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**Is the interval I of an autonomous diff closed???**

## Homework Statement

Given this autonomous diff.eqn

Where we have an open set E defined on R^n and [tex]f \in \mathcal{C}^1(E)[/tex]

[tex]x' = f(x)[/tex] where [tex]x(t_a) = x(t_b)[/tex] and [tex]t_a,t_b \in I[/tex] and where [tex]t_a < t_b[/tex].

Show for n = 1, that the solution x is constant.

## The Attempt at a Solution

According to the definition of the autonomous differential equation. Then for x to be a solution to the diff.eqn then certain conditions must upheld.

1) x(t) is differentiable on I and [tex]\forall t \in I[/tex], [tex]x(t) \in E[/tex] and

2) [tex]x'(t) = f(x(t))[/tex].

according to how f is defined its continous on E and have first derivatives on E. E is open on subset on [tex]\mathbb{R}^n[/tex] and in our case it must mean that if E is defined as [tex]E \subset \mathbb{R} \times \mathbb{R}[/tex] and then [tex]I \in \mathbb{R}[/tex] which is the interval of all solutions of the original problem. Then if [tex]x(t_a)[/tex] and [tex]x(t_b)[/tex] are both solution of the original equation. Then it must mean (as I see it!) that x(t_a) = x(t_b) = 0. Hence a t_c defined on I will result in [tex]x'(t_c) = f(x(t_c)) = 0[/tex]. Thusly for n = 1 x is a constant solution of of the autonomous diff.eqn x' = f(x).

maybe I have misunderstood something but isn't if E is a subset of R x R and I is subset R doesn't mean that both sets E and I are subset of the same set R?

Have expressed this satisfactory?

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