设$a_1,a_2,\cdots ,a_n$是非零常数.$f(x_1,x_2,\cdots ,x_n)$在$\mathbb{R}^n$上可微.求证：存在$\mathbb{R}$上一元可微函数$F(s)$使得$f(x_1,x_2,\cdots ,x_n)=F(a_1x_1+a_2x_2+\cdots +a_nx_n)$的充分必要条件是$a_j\frac{\partial f}{\partial x_i}=a_i\frac{\partial f}{\partial x_j},i,j=1,2,\cdots ,n$
必要性：
$$\frac{\partial F}{\partial x_i}=a_iF’=\frac{\partial f}{\partial x_i}\therefore a_j\frac{\partial f}{\partial x_i}=a_i\frac{\partial f}{\partial x_j}$$
充分性：
设
$$
\left\lbrace
\begin{aligned}
u_1=&a_1x_1+a_2x_2+\cdots +a_nx_n \\
u_2=&\phantom{a_1x_1+ } a_2x_2\phantom{+\cdots +a_nx_n}\\
\cdots \\
u_n=&\phantom{a_1x_1+a_2x_2+\cdots + }a_nx_n
\end{aligned}
\right.
$$
则
$$
\left\lbrace
\begin{aligned}
x_1&=\frac{1}{a_1}(u_1-u_2-\cdots-u_n)\\
x_2&=\frac{u_2}{a_2}\\
\cdots \\
x_n&=\frac{u_n}{a_n}\\
\end{aligned}
\right.
$$
$$\therefore f=f(x_1(u_1,u_2,\cdots,u_n),x_2(u_1,u_2,\cdots,u_n),\cdots,x_n(u_1,u_2,\cdots,u_n))$$
$$\frac{\partial f}{\partial u_i}=
\left\lbrace
\begin{aligned}
\frac{1}{a_1}f_{x_1}’,i=1\\
-\frac{1}{a_1}f_{x_1}’+\frac{1}{a_i}f_{x_i}’,i\neq 1
\end{aligned}
\right.
$$
$$
\because a_j\frac{\partial f}{\partial x_i}=a_i\frac{\partial f}{\partial x_j},i,j=1,2,\cdots ,n$$
$$
\therefore \frac{1}{a_i}\frac{\partial f}{\partial x_i}=\frac{1}{a_j}\frac{\partial f}{\partial x_j},i,j=1,2,\cdots ,n
$$
$$
\therefore
\frac{\partial f}{\partial u_i}=
\left\lbrace
\begin{aligned}
\frac{1}{a_1}f_{x_1}’,i=1\\
0,i\neq 1
\end{aligned}
\right.
$$
$$
\therefore
f=F(u_1)=F(a_1x_1+a_2x_2+\cdots +a_nx_n)
$$
若函数$u=f(x,y,z)$满足恒等式$f(tx,ty,tz)=t^kf(x,y,z)(t>0)$,则称$f(x,y,z)$为$k$次齐次函数。试证下述关于齐次函数的欧拉定理：可微函数$f(x,y,z)$为$k$次齐次函数齐次函数的充要条件是：
$$xf’_x(x,y,z)+yf’_y(x,y,z)+zf’_z(x,y,z)=kf(x,y,z)$$
必要性：
$$\because f(tx,ty,tz)=t^kf(x,y,z)(t>0)$$
$$\therefore \frac{\mathop{}\mathrm{d}{f}}{\mathop{}\mathrm{d}{t}}
=xf’_x(tx,ty,tz)+yf’_y(tx,ty,tz)+zf’_z(tx,ty,tz)=kt^{k-1}f(x,y,z)$$
令$t=1$
$$xf’_x(x,y,z)+yf’_y(x,y,z)+zf’_z(x,y,z)=kf(x,y,z)$$
充分性：
设$$\phi (t)=\frac{f(tx,ty,tz)}{t^k}$$
$$
\begin{aligned}
\phi’ &=\frac{1}{t^{2k}}(t^k(xf’_x(tx,ty,tz)+yf’_y(tx,ty,tz)+zf’_z(tx,ty,tz))-kt^{k-1}f(tx,ty,tz))\\
&=\frac{1}{t^{k+1}}(t(xf’_x(tx,ty,tz)+yf’_y(tx,ty,tz)+zf’_z(tx,ty,tz))-kf(tx,ty,tz))
&=0
\end{aligned}
$$
$$\therefore \phi(t)=\phi(1)=f(x,y,z)$$
$$\therefore f(tx,ty,tz)=t^kf(x,y,z)$$