This post has a few tricks for deriving state transitions matrices. These matrices arise when studying linear ODEs, for example: $$ \frac{d}{dt} x(t) = F(t)x(t) + G(t) u(t), \qquad x(t_0) = x_0 $$ where

• $F: [t_0, t_f] \rightarrow \mathbb{R}^{n\times n}$ is the state matrix,
• $G: [t_0, t_f] \rightarrow \mathbb{R}^{n \times m}$ is the input matrix,
• $u: [t_0, t_f] \rightarrow \mathbb{R}^m $ is the input,
• $x: [t_0, t_f] \rightarrow \mathbb{R}^n$ is the solution of the ODE.

Time-varying linear ODEs of this sort generally arise from linearizing a non-linear system of equations around a nominal solution.

The solution of this linear ODE can be written as: $$ x(t) = \Phi(t, t_0) x(t_0) + \int_{t_0}^t \Phi(t, s) G(s) u(s)\; ds $$

where $\Phi: [t_0, t_f] \times [t_0, t_f] \rightarrow \mathbb{R}^{n \times n}$ is the state transition matrix. This matrix itself is given by the family of solutions of the autonomous linear ODEs: $$ \begin{equation} \frac{d}{dt} \Phi(t, \tau) = F(t) \Phi(t, \tau), \qquad \Phi(\tau, \tau) = I, \qquad \tau \in [t_0, t_f] \label{eq:fund_mat} \end{equation} $$

Solving for $\Phi$

In general there won’t be a closed form solution for the ODE $\eqref{eq:fund_mat}$ but in many navigation problems such a solution does exist. When this solution is not available or is too expensive to implement one can resort to numerical integration of the ODE. Another approach is to apply an expansion such as the Peano-Baker Series: $$ \Phi(t, \tau) = \sum_{n=0}^\infty \mathcal{I}_n(t, \tau), \qquad \mathcal{I}_0(t, \tau) = I, \qquad \mathcal{I}_{n+1}(t, \tau) = \int_\tau^t A(s) \mathcal{I}_{n}(s) \; ds. $$ This series can also be used to derive exact formulas in some circumstances.

Linear Time-Invariant Case

In the case where $F$ is constant (i.e. $F(t) \equiv F_0$) the solution is given simply by: $$ \Phi(t, \tau) = \exp((t-\tau) F_0) $$ so that the overall solution reads: $$ x(t) = \exp((t-\tau)F_0) x(t_0) + \int_{t_0}^t \exp((t-s)F_0) G(s) u(s)\; ds $$

Solution for Block Triangular $F$

Say $F$ is of the form: $$F(t) = \begin{bmatrix} F_{11}(t) & F_{12}(t) \\ 0 & F_{22}(t)\end{bmatrix}.$$ Then $\Phi$ can also be written as: $$\Phi(t, \tau) = \begin{bmatrix} \Phi_{11}(t, \tau) & \Phi_{12}(t, \tau) \\ 0 & \Phi_{22}(t, \tau)\end{bmatrix},$$ with $\Phi_{11}$ and $\Phi_{22}$ being the state-transition matrices associated with $F_{11}$ and $F_{22}$ independently: $$ \begin{align} \frac{d}{dt} \Phi_{ii}(t, \tau) &= F_{ii}(t) \Phi_{ii}(t, \tau), \qquad \Phi_{ii}(\tau, \tau) = I \qquad i \in \{1, 2\}. \\
\end{align} $$ and $$ \Phi_{12}(t, \tau) = \int_{\tau}^t \Phi_{11}(t, s) F_{12}(s) \Phi_{22}(s, \tau) ; ds. $$

Proof: This result can be verified by noting that $\Phi(t, t) = I$ and differentiating $\Phi$: $$ \frac{d}{dt} \Phi(t, \tau) = \begin{bmatrix} F_{11}(t) \Phi_{11}(t, \tau) & F_{12}(t)\Phi_{22}(t, \tau) + F_{11}(t) \Phi_{12}(t, \tau) \\
0 & F_{11}(t) \Phi_{22}(t, \tau) \end{bmatrix} = F(t) \Phi(t, \tau). $$