下册 7.3 微分方程的验证及变量替换 第3题

\section*{解题过程：} （1）对方程 $z=x+y \varphi(z)$ 两边求微分得 $$ \mathrm{d} z=\mathrm{d} x+\varphi(z) \mathrm{d} y+y \varphi^{\prime}(z) \mathrm{d} z . $$ 于是 $$ z_{x}=\frac{1}{1-y \varphi^{\prime}}, z_{y}=\frac{\varphi}{1-y \varphi^{\prime}}, z_{y}=\frac{\varphi}{1-y \varphi^{\prime}}=\varphi z_{x} . $$ 所以 $$ u_{x}=f^{\prime}(z) z_{x}=f^{\prime}(z) \frac{-1}{1-y \varphi^{\prime}}, u_{y}=f^{\prime}(z) z_{y}=f^{\prime}(z) \frac{\varphi}{1-y \varphi^{\prime}} . $$ 故 $\displaystyle \frac{\partial u}{\partial y}=\varphi(z) \frac{\partial u}{\partial x}$ ． （2）由 $u_{x}=\varphi+x \varphi^{\prime} u_{x}$ 有 $\displaystyle u_{x}=\frac{\varphi}{1-x \varphi^{\prime}}$ ．由 $u_{y}=1+x \varphi^{\prime} u_{y}$ 有 $\displaystyle u_{y}=\frac{1}{1-x \varphi^{\prime}}$ ，所以 $$ \begin{aligned} & u_{x x}=\frac{\varphi^{\prime} u_{x}\left(1-x \varphi^{\prime}\right)-\varphi\left(-\varphi^{\prime}-x \varphi^{\prime \prime} u_{x}\right)}{\left(1-x \varphi^{\prime}\right)^{2}}=\frac{1}{\left(1-x \varphi^{\prime}\right)^{2}}\left(2 \varphi \varphi^{\prime}+\frac{x \varphi^{2} \varphi^{\prime \prime}}{1-x \varphi^{\prime}}\right) . \\ & \frac{\partial}{\partial y}\left(\varphi^{2}(u) u_{y}\right)=\frac{\partial}{\partial y}\left(\frac{\varphi^{2}}{1-x \varphi^{\prime}}\right)=\frac{2 \varphi \varphi^{\prime} u_{y}\left(1-x \varphi^{\prime}\right)-\varphi^{2}\left(-x \varphi^{\prime \prime} u_{y}\right)}{\left(1-x \varphi^{\prime}\right)^{2}}=\frac{1}{\left(1-x \varphi^{\prime}\right)^{2}}\left(2 \varphi \varphi^{\prime}+\frac{x \varphi^{2} \varphi^{\prime \prime}}{1-x \varphi^{\prime}}\right) . \end{aligned} $$ 所以 $\displaystyle \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial y}\left(\varphi^{2}(u) \frac{\partial u}{\partial y}\right)$ ． （3）对方程 $z=x+y \varphi(z)$ 求微分得 $$ \mathrm{d} z=\mathrm{d} x+\varphi(z) \mathrm{d} y+y \varphi^{\prime}(z) \mathrm{d} z . $$ 于是 $$ z_{x}=\frac{1}{1-y \varphi^{\prime}}, z_{y}=\frac{\varphi}{1-y \varphi^{\prime}}, z_{y}=\frac{\varphi}{1-y \varphi^{\prime}}=\varphi z_{x} . $$ 所以 $$ u_{x}=f^{\prime}(z) z_{x}=f^{\prime}(z) \frac{1}{1-y \varphi^{\prime}}, u_{y}=f^{\prime}(z) z_{y}=f^{\prime}(z) \frac{\varphi}{1-y \varphi^{\prime}} . $$ 于是 $\displaystyle \frac{\partial u}{\partial y}=\varphi(z) \frac{\partial u}{\partial x}$ ． 假设 $n=k-1$ 时成立 $\displaystyle \frac{\partial^{k-1} u}{\partial y^{k-1}}=\frac{\partial^{k-2}}{\partial x^{k-2}}\left(\varphi^{k-1}(z) \frac{\partial u}{\partial x}\right)$ ．下证：当 $n=k$ 时，$\displaystyle \frac{\partial^{k} u}{\partial y^{k}}=\frac{\partial^{k-1}}{\partial x^{k-1}}\left(\varphi^{k}(z) \frac{\partial u}{\partial x}\right)$ 成立． $$ \begin{aligned} \frac{\partial^{k} u}{\partial y^{k}} & =\frac{\partial}{\partial y}\left(\frac{\partial^{k-1} u}{\partial y^{k-1}}\right)=\frac{\partial}{\partial y}\left(\frac{\partial^{k-2}}{\partial x^{k-2}}\left(\varphi^{k-1}(z) \frac{\partial u}{\partial x}\right)\right)=\frac{\partial^{k-2}}{\partial x^{k-2} \partial y}\left(\varphi^{k-1}(z) \frac{\partial u}{\partial x}\right) \\ & =\frac{\partial^{k-2}}{\partial x^{k-2}}\left[(k-1) \varphi^{k-2}(z) \varphi^{\prime}(z) \frac{\partial z}{\partial y} \frac{\partial u}{\partial x}+\varphi^{k-1}(z) \frac{\partial^{2} u}{\partial x \partial y}\right] . \end{aligned} $$ 由 $\displaystyle \frac{\partial u}{\partial y}=\varphi(z) \frac{\partial u}{\partial x}$ 知 $$ \frac{\partial^{k-1}}{\partial x^{k-1}}\left(\varphi^{k}(z) \frac{\partial u}{\partial x}\right)=\frac{\partial^{k-1}}{\partial x^{k-1}}\left(\varphi^{k-1}(z) \frac{\partial u}{\partial y}\right)=\frac{\partial^{k-2}}{\partial x^{k-2}}\left[(k-1) \varphi^{k-2}(z) \varphi^{\prime}(z) \frac{\partial z}{\partial x} \frac{\partial u}{\partial y}+\varphi^{k-1}(z) \frac{\partial^{2} u}{\partial x \partial y}\right] . $$ 因 $\displaystyle z_{y}=\frac{\varphi}{1-y \varphi^{\prime}}=\varphi z_{x}$ 及 $\displaystyle \frac{\partial u}{\partial y}=\varphi(z) \frac{\partial u}{\partial x}$ 得 $\displaystyle z_{y} \frac{\partial u}{\partial x}=z_{x} \varphi(z) \frac{\partial u}{\partial x}=z_{x} \frac{\partial u}{\partial y}$ 。于是 $\displaystyle \frac{\partial^{k} u}{\partial y^{k}}=\frac{\partial^{k-1}}{\partial x^{k-1}}\left(\varphi^{k}(z) \frac{\partial u}{\partial x}\right)$ ． 由数学归纳法得证。