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I'm discretizing the following Poisson equation using FVM where the domain $\Omega$ of the solution is a regular hexagon of side $1$ centered about the origin. $$\Delta u =k,\text{ $k$ constant}\\ \partial_n u = 0\\ \Omega \subset \mathbb{R^2}$$

I'm not really quite sure if my discretization and especially the argument for existence and uniqueness are correct. So I would appreciate some feedback.

Let $V_i\subset \Omega$, $i=1,\dots,6$, be one of the six equilateral triangles comprising the hexagon; so $\bigcup\limits_{i=1}^6 V_i=\Omega$.

Then we obtain a system of six equations, that is a linear system after the following steps: $$\int_{V_i} \Delta udx = \int_{\partial V_i} \nabla u(x)\cdot n ds =\int_{V_i}kdx$$ $$\int_{partial V_i} \nabla u(x)\cdot nds=\int_{\partial V_i}\partial_n u ds$$ $$\approx \sum\limits_{j\in n_i}\frac{u_j-u_i}{h_{i,j}}l_{i,j}+\int_{\Gamma_i}0ds=\sum\limits_{j\in n_i} \frac{u_j-u_i}{h_{i,j}}l_{i,j}+\text{constant $C$}$$ $$=|V_i|k$$

Where $|V_i|$ means the area of $V_i$ and $n_i$ is the set of cells immediately neighboring $V_i$ (two such cells for a hexagon).

Now, we can use some geometry to deduce that

$$|V_i| = \frac{\sqrt{3}}{4}; h_{i,j} = \frac{\sqrt{3}}{2}; l_{i,j}=1$$

So the system becomes:

$$\frac{2}{\sqrt3}\sum_{j\in n_i} (u_j-u_i) + C = \frac{\sqrt3}{4}k$$

Which can be represented in matrix-vector form as

$$B\vec{u}=\vec{c}$$

where $B$ is a square $6\times 6$ matrix with the following components

$$b_{ij}=\frac{2}{\sqrt3}, i\ne j; b_{ii}=-\frac{2\sqrt2}{3}$$ $$c_i = \frac{\sqrt3}{4}k-C$$

Now, the "most interesting" part: is it enough to argue that there is a unique solution and it exists because $B$ is non-singular?

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    $\begingroup$ I don't think that your differential equation has a unique solution. If $u$ is a solution $u + c$, being $c$ a constant, is also a solution. $\endgroup$ – nicoguaro Mar 19 at 15:44
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    $\begingroup$ Integrating over whole domain $$ \int_\Omega k = \int_\Omega \Delta u = \int_{\partial\Omega} \frac{\partial u}{\partial n} = 0 $$ and if $k$ is constant, then we need $k=0$. $\endgroup$ – cpraveen Mar 19 at 16:53
  • $\begingroup$ @nicoguaro I understand that for the equation $\Delta u = k$ any solution $u+c$ is also a solution. But in this case the matrix $B$ can be derived explicitly, and it is non-singular. So, if we substitute a solution $u+c$ into the system we will not get the same vector $b$ on the RHS ($B(u+c) \ne b$ as opposed to $Bu=b$). Can you please clarify this peculiarity? $\endgroup$ – sequence Mar 24 at 21:38

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