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$C^{1}$ dependence of solutions on initial data

Statement

Let $F : D \subseteq R \times R^{n} \to R^{n}$ be continuous and $C^{1}$ in the state variable. Fix data $(t_{0}, x_{0}) \in D$ and a compact time interval $I = [t_{0} - h, t_{0} + h]$ on which the corresponding solutions through nearby initial states all exist. Then the solution map

$Φ : I \times U_{0} \to R^{n}, Φ (t, y) = x (t; y),$

is $C^{1}$ in the initial-state variable $y$ on some neighbourhood $U_{0}$ of $x_{0}$ . For each $y \in U_{0}$ , the derivative matrix $D_{y} Φ (t, y)$ is the solution of the variational equation along $t \mapsto x (t; y)$ .

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 $A^{'} (t) = D_{x} F (t, x (t)) A (t)$ with initial condition $A (t_{0}) = I_{n}$ (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).

[L3]

The norm of a vector-valued integral is at most the integral of the norm (For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable).

[L4]

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

Proof

technique · direct

1.1

By [L1], after shrinking $U_{0}$ if needed, every solution graph $(t, Φ (t, y))$ with $y \in U_{0}$ lies in one compact cylinder $K \subseteq D$ . Fix $y \in U_{0}$ , and let $A_{y} : I \to M_{n} (R)$ be the unique solution of the variational equation below.

F1L1L2

$A_{y}^{'} (t) = D_{x} F (t, Φ (t, y)) A_{y} (t), A_{y} (t_{0}) = I_{n},$

whose existence on $I$ is given by [L2]. Because $D_{x} F$ is continuous on the compact set $K$ , it is bounded and uniformly continuous there.

2.1

Fix $y \in U_{0}$ and an increment $u \in R^{n}$ with $y + u \in U_{0}$ . By [F2], the difference $z_{u} (t) : = Φ (t, y + u) - Φ (t, y)$ satisfies the Volterra equation below, and the state-variable mean-value formula gives the matrix field $B_{u}$ .

F2L1step 1.1algebra

$z_{u} (t) = u + \int_{t_{0}}^{t} (F (s, Φ (s, y + u)) - F (s, Φ (s, y))) d s .$

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

$F (s, Φ (s, y + u)) - F (s, Φ (s, y)) = B_{u} (s) z_{u} (s),$

where

$B_{u} (s) : = \int_{0}^{1} D_{x} F (s, Φ (s, y) + θ z_{u} (s)) d θ .$

By [L1], $z_{u} \to 0$ uniformly on $I$ as $u \to 0$ , so the uniform continuity of $D_{x} F$ on $K$ gives

$ε (u) : = \sup_{s \in I} ∥ B_{u} (s) - D_{x} F (s, Φ (s, y)) ∥ ⟶ 0.$

3.1

Put $ρ_{u} (t) : = z_{u} (t) - A_{y} (t) u$ . Subtracting the Volterra equations for $z_{u}$ and $A_{y} u$ gives the identity below.

L3L4step 1.1step 2.1algebra

$ρ_{u} (t) = \int_{t_{0}}^{t} B_{u} (s) ρ_{u} (s) d s + \int_{t_{0}}^{t} (B_{u} (s) - D_{x} F (s, Φ (s, y))) A_{y} (s) u d s .$

Let $M : = \sup_{K} ∥ D_{x} F ∥$ and $C_{y} : = \sup_{s \in I} ∥ A_{y} (s) ∥$ . Then $∥ B_{u} (s) ∥ \leq M$ , and [L3] gives, for $t \geq t_{0}$ ,

$∥ ρ_{u} (t) ∥_{2} \leq M \int_{t_{0}}^{t} ∥ ρ_{u} (s) ∥_{2} d s + C_{y} ε (u) ∥ u ∥ (t - t_{0}) .$

Applying [L4] on $[t_{0}, t]$ and its time-reflected form on $[t, t_{0}]$ yields a constant $C_{y}^{'}$ independent of $u$ such that $\sup_{t \in I} ∥ ρ_{u} (t) ∥_{2} \leq C_{y}^{'} ε (u) ∥ u ∥$ . Since $ε (u) \to 0$ , this proves

$\sup_{t \in I} \frac{∥ Φ (t, y + u) - Φ (t, y) - A_{y} (t) u ∥_{2}}{∥ u ∥} ⟶ 0.$

Therefore $A_{y} (t) = D_{y} Φ (t, y)$ for every $t \in I$ .

4.1

For $y, y^{'} \in U_{0}$ , write $E (t) : = A_{y^{'}} (t) - A_{y} (t)$ and $C_{y} (t) : = D_{x} F (t, Φ (t, y))$ . Then the continuity estimate below, together with the same Gronwall argument as in step 3.1, proves continuity of the derivative matrix.

L1L3L4step 1.1step 3.1algebra

$E (t) = \int_{t_{0}}^{t} C_{y^{'}} (s) E (s) d s + \int_{t_{0}}^{t} (C_{y^{'}} (s) - C_{y} (s)) A_{y} (s) d s .$

By [L1], $Φ (\cdot, y^{'}) \to Φ (\cdot, y)$ uniformly on $I$ as $y^{'} \to y$ , so the uniform continuity of $D_{x} F$ on $K$ gives $\sup_{s \in I} ∥ C_{y^{'}} (s) - C_{y} (s) ∥ \to 0$ . Applying [L3] and [L4] exactly as in step 3.1 shows $\sup_{t \in I} ∥ E (t) ∥_{2} \to 0$ . Hence $y \mapsto D_{y} Φ (t, y) = A_{y} (t)$ is continuous for each $t$ , and the derivative matrix is exactly the variational-equation solution.

5.1

Steps 3.1 and 4.1 prove that $Φ$ is $C^{1}$ in the initial-state variable and that its derivative matrix is the solution of the variational equation.

step 3.1step 4.1∎

Depends on

Used by

Continuous dependence does not by itself imply differentiable dependence False statement
Smooth dependence of solutions on initial data Theorem

Dependency tree · two levels

47 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Nigel Hitchin, Differentiable Manifolds, Appendix §10.3, Lemma 10.6 (standard reference, not scraped)
Chin-Lung Wang, Banach Calculus, §4.4 (standard reference, not scraped)

C1C1 dependence of solutions on initial data