Non Linear Eigenvalue problem

2018 7 24
2 个回答
更新时间：2025 1 8
15 次查看（30 天）

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I have been trying to solve a Non linear Eigenvalue problem using fsolve and Newton's iteration method and have not been successful. The matrix which I am looking to solve:
A=[2*w -300 0 0;sin(w/2) cos(w/2) -sin(w/2) cos(w/2); 2*cos(w/2) -2*sin(w/2) -cos(w/2) sin(w/2); 0 0 cos(w) -sin(w)]; (Found in a paper, using it as a practice case)
My Newton method code:
%Define intial values and tolerances for the variable
w0=0.1;
tol=2;
maxiter=1000;
w=w0;
wold=w0;
lambda=0.1;
%Start Iteration
for i=1:maxiter
%Define A and B
A=[2*w -300 0 0;sin(w/2) cos(w/2) -sin(w/2) cos(w/2); 2*cos(w/2) -2*sin(w/2) -cos(w/2) sin(w/2);
0 0 cos(w) -sin(w)];
B=[-2 0 0 0;-0.5*cos(w/2) 0.5*sin(w/2) 0.5*cos(w/2) -0.5*sin(w/2); sin(w/2) cos(w/2) -0.5*sin(w/2) -0.5*cos(w/2);
0 0 sin(w) cos(w)];
C=inv(B);
%Find Eigen value for the intermediate step
beta=eig(C*A);
epsilon=min(abs(beta));
%Update the variable
w=w0+epsilon;
err=abs(epsilon);
wold=w;
if(err<tol)
break;
end
end
Fsolve code
function fval=fun4evp(w)
A=[2*w -300 0 0;sin(w/2) cos(w/2) -sin(w/2) cos(w/2); 2*cos(w/2) -2*sin(w/2) -cos(w/2) sin(w/2);
0 0 cos(w) -sin(w)];
fval=det(A);
end
wsol=fsolve(@(w)fun4evp,0.1);
Thanks

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Gyan Swarup Nag
Gyan Swarup Nag 2019-3-16
clear all;
% One root at a time
% Number of iteration
k=0; % Number of iteration
ks=300;
epsi=1*0.01;
w0=-1
while 1e-14<abs(epsi)
k=k+1;
A=[2*w0 -ks 0 0;
sin(w0/2) cos(w0/2) -sin(w0/2) -cos(w0/2);
2*cos(w0/2) -2*sin(w0/2) -cos(w0/2) sin(w0/2);
0 0 cos(w0) -sin(w0)];
B=-[2 0 0 0;
cos(w0/2)/2 -sin(w0/2)/2 -cos(w0/2)/2 sin(w0/2)/2;
-sin(w0/2) -cos(w0/2) sin(w0/2)/2 cos(w0/2)/2;
0 0 -sin(w0) -cos(w0)];
e = eig(A, B);
[dum,jj]=min(abs(e));
epsi=e(jj)
w0=w0+epsi
end
[w0 k]

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回答（2 个）

Matt J
Matt J 2018-7-24

0 个投票

wsol=fsolve(@fun4evp,0.1),

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Matt J
Matt J 2018-7-24
As for "Newton's iterations", I don't see how
w=w0+min(abs(beta));
is a meaningful update. For one thing, it doesn't utilize wold. If you really meant to have
w=wold+min(abs(beta))
it still doesn't look right, because w will always be increasing with iterations. What if the algorithm needs to take a step backwards?
Abhishek Gaikwad
Abhishek Gaikwad 2018-7-24
编辑：Matt J 2018-7-24
!!!That's a bad mistake.Thanks. What about the Newton's method I am not sure about setting the tolerance. Currently, it is equal to the eigenvalue of C*A matrix which never goes to q very low value so i am able to solve the equation but the solution is incorrect.
Abhishek Gaikwad
Abhishek Gaikwad 2018-7-24
编辑：Abhishek Gaikwad 2018-7-24
The idea was to use the taylor series expansion and expand the matrix in the vicinity of w0 giving rise to epsilon which I found equal to the eigenvalue of the equation (A-lamba*B).

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Christine Tobler
Christine Tobler 2025-1-8
编辑：Christine Tobler 2025-1-8

0 个投票

Sorry for answering so long after the initial post. One option you could consider for solving a nonlinear eigenvalue problem is the Rational Krylov Toolbox for MATLAB: http://guettel.com/rktoolbox/examples/html/example_nlep.html.
(edit to fix the link)

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