竖直上抛问题，边值问题、变分问题与最小作用量
重力加速度为 $g$, 小球质量 $m=1$, 以初速度为 $v_0 = g/2$ 竖直上抛, $t$ 时刻动能为 $T(\dot{x}) = \dot{x}^2/2$, 势能为 $V(x) = gx$, 求
- $$\int_0^1 dt \ 2T -V$$
- $$\int_0^1 dt \ T -V$$
边值问题 (P)
$$ \left\{\begin{aligned} & - \ddot{x} = F \qquad 0 < t < 1 \\ & x(0) = x(1) = 0 \end{aligned}\right. $$$F(t)$ 是光滑函数, 求 $x$.
变分问题 (w)
找 $x_{\text{真}} \in \mathcal{X} = \lbrace x | x \in [0,1], x(0) = x(1) = 0 \rbrace$, $\dot{x}$ 分片有界, s.t. $$\int_0^1 dt \ \dot{x}_{\text{真}} \dot{x} = \int_0^1 dt \ F x$$
最小作用量问题(M)
$$S(x) = \int_0^1 dt \ T-V$$ 找 $x_{\text{真}} \in \mathcal{X}$, s.t. $S(x_{\text{真}}) \leq S(x),\forall x \in \mathcal{X}$
三个形式的解是等价的
$$(P) \iff (W) \iff (M)$$
$(W) \implies (M)$
假设 $x_{\text{真}} \in \mathcal{X}$ 是 (W) 的解, $\forall x \in \mathcal{X}, \Delta x = x - x_{\text{真}} \in \mathcal{X}$, $$ \begin{aligned} S(x) &= S(x_{\text{真}} + \Delta x) + \frac 12 \int_0^1 dt \ (\dot x_\text{真} + \Delta \dot x)^2- \int_0^1 dt \ F (x_\text{真} + \Delta x) \\ & = S(x_{\text{真}}) + \frac 12 \int_0^1 dt \ \Delta \dot x^2 \qquad \text{利用 (W)}\\ &\geq S(x_{\text{真}}) \end{aligned} $$
$(M) \implies (W)$
假设 $x \in \mathcal{X}$ 是 (M) 的解, $\forall x \in \mathcal X, \forall \epsilon > 0$, $x_\text{真} + \epsilon x \in \mathcal X$ $$S(x_\text{真}) \leq S(x_\text{真} + \epsilon x)$$.
设 $$ g(\epsilon) \equiv S(x_\text{真} + \epsilon x)$$ $$g(\epsilon) = \frac 12 \int_0^1 dt \ \dot x_\text{真}^2 - \left[ \int_0^1 dt \ F x_\text{真} + \epsilon \left( \int_0^1 dt \ \dot x_\text{真} \dot x - \int_0^1 dt \ F x \right) \right] + \frac{\epsilon^2}{2} \int_0^1 dt \ \dot x^2$$ $$\frac{\mathrm{d} g(0)}{\mathrm{d} \epsilon}= 0 \implies \int_0^1 dt \ \dot x_\text{真} x - \int_0^1 dt \ Fx = 0$$
$(W) \iff (P)$
$$\int_0^1 dt \ \dot x_\text{真} \dot x - \int_0^1 dt \ F x = 0$$ 分部积分 $$-\int_0^1 dt \ \ddot x_\text{真} x - \int_0^1 dt \ F x = 0$$ $$\int_0^1 dt \ (\ddot x_\text{真} + F )x = 0$$ $$\ddot x_\text{真} + F=0$$