[家里蹲大学数学杂志]第080期丘成桐大学生数学竞赛2010年分析与方程个人赛试题参考解答

 

 

1 (1)Let $\sed{x_k}_{k=1}^n \subset (0,\pi)$, and define $$\bex x=\frac{1}{n}\sum_{k=1}^n x_i. \eex$$ Show that $$\bex \prod_{k=1}^n \frac{\sin x_k}{x_k}\leq \sex{\frac{\sin x}{x}}^n. \eex$$

Proof. Direct computations show $$\bex \sex{\ln\frac{\sin x}{x}}'' =\sex{\ln \sin x-\ln x}'' =\frac{-1}{\sin^2 x}+\frac{1}{x^2} <0, \eex$$ for all $x\in (0,\pi)$. Thus $\dps{\ln\frac{\sin x}{x}}$ is a concave function in $(0,\pi)$. Jensen's inequality then yields $$\bex \frac{1}{n}\sum_{k=1}^n\ln\frac{\sin x_k}{x_k} \leq \ln\frac{\sin x}{x}. \eex$$ The exponential of this above inequality is the desired result.

(2)From $$\bex \int_0^\infty e^{-x^2}\rd x=\frac{\sqrt{\pi}}{2}, \eex$$ calculate the integral $\dps{\int_0^\infty \sin\sex{x^2}\rd x}$.

Proof. Consider the sector in $\bbR^2$ enclosed by the following three curves $$\bex \left\{\ba{lll} I:& 0\leq z\leq R,\\ II:&Re^{i\theta},\ 0\leq \theta\leq \frac{\pi}{4},\\ III:& re^{i\frac{\pi}{4}},\ 0\leq r\leq R. \ea\right. \eex$$ Cauchy's integration theorem then yields $$\bee\label{Yau_DE_1:eq} 0=\sez{\int_I+\int_{II}+\int_{III}}e^{iz^2}\rd z. \eee$$ Noticing

(a)$\dps{\int_I e^{iz^2}\rd z =\int_0^R e^{ix^2}\rd x}$,

(b)$$\bex \sev{\int_{II}e^{iz^2}\rd z} &=&\sev{\int_0^\frac{\pi}{4} e^{iR^2e^{2i\theta}}\cdot iRe^{i\theta} \rd \theta}\\ &\leq&R\int_0^\frac{\pi}{4} e^{-R^2\sin 2\theta}\rd \theta\\ &\leq&R\int_0^\frac{\pi}{4} e^{-R^2\cdot \frac{2}{\pi}\cdot 2\theta} \rd \theta\\ &=&\frac{\pi}{4R}\sex{1-e^{-R^2}}\\ &\to&0,\mbox{ as }R\to\infty, \eex$$

(c)$\dps{\int_{III}e^{iz^2}\rd z =-\int_0^R e^{ir^2e^{i\frac{\pi}{2}}}\cdot e^{i\frac{\pi}{4}}\rd r =e^{i\frac{\pi}{4}}\int_0^Re^{-r^2}\rd r}$,

we have, by sending $R\to\infty$ in \eqref{Yau_DE_1:eq}, that $$\bex \int_0^\infty e^{ix^2}\rd x =e^{i\frac{\pi}{4}}\int_0^\infty e^{-r^2} \rd r. \eex$$ Taking the imaginary part of this above equality gives $$\bex \int_0^\infty \sin(x^2)\rd x=\frac{\sqrt{\pi}}{2\sqrt{2}}. \eex$$

 

2 Let $f:\bbR\to \bbR$ be any function. Prove that the set $$\bex C=\sed{x_0\in \bbR;\ f(x_0)=\lim_{x\to x_0}f(x)} \eex$$ is a $G_\delta$-set.

Proof. By definition, $$\bex C=\cap_{k=1}^\infty C_k, \eex$$ where $$\bex C_k=\sed{x_0\in \bbR;\ \exists\ \delta_{x_0}>0,\ s.t.\ \sev{x-x_0}<\delta_{x_0}\ra \sev{f(x)-f(x_0)}<\frac{1}{k}} \eex$$ is an open set. In fact, $$\bex x_0\in C_k \ra U(x_0,\delta_{x_0})\subset C_k. \eex$$

 

3 Consider the ODE $$\bex \dot x=-x+f(t,x), \eex$$ where $$\bex \left\{\ba{ll} \sev{f(t,x)}\leq \varphi(t)\sev{x},\ (t,x)\in \bbR\times\bbR,\\ \int^\infty \varphi(t)\rd t<\infty. \ea\right. \eex$$ Prove that every solution approaches zero as $t\to\infty$.

Proof. For all $t\in [0,\infty)$, we have $$\bex \infty&>&\int_0^t \varphi(s)\rd s \geq \int_0^t \sev{\frac{\dot x(s)+x(s)}{x(s)}}\rd s =\int_0^t\sev{ \frac{\sex{e^sx(s)}'}{e^sx(s)}}\rd s\\ &\geq&\sev{\int_0^t \rd \sex{e^s x(s)}} =\sev{e^tx(t)-x(0)}. \eex$$ Thus $$\bex \lim_{t\to\infty}x(t) =\lim_{t\to\infty}e^{-t}\cdot\sez{e^tx(t)}=0. \eex$$

 

4 Solve the PDE $$\bex \left\{\ba{ll} \lap u=0,&\mbox{in }\bbR^+\times \bbR,\\ u=g,&\mbox{on }\sed{x_1=0}\times \bbR, \ea\right. \eex$$ where $$\bex g(x_2)=\left\{\ba{ll} 1,&\mbox{if }x_2>0,\\ -1,&\mbox{if }x_2<0. \ea\right. \eex$$

Proof. It is standard (easy to verfiy) that $$\bex u(x)=\int_{\sed{y_1=0}\times \bbR} u(y)\frac{\p G}{\p n}(x,y)\rd S(y), \eex$$ where $$\bex G(x,y) =\frac{1}{2\pi} \sez{\ln\sev{y-x} -\ln\sev{y-\tilde x}} \eex$$ is the Green's function for $\sed{x_1>0}$, with $\tilde x$ the reflection of $x$ in the plane $\sed{x_1=0}$. Direct computations show $$\bex \frac{\p G}{\p n} (x,y) &=&-\frac{\p G}{\p y_1}(x,y) =-\frac{1}{2\pi}\sez{\frac{y_1-x_1}{\sev{y-x}^2} -\frac{y_1+x_1}{\sev{y-\tilde x}}}\\ &=&-\frac{1}{2\pi}\frac{-2x_1}{\sev{y-x_1}^2}\ \sex{\sev{y-x}=\sev{y-\tilde x}}\\ &=&\frac{x_1}{\pi\sev{y-x}^2}. \eex$$ Thus $$\bex u(x) &=&\int_{\sed{y_1=0}\times \bbR} u(y)\frac{x_1}{\pi\sev{y-x}^2}\rd S(y)\\ &=&-\frac{x_1}{\pi}\int_{-\infty}^\infty \frac{g(y_2)}{x_1^2+(y_2-x_2)^2}\rd y_2\\ &=&-\frac{x_1}{\pi} \sez{ \frac{1}{x_1}\int_{-\infty}^0 \frac{-1}{1+\sev{\frac{y_2-x_2}{x_1}}^2}\rd \frac{y_2-x_2}{x_1} +\frac{1}{x_1}\int_0^\infty \frac{1}{1+\sex{\frac{y_2-x_2}{x_1}}^2}\rd \frac{y_2-x_2}{x_1} }\\ &=&-\frac{1}{\pi}\sez{ \left.-\arctan\frac{y_2-x_2}{x_1}\right| ^{y_2=0}_{y_2=-\infty} +\left.\arctan\frac{y_2-x_2}{x_1}\right| ^{y_2=\infty}_{y_2=0} }\\ &=&\frac{2}{\pi}\arctan\frac{x_2}{x_1}, \ x=(x_1,x_2)\in \bbR^2. \eex$$

 

5 Let $K\in C\sex{[0,1]\times [0,1]}$. For $f\in C[0,1]$, the space of continuous functions on $[0,1]$, define $$\bex Tf(x)=\int_0^1 K(x,y)f(y)\rd y. \eex$$ Prove that $Tf\in C[0,1]$. Moreover, $$\bex \Omega=\sed{Tf;\ \sen{f}_{sup}\leq 1} \eex$$ is precompact in $C[0,1]$.

Proof.

(1)$Tf\in C[0,1]$. $$\bee\label{Yau_DE_5:eq}\bea \sev{Tf(x_1)-Tf(x_2)} &\leq \int_0^1 \sev{K(x_1,y)-K(x_2,y)}\sev{f(y)} \rd y\\ &\to 0,\mbox{ as }\sev{x_1-x_2}\to 0, \eea\eee$$by the uniform continuity of $K$ in $x$ and $y$.

(2)$\Omega$ is precompact in $C[0,1]$. This follows readily from

(a)the unform boundedness of $f\in \Omega$: $$\bex \sen{f}_{\sup}\leq 1, \eex$$

(b)the equicontinuity of $f\in \Omega$, that is, \eqref{Yau_DE_5:eq},

(c)and the Ascoli-Azer\'a theorem.

 

6 Prove the Poisson summation formula $$\bex \sum_{n=-\infty}^\infty f(x+2n\pi) =\frac{1}{2\pi} \sum_{k=-\infty}^\infty \hat f(k)e^{ikx}, \eex$$ for $$\bex f\in \calS(\bbR) =\sed{f\in L^1_{loc}(\bbR);\ \sex{1+\sev{x}^m}\sev{f^{(n)}(x)}\leq C_{m,n},\ \forall\ m,n\geq 0}. \eex$$ Here $$\bex \hat f(\xi)=\int_{\bbR} f(x)e^{-ix\xi}\rd x. \eex$$

Proof. Define $$\bex h(x)=\sum_{n=-\infty}^\infty f(x+2n\pi). \eex$$ Then $h$ is periodic with periodical $2\pi$. And hence the coefficients of its Fourier series are $$\bex a_k &=&\frac{1}{2\pi}\int_0^{2\pi} h(x)e^{-ikx}\rd x =\frac{1}{2\pi}\sum_{n=-\infty}^\infty \int_0^{2\pi} f(x+2n\pi)e^{-ikx}\rd x\\ &=&\frac{1}{2\pi} \sum_{n=-\infty}^\infty \int_{2n\pi}^{2(n+1)\pi} f(y)e^{-ik(y-2n\pi)}\rd y\\ &=&\int_{-\infty}^\infty f(x)e^{-ikx}\rd x =\hat f(k). \eex$$ Consequently, $$\bex \sum_{n=-\infty}^\infty f(x+2n\pi) =h(x) =\sum_{k=-\infty}^\infty a_ke^{ikx} =\sum_{k=-\infty}^\infty \hat f(k)e^{ikx}. \eex$$