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Chapter 4: Problem 4

Show that, for analytic systems with connected state space \(X\), if the accessibility rank condition holds at even one point \(x^{0} \in X\), then it must also hold on an open dense subset of \(\mathcal{X}\). (Hint: Use that if an analytic function vanishes in some open set, then it must be identically zero.)

Short Answer

Expert verified
Given an analytic system with connected state space \(X\) and accessibility at one point \(x^0\), we can leverage the accessibility and analyticity of the system. On an open set around \(x^0\), the controls don't simultaneously vanish; as per the properties of analytic functions, they don't vanish on the entire space \(X\). Therefore, the accessibility rank condition must hold on an open dense subset of \(X\).

Step by step solution

01

Understanding the Accessibility Rank Condition

The accessibility rank condition is a fundamental concept in control theory. It essentially says that for a control system defined on a manifold or space, the system is 'controllable' if we can change the system's state from any initial point \(x^{0}\) to a final point \(x_{f}\) in a finite time using the system's controls. More precisely, it requires that the Lie algebra generated by the system's vector fields span the tangent space at every point of the manifold.
02

Understanding Analytic Systems and Functions

An analytic system or function is infinitely differentiable and equal to its Taylor series. Properties of analytic functions play a crucial role in this proof, particularly the principle that if an analytic function vanishes (i.e., equals zero) in some open set, then it must be identically zero. Meaning, it has to be zero everywhere in its domain.
03

Beginning the Proof

We start by considering an accessible point \(x^{0}\) in analytic system with state space \(X\). This implies that the accessibility rank condition holds at \(x^0\), that is, we have an enough number of independent controls in the neighbourhood of \(x^{0}\) to change the system's state.
04

Leveraging Accessibility and Analyticity

Given the analyticity of the system and the accessibility at \(x^{0}\), since controls can be expressed as analytic functions, we can say that on some open set around \(x^{0}\), these controls don't simultaneously vanish. Now, as per the properties of analytic functions, if they don't vanish on this open set, they don't vanish on the entire space \(X\).
05

Conclusion

So, given the accessibility at one point, the analyticity forces the rank condition to hold at least on an open dense subset of \(X\), if not the entire space. This is because an open dense subset can span the whole set in the contexts of topology and differential geometry. This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Accessibility Rank Condition
The Accessibility Rank Condition is a concept in control theory, which helps us check how controllable a system is. Imagine you have a robot on a field. You control its movements, and you want it to travel from one location to another. The Accessibility Rank Condition tells us whether this is possible by examining the vector fields related to your controls. In more technical terms, it requires that the Lie algebra created by combining these vector fields can cover all possible directions (or span the tangent space) at every point of the field you're on. If it does, then you can control the system to reach any point you wish from wherever you start. This condition is vital as it allows us to understand whether a system can reach all states, which is a key aspect in designing effective and functional controls.
Analytic Systems
Analytic systems are special because they involve functions that are not just smooth, but are equal to their Taylor series around any point. This means you can count on them being predictable and having neat properties. One of the most important properties is that if an analytic function equals zero in even a small open area, it can only be zero everywhere in its domain. This plays a critical role in mathematical proofs. For instance, when examining control systems, this property helps establish where certain conditions must hold true, supporting conclusions about the system's overall behavior. When control systems are modeled as analytic systems, this opens up straightforward ways to predict how they will act and assures us that the systems are behaving in a reliable and expected manner across their entire operational space, contributing to more robust and effective design.
Lie Algebra
Lie Algebra comes in handy when dealing with control systems as it provides a structure to explore the behaviors of these systems. It's like having a mathematical toolkit to understand systems where symmetry and transformation play a big role. In our context here, Lie Algebra is built from the vector fields of the control system. Combining these vector fields through operations yields a new set of vectors. If the set of all possible combinations spans out to cover every possible direction at any point, then you've got a powerful toolkit indeed. By examining this structure, we acquire insights into whether the entire system鈥檚 states are accessible from any given starting point. Thus, Lie Algebra helps us decide if we can truly control a system in its entirety, a critical criterion in effective control theory design.

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Most popular questions from this chapter

Suppose that \(f_{1}, \ldots, f_{r}\) are smooth vector fields on an open set \(\mathcal{O} \subseteq \mathbb{R}^{n}\), and that \(f_{1}(x), \ldots, f_{r}(x)\) are linearly independent for each \(x \in \mathcal{O}\). Show that the following two properties are equivalent: \- \(\left[f_{i}, f_{j}\right]=0\) for each \(i, j \in\\{1, \ldots, r\\} .\) \- For each \(x^{0} \in \mathcal{O}\) there is an open subset \(\mathcal{O}_{0} \subseteq \mathcal{O}\) which contains \(x^{0}\) and a diffeomorphism \(\Pi: \mathcal{O}_{0} \rightarrow \mathcal{V}\) into some open subset \(\mathcal{V} \subseteq \mathbb{R}^{n}\) such that \(\left(\Pi_{*} f_{i}\right)(z)=e_{i}\) for each \(z \in \mathcal{V}\), where \(e_{i}\) is the \(i\) th canonical basis vector. (That is, the vector fields commute if and only if there is a local change of variables where they all become \(\left.f_{i} \equiv e_{i} \cdot\right)\)

Suppose \(X\) and \(Y\) are analytic vector fields defined on \(\mathcal{O}\). For any \(x^{0} \in \mathcal{O}\), let \(\mathcal{I}=\mathcal{I}_{X, x^{0}}:=\left\\{t \in \mathbb{R} \mid\left(t, x^{0}\right) \in \mathcal{D}_{X}\right\\}\). Then, the function \(\gamma: \mathcal{I} \rightarrow \mathbb{R}^{n}: t \mapsto \operatorname{Ad}_{t X} Y\left(x^{0}\right)\) is analytic. Proof. Let \(\alpha(t):=\left(e^{-t X}\right),\left(e^{t X} x^{0}\right)\), seen as a function \(\mathcal{I} \rightarrow \mathbb{R}^{n \times n}\). Note that \(\alpha(0)=I\), and that, by Equation (4.28), the vector \(\left(e^{t X} x^{0}, \alpha(t)\right)\) is the solution of the differential equation $$ \begin{array}{lll} \dot{x}(t) & =X(x(t)) & x(0) & =x^{0} \\ \dot{\alpha}(t) & =-\alpha(t) \cdot X_{*}(x(t)) & & \alpha(0)=1 . \end{array} $$ This is a differential equation with analytic right-hand side, so \(x(\cdot)\) and \(\alpha(\cdot)\) are both analytic (see, for instance, Proposition C.3.12). Then, \(\gamma(t)=\alpha(t) Y(x(t))\) is also analytic.

Consider a model for the "shopping cart" shown in Figure 4.2 ("knife-edge" or "unicycle" are other names for this example). The state is given by the orientation \(\theta\), together with the coordinates \(x_{1}, x_{2}\) of the midpoint between the back wheels. Figure 4.2: Shopping cart. The front wheel is a castor, free to rotate. There is a non-slipping constraint on movement: the velocity \(\left(\dot{x}_{1}, \dot{x}_{2}\right)^{\prime}\) must be parallel to the vector \((\cos \theta, \sin \theta)^{\prime} .\) This leads to the following equations: $$ \begin{aligned} \dot{x}_{1} &=u_{1} \cos \theta \\ \dot{x}_{2} &=u_{1} \sin \theta \\ \dot{\theta} &=u_{2} \end{aligned} $$ where we may view \(u_{1}\) as a "drive" command and \(u_{2}\) as a steering control (in practice, we implement these controls by means of differential forces on the two back corners of the cart). We view the system as having state space \(\mathbb{R}^{3}\) (a more accurate state space would be the manifold \(\mathbb{R}^{2} \times \mathbb{S}^{1}\) ). (a) Show that the system is completely controllable. (b) Consider these new variables: \(z_{1}:=\theta, z_{2}:=x_{1} \cos \theta+x_{2} \sin \theta, z_{3}:=\) \(x_{1} \sin \theta-x_{2} \cos \theta, v_{1}:=u_{2}\), and \(v_{2}:=u_{1}-u_{2} z_{3}\). (Such a change of variables is called a "feedback transformation".) Write the system in these variables, as \(\dot{z}=\widetilde{f}(z, v) .\) Note that this is one of the systems \(\Sigma_{i}\) in Exercise 4.3.14. Explain why controllability can then be deduced from what you already concluded in that previous exercise.

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) is a distribution of constant rank \(r\) on an open set \(\mathcal{O} \subseteq \mathbb{R}^{n}\). Show that the following two properties are equivalent: \- \(\Delta\) is involutive. 176 4\. Nonlinear Controllability \- For each \(x^{0} \in \mathcal{O}\) there is an open subset \(\mathcal{O}_{0} \subseteq \mathcal{O}\) which contains \(x^{0}\) and a diffeomorphism \(\Pi: \mathcal{O}_{0} \rightarrow \mathcal{V}\) into some open subset \(\mathcal{V} \subseteq \mathbb{R}^{n}\) such that \(\left(\Pi_{*} \Delta\right)(z)=\operatorname{span}\left\\{e_{1}, \ldots, e_{r}\right\\}\) for each \(z \in \mathcal{V}\), where \(e_{i}\) is the \(i\) th canonical basis vector. (A distribution of the form \(\Delta_{e_{1}, \ldots, e_{r}}\) is said to be a flat distribution. So, the claim is that a distribution is involutive if and only if it is locally flat in appropriate coordinates.)

A distribution \(\Delta\) of constant rank \(r\) is completely integrable if, for each \(x^{0} \in \mathcal{O}\) there exists some neighborhood \(\mathcal{O}_{0}\) of \(x^{0}\) and a smooth function $$ \Psi: \mathcal{O}_{0} \rightarrow \mathbb{R}^{n-r} $$ such that $$ \operatorname{ker} \Psi_{\star}(x)=\Delta(x) $$ for all \(x \in \mathcal{O}_{0}\).
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