Hi Willie, thanks again. But on the isocontour of constant $| \nabla z |$, when you specify the ODE $w \cdot \nabla z = \pm | \nabla z | \sqrt{1 - (v \cdot w)^2}$, isn't the sign above unknown at every point on that curve? Which means that the ODE to solve for $z$ over the isocontour of constant $| \nabla z |$ is undefined due to that sign ambiguity?

Hi Willie, thanks for your response. I have added a couple of edits to the problem. In particular, I should have mentioned that the surface is a "general" one, so the ambiguities of special cases like solids of revolution are not an issue (that is, we know that at least some members of the two types of isocontours intersect). Also, a surface is characterized if we recover $z'(x,y) = a z(x,y) + b$, where $a$ and $b$ are global constants. So, I was seeking a characterization of any ambiguity over and above the multiplicative scale and additive offset.

It will vary with each instance. If it helps with the solution, here are some details of the physical setup. The surface z can be understood as the "visible" portion of a smooth object, observed by a fixed viewer. So, the boundary is where the surface curves away from the viewer. The shape of the boundary will depend on the shape of the object, but we know that at the boundary, the normal to the surface will be parallel to the retinal plane of viewer. So, while $z_x$ and $z_y$ go to infinity, the ratio ${z_y}/{z_x}$ is simply the direction of the normal to the silhouette in the retinal plane.

Thanks for the answer, Denis and everyone else. I can know a couple of additional things at a closed boundary contour. Along that contour, both $p$ and $q$ tend to infinity. But the ratio $q/p$ is finite and known. So, in your solution, I can specify $\theta$ at the boundary, but $\rho$ is infinity. I have also posed this as a second order PDE by substituting $p_y = q_x$ in PDE1. I have asked it here: mathoverflow.net/questions/41050/… For this second-order PDE, we may ignore the fact that it arises from a coupled first order system.