反函数与隐函数

隐函数定理

\(\bf Theorem\;1.\quad\)设 \(\Omega\in\mathbb R^m\times\mathbb R^n\) 为开集，\(\boldsymbol F(\boldsymbol x,\boldsymbol y)\)（其中 \(\boldsymbol x\in\mathbb R^m,\boldsymbol y\in\mathbb R^n\)）满足 \(\boldsymbol F(\boldsymbol x,\boldsymbol y)\in C^1(\Omega,\mathbb R^n)\)。设点 \((\boldsymbol x_0,\boldsymbol y_0)\in\Omega\)，若 \(\boldsymbol F(\boldsymbol x_0,\boldsymbol y_0)=\boldsymbol 0\)，且

\[\det\text D_y\boldsymbol F(\boldsymbol x_0,\boldsymbol y_0)=\frac{\partial(F_1,\cdots,F_n)}{\partial(y_1,\cdots,y_n)}\Bigg|_{(\boldsymbol x_0,\boldsymbol y_0)}\neq0, \]

则存在一个邻域 \(U(\boldsymbol x_0)\in\mathbb R^m\) 及唯一 \(\boldsymbol\varphi(\boldsymbol x)\) 满足

\[\boldsymbol\varphi(\boldsymbol x)\in C^1(U(\boldsymbol x_0),\mathbb R^n) \]

使 \(\forall\boldsymbol x_1\in U(\boldsymbol x_0),(\boldsymbol x_1,\boldsymbol\varphi(\boldsymbol x_1))\in\Omega\)，且

\[\begin{aligned} \boldsymbol F(\boldsymbol x_1,\boldsymbol\varphi(\boldsymbol x_1))=\text D_x\boldsymbol F(\boldsymbol x_1,\boldsymbol\psi(\boldsymbol x_1))+\text D_y\boldsymbol F(\boldsymbol x_1,\boldsymbol\psi(\boldsymbol x_1))\text D\boldsymbol\varphi(\boldsymbol x_1)=\boldsymbol 0, \end{aligned} \]

并且有

\[\text D\boldsymbol\varphi(\boldsymbol x_1)=-\frac{\text D_x\boldsymbol F(\boldsymbol x_1,\boldsymbol\varphi(\boldsymbol x_1))}{\text D_y\boldsymbol F(\boldsymbol x_1,\boldsymbol\varphi(\boldsymbol x_1))}. \]

复合依赖关系

记 \(\mathbb R(\!\!\{\!\mathcal X\!\}\!\!)\) 为 \(\mathbb R\) 上以 \(\mathcal X\) 为变元集的形式多元函数。对于 \(\mathbb R(\!\!\{\!\mathcal X\!\}\!\!)\) 定义“代入算符”\((\circ)\)，其中 \(\circ\) 属于 \(\prod_{j\leq|\mathcal X|}\mathbb R(\!\!\{\!\mathcal Y_J\!\}\!\!)\) 或 \(\mathbb R^{|\mathcal X|}\)，表示将 \(f\in\mathbb R(\!\!\{\!\mathcal X\!\}\!\!)\) 中的变元逐维用 \(\circ\) 代入，映至 \(\mathbb R\big(\!\!\big\{\!\coprod_{j\leq|\mathcal X|}\mathcal Y_J\!\big\}\!\!\big)\) 或 \(\mathbb R\)。

下面变元在 \(\mathcal Z=\{X_1,\cdots,X_n\}\) 中取，考虑 \(\mathbb R(\!\!\{\!\mathcal X_{i}\!\}\!\!):=\mathbb R(\!\!\{\!X_1,\cdots,X_{i-1},X_{i+1},\cdots,X_n\!\}\!\!)\) 为 \(\mathbb R\) 上以 \(\mathcal X_i\) 为变元集的形式 \(n-1\) 元函数的集合，记算子

\[\text D_{X_j}:=\frac{\partial}{\partial X_j}:\quad\mathbb R(\!\!\{\!\mathcal X_{k}\!\}\!\!)\to\mathbb R(\!\!\{\!\mathcal X_k\!\}\!\!),\;(k\neq j) \]

为对分量 \(X_j\) 的形式偏导数。

现有形式函数 \(f\in\mathbb R(\!\!\{\!\mathcal Z\!\}\!\!)\)，考虑商集

\[\mathbb R(\!\!\{\!\mathcal Z\!\}\!\!)/[f]:=\mathbb R(\!\!\{\!\mathcal Z\!\}\!\!)\big/\sim \;:\quad f\sim 0. \]

在此商集下，可解出 \(n\) 个形式 \(n-1\) 元函数

\[\widetilde{x_i}\in R(\!\!\{\!\mathcal X_{i}\!\}\!\!):X_i\sim\widetilde{x_i}. \]

从而商集下有等式

\[\widetilde{x_i}\;\sim\;\widetilde{x_i}(\widetilde{x_1},\cdots,\widetilde{x_{i-1}},\widetilde{x_{i+1}},\cdots,\widetilde{x_n}), \]

记 \(\boldsymbol x_i=(\widetilde{x_1},\cdots,\widetilde{x_{i-1}},\widetilde{x_{i+1}},\cdots,\widetilde{x_n})\in R(\!\!\{\!\mathcal Z\!\}\!\!)^{n-1},\boldsymbol x=(\widetilde{x_i})_{i\leq n}\)，两端求偏导数得

\[\frac{\partial\widetilde{x_i}}{\partial X_k}\sim\sum_{j=1}^{n}\frac{\partial\widetilde{x_i}}{\partial X_j}(\boldsymbol x_i)\frac{\partial\widetilde{x_j}}{\partial X_k}. \]

留意到 \(X_i\sim\widetilde{x_i}\) 蕴含 \(\text D_{X_i}\widetilde{x_i}=1\)，因此上式给出：在商集 \(\mathbb R(\!\!\{\!\mathcal Z\!\}\!\!)/[f]\) 中有

\[\sum_{j\neq k}\frac{\partial}{\partial X_j}(\boldsymbol x)\frac{\partial\widetilde{x_j}}{\partial X_k}\sim\sum_{j\neq i}\frac{\partial}{\partial X_j}(\boldsymbol x)\frac{\partial\widetilde{x_j}}{\partial X_k}\in[0]. \]

练习

\(\bf Example\;1.\quad\)证明：由方程 \(y=x\varphi(z)+\psi(z)\) 确定的二元函数 \(z:=z(x,y)\) 满足方程

\[\left(\frac{\partial z}{\partial y}\right)^2\frac{\partial^2z}{\partial x^2}-2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{\partial^2z}{\partial x\partial y}+\left(\frac{\partial z}{\partial x}\right)^2\frac{\partial^2z}{\partial y^2}=0. \]

\(\bf Example\;2.\quad\)解出关于 \(y:=y(x,z)\) 的方程

\[\left(\frac{\partial z}{\partial y}\right)^2\frac{\partial^2z}{\partial x^2}-2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{\partial^2z}{\partial x\partial y}+\left(\frac{\partial z}{\partial x}\right)^2\frac{\partial^2z}{\partial y^2}=0. \]