I've been reading through some notes on integrable systems/Hamiltonian dynamics, and got stuck on a problem relating to generating functions and canonical transformations.
Firstly, some background.
A coordinate change $(\vec{q},\vec{p})\to (\vec{Q},\vec{P})$ is called canonical if it leaves Hamilton's equations invariant, i.e. the equations in the original coordinates $$\dot{\vec{q}}=\frac{\partial H}{\partial\vec{p}},\quad \dot{\vec{p}}=-\frac{\partial H}{\partial\vec{q}}$$ are equivalent to $$\dot{\vec{Q}}=\frac{\partial\tilde{H}}{\partial\vec{P}}, \quad \dot{\vec{P}}=-\frac{\partial\tilde{H}}{\partial\vec{Q}}$$ where $\tilde{H}(\vec{Q},\vec{P})=H(\vec{q},\vec{p})$.
The notes also include the following proposition, but I don't know if it is intended to be used here.
A map $\vec{x}\to\vec{y}(\vec{x})$ is canonical iff $D\vec{y}$ is symplectic, i.e. $$D\vec{y}J(D\vec{y})^T=J$$ where $J$ is the symplectic matrix.
Generating function method for finding canonical transformations:
Suppose we have a function $S:\mathbb{R}^{2n}\to\mathbb{R}.$ Write its arguments $S(\vec{q},\vec{P})$. Now set $$\vec{p}=\frac{\partial S}{\partial \vec{q}}, \quad \vec{Q}=\frac{\partial S}{\partial \vec{P}}.$$ The first equation lets us to solve for $\vec{P}$ in terms of $\vec{q},\vec{p}$. The second equation lets us solve for $\vec{Q}$ in terms of $\vec{q},\vec{P}$, and hence in terms of $\vec{q},\vec{p}$. The new coordinates $\vec{Q}$, $\vec{P}$ we find this way will give a canonical transformation. Checking this is just a careful application of the chain rule.
My Problem:
So I decided to try and work out this 'careful application of the chain rule', i.e. prove that the transformation obtained via this generating function method is canonical. I have been unable to do so, and help with this problem would be greatly appreciated.
****my attempt****
Try to prove $\dot{\vec{P}}=-\frac{\partial\tilde{H}}{\partial\vec{Q}}$. Thinking of $H$ as $H(\vec{q}(\vec{Q},\vec{P}),\vec{p}(\vec{Q},\vec{P}))$, and using the chain rule, $$\frac{\partial\tilde{H}}{\partial Q_i}=\frac{\partial H}{\partial q_j}\frac{\partial q_j}{\partial Q_i}+\frac{\partial H}{\partial p_j}\frac{\partial p_j}{\partial Q_i}$$ which equals, using original Hamilton equations $$=-\dot{p}_j\frac{\partial q_j}{\partial Q_i}+\dot{q}_j\frac{\partial p_j}{\partial Q_i}$$ Meanwhile, $$-\dot{P}_i=-\frac{\partial P_i}{\partial q_j}\dot{q}_j-\frac{\partial P_i}{\partial p_j}\dot{p}_j,$$ so we want $$\frac{\partial P_i}{\partial p_j}=\frac{\partial q_j}{\partial Q_i},\quad \frac{\partial P_i}{\partial q_j}=-\frac{\partial p_j}{\partial Q_i}.$$ Not sure how to go about showing this using $S$.
You've restated the canonical transformation with two equations in matrices. It will help to invert them, viz. $\frac{\partial p_j}{\partial P_i}=\frac{\partial Q_i}{\partial q_j},\,\frac{\partial q_j}{\partial P_i}=-\frac{\partial Q_i}{\partial p_j}$. Let's derive this from $p_j=\frac{\partial S}{\partial q_j},\,Q_i=\frac{\partial S}{\partial P_i}$ viz.$$\frac{\partial p_j}{\partial P_i}=\frac{\partial^2S}{\partial P_i\partial q_j}=\frac{\partial Q_i}{\partial q_j},\,-\frac{\partial Q_i}{\partial p_j}=-\frac{\partial^2S}{\partial p_j\partial P_i}=\frac{\partial}{\partial P_i}\left(p_k\frac{\partial q_k}{\partial p_j}\right)=\frac{\partial q_j}{\partial P_i}.$$