# differential equations solve for exponents

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.

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 $1,1,5$.

• 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. Dec 3, 2017 at 9:21
• @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 $$e^{\int_0^t B(\tau)\,d\tau}=e^{\left(\begin{array}{cc}0& t-\cos(t)\\ -t+\cos(t)& 0 \end{array}\right)}=e^{\left(\begin{array}{cc}0& -\cos(t)\\ \cos(t)& 0 \end{array}\right)}e^{\left(\begin{array}{cc}0& 1\\ -1& 0 \end{array}\right)t}$$
– AVK
Dec 3, 2017 at 10:10
• Sorry, I meant to say that $B(t)B(s)=B(s)B(t)$
– AVK
Dec 3, 2017 at 10:22
• 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. Dec 3, 2017 at 19:37
• @James333 No, $\Lambda$ is a constant matrix; in this case, $\Lambda=\left(\begin{array}{rrr}-1&0&0\\0&0&1\\0&-1&0\end{array}\right)$.
– AVK
Dec 3, 2017 at 20:05