Invertibility of the Legendre Transformation

Question

The above image shows the Legendre Transformation in the context of an introduction to the Hamiltonian formalism.

My question is in 4.6, wherein $u(x, y)$ has been defined; what is the guarantee that we would be able to invert this relation in order to get $x(u, y)$? Or in other words, what is the guarantee that we would be able to write $\dot{q_{i}}$ as $\dot{q_{i}}(p_{i}, q_{i}, t)?$. And doesn't that affect the invertibility of the Legendre transformation as a whole as written in the lines following equation 4.9?

Another question: Is time supposed to be a spectator variable when it comes to a Legendre Transformation in Hamiltonian dynamics?

Could somebody possibly dwell on this rigorously? Thanks.

Please write equations yourself, rather than posting an image. The image cannot be searched and cannot be edited. — DanielSank, Apr 27 '20 at 16:40
Here's the simple answer without full derivation: in order for a Legendre transform $f(v)\rightarrow f(p)$ to be well-defined and invertible, you need $df/dv$ to be strictly concave in $v$.
For your second question, yes, time is a spectator variable in classical Hamiltonian dynamics. — Arturo don Juan, Apr 27 '20 at 16:46
Related : A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein. — Frobenius, May 01 '20 at 17:55

Invertibility of the Legendre Transformation

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