## 统计代写|最优控制作业代写optimal control代考|MAGIC106

2022年7月7日

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## 统计代写|最优控制作业代写optimal control代考|A Maximum Principle for Problems with Mixed Inequality Constraints

We will state the maximum principle for optimal control problems with mixed inequality constraints without proof. For further details see Pontryagin et al. (1962), Hestenes (1966), Arrow and Kurz (1970), Hadley and Kemp (1971), Bensoussan et al. (1974), Feichtinger and Hartl (1986), Seierstad and Sydsæter (1987), and Grass et al. (2008).

Let the system under consideration be described by the following vector differential equation
$$\dot{x}=f(x, u, t), x(0)=x_{0}$$
given the initial conditions $x_{0}$ and a control trajectory $u(t), t \in[0, T], T>0$, where $T$ can be the terminal time to be optimally determined or given as a fixed positive number. Note that in the above equation, $x(t) \in E^{n}$ and $u(t) \in E^{m}$, and the function $f: E^{n} \times E^{m} \times E^{1} \rightarrow E^{n}$ is assumed to be continuously differentiable.
Let us consider the following objective:
$$\max \left{J=\int_{0}^{T} F(x, u, t) d t+S[x(T), T]\right}$$
where $F: E^{n} \times E^{m} \times E^{1} \rightarrow E^{1}$ and $S: E^{n} \times E^{1} \rightarrow E^{1}$ are continuously differentiable functions and where $T$ denotes the terminal time. Depending on the situation being modeled, the terminal time $T$ may be given or to be determined. In the case when $T$ is given, the function $S(x(T), T)$ should be viewed as merely a function of the terminal state, and can be revised as $S(x(T))$.

Next we impose constraints on state and control variables. Specifically, for each $t \in[0, T], x(t)$ and $u(t)$ must satisfy
$$g(x, u, t) \geq 0, t \in[0, T],$$
where $g: E^{n} \times E^{m} \times E^{1} \rightarrow E^{q}$ is continuously differentiable in all its arguments and must contain terms in $u$. An important special case is that of controls having an upper bound that depends on the current state, i.e., $u(t) \leq M(x(t)), t \in[0, T]$,which can be written as $M(x)-u \geq 0$. Inequality constraints without terms in $u$ will be introduced later in Chap.

## 统计代写|最优控制作业代写optimal control代考|Sufficiency Conditions

In this section we will state, without proof, a number of sufficiency results. These results require the concepts of concave and quasiconcave functions.

Recall from Sect. $1.4$ that with $D \subset E^{n}$, a convex set, a function $\psi: D \rightarrow E^{1}$ is concave, if for all $y, z \in D$ and for all $p \in[0,1]$,
$$\psi(p y+(1-p) z) \geq p \psi(y)+(1-p) \psi(z)$$
The function $\psi$ is quasiconcave if $(3.24)$ is relaxed to
$$\psi(p y+(1-p) z) \geq \min {\psi(y), \psi(z)}$$
and $\psi$ is strictly concave if $y+z$ and $p \in(0,1)$ and (3.24) holds with a strict inequality. Furthermore, $\psi$ is convex, quasiconvex, or strictly convex if $-\psi$ is concave, quasiconcave, or strictly concave, respectively. Note that linearity implies both concavity and convexity, and concavity implies quasiconcavity. For further details on the properties of such functions, see Mangasarian (1969).

We can now state a sufficiency result concerning the problem with mixed constraints stated in (3.7). For this purpose, let us define the maximized Hamiltonian
$$H^{0}(x, \lambda, t)=\max _{{u \mid g(x, u, t) \geq 0}} H(x, u, \lambda, t)$$
Theorem 3.1 Let $\left(x^{}, u^{}, \lambda, \mu, \alpha, \beta\right)$ satisfy the necessary conditions in (3.12). If $H^{0}(x, \lambda(t), t)$ is concave in $x$ at each $t \in[0, T], S$ in $(3.2)$ is concave in $x, g$ in (3.3) is quasiconcave in $(x, u)$, a in (3.4) is quasiconcave in $x$, and $b$ in (3.5) is linear in $x$, then $\left(x^{}, u^{}\right)$ is optimal.

## 统计代写|最优控制作业代写optimal control代考|A Maximum Principle for Problems with Mixed Inequality Constraints

Hadley 和 Kemp (1971)、Bensoussan 等人。(1974)、Feichtinger 和 Hartl (1986)、
Seierstad 和 Sydsæter (1987)，以及 Grass 等人。（2008 年)。

$$\dot{x}=f(x, u, t), x(0)=x_{0}$$

\max Veft ${=\backslash$ int_{ ${0} \wedge{T} F(x, u, t) d t+S[x(T), T] \backslash$ right $}$

$$g(x, u, t) \geq 0, t \in[0, T]$$

## 统计代写|最优控制作业代写optimal control代考|Sufficiency Conditions

$$\psi(p y+(1-p) z) \geq p \psi(y)+(1-p) \psi(z)$$

$$\psi(p y+(1-p) z) \geq \min \psi(y), \psi(z)$$

$$H^{0}(x, \lambda, t)=\max _{u \mid g(x, u, t) \geq 0} H(x, u, \lambda, t)$$

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