Local nested transverse feedback linearization

Abstract

We study two local feedback equivalence problems for a nonlinear control-affine system with two nested, controlled-invariant, embedded submanifolds in its state space. The first, less restrictive, result gives necessary and sufficient conditions for the dynamics of the system restricted to the larger submanifold and transversal to the smaller submanifold to be linear and controllable. This normal form facilitates designing controllers that locally stabilize the smaller set relative to the larger set. The second, more restrictive, result additionally imposes that the transversal dynamics to the larger set be linear and controllable. This result can simplify designing controllers to locally stabilize the larger submanifold. This is illustrated by sufficient conditions under which these normal forms can be used to locally solve a nested set stabilization problem.

Appendix: Supporting results and proofs

Proof of Proposition 1

The proof that $\dim {(P(q))} = \nu (q)$ is obvious from their definitions and is omitted. Next, we have

$$\begin{aligned} \dim (Q(q))= & {} n-\dim (G_0(q)\cap T_qS_2)+\dim (G_0(q)\cap T_qS_1)\\&-\,\dim ([G_0(q)\cap T_qS_2]^\perp +[G_0(q)\cap T_qS_1])\\= & {} \dim (G_0(q)\cap T_qS_1)-\dim (G_0(q)\cap T_qS_2)\\= & {} \rho (q). \end{aligned}$$

Similar computations yield $\dim {(R(p))} = \sigma (p)$ on $S_1$. $\square$

Proof of Proposition 2

Let $U\subseteq {\mathbb {R}}^n$ be an open set containing $\bar{x}$ and set $V_1 = S_1\cap U$. If $\dim {(T_{x}S_1\cap G_0(x))}$ is constant on $V_1$ then, since $\dim {(T_xS_1)}$ and $\dim {(G_0(x))}$ are constant on $V_1$, the function $\sigma (x)$ in (11) is constant on $V_1$. If both $\dim {(T_{x}S_2\cap G_0(x))}$ and $\dim {(T_{x}S_1\cap G_0(x))}$ are constant on $V_2 = S_2\cap U$ then the functions $\nu$ and $\rho$ in (11) are constant on $V_2$.

Conversely, if the function $\sigma$ is constant on an open set $V_1 \subset S_1$ with $\bar{x}\in V_1$, then since $T_xS_1$ and $G_0(x)$ are constant dimensional and from the definition of $\sigma$ it follows that $\dim {(T_{x}S_1\cap G_0(x))}$ is constant on $V_1$. If $\nu$, $\rho$ are constant on an open set $V_2 \subset S_2$ with $\bar{x}\in V_2$ then from their definitions it follows that $\dim {(T_{x}S_2\cap G_0(x))}$ is constant on $V_2$. $\square$

Proof of Proposition 3

Let $\bar{x} \in S_2$ be a regular point of the distributions (10). Then, by Proposition 2 and Definition 4 P is non-singular in a neighbourhood $V_2=V_1\cap S_2$, with $V_1\subseteq S_1$ and containing $\bar{x}$. Lemma 3 proves that P is also smooth in a neighbourhood of $\bar{x}$, without loss of generality, $V_2$. Proposition 1 shows that Q is non-singular on $V_2$ and R is non-singular on $V_1$. Furthermore, by Proposition 2, the assumed non-singularity of $G_0$ and Lemma 3 we have, by possibly shrinking $V_1$, and hence $V_2$, that $G_0 \cap TS_1$ and $\left[ G_0\cap TS_1\right] ^\perp$ are smooth on $V_1$ and $\left[ G_0(x)\cap TS_2\right] ^\perp$ is smooth on $V_2$. Therefore, Q and R are the non-singular intersection of smooth non-singular distributions and by [14, Lemma 1.3.5] they are smooth themselves.

Conversely, suppose that the distribution R in (10) is smooth and non-singular in a neighbourhood $V_1 \subseteq S_1$ containing $\bar{x}$ and distributions P and Q in (10) are smooth and non-singular in $V_2=V_1\cap S_2$. By Proposition 1 and Definition 4 $\bar{x}$ is a regular point of (11). $\square$

Proof of Lemma 1

Let $x\in S_2$ be fixed but arbitrary and let $\varXi \in {{\mathrm{\text {Diff}}}}{(U)}$ be a diffeomorphism onto its image with domain U containing x. Let $\left( \alpha ,\beta \right)$ be a regular feedback transformation also defined on U and let $\tilde{g}(x) :=g(x)\beta (x)$, $\tilde{G}_0(x):={{\mathrm{\text {span}}}}\{\tilde{g}_1(x),\ldots ,\tilde{g}_m(x)\}$. Since each $\tilde{g}_i(x)$ is a linear combination of $g_1(x), \ldots , g_m(x)$, $\tilde{G}_0(x) \subseteq G_0(x)$. Furthermore, since $\beta :U \subseteq {\mathbb {R}}^n\rightarrow {{\mathrm{{\mathsf {GL}}}}}(m,{\mathbb {R}})$ is non-singular, $\tilde{G}_0(x) = G_0(x)$ and, therefore,

$$\begin{aligned} \nu (x) = \dim (T_{x}S_2\cap G_0(x)) = \dim (T_{x}S_2\cap \tilde{G}_0(x)). \end{aligned}$$

Next, let $\hat{g} :=\varXi _\star (g\beta ) = \varXi _\star (\tilde{g})$ and $\hat{G}_0 :={{\mathrm{\text {span}}}}\{\hat{g}_1,\ldots ,\hat{g}_m\}$. Since ${\mathrm {d}}\varXi _x$ is an isomorphism at each $x \in U$, we have

$$\begin{aligned} \begin{aligned} \dim (T_{x}S_2\cap \tilde{G}_0(x))&= \!\dim ({\mathrm {d}}\varXi _x (T_{x}S_2\cap \tilde{G}_0(x))) \!=\! \dim ({\mathrm {d}}\varXi _x (T_{x}S_2)\cap {\mathrm {d}}\varXi _x(\tilde{G}_0(x)))\\&= \dim (T_{\varXi (x)}\varXi (S_2\cap U)\cap \hat{G}_0(\varXi (x))) \end{aligned} \end{aligned}$$

where the next to last equality comes from the fact that ${{\mathrm{\text {Ker}}}}{({\mathrm {d}}\varXi _x)} = \{0\}$. From this it follows that the value $\nu (x)$ is unchanged under coordinate and feedback transformations. The same arguments hold for the other functions in (11). $\square$

Proof of Lemma 2

Let $\bar{x}\in S_2$ be arbitrary. Since $S_1\subseteq {\mathbb {R}}^n$ is an embedded submanifold there exist slice coordinates $\left( V_1,\psi \right)$ for ${\mathbb {R}}^n$ with $\bar{x}\in V _1$ such that

$$\begin{aligned} \psi (S_1\cap V_1)=\{x\in V_1: \psi _{s_1+1}(x)=c_{s_1+1},\ldots ,\psi _{n}(x)=c_{n}\} \end{aligned}$$

where, without loss of generality, we take the constants $c_i$ to be zero. Let $\pi _1:{\mathbb {R}}^n\rightarrow {\mathbb {R}}^{n-s_1}$ denote the projection onto the last $n-s_1$ factors, i.e, $\pi _1(x)=(x_{s_1+1},\ldots ,x_n)$. Define $\varPhi _1:V_1\rightarrow {\mathbb {R}}^{n-s_1},x\mapsto \pi _1\circ \psi (x)$. Then, $\varPhi _1$ is a submersion and

$$\begin{aligned} \psi (S_1\cap V_1)=\{x\in V_1:\varPhi _1(x)=0\}. \end{aligned}$$

This construction is summarized in the following commutative diagram

We now apply a similar construction to $S_2$. Let $\left( V_2, \varphi \right)$ be slice coordinates for ${\mathbb {R}}^n$ with $\bar{x}\in V_2$ and let $\pi _2:{\mathbb {R}}^n\rightarrow {\mathbb {R}}^{n-s_2}$ be the projection onto the last $n-s_2$ factors. Then, letting $\varPhi _2:=\pi _2\circ \phi$ we have

$$\begin{aligned} \varphi (S_2\cap V_2)=\{x\in V_2:\varPhi _2(x)=0\}. \end{aligned}$$

and the commutative diagram

Let $U:=V_1\cap V_2$ and note that $\bar{x}\in U$. Since $\varPhi _1$ and $\varPhi _2$ are submersions we have that, for all $x\in U$, ${{\mathrm{\text {rank}}}}({\text {d}}\varPhi _1)=n-s_1$ and ${{\mathrm{\text {rank}}}}({\text {d}}\varPhi _2)=n-s_2$. Furthermore, by [20, Lemma 8.15], for all $x\in S_2\cap U$, ${{\mathrm{\text {Ker}}}}{{\text {d}}\varPhi _{1,x}}=T_xS_1$ and ${{\mathrm{\text {Ker}}}}{ {\text {d}}\varPhi _{2,x}}=T_xS_2$. Therefore, ${{\mathrm{\text {Ker}}}}{\text {d}}\varPhi _{2,x}\subset {{\mathrm{\text {Ker}}}}{\text {d}}\varPhi _{1,x}$ and

$$\begin{aligned} {{\mathrm{\text {rank}}}}\left[ \begin{array}{c} {\text {d}}\varPhi _{1,x}\\ {\text {d}}\varPhi _{2,x} \end{array} \right] = {{\mathrm{\text {rank}}}}[ \begin{array}{c} {\text {d}}\varPhi _{2,x} \end{array} ]=n-s_2. \end{aligned}$$

(30)

This allows us to construct a submersion $\varPhi :U\rightarrow {\mathbb {R}}^{n-s_2}$. We take the last $n-s_1$ components of $\varPhi$ to be the function $\varPhi _1$. From (30) we conclude that in the set $\varPhi _2=\{\varphi _{s_2+1},\ldots ,\varphi _n\}$ it is possible to find $s_1-s_2$ functions, without of loss of generality $\{\varphi _{s_2+1},\ldots ,\varphi _{s_1}\}=:\bar{\varPhi }_2$, with the property that the $n-s_2$ differentials ${\text {d}}\varphi _{s_2+1},\ldots ,{\text {d}}\varphi _{s_1},{\text {d}}\psi _{s_1+1},\ldots ,{\text {d}}\psi _n$ are linearly independent at $\bar{x}$. Let $\varPhi :=\left( \bar{\varPhi }_2,\varPhi _1\right)$.

Since, ${\text {d}}\varPhi (\bar{x})$ has rank $n-s_2$ it has some $(n-s_2)\times (n-s_2)$ minor with non-zero determinant. By re-ordering the coordinates we assume that it is the minor corresponding to the first $n-s_2$ rows and columns of ${\mathrm {d}}\varPhi (\bar{x})$. Relabel the coordinates as $(y,z)=(x_1,\ldots ,x_{n-s_2},x_{n-s_2+1},\ldots ,x_n)$ in ${\mathbb {R}}^n$. Define $\varXi :U\rightarrow {\mathbb {R}}^n$ by $\varXi (y,z):=(z,\varPhi (y,z))$. Its total derivative at $\bar{x}$ is

$$\begin{aligned} {\text {d}}\varXi (\bar{x})=\left[ \begin{array}{c@{\quad }c} 0&{}I_{s_2}\\ \frac{\partial \varPhi _i}{\partial y_j}&{}\frac{\partial \varPhi _i}{\partial z_j} \end{array}\right] , \end{aligned}$$

which is non-singular because its columns are independent. Therefore, by the inverse function theorem [20, Theorem 7.6], by possibly shrinking U, $\varXi \in {{\mathrm{\text {Diff}}}}{(U)}$. In the chart $\left( U,\varXi \right)$ of ${\mathbb {R}}^n$ we have

$$\begin{aligned} \varXi (S_1\cap U)=\{x\in U: \varXi _{s_1+1}(x)=\cdots =\varXi _{n}(x)=0\} \end{aligned}$$

and

$$\begin{aligned} \varXi (S_2\cap U)=\{x\in U: \varXi _{s_2+1}(x)=\cdots =\varXi _{n}(x)=0\}. \end{aligned}$$

$\square$

Proof of Proposition 4

Let ${\mathcal {N}}(M)$ be a tubular neighbourhood of M. By [20, Proposition 10.20] there exists a smooth retraction $r : {\mathcal {N}}(M) \rightarrow M$. Let $U\subseteq {\mathcal {N}}(M)$ be an open set containing x. Then, restriction $\left. r \right| _U$ is a smooth retraction of U to $M\cap U$. $\square$

Proof of Proposition 5

Apply Lemma 2 to obtain an open set $U \subseteq {\mathbb {R}}^n$ containing $\bar{x}$ and maps $\varPhi _1$ and $\bar{\varPhi }_2$ such that $V_1=\varPhi _1^{-1}(0)$ and $V_2=(\bar{\varPhi }_2,\varPhi _1)^{-1}(0)$ where $V_1:=S_1\cap U$ and $V_2:=S_2\cap U$.

Since $S_1$ is a controlled-invariant submanifold there exists a smooth state feedback $\alpha _1:V_1\rightarrow {\mathbb {R}}^m$ such that

$$\begin{aligned} \left( \forall x\in V_1\right) \quad {\text {d}}\varPhi _1(x)\left( f(x)+g(x)\alpha _1(x)\right) =0. \end{aligned}$$

(31)

Similarly, since $S_2$ is a controlled-invariant submanifold there exists a smooth state feedback $\alpha _2:V_2\rightarrow {\mathbb {R}}^m$ such that

$$\begin{aligned} \left( \forall x\in V_2\right) \quad \left[ \begin{array}{c} {\text {d}}\bar{\varPhi }_2(x)\\ {\text {d}}{\varPhi }_1(x) \end{array}\right] \left( f(x)+g(x)\alpha _2(x)\right) =0. \end{aligned}$$

(32)

We now modify $\alpha _1$ so that the resulting state feedback simultaneously satisfies (31) and (32). We have that

$$\begin{aligned} \begin{aligned} \left( \forall x\in V_2\right) \quad&\left. {\text {d}}\varPhi _1(x)\left( f(x)+g(x) \alpha _2(x)\right) \right| _{V_2}-\left. {\text {d}}\varPhi _1(x)\left( f(x)+g(x)\alpha _1(x)\right) \right| _{V_2}=0\\&\Rightarrow \left. {\text {d}}\varPhi _1(x)g(x)\left( \alpha _2(x)-\alpha _1(x)\right) \right| _{V_2}=0. \end{aligned} \end{aligned}$$

Since $\alpha _1$ and $\alpha _2$ are both smooth, there exists a smooth $\hat{v}(x)\in {{\mathrm{\text {Ker}}}}(\left. {\text {d}}\varPhi _1(x)g(x)\right| _{V_2})$ such that, for all $x\in V_2$, $\alpha _2(x)=\left. \alpha _1(x)\right| _{V_2}+\hat{v}(x)$. We have that

$$\begin{aligned} \begin{aligned} (\forall x\in V_1)\quad {{\mathrm{\text {rank}}}}( {\text {d}}\varPhi _1(x)g(x))&={{\mathrm{\text {rank}}}}g(x)-\dim ({{\mathrm{\text {Ker}}}}{\text {d}}\varPhi _1(x)\cap {{\mathrm{\text {rank}}}}{g(x)})\\&=\dim G_0(x)-\dim (T_xS_1\cap G_0(x)). \end{aligned} \end{aligned}$$

By hypothesis, $\bar{x}$ is a regular point of (10) and by Proposition 2, by possibly shrinking $V_1$, $\dim (T_xS_1\cap G_0(x))$ is constant and $\dim G_0(x)$ is constant. Thus, ${{\mathrm{\text {rank}}}}( {\text {d}}\varPhi _1(x)g(x))$ is constant on $V_1$. It implies that $\dim \left( {{\mathrm{\text {Ker}}}}( {\text {d}}\varPhi _1(x)g(x))\right)$ is also constant on $V_1$. Assume that $\dim \left( {{\mathrm{\text {Ker}}}}( {\text {d}}\varPhi _1(x)g(x))\right) =q$. By [14, Lemma 1.3.1], there exists a set $\{v_1,\ldots ,v_q\}$ of smooth vector fields defined on $V_1$ such that at each $x\in V_1$, the vectors $v_1(x),\ldots ,v_q(x)$ are linearly independent and

$$\begin{aligned} \left( \forall x\in V_1\right) \quad {{\mathrm{\text {Ker}}}}({\text {d}}\varPhi _1(x)g(x))={{\mathrm{\text {span}}}}\{v_1(x),\ldots ,v_q(x)\}. \end{aligned}$$

Thus, we can write

$$\begin{aligned} \hat{v}(x)=\sum _{i=1}^{q}\hat{c}_i(x)v_i(x). \end{aligned}$$

where $\hat{c}_i:V_2\rightarrow {\mathbb {R}}$ are smooth real-valued functions. Apply Proposition 4 and, by possibly shrinking U , introduce a retraction $r_1:V_1\rightarrow V_2$ of $V_1$ onto $V_2$ and define

$$\begin{aligned} \begin{aligned} c_i:&V_1\rightarrow {\mathbb {R}}\\&x\mapsto \hat{c}_i\circ r_1(x). \end{aligned} \end{aligned}$$

and

$$\begin{aligned} v(x)=\sum _{i=1}^{q}c_i(x)v_i(x). \end{aligned}$$

Let $\alpha ^\prime :=\alpha _1+v$. It solves Eq. (31) since

$$\begin{aligned} \left( \forall x \in V_1\right) \; {\text {d}}\varPhi _1(x)(f(x)+g(x)\alpha ^\prime (x))= & {} {\text {d}}\varPhi _1(x)\left( f(x)+g(x)\alpha _1(x)\right) \\&+\,{\text {d}}\varPhi _1(x)g(x)v(x)=0. \end{aligned}$$

Similarly it can be verified that it solves Eq. (32). Again, applying Proposition 4 we introduce a retraction $r_2:U\rightarrow V_1$ of U into $V_1$ and define

$$\begin{aligned} \begin{aligned} \alpha :&U\rightarrow {\mathbb {R}}^m\\&x\mapsto \alpha ^\prime \circ r_2(x). \end{aligned} \end{aligned}$$

The state feedback $\alpha$ has the desired property. $\square$

Lemma 3

[24] Let $N \subset M$ be an n-dimensional submanifold of the m-dimensional manifold M. Let $p \in N$ be a regular point of a d-dimensional distribution D on M. Suppose there exists an open neighbourhood V of p in N such that $k = \dim (T_qN \cap D(q))$ is constant for all $q \in V$. Then, there exists a neighbourhood U of p in V such that $TN\cap D$ is smooth on U.

Proof

Let $(W, \psi )$ be a coordinate chart of M adapted to N, that is, such that $\psi (N \cap W) = \{x \in \psi (W): x_{n+1} = \cdots = x_m = 0\}$, and let $\{f_1, \ldots , f_d\}$ be a set of local generators of D around p. Let $\pi : (x_1, \ldots , x_m) \mapsto (x_1, \ldots , x_n)$ be the projection onto the first n factors. By making W smaller, we can assume that $f_1, \dots , f_d$ are linearly independent on W.

Recall that $\hat{\psi } := \pi \circ \psi : N \cap W \rightarrow {\mathbb {R}}^n$ is a diffeomorphism onto its image, and let

$$\begin{aligned} \hat{f}_i := \hat{\psi }_\star \left( f_i\right) ,\quad \; i \in \{1, \ldots , d\}. \end{aligned}$$

The vector fields $\hat{f}_i$ are defined on an open set of ${\mathbb {R}}^n$. Letting $\{e_1, \ldots , e_n\}$ denote the natural basis of ${\mathbb {R}}^n$, for each $q \in N \cap W$ we have

$$\begin{aligned} \begin{aligned} d\hat{\psi }_q(T_qN \cap D(q))&= d\hat{\psi }_q(T_qN) \cap d\hat{\psi }_q(D(q))\\&= {{\mathrm{\text {span}}}}\{e_1, \ldots , e_n\} \cap {{\mathrm{\text {span}}}}\{\hat{f}_1(\psi (q)), \ldots , \hat{f}_d(\psi (q))\}. \end{aligned} \end{aligned}$$

Hence, $d\hat{\psi }(TN\cap D)$ is a distribution on an open set of ${\mathbb {R}}^n$. By assumption, and since $d\hat{\psi }_q$ is an isomorphism at each $q \in N \cap W$, it is the intersection of two smooth non-singular distributions, and it has constant dimension near $\psi (p)$. Therefore, by [14, Lemma 1.3.5], it is smooth. This implies that $TN \cap D$ is also smooth on a neighbourhood V of p. $\square$