# Continuous non-autonomous vector field, behaviour of solution curves as initial condition changes

Suppose $F(x,t)$ is a continuous non-autonomous vector field on $\mathbb{R^n} \times \mathbb{R}$ such that $||F(x,t) - F(y,t)|| \leq L(t)||x-y||$ and let $\phi_t (x_0)$ be the solution of

$$x' = F, \quad x(0) = x_0$$

Show that $\phi_t$ satisfies the inequality $|| \phi_t (x_0) - \phi_t (y_0) || \leq ||x_0 - y_0|| \exp \bigg( | \int_0^t L(s) ds | \bigg)$

Suppose now that $n=1$, and that $F(x,t)$ is continuously differentiable in $x$. Show that

$$\frac{\partial \phi_t (x)}{\partial x} = \exp \bigg( \int_{0}^{t} \frac{\partial F(\phi_s (x),s)}{\partial x} ds \bigg)$$

I have no clue how to prove this, or even where to start in order to involve those exponentials. Any help/hints/solutions are welcome, thanks in advance!

## 1 Answer

Hint: $x(t)$ being the solution of $\dot{x} = F(x,t)$, $x(0) = x_0$ is equivalent to the statement that \begin{equation} x(t) = x_0 + \int_0^t F(x(\tau),\tau)\,\text{d}\tau. \end{equation} In your notation, we can just write $x(t) = \phi_t(x_0)$. Now, you can apply the inequality for $x(t) - y(t) = \phi_t(x_0) - \phi_t(y_0)$, and see what you get.