可方向导不一定连续的例子

可方向导不一定连续的例子

f ( x , y ) = { x 2 y x 4 + y 2 , ( 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 f f f 在两个坐标轴上恒为零,所以 f f f ( 0 , 0 ) (0,0) (0,0) 处的偏导数为零
  • 取单位向量 u = ( u 1 , u 2 ) u=(u_1,u_2) u=(u1,u2),当 u 2 ≠ 0 u_2\ne 0 u2=0 时,有:
    ∂ f ∂ u ( 0 , 0 ) = lim ⁡ t → 0 1 t t 3 u 1 2 u 2 t 4 u 1 4 + t 2 u 2 2 = u 1 2 u 2 t 2 u 1 4 + u 2 2 = u 1 2 u 2 \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} uf(0,0)=t0limt1t4u14+t2u22t3u12u2=t2u14+u22u12u2=u2u12
    这说明 f f f ( 0 , 0 ) (0,0) (0,0) 处方向导数都存在

f f f 不连续的证明:

lim ⁡ x → 0 y → k x 2 = lim ⁡ x → 0 k 2 x 4 x 4 + k 2 x 4 = lim ⁡ k 2 1 + k 2 \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} xy0kx2lim=x0limx4+k2x4k2x4=lim1+k2k2

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转载自blog.csdn.net/Infinity_07/article/details/109668493