How to find eigenvalues for a system of lenearized ordinary differential equations?
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I have a system of linearized ODEs with corresponding boundary conditions.
%----------------------------system of ODEs--------------------------------------%
y'(1)=y(2)
y'(2)=y(3)
y'(3)=(phi./Da).*y(2)+(2.*phi.*Fr./A1).*fd.*y(2)-(fd1.*1./A1).*y(3)-(fdd.*1./A1).*y(1)+(2.*fd.*1./A1).*y(2)-(e./A1).*y(2)-(phi.*Ra./(A1^2).*A2).*y(4)
y'(4)=y(5)
y'(5)=-(Pr./A2).*(fd.*y(5)+thd.*y(1)+e.*y(4))];
%---------------------------boundary conditions----------------------------------%
y(1)=y(2)=y(4)=0 at eta=0
y(2)=y(4)=0 at eta=0;
here Pr phi Ra Da Fr A1 A2 fd1 fd fdd thd are known quantities and 'e' is unknown.
I need to solve the system to find out the eigenvalues (e).
Thanks in advance.
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Pavel Osipov
2019년 10월 5일
Tanya, hi.
write so:
dyi/dt =...y1 (t)+...y2 (t)+...+y5(t);
let x (t)=[y1;y2;...;y5]; ->
((V/ve) x=Ah; A - matrix coeff. Your system. Let's formally denote d/dt=p
px-Ax=0; - > (p*E-A) x=0; since x is not 0, then
det(p*E-A)=0. This is the equation for the eigenvalues of p.
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Poly
2022년 3월 10일
Hello Tanya!! I don't know whether you get the code right or not? but can you share the code if possible
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Pavel Osipov
2019년 10월 9일
Hi!
det(p*E-A)=0. This is the equation for the eigenvalues of p. - -> The unkown "p" is solution det(p*E-A)=0. det - is the determinant with dimensions 5x5.
px-Ax=0 ->Ax=px, p is eigenvalues of A MATLAB command [V,D] = eig(A) returns diagonal matrix D of eigenvalues and matrix V whose columns are the corresponding right eigenvectors, so that A*V = V*D. (from MATLAB help).
eigenvalues p is 5x1 vector = liagonal elements D. eigenvectors of A see at columns V.
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