# Difference between revisions of "Bim package"

## 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

[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

pdemesh (mesh.p, mesh.e, mesh.t)
view (2)


Set the coefficients for the problem:

Get the node coordinates from the mesh structure

xu     = mesh.p(1,:).';
yu     = mesh.p(2,:).';


Get the number of elements and nodes in the mesh

nelems = columns (mesh.t);
nnodes = columns (mesh.p);

epsilon = .1;
phi     = xu + yu;


Construct the discretized operators

AdvDiff = bim2a_advection_diffusion (mesh, epsilon, 1, 1, phi);
Mass    = bim2a_reaction (mesh, 1, 1);
b       = bim2a_rhs (mesh,f,g);


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

GammaD = bim2c_unknowns_on_side (mesh, [1 2]); 	    ## DIRICHLET NODES LIST
GammaN = bim2c_unknowns_on_side (mesh, [3 4]); 	    ## NEUMANN NODES LIST

jn    = zeros (length (GammaN),1);           	    ## PRESCRIBED NEUMANN FLUXES
ud    = 3*xu;                                       ## DIRICHLET DATUM
Omega = setdiff (1:nnodes, union (GammaD, GammaN)); ## INTERIOR NODES LIST


Add = A(GammaD, GammaD);

And = A(GammaN, GammaD); ## shoud be all zeros hopefully!!
Ann = A(GammaN, GammaN);
Ani = A(GammaN, Omega);

Ain = A(Omega, GammaN);
Aii = A(Omega, Omega);

bn = b(GammaN);
bi = b(Omega);


Solve for the displacements

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

jd = [Add Adi Adn] * u([GammaD; Omega; GammaN]) - bd;


Compute the gradient of the solution

[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

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

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);