/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Consider a continuous-time syste... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 9

Consider a continuous-time system \(\dot{x}=f(x)\) with no controls and \(x=\mathbb{R}^{n}\). Suppose that \(V: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is proper and positive definite, and satisfies \(\dot{V}(x)=L_{f} V(x)<0\) for all \(x \neq 0\) (this is the Lyapunov condition in Lemma 5.7.4). Show that there exists a continuous function \(\alpha:[0, \infty) \rightarrow[0, \infty)\) which is positive definite (that is, \(\alpha(0)=0\) and \(\alpha(r)>0\) for all \(r>0\) ) such that the following differential inequality holds: \(\nabla V(x) \cdot f(x)=\dot{V}(x) \leq-\alpha(V(x))\) for all \(x \in \mathbb{R}^{n} .\) (Hint: Study the maximum of \(L_{f} V(x)\) on the set where \(V(x)=r_{.}\))

Short Answer

Expert verified
In summary, we define a continuous and positive definite function \(\alpha\) as \(\alpha(r) = -\beta(r)\) for \(r>0\) and \(\alpha(0) = 0\), where \(\beta(r) = \max_{x \in S_r} L_f V(x) < 0\) for all \(r > 0\), and \(S_r = \{x \in \mathbb{R}^n : V(x) = r\}\). The given differential inequality \(\nabla V(x) \cdot f(x) = \dot{V}(x) \leq -\alpha(V(x))\) holds for all \(x \in \mathbb{R}^n\), where \(\dot{V}(x) = L_f V(x)\).

Step by step solution

01

Study the maximum of \(L_f V(x)\) on the set where \(V(x) = r\).

Let \(S_r = \{x \in \mathbb{R}^n : V(x) = r\}\). Since \(V\) is proper and positive definite, \(S_r\) is compact and non-empty for any \(r > 0\). Also, by the Lyapunov condition, we have \(\dot{V}(x) = L_f V(x) < 0\) for all \(x \neq 0\). Thus, there exists \(\beta(r) = \max_{x \in S_r} L_f V(x) < 0\) for all \(r > 0\).
02

Define the function \(\alpha\).

Now, we define the function \(\alpha : [0, \infty) \rightarrow [0, \infty)\) as: \[\alpha(r) = -\beta(r) \ \ \text{for} \ \ r > 0,\] and \[\alpha(0) = 0.\]
03

Show that \(\alpha\) is continuous and positive definite.

We derived \(\alpha(r) = -\beta(r)\), where \(\beta(r) < 0\) for all \(r > 0\). Therefore, \(\alpha(r) > 0\) for all \(r > 0\). Since \(\alpha(0) = 0\), \(\alpha\) is positive definite. \(S_r\) is compact, and \(L_f V(x)\) is continuous. Thus, \(\beta(r) = \max_{x \in S_r} L_f V(x)\) is also continuous as a function of \(r\). Then, \(\alpha(r) = -\beta(r)\) is continuous as the composition of two continuous functions.
04

Prove the differential inequality.

Since \(L_f V(x) = \dot{V}(x)\), for every \(x \in \mathbb{R}^n\), we have \[\dot{V}(x) = L_f V(x) \leq -\alpha(V(x)).\] Hence, the given differential inequality holds: \[\nabla V(x) \cdot f(x) = \dot{V}(x) \leq -\alpha(V(x))\ \text{for all}\ x \in \mathbb{R}^n.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Differential Inequality
In the context of the given exercise, the concept of a differential inequality plays a crucial role. It involves analyzing the behavior of derivatives under certain inequalities.
Consider a function \( V(x) \) where its derivative \( \dot{V}(x) \) satisfies an inequality of the form \( \dot{V}(x) \leq -\alpha(V(x)) \). Here, \( \alpha \) is a continuous, positive definite function.
This inequality indicates that the derivative of \( V(x) \) is always negative or zero, suggesting that \( V(x) \) decreases over time.
  • The continuous function \( \alpha \) ensures that the rate of decrease of \( V(x) \) doesn鈥檛 stagnate, as it enforces \( \dot{V}(x) \) to be less than or equal to a non-positive value.
  • This provides key insights into the system's behavior without solving it directly.
By examining this differential inequality, we also secure information regarding the stability of the system, as it implies \( V(x) \)'s total decrease thus hinting at Lyapunov Stability.
Continuous-Time System
The exercise is primarily concerned with a continuous-time system. This refers to a system where changes with respect to time are continuous, as opposed to occurring in discrete steps.
Such a system is described mathematically by differential equations like \( \dot{x} = f(x) \). These equations provide information on how state variables change smoothly over time.
  • Continuous-time systems are essential models in various fields like control systems, physics, and economics, where reactions and changes do not happen in isolated steps but in a flowing manner.
  • The system dynamics described by \( f(x) \) dictate how the system evolves under the influence of its current state \( x \).
Understanding the role of these systems helps in predicting future states depending on current observations, thereby playing a fundamental role in control and stability analysis like the Lyapunov approach being explored here.
Lyapunov Function
Lyapunov functions are at the heart of analyzing stability within dynamic systems. They function as energy-like measures that help track whether a system's status is tending towards a stable point or diverging away.
In this specific scenario, \( V(x) \) is our Lyapunov function and it is positive definite, meaning it is always positive unless the system is in equilibrium (i.e., \( x = 0 \)).
  • A Lyapunov function, by design, decays over time such that \( \dot{V}(x) < 0 \) for all \( x eq 0 \), signifying that the system continues to approach steadiness.
  • This decline assures that as time progresses, the system inherits a trend towards a stable state or point.
  • The existence of such a function not only serves to confirm a system's potential stability but also provides a quantitative pathway to understanding its transient behavior.
The Lyapunov function is instrumental in proving stability through careful design and demonstration, like the setup we're examining in this exercise. By satisfying the condition \( \dot{V}(x) < 0 \), it supports the notion that the system is Lyapunov stable and further aids in constructing continuous functions like \( \alpha(r) \) to solidify this stability guarantee.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\diamond\) Show that, if the field \(\mathbb{K}\) is infinite, \((A, B)\) is controllable, and the matrix \(A\) is cyclic, then there exists some \(u\) so that \((A, B u)\) is controllable. Give counterexamples to show that this fails if the hypothesis that \(A\) is cyclic is dropped, and to show that even if \(A\) is cyclic the result fails over the field of two elements. (Hint: Use the fact that the geometric multiplicity of each eigenvalue of the transpose \(A^{\prime}\) must be one, and then solve the inequalities $$ v^{\prime} B u \neq 0 $$ for each eigenvector \(v\) of \(A^{\prime}\).)

For any pair \((A, B)\) and any \(F \in \mathbb{K}^{m \times n}\), $$ \mathcal{R}(A+B F, B)=\mathcal{R}(A, B) . $$ In particular, \((A+B F, B)\) is controllable if and only if \((A, B)\) is.

The function \(\varphi\) can be obtained as the solution of an optimization problem. For each fixed \(x \neq 0\), thought of as a parameter, not as a state, we may consider the pair \((a(x), B(x))\) as a \((1, m)\) pair describing a linear system of dimension \(\mathbf{1}\), with \(m\) inputs. The equations for this system are as follows, where we prefer to use " \(z\) " to denote its state, so as not to confuse with \(x\), which is now a fixed element of \(\mathbb{R}^{n} \backslash\\{0\\}\) : $$ \dot{z}=a z+\sum_{i=1}^{m} b_{i} u_{i}=a z+B u \text {. } $$ The control-Lyapunov function condition guarantees that this system is asymptotically controllable. In fact, the condition " \(B=0 \Rightarrow a<0^{\prime \prime}\) means precisely that this system is asymptotically controllable. A stabilizing feedback law \(k=k(x)\) for the original nonlinear system must have, for each fixed \(x\), the property that $$ a+\sum_{i=1}^{m} b_{i} k_{i}<0 \text {. } $$ This means that \(u=k z\) must be a stabilizing feedback for the linear system (5.59). Consider for this system the infinite-horizon linear-quadratic problem of minimizing (cf. Theorem 41 (p. 384)) $$ \int_{0}^{\infty} u^{2}(t)+\beta(x) z^{2}(t) d t $$ (For motivation, observe that the term \(u^{2}\) has greater relative weight when \(\beta\) is small, making controls small if \(x\) is small.) Prove that solving this optimization problem leads to our formula (5.56).

Assume that \(\Sigma\) is a time-invariant discrete-time system of class \(\mathcal{C}^{1}\), $$ x^{+}=\mathcal{P}(x, u), $$ with \(X \subseteq \mathbb{R}^{n}\) and \(\mathcal{U} \subseteq \mathbb{R}^{m}\) open, and let \(\left(x^{0}, u^{0}\right)\) be an equilibrium pair, i.e. $$ \mathcal{P}\left(x^{0}, u^{0}\right)=x^{0} . $$ Assume that the linearization of \(\Sigma\) at \(\left(x^{0}, u^{0}\right)\) is asymptotically controllable. Then \(\Sigma\) is locally asymptotically controllable (to \(x^{0}\) ). Moreover, there exists in that case a matrix \(F \in \mathbb{R}^{m \times n}\) such that the closed-loop system $$ x^{+}=\mathcal{P}_{c l}(x):=\mathcal{P}\left(x, u^{0}+F\left(x-x^{0}\right)\right) $$ is locally asymptotically stable.

Prove that we have an equivalence relation, in the following sense: (a) For each \(\Sigma\) and open \(\mathcal{O} \subseteq X,(\Sigma, \mathcal{O})\) is feedback equivalent to \((\Sigma, \mathcal{O})\). (b) \((\Sigma, \mathcal{O})\) feedback equivalent to \((\widetilde{\Sigma}, \widetilde{\mathcal{O}})\) implies \((\widetilde{\Sigma}, \widetilde{\mathcal{O}})\) feedback equivalent to \((\Sigma, \mathcal{O})\) (c) If \((\Sigma, \mathcal{O})\) is feedback equivalent to \((\tilde{\Sigma}, \tilde{\mathcal{O}})\) and \((\tilde{\Sigma}, \tilde{\mathcal{O}})\) is feedback equivalent to \((\hat{\Sigma}, \hat{\mathcal{O}})\), then \((\Sigma, \mathcal{O})\) is feedback equivalent to \((\hat{\Sigma}, \hat{\mathcal{O}})\).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.