For continuous coefficients on [a,b], the linear right-hand sides A(t)x+g(t) and A(t)X+XB(t)+G(t) are continuous, globally Lipschitz in the state with constant a bound of the coefficient norms, and their initial value problems have exactly one solution.
Let A:[a,b]→Mn(R) and g:[a,b]→Rn be maps. There is exactly one map F:[a,b]×Rn→Rn with F(t,x)=A(t)x+g(t) for all t∈[a,b] and x∈Rn. Likewise, let A:[a,b]→Mn(R), B:[a,b]→Mq(R) and G:[a,b]→Mn×q(R) be maps. There is exactly one map F:[a,b]×Mn×q(R)→Mn×q(R) with F(t,X)=A(t)X+XB(t)+G(t) for all t∈[a,b] and X∈Mn×q(R).
Let A:[a,b]→Mn(R) and g:[a,b]→Rn be continuous on [a,b], and let F:[a,b]×Rn→Rn be the map of (i) for A and g, F(t,x)=A(t)x+g(t). Then F is continuous on [a,b]×Rn; there is K∈R with K≥0 and ∥A(t)∥≤K for every t∈[a,b]; for every such K,
Let A:[a,b]→Mn(R), B:[a,b]→Mq(R) and G:[a,b]→Mn×q(R) be continuous on [a,b], and let F:[a,b]×Mn×q(R)→Mn×q(R) be the map of (i) for A, B and G, F(t,X)=A(t)X+XB(t)+G(t). Then F is continuous on [a,b]×Mn×q(R); there are KA,KB∈R with KA,KB≥0, ∥A(t)∥≤KA and ∥B(t)∥≤KB for every t∈[a,b]; for all such KA and KB,
Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.