Krdytkyu

Consider the following continuous-time state space representation of the form:

$frac{d}{dx}x(t) = Ax(t)+Bu(t), quad y(t)=Cx(t), quad tin mathbb{R}^{+}$

$A=begin{bmatrix}-1&3&0&0\-3&-1&0&0\0&0&0&3\0&0&-3&0 end{bmatrix} quad B = begin{bmatrix}0\1\0\0 end{bmatrix} quad C=begin{bmatrix}1&0&0&1 end{bmatrix}$

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?

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).