方企勤 第七章 典型综合题分析 第28题
📝 题目
例 28 设 $f\left( {x,y}\right)$ 在 ${x}^{2} + {y}^{2} < 1$ 上二次连续可微,且满足
$$ \frac{{\partial }^{2}f}{\partial {x}^{2}} + \frac{{\partial }^{2}f}{\partial {y}^{2}} = {\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }. $$
💡 答案解析
证明:
$$ {\iint }_{{x}^{2} + {y}^{2} < 1}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = \frac{\pi }{2\mathrm{e}}. $$
证法 1 令 $x = r\cos \theta ,y = r\sin \theta$ ,则 $\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x}\cos \theta + \frac{\partial f}{\partial y}\sin \theta$ ,因此 $r\frac{\partial f}{\partial r} = x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}$ ,所以
$$ {\iint }_{{x}^{2} + {y}^{2} < 1}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = {\int }_{0}^{1}\mathrm{\;d}r{\int }_{0}^{2\pi }r\frac{\partial f}{\partial r} \cdot r\mathrm{\;d}\theta . \tag{7.50} $$
再由格林公式有
$$ {\int }_{0}^{2\pi }\frac{\partial f}{\partial r}r\mathrm{\;d}\theta = {\int }_{{x}^{2} + {y}^{2} = {r}^{2}}\frac{\partial f}{\partial n}\mathrm{\;d}s $$
$$ = {\iint }_{{x}^{2} + {y}^{2} \leq {r}^{2}}\left( {\frac{{\partial }^{2}f}{\partial {x}^{2}} + \frac{{\partial }^{2}f}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y = {\iint }_{{x}^{2} + {y}^{2} \leq {r}^{2}}{\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}x\mathrm{\;d}y $$
$$ = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{r}t{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t = \pi \left( {1 - {\mathrm{e}}^{-{r}^{2}}}\right) \;\left( {0 < r < 1}\right) . $$
把所得结果代入 (7.50) 式, 我们得到
$$ {\iint }_{{x}^{2} + {y}^{2} < 1}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = {\int }_{0}^{1}\pi \left( {1 - {\mathrm{e}}^{-{r}^{2}}}\right) r\mathrm{\;d}r $$
$$ = \frac{\pi }{2} + \frac{\pi }{2}\left( {{\mathrm{e}}^{-1} - 1}\right) = \frac{\pi }{2\mathrm{e}}. $$
证法 2 利用广义格林公式
$$ {\int }_{{c}_{r}}v\frac{\partial u}{\partial n}\mathrm{\;d}s = {\iint }_{{\Delta }_{r}}\left( {\frac{\partial u}{\partial x}\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\frac{\partial v}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y + {\iint }_{{\Delta }_{r}}v\left( {\frac{{\partial }^{2}u}{\partial {x}^{2}} + \frac{{\partial }^{2}u}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y, $$
其中 ${c}_{r},{\Delta }_{r}$ 分别表示圆周 ${x}^{2} + {y}^{2} = {r}^{2}$ 与圆盘 ${x}^{2} + {y}^{2} \leq {r}^{2}\left( {0 < r < 1}\right)$ .
现在取 $u\left( {x,y}\right) = f\left( {x,y}\right) ,v\left( {x,y}\right) = \frac{1}{2}\left( {{x}^{2} + {y}^{2}}\right)$ ,则
$$ {\iint }_{{\Delta }_{r}}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = \mathop{\lim }\limits_{{r \rightarrow 1}}{\iint }_{{\Delta }_{r}}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y $$
$$ = \mathop{\lim }\limits_{{r \rightarrow 1}}\left\lbrack {{\int }_{{c}_{r}}\frac{{x}^{2} + {y}^{2}}{2}\frac{\partial f}{\partial n}\mathrm{\;d}s - {\iint }_{{\Delta }_{r}}\frac{{x}^{2} + {y}^{2}}{2}\left( {\frac{{\partial }^{2}f}{\partial {x}^{2}} + \frac{{\partial }^{2}f}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y}\right\rbrack $$
$$ = \mathop{\lim }\limits_{{r \rightarrow 1}}\left\lbrack {\frac{{r}^{2}}{2}{\int }_{{c}_{r}}\frac{\partial f}{\partial n}\mathrm{\;d}s - {\iint }_{{\Delta }_{r}}\frac{{x}^{2} + {y}^{2}}{2} \cdot {\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}x\mathrm{\;d}y}\right\rbrack $$
$$ = \mathop{\lim }\limits_{{r \rightarrow 1}}\left\lbrack {\frac{1}{2}{\iint }_{{\Delta }_{r}}\left( {\frac{{\partial }^{2}f}{\partial {x}^{2}} + \frac{{\partial }^{2}f}{\partial {y}^{2}}}\right) \mathrm{d}x\mathrm{\;d}y - {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{1}\frac{{t}^{2}}{2}{\mathrm{e}}^{-{t}^{2}} \cdot t\mathrm{\;d}t}\right\rbrack $$
$$ = \frac{1}{2}{\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{1}{\mathrm{e}}^{-{t}^{2}}t\mathrm{\;d}t - {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{1}\frac{{t}^{3}}{2}{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t = \frac{\pi }{2\mathrm{e}}. $$
证法 3 由证法 1 看出,若 $f\left( {x,y}\right)$ 是调和函数,即 $\frac{{\partial }^{2}f}{\partial {x}^{2}} + \frac{{\partial }^{2}f}{\partial {y}^{2}} =$ 0,则积分 $\displaystyle{\iint }_{{x}^{2} + {y}^{2} < 1}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = 0$ . 所以证明结果只要找一特解即成. 设要找的解为 $f\left( {x,y}\right) = f\left( \sqrt{{x}^{2} + {y}^{2}}\right) = f\left( r\right)$ ,则
$$ \frac{\partial f}{\partial x} = {f}^{\prime }\left( r\right) \frac{x}{r},\;\frac{{\partial }^{2}f}{\partial {x}^{2}} = {f}^{\prime \prime }\left( r\right) \frac{{x}^{2}}{{r}^{2}} + {f}^{\prime }\left( r\right) \frac{1}{r} - {f}^{\prime }\left( r\right) \frac{{x}^{2}}{{r}^{3}}. $$
同理
$$ \frac{{\partial }^{2}f}{\partial {y}^{2}} = {f}^{\prime \prime }\left( r\right) \frac{{y}^{2}}{{r}^{2}} + {f}^{\prime }\left( r\right) \frac{1}{r} - {f}^{\prime }\left( r\right) \frac{{y}^{2}}{{r}^{3}}. $$
把结果代入 $f$ 满足的方程,得
$$ {f}^{\prime \prime }\left( r\right) + {f}^{\prime }\left( r\right) \frac{1}{r} = {\mathrm{e}}^{-{r}^{2}}\text{ 或 }\;\frac{\mathrm{d}}{\mathrm{d}r}\left( {r\frac{\mathrm{d}f}{\mathrm{\;d}r}}\right) = r{\mathrm{e}}^{-{r}^{2}}. $$
积分一次得 $r\frac{\mathrm{d}f}{\mathrm{\;d}r} = - \frac{1}{2}{\mathrm{e}}^{-{r}^{2}} + C$ . 为了保证函数的连续性要求,任意常数 $C$ 必须取 $1/2$ ,即
$$ r\frac{\mathrm{d}f}{\mathrm{\;d}r} = \frac{1}{2}\left( {1 - {\mathrm{e}}^{-{r}^{2}}}\right) . $$
于是
$$ {\iint }_{{x}^{2} + {y}^{2} < 1}\left( {x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{1}\frac{1}{2}\left( {1 - {\mathrm{e}}^{-{r}^{2}}}\right) \cdot r\mathrm{\;d}r = \frac{\pi }{2\mathrm{e}}. $$