9.2 雨课堂作业

解 (1) 由弧长公式可得$$\begin{array}{}\text{(1)}& L={\int }_{\gamma }\sqrt{\mathrm{d}{x}^2+\mathrm{d}{y}^2}={\int }_a^b\sqrt{1+(f^{\prime }(x){)}^2}\mathrm{d}x\end{array}$$ 曲率为$$\begin{array}{}\text{(2)}& \kappa =\frac{\parallel {\mathit{r}}^{\prime }(x)\times {\mathit{r}}^{″}(x)\parallel }{\parallel {\mathit{r}}^{\prime }(x){\parallel }^3}=\frac{\parallel (1,f^{\prime }(x))\times (0,f^{″}(x))\parallel }{\parallel (1,f^{\prime }(x)){\parallel }^3}=\frac{|f^{″}(x)|}{{[1+(f^{\prime }(x){)}^2]}^{3/2}}\end{array}$$

(2) 由弧长公式可得$$\begin{array}{}\text{(3)}& L={\int }_{\gamma }\sqrt{\mathrm{d}{r}^2+(r\mathrm{d}\theta {)}^2}={\int }_{\alpha }^{\beta }\sqrt{(r^{\prime }(\theta ){)}^2+r(\theta {)}^2}\mathrm{d}\theta \end{array}$$ 曲率为$$\begin{array}{}\text{(4)}& \kappa =\frac{\parallel {\mathit{r}}^{\prime }(\theta )\times {\mathit{r}}^{″}(\theta )\parallel }{\parallel {\mathit{r}}^{\prime }(\theta ){\parallel }^3}=\frac{|r(\theta {)}^2+2(r^{\prime }(\theta ){)}^2-r(\theta )r^{″}(\theta )|}{{[(r^{\prime }(\theta ){)}^2+r(\theta {)}^2]}^{3/2}},\mathit{r}(\theta )=(r(\theta )\cos \theta ,r(\theta )\sin \theta )\end{array}$$ 直接求导计算曲率的运算量较大，另一种方法是利用单位向量。注意到$$\begin{array}{}\text{(5)}& \{\begin{array}{l}\hat{r}=(\cos \theta ,\sin \theta )\\ \hat{\theta }=(-\sin \theta ,\cos \theta )\end{array}⟹\frac{\mathrm{d}\hat{r}}{\mathrm{d}\theta }=\hat{\theta },\frac{\mathrm{d}\hat{\theta }}{\mathrm{d}\theta }=-\hat{r}\end{array}$$ 因此$$\begin{array}{}\text{(6)}& {\mathit{r}}^{\prime }={\left(r\hat{r}\right)}^{\prime }=r^{\prime }\hat{r}+r\hat{\theta },{\mathit{r}}^{″}={(r^{\prime }\hat{r}+r\hat{\theta })}^{\prime }=(r^{″}-r)\hat{r}+2r^{\prime }\hat{\theta }\end{array}$$ 由此可得$$\begin{array}{}\text{(7)}& \parallel {\mathit{r}}^{\prime }\times {\mathit{r}}^{″}\parallel =|r^{\prime }\cdot 2r^{\prime }-r(r^{″}-r)|=|r^2+2(r^{\prime }{)}^2-r{r}^{″}|,\parallel {\mathit{r}}^{\prime }\parallel =\sqrt{(r^{\prime }{)}^2+r^2}\end{array}$$ 故曲率为$$\begin{array}{}\text{(8)}& \kappa =\frac{\parallel {\mathit{r}}^{\prime }\times {\mathit{r}}^{″}\parallel }{\parallel {\mathit{r}}^{\prime }{\parallel }^3}=\frac{|r^2+2(r^{\prime }{)}^2-r{r}^{″}|}{{[(r^{\prime }{)}^2+r^2]}^{3/2}}\end{array}$$ $◻$

解 将椭圆方程化为极坐标形式可得$$\begin{array}{}\text{(9)}& A{r}^2{\cos }^2\theta +2B{r}^2\sin \theta \cos \theta +C{r}^2{\sin }^2\theta =1⟹r(\theta {)}^2=\frac{1}{A{\cos }^2\theta +2B\sin \theta \cos \theta +C{\sin }^2\theta }\end{array}$$ 由面积公式可得椭圆所围有界区域的面积为$$\begin{array}{}\text{(10)}& \begin{array}{rl}S& =\frac{1}{2}{\int }_{-\pi }^{\pi }r(\theta {)}^2\mathrm{d}\theta =\frac{1}{2}{\int }_{-\pi }^{\pi }\frac{\mathrm{d}\theta }{A{\cos }^2\theta +2B\sin \theta \cos \theta +C{\sin }^2\theta }={\int }_{-\pi }^{\pi }\frac{\mathrm{d}\theta }{A+C+(A-C)\cos 2\theta +2B\sin 2\theta }\\ & ={\int }_{-\pi }^{\pi }\frac{\mathrm{d}\theta }{A+C+\sqrt{(A-C{)}^2+4B^2}\cos (2\theta -\phi )}\underset{\mathrm{periodic}}{\overset{t=2\theta -\phi }{=}}{\int }_{-\pi }^{\pi }\frac{\mathrm{d}t}{A+C+\sqrt{(A-C{)}^2+4B^2}\cos t}\end{array}\end{array}$$ 由于$AC>B^2$，故$A+C>\sqrt{(A-C{)}^2+4B^2}$。利用以下积分$$\begin{array}{}\text{(11)}& {\int }_{-\pi }^{\pi }\frac{\mathrm{d}t}{a+b\cos t}=\frac{2\pi }{\sqrt{a^2-b^2}},|a|>|b|\end{array}$$ 可得$$\begin{array}{}\text{(12)}& S=\frac{2\pi }{\sqrt{(A+C{)}^2-[(A-C{)}^2+4B^2]}}=\frac{\pi }{\sqrt{AC-B^2}}\end{array}$$ $◻$

解 仅考虑第一象限的部分（即$x,y\ge 0$）的弧长、面积，随后乘以$4$即可。由题意可知$$\begin{array}{}\text{(13)}& y=a{[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{3/2},y^{\prime }(x)=-{\left(\frac{x}{a}\right)}^{-1/3}{[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{1/2},x\in [0,a]\end{array}$$ 计算可得$$\begin{array}{}\text{(14)}& L=4{\int }_0^a\sqrt{1+(y^{\prime }(x){)}^2}\mathrm{d}x=4{\int }_0^a{\left(\frac{x}{a}\right)}^{-1/3}\mathrm{d}x=4\cdot \frac{3a}{2}=6a\end{array}$$ $$\begin{array}{}\text{(15)}& A=4{\int }_0^{a}y(x)\mathrm{d}x=4a{\int }_0^a{[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{3/2}\mathrm{d}x\stackrel{t=\sqrt[3]{\frac{x}{a}}}{=}12a^2{\int }_0^1(1-t^2{)}^{3/2}{t}^2\mathrm{d}t=12a^2\cdot \frac{\pi }{32}=\frac{3\pi a^2}{8}\end{array}$$ $◻$

另解 另一种做法是利用星形线的参数方程：$$\begin{array}{}\text{(16)}& \{\begin{array}{ll}x(t)=a{\cos }^{3}t,& y(t)=a{\sin }^{3}t,\\ x^{\prime }(t)=-3a{\cos }^{2}t\sin t,& y^{\prime }(t)=3a{\sin }^{2}t\cos t,\end{array}t\in [0,\pi /2]\end{array}$$ 计算可得$$\begin{array}{}\text{(17)}& L=4{\int }_0^{\pi /2}\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}\mathrm{d}t=12a{\int }_0^{\pi /2}\sin t\cos t\mathrm{d}t=6a\end{array}$$ $$\begin{array}{}\text{(18)}& A=4\cdot \frac{1}{2}{\int }_0^{\pi /2}(x(t)y^{\prime }(t)-y(t)x^{\prime }(t))\mathrm{d}t=6a^2{\int }_0^{\pi /2}{\sin }^{2}t{\cos }^{2}t\mathrm{d}t=\frac{3\pi a^2}{8}\end{array}$$ $◻$

解 由题意可知$$\begin{array}{}\text{(19)}& x^{\prime }(t)=1-\cos t,y^{\prime }(t)=\sin t,t\in [0,2\pi ]\end{array}$$ 计算可得$$\begin{array}{}\text{(20)}& L={\int }_0^{2\pi }\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}\mathrm{d}t={\int }_0^{2\pi }\sqrt{2-2\cos t}\mathrm{d}t=8\end{array}$$ $$\begin{array}{}\text{(21)}& A={\int }_{\gamma }y\mathrm{d}x={\int }_0^{2\pi }{x}^{\prime }(t)y(t)\mathrm{d}t={\int }_0^{2\pi }(1-2\cos t+{\cos }^{2}t)\mathrm{d}t=3\pi \end{array}$$ $◻$

解 仅考虑第一象限的部分（即$\theta \in [0,\frac{\pi }{4}]$）的弧长、面积，随后乘以$4$即可。由题意可知$$\begin{array}{}\text{(22)}& r(\theta )=a\sqrt{2\cos 2\theta },r^{\prime }(\theta )=-\frac{\sqrt{2}a\sin 2\theta }{\sqrt{\cos 2\theta }},\theta \in [0,\frac{\pi }{4}]\end{array}$$ 计算可得$$\begin{array}{}\text{(23)}& \begin{array}{rl}L& =4{\int }_0^{\pi /4}\sqrt{(r^{\prime }(\theta ){)}^2+r(\theta {)}^2}\mathrm{d}\theta =4a{\int }_0^{\pi /4}\sqrt{\frac{2{\sin }^{2}2\theta }{\cos 2\theta }+2\cos 2\theta }\mathrm{d}\theta \\ & \stackrel{t=2\theta }{=}2\sqrt{2}a{\int }_0^{\pi /2}\frac{\mathrm{d}t}{\sqrt{\cos t}}=4aK\left(\frac{1}{2}\right)\end{array}\end{array}$$ $$\begin{array}{}\text{(24)}& A=4\cdot \frac{1}{2}{\int }_0^{\pi /4}r(\theta {)}^2\mathrm{d}\theta =2a^2{\int }_0^{\pi /4}2\cos 2\theta \mathrm{d}\theta =2a^2\end{array}$$ 其中$K(k)$为第一类完全椭圆积分，$K\left(\frac{1}{2}\right)\approx 1.85407$。 $◻$

解 (1) 由题意可知抛物线在圆内（包括边界）的部分满足$$\begin{array}{}\text{(25)}& x^2+(A{x}^2-1{)}^2\le 1⟹x^2(1-2A+A^2{x}^2)\le 0⟹|x|\le x_A:=\frac{\sqrt{2A-1}}{A}\end{array}$$ 曲线的弧长为$$\begin{array}{}\text{(26)}& L(A)=2{\int }_0^{x_A}\sqrt{1+(2Ax{)}^2}\mathrm{d}x=\sqrt{16-\frac{14}{A}+\frac{3}{A^2}}+\frac{1}{2A}\ln (2\sqrt{2A-1}+\sqrt{8A-3})\end{array}$$ 显然有$$\begin{array}{}\text{(27)}& \underset{A\to +\mathrm{\infty }}{lim}L(A)=4\end{array}$$

(2) 对$L$求导可得$$\begin{array}{}\text{(28)}& L^{\prime }(A)=\frac{1}{2A^2}[2\sqrt{\frac{8A-3}{2A-1}}-\ln (2\sqrt{2A-1}+\sqrt{8A-3})]\end{array}$$ 令$t=2\sqrt{2A-1}$，考虑$$\begin{array}{}\text{(29)}& f(t):=4\sqrt{1+\frac{1}{t^2}}-\ln (t+\sqrt{t^2+1}),t>0\end{array}$$ 显然$f$严格减，且有$f(0)=+\mathrm{\infty }$、$f(+\mathrm{\infty })=-\mathrm{\infty }$，故$\mathrm{\exists }{t}_0\in (0,+\mathrm{\infty })$使得$f(t_0)=0$。

令$A_0=\frac{1}{2}(1+\frac{t_0^2}{4})$，则$L$在$(\frac{1}{2},A_0)$上严格增，在$(A_0,+\mathrm{\infty })$上严格减，最终趋于$4$。因此$L(A_0)>L(+\mathrm{\infty })=4$，故存在$A>\frac{1}{2}$使得$L(A)>4$。

(3) 由于$L\left(\frac{1}{2}\right)=0<4$、$L(A_0)>4$且$L$在$(\frac{1}{2},A_0)$上严格增，故$\mathrm{\exists }!A_1\in (\frac{1}{2},A_0)$使得$L(A_1)=4$，即存在唯一$A$使得$L(A)=4$。 $◻$

注 以下内容与课程要求无关。我们尝试使用一些技巧探讨$f$的更多性质。记$L:=\underset{x\to +\mathrm{\infty }}{lim}f(x)$，由于直接求解$f(x)$较为困难，考虑利用反函数方法。设$y=f(x)$，其中$x>0$、$y>0$，则$x=f^{-1}(y)$满足$$\begin{array}{}\text{(37)}& \frac{\mathrm{d}x}{\mathrm{d}y}=x^2+y^2,x(1)=1\end{array}$$ 这是Ricatti方程，令$x(y)=\frac{u^{\prime }(y)}{u(y)}$，则$u(y)$满足二阶线性微分方程$$\begin{array}{}\text{(38)}& u^{″}(y)+y^{2}u(y)=0,u(1)+u^{\prime }(1)=0\end{array}$$ 考虑换元$(y,u)\mapsto (t,v)$，其中$t=\frac{1}{2}{y}^2$、$v=\frac{u}{\sqrt{y}}$，亦即$y=\sqrt{2t}$、$u=v\sqrt[4]{2t}$，由此可得$$\begin{array}{}\text{(39)}& \begin{array}{rl}\frac{\mathrm{d}u}{\mathrm{d}y}& =\frac{\mathrm{d}t}{\mathrm{d}y}\frac{\mathrm{d}(v\sqrt[4]{2t})}{\mathrm{d}t}=\sqrt{2t}(v^{\prime }(t)\sqrt[4]{2t}+\frac{v(t)}{2\sqrt[4]{(2t{)}^3}})=\frac{v(t)+4t{v}^{\prime }(t)}{2\sqrt[4]{2t}}\\ \frac{{\mathrm{d}}^{2}u}{\mathrm{d}{y}^2}& =\frac{\mathrm{d}t}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}t}\frac{v(t)+4t{v}^{\prime }(t)}{2\sqrt[4]{2t}}=\frac{16t^2{v}^{″}(t)+16t{v}^{\prime }(t)-v(t)}{4(2t{)}^{3/4}}\end{array}\end{array}$$ 代入计算可得$$\begin{array}{}\text{(40)}& \begin{array}{rl}{[\frac{\mathrm{d}u}{\mathrm{d}y}+u(y)]}_{y=1}& ={[\frac{v(t)+4t{v}^{\prime }(t)}{2\sqrt[4]{2t}}+v(t)\sqrt[4]{2t}]}_{t=1/2}=v^{\prime }\left(\frac{1}{2}\right)+\frac{3}{2}v\left(\frac{1}{2}\right)=0\\ \frac{{\mathrm{d}}^{2}u}{\mathrm{d}{y}^2}+y^{2}u& =\frac{16t^2{v}^{″}(t)+16t{v}^{\prime }(t)-v(t)}{4(2t{)}^{3/4}}+2tv(t)\sqrt[4]{2t}=\frac{4}{(2t{)}^{3/4}}[t^2{v}^{″}(t)+t{v}^{\prime }(t)+(t^2-\frac{1}{16})v(t)]\end{array}\end{array}$$ 亦即$v(t)$满足$\frac{1}{4}$阶Bessel方程$$\begin{array}{}\text{(41)}& t^2{v}^{″}(t)+t{v}^{\prime }(t)+(t^2-\frac{1}{16})v(t)=0⟹v(t)=c_1{J}_{1/4}(t)+c_2{J}_{-1/4}(t)\end{array}$$ 因此$$\begin{array}{}\text{(42)}& u(y)=\sqrt{y}[c_1{J}_{1/4}\left(\frac{y^2}{2}\right)+c_2{J}_{-1/4}\left(\frac{y^2}{2}\right)]\end{array}$$ 代入$t_0=\frac{1}{2}$处的初始条件，利用恒等式$J_{\nu }^{\prime }(z)=J_{\nu -1}(z)-\frac{\nu }{z}{J}_{\nu }(z)$可得$$\begin{array}{}\text{(43)}& \begin{array}{rl}0& =v^{\prime }\left(\frac{1}{2}\right)+\frac{3}{2}v\left(\frac{1}{2}\right)=c_1[J_{1/4}^{\prime }\left(\frac{1}{2}\right)+\frac{3}{2}{J}_{1/4}\left(\frac{1}{2}\right)]+c_2[J_{-1/4}^{\prime }\left(\frac{1}{2}\right)+\frac{3}{2}{J}_{-1/4}\left(\frac{1}{2}\right)]\\ & =c_1[J_{-3/4}\left(\frac{1}{2}\right)+J_{1/4}\left(\frac{1}{2}\right)]+c_2[J_{-5/4}\left(\frac{1}{2}\right)+2J_{-1/4}\left(\frac{1}{2}\right)]\end{array}\end{array}$$ 由题设可知$y(+\mathrm{\infty })=L\in (1,2)$，故$x(L)$发散，因此$u(L)=0$，由此可得$$\begin{array}{}\text{(44)}& u(L)=\sqrt{L}[c_1{J}_{1/4}\left(\frac{L^2}{2}\right)+c_2{J}_{-1/4}\left(\frac{L^2}{2}\right)]=0⟹\frac{J_{1/4}\left(\frac{L^2}{2}\right)}{J_{-1/4}\left(\frac{L^2}{2}\right)}=-\frac{c_2}{c_1}=\frac{J_{-5/4}\left(\frac{1}{2}\right)+2J_{-1/4}\left(\frac{1}{2}\right)}{J_{-3/4}\left(\frac{1}{2}\right)+J_{1/4}\left(\frac{1}{2}\right)}\end{array}$$ 因此$t=\frac{L^2}{2}$是函数$g(t):=\frac{J_{1/4}\left(t\right)}{J_{-1/4}\left(t\right)}$在区间$(\frac{1}{2},2)$内的（唯一）零点，数值计算可得$t\approx 1.49549$，因此$L\approx 1.72944$。

我们下面尝试求出$f$在$x=+\mathrm{\infty }$处的渐近展开。令$s=\frac{1}{x}\to 0^{+}$，代入可得$$\begin{array}{}\text{(45)}& \frac{\mathrm{d}f}{\mathrm{d}s}=\frac{\mathrm{d}x}{\mathrm{d}s}\frac{\mathrm{d}f}{\mathrm{d}x}=-\frac{1}{s^2}\cdot \frac{1}{\frac{1}{s^2}+f^2}=-\frac{1}{1+s^2{f}^2}⟹{\frac{\mathrm{d}f}{\mathrm{d}s}|}_{s=0}=-1\end{array}$$ 求导可得$$\begin{array}{}\text{(46)}& \frac{{\mathrm{d}}^{2}f}{\mathrm{d}{s}^2}=\frac{2s{f}^2+2s^{2}f\frac{\mathrm{d}f}{\mathrm{d}s}}{(1+s^2{f}^2{)}^2}⟹{\frac{{\mathrm{d}}^{2}f}{\mathrm{d}{s}^2}|}_{s=0}=0\end{array}$$ 继续求导、代入、计算可得$$\begin{array}{}\text{(47)}& {\frac{{\mathrm{d}}^{3}f}{\mathrm{d}{s}^3}|}_{s=0}=2L^2,{\frac{{\mathrm{d}}^{4}f}{\mathrm{d}{s}^4}|}_{s=0}=-12L,{\frac{{\mathrm{d}}^{5}f}{\mathrm{d}{s}^5}|}_{s=0}=24(1-L^4),{\frac{{\mathrm{d}}^{6}f}{\mathrm{d}{s}^6}|}_{s=0}=560L^3,\cdots \end{array}$$ 因此$f$在$x=+\mathrm{\infty }$处的渐近展开为$$\begin{array}{}\text{(48)}& f(x)=L-\frac{1}{x}+\frac{L^2}{3x^3}-\frac{L}{2x^4}+\frac{1-L^4}{5x^5}+\frac{7L^3}{9x^6}+\cdots \end{array}$$

注 第(2)问的第二个不等式可以加强为：设$f\in {\mathcal{C}}^1[a,b]$且满足$f(a)=0$，则有$$\begin{array}{}\text{(64)}& {\int }_a^{b}f(x{)}^2\mathrm{d}x\le \frac{4(b-a{)}^2}{{\pi }^2}{\int }_a^b(f^{\prime }(x){)}^2\mathrm{d}x\end{array}$$ 其中等号成立当且仅当$f(x)=C\sin \frac{\pi (x-a)}{2(b-a)}$或$f(x)=C\cos \frac{\pi (x-a)}{2(b-a)}$，$C$为常数。

不妨设$[a,b]=[0,\frac{\pi }{2}]$，证明可利用例10.2.1中的Wirtinger不等式：设$g\in {\mathcal{C}}^1[0,2\pi ]$且满足$g(0)=g(2\pi )$、${\int }_0^{2\pi }g(x)\mathrm{d}x=0$，则有$$\begin{array}{}\text{(65)}& {\int }_0^{2\pi }g(x{)}^2\mathrm{d}x\le {\int }_0^{2\pi }(g^{\prime }(x){)}^2\mathrm{d}x\end{array}$$ 只需令$$\begin{array}{}\text{(66)}& g(x)=\{\begin{array}{ll}f(x),& x\in [0,\frac{\pi }{2}]\\ f(\pi -x),& x\in (\frac{\pi }{2},\pi ]\\ -f(x-\pi ),& x\in (\pi ,\frac{3\pi }{2}]\\ -f(2\pi -x),& x\in (\frac{3\pi }{2},2\pi ]\end{array}\end{array}$$ 如有必要，可添加光滑子使得$g$在$0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$处可微，则$$\begin{array}{}\text{(67)}& {\int }_0^{\pi /2}f(x{)}^2\mathrm{d}x=\frac{1}{4}{\int }_0^{2\pi }g(x{)}^2\mathrm{d}x\le \frac{1}{4}{\int }_0^{2\pi }(g^{\prime }(x){)}^2\mathrm{d}x={\int }_0^{\pi /2}(f^{\prime }(x){)}^2\mathrm{d}x\end{array}$$