588 CHAPTER 30. LAPLACE TRANSFORM METHODS

Letting s→ ∞, this clearly converges to 0. Also,∥∥∥∥∫ δ

0se−st (Φ(t)−Φ(0))dt

∥∥∥∥≤ ∫ δ

0se−st

εdt < ε

and so for all s large enough, ∥(I−A/s)∫

∞

0 se−st (Φ(t)−Φ(0))dt∥ < ε showing that thisdoes indeed converge to 0. The last term in 30.6 converges to Φ(0) as s→ ∞ and so we doindeed have Φ(0) = I.

What about Φ′ (t) = AΦ(t)? For large s,integrate by parts using Φ(0) = I to obtain∫∞

0e−st

Φ′ (t)dt = −I +

∫∞

0se−st

Φ(t)dt =−I + sF (s)∫∞

0e−stAΦ(t)dt = AF (s)

Is −I + sF (s) = AF (s)? Yes because F (s) = (sI−A)−1 and so the Laplace transforms ofΦ′ (t) and AΦ(t) are the same. This means the two functions are the same because they areboth continuous, something which is shown later that the Laplace transform determines thefunctions from which it comes. This has shown the following important theorem.

Theorem 30.2.3 Let A be a p× p matrix and suppose F (s) = (sI−A)−1 = L (Φ(t)) forΦ(t) a matrix whose entries have exponential growth. Then Φ′ (t) = AΦ(t) ,Φ(0) = I.Conversely, if Φ′ (t) = AΦ(t) ,Φ(0) = I, then L (Φ(t)) = (sI−A)−1. Thus the fundamen-tal matrix is unique.

As noted above, one can ALWAYS find from the table of Laplace transforms an explicitsolution Φ(t) whose Laplace transform is (sI−A)−1 provided you can factor the polyno-mials in the denominators of the rational functions which are the entries of (sI−A)−1. Suchfactorizations always exist by the fundamental theorem of algebra and so the fundamentalmatrix always exists. Thus your ability to find an explicit formula for such a fundamentalmatrix is exactly as good as your ability to factor polynomials which occur as denominatorsin the formula for (sI−A)−1. Note that Φ(t) is unique, because if you have one then itsLaplace transform must be (sI−A)−1.

One other item is of interest in these fundamental matrices and this is the group prop-erty.

Theorem 30.2.4 The following hold:

Lemma 30.2.5 1. If Ψ′ (t) = AΨ(t) ,Ψ(0) = 0, then Ψ(t) = 0.

2. If Φ(t) is the fundamental matrix for A then Φ(t)A = AΦ(t) .

3. Also if Φ(t) is the fundamental matrix, then Φ(t +u) = Φ(t)Φ(u) . In particularΦ(t)−1 = Φ(−t).

Proof: 1. Consider the first claim. Letting G(s) be the Laplace transform, it followsthat

sG(s)−0 = AG(s)

for all s large enough. This is impossible unless G(s) = 0. Therefore, L (0) = L (Ψ(t))and so 0 = Ψ(t) from what is shown later about the Laplace transform determining thefunction.