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updated dynamical systems
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victorballester7 committed Oct 11, 2023
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\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 $\vf{\tilde{A}}$.
\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:
$$
\vf{H}\circ \vf{\tilde{A}}=\vf{B}\circ \vf{H}\iff
\vf{h}(\vf{\tilde{A}x})-\vf{\tilde{A}} \vf{h}(\vf{x})=\vf{b}(\vf{x}+\vf{h}(\vf{x}))
$$
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}}
\end{gather*}
Observe that:
$$
\sup_{\vf{x}\in T^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{x})}=\sup_{\vf{x}\in T^2}\norm{\vf{S}_{\vf{\tilde{A}}}\vf{h}(\vf{\tilde{A}}^{-1}\vf{x})}= \sup_{\vf{x}\in T^2}\norm{\vf{h}(\vf{x})}
$$
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:
$$
\vf{L}_{\vf{\tilde{A}}}\vf{h}=\vf{c}\iff \begin{cases}
\vf{L}_{\vf{\tilde{A}}}\vf{h}_+=\vf{S}_{\vf{\tilde{A}}}\vf{h}_+-\lambda_+\vf{h}_+=\vf{c}_+ \\
\vf{L}_{\vf{\tilde{A}}}\vf{h}_-=\vf{S}_{\vf{\tilde{A}}}\vf{h}_--\lambda_-\vf{h}_-=\vf{c}_-
\end{cases}
$$
where $\vf{h}=\vf{h}_++\vf{h}_-$, $\vf{c}=\vf{c}_++\vf{c}_-$ and $\vf{h}_\pm,\vf{c}_\pm\in \langle \vf{e}_\pm\rangle$. Now, note that $\norm{\frac{\vf{S}_{\vf{\tilde{A}}}}{\lambda_+}}<1$ and so
$$
(\vf{S}_{\vf{\tilde{A}}}-\lambda_+\vf{I})=\lambda_+\left(\frac{\vf{S}_{\vf{\tilde{A}}}}{\lambda_+}-\vf{I}\right)
$$
is invertible. Similarly, we have $\norm{\vf{S}_{\vf{\tilde{A}}}^{-1}\lambda_-}<1$ and so
$$
(\vf{S}_{\vf{\tilde{A}}}^{-1}-\lambda_-\vf{I})=\vf{S}_{\vf{\tilde{A}}}^{-1}\left(\vf{I}-\lambda_-\vf{S}_{\vf{\tilde{A}}}^{-1}\right)
$$
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:
\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.
\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 $d(f^n(x),f^n(y)) > \varepsilon$.
\end{definition}
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$$
\end{definition}
\begin{remark}
The Lyapunov exponent measures the exponential growth rate of tangent vectors along orbits.
The Lyapunov exponent measures the exponential growth rate of tangent vectors along orbits. It can rarely be computed explicitly, but if we can show that $\chi(x,\vf{v})>0$ for some $\vf{v}$, then we know that the system is \emph{chaotic}.
\end{remark}
\end{multicols}
\end{document}

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