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We are interested in characterizing a 2D surface $z(x,y)$, where $(x,y)$ is the regular 2D Cartesian grid. Let $\nabla z = (z_x, z_y)$ denote the gradient. The surface is a "general" one, that is, devoid of any special symmetries.

We know the isocontours of constant $z$ and the isocontours of constant $|| \nabla z ||$ at every point. To what extent does this information characterize the surface? (For example, we can the call the surface completely characterized if we know the values of $z$ at every point $(x,y)$ up to a global multiplicative scale and additive offset, or the exact values of the gradient $(z_x, z_y)$ at every point.)

Note that the surface is "general", so at least some isocontours of constant $z$ and constant $|| \nabla z ||$ can be assumed to intersect.

To look at the problem in another way, knowing the isocontours of constant $z$ is the same as knowing the direction of the gradient, that is the ratio $\displaystyle\frac{z_y}{z_x}$, at every point. The surface will be completely specified if, in addition, we also know the magnitude of the gradient, $|| \nabla z ||$, at every point. Instead, all we know are the isocontours of constant $|| \nabla z ||$. Is there an elegant characterization of the ambiguity to which we can infer the values of $z$?

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The two isocontours you know are invariant under scaling: let $\tilde{z} = \lambda z$ for a fixed constant. Then the isocontours of $\tilde{z}$ is the same of that of $z$. Ditto $\|\nabla \tilde{z}\|$. Furthermore, if the two isocontours agree (which happens, for example, if $z$ is a surface of revolution), then for any $C^1$ function $f$, the surface defined by $\hat{z} = f\circ z$ will have the same set of isocontours (in other words, any two surfaces of revolutions have the same set of isocontours). So you are missing quite a lot of information. –  Willie Wong Oct 22 '10 at 10:24
    
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. –  user9728 Oct 22 '10 at 23:14
    
... in view of the answer below, you may want to add [ap.analysis-of-pdes] to the tags of this question. =) –  Willie Wong Oct 23 '10 at 11:46
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1 Answer

Let $p$ be a generic point in the sense that there exists an open set $V\ni p$ such that the isocontours are nondegenerate (in that they are 1-dimensional curves, not 0-d points or 2-d regions) and intersect transversally. Then in a possibly smaller neighborhood $U\subset V$, the values of $z$ is uniquely determined by $(z, |\nabla z|)(p)$. These two degrees of freedom corresponds precisely to the trivial scaling and vertical translation freedoms. So in other words, in generic regions there are no more ambiguity.

The proof is simple: just observe that your conditions gives a hyperbolic PDE in 2 dimensions with initial data prescribed on characteristic curves.

Since in $V$ the isocontours are nondegenerate and intersects transversally, we can pick two unit vector fields $v,w$ tangent to the isocontours, and so they are transversal. Now, that $v$ is orthogonal to $\nabla z$ implies that

$$ w \cdot \nabla z = \pm |\nabla z| \sqrt{1 - (v\cdot w)^2} $$

The sign ambiguity is due to, again, the scaling ambiguity of the original setup. Now, we know that $|\nabla z|$ is constant along integral curves of $w$. So along the integral curve of $w$ that passes through $p$, the above equation uniquely determines $w\cdot\nabla z$. This means that we can integrate this first order ODE and solve for $z$ along the integral curve of $w$ through $p$.

Now, using that $z$ is constant along the integral curves of $v$, we can extend this solution off $p$ as long as $v$ is transversal to $w$. q.e.d.

This also illustrates why when the isocontours are parallel, you get a larger ambiguity: this corresponds to the two characteristics of the second order hyperbolic PDE coinciding. And therefore the data $(z,|\nabla z|)(p)$ at one point, when propagated, either gives you too little information (only the solution along one isocontour, and the solution remains free elsewhere), or too much (that the propagation equations rel. to $v$ and $w$ are not compatible.)

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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? –  user9728 Oct 26 '10 at 6:42
    
If you assume $z$ is continuously differentiable, and $\nabla z \neq 0$, then the sign ambiguity is global, in the sense that once you choose a sign it must be the same sign in a whole neighborhood. Otherwise the gradient vector field $\nabla z$ will point in one direction for a while and suddenly jump to a different direction, failing continuity. –  Willie Wong Oct 26 '10 at 9:17
    
To be more precise: $w$ is chosen to be smooth along its integral curve (norm 1), so if $z$ is continuously differentiable, $w\cdot \nabla z$ is a continuous function. On the other hand, $\sqrt{1-(v\cdot w)^2} \geq 0$ as long as $v,w$ are not tangent. And as long as $|\nabla z|\neq 0$, this means $w\cdot\nabla z \neq 0$. A continuous function that is never zero must have a fixed sign. –  Willie Wong Oct 26 '10 at 9:22
    
Ack, I meant $\sqrt{1-(v\cdot w)^2} > 0$ as long as $v,w$ not tangent, and only $= 0$ if they are tangent. –  Willie Wong Oct 26 '10 at 9:23
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