Can a Hamiltonian include explicitly the derivative of the conjugate momentum, especially after a canonical transformation?

Question

Can a Hamiltonian expression, say $H$ with $(q,p)$ as conjugate variable pair, include the total derivative of $p$ explicitly? That is, can we have $H=H(q,p,\dot{p})$?

And, if so, what does it imply in terms of the usual canonical equations? I can see that it will lead to second order derivatives now. Is this permissible in the Hamiltonian formulation, and what does it mean?

Finally, if a transformation $(Q,P,t) \rightarrow (q,p,t)$ is what had led a regular Hamiltonian $K(Q,P,t)$ to become a new one $H(q,p,\dot{p},t)$ with such dependence on $\dot{p}$, as described above, does that mean that such transformation is necessarily not canonical? Or can a regular canonical generating function $F$, for example, do so?

Answer 1

No, the Hamiltonian $H(q,p,t)$ depends by definition not on $\dot{q}$ or $\dot{p}$. Similarly for the Kamiltonian $K(Q,P,t)$.

For Legendre transformations involving $\dot{q}$ or $\dot{p}$, see e.g. this & this Phys.SE posts and links therein.