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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-29
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C1 dependence of solutions on initial data

Statement

Let F:D⊆R×Rn→Rn be continuous and C1 in the state variable. Fix data (t0,x0)∈D and a compact time interval I=[t0−h,t0+h] on which the corresponding solutions through nearby initial states all exist. Then the solution map

Φ:I×U0→Rn,Φ(t,y)=x(t;y),

is C1 in the initial-state variable y on some neighbourhood U0 of x0. For each y∈U0, the derivative matrix DyΦ(t,y) is the solution of the variational equation along t↦x(t;y).

Facts & Assumptions

Given: The field F, the compact interval I, and the common local family Φ(t,y)=x(t;y) of solutions through nearby initial states.

[F1]

The variational equation along a solution is A′(t)=DxF(t,x(t))A(t) with initial condition A(t0)=In (The variational equation along an ODE solution).

[F2]

Solutions satisfy the corresponding Volterra integral equation (A first-order initial value problem is equivalent to its Volterra integral equation).

[L1]

Nearby initial data and parameters give uniformly close solutions on one common compact time interval (Continuous dependence of ODE solutions on initial data and parameters).

[L2]

Linear matrix ODEs on a compact interval have unique solutions (Linear matrix ODEs have unique global solutions on a fixed interval).

[L4]

Gronwall's integral inequality controls difference equations of Volterra type (Gronwall's integral inequality with variable and constant coefficients).

Proof

technique · direct
1.1F1L1L2

By [L1], after shrinking U0 if needed, every solution graph (t,Φ(t,y)) with y∈U0 lies in one compact cylinder K⊆D. Fix y∈U0, and let Ay:I→Mn(R) be the unique solution of the variational equation below.

Ay′(t)=DxF(t,Φ(t,y))Ay(t),Ay(t0)=In,

whose existence on I is given by [L2]. Because DxF is continuous on the compact set K, it is bounded and uniformly continuous there.

2.1F2L1step 1.1algebra

Fix y∈U0 and an increment u∈Rn with y+u∈U0. By [F2], the difference zu(t):=Φ(t,y+u)−Φ(t,y) satisfies the Volterra equation below, and the state-variable mean-value formula gives the matrix field Bu.

zu(t)=u+∫t0t(F(s,Φ(s,y+u))−F(s,Φ(s,y))) ds.

For each s∈I, the one-variable mean-value formula in the state variable gives

F(s,Φ(s,y+u))−F(s,Φ(s,y))=Bu(s) zu(s),

where

Bu(s):=∫01DxF(s,Φ(s,y)+θzu(s)) dθ.

By [L1], zu→0 uniformly on I as u→0, so the uniform continuity of DxF on K gives

ε(u):=sup⁡s∈I∥Bu(s)−DxF(s,Φ(s,y))∥⟶0.

3.1L3L4step 1.1step 2.1algebra

Put ρu(t):=zu(t)−Ay(t)u. Subtracting the Volterra equations for zu and Ayu gives the identity below.

ρu(t)=∫t0tBu(s)ρu(s) ds+∫t0t(Bu(s)−DxF(s,Φ(s,y)))Ay(s)u ds.

Let M:=sup⁡K∥DxF∥ and Cy:=sup⁡s∈I∥Ay(s)∥. Then ∥Bu(s)∥≤M, and [L3] gives, for t≥t0,

∥ρu(t)∥2≤M∫t0t∥ρu(s)∥2 ds+Cyε(u)∥u∥(t−t0).

Applying [L4] on [t0,t] and its time-reflected form on [t,t0] yields a constant Cy′ independent of u such that sup⁡t∈I∥ρu(t)∥2≤Cy′ε(u)∥u∥. Since ε(u)→0, this proves

sup⁡t∈I∥Φ(t,y+u)−Φ(t,y)−Ay(t)u∥2∥u∥⟶0.

Therefore Ay(t)=DyΦ(t,y) for every t∈I.

4.1L1L3L4step 1.1step 3.1algebra

For y,y′∈U0, write E(t):=Ay′(t)−Ay(t) and Cy(t):=DxF(t,Φ(t,y)). Then the continuity estimate below, together with the same Gronwall argument as in step 3.1, proves continuity of the derivative matrix.

E(t)=∫t0tCy′(s)E(s) ds+∫t0t(Cy′(s)−Cy(s))Ay(s) ds.

By [L1], Φ( ⋅ ,y′)→Φ( ⋅ ,y) uniformly on I as y′→y, so the uniform continuity of DxF on K gives sup⁡s∈I∥Cy′(s)−Cy(s)∥→0. Applying [L3] and [L4] exactly as in step 3.1 shows sup⁡t∈I∥E(t)∥2→0. Hence y↦DyΦ(t,y)=Ay(t) is continuous for each t, and the derivative matrix is exactly the variational-equation solution.

5.1step 3.1step 4.1∎

Steps 3.1 and 4.1 prove that Φ is C1 in the initial-state variable and that its derivative matrix is the solution of the variational equation.

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