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I have an equation which is used for the finite volume method in partial differential equations. I can read it but I had to few vector analysis to really get it.

The equation is:

\begin{equation} \oint_{\partial \Gamma} \operatorname{grad} u(x, y) d \vec{n}=\int_{\Gamma} f(x, y) d x d y \end{equation}

It is used to calculate the boarders for the Voronoi method. The right hand side is clear to me. I have a minor insecurity depending the region $\Gamma$ but so far my solution to this part has been correct.

The left hand side troubles me. I have some speculations about it, which I tried to verify on to exercises. But it looks like I still haven't understood it correctly.

The first example goes like this:

enter image description here

I get the -8/3. Regarding my insecurity about the region $\Gamma$: Do I assume correctly that it from 0 to 1 because $\Omega$ gets divided into four regions with length 1x1?

However, the solution looks like this: \begin{equation} \left[\begin{array}{rrrr}{-4} & {1} & {} & {1} \\ {1} & {-4} & {1} & {} \\ {} & {1} & {-4} & {1} \\ {1} & {} & {1} & {-4}\end{array}\right] \cdot\left(\begin{array}{c}{\tilde{u}_{1}} \\ {\tilde{u}_{2}} \\ {\tilde{u}_{3}} \\ {\tilde{u}_{4}}\end{array}\right)=\left(\begin{array}{c}{-16 / 9} \\ {-16 / 9} \\ {-16 / 9} \\ {-16 / 9}\end{array}\right) \end{equation}

Because of the discretization there is a factor of $2$ included to get b. Here my problems of understanding the LHS begins. Why $2$?

The second example is this one: enter image description here

I got $\frac{1}{2}$ for the RHS (by integrating again from 0 to 1 in both directions ... (?) ). Based on the boundary conditions I got a b-vector $ \left(\begin{array}{c}{1} \\ {3} \\ {5} \\ {11}\end{array}\right) $ by simpling inserting a cells boundary middle-point into the boundary conditions, and adding adjacent boundaries for each cell. E.g. $[0.5 \quad 1]$ in $x-2$ => $-1.5$.

Yet I haven't used the integral from the RHS. There is again a factor of $2$ between my $b$ and the solutions $b$. Somewhere in there the $\frac{1}{2}$ and some factor regarding the discretization from the LHS should be multiplied.

My problem is that I don't get this Integral by Gauss-Theorem in this discrete case. Integrating / adding the gradient over the boundary regarding the normal vector should be the same as the RHS. Therefor I adjust the gradient somehow?

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