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Consider this nonlinear system, with \(X=\mathbb{R}^{2}\) and \(\mathcal{U}=[-1,1]\) : $$ \begin{aligned} &\dot{x}_{1}=\left(1-x_{2}^{2}\right) u^{2} \\ &\dot{x}_{2}=u . \end{aligned} $$ Show that the system is complete, that is, solutions exist for every initial state and every control. Show, however, that \(\mathcal{R}^{1}(0)\) is not a closed set. Why does this not contradict the discussion in Remark 10.1.10? (Hint: Consider the states \(x_{k}^{f}=\alpha\left(\omega_{k}\right)\), where \(\omega_{k}\) is a control that switches fast between \(+1\) and $\left.-1 .\right)$

Short Answer

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The given nonlinear system with state variables \(x_1\) and \(x_2\) and control variable \(u\) is complete, as solutions exist for every initial state and every control, due to the continuous functions in the system. However, \(\mathcal{R}^{1}(0)\) is not a closed set when using a control input \(\omega_k\) that switches fast between \(+1\) and \(-1\), which leads to states with no limit points inside the reachable set. This doesn't contradict Remark 10.1.10 because the control input used is not piecewise continuous, and therefore, the result from Remark 10.1.10 doesn't apply to this control input.
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Step 1: Analyzing the given system of equations

: We are given the following nonlinear system of equations: $$ \begin{aligned} &\dot{x}_{1}=\left(1-x_{2}^{2}\right) u^{2} \\ &\dot{x}_{2}=u . \end{aligned} $$ This system has two state variables, \(x_1\) and \(x_2\), and one control variable, \(u\). The control set \(\mathcal{U}=[-1,1]\), which means that the control variable can take any real value between -1 and 1, inclusive.

Step 2: Showing that the system is complete

: To show that the system is complete, we need to demonstrate that solutions exist for every initial state and every control. We will do this by showing that the right-hand sides of the differential equations are continuous functions of the state variables and the control variable. For the first equation, the right-hand side is given by \(\left(1-x_{2}^{2}\right) u^{2}\). This is a continuous function of \(x_2\) and \(u\), as the square of a real number is continuous, as is the product of continuous functions. For the second equation, the right-hand side is simply given by \(u\), which is also a continuous function of the control variable. Since both right-hand sides are continuous functions, and the control set \(\mathcal{U}\) is a closed and bounded interval, it follows that this nonlinear system has solutions for every initial state and every control, thereby proving that the system is complete.

Step 3: Showing that the reachable set is not closed

: Now we have to show that \(\mathcal{R}^{1}(0)\) is not a closed set. To do this, we'll consider states \(x_{k}^{f}=\alpha\left(\omega_{k}\right)\), where \(\omega_{k}\) is a control that switches fast between \(+1\) and \(-1\). As the switching frequency of the control \(\omega_k\) increases without bound, the state \(x_{k}^{f}\) will approach the origin, which is the element in the reachable set \(\mathcal{R}^{1}(0)\). However, the state trajectory will not converge to the origin, as it will continue oscillating with higher frequencies. This yields that the sequence of states \(\{x_k^f\}\) has no limit points inside the reachable set \(\mathcal{R}^{1}(0)\). Therefore, \(\mathcal{R}^{1}(0)\) is not a closed set.

Step 4: Explaining why this does not contradict Remark 10.1.10

: In Remark 10.1.10, it is mentioned that for a complete control system, the reachable set is closed. However, this statement assumes that the control inputs are piecewise continuous. In our case, the control input \(\omega_k\) that we've used to show that \(\mathcal{R}^{1}(0)\) is not closed is not piecewise continuous, as the switching frequency increases without bound. Thus, the result from Remark 10.1.10 does not apply to this particular control input, and there is no contradiction. In conclusion, we have shown that the given nonlinear system is complete, as solutions exist for every initial state and every control. We have also shown that \(\mathcal{R}^{1}(0)\) is not a closed set and explained why this does not contradict the discussion in Remark 10.1.10.

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Most popular questions from this chapter

Chapter 10

Let $\mathcal{R}^{\leq T}\left(x^{0}\right):=\bigcup_{0 \leq t \leq T} \mathcal{R}^{t}\left(x^{0}\right)\(. Show that, for any \)x\( and \)T, \mathcal{R}^{\leq T}\left(x^{0}\right)$ is: \- connected, \- compact, \- but not necessarily convex. (Hint: (For compactness.) If $x_{k}^{\mathrm{f}} \in \mathcal{R}^{t_{k}}\left(x^{0}\right)\(, you may assume that \)t_{k} \backslash T\( or \)t_{k} \backslash T\( for some \)T$ (why?). Then, restrict or extend controls \(\omega_{k}\) to \([0, T]\). Finally, use a compactness argument as in the proof that \(\mathcal{R}^{T}\left(x^{0}\right)\) is compact.)

Chapter 10

(a) Consider a controllable single-input system \(\dot{x}=A x+b u\). Show that there is some \(\delta>0\) with the following property: for each $\gamma \neq 0\(, the function \)t \mapsto \gamma^{\prime} e^{-t A} b\( has at most \)n-1$ zeros in the interval \([0, \delta]\). (b) Conclude that, for controllable single-input systems, with \(\mathcal{U}=[\underline{u}, \bar{u}]\), there is some \(\delta>0\) so that, whenever \(x^{0}\) and \(x^{\mathrm{f}}\) are so that \(x^{\mathrm{f}}\) can be reached in time at most \(\delta\) from \(x^{0}\), the time optimal control steering \(x^{0}\) to \(x^{f}\) is piecewise constant, taking values \(\underline{u}\) or \(\bar{u}\) in at most \(n\) intervals. (c) Conclude further that, if \(T\) is the minimum time for steering \(x^{0}\) to \(x^{\mathrm{f}}\), then the time optimal control steering a state \(x^{0}\) to a state \(x^{\mathrm{f}}\) is piecewise constant, with at most \(T n / \delta\) switches. (Hint: (For (a).) Assume the property is false on \([0,1 / k]\). Note that one may restrict attention to \(\gamma\) 's so that \(\|\gamma\|=1\). What can be said about the zeros of derivatives of \(\gamma^{\prime} e^{-t A} b\) ? Take limits \(k \rightarrow \infty\).)

Chapter 10

For any metric space \(M\), we use \(K(M)\) to denote the family of all nonempty compact subsets of \(M\), and define $$ D\left(C_{1}, C_{2}\right):=\max \left\\{\max _{x \in C_{1}} d\left(x, C_{2}\right), \max _{x \in C_{2}} d\left(x, C_{1}\right)\right\\} . $$ Show that \(D\) defines a metric on \(\mathbb{K}\) (usually called the Hausdorff metric). Now consider a linear system \(\dot{x}=A x+B u\) and a fixed initial state \(x^{0} \in \mathbb{R}^{n}\). Show that the mapping $T \mapsto \mathcal{R}^{T}\left(x^{0}\right)\( is continuous as a map from \)\mathbb{R}$ into \(\mathbb{K}\left(\mathbb{R}^{n}\right)\).

Chapter 10

Consider the nonlinear, affine in controls, system \(\dot{x}=x^{2}+u\), with \(\mathcal{U}=[-1,1]\) (any other compact convex set \(\mathcal{U}\) could be used). Show that there are states \(x^{0}\) and times \(T\) for which \(\mathcal{R}^{T}\left(x^{0}\right)\) is not compact. Why does this not contradict the discussion in Remark 10.1.10?

Chapter 10

Let \(\mathcal{V} \subseteq \mathcal{U}\) be so that the convex hull co \((\mathcal{V})=\mathcal{U} .\) Show that for each $\omega \in \mathcal{L}_{\mathrm{u}}(0, T)\( there is some sequence \)\nu_{k} \stackrel{\mathrm{w}}{\rightarrow} \omega\( so that \)\nu_{k}(t) \in \mathcal{V}\( for all \)t \in[0, T]$. (Suggestion: you may want to argue as follows. First, show that \(\omega\) can be weakly approximated by piecewise constant controls, i.e., of the form \(\sum_{\text {finite }} I_{J_{i}} u_{i}\), for intervals \(J_{i} \subseteq[0, T]\) and elements \(u_{i} \in \mathcal{U} .\) Next argue that, on each interval \(J\), a constant control with value \(u=\sum_{i=1}^{r} \rho_{i} v_{i}\), with \(\sum \rho_{i}=1\), $v_{i} \in \mathcal{V}, \rho_{i} \geq 0\( for all \)i$, can be in turn weakly approximated by controls with values in \(\mathcal{V}\). For this last approximation, you may think first of the special case \(r=2\) : in that case, the sequence \(\omega_{k}\left(v_{1}-v_{2}\right)+v_{2}\) converges to $\rho v_{1}+(1-\rho) v_{2}\(, if \)\omega_{k}$ is the sequence constructed in Example 10.1.7.)

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