If you a have a function of several variable like f(t,x,c), it is of class C^m if all partial derivative df/dt, df/dx and df/dc are of class C^m. If f is of class C^m, then the solution is of class C^m (cf. Arnol'd, Ordinary Differential Equations, part 2.7.5).

The answer is 'No', as the example x' = |c| shows.

By the way, I am not sure the statement you declared as known ("the solution is C^k...") is actually true. I don't think it is. The difficulty is this: The derivation of this kind of smoothness result involves mixed derivatives, e.g. w.r.t. x and w.r.t. c. The way you stated your assumptions ("for fixed t and x...") does not imply much about these mixed derivatives. It does not even follow that f is continuous! For more on the subject, you might want to consult the following books:

H Amann, Ordinary Differential Equations, de Gruyter;

P Hartman, Ordinary differential equations, SIAM

Thanks all.

Gunther, you are right about my "know" statement. For its truth f must be C^k in the variables (x,c) as Oliver said.

However in your example, x'=|c| is continuous (C^0) in c as well its solution x(t)=x(0)+|c| t. So it is still working for my real doubt (I have wrote wrong.) Of course the equation is C^infty in x and it solution is only continuous in c.

Let me rewrite better my doubt:

Suppose that f(t,x,c) is C^m in the variables (x,c) and the function c|-->f(t,x,c) C^n.

Again, if n>m can I say that the function c|-->x(t,z,c) is C^n (for fixed t and z)?

For your "modified doubt", I think the answer is 'No' again.

Assume f is independent of t, and let x(0) = 0. Assume further that for ||x|| > 1, f does not depend on c, and that for ||x|| 1 whenever t> t1. Then the dependence of x(t1) on c is of class C^n. Now consider some t > t1. Then x(t,t0=0,x(0),c) = x(t,t1,x(t1),c). Observe that x(t,t1,x(t1),c) does not depend on its fourth argument, and the dependence of x(t,t1,x(t1),c) on its third argument is only of class C^m, not C^n. Hence, in general, the dependence of x(t,t0=0,x(0),c) on c is only of class C^m.

Gunther,

You are taking x(t,t1,z,c) the solution of (1) such that x(t1,t1,z,c)=z, right?

In your "modified doubt example" z is depending on c. However I am asking about the differenciability of the function c|-->x(t,t1,z,c), note that the coordinates t, t1, z are fixed.

Douglas,

I do understand your question, and my example does answer it. I think that my explanation was just too sloppy. I am sorry. Below, I give a careful explanation in tiny steps. I will number all my arguments, and if my explanation does not convince you, please refer to the first of my arguments that is, in your opinion, flawed.

1. We may focus on the autonomous case, as we are looking for a counterexample to a conjecture. For the same reason, we may assume that both the state and the parameter are scalar (real), and that the initial value is 0. So the initial value problem is

x' = f(x,c), (1)

x(0) = 0, (2)

where c is a parameter and f maps R x R to R. Let \varphi be the flow of (1). As usual, \varphi takes three arguments, of which the third is the parameter.

2. Assume that the right hand side f of (1) is of class C^1, and that additionally, the second order partial derivative of f with respect to the second argument exists and is continuous on RxR. Then \varphi is of class C^1. You ask if for any given t and x0, the map $H \colon c \mapsto \varphi(t,x0,c)$ is even of class C^2, right? This is a question about the dependence of the flow on a parameter. To see that the answer is 'No', I use the group property of the flow to transform the question into one about the dependence on initial values.

3. We can find a right hand side f such that the following holds:

(3a) if |x| > 1/2, then f(x,c) does not depend on c,

(3b) f(x,c) >= 1 for all x and c,

(3c) for any t > 0, the derivative of \varphi(t,0,\cdot) does not vanish at the origin.

4. By assumption (3b), the map \varphi(.,0,0) is strictly increasing. Hence, there is a unique t1>0 such that \varphi(t1,0,0) = 1.

5. Let s >= 0 and x1 > 1/2.Then, by (3a) and (3b), \varphi(s,x1,c) does not depend on c.

6. Let t > t1 be arbitrary. Then, for x0 =0, the map H given in item 2 takes the following form:

H(c) = \varphi(t - t1, \varphi(t1, 0, c), c).

This is the group property of the flow. For c small enough, \varphi(t1,0,c) > 1/2 by item 4, so by item 5 and the chain rule we have:

H'(c) = D_2 \varphi(t-t1,\varphi(t1,0,c),c) D_3 \varphi(t1,0,c), (***)

where D_i denotes the partial derivative with respect to the i-th argument.

7. If the answer to your question were 'Yes', then both H and the map c \mapsto \varphi(t1,0,c) were of class C^2. Then, by (***) and (3c), the map c \mapsto D_2 \varphi(t-t1,\varphi(t1,0,c),c) was of class C^1. By (3c) and item 5, this implies that the map x1 \mapsto \varphi(t-t1,x1,0) is of class C^2 in a neigborhood of 1. However, for any prescribed value of the parameter, the solution is only of class C^1 with respect to initial conditions, as f is only C^1. This is a contradiction, so the map H is, in general, not of class C^2.

Let me know if you agree.

Dear Gunther, thanks for your very detailed answer. I still have some doubts.

In (3c) the derivative of \varphi(t,0,\cdot) is the derivative of the map c|--> \varphi(t,0,c)?

I followed your argument until concluding, in 7, that the conjecture implies that the map: c |---> D_2 \varphi(t-t1,\varphi(t1,0,c),c) was of class C^1.

However I did not realize how you concluded that the map x1 \mapsto \varphi(t-t1,x1,0) is of class C^2.

Thanks again.

Douglas,

thank you for your comments.

> In (3c) the derivative of \varphi(t,0,\cdot) is the derivative of the map

> c|--> \varphi(t,0,c)?

Yes.

>I followed your argument until concluding, in 7, that the conjecture implies that the >map: c |---> D_2 \varphi(t-t1,\varphi(t1,0,c),c) was of class C^1.

>However I did not realize how you concluded that the map

>x1 \mapsto \varphi(t-t1,x1,0) is of class C^2.

Let us define the map G by

G(c) = D_2 \varphi(t-t1,\varphi(t1,0,c),c).

For c small enough, \varphi(t1,0,c) > 1/2 by item 4, so by item 5, \varphi(t-t1,\varphi(t1,0,c),d) does not depend on d. Then D_2 \varphi(t-t1,\varphi(t1,0,c),d) does not depend on d either, hence

G(c) = D_2 \varphi(t-t1,\varphi(t1,0,c),0)

for all c small enough, and you have already accepted that the latter map is of class C^1. Now we rewrite that map as

G(c) = f(g(c)),

where

f(x1) = D_2 \varphi(t-t1,x1,0)

and

g(c) = \varphi(t1,0,c).

The map g is of class C^1 (C^2 if the answer to your question were 'Yes', but we do not use this fact here), and by (3c) and the inverse function theorem, g is a local C^1-diffeomorphism around the origin. Hence,

G( g^{-1} ( z ) ) = f(z)

for all z in a neighborhood of 1. As the maps on the left hand side are of class C^1, so is f, which shows that the map x1 \mapsto \varphi(t-t1,x1,0) is of class C^2 in a neighborhood of 1.