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
\[\text D\boldsymbol\varphi(\boldsymbol x_1)=-\frac{\text D_x\boldsymbol F(\boldsymbol x_1,\boldsymbol\varphi(\boldsymbol x_1))}{\text D_y\boldsymbol F(\boldsymbol x_1,\boldsymbol\varphi(\boldsymbol x_1))}. \]练习
\(\bf Example\;1.\quad\)证明：由隐函数方程 \(y=x\varphi(z)+\psi(z)\) 确定的二元函数 \(z:=z(x,y)\) 满足方程
\[\left(\frac{\partial z}{\partial y}\right)^2\frac{\partial^2z}{\partial x^2}-2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{\partial^2z}{\partial x\partial y}+\left(\frac{\partial z}{\partial x}\right)^2\frac{\partial^2z}{\partial y^2}=0. \]\(\bf Example\;2.\quad\)解方程（只需给出隐函数形式）
\[\left(\frac{\partial z}{\partial y}\right)^2\frac{\partial^2z}{\partial x^2}-2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\frac{\partial^2z}{\partial x\partial y}+\left(\frac{\partial z}{\partial x}\right)^2\frac{\partial^2z}{\partial y^2}=0. \]\[\begin{aligned} \frac{\partial}{\partial x_i}&=\sum_{j=1}^n\frac{\partial}{\partial x_j} \frac{\partial x_j}{\partial x_i}\\ 0&=\sum_{j\neq i}\frac{\partial}{\partial x_j} \frac{\partial x_j}{\partial x_i}\\ 0&= \frac{\partial x_k}{\partial x_i}+\sum_{j\neq i,j\neq k}\frac{\partial x_k}{\partial x_j} \frac{\partial x_j}{\partial x_i} \end{aligned} \] 标签：mathbb,frac,函数,text,boldsymbol,varphi,partial,反函数 From： https://www.cnblogs.com/SalomeJLQ/p/18337725