﻿ calculus - 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 $I$. - oipapio - oipapio.com oipapio

# 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?

1. 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 $$C^{k+1}$$.