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Solve PDE with Coefficients in Functional Form

This example shows how to write PDE coefficients in string form and in functional form for 2-D geometry.

Geometry

The geometry is a rectangle with a circular hole. Create a PDE model container, and incorporate the geometry into the container.

model = createpde(1);

% Rectangle is code 3, 4 sides,
% followed by x-coordinates and then y-coordinates
R1 = [3,4,-1,1,1,-1,-.4,-.4,.4,.4]';
% Circle is code 1, center (.5,0), radius .2
C1 = [1,.5,0,.2]';
% Pad C1 with zeros to enable concatenation with R1
C1 = [C1;zeros(length(R1)-length(C1),1)];
geom = [R1,C1];
% Names for the two geometric objects
ns = (char('R1','C1'))';
% Set formula
sf = 'R1-C1';
% Create geometry
gd = decsg(geom,sf,ns);

% Include the geometry in the model
geometryFromEdges(model,gd);
% View geometry
pdegplot(model,'EdgeLabels','on')
xlim([-1.1 1.1])
axis equal

PDE Coefficients

The PDE is parabolic,

dut(cu)+au=f,

with the following coefficients:

  • d = 5

  • a = 0

  • f is a linear ramp up to 10, holds at 10, then ramps back down to 0:

    f=10*{10t0t0.110.1t0.91010t0.9t1

  • c = 1 +.x2 + y2

Write a function for the f coefficient.

function f = framp(t)

if t <= 0.1
    f = 10*t;
elseif t <= 0.9
    f = 1;
else
    f = 10-10*t;
end
f = 10*f;

Boundary conditions

The boundary conditions on the outer boundary (segments 1 through 4) are Dirichlet, with the value u(x,y) = t(x – y), where t is time. Suppose the circular boundary (segments 5 through 8) has a generalized Neumann condition, with q = 1 and g = x2 + y2.

myufun = @(region,state)state.time*(region.x - region.y);
mygfun = @(region,state)(region.x.^2 + region.y.^2);
applyBoundaryCondition(model,'Edge',1:4,'u',myufun,'Vectorized','on');
applyBoundaryCondition(model,'Edge',5:8,'q',1,'g',mygfun,'Vectorized','on');

The boundary conditions are the same as in Boundary Conditions for Scalar PDE. That description uses the older function form for specifying boundary conditions, which is no longer recommended. This description uses the recommended object form.

Initial Conditions

The initial condition is u(x,y) = 0 at t = 0.

u0 = 0;

Mesh

Create the mesh.

generateMesh(model);

Parabolic Solution Times

Set the time steps for the parabolic solver to 50 steps from time 0 to time 1.

tlist = linspace(0,1,50);

Solution

Solve the parabolic PDE.

d = 5;
a = 0;
f = 'framp(t)';
c = '1+x.^2+y.^2';
u = parabolic(u0,tlist,model,c,a,f,d);

View an animation of the solution.

for tt = 1:size(u,2) % number of steps
    pdeplot(model,'xydata',u(:,tt),'zdata',u(:,tt),'colormap','jet')
    axis([-1 1 -1/2 1/2 -1.5 1.5 -1.5 1.5]) % use fixed axis
    title(['Step ' num2str(tt)])
    view(-45,22)
    drawnow
    pause(.1)
end

Alternative Coefficient Syntax

Equivalently, you can write a function for the coefficient f in the syntax described in Specify 2-D Scalar Coefficients in Function Form.

function f = framp2(p,t,u,time)

if time <= 0.1
    f = 10*time;
elseif time <= 0.9
    f = 1;
else
    f = 10-10*time;
end
f = 10*f;

Call this function by setting

f = @framp2;
u = parabolic(u0,tlist,model,c,a,f,d);

You can also write a function for the coefficient c, though it is more complicated than the string formulation.

function c = cfunc(p,t,u,time)

% Triangle point indices
it1 = t(1,:);
it2 = t(2,:);
it3 = t(3,:);

% Find centroids of triangles
xpts = (p(1,it1)+p(1,it2)+p(1,it3))/3;
ypts = (p(2,it1)+p(2,it2)+p(2,it3))/3;

c = 1 + xpts.^2 + ypts.^2;

Call this function by setting

c = @cfunc;
u = parabolic(u0,tlist,model,c,a,f,d);

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