---
tag: 6J83GN4KB8
kind: proof
name: Proof that time translates of solutions of an autonomous system are solutions
msc: [34A34]
proves: 0WA7ZR95V4
uses: [07BNRFQ0PJ, 04SW7WECS8]
sources:
  - url: https://en.wikipedia.org/wiki/Autonomous_system_(mathematics)
    locator: Properties
---

## Proof

We compute the derivative of $\delta$ from its difference quotients.

1. The set $I - c$ is an interval, since it is the image of the interval $I$ under the shift $t \mapsto t - c$.
2. For $t \in I - c$ we have $t + c \in I$. So $\delta(t) = \gamma(t + c)$ is defined, and $\delta(t) \in U$ because $\gamma$ takes values in $U$.
3. Fix $t \in I - c$. For each real $h \ne 0$ with $t + h \in I - c$, we have $\dfrac{\delta(t + h) - \delta(t)}{h} = \dfrac{\gamma(t + c + h) - \gamma(t + c)}{h}$.
4. As $h \to 0$, the right side tends to $\gamma'(t + c)$, because $\gamma$ is differentiable at $t + c \in I$. Here the admissible $h$ are the same on both sides, so one-sided derivatives at endpoints also match.
5. So $\delta$ is differentiable at $t$ and $\delta'(t) = \gamma'(t + c)$.
6. Since $\gamma$ is a [[solution|04SW7WECS8]] of the [[autonomous system|07BNRFQ0PJ]], $\gamma'(t + c) = f(\gamma(t + c)) = f(\delta(t))$.
7. Hence $\delta'(t) = f(\delta(t))$ for each $t \in I - c$, and $\delta$ is a solution of $x' = f(x)$ on $I - c$.

## Idea

The right side $f(x)$ does not depend on time. Shifting the time variable does not change the derivative of a curve, so the shifted curve still has velocity $f$ at its own position.