可方向导不一定连续的例子
f(x,y)={x2yx4+y2,(x,y)≠(0,0)0,(x,y)=(0,0) f(x,y)=\left\{\begin{aligned} &\frac{x^2y}{x^4+y^2},\quad &(x,y)\ne(0,0)\\ &0,&(x,y)=(0,0) \end{aligned}\right. f(x,y)=⎩⎪⎨⎪⎧x4+y2x2y,0,(x,y)=(0,0)(x,y)=(0,0)
可方向导的证明:
- 因为 fff 在两个坐标轴上恒为零,所以 fff 在 (0,0)(0,0)(0,0) 处的偏导数为零
- 取单位向量 u=(u1,u2)u=(u_1,u_2)u=(u1,u2),当 u2≠0u_2\ne 0u2=0 时,有:
∂f∂u(0,0)=limt→01tt3u12u2t4u14+t2u22=u12u2t2u14+u22=u12u2 \frac{\partial f}{\partial u}(0,0)=\lim_{t\to 0}\frac{1}{t}\frac{t^3u_1^2u_2}{t^4u_1^4+t^2u_2^2}=\frac{u_1^2u_2}{t^2u_1^4+u_2^2}=\frac{u_1^2}{u_2} ∂u∂f(0,0)=t→0limt1t4u14+t2u22t3u12u2=t2u14+u22u12u2=u2u12
这说明 fff 在 (0,0)(0,0)(0,0) 处方向导数都存在
fff 不连续的证明:
limx→0y→kx2=limx→0k2x4x4+k2x4=limk21+k2\lim_{\tiny \begin{aligned} x&\to0\\y&\to kx^2\end{aligned}}=\lim_{x\to 0}\frac{k^2x^4}{x^4+k^2x^4}=\lim\frac{k^2}{1+k^2} xy→0→kx2lim=x→0limx4+k2x4k2x4=lim1+k2k2