下册 7.3 微分方程的验证及变量替换 第4题

\section*{解题过程：} （1）在等式 $f(t x, t y)=t^{n} f(x, y)$ 两边对 $t$ 求导得 $$ \frac{\partial f(t x, t y)}{\partial t}=x f_{1}(t x, t y)+y f_{2}(t x, t y)=n t^{n-1} f(x, y) . $$ 将 $t=1$ 代人即得 $\displaystyle x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=n f$ ． （2）必要性．设可微函数 $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ 为 $k$ 齐次的，则 $\forall t>0$ ， $$ f\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)=t^{k} f\left(x_{1}, x_{2}, \cdots, x_{n}\right) . $$ 在等式两边对 $t$ 求导得 $\displaystyle \frac{\partial f\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)}{\partial t}=x_{1} f_{1}\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)+x_{2} f_{2}\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)+\cdots+x_{n} f_{n}\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)=k t^{k-1} f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. 将 $t=1$ 代人即得 $\displaystyle \sum_{i=1}^{n} x_{i} \frac{\partial f}{\partial x_{i}}=k f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ ． 充分性：任意固定定义域中一点 $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ ，记 $\displaystyle F(t)=\frac{f\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)}{t^{k}},(t>0)$ ，则 $F(t)$ 在 $t>0$ 时是可微的，且 $$ F^{\prime}(t)=\frac{1}{t^{k}}\left\{x_{1} f_{1}\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)+x_{2} f_{2}\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)+\cdots+x_{n} f_{n}\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)\right\}-\frac{k}{t^{k+1}} f\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right) \equiv 0 $$ 从而当 $t>0$ 时 $F(t)=c$（与 $t$ 无关的常数）．在函数 $F(t)$ 的等式中令 $t=1$ 得 $c=F(1)=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ ．于是 $$ F(t)=\frac{f\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)}{t^{k}}=f\left(x_{1}, x_{2}, \cdots, x_{n}\right), $$ 即 $\forall t>0, f\left(t x_{1}, t x_{2}, \cdots, t x_{n}\right)=t^{k} f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ ，即可微函数 $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ 为 $k$ 次齐次函数。 （3）须证 $\forall t>0, g(t x, t y)=\operatorname{tg}(x, y)$ 或 $x g_{x}+y g_{y}=g(x, y)$ ． 在 $f(t x, t y, t z)=t^{n} f(x, y, z)$ 中分别对 $x, y, z, t$ 求导得 于是 $$ \begin{aligned} & f_{1}(t x, t y, t z)=t^{n-1} f_{1}(x, y, z), f_{2}(t x, t y, t z)=t^{n-1} f_{2}(x, y, z), f_{3}(t x, t y, t z)=t^{n-1} f_{3}(x, y, z) . \\ & x f_{1}(t x, t y, t z)+y f_{2}(t x, t y, t z)+z f_{3}(t x, t y, t z)=(n-1) t^{n-1} f(x, y, z) . \end{aligned} $$ 即 $$ x t^{n-1} f_{1}(x, y, z)+y t^{n-1} f_{2}(x, y, z)+z t^{n-1} f_{3}(x, y, z)=(n-1) t^{n-1} f(x, y, z), $$ 由此得 $$ x f_{1}(x, y, z)+y f_{2}(x, y, z)+z f_{3}(x, y, z)=(n-1) f(x, y, z) . $$ 由此得 $\displaystyle \frac{x f_{1}+y f_{2}}{f_{3}}=(n-1) \frac{f(x, y, z)}{f_{3}}-z$ ． 由 $f(x, y, z)=0$ 确定的隐函数 $z=g(x, y)$ 满足： $$ g_{x}(x, y)=-\frac{f_{1}}{f_{3}}, g_{y}(x, y)=-\frac{f_{2}}{f_{3}}, f(x, y, g(x, y))=0 . $$ 于是 $\displaystyle x g_{x}(x, y)+y g_{y}(x, y)=-\left(x \frac{f_{1}}{f_{3}}+y \frac{f_{2}}{f_{3}}\right)=-\frac{x f_{1}+y f_{2}}{f_{3}}=-(n-1) \frac{f(x, y, g(x, y))}{f_{3}}+g(x, y)=g(x, y)$ ． 由（2）知，$z=g(x, y)$ 为一次齐次函数。