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If $\mathbb{R}_{+}^{n}$ is the half space $\{x\in\mathbb{R}^n\vert\ x_n>0\}$ and $u$ is a twice differentiable function in $\mathbb{R}^{n}$. If we write: \begin{equation} Du(x)=\left(\frac{\partial u(x)}{\partial x_1},\dots,\frac{\partial u(x)}{\partial x_n}\right)^T\quad\text{for} \ x\in\partial\mathbb{R}_{+}^{n}, \end{equation}

then what does it mean to say that $\left(\frac{\partial u(x)}{\partial x_1},\dots,\frac{\partial u(x)}{\partial x_{n-1}}\right)^T$ is the tangential gradient?

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As written your question doesn't make sense. You want: " If $\mathbb{R}_{+}^{n}$ is the half space $\{x\in\mathbb{R}^n\vert\ x_n\geq 0\}$ $\cdots$ ", and also you must locate yourself at a point where $x_n=0$ and compute derivatives there. – Georges Elencwajg Mar 20 '13 at 13:04
I made the specification that $x\in\partial\mathbb{R}_{+}^{n}$ when computing the derivatives but I've left the half-space notation as $\{x\in\mathbb{R}^{n}\ \vert\ x_n>0\}$ since I'm getting this from Gilbarg-Trudinger's book and that's their convention. – Nirav Mar 21 '13 at 7:28
up vote 1 down vote accepted

In $\mathbb{R}^n$, the boundary of $\mathbb{R}^n_+$ is the set $V=\{(x_1,...,x_{n-1},0):\ (x_1,...,x_{n-1})\in \mathbb{R}^{n-1}\}$. Note that the set $V$ is isomorphic to $\mathbb{R}^{n-1}$. In general, you can consider the set $V_a=V+\{(0,...,0,a)\}$, where $a>0$, note that $V_a\subset \mathbb{R}^n_+$ and $V_a$ is ismorphic within $V$. When you fix $a>0$ and take the derivative $u$ in the direction of an element in $V_a$, then you are taking the tangential derivative. In a similar way, for the same fixed $a$, if you derivate in the direction $(0,...,0,a)$, you are taking the normal derivative. Because of the ismorphism, you can use the notation $(\frac{\partial u}{\partial x_1},...,\frac{\partial u}{\partial x_{n-1}})$ for tangential derivative and $\frac{\partial u}{\partial x_n}$ for normal derivative.

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In your last line you probably mean $\left(\frac{\partial u}{\partial x_1},\dots, \frac{\partial u}{x_{n-1}}\right)$ for the tangential derivative, right? – Nirav Mar 21 '13 at 7:51
Yes, you are right, let me fix it. Thank you. – Tomás Mar 21 '13 at 12:55

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