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A C2 vector field has a C2 flow on a common compact tube

Statement

Let f:U→Rn be C2 on an open set U. Let T+>0, let W⊂U be open, and suppose every solution from x∈W exists on an open time interval containing [0,T+] and stays in one compact convex set K⊂U for all t∈[0,T+]. Write Φ(t,x)=ϕt(x). Then Φ is C2 jointly in (t,x) on (0,T+)×W, locally on every common-tube neighborhood in W, and its derivatives extend continuously in time to [0,T+]. Its first state derivative S(t,x)=DxΦ(t,x) solves

∂tS=Df(Φ)S,S(0,x)=In,

and its second state derivative is the bilinear map H(t,x) determined by

∂tH[u,v]=Df(Φ)H[u,v]+D2f(Φ)[Su,Sv],H(0,x)[u,v]=0.

In particular ∂tΦ=f(Φ), ∂txΦ=Df(Φ)S, and ∂ttΦ=Df(Φ)f(Φ). The assertion is local to the stated common tube.

Facts & Assumptions

Given: The C2 vector field, T+, open initial-state set W, and a common compact convex tube containing all solution graphs on [0,T+].

[F1]

On a common compact tube the flow is C1 in its initial state, and S=DxΦ is the unique solution of the variational equation (Initial-data sensitivity of an ODE flow on a common compact tube). Its proof's difference-quotient estimate is uniform over the compact time interval.

[F2]

A classical solution of the autonomous IVP is differentiable in time and satisfies ∂tΦ=f(Φ) (Autonomous ordinary differential equations, First-order systems, initial value problems, and solutions on intervals).

[F3]

The C2 convention supplies continuous first and second total derivatives; on compact K, Df and D2f are bounded and D2f is uniformly continuous (Ck Euclidean maps and diffeomorphisms, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[F4]

A continuous-coefficient linear matrix IVP has a unique solution on the compact interval; adjoining a constant coordinate also solves an affine linear equation (Linear matrix ODEs have unique global solutions on a fixed interval).

[F5]

A nonnegative solution of u(t)≤A+L∫atu(s) ds is bounded by AeL(t−a) (Gronwall's integral inequality with variable and constant coefficients).

Proof

1.1F1F2F3F5F6given

Fix a base state x∈W. By [F1], Φ is C1 in x and S solves S′=AS, where A(t)=Df(Φ(t,x)). The coefficient is bounded on [0,T+] by [F3]; integrating the variational equation by [F6] and applying [F5] bounds S uniformly in t. The flow itself is jointly continuous in (t,x) on the common tube by [F1] and the ODE relation [F2].

2.1F3F4F5F6step 1.1algebra

For a direction h∈Rn, define a matrix Bh(t) by Bh(t)v=D2f(Φ(t,x))[S(t,x)h,S(t,x)v]. It is continuous and has norm at most C∥h∥2 for a finite constant C, by [F3] and the bound on S. Let Hh solve Hh′=AHh+Bh, Hh(0)=0. For each column, adjoining a constant coordinate turns its inhomogeneous linear IVP into a homogeneous linear system; applying [F4] to the augmented matrix gives a unique solution on [0,T+]. Its integral equation from [F6] and [F5] give ∥Hh∥∞≤C′∥h∥2 for a finite constant C′.

2.2F1F3F6F7step 1.1algebra

For small h with x+h∈W, put Δ(t)=Φ(t,x+h)−Φ(t,x), Sh(t)=S(t,x+h), and Ah(t)=Df(Φ(t,x+h)). The uniform remainder in [F1] gives Δ=Sh+o(∥h∥2) uniformly on [0,T+]. On each segment between Φ(t,x) and Φ(t,x+h), the chain rule [F7] and [F6] applied coordinatewise to Df give Ah−A=∫01D2f(Φ+sΔ)[Δ,⋅]ds. Uniform continuity of D2f on K [F3] and Δ=Sh+o(∥h∥) therefore imply Ah−A=D2f(Φ)[Sh,⋅]+o(∥h∥2) uniformly in t.

3.1F1F3F5F6step 2.1step 2.2algebra

The variational equations give Sh′=AhSh and S′=AS. Subtract these and the equation for Hh from step 2.1, setting Rh=Sh−S−Hh, to obtain Rh′=AhRh+(Ah−A)S+(Ah−A)Hh−Bh,Rh(0)=0. By step 2.2, (Ah−A)S−Bh=o(∥h∥) uniformly; also Ah−A=O(∥h∥) and Hh=O(∥h∥) by steps 2.2 and 2.1. Thus the forcing after AhRh is o(∥h∥) uniformly. The coefficients Ah are uniformly bounded on K, so [F5] applied to the integral equation for Rh from [F6] gives ∥Rh∥∞=o(∥h∥2). Therefore DxS(t,x)[h]=Hh(t), proving existence of the second state derivative.

4.1F1F3F5F6step 2.1step 3.1

If xj→x, [F1] gives uniform convergence of Φ(⋅,xj) and S(⋅,xj) on [0,T+]. Continuity of D2f on K then gives uniform convergence of the coefficients and forcing in the linear equations for Hh; subtracting their integral equations and applying [F5] proves Hh(⋅,xj)→Hh(⋅,x) uniformly. Thus the second state derivative is continuous in (t,x).

5.1F2F7step 1.1step 4.1∎

By [F2], ∂tΦ=f(Φ). Differentiating this identity in x and in t using [F7] gives ∂txΦ=Df(Φ)S and ∂ttΦ=Df(Φ)f(Φ). These derivatives and Dx2Φ=H are continuous by steps 1.1 and 4.1, so Φ is jointly C2 on the stated tube.

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