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Consider $x' = A(t)x$, $x \in \mathbb{R}^n$ where $A$ is $2\pi$-periodic.

$$A(t) = \begin{bmatrix} 1+\sin(t)&0&0\\ 0&3&4\\ 0&1&3\end{bmatrix}$$

The question is to find the Floquet exponents and it also asked to find Lyapunov exponents. I am kind of new to ordinary differential equations, and it would be very nice to show step by steps to solve this problem to firmly understand the concept.

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1 Answer 1

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The original system consists of 2 parts: $$\tag{1} \dot x=(1+\sin t)x $$ and $$\tag{2} \frac{d}{dt}\left(\begin{array}{c}y\\z\end{array}\right)= B\left(\begin{array}{c}y\\z\end{array}\right),\qquad B= \left(\begin{array}{cc} 3&4\\1&3 \end{array}\right). $$ We can solve them separately, obtaining $$ x(t)=C_1e^{t-\cos t} $$ $$ \left(\begin{array}{c}y\\z\end{array}\right)= e^{Bt}\left(\begin{array}{c}C_2\\C_3\end{array}\right)= \left(\begin{array}{cc} e^{5t}/2 + e^t/2& e^{5t} - e^t\\ e^{5t}/4 - e^t/4& e^{5t}/2 + e^t/2 \end{array}\right)\left(\begin{array}{c}C_2\\C_3\end{array}\right); $$ or $$ \left(\begin{array}{c}x\\y\\z\end{array}\right)= \left(\begin{array}{ccc} e^{t-\cos t}&0&0\\ 0& e^{5t}/2 + e^t/2& e^{5t} - e^t\\ 0& e^{5t}/4 - e^t/4& e^{5t}/2 + e^t/2 \end{array}\right)\left(\begin{array}{c}C_1\\C_2\\C_3\end{array}\right)=X(t) \left(\begin{array}{c}C_1\\C_2\\C_3\end{array}\right). $$ The fundamental matrix $X(t)=\left(\begin{array}{c|c} e^{t-\cos t}&0\\ \hline 0&e^{Bt} \end{array}\right)$ can be written in the Floquet's canonical form $$ X(t)=\Phi(t)e^{\Lambda t}, $$ where $$ \Phi(t)= \left(\begin{array}{ccc} e^{-\cos t}&0&0\\ 0& 1& 0\\ 0& 0& 1 \end{array}\right) $$ is a $2\pi$ periodic matrix, $$ e^{\Lambda t}= \left(\begin{array}{c|c} e^t&0\\ \hline 0&e^{Bt} \end{array}\right)= e^{ \left(\begin{array}{c|c} t&0\\ \hline 0&Bt \end{array}\right)}= e^{ \left(\begin{array}{c|c} 1&0\\ \hline 0&B \end{array}\right)t}, $$ hence, $$ \Lambda= \left(\begin{array}{ccc} 1&0&0\\ 0& 3& 4\\ 0& 1& 3 \end{array}\right). $$ $\Lambda$ has the eigenvalues $1,1,5$, thus, the Floquet multipliers of the system are the eigenvalues of $e^{2\pi\Lambda}$, i.e. $e^{2\pi}$, $e^{2\pi}$, $e^{10\pi}$; the Floquet exponents are $1,1,5$; The Lyapunov exponents are the real parts of the Floquet exponents, so they are equal to $1,1,5$.

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  • $\begingroup$ Thank you for the answer!! Just got an one more question if possible, what if the A matrix is [-1, 0, 0; 0, 0, 1+sin(t); 0, -1-sin(t), 0]? Thank you for your help again. $\endgroup$
    – James333
    Dec 3, 2017 at 9:21
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    $\begingroup$ @James333 This system again consists of 2 systems: $\dot x=-x$ and the linear system with $B(t)=\left(\begin{array}{cc}0& 1+\sin(t)\\ -1-\sin(t)& 0 \end{array}\right)$. The matrix $B(t)$ has the property: $\forall t,s\in\mathbb R$ $A(t)A(s)=A(s)A(t)$. It implies that the fundamental matrix of the second system is $$e^{\int_0^t B(\tau)\,d\tau}=e^{\left(\begin{array}{cc}0& t-\cos(t)\\ -t+\cos(t)& 0 \end{array}\right)}=e^{\left(\begin{array}{cc}0& -\cos(t)\\ \cos(t)& 0 \end{array}\right)}e^{\left(\begin{array}{cc}0& 1\\ -1& 0 \end{array}\right)t}$$ $\endgroup$
    – AVK
    Dec 3, 2017 at 10:10
  • $\begingroup$ Sorry, I meant to say that $B(t)B(s)=B(s)B(t)$ $\endgroup$
    – AVK
    Dec 3, 2017 at 10:22
  • $\begingroup$ Thank you again for the comment! So, the Λ matrix will now have [1, 0, 0; 0, 0, t; 0, -t, 0], Then, how would you calculate for the eigenvalues if it has t in the matrix to get the Floquet multipliers. $\endgroup$
    – James333
    Dec 3, 2017 at 19:37
  • $\begingroup$ @James333 No, $\Lambda$ is a constant matrix; in this case, $\Lambda=\left(\begin{array}{rrr}-1&0&0\\0&0&1\\0&-1&0\end{array}\right)$. $\endgroup$
    – AVK
    Dec 3, 2017 at 20:05

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