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I am interested in finding out the stability of the system

$\dot{x} = -a \begin{bmatrix} \cos^2(t) &\cos(t)\sin(t) \\\cos(t)\sin(t) &\sin^2(t) \end{bmatrix}x$

with $a > 0$, via Lyapunov theory.

Now, I know that a Lyapunov function of the form $V = x^T P(t) x$, where $P(t) \in \mathbb{R}^{2 \times 2}$ is positive definite and symmetric that satisfies the matrix differential equation

$-\dot{P} = P(t) A(t) + A^T(t) P(t) + Q(t)$, here $A(t) = -a \begin{bmatrix} \cos^2(t) &\cos(t)\sin(t) \\ \cos(t)\sin(t) &\sin^2(t) \end{bmatrix}$ and $Q(t)$ is positive definite matrix.

does prove the stability of the above system. However, the question is how do I explicitly find such a matrix $P(t)$?

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  • $\begingroup$ (1) Please don't put multilinear math equations in subject lines. It messes up some of the displays. (2) Use \cos and \sin to get $\cos$ and $\sin$. $\endgroup$ Mar 14, 2023 at 18:12

1 Answer 1

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It is difficult in general to find a solution to this equation. One way would be to use Kronecker calculus, but in this case that would be of no help.

First note that the period of the system is $\pi$ and so the stability condition is given by the existence of positive definite matrix-valued functions $P,Q:[0,2\pi\mapsto\mathbb{R}^{2\times 2}$, $P$ differentiable, $Q$ continuous, verifying $P(0)=P(k\pi)$ and $Q(0)=Q(k\pi)$, for some $k\in\mathbb{N}$, $k\ne 0$, such that

$$ \dot{P}(t)+P(t) A(t) + A(t)^T P(t) + Q(t)=0 $$ holds for all $t\in[0,k\pi]$.

This is not analytically solvable in most cases and one often has to rely on algorithms. However, in the present case, one can exploit the nice structure of the matrix $A(t)$ to parametrize all the possible solutions.

To this aim, define $$ Z(t)=\begin{bmatrix} \sin(t) & \cos(t)\\ -\cos(t) & \sin(t) \end{bmatrix} $$ we see that $Z(t)^TZ(t)=I$. Define also $$ J=\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} $$ and observe that $\dot{Z}(t)=Z(t)J^T$.

We now observe that if we use the change of variables $z(t)=Z(t)^Tx(t)$ we get that

$$\dot{z}(t)=\dot{Z}(t)^Tx(t)+Z(t)^T\dot{x}(t)=(\dot{Z}(t)^TZ(t)+Z(t)^TA(t)Z(t))z(t).$$

Using now the fact that $\dot{Z}(t)=Z(t)J^T$, we get that $\dot{Z}(t)^TZ(t)=JZ(t)^TZ(t)=J$ and, so,

$$\dot{z}(t)=J_0z(t)$$

where $$ J_0=J+Z(t)^TA(t)Z(t)=\begin{bmatrix} 0 & 1\\ -1 & -a \end{bmatrix}. $$

So, this shows that by a proper change of variables, one can express the system into LTI form, for which stability is easy to assess. In fact, the eigenvalues of $J_0$ have all negative real part for all $a>0$.

Therefore, there exist some positive definite matrices $P_0$ and $Q_0$ such that

$$J_0^TP_0+P_0J_0+Q_0=0$$

holds.

From that, it is possible to show that $(P_0,Q_0)$ is a solution of the above Lyapunov equation if and only if $(P(t),Q(t))=(Z(t)P_0Z(t)^T,Z(t)Q_0Z(t)^T)$ is a solution to the time-varying Lyapunov equation.

The proof of this result follows from direct substitution of the expressions and very tedious algebra.

This allows us to state that the following statements are equivalent

  1. The system $\dot{x}(t)=A(t)x(t)$ is asympotically stable.
  2. The system $\dot{z}(t)=J_0z(t)$ is asympotically stable.
  3. There exist some positive definite matrices $P_0$ and $Q_0$ such that

$$J_0^TP_0+P_0J_0+Q_0=0$$ holds.

  1. There exist some positive definite matrix-valued functions $P,Q:[0,2\pi\mapsto\mathbb{R}^{2\times 2}$, $P$ differentiable, $Q$ continuous, verifying $P(0)=P(2\pi)$ and $Q(0)=Q(2\pi)$ such that

$$ \dot{P}(t)+P(t) A(t) + A(t)^T P(t) + Q(t)=0 $$ holds for all $t\in[0,2\pi]$.

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  • $\begingroup$ First of all, an amazing answer. Just some clarifications: 1. The transformations you are using are $P(t) = Z(t) P_0 Z(t)^T$ and $Q(t) = Z(t) Q_0 Z(t)^T$ and then you plug them in $\dot{P}(t) + .....$ and obtain this answer? 2. Since the matrix $J_0$ has negative real eigenvalues for some $a > 0$, does that mean that $P_0$ and $Q_0$ exist and they are positive definite and symmetric. Since I only want to prove global asymptotic stability, I only care whether $P_0, Q_0$ exist and not what their concrete value is, right? $\endgroup$ Mar 14, 2023 at 22:10
  • $\begingroup$ @peacecatfrog I have edited my answer. Let me know if this is still unclear. $\endgroup$
    – KBS
    Mar 15, 2023 at 5:36
  • $\begingroup$ Everything clear, sir! $\endgroup$ Mar 15, 2023 at 10:09
  • $\begingroup$ This approach works because the system has nice properties, i.e. the eigenvalues of $A(t)$ are 0 and $-a$ at all times. $\endgroup$
    – KBS
    Mar 15, 2023 at 10:53
  • $\begingroup$ This approach works because the system has nice properties, i.e. the eigenvalues of $A(t)$ are 0 and$-a$ at all times. If you are happy with an answer, please consider accepting it then. It is recommended to do so as reward. $\endgroup$
    – KBS
    Mar 15, 2023 at 11:01

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