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Consider the non-autonomous dynamical system

$$ \dot x = f(x,t) $$

with $x \in \mathbb{R}^n$. This may be converted to an autonomous system of dimension $n+1$ with $t = x_{n+1}$ and $\dot x_{n+1} = 1$.

Question: how can one prove that the new system has no fixed points?

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Hint: A fixed point $y_0 \in \mathbb R^{n+1}$ of the new system $\dot y = F(y)$ with $F(y) = \binom{f((y_1, \ldots, y_n), y_{n+1})}{1}$ has $F(y_0) = 0$.

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Maybe it's my understanding of fixed points that is somewhat lacking. Is it a necessary condition for a fixed point to occur that $\dot x_1 = 0, \dot x_2 = 0, ... \dot x_{n+1} = 0$? – trolle3000 Oct 28 '13 at 13:22
$y_0$ being a fixed point of $\dot y = F(y)$ means that $y(t) = y_0$ is a solution. That gives $f(y_0) = f(y(t)) = y'(t) = 0$ (as $y$ is constant). – martini Oct 28 '13 at 20:00

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