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A periodic C1 velocity has a global characteristic flow on a finite horizon

Statement

Let d≥1, T>0, ℓj>0, and let J⊆R be an open interval containing [0,T]. Suppose b:J×Rd→Rd is C1 and periodic in xj with period ℓj for every j. Choose a compact interval K⊂J with [0,T]⊂int⁡K, and put MK:=sup⁡K×[0,ℓ1]×⋯×[0,ℓd]∥b∥2. For every s∈[0,T] and x∈Rd, there is a unique solution X(⋅;s,x) of X′(t)=b(t,X(t)),X(s;s,x)=x, on the whole interval [0,T]. The lifted flow is periodic-equivariant: X(t;s,x+kℓ)=X(t;s,x)+kℓ(k∈Zd), where kℓ=(k1ℓ1,…,kdℓd). If MT:=sup⁡[0,T]×[0,ℓ1]×⋯×[0,ℓd]∥b∥2, then ∥X(t;s,x)−x∥2≤MT∣t−s∣(t∈[0,T]). Thus each characteristic stays in a compact tube [0,T]×B‾(x,MTT+1).

Facts & Assumptions

Given: d,T,ℓ,J,K,b,s,x as in the Statement.

[F1]

Linear transport equations and their characteristic flow: a characteristic solves X′(t)=b(t,X(t)); a single-valued flow requires existence and uniqueness for each initial value.

[F2]

Picard-Lindelöf local existence and uniqueness for first-order systems: a continuous vector field locally Lipschitz in the state has a unique local solution; on a compact cylinder with field bound M0 and state Lipschitz constant L0, it applies on a symmetric interval of radius δ when δM0≤r and L0δ<1 for the cylinder state radius r.

[F3]

A first-order initial value problem is equivalent to its Volterra integral equation: a solution satisfies X(t)=X(s)+∫stb(r,X(r)) dr with the oriented integral.

[F4]

For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable: ∥∫abf∥2≤∫ab∥f∥2 for integrable vector-valued f and ordered endpoints a≤b; a reversed oriented integral is first negated and rewritten with ordered endpoints.

[F5]

On a convex open set, a uniform bound ∥Df(z)v∥2≤M∥v∥2 implies ∥f(y)−f(x)∥2≤M∥y−x∥2: on a convex open set, a uniform derivative bound ∥Df(z)v∥2≤L∥v∥2 gives the corresponding L-Lipschitz bound.

Proof

1.1givenF2F5choose

Periodicity and continuity on the compact cylinder K×[0,ℓ1]×⋯×[0,ℓd] give finite bounds MK:=sup⁡∥b∥2 and L:=sup⁡∥Dxb∥ there. The same bounds hold on all spatial translates. By [F5] each time slice is globally L-Lipschitz in x. Since [0,T]⊂int⁡K, choose δ>0 small enough that every radius-δ time interval centered in [0,T] lies in K, δ(MK+1)≤1, and δ(L+1)<1. On each corresponding unit-radius state cylinder, [F2] therefore gives a unique local solution through its center.

2.1givenF2step 1.1construct

Partition the finite interval between s and any t∈[0,T] into finitely many subintervals of length at most δ. Apply [F2] successively at each endpoint, using the preceding endpoint value as the next initial state. The resulting curves agree on their shared endpoints and have the same derivative b(r,X(r)) there, so they concatenate to a solution on [0,T]. Repeated local uniqueness from [F2] proves uniqueness on the concatenated interval. This is a finite construction and selects no family of local solutions over an infinite index set.

3.1givenF1F3F4step 2.1algebra∎

For the constructed solution on [0,T], [F3] and [F4] give ∥X(t;s,x)−x∥2≤∫min⁡(s,t)max⁡(s,t)∥b(r,X(r))∥2 dr≤MT∣t−s∣. Thus its graph lies in the stated compact tube. For a period vector kℓ, the curve X(t;s,x+kℓ)−kℓ solves the same IVP as X(t;s,x) by periodicity of b; uniqueness gives periodic equivariance.

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