Bourbaki

Exercise 2, § 2

Functions of a Real Variable, IV, 204

Let t↦A(t)t \mapsto A(t) be a regulated map from J into L(E)\mathscr{L}(E) such that, for any two points s, t of J, A(s)A(s) and A(t)A(t) commute. Put B(t)=∫t0tA(s) dsB(t) = \int_{t_0}^t A(s)\,ds. Show that the resolvent C(t,t0)C(t, t_0) of the equation dx/dt=A(t).xd\mathbf{x}/dt = A(t).\mathbf{x} is equal to exp⁡(B(t))\exp(B(t)).

If t↦A1(t),t↦A2(t)t \mapsto A_1(t), t \mapsto A_2(t) are two regulated functions from J into L(E)\mathscr{L}(E) such that, for any two points s, t of J, A1(s),A1(t),A2(s),A2(t)A_1(s), A_1(t), A_2(s), A_2(t) commute pairwise, show that

exp⁡(∫t0t(A1(s)+A2(s))ds)=exp⁡(∫t0tA1(s) ds)exp⁡(∫t0tA2(s) ds). \exp \left( \int_{t_0}^t \left( A_1(s) + A_2(s) \right) ds \right) = \exp \left( \int_{t_0}^t A_1(s)\,ds \right) \exp \left( \int_{t_0}^t A_2(s)\,ds \right).

Solution

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Transcribed from the printed volume. Not translated.