# Existence and uniquness of solution of FVM for Poisson equation

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?

• 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. – nicoguaro Mar 19 '19 at 15:44
• 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 $k=0$. – cfdlab Mar 19 '19 at 16:53
• @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? – sequence Mar 24 '19 at 21:38