calculus - If $f \in \mathcal{C}^k (I \times \mathbb{R}^m)$, show that each solution $x(t)$ of $x'(t)=f(t,x(t))$ is of class $\mathcal{C}^{k+1}$ over $I$.

I have no idea on how to start this.

$x'(t)=f(t,x(t))$ is short for $$ \left\{\begin{align} x_1'(t)&=f_1(t,x_1(t),x_2(t),\dots,x_m(t)) \\ x_2'(t)&=f_2(t,x_1(t),x_2(t),\dots,x_m(t)) \\ &\vdots \\ x_m'(t)&=f_m(t,x_1(t),x_2(t),\dots,x_m(t)) \end{align} \right.$$

I feel like it has something to do with the fact that a differential equation of order $n$ can be rewritten as a system of $n$ first order differential equations (change of variables), but I can't really see where to start.

Any clues?

Thanks in advance.

1 Answer

  1. Frank- Reply

    2019-11-14

    The composition of a function of class $C^k$ with another of class $C^p$ is of class $C^{\min(k,p)}$. Also, any function of class $C^k$ is of class $C^p$ for $p\leq k$.

    The solution $x$ of your equation $x'=f(\cdot,x(\cdot))$ is at least continuous, so $x'=f(\cdot,x(\cdot))$ is also continuous, so $x$ is actually of class $C^1$ (because $k\geq 0$).

    More generally, if $x$ is of class $C^p$ where $p\leq k$, then $x'=f(\cdot,x(\cdot))$ is the composition of two $C^p$ functions, so $x'$ is of class $C^p$ and thus $x$ is $C^{p+1}$.

    So just iterate the paragraph above for $p=0,\ldots,k$ and conclude that $x$ is $C^{k+1}$.

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