第六章 多元函数积分学 · 第6.6题

练习题

📝 题目

6. 6.9 设 $f\left( {x,y,z}\right)$ 是一次齐次函数, $F = \frac{1}{4}f\left( {x,y,z}\right) r$ . 试证:

$$ \operatorname{div}\mathbf{F} = f\left( {x,y,z}\right) . $$

💡 答案与解析

解答:

已知 $f(x,y,z)$ 是一次齐次函数,即对任意实数 $t>0$ 有 $$ f(tx,ty,tz)=t\,f(x,y,z). $$ 并且定义 $$ \mathbf{F} = \frac{1}{4} f(x,y,z)\, \mathbf{r}, $$ 其中 $\mathbf{r} = (x,y,z)$,即位置向量。需要证明 $$ \operatorname{div} \mathbf{F} = f(x,y,z). $$

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**步骤 1:写出散度的表达式**

设 $\mathbf{F} = (F_1, F_2, F_3)$,则 $$ F_1 = \frac{1}{4} f(x,y,z)\, x,\quad F_2 = \frac{1}{4} f(x,y,z)\, y,\quad F_3 = \frac{1}{4} f(x,y,z)\, z. $$

散度定义为 $$ \operatorname{div} \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}. $$

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**步骤 2:计算偏导数**

先计算 $\frac{\partial F_1}{\partial x}$: $$ \frac{\partial F_1}{\partial x} = \frac{1}{4} \left( \frac{\partial f}{\partial x} \cdot x + f \cdot 1 \right) = \frac{1}{4} \left( x \frac{\partial f}{\partial x} + f \right). $$

类似地, $$ \frac{\partial F_2}{\partial y} = \frac{1}{4} \left( y \frac{\partial f}{\partial y} + f \right), $$ $$ \frac{\partial F_3}{\partial z} = \frac{1}{4} \left( z \frac{\partial f}{\partial z} + f \right). $$

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**步骤 3:求和得到散度**

将三个偏导数相加: $$ \operatorname{div} \mathbf{F} = \frac{1}{4} \left( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} + 3f \right). $$

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**步骤 4:利用齐次函数的欧拉定理**

对于一次齐次函数 $f$,欧拉定理指出: $$ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = f(x,y,z). $$ (说明:因为齐次次数为1,所以左边等于次数乘以函数本身。)

代入上式: $$ \operatorname{div} \mathbf{F} = \frac{1}{4} \left( f + 3f \right) = \frac{1}{4} \cdot 4f = f. $$

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**结论**: $$ \boxed{\operatorname{div}\mathbf{F} = f(x,y,z)}. $$