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I need help how to interpret this formula and how to write it in explicit form.

Let $\Omega \subset \mathbb{R}^2, u \in C^2(\Omega)$ and $v\in C^1(\Omega)$, then \begin{align} \int_{\Omega}(\nabla^2u)v \, \mathrm{d}x\,\mathrm{d}y=\int_{\partial \Omega} (\nabla^2u \cdot \mathbf{n})v\, \mathrm{d}s-\int_{\Omega}\nabla^2u\cdot\nabla^2u\, \mathrm{d}x \, \mathrm{d}y \end{align}

What is ds in this formula? Is it a abbreviation?

Is $u=u(x,y)$ and $v=v(x,y)$?

Is $\mathbf{n}$ a contant vector or a function, i.e. $\mathbf{n}(x,y)$?

Thanks!

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    $ds$ is the infinitesimal increment of the surface $\partial \Omega$, and $\mathbb{n}$ is the outwards pointing unit normal vector of $\partial \Omega$. – 2017-02-21

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Yes, $u$ and $v$ are functions so they depend in general on their arguments, i.e. $u = u(x,y)$ and $v = v(x,y)$.

$n$ is the outward normal vector and depends on the point of the boundary, so yes $n = n(x,y)$.

$ds$ can be interpreted as an infinitesimal increment of the surface $\partial \Omega$ as mentioned in a comment already. On the other hand you can just view it as a notation for integrating with respect to the surface measure of $\partial \Omega$.

What you call the "explicit form" is also valid, but get rid of the second integral symbol, it is not needed here. The first form of the statement is just an abbreviation by not denoting the arguments of the functions which might be confusing for beginners but saves you time later on.

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    Thanks for the answer! But $ds$, is it a abbreviation? I thought the integral with $ds$ was a double integral. – 2017-02-21
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    As i mentioned in my response there is more than one way to go about this, but you need to understand surface integrals and therefore manifolds first. I would write it in the following way:$ \int_{\partial \Omega} (\nabla^2u(x,y) \cdot \mathbf{n}(x,y))v(x,y)\, \mathrm{d}\mu(x,y)$, where $\mu$ is the surface measure of $\partial \Omega$. – 2017-02-21