Finding all the eigen values through bvp4c

Dear members,

I am trying to find all the eigen values of y''+lambda*y=0

I know that y=sin(kx) with lambda=k^2, k=1,2,3.. are the solutions.

When I try to find this solution numerically, I miss some of the roots. Is there a way to get all the solutions

Here is the code:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
lambda = 0;
k= 1;
hold on
for i=1:5
    lambda=lambda+1;
    k=k+1;
    solinit = bvpinit(linspace(-pi,pi,20),@matinit,k,lambda);
    sol = bvp4c(@matode, @matbc, solinit);
    fprintf(' eigenvalue is approximately %7.3f.\n',...
        sol.parameters)
    xint = linspace(-pi,pi);
    Sxint = deval(sol,xint);
    plot(xint,Sxint(1,:))
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end

 eigenvalue is approximately   1.000.
 eigenvalue is approximately   4.000.
 eigenvalue is approximately   4.000.
 eigenvalue is approximately  16.001.
 eigenvalue is approximately   4.000.

hold off

    function yinit = matinit(x,k) % initial guess function
    yinit = [sin(k*x)
        k*cos(k*x)];
    end
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    function res = matbc(ya,yb,lambda) % boundary conditions
    res = [ya(1)
        yb(1)
        ya(2)-1];
    end
    %%%%%%%%%%%%%%%%
    function dydx = matode(x,y,lambda) % equation being solved
    dydx = [y(2)
        -(lambda)*y(1)];
    end

%%Output:

eigenvalue is approximately 1.000.

eigenvalue is approximately 4.000.

eigenvalue is approximately 4.000. ** This is repeated I expect 9 here

eigenvalue is approximately 16.001.

eigenvalue is approximately 4.000. ** I expect 25 here.

PS: If I change my initial guess to more points I see more roots appearing but then again I miss some of them.

[Torsten]

Why is k*cos(k*(-pi)) = 1 as you claim in your boundary condition ya(2) - 1 = 0 ?

[Gaurav Singh]

As Lamdda is a parameter, so by using this BC, I am fixing the amplitude.

Answer 1

[Saarthak Gupta]

Hi Gaurav,

I understand you are trying to find all eigenvalues of the second order differential equation y” + lambda*y = 0 using ‘bvp4c’.

The ‘bvp4c’ code is for general BVPs, so all it can do is compute the eigenvalue closest to a guess. This BVP can be solved with a constant guess for the eigenfunction, but we can make it much more likely that we compute the desired eigenvalue by supplying a guess for the eigenfunction that has the correct qualitative behaviour.

You can make the following modifications to the code to achieve the desired result:

1. Parametrize the ‘matinit’ function w.r.t. k, and supply the function handle in the call to ‘bvpinit’. Please refer to the MATLAB documentation for Parameterizing Functions for more details: https://in.mathworks.com/help/matlab/math/parameterizing-functions.html

2. Give the initial guess for lambda as k^2, to increase the likelihood of getting the desired eigenvalue (as per the analytical solution).

The following code computes all eigenvalues for the BVP:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
lambda = 0;
k= 1;
hold on
for i=1:5
    lambda=lambda+1;
    k=k+1;
    matinit = @(x) [sin(k*x) k*cos(k*x)];
    solinit = bvpinit(linspace(-pi,pi,20),matinit,lambda.^2);
    sol = bvp4c(@matode, @matbc, solinit);
    fprintf(' eigenvalue is approximately %7.3f.\n',...
        sol.parameters)
    xint = linspace(-pi,pi);
    Sxint = deval(sol,xint);
    plot(xint,Sxint(1,:))
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
hold off
% function yinit = matinit(x,k) % initial guess function
% yinit = [sin(k*x)
%     k*cos(k*x)];
% end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function res = matbc(ya,yb,lambda) % boundary conditions
res = [ya(1)
    yb(1)
    ya(2)-1];
end
%%%%%%%%%%%%%%%%
function dydx = matode(x,y,lambda) % equation being solved
dydx = [y(2)
    -(lambda)*y(1)];
end

Please refer to the following MATLAB documentation for more details:

• https://in.mathworks.com/help/matlab/math/solve-bvp-with-unknown-parameter.html