4.5 When is the fundamental matrix an exponential?
For a scalar equation \(x' = a(t)x\) the solution is \(\exp \bigl (\int _{t_0}^ta\bigr )x_0\). It is tempting to write \(\Phi (t, t_0) = \exp \bigl (\int _{t_0}^tA(s)\,ds\bigr )\) for systems, using the matrix exponential \(e^M = \sum _kM^k/k!\) of the next section.
Proposition 4.5.1. If \(A(t)\) commutes with \(B(t) = \int _{t_0}^tA(s)\,ds\) for all \(t\), then \(\Phi (t, t_0) = e^{B(t)}\). Without commutativity the formula is false in general.
Proof. If \(A\) commutes with \(B\), it commutes with every power of \(B\), and differentiating the series term by term, \(\frac {d}{dt}B^k = \sum _{j=0}^{k-1}B^jB'B^{k-1-j} = kB^{k-1}A\), using \(B' = A\) and commutativity. Hence \(\frac {d}{dt}e^B = \sum _k\frac {kB^{k-1}A}{k!} = e^BA = Ae^B\), and \(e^{B(t_0)} = I\). Without commutativity the step \(B^jAB^{k-1-j} = B^{k-1}A\) fails; Software 4.6.3 gives a counterexample. □
The condition holds when \(A\) is constant, and when \(A(t) = a(t)A_0\) for a fixed matrix \(A_0\). It fails for most variable-coefficient systems, which is why they rarely have closed-form solutions, and why the theory of this section is structural rather than computational.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.