### System equivalent to 5 point finite difference stencil

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What kind of element and cell type for the finite element method in Fenics with Dirichlet boundary condition leads to an equivalent system of 5 point stencil with the Finite difference method for the Poisson equation ?

Consider the code below:

The MIT matlab code here shows the possibility of obtaining such a matrix using Finite element methods. I would like to get this with Fenics. If I can obtain the same matrix with any of Fenics' capabilities as in the example MIT MATLAB code, that would be great as well.

Thank you.

Consider the code below:

```
import numpy as np
from fenics import *
mesh = UnitSquareMesh.create(3, 3,CellType.Type_Option1)
V = FunctionSpace(mesh, 'Option 2', 1)
u = TrialFunction(V)
v = TestFunction(V)
a = (inner(grad(u), grad(v)))*dx
matA = assemble(a)
```

with the different parameters(Option1 and Option 2) available as in Functionspace to get the same stiffness matrix as for the 5 point stencil using the finite difference method ? The MIT matlab code here shows the possibility of obtaining such a matrix using Finite element methods. I would like to get this with Fenics. If I can obtain the same matrix with any of Fenics' capabilities as in the example MIT MATLAB code, that would be great as well.

Thank you.

Community: FEniCS Project

### 1 Answer

6

The short answer is that to get the standard 5-point finite difference mesh for the Laplacian with Dirichlet boundary conditions by assembling a finite element in FEniCS, you can use

```
from fenics import *
n = 3
mesh = UnitSquareMesh(n, n)
V = FunctionSpace(mesh, 'Lagrange', 1)
u = TrialFunction(V)
v = TestFunction(V)
a = -inner(grad(u), grad(v))*dx
bc = DirichletBC(V, 0., DomainBoundary())
A = assemble(a)
bc.apply(A)
print(A.array())
```

The long answer is a bit more complicated. This uses a mesh of (n+1)^2 = 16 vertices, so the FE matrix will be 16 x 16. (n-1)^2 = 4 of these vertices are interior to the unit square, and the other 12 are on the boundary. The standard FD matrix is only 4 x 4 in this case: the values at the boundary mesh points are not considered as unknowns. In the FE matrix, 12 rows will be all zeros except for a one on the diagonal. These rows correspond to the boundary nodes and simply impose the boundary condition. If you cross out these rows and also the corresponding columns, you will get the FD matrix. Well, except for one thing: you may have to reorder the vertices (apply a permutation to the rows and columns) in order to make the FE matrix identical to the FD matrix.
Great, Thanks! That makes a lot of sense.

written
5 months ago by
dommath

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