Let $E$ be a Banach space and $f,g:E\to E$ diffeomorphism of class $C^1$. $f$ is linear and $f\circ g=g\circ f$ which imply: $Dg(f(x))=f(g(x))\:,\:\forall x\in E$
i) Prove the if x(t) is the maximal solution of $x'=f(x)$ with $x(0)=x_0 $then $g(x(t))$ is the maximal solution of the initial problem $x'=f(x)$ with $x(0)=g(x_0)$.
ii) Deduce that $\phi_t$ is the flux of $x'=f(x)$ then $\phi_t\circ g=g\circ\phi_t$ for all $t$ in the respective maximal interval.
i) For the first part I assumed $g(x(t))$ to be the solution then I tried to verify if it was the solution for the initial problem. $(g(x(t)))'=g'(x(t))x'(t)=g'(x(t))f(x(t))=g'(f(x(t)))=f(g(x(t)))$ which is the vector field for $\begin{cases} \dot{x}=f(x)\\ x(0)=g(x_0) \end{cases}$
However I am struggling to prove that $x(0)=g(x_0)$.
ii)
By part i) I am wondering if I could consider $\phi_t=g(x(t))$ so that $\phi_t\circ g=g\circ g=g\circ\phi_t$.
Question:
Are my answers right? How should I finish i)? If my answers are not right. How should I solve the problem?
Thanks in advance!
In your question to i), the formulation of the task is unfortunate. What they mean is that $\tilde x=g(x(t))$ solves $\tilde x'=f(\tilde x)$ with $\tilde x(0)=g(x(0))=g(x_0)$. Which means that the second time $x$ is used, here $\tilde x$, it is meant as general, abstract $x$ as subject of the ODE, with no relation to the earlier specified solution $x$.
Your approach to ii) has notational misunderstandings. The flux or flow of the ODE is a function $\phi(t;x_0)$ which as function of $t$ solves the IVP with $\phi(0;x_0)=x_0$. An alternative notation is $\phi_t(x_0)$ as family of isomorphisms so that one more easily formulate the group action $\phi_{s+t}=\phi_s\circ\phi_t$. What you have to show is $g(\phi(t;x_0))=\phi(t;g(x_0))$. This is essentially what was proven in i), only in a more complicated notation.