# Difference between revisions of "Bim package"

Package for solving Diffusion Advection Reaction (DAR) Partial Differential Equations based on the Finite Volume Scharfetter-Gummel (FVSG) method a.k.a Box Integration Method (BIM).

## Tutorials

### 2D Diffusion Advection Reaction example

This is a short example on how to use bim to solve a 2D Diffusion Advection Reaction problem. The coplete code for this example can is on Agora at this link.

We want to solve the equation

${\displaystyle -\mathrm {div} \ (\varepsilon \ \nabla u(x,y)-\nabla \varphi (x,y)\ u(x,y)))+u(x,y)=1\qquad {\mbox{ in }}\Omega }$

${\displaystyle \varphi (x,y)\ =\ x+y}$

with mixed Dirichlet / Neumann boundary conditions

${\displaystyle u(x,y)=u_{d}(x,y)\qquad {\mbox{ on }}\Gamma _{D}}$

${\displaystyle -(\varepsilon \ \nabla u(x,y)-\nabla \varphi (x,y)\ u(x,y))\cdot \mathbf {n} =j_{N}(x,y)\qquad {\mbox{ on }}\Gamma _{N}}$

Create the mesh and precompute the mesh properties

To define the geometry of the domain we can use gmsh.

the following gmsh input

Point (1)  = {0, 0, 0, 0.1};
Point (2)  = {1, 1, 0, 0.1};
Point (3)  = {1, 0.9, 0, 0.1};
Point (4)  = {0, 0.1, 0, 0.1};
Point (5) = {0.3,0.1,-0,0.1};
Point (6) = {0.4,0.4,-0,0.1};
Point (7) = {0.5,0.6,0,0.1};
Point (8) = {0.6,0.9,0,0.1};
Point (9) = {0.8,0.8,0,0.1};
Point (10) = {0.2,0.2,-0,0.1};
Point (11) = {0.3,0.5,0,0.1};
Point (12) = {0.4,0.7,0,0.1};
Point (13) = {0.5,1,0,0.1};
Point (14) = {0.8,0.9,0,0.1};

Line (1)  = {3, 2};
Line (2) = {4, 1};

CatmullRom(3) = {1,5,6,7,8,9,3};
CatmullRom(4) = {4,10,11,12,13,14,2};
Line Loop(15) = {3,1,-4,2};
Plane Surface(16) = {15};

will produce the geometry below

we need to load the mesh into Octave and precompute mesh properties check out the tutorial for the msh package for info on the mesh structure

 Code: Meshing the 2D problem [mesh] = msh2m_gmsh ("fiume","scale",1,"clscale",.1); [mesh] = bim2c_mesh_properties (mesh);

to see the mesh you can use functions from the fpl package

 Code: Plot mesh in the 2D problem pdemesh (mesh.p, mesh.e, mesh.t) view (2)

Set the coefficients for the problem:

Get the node coordinates from the mesh structure

 Code: Get mesh coordinates in the 2D problem xu = mesh.p(1,:).'; yu = mesh.p(2,:).';

Get the number of elements and nodes in the mesh

 Code: Get number of elements in the 2D problem nelems = columns (mesh.t); nnodes = columns (mesh.p);
epsilon = .1;
phi     = xu + yu;

Construct the discretized operators

 Code: Discretized operators for the 2D problem AdvDiff = bim2a_advection_diffusion (mesh, epsilon, 1, 1, phi); Mass = bim2a_reaction (mesh, 1, 1); b = bim2a_rhs (mesh,f,g); A = AdvDiff + Mass;

To Apply Boundary Conditions, partition LHS and RHS

The tags of the sides are assigned by gmsh we let ${\displaystyle \Gamma _{D}}$ be composed by sides 1 and 2 and ${\displaystyle \Gamma _{D}}$ be the rest of the boundary

Solve for the displacements

 Code: Displacement in the 2D problem temp = [Ann Ani ; Ain Aii ] \ [ jn+bn-And*ud(GammaD) ; bi-Aid*ud(GammaD)]; u = ud; u(GammaN) = temp(1:numel (GammaN)); u(Omega) = temp(length(GammaN)+1:end);

Compute the fluxes through Dirichlet sides

Compute the gradient of the solution

 Code: Gradient of solution in the 2D problem [gx, gy] = bim2c_pde_gradient (mesh, u);

[jxglob, jyglob] = bim2c_global_flux (mesh, u, epsilon*ones(nelems, 1), ones(nnodes, 1), ones(nnodes, 1), phi);

Export data to VTK format

The resut can be exported to vtk format to visualize with [[1]] or [[2]]

fpl_vtk_write_field ("vtkdata", mesh, {u, "Solution"}, {[gx; gy]', "Gradient"}, 1);

you can also plot your data directly in Octave using pdesurf

pdesurf (mesh.p, mesh.t, u)

it will look like this

### 3D Time dependent problem

Here is an example of a 3D time-dependent Advection-Diffusion equation that uses lsode for time-stepping.

The equation being solved is

${\displaystyle {\frac {\partial u}{\partial t}}-\mathrm {div} \left(.01\nabla u-u\nabla \varphi \right)=0\qquad {\mbox{ in }}\Omega \times [0,T]=[0,1]^{3}\times [0,1]}$

${\displaystyle ~\varphi =x+y-z}$

${\displaystyle -\left(.01\nabla u-u\nabla \varphi \right)\cdot \mathbf {n} =0\qquad {\mbox{ on }}\partial \Omega }$

The initial condition is

${\displaystyle u=\exp(-\left({\frac {x-.2}{.2}}\right)^{2}-\left({\frac {y-.2}{.2}}\right)^{2}-\left({\frac {z-.2}{.2}}\right)^{2})}$

 Code: Define the 3D problem pkg load bim x = linspace (0, 1, 40); y = z = linspace (0, 1, 20); msh = bim3c_mesh_properties (msh3m_structured_mesh (x, y, z, 1, 1:6)); nn = columns (msh.p); ne = columns (msh.t); x = msh.p(1, :).'; y = msh.p(2, :).'; z = msh.p(3, :).'; x0 = .2; sx = .1; y0 = .2; sy = .1; z0 = .8; sz = .1; u = exp (- ((x-x0)/(2*sx)) .^2 - ((y-y0)/(2*sy)) .^2 - ((z-z0)/(2*sz)) .^2); A = bim3a_advection_diffusion (msh, .01*ones(ne, 1), 100*(x+y-z)); M = bim3a_reaction (msh, 1, 1); function du = f (u, t, A, M) du = - M \ (A * u); endfunction time = linspace (0, 1, 30); lsode_options ("integration method", "adams"); U = lsode (@(u, t) f(u, t, A, M), u, time); for ii = 1:1:numel (time) name = sprintf ("u_%3.3d", ii); delete ([name ".vtu"]); fpl_vtk_write_field (name, msh, {U(ii,:)', 'u'}, {}, 1); endfor

This is a video showing the .3 isosurface of the solution.