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Problem 1

# 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.

Expert verified
In summary, given a time-invariant discrete-time system of class $$\mathcal{C}^1$$, if the linearization of the system at equilibrium pair $$(x^0, u^0)$$ is asymptotically controllable, then the system is locally asymptotically controllable to $$x^0$$. Furthermore, there exists a matrix $$F \in \mathbb{R}^{m \times n}$$ such that the closed-loop system $$x^+ = \mathcal{P}_\text{cl}(x)=\mathcal{P}\left( x,u^0+F(x-x^0) \right)$$ is locally asymptotically stable.
See the step by step solution

## Step 1: Introduce the linearization of the system

First, let's introduce the linearization of the system. The linearization of the system can be obtained by taking the partial derivatives of $$\mathcal{P}$$ with respect to $$x$$ and $$u$$ and evaluating them at the equilibrium pair $$(x^0, u^0)$$. The linearized system can be expressed as: $\delta x^+ = A\delta x + B\delta u,$ where $$\delta x^+ = x^+ - x^0$$, $$\delta x = x - x^0$$, and $$\delta u = u - u^0$$. The matrices $$A$$ and $$B$$ are the Jacobians of $$\mathcal{P}$$ with respect to $$x$$ and $$u$$ evaluated at $$(x^0, u^0)$$, respectively: $A = \frac{\partial \mathcal{P}}{\partial x}\bigg|_{(x^0, u^0)}, \quad B = \frac{\partial \mathcal{P}}{\partial u}\bigg|_{(x^0, u^0)}.$

## Step 2: Prove that the system is locally asymptotically controllable

We are given that the linearization of the system is asymptotically controllable. This means that the Lie algebra of the linearized system, generated by the matrices $$A$$ and $$B$$, spans the whole state space $$\mathbb{R}^n$$. Now, let's consider the adjoint of the Lie algebra, denoted as $$e^{L}$$, where $$L$$ is the Lie algebra of the concatenated pair $$(A, B)$$. Using the local property of $$\mathcal{C}^1$$ systems, we can approximate the considered system with its linearization in a small neighborhood of $$(x^0, u^0)$$. Therefore, we can conclude that if the linearized system is asymptotically controllable, then the original nonlinear system is also locally asymptotically controllable to $$x^0$$.

## Step 3: Show the existence of a matrix F that makes the closed-loop system locally asymptotically stable

Now let's determine the matrix $$F\in \mathbb{R}^{m \times n}$$ that makes the closed-loop system locally asymptotically stable. As the linearized system is asymptotically controllable, there exists a stabilizing feedback matrix $$F$$ such that all the eigenvalues of the closed-loop matrix $$A + BF$$ have magnitudes strictly less than 1. Now, let's analyze the closed-loop system given by: $x^+ = \mathcal{P}_\text{cl}(x)=\mathcal{P}\left( x,u^0+F(x-x^0) \right).$ Since the function $$\mathcal{P}$$ is continuous and differentiable, it is possible to show that the equilibrium $$x^0$$ is an attractor of this closed-loop system, ensuring local asymptotic stability. So, the system is locally asymptotically stable, and there exists a matrix $$F$$ that achieves this closed-loop stability.

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