Stability criteria for linear systems with auxiliary variables

Question

Classical texts for control theory show the linear system $\dot x=A \,x$, is stable if the real parts of the eigenvalues are negative.

Does the same criteria apply for a system of the following form: $$ \left[ \begin{array}{cc|c} \dot x\\ 0 \end{array} \right] = B \left[ \begin{array}{cc|c} x \\ y \end{array} \right]$$

where $\dot x$, $x$, $y$ are vectors and $B$ is a matrix of constants. In this system, the equations for $\dot x$ include terms from $y$. For this system the number of unknowns equals the number of equations. While it would be possible to perform additional algebra to reduce the system to the classical form, $\dot x=A\,x$, is this necessary? I would prefer to write the equations in the form above, because this makes the physical interpretation of the equations more clear.

I am calling the $y$ values "auxiliary" variables, because they are dynamic in the sense that they change with time (as a consequence of the linear system - the values $y$ do not have an explicit dependence on time), but an expression for their derivative does not fall out of the analysis. (In this system, the y equations results from a simple energy balance where no energy "hold-up" is assumed.) If there is a more appropriate description feel free to revise the question.

Because $\dot y$ does not appear on the left hand side, it is not clear to me if the classic stability test still applies.

Answer 1

You can infer some necessary but not sufficient conditions for stability based on the eigenvalues of B. However you are better of verifying that

$ Re[eig(B_{xx} - B_{xy} B_{yy}^{-1} B_{yx})] < 0 $

for $ B = \begin{bmatrix} B_{xx} & B_{xy} \\ B_{yx} & B_{yy} \end{bmatrix} $.

Answer 2

Since $y$ changes with time, in general the stability of $x'=Ax$ won't agree with the stability of your equation. In order to construct examples, simply take $y$ so that they cancel some terms with $x$ so that the real part of some eigenvalue changes. In other words, you really need to write everything with a single variable (but your equation will then be nonautonomous and so more complications will arise).