下册 7.3 微分方程的验证及变量替换 第7题

\section*{解题过程：} （1）必要性：微分方程的特征方程为 $\displaystyle \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^{2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+1=0,\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}+1\right)^{2}=0$ ，特征线为 $x+y=C$ ． 作变换 $\xi=x+y, \eta=x$ ，则 $$ \begin{gathered} u_{x}=u_{\xi}+u_{\eta}, u_{y}=u_{\xi} \\ u_{x x}=u_{\xi \xi}+u_{\xi \eta}+u_{\eta \xi}+u_{\eta \eta}=u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}, u_{x y}=u_{\xi \xi}+u_{\eta \xi}, u_{y y}=u_{\xi \xi} \end{gathered} $$ 于是原方程化为 $u_{\eta \eta}=0$ ，从而 $u_{\eta}=f(\xi), u=\eta f(\xi)+g(\xi)$ ．故 $$ u(x, y)=x f(x+y)+g(x+y)=x f(x+y)+(x+y) \frac{g(x+y)}{x+y}=x \varphi(x+y)+y \psi(x+y) $$ 充分性：若 $u(x, y)=x \varphi(x+y)+y \psi(x+y)$ ，则 $$ \begin{gathered} u_{x}=\varphi+x \varphi^{\prime}+y \psi^{\prime}, u_{y}=x \varphi^{\prime}+\psi+y \psi^{\prime} \\ u_{x x}=2 \varphi^{\prime}+x \varphi^{\prime \prime}+y \psi^{\prime \prime}, u_{x y}=\varphi^{\prime}+x \varphi^{\prime \prime}+\psi^{\prime}+y \psi^{\prime \prime}, u_{y y}=x \varphi^{\prime \prime}+2 \psi^{\prime}+y \psi^{\prime \prime} \\ \text { 左 }=2 \varphi^{\prime}+x \varphi^{\prime \prime}+y \psi^{\prime \prime}-2\left(\varphi^{\prime}+x \varphi^{\prime \prime}+\dot{\psi}^{\prime}+y \psi^{\prime \prime}\right)+x \varphi^{\prime \prime}+2 \psi^{\prime}+y \psi^{\prime \prime}=0=\text { 右. } \end{gathered} $$ （2）由 $\displaystyle u \frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial u}{\partial x} \frac{\partial u}{\partial y}$ 得 $\displaystyle \frac{\partial}{\partial y}\left(\frac{1}{u} \frac{\partial u}{\partial x}\right)=\frac{1}{u^{2}}\left(u \frac{\partial^{2} u}{\partial x \partial y}-\frac{\partial u}{\partial x} \frac{\partial u}{\partial y}\right)=0$ ，于是 $$ \frac{1}{u} \frac{\partial u}{\partial x}=f(x) $$ 解之得 $\displaystyle \frac{\partial \ln u}{\partial x}=f(x)$ ．于是 $\ln u=\int f(x) \mathrm{d} x=F(x)+G(y)$ ，即 $u=\mathrm{e}^{F(x)} \mathrm{e}^{G(y)}$ ． 反之，若 $u(x, y)=f(x) g(y)$ ，则 $$ u_{x}(\dot{x}, y)=f^{\prime}(x) g(y), u_{y}(x, y)=f(x) g^{\prime}(y), u_{x y}(x, y)=f^{\prime}(x) g^{\prime}(y) $$ 于是 $\displaystyle u \frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial u}{\partial x} \frac{\partial u}{\partial y}$ ． （3）若 $f(x, y, z)=g(a x+b y+c z)$ ，则 $$ f_{x}=a g^{\prime}(a x+b y+c z), f_{y}=b g^{\prime}(a x+b y+c z), f_{z}=c g^{\prime}(a x+b y+c z) $$ 于是 $$ \frac{1}{a} \frac{\partial f}{\partial x}=\frac{1}{b} \frac{\partial f}{\partial y}=\frac{1}{c} \frac{\partial f}{\partial z} $$ 反之，记 $\displaystyle \frac{1}{a} \frac{\partial f}{\partial x}=\frac{1}{b} \frac{\partial f}{\partial y}=\frac{1}{c} \frac{\partial f}{\partial z}=g$ ，则 $\displaystyle \frac{\partial f}{\partial x}=a g, \frac{\partial f}{\partial y}=b g, \frac{\partial f}{\partial z}=c g$ ．于是 $$ \mathrm{d} f=\frac{\partial f}{\partial x} \mathrm{~d} x+\frac{\partial f}{\partial y} \mathrm{~d} y+\frac{\partial f}{\partial z} \mathrm{~d} z=g \mathrm{~d}(a x+b y+c z) . $$ 所以存在 $g(u) \in C^{1}(R)$ 使得 $f(x, y, z)=g(a x+b y+c z)$ ．