# 统计代写|最优控制作业代写optimal control代考|MA30061

#### Doug I. Jones

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

So far, we have shown the necessity of the maximum principle conditions for optimality. Next we prove a theorem that gives qualifications under which the maximum principle conditions are also sufficient for optimality. This theorem is important from our point of view since the models derived from many management science applications will satisfy conditions required for the sufficiency result. As remarked earlier, our technique for proving existence will be to display for any given model, a solution that satisfies both necessary and sufficient conditions. A good reference for sufficiency conditions is Seierstad and Sydsæter (1987).

We first define a function $H^{0}: E^{n} \times E^{m} \times E^{1} \rightarrow E^{1}$ called the derived Hamiltonian as follows:
$$H^{0}(x, \lambda, t)=\max _{u \in \Omega(t)} H(x, u, \lambda, t)$$
We assume that by this equation a function $u^{}(x, \lambda, t)$ is implicitly and uniquely defined. Given these assumptions we have by definition, $$H^{0}(x, \lambda, t)=H\left(x, u^{}, \lambda, t\right)$$

For our proof of the sufficiency of the maximum principle, we also need the derivative $H_{x}^{0}(x, \lambda, t)$, which by use of the Envelope Theorem can be given as
$$H_{x}^{0}(x, \lambda, t)=H_{x}\left(x, u^{}, \lambda, t\right):=\left.H_{x}(x, u, \lambda, t)\right|{u=u^{}}$$
To see this in the case when $u^{}(x, \lambda, t)$ is differentiable in $x$, let us differentiate (2.62) with respéct to $x$ : $$H{x}^{0}(x, \lambda, t)=H_{x}\left(x, u^{}, \lambda, t\right)+H_{u}\left(x, u^{}, \lambda, t\right) \frac{\partial u^{}}{\partial x}$$
To obtain (2.63) from (2.64), we need to show that the second term on the right-hand side of (2.64) vanishes, i.e.,
$$H_{u}\left(x, u^{}, \lambda, t\right) \frac{\partial u^{}}{\partial x}=0$$ for each $x$. There are two cases to consider. If $u^{}$ is in the interior of $\Omega(t)$, then it satisfies the first-order condition $H_{u}\left(x, u^{}, \lambda, t\right)=0$, thereby implying (2.65). Otherwise, $u^{}$ is on the boundary of $\Omega(t)$. Then, for each $i, j$, either $H_{u_{i}}=0$ or $\partial u_{i}^{} / \partial x_{j}=0$ or both. Once again, (2.65) holds. Exercise $2.25$ gives a specific instance of this case.

## 统计代写|最优控制作业代写optimal control代考|Solving a TPBVP by Using Excel

A number of examples and exercises found throughout this book involve finding a numerical solution to a two-point boundary value problem (TPBVP). In this section we will show how the GOAL SEEK function in Excel can be used for this purpose. We will solve the following example.
Example $2.8$ Consider the problem:
$$\max \left{J=\int_{0}^{1}-\frac{1}{2}\left(x^{2}+u^{2}\right) d t\right}$$

subject to
$$\dot{x}=-x^{3}+u, x(0)=5 .$$
Solution We form the Hamiltonian
$$H=-\frac{1}{2}\left(x^{2}+u^{2}\right)+\lambda\left(-x^{3}+u\right)$$
where the adjoint variable $\lambda$ satisfies the equation
$$\dot{\lambda}=x+3 x^{2} \lambda, \lambda(1)=0$$
Since $u$ is unconstrained, wee set $H_{u}=0$ to obtaiin $u^{*}=\lambda$. With this, the stateé equation (2.76) becomes
$$\dot{x}=-x^{3}+\lambda, x(0)=5$$
Thus, the TPBVP is given by the system of Eqs. (2.77) and (2.78).
A simple method to solve the TPBVP uses what is known as the shooting method, explained in the flowchart in Fig. 2.6.

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

$$H^{0}(x, \lambda, t)=\max {u \in \Omega(t)} H(x, u, \lambda, t)$$ 我们假设通过这个方程一个函数 $u(x, \lambda, t)$ 是隐含且唯一定义的。鉴于这些假设，我们根据 定义, $$H^{0}(x, \lambda, t)=H(x, u, \lambda, t)$$ 为了证明最大原理的充分性，我们还需要导数 $H{x}^{0}(x, \lambda, t)$, 通过使用包絡定理可以给出
$$H_{x}^{0}(x, \lambda, t)=H_{x}(x, u, \lambda, t):=H_{x}(x, u, \lambda, t) \mid u=u$$

$$H x^{0}(x, \lambda, t)=H_{x}(x, u, \lambda, t)+H_{u}(x, u, \lambda, t) \frac{\partial u}{\partial x}$$

$$H_{u}(x, u, \lambda, t) \frac{\partial u}{\partial x}=0$$

## 统计代写|最优控制作业代写optimal control代考|Solving a TPBVP by Using Excel

\max \eft $\left{J=\backslash\right.$ int_{ ${0} \wedge{1}-\backslash \operatorname{frac}{1}{2} \backslash \operatorname{lft}\left(\mathrm{x}^{\wedge}{2}+\mathrm{u}^{\wedge}{2} \backslash\right.$ right) $\mathrm{d} t \backslash$ right $}$

$$\dot{x}=-x^{3}+u, x(0)=5 .$$

$$H=-\frac{1}{2}\left(x^{2}+u^{2}\right)+\lambda\left(-x^{3}+u\right)$$

$$\dot{\lambda}=x+3 x^{2} \lambda, \lambda(1)=0$$

$$\dot{x}=-x^{3}+\lambda, x(0)=5$$

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