Stability analysis of a time dependent Lyapunov equation

4 次查看（过去 30 天）
Anil
Anil 2024-2-29
评论： Anil 2024-3-2
I want solution of coupled linear equations for which I am using time dependnet Lyapunov equation:
dP/dt=A*P + P*A'+ D,
where A (A' is transpose) is a system matrix, P is covariance matrix to be found and D is diffusion matrix. I solve it staright-forward as coupled equations extracted from a matrix in terms of a col. vector but the solution blows up i.e. "instabilities", even though the real eignevalues of matrix A are negative. Whie seraching literatures for this, an options is given, using vectorization: A*P+P*A'=(A⊗I+I⊗A)*P,. using this, giving me an error of incompatibel sizes. I am not sure if this vectorization is enough to make sure the stability of the system but I dont know other options, please suggets if any. For that vectorization, I am using a function:
% showing just a stucture of the function file
function dpdt=(t,p0)
dpdt=zero(m,1)% m is number of rows
[m,n]=size(A);
P1=reshape(P0, m,1);
D1=reshape(D,m,1);% D is some matrix
b=eye(m);
a=real(eig(A))
if all(a<0)
dprdt=(tensorprod(A.',b)+tensorprod(b,A.')).*P1 + D1;% error appears here;
dpdt(1)=dprdt(1);
.
.
.
.
dpdt(m)=dprdt(m)
end
end
is the above right way of doing the vectorization? Any other suggestions for the stability of this time depedent Lyapunov equation, I shall be grateful. Thanks.

请先登录，再回答此问题。

回答（1 个）

Christine Tobler
Christine Tobler 2024-2-29
The tensorprod function doesn't do what you are looking for here, that would be kron (short for Kronecker product).
It also seems like P being a column vector of size mx1 would have the wrong size, are you looking for it to be a m x m matrix that you reshape to be m*m x 1?
In terms of performance, you should expect A*P + P*A' (and then possibly reshaping the result of this) to be much more efficient than forming the (A' kron I + I kron A') matrix explicitly.
  3 个评论
显示 1更早的评论
Christine Tobler
Christine Tobler 2024-3-1
The equation A*B + P*A.' implies that P is an m x m matrix. That means that if it is reshaped, it must still have m^2 elements, so mx1 would not be a possible result of that reshaping.
You might start out to check your code is doing the right thing by using diagonal matrices A - then eigenvalues being negative should be enough. For non-symmetric A, it's possible for the values to grow before starting to stabilize.
Anil
Anil 2024-3-2
Dear @Christine Tobler, Yes, that's true, reshaped P is m*m x1. I missunderstood that. Thanks for the later part, I am checking it.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Computations 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by