I faced such a question. Consider the Cauchy problem for an ODE: $$ \begin{cases} y'=F(t,y)\\ y(0)=y_0. \end{cases} $$ Assume $F$ has the regularity $C^{k,\alpha}$ (i.e. it has partial derivatives of order up to $k$, and these derivatives are $\alpha$-Hölder). Can one conclude that the solution function also has the $C^{k,\alpha}$ dependence on $t$ and the initial data $y_0$? For nonzero $\alpha$ I couldn't find it in the literature. Thanks!
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$\begingroup$ For $k=1$ it seems a routine work (but always a work) since we know the expression for the differential of the general solution $\eta(t, x)$ in terms of $dF$, which is a composition of maps between various function spaces. The case $k>1$ I think should follow by induction. $\endgroup$– Pietro MajerCommented Jan 9 at 22:00
2 Answers
There are some papers that discuss the $C^{k,\alpha}$ dependence of solutions to ODEs or related systems. For example, this paper shows the continuous dependence of solutions to algebraic differential systems on the initial data, using the concept of $C^{k,\alpha}$-smoothness.
This other paper studies the dependence of the maximal interval of existence of solutions to a nonlinear differential equation on the initial data, assuming that the right-hand side is $C^{k,\alpha}$-smooth.
Perhaps you can try to generalize the results from the papers I mentioned
I think you might run into uniqueness problems here. For example, if we consider the ODE $y' = \sqrt{|y|}$, with $y(0) = y_0$, then this has the solution
$$ \hat{y}(t)= ( t^2 + 2 \sqrt{y_0} )^2/4 $$
But also, if $y_0 =0$, then there is another solution given by simply $ \tilde{y} (t) = 0 $.
Since we have two solutions for the same initial data, it seems that you'd have an infinite Holder constant if you try to compare the $\hat{y}$ solutions to the $\tilde{y}$ solutions.
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1$\begingroup$ Note that this only happens for k=0. For k greater or equal than 1, F is Lipschitz, which is enough to ensure continuity. $\endgroup$ Commented Jan 9 at 20:26