 Surjectivity of the operator $(ad_{Y_0})^2+b.ad_{Y_0}+cI$

Lemma 6.1

For all numbers $b,c \in {\mathbb {R}}$ such that $b^2-4c<0$, the operator $\varphi _1^2 + b \varphi _1 + c.id_{{\mathbb {R}}^k}$, (respectively, $\varphi _2^2 + b \varphi _2 + c.id_{{\mathbb {R}}^l}$) is surjective on $U_1^{\varepsilon }$ (respectively, on $U_2^{\varepsilon }$).

Proof

Let $Y^i \in U_i$ such that

$$\begin{aligned} Y^1&= \overset{}{\underset{i=1}{\overset{k}{\sum }}f_{i}\left( x\right) \frac{ \partial }{\partial x_{i}}} \in U_1\\ Y^2&= \overset{}{\underset{i=1}{\overset{l}{\sum }}g_{i}\left( y\right) \frac{ \partial }{\partial y_{i}}} \in U_2, \end{aligned}$$

such that $(\varphi _1^2 + b \varphi _1 + c id_{{\mathbb {R}}^k}) W_1=[-Y_0^-,[-Y_0^-,W_1]]+b[-Y_0^-,W_1]+cW_1$ and $(\varphi _2^2 + b \varphi _2 + c id_{{\mathbb {R}}^l}) W_2= [-Y_0^+,[-Y_0^+,W_2]]+b[-Y_0^+,W_2]+cW_2$.

$$\begin{aligned} {\left\{ \begin{array}{ll} W_1=\int _{0}^{\infty } R(t)(\psi _t^{-} )_*Y^1 {\text {d}}t \in U_1^\varepsilon \\ W_2=\int _{0}^{\infty } R(t)(\psi _{-t}^{+})_*Y^2 {\text {d}}t \in U_2^\varepsilon . \end{array}\right. } \end{aligned}$$

With

$$\begin{aligned} R(t)= \dfrac{2exp(\dfrac{b}{2}t)}{\sqrt{4c-b^2}}.\sin (\sqrt{4c-b^2}/2)t. \end{aligned}$$

- First stage: Let us show that $W_i$ is a solution of the equation $(\varphi _i^2 + b \varphi _i + c id) W_i=Y^i$.

If we put $Z_1=[-Y_0^-,W_1] =ad_{-Y_0^-}(W_1)$, then

$$\begin{aligned} Z_1&=\lim _{s \rightarrow 0 } \frac{{\text {d}}}{{\text {d}}s} (\psi _{s}^{-})_* \left( \int _{0}^{+\infty }R(t) (\psi _{t} ^{-})_* Y^1 {\text {d}}t \right) \\&=R(\tau ) (\psi _{\tau } ^{-})_*\cdot Y^1 |_0^{+\infty }-\int _{0}^{+\infty } (\psi _{\tau }^{-})_* Y^1 \frac{d}{d\tau } R(\tau ) d\tau \\ Z_1&=R(\tau ) (\psi _{\tau } ^-)_{*} Y^1|_0^{+\infty }- \int _{0}^{+\infty } R'(\tau ) (\psi _{\tau }^{-})_* Y^1 d\tau . \end{aligned}$$

As $R(t)= \dfrac{2exp(\dfrac{b}{2}t)}{\sqrt{4c-b^2}}.\sin (\sqrt{4c-b^2}/2)t$ and

$$\begin{aligned} \parallel R(\tau ) (\psi _{\tau }^-)_*\cdot Y^1\parallel _r^{\Omega _1}&\le C' e^{\frac{b}{2} \tau }. e^{-\tau \rho '_1 a_R}. \left\| Y^1 \right\| _{r+m_1}^{\Omega _1^{\varepsilon }} \\&\le C' e^{-\tau [-b/2+\rho '_1 a_R]}. \left\| Y^1 \right\| _{r+m_1}^{\Omega _1^{\varepsilon }} \end{aligned}$$

$\rho '_1$ being arbitrary; therefore, $\forall b \in {\mathbb {R}}, \exists \rho '_{1,b}> 0 / -b/2+ \rho '_{1,b}. a_R > 0$, where ultimately

$$\begin{aligned} \lim _{\tau \rightarrow +\infty } \parallel R(\tau ) (\psi _{\tau }^-)_*. Y^1\parallel _r = 0, \end{aligned}$$

then the first member vanishes and $Z_1$ becomes

$$\begin{aligned} Z_1= - \int _{0}^{+\infty } R'(\tau ) (\psi _{\tau }^{-})_* Y^1 d\tau . \end{aligned}$$

We will have

$$\begin{aligned} \varphi _1^2(W_1) =[-Y_0^-,[-Y_0^-,W_1]]&=[-Y_0^-,Z_1]\\&= -\lim _{s \rightarrow 0 } \frac{{\text {d}}}{{\text {d}}s} (\psi _s^{-})_* \int _{0}^{+\infty } R'(\tau ) (\psi _{\tau }^{-})_* Y^1 d\tau \\&= -\lim _{s \rightarrow 0 } \int _{0}^{+\infty } R'(\tau ) \frac{{\text {d}}}{{\text {d}}s} (\psi _{s+\tau }^{-})_* Y^1 d\tau \\&= -\left[ R'(t) (\psi _{t}^{-})_* Y^1 \right] _{0}^{\infty }+\int _{0}^{+\infty } R''(t) (\psi _{t}^{-})_* Y^1 {\text {d}}t \qquad \text {where } t=s+\tau \\&= Y^1+\int _{0}^{+\infty } R''(t) (\psi _{t}^{-})_* Y^1 {\text {d}}t. \end{aligned}$$

Consequently,

$$\begin{aligned} (\varphi _1^2 + b \varphi _1 + c id_{{\mathbb {R}}^k}) W_1&= [-Y_0^-,[-Y_0^-,W_1]] + b [-Y_0^-,W_1] + c W_1\\&= Y^1+\int _{0}^{+\infty } (R''(t)-bR'(t)+cR(t) ) (\psi _{t}^{-})_* Y {\text {d}}t \\&=Y^1, \end{aligned}$$

where $R(t)= \dfrac{2exp(\dfrac{b}{2}t)}{\sqrt{4c-b^2}}.\sin (\sqrt{4c-b^2}/2)t$ is solution of the Cauchy problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} R''(t)-b R'(t) +c R(t) =0\\ R(0)= 0\\ R'(0)=1. \end{array}\right. } \end{aligned}$$

The operator $\varphi _1^2 + b \varphi _1 + c.id_{{\mathbb {R}}^k}$ is, therefore, surjective on $U_1^{\varepsilon }$. The proof of the surjectivity of the operator $\varphi _2^2 + b \varphi _2 + c. id_{{\mathbb {R}}^l}$ on $U_2^{\varepsilon }$ is made in the same way.

- Second stage: According to the previous lemma, $W_i, (i=1,2)$ are of class $C^\infty$ on all compact. $\square$

Lemma 6.2

For all $b,c \in {\mathbb {R}}$, $b^2-4.c <0$ the operator $\varphi ^2+b.\varphi + c.id$ is surjective in U.

Proof

Let $Y \in U$, seeking a $W \in U$ such that

$$\begin{aligned} \forall (x,y) \in {\mathbb {R}}^n, Y(x,y) = (\varphi ^2+b.\varphi + c.id) W(x,y), \text { for all } b,c \in {\mathbb {R}}, b^2-4.c <0. \end{aligned}$$

As $Y \in U= U_1 \oplus U_2$, $\exists ! Y^i \in U_i$ such that

$$\begin{aligned} Y(x,y) = ( Y^1(x) , Y^2(y)) = \Big ( \varphi _1^2 W_1(x)+b.\varphi _1W_1(x) + c.W_1(x), \varphi _2^2 W_2(y)+b.\varphi _2W_2(y) + c.W_2(y) \Big ). \end{aligned}$$

We project on $U_i$:

- On $U_1$, we have $Y(x,y) = (Y^1(x),0)= \Big ( \varphi _1^2 W_1(x)+b.\varphi _1W_1(x) + c.W_1(x), 0 \Big )$, and according to Lemma 6.1, $\exists \varepsilon > 0$ and $W_1$, such that $W_1(x)=\int _{0}^{+\infty }R(t) (\psi _t ^{-})_* Y^1 (x){\text {d}}t \in \Omega _1^{\varepsilon }$ with

$$\begin{aligned} R(t)= \dfrac{2exp(\dfrac{b}{2}t)}{\sqrt{4c-b^2}}.\sin (\sqrt{4c-b^2}/2)t \end{aligned}$$

- On $U_2$, we have $Y(x,y)=(0,Y^2(y)) = \Big ( 0, \varphi _2^2 W_2(y)+b.\varphi _2W_2(y) + c.W_2(y) \Big )$, and according to Lemma 6.1, $\exists \varepsilon > 0$ and $W_2$ such that $W_2(y)=\int _{0}^{+\infty }R(t) (\psi _{-t} ^{+})_* Y^2 (y){\text {d}}t \in \Omega _2^{\varepsilon } \subset \Omega _2$ with

$$\begin{aligned} R(t)= \dfrac{2exp(\dfrac{b}{2}t)}{\sqrt{4c-b^2}}.\sin (\sqrt{4c-b^2}/2)t. \end{aligned}$$

It follows that

$$\begin{aligned} W_i \in U_i^{\varepsilon }. \end{aligned}$$

We put $W(x,y)=(W_1(x),W_2(y)) \in {\mathbb {R}}^k \times {\mathbb {R}}^l={\mathbb {R}}^n$. Since $W=W_1 + W_2 \in U_1^\varepsilon + U_2^\varepsilon = U$. Consequently, $\varphi ^2 + b.\varphi +c.id$ is surjective on U, for all $b,c \in {\mathbb {R}}$, $b^2-4.c <0$. $\square$

7 Ideals of finite codimension in hyperbolic-type algebra

Theorem Let U be the admissible algebra having a hyperbolic structure for the flow:

$(\psi _t)_*=(exptY_0)_*=(expt(X_0 + Z_0))_*$, if $dim (E-U)$ is finite, and $\varphi = (ad_{-Y_0^-} , ad_{Y_0^+})$ an endomorphism on U, such that

(i):

$\varphi (U) \subset U$,

(ii):

$\varphi + b.id_{{\mathbb {R}}^n}$ is surjective on U; $\forall b \in {\mathbb {R}}$,

(iii):

$\varphi ^2 + b \varphi + c .id_{{\mathbb {R}}^n}$ is surjective on U; $\forall b,c \in {\mathbb {R}} / b^2-4c < 0$ Then $U = E$.

Proof

The hypotheses of the fundamental lemma is satisfied thanks to Lemmas 4.2, 5.2 and 6.2, and, if in addition we have $dim (E-U)$ finite, then we deduce

$$\begin{aligned} U=E. \end{aligned}$$

$\square$