Consider the following continuous-time state space representation of the form:
$\frac{d}{dx}x(t) = Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t\in \mathbb{R}^{+}$
$A=\begin{bmatrix}-1&3&0&0\\-3&-1&0&0\\0&0&0&3\\0&0&-3&0 \end{bmatrix} \quad B = \begin{bmatrix}0\\1\\0\\0 \end{bmatrix} \quad C=\begin{bmatrix}1&0&0&1 \end{bmatrix}$
The corresponding eigenvalues are: $-1+3i, \ -1-3i, \ 0+3i \ \text{and} \ 0-3i$.
The answer states that this system is Lyaponov stable.
But I'm wondering why.
Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?
Best Answer
Write $I_n$ for a $n \times n$ identity matrix.
Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:
$$ \begin{align} \dot{V}(x) &= \dot{x}^T P x + x^T P \dot{x} \\ &= x^T A^T P x + x^T P A x \\ &= x^T(A^T P + P A) x \\ &= x^T Q x, \end{align} $$
insert $A$ and $P$ and derive $Q$:
$$ \begin{align} Q = A^T P + P A &= \begin{bmatrix} -1 & -3 & 0 & 0 \\ 3 & -1 & 0 & 0 \\ 0 & 0 & 0 & -3 \\ 0 & 0 & 3 & 0 \end{bmatrix} \frac{1}{2} I_4 + \frac{1}{2} I_4 \begin{bmatrix} -1 & 3 & 0 & 0 \\ -3 & -1 & 0 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & -3 & 0 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} . \end{align} $$
So your directional derivative is $\dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).