de-CH
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math math-format
Determine the Stationary State in the Inhomogeneous Case
linsys-03-01
multiple
65536
randRangeNonZero(-8,8) randRangeNonZero(-8,8) randRangeNonZero(-8,8) randRangeNonZero(-8,8)
A*D-B*C randRangeNonZero(-8,8) randRangeNonZero(-8,8) fractionReduce(-D*X+B*Y,det) fractionReduce(-A*Y+C*X,det)

The matrix A= \begin{pmatrix} A & B \\ C & D \end{pmatrix} and the vector \begin{pmatrix}X \\ Y \end{pmatrix} define y' = A \cdot y + \begin{pmatrix}X \\ Y \end{pmatrix}.

Determine {\color{red}X} and {\color{blue}Y} such that {\color{orange}y_{\infty}} = \begin{pmatrix} {\color{red}X} \\ {\color{blue}Y} \end{pmatrix} is a stationary state of the inhomogeneous system.

a \color{red} X = (-D*X+B*Y)/det
b \color{blue} Y = (-A*Y+C*X)/det

We are looking for a solution \color{orange}y_{\infty} with {\color{orange}y'_{\infty}} = 0 = A \cdot {\color{orange}y_{\infty}} + \begin{pmatrix}X \\ Y \end{pmatrix}.

Since A= \begin{pmatrix} A & B \\ C & D \end{pmatrix} is invertible, with \det(A) = det and A^{-1}= \frac 1{det}\begin{pmatrix} D & -B \\ -C & A \end{pmatrix}, we can solve the equation 0 = A \cdot {\color{orange}y_{\infty}} + \begin{pmatrix}X \\ Y \end{pmatrix} for \color{orange}y_{\infty} .

It follows that - \begin{pmatrix}X \\ Y \end{pmatrix}= A \cdot {\color{orange}y_{\infty}} and for multiplication with A^{-1} from the left A^{-1} \left(- \begin{pmatrix}X \\ Y \end{pmatrix}\right)= {\color{orange}y_{\infty}} .

Substituting this yields {\color{orange}y_{\infty}} = \begin{pmatrix} {\color{red}SX} \\ {\color{blue}SY} \end{pmatrix} .