Let $\Omega \subset \mathbb{R}^n$ be a limited domain of class $C^\infty$ (open, connected) and $\varepsilon > 0$ small such that $$ \Omega_\varepsilon = \{y \in \overline{\Omega} : d(y, \partial \Omega) \leq \varepsilon\} $$ is a (closed) tubular neighborhood of $\partial \Omega$ in $\Omega$. This is diffeomorphic to the product $\partial \Omega \times [0, \varepsilon]$ via the map $$ (x,t) \mapsto x + t \eta(x), $$ where $\eta : \partial \Omega \to \mathbb{R}^n$ is the inward unit normal to the boundary. Denote by $\pi : \Omega_\varepsilon \to \partial \Omega$ the projection, given by $$ \pi(x + t\eta(x)) = x, \quad (x, t) \in \partial \Omega \times [0, \varepsilon]. $$

Is there a way to compute the derivative of $\pi$? I tried unsuccessfully. Thanks!

PS.: my final objective is the following. Suppose $\varphi : \partial \Omega \to \mathbb{R}$ is a differentiable function. I want to calculate the gradient of $\varphi \circ \pi : \Omega_\varepsilon \to \mathbb{R}$.

  • $\begingroup$ Are you seeking something more specific than $D\pi(x, t) = \left[\begin{array}{cc}I_{n-1} & 0\end{array}\right]$, from $\pi(x, t) = x$? $\endgroup$ – Andrew D. Hwang Jun 29 '16 at 0:39
  • $\begingroup$ I wanted something in the original coordinates, if possible. But this looks nice. $\endgroup$ – Eduardo Longa Jun 29 '16 at 0:43
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    $\begingroup$ I added my goal. $\endgroup$ – Eduardo Longa Jun 29 '16 at 0:47
  • $\begingroup$ Since $\varphi \circ \pi$ is constant on the fibres of the projection, $D(\varphi \circ \pi)(x, t) = [D\varphi(x)\ 0]$ (again, modulo the identification of $\Omega_{\epsilon}$ with a product). $\endgroup$ – Andrew D. Hwang Jun 29 '16 at 0:54
  • $\begingroup$ @AndrewD.Hwang, the differential of the projection map is not the identity on hyperplanes parallel to the tangent spaces at the projection point. To convince yourself of this, consider $\partial\Omega$ a circle in the plane of radius $r$, and consider a concentric circle of radius $R$ and parameterized by arc length $\alpha: I \to \mathbb{R}^2$. If $\beta = \pi \circ \alpha$, the length of the tangent vector $\beta'$ is going to be $r/R$. $\endgroup$ – yasmar Apr 19 '17 at 0:20

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