Eigenvalues and Eigenmodes of Square

This example shows how to compute the eigenvalues and eigenmodes of a square domain.

The eigenvalue PDE problem is -Δu=λu. This example finds the eigenvalues smaller than 10 and the corresponding eigenmodes.

Create a model. Import and plot the geometry. The geometry description file for this problem is called squareg.m.

model = createpde();
geometryFromEdges(model,@squareg);

pdegplot(model,'EdgeLabels','on')
ylim([-1.5,1.5])
axis equal

Specify the Dirichlet boundary condition u=0 for the left boundary.

applyBoundaryCondition(model,'dirichlet','Edge',4,'u',0);

Specify the zero Neumann boundary condition for the upper and lower boundary.

applyBoundaryCondition(model,'neumann','Edge',[1,3],'g',0,'q',0);

Specify the generalized Neumann condition un-34u=0 for the right boundary.

applyBoundaryCondition(model,'neumann','Edge',2,'g',0,'q',-3/4);

The eigenvalue PDE coefficients for this problem are c = 1, a = 0, and d = 1. You can enter the eigenvalue range r as the vector [-Inf 10].

specifyCoefficients(model,'m',0,'d',1,'c',1,'a',0,'f',0);
r = [-Inf,10];

Create a mesh and solve the problem.

generateMesh(model,'Hmax',0.05);
results = solvepdeeig(model,r);
              Basis= 10,  Time=   0.19,  New conv eig=  0
              Basis= 11,  Time=   0.20,  New conv eig=  0
              Basis= 12,  Time=   0.20,  New conv eig=  1
              Basis= 13,  Time=   0.21,  New conv eig=  1
              Basis= 14,  Time=   0.21,  New conv eig=  1
              Basis= 15,  Time=   0.22,  New conv eig=  1
              Basis= 16,  Time=   0.22,  New conv eig=  1
              Basis= 17,  Time=   0.22,  New conv eig=  1
              Basis= 18,  Time=   0.23,  New conv eig=  2
              Basis= 19,  Time=   0.23,  New conv eig=  2
              Basis= 20,  Time=   0.24,  New conv eig=  2
              Basis= 21,  Time=   0.24,  New conv eig=  3
              Basis= 22,  Time=   0.25,  New conv eig=  3
              Basis= 23,  Time=   0.25,  New conv eig=  4
              Basis= 24,  Time=   0.25,  New conv eig=  6
End of sweep: Basis= 24,  Time=   0.25,  New conv eig=  3
              Basis= 13,  Time=   0.30,  New conv eig=  0
              Basis= 14,  Time=   0.31,  New conv eig=  0
              Basis= 15,  Time=   0.31,  New conv eig=  0
              Basis= 16,  Time=   0.32,  New conv eig=  0
              Basis= 17,  Time=   0.32,  New conv eig=  0
              Basis= 18,  Time=   0.33,  New conv eig=  0
              Basis= 19,  Time=   0.34,  New conv eig=  0
              Basis= 20,  Time=   0.34,  New conv eig=  0
              Basis= 21,  Time=   0.35,  New conv eig=  0
              Basis= 22,  Time=   0.36,  New conv eig=  1
              Basis= 23,  Time=   0.36,  New conv eig=  2
End of sweep: Basis= 23,  Time=   0.36,  New conv eig=  0
              Basis= 13,  Time=   0.41,  New conv eig=  1
End of sweep: Basis= 13,  Time=   0.41,  New conv eig=  1
              Basis= 14,  Time=   0.46,  New conv eig=  0
              Basis= 15,  Time=   0.46,  New conv eig=  0
              Basis= 16,  Time=   0.48,  New conv eig=  0
              Basis= 17,  Time=   0.48,  New conv eig=  0
              Basis= 18,  Time=   0.49,  New conv eig=  0
              Basis= 19,  Time=   0.50,  New conv eig=  0
              Basis= 20,  Time=   0.50,  New conv eig=  0
              Basis= 21,  Time=   0.51,  New conv eig=  0
              Basis= 22,  Time=   0.52,  New conv eig=  0
              Basis= 23,  Time=   0.52,  New conv eig=  1
End of sweep: Basis= 23,  Time=   0.52,  New conv eig=  0
              Basis= 14,  Time=   0.56,  New conv eig=  1
End of sweep: Basis= 14,  Time=   0.57,  New conv eig=  1
              Basis= 15,  Time=   0.61,  New conv eig=  0
              Basis= 16,  Time=   0.61,  New conv eig=  0
              Basis= 17,  Time=   0.62,  New conv eig=  0
End of sweep: Basis= 17,  Time=   0.62,  New conv eig=  0

There are six eigenvalues smaller than 10 for this problem.

l = results.Eigenvalues
l = 5×1

   -0.4146
    2.0528
    4.8019
    7.2693
    9.4550

Plot the first and last eigenfunctions in the specified range.

u = results.Eigenvectors;
pdeplot(model,'XYData',u(:,1));

pdeplot(model,'XYData',u(:,length(l)));

This problem is separable, meaning

u(x,y)=f(x)g(y).

The functions f and g are eigenfunctions in the x and y directions, respectively. In the x direction, the first eigenmode is a slowly increasing exponential function. The higher modes include sinusoids. In the y direction, the first eigenmode is a straight line (constant), the second is half a cosine, the third is a full cosine, the fourth is one and a half full cosines, etc. These eigenmodes in the y direction are associated with the eigenvalues

0,π24,4π24,9π24,...

It is possible to trace the preceding eigenvalues in the eigenvalues of the solution. Looking at a plot of the first eigenmode, you can see that it is made up of the first eigenmodes in the x and y directions. The second eigenmode is made up of the first eigenmode in the x direction and the second eigenmode in the y direction.

Look at the difference between the first and the second eigenvalue compared to π2/4:

l(2) - l(1) - pi^2/4
ans = 1.6752e-07

Likewise, the fifth eigenmode is made up of the first eigenmode in the x direction and the third eigenmode in the y direction. As expected, l(5)-l(1) is approximately equal to π2:

l(5) - l(1) - pi^2
ans = 6.2135e-06

You can explore higher modes by increasing the search range to include eigenvalues greater than 10.