# Differential equations – The regularity of the ODE with the Zygmund coefficients.

A function of zygmund $$f in mathscr C ^ 1$$ It is a continuous function that satisfies. $$| f (x + h) + f (x-h) -2f (x) | le C | h |$$ for all $$x, h in mathbb R ^ n$$ in the domain

According to Markus' article: A singularity theorem for ordinary differential equations involving smooth functions, a $$mathscr C ^ 1$$-Vector field $$X$$ can define a flow $$mathscr F_X (t, p): mathbb R times M to M$$, that is the only map that satisfies $$frac partial { partial t} mathscr F_X (t, p) = X circ mathscr F_X (t, p)$$, $$mathscr F_X (0, p) = p$$.

But unlike Lipschitz or Holder-$$mathscr C ^ gamma$$ for $$gamma> 1$$, where those functions are stable under composition, the composition of Zygmund's function may not be Zygmund.

For example, $$x log x circ x log x = x log ^ 2x log log x$$Y $$x log ^ 2x$$ It is no longer Zygmund.

I think the following is true, and my Question That is, is there any elementary example for:

Show that there is a $$mathscr C ^ 1$$-function $$f (u, s)$$ in $$mathbb R ^ 2$$ such that for the singularity solution $$frac partial { partial t} phi (t, s) = f ( phi (t, s), s)$$, $$phi (0, s) = s$$ It is not $$mathscr C ^ 1$$ in any neighborhood of $$0$$.

The original question I can think of is, if possible, to find an example to:

Show that there is a $$mathscr C ^ 1$$-Vector field $$X$$ so that if $$Phi (t, s)$$ is the parameterization (a continuous map homeomorphic to your image) near 0 such that $$partial_t Phi in C ^ 0$$ Y $$partial_t Phi (t, s) = X ( Phi (t, s))$$, so $$Phi notin mathscr C ^ 1$$ Y $$forall s, Phi ( cdot, s) notin mathscr C ^ 2_t$$ in any neighborhood of $$0$$.

The idea that I have is to do $$u$$-variable of $$f$$ behave to oscillate like $$u log u$$. I try $$f (u, s) = u log u$$ Y $$phi (1, s) = – s log s$$. I get $$partial_t phi (1,0) notin mathscr C ^ 1$$but this case $$phi$$ it seems still $$mathscr C ^ 1$$ near $$(1,0)$$, and of course not locally restricted near 0.

The obstruction also comes from verifying how the second differentiation explodes: $$frac { phi (t, s) + phi (t, s & # 39;)} 2- phi (t, frac {s + s & # 39;} 2) = int_0 ^ t frac {f ( phi (u, s), s) + f ( phi (u, s & # 39;), s & # 39;)} 2-f ( phi (t, frac {s + s & # 39;} 2), frac {s + s & # 39;} 2) du$$. Just consider $$int_0 ^ tf ( frac { phi (u, s) + phi (u, s & # 39;)} 2, frac {s + s & # 39;} 2) -f ( phi (u , frac {s + s & # 39;} 2), frac {s + s & # 39;} 2) du >> | s-s & # 39; |$$.

But the fact $$phi (0, s) = s$$ Y $$phi$$ is $$C ^ 1$$ shows that the integral is bounded by $$| int_0 ^ tf ( frac { phi (u, s) + phi (u, s & # 39;)} 2, frac {s + s & # 39;} 2) -f ( phi (u, frac {s + s & # 39;} 2), frac {s + s & # 39;} 2) du | lesssim_ epsilon | s-s & # 39; | ^ {1- epsilon} t ^ 2$$ (as $$f in mathscr C ^ {1- epsilon}$$). The calculation shows that we are different from having the idea if we arrange $$s$$ or $$s & # 39;$$ be $$0$$.