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Continuous and Lipschitz dependence on the initial condition on a Picard-Lindelof cylinder

Statement

Let n≥1, let U⊆Rn be open and let F:U→Rn be continuous and locally Lipschitz in the state variable (Local Lipschitz continuity in the state variable, locally uniform in time and parameters). Let a0∈U, let h>0 and R>0 satisfy B‾(a0,R)⊆U (Open ball, closed ball and sphere in a metric space), put

M:=sup⁡x∈B‾(a0,R)∥F(x)∥2,L:=a state-Lipschitz constant of F on B‾(a0,R),

so that ∥F(x)−F(y)∥2≤L∥x−y∥2 for all x,y∈B‾(a0,R) (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn), and let 0≤ρ<R satisfy

hM≤R−ρ,Lh<1.

Then for every a∈B‾(a0,ρ) the initial value problem

x′(t)=F(x(t)),x(0)=a

has exactly one solution x( ⋅ ;a) on [−h,h] whose graph lies in [−h,h]×B‾(a0,R), and for all a,b∈B‾(a0,ρ) and all s,t∈[−h,h]:

  1. ∥x(t;a)−x(t;b)∥2≤eL∣t∣∥a−b∥2;
  2. ∥x(t;a)−x(s;a)∥2≤M ∣t−s∣;
  3. ∥x(t;a)−x(s;b)∥2≤eLh∥a−b∥2+M ∣t−s∣; in particular the map (t,a)↦x(t;a) is Lipschitz, hence continuous, on the compact set [−h,h]×B‾(a0,ρ).

Facts & Assumptions

Given: The open set U, the continuous and locally Lipschitz field F, the point a0∈U, the numbers h,R>0 with B‾(a0,R)⊆U, the constants M,L and the radius ρ with hM≤R−ρ and Lh<1.

[F1]

Picard-Lindelöf: if [t0−h,t0+h]×B‾(x0,r)⊆R×U, if ∥F∥2≤M and F has state-Lipschitz constant L on that cylinder, and if hM≤r and Lh<1, then exactly one solution of x′=F(x), x(t0)=x0 exists on [t0−h,t0+h] with graph in the cylinder; any two solutions through the same initial data agree on every common subinterval containing the initial time (Picard-Lindelöf local existence and uniqueness for first-order systems, First-order systems, initial value problems, and solutions on intervals).

[F2]

F is locally Lipschitz in the state variable: every compact time-state cylinder carries a finite state-Lipschitz constant (Local Lipschitz continuity in the state variable, locally uniform in time and parameters); in particular L as chosen in the statement is finite and the inequality ∥F(x)−F(y)∥2≤L∥x−y∥2 holds on B‾(a0,R).

[F3]

For ρ=0 the notation B‾(a0,ρ) means the singleton {a0}, as follows from the inequality defining a closed ball. B‾(a0,R) is closed and bounded, hence compact in Rn (Open ball, closed ball and sphere in a metric space, Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), so the continuous function ∥F∥2 attains its maximum there and M<+∞ (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F6]

Gronwall: if continuous u:[0,T]→R and constants A≥0, L≥0 satisfy u(t)≤A+L∫0tu for t∈[0,T], then u(t)≤AeLt there; the reflected form for an integral oriented as ∫t0 gives u(t)≤AeL∣t∣ for t≤0 (Gronwall's integral inequality with variable and constant coefficients).

Proof

technique · direct; construct the solutions on a common cylinder and estimate their difference with Gronwall's inequality
1.1givenF1F2F3algebra

Fix a∈B‾(a0,ρ) and put ra:=R−∥a−a0∥2, so that ra≥R−ρ>0. Then [−h,h]×B‾(a,ra)⊆[−h,h]×B‾(a0,R)⊆R×U: if ∥x−a∥2≤ra then ∥x−a0∥2≤∥x−a∥2+∥a−a0∥2≤ra+∥a−a0∥2=R. Moreover hM≤R−ρ≤ra and Lh<1, and the bounds M and L hold on the larger cylinder [−h,h]×B‾(a0,R) by [F2] and [F3]. By [F1] with t0=0, x0=a, r=ra, exactly one solution x( ⋅ ;a) on [−h,h] exists whose graph lies in [−h,h]×B‾(a,ra)⊆[−h,h]×B‾(a0,R).

2.1F4F5step 1.1algebra

For a,b∈B‾(a0,ρ) put e:=x( ⋅ ;a)−x( ⋅ ;b). By [F4], for t∈[0,h], e(t)=a−b+∫0t(F(x(r;a))−F(x(r;b)))dr. The integrand is continuous because F and the two solutions are continuous, so [F5] gives ∥e(t)∥2≤∥a−b∥2+L∫0t∥e(r)∥2 dr: the Lipschitz estimate applies on B‾(a0,R) since both graphs lie there by step 1.1.

2.2step 1.1F4F5algebra

For s,t∈[−h,h], [F4] gives x(t;a)−x(s;a)=∫stF(x(r;a)) dr, and the graph of x( ⋅ ;a) lies in B‾(a0,R) by step 1.1, where ∥F∥≤M; hence ∥x(t;a)−x(s;a)∥2≤M∣t−s∣ by [F5]. This is clause 2.

3.1step 2.1F1F4F5F6algebra

Applying [F6] with u=∥e∥2, A=∥a−b∥2 gives ∥e(t)∥2≤eLt∥a−b∥2 for t∈[0,h]. For t∈[−h,0] apply the same argument to the reflected curves x~(s):=x(−s;a), x~(s):=x(−s;b) on [0,−t]: their difference obeys the same integral inequality, so ∥e(t)∥2≤e−Lt∥a−b∥2=eL∣t∣∥a−b∥2. This is clause 1.

4.1step 3.12.2algebra∎

Combining clauses 1 and 2 with the triangle inequality, ∥x(t;a)−x(s;b)∥2≤∥x(t;a)−x(s;a)∥2+∥x(s;a)−x(s;b)∥2≤M∣t−s∣+eLh∥a−b∥2 for all (t,a),(s,b)∈[−h,h]×B‾(a0,ρ); this is clause 3. The domain of the map is a closed bounded subset of R×Rn, hence compact (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and the displayed inequality exhibits it as Lipschitz.

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