Posted on Categories:Convex Analysis and Optimal Control, 凸分析和最优控制, 数学代写

数学代写|凸分析和最优控制代写Convex Analysis and Optimal Control代考|EECS598 Preliminaries

avatest™

avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

数学代写|凸分析和最优控制代写Convex Analysis and Optimal Control代考|Preliminaries

We recall first an abstract result which will be used in the proof of Theorem 1 . Let $X$ be a Banach space, let $Y$ and $W$ be linear normed spaces, and let $Z=W \times Y$. Consider a map $\Sigma$ acting from $Z$ to the subsets of $X$ and defined by
$$\Sigma(w, y)={x \in X \mid y \in \psi(x, w)+\mathcal{F}(x)}$$
where $\psi: X \times W \rightarrow Y$ is a function and $\mathcal{F}: X \rightarrow Y$ is a set-valued map with closed graph. Let $\left(w^, y^, x^\right) \in$ graph $\Sigma$. Assume that $\psi$ is Fréchet differentiable with respect to $x$ in a neighborhood of $\left(x^, w^\right)$ and its derivative $\nabla_x \psi$ is continuous around $\left(x^, w^\right)$; moreover, $\psi$ is Lipschitz continuous in $w$, uniformly in $x$, around $\left(x^, w^\right)$. Along with (2.1) consider the following map obtained by a linearization of $\psi$ : $$\Lambda(y)=\left{x \in X \mid y \in \psi\left(x^, w^\right)+\nabla_x \psi\left(x^, w^\right)\left(x-x^\right)+\mathcal{F}(x)\right} .$$
Our analysis is based on the following result from [1], for another proof see [4]:
Proposition 1. Let $L$ be a positive number. Then for the maps $\Sigma$ and $\Lambda$ defined in (2.1) and (2.2) the following are equivalent:
(j) $\Lambda$ is locally single-valued and lipschitzian at the point $\left(y^, x^\right)$ with a Lipschitz constant any number $\widehat{L}>L$
(jj) $\Sigma$ is locally single-valued and lipschitzian at the point $\left(w^, y^, x^*\right)$ with a Lipschitz constant any number $\widehat{L}>L$.

We apply Proposition 1 to the variational inequality $(\mathrm{VI})_p$. The linearization of (1.1) – (1.4) defining the map $\Lambda$ is given by the following linear variational inequality which we denote by $(\mathrm{LVI})\delta$ : \begin{aligned} &\dot{y}(t)-A(t) y(t)-B(t) v(t)+a_1(t)=0, \quad y(0)=0, \ &\Theta(t) v(t)+a_2(t) \in \mathcal{N}{\boldsymbol{R}{+}^k}(\mu(t)), \ &\dot{r}(t)+A^T(t) r(t)+Q{11}(t) y(t)+Q_{12}(t) v(t)+a_3(t)=0, \quad r(1)=0, \ &B(t)^T r(t)+Q_{22}(t) v(t)+Q_{12}(t)^T y(t)+\Theta(t)^T \mu(t)+a_4(t)=0, \end{aligned}
for a.e. $t \in[0,1]$, where the parameter $\delta:=\left(a_1, a_2, a_3, a_4\right) \in \Delta:=L^{\infty}\left(0,1 ; \mathbb{R}^n\right) \times$ $L^{\infty}\left(0,1 ; \mathbb{R}^k\right) \times L^{\infty}\left(0,1 ; \mathbb{R}^n\right) \times L^{\infty}\left(0,1 ; \mathbb{R}^m\right)$. The reference value $\delta_0=\left(a_{01}, a_{02}, a_{03}, a_{04}\right)$ is:
\begin{aligned} &a_{01}(t)=-\left(\dot{x}0(t)-A x_0(t)-B u_0(t)\right) \ &a{02}(t)=g\left(u_0(t), 0\right)-\Theta(t) u_0(t) \ &a_{03}(t)=-\left(\dot{q}0(t)+A^T(t) q_0(t)+Q{11}(t) x_0(t)+Q_{12}(t) u_0(t)\right) \ &a_{04}(t)=-\left(B(t)^T q_0(t)+Q_{22}(t) u_0(t)+Q_{12}(t)^T x_0(t)+\Theta(t)^T \nu_0(t)\right) \end{aligned}

数学代写|凸分析和最优控制代写Convex Analysis and Optimal Control代考|Proof of the implication (ii) =^ (i)

Let us assume that the Independence and the Coercivity conditions hold. We start with a proof of (I). Expressing $y_\delta$ and $r_\delta$ as functions of $\delta, v_\delta$ and $\mu_\delta$ from (2.3) and (2.5), respectively, substituting in (2.6) and using the definition of the map $\mathcal{M}$ in (2.8), we obtain that, for a given $\delta \in \Delta$, if $\left(y_\delta, v_\delta, r_\delta, \mu_\delta\right)$ is a solution of $(\mathrm{LVI})\delta$, then $\left(v\delta, \mu_\delta\right)$ is an $L^{\infty}$ in $t$ solution of the following variational inequality:
\begin{aligned} &(\mathcal{M} v)(t)+Q_{22}(t) v(t)+\Theta(t)^T \mu(t)+\zeta\left(a_1, a_3\right)(t)+a_4(t)=0, \ &\Theta(t) v(t)+a_2(t) \in \mathcal{N}{\boldsymbol{R}{+}^k}(\mu(t)) \end{aligned}
for a.e. $t \in[0,1]$, where $\zeta: L^{\infty}\left(0, T ; \mathbb{R}^n\right) \times L^{\infty}\left(0, T ; \mathbb{R}^n\right) \rightarrow L^{\infty}\left(0, T ; \mathbb{R}^m\right)$ is a linear map and for any $a_1, a_3$ in $L^{\infty}, \zeta\left(a_1, a_3\right)(\cdot)$ is a $L^{\infty}$ function in $[0,1]$. Observe that (3.1) can be equivalently written as a variational inequality of the form
\begin{aligned} &\left\langle(\mathcal{M} v)(t)+Q_{22}(t) v(t)+\zeta\left(a_1, a_3\right)(t)+a_4(t), w-v(t)\right\rangle \geq 0 \ &\text { for all } w \text { in the set } D(t):=\left{w \in \mathbb{R}^m \mid \Theta(t) w+a_2(t) \leq 0\right} . \end{aligned}
67

数学代写|凸分析和最优控制代写Convex Analysis and Optimal Control代 考|Preliminaries

$$\Sigma(w, y)=x \in X \mid y \in \psi(x, w)+\mathcal{F}(x)$$

〈left 的分隔符缺失或无法识别

(jj) $\Sigma$ 在该点是局部单值和lipschitzian $\left(w^{\prime} y^{\prime} x^*\right)$ 具有 Lipschitz 常数的任意数 $\widehat{L}>L$.

$L^{\infty}\left(0,1 ; \mathbb{R}^k\right) \times L^{\infty}\left(0,1 ; \mathbb{R}^n\right) \times L^{\infty}\left(0,1 ; \mathbb{R}^m\right)$. 参考值 $\delta_0=\left(a_{01}, a_{02}, a_{03}, a_{04}\right)$ 是:
$$a_{01}(t)=-\left(\dot{x} 0(t)-A x_0(t)-B u_0(t)\right) \quad a 02(t)=g\left(u_0(t), 0\right)-\Theta(t) u_0(t) a_{03}(t)=-\left(\dot{q} 0(t)+A^T(t) q_0(t)+Q 11(t) x_0(t)+Q_{12}(t) u_0(t)\right) \quad a_{04}(t)=$$

数学代写|凸分析和最优控制代写Convex Analysis and Optimal Control代考|Proof of the implication (ii) =^ (i)

(2.6) 并使用映射的定义 $\mathcal{M}$ 在 (2.8) 中，我们得到，对于给定的 $\delta \in \Delta$ ，如果 $\left(y_\delta, v_\delta, r_\delta, \mu_\delta\right)$ 是一个解䦼方宓(LVI) $\delta$ ，然后
$$(\mathcal{M} v)(t)+Q_{22}(t) v(t)+\Theta(t)^T \mu(t)+\zeta\left(a_1, a_3\right)(t)+a_4(t)=0, \quad \Theta(t) v(t)+a_2(t) \in \mathcal{N} \boldsymbol{R}+{ }^k(\mu(t))$$

〈left 的分隔符缺失或无法识别

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。