How do I find a D matrix that satisfies [K]*inv(M)​*[D]==[D]*​inv(M)*[K] ?

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ikram ros salamat
ikram ros salamat 2021 年 1 月 25 日
コメント済み: David Goodmanson 2021 年 1 月 26 日
M and K are respectively a mass and stiffness matrices 12x12, I need to find a matrix D 12x12 that satisfies this equality.
Thank you for your help
  4 件のコメント
Bruno Luong
Bruno Luong 2021 年 1 月 25 日
編集済み: Bruno Luong 2021 年 1 月 25 日
So you are saying the damping matrix can be entirely computed from K and M? Physically it doesn't seem right.
David Goodmanson
David Goodmanson 2021 年 1 月 26 日
Hi ikram,
I agree that the idea of D being the damping matrix on the basis of satisfying
K*inv(M)*D = D* inv(M)*K
does not make sense physically. For one thing I was remiss in not mentioning that if D satisfies the equation above, so does constant*D, which means if D were damping, then its overall size could be anything. Bruno pointed out what I missed, that w*X*inv(V) is also a solution for any diagonal X.
Could you provide a reference or expanation of what is meant by the first theorem of Rayleigh?

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David Goodmanson
David Goodmanson 2021 年 1 月 25 日
編集済み: David Goodmanson 2021 年 1 月 25 日
Hi Ikram,
Using N in place of inv(M) for simplicity, you are looking for a D such that
K*N*D = D*N*K
Let K*N and N*K have eigenvalue expansions
(K*N)*W = W*lambda
(N*K)*V = V*lambda
for matrices W and V and diagonal eigenvalue matrix lambda (although the code outputs lambda as a vector for convenience). It's the same lambda in both cases since the eigenvalues of K*N and N*K are identical. Matlab does not put eigenvalues in any particular order, so the code sorts the lambdas to make sure they are in the same order in each case. For simplicity I also assume no repeated eigenvalues because things are more complicated in that situation. It's easy to show that
D = W*inv(V)
is the solution.
n = 7;
K = rand(n,n);
N = rand(n,n);
[W lam] = eig(K*N,'vector');
[~,ind] = sort(lam);
W = W(:,ind);
[V lam] = eig(N*K,'vector');
[~,ind] = sort(lam);
V = V(:,ind);
D = real(W/V); % remove tiny imaginary part down in the numerical noise
K*N*D - D*N*K % check, should be small
  2 件のコメント
ikram ros salamat
ikram ros salamat 2021 年 1 月 25 日
thank you so much
Bruno Luong
Bruno Luong 2021 年 1 月 25 日
編集済み: Bruno Luong 2021 年 1 月 25 日
More generally
D = W*X*inv(V)
with X any diagonal matrix, is also a solution

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その他の回答 (1 件)

Bruno Luong
Bruno Luong 2021 年 1 月 25 日
編集済み: Bruno Luong 2021 年 1 月 25 日
You have homogeneous linear equation, the entire null space of operator (D considered as input)
K*inv(M)*D-D*inv(M)*K
(dimension n) contain all solutions:
K = rand(7);
M = rand(7);
I = eye(size(K));
A = K*inv(M);
B = inv(M)*K;
N = null(kron(I,A)-kron(B.',I)); % size(N,2) >= n
r = randn(size(N,2),1); % here we randomize the solution
D = reshape(N*r,size(K));
% Check, it should be small
norm(K*inv(M)*D-D*inv(M)*K)
Note that usually stiffnes and mass K and M are symmetric (definite positive). Thus B = A.'. Your equation is homegenuous Lyapunov equation

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