The Frobenius Theory

posted Nov 10, 2014, 9:47 PM by Javad Taghia   [ updated Nov 10, 2014, 9:47 PM ]

15.4.2.4 The Frobenius Theorem

The Lie bracket is the only tool needed to determine whether a system is completely integrable (holonomic) or nonholonomic (not integrable). Suppose that a system of the form (15.53) is given. Using the $ m$ system vector fields $ h_1$$ \ldots $$ h_m$ there are$ {(^{m}_{2})}$ Lie brackets of the form $ [h_i,h_j]$ for $ i < j$ that can be formed. A distribution $ {\triangle }$ is called involutive [133] if for each of these brackets there exist $ m$ coefficients $ c_k \in {\mathbb{R}}$ such that

$\displaystyle [h_i,h_j] = \sum_{k = 1}^m c_k h_k .$(15.86)

In other words, every Lie bracket can be expressed as a linear combination of the system vector fields, and therefore it already belongs to $ {\triangle }$. The Lie brackets are unable to escape $ {\triangle }$ and generate new directions of motion. We did not need to consider all $ n^2$ possible Lie brackets of the system vector fields because it turns out that $ [h_i,h_j]=-[h_j,h_i]$ and consequently $ [h_i,h_i] = 0$. Therefore, the definition of involutive is not altered by looking only at the $ {(^{m}_{2})}$ pairs.

If the system is smooth and the distribution is nonsingular, then the Frobenius theorem immediately characterizes integrability:

A system is completely integrable if and only if it is involutive.
Proofs of the Frobenius theorem appear in numerous differential geometry and control theory books [133,156,478,846]. There also exist versions that do not require the distribution to be nonsingular.

Determining integrability involves performing Lie brackets and determining whether (15.86) is satisfied. The search for the coefficients can luckily be avoided by using linear algebra tests for linear independence. The $ n \times m$ matrix $ H(x)$, which was defined in (15.56), can be augmented into an $ n \times (m+1)$ matrix $ H'(x)$ by adding $ [h_i,h_j]$ as a new column. If the rank of $ H'(x)$ is $ m+1$ for any pair $ h_i$ and $ h_j$, then it is immediately known that the system is nonholonomic. If the rank of $ H'(x)$ is $ m$ for all Lie brackets, then the system is completely integrable. Driftless linear systems, which are expressed as $ {\dot x}= B u$ for a fixed matrix $ B$, are completely integrable because all Lie brackets are zero.

Example 15..11 (The Differential Drive Is Nonholonomic)   For the differential drive model in (15.54), the Lie bracket $ [f,g]$ was determined in Example 15.9 to be $ [ \sin\theta \;\; -\cos\theta \;\; 0]^T$. The matrix $ H'(q)$, in which $ q = (x,y,\theta)$, is
$\displaystyle H'(q) = \begin{pmatrix}\cos \theta & 0 & \sin\theta  \sin \theta & 0 & -\cos\theta  0 & 1 & 0  \end{pmatrix} .$(15.87)

The rank of $ H'(q)$ is $ 3$ for all $ q \in {\cal C}$ (the determinant of $ H'(q)$ is $ 1$). Therefore, by the Frobenius theorem, the system is nonholonomic. $ \blacksquare$ 
Example 15..12 (The Nonholonomic Integrator Is Nonholonomic)   We would hope that the nonholonomic integrator is nonholonomic. In Example 15.10, the Lie bracket was determined to be $ [ 0 \;\; 0 \;\; 2]^T$. The matrix $ H'(q)$ is
$\displaystyle H'(q) = \begin{pmatrix}1 & 0 & 0  0 & 1 & 0  -x_2 & x_1 & 2  \end{pmatrix} ,$(15.88)

which clearly has full rank for all $ q \in {\cal C}$$ \blacksquare$ 
Example 15..13 (Trapped on a Sphere)   Suppose that the following system is given:
$\displaystyle \begin{pmatrix}{\dot x}_1  {\dot x}_2  {\dot x}_3 \end{pmatri... ... 0  \end{pmatrix} u_1 + \begin{pmatrix}x_3  0  -x_1  \end{pmatrix} u_2,$(15.89)

for which $ X = {\mathbb{R}}^3$ and $ U = {\mathbb{R}}^2$. Since the vector fields are linear, the Jacobians are constant (as in Example 15.10):
$\displaystyle \frac{\partial f}{\partial x} = \begin{pmatrix}0 & 1 & 0  -1 & 0 & 0  0 & 0 & 0 \end{pmatrix}$$\displaystyle \mbox {\;\;\; and \;\;\;} \frac{\partial g}{\partial x} = \begin{pmatrix}0 & 0 & 1   0 & 0 & 0   -1 & 0 & 0 \end{pmatrix} .$(15.90)

Using (15.80),
$\displaystyle \small \frac{\partial g}{\partial x} f - \frac{\partial f}{\parti... ...0  -x_1  \end{pmatrix} = \begin{pmatrix}0  x_3  -x_2  \end{pmatrix} .$(15.91)

This yields the matrix
$\displaystyle H'(x) = \begin{pmatrix}x_2 & -x_1 & 0  x_3 & 0 & -x_1  0 & x_3 & -x_2  \end{pmatrix} .$(15.92)

The determinant is zero for all $ x \in {\mathbb{R}}^3$, which means that $ [f,g]$ is never linearly independent of $ f$ and $ g$. Therefore, the system is completely integrable.15.10

The system can actually be constructed by differentiating the equation of a sphere. Let

$\displaystyle f(x) = x_1^2 + x_2^2 + x_3^2 - r^2 = 0 ,$(15.93)

and differentiate with respect to time to obtain
$\displaystyle x_1 {\dot x}_1 + x_2 {\dot x}_2 + x_3 {\dot x}_3 = 0,$(15.94)

which is a Pfaffian constraint. A parametric representation of the set of vectors that satisfy (15.94) is given by (15.89). For each $ (u_1,u_2) \in {\mathbb{R}}^2$, (15.89) yields a vector that satisfies (15.94). Thus, this was an example of being trapped on a sphere, which we would expect to be completely integrable. It was difficult, however, to suspect this using only (15.89). $ \blacksquare$ 

Steven M LaValle 2012-04-20

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Javad Taghia,
Jul 14, 2015, 1:14 AM
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