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Problem 4

# 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 $\left.-1 .\right)$

Expert verified
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.
See the step by step solution

## 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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