变分法概览

1.泛函数

泛函数指的是以函数集为原集，数集为像集的函数。 e.g.

I_{e g} (y) = y (x_{0}), w h e r e x_{0} = C o n s t .

$y(x)$ $I(y)$ 给出一个数值。另一种常见的泛函数为

I (y) = \int_{x_{1}}^{x_{2}} F (y, y^{'}; x) d x

$F(y,y';x)$ $F$ $y,y'$ $y$ $x$ 的函数。

令：

\begin{matrix} \tilde{y} (x, ϵ) \equiv y (x) + ϵ η (x) \end{matrix}

\begin{aligned} \tilde{I} \equiv & I (\tilde{y}) = I (y + ϵ η) \\ = & I (y) + ϵ \frac{\partial I (\tilde{y})}{\partial ϵ} + o (ϵ) \end{aligned}

$I$ 的一阶变分为

δ I = ϵ \frac{\partial I (\tilde{y})}{\partial ϵ}

$I_{eg}=y(x_0)$ $\delta I=\epsilon \eta(x_0)$

注意：不会出现形如 $\color{red} \delta I=\alpha \delta y$ 的表达式，因为对于I，x从x₁取到x₂，而对于y，x为某一固定的x₀。

$I(y)=\int_{x_1}^{x_2}F(y,y';x)\dd x$ $I(y)$ $\delta I=0$ . 我们有：

\begin{aligned} δ I = & ϵ \frac{\partial I (\tilde{y})}{\partial ϵ} \\ = & ϵ \int_{x_{1}}^{x_{2}} \frac{\partial}{\partial ϵ} F (\tilde{y}, {\tilde{y}}^{'}; x) d x \\ = & ϵ \int_{x_{1}}^{x_{2}} (\frac{\partial}{\partial \tilde{y}} \frac{\partial \tilde{y}}{\partial ϵ} + \frac{\partial}{\partial {\tilde{y}}^{'}} \frac{\partial {\tilde{y}}^{'}}{\partial ϵ}) F (\tilde{y}, {\tilde{y}}^{'}; x) d x \end{aligned}

显然有

\begin{matrix} \tilde{y} = y + ϵ η \Rightarrow \frac{\partial \tilde{y}}{\partial ϵ} = η \\ \tilde{y} = y^{'} + ϵ η^{'} \Rightarrow \frac{\partial {\tilde{y}}^{'}}{\partial ϵ} = η^{'} \end{matrix}

带入得

δ I = ϵ \int_{x_{1}}^{x_{2}} (η \frac{\partial}{\partial \tilde{y}} + η^{'} \frac{\partial}{\partial {\tilde{y}}^{'}}) F (\tilde{y}, {\tilde{y}}^{'}; x) d x

$I_{eg}=y(x_0)$ $\delta I=\epsilon \eta(x_0)$ $\delta y=\epsilon \eta$ $\delta y'=\epsilon \eta'$ . 代入得：

δ I = \int_{x_{1}}^{x_{2}} (δ y \frac{\partial}{\partial \tilde{y}} + δ y^{'} \frac{\partial}{\partial {\tilde{y}}^{'}}) F (\tilde{y}, {\tilde{y}}^{'}; x) d x

$\epsilon\rightarrow 0,~\tilde{y}\rightarrow 0$ ，带入得

δ I = \int_{x_{1}}^{x_{2}} (δ y \frac{\partial}{\partial y} + δ y^{'} \frac{\partial}{\partial y^{'}}) F (y, y^{'}; x) d x

对上式的第二项经行分部积分

\begin{aligned} 0 = δ I = & \int_{x_{1}}^{x_{2}} (δ y \frac{\partial F}{\partial y} + δ y^{'} \frac{\partial F}{\partial y^{'}}) d x \\ = & \int_{x_{1}}^{x_{2}} δ y \frac{\partial F}{\partial y} d x - \int_{x_{1}}^{x_{2}} δ y \frac{d}{d x} \frac{\partial F}{\partial y^{'}} d x + (δ y \frac{\partial F}{\partial y^{'}}) |_{x_{1}}^{x_{2}} \\ = & \int_{x_{1}}^{x_{2}} δ y (\frac{\partial F}{\partial y} - \frac{d}{d x} \frac{\partial F}{\partial y^{'}}) d x + (δ y \frac{\partial F}{\partial y^{'}}) |_{x_{1}}^{x_{2}} \end{aligned}

若我们有边界条件：

(δ y \frac{\partial F}{\partial y^{'}}) |_{x_{1} o r x_{2}} = 0

$\epsilon \eta(x)$ 的随意性可得

0 = \frac{\partial F}{\partial y} - \frac{d}{d x} \frac{\partial F}{\partial y^{'}}

是为欧拉-拉格朗日方程。

由上节，边界条件为：

(δ y \frac{\partial F}{\partial y^{'}}) |_{x_{1} o r x_{2}} = 0

(\frac{\partial F}{\partial y^{'}}) |_{x_{1} o r x_{2}} = 0

Essential boundaty condition:
$\begin{matrix} δ y (x_{1} o r x_{2}) = 0 \\ i . e . y (x_{1} o r x_{2}) = C o n s t . \end{matrix}$

$I(y)$ $y$ 的更高阶导数时

I (y) = \int_{x_{1}}^{x_{2}} F (y, y^{'}, y^{″}, . . .; x)

此时有：

δ I = \int_{x_{1}}^{x_{2}} (δ y \frac{\partial}{\partial y} + δ y^{'} \frac{\partial}{\partial y^{'}} + δ y^{″} \frac{\partial}{\partial y^{″}} + . . .) y d x

同理可以用分部积分获得其欧拉-拉格朗日方程以及边界条件。

在有约束

J (y) = \int_{x_{1}}^{x_{2}} G (y, y^{'}, . . .; x) d x = C o n s t .

下求如下泛函的极值

I (y) = \int_{x_{1}}^{x_{2}} F (y, y^{'}, . . .; x) d x = C o n s t .

$\lambda$ ，转换求如下新泛函的极值

I^{*} (y) = \int_{x_{1}}^{x_{2}} (F (y, y^{'}, . . .; x) + λ G (y, y^{'}, . . .; x)) d x

之后带入欧拉-拉格朗日方程，并上约束即可解得极值。

直观上理解，泛函的函数自变量相当于可数个数值自变量，那么对一般函数可以使用的拉格朗日乘子法对泛函也理应能用

$(x_1,y_1),(x_2,y_2)$ 。两点坐标不会改变，对应Essential boundary condition。

\begin{aligned} L (y) = \int_{x_{1}}^{x_{2}} \sqrt{1 + (y^{'})^{2}} d x \\ \Rightarrow & \frac{\partial F}{\partial y} = 0 \\ \frac{\partial F}{\partial y^{'}} = \frac{y^{'}}{\sqrt{1 + (y^{'})^{2}}} \end{aligned}

带入欧拉-拉格朗日方程

\begin{aligned} 0 = 0 - \frac{d}{d x} \frac{y^{'}}{\sqrt{1 + (y^{'})^{2}}} \\ \Rightarrow & \frac{y^{'}}{\sqrt{1 + (y^{'})^{2}}} = C o n s t . \\ i . e . & y^{'} = C o n s t . \end{aligned}

导数为定值，对应直线。

$(x_1,y_1),(x_2,y_2)$ 间的以定长曲线为上曲线的积分最大值在曲线为圆弧时去到。

\begin{matrix} A (y) = \int_{x_{1}}^{x_{2}} y d x \\ L (y) = \int_{x_{1}}^{x_{2}} \sqrt{1 + (y^{'})^{2}} d x = L = C o n s t . \end{matrix}

构造新函数：

\begin{matrix} I (x) = \int_{x_{1}}^{x_{2}} (y + λ \sqrt{1 + (y^{'})^{2}}) d x \end{matrix}

代入欧拉-拉格朗日方程：

\begin{aligned} \frac{\partial F}{\partial y} = 1 \\ \frac{\partial F^{'}}{\partial y} = \frac{λ y^{'}}{\sqrt{1 + (y^{'})^{2}}} \\ \Rightarrow & 0 = 1 - \frac{d}{d x} \frac{λ y^{'}}{\sqrt{1 + (y^{'})^{2}}} \\ \Rightarrow & \frac{y^{″}}{(1 + (y^{'})^{2})^{3 / 2}} = \frac{1}{λ} \end{aligned}

最后一个式子就是曲率为定值，即为圆弧。

$I[y]=\int F(y,y',...;x)\dd x$ ${\delta I\over \delta y}$ 如下：

δ I = I [y + δ y] - I [y] = \int (\frac{δ I}{δ y}) δ y d x

${\delta I\over \delta y}=0$ 时。