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全微分

1 全微分的严格定义

有了直观之后,下面来看看数学家给出的全微分定义:

定义 .设函数z=f(x,y)(x_0,y_0)点的某邻域内有定义,令:

\Delta x=\mathrm{d}x=x-x_0,\quad \Delta y=\mathrm{d}y=y-y_0

若函数z=f(x,y)(x_0,y_0)点的 全增量 (Total increment):

\Delta z=f(x_0+\Delta x, y_0+\Delta y)-f(x_0, y_0)

可表示为:

\Delta z=A\Delta x+B\Delta y+o(\rho)

其中AB不依赖于\Delta x\Delta y而仅与xy相关,且:

\rho=\sqrt{(\Delta x)^2+(\Delta y)^2}

那么称z=f(x,y)(x_0,y_0) 可微分 (Differentiable),而A\Delta x+B\Delta y称为z=f(x,y)(x_0,y_0)点的 全微分 (Total differential),此时通常改写为A\mathrm{d}x+B\mathrm{d}y,并记作\mathrm{d}z,即:

\mathrm{d}z=A\mathrm{d}x+B\mathrm{d}y

上述定义看起来很复杂,但重点是其中的三个式子:

  • \Delta z=f(x_0+\Delta x, y_0+\Delta y)-f(x_0, y_0),这是曲面的表达式
  • \mathrm{d}z=A\mathrm{d}x+B\mathrm{d}y,这是平面的表达式,也就是曲面在(x_0,y_0)点的全微分的表达式
  • \Delta z=A\Delta x+B\Delta y+o(\rho),该式可改写为o(\rho)=\Delta z-(A\Delta x+B\Delta y),所以该式说的是曲面\Delta z和全微分\mathrm{d} z相差o(\rho)

上述三个式子以及彼此的关系,如下图所示。

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mapython only code

所以上述定义其实说的就是:

\text{若}\ \text{曲面}-\text{平面}=o(\rho)\implies\text{该平面就是曲面的全微分}

2 全微分的计算方法
定理 .若函数z=f(x,y)(x_0,y_0)可微分,则偏导数f_x(x_0,y_0)f_y(x_0,y_0)必定存在,且z=f(x,y)在点(x_0,y_0)全微分为:

\mathrm{d}z=f_x(x_0,y_0)\mathrm{d}x+f_y(x_0,y_0)\mathrm{d}y

证明 .因为z=f(x,y)(x_0,y_0)可微分,所以对于(x_0,y_0)点某邻域内任意的(x_0+\Delta x,y_0+\Delta y)点,有:

\Delta z=A\Delta x+B\Delta y+o(\rho)=A\Delta x+B\Delta y+o(\sqrt{(\Delta x)^2+(\Delta y)^2})

\Delta y=0时,上式也成立,即有:

\Delta z=A\Delta x+o(|\Delta x|)

两边同时除以\Delta x。再令\Delta x\to 0,可得:


\begin{aligned}
    \frac{\Delta z}{\Delta x}=A+\frac{o(|\Delta x|)}{\Delta x}
        &\implies A=\frac{\Delta z}{\Delta x}-\frac{o(|\Delta x|)}{\Delta x}\implies A=\lim_{\Delta x\to 0}\frac{\Delta z}{\Delta x}-\lim_{\Delta x\to 0}\frac{o(|\Delta x|)}{\Delta x}\\
        &\implies A=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=f_x(x_0,y_0)\\
\end{aligned}

同理可得:

B=\lim_{\Delta y\to 0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}=f_y(x_0,y_0)

所以偏导数f_x(x_0,y_0)f_y(x_0,y_0)存在,且这两个偏导数就是AB,所以:

\mathrm{d}z=f_x(x_0,y_0)\mathrm{d}x+f_y(x_0,y_0)\mathrm{d}y

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3 可微分的充分条件
定理 .已知函数z=f(x,y)偏导数\frac{\partial f}{\partial x}\frac{\partial f}{\partial y},那么:

\frac{\partial f}{\partial x}、\frac{\partial f}{\partial y}\ 在\ (x_0,y_0)\ 连续\implies f(x,y)\ 在\ (x_0,y_0)\ 点可微分

证明 .根据偏导数\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}(x_0,y_0)连续,可知偏导数\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}(x_0,y_0)点的某邻域内存在。设点(x_0+\Delta x,y_0+\Delta y)为该邻域内任意一点,则:


\begin{aligned} 
    \Delta z
        &=f(x_0+\Delta x, y_0+\Delta y)-f(x_0, y_0) \\ 
        &=[f(x_0+\Delta x, y_0+\Delta y)-f(x_0, y_0+\Delta y)]+[f(x_0, y_0+\Delta y)-f(x_0, y_0)] 
\end{aligned}

第一个方括号内的表达式,由于y_0+\Delta y不变,所以可看作是x的一元函数f(x, y_0+\Delta y)的增量,应用拉格朗日中值定理,得到:

f(x_0+\Delta x, y_0+\Delta y)-f(x_0, y_0+\Delta y)=f_{x}\left(x_0+\theta_{1} \Delta x, y_0+\Delta y\right) \Delta x, \quad 0<\theta_{1}<1

因为偏导数f_x(x,y)(x_0,y_0)连续,又多元函数的极限也有类似于极限与无穷小的关系,即类似有\lim f(x)=L\implies f(x)=L+\alpha,其中\alpha无穷小,所以:

\lim_{(x,y)\to (x_0,y_0)}f_{x}(x_0+\theta_{1} \Delta x, y_0+\Delta y)=f_x(x_0,y_0)\implies f_{x}(x_0+\theta_{1} \Delta x, y_0+\Delta y)=f_x(x_0,y_0)+\varepsilon_1

其中\varepsilon_1满足\lim_{(x,y)\to (x_0,y_0)}\varepsilon_1=0,所以第一个方括号的式子最终可以写作:

f(x_0+\Delta x, y_0+\Delta y)-f(x_0, y_0+\Delta y)=f_{x}(x_0, y_0) \Delta x+\varepsilon_{1} \Delta x

同样的道理,第二个方括号中的表示式可以写作:

f(x_0, y_0+\Delta y)-f(x_0, y_0)=f_{y}(x_0, y_0) \Delta y+\varepsilon_{2} \Delta y

其中\varepsilon_2满足\lim_{y\to y_0}\varepsilon_2=0,所以:

\Delta z=f_{x}(x_0, y_0) \Delta x+f_y(x_0, y_0) \Delta y+\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y

\rho=\sqrt{(\Delta x)^2+(\Delta y)^2},容易看出:

\left|\frac{\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y}{\rho}\right|=\left|\frac{\varepsilon_{1} \Delta x}{\rho}+\frac{\varepsilon_{2} \Delta y}{\rho}\right|\le|\frac{\varepsilon_{1} \Delta x}{\rho}|+|\frac{\varepsilon_{2} \Delta y}{\rho}|\le|\varepsilon_{1}|+|\varepsilon_{2}|

类似于单变量微积分中的夹逼定理,根据上式可推出:


\begin{aligned}
    \left|\frac{\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y}{\rho}\right|\le|\varepsilon_{1}|+|\varepsilon_{2}|
        &\implies -|\varepsilon_{1}|-|\varepsilon_{2}|\le\frac{\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y}{\rho}\le|\varepsilon_{1}|+|\varepsilon_{2}|\\
        &\implies\lim_{\rho\to 0}\frac{\varepsilon_{1}\Delta x+\varepsilon_{2}\Delta y}{\rho}=0\implies \varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y=o(\rho)
\end{aligned}

所以:

\Delta z=f_{x}(x_0, y_0) \Delta x+f_y(x_0, y_0) \Delta y+\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y=f_{x}(x_0, y_0) \Delta x+f_y(x_0, y_0) \Delta y+o(\rho)

这也就证明了f(x,y)(x_0,y_0)可微分

blanksquare
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