下册 9.2 第二型曲线积分 第41题
📝 题目
41．证明下列结论．
（1）设函数 $u(x, y)$ 在光滑闭合曲线 $L$ 所围成的区域 $D$ 上有二阶连续偏导数，试证明：
$\displaystyle \iint_{D}\left(\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D} u \Delta u \mathrm{~d} x \mathrm{~d} y+\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s$ ，其中 $\displaystyle \Delta u=\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}, \frac{\partial u}{\partial n}$ 是 $u(x, y)$ 沿曲线 $L$ 的外法线 $\boldsymbol{n}$ 的方向导数。
（2）设函数 $u(x, y)$ 在由封闭的光滑曲线 $L$ 所围的区域 $D$ 上具有二阶连续偏导
数，$\displaystyle \Delta u=\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0$ 。证明： $\displaystyle \iint_{D}\left[\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right] \mathrm{d} x \mathrm{~d} y=\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s$ ，其中 $\displaystyle \frac{\partial u}{\partial n}$ 是沿 $L$ 的外法线的方向导数．
（3）设 $u(x, y), v(x, y)$ 是具有二阶连续偏导数的函数．证明：
$$
\iint_{D} v\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D}\left(\frac{\partial u}{\partial x} \cdot \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} \cdot \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{~d} y+\oint_{L} v \frac{\partial u}{\partial n} \mathrm{~d} s
$$
（4）设 $\Omega$ 为 $x y$ 平面上具有光滑边界 $\partial \Omega$ 的有界区域，$u \in C^{2}(\Omega) \cap C^{1}(\bar{\Omega})$ ，且 $u$ 为非常值函数及 $\left.u\right|_{\partial \Omega}=0$ ，证明： $\displaystyle \iint_{\Omega} u\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y<0$ ．
💡 答案解析
\section*{解题过程：}
（1）因 $\displaystyle \frac{\partial u}{\partial \vec{n}}=\frac{\partial u}{\partial x} \cos (\boldsymbol{n}, \boldsymbol{x})+\frac{\partial u}{\partial y} \cos (\boldsymbol{n}, \boldsymbol{y}), \cos (\boldsymbol{n}, \boldsymbol{x}) \mathrm{d} s=\mathrm{d} y, \cos (\boldsymbol{n}, \boldsymbol{y}) \mathrm{d} s=-\mathrm{d} x$ ，所以
$$
\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s=\oint_{L}\left(u \frac{\partial u}{\partial x} \cos (n, x)+u \frac{\partial u}{\partial y} \cos (n, y)\right) \mathrm{d} s=\oint_{L}-u \frac{\partial u}{\partial y} \mathrm{~d} x+u \frac{\partial u}{\partial x} \mathrm{~d} y
$$
由格林公式得
$$
\begin{aligned}
& \oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s=\oint_{L} u \frac{\partial u}{\partial x} \mathrm{~d} y-u \frac{\partial u}{\partial y} \mathrm{~d} x=\iint_{D}\left[\left(u \frac{\partial u}{\partial x}\right)_{x}+\left(u \frac{\partial u}{\partial y}\right)_{y}\right] \mathrm{d} x \mathrm{~d} y \\
&=\iint_{D}\left[\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right] \mathrm{d} x \mathrm{~d} y+\iint_{D} u\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y \\
& \iint_{D}\left(\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D} u \Delta u \mathrm{~d} x \mathrm{~d} y+\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s
\end{aligned}
$$
即
$$
\iint_{D}\left[\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right] \mathrm{d} x \mathrm{~d} y=-\iint_{D} u \Delta u \mathrm{~d} x \mathrm{~d} y+\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s
$$
又 $\displaystyle \Delta u=\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0$ ，所以 $\displaystyle \iint_{D}\left[\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right] \mathrm{d} x \mathrm{~d} y=\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s$ ．
（3）因 $\displaystyle \frac{\partial u}{\partial n}=\frac{\partial u}{\partial x} \cos (n, x)+\frac{\partial u}{\partial y} \cos (n, y), \cos (n, x) \mathrm{d} s=\mathrm{d} y, \cos (n, y) \mathrm{d} s=-\mathrm{d} x$ ，所以
$$
\oint_{L} v \frac{\partial u}{\partial n} \mathrm{~d} s=\oint_{L}\left(v \frac{\partial u}{\partial x} \cos (n, x)+v \frac{\partial u}{\partial y} \cos (n, y)\right) \mathrm{d} s=\oint_{L}-v \frac{\partial u}{\partial y} \mathrm{~d} x+v \frac{\partial u}{\partial x} \mathrm{~d} y
$$
由格林公式得
$$
\oint_{L} v \frac{\partial u}{\partial n} \mathrm{~d} s=\oint_{L} v \frac{\partial u}{\partial x} \mathrm{~d} y-v \frac{\partial u}{\partial y} \mathrm{~d} x=\iint_{D} v\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y+\iint_{D}\left(\frac{\partial u}{\partial x} \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{~d} y
$$
即
$$
\iint_{D} v\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D}\left(\frac{\partial u}{\partial x} \cdot \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} \cdot \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{~d} y+\oint_{L} v \frac{\partial u}{\partial n} \mathrm{~d} s
$$
（4）由（3）知
$$
\iint_{D} u\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D}\left(\frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \cdot \frac{\partial u}{\partial y}\right) \mathrm{d} x \mathrm{~d} y+\oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s
$$
又 $\left.u\right|_{\partial \Omega}=0$ ，于是 $\displaystyle \oint_{L} u \frac{\partial u}{\partial n} \mathrm{~d} s=0$ ．从而
$$
\iint_{D} u\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D}\left(\frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \cdot \frac{\partial u}{\partial y}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D}\left[\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}\right] \mathrm{d} x \mathrm{~d} y<0
$$
📋 详细解题步骤
步骤 1/6
目标：将外法向导数转化为坐标形式
由方向导数定义，$\frac{\partial u}{\partial n} = \frac{\partial u}{\partial x} \cos(n,x) + \frac{\partial u}{\partial y} \cos(n,y)$。对于外法线，有 $\cos(n,x) ds = dy$，$\cos(n,y) ds = -dx$。因此，$\oint_L u \frac{\partial u}{\partial n} ds = \oint_L (u \frac{\partial u}{\partial x} dy - u \frac{\partial u}{\partial y} dx)$。
公式：$\frac{\partial u}{\partial n} = \frac{\partial u}{\partial x} \cos(n,x) + \frac{\partial u}{\partial y} \cos(n,y)$
提示：注意外法线方向与坐标轴夹角的关系，确保符号正确。
步骤 2/6
目标：应用格林公式将曲线积分转化为二重积分
令 $P = -u \frac{\partial u}{\partial y}$，$Q = u \frac{\partial u}{\partial x}$，则 $\oint_L u \frac{\partial u}{\partial n} ds = \oint_L P dx + Q dy$。由格林公式，$\oint_L P dx + Q dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dxdy$。计算偏导数：$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(u \frac{\partial u}{\partial x}) = (\frac{\partial u}{\partial x})^2 + u \frac{\partial^2 u}{\partial x^2}$，$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(-u \frac{\partial u}{\partial y}) = -(\frac{\partial u}{\partial y})^2 - u \frac{\partial^2 u}{\partial y^2}$。因此，$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2 + u(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2})$。
公式：格林公式：$\oint_L P dx + Q dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dxdy$
提示：注意格林公式中 $P$ 和 $Q$ 的对应关系，以及求偏导时的链式法则。
步骤 3/6
目标：整理得到第一问的结论
由格林公式结果，$\oint_L u \frac{\partial u}{\partial n} ds = \iint_D [(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2] dxdy + \iint_D u \Delta u dxdy$。移项即得 $\iint_D [(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2] dxdy = -\iint_D u \Delta u dxdy + \oint_L u \frac{\partial u}{\partial n} ds$。
提示：移项时注意符号变化。
步骤 4/6
目标：第二问：利用调和函数条件简化
由第一问结论，若 $\Delta u = 0$，则 $\iint_D u \Delta u dxdy = 0$，代入即得 $\iint_D [(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2] dxdy = \oint_L u \frac{\partial u}{\partial n} ds$。
提示：直接代入即可，注意条件 $\Delta u = 0$。
步骤 5/6
目标：第三问：推广到两个函数的情形
类似第一问，将 $u$ 替换为 $v$，但注意 $\frac{\partial u}{\partial n}$ 不变。由方向导数定义，$\oint_L v \frac{\partial u}{\partial n} ds = \oint_L (v \frac{\partial u}{\partial x} dy - v \frac{\partial u}{\partial y} dx)$。令 $P = -v \frac{\partial u}{\partial y}$，$Q = v \frac{\partial u}{\partial x}$，应用格林公式：$\frac{\partial Q}{\partial x} = \frac{\partial v}{\partial x} \frac{\partial u}{\partial x} + v \frac{\partial^2 u}{\partial x^2}$，$\frac{\partial P}{\partial y} = -\frac{\partial v}{\partial y} \frac{\partial u}{\partial y} - v \frac{\partial^2 u}{\partial y^2}$。因此 $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial v}{\partial y} + v \Delta u$。于是 $\oint_L v \frac{\partial u}{\partial n} ds = \iint_D (\frac{\partial u}{\partial x} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial v}{\partial y}) dxdy + \iint_D v \Delta u dxdy$，移项即得结论。
提示：注意 $v$ 和 $u$ 的偏导顺序，格林公式中 $P$ 和 $Q$ 的构造与第一问类似。
步骤 6/6
目标：第四问：利用边界条件证明不等式
在第三问中取 $v = u$，得 $\iint_D u \Delta u dxdy = -\iint_D [(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2] dxdy + \oint_L u \frac{\partial u}{\partial n} ds$。由于 $u$ 在边界 $\partial \Omega$ 上为 $0$，故 $\oint_L u \frac{\partial u}{\partial n} ds = 0$。因此 $\iint_D u \Delta u dxdy = -\iint_D [(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2] dxdy$。由于 $u$ 非常值，梯度平方积分大于 $0$，故 $\iint_D u \Delta u dxdy < 0$。
提示：注意边界条件 $u|_{\partial \Omega}=0$ 导致曲线积分为零，且非常值函数保证梯度不恒为零。
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