improvements dyn syst

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}

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
     $$
@@ -306,10 +313,13 @@
     \end{multline*}
   \end{proof}
   \begin{lemma}\label{ADS:lema1}
-    Let $f\in \mathcal{D}^0(\TT^1)$ be such that $f=\id +\varphi$, with $\varphi$ 1-periodic. Let $m:=\min_{x\in\RR}\varphi$ and $M:=\max_{x\in\RR}\varphi$. Then, we have $m\leq M< m+1$.
+    Let $f=\id+\varphi\in\mathcal{D}^0(\TT^1)$ with $\varphi$ 1-periodic. Let $m:=\min_{x\in\RR}\varphi$ and $M:=\max_{x\in\RR}\varphi$. Then, we have $m\leq M< m+1$.
   \end{lemma}
   \begin{proof}
-    Let $0\leq x\leq y<1\leq x+1$. Then, $f(y)<f(x+1)=f(x)+1$. Thus, $f(y)+x< f(x)+1+y$, and so $\varphi(y) < \varphi(x)+1$. Now take supremum in $y$ and infimum in $x$.
+    By the periodicity and continuity of $\varphi$, we have that $\exists x_m, x_M\in\RR$ such that $\varphi(x_m)=m$, $\varphi(x_M)=M$ and $0\leq x_M-x_m<1$. Since $f\in \mathcal{D}^0(\TT^1)$, we must have $f(x_M)-f(x_m)<1$. Thus:
+    $$
+      M-m=f(x_M)-f(x_m)-(x_M-x_m)<1
+    $$
   \end{proof}
   \begin{definition}
     Let ${(u_n)}\in\RR$ be a sequence. We say that $(u_n)$ is \emph{subadditive} if $u_{n+m}\leq u_n+u_m$ for all $n,m\in\NN$. We say that $(u_n)$ is \emph{superadditive} if $u_{n+m}\geq u_n+u_m$ for all $n,m\in\NN$, that is, if $(-u_n)$ is subadditive.
@@ -363,6 +373,20 @@
       {f_k}^n(x)=f(f(\cdots f(f(x+k)+k)+\cdots )+k)=f^n(x)+nk
     $$
   \end{sproof}
+  \begin{proposition}
+    The function
+    $$
+      \function{R}{\mathcal{D}^0(\TT^1)}{\RR}{f}{\rho(f)}
+    $$
+    is continuous with respect to the $\mathcal{C}^0$-topology.
+  \end{proposition}
+  \begin{proof}
+    Let $\varepsilon >0$ 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}.