2 Totally revised

(Sorry if you saw my original answer but it was completely wrong)

I don't know the answer if the thickness $\epsilon$ to this question, because I believe that it is equivalent to an initial value problem for a system of PDE's, where the tubular neighborhood initial data is specified posed on a characteristic surface. My guess is that the answer is in advancegeneral no, but there does exist some $\epsilon > 0$ for which the answer it is yesjust a guess.

However, if you pose the initial data differently, then there is the following theorem: If instead of specifying the function $f$, you specify an 3-d orthonormal frame along the surface is real analyticso that none of the vectors in the frame are ever tangent to the surface, then it there exist triply orthogonal co-ordinates (i.e., co-ordinates such that the metric tensor is easily proved using Cauchy-Kovalevskidiagonal) in a neighborhood of the surface such that their gradients point in the same directions as the given orthonormal frame. If it This is only smooth, then this was proved inthis paper: DeTurck, Dennis M.; Yang, Deane Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243--260.

Any chance you have some use for this? I was quite disappointed that I couldn't find any use for co-ordinates where the metric tensor is diagonal.

1

I don't know the answer if the thickness $\epsilon$ of the tubular neighborhood is specified in advance, but there does exist some $\epsilon > 0$ for which the answer is yes. If the surface is real analytic, then it is easily proved using Cauchy-Kovalevski. If it is only smooth, then this was proved in this paper: DeTurck, Dennis M.; Yang, Deane Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243--260.

Any chance you have some use for this? I was quite disappointed that I couldn't find any use for co-ordinates where the metric tensor is diagonal.