[家里蹲大学数学杂志]第279期丘成桐大学生数学竞赛2011年分析与方程团体赛试题参考解答

S.-T.YauCollege Student Mathematics Contests 2012

Analysis and Differential Equations Team

Please solve $5$ out of the following $6$ problems.

 

 

 

 

 1. let $H^2(\lap)$ be the space of holomorphic function in the unit disk $\lap=\sed{|z|<1}$ such that $$\bex \int_\lap |f|^2|\rd z|^2<\infty. \eex$$ Prove that $H^2(\lap)$ is a Hilbert space and that for any $r<1$, the map $T:H^2(\lap)\to H^2(\lap)$ given by $Tf(z):=f(rz)$ is a compact operator.

 

 Solution: Setting $$\bex \sen{f}:=\sex{\int_\lap |f|^2|\rd z|^2}^\frac{1}{2}, \eex$$ we may easily conclude that $(H^2(\lap),\sen{\cdot})$ is a Hilbert space ($f_n$ holomorphic, $f_n\to f\ra f$ holomorphic by Cauchy integral theorem).

 

 

 2. For any continuous function $f(z)$ of period $1$, show that the equation $$\bee\label{2:eq} \frac{\rd \varphi}{\rd t}=2\pi \varphi+f(t) \eee$$has a unique solution of period $1$.

 

 Solution: Invoking method of variation of constant, we see the solution of \eqref{2:eq} is given by $$\bex \varphi(t)=e^{2\pi t}\sez{\varphi(0)+\int_0^1 e^{-2\pi s}f(s)\rd s}. \eex$$ Suppose $\varphi$ is a $1$ periodical solution, then $$\bex \varphi(0)=\varphi(1)=e^{2\pi}\sez{\varphi(0)+\int_0^1 e^{-2\pi s}f(s)\rd s}. \eex$$ Thus only if $$\bex \varphi(0)=\cfrac{e^{2\pi}\dps{\int_0^1 e^{-2\pi s}f(s)\rd s}}{1-e^{2\pi}}, \eex$$ \eqref{2:eq} has a $1$ periodical solution. We now show the if part. Consider $\psi(t)=\varphi(t+1)$, then $$\bex \sedd{\ba{ll} \cfrac{\rd \psi}{\rd t}&=2\pi \psi+f(t+1)=2\pi \psi+f(t),\\ \psi(0)&=\varphi(1)=\varphi(0). \ea} \eex$$ The uniqueness theorem then yields that $\psi\equiv \varphi$.

 

 

 3. Let $h(x)$ be a $C^\infty$ function on the real line $\bbR$. Find a $C^\infty$ function $u(x,y)$ on an open subset of $\bbR^2$ containing the $x$-axis such that $$\bee\label{3:eq} \sedd{\ba{ll} u_x+2u_y&=u^2,\\ u(x,0)&=h(x). \ea} \eee$$

 

 Solution: The characteristic ODE of \eqref{3:eq} is given by $$\bex \sedd{\ba{lll} \cfrac{\rd x}{\rd t}=1,&\cfrac{\rd y}{\rd t}=2,&\cfrac{\rd u}{\rd t}=u^2,\\ x(0)=x_0,&y(0)=0,&u(0)=h(x_0). \ea} \eex$$ Solving it yields that $$\bex u(x,y)=\cfrac{h\sex{x-\cfrac{y}{2}}}{1-\cfrac{y}{2}\cdot h\sex{x-\cfrac{y}{2}}}. \eex$$

 

 

 4. Let $c>0$ and $$\bex S=\sed{x\in\bbR;\ \sev{x-\frac{p}{q}}\leq \frac{c}{q^3}\mbox{ for all }p,q\in\bbZ,\ q>0}. \eex$$ Show that $S$ is uncountable and its measure is zero.

 

 Solution: This title is problematic.

 

 

 5. Let $sl(n)$ denote the set of all $n\times n$ real matrices with trace equal to zero and let $SL(n)$ be the set of all $n\times n$ matrices with determinant equal to one. Let $\varphi(z)$ be a real analytic function defined in a neighborhood of $z=0$ of the complex plane $\bbC$ satisfying the conditions $\varphi(0)=1$ and $\varphi'(0)=1$. (1) If $\varphi$ maps any near zero matrix in $sl(n)$ into $SL(n)$ for some $n\geq 3$, show that $\varphi(z)=\exp(z)$. (2) Is the conclusion of (1) still true in the case $n=2$? If it is true, prove it. If not, give a counterexample.

 

 Solution: (1) a. By the uniqueness result for the holomorphic functions, we need only to show $\varphi(z)=\exp(z)$ for $z$ small. b. let $A$ be an $n\times n$ matrix with eigenvalues $\lm_k, k=1,\cdots,n$ and with Jordan carnonical form $J$: $$\bex A=T^{-1}JT. \eex$$ Then for any polynomial $p(z)$, $$\bex p(A)=T^{-1}p(J)T,\quad \det p(A)=\prod_{k=1}^n p(\lm_k). \eex$$ And consequently, by analyticity and continuity, $$\bex \varphi(A)=T^{-1}\varphi (J)T,\quad \det \varphi(A)=\prod_{k=1}^n \varphi(\lm_k). \eex$$ (2) By assumptions, for $s$ small, $$\beex \bea 1&=\det \varphi\sex{\sex{\ba{ccc}s&&\\ &-s&\\ &&{\bf 0}\ea}}\\ &=\varphi(s)\cdot \varphi(-s)\cdot \varphi(0),\\ \varphi(-s)&=\frac{1}{\varphi(s)},\\ 1&=\varphi\sex{\sex{\ba{cccc}s&&&\\ &t&&\\ &&-s-t&\\ &&&{\bf 0}\ea}}\\ &=\varphi(s)\cdot \varphi(t)\cdot \varphi(-s-t)\\ &=\frac{\varphi(s)\varphi(t)}{\varphi(s+t)},\\ \varphi(s+t)&=\varphi(s)\cdot \varphi(t),\\ \varphi'(s)&=\lim_{t\to 0}\frac{\varphi(s+t)-\varphi(s)}{t}\\ &=\varphi(s)\lim_{t\to 0}\frac{\varphi(t)-1}{t}\\ &=\varphi(s)\cdot \varphi'(0)\\ &=\varphi(s),\\ \varphi(s)&=\exp(s). \eea \eeex$$ (2) If $n=2$, the conclusion of (1) does not hold true. For example, $\varphi(z)=\exp(z^3)$ satisfies $$\beex \bea \tr A=0&\ra \lm_1+\lm_2=\tr A=0\\ &\ra \det \varphi(A)=\varphi(\lm_1)\cdot \varphi(\lm_2) =1\quad\sex{\mbox{by }(1)c}. \eea \eeex$$

 

 

 6. Use mathematical analysis to show that (1) $e$ and $\pi$ are irrational numbers; (2) $e$ and $\pi$ are transcendental numbers.

 

 Solution: (1) By Taylor expansion, $$\bex e=1+1+\frac{1}{2!}+\cdots+\frac{1}{n!}+\frac{e^\xi}{(n+1)!}\quad (0<\xi<1). \eex$$ If $e$ is rational, and equals $\dps{\frac{p}{q}\ (p,q\in\bbZ_+,\ (p,q)=1)}$, then $$\bex n!\frac{p}{q}=n!\sex{1+1+\frac{1}{2!}+\cdots+\frac{1}{n!}} +\frac{e^\xi}{n+1}. \eex$$ Once $n$ is large, we find $\dps{\frac{e^\xi}{n+1}}$ is a natural number, which is a contradiction.

(2) Argue similarly with $$\bex \frac{\pi}{4}=\arctan 1=\sum_{n=0}^\infty (-1)^n\frac{1}{2n+1}, \eex$$ we see $\pi$ is also irrational.

(3) For the proof of the of $e$ and $\pi$, please refer to the transcendency of $e$ and $\pi$.