Controllability of Linear Systems on Low Dimensional Nilpotent and Solvable Lie Groups

Abstract

This paper is devoted to the study of controllability of linear systems on solvable and nilpotent Lie groups. Some general results are stated and used to completely characterize the controllable systems on the nilpotent Heisenberg group and the solvable two-dimensional affine group.

1 Introduction

In this paper, we are interested in finding some necessary and sufficient controllability conditions for linear systems on nilpotent and solvable Lie groups, in particular on the Heisenberg group and the two-dimensional affine group A f f ₊(2). By linear system is meant a controlled system

$$({\Sigma}) \qquad \dot{g}=\mathcal{X}_{g}+{\sum}_{j=1}^{m} u_{j}{Y^{j}_{g}} $$

on a connected Lie group G where $\mathcal {X}$ is a linear vector field, that is a vector field whose flow is a one-parameter group of automorphisms, and the Y _j ’s are right-invariant.

Up to now, the known controllability properties of these systems were either of local nature (see [3, 5]), or restricted to linear systems on compact or semisimple groups (see [8, 10, 14]), but few results were known in the nilpotent and solvable cases, excepted the following particular one: ([10]) if the derivation associated to the linear field is inner (see Section 2 for the definition), then system (Σ) is controllable if and only if the Lie algebra generated by the controllable vector fields $Y^{1},\dots ,Y^{m}$ is equal to the Lie algebra of G (some controllability results on compact homogeneous spaces involve also solvable groups, see [9]).

This statement is analog to the known one about invariant systems on nilpotent Lie groups (see [2]), but the results we present here show that in the case where the derivation is not inner, the behavior of linear systems is completely different from the one of invariant systems. For instance on the Heisenberg group, a one-input linear system can be controllable (see Theorem 6), which is impossible for an invariant one.

The present investigation makes use of subgroups and quotients. As a matter of fact, the nilpotent and solvable groups have subgroups that are invariant under the flow of linear vector fields, that is the derived groups and the elements of the lower central series. If these subgroups are moreover closed, the linear system can be projected to the quotient group, and our approach is to rely the controllability properties of the system to the ones of the systems induced on the involved subgroup and on the related quotient group.

The paper is organized as follows.

Firstly, the basic definitions and facts concerning linear systems on Lie groups are recalled in Section 2.

Then Section 3 is devoted to general results of two kinds. In Propositions 5 and 6, the relations between the controllability properties of the system and the ones of the systems induced on some subgroup and on the related quotient group are stated. The study of the action of group automorphisms on linear systems is done in Proposition 7.

These general results are used to completely characterize the one-input controllable systems on the two-dimensional affine group in Section 4. The main result of that section is Theorem 3; it is preceded by Theorem 2 which deals with systems transformed by automorphism into what we call a normal form.

In Section 5, the same work is done on the Heisenberg group, first for two-input systems (Theorem 4) and then for one-input systems (Theorems 5 and 6). The characterization of the controllable one-input systems on the group Heisenberg is the main result of this paper. It enlightens the kind of problems that one can encounter in the analysis of linear systems on general nilpotent groups.

The results are then discussed in Section 6 in order to try to identify controllability conditions that could hold on general nilpotent and solvable groups.

2 Basic Definitions and Notations

More details about linear vector fields and linear systems can be found in [7] and [10].

2.1 Linear Vector Fields

Let G be a connected Lie group and $\mathfrak {g}$ its Lie algebra (the set of right-invariant vector fields, identified with the tangent space at the identity). A vector field on G is said to be linear if its flow is a one-parameter group of automorphisms. Actually, the linear vector fields are nothing else than the so-called infinitesimal automorphisms in the Lie group literature (see [4] for instance). The following characterization will be useful in the sequel.

A vector field $\mathcal {X}$ on a connected Lie group G is linear if and only if it belongs to the normalizer of $\mathfrak {g}$ in the algebra of analytic vector fields of G, that is

$$\forall Y\in \mathfrak{g} \qquad [\mathcal{X},Y]\in\mathfrak{g}, $$

and verifies $\mathcal {X}(e)=0$.

On account of this characterization, one can associate to a linear vector field $\mathcal {X}$ the derivation $D=-ad(\mathcal {X})$ of the Lie algebra $\mathfrak {g}$ of G.

In the case where this derivation is inner, that is D=−a d(X) for some right-invariant vector field X on G, the linear vector field splits into $\mathcal {X}=X+\mathcal {I}_{*}X$, where $\mathcal {I}$ stands for the diffeomorphism $g\in G\longmapsto \mathcal {I}(g)=g^{-1}$. Thus, $\mathcal {X}$ is the sum of the right-invariant vector field X and the left-invariant one $\mathcal {I}_{*}X$.

About the existence of linear vector fields, we have:

Theorem 1 ([7])

The group G is assumed to be (connected and) simply connected. Let D be a derivation of its Lie algebra g. Then there exists one and only one linear vector field on G whose associated derivation is D.

Throughout the paper, the flow of a linear vector field $\mathcal {X}$ will be denoted by $(\varphi _{t})_{t\in \mathbb {R}}$.

2.2 Linear Systems

Definition 1

A linear system on a connected Lie group G is a controlled system

$$({\Sigma}) \qquad \dot{g}=\mathcal{X}_{g}+{\sum}_{j=1}^{m} u_{j}{Y^{j}_{g}}$$

where $\mathcal {X}$ is a linear vector field and the Y ^j’s are right-invariant ones. The control $u=(u_{1},\dots ,u_{m})$ takes its values in $\mathbb {R}^{m}$.

An input u being given (measurable and locally bounded); the corresponding trajectory of (Σ) starting from the identity e will be denoted by e _u (t), and the one starting from the point g by g _u (t). A straightforward computation shows that

$$g_{u}(t)=e_{u}(t)\varphi_{t}(g). $$

Remark

If the vector fields Y ^j were left-invariant, we would have g _u (t)=φ _t (g)e _u (t).

Notations. We denote by $\mathcal {A}(g,t)=\{g_{u}(t);\ u\in L^{\infty }[0,t]\}$ (resp. $\mathcal {A}(g,\leq t)$) (resp. $\mathcal {A}(g)$) the reachable set from g in time t (resp. in time less than or equal to t) (resp. in any time). In particular, the reachable sets from the identity e are denoted by

$$\mathcal{A}_{t}=\mathcal{A}(e,t) \qquad\text{ and }\qquad \mathcal{A}=\mathcal{A}(e). $$

These sets are related by the following equalities:

Proposition 1

• 1. $\forall g\in G \quad \mathcal {A}(g,t)=\mathcal {A}_{t}\varphi _{t}(g)$.

• 2. $\forall \ s,t\geq 0 \quad \mathcal {A}_{t+s}=\mathcal {A}_{t}\varphi _{t}(\mathcal {A}_{s})=\mathcal {A}_{s}\varphi _{s}(\mathcal {A}_{t})$.

We also denote by $\mathcal {A}^{-}=\{ g \in G; \ e \in \mathcal {A}(g)\}$ the set of points from which the identity can be reached. It is equal to the attainability set from the identity for the time-reversed system.

2.3 The System Lie Algebra and the Rank Condition

Let 𝔥 be the subalgebra of 𝔤 generated by $\{Y^{1},\dots ,Y^{m}\}$, let us denote by $D\mathfrak {h}$ the smallest D-invariant subspace of $\mathfrak {g}$ that contains $\mathfrak {h}$, i.e., $D\mathfrak {h}=\text {Sp}\{D^{k}Y;\ Y\in \mathfrak {h} \text { and } k\in \mathbb {N}\}$, and let $\mathcal {LA}(D\mathfrak {h})$ be the $\mathfrak {g}$ subalgebra generated by $D\mathfrak {h}$ (as previously $D=-\text {ad}(\mathcal {X})$).

Proposition 2 ([10])

The subalgebra $\mathcal {LA}(D\mathfrak {h})$ is D-invariant. It is therefore equal to the zero-time ideal $\mathcal {L}_{0}$ , and the system Lie algebra $\mathcal {L}$ is equal to

$$\mathbb{R}\mathcal{X}\oplus \mathcal{LA}(D\mathfrak{h})=\mathbb{R}\mathcal{X}\oplus \mathcal{L}_{0}. $$

The rank condition is satisfied by (Σ) if and only if $\mathcal {L}_{0}=\mathfrak {g}$.

2.4 The Lie Saturate

The Lie saturate $\mathcal {LS}({\Sigma })$ of (Σ) (resp. the strong Lie saturate $\mathcal {LSS}({\Sigma })$ of (Σ)) is the set of vector fields f belonging to the system Lie algebra $\mathcal {L}$ and whose flow $(\phi _{t})_{t\in \mathbb {R}}$ satisfies

$$\forall g\in G,\ \forall t\geq 0\qquad \phi_{t}(g)\in\overline{\mathcal{A}(g)} \qquad (\text{resp. } \phi_{t}(g)\in\overline{\mathcal{A}(g,\leq t)}) $$

as soon as ϕ _t (g) is defined (see [12]). Here, $\overline {\mathcal {A}(g)}$ stands for the closure of $\mathcal {A}(g)$.

As a first consequence of the enlargement technics that consists to add to a system some vector fields of the (strong) Lie saturate, we have the following proposition:

Proposition 3

The algebra $\mathfrak {h}$ is included in $\mathcal {LSS}({\Sigma })$ , so that (Σ) can be enlarged to the system

$$(\widetilde{\Sigma}) \qquad \dot{g}=\mathcal{X}_{g}+{\sum}_{j=1}^{p} u_{j}\widetilde{Y}^{j}_{g}, $$

where $\widetilde {Y}^{1},\dots ,\widetilde {Y}^{p}$ is a basis of $\mathfrak {h}$ , without modifying the sets $\overline {\mathcal {A}(g,\leq t)}.$

2.5 Local Controllability and the Ad-Rank Condition

It is well known that a system is locally controllable at an equilibrium point as soon as the linearized system is controllable (see [13] for instance). In this assertion, “locally controllable” at a point g means that the set $\mathcal {A}(g,t)$ is a neighborhood of g for all t>0.

According to Proposition 3, we can consider the linearization of the extended system instead of the one of the system itself. This leads to the following definition:

Definition 2

System (Σ) is said to satisfy the ad-rank condition if $D\mathfrak {h}=\mathfrak {g}$, in other words, if the linearized system of $(\widetilde {\Sigma })$ is controllable.

As the rank condition is implied by the ad-rank one, we obtain at once:

Proposition 4

If the ad-rank condition is satisfied then for all t>0, the reachable set $\mathcal {A}_{t}$ is a neighborhood of e.

This proposition was first stated in [3] with a different proof.

3 General Results

In this section, we state some general results that will be applied to the Heisenberg and affine groups. Proposition 7 deals with the action of group automorphisms on linear systems, but firstly, we will rely the controllability of a linear system (Σ) on G to the controllability on some quotient group G/H of the system induced by (Σ). Here, the subgroup H of G is assumed to be normal and closed, but also globally invariant under the flow of $\mathcal {X}$, for the induced linear field to exist (see [7]).

In the present paper, we are interested in nilpotent and solvable Lie groups ,and it is a well-known fact that the derived subalgebras, as well as the subalgebras of the lower central series, are characteristic ideals of $\mathfrak {g}$, hence invariant under all derivations. The corresponding connected subgroups are therefore normal and invariant under the flow of $\mathcal {X}$ (see [8]). Moreover, these subgroups are closed when G is simply connected (see [5]). Notice also that in the case when G is simply connected, the group $G/\mathcal {D}^{1}G$ is abelian and simply connected, hence diffeomorphic to $\mathbb {R}^{n}$ for some n, and that the induced system on $G/\mathcal {D}^{1}G$ is linear in the classical sense. For these reasons, the two forthcoming propositions are suitable to our study.

We will denote by $\mathcal {A}_{G/H}(gH)$ the reachable set from g H for the induced system. The condition $g'H \in \mathcal {A}_{G/H}(gH)$ means that there exists a time t≥0 and a control $u\in L^{\infty } [0,t]$ such that g ^′ H=g _u (t)H=e _u (t)φ _t (g)H, and is equivalent to the existence of h∈H such that $g'\in \mathcal {A}(gh)$. Indeed, the above condition is equivalent to the existence of h ^′∈H such that g ^′=e _u (t)φ _t (g)h ^′=e _u (t)φ _t (g φ _−t (h ^′))=e _u (t)φ _t (g h), where h=φ _−t (h ^′)∈H, thanks to the invariance of H under the flow of $\mathcal {X}$.

Proposition 5

Let H be a normal and closed subgroup of G, globally invariant under the flow of $\mathcal {X}$.

The linear system (Σ) on G, assumed to satisfy the rank condition, is controllable if and only if both conditions hold:

• 1. The system induced on G/H is controllable

• 2. The subgroup H is included in the closures $\overline {\mathcal {A}}$ and $\overline {\mathcal {A}^{-}}$ of $\mathcal {A}$ and $\mathcal {A}^{-}$.

Proof

These conditions are clearly necessary. To prove that they are sufficient, we have to show that they imply $\mathcal {A} = \mathcal {A}^{-} = G$.

Let g∈G. By the first condition $gH \in \mathcal {A}_{G/H}(H)$, and consequently there exists h belonging to H such that g belongs to $\mathcal {A}(h)$. By the second one h belongs to $\overline {\mathcal {A}}$; together with $g\in \mathcal {A}(h)$, this implies $g \in \overline {\mathcal {A}}$. Thus, we obtain $\overline {\mathcal {A}}=G$, and consequently $\mathcal {A}=G$ because the rank condition holds (see [11]).

In order to show that $\mathcal {A}^{-}=G$, we consider the time-reversed system (Σ⁻). Obviously, the first condition implies that the system induced by (Σ⁻) on G/H is controllable. The second condition implies as above that $\mathcal {A}^{-}=G$.

In conclusion, (Σ) is controllable. □

Proposition 6

Let us assume that H is a connected and closed subgroup of G and that the restriction of $\mathcal {X}$ to H vanishes. Then H is included in $\mathcal {A}$ (resp. in $\mathcal {A}^{-}$ ) if and only if $\mathcal {A} \cap H$ (resp. $\mathcal {A}^{-}\cap H$ ) is a neighborhood of e in H.

Proof

Let us first show that $\mathcal {A}\cap H$ is a semigroup. If h, h ^′ are two elements of $\mathcal {A}\cap H$, there exist nonnegative real numbers t and s such that $h\in \mathcal {A}_{t}$ and $h'\in \mathcal {A}_{s}$. As the restriction to H of the flow φ _t of $\mathcal {X}$ is the identity, we get $hh'=h\varphi _{t}(h') \in \mathcal {A}_{t}\varphi _{t}(\mathcal {A}_{s})= \mathcal {A}_{t+s}$. Consequently, $\mathcal {A}\cap H$ is a semigroup. If it contains a neighborhood of e in H which is connected, then $H \subset \mathcal {A}\cap H$, and $H\subset \mathcal {A}$.

Similarly, let us show that $\mathcal {A}^{-}\cap H$ is a semigroup. Notice first that for any h∈H and t≥0, we have $\mathcal {A}_{t}(h) = \mathcal {A}_{t}\varphi _{t}(h)= \mathcal {A}_{t}h$. Let h,h ^′ belonging to $\mathcal {A}^{-}\cap H$. From $e \in \mathcal {A}_{t}h$, we deduce $h' \in \mathcal {A}_{t}hh'= \mathcal {A}_{t}(hh')$. Together with $e \in \mathcal {A}(h')$, this shows $e \in \mathcal {A}(hh')$ and $hh' \in \mathcal {A}^{-}\cap H$. For the same reasons as above $\mathcal {A}^{-}\subset H$. □

Singular and regular systems.

The linear systems for which $\mathcal {X}$ vanishes on a connected, closed, normal, and $\mathcal {X}$-invariant (nontrivial) subgroup will be referred to as singular systems, the other ones being regular. The previous Proposition 6 is crucial in the singular case.

Proposition 7

Let (Σ): $\dot {g}=\mathcal {X}_{g}+{\sum }_{j=1}^{m} u_{j}{Y^{j}_{g}}$ be a linear system on a connected and simply connected Lie group G, D the derivation associated to $\mathcal {X}$ , and P an automorphism of $\mathfrak {g}$ . Then (Σ) is equivalent by group automorphism to

$$(\widetilde{\Sigma})\qquad \dot{g}=\widetilde{\mathcal{X}_{g}}+{\sum}_{j=1}^{m} u_{j}P{Y^{j}_{g}} $$

where $\widetilde {\mathcal {X}}$ is the linear vector field whose associated derivation is $\widetilde {D} = PDP^{-1}$.

Proof

• 1. Let us first show that P D P ⁻¹ is a derivation of $\mathfrak {g}$. Indeed for all $X,Y\in \mathfrak {g}$:

  $$\begin{array}{ll} PDP^{-1}[X,Y] & =PD[P^{-1}X,P^{-1}Y]\\ & =P([DP^{-1}X,P^{-1}Y]+[P^{-1}X,DP^{-1}Y])\\ & =[PDP^{-1}X,Y]+[X,PDP^{-1}]. \end{array} $$

• 2. Then the group G being simply connected there exists an automorphism ψ of G, unique, such that T _e ψ=P (see [4]). Let us show that for any linear vector field $\mathcal {X}$ on G, the vector field $\psi _{*}\mathcal {X}$ is linear and the associated derivation is P D P ⁻¹. If the flow of $\mathcal {X}$ is denoted by φ _t , the one of the vector field $\psi _{*}\mathcal {X}$ is ψ∘φ _t ∘ψ ⁻¹. It is a one-parameter group of automorphisms, hence the flow of a linear vector field.

  It remains to show that P D P ⁻¹ is the derivation $\widetilde {D}$ associated to $\psi _{*} \mathcal {X}$. According to the formula T _e φ _t =e ^tD (see [7]) applied to ψ∘φ _t ∘ψ ⁻¹, we get

  $$e^{t \widetilde{D}}= T_{e}(\psi \circ \varphi_{t}\circ\psi^{-1}) = PT_{e}\varphi_{t}P^{-1}=Pe^{tD}P^{-1}=e^{tPDP^{-1}}, $$

  so that $\widetilde {D}=PDP^{-1}$.

• 3. Finally, the vector field ψ _∗ Y ^j is equal to P Y ^j. Indeed, the mapping ψ being a group morphism, it is a basic fact that ψ _∗ Y ^j is the right-invariant field T _e ψ Y ^j.

The proof is complete. □

4 Controllability of Linear Systems on A f f ₊(2)

4.1 Linear Vector Fields and Linear Systems on A f f ₊(2)

Let G be the connected component of e in the two-dimensional affine group:

$$G=Aff_{+}(2)=\left\{\left(\begin{array}{ll}x&y\\0&1 \end{array}\right);\ \ (x,y)\in\mathbb{R}_{+}^{*}\times \mathbb{R}\right\}. $$

Its Lie algebra $\mathfrak {g}=\mathfrak {aff}(2)$ is identified with the set of left-invariant vector fields. Previously, we had considered right-invariant vector fields only, but the left-invariant ones are more convenient on A f f ₊(2) and, of course, the theory is the same. This two-dimensional algebra is solvable¹ and generated by the left-invariant vector fields whose value at the group identity is given by

$$X=\left(\begin{array}{ll}1&0\\0&0 \end{array}\right) \quad \text{and}\quad Y=\left(\begin{array}{ll}0&1\\0&0 \end{array}\right) $$

with [X,Y]=X Y−Y X=Y. The matrices X and Y are identified with the left-invariant vector fields:

$$gX=\left(\begin{array}{ll}x&0\\0&0 \end{array}\right) \quad \text{ and }\quad gY=\left(\begin{array}{ll}0&x\\0&0 \end{array}\right)\quad \text{ where} \quad g=\begin{pmatrix}x&y\\0&1\end{pmatrix}. $$

The three followings facts are straightforward:

• 1. In the basis (X,Y) all derivations D on $\mathfrak {aff}(2)$ are of the form $D=\left (\begin {array}{ll}0&0\\a&b \end {array}\right )$ where a and b are real numbers.

  Indeed, such an endomorphism of $\mathfrak {aff}(2)$ is clearly a derivation. Conversely, the derived ideal $\mathcal {D}^{1}\mathfrak {g}$ is characteristic, hence invariant with respect to D, so that D Y=b Y for some real number b. Derivating Y=[X,Y], we get b Y=D Y=[D X,Y]+[X,D Y]=[D X,Y]+b Y, hence [D X,Y]=0 and D X=a Y for some real number a.

• 2. All derivations on A f f ₊(2) are inner. Actually, D=a d(b X−a Y)=−a d(a Y−b X).

• 3. The linear vector field $\mathcal {X}$ associated to such a derivation is $\mathcal {X}(g)=\left (\begin {array}{lc}0 &a(x-1)+by\\0 &0 \end {array}\right )$.

  Indeed, $D=-ad(\mathcal {X})= -ad(aY-bX)$ and following Section 2.1:

  $$\mathcal{X}(g)= g(aY-bX)-(aY-bX)g= \left(\begin{array}{lc}0 &a(x-1)+by\\0 &0 \end{array}\right) $$

  at the point $g=\left (\begin {array}{ll}x&y\\0&1 \end {array}\right )$.

Let B=α X+β Y be a element of $\mathfrak {aff}(2)$, and (Σ) the one-input linear system defined by:

$$({\Sigma}):\dot{g}=\mathcal{X}(g)+ugB. $$

In coordinates, (Σ) writes:

$$({\Sigma})\qquad\left\{\begin{array}{ll} \dot{x} =& u\alpha x\\ \dot{y} =& a(x-1)+by+u\beta x \end{array} \right. $$

and the induced system on $G/\mathcal {D}^{1}G$ is equivalent to

$$\dot{x} = u \alpha x.$$

4.2 Necessary and Sufficient Controllability Conditions

Before stating the main result of this section, that is Theorem 3, we have first to exhibit what we call the normal form of (Σ), and to analyze the controllability properties of that particular form (Theorem 2).

Proposition 8

If (Σ) satisfies the rank condition then it is equivalent by group automorphism to (Σ _N ):

$$({\Sigma}_{N})\qquad \left\{\begin{array}{ll} \dot{x}=& u \alpha x\\ \dot{y}=& x-1+by \qquad\text{ with}\qquad \alpha \neq 0. \end{array} \right. $$

(Σ _N ) is called the normal form of (Σ).

Proof

We have

$$DB=D(\alpha X+\beta Y)=\alpha DX+\beta DY=(a\alpha +b \beta )Y .$$

Therefore, the 2×2-matrix whose first (resp. second) column contains the coefficients of B (resp. DB) in the basis (X,Y) is given by

$$(B\ DB)=\left(\begin{array}{cc} \alpha & 0\\ \beta & a\alpha + b\beta \end{array} \right), $$

and the rank condition is satisfied if and only if α(a α+b β)≠0. As a matter of fact, if B and DB were linearly dependent, then the system zero-time ideal would be the one-dimensional space $\mathbb {R} B$. Notice that here the rank condition is equivalent to the ad-rank one.

Since B and DB are linearly independent, we can define an isomorphism P of the vector space $\mathfrak {g}$ by

$$P(B)=\alpha X \qquad\text{ and}\qquad P(DB)=\alpha Y . $$

This isomorphism turns out to be an automorphism of the Lie algebra $\mathfrak {g}$ because

$$[B,DB]= [\alpha X+\beta Y,(a\alpha +b\beta )Y ] = \alpha (a\alpha +b\beta )Y = \alpha DB, $$

implies

$$P[B,DB]=P(\alpha DB)=\alpha^{2} Y = \alpha^{2}[X,Y]=[\alpha X , \alpha Y]=[PB,P(DB)]. $$

By this automorphism, the derivation D is transformed into the derivation P D P ⁻¹ (see Proposition 7) which is characterized by:

$$PDP^{-1}X = Y \qquad\text{ and}\qquad PDP^{-1}Y = bY. $$

Consequently, we get in the basis (X,Y)

$$PDP^{-1}=\left(\begin{array}{cc} 0 & 0\\ 1 & b \end{array}\right)\qquad\ \qquad\text{and}\qquad PB= \alpha X .$$

Finally (Σ) is equivalent to : $({\Sigma }_{N})\ \ \left \{\begin {array}{cc} \dot {x}= & u\alpha x\\ \dot {y}=&x-1+by \end {array} \right .$ , with α≠0. □

Theorem 2

The system Σ _N is controllable if and only if b=0.

Proof

For b=0, the restriction of $\mathcal {X}$ to $\mathcal {D}^{1}G$ vanishes, and the system is singular. Since the ad-rank condition is satisfied, the sets $\mathcal {A}$ and $\mathcal {A}^{-}$ are neighborhoods of e, and $\mathcal {D}^{1}G \cap \mathcal {A}$ and $\mathcal {D}^{1}G \cap \mathcal {A}^{-}$ are neighborhoods of e in $\mathcal {D}^{1}G$. According to Proposition 6, the derived group $\mathcal {D}^{1}G$ is included in $\mathcal {A}$ and $\mathcal {A}^{-}$. The induced system on $G/\mathcal {D}^{1}G$, that is $\dot {x}= u\alpha x$, being clearly controllable whenever α≠0 (which is guaranteed by the rank condition), we conclude that (Σ _N ) is controllable by Proposition 5.

Let us now show that the system is not controllable as soon as b≠0, and let us first assume b>0. Since x>0, we have

$$\left.\begin{array}{ll} \dot{y}<0 &\Longleftrightarrow x-1+by < 0 \\ &\Longrightarrow y < \frac{1-x}{b} < \frac{1}{b}. \end{array}\right.$$

Consequently,

$$y\geq\frac{1}{b}\Longrightarrow \forall x >0 \qquad \dot{y} > 0 $$

and the system is not controllable to e=(1,0). In the same way, if b<0 we have

$$ y \leq \frac{1}{b} \qquad \Longrightarrow \qquad \forall x>0 \qquad \dot{y} > 0, $$

and the system is not controllable from the identity. □

We are now in a position to go back to the original system and to state the main result of this section.

Theorem 3

The linear system $({\Sigma }): \dot {g}=\mathcal {X}_{g} + ugB$ on Aff ₊ (2), where in the basis (X,Y)

$$D=\begin{pmatrix}0&0\\a&b\end{pmatrix} \qquad\text{and}\qquad B=\alpha X +\beta Y, $$

is controllable if and only if b=0 and aα≠0.

In other words, (Σ) is controllable if and only if it is singular and satisfies the rank condition.

4.3 Controllability of the Associated Invariant System

Since the derivations of $\mathfrak {aff}(2)$ are inner, a linear vector field $\mathcal {X}(g)= g(aY-bX)-(aY-bX)g$ on the affine group is the sum of a left-invariant vector field and a right-invariant one. Consequently, we can associate to (Σ) the left-invariant system

$$(I)\qquad \dot{g}=g(aY-bX)+ugB. $$

In the semisimple case, all the derivations are inner, and an invariant system can always be associated to a linear one (see [10]). Moreover, in that case, no example of controllable linear system whose associated invariant system is not controllable is known, and we conjecture that on a semisimple Lie group (Σ) is controllable if and only if (I) is.

The next proposition states that a one-input left-invariant system on A f f ₊(2) is never controllable, showing that this conjecture is not true on solvable Lie groups. It can be proved using the hypersurface principle, see for instance [13], but the proof given here is closer to the one of the linear case.

Proposition 9

A one-input invariant system on Aff ₊ (2) is not controllable.

Proof

The invariant system (I) satisfies the rank condition if and only if the vector fields a Y−b X and B=α X+β Y are linearly independent, that is if and only if a α+b β≠0.

In particular, (I) satisfies the rank condition when (Σ) does. We can in that case consider the normalized system, that is a=1 and β=0. Then (I) writes

$$(I)\qquad\left\{\begin{array}{ll} \dot{x}= &-bx +u\alpha x\\ \dot{y}= & x \end{array} \right.$$

but this system is not controllable because $\dot {y}>0$.

For α=0 and b β≠0, the rank condition is still satisfied by (I), but controllability does not hold because $\dot {x}$ does not depend neither on u nor on y. □

5 Controllability of Linear Systems on the Heisenberg Group

5.1 Linear Vector Fields on the Heisenberg Group

The Heisenberg group is the matrix group

$$G=\left\{\left(\begin{array}{lll}1&y&z\\0&1&x\\0&0&1 \end{array}\right);\ \ (x,y,z)\in\mathbb{R}^{3}\right\}. $$

Its Lie algebra $\mathfrak {g}$ is generated by the right-invariant vector fields whose value at the group identity is given by

$$X=\left(\begin{array}{lll}0&0&0\\0&0&1\\0&0&0 \end{array}\right),\quad Y=\left(\begin{array}{lll}0&1&0\\0&0&0\\0&0&0 \end{array}\right), \quad Z=\left(\begin{array}{lll}0&0&1\\0&0&0\\0&0&0 \end{array}\right). $$

They verify² [X,Y]=Y X−X Y=Z, and can be written in natural coordinates

$$X=\frac{\partial}{\partial x}, \qquad Y=\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}, \qquad\ Z=\frac{\partial}{\partial z}. $$

A straightforward computation using the equality D Z=[D X,Y]+[X,D Y] shows that the derivations of $\mathfrak {g}$ are the endomorphisms whose matrix in the basis (X,Y,Z) writes

$$D=\left(\begin{array}{llc}a&b&0\\c&d&0\\e&f&a+d \end{array}\right). $$

Then the linear vector field $\mathcal {X}$ associated to this derivation is the unique vector field on G that vanishes at the identity and verifies $\text {ad}(\mathcal {X})=-D$. A computation making use of the equalities $[X,\mathcal {X}]=DX$ and so on shows that $\mathcal {X}$ writes in natural coordinates

$$\mathcal{X}=(ax+by)\frac{\partial}{\partial x}+(cx+dy)\frac{\partial}{\partial y}+(ex+fy+(a+d)z+\frac{1}{2}by^{2}+\frac{1}{2}cx^{2})\frac{\partial}{\partial z}. $$

Notations. The center of the Heisenberg group, which is equal to the derived group $\mathcal {D}^{1}G$, will be denoted by $\mathcal {Z}(G)$. Similarly, the center of its Lie algebra $\mathfrak {g}$ will be denoted by $\mathcal {Z}(\mathfrak {g})=\mathcal {D}^{1}\mathfrak {g}$.

5.2 Controllability of Two-Input Linear Systems

In this section, we consider the two-input linear system

$$({\Sigma}_{2}) \quad \dot{g}=\mathcal{X}_{g}+u_{1}{B^{1}_{g}}+ u_{2}{B^{2}_{g}} $$

where B ¹ and B ² are right-invariant vector fields assumed to be linearly independent.

Theorem 4

The two-input linear system (Σ ₂ ), where B ¹ and B ² are linearly independent, is controllable if and only if the rank condition is satisfied. In that case, it is exact time controllable in time T for every T>0.

Proof

We have only to prove that the system is controllable under the rank condition.

• 1. If the bracket of B ¹ and B ² does not not vanish, that is if [B ¹,B ²]=a Z with a≠0, then $\mathbb {R} Z$ is included in $\mathcal {LSS}({\Sigma }_{2})$ which therefore contains $\mathfrak {g}$, and (Σ₂) is controllable in exactly T unit of time for all T>0.

• 2. If [B ¹,B ²] vanishes, we can assume without loss of generality that one of the vector fields B ¹ or B ² belongs to $\mathcal {Z}(\mathfrak {g})$, say $B^{2}\in \mathcal {Z}(g)$ (to be more precise B ¹ or B ² can be replaced by a linear combination of B ¹ and B ² that belongs to $\mathcal {Z}(G)$). Then the rank condition implies that (B ¹,D B ¹,B ²) is a basis of $\mathfrak {g}$ (otherwise, the system zero-time ideal would be contained in the two-dimensional space $\mathbb {R} B^{1}+\mathcal {Z}(\mathfrak {g})$). As a consequence, B ¹ and D B ¹ are independant on $\mathcal {Z}(\mathfrak {g})$ and the induced system on $G/\mathcal {Z}(G)$, which is a classical linear system, is controllable since the Kalman rank condition is satisfied.

  As B ²=α Z, for some α≠0, the dynamic on $\mathcal {Z}(G)$ is $\dot {z} = (a+d)z +\alpha u_{2}$ for x=y=u ₁=0. This subsystem is clearly controllable, hence the center $\mathcal {Z}(G)$ of G is included in $\mathcal {A}$ and $\mathcal {A}^{-}$, and (Σ₂) is controllable by Proposition 5.

  The system is actually exact time controllable in any time T>0. Indeed, let p ₁=(x ₁,y ₁,z ₁)∈G. There exists a control that steers (0,0) to (x ₁,y ₁) in $G/\mathcal {Z}(G)$ in T/2 unit of time. In G, this control steers (0,0,z ₂) to (x ₁,y ₁,z ₁) for some z ₂. There exists as well a control that steers (0,0,0) to (0,0,z ₂) in T/2 unit of time, hence p ₁ can be reached from the identity at any time T>0. Clearly, the identity can also be reached from p ₁ at any time T>0.

□

5.3 Controllability of Single-Input Linear Systems

We turn now our attention to the single-input case, which is the most pleasant one because it enlights in an interesting way the controllability problem on nilpotent groups. It is also the most difficult. The system under consideration will be

$$({\Sigma}_{1})\qquad \dot{g}=\mathcal{X}_{g}+ u B_{g}, $$

and as well as on the affine group, we begin by exhibiting a normal form that will greatly simplify the forthcoming proofs.

5.3.1 Normal Form

Proposition 10

If (Σ ₁ ) verifies the rank condition then it is equivalent by automorphism to

$$({\Sigma}_{N})\qquad\dot{g}=\widetilde{\mathcal{X}}_{g}+ u X_{g} \qquad\text{where}\qquad \widetilde{D}=\left(\begin{array}{lll}0&b&0\\1&d&0\\0&f&d \end{array}\right) $$

is the derivation associated to $\widetilde {\mathcal {X}}$ in the basis (X,Y,Z).

(Σ _N ) will be referred to as a system in normal form.

Moreover:

• 1. The application (x,y,z)→(λx,λy,λ ² z) where λ≠0 is a group automorphism whose action on a normal form is to change f into λf and X into λX.

• 2. The time-reversed system

  $$({\Sigma}^{-}_{1})\qquad \dot{g}=-\mathcal{X}_{g}- u X_{g}$$

  is equivalent by automorphism to the system in normal form whose associated derivation is $\begin {pmatrix}0&b&0\\1&-d&0\\0&f&-d\end {pmatrix}$.

Proof

• 1. Let us first show that the rank condition is satisfied if and only if (B,D B,[B,D B]) is a basis of $\mathfrak {g}$. This condition is clearly sufficient. For the necessity, we know by Proposition 2 that the rank condition is satisfied if and only if $\mathcal {L}_{0} = \mathfrak {g}$, where $\mathcal {L}_{0} = \mathcal {L} \mathcal {A} \lbrace D^{k}B, k \geq 0 \rbrace$.

  If B and DB are linearly dependent then $\mathcal {L}_{0} \subseteq \mathbb {R} B$. If $DB \in \mathcal {Z}(g)$, then $\mathcal {L}_{0}\subseteq \mathbb {R} B + \mathcal {Z}(g)$. In both cases, the rank of $\mathcal {L}_{0}$ is not full. Consequently, (B,D B,[D,D B]) must be a basis of $\mathfrak {g}$ for the rank condition to hold, and we can define an isomorphism P of the vector space $\mathfrak {g}$ by P(B)=X, P(D B)=Y, P([B,D B])=Z. But [P(B),P(D B)]=[X,Y]=Z=P[B,D B], the other brackets vanish, and P is actually a Lie algebra automorphism.

  According to Proposition 7, there exists a group automorphism that transforms (Σ₁) into $({\Sigma }_{N}):\quad \dot {g}=\widetilde {\mathcal {X}}_{g}+ u X_{g}$. The derivation $\widetilde {D}= PDP^{-1}$ associated to $\widetilde { \mathcal {X}}$ verifies $\widetilde {D}X =PDP^{-1}X =PDB =Y$, hence its matrix in the basis (X,Y,Z) has the form $\begin {pmatrix}0&b&0\\1&d&0\\0&f&d\end {pmatrix}$.

• 2. The transformations ϕ:(x,y,z)→(λ x,λ y,λ ² z), where λ≠0, and ψ:(x,y,z)→(−x,y,−z) are automorphisms of the group G, whose law is (x,y,z).(x ^′,y ^′,z ^′)=(x+x ^′,y+y ^′,z+z ^′+y x ^′) in the natural coordinates. An easy computation gives:

  $T_{e}\phi = P = \begin {pmatrix}\lambda &0 & 0\\0 & \lambda & 0\\0 & 0 &\lambda ^{2}\end {pmatrix}$ hence $P\widetilde {D}P^{-1} = \begin {pmatrix}0&b&0\\1&d&0\\0&\lambda f&d\end {pmatrix}$, and

  $T_{e}\psi = P' =\begin {pmatrix}-1&0 & 0\\0 & 1 & 0\\0 & 0 &-1\end {pmatrix}$ hence $P'(-\widetilde {D})P'^{-1} = \begin {pmatrix}0&b&0\\1&-d&0\\0&f&-d\end {pmatrix}$.

□

In the forthcoming proof of Theorem 5, we will assume that the system under consideration is in normal form (hence satisfies the rank condition), that is B=X and $D=\begin {pmatrix}0&b&0\\1&d&0\\0&f&d\end {pmatrix}$.

Remark

• 1. In the previous proof the coefficients b,d,f of the derivation $\widetilde {D} =PDP^{-1}$ are not the same as the coefficients b,d,f of the original derivation $D=\left (\begin {array}{llc}a&b&0\\c&d&0\\e&f&a+d \end {array}\right )$. They can of course be related by some formulas, but we do not need them herein.

• 2. The controllability matrix of the linear part of (Σ _N ) is

  $$[X,\widetilde{D}X,\widetilde{D}^{2}X]=\left(\begin{array}{ccc} 1 & 0 & b\\ 0 & 1 & d \\ 0 & 0 & f \end{array}\right)$$

  and the ad-rank condition is equivalent to f≠0.

• 3. A system in normal form writes in natural coordinates

  $$({\Sigma}_{N})\ \ \left\{\begin{array}{lll} \dot{x}=& by + u\\ \dot{y}=& x +dy\\ \dot{z}=& fy+dz+\frac{1}{2}x^{2}+\frac{1}{2}by^{2} \end{array} \right. $$

  The induced system on $G/\mathcal {Z}(G)$, that is $\left \{\begin {array}{ll} \dot {x}=& by + u\\ \dot {y}=& x +dy \end {array} \right .$ , is always controllable.

5.3.2 The Lie Saturate

Proposition 11

System (Σ ₁ ) can be enlarged to the system

$$({\Sigma}_{E})\qquad \dot{g}=\mathcal{X}_{g}+u X_{g} +vY_{g} + \frac{v^{2}}{2}Z_{g} $$

without modifying the closures of the attainable sets. In particular, the right-invariant vector field Z belongs to the Lie saturate $\mathcal {LS}({\Sigma }_{1})$.

Proof

As the vector field v X belongs to $\mathcal {LS} ({\Sigma }_{1})$ for all $v\in \mathbb {R}$, the vector field $\exp (v\text {ad}(X))\mathcal {X}$ belongs also to $\mathcal {LS}({\Sigma }_{1})$ (see [11]). But

$$\begin{array}{ll} \exp(v\text{ad}(X))\mathcal{X}&=\mathcal{X}+(v\text{ad}(X))\mathcal{X} + \frac{1}{2}(v\text{ad}(X))^{2} \mathcal{X} \\ &= \mathcal{X} + v DX +\frac{v^{2}}{2}[X,DX]\\ &= \mathcal{X}+ vY + \frac{v^{2}}{2}Z \end{array} $$

The Lie saturate $\mathcal {LS}({\Sigma }_{1})$ being a closed convex cone we can multiply that vector field by $\frac {2}{v^{2}}$ and then let v go to infinity. We obtain $Z \in \mathcal {LS}({\Sigma }_{1})$. □

Remark

• 1. System (Σ _E ) will be referred to as the extended system. Assume (Σ₁) to be in normal form, then (Σ _E ) writes in coordinates

  $$({\Sigma}_{E})\qquad \left\{\begin{array}{lll} \dot{x}=&by +u\\ \dot{y}=&x + dy+ v\\ \dot{z}=&fy+dz+\frac{1}{2}x^{2}+ \frac{1}{2}by^{2} +\frac{1}{2}v^{2} \end{array}\right.$$

• 2. In the forthcoming controllability proof, we will have to show $\mathcal {Z}(G) \subset \overline {\mathcal {A}}$ and $\mathcal {Z}(G)\subset \overline {\mathcal {A}^{-}}$ (here $\mathcal {A}$ and $\mathcal {A}^{-}$ refer to (Σ _N )). For this purpose, we will set x=0 and u(t)=−b y(t). Eliminating the variable x, we obtain in the (y,z)-plane the system

  $$({\Sigma}_{RE})\quad \left\{\begin{array}{ll} \dot{y}=dy+v\\ \dot{z}=fy + dz+\frac{1}{2}by^{2} + \frac{1}{2}v^{2} \end{array}\right. $$

  It will be referred to as the reduced-extended systemin the sequel.

• 3. More accurately, we will show that for all $z_{0}\in \mathbb {R}$, the center $\mathcal {Z}(G)$ of G is contained in the attainable set from (0,0,z ₀) for the extended system (Σ _E ) by means of piecewise smooth controls. However, the attainable sets of (Σ _E ) from an initial point p ₀ for piecewise constant inputs and for measurable locally bounded ones have the same closures, which is also the closure of the attainable set from p ₀ for (Σ _N ). This will therefore prove the inclusions $\mathcal {Z}(G) \subset \overline {\mathcal {A}}$ and $\mathcal {Z}(G)\subset \overline {\mathcal {A}^{-}}$ (About the attainable sets for different classes of inputs see for instance [6] or [15]).

5.3.3 Main Result

Theorem 5

A system in normal form is controllable if and only if one of the conditions

• (i) $b < -\frac {d^{2}}{4}$,

• (ii) d = 0 and f ≠ 0,

holds.

Proof

The proof of this theorem is divided into three lemmas. □

Lemma 1

Let us assume that d=0. The system is controllable if and only if f≠0, or f=0 and b<0.

Proof

For d=0, the linear vector field $\mathcal {X}$ vanishes on $\mathcal {Z}(G)$: it is the singular case. As the induced system on $G/\mathcal {Z}(G)$ is controllable, and according to Proposition 5, we have only to show that $\mathcal {Z}(G)=\{(0,0,z);\ z\in \mathbb {R}\}$ is contained in the closures of $\mathcal {A}$ and $\mathcal {A}^{-}$.

• 1. If f≠0 the ad-rank condition is satisfied, the linearized system is controllable, and $\mathcal {A}$ and $\mathcal {A}^{-}$ are neighborhoods of the identity e=(0,0,0). According to Propositions 6 and 5, the system is controllable.

• 2. If f=0 and b≥0, then $\dot {z} = \frac {1}{2}x^{2}+ \frac {1}{2}by^{2} \geq 0$, therefore the system is not controllable.

• 3. Let us show that (Σ _N ) is controllable if f=0 and b<0. As the right-invariant vector field Z belongs to the Lie saturate, starting from a point (0,0,z ₀), any point (0,0,z ₁) with z ₁>z ₀ can be reached by the extended system.

  Since b<0, we can set b=−β ² so that the reduced-extended system writes

  $$\left\{\begin{array}{ll} \dot{y}=v\\ \dot{z}=\frac{1}{2}(v^{2}-\beta^{2}y^{2}) \end{array}\right. $$

  Let us start from a point (0,z ₀). By applying the control v=1, and then the control v=0, the reduced-extended system reaches a point (y ₁,z ₁) where y ₁>0 and z ₁<z ₀ with z ₀−z ₁ arbitrary. To finish the trajectory obtained for the control v=−β y tends to the point (0,z ₁).

  This shows that $\mathcal {Z}(G)$ is included in the closures of $\mathcal {A}$ and $\mathcal {A}^{-}$, hence that the system is controllable.

□

Lemma 2

Let us assume that d≠0. If $b \geq -\frac {d^{2}}{4}$ then the system is not controllable.

Proof

To prove this lemma, two changes of variable will be applied to z.

Let us first set $w=z+\frac {d}{4} y^{2}$. Using the expression of (Σ _N ) in coordinates, we obtain

$$\dot{w}=fy +dw+\frac{1}{2}(x+\frac{1}{2}dy)^{2}+\frac{1}{2}y^{2}(\frac{d^{2}}{4}+ b) $$

(the value $\alpha = \frac {d}{4}$ corresponds to the worst case in the change of variable w=z+α y ²).

If $b>-\frac {d^{2}}{4}$ then the polynomial $fy +\frac {1}{2}(x+\frac {1}{2}dy)^{2}+\frac {1}{2}y^{2}(\frac {d^{2}}{4}+ b)$ has a minimum m because $\frac {1}{2}(x+\frac {1}{2}dy)^{2}+\frac {1}{2}y^{2}(\frac {d^{2}}{4}+ b)$ is a positive definite quadratic form. We get $\dot {w}> 0$ for all x,y as soon as d w>−m. Consequently, if d>0 then w>−d ⁻¹ m implies $\dot {w} > 0$ (if d<0 then w<−d ⁻¹ m implies $\dot {w}>0$), and the system is not controllable.

It remains to prove the noncontrollability in the case $b = \frac {-d^{2}}{4}$. For that purpose, we make the second change of variable ${\Theta } = w+2\frac {f}{d} y$ (the value $\gamma =2\frac {f}{d}$ corresponds to the worst case in the change of variable Θ=w+γ y). We obtain

$$\left.\begin{array}{ll} \dot{\Theta}&= fy + d{\Theta} + 2\frac{f}{d} x+ \frac{1}{2}(x+\frac{1}{2}dy)^{2}. \end{array}\right. $$

The minimum of the expression $2\frac {f}{d} x+ \frac {1}{2}(x+\frac {1}{2}dy)^{2}$ is reached for $x = -2\frac {f}{d} - \frac {1}{2}dy$, and is equal to $-fy-2\frac {f^{2}}{d^{2}}$. Therefore, the inequality $\dot {\Theta } \geq d{\Theta } -2\frac {f^{2}}{d^{2}}$ holds for all $x, y \in \mathbb {R}$, and for the same reasons as above the system is not controllable. □

Lemma 3

Let us assume that d≠0. If $b < -\frac {d^{2}}{4}$ then the system is controllable.

Proof

According to Proposition 10, we can assume d>0. Indeed if d<0, then the time-reversed system satisfies d>0 and $b < -\frac {d^{2}}{4}$, it is controllable by the forthcoming proof and so is (Σ _N ). Moreover, f is defined up to a nonvanishing constant and we can also assume f≤0 (in order to keep the system in normal form the feedback transformation u↦λ ⁻¹ u is necessary, but of course it does not change anything to controllability).

The system induced on $G/\mathcal {Z}(G)$ being controllable we have only to show that $\mathcal {Z}(G)$ is included in the closures of $\mathcal {A}$ and $\mathcal {A}^{-}$, or that for any pair z ₀,z ₁ of real numbers, the point (0,0,z ₁) can be reached from (0,0,z ₀) by a trajectory of the extended system.

Since the vector field Z belongs to the Lie saturate, this can be achieved as soon as z ₁≥z ₀. Moreover, the trajectory starting from (0,0,z ₀) for the control u=0 being t↦(0,0,z ₀ e ^dt) with d>0, any point (0,0,z ₁) can be reached from the point (0,0,z ₀) whenever z ₀<0.

Consequently, it remains to show that starting from a given point (0,0,z ₀) with z ₀≥0, we can reach at least one point (0,0,z) with z<0.

For this purpose, we set b=−β ² with $\beta > \frac {d}{2}$, and we consider the reduced-extended system

$$\left\{\begin{array}{ll} \dot{y} & =dy+v\\ \dot{z} & =fy + dz-\frac{1}{2}\beta^{2} y^{2} + \frac{1}{2}v^{2} \end{array}\right. $$

up to the end of this proof.

To begin with, we define two areas $\mathcal {Z}_{1}$ and $\mathcal {Z}_{2}$. The first one $\mathcal {Z}_{1}$ is the set of points $(y , z)\in \mathbb {R}^{2}$ for which there exists at least one value of the control v such that $\dot {z} < 0$, that is

$$\mathcal{Z}_{1}=\lbrace(y,z)\in \mathbb{R}^{2};\ dz+fy -\frac{1}{2}\beta^{2} y^{2} < 0 \rbrace. $$

Since f≤0 the set $\lbrace (y,z)\in \mathbb {R}^{2};\ y\geq 0\,\, \text {and}\,\, dz -\frac {1}{2}\beta ^{2} y^{2}< 0 \rbrace$ is included in $\mathcal {Z}_{1}$.

The second area denoted by $\mathcal {Z}_{2}$ is the set of points $(y , z)\in \mathbb {R}^{2}$ for which there exists at least one value of the control v such that $\dot {z} < 0$ and $y \dot {y} < 0$.

The second condition entails v ²>d ² y ², hence $\dot {z} > fy+dz-\frac {1}{2}\beta ^{2}y^{2}+\frac {1}{2}d^{2}y^{2}$. Consequently, $\dot {z}\leq 0$ implies $fy+dz+\frac {1}{2}(d^{2}-\beta ^{2})y^{2}< 0$, and one obtains

$$ \mathcal{Z}_{2}=\{fy+dz+\frac{1}{2}(d^{2}-\beta^{2})y^{2}< 0\}. $$

Starting from (0,z ₀), with z ₀>0, we will successively reach $\mathcal {Z}_{1}$ and $\mathcal {Z}_{2}$, and finally a point (0,z) with z<0 in five steps.

• 1. The constant control v>0 on an arbitrary small time steers (0,z ₀) to a point (y ₁,z ₁) with y ₁>0,z ₁>0.

• 2. The control v=β y allows to enter $\mathcal {Z}_{1}$. Indeed, for v=β y the system is

  $$ \left\{\begin{array}{ll} \dot{y}=(d+\beta)y\\ \dot{z}=fy + dz \leq dz \qquad(\text{since}\,\, f\leq 0) \end{array}\right. $$

  and we get $y(t)=y_{1}\exp ((d+\beta )t)$, $z(t)\leq z_{1}\exp (dt)$. An easy computation shows that

  $$ z(t)-\frac{\beta^{2}}{2d}y^{2}(t)\longrightarrow_{t\rightarrow +\infty}-\infty $$

  and that the trajectory reaches a point (y ₂,z ₂) with y ₂>0 and $z_{2}<\frac {\beta ^{2}}{2d}{y_{2}^{2}}$, hence belonging to $\mathcal {Z}_{1}$, for some t large enough.

• 3. If z ₂≥0 by applying the control v=0, we get

  $$\left\{\begin{array}{ll} \dot{y}=dy\\ \dot{z}=fy + dz-\frac{1}{2}\beta^{2} y^{2} \end{array}\right.$$

  It is clear that $\dot {z}< 0$ because on the one hand, it is true at t=0, and on the other hand y(t) is increasing. Consequently, we get a trajectory inside the area $\mathcal {Z}_{1}$ that reaches a point $(y_{3},z_{3})\in \mathcal {Z}_{1}$ with y ₃>0 and z ₃<0.

• 4. We must now reach $\mathcal {Z}_{2}$. Whenever β ²≥d ², the point (y ₃,z ₃) is already in $\mathcal {Z}_{2}$. If not, that is if $\frac {d^{2}}{4} < \beta ^{2}< d^{2}$, we apply the control v=−a y with $\frac {d}{2}< a \leq \beta$. We have $y(t)=y_{3}\exp (d-a)t$ and

  $$\begin{array}{ll} \dot{z} & =dz+fy -\frac{1}{2}\beta^{2} y^{2}+\frac{1}{2}a^{2} y^{2}\\ & \leq dz-\frac{1}{2}(\beta^{2}-a^{2}) y^{2} \\ & \leq dz \qquad\text{ because}\qquad\beta^{2}-a^{2}\geq 0 \end{array} $$

  consequently $z(t)\leq z_{3}\exp (dt)$ and

  $$\begin{array}{ll} fy+dz+\frac{1}{2}(d^{2}-\beta^{2})y^{2} & \leq dz+\frac{1}{2}(d^{2}-\beta^{2})y^{2}\\ & \leq dz_{3}\exp(dt)+\frac{1}{2}(d^{2}-\beta^{2}){y_{3}^{2}}\exp(2(d-a)t)\\ & \longrightarrow_{t\rightarrow +\infty}-\infty \end{array} $$

  because $2(d-a)<2(d-\frac {d}{2})=d$. Thus, the trajectory enters $\mathcal {Z}_{2}$ and reaches a point $(y_{4},z_{4})\in \mathcal {Z}_{2}$ with y ₄>0 and z ₄<0.

• 5. We are now in a position to reach a point (0,z) with z<0. Firstly, we apply the control v=−d y, for which $\dot {y} = 0$ and $\dot {z} < 0$, and that steers (y ₄,z ₄) to (y ₅,z ₅) with y ₅=y ₄ and $z_{5}\leq \frac {-1}{2d}(dy_{5}+y_{5})^{2}$.

  To finish the control v=−d y−y ₅ steers at the time t=1, the point (y ₅,z ₅) to the point (0,z ₆) where z ₆<0 because

  $$\begin{array}{ll} \dot{z}&=dz+fy -\frac{1}{2}\beta^{2} y^{2}+\frac{1}{2}(dy+y_{5})^{2}\\ &\leq dz+\frac{1}{2}(dy+y_{5})^{2}\\ &\leq dz + \frac{1}{2}(dy_{5}+y_{5})^{2} \qquad\text{because} \,y>0\ \text{decreases}\\ & \leq 0. \end{array}$$

The proof is complete. □

Let us go back to the system induced on the quotient group $G/\mathcal {Z}(G)$. This classical linear system writes

$$\left\{\begin{array}{ll} \dot{x}=& by + u\\ \dot{y}=& x +dy \end{array} \right. $$

and we will denote it by (L). A standard computation shows that the eigenvalues of the matrix $\left (\begin {array}{ll}0&b\\1&d \end {array}\right )$are real if and only if $b\geq -\frac {d^{2}}{4}$. These eigenvalues are invariant under the group automorphisms because their action on $G/\mathcal {Z}(G)$ is linear in the classical meaning. Consequently, we can forget the normal form and state:

Theorem 6

The one-input system (Σ ₁ ) on the Heisenberg group is controllable if and only if it satisfies the rank condition and

• (i) In the regular case: the eigenvalues of (L) are not real

• (ii) In the singular case: the eigenvalues of (L) are not real or the ad-rank condition is satisfied.

6 Discussion of the Results

6.1 Eigenvalues

On the Heisenberg group, and in the non-singular case, we have found that a one-input system that verifies the rank condition is controllable if and only if the eigenvalues of the system induced on $G/\mathcal {Z}(G)$ are not real.

This condition is coherent with the results obtained on G=A f f ₊(2). Indeed, the change of variables $(x,y)\longmapsto (\zeta =\ln (x),y)$ transforms the system into:

$$\left\{\begin{array}{ll} \dot{\zeta}=& \alpha u\\ \dot{y}=& e^{\zeta}-1+y \end{array} \right. $$

The linear part of the system induced on $G/\mathcal {D}^{1}G$ vanishes, and its only eigenvalue, that is 0, is real. On the other hand, the system is controllable only in the singular case.

6.2 The Ad-Rank Condition

The results obtained herein, in particular Theorem 6, show also that the ad-rank condition arises only in the singular case because then the controllability of the linearized system allows to conclude to the controllability of the system. In the regular case, it does not matter if the ad-rank condition is satisfied or not.

6.3 Controllability from the Identity and Controllability

Theorem 6 allows also to answer to the following open question: Is it true that a system controllable from the identity is controllable? In other words: is the condition $\mathcal {A}=G$ sufficient for controllability?

In [10], it is proved that the answer is positive for linear systems on semisimple groups with finite center. On the other hand, it is no longer positive in the general case. As a matter of fact, consider on the Heisenberg group a system in normal form

$$({\Sigma}_{N})\ \ \left\{\begin{array}{lll} \dot{x}=& by + u\\ \dot{y}=& x +dy\\ \dot{z}=& fy+dz+\frac{1}{2}x^{2}+\frac{1}{2}by^{2} \end{array} \right. $$

with d>0 and $-\frac {d^{2}}{4}<b<0$. The system is not controllable since it is regular with $b\geq -\frac {d^{2}}{4}$. However, it is controllable from e if f≠0. To see that consider the eigenvalues of $\left (\begin {array}{ll}0&b\\1&d \end {array}\right )$. There are equal to

$$ \frac{d\pm\sqrt{d^{2}+4b}}{2} $$

hence positive, so that the three eigenvalues of the linearized system are positive, the third one being d.

According to f≠0, the linearized system is controllable and the attainable set $\mathcal {A}$ from the identity contains a neighborhood W of the identity. But for a simply connected nilpotent group, the exponential mapping is a diffeomorphism and W is the image by $\exp$ of some neighborhood V of 0 in $\mathfrak {g}$. For the control u=0, the flow of the system is φ _t , the one of $\mathcal {X}$, and for all t≥0 the set

$$\varphi_{t}(W)=\varphi_{t}(\exp(V))=\exp(e^{tD}V) $$

is included in $\mathcal {A}$. But the eigenvalues of D being positive the set $\cup _{t\geq 0}e^{tD}V$ is equal to $\mathfrak {g}$ and $\mathcal {A}$ is equal to G (see [1] for this idea).

In conclusion:

Theorem 7

For d>0, $-\frac {d^{2}}{4}<b<0$ , and f≠0, the system is controllable from e but not controllable.

One could think that the difference between the behavior of linear systems on semisimple Lie groups and the Heisenberg group is due to the fact that the derivations are inner on semisimple Lie algebras, while on the Heisenberg group the derivation associated to a one-input linear system that verifies the rank condition cannot be (see [10]).

But actually, the same phenomenon appears on the two-dimensional affine group, though all derivations are inner. Consider on A f f ₊(2) the system in normal form

$$({\Sigma}_{N})\qquad \left\{\begin{array}{ll} \dot{x}=& u \alpha x\\ \dot{y}=& x-1+by \qquad\text{ with}\qquad \alpha \neq 0, \end{array} \right. $$

and assume b>0. As the ad-rank condition holds, the set $\mathcal {A}$ is a neighborhood of the identity e=(1,0) and it is easy to see that $\mathcal {D}^{1}G$ is included in $\mathcal {A}$. Following the same lines as in the proof of Proposition 5, we get $\mathcal {A}=Aff_{+}(2)$, so that (Σ _N ) is controllable from e, but not controllable since b>0.

Proposition 12

For b > 0, the linear system in normal form (Σ _N ) on Aff ₊ (2) is controllable from the identity but not controllable.