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## Liapunov-Floquet transformation with worked examples

May 26, 2020   Compiled on September 9, 2023 at 2:24pm

2.1 Example 1

### 1 Introduction

Given vector system of linear time variant diﬀerential equations $x^{\prime }\left ( t\right ) =A(t)x\left ( t\right )$ where $$x,x^{\prime }(t)$$ are each $$n\times 1$$ vectors and $$A(t)$$ is an $$n\times n$$ which is periodic in $$T$$, which means $$A(t)=A(t+T)$$ and given that the matrix $$A(t)$$ is commutative, meaning there exists a matrix $$C(t)$$ such that $$A(t)C(t)=C(t)A(t)$$ then it is possible to ﬁnd closed form Matrix $$P(t)$$ which is $$n\times n$$ and periodic in $$T$$ and a constant matrix $$B$$ such that$\Phi (t,0)=P(t)e^{Bt}$ where $$\Phi (t,\tau )$$ is the state transition matrix (STM) of $$x^{\prime }=A(t)x$$

Finding $$P(t)$$ and $$B$$ allows one to convert the time varying system $$x^{\prime }\left ( t\right ) =A(t)x\left ( t\right )$$ to non-time varying system using the so called Liapunov-Floquet transformation $x(t)=P(t)y(t)$ And obtain the system $y^{\prime }(t)=By(t)$ To solve. Since this is now no longer time varying, it easier to solve. Then $$x(t)$$ is found by using the above Liapunov-Floquet transformation.

The method is best illustrated by worked examples. In each example, the solution found using Liapunov-Floquet transformation is then compared to the solution found by solving $$x^{\prime }\left ( t\right ) =A(t)x\left ( t\right )$$ using computer algebra software to verify the result.

### 2 Examples

#### 2.1 Example 1

Given $$x^{\prime }\left ( t\right ) =A(t)x\left ( t\right )$$ as$x^{\prime }\left ( t\right ) =\begin {pmatrix} \cos t & \sin t\\ -\sin t & \cos t \end {pmatrix} x\left ( t\right )$ Let the period of $$A\left ( t\right )$$ be $$T$$ (which is $$2\pi$$ in this case). The ﬁrst step is to ﬁnd the state transition matrix. Since the system is time varying, then \begin {equation} \Phi (t,t_{0})=e^{\int _{t_{0}}^{t}A\left ( s\right ) ds} \tag {1} \end {equation} Calculating the matrix exponential above gives\begin {equation} \Phi (t,t_{0})=e^{\sin t-\sin t_{0}}\begin {pmatrix} \cos \left ( \cos t-\cos t_{0}\right ) & -\sin \left ( \cos t-\cos t_{0}\right ) \\ \sin \left ( \cos t-\cos t_{0}\right ) & \cos \left ( \cos t-\cos t_{0}\right ) \end {pmatrix} \tag {2} \end {equation} Since the period is $$2\pi$$ then replacing $$t_{0}$$ by $$2\pi$$ in the above gives (this is Eq. 16 in ﬁrst reference below but not normalized)$\Phi (t,2\pi )=e^{\sin t}\begin {pmatrix} \cos \left ( \cos t-1\right ) & -\sin \left ( \cos t-1\right ) \\ \sin \left ( \cos t-1\right ) & \cos \left ( \cos t-1\right ) \end {pmatrix}$ Now $$\Phi (t,2\pi )=e^{Bt}$$. Which is valid for any $$t$$. Letting $$t=2\pi$$ gives\begin {align*} \Phi (2\pi ,2\pi ) & =e^{2\pi B}\\\begin {pmatrix} 1 & 0\\ 0 & 1 \end {pmatrix} & =e^{2\pi B} \end {align*}

Hence \begin {align*} B & =\frac {1}{2\pi }\begin {pmatrix} \ln 1 & 0\\ 0 & \ln 1 \end {pmatrix} \\ & =\begin {pmatrix} 0 & 0\\ 0 & 0 \end {pmatrix} \end {align*}

Therefore the transformed system becomes\begin {align*} y^{\prime }\left ( t\right ) & =By\left ( t\right ) \\ y^{\prime }\left ( t\right ) & =\begin {pmatrix} 0 & 0\\ 0 & 0 \end {pmatrix}\begin {pmatrix} y_{1}\left ( t\right ) \\ y_{2}\left ( t\right ) \end {pmatrix} \\ & =\begin {pmatrix} 0\\ 0 \end {pmatrix} \end {align*}

Which has the solution$\begin {pmatrix} y_{1}\left ( t\right ) \\ y_{2}\left ( t\right ) \end {pmatrix} =\begin {pmatrix} c_{1}\\ c_{2}\end {pmatrix}$ Now we need to ﬁnd $$P\left ( t\right )$$ to go back to $$x$$ space. Since $$\Phi (t,0)=P(t)e^{Bt}$$ then \begin {align*} P(t) & =\Phi (t,0)e^{-Bt}\\ & =\Phi (t,0) \end {align*}

In this case. This is because $$B=\begin {pmatrix} 0 & 0\\ 0 & 0 \end {pmatrix}$$. Hence, from (2) $P(t)=e^{\sin t}\begin {pmatrix} \cos \left ( \cos t-1\right ) & -\sin \left ( \cos t-1\right ) \\ \sin \left ( \cos t-1\right ) & \cos \left ( \cos t-1\right ) \end {pmatrix}$ Therefore\begin {align*} x\left ( t\right ) & =P\left ( t\right ) y\left ( t\right ) \\ & =e^{\sin t}\begin {pmatrix} \cos \left ( \cos t-1\right ) & -\sin \left ( \cos t-1\right ) \\ \sin \left ( \cos t-1\right ) & \cos \left ( \cos t-1\right ) \end {pmatrix}\begin {pmatrix} c_{1}\\ c_{2}\end {pmatrix} \\ & =e^{\sin t}\begin {pmatrix} c_{1}\cos \left ( \cos t-1\right ) -c_{2}\sin \left ( \cos t-1\right ) \\ c_{1}\sin \left ( \cos t-1\right ) +c_{2}\cos \left ( \cos t-1\right ) \end {pmatrix} \end {align*}

Hence $\begin {pmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {pmatrix} =\begin {pmatrix} e^{\sin t}\left ( c_{1}\cos \left ( \cos t-1\right ) -c_{2}\sin \left ( \cos t-1\right ) \right ) \\ e^{\sin t}\left ( c_{1}\sin \left ( \cos t-1\right ) +c_{2}\cos \left ( \cos t-1\right ) \right ) \end {pmatrix}$ There is something wrong. The correct solution should be$\begin {pmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end {pmatrix} =\begin {pmatrix} e^{\sin t}\left ( c_{1}\cos \left ( \cos t\right ) -c_{2}\sin \left ( \cos t\right ) \right ) \\ e^{\sin t}\left ( c_{1}\sin \left ( \cos t\right ) +c_{2}\cos \left ( \cos t\right ) \right ) \end {pmatrix}$

I need to ﬁnd out what is wrong.

1. Paper. Liapunov-Floquet Transformation: Computation and Applications to Periodic Systems. Article in Journal of Vibration and Acoustics April 1996. S.C.Sinha, R.Pandiyan, J.S.Bibb. Here is a copy of the paper copy of PDF
2. Floquet’s Theorem, Bachelor’s Project Mathematics. By E. Folkers. 2018. copy of PDF