同济高数 第11章 第11-3-8题

[AI解答]

[AI解答]

证明
设曲线 $L$ 是区域 $D$ 的边界，取正向（即逆时针方向）。记外法向量为 $\mathbf{n}$，切向量为 $\mathbf{T}$，且外法向量与切向量的关系为：若切向量方向为 $(\mathrm{d}x,\mathrm{d}y)$，则外法向量方向为 $(\mathrm{d}y, -\mathrm{d}x)$（因为旋转 $90^\circ$ 且指向外侧）。

沿曲线 $L$ 的外法向导数定义为：
$$
\frac{\partial u}{\partial n} = \nabla u \cdot \mathbf{n} = \frac{\partial u}{\partial x} n_x + \frac{\partial u}{\partial y} n_y.
$$

对于正向曲线，外法向量的方向余弦为：
$$
n_x = \frac{\mathrm{d}y}{\mathrm{d}s}, \quad n_y = -\frac{\mathrm{d}x}{\mathrm{d}s}.
$$
因此，
$$
\frac{\partial u}{\partial n} = \frac{\partial u}{\partial x} \frac{\mathrm{d}y}{\mathrm{d}s} - \frac{\partial u}{\partial y} \frac{\mathrm{d}x}{\mathrm{d}s}.
$$

于是，
$$
\oint_L \frac{\partial u}{\partial n} \, \mathrm{d}s = \oint_L \left( \frac{\partial u}{\partial x} \mathrm{d}y - \frac{\partial u}{\partial y} \mathrm{d}x \right).
$$

利用格林公式（Green's theorem），对于函数 $P = -\frac{\partial u}{\partial y}$，$Q = \frac{\partial u}{\partial x}$，有：
$$
\oint_L P\,\mathrm{d}x + Q\,\mathrm{d}y = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathrm{d}\sigma.
$$

代入 $P = -\frac{\partial u}{\partial y}$，$Q = \frac{\partial u}{\partial x}$，得：
$$
\oint_L \left( \frac{\partial u}{\partial x} \mathrm{d}y - \frac{\partial u}{\partial y} \mathrm{d}x \right) = \iint_D \left( \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} \right) - \frac{\partial}{\partial y}\left( -\frac{\partial u}{\partial y} \right) \right) \mathrm{d}\sigma.
$$

计算被积函数：
$$
\frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} \right) = \frac{\partial^2 u}{\partial x^2}, \quad
\frac{\partial}{\partial y}\left( -\frac{\partial u}{\partial y} \right) = -\frac{\partial^2 u}{\partial y^2},
$$
所以
$$
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}.
$$

因此，
$$
\oint_L \frac{\partial u}{\partial n} \, \mathrm{d}s = \iint_D \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) \mathrm{d}\sigma.
$$

证毕。