From ad1a61d224b5f7b04e62b92d9c80885a439eb151 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Sat, 16 Dec 2023 21:30:26 +0100 Subject: [PATCH] improvements dyn syst --- .../Functions_of_several_variables.tex | 2 +- .../Advanced_dynamical_systems.tex | 70 ++++++++++++++----- 2 files changed, 53 insertions(+), 19 deletions(-) diff --git a/Mathematics/2nd/Functions_of_several_variables/Functions_of_several_variables.tex b/Mathematics/2nd/Functions_of_several_variables/Functions_of_several_variables.tex index 204757b..3ea0c78 100644 --- a/Mathematics/2nd/Functions_of_several_variables/Functions_of_several_variables.tex +++ b/Mathematics/2nd/Functions_of_several_variables/Functions_of_several_variables.tex @@ -304,7 +304,7 @@ \begin{theorem}[Mean value theorem]\label{FSV:meanvaluethm} Let $f:B\rightarrow\RR $ be a function of class $\mathcal{C}^1$ in an open connected set $B\subseteq\RR^n$ and $x,y\in B$. Then: $$f(x)-f(y)=\grad f(z)\cdot(x-y)$$ for some $z\in[x,y]$. \end{theorem} - \begin{theorem}[Mean value theorem for vector-valued functions] + \begin{theorem}[Mean value theorem for vector-valued functions] \label{FSV:meanvaluethmvector} Let $\vf{f}:B\rightarrow\RR^m$ be a function of class $\mathcal{C}^1$ in an open connected set $B\subseteq\RR^n$ and $x,y\in B$. Then: $$\|\vf{f}(x)-\vf{f}(y)\|\leq\|\vf{Df}(z)\|\|x-y\|$$ for some $z\in[x,y]$. \end{theorem} \subsubsection{Higher order derivatives} diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex index 75861b1..62554c5 100644 --- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex +++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex @@ -149,7 +149,7 @@ $$ \end{definition} \begin{theorem}[Structal stability] - Let $\vf{B}$ be a diffeomorphism on $\TT^2$ $\mathcal{C}^1$-close to $\vf{\tilde{A}}$. Then, $\vf{B}$ is $\mathcal{C}^0$-conjugate to $\vf{\tilde{A}}$. + Let $\vf{B}$ be a diffeomorphism on $\TT^2$ which is $\mathcal{C}^1$-close to $\vf{\tilde{A}}$. Then, $\vf{B}$ is $\mathcal{C}^0$-conjugate to $\vf{\tilde{A}}$. \end{theorem} \begin{proof} We need to find a $\mathcal{C}^0$-conjugacy $\vf{H}$ between $\vf{B}$ and $\vf{\tilde{A}}$. Since, $\vf{B}$ is $\mathcal{C}^1$-close to $\vf{\tilde{A}}$, we may expect that both $\vf{H}$ and $\vf{B}$ are small perturbations of the identity and $\vf{\tilde{A}}$ respectively. So set $\vf{H}=\vf{I}+\vf{h}$ and $\vf{B}=\vf{\tilde{A}}+\vf{b}$. Then, we want to find $\vf{h}$ and $\vf{b}$ such that: @@ -159,12 +159,12 @@ $$ This equation is called \emph{conjugacy equation}. Consider the operators \begin{gather*} - \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}} + \function{\vf{S}_{\vf{\tilde{A}}}}{\mathcal{C}^0(\RR^2,\RR^2)}{\mathcal{C}^0(\RR^2,\RR^2)}{\vf{h}}{\vf{h}(\vf{\tilde{A}}(\vf{x}))}\\ + \function{\vf{L}_{\vf{\tilde{A}}}}{\mathcal{C}^0(\RR^2,\RR^2)}{\mathcal{C}^0(\RR^2,\RR^2)}{\vf{h}}{\vf{S}_{\vf{\tilde{A}}}\vf{h}-\vf{\tilde{A}}\vf{h}} \end{gather*} - Observe that: + where we consider the diffeomorphisms $\vf{\tilde{A}}$ and $\vf{B}$ as operators \textit{lifted} to $\mathcal{C}^1(\RR^2,\RR^2)$. Observe that: $$ - \sup_{\vf{x}\in \TT^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{x})}=\sup_{\vf{x}\in \TT^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{\tilde{A}}^{-1}\vf{x})}= \sup_{\vf{x}\in \TT^2}\norm{\vf{h}(\vf{x})} + \sup_{\vf{x}\in \RR^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{x})}=\sup_{\vf{x}\in \RR^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{\tilde{A}}^{-1}\vf{x})}= \sup_{\vf{x}\in \RR^2}\norm{\vf{h}(\vf{x})} $$ 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: $$ @@ -181,12 +181,12 @@ $$ (\vf{S}_{\vf{\tilde{A}}}^{-1}-\lambda_-\vf{I})=\vf{S}_{\vf{\tilde{A}}}^{-1}\left(\vf{I}-\lambda_-\vf{S}_{\vf{\tilde{A}}}^{-1}\right) $$ - is invertible because it is a product of invertible operators. Thus, $\vf{L}_{\vf{\tilde{A}}}$ is invertible. Now, we return to our initial problem. Find $\vf{h}$ such that $\vf{h}= {\vf{L}_{\vf{\tilde{A}}}}^{-1}(\vf{b}(\vf{x}+\vf{h}(\vf{x})))=:\vf\Psi(\vf{h})$, which is a fixed-point problem. Note that $\vf\Psi$ is a contraction. Indeed: + is invertible because it is a product of invertible operators. Thus, $\vf{L}_{\vf{\tilde{A}}}$ is invertible and its inverse is linear because $\vf{L}_{\vf{\tilde{A}}}$ is linear. Now, we return to our initial problem. Find $\vf{h}$ such that $\vf{h}= {\vf{L}_{\vf{\tilde{A}}}}^{-1}(\vf{b}(\vf{x}+\vf{h}(\vf{x})))=:\vf\Psi(\vf{h})$, which is a fixed-point problem. Note that $\vf\Psi$ is a contraction. Indeed: \begin{align*} \norm{\vf\Psi(\vf{h})-\vf\Psi(\vf{h}')} & \leq \norm{{\vf{L}_{\vf{\tilde{A}}}}^{-1}}\!\norm{\vf{b}(\vf{x}+\vf{h}(\vf{x}))\!-\!\vf{b}(\vf{x}+\vf{h}'(\vf{x}))} \\ & \leq \norm{{\vf{L}_{\vf{\tilde{A}}}}^{-1}}\norm{\vf{Db}}\norm{\vf{h}-\vf{h}'} \end{align*} - which is arbitrarily small ($\norm{\vf{Db}}$ is arbitrarily small) because $\vf{B}$ is $\mathcal{C}^1$-close to $\vf{\tilde{A}}$. Thus, $\vf{h}$ exists and it's unique. + because of \mnameref{FSV:meanvaluethmvector}. This last term is arbitrarily small ($\norm{\vf{Db}}$ is arbitrarily small) because $\vf{B}$ is $\mathcal{C}^1$-close to $\vf{\tilde{A}}$. Thus, $\vf{h}$ exists and it's unique. \end{proof} \begin{definition} A dynamical system $f : X\rightarrow X$ has \emph{sensitive dependence on initial conditions} on $X$ if $\exists\varepsilon >0$ such that for each $x\in X$ and any neighborhood $N_x$ of $x$, exists $y \in N_x$ and $n \geq 0$ such that $d(f^n(x),f^n(y)) > \varepsilon$. @@ -218,11 +218,18 @@ \end{lemma} \begin{proof} Let $W_t:=\vf{\phi}_t(W)$, where $\phi_t$ is the flow of \mcref{ADS:ham_system}. Then: + \begin{equation*} + \dv{}{t}\vol(W_t)=\dv{}{t}\int_{\phi_t(W)}\dd{\vf{x}}=\int_W\dv{}{t}\det \vf{D\phi}_t=\int_W\div \vf{X}_H + \end{equation*} + where $\vf{X}_H$ is the vector field of \mcref{ADS:ham_system}. But an easy computation shows that $\div \vf{X}_H=0$. Let's make the last step of the computation of above more explicit. Given $\vf{A}\in \GL_n(\RR)$, we have that: + \begin{align*} + \det(\vf{A} + \varepsilon \vf{T}) & = \varepsilon^n\det(\vf{A})\det\left(\frac{1}{\varepsilon} \vf{I} + \vf{T} \vf{A}^{-1}\right) \\ + & =\det(\vf{A})\left(1+\varepsilon\trace(\vf{T}\vf{A}^{-1})+\O{\varepsilon^2}\right) + \end{align*} + And so, $\det'(\vf{A})\vf{T}=\trace(\vf{T}\vf{A}^{-1})$. Finally, taking $\vf{A}=\vf{D\phi}_t$ and using that $\dv{}{t}\vf{D\phi}_t=\vf{DX}_H \vf{D\phi}_t$ we get: \begin{multline*} - \dv{}{t}\vol(W_t)=\dv{}{t}\int_{\phi_t(W)}\dd{\vf{x}}=\int_W\dv{}{t}\det \vf{D\phi}_t=\\ - =\int_W\trace\left(\dv{}{t}\vf{D\phi}_t\right)=\int_W\div \vf{X}_H + \dv{}{t}\det \vf{D\phi}_t=\det\empty{}'(\vf{D\phi}_t)\dv{}{t}\vf{D\phi}_t=\trace\!\left(\!\dv{}{t}\vf{D\phi}_t {(\vf{D\phi}_t)}^{-1}\!\right)=\\=\trace(\vf{DX}_H) = \div \vf{X}_H \end{multline*} - where $\vf{X}_H$ is the vector field of \mcref{ADS:ham_system}, and we used that the derivative of the determinant map is the trace. But an easy computation shows that $\div \vf{X}_H=0$. \end{proof} \subsection{Dynamics on the circle} \subsubsection{Generalities} @@ -242,7 +249,7 @@ We only prove the second property. Since, $f$, $g$ are both lifts of $F$, they belong to the same equivalence class. Thus, $f-g\in\ZZ$. And now use the continuity of $f-g$. \end{proof} \begin{remark} - Recall that $f:\RR\to\RR$ is a homeomorphism if and only if $f$ is monotone. + Recall that a continuous function $f:\RR\to\RR$ is a homeomorphism if and only if $f$ is strictly monotonous. \end{remark} \begin{definition} We say that a homeomorphism $F$ \emph{preserves orientation} if and only if $f$ is strictly increasing. We define the set of $\Homeoplus(\TT^1)$ as the set of homeomorphisms of $\TT^1$ that preserve orientation. @@ -293,7 +300,7 @@ \end{proof} \subsubsection{Rotation number}\label{ADS:rotation_number_section} \begin{lemma}\label{ADS:lema_sum} - Recall that $f=\id+\varphi$ with $\varphi$ 1-periodic. And thus: + Let $f=\id+\varphi\in\mathcal{D}^0(\TT^1)$ with $\varphi$ 1-periodic. Thus: $$ f^n=\id + \sum_{i=0}^{n-1} \varphi\circ f^i=: \id + \varphi_n $$ @@ -306,10 +313,13 @@ \end{multline*} \end{proof} \begin{lemma}\label{ADS:lema1} - Let $f\in \mathcal{D}^0(\TT^1)$ be such that $f=\id +\varphi$, with $\varphi$ 1-periodic. Let $m:=\min_{x\in\RR}\varphi$ and $M:=\max_{x\in\RR}\varphi$. Then, we have $m\leq M< m+1$. + Let $f=\id+\varphi\in\mathcal{D}^0(\TT^1)$ with $\varphi$ 1-periodic. Let $m:=\min_{x\in\RR}\varphi$ and $M:=\max_{x\in\RR}\varphi$. Then, we have $m\leq M< m+1$. \end{lemma} \begin{proof} - Let $0\leq x\leq y<1\leq x+1$. Then, $f(y)0$ and $N>0$ such that $\frac{1}{N}<\varepsilon$. Let $f,g\in \mathcal{D}^0(\TT^1)$ be close enough such that $\abs{f^N(x) - g^N(x)}<\varepsilon$ for all $x\in\RR$. In particular, $f^N(0)< g^N(0)+\varepsilon$. A straightforward induction shows that in fact we have $f^{kN}(0)< g^{kN}(0)+k-1+\varepsilon$ for all $k\in\NN$. Thus: + $$ + \rho(f)-\rho(g)=\lim_{k\to\infty}\frac{f^{kN}(0)-g^{kN}(0)}{kN}\leq \frac{1}{N}<\varepsilon + $$ + Exchanging the roles of $f$ and $g$ we get the other inequality. + \end{proof} \begin{definition} Let $F\in \Homeoplus(\TT^1)$ with lift $f$. We define the \emph{rotation number} of $F$ as $\rho(F):=[\rho(f)]\in \TT^1$. \end{definition} @@ -373,7 +397,7 @@ Let $F,G\in\Homeoplus(\TT^1)$ be such that $G$ is semi-conjugate to $F$. Then, if $F$ has a periodic point, then $G$ has a periodic point. \end{lemma} \begin{proof} - Let $p\in \TT^1$ be a periodic point of $F$ with period $n$. Then, $H(p)$ is a periodic point of $G$ with period $n$. Indeed: + Let $p\in \TT^1$ be a periodic point of $F$ with period $n$. Then, $H(p)$ is a periodic point of $G$ with period at most $n$. Indeed: $$ G^n(H(p))=H(F^n(p))=H(p) $$ @@ -434,7 +458,7 @@ Let $F\in\Homeoplus(\TT^1)$. Then, $\mathcal{M}_F(\TT^1)\ne\varnothing$. \end{theorem} \begin{proposition} - Let $F\in\Homeoplus(\TT^1)$ and $f=\id+\varphi$ be a lift of $F$, with $\varphi\in\mathcal{C}(\TT^1)$. Then, $\forall\mu\in\mathcal{M}_F(\TT^1)$, $\rho(f)=\mu(\varphi)$. Moreover: + Let $F\in\Homeo(\TT^1)$ and $f=\id+\varphi$ be a lift of $F$, with $\varphi\in\mathcal{C}(\TT^1)$. Then, $\forall\mu\in\mathcal{M}_F(\TT^1)$, $\rho(f)=\mu(\varphi)$. Moreover: \begin{enumerate} \item $\norm{f^n-\id -n\rho(f)}_{\mathcal{C}(\RR)}<1$ for all $n\in\NN$. \item $\forall n\in\NN$, $\exists x_n\in\RR$ such that $f^n(x_n)-x_n=n\rho(f)$. @@ -443,7 +467,7 @@ \begin{proof} Let $\psi_n:= f^n-\id -n\mu(\varphi)$ with $\mu\in\mathcal{M}_F(\TT^1)$. We have that: \begin{multline*} - \mu(\psi_n)=\sum_{i=0}^{n-1}\mu(\varphi\circ f^i)-n\varphi(n)=\sum_{i=0}^{n-1}\mu(\varphi\circ F^i)-n\varphi(n)=0 + \mu(\psi_n)=\sum_{i=0}^{n-1}\mu(\varphi\circ f^i)-n\mu(\varphi)=\sum_{i=0}^{n-1}\mu(\varphi\circ F^i)-n\mu(\varphi)=0 \end{multline*} where we have used \mcref{ADS:lema_sum}. Now we must have that $\psi_n$ change their sign in $[0,1]$ because otherwise that would contradict $\mu(\psi_n)=0$. So $\exists x_n\in[0,1]$ such that $\psi_n(x_n)=0$. So: $$ @@ -484,6 +508,16 @@ \end{align*} If the homeomorphism is not specified, we will omit the subscript. \end{definition} + \begin{definition} + Let $F\in\Homeoplus(\TT^1)$ and $x\in \TT^1$. We define the \emph{omega limit} of $x$ as the set of limit points of $\mathcal{O}_F^+(x)$, i.e.: + $$ + \omega(x):=\{y\in\TT^1:\exists (n_k)\nearrow+\infty\text{ such that }F^{n_k}(x)\to y\} + $$ + We define the \emph{alpha limit} of $x$ as the set of limit points of $\mathcal{O}_F^-(x)$, i.e.: + $$ + \alpha(x):=\{y\in\TT^1:\exists (n_k)\searrow-\infty\text{ such that }F^{n_k}(x)\to y\} + $$ + \end{definition} \begin{definition} Let $F\in\Homeoplus(\TT^1)$ and $X\subset \TT^1$. We say that $X$ is \emph{positively invariant} if $F(X)\subseteq X$ and \emph{negatively invariant} if $F^{-1}(X)\subseteq X$. We say that $X$ is \emph{invariant} if $F(X)=X$. \end{definition} @@ -524,7 +558,7 @@ $$ \end{lemma} \begin{proof} - All the inclusions are clear except for maybe $R^\pm(F)\subseteq \Omega(F)$. Let $x\in R^\pm(F)$. Then, $\exists (n_k)\in\NN$ with $n_k \nearrow \infty$ such that $F^{n_k}(x)\to x$. Now, let $U$ be a neighborhood of $x$. Then, $x\in U$ and thus $F^{n_1}(U)\cap U\ne\varnothing$. So $x\in \Omega(F)$. + All the inclusions are clear except for maybe $R^\pm(F)\subseteq \Omega(F)$. Let $x\in R^\pm(F)$. Then, $\exists (n_k)\in\NN$ with $n_k \nearrow \infty$ such that $F^{n_k}(x)\to x$. Now, let $U$ be a neighborhood of $x$. Then, $x\in U$ and for $k$ large enough, by the continuity of $F$, we must have $x\in F^{n_k}(U)$. So $F^{n_k}(U)\cap U\ne\varnothing$ and thus $x\in \Omega(F)$. \end{proof} \begin{definition} Let $F\in \Homeoplus(\TT^1)$ and $X\subseteq \TT^1$ be a non-empty closed invariant set. We say that $X$ is \emph{minimal} if $\forall x\in X$, $\overline{\mathcal{O}(x)}=X$. If $X=\TT^1$, we say that $F$ is \emph{minimal}.