MA102a 2024春 期末 答案

2024春数学分析2(H)期末考试答案-马嘉麒

一

设 $F_1(x,y,z)=x^3+y^3+z^3-3xyz,F_2(x,y,z)=x+y+z-a$ 则

$$\left(\begin{array}{c}\frac{dy}{dx}\\ \frac{dz}{dx}\end{array}\right)=-{\left(\begin{array}{ll}\frac{\partial F_1}{\partial y}& \frac{\partial F_1}{\partial z}\\ \frac{\partial F_2}{\partial y}& \frac{\partial F_2}{\partial z}\end{array}\right)}^{-1}\left(\begin{array}{c}\frac{\partial F_1}{\partial x}\\ \frac{\partial F_2}{\partial x}\end{array}\right)$$$$=-{\left(\begin{array}{cc}3y^2-3xz& 3z^2-3xy\\ 1& 1\end{array}\right)}^{-1}\left(\begin{array}{c}3x^2-3yz\\ 1\end{array}\right)$$$$=-\frac{1}{3(y+z)(y-z-x)}\left(\begin{array}{cc}1& 3xy-3z^2\\ -1& 3y^2-3xz\end{array}\right)\left(\begin{array}{c}3x^2-3yz\\ 1\end{array}\right)$$$$=\left(\begin{array}{c}\frac{(z-x)(z+x+y)}{(y+z)(y-z-x)}\\ \frac{(x-y)(x+y+z)}{(y+z)(y-z-x)}\end{array}\right)$$$$\Rightarrow \frac{dy}{dx}=\frac{(z-x)(z+x+y)}{(y+z)(y-z-x)},\frac{dz}{dx}=\frac{(x-y)(x+y+z)}{(y+z)(y-z-x)}$$

二

2-1

设 $\mathrm{C}=\{(x,y)\mid x⩾0,y⩾0,x+y⩽1\}$

$${\iiint }_D\frac{dxdydz}{(2+x+y{)}^2}={\iint }_{C}dxdy{\int }_0^{1-x-y}\frac{dz}{(2+x+y{)}^2}={\iint }_C\frac{1-x-y}{(2+x+y{)}^2}dxdy$$$$={\int }_0^{1}dx{\int }_0^{1-x}\frac{1-x-y}{(2+x+y{)}^2}dy$$$$={\int }_0^1\left({\int }_0^{1-x}(\frac{3}{(2+x+y{)}^2}-\frac{1}{2+x+y})dy\right)dx$$$$={\int }_0^1(3{\int }_{x+2}^3\frac{dy}{y^2}-{\int }_{x+2}^3\frac{dy}{y})dx$$$$={\int }_0^1(\frac{3}{x+2}-\ln \frac{3}{x+2}-1)dx$$$$={\int }_2^3(\frac{3}{x}-\ln \frac{3}{x}-1)dx$$$$=3\ln \frac{3}{2}-1-({x\ln \frac{3}{x}|}_2^3-{\int }_2^{3}xd(\ln \frac{3}{x}))=5\ln \frac{3}{2}-2$$

2-2

设 $C_z=\{(x,y)\mid x⩾0,y⩾0,x+y⩽1-z\}$

$${\iiint }_D\frac{dxdydz}{(2+x+y{)}^2}={\int }_0^{1}dz{\iint }_{C_z}\frac{dxdy}{(2+x+y{)}^2}$$$$={\int }_0^{1}dz{\int }_0^{1-z}dx{\int }_0^{1-z-x}\frac{dy}{(2+x+y{)}^2}$$$$={\int }_0^{1}dz{\int }_0^{1-z}(\frac{1}{z-3}+\frac{1}{x+2})dx$$$$={\int }_0^1(\frac{1-z}{z-3}+\ln \frac{3-z}{2})dz=5\ln \frac{3}{2}-2$$

三

3-(1)

$${\int }_1^{2}dx{\int }_{2-x}^{\sqrt{2-x}}f(x,y)dy={\int }_0^{1}dy{\int }_{2-y}^{2}f(x,y)dx+{\int }_1^{\sqrt{3}}dy{\int }_{\frac{y^2+1}{2}}^{2}f(x,y)dy$$

3-(2)

$${\int }_0^{1}dx{\int }_0^{1-x}dy{\int }_0^{x+y}f(x,y,z)dz={\int }_0^{1}dy{\int }_0^{1-y}dx{\int }_0^{x+y}f(x,y,z)dz$$$$={\int }_0^{1}dy{\int }_0^{y}dz{\int }_0^{1-y}f(x,y,z)dx+{\int }_0^{1}dy{\int }_y^{1}dz{\int }_{z-y}^{1-y}f(x,y,z)dx$$$$={\int }_0^{1}dz{\int }_z^{1}dy{\int }_0^{1-y}f(x,y,z)dx+{\int }_0^{1}dz{\int }_0^{z}dy{\int }_{z-y}^{1-y}f(x,y,z)dx$$

四

设 $P=x^2\ln y,Q=\frac{x^3}{3y},{\mathrm{\Gamma }}^{\prime }:y=-x+5(1\le x\le 4)$

由 $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=\frac{x^2}{y}$ 知

$${\int }_{\mathrm{\Gamma }}{x}^2\ln ydx+\frac{x^3}{3y}dy={\int }_{{\mathrm{\Gamma }}^{{}^{\prime }}}{x}^2\ln ydx+\frac{x^3}{3y}dy={\int }_1^4(x^2\ln (5-x)-\frac{x^3}{3(5-x)})dx$$$$={\frac{1}{3}{x}^3\ln (5-x)|}_1^4=-\frac{2}{3}\ln 2$$

五

设 $F(x,y,z)=f(x,y,z)+{\lambda }_1(x+y+z-12)+{\lambda }_2(x^2+y^2+z^2-56)$

$$=({\lambda }_1+1)x+({\lambda }_1+3)y+({\lambda }_1+5)z+{\lambda }_2{x}^2+{\lambda }_2{y}^2+{\lambda }_2{z}^2-2{\lambda }_1-56{\lambda }_2$$

令

$$\begin{array}{l}\frac{\partial F}{\partial x}={\lambda }_1+1+2{\lambda }_{2}x=0(1)\\ \frac{\partial F}{\partial y}={\lambda }_1+3+2{\lambda }_{2}y=0(2)\\ \frac{\partial F}{\partial z}={\lambda }_1+5+2{\lambda }_{2}z=0(3)\end{array}$$

$(1)+(2)+(3)$ 得 ${\lambda }_1+8{\lambda }_2+3=0(4)$

$(4)$ 带入 $(1)、(2)、(3)$得

$$\begin{array}{l}x=\frac{4{\lambda }_2+1}{{\lambda }_2}(5)\\ y=4,(6)\\ z=\frac{4{\lambda }_2-1}{{\lambda }_2}(7)\end{array}$$

$(5)(6)(7)$ 带入 $x^2+y^2+z^2=56$ 得 $\{\begin{array}{l}{\lambda }_2=\frac{1}{2}\\ x=6\\ y=4\\ z=2\end{array}$ 或 $\{\begin{array}{l}{\lambda }_2=-\frac{1}{2}\\ x=2\\ y=4\\ z=6\end{array}$

由连续函数在有界闭集上必然取到最小值和最大值，$f(6,4,2)=28$ 和 $f(2,4,6)=44$ 知

$f(x,y,z)$ 在前述条件下的最大值为 44

六

证明：由Gauss 公式， ${\iiint }_{\mathrm{\Omega }}(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})(x,y,z)dxdydz=0$

设 $g(x,y,z)=(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}-f)(x,y,z)$,

由积分平均值定理得在在 $(x_0,y_0,z_0)\in \overline{\mathrm{\Omega }}$ 使得

${\iiint }_{\mathrm{\Omega }}g(x,y,z)dxdydz=g(x_0,y_0,z_0)V(\mathrm{\Omega })$ ，即

$(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})(x_0,y_0,z_0)=f(x_0,y_0,z_0)-\frac{1}{V(\mathrm{\Omega })}{\iiint }_{\mathrm{\Omega }}f(x,y,z)dxdydz$

七 证明

7-(1)

$f(x+h)-f(x)=Jf(x)h+r(h)$ ，其中 $\underset{h\to 0}{lim}\frac{\parallel r(h)\parallel }{\parallel h\parallel }=0$

$⟨f(x+h)-f(x),h⟩⩾\alpha \parallel h{\parallel }^2$

$\Rightarrow ⟨Jf(x)h,h⟩+⟨r(h),h⟩⩾\alpha \parallel h{\parallel }^2$

利用Cauchy-Schwarz不等式得

$$\begin{array}{c}⟨Jf(x)h,h⟩⩾\alpha \parallel h{\parallel }^2-⟨r(h),h⟩\\ ⩾\alpha \parallel h{\parallel }^2-\parallel r(h)\parallel \parallel h\parallel \end{array}$$

对于 $\epsilon =\frac{1}{2}\alpha ,\mathrm{\exists }\delta >0,h\in B_{\delta }(0)$ 时 $\frac{\parallel r(h)\parallel }{\parallel h\parallel }<\epsilon$

$\Rightarrow ⟨Jf(x)h,h⟩⩾(\alpha -\epsilon )\parallel h{\parallel }^2=\frac{\alpha }{2}\parallel h{\parallel }^2>0$

若 $\det Jf(x)=0$ ，则 $Jf(x)$ 有特征值 0 ，设 $h\ne 0$ 为对应的特征向量，则 $Jf(x)h=0$ ，

取 $x>0$ 充分大使得 $\frac{h}{x}\in B_{\delta }(0)$ ，则 $⟨Jf(x)\frac{h}{x},\frac{h}{x}⟩=0$ ，矛盾

$⟹\det Jf(x)\ne 0$

7-(2)

由 $f\in C^{\prime }\left({\mathbb{R}}^n\right)$ 及det $Jf(x)\ne 0$ 可得 $f\left({\mathbb{R}}^n\right)$ 为开集

对 $\mathrm{\forall }x\ne 0\in {\mathbb{R}}^n,\alpha \parallel x{\parallel }^2⩽⟨f(x)-f(0),x⟩⩽\parallel f(x)-f(0)\parallel \parallel x\parallel$

$\Rightarrow \alpha \parallel x\parallel \le \parallel f(x)-f(0)\parallel \le \parallel f(x)\parallel +\parallel f(0)\parallel$

$\Rightarrow$ 若点列 $\left\{f\left(x_i\right)\right\}$ 有界，即存在 $r>0$ 使得 $\Vert f\left(x_i\right)\Vert <r$ ，则 $\Vert x_i\Vert <\frac{r+\parallel f(0)\parallel }{\alpha }$ ，则点列 $\left\{x_i\right\}$ 有界

取 $f\left({\mathbb{R}}^n\right)$ 中任一收敛点列 $\left\{f\left(x_i\right)\right\}$ ，由于 $\left\{f\left(x_i\right)\right\}$ 收敛，故 $f\left(x_i\right)$ 有界，则 $\left\{x_i\right\}$ 有界，由

Bolzano-Weierstrass定理，取出 $\left\{x_i\right\}$ 中收敛子列 $\left\{x_{k_i}\right\}$ ，设 $\underset{i\to \mathrm{\infty }}{lim}{x}_{k_i}=x$ ，由 $f$ 的连续性

知 $\underset{i\to \mathrm{\infty }}{lim}f\left(x_{k_i}\right)=f(x)=\underset{i\to \mathrm{\infty }}{lim}f\left(x_i\right)$

$\Rightarrow \underset{i\to \mathrm{\infty }}{lim}f\left(x_i\right)\in f\left({\mathbb{R}}^n\right)$

$\Rightarrow f\left({\mathbb{R}}^n\right)$ 是闭集

因此 $f{\left({\mathbb{R}}^n\right)}^c$ 是开集，而 ${\mathbb{R}}^n=f\left({\mathbb{R}}^n\right)\cup f{\left({\mathbb{R}}^n\right)}^c$ ，但是 ${\mathbb{R}}^n$ 是连通集，不能分解成两个不相交的非空开集之并，故 $f\left({\mathbb{R}}^n\right)$ 与 $f{\left({\mathbb{R}}^n\right)}^c$ 中有一个是空集，故 $f\left({\mathbb{R}}^n\right)={\mathbb{R}}^n$

八

证明：设 $\phi :{\mathbb{R}}^2\to {\mathbb{R}}^2,(r,\theta )\mapsto (r\cos \theta ,r\sin \theta )$ ，则 $\frac{\partial (f\circ \phi )}{\partial r}=\frac{\partial f}{\partial x}\cos \theta +\frac{\partial f}{\partial y}\sin \theta$

$$\underset{\epsilon \to 0^{+}}{lim}\underset{{\epsilon }^2\le x^2+y^2\le 1}{\iint }\frac{1}{x^2+y^2}(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y})(x,y)dxdy$$$$=\underset{\epsilon \to 0^{+}}{lim}{\int }_0^{2\pi }d\theta {\int }_{\epsilon }^1\frac{1}{r^2}(r\cos \theta \frac{\partial f}{\partial x}+r\sin \theta \frac{\partial f}{\partial y})rdr$$$$=\underset{\epsilon \to 0^{+}}{lim}{\int }_0^{2\pi }d\theta {\int }_{\epsilon }^1\frac{\partial (f\circ \phi )}{\partial r}dr$$$$=\underset{\epsilon \to 0^{+}}{lim}{\int }_0^{2\pi }[(f\circ \phi )(1,\theta )-(f\circ \phi )(\epsilon ,\theta )]d\theta$$$$=\underset{\epsilon \to 0^{+}}{lim}{\int }_0^{2\pi }[-(f\circ \phi )(\epsilon ,\theta )]d\theta$$

由积分平均值定理知存在 ${\theta }_0\in [0,2\pi ]$ 使得 ${\int }_0^{2\pi }[-(f\circ \phi )(\epsilon ,\theta )]d\theta =-2\pi (f\circ \phi )(\epsilon ,{\theta }_0)$

由 $f$ 的连续性知 $\underset{\epsilon \to 0^{+}}{lim}{\int }_0^{2\pi }[-(f\circ \phi )(\epsilon ,\theta )]d\theta =-2\pi \underset{\epsilon \to 0^{+}}{lim}(f\circ \phi )(\epsilon ,{\theta }_0)=-2\pi f(0,0)$