Finding all the eigen values through bvp4c

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Gaurav Singh
Gaurav Singh 2022-11-17
回答： Saarthak Gupta 2023-9-7
Ran in:
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.
  2 个评论
Torsten
Torsten 2022-11-17
Why is k*cos(k*(-pi)) = 1 as you claim in your boundary condition ya(2) - 1 = 0 ?
Gaurav Singh
Gaurav Singh 2022-11-17
As Lamdda is a parameter, so by using this BC, I am fixing the amplitude.

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回答（1 个）

Saarthak Gupta
Saarthak Gupta 2023-9-7
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:

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