Reversing time by u ↦ a+b-u turns the terminal value problem x' = F(t,x), x(b) = ξ on [a,b] into the initial value problem y' = -F(a+b-u,y), y(a) = ξ, for vectors and matrices, so that a continuous globally Lipschitz F gives exactly one solution of the terminal value problem.
Let F:[a,b]×Rn→Rn be a function and let ξ∈Rn. Let G:[a,b]×Rn→Rn be the function with G(u,z)=−F(a+b−u,z) for all u∈[a,b] and z∈Rn, and, for a function x:[a,b]→Rn, let y:[a,b]→Rn be the function with y(u)=x(a+b−u) for every u∈[a,b]. Then x is differentiable with
Let F:[a,b]×Mp×q(R)→Mp×q(R) be a function and let Ξ∈Mp×q(R). Let G:[a,b]×Mp×q(R)→Mp×q(R) be the function with G(u,Z)=−F(a+b−u,Z) for all u∈[a,b] and Z∈Mp×q(R), and, for a function X:[a,b]→Mp×q(R), let Y:[a,b]→Mp×q(R) be the function with Y(u)=X(a+b−u) for every u∈[a,b]. Then X is differentiable with
Let F:[a,b]×Rn→Rn be continuous on [a,b]×Rn, and let L∈R with L≥0 satisfy ∥F(t,v)−F(t,w)∥≤L∥v−w∥ for all t∈[a,b] and v,w∈Rn. Then for every ξ∈Rn there is exactly one function x:[a,b]→Rn that is differentiable with x′(t)=F(t,x(t)) for every t∈[a,b] and x(b)=ξ. Likewise, let F:[a,b]×Mp×q(R)→Mp×q(R) be continuous on [a,b]×Mp×q(R), and let L∈R with L≥0 satisfy ∥F(t,V)−F(t,W)∥≤L∥V−W∥ for all t∈[a,b] and V,W∈Mp×q(R). Then for every Ξ∈Mp×q(R) there is exactly one function X:[a,b]→Mp×q(R) that is differentiable with X′(t)=F(t,X(t)) for every t∈[a,b] and X(b)=Ξ.
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.