Let T>0 be a real number and k≥1 a natural number. Let A assign to each t∈[0,T] a real k×k matrix A(t) (real matrix) all of whose entries are continuous functions of t on [0,T], the interval being regarded as a subset of the real line with the absolute value metric and R carrying the same metric. Let Φ be the fundamental solution of A on [0,T] and Ψ(t)=Φ(t)−1 its inverse (claims 1 and 2 of that theorem), and define the two-parameter fundamental solution by
Φ(t,u)=Φ(t)Ψ(u)(t,u∈[0,T]),
with the matrix product. For a real matrix M with k rows and k columns put ∥M∥rs=(∑i=1k(∑j=1k∣Mij∣)2)1/2 (the Euclidean norm of the vector of absolute row sums, an ad hoc norm used only in this lemma and called the row-sum vector norm here), so that ∣Mv∣≤∥M∥rs∣v∣ for every v∈Rk by claim 3 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product, where ∣⋅∣ is the Euclidean norm on Euclidean space Rk and Mv the matrix-vector product. Let Aˉ≥0 and Φˉ≥0 be real numbers with ∥A(t)∥rs≤Aˉ, ∥Φ(t)∥rs≤Φˉ and ∥Ψ(t)∥rs≤Φˉ for all t∈[0,T]; such numbers exist because every entry of A, Φ and Ψ is continuous on [0,T], hence bounded by Continuous Real-Valued Functions on a Compact Interval are Bounded, and a matrix whose entries are bounded by C has row-sum vector norm at most k3/2C (each absolute row sum being at most kC). A map [0,T]→Rk is called bounded measurable when its components are bounded and measurable with respect to the trace Borel σ-algebra on [0,T] and the Borel σ-algebra of the real line; for such maps, ∫[a,b]⋅du denotes the componentwise Lebesgue integral over the compact interval [a,b]⊆[0,T], equal to 0 when a=b. Let Ik be the identity matrix.
1. (Two-parameter fundamental solution.) Φ(t,t)=Ik and ∥Φ(t,u)∥rs≤Φˉ2 for all t,u∈[0,T]; for fixed t every entry of u↦Φ(t,u) is continuous on [0,T]; and for all 0≤τ≤t≤T and every v∈Rk,
Φ(t,τ)v−v=∫[τ,t]Φ(t,u)A(u)vdu.
2. (Existence.) Let ξ∈Rk and let g:[0,T]→Rk be bounded measurable. Then u↦Φ(t,u)g(u) is bounded measurable for every t, the map
x(t)=Φ(t)ξ+∫[0,t]Φ(t,u)g(u)du(t∈[0,T])
has continuous, hence bounded measurable, components, ∣x(t)∣≤Φˉ∣ξ∣+Φˉ2∫[0,t]∣g(u)∣du for every t, and
x(t)=ξ+∫[0,t](A(u)x(u)+g(u))du(t∈[0,T]).
3. (Uniqueness.) If z:[0,T]→Rk is bounded measurable and z(t)=ξ+∫[0,t](A(u)z(u)+g(u))du for every t∈[0,T], then z=x on [0,T].