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Let $t\in I=[a,b]\subset\mathbb{R}$, $a<b$, and \begin{equation} \frac{\mathrm{d} x_i}{\mathrm{d}t} = p_i(t,x_1,...,x_n), \quad i\in \{1,...,n\}, \end{equation} be a normal system of first-order ODEs, where $p_i$'s are polynomial functions.

Since $p_i$'s are polynomial, we can always apply the chain rule to get a higher order of derivative with respect to $t$. Thus $x_i(t)\in \mathcal{C}^\infty (I),i \in\{1,...,n\}$.

My question is: does $x_i(t)\in \mathcal{C}^\omega(I)$?

(i.e. is $x_i(t)$ real-analytic?)

If yes, why?

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  • $\begingroup$ What difference do you make between $\mathcal{C}^\infty (I)$ and $\mathcal{C}^\omega(I)$? $\endgroup$ Dec 1, 2015 at 20:45
  • $\begingroup$ @mathcounterexamples.net $\mathcal{C}^\omega(I)$ means real-analytic. en.wikipedia.org/wiki/Analytic_function When the class is $\mathcal{C}^\infty(I)$, then it's not necessarily true that the Taylor expansion converges for any neighborhood of $x_0\in I$. $\endgroup$ Dec 1, 2015 at 20:47
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    $\begingroup$ The answer is "yes", if $\dot x=f(x)$ has $f$ analytic then the solutions are also analytic. This is often called Cauchy's theorem. $\endgroup$
    – Artem
    Dec 1, 2015 at 21:08
  • $\begingroup$ @Artem Do you have a reference for this theorem? $\endgroup$ Dec 1, 2015 at 21:13
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    $\begingroup$ Yup. But in general Cauchy-Kowalevski theorem is the name for a theorem about PDEs. Look up Theorem 4.1 here $\endgroup$
    – Artem
    Dec 2, 2015 at 2:27

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