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The maximal solution domain is open

Statement

Let V:URn be a smooth vector field on an open set U, and for each x0U let x(;x0) denote the unique maximal solution of x=V(x) with x(0)=x0. Then the maximal solution domain

Ω:={(t,x0)R×U:t lies in the maximal interval of x(;x0)}

is open in R×U. On Ω, the evaluation map Φ(t,x0):=x(t;x0) is smooth in the state variable and continuous jointly in (t,x0).

Facts & Assumptions

Given: A smooth vector field V:URn and its maximal solutions.

[L1]

Autonomous smooth ODEs have local smooth flows near every point (The fundamental theorem for autonomous smooth ODEs).

[L3]

Solutions depend continuously on nearby initial data on common compact local intervals (Continuous dependence of ODE solutions on initial data and parameters).

Proof

technique · direct
1.1

Let (t,x)Ω and put p:=Φ(t,x). By [L1], applied at the [L1, choose] state point p, there exist δ>0 and an open neighbourhood WU of p such that every yW has a unique solution on [δ,δ], and these solutions vary smoothly with y.

L1choose
2.1

By [L3], for initial states x sufficiently close to x, the solution [L3, step 1.1] Φ(,x) is defined at least on a compact interval around t and its value at time t lies in W. Therefore for every such x and every s<δ, the solution continued from time t by the local flow of step 1.1 is defined at time t+s. Hence all pairs (t+s,x) with s<δ and x near x lie in Ω.

L3step 1.1
3.1

Step 2.1 gives an open neighbourhood of (t,x) contained in Ω, [L1, L2, L3, step 2.1] so Ω is open. On that neighbourhood, the evaluation map is the composite of the continuous time-t map with the local smooth flow from step 1.1, hence is jointly continuous and smooth in the state variable. Since (t,x) was arbitrary, the same holds on all of Ω.

L1L2L3step 2.1

Depends on

Used by

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