The Principle of Least Action
定义 \(T\) 为动能，\(V\) 为势能
Lagrangian
\[L = T - V = L(x, \dot{x}, t)\]
书里面写的是 \(L(x, \dot{x})\)，感觉像是假设势能场是不变的。加上 \(t\) 的话 \(V\) 就可以写成 \(V(x, t)\) 了。
The action of a trajectory is written
\[\mathcal{A} = \int_{t_0}^{t_1}L(x, \dot{x}, t)\mathrm{d}t\]
这里的 Least Action 其实不是 least，是 stationary，也就是 \(\delta \mathcal{A}=0\)。
然后根据 Euler-Lagrange equation
\[\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x}=0\]
具体证明的话，The Theoretical Minimum 和 The Feynman Lectures on Physics 里的证明不太一样。TM 是把 \(L\) 看成 \(t\) 上的黎曼和，然后对每一小段进行求导。FL 那本书的证明挺标准化的，我们不如再标准化一点，去复习一下 Bishop 那本 Deep Learning 里的证明。
The Theoretical Minimum 的证明
他是把积分看成求和
\[\begin{aligned}
\mathcal{A} &= \int_{t_0}^{t_1}L(x, \dot{x}, t)\mathrm{d}t\\
&=\sum_{n}L(x_n, \dot{x}_n, t)\Delta t\\
&=\sum_{n}L\left(x_n, \frac{x_{n+1}-x_n}{\Delta t}, t\right)\Delta t
\end{aligned}\]
酱紫
\[\frac{\partial \mathcal{A}}{\partial x_n} = \left(\frac{\partial}{\partial x_n}L\left(x_n, \frac{x_{n+1}-x_n}{\Delta t}, t\right)+\frac{\partial}{\partial x_n}L\left(x_{n-1}, \frac{x_n-x_{n-1}}{\Delta t}, t\right)\right)\Delta t\]
其中
\[\frac{\partial}{\partial x_n}L\left(x_n, \frac{x_{n+1}-x_n}{\Delta t}, t\right)
=
\left.\frac{\partial L}{\partial x}\right|_{n}-\frac{1}{\Delta t}\cdot\left.\frac{\partial L}{\partial\dot{x}}\right|_{n}\]
然后
\[\frac{\partial}{\partial x_n}L\left(x_{n-1}, \frac{x_n-x_{n-1}}{\Delta t}, t\right)
=
\frac{1}{\Delta t}\cdot\left.\frac{\partial L}{\partial \dot{x}}\right|_{n-1}\]
所以说
\[\frac{\partial \mathcal{A}}{\partial x_n}
=
\Delta t \left.\frac{\partial L}{\partial x}\right|_{n}-\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{n}-\left.\frac{\partial L}{\partial \dot{x}}\right|_{n-1}\right)\]
考虑到 \(\mathcal{A}\) 要 stationary，也就是 \(\frac{\partial\mathcal{A}}{\partial x_n}=0\)：
\[\left.\frac{\partial L}{\partial x}\right|_{n}-\frac{1}{\Delta t}\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{n}-\left.\frac{\partial L}{\partial \dot{x}}\right|_{n-1}\right)=0\]
也就是
\[\frac{\partial L}{\partial x}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{x}}=0\]
Bishop Deep Learning 的证明
Bishop 书里详细定义了 \(\delta F\)，对于 funcional \(F(y(x))\)，我们定义 \(\frac{\delta F}{\delta y(x)}\)：
\[F(y(x)+\epsilon \eta(x))=F(y(x))+\epsilon\int_l^r\frac{\delta F}{\delta y(x)}\eta(x)\mathrm{d}x+\mathcal{O}(\epsilon^2)\]
其中 \(\eta(l)=\eta(r)=0\)。
那我们假设
\[F(y(x)) = \int_l^r G(y(x), y'(x), x)\mathrm{d}x\]
我们有
\[\begin{aligned}
F(y(x) + \epsilon \eta(x))
&=\int_l^r G(y(x) + \epsilon\eta(x), y'(x)+\epsilon\eta'(x), x)\mathrm{d}x\\
&=\int_l^r \left(G(y(x), y'(x),x)+\frac{\partial G}{\partial y}\cdot \epsilon\eta(x)+\frac{\partial G}{\partial y'}\cdot\epsilon\eta'(x)+\mathcal{O}(\epsilon^2)\right)\mathrm{d}x\\
&=F(y(x))+\epsilon\int_l^r\left(\frac{\partial G}{\partial y}\eta(x)+\frac{\partial G}{\partial y'}\eta'(x)\right)\mathrm{d}x+\mathcal{O}(\epsilon^2)
\end{aligned}\]
我们之后想让 \(\frac{\delta F}{\delta y(x)}=0\) 然后去解，所以我们希望把 \(\eta(x)\) 消掉，我们考虑对 \(\frac{\partial G}{\partial y'}\eta'\) 做分步积分：
\[\int_l^r\frac{\partial G}{\partial y'}\eta'(x)\mathrm{d}x
=\int_l^r\frac{\partial G}{\partial y'}\mathrm{d}\eta
=\left.\left(\eta\frac{\partial G}{\partial y'}\right)\right|_l^r-\int_l^r\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial G}{\partial y'}\right)\]
考虑到 \(\eta(l)=\eta(r)=0\)，我们有
\[\int_l^r\frac{\partial G}{\partial y'}\eta'(x)\mathrm{d}x
=-\int_l^r\eta(x)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial G}{\partial y'}\right)\mathrm{d}x\]
所以
\[F(y(x) + \epsilon \eta(x))
=F(y(x))+\epsilon\int_l^r\left(\frac{\partial G}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial G}{\partial y'}\right)\right)\eta(x)\mathrm{d}x+\mathcal{O}(\epsilon^2)\]
所以我们就能推出 Euler-Lagrange 方程：
\[\frac{\partial G}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial G}{\partial y'}\right)=0\]
如果有多个物体的话，我们可以把他们的集合看成整体
\[L(\{x\}, \dot{\{x\}}, t) = \sum_{i}\left(\frac{1}{2}m_i \dot{x}_i^2\right)-V(\{x\}, t)\]