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I have a question regarding the weak formulation of the poisson equaiton:

$-\nabla^2 u = f$ on $\Omega \subset \mathbb{R}^n$, $u_{\delta\omega}=0$

In my notes as well as on wiki, it says that the weak formulation can be derived using integration by parts and greens identity. Can someone do this explicit? Where does the use of greens identity come in?

Regards

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Take a test function $v$ such that $v=0$ on $\partial \Omega$, multiply both sides by this $v$ and integrate over $\Omega$, we get (I implicitly suppose that all fields and functions are sufficiently regular to apply those operators involved):

$$-\int_{\Omega} (\Delta u) v = \int_{\Omega} fv $$

Now, remember the identity $ \text{div} (v \nabla u )= (\nabla u )(\nabla v) + v \Delta u $ to get:

$$\int_{\Omega} \nabla u \nabla v - \int_{\Omega} \text{div} (v \nabla u ) = \int_{\Omega} fv $$

By Green's theorem we can write (if $\partial \Omega$ is sufficiently regular):

$$ \int_{\Omega} \nabla u \nabla v - \int_{\partial \Omega} (v \nabla u) \cdot \mathbf{n} \, dS = \int_{\Omega} fv $$

Finally, because $v=0$ on $\partial \Omega$ we conclude:

$$ \int_{\Omega} \nabla u \nabla v = \int_{\Omega} fv $$

Note that the identity $\text{div} (f \mathbf{F})= \nabla f \cdot \mathbf{F} + f \text{div} \mathbf{F}$ plays the role of "integration by parts", as Wikipedia states.

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    Thanks! Yeah I have one further question wrt. integration by parts. If you just do it clasically isn't that enough then. You get the surface integral as well right? – 2017-02-21
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    Edited it:) I think this is good as a general approach, if you want something more similar to the integration by parts in one dimension, we should work in a Sobolev space and test on compactly supported functions. – 2017-02-21
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    Well, in one dimension the gradient of a scalar function is just its derivative and the Laplacian is just the second derivative! – 2017-02-21
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    I can add the example in one dimension if you need it ! – 2017-02-21
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    The problem with the one dimensional example, is that then classical integration by parts just does it, you don't get the insight with the boundary integral, since it's just two points you have to substract. – 2017-02-21
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    By classical integration by parts I mean: $\int_{\Omega}\nabla^2uv = [\nabla uv]_{\delta\Omega}-\int_{\Omega}\nabla u \nabla v$ – 2017-02-21
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    @1233023 Yes, that's it :) that's why I think the general one is more interesting. – 2017-02-21