Let's say we have a sobolev space $H^1(\Omega) $

Therefore if $u\in H^1(\Omega) $ ,$ \frac{\partial u}{\partial x_i} $ exists weakly. i.e

There exists a $g\in L^2(\Omega)$, such that for any $\phi\in C^\infty_{c}(\Omega) $ $$ \int_{\Omega}g \phi \ dx =-\int_{\Omega} u \frac{\partial \phi}{\partial x_i} \ dx $$ So bascially $\frac{\partial u}{\partial x_i}$ can be represented as $g\in L^2 $ distributionally (weakly)

But I am confused that for the Green's formula , we have for $u,v\in H^1(\Omega)$ $$ \int_{\Omega}\frac{\partial u}{\partial x_i}v\ dx=-\int_{\Omega}\frac{\partial v}{\partial x_i}u\ dx \ + \ \int_{\partial\Omega}uv\nu_i\ dS$$ Since the derivative of $u$ in $ H^1(\Omega)$ only exists distributionally (weakly) when we intergrate with a $C^\infty_c$ function , How does the derivative of $u,v$ in the above Green's formula makes sense for $u,v\in H^1(\Omega)$. I am studying analysis of PDE for the first time so the question maybe funny but please help! I am really confused.

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    $\begingroup$ The derivatives of $u$ and $v$ make sense as functions in $L^2(\Omega)$, so it makes sense to consider integrals such as $\int_\Omega \frac{\partial u}{\partial x_i} v$. I would be more concerned about the boundary integral, since you need to make sense of $u|_{\Omega}$ as a function in $L^2(\partial \Omega)$. For this, you need the so-called trace (look it up!). $\endgroup$ – Michał Miśkiewicz Apr 23 '18 at 21:37

The idea is that you can approximate any $H^1(\Omega)$ function by a test function in the $\lVert \cdot \rVert_{H^1}$-norm. Thus you can find a sequence $\phi_n \in C^\infty_c(\Omega)$ (assuming the boundary of $\Omega$ is nice enough) such that $\lVert \phi_n -v \rVert_{H^1} \to 0$ as $n \to \infty$. Use this and the definition of the weak derivative and the formula will make sense


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