Differentiable dependence on parameters

Remarks

Proof idea: the parameter $\lambda$ can be absorbed into the state by adding the equation $\lambda' = 0$, so it suffices to treat initial values. One shows that the difference quotients $(\varphi(t; t_0, x_0 + h e_i) - \varphi(t; t_0, x_0))/h$ converge, using the mean value theorem and Grönwall's inequality to compare them with the solution of the variational equation. Example: for $x' = \lambda x$, $\varphi(t; 0, x_0, \lambda) = x_0 e^{\lambda t}$, and $K = \partial \varphi / \partial \lambda = t x_0 e^{\lambda t}$ indeed solves $K' = \lambda K + x_0 e^{\lambda t}$ with $K(0) = 0$. If $f$ is only Lipschitz in $x$, the solution depends continuously on the data but need not be differentiable. For $x' = \abs{x}$, the map $x_0 \mapsto \varphi(1; 0, x_0)$ is $e x_0$ for $x_0 \ge 0$ and $e^{-1} x_0$ for $x_0 < 0$, not differentiable at $0$. If $f$ is of class $C^r$, then $\varphi$ is of class $C^r$.

Going deeper

For a student

The variational equation is the linearization of the system along a given solution. It is the basic tool for stability of periodic orbits, for Lyapunov exponents and for perturbation methods. The result also shows that the flow of a $C^1$ vector field is a $C^1$ map, with derivative a fundamental matrix of the variational equation.