Set Up and Solve the PDE

Define a function for the nonlinear f coefficient in the syntax as given in f Coefficient for specifyCoefficients.

function bcMatrix = myfffun(region,state)

bcMatrix = 1+10*region.z.^2+region.y;

Import the geometry and examine the face labels.

model = createpde;
g = importGeometry(model,"Plate10x10x1.stl");
pdegplot(g,"FaceLabels","on","FaceAlpha",0.5)

The faces are numbered 1 through 6.

Create the coefficients and boundary conditions.

c = 1;
a = 0;
d = 1;
f = @myfffun;
specifyCoefficients(model,"m",0,"d",d,"c",c,"a",a,"f",f);

applyBoundaryCondition(model,"dirichlet","Face",1:6,"u",0);

Set a zero initial condition.

setInitialConditions(model,0);

Create a mesh with sides no longer than 0.3.

generateMesh(model,"Hmax",0.3);

Set times from 0 through 0.2 and solve the PDE.

tlist = 0:0.02:0.2;
results = solvepde(model,tlist);

Plot Slices Through the Solution

Create a grid of (x,y,z) points, where x = 5, y ranges from 0 through 10, and z ranges from 0 through 1. Interpolate the solution to these grid points and all times.

yy = 0:0.5:10;
zz = 0:0.25:1;
[YY,ZZ] = meshgrid(yy,zz);
XX = 5*ones(size(YY));
uintrp = interpolateSolution(results,XX,YY,ZZ,1:length(tlist));

The solution matrix uintrp has 11 columns, one for each time in tlist. Take the interpolated solution for the second column, which corresponds to time 0.02.

usol = uintrp(:,2);

The elements of usol come from interpolating the solution to the XX, YY, and ZZ matrices, which are each 5-by-21, corresponding to z-by-y variables. Reshape usol to the same 5-by-21 size, and make a surface plot of the solution. Also make surface plots corresponding to times 0.06, 0.10, and 0.20.

figure
usol = reshape(usol,size(XX));
subplot(2,2,1)
surf(usol)
title("t = 0.02")
zlim([0,1.5])
xlim([1,21])
ylim([1,5])

usol = uintrp(:,4);
usol = reshape(usol,size(XX));
subplot(2,2,2)
surf(usol)
title("t = 0.06")
zlim([0,1.5])
xlim([1,21])
ylim([1,5])

usol = uintrp(:,6);
usol = reshape(usol,size(XX));
subplot(2,2,3)
surf(usol)
title("t = 0.10")
zlim([0,1.5])
xlim([1,21])
ylim([1,5])

usol = uintrp(:,11);
usol = reshape(usol,size(XX));
subplot(2,2,4)
surf(usol)
title("t = 0.20")
zlim([0,1.5])
xlim([1,21])
ylim([1,5])

Set Up and Solve the PDE

The problem is to solve Poisson's equation with zero Dirichlet boundary conditions for a complicated geometry. Poisson's equation is

Partial Differential Equation Toolbox™ solves equations in the form

So you can represent the problem by setting $$c=1$$ and $$a=0$$. Arbitrarily set $$f=10$$.

c = 1;
a = 0;
f = 10;

The first step in solving any 3-D PDE problem is to create a PDE Model. This is a container that holds the number of equations, geometry, mesh, and boundary conditions for your PDE. Create the model, then import the "ForearmLink.stl" file and view the geometry.

N = 1;
model = createpde(N);
importGeometry(model,"ForearmLink.stl");
pdegplot(model,"FaceAlpha",0.5)
view(-42,24)

Specify PDE Coefficients

Include the PDE coefficients in model.

specifyCoefficients(model,"m",0,"d",0,"c",c,"a",a,"f",f);

Create zero Dirichlet boundary conditions on all faces.

applyBoundaryCondition(model,"dirichlet", ...
                             "Face",1:model.Geometry.NumFaces, ...
                             "u",0);

Create a mesh and solve the PDE.

generateMesh(model);
result = solvepde(model);

Plot the Solution as Contour Slices

Because the boundary conditions are $$u=0$$ on all faces, the solution $$u$$ is nonzero only in the interior. To examine the interior, take a rectangular grid that covers the geometry with a spacing of one unit in each coordinate direction.

[X,Y,Z] = meshgrid(0:135,0:35,0:61);

For plotting and analysis, create a PDEResults object from the solution. Interpolate the result at every grid point.

V = interpolateSolution(result,X,Y,Z);
V = reshape(V,size(X));

Plot contour slices for various values of $$z$$.

figure
colormap jet
contourslice(X,Y,Z,V,[],[],0:5:60)
xlabel("x")
ylabel("y")
zlabel("z")
colorbar
view(-11,14)
axis equal

Plot contour slices for various values of $$y$$.

figure
colormap jet
contourslice(X,Y,Z,V,[],1:6:31,[])
xlabel("x")
ylabel("y")
zlabel("z")
colorbar
view(-62,34)
axis equal

Save Memory by Evaluating As Needed

For large problems you can run out of memory when creating a fine 3-D grid. Furthermore, it can be time-consuming to evaluate the solution on a full grid. To save memory and time, evaluate only at the points you plot. You can also use this technique to interpolate to tilted grids, or to other surfaces.

For example, interpolate the solution to a grid on the tilted plane $$0\le x\le 135$$, $$0\le y\le 35$$, and $$z=x/10+y/2$$. Plot both contours and colored surface data. Use a fine grid, with spacing 0.2.

[X,Y] = meshgrid(0:0.2:135,0:0.2:35);
Z = X/10 + Y/2;
V = interpolateSolution(result,X,Y,Z);
V = reshape(V,size(X));
figure
subplot(2,1,1)
contour(X,Y,V);
axis equal
title("Contour Plot on Tilted Plane")
xlabel("x")
ylabel("y")
colorbar
subplot(2,1,2)
surf(X,Y,V,"LineStyle","none");
axis equal
view(0,90)
title("Colored Plot on Tilted Plane")
xlabel("x")
ylabel("y")
colorbar

Import and View the Geometry

model = createpde;
importGeometry(model,"Block.stl");
pdegplot(model,"FaceLabels","on","FaceAlpha",0.5)
view(-42,24)

Set Boundary Conditions

Set faces 1, 2, and 5 to be the places where the particle can escape. On these faces, the solution $$u=0$$. Keep the default reflecting boundary conditions on faces 3, 4, and 6.

applyBoundaryCondition(model,"dirichlet","Face",[1,2,5],"u",0);

Create PDE Coefficients

The PDE is

In Partial Differential Equation Toolbox™ syntax,

This equation translates to coefficients c = 1, a = 0, and f = 2. Enter the coefficients.

c = 1;
a = 0;
f = 2;
specifyCoefficients(model,"m",0,"d",0,"c",c,"a",a,"f",f);

Create Mesh and Solve PDE

Initialize the mesh.

generateMesh(model);

Solve the PDE.

results = solvepde(model);

Examine the Solution in a Contour Slice Plot

Create a grid and interpolate the solution to the grid.

[X,Y,Z] = meshgrid(0:135,0:35,0:61);
V = interpolateSolution(results,X,Y,Z);
V = reshape(V,size(X));

Create a contour slice plot for five fixed values of the y-coordinate.

figure
colormap jet
contourslice(X,Y,Z,V,[],0:4:16,[])
xlabel("x")
ylabel("y")
zlabel("z")
xlim([0,100])
ylim([0,20])
zlim([0,50])
axis equal
view(-50,22)
colorbar

The particle has the largest mean exit time near the point $$(x,y,z)=(100,0,0)$$.

Use Gradients for Quiver and Streamline Plots

Examine the solution in more detail by evaluating the gradient of the solution. Use a rather coarse mesh so that you can see the details on the quiver and streamline plots.

[X,Y,Z] = meshgrid(1:9:99,1:3:20,1:6:50);
[gradx,grady,gradz] = evaluateGradient(results,X,Y,Z);

Plot the gradient vectors. First reshape the approximate gradients to the shape of the mesh.

gradx = reshape(gradx,size(X));
grady = reshape(grady,size(Y));
gradz = reshape(gradz,size(Z));

figure
quiver3(X,Y,Z,gradx,grady,gradz)
axis equal
xlabel("x")
ylabel("y")
zlabel("z")
title("Quiver Plot of Estimated Gradient of Solution")

Plot the streamlines of the approximate gradient. Start the streamlines from a sparser set of initial points.

hold on
[sx,sy,sz] = meshgrid([1,46],1:6:20,1:12:50);
streamline(X,Y,Z,gradx,grady,gradz,sx,sy,sz)
title("Quiver Plot with Streamlines")
hold off

The streamlines show that small values of y and z give larger mean exit times. They also show that the x-coordinate has a significant effect on u when x is small, but when x is greater than 40, the larger values have little effect on u. Similarly, when z is less than 20, its values have little effect on u.