Indeed $H$ does not identically vanish. The space of solutions of the system $H_u=H_v=0$ plus $(1)$ is 'bigger' than the original system, because we are essentially setting $H$ to be any constant whereas it has to be zero, is it ? Besides that, that's a 'doubly overdetermined' system - 4 equations for a single function $\gamma$, which seems even more difficult to handle than just $(1)$. Is there in the literature a similar case-study to educate myself about what to do ?

In this case and similar ones of interest indeed $A_0=A_3=0$ identically, but $H$ does vanish and $(1)$ cannot be written as linear combinations of its derivatives. So $H=0$ is a second order determined equation for $\gamma$ but it does not necessarily imply $(1)$.

@IgorKhavkine Thank you. It's exactly as you mention.

Thank you. Also, in $(2)$ shouldn't we have some $A_4$ containing (only) second and lower derivatives ?

Thanks for the guidance. A preliminary question is, shouldn't we consider a third equation? I mean, in the original system we take $\tau$ from the Gauss (third equation) and replace in the first two (Codazzi) equations, thus obtaining the two in $(1)$. But taking cross derivatives of the Codazzi's, $(\tau_u)_v=(\tau_v)_u$, puts into another equation containing $\gamma_{uvv}$ and $\gamma_{uuv}$.

@RobertBryant Similarly, in the case $\alpha_v=\beta_u=0$ one can write down a single equation involving only $\tau$, but the generic case of $\alpha$ and $\beta$ with non-vanishing derivatives is what we aim to understand. The problem can be rephrased as follows: The displayed system of PDEs are the Gauss-Codazzi equations associated with the fundamental forms $I=du^2+dv^2+\cos\gamma\: du dv$, $II=\alpha \gamma_u \:du^2-\beta\gamma_v\: dv^2+\tau \:\sin\gamma \:du dv$. The question is, what can $\gamma$ and $\tau$ be, or which surfaces admit such fundamental forms.

@RobertBryant Thank you. $\alpha$ and $\beta$ are know functions, specified in advance. They are part of an experimental input, and as you rightly say, when they vanish or when they are identical constants one can immediately solve for $\gamma$ and $\tau$.

@DeaneYang Thank you. Please see edit.

@MattF. This is for unit length edges. The corresponding sphere to compare would have radius 1/2 so the volume is $\pi/6\approx 0.5$, and we consider only half of it so the bound is $0.25$.

@M.Winter Thank you. Pak considers submetric mappings, meaning that the lengths of the geodesics are smaller than the lengths of their corresponding pre-images . Here we require that not the geodesics but some other curves (the ones aligned with the axes, that is, the warp and weft ) preserve their lengths. In that sense this mapping is less constrained.