黎曼流形上具有凸约束最优控制问题的必要条件

Necessary Conditions for Optimal Control Problems with Convex Constraints on Riemannian Manifolds

摘要: 本文讨论黎曼流形上的最优控制问题，主要考虑的是初始状态固定，终止状态无约束，控制具有凸约束的情况。对于这种情况，通过引入适当的对偶方程，且它依赖于流形中的曲率张量，从而建立二阶必要条件。本文给出了黎曼流形上具有凸约束最优控制问题的一阶和二阶必要条件。

Abstract: This paper mainly discusses the optimal control problem on the Riemannian manifold. The main considerations are the fixed initial state, the endpoint is free, and the control set is assumed to be convex in a Euclidian space. In this case, a second-order necessary condition is established by introducing an appropriate dual equation, which depends on the curvature tensor in the manifold. In this paper, first- and second-order necessary conditions for optimal control problems with convex constraints on Riemannian manifolds are given.

1. 引言

设 $n\in N,M$ 是一个具有黎曼度量g的完全单连通的黎曼流形。设 $\nabla$ 是M上与g相关的Levi-Civita

联络， $\rho \left(\cdot ,\cdot \right)$ 是M上的距离函数， $T_{x}M$ 是M在 $x\in M$ 处的切空间， $T_x^{*}M$ 为该余切空间。我们用 $\langle \cdot ,\cdot \rangle$ 和 $\left|\cdot \right|$ 分别表示与g相关的内积和 $T_{x}M$ 上的范数。另外，分别用 $TM={\displaystyle \underset{x\in M}{\cup }{T}_{x}M},T_x^{*}M\equiv {\displaystyle \underset{x\in M}{\cup }{T}_x^{*}M}$ 和 $C^{\infty }\left(M\right)$ 表示

流形上M的切丛，余切丛以及光滑函数集。

给定 $T>0$，凸集U， $y_0\in M$ 以及两个函数：

$f:\left[0,T\right]\times M\times U\to TM,f^0:\left[0,T\right]\times M\times U\to R$

我们来考虑如下控制系统：

$\{\begin{array}{l}\stackrel{\dot{}}{y}\left(t\right)=f\left(t,y\left(t\right),u\left(t\right)\right)t\in \left[0,T\right],y\left(t\right)\in M\hfill \\ y\left(0\right)=y_0\hfill \end{array}$ (0.1)

这里 $y(\cdot )$ 和 $u(\cdot )$ 分别表示状态变量和控制变量。与(0.1)相关的性能指标值：

$J\left(u(\cdot )\right)={\displaystyle {\int }_0^T{f}^0\left(t,y\left(t\right),u\left(t\right)\right)\text{d}t}$ (0.2)

其中(0.2)里的 $u(\cdot )$ 属于以下允许控制集合：

$\Omega \equiv \left\{u(\cdot ):\left[0,T\right]\to U;u(\cdot )是可测的\right\}$ (0.3)

问题(P)

找到一个 $\bar{u}(\cdot )\in \Omega$，使得：

$J\left(\bar{u}(\cdot )\right)=\underset{u(\cdot )\in \mathfrak{A}}{\inf }J\left(u(\cdot )\right)$ (0.4)

成立。

对于以上问题(P)，我们称 $\bar{u}(\cdot )$ 为最优控制， $\bar{y}(\cdot )$ 是(0.1)中以 $\bar{u}(\cdot )$ 为控制的解，称为最优轨迹， $\left(\bar{u}(\cdot ),\bar{y}(\cdot )\right)$ 为最优对。显然，该最优控制问题可以被视为状态被约束在欧氏空间的子流形上最优控制问题。控制问题的核心问题之一是找到最优控的必要和充分性条件，本文主要讨论其必要条件。

有许多工作讨论了状态变量取值于欧氏空间中的最优控制问题。例如文献( [1] [2] [3] )给出了一阶必要条件的例子，文献( [1] [4] [5] [6] [7] )给出了二阶必要条件的例子。正如微积分中的那样，即使对于某些状态约束的情况，也可以得出经典控制论 [1] 中所做的最优控制的一阶必要条件。然而，对于某些最优控制问题，一阶必要条件不能体现最优控制的足够信息。在这种情况下，一阶必要条件不能为理论分析和数值计算提供足够的信息，因此需要研究二阶(或更高阶)必要条件条件。

对于一些状态被约束在流形上的控制系统，也有一些文献讨论关于最优控制问题的一阶和二阶必要条件。然而，与平坦空间的情况相比，我们认为弯曲空间中的还有一些最优条件是非常不完整的。将平坦空间中的相关结果推广到弯曲空间的假设将是非常有趣的(例如，在研究 [8] 和 [9] 中针对二阶最佳条件)。在参考文献 [9] 中，所提的控制问题的控制集为 ${ℝ}^m$ 中的一般集合，对控制作针状变分，而本文是一个凸集，因此做的是凸变分， [9] 中的二阶必要条件是针对使得庞特里亚金形最大值原理退化的控制所满足的条件。本文所得到的二阶必要条件，是针对庞特里亚金型最大值原理在古典意义下退化的控制所满足的条件。参考文献 [9] 所考虑的控制集合是开集的情况。本文结果对文献 [9] 和文献 [10] 有重要补充。对不同控制给予不同的条件得到的结果的强弱也是不同的，每种方法各有优势。因此在弯曲空间中依然还有事情要做。

2. 准备

2.1. 指数映射

读者可以在参考文献( [8] [9] [11] [12] [13] [14] )看到更多的细节。

一个流形M上的可导曲线 $\gamma \left(t\right),t\in \left[0,\alpha \right)\left(\alpha >0\right)$ 如果满足

${\nabla }_{\stackrel{\dot{}}{\gamma }\left(t\right)}\stackrel{\dot{}}{\gamma }\left(t\right)=0,t\in \left[0,\alpha \right)$ (1.1)

则被称为测地线。

假设 $x\in M$ 是固定的。对任意的 $v\in T_{x}M$，这里存在唯一的测地线 ${\gamma }_v(\cdot )$ 满足 ${\gamma }_v\left(0\right)=x,{\stackrel{\dot{}}{\gamma }}_v\left(0\right)=v$。设 $\left[0,L_v\right]$ 是定义在 ${\gamma }_v(\cdot )$ 上的最大间距。设 $O_x\subset T_{x}M$ 为向量v所在集合使得 $L_v>1$。则定义指数映射：

$ex{p}_x:O_x\to M,ex{p}_v={\gamma }_v(1)$

很显然 $O_x$ 是原点 $O\in T_{x}M$ 的一个邻域， $ex{p}_x$ 是将经过原点O， $T_{x}M$ 中的直线段映射到M中过x点的测地线上。对于任意的 $v\in T_{x}M$，定义 $ex{p}_x$ 在点v的导数，表示为：

${dex{p}_x|}_v:T_v{T}_{x}M\to T_{ex{p}_{x}v}M$ (1.2)

这里 $T_v{T}_{x}M$ 表示流形 T x M $T_{x}M$ 在点 $v\in T_{x}M$ 处的切空间。

给定一个 $\epsilon >0$，

$\begin{array}{l}B\left(0,\epsilon \right)\equiv \left\{v\in T_{x}M;\left|v\right|<\epsilon \right\}\\ B_x\left(\epsilon \right)\equiv \left\{y\in M;\rho \left(x,y\right)<\epsilon \right\}\end{array}$ (1.3)

我们称 $i\left(x\right)\equiv \sup \left\{\epsilon >0,映射ex{p}_x:B\left(O,\epsilon \right)\to B_x\left(\epsilon \right)是微分同胚的\right\}$ 为点x的单射半径。

2.2. 平行移动

对于一些细节的定义可以参考文献( [8] [9] [13] [14] )。

对于任意的 $x\in M$ 及 $r,s\in N$，所谓M在x处的一个 $\left(r,s\right)$ 型张量 $\Gamma$ 是指： $r+s$ 重线性映射：

$F:\underset{r}{\underset{︸}{T_x^{*}M\times \cdots \times T_x^{*}M}}\times \underset{s}{\underset{︸}{T_{x}M\times \cdots \times T_{x}M}}\to R$

其中r称为 $\Gamma$ 的反变阶数，s称为 $\Gamma$ 的协变阶数。 ${\Gamma }_s^r\left(x\right)$ 表示M在x点的所有 $\left(r,s\right)$ 型张量构成的集合，其中 $\Gamma \left(x\right)\in {\Gamma }_s^r\left(x\right)$ 为该集合中的元素。 $\Gamma$ 在 $x\in M$ 点的范数的定义为：

$\begin{array}{l}\left|\Gamma \left(x\right)\right|=\sup \left\{\Gamma \left(x\right)\left(Y_1,\cdots ,Y_r,{\lambda }_1,\cdots ,{\lambda }_s\right)\right\};\\ Y_j\in T_x^{*}M,{\lambda }_l\in T_{x}M,\left|Y_j\right|\le 1,\left|{\lambda }_l\right|\le 1,j=1,\cdots ,r,l\in \left\{1,\cdots ,s\right\}\end{array}$ (1.4)

设 $\gamma :\left[0,L_v\right)\to M$ 是一个可导曲线，有 $\gamma \left(0\right)=x\in M,\gamma \left(L\right)=y$，且 $L>0$。给定一个向量 $v\in T_{x}M$，这里存在一个唯一沿着 $\gamma$ 的向量场X，它满足：

${\nabla }_{\stackrel{\dot{}}{\gamma }\left(t\right)}X=0,\forall s\in \left[0,l_v\right],X\left(0\right)=v$

映射 $T_{x}M∍v\mapsto X\left(\gamma \left(l\right)\right)\in T_{y}M$ 是 $T_{x}M$ 和 $T_{y}M$ 之间的线性等距同构。我们称这个映射沿着 $\gamma$ 的平行移动，表示为 $L_{xy}^{\gamma }v$。

2.3. 主要引理

(该引力主要引用参考文献 [9]，具体内容可参考( [9] , p.34)。

引理1 假设满足猜想(A1)-(A3)，对任何 $\bar{u}(\cdot )$ 和 $v(\cdot )$ 满足 ${\Vert v(\cdot )-\bar{u}(\cdot )\Vert }_{\infty }<\infty$，设：

$u^{\epsilon }(\cdot )=\bar{u}(\cdot )+\epsilon \left(v(\cdot )-\bar{u}(\cdot )\right),\epsilon >0$

其中 $y^{\epsilon }(\cdot )$ 是(0.1)中与 $u^{\epsilon }(\cdot )$ 相关的解。特别的， $\bar{y}(\cdot )$ 是(0.1)中与 $\bar{u}(\cdot )$ 相关的解。对于任意充分小的 $\epsilon$，我们可以定义一个沿着 $\bar{y}(\cdot )$ 的向量场：

$V^{\epsilon }\left(t\right)={\exp }_{\bar{y}\left(t\right)}^{-1}{y}^{\epsilon }\left(t\right),t\in \left[0,T\right]$ (1.5)

设 $V(\cdot )$ 和 $Y(\cdot )$ 分别表示沿着 $\bar{y}(\cdot )$ 的向量场分别满足以下方程组：

$\{\begin{array}{l}{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}V\left(Z\right)={\nabla }_{x}f\left[t\right]\left(Z,V\left(t\right)\right)+{\nabla }_{u}f\left[t\right]\left(Z,v\left(t\right)-\bar{u}\left(t\right)\right)\hfill \\ V\left(0\right)=0\hfill \end{array}$ (1.6)

以及

$\{\begin{array}{l}\begin{array}{l}{\nabla }_{\stackrel{\dot{}}{\bar{y}}}Y\left(Z\right))={\nabla }_{x}f\left[t\right]\left(Z\left(t\right),Y\left(t\right)\right)+{\nabla }_x{\nabla }_{u}f\left[t\right]\left(Z\left(t\right),v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\\ -\frac{1}{2}R\left(\tilde{Z}\left(t\right),V\left(t\right),\bar{y}\left(t\right),V\left(t\right)\right)+\frac{1}{2}{\nabla }_x^{2}f\left[t\right]\left(Z\left(t\right),V\left(t\right),V\left(t\right)\right)\\ +\frac{1}{2}{\nabla }_u^{2}f\left[t\right]\left(Z\left(t\right),v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)a.e.t\in \left(0,T\right],\forall Z\in T^{*}M\end{array}\hfill \\ Y\left(0\right)=0\hfill \end{array}$ (1.7)

则我们有：

$V^{\epsilon }\left(t\right)=\epsilon V\left(t\right)+{\epsilon }^{2}Y\left(t\right)+o\left({\epsilon }^2\right)\forall t\in \left(0,T\right]$ (1.8)

3. 主要结论

对于问题(P)给出如下假设：

(A1)映射： $f:\left[0,T\right]\times M\times U\to TM$， $f^0:\left[0,T\right]\times M\times U\to R$ 关于t可测，关于u连续，且关于x是 $C^1$ 的。除此之外存在一个常数 $L>1$，使得对所有 $s\in \left[0,T\right]$， $u\in U$ 以及 $x_1,x_2\in M$ 有 $\rho \left(x_1,x_2\right)<\min \left\{i\left(x_1\right),i\left(x_2\right)\right\},i\left(x\right)$ 为x点的单射半径。这里 $x_0\in M$ 是一个固定点，使得以下等式成立：

$\begin{array}{l}\left|f^0\left(s,x_1,u\right)-f^0\left(s,x_2,u\right)\right|\le L\rho \left(x_1,x_2\right)\\ \left|L_{x_1{x}_2}f\left(s,x_1,u\right)-f\left(s,x_2,u\right)\right|\le L\rho \left(x_1,x_2\right)\\ \left|f^0\left(s,x_0,u\right)\right|\le L\\ \left|f\left(s,x_0,u\right)\right|\le L\end{array}$ (2.1)

(A2) 映射 $f\left(s,x,u\right)$ 和 $f^0\left(s,x,u\right)$ 在x是 $C^2$ 的。进一步，对于所有的 $x_1,x_2\in M$ 有 $\rho \left(x_1,x_2\right)<\min \left\{i\left(x_1\right),i\left(x_2\right)\right\},i\left(x\right)$，有：

$\begin{array}{l}\left|{\nabla }_x{f}^0\left(t,x_1,u\right)-L_{x_2{x}_1}{\nabla }_x{f}^0\left(t,x_2,u\right)\right|\le L\rho \left(x_1,x_2\right)\\ \left|{\nabla }_{x}f\left(t,x_1,u\right)-L_{x_2{x}_1}{\nabla }_{x}f\left(t,x_2,u\right)\right|\le L\rho \left(x_1,x_2\right)\end{array}$ (2.2)

其中 ${\nabla }_x{f}^0\left(t,x,u\right)$， ${\nabla }_{x}f\left(t,x,u\right)$ 分别为 $f^0\left(t,x,u\right)$， $f\left(t,x,u\right)$ 关于状态变量的协变导数。

(A3) $U\subset R^m$，是凸集。映射 $f\left(t,x,u\right)$ 和 $f^0\left(t,x,u\right)$ 在x是 $C^2$ 的。进一步，对所有的， $u_1,u_2\in U,\left(t,x\right)\in \left[0,T\right]\times M$。这里 ${\nabla }_u{f}^0\left(t,x,\cdot \right)$， ${\nabla }_{u}f\left(t,x,\cdot \right)$ 分别为 $f^0\left(t,x,u\right)$， $f\left(t,x,u\right)$ 关于控制变量的导数，有：

$\begin{array}{l}\left|f^0\left(t,x,u_1\right)-f^0\left(t,x,u_2\right)\right|\le L\left|u_1-u_2\right|\\ \left|f\left(t,x,u_1\right)-f\left(t,x,u_2\right)\right|\le L\left|u_1-u_2\right|\\ \left|{\nabla }_u{f}^0\left(t,x,u_1\right)-{\nabla }_u{f}^0\left(t,x,u_2\right)\right|\le L\left|u_1-u_2\right|\\ \left|{\nabla }_{u}f\left(t,x,u_1\right)-{\nabla }_{u}f\left(t,x,u_2\right)\right|\le L\left|u_1-u_2\right|\\ \left|{\nabla }_x{f}^0\left(t,x,u_1\right)-{\nabla }_x{f}^0\left(t,x,u_2\right)\right|\le L\left|u_1-u_2\right|\\ \left|{\nabla }_{x}f\left(t,x,u_1\right)-{\nabla }_{x}f\left(t,x,u_2\right)\right|\le L\left|u_1-u_2\right|\end{array}$ (2.3)

本文对于 $X\in TM$ (或者 $X\in T^{\ast }M$ )，我们用 $\tilde{X}$ 表示X的对偶切向量，Hamiltonian函数： $H:\left[0,T\right]\times T^{*}M\times U\to R$，定义为：

$H\left(t,x,\psi ,u\right)\equiv \psi \left(f\left(t,x,u\right)\right)-f^0\left(t,x,u\right)$

对任意的 $\left(t,x,\psi ,u\right)\in \left[0,T\right]\times T^{*}M\times U$。

我们固定一个控制 $\bar{u}(\cdot )\in \Omega$，其中 $\Omega$ 是在引言中的公式(0.3)给定。设是 $\bar{y}(\cdot )$ 是(0.1)中与 $\bar{u}(\cdot )$ 相关的解。为了简洁，我们表示：

$\left[t\right]=\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)$ (2.4)

其中对任意的 $t\in \left[0,T\right]$。

假设 $\psi \left(t\right)\in T_{\bar{y}\left(t\right)}^{*}M\left(t\in \left[0,T\right]\right)$ 是以下一阶对偶方程的解：

$\{\begin{array}{l}{\nabla }_{\stackrel{\dot{}}{\bar{y}}}\psi \left(t\right)=-{\nabla }_{x}f\left[t\right]\left(\psi \left(t\right),\cdot \right)+d_x{f}^0\left[t\right],a.e.t\in \left[0,T\right]\hfill \\ \psi \left(T\right)=0\hfill \end{array}$ (2.5)

其中 $d_x{f}^0$ 表示 $f^0$ 关于状态变量x的导数， ${\nabla }_{x}f\left[t\right]\left(\psi \left(t\right),\cdot \right)$ 是一个张量，定义为：

${\nabla }_{x}f\left[t\right]\left(\psi \left(t\right),X\left(\bar{y}\left(t\right)\right)\right)\equiv {\nabla }_{X\left(\bar{y}\left(t\right)\right)}f\left(t,\cdot ,\bar{u}(\cdot )\right)\psi (t)$

其中 $\forall X\in TM$。

3.1. 一阶必要条件

定理一 假设(A1)~(A3)成立， $U\subset R^m$ 是凸集， $\left(\bar{y}(\cdot ),\bar{u}(\cdot )\right)$ 为问题(P)的最优对，则：

${\nabla }_{u}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\le 0$ (2.6)

在 $\left[0,T\right]$ 上处处成立，其中 $\psi (\cdot )$ 是(2.5)的解， ${\nabla }_{u}H$ 表示H关于u求偏导。

3.2. 二阶必要条件

定理二 假设(A1)~(A3)成立， $U\subset R^m$ 是凸集， $\left(\bar{y}(\cdot ),\bar{u}(\cdot )\right)$ 为问题(P)的最优对，则使得(2.6)式成立的 $v(\cdot )$，下列不等式成立：

$\begin{array}{l}0\ge {\displaystyle {\int }_0^T[{\nabla }_x^{2}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(V\left(t\right),V\left(t\right)\right)}\\ +2{\nabla }_x{\nabla }_{u}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\\ +{\nabla }_u^{2}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\\ -R\left(\hat{\psi }\left(t\right),V\left(t\right),\bar{y}\left(t\right),V\left(t\right)\right)]\text{d}t\end{array}$ (2.7)

其中， $V(\cdot )$ 满足(1.6)。

4. 主要结论证明

4.1. 定理一的证明

任取 $v(\cdot )\in L^{\infty }\left(0,T;R^m\right)\cap \Omega$，则对任意的 $\epsilon >0$，定义：

$u^{\epsilon }(\cdot )\equiv \bar{u}(\cdot )+\epsilon \left(v(\cdot )-\bar{u}(\cdot )\right)\in \Omega$

以及记 $u^{\epsilon }(\cdot )$ 为控制(0.1)的解为 $y^{\epsilon }(\cdot )$，由引理1，当 $\epsilon >0$ 充分小，可定义：

$V^{\epsilon }\left(t\right)={\exp }_{\bar{y}\left(t\right)}^{-1}{y}^{\epsilon }\left(t\right),\forall t\in \left[0,T\right]$

现单位化 $V^{\epsilon }(\cdot )$ ：

${\hat{V}}^{\epsilon }(\cdot )=\{\begin{array}{l}\frac{V^{\epsilon }\left(t\right)}{\rho \left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)},\left|V^{\epsilon }\left(t\right)\right|\ne 0\\ 0\left|V^{\epsilon }\left(t\right)\right|=0\end{array}$ (3.1)

$\beta \left(\theta ;t\right)={\exp }_{\bar{y}\left(t\right)}\theta V^{\epsilon }\left(t\right),\theta \in \left[0,\rho \left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)\right]$ (3.2)

${\frac{\partial }{\partial \theta }|}_{\theta =0}\beta \left(\theta ;t\right)={dex{p}_{\bar{y}\left(t\right)}|}_O{\hat{V}}^{\epsilon }\left(t\right)={\hat{V}}^{\epsilon }\left(t\right)$ (3.3)

由 $\bar{u}(\cdot )$ 的最优性有：

$\begin{array}{c}0\le J\left(u^{\epsilon }(\cdot )\right)-J\left(\bar{u}(\cdot )\right)\\ ={\displaystyle {\int }_0^T\left[f^0\left(t,y^{\epsilon }\left(t\right),u^{\epsilon }(\cdot )\right)-f^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)\right]\text{d}t}\\ ={\displaystyle {\int }_0^T[f^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)-f^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)}\\ +f^0\left(t,y^{\epsilon }\left(t\right),u^{\epsilon }(\cdot )\right)-f^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)]\text{d}t\end{array}$

令：

$\begin{array}{l}{I}_1={\displaystyle {\int }_0^T\left[f^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)-f^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)\right]\text{d}t}\\ I_2={\displaystyle {\int }_0^T\left[f^0\left(t,y^{\epsilon }\left(t\right),u^{\epsilon }(\cdot )\right)-f^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)\right]\text{d}t}\end{array}$

则有：

$\begin{array}{c}{I}_1={\displaystyle {\int }_0^T\left[f^0\left(t,\beta \left(\theta ;t\right),\bar{u}\left(t\right)\right)-f^0\left(t,\beta \left(0;t\right),\bar{u}\left(t\right)\right)\right]\text{d}t}\\ ={\displaystyle {\int }_0^T\{{\frac{\partial }{\partial \theta }|}_{\theta =0}{f}^0\left(t,\beta \left(\theta ;t\right),\bar{u}\left(t\right)\right)\rho \left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)}\\ +\frac{1}{2}{\frac{{\partial }^2}{\partial {\theta }^2}|}_{\theta =0}{f}^0\left(t,\beta \left(\theta ;t\right),\bar{u}\left(t\right)\right){\rho }^2\left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)+o\left({\rho }^2\left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)\right)\}\text{d}t\\ ={\displaystyle {\int }_0^T\left\{{\nabla }_x{f}^0\left(t,\beta \left(\text{0};t\right),\bar{u}\left(t\right)\right)\left({\frac{\partial }{\partial \theta }|}_{\theta =0}\beta \left(\theta ;t\right)\right)\rho (\bar{y}\left(t\right),y^{\epsilon }(t)\right)}\end{array}$

$\begin{array}{l}+\frac{1}{2}{\nabla }_x^2{f}^0\left(t,\beta \left(0;t\right),\bar{u}\left(t\right)\right)\left({\frac{\partial }{\partial \theta }|}_{\theta =0}\beta \left(\theta ;t\right),{\frac{\partial }{\partial \theta }|}_{\theta =0}\beta \left(\theta ;t\right)\right){\rho }^2\left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)\\ \begin{array}{c}\\ \end{array}+o\left({\rho }^2\left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)\right)\}\text{d}t\\ ={\displaystyle {\int }_0^T\{{\nabla }_x{f}^0\left[t\right]\left({\hat{V}}^{\epsilon }\left(t\right)\right)\rho \left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)\begin{array}{c}\\ \end{array}}\\ +\frac{1}{2}{\nabla }_x^2{f}^0\left[t\right]\left({\hat{V}}^{\epsilon }\left(t\right),{\hat{V}}^{\epsilon }\left(t\right)\right){\rho }^2\left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)+o\left({\rho }^2\left(\bar{y}\left(t\right),y^{\epsilon }\left(t\right)\right)\right)\}\text{d}t\end{array}$

则：

$\begin{array}{c}{I}_2={\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }\frac{\partial }{\partial \theta }{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+\theta \left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)\text{d}\theta \text{d}t}}\\ ={\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+\theta \left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)}}\text{d}\theta \text{d}t\\ =\epsilon {\displaystyle {\int }_0^T{\nabla }_u}{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }[{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+\theta \left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)}}\\ -{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}\theta \text{d}t\end{array}$

$\begin{array}{l}=\epsilon {\displaystyle {\int }_0^T{\nabla }_u{f}^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t}\\ +\epsilon {\displaystyle {\int }_0^T\left[{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)-{\nabla }_u{f}^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)\right]}\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }{\displaystyle {\int }_0^{\theta }\frac{\partial }{\partial \tau }}}}{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+\tau \left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}\tau \end{array}$

记：

$I_2=\epsilon {\displaystyle {\int }_0^T{\nabla }_u{f}^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t}+I_3+I_4$

其中，

$\begin{array}{l}{I}_3=\epsilon {\displaystyle {\int }_0^T\left[{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)-{\nabla }_u{f}^0\left(t,\bar{y}\left(t\right),\bar{u}\left(t\right)\right)\right]}\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\\ I_4={\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }{\displaystyle {\int }_0^{\theta }\frac{\partial }{\partial \tau }}}}{\nabla }_u{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+\tau \left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}\tau \text{d}\theta \text{d}t\end{array}$

则：

$\begin{array}{c}{I}_4={\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }{\displaystyle {\int }_0^{\theta }{\nabla }_u^2{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+\tau \left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}\tau \text{d}\theta \text{d}t}}}\\ =\frac{{\epsilon }^2}{2}{\displaystyle {\int }_0^T{\nabla }_u^2{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\left(\epsilon -\tau \right)\text{d}\tau \text{d}t}\\ +{\displaystyle {\int }_0^T{\displaystyle {\int }_0^{\epsilon }{\nabla }_u^2{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)}}\\ -\frac{{\epsilon }^2}{2}{\displaystyle {\int }_0^T{\nabla }_u^2}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\end{array}$

由积分中值定理：

$\begin{array}{l}=\frac{{\epsilon }^2}{2}{\displaystyle {\int }_0^T{\nabla }_u^2{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t}\\ +\frac{{\epsilon }^2}{2}{\displaystyle {\int }_0^T[{\nabla }_u^2{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)+{\tau }_{\epsilon ,t}\left(v\left(t\right)-\bar{u}\left(t\right)\right)\right)}\\ +{\nabla }_u^2{f}^0\left(t,y^{\epsilon }\left(t\right),\bar{u}\left(t\right)\right)]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\end{array}$

再由Lebesgue控制收敛定理：

$=\frac{{\epsilon }^2}{2}{\displaystyle {\int }_0^T{\nabla }_u^2{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t}+o(\epsilon 2)$

对于 $I_3$，记：

$\hat{\beta }\left(\theta ;t\right)=ex{p}_{\bar{y}\left(t\right)}\theta V^{\epsilon }(t)$

其中： $\frac{\partial }{\partial \theta }\hat{\beta }\left(\theta ;t\right)=L_{\hat{\beta }\left(0\right)\hat{\beta }\left(\theta \right)}{\frac{\partial }{\partial \theta }|}_O\hat{\beta }\left(\theta ;t\right)=L_{\bar{y}\left(t\right)\hat{\beta }\left(\theta \right)}{V}^{\epsilon }\left(t\right)$。

则：

$\begin{array}{c}{I}_3=\epsilon {\displaystyle {\int }_0^T{\displaystyle {\int }_0^1\frac{\partial }{\partial \theta }{\nabla }_u{f}^0\left(t,\hat{\beta }\left(\theta ;t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}\theta \text{d}t}}\\ =\epsilon {\displaystyle {\int }_0^T{\displaystyle {\int }_0^T{\nabla }_x{\nabla }_u}}{f}^0\left(t,\hat{\beta }\left(\theta ;t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right),L_{\bar{y}\left(t\right)\hat{\beta }\left(\theta ;t\right)}{V}^{\epsilon }\left(t\right)\right)\text{d}\theta \text{d}t\\ =\epsilon {\displaystyle {\int }_0^T{\displaystyle {\int }_0^T{L}_{\hat{\beta }\left(\theta ;t\right)\bar{y}\left(t\right)}}}{\nabla }_x{\nabla }_u{f}^0\left(t,\hat{\beta }\left(\theta ;t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right),V^{\epsilon }\left(t\right)\right)\text{d}\theta \text{d}t\end{array}$

由Lebesgue控制收敛定理且将(1.8)带入上式得：

$\begin{array}{l}=\epsilon {\displaystyle {\int }_0^T{\nabla }_x{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V^{\epsilon }\left(t\right)\right)}\\ +\epsilon {\displaystyle {\int }_0^T{\displaystyle {\int }_0^1\left[L_{\hat{\beta }\left(\theta ;t\right)\bar{y}\left(t\right)}{\nabla }_x{\nabla }_u{f}^0\left(t,\hat{\beta }\left(\theta ;t\right)\bar{u}\left(t\right)\right)-{\nabla }_x{\nabla }_u{f}^0\left[t\right]\right]}}\left(v\left(t\right)-\bar{u}\left(t\right),V^{\epsilon }\left(t\right)\right)\text{d}\theta \text{d}t\\ ={\epsilon }^2{\displaystyle {\int }_0^T{\nabla }_x{\nabla }_u}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)+o(\epsilon 2)\end{array}$

综上：

$\begin{array}{c}0\le J\left(u^{\epsilon }(\cdot )\right)-J\left(\bar{u}(\cdot )\right)\\ =\epsilon {\displaystyle {\int }_0^T{\nabla }_x{f}^0\left[t\right]V\left(t\right)\text{d}t}\\ +{\epsilon }^2{\displaystyle {\int }_0^T\left[{\nabla }_x{f}^0\left[t\right]\left(Y\left(t\right)\right)+\frac{1}{2}{\nabla }_x^2{f}^0\left[t\right]\left(V\left(t\right),V\left(t\right)\right)\right]\text{d}t}+o\left({\epsilon }^2\right)\\ +\epsilon {\displaystyle {\int }_0^T{\nabla }_u}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\\ +{\epsilon }^2{\displaystyle {\int }_0^T{\nabla }_x{\nabla }_u}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\text{d}t\\ +\frac{{\epsilon }^2}{2}{\displaystyle {\int }_0^T{\nabla }_u^2}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t+o(\epsilon 2)\end{array}$

进一步有：

$0\le \frac{J\left(u^{\epsilon }(\cdot )\right)-J\left(\bar{u}(\cdot )\right)}{\epsilon }$

当 $\epsilon \to 0^{+}$ 时，

$0\le {\displaystyle {\int }_0^T\left[{\nabla }_x{f}^0\left[t\right]\left(V\left(t\right)\right)+{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\right]\text{d}t}$

将对偶方程(2.5)带入得到：

$0\le {\displaystyle {\int }_0^T\left({\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}\psi \left(t\right)+{\nabla }_{x}f\left[t\right]\left(\psi \left(t\right),V\left(t\right)\right)\right)\text{d}t+{\displaystyle {\int }_0^T{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\ge 0}}$

再由(1.6)得：

$\begin{array}{c}0\le {\displaystyle {\int }_0^T\left({\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}\psi \left(t\right)+{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}V\left(\psi \left(t\right)\right)\right)\text{d}t}-{\displaystyle {\int }_0^T{\nabla }_{u}f\left[t\right]\left(\psi \left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t}\\ +{\displaystyle {\int }_0^T{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t}\\ ={\displaystyle {\int }_0^T{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}\psi \left(t\right)\text{d}t}+\langle V\left(T\right),\psi \left(T\right)\rangle -\langle V\left(0\right),\psi \left(0\right)\rangle \\ -{\displaystyle {\int }_0^T{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}\psi \left(t\right)\text{d}t}-{\displaystyle {\int }_0^T{\nabla }_u}f\left[t\right]\left(\psi \left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\\ +{\displaystyle {\int }_0^T{\nabla }_u}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\end{array}$

$\begin{array}{l}={\displaystyle {\int }_0^T\left[-{\nabla }_{u}f\left[t\right]\left(\psi \left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)+{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right)\right)\right]}\\ ={\displaystyle {\int }_0^T-}{\nabla }_{u}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\end{array}$

由参考文献( [1] , p.105)中的方法将以上积分型极大值原理变为(2.6)，证毕。

4.2. 定理二的证明

若 $v(\cdot )\in \Omega$ 满足(2.6)有：

$\begin{array}{c}0\le \frac{J\left(u^{\epsilon }(\cdot )\right)-J\left(\bar{u}(\cdot )\right)}{{\epsilon }^2}\\ ={\displaystyle {\int }_0^T\left[{\nabla }_x{f}^0\left[t\right]\left(Y\left(t\right)\right)+\frac{1}{2}{\nabla }_x^2{f}^0\left[t\right]\left(V\left(t\right),V\left(t\right)\right)\right]}\\ +{\displaystyle {\int }_0^T{\nabla }_x{\nabla }_u}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\text{d}t\\ +\frac{1}{2}{\displaystyle {\int }_0^T{\nabla }_u^2}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t+\frac{o\left({\epsilon }^2\right)}{{\epsilon }^2}\end{array}$

当 $\epsilon \to 0^{+}$ 时，

$\begin{array}{c}0\le \frac{J\left(u^{\epsilon }(\cdot )\right)-J\left(\bar{u}(\cdot )\right)}{{\epsilon }^2}\\ ={\displaystyle {\int }_0^T\left[{\nabla }_x{f}^0\left[t\right]\left(Y\left(t\right)\right)+\frac{1}{2}{\nabla }_x^2{f}^0\left[t\right]\left(V\left(t\right),V\left(t\right)\right)\right]}\\ +{\displaystyle {\int }_0^T{\nabla }_x{\nabla }_u}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\text{d}t\\ +\frac{1}{2}{\displaystyle {\int }_0^T{\nabla }_u^2}{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)\text{d}t\end{array}$

将对偶方程(2.5)带入得到：

$\begin{array}{c}0\le {\displaystyle {\int }_0^T[{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}\psi \left(t\right)+{\nabla }_{x}f\left[t\right]\left(\psi \left(t\right),Y\left(t\right)\right)+\frac{1}{2}{\nabla }_x^2{f}^0\left[t\right]\left(V\left(t\right),V\left(t\right)\right)}\\ +\frac{1}{2}{\nabla }_u^2{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)+{\nabla }_x{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)]\text{d}t\end{array}$

再将(1.7)带入其中，利用与证明定理一相同的方法得到：

$\begin{array}{c}0\le {\displaystyle {\int }_0^T[{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}\psi \left(t\right)+{\nabla }_{\stackrel{\dot{}}{\bar{y}}\left(t\right)}Y\left(\psi \left(t\right)\right)-{\nabla }_x{\nabla }_{u}f\left[t\right]\left(\psi \left(t\right),v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\begin{array}{c}\\ \end{array}}\\ +\frac{1}{2}R\left(\tilde{\psi }\left(t\right),V\left(t\right),\stackrel{\dot{}}{\bar{y}}\left(t\right),V\left(t\right)\right)-\frac{1}{2}{\nabla }_x^{2}f\left[t\right]\left(\psi \left(t\right),V\left(t\right),V\left(t\right)\right)\\ -\frac{1}{2}{\nabla }_u^{2}f\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)+\frac{1}{2}{\nabla }_x^2{f}^0\left[t\right]\left(V\left(t\right),V\left(t\right)\right)\\ +\frac{1}{2}{\nabla }_u^2{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),v\left(t\right)-\bar{u}\left(t\right)\right)+{\nabla }_x{\nabla }_u{f}^0\left[t\right]\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)]\text{d}t\end{array}$

$\begin{array}{l}=\frac{1}{2}{\displaystyle {\int }_0^T-\left[{\nabla }_x^{2}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(V\left(t\right),V\left(t\right)\right)\right]}\\ +2{\nabla }_x{\nabla }_{u}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)\left(v\left(t\right)-\bar{u}\left(t\right),V\left(t\right)\right)\\ +{\nabla }_u^{2}H\left(t,\bar{y}\left(t\right),\psi \left(t\right),\bar{u}\left(t\right)\right)-R\left(\tilde{\psi }\left(t\right),V\left(t\right),\stackrel{\dot{}}{\bar{y}}\left(t\right),V\left(t\right)\right)]\text{d}t\end{array}$

证毕。