下册 7.3 微分方程的验证及变量替换 第22题

\section*{解题过程：} （1）$z_{x}=z_{u} u_{x}+z_{v} v_{x}=z_{u}+z_{v}, z_{y}=z_{u} u_{y}+z_{v} v_{y}=-2 z_{u}+a z_{v}$. 进一步计算二阶偏导数得 $$ z_{x x}=z_{u u}+2 z_{u v}+z_{v v}, z_{x y}=-2 z_{u u}+(a-2) z_{u v}+a z_{v v}, z_{y y}=4 z_{u u}-4 a z_{u v}+a^{2} z_{v v} $$ 将其代人方程 $6 z_{x x}+z_{x y}-z_{y y}=0$ 得 $$ (10+5 a) z_{u v}+\left(6+a-a^{2}\right) z_{v v}=0 $$ 由题设知， $10+5 a \neq 0,6+a-a^{2}=0$ ．由此解得 $a=3$ ． （2）$z_{x}=z_{u} u_{x}+z_{v} v_{x}=z_{u}+z_{v}, z_{y}=z_{u} u_{y}+z_{v} v_{y}=2 z_{u}+a z_{v}$. 进一步计算二阶偏导数得 $$ z_{x x}=z_{u u}+2 z_{u v}+z_{v v}, z_{x y}=2 z_{u u}+(a+2) z_{u v}+a z_{v v}, z_{y y}=4 z_{u u}+4 a z_{u v}+a^{2} z_{v v} . $$ 将其代人方程 $2 z_{x x}+z_{x y}-z_{y y}=0$ 得 $$ (6-3 a) z_{u v}+\left(2+a-a^{2}\right) z_{v v}=0 $$ 由题设知， $6-3 a \neq 0,2+a-a^{2}=0$ ．由此解得 $a=-1$ ． （3）经计算得 $$ z_{x x}=z_{\xi \xi}+2 z_{\xi \eta}+z_{\eta \eta}, z_{x y}=\lambda z_{\xi \xi}+(\lambda-2) z_{\xi \eta}-2 z_{\eta \eta}, z_{y y}=\lambda^{2} z_{\xi \xi}-4 \lambda z_{\xi \eta}+4 z_{\eta \eta} . $$ 于是由 $14 z_{x x}+5 z_{x y}-z_{y y}=0$ 得 $$ \left(14+5 \lambda-\lambda^{2}\right) z_{\xi \xi}+(9 \lambda+18) z_{\xi \eta}=0 $$ 于是 $\lambda=-2, \lambda=7$ ．所以 $z_{\xi}=f(\xi)$ ，故 $$ z=\int f(\xi) \mathrm{d} \xi=F(\xi)+G(\eta)=F(x+7 y)+G(x-2 y) $$ （4）$z_{x}=z_{u} u_{x}+z_{v} v_{x}=z_{u}+z_{v}, z_{y}=z_{u} u_{y}+z_{v} v_{y}=a z_{u}+b z_{v}$ ． 进一步计算二阶偏导数得 $$ z_{x x}=z_{u u}+2 z_{u v}+z_{v v}, z_{x y}=a z_{u u}+(a+b) z_{u v}+b z_{v v}, z_{y y}=a^{2} z_{u u}+2 a b z_{u v}+b^{2} z_{v v} . $$ 将其代人方程 $z_{x x}+4 z_{x y}+3 z_{y y}=0$ 得 $$ \left(1+4 a+3 a^{2}\right) z_{u u}+(2+4 a+4 b+6 a b) z_{u v}+\left(1+4 b+3 b^{2}\right) z_{v v}=0 $$ 欲使变换 $u=x+a y, v=x+b y$ 将微分方程 $z_{x x}+4 z_{x y}+3 z_{y y}=0$ 化为 $\displaystyle \frac{\partial^{2} z}{\partial u \partial v}=0$ ，必须 $$ 1+4 a+3 a^{2}=0,1+4 b+3 b^{2}=0,2+4 a+4 b+6 a b \neq 0 $$ 解之得 $\displaystyle a=-\frac{1}{3}, b=-1$ 或 $\displaystyle a=-1, b=-\frac{1}{3}$ ． $$ \text { (5) } \begin{aligned} u_{x} & =u_{\xi} \xi_{x}+u_{\eta} \mu_{x}=u_{\xi}+u_{\eta}, u_{y}=u_{\xi} \xi_{y}+u_{\eta} \mu_{y}=a u_{\xi}+b u_{\eta} \\ u_{x x} & =u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}, u_{x y}=a u_{\xi \xi}+(a+b) u_{\xi \eta}+b u_{\eta \eta}, u_{y y}=a^{2} u_{\xi \xi}+2 a b u_{\xi \eta}+b^{2} u_{\eta \eta} \end{aligned} $$ 将其代人方程 $2 z_{x x}-3 z_{x y}+z_{y y}=0$ 得 $$ \left(2-3 a+a^{2}\right) u_{\xi \xi}+\left(2-3 b+b^{2}\right) u_{\eta \eta}+[4-3(a+b)+2 a b] u_{\xi \eta}=0 $$ 因为变换 $\xi=x+a y, \eta=x+b y$ 将方程 $2 z_{x x}-3 z_{x y}+z_{y y}=0$ 化为 $z_{\xi \eta}=0$ ，所以 $$ 2-3 a+a^{2}=0,2-3 b+b^{2}=0,4-3(a+b)+2 a b \neq 0 $$ 解之得 $a=1, b=2$ 或 $a=2, b=1$ ． （6）$u_{x}=u_{\xi} \xi_{x}+u_{\eta} \eta_{x}=u_{\xi}+u_{\eta}, u_{t}=u_{\xi} \xi_{t}+u_{\eta} \eta_{t}=a u_{\xi}-a u_{\eta}$ ． $$ u_{x x}=u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}, u_{n}=a^{2}\left(u_{\xi \xi}-2 u_{\xi \eta}+u_{\eta \eta}\right) . $$ 由方程 $u_{t t}=a^{2} u_{x x}$ 得 $u_{t t}=a^{2}\left(u_{\xi \xi}-2 u_{\xi \eta}+u_{\eta \eta}\right)=a^{2}\left(u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}\right)=a^{2} u_{x x}$ ．所以 $u_{\xi \eta}=0$ ． 由 $\displaystyle u_{\xi \eta}=\frac{\partial}{\partial \eta}\left(\frac{\partial u}{\partial \xi}\right)=0$ 得 $$ \frac{\partial u}{\partial \xi}=\varphi\left(\xi, c_{1}\right), u=u(\xi, \eta)=\int \varphi\left(\xi, c_{1}\right) \mathrm{d} \xi+g\left(\eta, c_{2}\right) $$ 即方程 $u_{t t}=a^{2} u_{x x}$ 的通解为 $u=u(\xi, \eta)=\int \varphi\left(x+a t, c_{1}\right)(\mathrm{d} x+a \mathrm{~d} t)+g\left(x-a t, c_{2}\right)$ ．