Hessian condition for solving Hamilton-Jacobi equation I am reading the book Classical dynamics: a contemporary approach by J. V. José and E. J. Saletan. In chapter 6.1, after deducing Hamilton-Jacobi (HJ) equation, the following is stated:

This (obtaining the solution of the HJ equation) can be done only if the Hessian satisfies the condition $|\partial^2S/\partial q^\alpha\partial Q^\beta|\neq 0$, but recall that this is a condition already imposed on a Type 1 generator.

A Type 1 generating function is $F_1(q,Q,t)=S(q,Q,t)$. I don't understand why such condition is already imposed in this kind of generating function, this only happens to type 1, or it also works in type 2 and others? Also, why is this condition even needed? 
The book doesn't explain any further, so I would like to understand this mathematical needs for solving HJ equation.
 A: *

*The full quote in Ref. 1 reads

[...] In the meantime, assume that a solution for $S(q, Q, t)$ has been found. Because $K = 0$ by construction, the motion in the $(Q, P)$ coordinates is trivial:
  $$ Q^i(t) ~=~ \text{const}., \qquad P_i(t) = \text{const}.\tag{6.1b}$$
  To find the motion in the $(q, p)$ coordinates, use Eqs. (5.121):
  $$p_i~=~\frac{\partial S}{\partial q^i}, \qquad P_i~=~-\frac{\partial S}{\partial Q^i}. \tag{6.2} $$
  Invert the second of these to find the $q^i$ in terms of $(Q, P, t)$ and insert that into the first to find the $p_i$ in terms of $(Q, P, t).$ This can be done only if the Hessian satisfies the condition 
  $$\det\frac{\partial^2 S}{\partial q^i\partial Q^j}~\neq~ 0,\tag{6.2c}$$ 
  but recall that this is a condition already imposed on a Type 1 generator. The problem is thereby solved, since $q(t)$ and $p(t)$ have been obtained
  in terms of the $2n$ constants of the motion $(Q, P).$ [...]


*Note that Ref. 1 is not claiming that condition (6.2c) is needed to solve the Hamilton-Jacobi (HJ) equation wrt. $S$. Instead condition (6.2c) is needed to solve eq. (6.2b) wrt. $q^i$, cf. the inverse function theorem. 

*In the rest of this answer, we would like to show that condition (6.2c) is automatically imposed on a Type 1 generator. To this end, let us introduce the collective notation 
$$z^I~\equiv~(q^i,p_i), \qquad Z^I~\equiv~(Q^i,P_i), \qquad \mathbb{Q}^I~\equiv~(q^i,Q^i), \qquad \mathbb{P}_I~\equiv~(p_i,P_i), $$ 
$$ i~\in~\{1,\ldots,n\},\qquad I~\in~\{1,\ldots,2n\}. \tag{1}$$

*For notational simplicity, the time parameter $t$ is implicitly implied in what follows. So for instance a canonical transformation (CT)
$$Z^I~=~G^I(z,t)\tag{2}$$ is written as $$Z^I~=~G^I(z)
~=~\begin{pmatrix} g^i(z) \\ \star \end{pmatrix}, \qquad Q^I~=~g^i(z),\tag{3}$$ 
with the parametric $t$-dependence implicitly implied, and so forth. (The star $\star$ indicates that the precise form of this sector will play no role in what follows.)

*If the CT has a type 1 generating function $F_1(\mathbb{Q})$, it means that
$$\mathbb{P}_I~=~ H_I(\mathbb{Q})
~=~\begin{pmatrix} h_i(\mathbb{Q}) \\ \star \end{pmatrix},\qquad 
p_i~=~h_i( \mathbb{Q})~=~\frac{\partial F_1(\mathbb{Q})}{\partial q^i}.\tag{4}$$

*For fixed $q$, we conclude that the functions $g$ and $h$ are each other's inverse functions. More precisely
$$ Q^i~=~g^i(q, h(q,Q)), \qquad p_i~=~h_i(q, g(q,p)).\tag{5}$$
We conclude that the matrices
$$ \frac{\partial g^i(z)}{\partial p_j} \quad \text{and}\quad 
\frac{\partial h_i(\mathbb{Q})}{\partial Q^j}
~=~\frac{\partial^2 F_1(\mathbb{Q})}{\partial q^i\partial Q^j} \tag{6}$$
are each other's inverse matrices, and in particular have non-zero determinants, which is the sought-for eq. (6.2c).
References:


*

*J.V. Jose & E.J. Saletan, Classical dynamics: a contemporary approach; p.244 eq. (5.122b) & p. 286.

