From 15e1bf0c46f0e13b85b0f84c89b5f30e6a2857a2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Fri, 15 Dec 2023 18:45:20 +0100 Subject: [PATCH] few changes to dyn sym --- .../Advanced_dynamical_systems.tex | 70 ++++++++++--------- 1 file changed, 37 insertions(+), 33 deletions(-) diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex index 0470aad..75861b1 100644 --- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex +++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex @@ -12,7 +12,7 @@ We identify the elements of $\TT^1$ as $\quot{\RR}{\ZZ}$. Let $x\in \TT^1$. Then, ${R_\alpha}^q x=x+\alpha q=x+p=x$. And $q$ is the smallest integer such that ${R_\alpha}^q x=x$ because we assume that $p$ and $q$ are coprime. \end{proof} \begin{proposition} - Let $\alpha\in\RR\setminus\QQ$ and let $R_\alpha:\TT^1\to \TT^1$ be the rotation of angle $\alpha$. Then, all the points of $\TT^1$ are dense in $\TT^1$. + Let $\alpha\in\RR\setminus\QQ$ and let $R_\alpha:\TT^1\to \TT^1$ be the rotation of angle $\alpha$. Then, all the orbits of $R_\alpha$ are dense in $\TT^1$. \end{proposition} \begin{proof} Let $\varepsilon>0$, $x,y\in \TT^1$. Discretize $\TT^1$ in intervals of length at most $\frac{1}{\varepsilon}$. Then, $\exists m,n\in \NN$ with $m< n\leq \frac{1}{\varepsilon}+1$ such that ${R_\alpha}^m x$ and ${R_\alpha}^nx$ are in the same interval. Thus, $\abs{{R_\alpha}^{n-m}x-x}<\varepsilon$. Now, concatenating ${R_\alpha}^{n-m}x$ repeatedly, we will eventually have $\abs{{R_\alpha}^{k(n-m)}x - y}<\varepsilon$ for some $k\in \NN$. @@ -21,7 +21,7 @@ Let $\alpha\in\RR\setminus\QQ$ and $A\subset \TT^1$ be a non-empty closed invariant set for $R_\alpha$. Then, $A=\TT^1$. \end{corollary} \begin{proof} - Let $x\in \TT^1$ and $y\in A$. Then, $\forall k\in\NN$ $\exists n_k\in\NN$ such that $R_\alpha^{n_k}y\in(x-\frac{1}{k},x+\frac{1}{k})$. Thus, $R_\alpha^{n_k}y\to x$ and $x\in A$ because $A$ is closed and $R_\alpha^{n_k}y\in A$ $\forall k\in\NN$. + Let $x\in \TT^1$ and $y\in A$. Then, $\forall k\in\NN$ $\exists n_k\in\NN$ such that $R_\alpha^{n_k}y\in(x-\frac{1}{k},x+\frac{1}{k})$. Thus, $R_\alpha^{n_k}y\overset{k\to\infty}{\longrightarrow}x$ and $x\in A$ because $A$ is closed and $R_\alpha^{n_k}y\in A$ $\forall k\in\NN$ because $A$ is invariant. \end{proof} \begin{definition} Consider the set $$\Sigma_m @@ -32,7 +32,7 @@ $$ \end{definition} \begin{remark} - Note that some elements in $[0,1]$ have two different representations in base-$m$ identified as elements of $\Sigma_m$. So we can think of $\Sigma_m$ a the quotient space $\quot{\Sigma_m}{\sim}$ where $(x_1,x_2,\ldots)\sim (y_1,y_2,\ldots)$ if and only if $\sum_{i=1}^\infty \frac{x_i}{m^i}=\sum_{i=1}^\infty \frac{y_i}{m^i}$. + Note that some elements in $[0,1]$ have two different representations in base-$m$ identified as elements of $\Sigma_m$. So we can think of $\Sigma_m$ as the quotient space $\quot{\Sigma_m}{\sim}$, where $(x_1,x_2,\ldots)\sim (y_1,y_2,\ldots)$ if and only if $\sum_{i=1}^\infty \frac{x_i}{m^i}=\sum_{i=1}^\infty \frac{y_i}{m^i}$. \end{remark} \begin{proposition} Let $m\in\NN$. Consider the \emph{expansion map} @@ -43,9 +43,9 @@ \end{proposition} \begin{proof} Let $x=(x_1,x_2,\ldots)\in \Sigma_m$. Then, $\phi\circ \sigma_m(x)=\sum_{i=1}^\infty \frac{x_{i+1}}{m^i}$. Moreover: - \begin{multline*} - E_m\circ \phi(x)=E\left(\sum_{i=1}^\infty \frac{x_i}{m^i}\right)=\sum_{i=1}^\infty \frac{x_i}{m^{i-1}}=\\=x_i+\sum_{i=1}\frac{x_{i+1}}{m^i}\equiv\sum_{i=1}\frac{x_{i+1}}{m^i} - \end{multline*} + \begin{equation*} + E_m\circ \phi(x)=m\sum_{i=1}^\infty \frac{x_i}{m^i}=x_i+\sum_{i=1}\frac{x_{i+1}}{m^i}\equiv\sum_{i=1}\frac{x_{i+1}}{m^i} + \end{equation*} \end{proof} \begin{remark} Note that $E$ preserves the Lebesgue measure \textit{backwards}: $\abs{{E_m}^{-1}(A)}=\abs{A}$ for all $A\subseteq \TT^1$, but $\abs{E_m(A)}\ne \abs{A}$ in general. @@ -67,39 +67,39 @@ is periodic and $d(x,y)<\varepsilon$. So periodic points of $\sigma_m$ are dense in $\Sigma_m$. \end{proof} \begin{proposition} - Le $x\in \TT^1$. Then, the positive orbit of $x$ for $E_m$ is dense in $\TT^1$. + There exists $x\in \TT^1$ such that its orbit under $E_m$ is dense in $\TT^1$. \end{proposition} \begin{proof} - By conjugacy, we only prove it for $\sigma_m$. But this is clear by taking: - $$ - x=(0,1,\ldots,m-1,10,\ldots,1(m-1),20,\ldots,2(m-1),\ldots) - $$ + By conjugacy, we only prove it for $\sigma_m$. But this is clear by taking he sequence of \textit{all sequences}: + \begin{multline*} + x=(0,1,\ldots,m-1,0,0,1,0,0,1,1,1,0,2,2,0,1,2,2,1,\\2,2,\ldots,(m-1), (m-1),0,0,0,\ldots) + \end{multline*} \end{proof} - \subsubsection{A hyperbolic automorphism of \texorpdfstring{$T^2$}{T2}} + \subsubsection{A hyperbolic automorphism of \texorpdfstring{$\TT^2$}{T2}} \begin{proposition} Consider $\vf{A}=\begin{pmatrix} 2 & 1 \\ 1 & 1 - \end{pmatrix}\in \GL_2(\RR)$. Then, $\vf{A}(\ZZ^2)=\ZZ^2$ and this induces an automorphism $\vf{\tilde{A}}$ of $T^2=\quot{\RR^2}{\ZZ^2}$. + \end{pmatrix}\in \GL_2(\RR)$. Then, $\vf{A}(\ZZ^2)=\ZZ^2$ and this induces an automorphism $\vf{\tilde{A}}$ of $\TT^2=\quot{\RR^2}{\ZZ^2}$. \end{proposition} \begin{definition} We define the set of periodic points of $\vf{\tilde{A}}$ as $\Per\vf{\tilde{A}}$. \end{definition} \begin{lemma} - $\Per\vf{\tilde{A}}=\quot{\QQ^2}{\ZZ^2}$. Thus, $\Per\vf{\tilde{A}}$ is dense in $T^2$. + $\Per\vf{\tilde{A}}=\quot{\QQ^2}{\ZZ^2}$. Thus, $\Per\vf{\tilde{A}}$ is dense in $\TT^2$. \end{lemma} \begin{proof} Let $\vf{x}\in \Per\vf{\tilde{A}}$. Then, $\exists k\in\NN$ and $\vf{n}\in\ZZ^2$ such that $\vf{A}^k\vf{x}=\vf{x}+\vf{n}$. One can easily check that $\sigma(\vf{\tilde{A}})=\left\{\frac{3}{2}\pm \frac{\sqrt{5}}{2}\right\}=:\{\lambda_{\pm}\}$ with $\lambda_-<1<\lambda_+$. Thus, $$ \det(\vf{A}^k-\vf{I})=({\lambda_+}^k-1)({\lambda_-}^k-1)\ne 0 $$ - and so the equation $\vf{A}^k\vf{x}=\vf{x}+\vf{n}$ has a unique (rational) solution. Now let $(\frac{p_1}{q_1},\frac{p_2}{q_2})\in \quot{\QQ^2}{\ZZ^2}$ and $N\geq 1$ left to be chosen. We define the set $Q_N:=\frac{\ZZ^2}{N} \mod{\ZZ^2}$, which is a subset finite set of $T^2$. But observe that $Q_N$ is invariant under $\vf{\tilde{A}}$, and thus, all of its points are periodic. For the above rational numbers, just choose $N=q_1q_2$. + and so the equation $\vf{A}^k\vf{x}=\vf{x}+\vf{n}$ has a unique (rational) solution. Now let $(\frac{p_1}{q_1},\frac{p_2}{q_2})\in \quot{\QQ^2}{\ZZ^2}$ and $N\geq 1$ left to be chosen. We define the set $Q_N:=\frac{\ZZ^2}{N} \mod{\ZZ^2}$, which is a subset finite set of $\TT^2$. Observe that $Q_N$ is invariant under $\vf{\tilde{A}}$, and thus, all of its points are periodic because the set is finite. For the above rational numbers, just choose $N=q_1q_2$. \end{proof} \begin{remark} The \emph{hyperbolicity} comes from the fact that there is one eigenvector with eigenvalue greater than $1$ and another with eigenvalue less than $1$, both eigenvalues being positive. \end{remark} \begin{theorem} - The iterates of $\vf{\tilde{A}}$ smear every domain $F\subseteq T^2$ uniformly over $T^2$, that is, for every domain $G\subseteq T^2$, we have that the following limit exists: + The iterates of $\vf{\tilde{A}}$ smear every domain $F\subseteq \TT^2$ uniformly over $\TT^2$, that is, for every domain $G\subseteq \TT^2$, we have that the following limit exists: $$ \abs{(\vf{\tilde{A}}^{-n} F)\cap G}\overset{n\to\infty}{\longrightarrow} \abs{F}\abs{G} $$ @@ -108,44 +108,48 @@ \begin{proof} We can prove a more general property in terms of functions in the torus (and then apply it to $f=\indi{F}$ and $g=\indi{G}$): $$ - \lim_{n\to\infty}\int_{T^2} f(\vf{\tilde{A}}^n \vf{x}) g(\vf{x})\dd{\vf{x}}=\int_{T^2} f(\vf{x})\dd{\vf{x}}\int_{T^2} g(\vf{x})\dd{\vf{x}} + \lim_{n\to\infty}\int_{\TT^2} f(\vf{\tilde{A}}^n \vf{x}) g(\vf{x})\dd{\vf{x}}=\int_{\TT^2} f(\vf{x})\dd{\vf{x}}\int_{\TT^2} g(\vf{x})\dd{\vf{x}} $$ We will prove this for the orthonormal basis of Fourier series $\{\exp{2\pi i \vf{p}\cdot \vf{x}}\}_{\vf{p}\in\ZZ^2}$. Note that: $$ - \int_{T^2} \exp{2\pi i (\transpose{(\vf{\tilde{A}}^n)}\vf{p})\cdot \vf{x}}\dd{\vf{x}}=\begin{cases} + \int_{\TT^2} \exp{2\pi i (\transpose{(\vf{\tilde{A}}^n)}\vf{p})\cdot \vf{x}}\dd{\vf{x}}=\begin{cases} 1 & \text{if }\vf{p}=\vf{0} \\ 0 & \text{if }\vf{p}\ne \vf{0} \end{cases} $$ - Therefore, since $\transpose{(\vf{\tilde{A}}^n)}\vf{p}$ takes infinitely many values for $\vf{p}\ne \vf{0}$, we have that if $g=\exp{2\pi i \vf{q} \cdot \vf{x}}$ then: + Now for large $n$, the norm of the vector $\transpose{(\vf{\tilde{A}}^n)}\vf{p}$ is large for $\vf{p}\ne \vf{0}$ as we have: + $$ + \vf{\tilde{A}}^n\vf{p} \simeq \lambda_+^n \langle \vf{p}, \vf{e}_+\rangle \vf{e}_+ + $$ + And so its transpose will eventually be different from $-\vf{q}$. Therefore, we have that if $g=\exp{2\pi i \vf{q} \cdot \vf{x}}$ then: $$ - \lim_{n\to\infty}\int_{T^2} \exp{2\pi i(\transpose{(\vf{\tilde{A}}^n)}\vf{p}+\vf{q})\cdot \vf{x}}\dd{\vf{x}}=0 + \lim_{n\to\infty}\int_{\TT^2} \exp{2\pi i(\transpose{(\vf{\tilde{A}}^n)}\vf{p}+\vf{q})\cdot \vf{x}}\dd{\vf{x}}=0 $$ So for any $\vf{p}, \vf{q}\in\ZZ^2$ we have the equality. Then, we use that any function nice enough can be approximated with its Fourier series. \end{proof} \begin{theorem} - On the torus $T^2$ there exist two direction fields invariant with respect to the automorphism $\vf{\tilde{A}}$. The integral curves of each of these directions fields are everywhere dense on the torus. The automorphism $\vf{\tilde{A}}$ converts the integral curves of each field into integral curves of the same field, expanding by $\lambda_+$ for the first field and contracting by $\lambda_-$ for the second. + On the torus $\TT^2$ there exist two direction fields invariant with respect to the automorphism $\vf{\tilde{A}}$. The integral curves of each of these directions fields are everywhere dense on the torus. The automorphism $\vf{\tilde{A}}$ converts the integral curves of each field into integral curves of the same field, expanding by $\lambda_+$ for the first field and contracting by $\lambda_-$ for the second. \end{theorem} \begin{proof} - Let $\vf{e}_+$ and $\vf{e}_-$ be the eigenvectors of $\vf{A}$ with eigenvalues $\lambda_+$ and $\lambda_-$ respectively. Let $\vf{x}\in T^2$ and + Let $\vf{e}_+$ and $\vf{e}_-$ be the eigenvectors of $\vf{A}$ with eigenvalues $\lambda_+$ and $\lambda_-$ respectively. Let $\vf{x}\in \TT^2$ and $$ - \function{\vf\gamma_+}{\RR}{T^2}{t}{\vf{x}+t \vf{e}_+}\quad - \function{\vf\gamma_-}{\RR}{T^2}{t}{\vf{x}+t \vf{e}_-} + \function{\vf\gamma_+}{\RR}{\TT^2}{t}{\vf{x}+t \vf{e}_+}\quad + \function{\vf\gamma_-}{\RR}{\TT^2}{t}{\vf{x}+t \vf{e}_-} $$ - be the expanding and contracting curves and let $\vf{\xi}_{\vf{x}}=\im(\vf\gamma_+)$, $\vf{\eta}_{\vf{x}}=\im(\vf\gamma_-)$. + be the expanding and contracting curves and let $\vf{\xi}_{\vf{x}}=\im(\vf\gamma_+)$, $\vf{\eta}_{\vf{x}}=\im(\vf\gamma_-)$ be the corresponding direction fields. The density of the curves is a consequence of the density of orbits in rotation maps in the circle with irrational angle. \end{proof} \begin{definition} - Let $\vf{A},\vf{B}:T^2\rightarrow T^2$ be $\mathcal{C}^1$ functions. We say that $B$ is \emph{$\mathcal{C}^0$-close} to $\vf{A}$ if for all $\varepsilon>0$: + Let $\vf{A},\vf{B}:\TT^2\rightarrow \TT^2$ be $\mathcal{C}^1$ functions. We say that $B$ is \emph{$\mathcal{C}^0$-close} to $\vf{A}$ if for all $\varepsilon>0$: $$ - \sup_{\vf{x}\in T^2}\norm{\vf{A}(\vf{x})-\vf{B}(\vf{x})}<\varepsilon + \sup_{\vf{x}\in \TT^2}\norm{\vf{A}(\vf{x})-\vf{B}(\vf{x})}<\varepsilon $$ - We say that $\vf{B}$ is \emph{$\mathcal{C}^1$-close} to $\vf{A}$ if for all $\varepsilon>0$, $\vf{B}$ is $\mathcal{C}^0$-close to $\vf{A}$ and: + We say that $\vf{B}$ is \emph{$\mathcal{C}^1$-close} to $\vf{A}$ if $\vf{B}$ is $\mathcal{C}^0$-close to $\vf{A}$ and for all $\varepsilon>0$: $$ - \sup_{\vf{x}\in T^2}\norm{\vf{D}\vf{A}(\vf{x})-\vf{D}\vf{B}(\vf{x})}<\varepsilon + \sup_{\vf{x}\in \TT^2}\norm{\vf{D}\vf{A}(\vf{x})-\vf{D}\vf{B}(\vf{x})}<\varepsilon $$ \end{definition} \begin{theorem}[Structal stability] - Let $\vf{B}$ be a diffeomorphism on $T^2$ $\mathcal{C}^1$-close to $\vf{\tilde{A}}$. Then, $\vf{B}$ is $\mathcal{C}^0$-conjugate to $\vf{\tilde{A}}$. + Let $\vf{B}$ be a diffeomorphism on $\TT^2$ $\mathcal{C}^1$-close to $\vf{\tilde{A}}$. Then, $\vf{B}$ is $\mathcal{C}^0$-conjugate to $\vf{\tilde{A}}$. \end{theorem} \begin{proof} We need to find a $\mathcal{C}^0$-conjugacy $\vf{H}$ between $\vf{B}$ and $\vf{\tilde{A}}$. Since, $\vf{B}$ is $\mathcal{C}^1$-close to $\vf{\tilde{A}}$, we may expect that both $\vf{H}$ and $\vf{B}$ are small perturbations of the identity and $\vf{\tilde{A}}$ respectively. So set $\vf{H}=\vf{I}+\vf{h}$ and $\vf{B}=\vf{\tilde{A}}+\vf{b}$. Then, we want to find $\vf{h}$ and $\vf{b}$ such that: @@ -155,12 +159,12 @@ $$ This equation is called \emph{conjugacy equation}. Consider the operators \begin{gather*} - \function{\vf{S}_{\vf{\tilde{A}}}}{\mathcal{C}^0(T^2,\RR^2)}{\mathcal{C}^0(T^2,\RR^2)}{\vf{h}}{\vf{h}(\vf{\tilde{A}}(\vf{x}))}\\ - \function{\vf{L}_{\vf{\tilde{A}}}}{\mathcal{C}^0(T^2,\RR^2)}{\mathcal{C}^0(T^2,\RR^2)}{\vf{h}}{\vf{S}_{\vf{\tilde{A}}}\vf{h}-\vf{\tilde{A}}\vf{h}} + \function{\vf{S}_{\vf{\tilde{A}}}}{\mathcal{C}^0(\TT^2,\RR^2)}{\mathcal{C}^0(\TT^2,\RR^2)}{\vf{h}}{\vf{h}(\vf{\tilde{A}}(\vf{x}))}\\ + \function{\vf{L}_{\vf{\tilde{A}}}}{\mathcal{C}^0(\TT^2,\RR^2)}{\mathcal{C}^0(\TT^2,\RR^2)}{\vf{h}}{\vf{S}_{\vf{\tilde{A}}}\vf{h}-\vf{\tilde{A}}\vf{h}} \end{gather*} Observe that: $$ - \sup_{\vf{x}\in T^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{x})}=\sup_{\vf{x}\in T^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{\tilde{A}}^{-1}\vf{x})}= \sup_{\vf{x}\in T^2}\norm{\vf{h}(\vf{x})} + \sup_{\vf{x}\in \TT^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{x})}=\sup_{\vf{x}\in \TT^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{\tilde{A}}^{-1}\vf{x})}= \sup_{\vf{x}\in \TT^2}\norm{\vf{h}(\vf{x})} $$ Hence, $\norm{\vf{S}_{\vf{\tilde{A}}}}=1$ and similarly $\norm{\vf{S}_{\vf{\tilde{A}}}^{-1}}=1$, where $\vf{S}_{\vf{\tilde{A}}}^{-1}:\vf{h}\mapsto \vf{h}(\vf{\tilde{A}}^{-1}(\vf{x}))$. We'll now prove that $\vf{L}_{\vf{\tilde{A}}}$ is invertible. Note that $\RR^2=\langle \vf{e}_+\rangle \oplus \langle \vf{e}_-\rangle$ because $\vf{\tilde{A}}$ is invertible. Thus: $$