Liapunov stability test (G = [0 1; −0.5 −1])

Liapunov stability theorem (discrete time): the equilibrium $\mathbf{x} = \mathbf{0}$ of $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k)$ is asymptotically stable if and only if, for any positive-definite Hermitian matrix $\mathbf{Q}$, there is a positive-definite matrix $\mathbf{P}$ satisfying

$$\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{Q}$$

Then $V(\mathbf{x}) = \mathbf{x}^T\mathbf{P}\mathbf{x}$ is a Liapunov function, and $\Delta V = \mathbf{x}^T(\mathbf{G}^T\mathbf{P}\mathbf{G}-\mathbf{P})\mathbf{x} = -\mathbf{x}^T\mathbf{Q}\mathbf{x} < 0$.

Choose $\mathbf{Q} = \mathbf{I}$ and $\mathbf{P} = \begin{bmatrix} p_{11} & p_{12} \\ p_{12} & p_{22} \end{bmatrix}$.

Forming the equations

$$\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.5 & -1 \end{bmatrix},\qquad \mathbf{G}^T\mathbf{P}\mathbf{G} = \begin{bmatrix} 0.25p_{22} & -0.5(p_{12}-p_{22}) \\ -0.5(p_{12}-p_{22}) & p_{11}-2p_{12}+p_{22} \end{bmatrix}$$

Setting $\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{I}$:

$$\begin{aligned} (1,1):&\quad 0.25p_{22} - p_{11} = -1\\ (1,2):&\quad -0.5p_{12} + 0.5p_{22} - p_{12} = 0 \;\Rightarrow\; p_{22} = 3p_{12}\\ (2,2):&\quad p_{11} - 2p_{12} = -1 \;\Rightarrow\; p_{11} = 2p_{12} - 1 \end{aligned}$$

Solving

Substitute into (1,1): $0.75p_{12} - 2p_{12} + 1 = -1$, so $p_{12} = 1.6$. Then

$$p_{22} = 4.8,\qquad p_{11} = 2.2$$ $$\mathbf{P} = \begin{bmatrix} 2.2 & 1.6 \\ 1.6 & 4.8 \end{bmatrix}$$

Positive definiteness

• $\Delta_1 = 2.2 > 0$
• $\Delta_2 = 2.2\times4.8 - 1.6^2 = 8 > 0$

So $\mathbf{P}$ is positive definite.

Conclusion

The equilibrium state $\mathbf{x} = \mathbf{0}$ is asymptotically stable, with Liapunov function $V(\mathbf{x}) = 2.2x_1^2 + 3.2x_1x_2 + 4.8x_2^2$ and $\Delta V(\mathbf{x}) = -(x_1^2 + x_2^2) < 0$.

Cross-check: the characteristic equation $z^2 + z + 0.5 = 0$ gives $z = -0.5 \pm j0.5$, with $|z| = 0.707 < 1$.