5.5 动态优化

5.5.1 Euler-Lagrange方程

在静态优化问题中，我们研究的是目标函数在可行域中的最值问题，其中自变量是可行域中的点。在动态优化问题中，我们研究的仍然是最值问题，但是自变量不再是点、而是函数，亦即在某个函数空间中找到使得目标“函数”取最大值的函数。这种将函数映射为实数的映射称为泛函，动态优化问题也称为泛函优化问题。

为了将函数映射为实数，泛函通常以积分的形式呈现。我们首先给出最常见的动态优化问题的定义：

解决该类问题的基本方法是变分法，其核心思想是：假设${\mathit{x}}^{\ast }$为最优解，考虑在${\mathit{x}}^{\ast }$附近的函数$\mathit{x}={\mathit{x}}^{\ast }+\epsilon \mathit{h}$，其中$\epsilon$为足够小的正数，$\mathit{h}$为任意满足边界条件的扰动函数，即$$\begin{array}{}\text{(3)}& \mathit{h}\in \mathcal{H}:=\{\mathit{h}=\tilde{\mathit{x}}-\mathit{x}\mid \mathit{x},\tilde{\mathit{x}}\in \mathcal{F}\}\end{array}$$ 对于本例而言，实际上有$\mathcal{H}:=\{\mathit{h}\in \mathcal{F}\mid \mathit{h}(a)=\mathit{h}(b)=\mathbf{0}\}$。考虑扰动后的目标泛函$$\begin{array}{}\text{(4)}& F(\epsilon ):=J[{\mathit{x}}^{\ast }+\epsilon \mathit{h}]={\int }_a^{b}L(t,{\mathit{x}}^{\ast }(t)+\epsilon \mathit{h}(t),{\dot{\mathit{x}}}^{\ast }(t)+\epsilon \dot{\mathit{h}}(t))\mathrm{d}t\end{array}$$ 对$\epsilon$求导可得$$\begin{array}{}\text{(5)}& \frac{\mathrm{d}F}{\mathrm{d}\epsilon }={\int }_a^b[\frac{\partial L}{\partial \mathit{x}}\mathit{h}(t)+\frac{\partial L}{\partial \dot{\mathit{x}}}\dot{\mathit{h}}(t)]\mathrm{d}t\end{array}$$

为了消除$\dot{\mathit{h}}$，我们对上式中的第二项进行分部积分，结合边界条件$\mathit{h}(a)=\mathit{h}(b)=\mathbf{0}$可得$$\begin{array}{}\text{(6)}& {\int }_a^b\frac{\partial L}{\partial \dot{\mathit{x}}}\dot{\mathit{h}}(t)\mathrm{d}t={\frac{\partial L}{\partial \dot{\mathit{x}}}\mathit{h}(t)|}_a^b-{\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)\mathit{h}(t)\mathrm{d}t=-{\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)\mathit{h}(t)\mathrm{d}t\end{array}$$ 代入原式可得$$\begin{array}{}\text{(7)}& \frac{\mathrm{d}F}{\mathrm{d}\epsilon }={\int }_a^b{[\frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)]}_{\mathit{x}={\mathit{x}}^{\ast }+\epsilon \mathit{h}}\mathit{h}(t)\mathrm{d}t\end{array}$$

令$\epsilon =0$可得$$\begin{array}{}\text{(8)}& {\frac{\mathrm{d}F}{\mathrm{d}\epsilon }|}_{\epsilon =0}={\int }_a^b{[\frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)]}_{\mathit{x}={\mathit{x}}^{\ast }}\mathit{h}(t)\mathrm{d}t\end{array}$$ 由于${\mathit{x}}^{\ast }$为最优解，故$J[\mathit{x}+\epsilon \mathit{h}]$在$\epsilon =0$处取得极大值，亦即$F$在$\epsilon =0$处取得极大值，由此可得$$\begin{array}{}\text{(9)}& {\int }_a^b{[\frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)]}_{\mathit{x}={\mathit{x}}^{\ast }}\mathit{h}(t)\mathrm{d}t=0,\mathrm{\forall }\mathit{h}\in \mathcal{H}\end{array}$$ 结合$\mathit{h}$的任意性，可得著名的Euler-Lagrange方程，即动态优化问题的一阶必要条件：

5.5.2 变分法

为了更好地解决动态优化问题，我们引入变分的概念，即满足边界条件的函数值的微小变化（扰动函数），定义为

对于多元泛函而言，容易证明$$\begin{array}{}\text{(13)}& \delta J[{\mathit{x}}_1,\cdots ,{\mathit{x}}_n,\delta {\mathit{x}}_1,\cdots ,\delta {\mathit{x}}_n]=\sum _{i=1}^n\frac{\partial J}{\partial {\mathit{x}}_i}[{\mathit{x}}_1,\cdots ,{\mathit{x}}_n]\delta {\mathit{x}}_i\end{array}$$ 需要特别注意的是，$\delta \mathit{x}$和$\delta \dot{\mathit{x}}$通常视作两个独立的变分。

类似方向导数形式的多元一阶Taylor展开式：$$\begin{array}{}\text{(14)}& J(\mathit{x}+\mathrm{\Delta }\mathit{x})=J(\mathit{x})+\underset{\begin{array}{c}=\mathrm{d}J(\mathit{x})(\mathrm{\Delta }\mathit{x})\\ =\mathrm{\nabla }J(\mathit{x})\cdot \mathrm{\Delta }\mathit{x}\end{array}}{\underbrace{\frac{\partial J}{\partial \mathrm{\Delta }\mathit{x}}(\mathit{x})}}+o(\mathrm{\Delta }\mathit{x})\end{array}$$ 泛函$J[\mathit{x}]$的变分也可以表示为$$\begin{array}{}\text{(15)}& J[\mathit{x}+\delta \mathit{x}]=J[\mathit{x}]+\delta J[\mathit{x},\delta \mathit{x}]+o(\delta \mathit{x})\end{array}$$ 因此，函数$\mathit{x}$的变分$\delta \mathit{x}$的几何意义类似于方向$\mathrm{\Delta }\mathit{x}$，表示“沿着$\delta \mathit{x}$方向移动的函数自变量”；而泛函$J[\mathit{x}]$的变分$\delta J[\mathit{x},\delta \mathit{x}]$的几何意义类似于方向导数$\frac{\partial J}{\partial \mathrm{\Delta }\mathit{x}}(\mathit{x})$的值，表示“沿着$\delta \mathit{x}$方向移动的泛函值变化率”。

可以证明，变分具有以下性质：

对于上例中的动态优化问题，我们可以用变分重新推导Euler-Lagrange方程：$$\begin{array}{}\text{(18)}& \begin{array}{rl}\delta J[\mathit{x},\delta \mathit{x}]& =\delta {\int }_a^{b}L(t,\mathit{x}(t),\dot{\mathit{x}}(t))\mathrm{d}t={\int }_a^b\delta L(t,\mathit{x}(t),\dot{\mathit{x}}(t))\mathrm{d}t\\ & ={\int }_a^b[\frac{\partial L}{\partial \mathit{x}}\delta \mathit{x}(t)+\frac{\partial L}{\partial \dot{\mathit{x}}}\delta \dot{\mathit{x}}(t)]\mathrm{d}t={\int }_a^b[\frac{\partial L}{\partial \mathit{x}}\delta \mathit{x}(t)+\frac{\partial L}{\partial \dot{\mathit{x}}}\frac{\mathrm{d}(\delta \mathit{x}(t))}{\mathrm{d}t}]\mathrm{d}t\\ & ={\int }_a^b\frac{\partial L}{\partial \mathit{x}}\delta \mathit{x}(t)\mathrm{d}t+{\frac{\partial L}{\partial \dot{\mathit{x}}}\delta \mathit{x}(t)|}_a^b-{\int }_a^b\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)\delta \mathit{x}(t)\mathrm{d}t\\ & ={\int }_a^b[\frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)]\delta \mathit{x}(t)\mathrm{d}t\end{array}\end{array}$$ 当${\mathit{x}}^{\ast }$为$J$的极大（小）值时，在${\mathit{x}}^{\ast }$的邻域内应成立$\delta J[{\mathit{x}}^{\ast },\delta \mathit{x}]\le 0(\ge 0)$，其中$\delta \mathit{x}$为满足边界条件的任意变分。不妨设${\mathit{x}}^{\ast }$为$J$的极大值点（否则考虑$-J$），设$\epsilon >0$，取特殊的$\delta \mathit{x}$可得$$\begin{array}{}\text{(19)}& \delta \mathit{x}=\epsilon {[\frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)]}_{\mathit{x}={\mathit{x}}^{\ast }}⟹\delta J[{\mathit{x}}^{\ast },\delta \mathit{x}]=\epsilon {\int }_a^b{\Vert \frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)\Vert }_{\mathit{x}={\mathit{x}}^{\ast }}^2\mathrm{d}t\le 0\end{array}$$ 因此$$\begin{array}{}\text{(20)}& {[\frac{\partial L}{\partial \mathit{x}}-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{\mathit{x}}}\right)]}_{\mathit{x}={\mathit{x}}^{\ast }}=\mathbf{0}\end{array}$$

5.5.3 变分法的应用

我们介绍两个经典的例子。

解 设曲线方程为$y=y(x)$，则曲线长度为$$\begin{array}{}\text{(21)}& S[y]={\int }_{x_a}^{x_b}\sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{d}x=:{\int }_{x_a}^{x_b}L(y,y^{\prime })\mathrm{d}x\end{array}$$ 由Euler-Lagrange方程可得$$\begin{array}{}\text{(22)}& \frac{\partial L}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial L}{\partial y^{\prime }}\right)=0⟹-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}\right)=0\end{array}$$ 故${\displaystyle \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}}=C$，其中$C$为常数。解得$y^{\prime }=k$，其中$k$为常数，故最短曲线为直线段。 $◻$

解 设曲线方程为$y=y(x)$，则质点在曲线上的速度为$$\begin{array}{}\text{(23)}& v=\sqrt{2gy}=\frac{\mathrm{d}s}{\mathrm{d}t}=\frac{\sqrt{1+y^{\prime 2}}}{\frac{\mathrm{d}t}{\mathrm{d}x}}⟹\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}\end{array}$$ 故质点从$A$到$B$所需时间为$$\begin{array}{}\text{(24)}& T[y]={\int }_0^l\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}\mathrm{d}x=:{\int }_0^{l}L(y,y^{\prime })\mathrm{d}x\end{array}$$ 由于$L$不显含$x$，由Euler-Lagrange方程可得$$\begin{array}{}\text{(25)}& \frac{\partial L}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial y}-\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\partial }{\partial y}\left(\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial }{\partial y}(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }})=0\end{array}$$ 故有$$\begin{array}{}\text{(26)}& L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}-y^{\prime }\cdot \frac{y^{\prime }}{\sqrt{2gy(1+y^{\prime 2})}}=\frac{1}{\sqrt{2gy(1+y^{\prime 2})}}=C\end{array}$$ 其中$C$为常数，解得$$\begin{array}{}\text{(27)}& y(1+y^{\prime 2})=k,k=\frac{1}{2g{C}^2}⟹\frac{\mathrm{d}y}{\mathrm{d}x}=\pm \sqrt{\frac{k}{y}-1}⟹x=\pm \int \sqrt{\frac{y}{k-y}}\mathrm{d}y\end{array}$$ 作变量代换$y=k{\sin }^2\frac{\theta }{2}$，则有$$\begin{array}{}\text{(28)}& \begin{array}{rl}x& =\pm k\int {\sin }^2\frac{\theta }{2}\mathrm{d}\theta =\pm k(\frac{\theta }{2}-\frac{\sin \theta }{2})+C_1\\ y& =k{\sin }^2\frac{\theta }{2}=\frac{k}{2}(1-\cos \theta )\end{array}\end{array}$$ 其中$C_1$为常数。取正号并结合边界条件可得$$\begin{array}{}\text{(29)}& \{\begin{array}{l}x=\frac{k}{2}(\theta -\sin \theta )\\ y=\frac{k}{2}(1-\cos \theta )\end{array}\theta \in [0,{\theta }_a],\{\begin{array}{l}l=\frac{k}{2}({\theta }_a-\sin {\theta }_a)\\ h=\frac{k}{2}(1-\cos {\theta }_a)\end{array}\end{array}$$ 上式即为最速降线的参数方程，其为摆线的一部分。 $◻$