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Given known functions $a(x,y)$ and $b(x,y)$, I have a linear, second-order, hyperbolic PDE for a surface $z(x,y)$. The PDE is of the form:
$z_{xx} - a^2 z_{yy} + ab z_x - b z_y = 0$.

The PDE does not have $z_{xy}$ term, $z$ term and constant term.

Along a closed boundary contour, it is known that both $z_x$ and $z_y$ approach infinity. But the limiting ratio $c(x,y) = {z_y}/{z_x}$ is known at the boundary.

The boundary conditions are somewhat unusual, but they arise naturally in my application.

Can the above PDE be solved with such BCs? If I need to specify additional conditions, what should they be? (I suppose at the least I will need to specify the value of z at some points in the domain.)

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Do you have a specific boundary contour, or will that vary with each instance? – Ricky Demer Oct 4 '10 at 21:02
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. – user9728 Oct 4 '10 at 22:17

A boundary value problem with one data at every point of the boundary is not suitable for a second-order hyperbolic equation, although it would be for an elliptic equation like $\Delta z+\cdots=0$. Just try $z_{xx}-z_{yy}=0$ in a disk, and you will find severe obstruction to the solvability of the BVP.

Instead, two kinds of problems can be solved in full generality. 1- the Cauchy problem, where both $z$ and its normal derivative are given on a non-characteristic curves (here, a curve for which $dy^2\ne a^2dx^2$). 2- the Goursat problem, where $z$ is given on two interesecting characteristic curves (here, one curve of each family $dy=\pm adx$).

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I know you already said it in your answer, but it is worth emphasizing that for the Goursat problem, only $z$, not the normal derivative, is given. The values of the normal derivatives are "solved for" by propagating along the characteristic with the equation. – Willie Wong Oct 5 '10 at 11:39

Just to make sense of your problem, think of the following special case: $a=1$, $b=0$ are both constant functions. Then every solution can be written as $z(x,y)=p(x+y)+q(x-y)$, and you have $$ z_x(x,y) = p'(x+y)+q'(x-y), $$ $$ z_y(x,y) = p'(x+y)-q'(x-y). $$ Thus you get $$ 2 p'(x+y) = z_x(x,y)+z_y(x,y) $$ which means, for all $s,t$ such that the points are in the domain you are working in, $$2p'(2s)=z_x(s+t,s-t)+z_y(s+t,s-t). $$ Now fix $s$ and move $t$ so that the line $(s+t,s-t)$ hits the boundary. You see that many problems arise. First of all, if it hits at two different points, the values must match (which is connected to Denis' remark that the problem is ill posed in a closed domain). But, more important for your question, the ratio $z_x/z_y$ has no role here, one needs that the sum $z_x+z_y$ remains bounded, so some cancelation is required. Unfortunately, if you do the same computation to calculate $q$, you get that also the difference $z_x-z_y$ must stay bounded. So I feel that there is no way to make sense of your problem...

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