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If I have a generating function, say, $$G(q,p,P,Q)= qp - e^Q e^P\tag{1}$$

what are the equations that give me the transformations $Q=Q(p,q)$ and $P=P(q,p)$?

I have only seen generating functions depending on only two variables: $(q,Q)$, $(q,P)$, $(p,Q)$ and $(p,P)$, these cases I know are related by a Legendre transformation and then it is straightforward to get the canonical transformations. But what changes when we allow the generating function to depend on more than two variables (disconsidering time $t$)?

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  • $\begingroup$ Where does your eq. (1) come from? To me that "generating function" looks nonsense. $\endgroup$ – Elio Fabri Apr 6 '19 at 13:57
  • $\begingroup$ I was just wondering if it can be done, never seen it anywhere. $\endgroup$ – Slayer147 Apr 6 '19 at 14:10
  • $\begingroup$ Since canonical transformations don't have units - and $G(p,q,Q.P)$ would look like a Legendre transformation if you set $Q=e^P$ and $P=e^Q$, i.e., $G(p,q,Q.P)=qp-QP$. And it would satisfy the form of $dG$ - but it seems to easy. $\endgroup$ – Cinaed Simson Apr 8 '19 at 2:33
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FWIW, in the time-independent case, besides OP's equation (1), the equations are $$ p\mathrm{d}q-P\mathrm{d}Q~=~\mathrm{d}G $$ together with $$ Q~=~Q(q,p)\qquad\text{and}\qquad P~=~P(q,p).$$

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