I'm studying classical mechanics reading Mathematical Methods of classical mechanics, by Arnold and doing some exercises lists. As I'm studyng by myself, I got stuck on this exercise.
Consider the system $$\dot q=F_1(q,p), \dot p= F_2(q,p)$$ with $q,p$ in $\mathbb{R}$. Assuming that the vector $(F_1(q1,q2),F_2(q1,q2))$ is perpendicular to $(q1,q2)$ for all $(q,p)\in \mathbb{R}^2$, and that the functions $F_1,F_2$ are class $C^2$. Show that there is a Hamiltonian, such that the ODEs above are equal to Hamilton equations if, and only if, the function $G(q,p)=G_1(q,p)^2 + G_2(q,p)^2$ depends only on the norm of $(q,p)$.
Thoughts: So, if $G$ only depends on the norm, then the vector field $(F1,F2)$ is tangent to circles around the origin. Is this correct? To be honest I don't know how to continue, I'm a little lost.
Given $qF_1+pF_2=0$, we want to prove some $H$ satisfies $F_1=H_p,\,F_2=-H_q$ iff some $f$ satisfies $F_1^2+F_2^2=f(q^2+p^2)$. Given such an $H$, can we get such an $f$; given such an $f$, can we get such an $H$?
The constraint on $H$ is equivalent to $H_{q^2}=H_{p^2}$, i.e. $H$ is a function of $q^2+p^2$, say $H=g(q^2+p^2)$. Then $F_1=2pg^\prime,\,F_2=-2qg^\prime,\,f=4(p^2+q^2)g^{\prime2}$.
So the real question is, can we get from $f$ to $g$ or vice versa (not necessarily uniquely)? Yes, viz. $f(x)=4xg^{\prime2}(x)$ with $x:=q^2+p^2$.
As a famous example, take $g(x)=x/2,\,f(x)=x$.