4.3 习题课讲解

解 将函数代入方程得到恒等式⁴$$\begin{array}{}\text{(1)}& F(x,y(x))=0\Rightarrow F_x(x,y(x))+F_y(x,y(x))y^{\prime }(x)=0\Rightarrow y^{\prime }(x)=-\frac{F_x(x,y(x))}{F_y(x,y(x))}\end{array}$$ $F_y(x,y(x))\ne 0$是保证隐函数$y=y(x)$存在的充分条件。

采用全微分的过程与上面类似。 $◻$

解 特征方程组为$$\begin{array}{}\text{(5)}& \frac{\mathrm{d}x}{y}=\frac{\mathrm{d}y}{-x}=\frac{\mathrm{d}z}{x^2-y^2}\end{array}$$ 亦即$$\begin{array}{}\text{(6)}& x\mathrm{d}x+y\mathrm{d}y=0,\frac{\mathrm{d}(x+y)}{y-x}=\frac{\mathrm{d}z}{x^2-y^2}\Rightarrow \mathrm{d}z+(x+y)\mathrm{d}(x+y)=0\end{array}$$ 解得特征线为$$\begin{array}{}\text{(7)}& x^2+y^2=h_1,2z+(x+y{)}^2=2z+2xy+x^2+y^2=\mathrm{const}\Rightarrow z+xy=h_2\end{array}$$ 设$g\in {\mathcal{C}}^1({\mathbb{R}}^2)$，则原方程的通解为$$\begin{array}{}\text{(8)}& u(x,y,z)=g(x^2+y^2,z+xy)\end{array}$$ $◻$

注1 如果${\lambda }_0$不是$A_0$的单重特征值，上面的结论还成立吗？

从证明中不难看出，如果${\lambda }_0$的代数重数不为1，即便其几何重数为1（见注2），仍有$B$不可逆，亦即$J$不可逆，故隐函数定理不适用。我们可以举一个反例：$$\begin{array}{}\text{(14)}& \left(\begin{array}{cc}1+a& 1+b\\ c& 1+d\end{array}\right)⟶{(\frac{a-d\pm \sqrt{a^2+4c+4bc-2ad+d^2}}{2c},1)}^{\mathrm{T}}\end{array}$$ 显然特征向量的方向是不确定的，我们可以选择合适的$a,b,c,d$使其朝向任意方向。

如果仅仅考虑特征值$\lambda (A)$，则本题会更加简单：构造函数$p(\lambda ,A)=det(\lambda I-A)$，隐函数定理适用的充分条件是$\frac{\partial p}{\partial \lambda }({\lambda }_0,A_0)\ne 0$，即${\lambda }_0$是$A_0$的单重特征值。

注2 设${\lambda }_1$为$n$阶方阵$A$的特征值，几何重数$>1$，则其特征子空间中存在垂直于${\mathit{x}}_1$的非零向量${\mathit{x}}_2$使得$B{\mathit{x}}_2=\mathbf{0}$，此时$detB=0$，故$J$不可逆。

设${\lambda }_1$为$n$阶方阵$A$的特征值，代数重数$>1$、几何重数$=1$，相应的特征向量为${\mathit{x}}_1$。设${\mathit{x}}_1,\cdots ,{\mathit{x}}_r$是$A$的所有单位特征向量，相应的特征值为${\lambda }_1,\cdots ,{\lambda }_r$，$V=\mathrm{span}\{{\mathit{x}}_1,\cdots ,{\mathit{x}}_r\}$，补充单位正交基$V^{\perp }=\mathrm{span}\{{\mathit{x}}_{r+1},\cdots ,{\mathit{x}}_n\}$。由于${\lambda }_1$的几何重数为1，故$i\ne 1\Rightarrow {\lambda }_i\ne {\lambda }_1$。

设$A$在单位正交基$\{{\mathit{x}}_1,\cdots ,{\mathit{x}}_n\}$上的表示矩阵为$$\begin{array}{}\text{(15)}& A^{\prime }={\left(\begin{array}{cc}\mathrm{diag}\{{\lambda }_1,\cdots ,{\lambda }_r\}& 0\\ U& W\end{array}\right)}_{n\times n}\end{array}$$ 假设$W$的所有特征值都不等于${\lambda }_1$，则$$\begin{array}{}\text{(16)}& p(\lambda ):=det(\lambda I-A)=det(\lambda I-A^{\prime })=det(\lambda I-W)\prod _{i=1}^r(\lambda -{\lambda }_i)\end{array}$$ 此时${\lambda }_1$的代数重数为1，与题设矛盾！故$W$至少有一个特征值等于${\lambda }_1$。

设$W$的特征值${\lambda }_1$对应的特征向量为$(y_{r+1},\cdots ,y_n)$，则对于向量$\mathit{x}=\sum _{i=r+1}^n{y}_i{\mathit{x}}_i$，设$$\begin{array}{}\text{(17)}& A\mathit{x}=\sum _{i=1}^r{a}_i{\mathit{x}}_i+\sum _{i=r+1}^n{\lambda }_1{y}_i{\mathit{x}}_i\end{array}$$ 选择${\mathit{x}}^{\prime }=\sum _{i=1}^r{y}_i{\mathit{x}}_i$使得$$\begin{array}{}\text{(18)}& B(\mathit{x}+{\mathit{x}}^{\prime })=[{\mathit{x}}_1{\mathit{x}}_1^{\mathrm{T}}+({\lambda }_{1}I-A)]\sum _{i=1}^n{y}_i{\mathit{x}}_i=\sum _{i=1}^r[({\lambda }_1-{\lambda }_i+{\delta }_{i1})y_i-a_i]{\mathit{x}}_i\stackrel{?}{=}\mathbf{0}\end{array}$$ 故${\mathit{x}}^{\prime }$需满足$$\begin{array}{}\text{(19)}& y_1=a_1,y_i=\frac{a_i}{{\lambda }_1-{\lambda }_i},i=2,\cdots ,r\end{array}$$ 因此$B$不可逆，故隐函数定理不适用。

解 (1) 设$(X,Y,Z)=(tx,ty,tz)$，则$$\begin{array}{}\text{(20)}& 1=X^2+Y^2=t^2(x^2+y^2)=t^2\Rightarrow t=\frac{1}{\sqrt{x^2+y^2}}\end{array}$$ 因此$$\begin{array}{}\text{(21)}& X=\frac{x}{\sqrt{x^2+y^2}},Y=\frac{y}{\sqrt{x^2+y^2}},Z=\frac{z}{\sqrt{x^2+y^2}}\end{array}$$

(2) 由题可得$(x,y,z)=(\cos \theta \cos \phi ,\cos \theta \sin \phi ,\sin \theta )$，故$$\begin{array}{}\text{(22)}& X=\cos \phi ,Y=\sin \phi ,Z=\tan \theta \end{array}$$

(3) 设$\gamma :(\cos \theta (t)\cos \phi (t),\cos \theta (t)\sin \phi (t),\sin \theta (t))$是球面上过点$P_0=\gamma (t_0)$的一条光滑曲线，则$F$把$\gamma$映为柱面上过点$Q_0=F(P_0)$的一条光滑曲线$$\begin{array}{}\text{(23)}& F(\gamma ):(\cos \phi (t),\sin \phi (t),f(\tan \theta (t)))\end{array}$$

设$\xi ={\theta }^{\prime }(t_0)$、$\eta ={\phi }^{\prime }(t_0)$，计算可知$\gamma$的速度向量为$$\begin{array}{}\text{(24)}& \mathit{v}=\xi (-\sin \theta \cos \phi ,-\sin \theta \sin \phi ,\cos \theta )+\eta (-\cos \theta \sin \phi ,\cos \theta \cos \phi ,0)\end{array}$$ 球面上$P_0$处的两个切向量${\mathit{v}}_1,{\mathit{v}}_2$的内积为$$\begin{array}{}\text{(25)}& {\mathit{v}}_1\cdot {\mathit{v}}_2={\xi }_1{\xi }_2+{\eta }_1{\eta }_2{\cos }^2\theta \end{array}$$ 夹角${\alpha }_1$的余弦为$$\begin{array}{}\text{(26)}& \cos {\alpha }_1=\frac{{\mathit{v}}_1\cdot {\mathit{v}}_2}{|{\mathit{v}}_1||{\mathit{v}}_2|}=\frac{{\xi }_1{\xi }_2+{\eta }_1{\eta }_2{\cos }^2\theta }{\sqrt{{\xi }_1^2+{\eta }_1^2{\cos }^2\theta }\sqrt{{\xi }_2^2+{\eta }_2^2{\cos }^2\theta }}=\frac{{\xi }_1{\xi }_2{\sec }^2\theta +{\eta }_1{\eta }_2}{\sqrt{{\xi }_1^2{\sec }^2\theta +{\eta }_1^2}\sqrt{{\xi }_2^2{\sec }^2\theta +{\eta }_2^2}}\end{array}$$

$F(\gamma )$上相应的速度向量为$$\begin{array}{}\text{(27)}& \mathit{w}=\xi (0,0,f^{\prime }(\tan \theta ){\sec }^2\theta )+\eta (-\sin \phi ,\cos \phi ,0)\end{array}$$ 柱面上$Q_0$处的两个切向量${\mathit{w}}_1,{\mathit{w}}_2$的内积为$$\begin{array}{}\text{(28)}& {\mathit{w}}_1\cdot {\mathit{w}}_2={\xi }_1{\xi }_2[f^{\prime }(\tan \theta ){\sec }^2\theta {]}^2+{\eta }_1{\eta }_2\end{array}$$ 夹角${\alpha }_2$的余弦为$$\begin{array}{}\text{(29)}& \cos {\alpha }_2=\frac{{\mathit{w}}_1\cdot {\mathit{w}}_2}{|{\mathit{w}}_1||{\mathit{w}}_2|}=\frac{{\xi }_1{\xi }_2[f^{\prime }(\tan \theta ){\sec }^2\theta {]}^2+{\eta }_1{\eta }_2}{\sqrt{{\xi }_1^2[f^{\prime }(\tan \theta ){\sec }^2\theta {]}^2+{\eta }_1^2}\sqrt{{\xi }_2^2[f^{\prime }(\tan \theta ){\sec }^2\theta {]}^2+{\eta }_2^2}}\end{array}$$

由题可知$\cos {\alpha }_1=\cos {\alpha }_2$对任意${\xi }_1,{\xi }_2,{\eta }_1,{\eta }_2$均成立，当且仅当$$\begin{array}{}\text{(30)}& f^{\prime }(\tan \theta )=\pm \cos \theta =\pm \frac{1}{\sqrt{1+{\tan }^2\theta }}\end{array}$$ 因此$$\begin{array}{}\text{(31)}& f^{\prime }(x)=\pm \frac{1}{\sqrt{1+x^2}}\Rightarrow f(x)=\pm \ln (x+\sqrt{1+x^2})+C=\pm \mathrm{arsinh}x+C\end{array}$$ 在实际使用中考虑到对称性等因素，我们选择$f(x)=\mathrm{arsinh}x$。

将球面上的所有点通过$F$映射到柱面上，沿柱面的母线剪开就得到一张地图，这就是Mercator地图。我们实际上证明了：Mercator地图上所画的两条相交道路的夹角与实际道路的夹角大小相等。 $◻$

另解 (3) 设曲线$\gamma$在$\gamma (t_0)=P_0$处的速度向量为$\mathit{v}={\gamma }^{\prime }(t_0)$。构造如下映射：$$\begin{array}{}\text{(32)}& t\stackrel{g}{\mapsto }(\phi ,\theta )\stackrel{h}{\mapsto }(x,y,z),\gamma =h\circ g\end{array}$$ 故有$$\begin{array}{}\text{(33)}& \mathit{v}={\gamma }^{\prime }(t_0)=\frac{\partial (x,y,z)}{\partial (\phi ,\theta )}(\genfrac{}{}{0ex}{}{{\phi }^{\prime }(t_0)}{{\theta }^{\prime }(t_0)})\end{array}$$

由于$(\phi ,\theta )$为球面或柱面的正交参数化，故可设$\phi ,\theta$方向的单位向量分别为$\mathit{e},{\mathit{e}}^{\prime }$，定义曲线的倾角$\alpha \in (-\pi ,\pi ]$满足下列方程的唯一解：$$\begin{array}{}\text{(34)}& \mathit{v}=\parallel \mathit{v}\parallel (\cos \alpha \mathit{e}+\sin \alpha {\mathit{e}}^{\prime })\end{array}$$ 容易验证有下式成立$$\begin{array}{}\text{(35)}& \cos ({\alpha }_1-{\alpha }_2)=\frac{(\mathit{e}\cdot {\mathit{v}}_1)(\mathit{e}\cdot {\mathit{v}}_2)+({\mathit{e}}^{\prime }\cdot {\mathit{v}}_1)({\mathit{e}}^{\prime }\cdot {\mathit{v}}_2)}{\parallel {\mathit{v}}_1\parallel \parallel {\mathit{v}}_2\parallel }=\frac{{\mathit{v}}_1\cdot {\mathit{v}}_2}{\parallel {\mathit{v}}_1\parallel \parallel {\mathit{v}}_2\parallel }\end{array}$$ 因此我们只需要证明$F$保倾角$\alpha$即可证明$F$保角。

保倾角$\alpha$的必要条件为保其正切值$\tan \alpha$。注意到$$\begin{array}{}\text{(36)}& \mathit{v}={\gamma }^{\prime }(t_0)=\frac{\partial (x,y,z)}{\partial (\phi ,\theta )}(\genfrac{}{}{0ex}{}{{\phi }^{\prime }(t_0)}{{\theta }^{\prime }(t_0)})=\Vert \frac{\partial (x,y,z)}{\partial \phi }\Vert {\phi }^{\prime }(r_0)\mathit{e}+\Vert \frac{\partial (x,y,z)}{\partial \theta }\Vert {\theta }^{\prime }(t_0){\mathit{e}}^{\prime }\end{array}$$ 则有$$\begin{array}{}\text{(37)}& \cos \alpha =\frac{\mathit{e}\cdot \mathit{v}}{\parallel \mathit{v}\parallel }=\frac{{\phi }^{\prime }(t_0)}{\parallel \mathit{v}\parallel }\Vert \frac{\partial (x,y,z)}{\partial \phi }\Vert ,\sin \alpha =\frac{{\mathit{e}}^{\prime }\cdot \mathit{v}}{\parallel \mathit{v}\parallel }=\frac{{\theta }^{\prime }(t_0)}{\parallel \mathit{v}\parallel }\Vert \frac{\partial (x,y,z)}{\partial \theta }\Vert \end{array}$$ 故有$$\begin{array}{}\text{(38)}& \frac{{\phi }^{\prime }(t_0)}{{\theta }^{\prime }(t_0)}\tan \alpha =\frac{\Vert \frac{\partial (x,y,z)}{\partial \theta }\Vert }{\Vert \frac{\partial (x,y,z)}{\partial \phi }\Vert }\end{array}$$ 因此$$\begin{array}{}\text{(39)}& \frac{1}{\sqrt{1+{\tan }^2\theta }}=\frac{1}{\cos \theta }=\frac{\Vert \frac{\partial (x,y,z)}{\partial \theta }\Vert }{\Vert \frac{\partial (x,y,z)}{\partial \phi }\Vert }=\frac{\Vert \frac{\partial (X,Y,Z)}{\partial \theta }\Vert }{\Vert \frac{\partial (X,Y,Z)}{\partial \phi }\Vert }=\frac{|f^{\prime }(\tan \theta )|}{1}\end{array}$$ 亦即$$\begin{array}{}\text{(40)}& f^{\prime }(x)=\pm \frac{1}{\sqrt{1+x^2}}\Rightarrow f(x)=\pm \ln (x+\sqrt{1+x^2})+C=\pm \mathrm{arsinh}x+C\end{array}$$ 为了保倾角，应选择正号⁵；考虑到对称性，应选择$C=0$。故$f(x)=\mathrm{arsinh}x$。