Let $v,w\in C^1(\bar{\Omega})$ and $\partial\Omega$ be smooth. Then Green's formula in $\mathbb{R}^2$, which is some integration by parts analogon to $\mathbb{R}^1$, is given to be $$ \int_{\Omega}v_{x_i}w\, dx=-\int_{\Omega}vw_{x_i}\, dx+\int_{\partial\Omega}vwn_i\, d\sigma,\qquad i=1,2,~~~~(*) $$ where $n=(n_1,n_2)$ is the outer normal on $\partial\Omega$.

I have two problems with this.

Problem 1: I get something different!

I think one can use Gauß-formula in $\mathbb{R}^2$ which is $$ \int_{\partial\Omega}W dx=\int_{\Omega}\left(\frac{\partial W_2}{\partial x_1}-\frac{\partial W_1}{\partial x_2}\right)\, d(x_1,x_2) $$

So setting $W=(W_1,W_2)$ with $W_1:=vw, W_2:=0$ what I get is, e.g. for $i=1$, $$ \int_{\Omega}v_{x_2} w\, dx = -\int_{\partial\Omega}v w n_1\, d\sigma - \int_{\Omega}vw_{x_2}\, dx.~~~~~(**) $$

This is not the same as claimed above!

Problem 2: In analysis in $\mathbb{R}^1$, the integration by parts formula is $$ \int_a^b vw'\, dx = -\int_a^b v'w + [vw]_a^b. $$

I would like to deduce this as a special case of the Green's formula $(*)$ resp. $(**)$.

In $\mathbb{R}^1$, we of course only have $i=1$. So let $u,v\in C^1[a,b]$.

Hence using Greens' formula, we simply should have that $$ \int_{\Omega}v_{x_i}w\, dx=\int_{a}^b v'w\, dx $$ and $$ -\int_{\Omega}vw_{x_i}\, dx=-\int_a^b vw'\, dx $$

where now this are not double integrals anymore.

The only point where I am struggling is why here $$ \int_{\partial\Omega}vwn\, d\sigma=[vw]_a^b $$ where $\partial\Omega=\left\{a\right\}\cup\left\{b\right\}$.

We should have $n(a)=-1, n(b)=1$.


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