I have two $n$-dimensional ODEs:
$$\begin{cases} \ddot x + K_1 x = 0 & \text{for $t$ in $[0,t_1)$}\\ \ddot x + K_2 x = v & \text{for $t$ in $[t_1,t_1+t_2)$} \end{cases}$$
where $K_1, K_2$ are symmetric positive definite matrices with $K_2=K_1+e_ne_n^\top$ ($e_n=[0 \ \dots \ 0 \ 1]^\top$). I am looking for $(x_0, \dot x_0)$ and $t_1,t_2>0$ that lead to periodic solutions. I managed to find some numerical solutions, but I am wondering if the following system of equations can be further simplified.
Since $K_1$, $K_2$ are spd, there have spd square roots $L_1$, $L_2$. On each of the two intervals, the solutions can be expressed as functions of $L_1$, $L_2$:
$$\begin{bmatrix} x(t) \\ \dot x(t)\end{bmatrix} = S_1(t) \begin{bmatrix} x_0 \\ \dot x_0\end{bmatrix} \quad \text{for $t$ in $[0,t_1)$}$$
$$\begin{bmatrix} x(t) \\ \dot x(t)\end{bmatrix} = S_2(t) \begin{bmatrix} x_0 \\ \dot x_0\end{bmatrix} +\begin{bmatrix} -L_2^{-2}v \\ 0 \end{bmatrix}\quad \text{for $t$ in $[t_1,t_1+t_2)$}$$
with
$$S_i(t)=\begin{bmatrix} \cos(tL_i) & t\operatorname{sinc}(tL_i) \\ -L_i\sin(tL_i) & \cos(tL_i)\end{bmatrix}\in\mathbb{R}^{2n\times 2n}.$$
The problem of finding periodic solutions is reduced to finding $X_0=(x_0,\dot x_0)\in\mathbb{R}^{2n}$ and $(t_1,t_2)$ such that:
$$\boxed{(S_2(t_2)S_1(t_1)-I)X_0 = (S_2(t_2) -I)V}$$ with $V^\top = [-(L_2^{-2}v)^\top,\ 0]$. I add the following two equations: $X_0 e_n = 1$ and $S_1(t_1)X_0e_n=1$, so there are $2n$ unknowns and $2n$ equations.
The question is, can this problem be simplified by the choice of a more appropriate basis?
Remarks:
- The solutions I found numerically all satisfy $\det(S_2(t_2)S_1(t_1)-I)=0$, the reason is not clear to me.
- I don't use the fact that $K_2=K_1+e_ne_n^\top$; could it be taken advantage of?