2.1二阶方程的特征理论与分类习题参考解答

zhangzujin

# 二阶方程的特征习题参考解答 --- 1、 求下列方程的特征线: (1)、 $\displaystyle u\_\{xy\}+au\_x+bu\_y+cu=f$. [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) / 特征方程为 \begin\{aligned\} -\mathrm\{ d\} x\mathrm\{ d\} y=0\Leftrightarrow \mathrm\{ d\} x=0\mbox\{或\} \mathrm\{ d\} y=0\Leftrightarrow x=C\mbox\{或\} y=C. \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)/ (2)、 $\displaystyle yu\_\{xx\}+u\_\{yy\}=0$. [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) / 特征方程为 \begin\{aligned\} y(\mathrm\{ d\} y)^2+(\mathrm\{ d\} x)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-1)、 若 $\displaystyle y < 0$, 则 $\displaystyle \sqrt\{-y\}\mathrm\{ d\} y=\pm\mathrm\{ d\} x \Leftrightarrow -\frac\{2\}\{3\}(-y)^\frac\{3\}\{2\}=\pm x+C$. (2-2)、 若 $\displaystyle y=0$, 则 $\displaystyle \mathrm\{ d\} x=0\Leftrightarrow x=C$. (2-3)、 若 $\displaystyle y > 0$, 则原 pde 没有实特征线. 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)、 $\displaystyle x^2u\_\{xx\}-y^2u\_\{yy\}=0$. [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) / 特征方程为 \begin\{aligned\} &x^2(\mathrm\{ d\} y)^2-y^2(\mathrm\{ d\} x)^2=0 \Leftrightarrow x\mathrm\{ d\} y=\pm y\mathrm\{ d\} x \Leftrightarrow \frac\{\mathrm\{ d\} y\}\{y\}=\pm\frac\{\mathrm\{ d\} x\}\{x\}\\\\ \Leftrightarrow& \ln |y|=\pm \ln |x|+C\_1 \Leftrightarrow y=Cx\mbox\{或\} xy=C. \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)、 $\displaystyle u\_\{xx\}+4u\_\{xy\}+13u\_\{yy\}=u^2$. [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) / 特征方程为 \begin\{aligned\} (\mathrm\{ d\} y)^2-4\mathrm\{ d\} x\mathrm\{ d\} y+13(\mathrm\{ d\} x)^2=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle (-4)^2-3\cdot 13 < 0$ 知原 pde 没有实特征线. 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、 写出下列方程的特征方程及特征方向: (1)、 $\displaystyle u\_\{x\_1x\_1\}+u\_\{x\_2x\_2\}=u\_\{x\_3x\_3\}u\_\{x\_4x\_4\}$; [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) / 特征方程为 \begin\{aligned\} \alpha\_1^2+\alpha\_2^2=\alpha\_3^2+\alpha\_4^2 \Leftrightarrow \alpha\_1^2+\alpha\_2^2=\alpha\_3^2+\alpha\_4^2=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而满足上述方程的任一单位向量 $\displaystyle (\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)$ 都是特征方向. 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)、 $\displaystyle u\_t=u\_\{xx\}-u\_\{yy\}$. [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) / 特征方程为 \begin\{aligned\} \alpha\_1^2-\alpha\_2^2=0\Leftrightarrow \alpha\_0^2+2\alpha\_1^2=1, \alpha\_1=\pm \alpha\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而满足上述方程的任一单位向量 $\displaystyle (\alpha\_0,\alpha\_1,\alpha\_2)$ 都是特征方向. 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)、 $\displaystyle u\_\{xy\}+u\_\{yz\}+u\_\{xz\}=0$. [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) / 特征方程为 \begin\{aligned\} &\alpha\_1\alpha\_2+\alpha\_2\alpha\_3+\alpha\_1\alpha\_3=0\\\\ \Leftrightarrow& (\alpha\_1+\alpha\_2+\alpha\_3)^2 =\alpha\_1^2+\alpha\_2^2+\alpha\_3^2+2(\alpha\_1\alpha\_2+\alpha\_2\alpha\_3+\alpha\_1\alpha\_3)=1\\\\ \Leftrightarrow& \alpha\_1+\alpha\_2+\alpha\_3=\pm 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而满足上述方程的任一单位向量 $\displaystyle (\alpha\_0,\alpha\_1,\alpha\_2)$ 都是特征方向. 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、 求波动方程 $\displaystyle u\_\{tt\}=u\_\{xx\}+u\_\{yy\}$ 过直线 $\displaystyle l: x=2t, y=0$ 的特征曲面. [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) / 波动方程的特征方程为 \begin\{aligned\} \alpha\_0^2=\alpha\_1^2+\alpha\_2^2\Leftrightarrow \alpha\_0^2=\alpha\_1^2+\alpha\_2^2=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由特征曲面过直线 $\displaystyle l: \frac\{t\}\{1\}=\frac\{x\}\{2\}=\frac\{y\}\{0\}$, 而 \begin\{aligned\} 0=\left\\{\alpha\_0,\alpha\_1,\alpha\_2\right\\}\cdot \left\\{1,2,0\right\\} =\alpha\_0+2\alpha\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 解得 \begin\{aligned\} \left\\{\alpha\_0,\alpha\_1,\alpha\_2\right\\}=\left\\{\frac\{1\}\{\sqrt\{2\}\},-\frac\{1\}\{2\sqrt\{2\}\},\pm\frac\{\sqrt\{6\}\}\{4\}\right\\} \mbox\{或\}\left\\{-\frac\{1\}\{\sqrt\{2\}\},\frac\{1\}\{2\sqrt\{2\}\},\pm\frac\{\sqrt\{6\}\}\{4\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是波动方程过直线 $\displaystyle l$ 的特征平面就是任一以上述单位向量为单位法向量的平面. 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、 对每个 $\displaystyle \alpha\in\mathbb\{R\}$, 求方程 \begin\{aligned\} u\_\{xx\}+2u\_\{yy\}+2\alpha u\_\{yz\}+\alpha^2u\_\{zz\}+u\_z+u=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的所有特征. [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) / 特征方程为 \begin\{aligned\} &=\alpha\_1^2+2\alpha\_2^2+2\alpha\alpha\_2\alpha\_3+\alpha^2\alpha\_3^2 =\alpha\_1^2+\alpha\_2^2+(\alpha\_2+\alpha \alpha\_3)^2\\\\ \Rightarrow&\alpha\_1=\alpha\_2=0, \alpha\_2+\alpha\alpha\_3=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (1)、 若 $\displaystyle \alpha=0$, 则 $\displaystyle \alpha\_3=\pm 1$, 而方程有特征 $\displaystyle \left\\{0,0,\pm 1\right\\}$. (2)、 若 $\displaystyle \alpha\neq 0$, 则 $\displaystyle \alpha\_3=0$, 这与 $\displaystyle \alpha\_1^2+\alpha\_2^2+\alpha\_3^2=1$ 矛盾. 故方程没有实特征. 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、 证明经过可逆的自变量变换后, 特征曲面仍变为特征曲面. [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) / 二阶半线性 pde \begin\{aligned\} \sum\_\{ij\}a\_\{ij\}(x\_1,\cdots,x\_n)u\_\{x\_ix\_j\}+F(x\_1,\cdots,x\_n,u,Du)=0\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的特征曲面 $\displaystyle S: G(x\_1,\cdots,x\_n)=0$ 应满足 \begin\{aligned\} \sum\_\{ij\}a\_\{ij\} G\_\{x\_i\}G\_\{x\_j\}=0.\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 经过可逆自变量变换 \begin\{aligned\} y\_i=\phi\_i(x\_1,\cdots,x\_n), 1\leq i\leq n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [$J=\det\left(\frac\{\partial \phi\_i\}\{\partial x\_j\}\right)\neq 0$] 后, \begin\{aligned\} u\_\{x\_i\}=\sum\_k u\_\{y\_k\}\frac\{\partial \phi\_k\}\{\partial x\_i\}, u\_\{x\_ix\_j\}=\sum\_\{kl\}u\_\{y\_ky\_l\} \frac\{\partial \phi\_l\}\{\partial x\_j\}\cdot \frac\{\partial \phi\_k\}\{\partial x\_i\} +\sum\_k u\_\{y\_k\}\frac\{\partial^2\phi\_k\}\{\partial x\_i\partial x\_j\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 $\displaystyle (I)$ 化为 \begin\{aligned\} &\sum\_\{ij\}a\_\{ij\}\sum\_\{kl\}u\_\{y\_ky\_l\} \frac\{\partial \phi\_l\}\{\partial x\_j\}\cdot \frac\{\partial \phi\_k\}\{\partial x\_i\} +\sum\_k u\_\{y\_k\}\frac\{\partial^2\phi\_k\}\{\partial x\_i\partial x\_j\}+\cdots\\\\ =&\sum\_\{kl\}\left(\sum\_\{ij\}a\_\{ij\}\frac\{\partial \phi\_k\}\{\partial x\_i\}\frac\{\partial \phi\_l\}\{\partial x\_j\}\right) u\_\{y\_ky\_l\}+\cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} $\displaystyle (II)$ 化为 \begin\{aligned\} 0=&\sum\_\{ij\}a\_\{ij\}G\_\{x\_i\}G\_\{x\_j\} =\sum\_\{ij\}a\_\{ij\} \sum\_k G\_\{y\_k\}\frac\{\partial \phi\_k\}\{\partial x\_i\} \sum\_l \frac\{\partial \phi\_l\}\{\partial x\_j\}\\\\ =&\sum\_\{kl\}\left(\sum\_\{ij\}a\_\{ij\}\frac\{\partial \phi\_k\}\{\partial x\_i\}\frac\{\partial \phi\_l\}\{\partial x\_j\}\right)G\_\{y\_k\}G\_\{y\_l\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这恰好就是变换后的二阶半线性 pde $\displaystyle (II)$ 的特征方程. 故原特征曲面经过可逆的自变量变换后还是特征曲面. 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)/