1. 求解下列方程的通解:
(1). $\dps{z\frac{\p z}{\p x} -y\frac{\p z}{\p y}=0}$.
解答: 全特征线为 $$\bex \frac{\rd x}{z} =\frac{\rd y}{-y}=\frac{\rd z}{0}, \eex$$ 由 $$\beex \bea \rd\frac{x}{z} =\frac{\rd x}{z}-\frac{x}{z^2}\rd z =\frac{\rd x}{z} =\frac{\rd y}{-y}&\ra \frac{x}{z}+\ln |y|=\tilde C_1\ra e^\frac{x}{z}y=C_1,\\ \rd z=0&\ra z=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex \varPhi\sex{z,e^\frac{x}{z}y}=0. \eex$$
(2). $\dps{\frac{\p z}{\p x}+\frac{\p z}{\p y}=2z}$.
解答: 全特征线为 $$\bex \frac{\rd x}{1}=\frac{\rd y}{1}=\frac{\rd z}{2z}, \eex$$ 其由首次积分 $$\bex x-y=C_1,\quad \frac{e^{2y}}{z}=C_2. \eex$$ 故原方程的通解为 $$\bex \varPhi\sex{x-y,\frac{e^{2y}}{z}}=0. \eex$$
(3). $\dps{x\frac{\p z}{\p y}=z}$.
解答: 全特征线为 $$\bex \frac{\rd x}{0}=\frac{\rd y}{x}=\frac{\rd z}{z}. \eex$$ 由 $$\beex \bea \rd \frac{y}{x}=\frac{\rd y}{x}-\frac{y}{x^2}\rd x =\frac{\rd y}{x}=\frac{\rd z}{z} &\ra \frac{y}{x}=\ln |z|+\tilde C_1 \ra \frac{e^\frac{y}{x}}{z}=C_1,\\ \rd x=0&\ra x=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex \varPhi\sex{\frac{e^\frac{y}{x}}{z},x}=0. \eex$$
(4). $\dps{\frac{\p u}{\p x} +b\frac{\p u}{\p y}+c\frac{\p u}{\p z}=xyz}$ ($b,c$ 是常数).
解答: 全特征线为 $$\bex \frac{\rd x}{1}=\frac{\rd y}{b}=\frac{\rd z}{c}=\frac{\rd u}{xyz}. \eex$$ 由 $$\bex \rd y=b\rd x\ra y-bx=C_1,\quad \rd z=c\rd x\ra z-cx=C_2 \eex$$ 及 $$\beex \bea c\rd u&=xyz\rd z =xy\rd \frac{z^2}{2} =\rd \frac{xyz^2}{2}-\frac{z^2}{2}(x\rd y+y\rd x)\\ &=\rd \frac{xyz^2}{2} -\frac{z^2}{2}\sex{x\frac{b}{c}\rd z +y\frac{1}{c}\rd z}\\ &=\rd \sex{\frac{xy z^2}{2}-\frac{bxz^3}{6c} -\frac{yz^3}{6c}} +\frac{bz^3}{6c}\rd x +\frac{z^3}{6c}\rd y\\ &=\rd \sex{\frac{xy z^2}{2}-\frac{bxz^3}{6c} -\frac{yz^3}{6c}} +\frac{bz^3}{6c}\frac{\rd z}{c} +\frac{z^3}{6c}\frac{b}{c}\rd z\\ &=\rd \sex{\frac{xy z^2}{2}-\frac{bxz^3}{6c} -\frac{yz^3}{6c}+\frac{bz^4}{12c^2}}\\ &\ra 12c^3u -6c^2xyz^2+2bc xz^3 +2cyz^3-bz^4=C_3 \eea \eeex$$ 知原方程的通解为 $$\bex \varPhi\sex{y-bx,z-cx,12c^3u -6c^2xyz^2+2bc xz^3 +2cyz^3-bz^4}=0. \eex$$
(5). $\dps{(xy^3-2x^4)\frac{\p u}{\p x} +(2y^4-x^3y)\frac{\p u}{\p y} =9u(x^3-y^3)}$.
解答: 全特征方程为 $$\bex \frac{\rd x}{xy^3-2x^4} =\frac{\rd y}{2y^4-x^3y} =\frac{\rd u}{9u(x^3-y^3)}. \eex$$ 由 $$\beex \bea &\quad \frac{\rd y}{\rd x}=\frac{2y^4-x^3y}{xy^3-2x^4} =\frac{2\sex{\frac{y}{x}}^4-\frac{y}{x}}{\sex{\frac{y}{x}}^3-2}\\ &\ra s+x\frac{\rd s}{\rd x}=\frac{2s^4-s}{s^3-2}\quad\sex{s=\frac{y}{x}}\\ &\ra x\frac{\rd s}{\rd x}=\frac{s^4+s}{s^3-2}\\ &\ra \frac{s^3-2}{s^4+s}\rd s =\frac{\rd x}{x}\\ &\ra \sex{-\frac{2}{s}+\frac{3s^2}{s^3+1}}\rd s =\frac{\rd x}{x}\\ &\ra \frac{s^3+1}{s^2x}=C_1\\ &\ra \frac{x^3+y^3}{x^2y^2}=C_1,\\ &\quad \frac{\rd (x^3)}{3x^3(y^3-2x^3)} =\frac{\rd (y^3)}{3y^3(2y^3-x^3)} =\frac{\rd (x^3-y^3)}{-6(x^6-y^6)}=\frac{\rd u}{9u(x^3-y^3)}\\ &\ra \frac{\rd t}{-2t}=\frac{\rd u}{3u}\quad\sex{t=x^3+y^3}\\ &\ra u^2t^3=C_2\\ &\ra u^2(x^3+y^3)^3=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex \varPhi\sex{ \frac{x^3+y^3}{x^2y^2},u^2(x^3+y^3)^3 }=0. \eex$$
2. 求解下列初值问题:
(1). $\dps{\sedd{\ba{ll} yz\cfrac{\p z}{\p x}+\cfrac{\p z}{\p y}=0\\ z|_{x=0}=y^2 \ea}}$.
解答: 全特征方程为 $$\bex \frac{\rd x}{yz}=\frac{\rd y}{1}=\frac{\rd z}{0}. \eex$$ 由 $$\beex \bea \rd z=0&\ra z=C_1,\\ \rd x=yz\rd y=z\rd \frac{y^2}{2}=\rd \frac{y^2z}{2}-\frac{y^2}{2}\rd z =\rd \frac{y^2z}{2} &\ra x-\frac{y^2z}{2}=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex \varPhi\sex{z,x-\frac{y^2z}{2}}=0. \eex$$ 又 $$\beex \bea &\quad z|_{x=0}=y^2\\ &\ra y^2=C_1,\quad 0-\frac{y^2\cdot y^2}{2}=C_2\\ &\ra C_1^2+2C_2=0\\ &\ra z^2+2\sex{x-\frac{y^2z}{2}}=0\\ &\ra z^2+2x-y^2z=0\\ &\ra z=\frac{y^2+\sqrt{y^4-8z}}{2}\quad\sex{\mbox{另外一个因为初值为零而应舍去}}. \eea \eeex$$
(2). $\dps{\sedd{\ba{ll} (x^2-yu)\cfrac{\p u}{\p x} +(y^2-xu)\cfrac{\p u}{\p y}=u^2-xy\\ u|_{x=0}=2y \ea}}$.
解答: 全特征线为 $$\bex \frac{\rd x}{x^2-yu}=\frac{\rd y}{y^2-xu} =\frac{\rd u}{u^2-xy}. \eex$$ 由 $$\beex \bea &\quad \frac{\rd x-\rd y}{x^2-yu-y^2+xu} =\frac{\rd y-\rd u}{y^2-xu-u^2+xy}\\ &\ra \frac{\rd (x-y)}{(x-y)(x+y+u)} =\frac{\rd (y-u)}{(y-u)(x+y+u)}\\ &\ra \frac{\rd (x-y)}{x-y}=\frac{\rd (y-u)}{y-u}\\ &\ra \frac{x-y}{y-u}=C_1 \eea \eeex$$ 及对称性, $$\bex \frac{y-u}{u-x}=C_2 \eex$$ 知原方程的通解为 $$\bex \varPhi\sex{\frac{x-y}{y-u},\frac{y-u}{u-x}}=0. \eex$$ 又当 $x=0$ 时, $u=2y$, $$\beex \bea &\quad C_1=1,\quad C_2=-\frac{1}{2}\\ &\ra C_1+2C_2=0\\ &\ra \frac{x-y}{y-u}+2\frac{y-u}{u-x}=0\\ &\ra u=-x+2y\quad\sex{u=\frac{x+y}{2}\mbox{ 应舍去}}. \eea \eeex$$
(3). $\dps{\sedd{\ba{ll} x\cfrac{\p v}{\p x}+y\cfrac{\p v}{\p y}=hv\quad\sex{h\mbox{ 是正整数}}\\ v|_{x=1}=e^y \ea}}$.
解答: 全特征线为 $$\bex \frac{\rd x}{x}=\frac{\rd y}{y}=\frac{\rd v}{hv}, \eex$$ 而有首次积分 $$\bex \frac{x}{y}=C_1,\quad \frac{y^h}{v}=C_2. \eex$$ 又当 $x=1$ 时, $v=e^y$, $$\beex \bea &\quad \frac{1}{y}=C_1,\quad \frac{y^h}{e^y}=C_2\\ &\ra \frac{1}{C_1^h e^\frac{1}{C_1}} =C_2\\ &\ra C_1^h e^\frac{1}{C_1} C_2=1\\ &\ra \sex{\frac{x}{y}}^h e^\frac{y}{x} \frac{y^h}{v}=1\\ &\ra v=\sex{\frac{x}{y}}^he^{\frac{y}{x}} y^h. \eea \eeex$$
(4). $\dps{\sedd{\ba{ll} \sum_{k=1}^n x_k\cfrac{\p u}{\p x_k}=3u\\ u(x_1,\cdots,x_{n-1},1)=\varphi(x_1,\cdots,x_{n-1}) \ea}}$.
解答: 全特征线为 $$\bex \frac{\rd x_1}{x_1}=\cdots=\frac{\rd x_n}{x_n}=\frac{\rd u}{3u}, \eex$$ 而有首次积分 $$\bex \frac{x_1}{x_n}=C_1,\quad \cdots,\quad \frac{x_{n-1}}{x_n}=C_{n-1},\quad \frac{u}{x_n^3}=C_n. \eex$$ 又 $x_n=1$ 时, $$\beex \bea &\quad x_1=C_1,\cdots,x_{n-1}=C_{n-1},\varphi(x_1,\cdots,x_{n-1})=u=C_n\\ &\ra \varphi(C_1,\cdots,C_{n-1})=C_n\\ &\ra \varphi\sex{\frac{x_1}{x_n},\cdots,\frac{x_{n-1}}{x_n}}=\frac{u}{x_n^3}\\ &\ra u=x_n^3 \varphi\sex{\frac{x_1}{x_n},\cdots,\frac{x_{n-1}}{x_n}}. \eea \eeex$$
3. 求解: $\dps{(x^2+y^2)\frac{\p u}{\p x} +2xy\frac{\p u}{\p y}=xu}$, $u=u(x,y)$ 过曲线 $x=a$, $y^2+u^2=a^2$.
解答: 全特征线为 $$\bex \frac{\rd x}{x^2+y^2}=\frac{\rd y}{2xy}=\frac{\rd u}{xu}. \eex$$ 而 $$\beex \bea \frac{\rd (x+y)}{(x+y)^2}=\frac{\rd (x-y)}{(x-y)^2} &\ra \frac{1}{x+y}-\frac{1}{x-y}=C_1,\\ \frac{\rd y}{2y}=\frac{\rd u}{u}&\ra \frac{y}{u^2}=C_2. \eea \eeex$$ 又当 $x=a$ 时, $y^2+u^2=a^2$, $$\beex \bea &\quad \frac{-2y}{a^2-y^2}=C_1,\quad \frac{y}{a^2-y^2}=C_2\\ &\ra C_1+2C_2=0\\ &\ra -\frac{2y}{x^2-y^2}+2\frac{y}{u^2}=0\\ &\ra x^2+u^2-y^2=0. \eea \eeex$$
4. 求与曲面 $v^2=axy$ 正交的曲面.
解答: 曲面 $v^2=axy$ 在 $(x,y,v)$ 处的法向量为 $$\bex (ay,ax,-2v), \eex$$ 而所求曲面为 $$\bex ay(X-x) +ax(Y-y) -2v(V-v)=0\lra ayX+axY-2vV=0. \eex$$
5. 证明偏微分方程 $\dps{a(x,y)\frac{\p u}{\p x}+b(x,y)\frac{\p u}{\p y}=0}$ 的特征曲线都是平面曲线.
解答: 全特征方程为 $$\bex \frac{\rd x}{a(x,y)}=\frac{\rd y}{b(x,y)}=\frac{\rd u}{0}, \eex$$ 其积分曲线在 $u=C$ 上, 而为平面曲线.
6. 证明拟线性偏微分方程 $$\bex u_t+a(u)u_x=0 \eex$$ 满足初值条件 $u(x,0)=h(x)$ 的解 $u(x,t)$ 由 $$\bex u(x,t)=h(x-a(u(x,t))t) \eex$$ 隐式地确定.
解答: 全特征方程为 $$\bex \frac{\rd t}{1}=\frac{\rd x}{a(u)}=\frac{\rd u}{0}. \eex$$ 由 $$\beex \bea \rd u=0&\ra u=C_1,\\ \rd (x-a(u)t)=\rd x-ta‘(u)\rd u-a(u)\rd t=0&\ra x-a(u)t=C_2 \eea \eeex$$ 及 $u(x,0)=h(x)$ 知 $$\bex h(x)=C_1,\ x=C_2 \ra C_1=h(C_2)\ra u=h(x-a(u)t). \eex$$