# Hamilton Flow of Homogeneous Hamiltonian

The context comes from trying to understand that Hamiltonian flow of a Homogeneous Hamiltonian. Let $$p(x,\xi) \in C^\infty\big(\mathbb{R}^{2n}\big)$$ be homogeneous of degree $$2$$ in $$\xi$$ and let $$X(t,x,\xi)$$ and $$\Xi(t,x,\xi)$$ solve the Hamilton flow equations

\begin{align*} \partial_t X &= (\partial_\xi p)(X, \Xi), \quad \ \ \ X(0,x,\xi) = x \\ \partial_t \Xi &= -(\partial_xp)(X, \Xi), \quad \Xi(0,x,\xi) = \xi \end{align*}

How do we show that from the degree $$2$$ homogeneity of $$p$$ in $$\xi$$ that for $$\lambda > 0$$ we have \begin{align} X(\lambda t, x , \xi/\lambda) = X(t,x,\xi), \quad \Xi(\lambda t,x,\xi/\lambda) = \lambda^{-1}\Xi(t,z,\xi) \end{align}

The first set of equations tell us that $$X$$ is homogeneous of degree $$1$$ in $$\xi$$ and that $$\Xi$$ is homogeneous of degree $$2$$ in $$\xi$$. I've tried to exploit this fact and the chain rule to differentiate the rescaled equations in $$t$$ and show they solve the same equations as $$X(t,x,\xi), \lambda^{-1}\Xi(t,x,\xi)$$ but I can't seem to get exactly what I want.

Edit: The bolded statement is not necessarily correct. The answer to the initial question is immediate once you noticed the same system is solved with the same initial conditions.

• Oh, I think you can show that $X$ and $\Xi$ are homogeneous of degree $1$ in $t$ and then this is immediate? Aug 12 '19 at 13:16

I don't think my statement about any homogeneity of $$Z$$ and $$\Xi$$ is necessarily correct. In any case this is immediate. Let $$Z_\lambda$$ and $$\Xi_\lambda$$ denote $$Z(\lambda t,x, \xi/\lambda), \ \Xi(\lambda t,x, \xi/\lambda)$$ respectively. By definition \begin{align} \partial_t Z_\lambda &= \lambda (\partial_\xi p)(Z_\lambda, \Xi_\lambda)\\ \partial_t \Xi_\lambda &= - \lambda (\partial_zp)(Z_\lambda, \Xi_\lambda) \end{align}
Additionally we see using the Homogeneity of $$p$$ that \begin{align} \partial_t Z &= \lambda (\partial_\xi p)(Z, \frac{\Xi}{\lambda})\\ \partial_t \frac{\Xi}{\lambda} &= -\lambda(\partial_zp)(Z, \frac{\Xi}{\lambda}) \end{align}
Therefore these solve the same system of equations with the same initial conditions and it follows that $$Z_\lambda = Z$$ and $$\Xi_\lambda = \frac{\Xi}{\lambda}$$