全微分形式的不变性

日期: 2022-12-31 09:36 查看: 98 次

设函数 $z=f(u, v)$ 可微，
则有 $\mathrm{d} z=\frac{\partial z}{\partial u} \mathrm{~d} u+\frac{\partial z}{\partial v} \mathrm{~d} v$ ；
如果 $u=\varphi(x, y) ， v=\psi(x, y)$ 可微，则 $z=f(u(x, y), v(x, y))=f(x, y)$ 也可微，则

$$
\mathrm{d} z=\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y \text {, }
$$

$$
\begin{gathered}
\text { 由 }\left\{\begin{array}{l}
\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \\
\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}
\end{array}\right. \text { ，知 } \\
\mathrm{d} z=\left(\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}\right) \mathrm{d} x+\left(\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}\right) \mathrm{d} y \\
=\frac{\partial z}{\partial u}\left(\frac{\partial u}{\partial x} \mathrm{~d} x+\frac{\partial u}{\partial y} \mathrm{~d} y\right)+\frac{\partial z}{\partial v}\left(\frac{\partial v}{\partial x} \mathrm{~d} x+\frac{\partial v}{\partial y} \mathrm{~d} y\right)=\frac{\partial z}{\partial u} \mathrm{~d} u+\frac{\partial z}{\partial v} \mathrm{~d} v
\end{gathered}
$$

可以看出，无论 $z$ 是自变量 $u 、 v$ 的函数，还是中间变量 $u 、 v$ 的函数，它 的全微分形式是一样的，这个性质就叫做全微分形式的不变性.
例 10 设 $z=(x-y)^{x^2+y^2}$ ，求 $\frac{\partial z}{\partial x} ， \frac{\partial z}{\partial y}$.
解 $z=(x-y)^{x^2+y^2}=u^v$ ，

$$
\mathrm{d} z=\mathrm{d} u^v=\frac{\partial}{\partial u}\left(u^v\right) \mathrm{d} u+\frac{\partial}{\partial v}\left(u^v\right) \mathrm{d} v=v u^{v-1} \mathrm{~d} u+u^v \ln u \mathrm{~d} v ，
$$

$$
\mathrm{d} u=\mathrm{d}(x-y)=\mathrm{d} x-\mathrm{d} y, \quad \mathrm{~d} v=\mathrm{d}\left(x^2+y^2\right)=2 x \mathrm{~d} x+2 y \mathrm{~d} y
$$

例 10 设 $z=(x-y)^{x^2+y^2}$ ，求 $\frac{\partial z}{\partial x} ， \frac{\partial z}{\partial y}$.

$$
\mathrm{d} z=v u^{v-1} \mathrm{~d} u+u^v \ln u \mathrm{~d} v ， \quad \mathrm{~d} u=\mathrm{d} x-\mathrm{d} y ， \quad \mathrm{~d} v=2 x \mathrm{~d} x+2 y \mathrm{~d} y ，
$$

代入并归并含 $\mathrm{d} x$ 及 $\mathrm{d} y$ 的项，得

$$
\mathrm{d} u=(x-y)^{x^2+y^2}\left[\left(2 x \ln (x-y)+\frac{x^2+y^2}{x-y}\right) \mathrm{d} x+\left(2 y \ln (x-y)-\frac{x^2+y^2}{x-y}\right) \mathrm{d} y\right],
$$

故 $\frac{\partial u}{\partial x}=(x-y)^{x^2+y^2}\left(2 x \ln (x-y)+\frac{x^2+y^2}{x-y}\right)$ ，

$$
\frac{\partial u}{\partial y}=(x-y)^{x^2+y^2}\left(2 y \ln (x-y)-\frac{x^2+y^2}{x-y}\right) .

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