Find all stable dynamical systems in 2 variables

Question

- To set jargon: The set of all $2\times2$ real matrices is $\mathbb{R}^4$. The 4 matrix elements fully parametrise this set: if we see this set as a manifold, then the 4 parameters are coordinates (we can always change coordinates and find another equivalent parametrisation of the matrix set).

- Motivation: I was wondering how to classify and write down all stable linear dynamical systems of the first order.

Let's start from the minimal case where the system has only two real variables, which leads us to consider $2\times2$ real Hurwitz-stable matrices. They should correspond to a union of submanifolds embedded in $\mathbb{R}^4$. I want to find these submanifolds, together with a coordinate chart on them (i.e., I would like to find a parameterization of families of Hurwitz matrices belonging to each submanifold).

- Attempt and further explanation: First, the real part of the two eigenvalues must be negative (see this): for $2\times2$ real matrices, this is equivalent to demanding that the trace is negative, while the determinant is positive. It seems to me that we already have 3 submanifolds (sink, degenerate sink, spiral sink, see e.g. stability theory):

1. Two real distinct eigenvalues $\lambda_1<0$ and $\lambda_2<0$ ("sink").

2. An eigenvalue $\lambda<0$ with multiplicity 2, meaning that the Jordan form is $ \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix} $ ("degenerate sink") or $ \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} $ ("sink" but with $\lambda_1=\lambda_2$).

3. Two complex eigenvalues $\lambda = a+i b$ and $\lambda^* = a-i b$, with $a<0$ and real non-zero $b$ (this case would be the "spiral sink").

Provided that I am not missing anything, how to parametrise the most general real $2 \times 2$ Hurwitz-stable matrix?

- We have different submanifolds, so different parameterizations: I believe there is no single form that is good for all 3 cases (i.e., each class is a distinct submanifold of dimension less or equal to 4 embedded into $\mathbb{R}^4$). Hence, we have to parametrise (at least) 3 distinct submanifolds.

Answer 1

Let the state-space equation of a system be of the form $\dot{\mathbf{x}}=A\mathbf{x}+Bu$, then its controllable canonical form is $$\begin{split}\begin{eqnarray*} \dot{\mathbf{x}} &=& \left[\begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_{0} & -a_{1} & -a_{2} & \cdots & -a_{n-1} \end{array}\right]\mathbf{x}+\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ \vdots \\ b_0 \end{array}\right]u\end{eqnarray*}\end{split}$$ If $b_0\neq0$ then the system's stability is a sufficient condition for its controllability. So continuing on @XegaXam's comment, a Hurwitz stable $2\times 2$ matrix in its canonical form is $$A=\left[\matrix{0 & 1\\ -a_0 & -a_1}\right]=\left[\matrix{0 & 1\\ -\lambda_{1} \lambda_{2} & \lambda_{1} + \lambda_{2}}\right]$$ where $a_0,\;a_1$ are positive real values.

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We can later show that almost all stable $2\times2$ matrices are similar to this canonical form, i.e. for almost every Hurwitz stable $H=\left[\matrix{h_1&h_2\\h_3&h_4}\right]$ there exists a similarity transformation so that $$P^{-1}HP=A$$ the only exception is when $h_2=h_3=0$ and $h_1=h_4$. In other words, stable matrices of the form $H=hI$ are not similar to $A$. Now what does this imply?

In $\mathbb R^4$ the matrix $A$ represents a quarter-plane (since $x_1=0, \;x_2=1$ are constant):

This can be analogous to the 1D case, where all stable $1\times1$ matrices lie on the left side of the real line. Anyway, those exceptional cases that are not similar to $A$ can be represented by a half-line:

Thus a combination of a quarter-plane and a half-line in $\mathbb R^4$ could be the basis for all Hurwitz-stable matrices under a similarity transform. But how does this similarity transform affect those so-called submanifolds? Well, the half-line is invariant under such transformation but for the plane it can't be answered easily (or maybe it's unanswerable; I don't know ‍♂️)

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Proof of similarity transform's existence: If $h_2\ne 0$ then let $P=\left[\matrix{h_2 & 0\\ -h_1 & 1}\right]$, else if $h_3\ne 0$ let $P=\left[\matrix{-h_4 & 1\\ h_3 & 0}\right]$ so that $$P^{-1}\left[\matrix{h_1 & h_2\\ h_3 & h_4}\right]P=\left[\matrix{0 & 1\\ -\det{H} & \text{tr }H}\right]$$ Otherwise if $h_2=h_3=0$, let $P=\left[\matrix{-h_4 & 1\\ -h_1 & 1}\right]$. It will be a singular matrix if $h_1=h_4$, which means no similarity transform exists in this special case.