2.2二阶方程的分类习题参考解答

zhangzujin

# 二阶方程的分类习题参考解答 --- 1、 将下列方程分类, 并化成标准型: (1)、 $\displaystyle u\_\{xx\}+2u\_\{xy\}-3u\_\{yy\}+2u\_x+6u\_y=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle a=1, b=1, c=-3$ 知 $\displaystyle \Delta=b^2-ac=4 > 0$, 而方程为双曲型. 作自变量变换 $\displaystyle \xi=y-3x, \eta=y+x$, 则原方程化为标准形: $\displaystyle u\_\{\eta\eta\}-\frac\{1\}\{2\}u\_\eta=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle u\_\{xx\}+4u\_\{xy\}+5u\_\{yy\}+u\_x+2u\_y=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle a=1, b=2, c=5$ 知 $\displaystyle \Delta=b^2-ac=-1 < 0$, 而方程为椭圆型. 作自变量变换 $\displaystyle \xi=y-2x, \eta=y$, 则原方程化为标准形: $\displaystyle u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\eta=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle x^2u\_\{xx\}+4xyu\_\{xy\}+4y^2u\_\{yy\}+5xu\_x+12yu\_y=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle a=x^2, b=2xy, c=4y^2$ 知 $\displaystyle \Delta=b^2-ac=0$, 而方程为抛物型. 作自变量变换 $\displaystyle \xi=\frac\{y\}\{x^2\}, \eta=y$, 则原方程化为标准形: $\displaystyle u\_\{\eta\eta\}+\frac\{3\}\{\eta\}u\_\eta=0$, 当 $\displaystyle \eta\neq 0$ 时. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (4)、 $\displaystyle u\_\{xx\}-2\cos x u\_\{xy\}-(3+\sin^2x)u\_\{yy\}-yu\_y+u\_x=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} a=1, b=-\cos x, c=-(3+\sin^2x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle \Delta=b^2-ac=4 > 0$, 而方程为双曲型. 特征方程为 \begin\{aligned\} &(\mathrm\{ d\} y)^2+2\cos x\mathrm\{ d\} x\mathrm\{ d\} y-(3+\sin^2x)(\mathrm\{ d\} x)^2=0\\\\ \Leftrightarrow& \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}=-\cos x\pm 2 \Leftrightarrow y=-\sin x\pm 2x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 作自变量变换 $\displaystyle \xi=y+\sin x+2x, \eta=y+\sin x-2x$, 则原方程化为标准形 [利用 $\displaystyle 4x=\xi-\eta$]: \begin\{aligned\} u\_\{\xi\eta\}-\frac\{2+\cos\frac\{\xi-\eta\}\{4\}-\frac\{\xi+\eta\}\{2\}\}\{16\}u\_\xi +\frac\{2+\frac\{\xi+\eta\}\{2\}-\cos \frac\{\xi-\eta\}\{4\}\}\{16\}u\_\eta=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (5)、 $\displaystyle u\_\{xx\}+yu\_\{yy\}+\frac\{1\}\{2\}u\_y=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle a=1, b=0, c=y$ 知 $\displaystyle \Delta=b^2-ac=-y$, 而当 $\displaystyle y < 0$ 时方程为双曲型, 特征方程为 \begin\{aligned\} \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}=\mp \sqrt\{-y\}\Rightarrow x=\mp 2\sqrt\{-y\}+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 作自变量变换 $\displaystyle \xi=x-2\sqrt\{-y\}, \eta=x+2\sqrt\{-y\}$, 则原方程化为标准形: $\displaystyle u\_\{\xi\eta\}=0$. 当 $\displaystyle y=0$ 时, 方程为抛物型, 不用作任何变换, 题中方程就是标准形: $\displaystyle u\_\{xx\}+\frac\{1\}\{2\}u\_y=0$. 当 $\displaystyle y > 0$ 时方程为椭圆型, 作自变量变换 $\displaystyle \xi=x, \eta=2\sqrt\{y\}$, 则原方程化为标准形: $\displaystyle u\_\{\xi\xi\}+u\_\{\eta\eta\}=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (6)、 $\displaystyle y^m u\_\{xx\}+u\_\{yy\}+a u\_x+bu\_y+cu=0, m$ 为正常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle a=y^m, b=0, c=1$ 知 $\displaystyle \Delta=b^2-ac=-y^m$. 故当 $\displaystyle y^m < 0$ 时方程为双曲型; 当 $\displaystyle y=0$ 时方程为抛物型; 当 $\displaystyle y^m > 0$ 时方程为椭圆型. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (7)、 $\displaystyle yu\_\{xx\}+(x+y)u\_\{xy\}+x u\_\{yy\}+2u\_y=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} a=y, b=\frac\{x+y\}\{2\}, c=x\Rightarrow \Delta=b^2-ac=\frac\{1\}\{4\}(x-y)^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知当 $\displaystyle x\neq y$ 时方程为双曲型; 当 $\displaystyle x=y$ 时, 方程为抛物型. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (8)、 $\displaystyle u\_\{xx\}+(x+y)^2u\_\{yy\}+xu\_x=0$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} a=1, b=0, c=(x+y)^2\Rightarrow \Delta=b^2-ac=-(x+y)^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知当 $\displaystyle x\neq y$ 时方程为椭圆型; 当 $\displaystyle x=y$ 时, 方程为抛物型. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 2、 讨论下列方程属于哪种类型, 并化成标准型: (1)、 $\displaystyle u\_\{xx\}+2u\_\{xy\}+2u\_\{yy\}+2u\_\{yz\}+2u\_\{zz\}+u\_x+u\_z=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程的特征二次型 $\displaystyle \mathcal\{D\}$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\} 1&1&0\\\\ 1&2&1\\\\ 0&1&2\end\{array\}\right)$. 由 \begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right), P\_1^\mathrm\{T\} AP\_1&=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&1\\\\ 0&1&2\end\{array\}\right),\\\\ P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&-1\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2&=E\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} P=P\_1P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&1&-1\\\\ 0&0&1\end\{array\}\right)\Rightarrow P^\mathrm\{T\} AP=E\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\xi\\\\\eta\\\\\zeta\end\{array\}\right)=P^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right)$ 后, 方程化为 \begin\{aligned\} 0&=u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\{\zeta\zeta\} +(1,0,1)P\left(\begin\{array\}\{cccccccccccccccccccc\}u\_\xi\\\\u\_\eta\\\\u\_\zeta\end\{array\}\right)\\\\ &=u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\{\zeta\zeta\} +u\_\xi-u\_\eta+2u\_\zeta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故原方程为椭圆型方程. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle u\_\{xx\}+2u\_\{xy\}+2u\_\{yy\}+4u\_\{yz\}+5u\_\{zz\}+3u\_x+u\_y=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程的特征二次型 $\displaystyle \mathcal\{D\}$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\} 1&1&0\\\\ 1&2&2\\\\ 0&2&5\end\{array\}\right)$. 由 \begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right), P\_1^\mathrm\{T\} AP\_1&=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&2\\\\ 0&2&5\end\{array\}\right),\\\\ P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&-2\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2&=E\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} P=P\_1P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&2\\\\ 0&1&-2\\\\ 0&0&1\end\{array\}\right)\Rightarrow P^\mathrm\{T\} AP=E\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\xi\\\\\eta\\\\\zeta\end\{array\}\right)=P^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right)$ 后, 方程化为 \begin\{aligned\} 0&=u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\{\zeta\zeta\} +(3,1,0)P\left(\begin\{array\}\{cccccccccccccccccccc\}u\_\xi\\\\u\_\eta\\\\u\_\zeta\end\{array\}\right)\\\\ &=u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\{\zeta\zeta\} +3u\_\xi-2u\_\eta+4u\_\zeta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故原方程为椭圆型方程. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle u\_\{xx\}+2u\_\{yy\}+3u\_\{zz\}+u\_\{xy\}+u\_\{yz\}+u\_\{xz\}-u\_x=0$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程的特征二次型 $\displaystyle \mathcal\{D\}$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\} 1&\frac\{1\}\{2\}&\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&2&\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&\frac\{1\}\{2\}&3\end\{array\}\right)$. 由 \begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-\frac\{1\}\{2\}&-\frac\{1\}\{2\}\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right), P\_1^\mathrm\{T\} AP\_1&=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&\frac\{7\}\{4\}&\frac\{1\}\{4\}\\\\ 0&\frac\{1\}\{4\}&\frac\{11\}\{4\}\end\{array\}\right),\\\\ P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&-\frac\{1\}\{7\}\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2&=A\_2=\mathrm\{diag\}\left(1,\frac\{7\}\{4\},\frac\{19\}\{7\}\right),\\\\ P\_3=\mathrm\{diag\}\left(1,\frac\{2\}\{\sqrt\{7\}\},\sqrt\{\frac\{7\}\{19\}\}\right) \Rightarrow P\_3^\mathrm\{T\} A\_2P\_3=E\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} P=P\_1P\_2P\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-\frac\{1\}\{\sqrt\{7\}\}&-\frac\{3\}\{\sqrt\{133\}\}\\\\ 0&\frac\{2\}\{\sqrt\{7\}\}&-\frac\{1\}\{\sqrt\{133\}\}\\\\ 0&0&\sqrt\{\frac\{7\}\{19\}\}\end\{array\}\right)\Rightarrow P^\mathrm\{T\} AP=E\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\xi\\\\\eta\\\\\zeta\end\{array\}\right)=P^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\\\\z\end\{array\}\right)$ 后, 方程化为 \begin\{aligned\} 0&=u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\{\zeta\zeta\} +(-1,0,0)P\left(\begin\{array\}\{cccccccccccccccccccc\}u\_\xi\\\\u\_\eta\\\\u\_\zeta\end\{array\}\right)\\\\ &=u\_\{\xi\xi\}+u\_\{\eta\eta\}+u\_\{\zeta\zeta\} -u\_\xi+\frac\{1\}\{\sqrt\{7\}\}u\_\eta+\frac\{3\}\{\sqrt\{133\}\}u\_\zeta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故原方程为椭圆型方程. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (4)、 $\displaystyle u\_\{x\_1x\_1\}+2\sum\_\{k=2\}^n u\_\{x\_kx\_k\} -2\sum\_\{k=1\}^p u\_\{x\_kx\_\{k+1\}\}=0, p=n-1$ 或 $\displaystyle n$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (4-1)、 当 $\displaystyle p=n$ 时, 特征二次型 $\displaystyle \mathcal\{D\}$ 的矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&\cdots&-1\\\\ -1&2&\cdots&-1\\\\ \vdots&\vdots&&\vdots\\\\ -1&-1&\cdots&2\end\{array\}\right)$. 由 \begin\{aligned\} &|\lambda E-A|=|\varLambda +ee^\mathrm\{T\}|\left(\left\\{\begin\{array\}\{llllllllllll\}\varLambda=\mathrm\{diag\}(\lambda-2,\lambda-3,\cdots,\lambda-3)\\\\ e=(1,\cdots,1)^\mathrm\{T\}\end\{array\}\right.\right)\\\\ =&|\varLambda|\cdot |E+\varLambda^\{-1\}ee^\mathrm\{T\}| =|\varLambda|(1+e^\mathrm\{T\} \varLambda^\{-1\}e)\\\\ =&(\lambda-2)(\lambda-3)^\{n-1\}\left(1+\frac\{1\}\{\lambda-2\}+\frac\{n-1\}\{\lambda-3\}\right)\\\\ =&(\lambda-3)^\{n-2\}\left\[\lambda^2+(n-5)\lambda+5-2n\right\] \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} n= 2\Rightarrow&A\mbox\{有 $\displaystyle 2$ 个正特征值, 而方程是椭圆型的\},\\\\ n\geq 3\Rightarrow&A\mbox\{有 $\displaystyle n-1$ 个正特征值, $\displaystyle 1$ 个负特征值, 而方程是双曲型的\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (4-2)、 当 $\displaystyle p=n-1$ 时, \begin\{aligned\} |\lambda E-A|=(\lambda-2)(\lambda-3)^\{n-3\}\left\[\lambda^2+(n-6)\lambda+7-2n\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} n= 3\Rightarrow&A\mbox\{有 $\displaystyle 2$ 个正特征值, 而方程是椭圆型的\},\\\\ n\geq 4\Rightarrow&A\mbox\{有 $\displaystyle n-1$ 个正特征值, $\displaystyle 1$ 个负特征值, 而方程是双曲型的\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (5)、 $\displaystyle u\_\{x\_1x\_1\}-2\sum\_\{k=2\}^n (-1)^k u\_\{x\_\{k-1\}x\_k\}=0$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程的特征二次型 \begin\{aligned\} \mathcal\{D\}=&\alpha\_1^2-2\alpha\_1\alpha\_2+2\alpha\_2\alpha\_3-2\alpha\_3\alpha\_4+\cdots+2(-1)^n \alpha\_\{n-1\}\alpha\_n\\\\ =&(\alpha\_1-\alpha\_2)^2-(\alpha\_2-\alpha\_3)^2 +(\alpha\_3-\alpha\_4)^2+\cdots\\\\ &+(-1)^\{n-2\}(\alpha\_\{n-1\}-\alpha\_n)^2+(-1)^\{n-1\}\alpha\_n^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故经过可逆线性变换 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1\\\\\vdots\\\\\beta\_n\end\{array\}\right) =\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&&&\\\\ &1&-1&&\\\\ &&\ddots&-1&\\\\ &&&1&-1\\\\ &&&&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1\\\\\vdots\\\\\alpha\_n\end\{array\}\right)$ 后, \begin\{aligned\} \mathcal\{D\}=\beta\_1^2-\beta\_2^2+\beta\_3^2-\beta\_4^2+\cdots+(-1)^\{n-2\}\beta\_\{n-1\}^2+(-1)^\{n-1\}\beta\_n^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故当 $\displaystyle n=2,3$ 时, 方程是双曲的; 当 $\displaystyle n\geq 4$ 时, 方程是超双曲的. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 3、 设 $\displaystyle \lambda$ 是参数, 试讨论方程 \begin\{aligned\} (\lambda+x)u\_\{xx\}+2xyu\_\{xy\}-y^2u\_\{yy\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的类型, 并指出它们对 $\displaystyle \lambda$ 的依赖关系. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} a=\lambda +x, b=xy, c=-y^2\Rightarrow \Delta =b^2-ac =(x^2+x+\lambda)y^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 (1)、 当 $\displaystyle y=0$ 时, 方程是抛物型的; (2)、 当 $\displaystyle y\neq 0$ 时, (2-1)、 当 $\displaystyle \lambda > \frac\{1\}\{4\}$ 时, $\displaystyle \Delta > 0$, 而方程是双曲型的; (2-2)、 当 $\displaystyle \lambda=\frac\{1\}\{4\}$ 时, \begin\{aligned\} x=-\frac\{1\}\{2\}\Rightarrow \mbox\{方程是抛物型的\}; x\neq -\frac\{1\}\{2\}\Rightarrow \mbox\{方程是双曲型的\}; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-3)、 当 $\displaystyle \lambda < -\frac\{1\}\{4\}$ 时, \begin\{aligned\} x < \frac\{-1-\sqrt\{1-4\lambda\}\}\{2\}\mbox\{或\} x > \frac\{-1+\sqrt\{1-4\lambda\}\}\{2\}\Rightarrow&\mbox\{方程是双曲型的\},\\\\ x=\frac\{-1\pm \sqrt\{1-4\lambda\}\}\{2\}\Rightarrow&\mbox\{方程是抛物型的\},\\\\ \frac\{-1-\sqrt\{1-4\lambda\}\}\{2\} < x < \frac\{-1+\sqrt\{1-4\lambda\}\}\{2\}\Rightarrow&\mbox\{方程是椭圆型的\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 4、 试找出一个未知函数变换, 将常系数线性方程 \begin\{aligned\} u\_\{xx\}+u\_\{yy\}+u\_\{zz\}+au\_x+bu\_y+cu\_z+du=f \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 中的所有一阶微商项消去. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由一般公式 \begin\{aligned\} (\mathrm\{e\}^\{Ax\}u)'=A\mathrm\{e\}^\{Ax\}u+\mathrm\{e\}^\{Ax\}u', (\mathrm\{e\}^\{Ax\}u)''=\mathrm\{e\}^\{Ax\}(u''+2Au'+A^2u) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知应令 $\displaystyle u=\mathrm\{e\}^\{-\frac\{ax+by+cz\}\{2\}\}v$, 则原 pde 化为 \begin\{aligned\} v\_\{xx\}+v\_\{yy\}+v\_\{zz\} +\left(d\mathrm\{e\}^\frac\{ax+by+cz\}\{2\}-\frac\{a^2+b^2+c^2\}\{4\}\right)v=\mathrm\{e\}^\frac\{ax+by+cz\}\{2\}f. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 5、 证明 $\displaystyle n$ 个自变量的常系数双曲型方程或椭圆型方程一定可以经过自变量及未知函数变换化成 \begin\{aligned\} \frac\{\partial^2u\}\{\partial \xi\_1^2\}-\sum\_\{k=2\}^n \frac\{\partial^2u\}\{\partial \xi\_k^2\}+cu=f, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 或 \begin\{aligned\} \sum\_\{k=1\}^n \frac\{\partial^2u\}\{\partial \xi\_k^2\}+cu=f. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 只需看椭圆型的情形. 此时, 已经知道可以通过可逆自变量变换 $\displaystyle y=P^\mathrm\{T\} x$ (这里 $\displaystyle P$: $\displaystyle P^\mathrm\{T\} AP=E\_n$, $\displaystyle A$ 是特征二次型的矩阵) 将椭圆型方程化为 \begin\{aligned\} \sum\_i u\_\{y\_iy\_i\}+\sum\_i B\_i u\_\{y\_i\}+Cu=F. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由一般公式 \begin\{aligned\} (\mathrm\{e\}^\{Ax\}u)'=A\mathrm\{e\}^\{Ax\}u+\mathrm\{e\}^\{Ax\}u', (\mathrm\{e\}^\{Ax\}u)''=\mathrm\{e\}^\{Ax\}(u''+2Au'+A^2u) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知应令 $\displaystyle u=\mathrm\{e\}^\{-\frac\{B\_1y\_1+\cdots+B\_ny\_n\}\{2\}\}v$ 后, 一阶微商项全部消去了, 而最终得到 \begin\{aligned\} \sum\_\{k=1\}^n \frac\{\partial^2v\}\{\partial y\_k^2\}+cu=f. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 6、 判断方程 \begin\{aligned\} \sum\_\{i,j=1\}^n \left(\delta\_\{ij\}-x\_ix\_j\right)\frac\{\partial^2u\}\{\partial x\_i\partial x\_j\} +2\sum\_\{i=1\}^n x\_i\frac\{\partial u\}\{\partial x\_i\}+cu=f \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的类型, 其中 $\displaystyle \delta\_\{ij\}=\left\\{\begin\{array\}\{llllllllllll\}1,&i=j,\\\\ 0,&i\neq j.\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程的特征二次型 $\displaystyle \mathcal\{D\}$ 的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1-x\_1^2&-x\_1x\_2&\cdots&-x\_1x\_n\\\\ -x\_1x\_2&1-x\_2^2&\cdots&-x\_2x\_n\\\\ \vdots&\vdots&&\vdots\\\\ -x\_1x\_n&-x\_2x\_n&\cdots&1-x\_n^2\end\{array\}\right)$. 由 \begin\{aligned\} &|\lambda E-A|=\left|\begin\{array\}\{cccccccccc\}\lambda-1+x\_1^2&x\_1x\_2&\cdots&x\_1x\_n\\\\ x\_1x\_2&\lambda-1+x\_2^2&\cdots&x\_2x\_n\\\\ \vdots&\vdots&&\vdots\\\\ x\_1x\_n&x\_2x\_n&\cdots&\lambda-1+x\_n^2\end\{array\}\right|\\\\ =&|(\lambda-1)E+\alpha\alpha^\mathrm\{T\}|\left(\alpha=(x\_1,\cdots,x\_n)^\mathrm\{T\}\right)\\\\ =&(\lambda-1)^n\left|E+\frac\{1\}\{\lambda-1\}\alpha\alpha^\mathrm\{T\}\right| =(\lambda-1)^n \left(1+\frac\{1\}\{\lambda-1\}\alpha^\mathrm\{T\}\alpha\right)\\\\ =&(\lambda-1)^\{n-1\} \left\[\lambda-\left(1-x\_1^2-\cdots-x\_n^2\right)\right\] \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的特征值为 $\displaystyle 1,\cdots,1,1-x\_1^2-\cdots-x\_n^2$. 故 (1)、 当 $\displaystyle x\_1^2+\cdots+x\_n^2 < 1$ 时, 方程是椭圆型的; (2)、 当 $\displaystyle x\_1^2+\cdots+x\_n^2=1$ 时, 方程是抛物型的; (3)、 当 $\displaystyle x\_1^2+\cdots+x\_n^2 > 1$ 时, 方程是双曲型的. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 7、 是否存在形如 \begin\{aligned\} \sum\_\{i,j=1\}^n a\_\{ij\}(x\_1,\cdots,x\_n)u\_\{x\_ix\_j\}=0, \quad a\_\{ij\}\in C(\mathbb\{R\}^n) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的方程, 它在非空集合 $\displaystyle D\subset \mathbb\{R\}^n$ 上是椭圆型的, 而在 $\displaystyle D$ 的余集 (含于 $\displaystyle \mathbb\{R\}^n$) 上是双曲型的? [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 不存在. 由题设, (1)、 当 $\displaystyle (x\_1,\cdots,x\_n)\in D$ 时, $\displaystyle (a\_\{ij\}(x))$ 是正定的或负定的. (2)、 当 $\displaystyle (x\_1,\cdots,x\_n)\in D^c$ 时, $\displaystyle (a\_\{ij\}(x))$ 有 $\displaystyle n-1$ 个正 (负) 的特征值 $\displaystyle 1$ 个负 (正) 的特征值. 对 $\displaystyle (x\_1,\cdots,x\_n)\in \partial D$, (1)、 若 $\displaystyle (x\_1,\cdots,x\_n)\in D$ 时, $\displaystyle (a\_\{ij\}(x))$ 是正定的. 由特征值是矩阵的连续函数知在 $\displaystyle (x\_1,\cdots,x\_n)$ 的某个邻域内 $\displaystyle (a\_\{ij\}(x))$ 都是正定的. 这与上述第 2 点矛盾. (2)、 若 $\displaystyle (x\_1,\cdots,x\_n)\in D^c$ 时, $\displaystyle (a\_\{ij\}(x))$ 有 $\displaystyle n-1$ 个正 (负) 的特征值 $\displaystyle 1$ 个负 (正) 的特征值. 由特征值是矩阵的连续函数知在 $\displaystyle (x\_1,\cdots,x\_n)$ 的某个邻域内 $\displaystyle (a\_\{ij\}(x))$ 有 $\displaystyle n-1$ 个正 (负) 的特征值 $\displaystyle 1$ 个负 (正) 的特征值. 这与上述第 1 点矛盾. 综上所述, 满足题设的方程是不存在的. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/