1
$\begingroup$

I'm looking on the ordinary differential equation

$$ \frac{\partial y}{\partial t}(t,\lambda)=f(t,y(t,\lambda),\lambda) $$ $$ y(t_0,\lambda)=y_0 $$ where $f$ is Lipschitz in $y$ and $C^p$ with respect to every variable. $\lambda$ is a parameter in some compact set in $R^n$.

For fixed $t$, let $y_t(\lambda)=y(t,\lambda)$. I want to know if $y_t$ is differentiable (with respect to $\lambda$).

Starting from $$ y(t,\lambda)=y(t_0,\lambda)+\int_{t_0}^{t}\frac{\partial y}{\partial t}(s,\lambda)\,ds =y_0+\int_{t_0}^{t} f(s,y(s,\lambda),\lambda)ds $$ then differentiating with respect to $\lambda$ and taking the derivative with respect to $t$ I get that $(dy_t)_{\lambda}$ satisfies an equation of the form

$$ \frac{\partial A}{\partial t}(t,\lambda)=F_{\lambda}(t,A) $$ $$ A(t_0)=id $$ where $F$ is some function made from the derivatives of $f$.

This proves that the differential if exists, satisfies that equation.

From the usual Picard–Lindelöf theorem, I know that this equation has unique continuous solution.

If I can prove that this solution is the differential of $y_t$, I'm done.

QUESTION : how to check that A(t) is the differential of $y_t$ ?

QUESTION : do you know an online document providing a precise statement and a proof of the fact that $(t,\lambda)\mapsto y(t,\lambda)$ is $C^p$ ?

This question is related to this one, but I'm asking more : first I want to know why the derivative with respecto to the parameter exists, and second, I want to have a $C^p$-regularity with respect to the parameters.

EDIT

John's answer is correct. Here are two other documents in which it is explaned with more details:

https://www.math.uni-bielefeld.de/%7Egrigor/odelec2009.pdf http://www.math.pitt.edu/%7Ebard/bardware/classes/2920/Grant_4july2007.pdf

1 Answers 1

2

All is quite simple provided that you add the equation $\lambda'=0$ (which means that you don't want $\lambda$ to change with $t$ as in fact we don't). Then the initial differential equation becomes $$ (y',\lambda')=(f(t,y,\lambda),0) $$ or if you prefer $z'=F(t,z)$, where $z=(y,\lambda)$ and $$ F(t,z)=(f(t,z),0). $$ From the (standard) results on the dependence of the solutions on the initial condition, the solution $z(t)$ with $z(t_0)=z_0$ is of class $C^p$ on $z_0$. Hence, the first component $y=y(t)$ is of class $C^p$ on $z_0=(y_0,\lambda_0)$. But since the second component of the differential equation is $\lambda'=0$ we have $\lambda(t)=\lambda_0$. As a consequence , the first component $y=y(t)$ is actually of class $C^p$ on $z_0=(y_0,\lambda)$ (look at the second component, which changed from $\lambda_0$ to $\lambda$).

For books with a more direct approach (which has some advantages, for example in case the regularity of $f$ is not the same in $y$ and $\lambda$), look at Hale's ODE's book (using uniform contractions) or at Chicone's ODE's book (using fiber contractions). Chicone's book is more complete at this particular place, but Hale's book is much better (in my opinion). In general the approach with uniform contractions tends to be more delicate though.

One can also proceed as you describe, making it rigorously using basically fiber contractions (otherwise it is really heavy the proof: you need to show that the primitive of the solution of the equation that you derived is a solution of the original equation, but often this requires additional regularity than the original one, unless we use fiber contractions).