Given the system $\dot{x} = f(x)$, $x(0) = x_0$, with $f \in C^{\infty}(\mathbb{R})$. I am trying to show that the flow map $\Phi(t, x_0)$ is invertible at any fixed point $x^{\ast}$ and at any finite time $t^{\ast}$. I was told that this would follow if I could show that $\Phi(t,x_0)$ is differentiable at $(t^{\ast}, x^{\ast})$ with respect to $x_0$ and $\frac{\partial \Phi(t^{\ast}, x^{\ast})}{\partial x_0} \neq 0$.

I am curious why differentiability of the flow map with respect to the initial condition at the fixed point guarantees invertibility and how one can show that in this case. My thought is that in order to use the smoothness of the vector field $f(x)$ I may want to expand $f(x)$ in a Taylor series about $x^{\ast}$.

  • $\begingroup$ Indeed it is neither related to differentiability nor to having a nonzero derivative. Also, do you want invertible in which variable? What have you tried anyways? $\endgroup$ – John B Oct 10 '18 at 19:59
  • $\begingroup$ The inverse flow map I am looking for is $\Phi_0(t, \Phi) = \Phi(-t,\Phi)$. I have tried plugging the flow map into the system and differentiating with respect to $x_0$. The reason I thought the inverse flow map was related to differentiability was because I have from lecture notes that $$\frac{\partial \Phi_0(t^{\ast},x)}{\partial x} = \frac{1}{\frac{\partial \Phi(t, \Phi_0(t^{\ast}, x)) }{\partial x_0}}.$$ $\endgroup$ – AMD Oct 10 '18 at 20:20
  • $\begingroup$ So you want to show that $\Phi(-t,\Phi(t,x_0))=x_0$. Best hint: use the uniqueness of the solutions of the differential equation (unless you already know that it is a flow). $\endgroup$ – John B Oct 10 '18 at 20:44

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