I have this Cauchy problem: $$Dy(x)=X(y(x))$$ $$y(0)=y_0.$$
I also have that $\Phi: \mathbb{R} \times\Omega\rightarrow\Omega$ is a continuously differentiable function such that, $\forall y_0 \in \Omega$,
- $\Phi(0,y_0)=y_0$,
- $\forall x_1,x_2 \in \mathbb{R}, \Phi(x_1,\Phi (x_2,y_0))= \Phi (x_1+x_2,y_0)$.
First, I had to show that if the function $x\rightarrow\Phi(x,y_0)$ is solution of this Cauchy problem, then $$X:\Omega\rightarrow\Omega : y_0\rightarrow\Phi'(0,y_0). \quad(1)$$ But now my question: How do I prove this the other way?
So: If I have $(1)$, then the function $x\rightarrow\Phi(x,y_0)$ is solution of this Cauchy problem?
Thanks in advance!
That's an interesting question, from which book is it from?
The assumptions on $\Phi$ means that $\Phi$ is a flow for $X$.
I would say that the best way to solve your problem is to define, for $x$ and $y_0$ fixed the function $$ F_{x,y_0}(\xi)=\Phi(\xi+x,y_0).$$
On one hand you have $F_{x,y_0}'(0)=\Phi'(x,y_0)$ and on the other hand, using your formula $\Phi(x_1+x_2,y_0)=\Phi(x_1,\Phi(x_2,y_0))$ then $$ F_{x,y_0}(\xi)=\Phi(\xi,\Phi(x,y_0))$$ and thus $$ F_{x,y_0}'(0)=\Phi'(0,\Phi(x,y_0)).$$ As $F_{x,y_0}$ has of course an unique derivative at $0$ this leads to $$ \Phi'(x,y_0)= \Phi'(0,\Phi(x,y_0)) = X(\Phi(x,y_0))$$