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.
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.