User:Margav06/sandbox/Click here to continue/LMIs in system and stability Theory/Positive Orthant Stabilizability

Positive Orthant Stabilizability

The positive orthant stability of a linear system refers to the property of the system states being real and positive for all $$ t \geq 0 $$ and decaying down to zero over time. In this section, the feasibility problem for systems to be positive orthant stable, and the stabilizability conditions to make the system positive orthant stable will be covered.

The System
Consider a linear state-space representation of a system as:



\begin{align} \dot x(t)&=Ax(t)+Bu(t) \end{align}$$

where $$x(t)\in \R^n$$ and $$u(t)\in \R^r$$ are the system state and the input vector respectively. A and B are system coefficient matrices of appropriate dimensions.

The Data
Number of states n and number of control inputs r need to be known. Moreover, the system matrices A,B are also required to be known.

The Feasibility LMI
An LTI system is positive orthant stable if $$ x(0) \in \R^n_+ $$ implies that $$ \forall t \geq 0, x(t) \in \R^n_+ $$. Moreover, as $$t \rightarrow \infty$$, $$ x(t) \rightarrow 0 $$. This is possible if and only if the following conditions hold:



\begin{align} A_{ij} & \geq 0, \forall i \neq j, \\ \exists P & > 0 \text{ s.t. } PA^\top + AP < 0, \end{align} $$

The above LMI feasibility is the positive orthant stability criteria. To convert it into a positive orthant stabilizability check, the problem can be modified so that we check if $$ \dot{x}=(A+BK)x $$ is positive orthant stable. As $$ K $$ is also a design variable here, the second inequality in the above LMI will result in bilinearity. A simple change of variables can overcome that to result in the following LMI feasibility problem for checking positive orthant stabilizability of the LTI system:



\begin{align} \text{Find } Q,Y & \text{ subj. to:} \\ & Q > 0, (AQ+BY)_{ij} \geq 0, \forall i \neq j, \\ & \text{ s.t. } QA^\top + AQ + BY + Y^\top B^\top < 0, \end{align} $$

If the above LMI is feasible, the LTI system is stabilizable with controller $$ K = YQ^{-1} $$.

Conclusion:
The feasibility of the above LMIs guarantees that the system is positive orthant stable if the first LMI is feasible or stabilizable with a controller if the second LMI holds.

Implementation
To solve the feasibility LMI, YALMIP toolbox is required for setting up the feasibility problem, and SeDuMi is required to solve the problem. The following link showcases an example of the feasibility problem:

https://github.com/smhassaan/LMI-Examples/blob/master/Positive_Orthant_LMI.m