4.1 第3次作业评讲

4.1.1 概念和计算部分

解 (1) 错。沿从$(x,y)$经$(x,b)$到$(a,b)$的折线趋于$(a,b)$时$f(x,y)$的极限，实际上就是$\underset{x\to a}{lim}f(x,b)$，只取一次极限；累次极限$\underset{x\to a}{lim}\underset{y\to b}{lim}f(x,y)$是两次极限，第一次极限是沿$(x,y)$到$(x,b)$取极限$\underset{y\to b}{lim}f(x,y)=g(x)$，然后再对一元函数$g(x)$取极限$\underset{x\to a}{lim}g(x)$。

(2) 错。混合偏导数未必总是相等的；但当这两个偏导函数都连续时，它们是相等的。 $◻$

解 (3) A。对梯度$\mathrm{\nabla }f(P_0)$的单位向量$\mathit{v}$，以及任意单位向量$\mathit{u}$，都有$$\begin{array}{}\text{(1)}& \mathrm{\nabla }f(P_0)\cdot \mathit{v}=\parallel \mathrm{\nabla }f(P_0)\parallel =\parallel \mathrm{\nabla }f(P_0)\parallel \parallel \mathit{u}\parallel \ge \mathrm{\nabla }f(P_0)\cdot \mathit{u}=\frac{\partial f}{\partial \mathit{u}}(P_0)\end{array}$$ $◻$

解 (4) $\mathrm{d}f=2x{y}^3\mathrm{d}x+3x^2{y}^2\mathrm{d}y$，$\mathrm{d}f(1,1)=2\mathrm{d}x+3\mathrm{d}y$。

(5) 由题知$z(x,y)=\frac{y}{x}+xyf\left(\frac{y}{x}\right)$，计算可得$$\begin{array}{}\text{(5)}& z_{xy}=-\frac{1-x^{2}f\left(\frac{y}{x}\right)+xy{f}^{\prime }\left(\frac{y}{x}\right)+y^2{f}^{″}\left(\frac{y}{x}\right)}{x^2}\end{array}$$ 故$z_{xy}(1,1)=-5$。

(6) 由题知$z=f(u,x,y)$，其中$u={\mathrm{e}}^y$，计算可得$$\begin{array}{}\text{(6)}& \begin{array}{rl}\frac{\partial z}{\partial x}& =f_1\frac{\partial (x{\mathrm{e}}^y)}{\partial x}+f_2={\mathrm{e}}^y{f}_1+f_2\\ \frac{{\partial }^{2}z}{\partial x\partial y}& =[f_{1,1}\frac{\partial (x{\mathrm{e}}^y)}{\partial y}+f_{1,3}]{\mathrm{e}}^y+f_1{\mathrm{e}}^y+[f_{2,1}\frac{\partial (x{\mathrm{e}}^y)}{\partial y}+f_{2,3}]\\ & =x{\mathrm{e}}^y{f}_{1,1}+x{\mathrm{e}}^y{f}_{1,2}+{\mathrm{e}}^y{f}_{1,3}+f_{2,3}+{\mathrm{e}}^y{f}_1\end{array}\end{array}$$ 故$(A,B,C)=(1,1,0)$，$(D,E,F,G,H,K,L,M,N)=(2,0,0,2,1,1,1,0,0)$。

(7) 计算可得Jacobi矩阵为$$\begin{array}{}\text{(7)}& J=\left(\begin{array}{cc}\frac{x}{x^2+y^2}& \frac{y}{x^2+y^2}\\ -\frac{y}{x^2+y^2}& \frac{x}{x^2+y^2}\end{array}\right)⟹detJ=\frac{1}{x^2+y^2}\end{array}$$ 故$detJ=1$。

(8) 计算可得Jacobi矩阵为$$\begin{array}{}\text{(8)}& J=\left(\begin{array}{ccc}\sin \theta \cos \varphi & r\cos \theta \cos \varphi & -r\sin \theta \sin \varphi \\ \sin \theta \sin \varphi & r\cos \theta \sin \varphi & r\sin \theta \cos \varphi \\ \cos \theta & -r\sin \theta & 0\end{array}\right)⟹detJ=r^2\sin \theta \end{array}$$ 故$detJ=2$。

(9) 归一化$\mathit{v}$得到$\mathit{u}=\frac{\mathit{v}}{\parallel \mathit{v}\parallel }={(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})}^{\mathrm{T}}$。由梯度和方向导数的关系可得$$\begin{array}{}\text{(9)}& \frac{\partial f}{\partial \mathit{u}}(2,3)=\mathrm{\nabla }f(2,3)\cdot \mathit{u}=\left(\begin{array}{c}3\\ 4\end{array}\right)\cdot \left(\begin{array}{c}-\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right)=\frac{\sqrt{2}}{2}\end{array}$$ $◻$

4.1.2 解答和证明部分

解 记$(u,v)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$，计算可得$$\begin{array}{}\text{(10)}& u_x=-v_y=v^2-u^2,u_y=v_x=-2uv\end{array}$$ 所以$$\begin{array}{}\text{(11)}& u_x^2+u_y^2=v_x^2+v_y^2,u_x{v}_x+u_y{v}_y=0,\mathrm{\Delta }u=\mathrm{\Delta }v=0\end{array}$$ 进一步计算可得$$\begin{array}{}\text{(12)}& \begin{array}{rl}\frac{\partial g}{\partial x}& =\frac{\partial f}{\partial u}{u}_x+\frac{\partial f}{\partial v}{v}_x\\ \frac{{\partial }^{2}g}{\partial x^2}& =\frac{{\partial }^{2}f}{\partial u^2}{u}_x^2+2\frac{{\partial }^{2}f}{\partial u\partial v}{u}_x{v}_x+\frac{{\partial }^{2}f}{\partial v^2}{v}_x^2+\frac{\partial f}{\partial u}{u}_{xx}+\frac{\partial f}{\partial v}{v}_{xx}\\ \frac{{\partial }^{2}g}{\partial y^2}& =\frac{{\partial }^{2}f}{\partial u^2}{u}_y^2+2\frac{{\partial }^{2}f}{\partial u\partial v}{u}_y{v}_y+\frac{{\partial }^{2}f}{\partial v^2}{v}_y^2+\frac{\partial f}{\partial u}{u}_{yy}+\frac{\partial f}{\partial v}{v}_{yy}\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(13)}& \mathrm{\Delta }g=(\frac{{\partial }^{2}f}{\partial u^2}+\frac{{\partial }^{2}f}{\partial v^2})(u_x^2+u_y^2)+2\frac{{\partial }^{2}f}{\partial u\partial v}(u_x{v}_x+u_y{v}_y)+\frac{\partial f}{\partial u}\mathrm{\Delta }u+\frac{\partial f}{\partial v}\mathrm{\Delta }v=0\end{array}$$ 即$g$满足Laplace方程。 $◻$

解 计算可得$$\begin{array}{}\text{(15)}& \begin{array}{rl}\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)& =\left(\begin{array}{cc}\frac{\partial }{\partial u}& \frac{\partial }{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}\frac{\partial }{\partial u}+\frac{\partial }{\partial v}& \frac{\partial }{\partial u}-\frac{\partial }{\partial v}\end{array}\right)\\ \left(\begin{array}{cc}\frac{{\partial }^2}{\partial x^2}& \frac{{\partial }^2}{\partial x\partial y}\\ \frac{{\partial }^2}{\partial y\partial x}& \frac{{\partial }^2}{\partial y^2}\end{array}\right)& =\left(\begin{array}{c}\frac{\partial }{\partial u}\\ \frac{\partial }{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial }{\partial u}& \frac{\partial }{\partial v}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}\frac{{\partial }^2}{\partial u^2}& \frac{{\partial }^2}{\partial u\partial v}\\ \frac{{\partial }^2}{\partial v\partial u}& \frac{{\partial }^2}{\partial v^2}\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)\\ & =\frac{1}{4}\left(\begin{array}{cc}\frac{{\partial }^2}{\partial u^2}+2\frac{{\partial }^2}{\partial u\partial v}+\frac{{\partial }^2}{\partial v^2}& \frac{{\partial }^2}{\partial u^2}-\frac{{\partial }^2}{\partial v^2}\\ \frac{{\partial }^2}{\partial u^2}-\frac{{\partial }^2}{\partial v^2}& \frac{{\partial }^2}{\partial v^2}-2\frac{{\partial }^2}{\partial u\partial v}+\frac{{\partial }^2}{\partial u^2}\end{array}\right)\end{array}\end{array}$$ 作换元$(x,y)\mapsto (u,v)$，则$z$满足的微分方程为$$\begin{array}{}\text{(16)}& z=\frac{1}{4}(z_{uu}+2z_{uv}+z_{vv})+\frac{1}{4}(z_{uu}-z_{vv})+\frac{1}{2}(z_u+z_v)=\frac{1}{2}(z_{uu}+z_{uv}+z_u+z_v)\end{array}$$ 再代入$z=w{\mathrm{e}}^{-y}=w{\mathrm{e}}^{v-u}$可得$$\begin{array}{}\text{(17)}& 2w{\mathrm{e}}^{v-u}=2z={\mathrm{e}}^{v-u}(w_{uu}+w_{uv})⟹w_{uu}+w_{uv}=2w\end{array}$$ $◻$

解 设$(u,v)=(x+y,x-y)$，计算可得$$\begin{array}{}\text{(18)}& \arctan \frac{1+u}{1-v}=\arctan [(1+u)(1+v+v^2+o(v^2))]=\arctan [1+u+v+uv+v^2+o(u^2+v^2)]\end{array}$$ 注意到$$\begin{array}{}\text{(19)}& \arctan (1+t)=\frac{\pi }{4}+\frac{1}{2}t-\frac{1}{4}{t}^2+o(t^2)\end{array}$$ 代入$t=u+v+uv+v^2$可得$$\begin{array}{}\text{(20)}& \begin{array}{rl}\arctan \frac{1+u}{1-v}& =\frac{\pi }{4}+\frac{1}{2}(u+v+uv+v^2)-\frac{1}{4}(u+v+uv+v^2{)}^2+o(u^2+v^2)\\ & =\frac{\pi }{4}+\frac{1}{2}(u+v+uv+v^2)-\frac{1}{4}(u+v{)}^2+o(u^2+v^2)\\ & =\frac{\pi }{4}+\frac{1}{2}\cdot 2x(x-y+1)-\frac{1}{4}\cdot (2x{)}^2+o(x^2+y^2)=\frac{\pi }{4}+x-xy+o(x^2+y^2)\end{array}\end{array}$$ $◻$

解 令$x=1+t$，计算可得$$\begin{array}{}\text{(21)}& x^y=\exp [y\ln (1+t)]=\exp \left[y(t+o(t))\right]=\exp (ty+o(ty))=1+(x-1)y+o((x-1{)}^2+y^2)\end{array}$$ $◻$

解 注意到$z$连续且满足$\underset{(x,y)\to \mathrm{\infty }}{lim}z(x,y)=+\mathrm{\infty }$，故$z$在${\mathbb{R}}^2$上有最小值、无最大值，且最小值点为极值点（之一）。计算可得$$\begin{array}{}\text{(24)}& \left(\begin{array}{c}\frac{\partial z}{\partial x}\\ \frac{\partial z}{\partial y}\end{array}\right)=4\left(\begin{array}{c}{x}^3-2x+2y\\ y^3-2y+2x\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$$ 解得所有驻点为$$\begin{array}{}\text{(25)}& (x,y)=(0,0),(2,-2),(-2,2)\end{array}$$ 继续计算可得$$\begin{array}{}\text{(26)}& H_f(x,y)=\left(\begin{array}{cc}\frac{{\partial }^{2}z}{\partial x^2}& \frac{{\partial }^{2}z}{\partial x\partial y}\\ \frac{{\partial }^{2}z}{\partial y\partial x}& \frac{{\partial }^{2}z}{\partial y^2}\end{array}\right)=4\left(\begin{array}{cc}3x^2-2& 2\\ 2& 3y^2-2\end{array}\right)\end{array}$$ 因此$$\begin{array}{}\text{(27)}& H_f(0,0)=\left(\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right),H_f(2,-2)=\left(\begin{array}{cc}10& 2\\ 2& 10\end{array}\right),H_f(-2,2)=\left(\begin{array}{cc}10& 2\\ 2& 10\end{array}\right)\end{array}$$ 故$(0,0)$为退化驻点，$(2,-2),(-2,2)$为极小值点，且$z(2,-2)=z(-2,2)=-32$，因此$z$的最小值为$-32$。注意到$$\begin{array}{}\text{(28)}& z(t,0)=t^4-4t^2,z(t,t)=2t^4\end{array}$$ 即在$0^{\circ }$方向$(0,0)$为极大值，在${45}^{\circ }$方向$(0,0)$为极小值，故$(0,0)$为鞍点。 $◻$

解 注意到$$\begin{array}{}\text{(29)}& \begin{array}{rl}f(0,y)& =y-{\mathrm{e}}^y\to -\mathrm{\infty },y\to \mathrm{\infty }\\ f(x,y)& =x-x^2+(x+y)-{\mathrm{e}}^{x+y}\le \frac{1}{4}+(-1)=-\frac{3}{4},\mathrm{\forall }(x,y)\in {\mathbb{R}}^2\end{array}\end{array}$$ 第二个式子在$(x,y)=(\frac{1}{2},-\frac{1}{2})$时取得等号。故$f$在${\mathbb{R}}^2$上有最大值$-\frac{3}{4}$、无最小值，值域为$(-\mathrm{\infty },-\frac{3}{4}]$，且最大值点为极值点（之一）。计算可得$$\begin{array}{}\text{(30)}& \left(\begin{array}{c}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\end{array}\right)=\left(\begin{array}{c}2-2x-{\mathrm{e}}^{x+y}\\ 1-{\mathrm{e}}^{x+y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$$ 解得$(x,y)=(\frac{1}{2},-\frac{1}{2})$为唯一驻点，其必为极大值点。 $◻$

解 注意到$$\begin{array}{}\text{(32)}& u(x,y)=(1+{\mathrm{e}}^y)\cos x-y{\mathrm{e}}^y\le 1+{\mathrm{e}}^y-y{\mathrm{e}}^y\le 2,\mathrm{\forall }(x,y)\in {\mathbb{R}}^2\end{array}$$ 当$(x,y)=(2k\pi ,0)$时取得等号。故$z$在${\mathbb{R}}^2$上有最大值$2$、无最小值，值域为$(-\mathrm{\infty },2]$，且最大值点为极值点（之一）。随后参见例3.3.9，计算驻点、分析得到极大值点和鞍点。 $◻$

注 在说明极值点是最值点时，一定要首先证明最值存在！