(1)

$$\begin{aligned} \int_{\boldsymbol c^{-1}} f \,ds &= \int_a^b f(\boldsymbol c^{-1}(t)) \|\boldsymbol c^{-1}\,'(t)\| \,dt \\ &= \int_a^b f(\boldsymbol c(a + b - t)) \|\boldsymbol c\,'(a + b - t)| \,dt \\ &= \int_b^a f(\boldsymbol c(u)) \|\boldsymbol c\,'(u)\| \,du \\ &= \int_a^b f(\boldsymbol c(t)) \|\boldsymbol c\,'(t)\| \,dt \end{aligned}$$

(2)

$$\begin{aligned} \int_{\boldsymbol c_1 \cdot \boldsymbol c_2} f \,ds &= \int_a^b f((\boldsymbol c_1 \cdot \boldsymbol c_2)(t)) \|(\boldsymbol c_1 \cdot \boldsymbol c_2)\,'(t)\| \,dt \\ &= \int_a^b f(\boldsymbol c_1(t)) \|\boldsymbol c_1\,'(t)\| \,dt + \int_b^c f(\boldsymbol c_2(t)) \|\boldsymbol c_2\,'(t)\| \,dt \\ &= \int_{\boldsymbol c_1} f \,ds + \int_{\boldsymbol c_2} f \,ds \end{aligned}$$

(3)

$$\begin{aligned} \int_{\boldsymbol c} (af + bg) \,ds &= \int_a^b (af(\boldsymbol c(t)) + bg(\boldsymbol c(t))) \|\boldsymbol c\,'(t)\| \,dt \\ &= a \int_a^b f(\boldsymbol c(t)) \|\boldsymbol c\,'(t)\| \,dt + b \int_a^b g(\boldsymbol c(t)) \|\boldsymbol c\,'(t)\| \,dt \\ &= a \int_{\boldsymbol c} f \,ds + b \int_{\boldsymbol c} g \,ds \end{aligned}$$

$\square$

$$\begin{aligned} \int_{\boldsymbol c \circ \varphi} f \,ds &= \int_c^d f((\boldsymbol c \circ \varphi)(t)) \|(\boldsymbol c \circ \varphi)\,'(t)\| \,dt \\ &= \int_c^d f(\boldsymbol c(\varphi(t))) \|\boldsymbol c\,'(\varphi(t))\| \cdot |\varphi'(t)| \,dt \\ &= \int_a^b f(\boldsymbol c(u)) \|\boldsymbol c\,'(u)\| \,du \\ &= \int_{\boldsymbol c} f \,ds \end{aligned}$$

$\square$

(1)

$$\begin{aligned} \int_{\boldsymbol c^{-1}} \boldsymbol F \cdot \mathrm{d\boldsymbol s} &= \int_a^b \boldsymbol F(\boldsymbol c^{-1}(t)) \cdot \boldsymbol c^{-1}\,'(t) \,dt \\ &= \int_a^b \boldsymbol F(\boldsymbol c(a + b - t)) \cdot (-\boldsymbol c\,'(a + b - t)) \,dt \\ &= -\int_b^a \boldsymbol F(\boldsymbol c(u)) \cdot \boldsymbol c\,'(u) \,du \\ &= -\int_a^b \boldsymbol F(\boldsymbol c(t)) \cdot \boldsymbol c\,'(t) \,dt \end{aligned}$$

(2)

$$\begin{aligned} \int_{\boldsymbol c_1 \cdot \boldsymbol c_2} \boldsymbol F \cdot \mathrm{d\boldsymbol s} &= \int_a^b \boldsymbol F((\boldsymbol c_1 \cdot \boldsymbol c_2)(t)) \cdot (\boldsymbol c_1 \cdot \boldsymbol c_2)\,'(t) \,dt \\ &= \int_a^b \boldsymbol F(\boldsymbol c_1(t)) \cdot \boldsymbol c_1\,'(t) \,dt + \int_b^c \boldsymbol F(\boldsymbol c_2(t)) \cdot \boldsymbol c_2\,'(t) \,dt \\ &= \int_{\boldsymbol c_1} \boldsymbol F \cdot \mathrm{d\boldsymbol s} + \int_{\boldsymbol c_2} \boldsymbol F \cdot \mathrm{d\boldsymbol s} \end{aligned}$$

(3)

$$\begin{aligned} \int_{\boldsymbol c} (a\boldsymbol F + b\boldsymbol G) \cdot \mathrm{d\boldsymbol s} &= \int_a^b (a\boldsymbol F(\boldsymbol c(t)) + b\boldsymbol G(\boldsymbol c(t))) \cdot \boldsymbol c\,'(t) \,dt \\ &= a \int_a^b \boldsymbol F(\boldsymbol c(t)) \cdot \boldsymbol c\,'(t) \,dt + b \int_a^b \boldsymbol G(\boldsymbol c(t)) \cdot \boldsymbol c\,'(t) \,dt \\ &= a \int_{\boldsymbol c} \boldsymbol F \cdot \mathrm{d\boldsymbol s} + b \int_{\boldsymbol c} \boldsymbol G \cdot \mathrm{d\boldsymbol s} \end{aligned}$$

$\square$

$$\begin{aligned} \int_{\boldsymbol c \circ \varphi} \boldsymbol F \cdot \mathrm{d\boldsymbol s} &= \int_c^d \boldsymbol F((\boldsymbol c \circ \varphi)(t)) \cdot (\boldsymbol c \circ \varphi)\,'(t) \,dt \\ &= \int_c^d \boldsymbol F(\boldsymbol c(\varphi(t))) \cdot \boldsymbol c\,'(\varphi(t)) \cdot \varphi'(t) \,dt \\ &= \int_a^b \boldsymbol F(\boldsymbol c(u)) \cdot \boldsymbol c\,'(u) \,du \\ &= \int_{\boldsymbol c} \boldsymbol F \cdot \mathrm{d\boldsymbol s} \end{aligned}$$

$\square$

根据梯度的定义和链式法则，有

$$\begin{aligned} \int_{\boldsymbol c} \mathrm{grad} f \cdot \mathrm{d\boldsymbol s} &= \int_a^b \mathrm{grad} f(\boldsymbol c(t)) \cdot \boldsymbol c\,'(t) \,dt \\ &= \int_a^b \frac{\mathrm{d}}{\mathrm{d}t} f(\boldsymbol c(t)) \,dt \\ &= f(\boldsymbol c(b)) - f(\boldsymbol c(a)) \end{aligned}$$

$\square$

我们在 $D$ 中构造一个递增列 $\{D_k\}_{k=1}^{\infty}$，使得

• 每一个 $D_k$ 均为有界闭的单连通区域
• 每一个 $D_k$ 都可以被分解为 $m_k$ 个子区域 $E_k^j$ 的并集，且每个子区域的边界均为正向的分段光滑闭合曲线，即

$$D_k = \bigcup_{j=1}^{m_k} E_k^j$$

• 当 $k \to \infty$ 时，$D_k$ 的覆盖范围趋近于 $D$，即 $\displaystyle \lim_{k \to \infty} \mathrm{Area}(D \setminus D_k) = 0$

不妨设各个 $D_k$ 上的矩形分割为

$$E_k^j = [a_k^j, b_k^j] \times [\overline a_k^j, \overline b_k^j]$$

那么矩形的两个边长则可以分别表示为

$$\ell_k^j = b_k^j - a_k^j, \quad \overline \ell_k^j = \overline b_k^j - \overline a_k^j$$

显然矩形的周长为 $2\ell_k^j + 2\overline \ell_k^j$，定义正向的分段光滑闭合曲线 $\boldsymbol c_k^j$ 为矩形 $E_k^j$ 的边界，如下

$$\boldsymbol c_k^j(t):[a_k^j, a_k^j + 2\ell_k^j + 2\overline \ell_k^j] \to \mathbb R^2,\quad t \mapsto \begin{cases} \binom{t}{\overline a_k^j}, & t \in [a_k^j, a_k^j + \ell_k^j] \\ \binom{b_k^j}{t - (a_k^j + \ell_k^j) + \overline a_k^j}, & t \in [a_k^j + \ell_k^j, a_k^j + \ell_k^j + \overline \ell_k^j] \\ \binom{a_k^j + b_k^j + \ell_k^j + \overline \ell_k^j - t}{\overline b_k^j}, & t \in [a_k^j + \ell_k^j + \overline \ell_k^j, a_k^j + 2\ell_k^j + \overline \ell_k^j] \\ \binom{a_k^j}{a_k^j + \overline b_k^j + 2\ell_k^j + \overline \ell_k^j - t}, & t \in [a_k^j + 2\ell_k^j + \overline \ell_k^j, a_k^j + 2\ell_k^j + 2\overline \ell_k^j] \end{cases}$$

以下开始证明对于各个矩形 $E_k^j$，Green 定理成立。令向量场 $\boldsymbol F(x_1, x_2) = \binom{F_1(x_1, x_2)}{F_2(x_1, x_2)}$，则有

$$\begin{aligned} \iint_{E_k^j} \left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}\right) \,dA &= \int_{a_k^j}^{b_k^j} \int_{\overline a_k^j}^{\overline b_k^j} \frac{\partial F_2}{\partial x_1} \,dx_2 \,dx_1 - \int_{\overline a_k^j}^{\overline b_k^j} \int_{a_k^j}^{b_k^j} \frac{\partial F_1}{\partial x_2} \,dx_1 \,dx_2 \\ &= \int_{a_k^j}^{b_k^j} \left[F_2(x_1, \overline b_k^j) - F_2(x_1, \overline a_k^j)\right] \,dx_1 - \int_{\overline a_k^j}^{\overline b_k^j} \left[F_1(b_k^j, x_2) - F_1(a_k^j, x_2)\right] \,dx_2 \\ &= \int_{a_k^j}^{b_k^j} F_2(x_1, \overline b_k^j) \,dx_1 - \int_{a_k^j}^{b_k^j} F_2(x_1, \overline a_k^j) \,dx_1 - \int_{\overline a_k^j}^{\overline b_k^j} F_1(b_k^j, x_2) \,dx_2 + \int_{\overline a_k^j}^{\overline b_k^j} F_1(a_k^j, x_2) \,dx_2 \end{aligned}$$

另一方面，有

$$\begin{aligned} \int_{\partial E_k^j} \boldsymbol F \cdot \mathrm{d\boldsymbol s} &= \int_{a_k^j}^{a_k^j + 2\ell_k^j + 2\overline \ell_k^j} \boldsymbol F(\boldsymbol c_k^j(t)) \cdot \boldsymbol c_k^j\,'(t) \,dt \\ &= \int_{a_k^j}^{a_k^j + \ell_k^j} \binom{F_1(t, \overline a_k^j)}{F_2(t, \overline a_k^j)} \cdot \binom{1}{0} \,dt \\ &\quad + \int_{a_k^j + \ell_k^j}^{a_k^j + \ell_k^j + \overline \ell_k^j} \binom{F_1(b_k^j, t - (a_k^j + \ell_k^j) + \overline a_k^j)}{F_2(b_k^j, t - (a_k^j + \ell_k^j) + \overline a_k^j)} \cdot \binom{0}{1} \,dt \\ &\quad + \int_{a_k^j + \ell_k^j + \overline \ell_k^j}^{a_k^j + 2\ell_k^j + \overline \ell_k^j} \binom{F_1(a_k^j + b_k^j + \ell_k^j + \overline \ell_k^j - t, \overline b_k^j)}{F_2(a_k^j + b_k^j + \ell_k^j + \overline \ell_k^j - t, \overline b_k^j)} \cdot \binom{-1}{0} \,dt \\ &\quad + \int_{a_k^j + 2\ell_k^j + \overline \ell_k^j}^{a_k^j + 2\ell_k^j + 2\overline \ell_k^j} \binom{F_1(a_k^j, a_k^j + \overline b_k^j + 2\ell_k^j + \overline \ell_k^j - t)}{F_2(a_k^j, a_k^j + \overline b_k^j + 2\ell_k^j + \overline \ell_k^j - t)} \cdot \binom{0}{-1} \,dt \\ &= \int_{a_k^j}^{b_k^j} F_1(t, \overline a_k^j) \,dt + \int_{\overline a_k^j}^{\overline b_k^j} F_2(b_k^j, x_2) \,dx_2 - \int_{a_k^j}^{b_k^j} F_1(x_1, \overline b_k^j) \,dx_1 - \int_{\overline a_k^j}^{\overline b_k^j} F_2(a_k^j, x_2) \,dx_2 \end{aligned}$$

由此可见，对于每个矩形 $E_k^j$，Green 定理成立，即

$$\int_{\partial E_k^j} \boldsymbol F \cdot \mathrm{d\boldsymbol s} = \iint_{E_k^j} \left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}\right) \,dA$$

现在回到 $D_k$，矩阵分割的内边界处的线积分由于方向相反会相互抵消，所以有

$$\begin{aligned} \int_{\partial D_k} \boldsymbol F \cdot \mathrm{d\boldsymbol s} &= \sum_{j=1}^{m_k} \int_{\partial E_k^j} \boldsymbol F \cdot \mathrm{d\boldsymbol s} \\ &= \sum_{j=1}^{m_k} \iint_{E_k^j} \left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}\right) \,dA \\ &= \iint_{D_k} \left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}\right) \,dA \end{aligned}$$

因此

$$\iint_D \left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}\right) \,dA = \lim_{k \to \infty} \iint_{D_k} \left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}\right) \,dA = \lim_{k \to \infty} \int_{\partial D_k} \boldsymbol F \cdot \mathrm{d\boldsymbol s} = \int_{\partial D} \boldsymbol F \cdot \mathrm{d\boldsymbol s}$$

$\square$